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<?xml version="1.0"?> <!DOCTYPE archimedes SYSTEM "../dtd/archimedes.dtd" > <archimedes> <info> <author>Guidobaldo del Monte</author> <title>Le Mechaniche</title> <date>1581</date> <place>Venezia</place> <translator/> <lang>it</lang> <cvs_file>monte_mecha_02_it_1581</cvs_file> <cvs_version/> <locator>monte-it.xml</locator> </info> <text id="id.0.0.0.0.3"> <front id="id.1.0.0.0.0"> <section> <pb id="p.0001" xlink:href="pageimg-it/004.jpg"/> <p type="head" id="id.2.1.2.0.0"> <s id="id.2.1.2.1.0"> LE <lb/>MECHANICHE <lb/>DELL'ILLVSTRISS SIG.<lb/>GVIDO VBALDO <lb/>DE' MARCHESI DEL <lb/>MONTE: <lb/>TRADOTTE IN VOL GARE <lb/>DAL SIG. FILIPPO PIGAFETTA: <lb/>Nellequali &longs;i contiene la vera Dottrina di rutti gli I&longs;trumenti <lb/>principali da mouer pe&longs;i grandis&longs;imi con <lb/>picciola forza. </s> </p> <p type="head" id="id.2.1.4.0.0"> <s id="id.2.1.4.1.0"><emph type="italics"/>A beneficio di chi &longs;i diletta di que&longs;ta nobili&longs;&longs;ima scionza; & ma&longs;&longs;imamente <lb/>di Capitani di guerra, Ingegnieri, Architetti, & d'ogni <lb/>Artefice, che intenda per via di Machine <lb/>far opre marauiglio&longs;e, e qua&longs;i <lb/>&longs;opra naturali.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.5.0.0"> <s id="id.2.1.5.1.0"> Et &longs;i dichiarano i vocaboli, & luoghi più difficili. </s> </p> <figure place="text" id="id.2.1.6.0.0" xlink:href="figures-it/004_01.jpg"></figure> <p type="head" id="id.2.1.7.0.0"> <s id="id.2.1.7.1.0"><emph type="italics"/>In Venetia, Appre&longs;&longs;o France&longs;co di France&longs;chi Sane&longs;e. </s> <s id="id.2.1.7.2.0"> MD LXXXI.<emph.end type="italics"/> </s> </p> </section> <section> <pb id="p.0002" xlink:href="pageimg-it/005.jpg"/> <pb id="p.0003" xlink:href="pageimg-it/006.jpg"/> <figure place="text" id="id.2.1.9.0.0" xlink:href="figures-it/006_01.jpg"></figure> <p type="head" id="id.2.1.10.0.0"> <s id="id.2.1.10.1.0"> ALL'ILLVSTRISSIMO <lb/>SIGNOR GIVLIO <lb/>SAVOR GNANO, <lb/>CONTE DI BELGRADO. &c. </s> </p> <p type="head" id="id.2.1.11.0.0"> <s id="id.2.1.11.1.0"> Signore o&longs;&longs;eruandi&longs;&longs;imo. </s> </p> <figure place="text" id="id.2.1.12.0.0" xlink:href="figures-it/006_02.jpg"></figure> <p type="main" id="id.2.1.13.0.0"> <s id="id.2.1.13.1.0"> C<emph type="italics"/>onciosia co&longs;a, che la &longs;cienza delle Mecha-<lb/>niche gioui &longs;ommamente à molte, & importan-<lb/>ti attioni della no&longs;tra vita, à gran ragione fu ella <lb/>da i Filo&longs;ofi, & da i Rè antichi &longs;timata degna di <lb/>laudi &longs;ingularißime; & i Matematici vi han-<lb/>no impiegato lo &longs;tudio, & l'opera più che meza-<lb/>namente, & i Principi fauoriti gl'ingegnieri ec <lb/>cellenti, & arricchiti. </s> <s id="id.2.1.13.2.0"> Ben è per certo di alti&longs;-<lb/>&longs;ima &longs;peculatione, & di &longs;ottile manifattura; imperoche tocca quella par-<lb/>te della Filo&longs;ofia, che tratta de gli elementi in vniuer&longs;ale, & del moto, & <lb/>della quiete de' corpi, &longs;econdo i luoghi &longs;uoi, a&longs;&longs;egnando la cagione in certo <lb/>modo de' loro mouimenti naturali; & anco sforzandoli, per via di machi-<lb/>ne à partir&longs;i da proprij &longs;iti, gli tra&longs;porta all'insù, & per ogni lato in mo-<lb/>uimenti contrari alla natura loro.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.14.0.0"> <s id="id.2.1.14.1.0"><emph type="italics"/>Mena ella ad effetto ambedue que&longs;te intentioni con le propo&longs;itioni che <lb/>na&longs;cono, & &longs;ono congiunte con la materia &longs;te&longs;&longs;a, & co' difici, & i&longs;trumen <lb/>ti, che forma artificialmente. </s> <s id="id.2.1.14.2.0"> La onde egli è dibi&longs;ogno con&longs;iderare que&longs;ta<emph.end type="italics"/> <pb id="p.0004" xlink:href="pageimg-it/007.jpg"/><emph type="italics"/>dottrina in due manìere; l'vna ìn quanto và &longs;peculando, & con ragione <lb/>di&longs;correndo &longs;opra le co&longs;e, che s'hanno à &longs;are, &longs;eruendo&longs;i dell' Arithmetica, <lb/>della Geometria, dell' A&longs;trologia, & della Filo&longs;ofia naturale: & l'altra che <lb/>po&longs;cia le manda ad e&longs;ecutione, & haue nece&longs;sita dell'e&longs;&longs;ercitio, & lauoro <lb/>delle mani, v&longs;ando l'Architettura, la Pittura, il di&longs;egno, l'arte de' fabri, <lb/>de'legnaiuoli, de'muratori, & d'altri me&longs;tieri tali, per modo che ella vie-<lb/>ne ad e&longs;&longs;ere me&longs;colata, & in parte compo&longs;ta della naturale Filo&longs;ofia, delle <lb/>Matematiche, & delle arti manuali. </s> <s id="id.2.1.14.3.0"> Per laqual co&longs;a chiunque &longs;i troua <lb/>dotato d'ingegno acuto, & da fanciullo hà incominciato ad apprendere le <lb/>già dette &longs;cienze, & &longs;a di&longs;egnare, & lauorare di &longs;ua mano, potrà nel vero <lb/>ottimo Mechanico, & <expan abbr="inuētore">inuentore</expan>, & facitore di opere marauiglio&longs;e riu&longs;cire.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.15.0.0"> <s id="id.2.1.15.1.0"><emph type="italics"/>Infinite parti, & vtilißime à gli huomini comprende que&longs;la noti-<lb/>tia, & in guerra, & in pace, ne i commodi della città, della villa, & della <lb/>mercatantia, & in altri; peroche la Medicina toglie da lei i difici per ri-<lb/>porre le o&longs;&longs;a &longs;mo&longs;&longs;e, & rotte ne i &longs;iti &longs;uoi. </s> <s id="id.2.1.15.2.0"> Onde pone Oriba&longs;io nel libro delle <lb/>Machine, diuer&longs;i i&longs;trumenti pre&longs;i dalla Mechanica, & <expan abbr="cõuertiti">conuertiti</expan> nell'v&longs;o del <lb/>la Medicina, come il Tri&longs;pa&longs;ton di Archimede: l'arte del nauigare ricono-<lb/>&longs;ce anco diuer&longs;i aiuti, come il timone, co'l quale, collocato di dietro, ouero <lb/>alle bande del nauilio ageuolmente lo moue, & dirizza, quantunque per <lb/>ri&longs;petto à tutto il corpo del va&longs;ello piccioli&longs;simo &longs;ia. </s> <s id="id.2.1.15.3.0"> I remi, che à gui&longs;a di <lb/>leua lo &longs;pingono innanzi, & l'arbore, & la vela &longs;ono pur di &longs;ua inuentio <lb/>ne. </s> <s id="id.2.1.15.4.0"> I molini, i quali &longs;i girano co'l vento, con l'acqua, & con la forza vi-<lb/>ua: & i pi&longs;trini, le carra, gli aratri, & altri ordigni di villa; il pe&longs;are con <lb/>la bilancia, & con la &longs;tadera; il cauare l'acqua da pozzi con le grù, ouero <lb/>cicogne, dette da latini to&longs;senoni, che &longs;ono come grandis&longs;ime bilancie, & <lb/>con le rote, & altre co&longs;e tali &longs;i riducono alla Mechanica. </s> <s id="id.2.1.15.5.0"> La ragione pa-<lb/>rimente del condurre le acque, & da profondis&longs;ime valli in alto farle &longs;ur <lb/>gere uà &longs;otto lei. </s> <s id="id.2.1.15.6.0"> Chiamarono gli antichi coloro Mechanici ancora, i quali <lb/>co'l fiato, ò vento, ouero acqua, ò corde, ò nerui faceuano vedere, & vdi <lb/>re effetti miracolo&longs;i; come &longs;uoni diuer&longs;i, & canti d'augelli, & fin ad e&longs;pri-<lb/>mere la voce humana in parole: & quelli che con horologi, i quali &longs;i mo-<lb/>uono da &longs;e &longs;tes&longs;i con rote, ò da acqua, ò da &longs;ole il tempo mi&longs;urarono, & di-<lb/>&longs;tin&longs;ero in hore. </s> <s id="id.2.1.15.7.0"> Appartengono alla Mechanica gli facitori delle Sfere <lb/>compartite ne'&longs;uoi cieli, co'l mouimento de'Pianeti, & di tutti i corpi <lb/>cele&longs;tiali à &longs;embianza dell'vniuer&longs;o mondo, & ciò mediante il mouimen-<lb/>to eguale, & in giro, che loro daua l'acqua, di cui la fama &longs;uona e&longs;&longs;ere <lb/>&longs;tato Archimede Siracu&longs;ano il primo mae&longs;tro. </s> <s id="id.2.1.15.8.0"> il mouere etiandio con poca<emph.end type="italics"/> <pb id="p.0005" xlink:href="pageimg-it/008.jpg"/><emph type="italics"/>forza pe&longs;i grandis&longs;imi con i&longs;trumenti, & ingegni diuer&longs;i è principale of-<lb/>ficio della Mechanica, come Bilancie, Stadere, Leue, Taglie, Cunei, Moli-<lb/>nelli, Rote co' denti & &longs;enza, Viti d'ogni &longs;orte, Argani, Mangani, Triuel <lb/>le, & altri molti, i quali da que&longs;ti &longs;i compongono: & &longs;econdo Ari&longs;totele <lb/>tutti &longs;iriducono alla Leua, & al cerchio, & alla machina ritonda, laquale <lb/>quanto è maggiore, tanto più velocemente &longs;i moue. </s> <s id="id.2.1.15.9.0"> L'arte del fortificare <lb/>le piazze, & i &longs;iti, & del difendergli, laquale acconciamente &longs;i puote chia <lb/>mare Architettur a militare, è pro&longs;es&longs;ione Mechanica: peroche per via di <lb/>Cortine, & di Baloardi, & d'altri ripari, qua&longs;i con machine, & i&longs;trumen-<lb/>ti s'ingegna l'huomo con po hi &longs;oldati di ributtarne in dietro molti, & <lb/>mantener&longs;i con vantaggio. </s> <s id="id.2.1.15.10.0"> Il fabricare, & adoprare oltre à ciò gli i&longs;tru <lb/>menti da guerra è proprio dono di que&longs;ta &longs;cienza, come Bali&longs;te, ò Bale&longs;tre, <lb/>Catapulte, Scorpioni, Fionde, & &longs;imili, che da lontano gittano foco, & &longs;aßi, <lb/>& ma&longs;&longs;e di ferro pe&longs;anti dugento cinquanta, & più libre, & Moli da <lb/>molino &longs;econdo Silio Italico, & Vitruuio, per di&longs;tanza di for&longs;e<emph.end type="italics"/> 300. <emph type="italics"/>pas&longs;i <lb/>à mi&longs;ura con ruin &longs;o colpo; & &longs;aette, & verettoni, & falariche grandi <lb/>à gui&longs;a di traui: & quelli che percoteudno con l'vrto da pre&longs;&longs;o, come Arie <lb/>ti, Onagri, Te&longs;tugini, & &longs;imili; & in altri v&longs;i, come <expan abbr="Sãbuche">Sambuche</expan>, Corui, Mani <lb/>di ferro, & gli altri maritimi, & Angoni, Monangoni, Tollenoni, &longs;cale &longs;no-<lb/>date, ponti, torri mobili, & &longs;imili difici antichi, i quali &longs;ono &longs;tati poi ri <lb/>fiutati, &longs;uccedendo in &longs;uo luogo le Artiglierie, da e&longs;&longs;ere anch'e&longs;&longs;e ordinate <lb/>nell' ampiezza della con&longs;ideratione Mechanica, facendo elle còn sì poca mæ <lb/>teria acce&longs;a, tanto horribile perco&longs;&longs;a.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.16.0.0"> <s id="id.2.1.16.1.0"><emph type="italics"/>Que&longs;ta &longs;cienza, che fuor di quanto &longs;i è detto, abbraccia innumerab ili <lb/>altri v&longs;i, & diletteuoli, & nece&longs;&longs;ari à mortali, in diuer&longs;i tempi hebbe in <lb/>&longs;orte vari &longs;tati, per ri&longs;petto à gli artefici, che la e&longs;ercitarono: peroche, <lb/>di là cominciando, ne gli antichis&longs;imi &longs;ecoli, che pa&longs;&longs;arono auanti la guer <lb/>ra di Troia vi&longs;&longs;e Dedalo Athenie&longs;e gran mae&longs;tro di Mechanica, ilquale <lb/>trouò il primiero la &longs;ega, l'a&longs;cia, il piombino da torre le diritture, la tri-<lb/>uella, l'albero, l'antenna, la vela, & altri or digni: di&longs;egnò in Creta poi <lb/>quell'intricato labirinto, & alla fine gli conuenne fabricare per &longs;e, & per <lb/>Icaro &longs;uo figlio due paia d'ali, & volar&longs;ene via per l'aere à gui&longs;a d'au-<lb/>gelli, come cantano i Poeti.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.17.0.0"> <s id="id.2.1.17.1.0"><emph type="italics"/>Nella fabrica del tempio di Salomone, che fu la maggiore per grandez <lb/>za, per mae&longs;tria d' Architettura, & ornamento, di quanie ne &longs;iano &longs;tate <lb/>fatte giamai; & delle piramidi, & di tanti altri difici di quei &longs;eco'i, che <lb/>hanno riempito il mondo di &longs;tupore, egli &longs;i può credere, che interueni&longs;&longs;ero<emph.end type="italics"/> <pb id="p.0006" xlink:href="pageimg-it/009.jpg"/><emph type="italics"/>eccellenti Mechanici, per leuare in alto le pietre &longs;mi&longs;urate, & per altre <lb/>opere, lequali à condurgli à fine &longs;i ricercauano. </s> <s id="id.2.1.19.4.0"> Nacquero dapoi Eudo&longs; <lb/>&longs;o, & Archita Tarentino, ambidue valenti ingegnieri; & di Archita &longs;i <lb/>legge, che lauorò di legno vna colomba con tanta mae&longs;tria temperata, & <lb/>gonfiata, che da &longs;e volaua per l'aria à gui&longs;a di viua colomba. </s> <s id="id.2.1.19.5.0"> Seguì co-<lb/>&longs;toro il Filo&longs;ofo Ari&longs;totele, ilquale certe poche, ma bellis&longs;ime que&longs;tioni Me-<lb/>chaniche, la&longs;ciò &longs;critte. </s> <s id="id.2.1.19.6.0"> A lui venne appre&longs;&longs;o Demetrio Rè, nominato il <lb/>pigliatore, ò di&longs;truggitore delle città, peroche fabricaua machine, & difi-<lb/>ci, co' quali per di&longs;opra vi montaua, & &longs;e ne faceua padrone, lequali per <lb/>auentura furono &longs;imiglianti alla machina detta Cauallo, con cui li Greci <lb/>pre&longs;ero la famo&longs;a Troia; di che ragionando Pau&longs;ania nell' Attica, dice che <lb/>giudica e&longs;pre&longs;&longs;a mattezza il credere, che fo&longs;&longs;e vn cauallo, & non machina <lb/>bellico&longs;a per acco&longs;tare alle muraglie, & prenderle. </s> <s id="id.2.1.19.7.0"> Que&longs;to Rè cominciò <lb/>ad aumentare la Mechanica in qualche honore. </s> <s id="id.2.1.19.8.0"> Ma Archimede, che fù <lb/>il megliore artefice di quanti fecero giamai que&longs;ta profes&longs;ione innanzi, & <lb/>dopo lui, & qua&longs;i vn lume, che poi ha illu&longs;trato tutto il mondo, accrebbe <lb/>in colmo la riputatione della Mechanica, & di pouera arte, & vile, che pri <lb/>ma era, come vuole Plutarco nella vita di Marcello, nel numero delle arti <lb/>nobili, & pregiate alla militia pertinenti la ripo&longs;e. </s> <s id="id.2.1.19.9.0"> Imperoche combat-<lb/>tendo Marcello Sir acu&longs;a patria &longs;ua per mare, & per terra con grande <lb/>ho&longs;te di Romani, egli co'&longs;uoi diuer&longs;i ingegni, & machine differenti, ribut-<lb/>tò &longs;empre gli sforzi, con graue lor danno, & vergogna; come Liuio, Plutar <lb/>co, & altri nominando i difici che v&longs;aua, diffu&longs;amente raccontano. </s> <s id="id.2.1.19.10.0"> Per-<lb/>cioche quando Marcello s'auicinaua aile muraglie per conqui&longs;tar le con la <lb/>Sambuca, il buon Archimede co'l Tollenone, & con le mani di ferro la al-<lb/>zaua di pe&longs;o in aere, & poi &longs;nodando quegli vncini &longs;uoi, la faceua cadere <lb/>da alto, in mare &longs;ommergendola; il mede&longs;imo effetto adoprando eontra gli <lb/>altri nauili, sì fattamente, che gli conuenne allontanare l'armata ben to <lb/>&longs;to dalle mura. </s> <s id="id.2.1.19.11.0"> Ne ce&longs;sò tuttauia d'infe&longs;tare il nemico: ma &longs;i come nota <lb/>Galeno nel terzo libro de' temper amenti, & Giouanni Zonara, & Tze&longs;es <lb/>confermano, allegando Diodoro, & Dione, compo&longs;e certi &longs;pecchi grandi <lb/>& concaui, &longs;econdo la proportione della di&longs;t anza di quei va&longs;elli dalla mu-<lb/>raglia, & opponendogli à raggi del Sole in diritta linea qua&longs;i per miraco-<lb/>lo, gli bru&longs;ciaua. </s> <s id="id.2.1.19.12.0"> Dalla parte della terra &longs;imilmente offendeua gli aduer-<lb/>&longs;ari con arme diuer&longs;e da gittare. </s> <s id="id.2.1.19.13.0"> Per laqual co&longs;a nè in mare, nè in terra <lb/>da gl'ingegni di quell'eccellente Mechanico &longs;i poteua egli &longs;chermire, nuoui <lb/>ripari, & horribili offe&longs;e apparecchiando &longs;empre. </s> <s id="id.2.1.19.14.0"> Pappo Ale&longs;&longs;andrino<emph.end type="italics"/> <pb id="p.0007" xlink:href="pageimg-it/010.jpg"/><emph type="italics"/>allega il quar ante&longs;imo trouato di Archimede, per dichiarare, che almeno <lb/>i &longs;uoi difici al numero di quaranta a&longs;cendeuano. </s> <s id="id.2.1.19.15.0"> La onde Marcello, veg-<lb/>gendo, che niuno profitto apportauano all'impre&longs;a gli a&longs;&longs;alti &longs;uoi, & che <lb/>erano vn mettere le genti ad euidente pericolo, per cagione di quel &longs;olo <lb/>valoro&longs;o vecchio, gli nacque vna tal opinione, & à tutto l'e&longs;ercito, che da <lb/>po&longs;&longs;anza diuina fo&longs;&longs;e gouernato in quella dife&longs;a, & mutò la ragione del <lb/>guerreggiare, dando&longs;i all'aßedio, & al vietare &longs;tret tißimamente le vitto-<lb/>uaglie a quella città.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.20.0.0"> <s id="id.2.1.20.1.0"><emph type="italics"/>Que&longs;te furono le cagioni, che la Mechanica &longs;alì in tanta gloria, & che <lb/>i Romani le a&longs;&longs;egnarono dapoi grado honoreuolißimo ne gli e&longs;erciti lo-<lb/>ro, come &longs;i legge nel primo libro della guerra ciuile, che Ce&longs;are &longs;e prigione <lb/>il Capitano de' fabri di Pompeio, nomato Magio Cremona, & Vitruuio fu <lb/>Capitano delle Bali&longs;te di Ce&longs;are Augu&longs;to, che &longs;arebbe nella militia moder-<lb/>na, come Capitano generale dell' artiglieria. </s> <s id="id.2.1.20.2.0"> La qual gloria &longs;ucceßiua-<lb/>mente le fu mantenuta poi da molti dottißimi &longs;crittori, & mae&longs;tri di <lb/>Mechani a, come da Cte&longs;ibio Ale&longs;&longs;andrino, da Herone Ale&longs;&longs;andrino, da <lb/>un'altro Herone, da Ateneo, da Bione, da Pappo Ale&longs;&longs;andrino, che allega <lb/>Carpo di Antiochia, da Eliodoro, da Oriba&longs;io, & da altri Greci, i quali fio <lb/>rirono in diuer&longs;i tempi, in&longs;egnando la ragione, la mi&longs;ura, & l'v&longs;o de gli <lb/>i&longs;trumenti bellico&longs;i non &longs;olo, ma di tutti gli altri, che le pertengono. </s> <s id="id.2.1.20.3.0"> Fra <lb/>Latini antichi Varrone &longs;cri&longs;&longs;e dell' Architettura, & per con&longs;eguente douet <lb/>te anco far mentione della Mechanica: & Vitruuio, & Vegetio, & qual-<lb/>che altro hanno fauellato d'intorno alla fabrica delle machine militari, <lb/>& da mouer pe&longs;i, & aiutato à con&longs;eruare fra gli huomini viua la digni-<lb/>tà della Mechanica.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.21.0.0"> <s id="id.2.1.21.1.0"><emph type="italics"/>Ma ruinando l'Imperio di Romani, & &longs;uccedendo i barbari in Italia, <lb/>in Grecia, in Egitto, & in ogni contrada, oue &longs;i e&longs;er cita&longs;&longs;ero le buone lette <lb/>re, caddero mi&longs;erabilmente, & &longs;i perderono qua&longs;i del tutto le &longs;cienze, & in <lb/>&longs;pecialità re&longs;tò la Mechanica lunghißimo tempo negletta, non cono&longs;cen-<lb/>do&longs;i in guerra altri difici, che Bricole, Trabucchi, Mangani, Martinelli, & <lb/>certi i&longs;trumenti tali, finche &longs;ouragiun&longs;e l'artiglieria, laquale à poco à po-<lb/>co gli fe di&longs;u&longs;are à fatto: & di quella parte altre&longs;i della Mechanica, laqua <lb/>le s'adopra al mouer pe&longs;i, ben picciolo intendimento rima&longs;e. </s> <s id="id.2.1.21.2.0"> Vera co&longs;a è, <lb/>che &longs;embra da vn tempo in quà le arti, & le dottrine più nobili, come le <lb/>belle lettere appellate humane, la Filo&longs;ofia, la Medicina, l'A&longs;trologia, l'A-<lb/>rithmetica con la Mu&longs;ica, la Geometria, l'Architettura, la Scoltura, la Pit <lb/>tura con molte altre: & &longs;pecialmente la Mechanica e&longs;&longs;ere dalle o&longs;cure te-<emph.end type="italics"/> <pb id="p.0008" xlink:href="pageimg-it/011.jpg"/><emph type="italics"/>nebre, oue giaceuano &longs;epolte, alla thiara luce ri&longs;u&longs;citate: Percioche ri-<lb/>&longs;tringendomi alle Mechaniche Giordano, che &longs;cri&longs;&longs;e de' pe&longs;i, la incominciò <lb/>à &longs;olleuare alquanto, & poi Leon Batti&longs;ta Alberti nella &longs;ua Architettu-<lb/>ra: il Tartaglia aper&longs;e anco la via à molte &longs;peculationi Mechaniche: Vitto<lb/>rio Fau&longs;to nell' Arzanà di Venetia mo&longs;trò d'e&longs;&longs;ere buon Mechanico: Mon-<lb/>&longs;ig. Reuerendi&longs;s. barbaro eletto d' Aquileia ne' Commentari del decimo di <lb/>Vitruuio nominò gli i&longs;trumenti da mouer pe&longs;i: Georgio Agricola nel &longs;e&longs;to <lb/>de' Metalli raccol&longs;e a&longs;&longs;ai machine da leuar pe&longs;i, & qualched'vn'altro: & <lb/>nuoua <expan abbr="m&etilde;te">mente</expan> l'Autoredi que&longs;t'opera, ilquale ben d'altra maniera in ciò pro <lb/>ce lette, che gli autori nominati, peroche con ordine ammir abile, & con <lb/>vere, & certe ragioni ha in&longs;egnato &longs;olo fra Latini ottimamente que&longs;tæ <lb/>&longs;cienza tutta da mouer pe&longs;i.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.22.0.0"> <s id="id.2.1.22.1.0"><emph type="italics"/>Ma &longs;i come i moderni da me ricordati, & principalmente l'Autore del <lb/>pre&longs;ente libro hanno ornata & e&longs;altata la Mechanica con le parole, & co i <lb/>volumi; co&longs;i V. S. Illu&longs;triß. </s> <s id="id.2.1.22.2.0"> l'hà celebrata, & magnificata co' di&longs;cor&longs;i, & <lb/>con le operationi i&longs;te&longs;&longs;e, & co' fattire&longs;a famigliare, & dome&longs;tica, diuer&longs;e <lb/>machine fabricando con profondißima dottrina, & facendone e&longs;perten-<lb/>ze nel mouere qualunque gran pe&longs;o, di cui &longs;i po&longs;&longs;a l'huomo in ogni bi&longs;ogno <lb/>&longs;eruire. </s> <s id="id.2.1.22.3.0"> Talche ben &longs;i puote con verità affermare, che per vna parte e&longs;&longs;a, <lb/>& l'Autore di que&longs;ti trattati per l'altra, habbiate alla Mechanica il pri&longs;ti <lb/>no honore re&longs;tituito, che da i tempi antichi in quà le cra &longs;marrito.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.23.0.0"> <s id="id.2.1.23.1.0"><emph type="italics"/>Sono for&longs;e quaranta anni gia &longs;cor&longs;i, che per i&longs;cherzare con Nicolò Tar<lb/>taglia, per&longs;ona à &longs;uoi tempi molto &longs;timata in que&longs;ta profeßone, & che &longs;i <lb/>dilettaua di andare &longs;oluendo que&longs;tioni &longs;ottili di Mechanica, & di Mathe <lb/>matica, & ne' &longs;uoi dialoghi introduceua à fauellare per&longs;onaggi grandi: <lb/>& alcuna fiata gli faceua dire qualche co&longs;a, di cui eßi prendeuano onta, <lb/>V. S. Illu&longs;triß. </s> <s id="id.2.1.23.2.0"> gliene propo&longs;e for&longs;e quaranta Mechaniche qua&longs;i tutte, & <lb/>difficili: alcune delle quali egli prouò di &longs;oluere, delle altre &longs;i &longs;cusò con di <lb/>re, che à cia&longs;cheduna di loro &longs;arebbe &longs;tato me&longs;tieri vn volume intero, co-<lb/>me &longs;i legge ne' &longs;uoi lib i &longs;tampati della noua &longs;cientia.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.24.0.0"> <s id="id.2.1.24.1.0"><emph type="italics"/>Hor non è punto di marauiglia, che ella habbia penetrato con l'inten-<lb/>dimento <expan abbr="tãto">tanto</expan> dentro, & &longs;aputo co&longs;i bene operare nelle Mechaniche, & &longs;ia <lb/>fatta padrona in tutto dell'arte del fortificare i &longs;iti, & d'ogni altra parte <lb/>della militia: peroche fu dall'ottimo &longs;uo padre alleuata in compagnia di <lb/>huomini &longs;cientiati, & d'alio affare, tra quali fu vn tempo Con&longs;tantino <lb/>La&longs;cari nobilißimo huomo Greco, & piono di dottrina, da cui &longs;ucceßi-<lb/>uamente imparò, oltra le altre lettere, Arithmetica, Geometria, A&longs;tro-<emph.end type="italics"/> <pb id="p.0009" xlink:href="pageimg-it/012.jpg"/><emph type="italics"/>logia, Geogra&longs;ia; à ài&longs;egnare, & lauorare manualmente in me&longs;tieri diuer-<lb/>&longs;i; à caualcare, à maneggiare le arme, à tirare d'archibugio, & d'artiglie <lb/>ria, & à <expan abbr="cõporre">comporre</expan> fochi artificiati, & l'arte per eccellenza detta del bom <lb/>bardiero; à viuere &longs;obriamente, & le fatiche rolerare al caldo, al freddo, <lb/>& ad ogni di&longs;agio, co&longs;e tutte, che di&longs;pongono l'animo, & indurano il corpo <lb/>alla militia. </s> <s id="id.2.1.24.2.0"> Giunta poi all' età di &longs;edici anni, fu inutata con dodici caual <lb/>li qua&longs;i tutti Turchi, & con prouedimento conueneuole di denari à vede-<lb/>re tutta quella guerra, che pa&longs;sò in Italia dalla pre&longs;ura del Rè France&longs;co <lb/>Primo di Francia, fin alla pace generale, che &longs;eguì l'anno<emph.end type="italics"/> 1529. <emph type="italics"/>Nella-<lb/>quale interuennero qua&longs;i tutti i mouimenti militari, che &longs;i po&longs;&longs;ano ima-<lb/>ginare, sì per gli e&longs;erciti grandi, che erano à fronte l'vn contra l'altro, sì <lb/>per la qualità, & quantità delle impre&longs;e fatte, & per mille altri acciden <lb/>ti importantißimi, & &longs;tratagemi auenuti, & sì principalmente; percio-<lb/>che nell'vn campo, & l'altro in varie &longs;tagioni militarono i primi guer-<lb/>rieri del mondo, & in gran numero, i quali con prudenza, a&longs;tutia, & <lb/>brauura contendeuano à gara, & per honore di &longs;oura&longs;tare, & e&longs;&longs;ere vinci <lb/>tori. </s> <s id="id.2.1.24.3.0"> Et veramente chi ben con&longs;idera, fin da i tempi antichi, rarißime vol <lb/>te è &longs;tato con numero maggiore di Capitani famo&longs;i, ò con più copia d'im-<lb/>pre&longs;e grandi guerreggiato, che in quegli anni: Peroche furono fatti prigio-<lb/>ni due de' maggiori Prencipi del mondo, &longs;i a&longs;&longs;ediò Milano, & per forza fu-<lb/>rono pre&longs;e tre città, Roma, Cremona, & Pauia; &longs;i videro più fatti d'ar-<lb/>me, & gli e&longs;erciti &longs;i andarono per&longs;eguitando da Milano à Roma; &longs;i che Pia <lb/>cenza, Parma, Bologna, & Fiorenza guardaròn&longs;i dalle armi nemiche.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.25.0.0"> <s id="id.2.1.25.1.0"><emph type="italics"/>Nello &longs;plendore dunque della &longs;cola del Duca France&longs;co Maria d'V rbino, <lb/>ilquale era Capitano generale della Lega, & di quegli altri valentißimi<lb/>Capitani, andaua V. S. Illu&longs;triß. </s> <s id="id.2.1.25.2.0"> come di &longs;ua libertà, & benißimo à ca-<lb/>uallo, con chi le piaceua, & &longs;i trouaua à quelle fattioni, che volea, &longs;eguen <lb/>do le più volte il Sig. </s> <s id="id.2.1.25.3.0"> Giouanni de' Medici, & Paulo Luzza&longs;co, che erano <lb/>&longs;empre de&longs;ti, & arditi, & come l'occhio dell'e&longs;ercito. </s> <s id="id.2.1.25.4.0"> Quì non è mia in-<lb/>tentione di narrare gli auenimenti di quella guerra, ma &longs;i bene di auerti <lb/>re, che chi la vide, & appre&longs;e da buon &longs;enno i &longs;uoi moti; & &longs;eppe manda-<lb/>re à memoria quei &longs;atti marauiglio&longs;i, ben puote meritamente vantar&longs;i <lb/>di hauer mirato ca&longs;i memorabili, i quali nè anche in migliaia d'anni &longs;o-<lb/>gliono accadere; come ella, che e&longs;&longs;endo giouine di viuace &longs;pirito, & am-<lb/>mae&longs;trata nclle arti nece&longs;&longs;arie al&longs;oldato, & volentern&longs;ißima d'imparare, <lb/>hebbe opportuna occa&longs;ione di far&longs;i prattica dell'ordinare, dell'e&longs;ercitare, <lb/>del far marciare in battaglia, dell'alloggiare in campagna gli e&longs;erciti &longs;i-<emph.end type="italics"/> <pb id="p.0010" xlink:href="pageimg-it/013.jpg"/><emph type="italics"/>curamente: & del pre&longs;entare al nemico il fatto d'arme con vantaggio: <lb/>Del fortificare, & difendere i &longs;iti, & offenderli con le mine, con le trin-<lb/>cee, con le artiglierie, con gli a&longs;&longs;alts, & con tutti gli altri sforzi; & d'o-<lb/>gni parte della militare &longs;cienza.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.18.0.0"> <s id="id.2.1.18.1.0"><emph type="italics"/>Ritornati in pace i Prencipi Chri&longs;tiani, &longs;i dedicò al &longs;eruigio de' Sereniß. <lb/>&longs;uoi Signori, oue ne i più importanti carichi, & maggiori, & in due guer <lb/>re haue e&longs;&longs;a aggiunto cinquanta anni di noua, & ottima &longs;eruitù all'an-<lb/>tica di qua&longs;i dugento anni, continua, & fedeli&longs;s. </s> <s id="id.2.1.18.2.0"> fattagli da i &longs;uoi pre-<lb/>dece&longs;&longs;ori Sauorgnani, fabricando nello &longs;patio di que&longs;to tempo in diuer&longs;e pro<lb/>uincie de' &longs;uoi &longs;tati pre&longs;&longs;o che cinquanta Baloardi, con eccellentißima ra-<lb/>gione inte&longs;i, & con vero magi&longs;terio lauorati, & notabilißimo ri&longs;parmio <lb/>del publico danaro.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.19.0.0"> <s id="id.2.1.19.1.0"><emph type="italics"/>Ma per tornare alle Mechaniche dico, che quando gli anni pa&longs;&longs;ati io <lb/>venni à vi&longs;it arla ad O&longs;opo &longs;ua fortezza, &longs;entì &longs;ommo piacere in &longs;corgere <lb/>quel monte, che circonda più d'un miglio, &longs;ituato alla foce del fiume Ta-<lb/>gliamento, oue dalle &longs;trettezze di quei gioghi s'allarga nelle pianure del <lb/>Friuli, d'ogn'intorno alto pre&longs;&longs;o che &longs;e&longs;&longs;anta paßi à mi&longs;ura, tutto di ma-<lb/>cigno duro, & di&longs;co&longs;ce&longs;e, & erto sì, che rende la &longs;alita impoßibile fornito <lb/>attorno di baloardi cauati nel &longs;a&longs;&longs;o, & di molti tagli, & canoniere per <lb/>ferire gli aduer&longs;ari, & di artiglierie, & d'arme d'ogni &longs;orte à &longs;u&longs;&longs;icienza, <lb/>da cui &longs;i hà vi&longs;ta di qua&longs;i tutto il Friuli, & è &longs;cudo, & riparo, come al-<lb/>tra volta fù, contra l'empito delle genti nemiche, lequali in Italia tenta&longs; <lb/>&longs;ero di &longs;oendere da quella parte; po&longs;to di co&longs;ta alla &longs;trada principale, che <lb/>conduce in Lamagna, per laqual vanno, & vengono Signori, & Principi, <lb/>& Amba&longs;ciadori, & infinite mercatantie; onde ella, che tiene &longs;empre le <lb/>guardie, & vedette sù quel monte, quando pa&longs;&longs;ano Signori principali, <lb/>hà per co&longs;tume di &longs;alutargli con le &longs;ue artiglterie, & conuitarg'i anco <lb/>nel &longs;uo alloggiamento d'O&longs;opo, oue tutto l'anno &longs;oggiorna, quantunque <lb/>habbia & Belgrado, & Aris, & Ca&longs;telnouo, & Sauorgnano, & villaggi <lb/>a&longs;&longs;ai: percioche l'aere vi è purißimo, & &longs;pende il &longs;uo tempo in ocio con ne <lb/>gocio, di continuo vi&longs;itata da Gentil'huomini, & Signori diuer&longs;i; ta che la <lb/>&longs;ua ca&longs;a viene ad e&longs;&longs;ere vn ridotto di per&longs;one virtuo&longs;e, & vn'albergo di &longs;ol <lb/>dati, & di dottori. </s> <s id="id.2.1.19.2.0"> lui &longs;i caualca, tenendo ella vna &longs;talla piena di buoni&longs;-<lb/>&longs;imi caualli, &longs;i armeggia, &longs;i và alla caccia, & in ogni attione &longs;i e&longs;ercita vi-<lb/>ta cauallere&longs;ca. </s> <s id="id.2.1.19.3.0"> Oltre à quanto hò diui&longs;ato, pre&longs;i anco diletto in vedere la <lb/>&longs;ua habitatione e&longs;&longs;ere à gui&longs;a d'vna bottega d'arme politamente à &longs;uoi <lb/>luoghi &longs;erbate: & vn magazino di machine bellico&longs;e, & da mouer pe&longs;i,<emph.end type="italics"/> <pb id="p.0011" xlink:href="pageimg-it/014.jpg"/><emph type="italics"/>hauendone ella fabricate di &longs;ua indu&longs;tria for&longs;e dodici di maniere differen <lb/>ti, parte da &longs;tra &longs;cinare, & parte da alzare con pochißima forza &longs;mi&longs;u-<lb/>rati pe&longs;i: come quella, che hà vna &longs;ola rota co' denti, & ali'erta tira cin-<lb/>que de' &longs;uoi canoni con la po&longs;&longs;anza di Grada&longs;&longs;o &longs;uo Nano: & quell'altra, la <lb/>quale con vna oncia di forza &longs;ola, po&longs;ta nel manico, che la volge, dà il me <lb/>to à quattordici mila libre di pe&longs;o: che &longs;e al detto manico &longs;i attribui&longs;ce <lb/>la forza, che comunalmente haue l'huomo con la mano, cioè libre cinquan <lb/>ta, egli è manife&longs;to la predetta machina hauere po&longs;&longs;anza di mouere, co&longs;a <lb/>incredibile, molto più di otto millioni di libre. </s> <s id="id.2.1.25.5.0"> Que&longs;te machine portabi-<lb/>li da vn mulo, & alcune anche da vn'huomo &longs;ono à diuer&longs;i affari nece&longs;&longs;a-<lb/>rijßime, & maßimamente à maneggiare, & condurrei pezzi großi del-<lb/>l'artiglieria. </s> <s id="id.2.1.25.6.0"> & per certo &longs;e l'anno<emph.end type="italics"/> 1529. <emph type="italics"/>il Conte di San Polo Capitano <lb/>France&longs;e nel ritirar&longs;i dall'a&longs;&longs;edio di Milano inuer&longs;o Piemonte con l'e&longs;er-<lb/>cito, & con l'artiglieria, haue&longs;&longs;e portato &longs;eco vno de' minimi i&longs;trumenti <lb/>d'O&longs;opo, non &longs;arcbbe &longs;cor&longs;o in quello &longs;tremo infortunio, percioche in mar <lb/>ciando fu da vn graue canone rotto il ponte, che trauer&longs;aua il &longs;o&longs;&longs;o della <lb/>&longs;trada, & il pezzo cadè nel fango. </s> <s id="id.2.1.25.7.0"> Onde formoßi il campo per non la-<lb/>&longs;ciarlo à dietro, & non hauendo ingegno da cauarlo fuori, &longs;i con&longs;umò tan-<lb/>to tempo, che &longs;opragiun&longs;e Antonio da Leua con le &longs;ue genti, & ritrouan <lb/>do l'e&longs;&longs;ercito nemico &longs;eparato, & in quel di&longs;ordine, lo mi&longs;e in rotta, & fè <lb/>preda delle bagaglie, delle artiglierie, & del Capitano mede&longs;mo. </s> <s id="id.2.1.25.8.0"> Non hà <lb/>troppo iempo, che il Duca France&longs;co di Gui&longs;a, allhor che di Francia guidò <lb/>l'e&longs;ercito in Abruzzo, douendo partire, volle &longs;piegare prima la fanteria, <lb/>& cauælleria &longs;ua in ordinanza à fronte del nemico, qua&longs;i à battaglia sfi-<lb/>dandolo; ma poi nel ritorno &longs;caualco&longs;si vn pezzo d'artiglieria, & s'arre-<lb/>&longs;tò tutta la ma&longs;&longs;a delle genti, & quei Prencipi France&longs;i &longs;montati da ca-<lb/>uallo, penarono buona pezza auanti, che lo ripone&longs;&longs;ero &longs;u le rote, con ri&longs;chio <lb/>di patir danno da gli aduer&longs;ari, che haue&longs;&longs;ero con quella occa&longs;ione &longs;pinto <lb/>innanzi. </s> <s id="id.2.1.25.9.0"> Di quc&longs;ti e&longs;empi non mancano per l'hi&longs;torie.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.26.0.0"> <s id="id.2.1.26.1.0"><emph type="italics"/>Hora che è pace V. S. Illu&longs;triß. è andata inue&longs;tigando per &longs;uo diporto <lb/>molte, & varie &longs;orti di ordigni da mouer pe&longs;i, affine di valer&longs;ene nelle <lb/>fabriche, & nell'argine di pietre, che fa per ritenere l'impeto del Taglia-<lb/>mento, che non gua&longs;ti i colti di O&longs;opo, & per douer&longs;ene anco &longs;eruire, quan<lb/>do che &longs;ia in guerra. </s> <s id="id.2.1.26.3.0"> Si come fece Archimede, ilquale, &longs;econdo Plutarco, &longs;tan<lb/>do in pace à petitione di Hierone Rè, compo&longs;e quelle <expan abbr="tãie">tanie</expan> Machine per giuo-<lb/>co, & i&longs;cherzo di Geometria, lequali poi &longs;oprauenendo la guerra, le &longs;eppe <expan abbr="cõ">com</expan> <lb/>uertire opportunamente contra Romani. </s> <s id="id.2.1.26.4.0"> Et &longs;e egli, come te&longs;tificano diuer&longs;i<emph.end type="italics"/> <pb id="p.0012" xlink:href="pageimg-it/015.jpg"/><emph type="italics"/>autori, &longs;edendo con certa machina detta, &longs;econdo Oriba&longs;io, Tri&longs;pa&longs;ton, per <lb/>che &longs;i maneggiaua con tre cor de, tirò dal mare in terra quella gran naue <lb/>del Rè &longs;uo; & con la forza della mano &longs;ini&longs;tra mo&longs;&longs;e mediante l'i&longs;tru-<lb/>mento vn pe&longs;o di cinque mila &longs;taia ò moggia, sì fattamente che diputan-<lb/>do à cia&longs;cuno &longs;taio quarantacinque libre di pe&longs;o, a&longs;cenderebbono alla &longs;om-<lb/>ma di dugento venticinque mila libre; & pre&longs;umeua&longs;i di hauer potuto <lb/>mouere la terra, trouando doue fermar&longs;i con la leua, ò con quella &longs;ua ma-<lb/>china de&longs;critta da Pappo nell'ottauo libro delle raccolte matematiche, la <lb/>quale hauea cinque rote co' &longs;uoi as&longs;i, & vna vite perpetua co'l manico: Io <lb/>mi rendo certo, che ella s'ingegnerebbe di &longs;ormare i&longs;trumenti per adoprare <lb/>altretanto.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.27.0.0"> <s id="id.2.1.27.1.0"><emph type="italics"/>Hauendo io dunque veduti, & i&longs;perimentati que&longs;ti vari difici ad O&longs;o-<lb/>po; & es&longs;endomi &longs;tato da lei mo&longs;trato la prima volta il pre&longs;ente libro, & <lb/>commendato &longs;ommamente, mi propo&longs;i nell' animo, che vtile &longs;arebbe il ri <lb/>durlo in volgare, accioche coloro i quali &longs;ono atti per altro ad intenderlo, <lb/>ma non hanno cono&longs;cenza del Latino, pote&longs;sero, farne &longs;uo profitto. </s> <s id="id.2.1.27.2.0"> Co&longs;i <lb/>compiuta l'opera, & fattala &longs;tampare, la mando à V. S. Illu&longs;tri&longs;s. </s> <s id="id.2.1.27.3.0"> che po&longs; <lb/>&longs;ede e&longs;qui&longs;it amente que&longs;ta materia, & &longs;econda i &longs;tudi delle buone lette-<lb/>re, i quali, &longs;e dopo Iddio, non vengono fauoriti da i gran Signori, nulla va <lb/>gliono. </s> <s id="id.2.1.27.4.0"> Che &longs;e in qualche parte haurò à gli amatori delle Mechaniche re-<lb/>cata ageuolezza, & vtilità con le mie fatiche, douranno eglino &longs;aper' à <lb/>lei buon grado, che di que&longs;ta fattura è &longs;tata cagione. <lb/>Di Venetia à<emph.end type="italics"/> 28. <emph type="italics"/>di Giugno<emph.end type="italics"/> 1581. </s> </p> <p type="main" id="id.2.1.28.0.0"> <s id="id.2.1.28.1.0"><emph type="italics"/>Di V. S. Illu&longs;tri&longs;s.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.29.0.0"> <s id="id.2.1.29.1.0"><emph type="italics"/>Affettionatiß. </s> <s id="id.2.1.29.2.0"> &longs;cruidore <lb/>Filippo Pigafetta.<emph.end type="italics"/> </s> </p> </section> <section> <pb id="p.0013" xlink:href="pageimg-it/016.jpg"/> <p type="head" id="id.2.1.31.0.0"> <s id="id.2.1.31.1.0"> AI LETTORI </s> </p> <p type="main" id="id.2.1.32.0.0"> <s id="id.2.1.32.1.0"> Il prefente libro contiene &longs;ei trattati, il primo de <lb/>quali è della Bilancia con la Stadera, l'altro della <lb/>Leua, il terzo della Taglia, il quarto dell' A&longs;&longs;e nel-<lb/>la rota, il quinto del Cuneo, & l'vltimo della Vite, <lb/>che tutti &longs;ono i&longs;trumenti Mechanici. </s> <s id="id.2.1.32.2.0"> Intitula&longs;i le <lb/>Mechaniche. </s> <s id="id.2.1.32.3.0"> Ma percioche que&longs;ta parola Mechaniche non ver <lb/>rà for &longs;e int e&longs;a da cia&longs;cheduno perlo &longs;uo vero &longs;ignificato, anzi <lb/>troueran&longs;i di quelli, che &longs;timeranno lei e&longs;&longs;ere voce d'ingiuria, <lb/>&longs;olendo&longs;i in molte parti d'Italia dire ad altrui Mechanico per <lb/>i&longs;cherno, & villania; & alcuni per e&longs;&longs;ere chiamati Ingegnieri &longs;i <lb/>prendono &longs;degno: non &longs;arà per auentura fuori di propo&longs;ito il <lb/>ricordare, che Mechanico è vocabolo honorati&longs;&longs;imo, dimo&longs;tran <lb/>te, &longs;econdo Plutarco, me&longs;tiero alla Militia pertinente, & conue <lb/>neuole ad huomo di alto affare, & che &longs;appia con le &longs;ue mani, <lb/>& co'l &longs;enno mandare ad e&longs;ecutione opre marauiglio&longs;e à &longs;ingu <lb/>lare vtilità, & diletto del viuere humano. </s> </p> <p type="main" id="id.2.1.33.0.0"> <s id="id.2.1.33.1.0"> Fù, per nomiuarne alcuno tra molti Filo&longs;ofi, & Prencipi de' <lb/>preteriti &longs;ecoli, Archita Tarentino, & Eudo&longs;&longs;o <expan abbr="cõpagni">compagni</expan> di Pla-<lb/>tone, & valenti&longs;&longs;imi Ingegnieri, & Mechanici, che &longs;ono vna me <lb/>de&longs;ma co&longs;a, di cui fa Plutarco mentione nella vita di Marcello: <lb/>& Demetrio Rè, inuentore &longs;ottili&longs;&longs;imo di Machine bellico&longs;e, <lb/>& ne lauoraua di &longs;ua mano ancora: & fra Greci di Sicilia Me-<lb/>chanico, & Ingegniere famo&longs;is&longs;imo Archimede Siracu&longs;ano, il <lb/>quale era di <expan abbr="grã">gram</expan> legnaggio, & parente di Hierone Rè di Sicilia. </s> </p> <p type="main" id="id.2.1.34.0.0"> <s id="id.2.1.34.1.0"> Et quantunque Plutarco nell'i&longs;te&longs;&longs;a vita affermi, che egli di <lb/>&longs;pregia&longs;&longs;e le Mechaniche, come bas&longs;i & vili, & materiali, nè di <lb/>loro degna&longs;&longs;e &longs;criuere giamai, & che non per opera principale, <lb/>ma per vn cotale &longs;ollazzo, & giuoco di Geometria impiegaua <lb/>la fatica nelle Mechaniche, pregato da quel Rè; sì leggiamo <lb/>noi tuttauia in altri autori, lui hauere dettato vn libro della mi <lb/>&longs;ura, & proportione d'ogni maniera di va&longs;ello, diui&longs;ando la for <lb/>ma della gran naue fabricata da Hierone, à cui nulla manca-<lb/>ua: & Pappo Ale&longs;&longs;andrino allega il libro della Bilancia di Ar-<lb/>chimede, che è pur Mechanico tutto: & l'i&longs;te&longs;&longs;o nell'ottauo del <lb/>le raccolte Matematiche pone vn'i&longs;trumento da mouer pe&longs;i, <pb id="p.0014" xlink:href="pageimg-it/017.jpg"/> mo&longs;trando e&longs;&longs;ere il quarante&longs;imo trouato d' Archimede, per cui <lb/>di&longs;&longs;e; Dami oue io mi fermi, ch'io mouerò la terra; & Carpo <lb/>Mechanico &longs;c ri&longs;&longs;e, che Archimede compo&longs;e vn libro del modo <lb/>del fare le Sfere, che è fattura Mechanica. </s> <s id="id.2.1.39.2.0"> Ma più il mede&longs;imo <lb/>Archimede, non vna &longs;ola volta cita &longs;e &longs;te&longs;&longs;o, nel libro della Qua <lb/>dratura della Parabola, con parole tali. </s> <s id="id.2.1.39.3.0"> Imperoche egli è dimo-<lb/>&longs;trato nelle Mechaniche; accennando alcune propo&longs;itioni del <lb/>&longs;uo libro delle co&longs;e, che egualmente pe&longs;ano, ilquale è tutto Me-<lb/>chanico. </s> <s id="id.2.1.39.4.0"> Oltre à ciò vna parte del libro della Quadratura della <lb/>Parabola, & il &longs;econdo delle co&longs;e, che &longs;tanno &longs;opra l'acqua, oue-<lb/>ro à galla &longs;ono Mechanici. </s> <s id="id.2.1.39.5.0"> Da que&longs;ti luoghi vede&longs;i e&longs;pre&longs;&longs;o, che <lb/>non &longs;olamente Archimede fece opre Mechaniche, ma ne &longs;cri&longs;&longs;e <lb/>anco molti trattati; & confe&longs;&longs;a Plutarco per niuna altra dottri-<lb/>na e&longs;&longs;ere tanto in riputatione &longs;alito Archimede, quanto per le <lb/>impre&longs;e Mechaniche; anzi veramente co'l mezo loro hauer&longs;i egli <lb/>all'hora procacciato fama non di &longs;cienza humana, ma di &longs;apien-<lb/>za diuina. </s> <s id="id.2.1.39.6.0"> Per la qual co&longs;a egli è ben da con&longs;iderare, come Plu-<lb/>tarco &longs;i &longs;ia la&longs;ciato tra&longs;correr' à dire, che Archimede le Mechani <lb/>che di&longs;preggia&longs;&longs;e, nè di loro degna&longs;&longs;e &longs;criuere: & per certo egli <lb/>forte d'opinione &longs;arebbe&longs;i <expan abbr="ingãnato">ingannato</expan>, &longs;e haue&longs;&longs;e poco &longs;timata quel <lb/>la facultà, che lo fè guadagnare gloria di gran lunga maggio-<lb/>re, che qualunque altra &longs;cienza &longs;i po&longs;&longs;ede&longs;&longs;e. </s> <s id="id.2.1.39.7.0"> Vitruuio de i <lb/>Latini fù buon Mechanico, & &longs;eruì per Capitano delle Bali&longs;te, <lb/>& delle altre machine da guerra Ottauiano Ce&longs;are, & gli intitu-<lb/>lò le &longs;ue fatiche dell' Architettura, & ne diuenne ricco. </s> </p> <p type="main" id="id.2.1.40.0.0"> <s id="id.2.1.40.1.0"> L'e&longs;&longs;ere Mechanico dunque, & Ingegniero con l'e&longs;empio di <lb/>tanti valent'huomini, è officio da per&longs;ona degna, & &longs;ignorile: <lb/>& Mechanica è voce Greca &longs;ignificante co&longs;a fatta con artificio <lb/>da mouere, come per miracolo, & fuori dell'humana po&longs;&longs;anza <lb/>grandis&longs;imi pe&longs;i con picciola forza, & in generale comprende <lb/>cia&longs;cun Dificio, Ordigno, I&longs;trumento, Argano, Mangano, oue-<lb/>ro ingegno mae&longs;treuolmente ritrouato, & lauorato per cotali ef <lb/>fetti, & &longs;imili altri infiniti in qual &longs;i voglia &longs;cienza, arte, & e&longs;er-<lb/>citio. </s> <s id="id.2.1.40.2.0"> Laquale hò de&longs;critta co&longs;i materialmente per darne vn cer <lb/>to &longs;aggio accommodato al gu&longs;to del più de gli huomini; trala-<lb/>&longs;ciando le accurate di&longs;&longs;initioni à miglior tempo. </s> </p> <p type="main" id="id.2.1.41.0.0"> <s id="id.2.1.41.1.0"> Aggiunga&longs;i, che &longs;otto que&longs;to vniuer&longs;ali&longs;&longs;imo titolo &longs;i è con-<pb id="p.0015" xlink:href="pageimg-it/018.jpg"/> tentato l'Autore di manife&longs;tare per hora, & il primo de Latini <lb/>con dimo&longs;trationi ageuoli, & piane, in&longs;egnare &longs;olamente la ra-<lb/>gion dello intendere, & maneggiare gli &longs;ei predetti l&longs;trumenti <lb/>Mechanici; à quali &longs;i riducono tutti gli altri, come à &longs;uoi prin-<lb/>cipil, & fondamenti; & da'quali &longs;i po&longs;&longs;ono comporne diuer&longs;e ma <lb/>niere, accozzandone in&longs;ieme due, tre, & più, come l'A&longs;&longs;e nella <lb/>rota con la Taglia, la Vite co'l detto A&longs;&longs;e, & con la Leua, & &longs;uc-<lb/>ces&longs;iuamente de gli altri ad arbitrio di chiunque in varie opre &longs;e <lb/>ne sà con giudicio valere, come nota l'Autore nel fine di que&longs;to <lb/>volume. </s> </p> <p type="main" id="id.2.1.35.0.0"> <s id="id.2.1.35.1.0"> Hor come che l'Autore con bella via, & chiara, & con ordine <lb/>ammirabile di que&longs;ti difici habbia ragionato, & la co&longs;a per &longs;e <lb/>molto o&longs;cura non &longs;ia ad intender&longs;i: nondimeno ben ricerca ella <lb/>tutto l'intelletto dell'huomo, & che con &longs;i&longs;&longs;a &longs;peculatione &longs;i leg-<lb/>gano attentis&longs;imamente più d'vna volta le dimo&longs;trationi. </s> </p> <p type="main" id="id.2.1.36.0.0"> <s id="id.2.1.36.1.0"> Doue &longs;i vede in alcuni luoghi di que&longs;ti trattati cotale &longs;orte di <lb/>lettere picciole, differente dalle altre, come la pre&longs;ente; auer-<lb/>ta&longs;i che non vi &longs;ono co&longs;e dettate dall' Autore di que&longs;to libro di <lb/>Mechaniche, ma notate da colui che l'hà volgarizato, à fine di <lb/>chiarire qualche pa&longs;&longs;o difficile, & ageuolare l'intendimento à' <lb/>Lettori non co&longs;i prattichi nelle Scole de' Filo&longs;ofi. </s> </p> <p type="main" id="id.2.1.37.0.0"> <s id="id.2.1.37.1.0"> Ponga&longs;i anco mente, che à carte 121. nel trattato della Vite, <lb/>è po&longs;to fra i detti dell' Autore il Problema di Pappo, ilquale do-<lb/>uea e&longs;&longs;ere &longs;tampato con lettere differenti dalle altre, ma per in-<lb/>auertenza è &longs;tato me&longs;&longs;o co' caratteri &longs;te&longs;&longs;i delle propo&longs;itioni del <lb/>l' Autore, che è difetto. </s> <s id="id.2.1.37.2.0"> Non è &longs;tato pos&longs;ibile &longs;chluare alcuni <lb/>falli nello &longs;tampare. </s> <s id="id.2.1.37.3.0"> Onde correggan&longs;i in que&longs;ta maniera. </s> <s id="id.2.1.37.4.0"> Nel <lb/>la Lettera à carte 1. faccia 2. ver&longs;i 25. to&longs;&longs;enoni, leggi tollenoni. <lb/>car. 43. ver. 22. dell'angolo, all'angolo. carte 48. f. 2. nella po-<lb/>&longs;tilla, per la 2. di que&longs;to; della 2. di que&longs;to. carte 87. f. 2. <lb/>ver. 14. dalla, alla. carte 93. ver. 32. cni, cui. carte 115. ver. 20. <lb/>Hlici, Helici. </s> <s id="id.2.1.37.16.0"> Gli altri errori di lettere meno importanti, & che <lb/>non mouono il &longs;en&longs;o alla, di&longs;cretione del giudicio&longs;o Lettore &longs;i ri <lb/>mettono. </s> </p> </section> <section> <pb id="p.0016" xlink:href="pageimg-it/019.jpg"/> <p type="head" id="id.2.1.68.0.0"> <s id="id.2.1.68.1.0"> TRATTATI IN QVEST'OPERA <lb/>CONTENVTI.</s> </p> <p type="main" id="id.2.1.69.0.0"> <s id="id.2.1.69.1.0"> I. <lb/>Della Bilancia, con la Stadera à carte 1<lb/> </s> <s id="id.2.1.69.2.0"> II. <lb/>Della Leua. 35<lb/> </s> <s id="id.2.1.69.3.0"> III. <lb/>Della Taglia. 56<lb/> </s> <s id="id.2.1.69.4.0"> IIII. <lb/>Dell'A&longs;&longs;e nella Rota. 102<lb/> </s> <s id="id.2.1.69.5.0"> V. <lb/>Del Cuneo. 107<lb/> </s> <s id="id.2.1.69.6.0"> VI. <lb/>Della Vite. 115<lb/> </s> </p> </section> </front> <body id="id.2.0.0.0.0"> <chap id="id.2.2.0.0.1"> <pb id="p.1" xlink:href="pageimg-it/020.jpg"/> <p type="head" id="id.2.1.43.0.0"> <s id="id.2.1.43.1.0"> LIBRO DI <lb/>MECHANICHE, <lb/>DELL'ILLVSTRISSIMO <lb/>SIGNORE, <lb/>II. S. GVIDO VBALDO DE' MARCHESI <lb/>DEL MONTE. </s> </p> <figure place="text" id="id.2.1.44.0.0" xlink:href="figures-it/020_01.jpg"></figure> <p type="head" id="id.2.1.45.0.0"> <s id="id.2.1.45.1.0"> Diffinitioni. </s> </p> <p type="main" id="id.2.1.46.0.0"> <s id="id.2.1.46.1.0"> Il centro della grauezza di cia&longs;cun corpo e vn <lb/>certo punto po&longs;to dentro, dal quale &longs;e con la <lb/>imaginatione s'intende e&longs;&longs;erui appe&longs;o il gra-<lb/>ue, mentre è portato &longs;ta fermo, & mantiene <lb/>quel &longs;ito, che egli hauea da principio, ne in <lb/>quel portamento &longs;i và riuolgendo. </s> </p> <p type="main" id="id.2.1.47.0.0"> <s id="id.2.1.47.1.0"><emph type="italics"/>Que&longs;ta diffinitione del centro della grauezza in&longs;egnò <lb/>Pappo Ale&longs;&longs;andrino nell'ottauo libro delle raccolte ma-<lb/>thematiche. </s> <s id="id.2.1.47.2.0"> Ma Federico Comandino nel libro del cen-<lb/>tro della grauezza de' corpi &longs;olidi dichiarò l'i&longs;te&longs;&longs;o centro in questa maniera de&longs;cri-<lb/>uendolo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.48.0.0"> <s id="id.2.1.48.1.0"> Il centro della grauezza di cia&longs;cuna figura &longs;olida è quel punto po&longs;to <lb/>dentro, d'intorno alquale le parti di momenti eguali da ogni parte <lb/>&longs;i fermano. </s> <s id="id.2.1.48.2.0"> Peroche &longs;e per tale centro &longs;arà condotto vn piano, che <lb/>&longs;eghi in qual &longs;i voglia modo la figura, &longs;empre la diuiderà in parti, <lb/>che pe&longs;eranno egualmente. </s> </p> <pb id="p.1v" xlink:href="pageimg-it/021.jpg"/> <p type="head" id="id.2.1.50.0.0"> <s id="id.2.1.50.1.0"> NOTITIE COMVNI. </s> </p> <p type="head" id="id.2.1.51.0.0"> <s id="id.2.1.51.1.0"> I. </s> </p> <p type="main" id="id.2.1.52.0.0"> <s id="id.2.1.52.1.0"> Se da co&longs;e egualmente pe&longs;anti &longs;i leneranno co&longs;e, che pur egualmente <lb/>pe&longs;ino, le re&longs;tanti pe&longs;eranno egualmente. </s> </p> <p type="head" id="id.2.1.53.0.0"> <s id="id.2.1.53.1.0"> II. </s> </p> <p type="main" id="id.2.1.54.0.0"> <s id="id.2.1.54.1.0"> Se à co&longs;e egualmente pe&longs;anti &longs;i aggiungeranno co&longs;e, che pur <expan abbr="egualm&etilde;">egualmen</expan> <lb/>te pe&longs;ino, tutte in&longs;ieme pe&longs;eranno egualmente. </s> </p> <p type="head" id="id.2.1.55.0.0"> <s id="id.2.1.55.1.0"> III. </s> </p> <p type="main" id="id.2.1.56.0.0"> <s id="id.2.1.56.1.0"> Le co&longs;e, che all'i&longs;te&longs;&longs;o &longs;ono eguali in pe&longs;o, &longs;ono tra loro anco gra-<lb/>ui egualmente. </s> </p> <p type="head" id="id.2.1.57.0.0"> <s id="id.2.1.57.1.0"> PRESVPPOSTE. </s> </p> <p type="head" id="id.2.1.58.0.0"> <s id="id.2.1.58.1.0"> I. </s> </p> <p type="main" id="id.2.1.59.0.0"> <s id="id.2.1.59.1.0"> Di vno corpo è vn &longs;olo centro della grauezza. </s> </p> <p type="head" id="id.2.1.60.0.0"> <s id="id.2.1.60.1.0"> II. </s> </p> <p type="main" id="id.2.1.61.0.0"> <s id="id.2.1.61.1.0"> Il centro della grauezza di vn corpo è &longs;empre nel mede&longs;imo &longs;ito per <lb/>ri&longs;petto al &longs;uo corpo. </s> </p> <p type="head" id="id.2.1.62.0.0"> <s id="id.2.1.62.1.0"> III. </s> </p> <p type="main" id="id.2.1.63.0.0"> <s id="id.2.1.63.1.0"> I Pe&longs;i &longs;ono portati in giu &longs;econdo il centro della grauezza. </s> </p> <p type="main" id="id.2.1.64.0.0"> <s id="id.2.1.64.1.0"> DIFFINITIONI. </s> <s id="id.2.1.64.2.0"> La diffinitione è vn breue parlare, che manife&longs;ta, & inte-<lb/>ramente dichiara la co&longs;a propo&longs;ta, &longs;i fattamente che non &longs;i po&longs;&longs;a trouare condi-<lb/>tione, ouero accidente alcuno principale in e&longs;&longs;a co&longs;a, &longs;e la diffinitione è buona, <lb/>che non &longs;ia in virtù compre&longs;a, & detta da lui; come per e&longs;empio l'Autore qui di <lb/>&longs;opra da ad inten dere che &longs;ia il centro della grauezza con due diffinitioni. </s> </p> <p type="main" id="id.2.1.65.0.0"> <s id="id.2.1.65.1.0"> Le Notitie comuni poi &longs;ono certe &longs;entenze manife&longs;te al &longs;en&longs;o comune de gli huomi-<lb/>ni, lequali pur che vi &longs;i ponga mente, &longs;ubito vdite, &longs;i intendono, & &longs;e le pre&longs;ta il <lb/>con&longs;entimento. </s> </p> <p type="main" id="id.2.1.66.0.0"> <s id="id.2.1.66.1.0"> Ma la Pre&longs;uppo&longs;ta è diuer&longs;a, peroche mette per vero la co&longs;a co&longs;i e&longs;&longs;ere, come &longs;i pro-<lb/>pone &longs;enza altro di&longs;cor&longs;o per farla cono&longs;cere. </s> </p> </chap> <chap id="id.2.2.0.0.2"> <pb id="p.2" xlink:href="pageimg-it/022.jpg"/> <p type="head" id="id.2.1.70.0.0"> <s id="id.2.1.70.1.0"> DELLA BILANCIA </s> </p> <p type="main" id="id.2.1.71.0.0"> <s id="id.2.1.71.1.0"> Avanti che &longs;i faccia mentione della Bilancia, accioche la <lb/>co&longs;a re&longs;ti più chiara, &longs;ia la Bilancia AB in linea diritta, & <lb/>CD la Truttina della Bilancia, laquale &longs;econdo la con&longs;uetu <lb/>dine comune &longs;tà &longs;empre à piombo dell'orizonte. </s> <s id="id.2.1.71.2.0"> & il punto C im <lb/>mobile, d'intorno alquale &longs;i volge la Bilancia, &longs;i chiami il centro del <lb/>la bilancia, &longs;ia pur collo-<lb/>cato di &longs;opra della bilan <lb/>cia, ò di &longs;otto, benche <lb/>non propriamente, che <lb/>non fa nulla Ma il CA, <lb/>& il CB &longs;iano le di&longs;tan <lb/>ze, & braccia della Bilan <lb/>cia, co&longs;i nomate. </s> <s id="id.2.1.71.3.0"> & &longs;e <lb/>dal centro della bilancia <lb/>collocato di &longs;opra, ò di <lb/>&longs;otto della Bilancia, &longs;arà <lb/>tirata vna linea à piom-<lb/>bo di AB, que&longs;ta &longs;i chia <lb/>merà perpendicolo, che <lb/>&longs;o&longs;terrà la Bilancia AB, <lb/>& &longs;empre &longs;tarà à piom-<lb/>bo di e&longs;&longs;a Bilancia, mo-<lb/>ua&longs;i ella in qual &longs;i voglia <lb/>modo. </s> </p> <figure place="text" id="id.2.1.72.0.0" xlink:href="figures-it/022_01.jpg"></figure> <p type="main" id="id.2.1.73.0.0"> <s id="id.2.1.73.1.0"> Concio&longs;ia che in que&longs;to trattato della Bilancia, & ne gli altri ancora l'Autore v&longs;i <lb/>alcune parole, lequali non &longs;i &longs;ono potute tra&longs;portare commodamente in v olga-<lb/>re, per non e&longs;&longs;ere e&longs;&longs;e anco &longs;tate accettate in que&longs;ta lingua, ne inte&longs;e da ognuno, <lb/>io le ho la&longs;ciate co&longs;i latine. </s> <s id="id.2.1.73.2.0"> Ma accioche non facciano difficultà à coloro, i quali <lb/>non intendono il latino, le andrò per tutto à fuoi luoghi dichiarando. </s> </p> <p type="main" id="id.2.1.74.0.0"> <s id="id.2.1.74.1.0"> Nel re&longs;to poi delle parole mi &longs;ono attenuto più al chiaro, & all'v&longs;ato, che &longs;ia pos&longs;i-<lb/>bile, & ho po&longs;to angolo retto, & linea retta in cambio di angolo diritto, & linea <lb/>diritta, & linea della direttione in lo co di linea della dirittura, & co&longs;i diretto per <lb/>diritto, & alcuna volta magnitudine in vece di grandezza, & angolo mi&longs;to per <lb/>me&longs;colato, & angolo curuilineo per di linee torte, & linea curua per torta, & &longs;oli-<lb/>do per &longs;odo, & for&longs;e qualche altro vocabolo poco v&longs;ato in que&longs;ta no&longs;tra fauella, <lb/>&longs;timando che cote&longs;te parole &longs;iano per dimo&longs;trare maggiormente la co&longs;a, & la in-<lb/>tentione dell' Autore: & etiandio de&longs;iderando, che &longs;i rendano famigliari, & dome <lb/>&longs;tiche in que&longs;ta &longs;cienza, talche ognuno le po&longs;&longs;a ageuolmente intendere. </s> </p> <p type="main" id="id.2.1.75.0.0"> <s id="id.2.1.75.1.0"> Trutina è quella co&longs;a, che &longs;o&longs;tiene tutta la Bilancia, laquale Trutina pigli a il Perno, <lb/>ouero l'A&longs;&longs;etto, & noma&longs;i in que&longs;ti pae&longs;i Gioa, altroue Giouola, ouero l'o recchie <lb/>della Bilancia, & in altre contrade Scocca, talche non &longs;i troua &longs;in hora vocabolo, <pb id="p.2v" xlink:href="pageimg-it/023.jpg"/>che in Italia communemente vi &longs;i confaccia, ne alcuno di qne&longs;ti &longs;arebbe inte&longs;o <lb/>per tutto. </s> <s id="id.2.1.75.2.0"> Onde io ho &longs;critto co&longs;i la Trutina, &longs;perando, che &longs;i habbia à fare termi <lb/>ne, & parola generale à tutte le nationi d'Italia. </s> </p> <p type="main" id="id.2.1.76.0.0"> <s id="id.2.1.76.1.0"> Perpendicolo vuol dire quella linea, che &longs;porge in fuori dal centro della Bilancia al <lb/>mezo di detta Bilancia, ilqual Perpendicolo è &longs;olamente nelle Bilancie, lequali han <lb/>no il centro di fuori della Bilancia, o &longs;ia di &longs;otto, ò &longs;ia di &longs;opra. </s> <s id="id.2.1.76.2.0"> Ma quando il cen-<lb/>tro della Bilancia è nel mezo di e&longs;&longs;a, all'hora non vi è que&longs;to Perpendicolo per e&longs; <lb/>&longs;ere il centro della Bilancia, & il mezo di e&longs;&longs;a vn'i&longs;te&longs;&longs;o punto. </s> <s id="id.2.1.76.3.0"> Et que&longs;to Perpen-<lb/>dicolo è co&longs;a imaginata dall' Autore &longs;olamente, & non da altri, per ageuolare al-<lb/>cune dimo&longs;trationi della Bilancia, che di nouo ha inue&longs;tigate: & non è la linguet-<lb/>ta, ne meno la linea della direttione, ò dirittura che &longs;i habbia à dire. </s> </p> <p type="head" id="id.2.1.77.0.0"> <s id="id.2.1.77.1.0"> LEMMA. </s> </p> <p type="main" id="id.2.1.78.0.0"> <s id="id.2.1.78.1.0"> Sia la linca AB à piombo dell'orizonte, & col diametro AB &longs;i de&longs;cri-<lb/>ua il cerchio AEBD, il cui centro &longs;ia C. Dico il punto B e&longs;&longs;ere <lb/>l'infimo luogo della circonferenza del cerchio AEBD, & il pun-<lb/>to A il piu alto, & quali &longs;i voglian punti, come DE, i quali &longs;iano <lb/>però egualmente di&longs;tanti da A e&longs;&longs;ere egualmente po&longs;ti di &longs;otto, & <lb/>quelli che &longs;tanno piu da pre&longs;&longs;o ad e&longs;&longs;o A, e&longs;&longs;ere più alti di quelli, che <lb/>&longs;ono più da lunge. </s> </p> <figure place="text" id="id.2.1.81.0.0" xlink:href="figures-it/023_01.jpg"></figure> <p type="main" id="id.2.1.79.0.0"> <s id="id.2.1.79.1.0"> <arrow.to.target n="note1"></arrow.to.target><emph type="italics"/>Allunghi&longs;i la linea AB fin al centro del mondo, <lb/>che &longs;ia F. Dapoi &longs;ia pre&longs;o nella circonferenza <lb/>del cerchio qual &longs;i voglia punto, come G, & &longs;i <lb/>congiungano le linee FG FD FE. Hor per-<lb/>cioche BF è la minima linea di tutte quelle, <lb/>che dal punto F &longs;ono tirate alla circonferenza <lb/>AEBD, &longs;arà la BF minore della FG. Per <lb/>laqual co&longs;a il punto B &longs;arà piu da pre&longs;&longs;o al pun-<lb/>to F, che il G. Et per cotesta ragione &longs;i dimo-<lb/>strerà, che il punto B &longs;ta più da pre&longs;&longs;o al centro <lb/>del mondo di qual &longs;i voglia altro punto della cir-<lb/>conferenza del cerchio AEBD. Sarà dunque <lb/>il punto B l'infimo luogo della circon&longs;erenza del <lb/>cerchio AEBD. Dapoi perche AF tirata <lb/>per lo centro è maggiore di GF, &longs;arà il punto A <lb/>più alto non &longs;olamente di G, ma etiandio di qual <lb/>&longs;i voglia altro punto della circon&longs;erenza del cer-<lb/>chio AEBD. Oltre à ciò perche DF, & FE <lb/>&longs;ono eguali, i punti DE &longs;aranno egualmente di <lb/>stanti dal centro del mondo. </s> <s id="id.2.1.79.2.0"> Et e&longs;&longs;endo DF <lb/>maggiore di FG, &longs;arà il punto D, che è più da <lb/>pre&longs;&longs;o al punto A, più alto del punto G, lequali <lb/>co&longs;e tutte erano da mo&longs;trar&longs;i.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.80.0.0"> <s id="id.2.1.80.1.0"> <margin.target id="note1"></margin.target><emph type="italics"/>Per la ottaua del terzo.<emph.end type="italics"/> </s> </p> <pb id="p.3" xlink:href="pageimg-it/024.jpg"/> <p type="main" id="id.2.1.83.0.0"> <s id="id.2.1.83.1.0"> Que&longs;to vocabolo Lemma greco v&longs;ato da tutti i volgarizatori di Euclide, & da gli <lb/>altri Scrittori di Mathematica ancora, hò accettato anch'io. </s> <s id="id.2.1.83.2.0"> Ma ben con tutto ciò <lb/>&longs;timo che egli habbia me&longs;tieri di vn poco di lume per e&longs;&longs;er inte&longs;o; & viene à dire, <lb/>&longs;i come nota Cicerone nel &longs;econdo della Diuinatione, co&longs;a che prima &longs;i prende <lb/>per render facile l'intendimento delle co&longs;e, lequali &longs;i hanno dapoi à mo&longs;trare, & <lb/> <expan abbr="nõ">non</expan> è Pre&longs;uppo&longs;ta, perche ella <expan abbr="nõ">non</expan> &longs;i proua <expan abbr="cõ">com</expan> ragione, ma &longs;uppon&longs;i; ma il Lemma <lb/>&longs;i dimo&longs;tra, come in que&longs;to luogo, che prende il punto B e&longs;&longs;ere po&longs;to nell'infimo <lb/>&longs;ito della circonferenza del cerchio, & lo proua per douer&longs;ene valere nelle &longs;eguen <lb/>ti dimo&longs;trationi. </s> </p> <p type="main" id="id.2.1.84.0.0"> <s id="id.2.1.84.1.0"> Doue in que&longs;to Lemma &longs;i dice, che la linea AB è à piombo dell'orizonte, intenda&longs;i <lb/>per orizonte il piano della campagna, & del terreno &longs;ottopo&longs;to, volendo dire ori <lb/>zonte parola greca vn cerchio, che termina la no&longs;tra veduta, & abbraccia & diui <lb/>de la metà della terra tutta. </s> <s id="id.2.1.84.2.0"> Quando dunque &longs;i troua in que&longs;ti libri vna linea, oue-<lb/>ro altra quantità e&longs;&longs;ere à piombo, ouero egualmente di&longs;tante, ò inchinata all'ori-<lb/>zonte, intenda&longs;i per l'orizonte il piano della campagna, ò del terreno. </s> </p> <p type="head" id="id.2.1.85.0.0"> <s id="id.2.1.85.1.0"> PROPOSITIONE I. </s> </p> <p type="main" id="id.2.1.86.0.0"> <s id="id.2.1.86.1.0"> Se il pe&longs;o &longs;arà &longs;o&longs;tenuto nel centro della &longs;ua grauezza da linea diritta <lb/>non &longs;i fermerà giamai, &longs;e quella i&longs;te&longs;&longs;a linea non &longs;arà à piombo del <lb/>l'orizonte. </s> </p> <figure place="text" id="id.2.1.87.0.0" xlink:href="figures-it/024_01.jpg"></figure> <p type="main" id="id.2.1.88.0.0"> <s id="id.2.1.88.1.0"><emph type="italics"/>Sia il pe&longs;o A, & il centro della &longs;ua <lb/>grauezza B, ilqual pe&longs;o venga &longs;o <lb/>&longs;tenuto dalla linea CB. Dico che <lb/>il pe&longs;o non è per fermar&longs;i giamai, <lb/>&longs;e CB non &longs;arà à piombo dell'o-<lb/>rizonte. </s> <s id="id.2.1.88.2.0"> Sia il punto C immobi-<lb/>le, e&longs;&longs;endo co&longs;i nece&longs;&longs;ario, accio il <lb/>pe&longs;o &longs;ia &longs;o&longs;tenuto: & e&longs;&longs;endo il pun <lb/>to C immobile, &longs;e il pe&longs;o A de-<lb/>ue&longs;i mouere, il punto B de&longs;criuerà <lb/>la circonferenza di vn cerchio, il <lb/>cui mezo diametro &longs;arà CB. Per <lb/>laqual co&longs;a &longs;u'l centro A & con <lb/>lo &longs;patio BC &longs;i de&longs;criua il cerchio <lb/>BFDE. & &longs;ia di prima BC à <lb/>piombo dell'orizonte, & &longs;ia tirata <lb/>&longs;in à D, & il punto C &longs;tia di &longs;ot <lb/>to al punto B. Hor percioche il pe&longs;o A &longs;i moue in giù &longs;econdo il centro della gra-<emph.end type="italics"/> <arrow.to.target n="note2"></arrow.to.target> <lb/><emph type="italics"/>uezza, il punto B &longs;i mouerà in giù, oue naturalmente inchina ver&longs;o il centro del mon <lb/>do per la linea diritta BD: tutto il pe&longs;o A dunque con B &longs;uo centro della gra-<lb/>uezza, grauerà &longs;opra la linea diritta BC, & concio&longs;ia che il pe&longs;o venga &longs;o&longs;tenuto <lb/>dalla linea CB, la linea CB &longs;o&longs;terrà tutto il pe&longs;o A, &longs;opra laquale non puote mo<emph.end type="italics"/> <pb id="p.3v" xlink:href="pageimg-it/025.jpg"/><emph type="italics"/>uer&longs;i in giù, e&longs;&longs;endogliene da e&longs;&longs;a <lb/>vietato. </s> <s id="id.2.1.88.3.0"> Per la diffinitione dun-<lb/>que del centro della grauezza, il <lb/>punto B & il pe&longs;o A &longs;taranno <lb/>in que&longs;to &longs;ito. </s> <s id="id.2.1.88.4.0"> & quantunque il <lb/>B &longs;ia piu alto di qual &longs;i voglia al-<lb/>tro punto del cerchio, tuttauia non <lb/>&longs;i mouerà in giù da que&longs;to &longs;ito per <lb/>la circonferenza del cerchio, pero-<lb/>che non &longs;i inchinerà più ver&longs;o lo F, <lb/>che ver&longs;o lo E, per e&longs;&longs;ere nell'vna <lb/>parte & nell'altra eguale la di&longs;ce-<lb/>&longs;a: ne il pe&longs;a A piu &longs;tà pendente <lb/>in vna parte che nell'altra, ilche <lb/>non auiene in qual &longs;i voglia altro <lb/>punto della circon&longs;erenza del cer-<lb/>chin, eccettuato il D. Sia il centro <lb/> <arrow.to.target n="fig1"></arrow.to.target> <lb/>della grauezza dell'i&longs;te&longs;&longs;o pe&longs;o, come in F, concio&longs;ia che la di&longs;ce&longs;a &longs;ia dal punto <lb/>F ver&longs;o il D, & la a&longs;ce&longs;a ver&longs;o il B, però il punto F mouera&longs;&longs;i in giù: & per-<lb/>cioche non &longs;i puote mouere al centro del mondo per linea diritta, per e&longs;&longs;ere impe-<lb/>dito dal punto C immobile per cau&longs;a della linea CF, ma ben &longs;i mouerà &longs;empre <lb/>in giù come richiede la &longs;ua natura: & e&longs;&longs;endo il D il luogo infimo, &longs;i mouerà per <lb/>la circonferenza FD finche peruenga in D, nelqual &longs;ito fermera&longs;&longs;i il pe&longs;o, & <lb/>re&longs;terà immobile, sì perche non &longs;i puote più mouere in giù per e&longs;&longs;ere attaccato al <lb/>punto C, sì anche percioche egli è &longs;o&longs;tenuto nel &longs;uo centro della grauezza. </s> <s id="id.2.1.88.5.0"> Et <lb/>quando F &longs;arà in D, &longs;arà &longs;imilmente la FC in DC, & in&longs;ieme à piombo <lb/>dell'orizonte. </s> <s id="id.2.1.88.6.0"> il pe&longs;o dunque non &longs;i fermerà giamai finche la linea CF non &longs;tia <lb/>à piombo dell'orizonte, che bi&longs;ognaua prouare.<emph.end type="italics"/> </s> </p> <figure id="fig1" place="text" xlink:href="figures-it/025_01.jpg"></figure> <p type="margin" id="id.2.1.90.0.0"> <s id="id.2.1.90.1.0"> <margin.target id="note2"></margin.target><emph type="italics"/>Per la terza pre&longs;upposta di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.91.0.0"> <s id="id.2.1.91.1.0"> Di quì &longs;i puote cauare, che il pe&longs;o &longs;ia pur &longs;o&longs;tenuto in vn dato punto <lb/>in qual &longs;i voglia modo, non &longs;tarà fermo giamai, &longs;e non quando la <lb/>linea tirata dal centro della grauezza del pe&longs;o à quel punto, &longs;tia à <lb/>piombo dell'orizonte. </s> </p> <pb id="p.4" xlink:href="pageimg-it/026.jpg"/> <p type="main" id="id.2.1.93.0.0"> <s id="id.2.1.93.1.0"><emph type="italics"/>Come, po&longs;te le co&longs;e i&longs;te&longs;&longs;e, &longs;ia &longs;o&longs;tenuto <lb/>il pe&longs;o dalle linee CG CH. Dico <lb/>che &longs;e la tirata linea BC &longs;arà à <lb/>piombo dell'orizonte, il pe&longs;o &longs;tarà <lb/>fermo: ma &longs;e la tirata linea CF <lb/>non &longs;arà à piombo dell'orizonte, il <lb/>punto F &longs;imouerà in giù fin al D, <lb/>nel qual &longs;ito &longs;tarà fermo il pe&longs;o, <lb/>& la tirata linea CD &longs;arà à piom-<lb/>bo dell'orizonte. </s> <s id="id.2.1.93.2.0"> Le quali co&longs;e <lb/>tutte con laragione mede&longs;ima &longs;i pro-<lb/>uerebbono.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.94.0.0" xlink:href="figures-it/026_01.jpg"></figure> <p type="head" id="id.2.1.95.0.0"> <s id="id.2.1.95.1.0"> PROPOSITIONE II. </s> </p> <p type="main" id="id.2.1.96.0.0"> <s id="id.2.1.96.1.0"> La bilancia egualmente di&longs;tante dall'orizonte, il cui centro &longs;tia &longs;opra <lb/>la detta bilancia, & che habbia i pe&longs;i eguali nelle &longs;tremità, & egual-<lb/>mente di&longs;tanti dal perpendicolo, &longs;e da cotale &longs;ito &longs;arà mo&longs;&longs;a, & <lb/>nell'i&longs;te&longs;&longs;o di nuouo la&longs;ciata, ritornerà, & iui re&longs;terà. </s> </p> <figure place="text" id="id.2.1.97.0.0" xlink:href="figures-it/026_02.jpg"></figure> <p type="main" id="id.2.1.98.0.0"> <s id="id.2.1.98.1.0"><emph type="italics"/>Sia la bilancia AB in <lb/>linea diritta egualmen <lb/>te di&longs;tante dall'orizon <lb/>te, il cui centro C &longs;ia <lb/>&longs;opra la bilancia, & <lb/>&longs;ia CD il perpendi-<lb/>colo, il quale &longs;arà à <lb/>piombo dell'orizonte: <lb/>& la di&longs;tanza DA <lb/>&longs;ia eguale alla di&longs;tan-<lb/>za DB: & &longs;iano i <lb/>pe&longs;i in AB eguali, <lb/>i centri della grauez-<lb/>za de' quali &longs;iano ne i <lb/>punti AB. Moua&longs;i <lb/>da que&longs;to &longs;ito la bi-<lb/>lancia AB come in EF, dapoi &longs;ia la&longs;ciata. </s> <s id="id.2.1.98.2.0"> Dico che la bilancia EF ritor-<lb/>neràin AB di&longs;tante egualmente dall'orizonte, & iui rimanerà. </s> <s id="id.2.1.98.3.0"> Hora percioche<emph.end type="italics"/> <pb id="p.4v" xlink:href="pageimg-it/027.jpg"/><emph type="italics"/>il punto C &longs;tà immobì <lb/>le mentre la bilancia &longs;i <lb/>moue, il punto D veni <lb/>rà à de&longs;criuere vna cir-<lb/>con&longs;erenza di cerchio, il <lb/>cui mezo diametro &longs;a-<lb/>rà CD. Per laqual <lb/>co&longs;a co'lcentro D, & <lb/>lo &longs;patio CD de&longs;cri-<lb/>ua&longs;i il cerchio DGH. <lb/>Et perche CD &longs;empre <lb/>&longs;tà à piombo della bi-<lb/>lancia, mentre la bilan <lb/>cia &longs;arà in EF, la li-<lb/>nea CD &longs;arà in CG <lb/>&longs;i fattamente, che CG <lb/> <arrow.to.target n="fig2"></arrow.to.target> <lb/>venga ad e&longs;&longs;ere à piombo di EF: & concio&longs;ia che AB &longs;ia diui&longs;a in due parti<emph.end type="italics"/> <lb/> <arrow.to.target n="note3"></arrow.to.target><emph type="italics"/>eguali nel punto D, & i pe&longs;iin AB &longs;iano eguali, &longs;arà etiandio il centro della <lb/>grauezza della magnitudine compo&longs;ta di que&longs;ti due corpi AB nel mezo, cioè in <lb/>D: & quando la bilancia in&longs;ieme co i pe&longs;i &longs;arà in EF, &longs;arà parimente G il cen <lb/>tro della grauezza della magnitudine compo&longs;ta di e&longs;&longs;i AB: & percioche CG <lb/>non è à piombo dell'orizonte, la grandezza compo&longs;ta de i pe&longs;i EF non rimarrà<emph.end type="italics"/> <lb/> <arrow.to.target n="note4"></arrow.to.target><emph type="italics"/>in questo &longs;ito, ma &longs;i mouerà in giù &longs;econdo il centro della grauezza &longs;ua, che è in <lb/>G, per la circonferenza GD, finche &longs;i faccia à piombo dell'orizonte, cioè finche <lb/>CG ritorni in CD. Et quando CG &longs;arà in CD, la linea EF (perche &longs;em-<lb/>pre &longs;tà ad angoli retti con CG) &longs;arà in AB, nelqual &longs;ito &longs;tarà &longs;erma. </s> <s id="id.2.1.98.4.0"> La bi-<emph.end type="italics"/> <lb/> <arrow.to.target n="note5"></arrow.to.target><emph type="italics"/>lancia dunque EF ritornerà in AB, laquale è di&longs;tante egualmente dall'orizon-<lb/>te, & iui rimarrà, che bi&longs;ognaua dimo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig2" place="text" xlink:href="figures-it/027_01.jpg"></figure> <p type="margin" id="id.2.1.100.0.0"> <s id="id.2.1.100.1.0"> <margin.target id="note3"></margin.target><emph type="italics"/>Per la quarta del primo di Archimede delle co&longs;e che pe&longs;ano egualmente.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.101.0.0"> <s id="id.2.1.101.1.0"> <margin.target id="note4"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.102.0.0"> <s id="id.2.1.102.1.0"> <margin.target id="note5"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.103.0.0"> <s id="id.2.1.103.1.0"> PROPOSITIONE III. </s> </p> <p type="main" id="id.2.1.104.0.0"> <s id="id.2.1.104.1.0"> La bilancia egualmente di&longs;tante dall'orizonte, che habbia nelle &longs;tre-<lb/>mità pe&longs;i eguali, & egualmente lontani dal perpendicolo, e&longs;&longs;endo <lb/>collocato il centro di &longs;otto, rimarrà in que&longs;to &longs;ito. </s> <s id="id.2.1.104.2.0"> Ma &longs;e indi &longs;arà <lb/>mo&longs;&longs;a, & la&longs;ciata à ba&longs;&longs;o, &longs;i mouerà &longs;econdo la parte piu ba&longs;&longs;a. </s> </p> <pb id="p.5" xlink:href="pageimg-it/028.jpg"/> <figure place="text" id="id.2.1.106.0.0" xlink:href="figures-it/028_01.jpg"></figure> <p type="main" id="id.2.1.107.0.0"> <s id="id.2.1.107.1.0"><emph type="italics"/>Sia la bilancia AB in <lb/>linea diritta, egual-<lb/>mente di&longs;tante dall'ori<lb/>zonte, il cui centro C <lb/>&longs;ia di &longs;otto alla bilan-<lb/>cia, & &longs;ia CD il per-<lb/>pendicolo, ilquale &longs;arà <lb/>à piombo dell'orizon-<lb/>te, & la di&longs;tanza AD <lb/>&longs;ia eguale alla distan-<lb/>za DB, & &longs;iano in <lb/>AB pe&longs;i eguali, i cen-<lb/>tri della grauezza de' <lb/>quali &longs;iano ne' punti <lb/>AB. Dico primiera-<lb/>mente che la bilancia <lb/>AB &longs;tarà &longs;erma in <lb/>que&longs;to &longs;ito. </s> <s id="id.2.1.107.2.0"> Hor percioche AB &longs;i diuide in parti eguali nel punto D, & i <lb/>pe&longs;i po&longs;ti in AB &longs;ono eguali, &longs;egue, che il punto D &longs;ia il centro della grauez-<lb/>za della magnitudine compo&longs;ta di ambedue i corpi me&longs;&longs;i in AB; & il CD che<emph.end type="italics"/> <arrow.to.target n="note6"></arrow.to.target> <lb/><emph type="italics"/>&longs;ostiene la bilancia &longs;tà à piombo dell'orizonte: Adunque la bilancia AB in <lb/>que&longs;to &longs;ito rimarrà ferma. </s> <s id="id.2.1.107.3.0"> Ma da que&longs;to &longs;ito moua&longs;i la bilancia AB come in <lb/>EF, & la&longs;ci&longs;i dapoi. </s> <s id="id.2.1.107.4.0"> Dico che la bilancia EF &longs;i mouerà dalla parte dello F. <lb/>Et percioche il CD &longs;tà &longs;empre à piombo della bilancia, mentre la bilancia &longs;arà <lb/>in EF verrà ad e&longs;&longs;ere anche il CD in CG à piombo di EF, & il punto<emph.end type="italics"/> <arrow.to.target n="note7"></arrow.to.target> <lb/><emph type="italics"/>G della magnitudine composta di EF &longs;arà il centro della grauezza, ilquale men <lb/>tre &longs;i moue de&longs;criuerà la circonferenza del cerchio DGH, il cui mezo diametro <lb/>è CD, & il centro C. Ma perche CG non &longs;tà à piombo dell'orizonte, la <lb/>grandezza compo&longs;ta de i pe&longs;i EF non rimarrà in questo &longs;ito, ma &longs;econdo il cen-<lb/>tro della &longs;ua grauezza &longs;i mouerà in giù per la circonferenza GH. La bilancia <lb/>dunque EF &longs;i mouerà in giù dalla parte dello F, che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.108.0.0"> <s id="id.2.1.108.1.0"> <margin.target id="note6"></margin.target><emph type="italics"/>Per la quar ta del primo d' Archimede delle co&longs;e che pe&longs;ano <expan abbr="egualm&etilde;te">egualmente</expan>.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.109.0.0"> <s id="id.2.1.109.1.0"> <margin.target id="note7"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.110.0.0"> <s id="id.2.1.110.1.0"> PROPOSITIONE IIII. </s> </p> <p type="main" id="id.2.1.111.0.0"> <s id="id.2.1.111.1.0"> La bilancia egualmente di&longs;tante dall'orizonte, & che habbia nelle &longs;tre <lb/>mità pe&longs;i eguali, & egualmente di&longs;tanti dal centro collocato in e&longs;&longs;a <lb/>bilancia. </s> <s id="id.2.1.111.2.0"> Se ella indi &longs;arà mo&longs;&longs;a, ò non, douunque ella &longs;arà la&longs;cia-<lb/>ta, rimarrà. </s> </p> <pb id="p.5v" xlink:href="pageimg-it/029.jpg"/> <figure id="fig3" place="text" xlink:href="figures-it/029_01.jpg"></figure> <p type="main" id="id.2.1.113.0.0"> <s id="id.2.1.113.1.0"><emph type="italics"/>Sia la bilancia nella linea <lb/>diritta AB egualmen <lb/>te di&longs;tante dall'orizon-<lb/>te, il cui centro C &longs;ia <lb/>nella i&longs;te&longs;&longs;a linea AB, <lb/>& la di&longs;tanza CA &longs;ia <lb/>eguale alla distanza <lb/>CB, & &longs;iano i pe&longs;i <lb/>AB eguali, i cui cen-<lb/>tri della grauezza &longs;tia <lb/>no ne i punti AB. Mo <lb/>ua&longs;i la bilancia come in <lb/>DE, & iui &longs;ia la&longs;cia-<lb/>ta. </s> <s id="id.2.1.113.2.0"> Dico primamen-<lb/> <arrow.to.target n="fig3"></arrow.to.target> <lb/>te che la bilancia DE non &longs;i mouerà, & rimarrà in quel &longs;ito. </s> <s id="id.2.1.113.3.0"> Hor percioche i <lb/>pe&longs;i AB &longs;ono eguali, &longs;arà il centro della grauezza della magnitudine compo&longs;ta <lb/>delli due pe&longs;i A & B in C. Per laqual co&longs;a l'i&longs;te&longs;&longs;o punto C &longs;arà il centro <lb/>della bilancia, & il centro della grauezza di tutto il pe&longs;o. </s> <s id="id.2.1.113.4.0"> Et percioche il centro <lb/>della bilancia che è C, mentre la bilancia AB in&longs;ieme co'pe&longs;i &longs;i moue in DE, <lb/>rimane immobile, non &longs;i mouerà ne anche il centro della grauezza, che è l'i&longs;te&longs;&longs;o C. <lb/>Adunque ne anche la bilancia DE &longs;i mouerà per la diffinitione del centro della <lb/>grauezza, e&longs;&longs;endo in e&longs;&longs;o appiccata. </s> <s id="id.2.1.113.5.0"> L'i&longs;te&longs;&longs;o accade parimente &longs;tando la bilancia <lb/>AB egualmente di&longs;tante dall'orizonte, ouero e&longs;&longs;endo in qual &longs;i voglia altro &longs;ito. <lb/>Rimarrà dunque la bilancia oue &longs;arà la&longs;ciata, che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.115.0.0"> <s id="id.2.1.115.1.0"><emph type="italics"/>Benche habbiamo con&longs;iderato nelle co&longs;e predette le grauezze &longs;olamente delle magni-<lb/>tudini, le quali &longs;ono po&longs;te nelle &longs;tremità della bilancia, &longs;enza la grauezza della bi-<lb/>lancia; niente di manco per e&longs;&longs;ere anche le braccia della bilancia eguali, auenir à lo <lb/>i&longs;te&longs;&longs;o alla bilancia, con&longs;iderata la &longs;ua grauezza in&longs;ieme co' pe&longs;i, ouero &longs;enza pe&longs;i, <lb/>percioche il centro iste&longs;&longs;o della grauezza &longs;enza pe&longs;i &longs;arà anco centro della grauez-<lb/>za della bilancia &longs;ola. </s> <s id="id.2.1.115.2.0"> Similmente &longs;e li pe&longs;i &longs;aranno appiccati nelle &longs;tremità del-<lb/>la bilancia, come &longs;uole far&longs;t, aùerrà l'iste&longs;&longs;o, purche le linee tirate da i punti oue &longs;o-<lb/>no attaccati i pe&longs;i ver&longs;o i centri delle grauezze, (moua&longs;i la bilancia in qual &longs;i vo-<lb/>gliamodo) vadano à concorrere nel centro del mondo, peroche doue &longs;ono attaccati <lb/>i pe&longs;i in questa maniera, iui grauano, come &longs;e in quegli &longs;te&longs;&longs;i punti baue&longs;&longs;ero i cen <lb/>tri delle grauezze. </s> <s id="id.2.1.115.3.0"> Oltre à ciò poßiamo con&longs;iderare le co&longs;e che &longs;eguono in tut-<lb/>to al modo i&longs;te&longs;&longs;o.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.116.0.0"> <s id="id.2.1.116.1.0"> <arrow.to.target n="note8"></arrow.to.target><emph type="italics"/>Ma percioche à que&longs;ta vltima conchiu&longs;ione molte co&longs;e dette da alcuni, che &longs;entono al-<lb/>tramente, paiono contra&longs;tare; però in cote&longs;ta parte egli &longs;arà bi&longs;ogno dimorare <lb/>alquanto, & &longs;econdo le mie forze non &longs;olo farò opra di difendere la propria <lb/>mia &longs;entenza, ma Archimede ancora, ilquale &longs;embra e&longs;&longs;ere &longs;tato in que&longs;to i&longs;te&longs;-<lb/>&longs;o parere.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.118.0.0"> <s id="id.2.1.118.1.0"> <margin.target id="note8"></margin.target><emph type="italics"/>Giord. </s> <s id="id.2.1.118.2.0"> de' pe&longs;i. </s> <s id="id.2.1.118.3.0"> Il Car dano della &longs;ottigliezza. </s> <s id="id.2.1.118.4.0"> Il Tartaglia de' que&longs;iti, & <expan abbr="inu&etilde;tioni">inuentioni</expan><emph.end type="italics"/> </s> </p> <pb id="p.6" xlink:href="pageimg-it/030.jpg"/> <figure id="fig4" place="text" xlink:href="figures-it/030_01.jpg"></figure> <p type="main" id="id.2.1.119.0.0"> <s id="id.2.1.119.1.0"><emph type="italics"/>Po&longs;te le co&longs;e i&longs;te&longs;&longs;e, &longs;ia <lb/>tirata la linea FCG <lb/>à piombo di AB, & <lb/>dell'orizonte: & col <lb/>centro C, & lo &longs;pa-<lb/>tio CA &longs;ia de&longs;crit-<lb/>to il cerchio ADFB <lb/>EG: &longs;aranno i punti <lb/>ADBE nella circon <lb/>ferenza del cerchio, <lb/>per e&longs;&longs;ere le braccia <lb/>della bilancia eguali. <lb/>& percioche conuen-<lb/>gono que&longs;ti autori in <lb/>vna &longs;entenza, affer-<lb/>mando, che la bilan-<lb/>cia DE non &longs;i moue <lb/>in FG, ne rimane in<emph.end type="italics"/> <lb/> <arrow.to.target n="fig4"></arrow.to.target> <lb/><emph type="italics"/>DE, maritornanellalinea AB egualmente di&longs;tante dall'orizonte, mo&longs;trerò que <lb/>&longs;ta loro opinione non potere à modo alcuno &longs;tare. </s> <s id="id.2.1.119.2.0"> Percioche &longs;e egli è vero quel <lb/>che dicono, ouero auenir à questo effetto per e&longs;&longs;ere il pe&longs;o D più graue del pe&longs;o E, <lb/>ouero &longs;e li pe&longs;i &longs;ono eguali, le di&longs;tanze nelle quali &longs;ono po&longs;ti, non &longs;aranno eguali, <lb/>cioè la CD non &longs;arà eguale alla CE, ma più grande. </s> <s id="id.2.1.119.3.0"> Ma che i pe&longs;i col-<lb/>locati in DE &longs;iano eguali, & la di&longs;tanza CD &longs;ia eguale alla di&longs;tanza CE, è <lb/>chiaro dalla pre&longs;uppo&longs;ta. </s> <s id="id.2.1.119.4.0"> Hor perche dicono che il pe&longs;o po&longs;to in D in quel &longs;i-<lb/>to è più graue del pe&longs;o po&longs;to in E nell altro &longs;ito da ba&longs;&longs;o: mentre i pe&longs;i &longs;ono in <lb/>DE, non &longs;arà il punto C piu centro della grauezza, imperoche non stanno fer-<lb/>mi &longs;e &longs;ono attaccati al C, ma &longs;arà nella linea CD per la terza del primo di At <lb/>chimede delle co&longs;e che pe&longs;ano egualmente. </s> <s id="id.2.1.119.5.0"> Non &longs;arà già nella CE per e&longs;&longs;ere il <lb/>pe&longs;o D più graue del pe&longs;o E: &longs;ia dunque in H, nelquale &longs;e &longs;aranno attacca-<lb/>ti, rimarranno. </s> <s id="id.2.1.119.6.0"> Et percioche il centro della grauezza de' pe&longs;i congiunti in AB <lb/>&longs;tà nel punto C: ma de' pe&longs;i po&longs;tiin DE il punto è H: mentre dunque i pe&longs;i <lb/>AB &longs;i muouono in DE, il centro della grauezza C moueraßi ver&longs;o D, & <lb/>s'appre&longs;&longs;er à più da vicino al D, ilche è impoßibile, per mantenere i pe&longs;i vname-<lb/>de&longs;ima di&longs;tanza fra loro: peroche il centro della grauezza di cia&longs;cun corpo &longs;tà &longs;em-<lb/>pre nel mede&longs;imo &longs;ito per ri&longs;petto al &longs;uo corpo. </s> <s id="id.2.1.119.7.0"> Et quantunque il punto C &longs;ia il<emph.end type="italics"/> <arrow.to.target n="note9"></arrow.to.target> <lb/><emph type="italics"/>centro della grauezza di due corpi A. & B, tuttauia per e&longs;&longs;ere mediante la bi-<lb/>lancia co&longs;i giunti in&longs;ieme, che &longs;empre &longs;i trouano nell'iste&longs;&longs;o modo; però il punto C<emph.end type="italics"/> <arrow.to.target n="note10"></arrow.to.target> <lb/><emph type="italics"/>&longs;arà co&longs;i centro della grauezza loro, come &longs;e fo&longs;&longs;e vna &longs;ola magnitudine; percio-<lb/>che la bilancia in&longs;ieme co' pe&longs;i fa vn &longs;olo corpo continuo, il cui centro della grauez <lb/>za &longs;empre &longs;tarà nel mezo. </s> <s id="id.2.1.119.8.0"> Non è dunque il pe&longs;o po&longs;to in D più graue del pe-<lb/>&longs;o po&longs;to in E. Che &longs;e dice&longs;&longs;ero il centro della grauezza non nella linea CD, ma<emph.end type="italics"/> <pb id="p.6v" xlink:href="pageimg-it/030.jpg"/><emph type="italics"/>nella CE douer e&longs;&longs;ere, auerrà l'i&longs;te&longs;&longs;o &longs;allo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.121.0.0"> <s id="id.2.1.121.1.0"> <margin.target id="note9"></margin.target><emph type="italics"/>Per la &longs;ecom da &longs;upposta di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.122.0.0"> <s id="id.2.1.122.1.0"> <margin.target id="note10"></margin.target><emph type="italics"/>Per la quar ta del primo di Archime de delle co&longs;e che pe&longs;ano egualmente.<emph.end type="italics"/> </s> </p> <figure id="fig5" place="text" xlink:href="figures-it/031_01.jpg"></figure> <p type="main" id="id.2.1.123.0.0"> <s id="id.2.1.123.1.0"><emph type="italics"/>Di più &longs;e il pe&longs;o D &longs;i mouer à in giù, mouer à il pe&longs;o E in sù. </s> <s id="id.2.1.123.2.0"> Adunque vn pe&longs;o <lb/>più graue di E nel mede&longs;imo &longs;ito pe&longs;erà tanto quanto il pe&longs;o D, & auerr à che <lb/>co&longs;e graui di&longs;uguali, po&longs;te in eguale distanza pe&longs;eranno egualmente. </s> <s id="id.2.1.123.3.0"> Aggiun-<lb/>ga&longs;i dunque al pe&longs;o E qualche co&longs;a graue, &longs;i &longs;attamente, che contrape&longs;i al D &longs;e <lb/>nel C &longs;aranno attac <lb/>cati. </s> <s id="id.2.1.123.4.0"> Ma e&longs;&longs;endo &longs;ta-<lb/>to di &longs;opra mo&longs;trato <lb/>il punto C e&longs;&longs;ere il cè-<emph.end type="italics"/> <lb/> <arrow.to.target n="note11"></arrow.to.target><emph type="italics"/>tro della grauezza di <lb/>pe&longs;i eguali po&longs;ti in <lb/>DE; &longs;e dunque il pe-<lb/>&longs;o. </s> <s id="id.2.1.123.5.0"> E &longs;arà più graue <lb/>del pe&longs;o D, &longs;arà anche <lb/>il centro della grauez <lb/>za nella linea CE. <lb/>& &longs;ia que&longs;to centro <lb/>il<emph.end type="italics"/> K. <emph type="italics"/>Ma per la diffi-<lb/>nitione del centro del <lb/>la grauezza, &longs;e li pe&longs;i <lb/>&longs;aranno appiccati al<emph.end type="italics"/> <lb/>K, <emph type="italics"/>staranno fermi. <lb/>Dunque &longs;e &longs;aranno<emph.end type="italics"/> <lb/> <arrow.to.target n="fig5"></arrow.to.target> <lb/><emph type="italics"/>appiccati al C, non &longs;taranno fermi, che è contra la pre&longs;uppo&longs;ta: ma il pe&longs;o E &longs;i<emph.end type="italics"/> <lb/> <arrow.to.target n="note12"></arrow.to.target><emph type="italics"/>mouer à in giù. </s> <s id="id.2.1.123.6.0"> Che &longs;e appiccati al C pe&longs;a&longs;&longs;ero ancora egualmente, na&longs;cerebbe <lb/>che di vna magnitudine, due &longs;arebbono i centri della grauezza, che è impo&longs;&longs;ibile. <lb/>Adunque il pe&longs;o po&longs;to in E più graue di quello che è in D, non pe&longs;er à tanto <lb/>quanto il D attaccando&longs;i al punto C. I pe&longs;i dunque eguali po&longs;ti in DE, attac-<lb/>cati nel centro della loro grauezza pe&longs;eranno egualmente, & &longs;taranno immobili, <lb/>che &longs;u proposto di mo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.125.0.0"> <s id="id.2.1.125.1.0"> <margin.target id="note11"></margin.target><emph type="italics"/>Per laterza del primo di Archimede delle co&longs;e che pe&longs;ano egual mente.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.126.0.0"> <s id="id.2.1.126.1.0"> <margin.target id="note12"></margin.target><emph type="italics"/>Per la prima &longs;upposta di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.127.0.0"> <s id="id.2.1.127.1.0"> <arrow.to.target n="note13"></arrow.to.target><emph type="italics"/>A que&longs;ta vltima &longs;conueneuolezza ri&longs;pondono, dicendo e&longs;&longs;ere impo&longs;&longs;ibile aggiungere al <lb/>lo E &longs;i picciolo pe&longs;o, che in ogni modo &longs;e ben &longs;i appiccano al C, il pe&longs;o E non <lb/>&longs;i moua &longs;empre in giù ver&longs;o il G. La qual co&longs;a habbiamo noi pre&longs;uppo&longs;to poter&longs;i <lb/>fare, & credeuamo poter&longs;i fare: Peroche quel che è di più del pe&longs;o D &longs;opra <lb/>il pe&longs;o E, hauendo ragione, & parte di quantità, &longs;i imaginauamo non &longs;olamente <lb/>e&longs;&longs;ere minimo, ma ancora poter&longs;i diuidere in infinito, il che eßi per certo non &longs;ola-<lb/>mente minimo, ma ne anche e&longs;&longs;ere minimo, non potendo&longs;i ritrouare, &longs;i s&longs;orzano di <lb/>mo&longs;tr are in que&longs;ta maniera.<emph.end type="italics"/> </s> </p> <pb id="p.7" xlink:href="pageimg-it/032.jpg"/> <p type="margin" id="id.2.1.129.0.0"> <s id="id.2.1.129.1.0"> <margin.target id="note13"></margin.target><emph type="italics"/>Il Tartaglia nella &longs;esta propo&longs;itione del quate libre.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.130.0.0"> <s id="id.2.1.130.1.0"><emph type="italics"/>Pongan&longs;i le co&longs;e iste&longs;&longs;e <lb/>& da i punti DE <lb/>&longs;iano tirate le linee <lb/>DHE<emph.end type="italics"/>K <emph type="italics"/>à piombo <lb/>dell'orizonte, & &longs;ia <lb/>vn'altro cerchio L <lb/>DM, il cui centro <lb/>&longs;ia N, ilquale toc<emph.end type="italics"/> <arrow.to.target n="note14"></arrow.to.target> <lb/><emph type="italics"/>chi FDG nel pun <lb/>to D, & &longs;ia eguale<emph.end type="italics"/> <arrow.to.target n="note15"></arrow.to.target> <lb/><emph type="italics"/>ad FDG. Sarà <lb/>NC linea retta: & <lb/>perche l'angolo<emph.end type="italics"/> K <lb/><emph type="italics"/>EC è eguale all'an-<lb/>golo HDN, & <lb/>l'angolo CEG è pa <lb/>rimente eguale al-<lb/>l'angolo NDM,<emph.end type="italics"/> <lb/> <arrow.to.target n="fig6"></arrow.to.target> <lb/><emph type="italics"/>peroche egli è contenuto da mezi diametri, & da circonferenze eguali: &longs;arà il re-<lb/>stante angolo & mi&longs;to<emph.end type="italics"/> K<emph type="italics"/>EG eguale al re&longs;tante angolo & mi&longs;to HDM. Et per-<lb/>cioche pre&longs;uppongono, che quanto è minore l'angolo contenuto dalla linea tirata à <lb/>piombo dell'orizonte, & dalla circonferenza, tanto in quel &longs;ito e&longs;&longs;ere anco più gra <lb/>ue il pe&longs;o. </s> <s id="id.2.1.130.2.0"> Talche &longs;i come l'angolo contenuto da HD, & dalla circonferenza <lb/>DG, è minore dell'angolo<emph.end type="italics"/> K<emph type="italics"/>EG, cioè dell'angolo HDM, co&longs;i &longs;econdo que&longs;ta <lb/>proportione il pe&longs;o po&longs;to in D &longs;ia più graue di quello che &longs;tà in E. Mala pro-<lb/>portione dell'angolo MHD all'angolo HDG è minore di qual &longs;i voglia altra <lb/>proportione, che &longs;i troui tra la maggiore, & minore quantità: Adunque la pro-<lb/>portione de i pe&longs;i DE &longs;arà la minima di tutte le proportioni, anzinon &longs;arà qua&longs;i <lb/>ne anche proportione, e&longs;&longs;endo la minima di tutte le proportioni. </s> <s id="id.2.1.130.3.0"> Che la propor-<lb/>tione di MDH ver&longs;o HDG &longs;ia di tutte la minima, mo&longs;trano con que&longs;ta ne-<lb/>ce&longs;&longs;aria ragione, peroche MHD &longs;upera HDG con angolo di linea curua, che <lb/>è MGD, ilquale angolo è il minimo di tutti gli angoli fatti di linee rette: ne po-<lb/>tendo&longs;i dare angolo minore di MGD &longs;arà la proportione di MDH ver&longs;o HDG <lb/>la minima di tutte le proportioni. </s> <s id="id.2.1.130.4.0"> Laqual ragione pare e&longs;&longs;ere grandemente friuo-<lb/>la, peroche quantunque l'angolo MDG &longs;ia di tutti gli angoli fatti di linee rette <lb/>il minore, non perciò &longs;egue totalmente egli e&longs;&longs;ere di tutti gli angoli il minimo, im-<emph.end type="italics"/> <arrow.to.target n="note16"></arrow.to.target> <lb/><emph type="italics"/>peroche &longs;ia dal punto D tirata la linea DO à piombo di NC, ambedue que-<lb/>ste toccberanno le circonferenze LDMFDG nel punto D. Ma percioche le <lb/>circonferenze &longs;ono eguali, &longs;arà l'angolo MDO misto eguale all'angolo ODG mi-<lb/>&longs;to. </s> <s id="id.2.1.130.5.0"> L'vno de gli angoli dunque, cioè ODG &longs;arà minore di MDG, cioè minore<emph.end type="italics"/> <arrow.to.target n="note17"></arrow.to.target> <lb/><emph type="italics"/>del minimo. </s> <s id="id.2.1.130.6.0"> Dapoi l'angolo ODH &longs;arà minore dell'angolo MDH. Per laqual co&longs;a <lb/>ODH haurà proportione minore all'angolo HDG, che MDH all'i&longs;te&longs;&longs;o<emph.end type="italics"/> <pb id="p.7v" xlink:href="pageimg-it/033.jpg"/><emph type="italics"/>HDG. Dara&longs;&longs;i dunque la proportione anco minore della minima, laquale mostre-<lb/>remo dauantaggio in in&longs;inito minore in questo modo. </s> <s id="id.2.1.130.7.0"> De&longs;criua&longs;i il cerchio DR, <lb/>il cui centro &longs;ia E, & il mezo diametro ED, la circonferentia DR tocche-<emph.end type="italics"/> <lb/> <arrow.to.target n="note18"></arrow.to.target><emph type="italics"/>rà la circonferenza <lb/>DG nel punto D, <lb/>& la linea DO nel<emph.end type="italics"/> <lb/> <arrow.to.target n="note19"></arrow.to.target><emph type="italics"/>punto D. Per laqual <lb/>co&longs;a minore &longs;arà l'an <lb/>golo RDG dell'an-<lb/>golo ODG, & &longs;i-<lb/>milmente l'angolo R <lb/>DH dell'angolo O <lb/>DH. Adunque ha-<lb/>uer à minore propor-<lb/>tione RDH ad HD <lb/>G di quel che haurà <lb/>ODH ad HDG. <lb/>Pigli&longs;i dapoi tra E <lb/>& C, come &longs;i vuo-<lb/>le, il punto P, dal <lb/>quale nella di&longs;tanza<emph.end type="italics"/> <lb/> <arrow.to.target n="fig7"></arrow.to.target> <lb/><emph type="italics"/>di PD &longs;i de&longs;criua vn'altra circonferenza DQ, laquale toccherà la circonferen-<lb/>tia DR, & la circonferentia DG nel punto D, & l'angolo QDH &longs;arà mi <lb/>nore dell'angolo RDH. Adunque QDH haurà proportione minore ad HDG <lb/>che RDH ad HDG, & nell'i&longs;te&longs;&longs;o modo in tutto, &longs;e tra il C & il P &longs;i tor-<lb/>rà vn'altro punto, & tra que&longs;to, & il C vn'altro, & co&longs;i &longs;ucceßiuamente &longs;i de-<lb/>&longs;criueranno infinite circonferentie tra DO, & la circonferenza DG: dalle quali <lb/>troueremo &longs;empre la proportione minore in infinito: & co&longs;i &longs;egue, che la propor-<lb/>tione del pe&longs;o po&longs;to in D al pe&longs;o po&longs;to in E non &longs;ia tanto picciola, che non &longs;i <lb/>po&longs;&longs;a ritrouarla &longs;empre minore in infinito. </s> <s id="id.2.1.130.8.0"> Et perche l'angolo MDG &longs;i puote <lb/>diuidere in infinito, &longs;i potrà anche diuidere quel più di grauezza che ha il D &longs;o-<lb/>pra lo E in infinito.<emph.end type="italics"/> </s> </p> <figure id="fig6" place="text" xlink:href="figures-it/032_01.jpg"></figure> <figure id="fig7" place="text" xlink:href="figures-it/033_01.jpg"></figure> <p type="margin" id="id.2.1.134.0.0"> <s id="id.2.1.134.1.0"> <margin.target id="note14"></margin.target><emph type="italics"/>Per la &longs;econda del terzo<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.135.0.0"> <s id="id.2.1.135.1.0"> <margin.target id="note15"></margin.target><emph type="italics"/>Per la vige&longs;imanona del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.136.0.0"> <s id="id.2.1.136.1.0"> <margin.target id="note16"></margin.target><emph type="italics"/>Per la decima ottaua del terzo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.137.0.0"> <s id="id.2.1.137.1.0"> <margin.target id="note17"></margin.target><emph type="italics"/>Per la ottaua del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.138.0.0"> <s id="id.2.1.138.1.0"> <margin.target id="note18"></margin.target><emph type="italics"/>Per la vnde cima del terzo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.139.0.0"> <s id="id.2.1.139.1.0"> <margin.target id="note19"></margin.target><emph type="italics"/>Per la decima ottaua del terzo.<emph.end type="italics"/> </s> </p> <pb id="p.8" xlink:href="pageimg-it/034.jpg"/> <p type="main" id="id.2.1.140.0.0"> <s id="id.2.1.140.1.0"><emph type="italics"/>Ne bi&longs;ogna trala&longs;ciare, che <lb/>eglino hanno pre&longs;upp o&longs;to <lb/>nella demo&longs;tratione l'ango <lb/>lo<emph.end type="italics"/> K<emph type="italics"/>EG e&longs;&longs;er maggiore del <lb/>l'angolo HDC, come co <lb/>&longs;a nota: il che ben è vero &longs;e <lb/>DHE<emph.end type="italics"/>K <emph type="italics"/>&longs;ono fra loro e-<lb/>gualmente di&longs;tanti. </s> <s id="id.2.1.140.2.0"> Ma <lb/>percioche, come eßi pari-<lb/>mente pre&longs;uppongono, le <lb/>linee DHE<emph.end type="italics"/>K <emph type="italics"/>&longs;i vanno à <lb/>trouare nel centro del mon <lb/>do, le linee DHE<emph.end type="italics"/>K <emph type="italics"/>non <lb/>&longs;aranno egualmente di&longs;tan <lb/>ti giamai, et <expan abbr="l'ãgolo">l'angolo</expan><emph.end type="italics"/> K<emph type="italics"/>EG <lb/>non &longs;olo non &longs;arà maggio-<lb/>re dall'angolo HDG, ma <lb/>minore. </s> <s id="id.2.1.140.3.0"> Come per gra-<lb/>tia di e&longs;&longs;empio, &longs;ia tirata la <lb/>linea FG &longs;in al centro del <lb/>mondo, che &longs;ia S, & con <lb/>giungan&longs;i DS ES. Egli <lb/>è da mostrare l'angolo SE <lb/>G e&longs;&longs;ere minore dell'ango <lb/>lo SDG. Tiri&longs;i dal punto <lb/>E la linea ET, che toc-<lb/>chi il cerchio DGEF, & <lb/>dall'i&longs;te&longs;&longs;o punto &longs;ia tirata <lb/>la EV egualmente di&longs;tan<emph.end type="italics"/> <lb/> <arrow.to.target n="fig8"></arrow.to.target> <lb/><emph type="italics"/>te da DS: Percioche dunque EVDS &longs;ono traloro egualmente di&longs;tanti, &longs;imil-<lb/>mente ET DO &longs;ono egualmente di&longs;tanti: &longs;arà l'angolo VET eguale all'ango-<lb/>lo SDO: & l'angolo TEG eguale all'angolo ODM, per e&longs;&longs;ere contenuto da <lb/>linee toccanti la circonferenza, & da circonferenze eguali. </s> <s id="id.2.1.140.4.0"> Tutto l'angolo dun-<lb/>que VEG &longs;arà eguale all'angolo SDM. Leui&longs;i via dall'angolo SDM l'ango <lb/>lo di linee curue MDG: & dall'angolo VEG leui&longs;i via l'angolo VES, & <lb/>l'angolo VES fatto di linee rette è maggiore dell'angolo MDG fatto di linee <lb/>curue; &longs;arà il re&longs;tante angolo SEG minore dell'angolo SDG. Per laqual co&longs;a <lb/>dalle pre&longs;uppo&longs;te loro non &longs;olo il pe&longs;o posto in D &longs;arà più graue del pe&longs;o po&longs;to <lb/>in E, ma per lo contrario il pe&longs;o E &longs;arà più graue dell'i&longs;te&longs;&longs;o D.<emph.end type="italics"/> </s> </p> <figure id="fig8" place="text" xlink:href="figures-it/034_01.jpg"></figure> <pb id="p.8v" xlink:href="pageimg-it/035.jpg"/> <p type="main" id="id.2.1.143.0.0"> <s id="id.2.1.143.1.0"><emph type="italics"/>Producono tutta via <lb/>ragioni con le quali <lb/>&longs;i sforzano di mo-<lb/>&longs;trare, che la bilan-<lb/>cia DE ritorna per <lb/>neceßità in AB e-<lb/>gualmente distante <lb/>dall'orizonte. </s> <s id="id.2.1.143.2.0"> Pri-<lb/>ma dimo&longs;trano l'i-<lb/>&longs;te&longs;&longs;o pe&longs;o e&longs;&longs;ere più <lb/>graue in A, che <lb/>in altro &longs;ito, che <lb/>chiamano &longs;ito della <lb/>egualità, e&longs;&longs;endo la <lb/>linea AB egual-<lb/>mente di&longs;tante dal-<lb/>l'orizonte. </s> <s id="id.2.1.143.3.0"> Da-<lb/>poi quanto è più da<emph.end type="italics"/> <lb/> <arrow.to.target n="fig9"></arrow.to.target> <lb/><emph type="italics"/>pre&longs;&longs;o allo A, tanto e&longs;&longs;ere piu graue di qual &longs;i voglia altro più da lontano, cioè <lb/>il pe&longs;o po&longs;to in A e&longs;&longs;ere più graue, che in D; & in D, che in L: & &longs;imil-<lb/>mente in A più graue, che in N; & in N più graue, che in M. Con&longs;ide-<emph.end type="italics"/> <lb/> <arrow.to.target n="note20"></arrow.to.target><emph type="italics"/>rando &longs;olamente vn pe&longs;o in vno delle braccia in sù, ouero in giù mo&longs;&longs;o. </s> <s id="id.2.1.143.4.0"> Percio-<lb/>che dicono, po&longs;ta la trutina della bilancia in CF, il pe&longs;o me&longs;&longs;o in A è più lunge <lb/>dalla trutina che in D; & in D più lunge, che in L: peroche tirate le linee DO<emph.end type="italics"/> <lb/> <arrow.to.target n="note21"></arrow.to.target><emph type="italics"/>LP à piombo di CF, la linea AC re&longs;ta maggiore di DO, & DO di e&longs;&longs;a LP, <lb/>& auiene l'i&longs;te&longs;&longs;o ne i punti NM. Dapoi dicono da qual luogo il pe&longs;o &longs;i mo-<lb/>ue più velocemente, iui è più graue: ma egli &longs;i moue più velocemente dallo<emph.end type="italics"/> <lb/> <arrow.to.target n="note22"></arrow.to.target><emph type="italics"/>A, che da altro &longs;ito; adunque egli è più graue nello A. Con &longs;imile mo-<lb/>do, quanto più egli è da pre&longs;&longs;o allo A, tanto più velocemente &longs;i moue: <lb/>adunque nel D &longs;arà più graue, che in L. L'altra cagione poi che cauano dal mo-<emph.end type="italics"/> <lb/> <arrow.to.target n="note23"></arrow.to.target><emph type="italics"/>uimento più diritto, & più torto è, che quanto il pe&longs;o di&longs;cende più diritto in archi <lb/>eguali, pare e&longs;&longs;er anco più graue; concio&longs;ia che il pe&longs;o e&longs;&longs;endo libero, & &longs;ciolto, &longs;i <lb/>moua di &longs;ua propria natura per lo diritto; ma in A egli di&longs;cende più dirittamen<emph.end type="italics"/> <lb/> <arrow.to.target n="note24"></arrow.to.target><emph type="italics"/>te; dunque in A &longs;arà più graue, & dimo&longs;trano ciò pigliando l'arco AN egua-<lb/>le all'arco LD. & da i punti NL &longs;iano tirate le linee NRLQ egualmente di-<lb/>&longs;tanti dalla linea FG, laquale chiamano anche della direttione; & quelle altre &longs;e-<lb/>gheranno le linee ABDO in QR, & dal punto N &longs;ia tirata la NT à piombo <lb/>di FG: Dimo&longs;trano veramente LQ e&longs;&longs;ere eguale à PO, & NR ad e&longs;&longs;a CT, <lb/>& la linea NR e&longs;&longs;er maggiore di LQ. Hor percioche la di&longs;ce&longs;a del pe&longs;o dallo A <lb/>fin ad N per la circonferentia di AN trapa&longs;&longs;a maggior parte della linea FG, <lb/>(che eßi chiamano pigliare di diritto) che la di&longs;ce&longs;a di L in D per la circonferenza <lb/>LD; concio&longs;ia che la di&longs;ce&longs;a AN trapaßi la linea CT, ma la di&longs;ce&longs;a LD la linea<emph.end type="italics"/> <pb id="p.9" xlink:href="pageimg-it/036.jpg"/><emph type="italics"/>PO, & CT è maggiore di PO, la di&longs;ce&longs;a di AN &longs;arà più diritta, che la di-<lb/>&longs;ce&longs;a di LD: &longs;arà dunque più graue il pe&longs;o po&longs;to in A, che in L, ouero in qual <lb/>&longs;i voglia altro &longs;ito, & nell'i&longs;te&longs;&longs;o modo dimo&longs;trano, che quanto il pe&longs;o è più vicino <lb/>allo A, è più graue; cioè &longs;iano le circonferenze LD DA traloro eguali, & <lb/>dal punto D &longs;ia tirata la linea DR à piombo di AB; &longs;arà la DR eguale al-<emph.end type="italics"/> <arrow.to.target n="note25"></arrow.to.target> <lb/><emph type="italics"/>la CO. & dimo-<lb/>&longs;trano po&longs;cia, che <lb/>la linea DR è mag<lb/>giore della LQ, & <lb/>dicono che la &longs;ce&longs;a <lb/>di DA prende più <lb/>di &longs;ce&longs;a diritta, che <lb/>non fa LD, pe-<lb/>roche è maggiore <lb/>la linea CO, che <lb/>la OT: Per la-<lb/>qual co&longs;a il pe&longs;o &longs;a <lb/>rà più graue in D, <lb/>che in L, ilche pa<lb/>rimente auiene ne <lb/>punti NM. & <lb/>co&longs;i il pre&longs;uppo&longs;to, <lb/>per loquale dimo-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig10"></arrow.to.target> <lb/><emph type="italics"/>&longs;trano la bilancia DE ritornare in AB a&longs;fermano come noto, & manife&longs;to; cioè<emph.end type="italics"/> <arrow.to.target n="note26"></arrow.to.target> <lb/><emph type="italics"/>che &longs;econdo il &longs;ito il pe&longs;o è tanto più graue, quanto nel mede&longs;imo &longs;ito manco tor-<lb/>ta è la &longs;ce&longs;a: & la cagione di cotal ritorno dicono e&longs;&longs;ere que&longs;ta; peroche la &longs;ce&longs;a del <lb/>pe&longs;o po&longs;to in D è più diritta della &longs;ce&longs;a del pe&longs;o po&longs;to in E, per pigliare il pe&longs;o<emph.end type="italics"/> <arrow.to.target n="note27"></arrow.to.target> <lb/><emph type="italics"/>di E manco della direttione in de&longs;cendendo che non fa il pe&longs;o di D pur nel di&longs;cen <lb/>dere: Come &longs;e l'arco EV &longs;ia eguale à DA, & &longs;iano tirate VHET à piom<lb/>bo di FG; &longs;arà maggiore DR di TH. Per laqual co&longs;a per la pre&longs;uppo&longs;ta il pe<emph.end type="italics"/> <arrow.to.target n="note28"></arrow.to.target> <lb/><emph type="italics"/>&longs;o me&longs;&longs;o in D per ri&longs;petto al &longs;ito &longs;arà più graue del pe&longs;o me&longs;&longs;o in E. Adunque <lb/>il pe&longs;o me&longs;&longs;o in D e&longs;&longs;endo più graue &longs;i mouerà in giù, & il pe&longs;o po&longs;to in E in <lb/>&longs;u fin che la bilancia DE ritorni in AB.<emph.end type="italics"/> </s> </p> <figure id="fig9" place="text" xlink:href="figures-it/035_01.jpg"></figure> <figure id="fig10" place="text" xlink:href="figures-it/036_01.jpg"></figure> <p type="margin" id="id.2.1.146.0.0"> <s id="id.2.1.146.1.0"> <margin.target id="note20"></margin.target><emph type="italics"/>Il Cardano nel primo della &longs;ottigliezza.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.147.0.0"> <s id="id.2.1.147.1.0"> <margin.target id="note21"></margin.target><emph type="italics"/>Giordano nella quarta propo&longs;itione<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.148.0.0"> <s id="id.2.1.148.1.0"> <margin.target id="note22"></margin.target><emph type="italics"/>Il Tartaglia nella quinta propo&longs;itione.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.149.0.0"> <s id="id.2.1.149.1.0"> <margin.target id="note23"></margin.target><emph type="italics"/>Il Cardano. </s> <s id="id.2.1.149.2.0"> Giordano al la propo&longs;itio ne quarta.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.150.0.0"> <s id="id.2.1.150.1.0"> <margin.target id="note24"></margin.target><emph type="italics"/>Il Tartaglia alla pro po&longs;itione<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.151.0.0"> <s id="id.2.1.151.1.0"> <margin.target id="note25"></margin.target><emph type="italics"/>Per la trige &longs;imaquarta del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.152.0.0"> <s id="id.2.1.152.1.0"> <margin.target id="note26"></margin.target><emph type="italics"/>Giordane nella quarta pre &longs;apoosta<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.153.0.0"> <s id="id.2.1.153.1.0"> <margin.target id="note27"></margin.target><emph type="italics"/>Giordano nella &longs;econda propo&longs;itione.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.154.0.0"> <s id="id.2.1.154.1.0"> <margin.target id="note28"></margin.target><emph type="italics"/>Il Tartaglia nella quinta propo&longs;icione.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.155.0.0"> <s id="id.2.1.155.1.0"><emph type="italics"/>L'altra ragione ancora di que&longs;to ritorno è, che <expan abbr="quãdo">quando</expan> la trutina della bilancia è &longs;opra<emph.end type="italics"/> <arrow.to.target n="note29"></arrow.to.target> <lb/><emph type="italics"/>dilei in CF; la linea CG è la meta: & percio che l'angolo GCD è maggiore <lb/>dell'angolo GCE, & l'angolo maggiore dalla meta rende più graue il pe&longs;o: adun-<lb/>que &longs;tando la trutina della bilancia di &longs;opra &longs;arà più graue il pe&longs;o in D, che in E, <lb/>& perciò il D ritorner à nello A, & lo E nel B.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.156.0.0"> <s id="id.2.1.156.1.0"> <margin.target id="note29"></margin.target><emph type="italics"/>Il Cardano.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.157.0.0"> <s id="id.2.1.157.1.0"> Meta è pur voce Latina co&longs;tumata da gli antichi ne i giuo chi, & conte&longs;e fatte ne i cer <lb/>chi murati, & ne i Theatri, percio che il principio, oue &longs;i dauano le mo&longs;&longs;e a' corri-<lb/>tori, &longs;i chiamaua Carcere, & il fine Meta; di modo, che meta viene à dire termine <lb/>& fine: & piu in altro &longs;ignificato il luogo piu ba&longs;&longs;o, & in&longs;imo. </s> <s id="id.2.1.157.2.0"> Hor qui &longs;i puote <pb id="p.9v" xlink:href="pageimg-it/037.jpg"/>intendere ad ambidue i modi, cioè che la linea CG &longs;ia la meta, cioè il termine <lb/>& fine, nelquale ha da peruenire il pe&longs;o collo cato nella bilancia; ouero il luogo <lb/>infimo della circonferenza, alquale capita il pe&longs;o per natura. </s> <s id="id.2.1.157.3.0"> Doue &longs;criue l'Auto <lb/>re l'angolo maggiore dalla Meta, vuol dire l'angolo, che fa il braccio della bilan-<lb/>cia con la Meta CG. </s> </p> <p type="main" id="id.2.1.158.0.0"> <s id="id.2.1.158.1.0"><emph type="italics"/>Et co&longs;i <expan abbr="cõ">com</expan> que&longs;te ragioni &longs;i sforzano dimo&longs;trare la bilancia DE ritornare in AB; le <lb/>quali al parer mio &longs;i po&longs;&longs;ono ageuolmente &longs;oluere.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.159.0.0"> <s id="id.2.1.159.1.0"><emph type="italics"/>Primieramente dunque in quanto s'appartiene alle ragioni, che dicono il pe&longs;o me&longs;&longs;o <lb/>in A e&longs;&longs;ere piu graue, che in altro &longs;ito, lequali cauano dalla di&longs;tanza piu da lonta-<lb/>no, & piu da pre&longs;&longs;o della linea FG, & dal mouimento piu veloce, & piu diritto <lb/>dal punto A. In prima non dimo&longs;trano veramente perche il pe&longs;o &longs;i moua piu velo-<lb/>cemente dallo A, che da al tro &longs;ito. </s> <s id="id.2.1.159.2.0"> ne perche &longs;ia maggiore CA di DO, & DO <lb/>di LP, per que&longs;to, come per vera cagione, &longs;egue il pe&longs;o po&longs;to in A e&longs;&longs;ere piu gra-<lb/>ue di quello, che è in D, & quello di D, di quel che &longs;tà in L, percioche non &longs;i queta <lb/>l'intelletto, &longs;e di ciò altra cagione non &longs;i dimo&longs;tra, parendo &longs;egno piu to&longs;to, che vera <lb/>cagione. </s> <s id="id.2.1.159.3.0"> Quello ste&longs;&longs;o accade parimente all'altra ragione, laquale adducono dal <lb/>mouimento piu diritto, & piu torto. </s> <s id="id.2.1.159.4.0"> Oltre à ciò tutte quelle co&longs;e, che per&longs;uadono <lb/>per via del <expan abbr="mouim&etilde;">mouimen</expan> <lb/>to piu veloce, & <lb/>piu tardo il pe&longs;o in <lb/>A e&longs;&longs;ere piu graue, <lb/>che in D, non per-<lb/>ciò dimo &longs;trano, che <lb/>il pe&longs;o in A, in <expan abbr="quã">quam</expan> <lb/>to è in A, &longs;ia piu <lb/>graue del pe&longs;o D, in <lb/>quanto è in D, ma <lb/>in quanto &longs;i parte <lb/>da i punti DA. <lb/>Onde, <expan abbr="auãti">auanti</expan> che piu <lb/>oltre &longs;i proceda, pri <lb/>ma dimo&longs;trerò, che <lb/>il pe&longs;o quanto egli <lb/>è piu da pre&longs;&longs;o ad <lb/>FG manco graua, <lb/>&longs;i in quanto egli &longs;tà <lb/>nel &longs;ito, oue &longs;i ritro<emph.end type="italics"/> <lb/> <arrow.to.target n="fig11"></arrow.to.target> <lb/><emph type="italics"/>ua, come anche in quanto &longs;i parte da quello: & in&longs;ieme, che egli è fal&longs;o il pe&longs;o e&longs;&longs;ere <lb/>piu graue in A, che in altro &longs;ito.<emph.end type="italics"/> </s> </p> <figure id="fig11" place="text" xlink:href="figures-it/037_01.jpg"></figure> <p type="main" id="id.2.1.161.0.0"> <s id="id.2.1.161.1.0"><emph type="italics"/>Tiri&longs;i la FG fin al centro del mondo, che &longs;ia in S, & dal punto S tiri&longs;i anco vna linea, <lb/>che tocchi il cerchio AFBG. non potrà già questa linea tirata dal punto S toc-<lb/>care il cerchio nel punto A; imperoche tirata la linea AS, il triangolo ACS ver<emph.end type="italics"/> <pb id="p.10" xlink:href="pageimg-it/038.jpg"/><emph type="italics"/>rebbe ad hauere due angoli retti, cioè SAC, & ACS, che è impoßibile: ne me <lb/>no toccherà &longs;opra il punto A nella circonferenza AF; peroche &longs;egherebbe il cer-<emph.end type="italics"/> 2<arrow.to.target n="note30"></arrow.to.target> <lb/><emph type="italics"/>chio. </s> <s id="id.2.1.161.2.0"> Toccherà dunque &longs;otto, & &longs;ia SO: &longs;iano dapoi congiunte le lince SD SL, <lb/>lequali &longs;eghino la circonferenza AOG ne' punti<emph.end type="italics"/> K<emph type="italics"/>H, & &longs;iano ancho congiunte le <lb/>linee C<emph.end type="italics"/>K <emph type="italics"/>CH. Et percioche il pe&longs;o, quanto egli è piu da pre&longs;&longs;o di F, tanto piu an-<lb/>co &longs;tà &longs;opra il centro; come il pe&longs;o in D preme, & &longs;tà piu &longs;opra il punto del volgi-<lb/>mento C, come à centro, cioè in D piu graua &longs;opra la linea CD, che &longs;e egli fo&longs;&longs;e in A <lb/>&longs;opra la linea CA: & dauantaggio piu in L &longs;opra la linea CL. imperoche e&longs;&longs;endo <lb/>li tre angoli di cia&longs;cun triangolo eguali à due angoli retti, & l'angolo DC<emph.end type="italics"/>K <emph type="italics"/>del <lb/>triangolo DC<emph.end type="italics"/>K, <emph type="italics"/>che è di due lati eguali &longs;ia <lb/>minore dell'angolo LCH del <expan abbr="triãgolo">triangolo</expan> LCH, <lb/>che è pur di due lati eguali: &longs;aranno gli altri <lb/>alla ba&longs;e, cioè CD<emph.end type="italics"/>K <emph type="italics"/>C<emph.end type="italics"/>K<emph type="italics"/>D in&longs;ieme pre&longs;i <lb/>maggiori de gli altri CLH CHL; & le <lb/>metà di que&longs;ti, cioè l'angolo CDS &longs;arà mag <lb/>giore dell'angolo CLS. E&longs;&longs;endo adunque <lb/>CLS minore, la linea CL piu &longs;i acco&longs;terà <lb/>al mouimento naturale del pe&longs;o me&longs;&longs;o in L <lb/>del tutto &longs;ciolto; cioè à dire alla linea LS, <lb/>che CD al mouimento DS: percioche il pe <lb/>&longs;o po&longs;to in L libero, & &longs;ciolto &longs;i mouerebbe <lb/>ver&longs;o il centro del mondo per LS, & il pe-<lb/>&longs;o po&longs;to in D per DS. Ma perche il pe&longs;o <lb/>me&longs;&longs;o in L graua tutto &longs;opra LS, & quello <lb/>che è in D &longs;opra DS, il pe&longs;o in L grauerà <lb/>pu &longs;opra la linea CL, che quello, che &longs;tà in <lb/>D &longs;opra la linea DC. Adunque la linea <lb/>CL &longs;o&longs;terrà piu il pe&longs;o, che lalinea CD, & <lb/>nel modo iste&longs;&longs;o quanto piu il pe&longs;o &longs;arà da <lb/>pre&longs;&longs;o ad F, &longs;i dimo strerà piu e&longs;&longs;er &longs;o&longs;tenuto <lb/>dalla linea CL per cotesta cagione, peroche <lb/>&longs;empre l'angolo CLS &longs;arebbe minore, la-<lb/>qual co&longs;a etiandio èmanife&longs;ta; perche &longs;e le li <lb/>nee CL, & LS s'incontra&longs;&longs;ero in vna li <lb/>nea, ilche auiene in FCS, all'hora la linea <lb/>CF &longs;o&longs;terrebbe tutto il pe&longs;o, che è in F, & <lb/>lo renderebbe immobile, nè haurebbe niuna <lb/>grauezza in tutto nella circonferenza del cer <lb/>chio. </s> <s id="id.2.1.161.3.0"> Li&longs;te&longs;&longs;o pe&longs;o dunque per la diuer&longs;ità<emph.end type="italics"/> <lb/> <arrow.to.target n="fig12"></arrow.to.target> <lb/><emph type="italics"/>de' &longs;iti &longs;arà piu graue, & piu lieue. </s> <s id="id.2.1.161.4.0"> & que&longs;to non già percio che per ragione del &longs;ito <lb/>alcuna volta egli acqui&longs;ti veramente grauezza maggiore, & alcuna volta la perda, <lb/>e&longs;&longs;endo &longs;empre della i&longs;te&longs;&longs;a grauezza, troui&longs;i douunque &longs;i voglia: ma percioche egli<emph.end type="italics"/> <pb id="p.10v" xlink:href="pageimg-it/039.jpg"/><emph type="italics"/>graua piu, & meno nella circonferenza, come in D piu graua &longs;opra la circonferenza <lb/>DA, che in L &longs;opra la circonferenza LD: cioè &longs;e il pe&longs;o &longs;arà &longs;o&longs;tenuto dalle circon <lb/>ferenze, & dalle linee diritte; la circonferenza AD &longs;o&longs;terrà piu il pe&longs;o po&longs;to in D, <lb/>che la circonferenza DL, &longs;tando il pe&longs;o in L; peroche meno aiuta CD, che CL. <lb/>Oltre à ciò quando il pe&longs;o è in L, &longs;e egli fo&longs;&longs;e del tutto libero & &longs;ciolto, &longs;i mouerebbe <lb/>in giu per LS, &longs;e non gliene fu&longs;&longs;e vietato dalla linea CL, laquale sforza il pe&longs;o po&longs;to <lb/>in L à mouer&longs;i oltre la linea LS per la circonferenza LD, & lo caccia in certo mo <lb/>do, & in cacciandolo viene in parte à &longs;o&longs;tenerlo; percioche &longs;e non lo &longs;o&longs;tene&longs;&longs;e, & <lb/>gli face&longs;&longs;e re&longs;i&longs;tenza, &longs;i mouerebbe in giu per la linea LS, ma non già per la cir-<lb/>conferenza LD. Similmente la CD fa re&longs;i&longs;tenza al pe&longs;o po&longs;to in D, sforzan-<lb/>dolo à mouer&longs;i per la circonferenza DA. Nell'iste&longs;&longs;o modo &longs;tando il pe&longs;o in A, <lb/>la linea CA con&longs;tringerà il pe&longs;o à mouer&longs;i <lb/>oltre la linea AS per la circoferenza AO; <lb/>peroche l'angolo CAS è acuto, e&longs;&longs;endo lo <lb/>angolo ACS retto. </s> <s id="id.2.1.161.5.0"> Adunque le linee <lb/>CA CD in qualche parte, ma non già e-<lb/>gualmente fanno re&longs;istenza al pe&longs;o. </s> <s id="id.2.1.161.6.0"> & qua <lb/>lunque volta l'angolo, che è nella circonfe-<lb/>renza del cerchio fatto dalle linee che e&longs;cono <lb/>dal centro del monde S, & dal centro C &longs;a-<lb/>rà acuto, dimo&longs;treremo auenire l'i&longs;te&longs;&longs;o. </s> <s id="id.2.1.161.7.0"> Hor <lb/>percioche l'angolo mi&longs;to CLD è eguale à <lb/>l'angolo CDA, per e&longs;&longs;ere conteuuto da <lb/>mezi diametri, & dall'i&longs;te&longs;&longs;a circonferenza; <lb/>& l'angolo CLS è minore dell'angolo <lb/>CDS; &longs;arà il reflante SLD maggiore <lb/>del re&longs;tante SDA. Per laqual co&longs;a la cir <lb/>conferenza DA, cioè la di&longs;ce&longs;a del pe&longs;o <lb/>in D &longs;ara piu da pre&longs;&longs;o al mouimento natu-<lb/>rale del pe&longs;o &longs;ciolto me&longs;&longs;o in D, cioè della li-<lb/>nea DS, che la circonferenza LD della <lb/>linea LS. Meno dunque farà re&longs;i&longs;tenza la <lb/>linea CD al pe&longs;o po&longs;to in D, che la linea <lb/>CL al pe&longs;o po&longs;to in L. Però la linea CD <lb/>&longs;o&longs;terrà meno, che CL, & il pe&longs;o &longs;arà <lb/>piu libero in D, che in L: mouendo&longs;i piu <lb/>naturalmente il pe&longs;o per DA, che per LD. <lb/>Per laqual co&longs;a piu graue &longs;arà in D, che in <lb/>L. Similmente dimo&longs;treremo, che CA man <lb/>co&longs;o&longs;tiene, che CD & che il pe&longs;o piu in A, <lb/>che in D è libero, & piu graue. </s> <s id="id.2.1.161.8.0"> Dopo dalla<emph.end type="italics"/> <lb/> <arrow.to.target n="fig13"></arrow.to.target> <lb/><emph type="italics"/>parte di &longs;otto per l'i&longs;te&longs;&longs;e cagioni, quanto il pe&longs;o &longs;arà piu da pre&longs;&longs;o al G, &longs;arà piu ri-<emph.end type="italics"/> <pb id="p.11" xlink:href="pageimg-it/040.jpg"/><emph type="italics"/>tenuto, come in H dalla linea CH, che in<emph.end type="italics"/> K <emph type="italics"/>dalla linea C<emph.end type="italics"/>K: <emph type="italics"/>percioche e&longs;&longs;en<emph.end type="italics"/> <arrow.to.target n="note31"></arrow.to.target> <lb/><emph type="italics"/>do l'angolo CHS maggiore dell'angolo C<emph.end type="italics"/>K<emph type="italics"/>S, le linee CH HS, &longs;i acco&longs;te-<lb/>ranno piu alla direttione, che C<emph.end type="italics"/>K K<emph type="italics"/>S. & per que&longs;to &longs;arà piu ritenuto il pe&longs;o da <lb/>CH, che da C<emph.end type="italics"/>K; <emph type="italics"/>percioche &longs;e CH HS &longs;i incontra&longs;&longs;ero in vna linea, come auie-<lb/>ne &longs;tando il pe&longs;o in G, allbora la linea CG &longs;o&longs;terrebbe tutto il pe&longs;o in G, per <lb/>modo che &longs;tarebbe immobile. </s> <s id="id.2.1.161.9.0"> Quanto minore dunque &longs;arà l'angolo contenuto dal <lb/>la linea CH, & dalla di&longs;ce&longs;a del pe&longs;o &longs;ciolto, cioè dalla linea HS, tanto meno <lb/>anco quella linea ritenirà il pe&longs;o, & doue &longs;arà manco ritenuto, iui &longs;arà piu libero, & <lb/>piu graue. </s> <s id="id.2.1.161.10.0"> Oltre à ciò &longs;e il pe&longs;o fo&longs;&longs;e libero in K, & &longs;ciolto, &longs;i mouerebbe per la li-<lb/>nea KS, ma egli è impedito dalla linea CK, laquale sforza il pe&longs;o a mouer&longs;i di <lb/>qua dalla linea KS per la circonferenza KH; percio che lo ritira in certo modo, <lb/>& in ritirandolo viene a &longs;o&longs;tenerlo, peroche &longs;e non lo &longs;o&longs;tene&longs;&longs;e, &longs;i mouerebbe il pe-<lb/>&longs;o in giu per la linea diritta KS, ma non per la circonferenza KH. Similmente <lb/>la CH ritiene il pe&longs;o, sforzandolo a mouer&longs;i per la circonferenza HG. Et percio-<lb/>che l'angolo CHS è maggiore dell'angolo CKS, leuati via gli angoli eguali <lb/>CHG, C<emph.end type="italics"/>K<emph type="italics"/>H, &longs;arà il re&longs;tante SHG maggiore del re&longs;tante S<emph.end type="italics"/>K<emph type="italics"/>H. Adunque <lb/>la circonferenza KH, cioè la di&longs;ce&longs;a del pe&longs;o po&longs;to in K &longs;arà piu da pre&longs;&longs;o al mo-<lb/>uimento naturale del pe&longs;o po&longs;to in K &longs;ciolto, cioè alla linea KS, che la circonfe-<lb/>renza HG alla linea HS. Per laqual co&longs;a meno ritiene la linea CK, che CH, <lb/>mouend o&longs;i il pe&longs;o piu naturalmente per KH, che per HG, Con ragione &longs;imile <lb/>anco &longs;i mo&longs;trerà, che quanto minore &longs;arà l'angolo SKH, la linea CK &longs;o&longs;terrà <lb/>meno. </s> <s id="id.2.1.161.11.0"> Stando dunque il pe&longs;o in O, percioche l'angolo SOC non &longs;olamente è <lb/>minore dell'angolo CKS, ma anco il minimo di tutti gli angoli, che e&longs;con da i pun <lb/>ti CS, & hanno la cima nella circonferenza OKG; &longs;arà l'angolo SOK il mi <lb/>nimo &longs;i dell'angolo SKH, come de tutti gli altri co&longs;i fatti. </s> <s id="id.2.1.161.12.0"> Adunque la di&longs;ce&longs;a <lb/>del pe&longs;o po&longs;to in O &longs;arà piu da pre&longs;&longs;o al mouimento naturale di e&longs;&longs;o pe&longs;o &longs;ciolto in <lb/>O, che in altro &longs;ito della circonferenza OKG: & la linea CO meno &longs;o&longs;tenirà <lb/>il pe&longs;o, che &longs;e egli fo&longs;&longs;e in qual &longs;i voglia altro &longs;ito della i&longs;te&longs;&longs;a circonferenza OG.<lb/> Similmente perche l'angolo del toccamento SOK è minore &longs;i dell'angolo SDA, <lb/>&longs;i dello SAO, & &longs;i di qual &longs;i voglia altro &longs;imile; &longs;arà la &longs;ce&longs;a del pe&longs;o me&longs;&longs;o in O <lb/>piu da pre&longs;&longs;o al mouimento naturale di e&longs;&longs;o pe&longs;o &longs;ciolto in O, che in altro &longs;ito del-<lb/>la <expan abbr="circõfer&etilde;za">circonferenza</expan> ODF. Oltre a ciò perche la linea CO no puote &longs;pingere il pe&longs;o po&longs;to <lb/>in O mentre egli &longs;i moue in giu, per modo che egli &longs;i moua oltre la linea OS, per <lb/>cioche la linea OS non taglia il cerchio, ma lo tocca; & l'angolo SOC è retto <lb/>& non acuto, il pe&longs;o po&longs;to in O non grauerà niente &longs;opra la linea CO, ne &longs;tarà <lb/>&longs;opra il centro, come accaderebbe in qual &longs;i voglia altro punto &longs;opra l'O. Sarà dun-<lb/>que il pe&longs;o po&longs;to in O per que&longs;te cagioni libero, & &longs;ciolto piu in que&longs;to &longs;ito, che in <lb/>qual &longs;i voglia altro della circonferenza FOG; & perciò in que&longs;to &longs;arà piu graue, <lb/>cioè a dire piu grauerà, che in altro &longs;ito. </s> <s id="id.2.1.161.13.0"> Et quanto &longs;arà piu da pre&longs;&longs;o ad O, &longs;arà <lb/>piu graue di quello, che &longs;e fo&longs;&longs;e piu da lunge: & la linea CO &longs;arà egualmente di-<lb/>&longs;tante dall'orizonte: non pero all'orizonte del punto C (come &longs;timano e&longs;&longs;i) ma <lb/>del pe&longs;o po&longs;to in O, douendo&longs;i prendere l'orizonte dal centro della grauezza del pe <lb/>&longs;o. </s> <s id="id.2.1.161.14.0"> Lequali co&longs;e tutte bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig12" place="text" xlink:href="figures-it/038_01.jpg"></figure> <figure id="fig13" place="text" xlink:href="figures-it/039_01.jpg"></figure> <p type="margin" id="id.2.1.165.0.0"> <s id="id.2.1.165.1.0"> <margin.target id="note30"></margin.target><emph type="italics"/>Per la decimaottaua del terzo<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.166.0.0"> <s id="id.2.1.166.1.0"> <margin.target id="note31"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 21. <emph type="italics"/>del prim.<emph.end type="italics"/> </s> </p> <pb id="p.11v" xlink:href="pageimg-it/041.jpg"/> <p type="main" id="id.2.1.167.0.0"> <s id="id.2.1.167.1.0"><emph type="italics"/>Ma &longs;e il braccio della bilancia fo&longs;&longs;e maggiore <lb/>di CO, come per la quantità di CD; <lb/>&longs;arà parimente il pe&longs;o me&longs;&longs;o in O piu gra-<lb/>ue. </s> <s id="id.2.1.167.2.0"> De&longs;criua&longs;i il cerchio OH, il cui <lb/>centro &longs;ia D, & il mezo diametro DO. <lb/>il cerchio OH toccherà il cerchio FOG <lb/>nel punto O, & toccherà anche la linea<emph.end type="italics"/> <lb/> <arrow.to.target n="note32"></arrow.to.target><emph type="italics"/>OS nel punto mede&longs;imo, laquale è la &longs;ce-<lb/>&longs;a naturale, & diritta del pe&longs;o po&longs;to in O.<emph.end type="italics"/> <lb/> <arrow.to.target n="note33"></arrow.to.target> </s> <s id="id.2.1.167.3.0"><emph type="italics"/>Et percioche l'angolo SOH è minore del <lb/>l'angolo SOG, &longs;arà la &longs;ce&longs;a del pe&longs;o po&longs;to <lb/>in O per la circonferenza OH piu dapre&longs; <lb/>&longs;o al mouimento naturale OS, che per la <lb/>circonferenza OG. </s> <s id="id.2.1.167.4.0"> Piu libero dunque <lb/>& &longs;ciolto, & per con&longs;equente piu graue &longs;a-<lb/>ràin O, &longs;tante il centro della bilancia in <lb/>D, che in C. </s> <s id="id2.1.167.5.0"> Similmente &longs;i mo&longs;trerà, <lb/>che quanto piu grande &longs;arà il braccio DO, <lb/>il pe&longs;o po&longs;to in O &longs;arà d'auantaggio piu <lb/>graue.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.168.0.0"> <s id="id.2.1.168.1.0"> <margin.target id="note32"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>del terzo<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.169.0.0"> <s id="id.2.1.169.1.0"> <margin.target id="note33"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 18. <emph type="italics"/>del terzo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.170.0.0" xlink:href="figures-it/041_01.jpg"></figure> <p type="main" id="id.2.1.171.0.0"> <s id="id.2.1.171.1.0"><emph type="italics"/>Ma &longs;e l'iste&longs;&longs;o cerchio AFBG co'l &longs;uo centro R &longs;arà piu da pre&longs;&longs;o ad S centro <lb/>del mondo, & dal punto S &longs;ia tirata vna linea, che tocchi il cerchio ST, il pun-<lb/>to T, (doue il pe&longs;o è piu graue) &longs;arà piu lontano dal punto A, che il punto O: <lb/>percioche &longs;iano tirate da i punti OT le linee OMTN à piombo di CS, & <lb/>congiungan &longs;i RT, & &longs;ia il centro R nella linea CS, & la linea ARB &longs;ia <lb/>egualmente di&longs;tante ad ACB. Percioche dunque i triangoli COS RTS &longs;ono<emph.end type="italics"/> <lb/> <arrow.to.target n="note34"></arrow.to.target><emph type="italics"/>di angoli retti, &longs;arà SC à CO, come CO à CM. Similmente SR ad<emph.end type="italics"/> <lb/> <arrow.to.target n="note35"></arrow.to.target><emph type="italics"/>RT, come RT ad RN. E&longs;&longs;endo dunque RT eguale à CO, & SC mag <lb/>giore di RS: haurà proportione maggiore SC à CO, che SR ad RT. on<emph.end type="italics"/> <lb/> <arrow.to.target n="note36"></arrow.to.target><emph type="italics"/>de baurà parimente proportione maggiore CO à CM, che RT ad RN. &longs;a <lb/>rà dunque minore CM, che RN. Tagli&longs;i dunque RN in P &longs;i fattamen-<emph.end type="italics"/> <pb id="p.12" xlink:href="pageimg-it/042.jpg"/><emph type="italics"/>te, che RP &longs;ia eguale à CM; & dal <lb/> <expan abbr="pũto">punto</expan> P &longs;ia tirata la linea PQ egual <lb/>mente di&longs;tante dalle linee MONT, <lb/>laquale tagli la <expan abbr="circõfer&etilde;za">circonferenza</expan> AT in Q, <lb/>& in fine <expan abbr="cõgiõgan&longs;i">congiongan&longs;i</expan> la RQ. Hor per <lb/>cioche le due CO CM &longs;ono eguali à <lb/>le due RQ RP, & l'angolo CMO<emph.end type="italics"/> <arrow.to.target n="note37"></arrow.to.target> <lb/><emph type="italics"/>è eguale all'angolo RPQ; &longs;arà an-<lb/>che l'angolo MCO eguale all'angolo <lb/>PRQ. Ma l'angolo MCA retto <lb/>è eguale all'angolo PRA retto; a-<lb/>dunque il re&longs;tante OCA al restante<emph.end type="italics"/> <arrow.to.target n="note38"></arrow.to.target> <lb/><emph type="italics"/>QRA &longs;arà eguale, & la circonferen-<lb/>za OA parimente eguale alla circon <lb/>ferenza QA. Però il punto T per <lb/>e&longs;&longs;ere piu di&longs;tante dal punto A, che <lb/>Q, &longs;arà anco piu di&longs;tante dal punto <lb/>A, che il punto O. Dimo&longs;trera&longs;&longs;i pa<lb/>rimente, che quanto piu il cerchio &longs;arà <lb/>vicino al centro del mondo, che egli &longs;a <lb/>rà anco piu lontano. </s> <s id="id.2.1.171.2.0"> Et co&longs;i come pri-<lb/>ma dimo&longs;trera&longs;&longs;i il pe&longs;o nella cir confe-<lb/>renza TAF &longs;tar &longs;opra il centro R, <lb/>ma nella circonferenza TG e&longs;&longs;ere ri-<lb/>tenuto dalla linea, & ritrouar&longs;i piu gra<lb/>ue nel punto T.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.172.0.0"> <s id="id.2.1.172.1.0"> <margin.target id="note34"></margin.target><emph type="italics"/>Per la ottaua del &longs;esto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.173.0.0"> <s id="id.2.1.173.1.0"> <margin.target id="note35"></margin.target><emph type="italics"/>Per la ottaua del quito<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.174.0.0"> <s id="id.2.1.174.1.0"> <margin.target id="note36"></margin.target><emph type="italics"/>Per la decima del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.175.0.0"> <s id="id.2.1.175.1.0"> <margin.target id="note37"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 7. <emph type="italics"/>del &longs;esto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.176.0.0"> <s id="id.2.1.176.1.0"> <margin.target id="note38"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 26. <emph type="italics"/>del terzo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.177.0.0" xlink:href="figures-it/042_01.jpg"></figure> <pb id="p.12v" xlink:href="pageimg-it/043.jpg"/> <p type="main" id="id.2.1.179.0.0"> <s id="id.2.1.179.1.0"><emph type="italics"/>Che &longs;e il punto G fo&longs;&longs;e nel centro del mondo; allhora quanto piu il pe&longs;o &longs;arà da pre&longs;&longs;o al <lb/>G, &longs;arà piu graue: & douunque &longs;ia po&longs;to il pe&longs;o, fuor che nel G &longs;empre &longs;tarà &longs;opra <lb/>il centro C, come in<emph.end type="italics"/> K<emph type="italics"/>: Imperoche tirata la linea G<emph.end type="italics"/>K; <emph type="italics"/>que&longs;ta (&longs;e condo laqua <lb/>le &longs;i fa il mouimento naturale del pe&longs;o) in&longs;ieme co'l braccio della bilancia<emph.end type="italics"/> K<emph type="italics"/>C <lb/>farà vn'angolo acuto, peroche <lb/>gli angoli posti alla ba&longs;e in<emph.end type="italics"/> K <lb/><emph type="italics"/>& G del triangolo di due la <lb/>ti eguali C<emph.end type="italics"/>K<emph type="italics"/>G &longs;ono &longs;empre <lb/>acuti. </s> <s id="id.2.1.179.2.0"> Hor &longs;iano paragonate <lb/>in&longs;ieme que&longs;te due co&longs;e, cioè il <lb/>pe&longs;o posto in<emph.end type="italics"/> K, <emph type="italics"/>& quello, <lb/>che è po&longs;to in D, &longs;arà il pe&longs;o <lb/>in K piu graue, che quello <lb/>in D; imperoche tirata la li-<lb/>nea DG, e&longs;&longs;endo che li tre an <lb/>goli di cia&longs;cuno triangolo &longs;iano <lb/>eguali à due angoli retti, & <lb/>l'angolo DCG del triangolo <lb/>CDG di due lati eguali &longs;ia <lb/>maggiore dell'angolo KCG <lb/>del triangolo CKG di due <lb/>lati eguali; &longs;aranno gli altri an <lb/>goli alla ba&longs;e DGC GDC <lb/>pre&longs;i in&longs;ieme minori de gli al-<lb/>tri KGC GKC pre&longs;i in&longs;ie<emph.end type="italics"/> <lb/> <arrow.to.target n="fig14"></arrow.to.target> <lb/><emph type="italics"/>me; & la metà di questi, cioè l'angolo CDG &longs;arà minore dell'angolo CKG: <lb/>Per laqual co&longs;a mouendo&longs;i il pe&longs;o po&longs;to in K &longs;ciolto naturalmente per KG, & <lb/>il pe&longs;o po&longs;to in D per DG come per &longs;patij, per i quali &longs;ono portati nel centro del <lb/>mondo; la linea CD, cioè il braccio della bilancia &longs;i acco&longs;terà piu al mouimento <lb/>naturale del pe&longs;o po&longs;to in D <expan abbr="totalm&etilde;te">totalmente</expan> &longs;ciolto, alla linea cioè DG, che CK al <lb/>mouimento &longs;atto &longs;econdo KG. So&longs;tenterà dunque piu la linea CD, che CK. <lb/>& perciò il pe&longs;o po&longs;to in K per le co&longs;e di &longs;opra dette &longs;arà piu graue, che in D. Ol-<lb/>tre à ciò, perche &longs;e il pe&longs;o po&longs;to in K fo&longs;&longs;e del tutto libero, & &longs;ciolto, &longs;i mouerebbe <lb/>in giu per KG, &longs;e egli non fo&longs;&longs;e impedito dalla linea CK, laquale sforza il pe&longs;o <lb/>à mouer&longs;i oltra la linea KG per la circonferenza KH; la linea KG &longs;o&longs;tente-<lb/>rà il pe&longs;o in parte, & gli farà re&longs;istenza, sforzandolo à mouer&longs;i per la circonferenza <lb/>KH. Et percioche l'angolo CDG è minore dell'angolo CKG, & l'angolo <lb/>CDK è eguale all'angolo CKH, &longs;arà l'angolo re&longs;tante GDK maggiore del re <lb/>&longs;tante GKH. Dunque la circonferenza KH &longs;arà piu da pre&longs;&longs;o al mouimento <lb/>naturale del pe&longs;o &longs;ciolto po&longs;to in K, cioè alla linea KG, che la circonferenza <lb/>DK alla linea DG. Per laqual co&longs;a la linea CD &longs;a piu re&longs;i&longs;tenza al pe&longs;o po&longs;to <lb/>in D, che la linea CK al pe&longs;o posto in K. Adunque il pe&longs;o po&longs;to in K &longs;arà<emph.end type="italics"/> <pb id="p.13" xlink:href="pageimg-it/044.jpg"/><emph type="italics"/>piu graue, che in D. Similmente mostrera&longs;&longs;i, che quanto il pe&longs;o &longs;arà piu da pre&longs;&longs;o <lb/>ad F, come in L manco grauerà; ma quanto piu da pre&longs;&longs;o &longs;i trouerà al G, co-<lb/>me in H, e&longs;&longs;ere piu graue.<emph.end type="italics"/> </s> </p> <figure id="fig14" place="text" xlink:href="figures-it/043_01.jpg"></figure> <p type="main" id="id.2.1.181.0.0"> <s id="id.2.1.181.1.0"><emph type="italics"/>Che &longs;e il centro del mondo fo&longs;&longs;e in S fra i punti CG; Primieramente &longs;i mo&longs;trerà nel <lb/>modo i&longs;te&longs;&longs;o, che il pe&longs;o in qualunque luogo po&longs;to starà &longs;opra il centro C, come in <lb/>H: peroche tirate le li-<lb/>nee HG HS, l'angolo <lb/>che è alla ba&longs;e GHC del <lb/> <expan abbr="triãgolo">triangolo</expan> di due lati eguali <lb/>CHG è &longs;empre acuto: <lb/>Perlaqual co&longs;a anco SHC <lb/>minor di lui &longs;arà parimen <lb/>te &longs;empre acuto. </s> <s id="id.2.1.181.2.0"> ma &longs;ia ti <lb/>rata dal punto S la linea <lb/>SK à piombo di CS. <lb/> Dico che il pe&longs;o è piu gra-<lb/>ue in<emph.end type="italics"/> K, <emph type="italics"/>che in alcun'al <lb/>tro &longs;ito della circonferen <lb/>za FKG; & quanto <lb/>piu da pre&longs;&longs;o &longs;arà allo F, <lb/>ouero al G meno graue-<lb/>rà. </s> <s id="id.2.1.181.3.0"> Prendan&longs;i ver&longs;o lo <lb/>F i punti DL, & con <lb/> <expan abbr="giungã&longs;i">giungan&longs;i</expan> le linee LC LS <lb/>DC DS, & &longs;iano al-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig15"></arrow.to.target> <lb/><emph type="italics"/>lungate le linee LS DS KS HS fin'alla <expan abbr="circõferenza">circonferenza</expan> del cerchio in EM NO; <lb/>& &longs;iano <expan abbr="cõgiunte">congiunte</expan> CE, CM, CN, CO. Hor percioche LE DM &longs;i taglia-<lb/>no in&longs;ieme in S, &longs;arà il rettangolo LSE eguale al rettangolo DSM. Onde &longs;i co<emph.end type="italics"/> <arrow.to.target n="note39"></arrow.to.target> <lb/><emph type="italics"/>me è la LS ver&longs;o la DS, co&longs;i &longs;arà la SM ver&longs;ola SE; ma è maggior la LS <lb/>della DS; & la SM di e&longs;&longs;a SE. Dunque LS SE pre&longs;e in&longs;ieme &longs;aranno mag-<emph.end type="italics"/> <arrow.to.target n="note40"></arrow.to.target> <lb/><emph type="italics"/>giori delle DS SM. & per la ragion i&longs;te&longs;&longs;a &longs;i mo&longs;trerà la KN e&longs;&longs;er minore di DM. <lb/> Di piu percioche il rettangolo OSH è eguale alrett'angolo KSN; per la mede&longs;i-<emph.end type="italics"/> <arrow.to.target n="note41"></arrow.to.target> <lb/><emph type="italics"/>ma ragione la HO &longs;arà maggiore della KN. & nell'i&longs;te&longs;&longs;o modo in tutto la<emph.end type="italics"/> <arrow.to.target n="note42"></arrow.to.target> <lb/><emph type="italics"/>KN &longs;i dimostrerà minore di tutte le altre linee, che pa&longs;&longs;ino per lo punto S. Et <lb/>percioche de i triangoli di due lati eguali CLE DCM i lati LC CE &longs;ono e-<lb/>guali a i lati DC CM; & la ba&longs;e LE è maggiore di DM: &longs;arà l'angolo <lb/>LCE maggiore dell'angolo DCM. Per laqual co&longs;a gli angoli CLE CEL po<emph.end type="italics"/> <arrow.to.target n="note43"></arrow.to.target> <lb/><emph type="italics"/>sti alla ba&longs;e tolti in&longs;ieme &longs;aranno minori de gli angoli CDM CMD; & le me-<lb/>tà di que&longs;ti, cioè l'angolo CLS &longs;arà minore dell'angolo CDS. Dunque il pe&longs;o po <lb/>&longs;to in L &longs;opra la linea LC grauerà piu, che po&longs;to in D &longs;opra la DC; & piu <lb/>&longs;tarà &longs;opra il centro in L, che in D. Similmente &longs;i mo&longs;trerà, che il pe&longs;o in D<emph.end type="italics"/> <pb id="p.13v" xlink:href="pageimg-it/045.jpg"/><emph type="italics"/>&longs;tarà piu &longs;opra il centro C, che in K. Adunque il pe&longs;o po&longs;to in K &longs;arà piu <lb/>graue, che in D, & in D, che in L. & con la mede&longs;ima ragione in tutto, pero-<lb/>che KN è minore di HO, &longs;arà l'angolo CKS maggiore dell'angolo CHS. <lb/> Per laqual co&longs;a il pe&longs;o posto in H &longs;tarà piu &longs;opra il centro C, che in<emph.end type="italics"/> K; <emph type="italics"/>& in que-<lb/>&longs;ta maniera &longs;i mostrerà, che douunque &longs;ia il pe&longs;o nella circonferenza FDG, manco <lb/>starà &longs;opra il centro quando &longs;arà po&longs;to in K, che in altro &longs;ito: & quanto piu da <lb/>pre&longs;&longs;o egli &longs;arà ad F, ouero à G piu &longs;tarà &longs;opra. </s> <s id="id.2.1.181.4.0"> Dopo percioche l'angolo CKS <lb/>è maggiore del CDS, & CDK è eguale à CKH: &longs;arà il re&longs;tante SKH mi-<lb/>nore del re&longs;tante SDK. Per laqual co&longs;a la circonferenza KH &longs;arà piu da pre&longs;&longs;o <lb/>al mouimento naturale <lb/>diritto del pe&longs;o po&longs;to in <lb/>K &longs;ciolto, cioè alla li-<lb/>nea KS, che la circon <lb/>ferenza DK al moui-<lb/>mento DS. & perciò <lb/>la linea CD &longs;a piu re&longs;i <lb/>&longs;tenza al pe&longs;o po&longs;to in D <lb/>che la CK al pe&longs;o me&longs;-<lb/>&longs;o in<emph.end type="italics"/> K. <emph type="italics"/>& per que&longs;ta <lb/>ragione &longs;i mo&longs;trera l'an-<lb/>golo SHG e&longs;&longs;er mag-<lb/>giore dello SKH; & <lb/>per con&longs;equente la linea <lb/>CH &longs;are piu re&longs;i&longs;tenza <lb/>al pe&longs;o po&longs;to in H, che <lb/>CK al pe&longs;o me&longs;&longs;o in K. <lb/> Similmente dimo&longs;trera&longs;&longs;i <lb/>che la linea CL piu &longs;o-<lb/>&longs;tenterà il pe&longs;o, che CD:<emph.end type="italics"/> <lb/> <arrow.to.target n="fig16"></arrow.to.target> <lb/><emph type="italics"/>& per le cagioni i&longs;te&longs;&longs;e &longs;i prouerà, che il pe&longs;o me&longs;&longs;o in K grauerà meno &longs;opra la li-<lb/>nea CK, che in qual &longs;i voglia altro &longs;ito della circon&longs;erenza FDG: & quanto <lb/>piu da pre&longs;&longs;o &longs;arà ad F, ouero à G, manco grauerà. </s> <s id="id.2.1.181.5.0"> dunque piu graue &longs;ara in K, <lb/>che in altro &longs;ito: & &longs;arà meno graue quanto piu da pre&longs;&longs;o &longs;tara ad F, ouero a G.<emph.end type="italics"/> </s> </p> <figure id="fig15" place="text" xlink:href="figures-it/044_01.jpg"></figure> <figure id="fig16" place="text" xlink:href="figures-it/045_01.jpg"></figure> <p type="margin" id="id.2.1.184.0.0"> <s id="id.2.1.184.1.0"> <margin.target id="note39"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 35. <emph type="italics"/>del terzo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.185.0.0"> <s id="id.2.1.185.1.0"> <margin.target id="note40"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 16. <emph type="italics"/>del &longs;esto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.186.0.0"> <s id="id.2.1.186.1.0"> <margin.target id="note41"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 7. <emph type="italics"/>del terzo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.187.0.0"> <s id="id.2.1.187.1.0"> <margin.target id="note42"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 25. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.188.0.0"> <s id="id.2.1.188.1.0"> <margin.target id="note43"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 25. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.189.0.0"> <s id="id.2.1.189.1.0"><emph type="italics"/>Se in fine il centro C fo&longs;&longs;e nel centro del mondo, egli è manife&longs;to, che il pe&longs;o po&longs;to doue<emph.end type="italics"/> <lb/> <arrow.to.target n="note44"></arrow.to.target><emph type="italics"/>&longs;i voglia &longs;tarà fermo. </s> <s id="id.2.1.189.2.0"> Come posto il pe&longs;o in D la linea CD &longs;o&longs;terrà tutto il pe&longs;o, <lb/>per e&longs;&longs;er a piombo dell'orizonte di e&longs;&longs;o pe&longs;o po&longs;to in D. Dunque &longs;tarà fermo <lb/>il pe&longs;o.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.190.0.0"> <s id="id.2.1.190.1.0"> <margin.target id="note44"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.191.0.0"> <s id="id.2.1.191.1.0"><emph type="italics"/>Hor percioche nelle co&longs;e, che fin qui &longs;ono &longs;tate dimostrate non habbiamo fatto mentio-<lb/>ne alcuna della grauezza del braccio della bilancia, però &longs;e vorremo anco con&longs;idera-<lb/>re la grauezza del detto braccio, &longs;i potrà ritrouare il centro della grauezza della ma<emph.end type="italics"/> <pb id="p.14" xlink:href="pageimg-it/046.jpg"/><emph type="italics"/>gnitudine fatta dal pe&longs;o, & dal braccio, & &longs;i <expan abbr="de&longs;criuerãno">de&longs;criueranno</expan> le circonferenze de' cerch; <lb/>&longs;econdo la di&longs;tanza dal centro della bilancia ad e&longs;&longs;o centro della grauezza, come &longs;e <lb/>in e&longs;&longs;o (come è veramente) fo&longs;&longs;e posto il pe&longs;o, Et le co&longs;e che &longs;enza la con&longs;ideratio <lb/>ne della grauezza del braccio della bilancia habbiamo trouato, tutte nell'i&longs;te&longs;&longs;o mo-<lb/>do con&longs;iderando ancora tal grauit à le ritrouaremo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.192.0.0"> <s id="id.2.1.192.1.0"><emph type="italics"/>Dalle co&longs;e dette dunque, <expan abbr="con&longs;iderãdo">con&longs;iderando</expan> la bilancia, <lb/>come ella è lontana dal centro del mondo <lb/>nel modo che e&longs;&longs;i hanno fatto, come etiandio <lb/>è in atto, appare la fal&longs;it à di coloro, che dico-<lb/>no il pe&longs;o po&longs;to in A e&longs;&longs;ere piu graue, che <lb/>in altro &longs;ito; & in&longs;ieme e&longs;&longs;er fal&longs;o, che quan-<lb/>to piu il pe&longs;o è lontano dalla linea FG, tan-<lb/>to e&longs;&longs;ere piu graue: imperoche il punto O <lb/>è piu da pre&longs;&longs;o alla FG, che il punto A; <lb/>percioche la linea tirata a piombo dal pun-<emph.end type="italics"/> <arrow.to.target n="note45"></arrow.to.target> <lb/><emph type="italics"/>to O ad FG è minore della CA. Da poi <lb/>egli è parimente fal&longs;o, che il pe&longs;o dal punto <lb/>A &longs;i moua piu velocemente, che da altro <lb/>&longs;ito. </s> <s id="id.2.1.192.2.0"> peroche dal punto O &longs;i mouerà piu ve <lb/>locemente, che dal punto A, concio&longs;ia che <lb/>in O &longs;ia piu libero e &longs;ciolto, che in altro &longs;ito; <lb/>& la &longs;ce&longs;a dal punto O &longs;ia piu da pre&longs;&longs;o al <lb/>mouimento naturale diritto, che qual &longs;i vo-<lb/>glia altra di&longs;ce&longs;a.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.193.0.0"> <s id="id.2.1.193.1.0"> <margin.target id="note45"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 15. <emph type="italics"/>del terzo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.194.0.0" xlink:href="figures-it/046_01.jpg"></figure> <pb id="p.14v" xlink:href="pageimg-it/047.jpg"/> <p type="main" id="id.2.1.196.0.0"> <s id="id.2.1.196.1.0"><emph type="italics"/>Oltre a ciò quando mo&longs;trano per via della piu diritta, & della piu torta di&longs;ce&longs;a, che il pe-<lb/>&longs;o è piu graue in A, che in D, & in D, che in L. Primieramente per certo e&longs;tima <lb/>no il fal&longs;o, che &longs;e alcun pe&longs;o &longs;arà collocato in qual &longs;i voglia &longs;ito della circonferenza, <lb/>come in D, la &longs;ua vera di&longs;ce&longs;a douer&longs;i fare per la linea diritta DR egualmente di-<lb/>&longs;tante da e&longs;&longs;a FG, come &longs;econdo il mouimento naturale, &longs;i come prima è &longs;tato det-<lb/>to. </s> <s id="id.2.1.196.2.0"> Percioche in qual &longs;i voglia &longs;ito &longs;i collochi alcun pe&longs;o, &longs;e riguardiamo il mouimen <lb/>to &longs;uo naturale al proprio luogo, alquale &longs;i moue dirittamente per &longs;ua natura, pre&longs;up <lb/>po&longs;ta tutta la figura dell'vniuer&longs;o mondo, &longs;arà tale, che &longs;empre lo &longs;patio, per lo qua-<lb/>le &longs;i moue naturalmente, parerà hauere ragione di linea tirata dalla circonferenza al <lb/>centro. </s> <s id="id.2.1.196.3.0"> Adunque le na <lb/>turali di&longs;ce&longs;e diritte di <lb/>qual &longs;i voglia pe&longs;o &longs;ciol <lb/>to non &longs;i po&longs;&longs;ono fare <lb/>per linee tra loro egual <lb/>mente di&longs;tanti, per an-<lb/>dar&longs;i à trouar tutte nel <lb/>centro del mondo. </s> <s id="id.2.1.196.4.0"> pre <lb/>&longs;uppongono da poi, che <lb/>il pe&longs;o mo&longs;&longs;o da D in <lb/>A per linea diritta ver <lb/>&longs;o il centro del mondo <lb/>&longs;ia della <expan abbr="quãtità">quantità</expan> i&longs;te&longs;&longs;a, <lb/>come &longs;e egli fo&longs;&longs;e da O <lb/>in C &longs;i fattamente, <lb/>che il <expan abbr="pũto">punto</expan> A &longs;ia egual <lb/>mente di&longs;tante dal cen-<lb/>tro del mondo, come C; <lb/>ilche è parimente fal&longs;o:<emph.end type="italics"/> <lb/> <arrow.to.target n="fig17"></arrow.to.target> <lb/> <arrow.to.target n="note46"></arrow.to.target><emph type="italics"/> Imperoche il punto A è piu da lontano dal centro del mondo, che C: percioche <lb/>maggior è la linea tirata dal centro del mondo fin ad A, che quella del centro del <lb/>mondo fin a C, concio&longs;ia che vna linea dal centro del mondo fin ad A &longs;i di&longs;tenda <lb/>&longs;otto vn'angolo retto contenuto dalle linee AC, & dal punto C al centro del <lb/>mondo. </s> <s id="id.2.1.196.5.0"> Dalle quali co&longs;e non &longs;olo rie&longs;ce vana quella pre&longs;uppo&longs;ta, laquale dimostra, <lb/>che la bilancia DE ritorna in AB, ma anco cadono tutte le loro dimo&longs;trationi; <lb/>&longs;e for&longs;e non dice&longs;&longs;ero, che que&longs;te co&longs;e tutte per la grandi&longs;&longs;ima di&longs;tanza, che è fra il cen <lb/>tro del mondo, & noi &longs;ono co&longs;i in&longs;en&longs;ibili, che per cagione di que&longs;ta in&longs;en&longs;ibilità, <lb/>&longs;i po&longs;&longs;ano pre&longs;upponere, come vere; concio&longs;ia, che tutti quelli, iquali hanno trattato <lb/>que&longs;te co&longs;e, le habbiano pre&longs;uppo&longs;te, come note; ma&longs;&longs;imamente, percioche quello <lb/>e&longs;&longs;ere in&longs;en&longs;ibile non fà, che la di&longs;ce&longs;a del pe&longs;o da L in D (per v&longs;are le loro paro-<lb/>le) non pigli meno del diretto, che la di&longs;ce&longs;a DA. Similmente l'arco DA piglie-<lb/>rà piu del diretto, che la circon&longs;erenza EV. onde &longs;arà vera la pre&longs;uppo&longs;ta, & le <lb/>altre dimo&longs;trationi rimarranno nella &longs;ua &longs;ua forza. </s> <s id="id.2.1.196.6.0"> Concediamo etiandio, che il pe<emph.end type="italics"/> <pb id="p.15" xlink:href="pageimg-it/048.jpg"/><emph type="italics"/>&longs;o po&longs;to in A &longs;ia piu graue, che in altro &longs;ito; & che la di&longs;ce&longs;a diritta del pe&longs;o &longs;i deb <lb/>ba fare per linea diritta egualmente di&longs;tante da FG, & quali&longs;i voglian punti pre&longs;i <lb/>nelle linee egualmente di&longs;tanti dall'orizonte e&longs;&longs;ere egualmente lontani dal centro <lb/>del mondo: non &longs;eguiter à gia per que&longs;to, che la loro dimostratione &longs;ia vera, con la-<lb/>quale vengono a dire, che il pe&longs;o posto in A è piu grane, che in altro &longs;ito, come in <lb/>L. Percioche &longs;e egli fo&longs;&longs;e vero, che quanto piu il pe&longs;o in que&longs;ta maniera di&longs;cende <lb/>piu al diritto, iui fo&longs;&longs;e piu graue; &longs;eguirebbe etiandio, che quanto l'iste&longs;&longs;o pe&longs;o de-<lb/>&longs;cende&longs;&longs;e egualmente in archi eguali al diritto, che ne i luoghi mede&longs;imi haue&longs;&longs;e gra-<lb/>uezza eguale, ilche in que&longs;to modo e&longs;&longs;er fal&longs;o &longs;i dimo&longs;tra.<emph.end type="italics"/> </s> </p> <figure id="fig17" place="text" xlink:href="figures-it/047_01.jpg"></figure> <p type="margin" id="id.2.1.198.0.0"> <s id="id.2.1.198.1.0"> <margin.target id="note46"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 18. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.199.0.0"> <s id="id.2.1.199.1.0"><emph type="italics"/>Siano le circonferenze AL AM tra loro eguali, & congiunga&longs;i LM, laquale ta-<lb/>gli AB in X; &longs;arà LM egualmente di&longs;tante da FG, & à piombo di AB,<emph.end type="italics"/> <arrow.to.target n="note47"></arrow.to.target> <lb/><emph type="italics"/>& XM &longs;arà eguale ad XL. Se dunque il pe&longs;o da L &longs;arà mo&longs;&longs;o in A per la cir-<lb/>conferenza LA, il mouimento &longs;uo diritto &longs;arà &longs;econdo la linea LX. Ma &longs;e egli &longs;i <lb/>mouerà da A in M per la circonferenza AM, il &longs;uo mouimento &longs;arà &longs;econdo <lb/>la linea diritta XM. Per laqual co&longs;a la &longs;ce&longs;a da L in A &longs;arà eguale alla &longs;ce&longs;a da <lb/>A in M, &longs;i per cau&longs;a delle circonferenze eguali, & &longs;i per le linee rette eguali, & à <lb/>piombo di e&longs;&longs;a AB. Adunque il pe&longs;o mede&longs;imo po&longs;to in L grauerà egualmente, <lb/>come in A, ilche è fal&longs;o, concio&longs;ia, che egli è di gran lunga piu graue in A, che in L.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.200.0.0"> <s id="id.2.1.200.1.0"> <margin.target id="note47"></margin.target><emph type="italics"/>Per la terza del terzo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.201.0.0"> <s id="id.2.1.201.1.0"><emph type="italics"/>Et benche AMLA prendano, &longs;econdo e&longs;&longs;i, egualmente del diretto, diranno for&longs;e, <lb/>nondimeno perche il principio della &longs;ce&longs;a da L, cioè LD piglia meno del diretto, che <lb/>il principio della &longs;ce&longs;a da A, cioè AN, il pe&longs;o in A &longs;arà piu graue, che in L. <lb/> Imperoche e&longs;&longs;endo (come è &longs;tato di &longs;opra po&longs;to) la circonferenza AN eguale ad <lb/>LD, laquale (&longs;econdo eßi) piglia di diretto CT; ma LD piglia di diretto PO, <lb/>però il pe&longs;o &longs;arà piu graue in A, che in L. ilche &longs;e fo&longs;&longs;e vero, &longs;eguirebbe, che l'i&longs;te&longs;-<lb/>&longs;o pe&longs;o nel mede&longs;imo &longs;ito, in diuer&longs;o modo &longs;olamente con&longs;iderato, ver&longs;o il mede&longs;imo <lb/>&longs;ito fo&longs;&longs;e & piu graue, & piu lieue; ilche è impo&longs;&longs;ibile. </s> <s id="id.2.1.201.2.0"> cioè &longs;e con&longs;ideriamo la &longs;ce&longs;a <lb/>del pe&longs;o po&longs;to in L in quanto egli de&longs;cende da L in A &longs;arà piu graue, che &longs;e con&longs;ide <lb/>reremo la &longs;ce&longs;a del pe&longs;o i&longs;te&longs;&longs;o da L in D &longs;olamente. </s> <s id="id.2.1.201.3.0"> ne po&longs;&longs;ono negare per i mede <lb/>&longs;imi detti &longs;uoi, che la di&longs;ce&longs;a del pe&longs;o da L in A non pigli del diretto LX, ouero PC. <lb/> Et che &longs;imilmente la &longs;ce&longs;a AM non prenda di diretto XM: pigliando cßi ancora <lb/>à que&longs;to modo, & co&longs;i nece&longs;&longs;ario &longs;ia di pigliare. </s> <s id="id.2.1.201.4.0"> percioche &longs;e vogliono dimo&longs;trare, <lb/>che la bilancia DE ritorni in AB paragonando la &longs;ce&longs;a del pe&longs;o po&longs;to in D con <lb/>la &longs;ce&longs;a del pe&longs;o posto in E, egli è nece&longs;&longs;ario, che mo&longs;trino, che la diritta &longs;ce&longs;a OC <lb/>ri&longs;pondente alla circonferenza DA &longs;ia maggiore della &longs;ce&longs;a diritta TH ri&longs;ponden-<lb/>te alla circonferenza EV. peroche &longs;e piglia&longs;&longs;ero &longs;olamente vna parte di tutta la &longs;ce <lb/>&longs;a da D in A, come D<emph.end type="italics"/>K, <emph type="italics"/>& dimo&longs;tra&longs;&longs;ero, che piu di diretto piglia la &longs;ce&longs;a D<emph.end type="italics"/>K, <lb/><emph type="italics"/>che la eguale portione della &longs;ce&longs;a dal punto E, &longs;eguirebbe il pe&longs;o po&longs;to in D, &longs;econ-<lb/>do eßi, e&longs;&longs;ere piu graue del pe&longs;o po&longs;to in E, & mouer&longs;i in giu fin al K &longs;olamente. <lb/>per modo che la bilancia &longs;ia mo&longs;&longs;a in KI. Similmente &longs;e vogliono mo&longs;trare, che la <lb/>bilancia KI ritorni in AB pigliando vna portione della &longs;ce&longs;a da K in A, cioè KS, <lb/>& mo&longs;tra&longs;&longs;ero, che KS pigli piu di diretto, che la &longs;ce&longs;a eguale, che è dirimpetto dal <lb/>punto I: &longs;eguirebbe con &longs;imile modo il pe&longs;o po&longs;to in K e&longs;&longs;ere piu graue, che in I, &<emph.end type="italics"/> <pb id="p.15v" xlink:href="pageimg-it/049.jpg"/><emph type="italics"/>mouer&longs;i &longs;olamente fin ad S. Et &longs;e di nouo mo&longs;tra&longs;&longs;ero vna portione della &longs;ce&longs;a da S <lb/>in A, & co&longs;i &longs;ucceßiuamente e&longs;&longs;ere piu diritta della &longs;ce&longs;a eguale del pe&longs;o oppo&longs;to; <lb/>&longs;empre &longs;eguirà, che la bilancia SI andarà piu da pre&longs;&longs;o ad AB, ma non <expan abbr="dimostre-<lb> rãno">dimostre-<lb/>ranno</expan> giamai che per <lb/>uenga in AB. Se <lb/>dunque vogliono di <lb/>mo&longs;trare, che la <expan abbr="bilã">bilam</expan> <lb/>cia DE ritorni in <lb/>AB, egli è nece&longs;&longs;a-<lb/>rio, che pre&longs;upponga <lb/>no, che la &longs;ce&longs;a del <lb/>pe&longs;o da D in A <expan abbr="pr&etilde;">prem</expan> <lb/>da di diretto la quan <lb/>tità della linea tira-<lb/>ta dal punto D ad <lb/>AB ad angoli ret-<lb/>ti; & co&longs;i, &longs;e para-<lb/>goneremo le &longs;ce&longs;e e-<lb/>guali di DA AN <lb/>fra loro, lequali <expan abbr="pr&etilde;">prem</expan> <lb/>dono di diretto OC <lb/>CT, accaderà, che<emph.end type="italics"/> <lb/> <arrow.to.target n="fig18"></arrow.to.target> <lb/><emph type="italics"/>il pe&longs;o i&longs;te&longs;&longs;o &longs;arà in D graue egualmente, come in A. Ma &longs;e le portioni &longs;olamente <lb/>piglieremo da DA, &longs;arà piu graue in A, che in D. Adunque dalla diuer&longs;ità &longs;o-<lb/>lamente del modo del con&longs;iderare, auerrà, che il pe&longs;o mede&longs;imo &longs;arà & piu graue, <lb/>& piu leggiero; & non per la natura della co&longs;a. </s> <s id="id.2.1.201.5.0"> Di piu la pre&longs;uppo&longs;ta loro non <lb/>afferma, che il pe&longs;o &longs;econdo il &longs;ito &longs;ia piu graue, quanto nel &longs;ito mede&longs;imo il principio <lb/>della &longs;ua di&longs;ce&longs;a è meno obliquo. </s> <s id="id.2.1.201.6.0"> La pre&longs;upposta dunque di &longs;opra addotta, cioè che <lb/>&longs;econdo il &longs;ito il pe&longs;o è piu graue quanto nell'i&longs;te&longs;&longs;o &longs;ito meno obliqua è la di&longs;ce&longs;a, non <lb/>&longs;olamente non &longs;i puote concedere à modo alcuno, per le co&longs;e, che habbiamo dette; <lb/>ma anco percioche non è co&longs;a difficile il dimo&longs;trare tutto l'oppo&longs;to, cioè il pe&longs;o mede&longs;i <lb/>mo in eguali circonferenze quanto meno obliqua è la di&longs;ce&longs;a, iui meno grauare.<emph.end type="italics"/> </s> </p> <figure id="fig18" place="text" xlink:href="figures-it/049_01.jpg"></figure> <p type="main" id="id.2.1.203.0.0"> <s id="id.2.1.203.1.0"><emph type="italics"/>Siano come prima le circonferenze AL AM tra loro eguali; & &longs;ia il punto L vici <lb/>no ad F, & congiunga&longs;i LM, la quale &longs;arà à piombo di AB & LX &longs;arà anco <lb/>eguale ad XM. Dapoi pre&longs;&longs;o ad M tra M & G &longs;ia pre&longs;o come &longs;i vuole, il pun<emph.end type="italics"/> <lb/> <arrow.to.target n="note48"></arrow.to.target><emph type="italics"/>to P, & &longs;ia fatta la circonferenza PO eguale alla circonferenza AM, &longs;arà il <lb/>punto O pre&longs;&longs;o ad A. & &longs;iano congiunte le linee CL, CO, CM, CP, OP.<emph.end type="italics"/> <lb/> <arrow.to.target n="note49"></arrow.to.target><emph type="italics"/>& dal punto P tiri&longs;i la PN a piombo di OC. & percioche la circonferenza. <lb/>AM è eguale alla circonferentia OP; &longs;arà l'angolo ACM eguale all'angolo<emph.end type="italics"/> <lb/> <arrow.to.target n="note50"></arrow.to.target><emph type="italics"/>OCP, & l'angolo CXM retto eguale al retto CNP, &longs;arà anco il re&longs;tante angolo <lb/>XMC del triangolo MXC eguale al re&longs;tante NPC del triangolo PCN.<emph.end type="italics"/> <pb id="p.16" xlink:href="pageimg-it/050.jpg"/> <emph type="italics"/>Ma il lato ancora CM è eguale al lato CP, dunque il triangolo MCX è egua <lb/>le al triangolo PCN, & il lato MX eguale al lato NP. Onde la linea PN <lb/>&longs;arà eguale ad LX. Tiri&longs;i oltre a ciò dal punto O la linea OT egualmente di-<lb/>&longs;tante da AC, laquale tagli NP in V. & &longs;ia anco tirata dal punto P vna <lb/>linea a piombo di OT, <lb/>la quale per certo non <lb/>puote cadere tra OV, <lb/>perche e&longs;&longs;endo l'angolo <lb/>ONV retto, &longs;arà acu<emph.end type="italics"/> <arrow.to.target n="note51"></arrow.to.target> <lb/><emph type="italics"/>to lo OVN. Per la <lb/>qualco&longs;a OVP &longs;arà <lb/>ottu&longs;o. </s> <s id="id.2.1.203.2.0"> Non caderà <lb/>dunque la linea tirata <lb/>dal punto P tra OV <lb/>à piombo di OT: pe-<lb/>roche due angoli d'uno <lb/> <expan abbr="triãgolo">triangolo</expan> &longs;arebbono l'u-<lb/>no retto, & l'altro ot-<lb/>tu&longs;o, che è impoßibile. <lb/>Caderà dun que nella li <lb/>nea OT nellaparte di <lb/>VT, et &longs;ia PT. &longs;arà &longs;e <lb/>condo e&longs;&longs;i, PT la di<emph.end type="italics"/> <lb/> <arrow.to.target n="fig19"></arrow.to.target> <lb/><emph type="italics"/>ritta &longs;ce&longs;a della circonferenza OP. Percioche dunque l'angolo ONV è retto,<emph.end type="italics"/> <arrow.to.target n="note52"></arrow.to.target> <lb/><emph type="italics"/>&longs;arà la linea OV maggiore della ON. Onde la OT &longs;arà parimente maggiore <lb/>della ON. & co&longs;i di&longs;tendendo&longs;i la linea OP &longs;otto gli angoli retti ONP, <lb/>OTP, &longs;arà il quadrato di OP eguale alli quadrati ON NP in&longs;ieme pre&longs;i, &longs;i<emph.end type="italics"/> <arrow.to.target n="note53"></arrow.to.target> <lb/><emph type="italics"/>milmente eguale a i quadrati di OT TP in&longs;ieme. </s> <s id="id.2.1.203.3.0"> per laqual co&longs;a li quadrati in&longs;ie-<lb/>me di ON NP &longs;aranno eguali a i quadrati in&longs;ieme di OT TP. Ma il quadrato <lb/>di OT è maggiore del quadrato di ON. per e&longs;&longs;ere maggiore la linea OT della <lb/>ON. Adunque il quadrato di NP &longs;ara maggiore del quadrato TP & perciò la <lb/>linea TP &longs;arà minore della linea PN, & della linea LX. Meno obliqua <lb/>dunque &longs;arà la &longs;ce&longs;a dell'arco LA, che dell'arco OP. Dunque il pe&longs;o po-<lb/>sto in L, per i loro detti, &longs;arà piu graue, che in O, il che, per le co&longs;e, che di <lb/>&longs;opra habbiamo detto, è manife&longs;tamente fal&longs;o. concio&longs;ia, che il pe&longs;o po&longs;to in O <lb/>&longs;ia piu graue, che in L. </s> <s id="id.2.1.203.4.0"> Non &longs;i puote dunque raccogliere dal piu diritto, & <lb/>piu torto mouimento in quel modo pigliato, e&longs;&longs;ere il pe&longs;o tanto piu graue &longs;econ-<lb/>do il &longs;ito, quanto nel mede&longs;imo &longs;ito è meno torta la &longs;ce&longs;a. </s> <s id="id.2.1.203.5.0"> & quinci na&longs;ce tutto <lb/>qua&longs;i il &longs;uo errore & inganno in cote&longs;ta co&longs;a. </s> <s id="id.2.1.203.6.0"> Imperoche quantunque per acciden-<lb/>te alle volte dalle co&longs;e fal&longs;e ne &longs;egua il vero, tutta via per &longs;e &longs;te&longs;&longs;e principalmente <lb/>dalle fal&longs;e ne &longs;egue il fal&longs;o, &longs;i come dalle vere &longs;empre il vero ne &longs;egue. </s> <s id="id.2.1.203.7.0"> Non è pero <lb/>da mar auigliar&longs;i, &longs;e mentre e&longs;&longs;i prendono co&longs;e fal&longs;e, & &longs;tanno &longs;opra quelle, come ve<emph.end type="italics"/> <pb id="p.16v" xlink:href="pageimg-it/051.jpg"/><emph type="italics"/>ri&longs;&longs;ime, raccolgono, & conchiudono co&longs;e in tutto fal&longs;i&longs;&longs;ime. </s> <s id="id.2.1.203.8.0"> &longs;ono oltre a ciò inganna-<lb/>ti, mentre pigliano a contemplare la bilancia &longs;emplicemente per via di matematica, <lb/>e&longs;&longs;endo la con&longs;ideratione &longs;ua mechanica affatto, ne di lei &longs;i po&longs;&longs;a ragionare a modo al <lb/>cuno &longs;enza il vero mouimento, & &longs;enza i pe&longs;i, che &longs;ono in tutto co&longs;e naturali, &longs;en-<lb/>za le quali non &longs;i po&longs;&longs;ono ritrouare per niuna maniera le vere cagioni di quelle co&longs;e, <lb/>che accadono alla bilancia.<emph.end type="italics"/> </s> </p> <figure id="fig19" place="text" xlink:href="figures-it/050_01.jpg"></figure> <p type="margin" id="id.2.1.205.0.0"> <s id="id.2.1.205.1.0"> <margin.target id="note48"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 27. <emph type="italics"/>del terzo<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.206.0.0"> <s id="id.2.1.206.1.0"> <margin.target id="note49"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 32. <emph type="italics"/>del primo<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.207.0.0"> <s id="id.2.1.207.1.0"> <margin.target id="note50"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 26. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.208.0.0"> <s id="id.2.1.208.1.0"> <margin.target id="note51"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 13. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.209.0.0"> <s id="id.2.1.209.1.0"> <margin.target id="note52"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 19. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.210.0.0"> <s id="id.2.1.210.1.0"> <margin.target id="note53"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 47. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.211.0.0"> <s id="id.2.1.211.1.0"><emph type="italics"/>Oltre a ciò &longs;e anche con <lb/>cederemo la pre&longs;up-<lb/>po&longs;ta, &longs;i partono tut<lb/>tauia molto <expan abbr="lũge">lunge</expan> dal <lb/>la <expan abbr="cõ&longs;ideratione">con&longs;ideratione</expan> della <lb/>bilancia, mentre di-<lb/>&longs;corrono; che in quel <lb/>la maniera debba la <lb/>bilancia DE ritor-<lb/>nare in AB: percio <lb/>che &longs;empre pigliano <lb/>vn di due pe&longs;i &longs;epara <lb/>tamente come D, <lb/>ouero E, come &longs;e hor <lb/>l'uno, hor l'altro fo&longs; <lb/>&longs;e po&longs;to nella bilan-<lb/>cia, non congiunti in <lb/>&longs;ieme ambidue in <lb/>modo veruno, e&longs;&longs;en-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig20"></arrow.to.target> <lb/><emph type="italics"/>doche nondimeno bi&longs;ogni fare tutto all'oppo&longs;ito di ciò, ne &longs;i puote con&longs;iderare dirit-<lb/>tamente l'uno &longs;enza l'altro, e&longs;&longs;endoche &longs;i ragiona di loro nella bilancia collocati. <lb/> Concio&longs;ia che quando dicono la di&longs;ce&longs;a del pe&longs;o po&longs;to in D e&longs;&longs;ere meno torta, che <lb/>la di&longs;ce&longs;a del pe&longs;o po&longs;to in E, co&longs;i &longs;arà il pe&longs;o in D, per la pre&longs;uppo&longs;ta, piu graue <lb/>del pe&longs;o po&longs;to in E; onde per e&longs;&longs;ere piu graue, eglie nece&longs;&longs;ario, che &longs;i moua in giu, <lb/>& che la bilancia DE ritorni in AB: Cote&longs;to di&longs;cor&longs;o non è di momento alcu-<lb/>no. </s> <s id="id.2.1.211.2.0"> Primieramente &longs;empre argomentano come &longs;e i pe&longs;i in DE debbano &longs;cende-<lb/>re, con&longs;iderando la &longs;ce&longs;a di vno &longs;olameute &longs;enza la compagnia, & congiungimen-<lb/>to dell'altro. </s> <s id="id.2.1.211.3.0"> Vltimamente nondimeno e&longs;&longs;i per la comparatione delle di&longs;ce&longs;e de'pe-<lb/>&longs;i conchiudono il pe&longs;o posto in D mouer&longs;i in giu, & il po&longs;to in E in &longs;u, prenden-<lb/>do l'uno, & l'altro pe&longs;o congiunti in&longs;ieme fra loro nella bilancia. </s> <s id="id.2.1.211.4.0"> Ma da &longs;uoi me-<lb/>de&longs;imi principij, i quali v&longs;ano, & dalle &longs;ue dimo&longs;trationi &longs;i puote cauare ageuoli&longs;&longs;i-<emph.end type="italics"/> <lb/> <arrow.to.target n="note54"></arrow.to.target><emph type="italics"/>mamente l'oppo&longs;ito di quel che &longs;i faticano di difendere. </s> <s id="id.2.1.211.5.0"> Imperoche &longs;e &longs;i paragona <lb/>la di&longs;ce&longs;a del pe&longs;o po&longs;to in D con la &longs;alita del pe&longs;o po&longs;to in E, come tirate le'linee <lb/>E<emph.end type="italics"/>K <emph type="italics"/>DH a piombo di AB, e&longs;&longs;endo l'angolo DCH eguale all'angolo ECK,<emph.end type="italics"/> <lb/> <arrow.to.target n="note55"></arrow.to.target><emph type="italics"/>& l'angolo DHC retto eguale al retto E<emph.end type="italics"/>K<emph type="italics"/>C, & il lato DC eguale al lato <lb/>CE; &longs;arà il triangolo CDH eguale al triangolo CEK, & il lato DH egua<emph.end type="italics"/> <pb id="p.17" xlink:href="pageimg-it/052.jpg"/><emph type="italics"/>le al lato EK: & e&longs;&longs;endo l'angolo DCA eguale all'angolo ECB, &longs;arà anco <lb/>la circonferenza DA eguale alla circonferenza BE. Mentre dunque il pe&longs;o po-<lb/>sto in D &longs;cende per la circonferenza DA, il pe&longs;o po&longs;to in E &longs;ale per la circon-<lb/>ferenza EB eguale a DA, & la &longs;ce&longs;a del pe&longs;o po&longs;to in D prenderà, (&longs;econdo <lb/>il co&longs;tume loro) di diretto DH: & la &longs;alita del pe&longs;o E prenderà di diretto EK <lb/>eguale a DH: &longs;arà dunque la &longs;ce&longs;a del pe&longs;o posto in D eguale alla &longs;alita del pe&longs;o <lb/>po&longs;to in E: & quale &longs;arà la inclinatione d'uno al mouimento in giù, tale &longs;arà etian <lb/>dio la re&longs;i&longs;tenza dell'altro al mouimento in sù, cioè la re&longs;istentia della violenza del <lb/>pe&longs;o po&longs;to in E nella a&longs;ce&longs;a, contra&longs;tando &longs;i oppone alla naturale po&longs;&longs;anza del pe-<lb/>&longs;o po&longs;to in D per e&longs;&longs;ere a lei eguale; percioche quanto il pe&longs;o po&longs;to in D per la na-<lb/>tural po&longs;&longs;anza de&longs;cende piu velocemente in giù, in tanto il pe&longs;o po&longs;to in E più tar-<lb/>do &longs;ale violentemente. </s> <s id="id.2.1.211.6.0"> Per laqual co&longs;a niuno di loro due pe&longs;era piu dell'altro, non <lb/>procedendo attione da eguale. </s> <s id="id.2.1.211.7.0"> il pe&longs;o po&longs;to in D dunque non mouerà il pe&longs;o po&longs;to <lb/>in E in &longs;u&longs;o, peroche &longs;e lo moue&longs;&longs;e, &longs;arebbe nece&longs;&longs;ario, che il pe&longs;o po&longs;to in D ha-<lb/>ue&longs;&longs;e virtu maggiore in di&longs;cendendo, che il pe&longs;o po&longs;to in E in &longs;alendo, ma que&longs;te co-<lb/>&longs;e &longs;ono eguali: adunque &longs;taranno &longs;ermi i pe&longs;i, & la grauezza del pe&longs;o po&longs;to in D &longs;a-<lb/>rà eguale alla grauezza del pe&longs;o po&longs;to in E. Oltre a ciò perche pre&longs;uppongono, che <lb/>quanto il pe&longs;o è piu di&longs;tante dalla linea FG della dirittura, tanto e&longs;&longs;ere piu graue. <lb/>però tirate parimente da i punti DE le linee DO, EI apiombo di FG, con <lb/>modo &longs;imile &longs;i dimostrerà il triangolo CDO e&longs;&longs;ere eguale al triangolo CEI: & <lb/>la linea DO e&longs;&longs;ere eguale ad EI. Tanto dunque è di&longs;tante il pe&longs;o po&longs;to in D <lb/>dalla linea FG, quanto il pe&longs;o po&longs;to in E. Dalle ragioni loro dunque, & dalle &longs;ue <lb/>pre&longs;uppo&longs;te li pe&longs;i me&longs;&longs;i in DE &longs;ono graui egualmente. </s> <s id="id.2.1.211.8.0"> Di piu, che vieta che non &longs;i di <lb/>mo&longs;tri la bi lancia DE mouer&longs;i per nece&longs;&longs;ità in FG con &longs;imile ragione? </s> <s id="id.2.1.211.9.0"> Primie-<lb/>ramente &longs;i puote raccogliere dalle loro mede&longs;ime dimo&longs;trationi, la &longs;alita del pe&longs;o po-<lb/>&longs;to in E ver&longs;o il B e&longs;&longs;ere piu diritta della &longs;alita del pe&longs;o po&longs;to in D ver&longs;o lo F, <lb/>cioè manco prendere di diretto la &longs;alita del pe&longs;o po&longs;to in D in archi eguali, che la <lb/>&longs;alita del pe&longs;o po&longs;to in E. Pre&longs;upponga&longs;i dunque, che il pe&longs;o &longs;ia piu leggiero &longs;econ-<lb/>do il &longs;ito tanto quanto nel &longs;ito mede&longs;imo meno diritta è la &longs;ua &longs;alita: Laqual pre-<lb/>&longs;upposta pare tanto manife&longs;ta, quanto l'altraloro. </s> <s id="id.2.1.211.10.0"> percioche dunque la &longs;alita del <lb/>pe&longs;o po&longs;to in E è piu diritta della &longs;alita del pe&longs;o po&longs;to in D, per la pre&longs;uppo&longs;ta il <lb/>pe&longs;o po&longs;to in D &longs;arà piu leggiero del pe&longs;o po&longs;to in E. Adunque il pe&longs;o po&longs;to in D <lb/>&longs;i mouerà in sù dal pe&longs;o po&longs;to in E, &longs;i &longs;attamente che la bilancia peruenga in FG, <lb/>& co&longs;i potra&longs;si dimo&longs;trare la bilancia DE mouer&longs;i in FG, laqual dimo&longs;tratio-<lb/>ne è del tutto veramente friuola, & pati&longs;ce le difficultà mede&longs;ime. </s> <s id="id.2.1.211.11.0"> Percioche quan-<lb/>tunque &longs;i conceda, come vero, che il pe&longs;o po&longs;to in E &longs;alendo &longs;ia piu graue del pe&longs;o <lb/>in D &longs;imilmente &longs;alendo, non perciò da que&longs;to &longs;egue, che il pe&longs;o po&longs;to in E de-<lb/>&longs;cendendo &longs;ia piu graue del pe&longs;o posto in D &longs;alendo. </s> <s id="id.2.1.211.12.0"> Niuna dunque di que&longs;te due <lb/>dimo&longs;trationi, che dicono la bilancia DE ritornare in AB, ouero mouer&longs;i in <lb/>FG, è vera.<emph.end type="italics"/> </s> </p> <figure id="fig20" place="text" xlink:href="figures-it/051_01.jpg"></figure> <p type="margin" id="id.2.1.213.0.0"> <s id="id.2.1.213.1.0"> <margin.target id="note54"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 15. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.214.0.0"> <s id="id.2.1.214.1.0"> <margin.target id="note55"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 25. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.215.0.0"> <s id="id.2.1.215.1.0"><emph type="italics"/>Oltre a ciò &longs;e e&longs;amineremo la loro pre&longs;uppo&longs;ta, & la &longs;orza delle loro parole, vedremo <lb/>per certo che altro &longs;entimento hanno. </s> <s id="id.2.1.215.2.0"> Imperoche e&longs;&longs;endo che &longs;empre lo &longs;patio per lo<emph.end type="italics"/> <pb id="p.17v" xlink:href="pageimg-it/053.jpg"/><emph type="italics"/>quale il pe&longs;o natur <expan abbr="alm&etilde;te">almente</expan> &longs;i moue, &longs;i deue prendere dal centro della grauezza di e&longs;-<lb/>&longs;o pe&longs;o ver&longs;o il centro del mondo à &longs;embianza di vna linea diritta tirata dal centro <lb/>della grauezza al centro del mondo, tanto &longs;i dir à que&longs;ta co&longs;i fatta di&longs;ce&longs;a del pe&longs;o <lb/>piu, & meno obliqua, quanto, &longs;econdo lo &longs;patio di&longs;&longs;egnato, a &longs;embianza della pre-<lb/>detta linea piu ò meno &longs;i mouerà, (andando pero &longs;empre a trouare il luogo &longs;uo natu <lb/>rale, & vie piu &longs;empre auicinandoui&longs;i.) talche tanto piu obliqua &longs;i dica la &longs;ce&longs;a <expan abbr="quã">quam</expan> <lb/>to &longs;i parte da cotale &longs;patio: & piu diritta quanto a lui &longs;i acco&longs;ta. </s> <s id="id.2.1.215.3.0"> & in que&longs;to <lb/>&longs;entimento quella pre&longs;upposta non deue partorire difficulta ad alcuno, percioche co-<lb/>&longs;i è la verita &longs;ua chiara, & conforme alla ragione, che non pare hauer me&longs;tieri di e&longs;-<lb/>&longs;er &longs;atta in alcun modo manife&longs;ta.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.216.0.0"> <s id="id.2.1.216.1.0"><emph type="italics"/>Se dunque il pe&longs;o &longs;ciolto, collocato nel &longs;i-<lb/>to di D &longs;i deue mouere al luogo pro-<lb/>prio, &longs;enza dubbio, po&longs;to S centro del <lb/>mondo, &longs;i mouerà per la linea DS, <expan abbr="&longs;i-<lb> milm&etilde;te">&longs;i-<lb/>milmente</expan> il pe&longs;o po&longs;to in E &longs;ciolto &longs;i mo <lb/>uerà per la linea ES. Per laqual co-<lb/>&longs;a &longs;e, (come è vero) la &longs;ce&longs;a del pe&longs;o &longs;i <lb/>dirà piu, ò meno obliqua, &longs;econdo lo al <lb/>lontanar&longs;i, ouero appre&longs;&longs;ar&longs;i a gli &longs;patij <lb/>di&longs;segnati per le linee DS ES, per ri <lb/>&longs;petto a'loro naturali mouimenti ver&longs;o <lb/>iproprij luoghi, egli è chiaro, che meno <lb/>obliqua è la &longs;ce&longs;a di E per EG, che <lb/>di D per DA, per e&longs;&longs;ere stato di <lb/>&longs;opra mo&longs;trato che l'angolo SEG è <lb/>minore dell'angolo SDA. Per laqual <lb/>co&longs;a piu grauer à il pe&longs;o in E, che in D, <lb/>il che totalmente è il contrario di quel-<lb/>lo, che e&longs;si &longs;i &longs;ono s&longs;orzati di prouare. <lb/>Leueran&longs;i per auuentura contra di noi <lb/>dicendo. </s> <s id="id.2.1.216.2.0"> Se dundue il pe&longs;o po&longs;to in E è <lb/>piu graue del pe&longs;o po&longs;to in D, la bi-<lb/>lancia DE non &longs;tar à giamai in que-<lb/>&longs;to &longs;ito, laqual co&longs;a noi habbiamo pro-<lb/>po&longs;to di mantenere, ma &longs;i mouer à in F <lb/>G. </s> <s id="id.2.1.216.3.0"> Allequali co&longs;e ri&longs;pondiamo. che im-<lb/>porta a&longs;&longs;ai, &longs;e noi con&longs;ideriamo i pe&longs;i o-<lb/>uero in quanto &longs;ono &longs;eparati l'uno dal-<lb/>l'altro, ouero in quanto &longs;ono traloro <lb/>congiunti: perche altra è la ragione del<emph.end type="italics"/> <lb/> <arrow.to.target n="fig21"></arrow.to.target> <lb/><emph type="italics"/>pe&longs;o po&longs;to in E &longs;enza il congiungimento del pe&longs;o po&longs;to in D, & altra di lui con <lb/>l'altro pe&longs;o congiunto, &longs;i fattamente che l'uno &longs;enza l'altro non &longs;i po&longs;&longs;a mouere. </s> <s id="id.2.1.216.4.0"> Im<emph.end type="italics"/> <pb id="p.18" xlink:href="pageimg-it/054.jpg"/><emph type="italics"/>peroche la diritta, & naturale di&longs;ce&longs;a dal pe&longs;o po&longs;to in E, inquanto egli è &longs;enza al-<lb/>tro congiungimento di pe&longs;o, &longs;i fa per la linea ES. ma inquanto egli è congiunto <lb/>col pe&longs;o D, la &longs;ua naturale di&longs;ce&longs;a non &longs;arà piu per la linea ES, ma per vna li-<lb/>nea egualmente di&longs;tante da CS. percioche la magnitudine compo&longs;ta de i pe&longs;i ED. <lb/>& della bilancia DE il cui centro della grauezza è C, &longs;e in ne&longs;&longs;un luogo non &longs;a-<lb/>rà &longs;o&longs;tenuta, &longs;i mouerà naturalmente in giu nel modo che &longs;i troua, &longs;econdo la gra-<lb/>uezza del centro per la linea diritta tirata dal centro della grauezza C al centro <lb/>del mondo S, finche il centro C peruenga nel centro S. La bilancia dunque DE <lb/>in&longs;ieme co'pe&longs;i, in quella maniera, che &longs;i troua &longs;i mouerà in giu per modo tale, che il <lb/>punto C &longs;i moua per la linea CS, fin che C peruenga in S, & la bilancia <lb/>DE in HK; & habbia la bilancia in HK la po&longs;itione i&longs;te&longs;&longs;a, che prima hauea; <lb/>cio è, che la HK &longs;ia egualmente distante da DE. Congiungan&longs;i dunque DH <lb/>EK. egli è manife&longs;to, che mentre la bilancia DE &longs;i moue in HK, mouer&longs;i an-<lb/>che ipunti DE per le linee DH EK, come quelle che &longs;ono & &longs;ra &longs;e, & ad<emph.end type="italics"/> <arrow.to.target n="note56"></arrow.to.target> <lb/><emph type="italics"/>e&longs;&longs;a CS eguali, & egualmente di&longs;tanti. </s> <s id="id.2.1.216.5.0"> Per la qual co&longs;a i pe&longs;i posti in DE, in <lb/>quanto &longs;ono &longs;ra loro congiunti, &longs;e riguarderemo il mouimento loro naturale &longs;imoue <lb/>ranno non &longs;econdo le linee DS, ES, ma &longs;econdo LDH MEK egualmente <lb/>di&longs;tanti da e&longs;&longs;a CS. Ma la naturale inclinatione del pe&longs;o po&longs;to in E libero, & <lb/>&longs;ciolto &longs;arà per ES, & del pe&longs;o po&longs;to in D <expan abbr="&longs;imilm&etilde;te">&longs;imilmente</expan> &longs;ciolto &longs;arà per DS. & per-<lb/>cio non è &longs;conueneuole, che il pe&longs;o mede&longs;imo hora in E, hora in D, &longs;ia piu graue <lb/>in E, che in D. Ma&longs;e i pe&longs;i po&longs;ti in ED &longs;ono l'un l'altro fra &longs;e congiunti, & gli <lb/>con&longs;idereremo in quanto &longs;ono congiunti, &longs;arà la naturale inclinatione del pe-<lb/>&longs;o po&longs;to in E per la linea MEK, percioche la grauezza dell'altro pe&longs;o po&longs;to <lb/>in D fa &longs;i, che il pe&longs;o po&longs;to in E non graui &longs;opra la linea ES, ma nella EK. <lb/> Ilche fa parimente la grauezza del pe&longs;o po&longs;to in E, cioè, che il pe&longs;o po&longs;to in D <lb/>non graui per la linea retta DS, ma &longs;econdo DH, per impedir&longs;i ambedue l'uno <lb/>l'altro che non vadino à propri luoghi. </s> <s id="id.2.1.216.6.0"> Concio&longs;ia dunque che la naturale &longs;ce&longs;a dirit-<lb/>ta de i pe&longs;i po&longs;ti in DE &longs;ia &longs;econdo LDH, MEK, &longs;arà &longs;imilmente la naturale <lb/>&longs;alita diritta loro &longs;econdo le i&longs;te&longs;&longs;e linee HDL KEM. & la naturale &longs;alita del <lb/>pe&longs;o po&longs;to in E &longs;i dirà più, & meno torta, quanto che &longs;econdo lo &longs;patio &longs;i mouerà <lb/>più, & meno pre&longs;&longs;o la linea MK. & a que&longs;to modo in tutto &longs;i ha da pigliare & la &longs;a <lb/>lita & la di&longs;ce&longs;a del pe&longs;o po&longs;to in D &longs;econdo la linea LH, &longs;e dunque il pe&longs;o po&longs;to <lb/>in E &longs;i moue&longs;&longs;e in giù per la linea EG, mouerebbe il pe&longs;o po&longs;to in D in sù per <lb/>DF. & percioche l'angolo CEK è eguale all'angolo CDL, & l'angolo CEG<emph.end type="italics"/> <arrow.to.target n="note57"></arrow.to.target> <lb/><emph type="italics"/>è eguale all'angolo CDF; &longs;arà il <expan abbr="re&longs;tãte">re&longs;tante</expan> angolo GEK al re&longs;tante LDF egua <lb/>le. </s> <s id="id.2.1.216.7.0"> & e&longs;&longs;endo quella pre&longs;uppo&longs;ta, che dice il pe&longs;o e&longs;&longs;er più graue &longs;econdo il &longs;ito, <lb/>quanto in quel mede&longs;imo &longs;ito la di&longs;ce&longs;a è meno obliqua per chiara, & manife&longs;ta ri-<lb/>ceuuta, &longs;arà anche da e&longs;&longs;ere accettata &longs;enza dubbio que&longs;t' altra, cioè, che il pe&longs;o &longs;arà <lb/>più graue &longs;econdo il &longs;ito, quanto nel &longs;ito mede&longs;imo meno obliqua &longs;arà la &longs;alita; per <lb/>non e&longs;&longs;ere manco manife&longs;ta, ne meno conforme alla ragione. </s> <s id="id.2.1.216.8.0"> &longs;arà dunque eguale <lb/>la &longs;ce&longs;a del pe&longs;o po&longs;to in E alla &longs;alita del pe&longs;o po&longs;to in D, percioche la &longs;ce&longs;a del pe <lb/>&longs;o po&longs;to in E tiene tanto di obliquo, quanto la &longs;alita del pe&longs;o po&longs;to in D. & quale <pb id="p.18v" xlink:href="pageimg-it/055.jpg"/>&longs;arà la inclinatione dell'vno al moui-<lb/>mento in giù, tale parimente &longs;arà la re <lb/>&longs;istenza dell'altro al mouimento in sù.<lb/> Adunque il pe&longs;o po&longs;to in E non mo-<lb/>uerà in sù il pe&longs;o po&longs;to in D: ne il pe&longs;o <lb/>po&longs;to in D: &longs;i mouerà in giù &longs;i fatta-<lb/>mente, che moua in sù il pe&longs;o po&longs;to in <lb/>E. imperoche e&longs;&longs;endo l'angolo CEB <lb/>egualea CDA, & l'angolo CEM <lb/>&longs;ia eguale all'angolo CDH; &longs;arà il <lb/>re&longs;tante MEB eguale al re&longs;tante<emph.end type="italics"/> <lb/> <arrow.to.target n="note116"></arrow.to.target><emph type="italics"/>HDA. La &longs;ce&longs;a dunque del pe&longs;o po-<lb/>&longs;to in D &longs;arà eguale alla &longs;alita del pe-<lb/>&longs;o po&longs;to in E. Adunque il pe&longs;o po&longs;to <lb/>in D non mouerà in sù il pe&longs;o po&longs;to <lb/>in E. Dalle quali co&longs;e &longs;egue che i pe&longs;i <lb/>po&longs;ti in DE, in quanto tra loro &longs;o-<lb/>no congiunti, &longs;ono egualmente graui.<emph.end type="italics"/> </s> </p> <figure id="fig21" place="text" xlink:href="figures-it/053_01.jpg"></figure> <p type="margin" id="id.2.1.219.0.0"> <s id="id.2.1.219.1.0"> <margin.target id="note56"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 35. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.220.0.0"> <s id="id.2.1.220.1.0"> <margin.target id="note57"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 29. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.379.0.0"> <s id="id.2.1.379.1.0"> <margin.target id="note116"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 29. <emph type="italics"/>del prime.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.380.0.0" xlink:href="figures-it/055_01.jpg"></figure> <p type="main" id="id.2.1.381.0.0"> <s id="id.2.1.381.1.0"><emph type="italics"/>L'altra ragione po&longs;cia, con laquale vorrebbono mo&longs;trare, che &longs;imilmente la bilancia <lb/>DE ritorna in AB, con dire, che e&longs;&longs;endo la trutina della bilancia CF, la méta <lb/>viene ad e&longs;&longs;er CG. & percioche l'angolo DCG è maggiore dell'angolo ECG, <lb/>il pe&longs;o po&longs;to in D &longs;arà più graue del po&longs;to in E; dunque la bilancia DE ritorne <lb/>ra in AB; non conchiude nulla al parer mio; & que&longs;ta fintione della trutina, & <lb/>della méta è più to&longs;to da trala&longs;ciare, & pa&longs;&longs;arla con &longs;ilentio, che farne pur vna paro <lb/>la per confonderla, e&longs;&longs;endo del tutto co&longs;a volontaria, percioche la nece&longs;&longs;aria ragione <lb/>per laquale il pe&longs;o po&longs;to in D dall' angolo maggiore &longs;ia più graue, & perche il mag <lb/>giore angolo &longs;ia cagione di grauezzamaggiore non appare in niun loco. </s> <s id="id.2.1.381.2.0"> che &longs;e gli <lb/>angoli &longs;aranno tra loro paragonati, e&longs;&longs;endo l'angolo GCD eguale all'angolo <lb/>FCE; &longs;e l'angolo GCD è cau&longs;a della grauezza, perche l'angolo FCE &longs;imil-<pb id="p.19" xlink:href="pageimg-it/056.jpg"/>mente non è della grauez <lb/>za cagione? </s> <s id="id.2.1.375.3.0"> Di questo ef <lb/>fetto mostrano di produ-<lb/>cere in mezo que&longs;ta cagio <lb/>ne, perche CG è la mé-<lb/>ta, & CF la trutina; <lb/>&longs;e (dicono e&longs;&longs;i) CG fo&longs; <lb/>&longs;e la trutina, & CF la <lb/>méta, all'hora l'angolo <lb/>FCE &longs;arebbe cagione <lb/>della grauezza, ma non <lb/>già il DCG ad e&longs;&longs;o e-<lb/>guale laquale ragione è al <lb/>tutto fatta con la imagi-<lb/>natione, & di voglia pro <lb/>pria. </s> <s id="id.2.1.375.4.0"> Peroche, che puote <lb/>importare che la trutina <lb/>&longs;ia ouero in CF, ouero <lb/>in CG, e&longs;&longs;endo la bilan <lb/>cia DE &longs;empre &longs;o&longs;ten-<lb/>tata nell'i&longs;te&longs;&longs;o punto C? Ma affine che l'inganno loro re&longs;ti più chiaro.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.376.0.0" xlink:href="figures-it/056_01.jpg"></figure> <p type="main" id="id.2.1.377.0.0"> <s id="id.2.1.377.1.0"><emph type="italics"/>Sia la mede&longs;ima bilancia AB, il cui mezo C. dapoitutta la FG &longs;ia la trutina, <lb/>laquale &longs;tia immobile, & &longs;o&longs;tenga la bilancia AB nel punto C. & moua&longs;i la <lb/>bilancia in DE. & per-<lb/>cioche la trutina è &longs;opra, & <lb/>&longs;otto la bilancia, quale ango <lb/>lo &longs;arà cagione della grauez <lb/>za, e&longs;&longs;endo &longs;o&longs;tenuta la bi-<lb/>lancia DE &longs;empre nel pun <lb/>to mede&longs;imo? </s> <s id="id.2.1.377.2.0"> Diranno for-<lb/>&longs;e &longs;e la trutina &longs;arà &longs;o&longs;tenu-<lb/>ta dalla po&longs;&longs;anza po&longs;ta in <lb/>F, allhora CG &longs;arà tan-<lb/>to quanto la méta, & l'an-<lb/>golo DCG &longs;arà della gra <lb/>uezza cagione. </s> <s id="id.2.1.377.3.0"> Ma&longs;e<emph.end type="italics"/> <lb/> <arrow.to.target n="fig35"></arrow.to.target> <lb/><emph type="italics"/>egli &longs;arà &longs;ostenuto in G, allhora FCE &longs;arà cagione della grauezza, & la CF <lb/>&longs;arà tanto quanto la méta. </s> <s id="id.2.1.377.4.0"> della qual co&longs;a niuna cagione pare poter&longs;i addurre, <lb/>&longs;e <expan abbr="nõ">non</expan> imaginata; peroche la méta (che dicono) non pare hauere à modo veruno nien <lb/>te di virtù che tiri dalla parte dell'angolo maggiore alcuna volta, & alcuna dalla <lb/>parte del minore. </s> <s id="id.2.1.377.5.0"> Ma &longs;ia &longs;o&longs;tenuta la trutina da due po&longs;&longs;anze in F cioè, & in G,<pb id="p.19v" xlink:href="pageimg-it/057.jpg"/>ilche &longs;i puote fare per nece&longs;&longs;ità, come &longs;e la po&longs;&longs;anza posta in F &longs;o&longs;&longs;e tanto debile, <lb/>che per &longs;e &longs;te&longs;&longs;a pote&longs;&longs;e &longs;o&longs;tentare &longs;olamente la metà del pe&longs;o & &longs;ia la po&longs;&longs;anza <lb/>posta in G eguale alla po&longs;&longs;anza po&longs;ta in F, & ambedue in&longs;ieme co' pe&longs;i &longs;o&longs;tenga-<lb/>no la bilancia. </s> <s id="id.2.1.381.3.0"> all'hora quale angolo &longs;arà cagione della grauezza? </s> <s id="id.2.1.381.4.0"> non gia <lb/>FCE, peroche la trutina è <lb/>in CF, & è &longs;o&longs;tentata in <lb/>F: ne meno il DCG, e&longs;&longs;en <lb/>do la trutina in CG, & pa <lb/>rimente &longs;o&longs;tentata in G. <lb/>Non &longs;aranno dunque gli an <lb/>goli della grauezza cagione.<lb/> Co&longs;i ne anche la bilancia <lb/>DE da que&longs;to &longs;ito per que <lb/>&longs;ta cagione &longs;i mouerà. </s> <s id="id.2.1.381.5.0"> Ma<emph.end type="italics"/> <lb/> <arrow.to.target n="note117"></arrow.to.target><emph type="italics"/>que&longs;ta loro &longs;entenza pare <lb/>e&longs;&longs;ere confermata da e&longs;&longs;i in <lb/>due modi. </s> <s id="id.2.1.381.6.0"> Primieramente<emph.end type="italics"/> <lb/> <arrow.to.target n="fig36"></arrow.to.target> <lb/><emph type="italics"/>dicono Ari&longs;totele nelle que&longs;tioni mecaniche hauere propo&longs;to que&longs;te due que&longs;tioni &longs;o <lb/>lamente, & le &longs;ue dimo&longs;trationi e&longs;&longs;ere fondate &longs;i nel maggiore, & nel minore <lb/>angolo, & &longs;i nella giacitura della trutina della bilancia. </s> <s id="id.2.1.381.7.0"> Affermano dapoi que&longs;to <lb/>iste&longs;&longs;o in&longs;egnare la e&longs;perientia ancora, cioè, che la bilancia DE, &longs;tando la &longs;ua <lb/>trutina in CF, ritorna in AB egualmente di&longs;tante dall'orizonte. </s> <s id="id.2.1.381.8.0"> & quando <lb/>la trutina &longs;tà in CG, mouer&longs;i in FG. Mane Ari&longs;totele, ne la e&longs;perienza fauo-<lb/>ri&longs;cono que&longs;ta loro opinione, anzi più to&longs;to le &longs;ono contrarij. </s> <s id="id.2.1.381.9.0"> Peroche in quan-<lb/>to appartiene alla e&longs;perienza &longs;i ingannano, e&longs;&longs;endo mani&longs;e&longs;to ciò per e&longs;perienza <lb/>accadere, all'hor che il centro ancora della bilancia &longs;arà collocato ò &longs;opra, ò &longs;ot-<lb/>to della bilancia, ma non già auenire que&longs;to stando la trutina ò &longs;opra &longs;olamente, <lb/>è &longs;otto.<emph.end type="italics"/> </s> </p> <pb id="p.20" xlink:href="pageimg-it/058.jpg"/> <figure id="fig35" place="text" xlink:href="figures-it/056_02.jpg"></figure> <figure id="fig36" place="text" xlink:href="figures-it/057_01.jpg"></figure> <p type="margin" id="id.2.1.384.0.0"> <s id="id.2.1.384.1.0"> <margin.target id="note117"></margin.target><emph type="italics"/>il Cardano.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.385.0.0"> <s id="id.2.1.385.1.0"><emph type="italics"/>Imperoche &longs;e la bilancia A <lb/>B haue&longs;&longs;e il centro C <lb/>&longs;opra la bilancia, & fo&longs;-<lb/>&longs;e la trutina CD &longs;otto <lb/>la bilancia, & &longs;i moue&longs;-<lb/>&longs;e la bilancia in EF, al <lb/>lhora EF di nouo ri-<lb/>tornerà in AB. egual-<lb/>mente di&longs;tante dall'o-<lb/>rizonte. </s> <s id="id.2.1.385.2.0"> &longs;imilmente &longs;e la <lb/>bilancia haue&longs;&longs;e il cen-<lb/>tro C &longs;otto la bilancia, <lb/>& &longs;o&longs;&longs;e la trutina CD <lb/>&longs;opra la bilancia, et &longs;i mo <lb/>ue&longs;&longs;e la bilancia in EF,<emph.end type="italics"/> <arrow.to.target n="note118"></arrow.to.target> <lb/><emph type="italics"/>egli è manife&longs;to, che la bi <lb/>lancia &longs;i mouerà in giu <lb/>dalla parte di F, &longs;tan-<lb/>do la trutina &longs;opra la bi-<lb/>lancia. </s> <s id="id.2.1.385.3.0"> & in qual &longs;i vo-<lb/>glia altro &longs;ito che &longs;ia la <lb/>trutina, auerrà &longs;empre il <lb/>mede&longs;imo. </s> <s id="id.2.1.385.4.0"> Adunque <expan abbr="nõ">non</expan> <lb/>è la trutina, ma il centro <lb/>della bilancia cagione di <lb/>cotali diuer&longs;i effetti.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.386.0.0"> <s id="id.2.1.386.1.0"> <margin.target id="note118"></margin.target><emph type="italics"/>Per la terza di questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.387.0.0" xlink:href="figures-it/058_01.jpg"></figure> <figure place="text" id="id.2.1.387.1.0" xlink:href="figures-it/058_02.jpg"></figure> <p type="main" id="id.2.1.388.0.0"> <s id="id.2.1.388.1.0"><emph type="italics"/>Egli è pero d'auertire in que&longs;ta parte che con di&longs;&longs;icultà &longs;i puote lauora re vna bilancia <lb/>materiale, che in vno punto &longs;olamente &longs;ia &longs;o&longs;tenuta, &longs;i come con la mente la imagi-<lb/>niamo, & habbia le braccia dal centro co&longs;i eguali non &longs;olamente in lunghezza, ma <lb/>in larghezza, & in profundità, ò gro&longs;&longs;ezza, che tutte le parti di quà, & di là pe&longs;i-<lb/>no a punto egualmente. </s> <s id="id.2.1.388.2.0"> percio che la materia di&longs;&longs;icili&longs;&longs;imam ente pati&longs;ce cotale giu-<lb/>&longs;ta mi&longs;ura. </s> <s id="id.2.1.388.3.0"> Per laqual co&longs;a &longs;e con&longs;idereremo il centro e&longs;&longs;ere in e&longs;&longs;a bilancia, non bi-<lb/>&longs;ogna ricorrere al &longs;en&longs;o, concio&longs;ia, che le co&longs;e artificiate non &longs;i po&longs;&longs;ano ridurre a quel <lb/>fommo grado di per&longs;ettione. </s> <s id="id.2.1.388.4.0"> Ma nelle altre co&longs;e la e&longs;perienza veramente potrà in&longs;e <lb/>gnare le co&longs;e che appaiono percioche <expan abbr="quãtunque">quantunque</expan> ilcentro della <expan abbr="bilãcia">bilancia</expan> &longs;empre &longs;ia vn <lb/>punto, nondimeno quando egli &longs;arà &longs;opra la bilancia, poco importa, &longs;e ben la bilancia <lb/>non &longs;ara &longs;o&longs;tenuta in quel punto co&longs;i puntalmente però che per e&longs;&longs;ere &longs;empre &longs;opra la <lb/>bilancia auerrà &longs;empre il mede&longs;imo. </s> <s id="id.2.1.388.5.0"> Con &longs;imile modo, quando egli anco è &longs;otto la bi-<lb/>lancia, ilche tuttauia non accade stando il centro in e&longs;&longs;a bilancia, per che &longs;e egli non <lb/>&longs;arà &longs;o&longs;tenuto &longs;empre in quel mezo accuratamente, &longs;ara differenza, e&longs;&longs;endo co&longs;a faci <lb/>li&longs;&longs;ima, che quel centro, muti il proprio &longs;ito, mentre &longs;i moue la bilancia.<emph.end type="italics"/> </s> </p> <pb id="p.20v" xlink:href="pageimg-it/059.jpg"/> <p type="main" id="id.2.1.390.0.0"> <s id="id.2.1.390.1.0"><emph type="italics"/>Ma che Ari&longs;totele habbia <lb/>propo&longs;to due que&longs;tioni &longs;o <lb/>lamente, cioè perche la <lb/>trutina &longs;tando &longs;opra, &longs;e <lb/>la bilancia <expan abbr="nõ">non</expan> &longs;arà egual <lb/>mente di&longs;tante dall'ori-<lb/>zonte in equilibrio, cioè <lb/>egualmente di&longs;tante dal <lb/>orizonte ritorna, ma &longs;e la <lb/>trutina &longs;ara po&longs;ta &longs;otto <lb/>non ritorna, ma di piu &longs;i <lb/>moue <expan abbr="&longs;ecõdo">&longs;econdo</expan> la parte ba&longs; <lb/>&longs;a: egli è verò per certo. <lb/>Ma non già per que&longs;to le <lb/>dimo&longs;trationi &longs;ue &longs;ono <lb/>&longs;ondate nell'angolo mag <lb/>giore, ò minore, & nella <lb/>giacitura della trutina, <lb/>come e&longs;&longs;i dicono: per cio-<lb/>che in questo non com-<lb/>prendono la <expan abbr="m&etilde;te">mente</expan> del filo <lb/>&longs;ofo, che a&longs;&longs;egna la ragio <lb/>ne de gli effetti diuer&longs;i <lb/>de'mouimenti della bilan <lb/>cia. </s> <s id="id.2.1.390.2.0"> peroche tanto è lon-<lb/>tano, che il filo&longs;o&longs;o attri <lb/>bui&longs;ca que&longs;ti diuer&longs;i effet<emph.end type="italics"/> <lb/> <arrow.to.target n="fig37"></arrow.to.target> <lb/><emph type="italics"/>ti à gli angoli, che piu to&longs;to dica e&longs;&longs;ere cagione l'ecce&longs;&longs;o, & quel &longs;opra più della gran <lb/>dezza che è dal perpendicolo dell'uno delle braccia della bilancia hor dall'una parte, <lb/>hora dall'altra.<emph.end type="italics"/> </s> </p> <figure id="fig37" place="text" xlink:href="figures-it/059_01.jpg"></figure> <figure id="fig37b" place="text" xlink:href="figures-it/059_02.jpg"></figure> <p type="main" id="id.2.1.392.0.0"> <s id="id.2.1.392.1.0"><emph type="italics"/>Come stando la trutina &longs;opra in CF, il perpendicolo &longs;arà FCG, il quale &longs;em-<lb/>pre inchina, &longs;econdo lui, ver&longs;o il centro del mondo, il quale anco diuide la bilancia mo&longs; <lb/>&longs;a in DE in parti di&longs;uguali: & la parte maggiore è ver&longs;o il D, & quel che è piu, <lb/>inchina in giu. </s> <s id="id.2.1.392.2.0"> Adunque dalla parte di D la bilancia &longs;i mouerà in giu fin che ri-<lb/>torni in AB. Ma &longs;e la trutina &longs;arà in CG di &longs;otto, &longs;arà GCF il perpendico-<lb/>lo, ilquale diuiderà parimente la bilancia DE in parte di&longs;uguali, & la parte mag <lb/>giore &longs;arà ver&longs;o E; Per laqual co&longs;a la bilancia &longs;i mouerà in giu dalla parte di E. <lb/>& accioche que&longs;to &longs;ia dirittamente compre&longs;o, &longs;appia&longs;i, che quando la trutina è &longs;o-<lb/>pra la bilancia, &longs;i ha da intendere, che anche il centro della bilancia &longs;ia &longs;opra la bi-<lb/>lancia, & &longs;e di &longs;otto, anche il centro deue &longs;tare di &longs;otto, come piu a ba&longs;&longs;o manife&longs;te-<lb/>ra&longs;&longs;i. Altramente la dimo&longs;tratione di Ari&longs;totele non conchiuderebbe nulla, pero <lb/>che stando il centro in e&longs;&longs;a bilancia, come in C moua&longs;i la bilancia in qual &longs;i voglia<emph.end type="italics"/> <pb id="p.21" xlink:href="pageimg-it/060.jpg"/><emph type="italics"/>modo, il perpendicolo FG non diuiderà giamai la bilancia &longs;e non nel punto C, et <lb/>in parti eguali. </s> <s id="id.2.1.392.3.0"> Onde la &longs;entenza di Ari&longs;totele non &longs;olamente non gli &longs;auori&longs;ce, ma <lb/>gli fa anche grandi&longs;sima <lb/>mente contra. il che <lb/>non &longs;olamente è chiaro <lb/>dalla &longs;econda & terza <lb/>propo&longs;itione di que&longs;to li <lb/>bro, ma anco percioche <lb/>&longs;tando il centro &longs;opra <lb/>la bilancia, il pe&longs;o alzato <lb/>acqui&longs;ta grauezza mag <lb/>giore per cau&longs;a del &longs;ito. <lb/> </s> <s id="id.2.1.392.4.0">Dalla qual co&longs;a accade il <lb/>ritorno della bilancia ad <lb/>eguale di&longs;tanza dall'ori-<lb/>zonte. </s> <s id="id.2.1.392.5.0"> Ma per lo con-<lb/>trario auiene quando il <lb/>centro è &longs;otto la bilan-<lb/>cia. </s> <s id="id.2.1.392.6.0"> Le quali co&longs;e tutte <lb/>&longs;i dimo&longs;treranno in que-<lb/>&longs;ta maniera, pre&longs;uppo-<lb/>nendo le co&longs;e, che di &longs;o-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig38"></arrow.to.target> <lb/><emph type="italics"/>pra furono dechiarate, cioè il pe&longs;o &longs;ar&longs;i più graue da quelloco dal quale &longs;cende piu <lb/>dirittamente, & da quello che egli &longs;ale piu dirittamente far&longs;i parimente piu <lb/>graue.<emph.end type="italics"/> </s> </p> <pb id="p.21v" xlink:href="pageimg-it/061.jpg"/> <figure id="fig38" place="text" xlink:href="figures-it/060_01.jpg"></figure> <p type="main" id="id.2.1.395.0.0"> <s id="id.2.1.395.1.0"><emph type="italics"/>Sia la bilancia AB egualmente di&longs;tante dall'orizonte, il cui centro C &longs;ia &longs;opra la <lb/>bilancia, & &longs;ia il perpendicolo CD: & &longs;iano i centri della grauezza di pe&longs;i eguali <lb/>po&longs;ti in AB: & la bilancia &longs;ia mo&longs;&longs;a in EF. Dico, che il pe&longs;o posto in E ha <lb/>grauezzamaggiore, che il <lb/>pe&longs;o posto in F. & per-<lb/>ciò la bilancia EF e&longs;&longs;e-<lb/>re per ritornare in AB. <lb/> &longs;ia allungata prima la linea <lb/>CD fin'al centro del mon <lb/>do, che &longs;ia S. Dapoi &longs;ia-<lb/>no congiunte le linee AC, <lb/>CB, EC, CF, HS; <lb/>& dai punti EF &longs;iano ti-<lb/>rate le linee EKGFL egual <lb/> <expan abbr="m&etilde;te">mente</expan> di&longs;tanti da HS. Per-<lb/>cioche dunque la di&longs;ce&longs;a na <lb/>turale diritta di tutta la <lb/>grandezza, cioè della bilan <lb/>cia EF co&longs;i di&longs;po&longs;ta in&longs;ie <lb/>me co'pe&longs;i è &longs;econdo la gra-<lb/>uezza del centro H per la <lb/>dirittalinea HS; &longs;arà pa <lb/> <expan abbr="rim&etilde;te">rimente</expan> la di&longs;ce&longs;a de'pe&longs;ime&longs; <lb/>&longs;i in EF co&longs;i di&longs;po&longs;ti &longs;econ <lb/>do le linee diritte E<emph.end type="italics"/>K <lb/><emph type="italics"/>FL egualmente distanti <lb/>da HS, &longs;i come di &longs;opra <lb/>habbiamo dimo&longs;trato. </s> <s id="id.2.1.395.2.0"> La <lb/>di&longs;ce&longs;a dunque, & la &longs;ali-<lb/>ta de i pe&longs;i po&longs;ti in EF &longs;i <lb/>dirà più, & meno obliqua <lb/>&longs;econdo la vicinanza, ò lon <lb/>tananza diputata &longs;econdo<emph.end type="italics"/> <lb/> <arrow.to.target n="note119"></arrow.to.target><emph type="italics"/>le linee EK FL. & per-<lb/>cioche li due lati AD DC <lb/>&longs;ono eguali a i due lati BD<emph.end type="italics"/> <lb/> <arrow.to.target n="fig39"></arrow.to.target> <lb/><emph type="italics"/>DC; & gli angoli al D &longs;ono retti, &longs;arà il lato AC eguale al lato CB. & e&longs;-<lb/>&longs;endo il punto C immobile; mentre, che i punti AB &longs;imoueranno, de &longs;criueran-<lb/>no la circonferenza di vno cerchio, il cui mezo diametro &longs;arà AC. Per laqual co<emph.end type="italics"/> <lb/> <arrow.to.target n="note120"></arrow.to.target><emph type="italics"/>&longs;a co'l centro C &longs;ia de&longs;critto il cerchio AE BF, i punti AB E<emph.end type="italics"/>F <emph type="italics"/>&longs;aranno nel <lb/>la circonferenza del cerchio. </s> <s id="id.2.1.395.3.0"> ma e&longs;&longs;endo EF eguale ad AB, &longs;arà la circonfe-<lb/>renza EAF eguale alla circonferenza AFB. Onde tolta via la comune AF<emph.end type="italics"/> <pb id="p.22" xlink:href="pageimg-it/062.jpg"/><emph type="italics"/>&longs;arà la circonferenza EA eguale alla cir conferenza FB. Hor percioche l'ango-<lb/>lo mi&longs;to CEA è eguale al mi&longs;to CFB, & HFB è maggiore di CFB, &<emph.end type="italics"/> <arrow.to.target n="note121"></arrow.to.target> <lb/><emph type="italics"/>l'angolo HEA è minore di CEA; &longs;arà l'angolo HFB maggiore dell'angolo <lb/>HEA. Da quali &longs;e &longs;aranno leuati via gli angoli HFG HEK eguali, &longs;arà l'an <lb/>golo GFB maggiore dell'angolo KEA. Adunque la di&longs;ce&longs;a del pe&longs;o po&longs;to in <lb/>E &longs;arà meno obliqua della &longs;alita del pe&longs;o po&longs;to in F. & <expan abbr="quātunque">quantunque</expan> il pe&longs;o po&longs;to in E <lb/>de&longs;cendendo, & il pe&longs;o po&longs;to in F &longs;alendo &longs;i mouino per eguali circonferenze, nondi <lb/>meno percioche il pe&longs;o po&longs;to in E da que&longs;to luogo di&longs;cende piu dirittamente di quel <lb/>che il pe&longs;o F <expan abbr="a&longs;c&etilde;de">a&longs;cende</expan>:pero la naturale po&longs;&longs;anza del pe&longs;o po&longs;to in E &longs;upererà la <expan abbr="re&longs;i&longs;t&etilde;">re&longs;i&longs;tem</expan> <lb/>za della violentia del pe&longs;o F. Onde grauezza maggiore hauerà il pe&longs;o posto in E, <lb/>che il pe&longs;o po&longs;to in F. Adunque il pe&longs;o po&longs;to in E &longs;i mouerà in giù & il pe&longs;o po&longs;to <lb/>in F in sù, fin che la bilancia EF ritorni in AB, che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig39" place="text" xlink:href="figures-it/061_01.jpg"></figure> <p type="margin" id="id.2.1.397.0.0"> <s id="id.2.1.397.1.0"> <margin.target id="note119"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 4. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.398.0.0"> <s id="id.2.1.398.1.0"> <margin.target id="note120"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 28. <emph type="italics"/>del terzo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.399.0.0"> <s id="id.2.1.399.1.0"> <margin.target id="note121"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 29. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.400.0.0"> <s id="id.2.1.400.1.0"><emph type="italics"/>La ragione di que&longs;to effetto po&longs;ta da Ari&longs;totele qui &longs;i puote vedere manife&longs;ta. </s> <s id="id.2.1.400.2.0"> Percio-<emph.end type="italics"/> <arrow.to.target n="note122"></arrow.to.target> <lb/><emph type="italics"/>che &longs;ia il punto N doue le linee CS EF &longs;i tagliano in&longs;ieme. </s> <s id="id.2.1.400.3.0"> & percioche HE <lb/>è eguale ad HF; &longs;arà NE maggiore di NF. adunque la linea CS, che no-<lb/>ma perpendicolo, diuiderà la bilancia EF in parti di&longs;uguali. </s> <s id="id.2.1.400.4.0"> concio&longs;ia dunque, che <lb/>la parte della bilancia NE &longs;ia maggiore della NF, & quel che è di più bi&longs;o-<lb/>gni, che &longs;ia portato in giù, la bilancia EF dalla parte di E &longs;i mouerà in giu finche <lb/>ritorni in AB.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.401.0.0"> <s id="id.2.1.401.1.0"> <margin.target id="note122"></margin.target><emph type="italics"/>Ragione de Aristotele.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.402.0.0"> <s id="id.2.1.402.1.0"><emph type="italics"/>Oltre à cio da quelle co&longs;e, che <lb/>fin hora &longs;ono &longs;tate dette, <lb/>&longs;i puote affermare, la bilan <lb/>cia EF da quel &longs;ito mo-<lb/>uer&longs;i piu velocemente in <lb/>AB; d'onde la linea EF <lb/>allungata a dirittura per-<lb/>uenga nel centro del mon-<lb/>do. </s> <s id="id.2.1.402.2.0"> come &longs;ia EFS vna <lb/>linea diritta. </s> <s id="id.2.1.402.3.0"> & percioche <lb/>CD CK &longs;ono tra loro <lb/>eguali. </s> <s id="id.2.1.402.4.0"> &longs;e dunque col cen-<lb/>tro C, & con lo &longs;patio <lb/>CD &longs;i de&longs;criuerà il cerchio <lb/>DHM, &longs;aranno i punti <lb/>DH nella circonferenza <lb/>del cerchio. </s> <s id="id.2.1.402.5.0"> Ma perche la <lb/>CH è à piombo di EF, <lb/>toccherà la EHS il cer-<lb/>chio DHM nel punto <lb/>H. il pe&longs;o dunque po&longs;to in <lb/>H, (&longs;i come di &longs;opra hab <lb/>biamo prouato) &longs;arà piu<emph.end type="italics"/> <lb/> <arrow.to.target n="fig40"></arrow.to.target> <pb id="p.22v" xlink:href="pageimg-it/063.jpg"/><emph type="italics"/>graue che in verun altro &longs;ito del cerchio DHM. Adunque la grandezza fatta de' <lb/>pe&longs;i EF, & della bilancia EF, il cui centro della grauczza sta in H, in cote&longs;to <lb/>&longs;ito grauerà più, che in qual &longs;i voglia altro &longs;ito del cerchio &longs;i troui il punto H. Da <lb/>que&longs;to &longs;ito adunque &longs;i mouera piu velocemente che da qualunque altro. </s> <s id="id.2.1.402.6.0"> & &longs;e lo H <lb/>&longs;arà piu da pre&longs;&longs;o al D <lb/>manco grauerà, & me-<lb/>no &longs;i mouerà da quel &longs;ito; <lb/>peroche &longs;empreè piu torta <lb/>la &longs;ce&longs;a, & meno diritta. <lb/> La bilancia dunque EF <lb/>&longs;i mouerà più velocemen-<lb/>te da que&longs;to &longs;ito, che da <lb/>altro &longs;ito, & &longs;e piu dapre&longs; <lb/>&longs;o acco&longs;teraßi ad AB, <lb/>d'indi &longs;i mouerà meno poi <lb/>quanto piu da lunge &longs;arà <lb/>di&longs;tante il punto H dal <lb/>punto C &longs;i mouerà più ve <lb/>locemente, il che non &longs;olo <lb/>da Ari&longs;totele nel principio <lb/>delle que&longs;tioni mecaniche, <lb/>& dai detti di &longs;opra è ma <lb/>nife&longs;to, ma ancora da quel <lb/>le co&longs;e, che di &longs;otto nella <lb/>&longs;e&longs;ta propo&longs;itione &longs;iamo <lb/>per dire, apparerà chiaro. <lb/> La bilancia dunque EF <lb/>quanto più &longs;arà lontana <lb/>dal &longs;uo centro, &longs;i mouerà anche piu velocemente.<emph.end type="italics"/> </s> </p> <figure id="fig40" place="text" xlink:href="figures-it/062_01.jpg"></figure> <figure place="text" id="id.2.1.404.0.0" xlink:href="figures-it/063_01.jpg"></figure> <pb id="p.23" xlink:href="pageimg-it/064.jpg"/> <p type="main" id="id.2.1.406.0.0"> <s id="id.2.1.406.1.0"><emph type="italics"/>Sia poi labilancia AB, il cui centro C stia &longs;otto la bilancia, & &longs;iano in AB <lb/>pe&longs;i eguali, & &longs;ia mo&longs;&longs;a la bilancia in EF. Dico che il pe&longs;o ha grauezza maggio-<lb/>re in F, che in E. & <lb/>perciò la bilancia EF <lb/>e&longs;&longs;ere per mouer&longs;i in giù <lb/>dalla parte di F. &longs;ia allun <lb/>gata la linea DC dall'una <lb/>parte, & dall'altra fin <lb/>nel centro del mondo S, <lb/>& fin ad O, & &longs;ia tira <lb/>tala linea HS, alla qua <lb/>le dai punti EF &longs;iano ti<lb/>rate le linee GEK FL <lb/>egualmente di&longs;tanti, & <lb/>&longs;iano congiunte le CE <lb/>CF: & dal centro C <expan abbr="cõ">com</expan> <lb/>lo &longs;patio CE de&longs;criua&longs;i <lb/>il cerchio AEO BF. <lb/>&longs;i dimo&longs;trerà &longs;imilmente <lb/>i punti AB EF e&longs;&longs;e-<lb/>re nella circonferenza del <lb/>cerchio, & che la di&longs;ce&longs;a <lb/>della bilancia EF in&longs;ie-<lb/>me co'pe&longs;i &longs;i fà diritta &longs;e <lb/>condo la linea HS: & <lb/>de ipe&longs;i po&longs;ti in EF &longs;e-<lb/>condo le linee GK FL <lb/>egualmente di&longs;tanti da <lb/>HS. Et percioche l'ango <lb/>lo CFP è eguale all'an <lb/>golo CEO &longs;arà l'ango-<lb/>lo HFP maggiore del-<lb/>l'angolo HEO. ma l'an<emph.end type="italics"/> <arrow.to.target n="note123"></arrow.to.target> <lb/><emph type="italics"/>golo HFL è eguale al-<lb/>l'angolo HEG. Da qua <lb/>li &longs;e &longs;aranno leuati via <lb/>gli angoli HFP HEO,<emph.end type="italics"/> <lb/> <arrow.to.target n="fig41"></arrow.to.target> <lb/><emph type="italics"/>&longs;arà l'angolo LFP minore dell' angolo GEO. Per laqual co&longs;a la &longs;ce&longs;a del pe&longs;o <lb/>po&longs;to in F &longs;arà piu diritta della a&longs;ce&longs;a del pe&longs;o po&longs;to in E. Adunque la po&longs;&longs;anza <lb/>naturale del pe&longs;o po&longs;to in F &longs;upererà la re&longs;i&longs;tenza della violentia del pe&longs;o po&longs;to in <lb/>E. & percio hauerà maggior grauezza il pe&longs;o di F, che il pe&longs;o di E. Adunque <lb/>il pe&longs;o di F &longs;imouer à in giù, & il pe&longs;o di E &longs;i mouerà in sù.<emph.end type="italics"/> </s> </p> <pb id="p.23v" xlink:href="pageimg-it/065.jpg"/> <figure id="fig41" place="text" xlink:href="figures-it/064_01.jpg"></figure> <p type="margin" id="id.2.1.409.0.0"> <s id="id.2.1.409.1.0"> <margin.target id="note123"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 29. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.410.0.0"> <s id="id.2.1.410.1.0"> <arrow.to.target n="note124"></arrow.to.target><emph type="italics"/>La ragione di Ari&longs;totele parimente qui è chiara. </s> <s id="id.2.1.410.2.0"> Percioche &longs;ia il punto N doue le <lb/>linee CO EF &longs;i tagliano in&longs;ieme. </s> <s id="id.2.1.410.3.0"> &longs;arà la NF maggiore della NE. & perche <lb/>il perpendicolo CO, &longs;e-<lb/>condo lui, diuide in parti <lb/>di&longs;uguali la bilancia, & <lb/>la parte maggiore è ver&longs;o <lb/>F, cioè NF; la bilan-<lb/>cia EF &longs;i mouerà in giù <lb/>dalla parte di F, concio <lb/>&longs;ia che quel che è di piu <lb/>venga portato à ba&longs;&longs;o.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.411.0.0"> <s id="id.2.1.411.1.0"> <margin.target id="note124"></margin.target><emph type="italics"/>Ragione di Aristotele.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.412.0.0"> <s id="id.2.1.412.1.0"><emph type="italics"/>Similmente dalle co&longs;e dette <lb/>caueremo, che <expan abbr="quāto">quanto</expan> piu <lb/>la bilancia EF tenente <lb/>il centro &longs;otto la bilancia, <lb/>&longs;arà <expan abbr="lõtana">lontana</expan> dal &longs;ito AB <lb/>&longs;i mouerà piu velocemen<lb/>te, percioche il centro del <lb/>la grauezza H, quanto <lb/>piu è di&longs;tante dal punto <lb/>D, tanto piu velocemen<lb/>te il pe&longs;o compo&longs;to dei pe<lb/>&longs;i EF, & della bilancia <lb/>EF &longs;i mouerà, finche <lb/>l'angolo CHS diuenga <lb/>retto. </s> <s id="id.2.1.412.2.0"> & dauantaggio &longs;i <lb/>mouerà anche piu veloce <lb/>mente quanto la bilancia <lb/>&longs;arà piu lontana dal cen-<lb/>tro C.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.413.0.0"> <s id="id.2.1.413.1.0"><emph type="italics"/>Oltre à ciò ne piace dalle &longs;ue <lb/>ragioni, & fal&longs;e pre&longs;uppo <lb/>&longs;te manife&longs;tare, & pro <lb/>durre gli effetti, & i moti <lb/>già dichiarati della bilan <lb/>cia, affine che appaia <expan abbr="quā">quam</expan> <lb/>ta &longs;ia la efficacia della ve-<lb/>rità, come quella, che dalle co&longs;e fal&longs;e ancora &longs;i sforza di ri&longs;plendere.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.414.0.0" xlink:href="figures-it/065_01.jpg"></figure> <p type="main" id="id.2.1.415.0.0"> <s id="id.2.1.415.1.0"><emph type="italics"/>Pongan&longs;i le co&longs;e iste&longs;&longs;e, cioè &longs;iail cerchio AE BF, & la bilancia AB, il cui cen-<lb/>tro C &longs;ia &longs;opra la bilancia, moua&longs;i in EF. Dico che il pe&longs;o po&longs;to in E hà iui <lb/>grauezza maggiore, che il pe&longs;o po&longs;to in F; & che la <expan abbr="bilācia">bilancia</expan> EF ritornerà in AB<emph.end type="italics"/> <pb id="p.24" xlink:href="pageimg-it/066.jpg"/><emph type="italics"/>&longs;iano tirate dai punti EF le linee EL FM à piombo di AB, le quali &longs;aran-<emph.end type="italics"/> <arrow.to.target n="note125"></arrow.to.target> <lb/><emph type="italics"/>no tra loro egualmente di&longs;tanti, & &longs;ia il punto N doue la AB, & la EF &longs;i <lb/>tagliano fra loro. </s> <s id="id.2.1.415.2.0"> Percioche dunque l'angolo FNM è eguale all'angolo ENL,<emph.end type="italics"/> <arrow.to.target n="note126"></arrow.to.target> <lb/><emph type="italics"/>& l'angolo FMN ret<lb/>to è eguale ad ELN <lb/>retto, & il re&longs;tante <lb/>NFM al re&longs;tante<emph.end type="italics"/> <arrow.to.target n="note127"></arrow.to.target> <lb/><emph type="italics"/>NEL è etiandio egua-<lb/>le; &longs;arà il triangolo NLE <lb/>&longs;imile al triangolo NMF.<emph.end type="italics"/> <arrow.to.target n="note128"></arrow.to.target> <lb/><emph type="italics"/>Si come dunque è la NE <lb/>ver&longs;o la EL, co&longs;i NF<emph.end type="italics"/> <arrow.to.target n="note129"></arrow.to.target> <lb/><emph type="italics"/>ad FM; & permut an-<lb/>do, &longs;i come EN ad NF, <lb/>co&longs;i EL ad FM. Ma <lb/>e&longs;&longs;endo HE eguale ad <lb/>HF, &longs;arà EN mag-<lb/>gior di NF. Per laqual <lb/>co&longs;a anco EL &longs;arà mag<emph.end type="italics"/> <lb/> <arrow.to.target n="fig42"></arrow.to.target> <lb/><emph type="italics"/>giore di FM. & percioche mentre il pe&longs;o po&longs;to in E de&longs;cende per la circonferen-<lb/>za EA, il pe&longs;o po&longs;to in F &longs;ale per la circon&longs;erenza FB eguale alla circonferen-<lb/>za EA, & la di&longs;ce&longs;a del pe&longs;o po&longs;to in E piglia (come e&longs;&longs;i dicono) di diretto EL: <lb/>& la &longs;alita del pe&longs;o po&longs;to in F piglia di diretto FM, meno di diretto verrà a pi-<lb/>gliare la &longs;alita del pe&longs;o po&longs;to in F, che la di&longs;ce&longs;a del pe&longs;o po&longs;to in E. Dunque il pe <lb/>&longs;o po&longs;to in E haurà grauezza maggiore, che il pe&longs;o po&longs;to in F.<emph.end type="italics"/> </s> </p> <figure id="fig42" place="text" xlink:href="figures-it/066_01.jpg"></figure> <p type="margin" id="id.2.1.417.0.0"> <s id="id.2.1.417.1.0"> <margin.target id="note125"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 28. <emph type="italics"/>d l primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.418.0.0"> <s id="id.2.1.418.1.0"> <margin.target id="note126"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 15. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.419.0.0"> <s id="id.2.1.419.1.0"> <margin.target id="note127"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 29. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.420.0.0"> <s id="id.2.1.420.1.0"> <margin.target id="note128"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 4.<emph type="italics"/>del &longs;esto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.421.0.0"> <s id="id.2.1.421.1.0"> <margin.target id="note129"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 16. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.422.0.0"> <s id="id.2.1.422.1.0"><emph type="italics"/>Sia allungata la linea CD dall una parte, & dall'altra in OP, laquale tagli la linea <lb/>EF nel punto S. & percioche (come dicono) quanto piu è lontano il pe&longs;o dalla <lb/>linea della direttione OP, tanto &longs;i fa piu graue; però con que&longs;to mezo ancora pro-<lb/>uera&longs;&longs;i il pe&longs;o po&longs;to in E hauer grauezza maggiore del pe&longs;o po&longs;to in F. Siano dai <lb/>punti EF tirate le linee EQ FR a piombo di OP. Con &longs;imile ragione mo&longs;tre <lb/>ra&longs;&longs;i, che il triangolo QES è &longs;imile al triangolo RFS; & che la linea EQ è <lb/>maggiore di RF. & co&longs;i il pe&longs;o po&longs;to in E &longs;arà piu lontano dalla linea OP, che <lb/>il pe&longs;o po&longs;to in F; & per ciò il pe&longs;o po&longs;to in E hauerà grauezza maggiore del pe <lb/>&longs;o po&longs;to in<emph.end type="italics"/> F. <emph type="italics"/>Dallequali co&longs;e appare euidente il ritorno della bilancia E<emph.end type="italics"/>F <emph type="italics"/>in AB.<emph.end type="italics"/> </s> </p> <pb id="p.24v" xlink:href="pageimg-it/067.jpg"/> <p type="main" id="id.2.1.424.0.0"> <s id="id.2.1.424.1.0"><emph type="italics"/>Ma &longs;e il centro della bilancia &longs;arà &longs;otto la bilancia, allhora &longs;i mo&longs;trerà con gli i&longs;te&longs;&longs;i me <lb/>zi, che il pe&longs;o abba&longs;&longs;ato hauerà grauezza maggiore dall'alzato. </s> <s id="id.2.1.424.2.0"> &longs;iano tirate dapun-<lb/>ti EF le linee EL FM <lb/>a piombo di AB. &longs;imil <lb/>mente&longs;i prouerà EL e&longs; <lb/>&longs;ere maggiore di FM; et <lb/>perciò la &longs;ce&longs;a del pe&longs;o po <lb/>sto in F prenderà meno <lb/>di dirittura, che la &longs;alita <lb/>del pe&longs;o po&longs;to in E. On-<lb/>de la re&longs;i&longs;tenza della vio-<lb/>lentia del pe&longs;o po&longs;to in E <lb/>&longs;upererà la naturale incli-<lb/>natione del pe&longs;o po&longs;to in <lb/>F. Adunque il pe&longs;o po&longs;to <lb/>in E &longs;arà piu graue del <lb/>pe&longs;o posto in F.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.425.0.0"> <s id="id.2.1.425.1.0"><emph type="italics"/>Sia <expan abbr="allūgata">allungata</expan> etiandio la CD <lb/>dall'una parte & l'altra<emph.end type="italics"/> <lb/> <arrow.to.target n="fig43"></arrow.to.target> <lb/><emph type="italics"/>in OP, & &longs;iano tirate dai punti EF le linee EQ FR à piombo dilei. </s> <s id="id.2.1.425.2.0"> &longs;i pro <lb/>verà con l'i&longs;te&longs;&longs;o modo in tutto, che la linea EQ è maggiore di FR. & percio il <lb/>pe&longs;o po&longs;to in E &longs;arà piu lontano dalla linea della dirittura OP, che il pe&longs;o po&longs;to <lb/>in F. Adunque il pe&longs;o po&longs;to in E haurà grauezza maggiore del pe&longs;o po&longs;to in F.<lb/> Dalle quali co&longs;e &longs;egue, che la bilancia EF &longs;i moue in giù dalla parte di E.<emph.end type="italics"/> </s> </p> <figure id="fig43" place="text" xlink:href="figures-it/067_01.jpg"></figure> <p type="main" id="id.2.1.427.0.0"> <s id="id.2.1.427.1.0"><emph type="italics"/>Si che Aristotele propo&longs;e que&longs;te due que&longs;tioni &longs;olamente, & la&longs;ciò la terza, cioè quando <lb/>il centro della bilancia &longs;tà nella bilancia i&longs;te&longs;&longs;a. </s> <s id="id.2.1.427.2.0"> Que&longs;ta però trala&longs;ciò egli, co-<lb/>me nota, &longs;i come egli &longs;ole trala&longs;ciare le co&longs;e molto note. </s> <s id="id.2.1.427.3.0"> Imperoche à chi puote <lb/>far dubbio, che &longs;e il pe&longs;o &longs;arà &longs;o&longs;tentato nel centro della grauezza&longs;ua, che non i&longs;tia <lb/>fermo? </s> <s id="id.2.1.427.4.0"> Mapotrebbe for&longs;e alcuno riprendere quelle co&longs;e che per &longs;ua &longs;ententia hab-<lb/>biamo propo&longs;to, affermando noi non hauere prodotto in mezo tutta la intera &longs;enten <lb/>za &longs;ua. </s> <s id="id.2.1.427.5.0"> Imperoche proponendo egli nella &longs;econda parte della que&longs;tione &longs;econda. <lb/> “Perche la bilancia e&longs;&longs;endo posta la trutina di &longs;otto, quando, portato il pe&longs;o in giu, al <lb/>cuno lo rimoue, non a&longs;cende, ma rimane?” non afferma perciò la bilancia mouer&longs;i in <lb/>giù, ma rimanere, il che pare &longs;imilmente hauere nella vltima conclu&longs;ione raccolto. <lb/> Ma que&longs;to non &longs;o larnente non ci &longs;a contra, ma&longs;e egli è ben' inte&longs;o grandi&longs;&longs;imamen-<lb/>te aiuta.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.428.0.0"> <s id="id.2.1.428.1.0"><emph type="italics"/>Percioche &longs;ia la bilancia AB egualmente di&longs;tante dall'orizonte, il cui centro E &longs;ia <lb/>&longs;otto la bilancia. </s> <s id="id.2.1.428.2.0"> & perche Ari&longs;totele con&longs;idera la bilancia come ella è in fatto, però <lb/>egli è nece&longs;&longs;ario collocare la trutina, ouero qualche altra co&longs;a &longs;otto il centro E, co-<lb/>me EF, che in ogni modo &longs;arà trutina, per modo, che &longs;o&longs;tengail centro E. & &longs;ia <lb/>ECD il perpendicolo. </s> <s id="id.2.1.428.3.0"> & accioche la bilancia AB &longs;i moua da que&longs;to &longs;ito, dice<emph.end type="italics"/> <pb id="p.25" xlink:href="pageimg-it/068.jpg"/><emph type="italics"/>Ari&longs;totele, ponga&longs;i il pe&longs;o in B, ilquale e&longs;&longs;endo graue mouerà la bilancia dalla par-<lb/>te B in giù, come in G, talche per l'impedimento non potrà egli piu mouer&longs;i in <lb/>giu, ma non dice gia Ari&longs;totele, che &longs;i moua la bilancia in giu dalla parte di B fin <lb/>tanto che parerà, da <lb/>poi &longs;i la&longs;ci, come noi <lb/>dicemmo: ma ordina <lb/>che &longs;ia posto il pe&longs;o <lb/>in B, il quale di &longs;ua <lb/>natura &longs;i mouera <lb/>&longs;empre in giù finche <lb/>la bilancia &longs;i appog-<lb/>gi alla trutina, ouerò <lb/>a qualche altra co&longs;a. <lb/>& quando il B &longs;a-<lb/>rà nel G, la bilan-<lb/>cia &longs;arà in GH, nel <lb/>qual &longs;ite leuato via <lb/>il pe&longs;o, rimarrà: per <lb/>e&longs;&longs;ere la maggior par <lb/>te della bilancia dal <lb/>perpendicolo uer&longs;o il<emph.end type="italics"/> <lb/> <arrow.to.target n="fig44"></arrow.to.target> <lb/><emph type="italics"/>G, che è DG, che DH. ne piu mouera&longs;&longs;i in giu, imperoche la bilancia &longs;tar à &longs;opra <lb/>la trutina, ouero qualche altra co&longs;a, che &longs;o&longs;tenga il centro della bilancia. </s> <s id="id.2.1.428.4.0"> peroche &longs;e a <lb/>cote&longs;ta non &longs;i appoggia&longs;&longs;e, verrebbe la bilancia à mouer&longs;i, &longs;econdo la &longs;ua opinione, <lb/>in giù dalla parte di G, concio&longs;ia, che quello che è di piu, cioè DG debba e&longs;&longs;ere <lb/>per nece&longs;&longs;ità in giu portato.<emph.end type="italics"/> </s> </p> <figure id="fig44" place="text" xlink:href="figures-it/068_01.jpg"></figure> <p type="main" id="id.2.1.430.0.0"> <s id="id.2.1.430.1.0"><emph type="italics"/>Ma potrebbe dauantagio dire alcuno, &longs;e in B &longs;arà collocato vn pe&longs;o picciolo, &longs;i mo-<lb/>uerà ben la bilancia in giu, ma non gia fin al G; nel qual &longs;ito, &longs;econdo Aristo-<lb/>tele, leuato via il pe&longs;o, deue remanere. </s> <s id="id.2.1.430.2.0"> ilche è manife&longs;to per la e&longs;perientia, inchi-<lb/>nando&longs;i la <expan abbr="bilãcia">bilancia</expan> più, & meno, quando in vna e&longs;tremita della bilancia &longs;olamente <lb/>vi è po&longs;to il pe&longs;o, che &longs;ia ò maggiore, ò minore. </s> <s id="id.2.1.430.3.0"> ilche è veri&longs;&longs;imo allhora che il centro <lb/>è collocato &longs;opra la bilancia, ma non già &longs;otto, ne in e&longs;&longs;a bilancia, come per gratia <lb/>di e&longs;empio.<emph.end type="italics"/> </s> </p> <pb id="p.25v" xlink:href="pageimg-it/069.jpg"/> <p type="main" id="id.2.1.432.0.0"> <s id="id.2.1.432.1.0"><emph type="italics"/>Sia la bilancia AB egualmente di&longs;tante dall'orizonte, il cui centro C &longs;ia &longs;opra la bi <lb/>lancia, & il perpendicolo <lb/>CD a piombo dell' ori-<lb/>zonte, il quale da la par-<lb/>te D &longs;ia allungato in H. <lb/> Hor percioche con&longs;idera <lb/>ta la grauezza della bi-<lb/>lancia, &longs;arà il punto D <lb/>il centro della grauezza <lb/>della bilancia. </s> <s id="id.2.1.432.2.0"> &longs;e dunque <lb/>vn piccolo pe&longs;o &longs;arà po-<lb/>&longs;to nel B, il cui centro <lb/>della grauezza &longs;ia nel <expan abbr="pũ">pum</expan> <lb/>to B; gia piu non &longs;arà <lb/>il centro della grauezza <lb/>D della magnitudine<emph.end type="italics"/> <lb/> <arrow.to.target n="note130"></arrow.to.target><emph type="italics"/>compo&longs;ta della bilancia<emph.end type="italics"/> <lb/> <arrow.to.target n="fig45"></arrow.to.target> <lb/><emph type="italics"/>AB, & del pe&longs;o po&longs;to in B, ma &longs;arà nella linea DB, come in K: per modo <lb/>che DE ad EB &longs;ia come il pe&longs;o po&longs;to in B alla grauezza della bilancia AB. <lb/>congiunga&longs;i la CE. & percioche il punto C è immobile, mentre la bilancia &longs;i <lb/>moue, il punto E de&longs;criuerà la circonferenza del cerchio EFG, il cui mezo dia-<lb/>metro è CE, & il centro C. Ma perche CD &longs;tà a piombo dell' orizonte, la li <lb/>nea CE non &longs;arà gia ella à piombo dell' orizonte. </s> <s id="id.2.1.432.3.0"> Per laqual co&longs;a la grandez-<lb/>za composta di AB, & del pe&longs;o po&longs;to in B non rimarrà in questo &longs;ito; ma &longs;i <lb/>mouerà in giu &longs;econdo il centro E della &longs;ua grauezza per la circonferenza EFG,<emph.end type="italics"/> <lb/> <arrow.to.target n="note131"></arrow.to.target><emph type="italics"/>finche CE diuenti a piombo dell' orizonte, cioè finche la CE peruenga in CDF. <lb/>& allhora la bilancia AB &longs;arà mo&longs;&longs;ain KL, nel qual &longs;ito la bilancia rimarrà <lb/>in&longs;ieme co'l pe&longs;o, ne d'auantaggio &longs;i mouerà in giù. </s> <s id="id.2.1.432.4.0"> che &longs;e in B &longs;arà po&longs;to vn pe&longs;o <lb/>piu graue, il centro'della grauezza di tutta la magnitudine &longs;arà piu dappre&longs;&longs;o al B, <lb/>come in M. & allhora la bilancia &longs;i mouerà in giu, finche la congiunta linea CM <lb/>peruenga nella linea CDH. Dal por&longs;i dunque pe&longs;o maggiore ò minore in B, la <lb/>bilancia &longs;i inchinerà piu ò meno. </s> <s id="id.2.1.432.5.0"> Da che &longs;egue che il pe&longs;o B de&longs;criuerà &longs;empre vna <lb/>circonferenza minore della quarta parte d'un cerchio, per e&longs;&longs;ere l'angolo FCE &longs;em <lb/>pre acuto:ne il punto B peruenirà gia mai fin alla linea CH, percioche &longs;empre il <lb/>centro della grauezza del pe&longs;o, & dalla bilancia in&longs;ieme &longs;arà fra BD. tuttauia <expan abbr="quā">quam</expan> <lb/>to &longs;arà il pe&longs;o po&longs;to in B piu graue, de&longs;criuerà anche circonferenza maggiore, ve-<lb/>nendo&longs;i per que&longs;to il punto B ad acco&longs;tare piu alla linea CH.<emph.end type="italics"/> </s> </p> <figure id="fig45" place="text" xlink:href="figures-it/069_01.jpg"></figure> <p type="margin" id="id.2.1.434.0.0"> <s id="id.2.1.434.1.0"> <margin.target id="note130"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>del primo. </s> <s id="id.2.1.434.2.0"> di. </s> <s id="id.2.1.434.3.0"> drch. </s> <s id="id.2.1.434.4.0"> del le co&longs;e cgual <expan abbr="m&etilde;se">mense</expan> <expan abbr="pesāti">pesanti</expan>.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.435.0.0"> <s id="id.2.1.435.1.0"> <margin.target id="note131"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.436.0.0"> <s id="id.2.1.436.1.0"><emph type="italics"/>Ma habbia la bilancia AB il centro C nella i&longs;te&longs;&longs;a bilancia, & nel &longs;uo mezo, <lb/>&longs;arà il C centro ancora della grauezza della bilancia, dal quale &longs;ia tirata la li-<lb/>nea FCG a piombo di e&longs;&longs;a AB, & dell' orizonte. </s> <s id="id.2.1.436.2.0"> Ponga&longs;i dapoi in B qual <lb/>pe&longs;o &longs;i voglia; &longs;arà il centro di tutta la grauezza, come in E; &longs;i fattamente che <lb/>la CE ver&longs;o EB &longs;ia come il pe&longs;o po&longs;to in B alla grauezza della bilancia. </s> <s id="id.2.1.436.3.0"> & per<emph.end type="italics"/> <pb id="p.26" xlink:href="pageimg-it/070.jpg"/><emph type="italics"/>cioche la CE non è a piombo dell' orizonte, la bilancia AB, & il pe&longs;o po&longs;to in <lb/>B non rimaranno in que-<lb/>&longs;to &longs;ito gia mai; ma &longs;i mo-<lb/>ueranno in giu dalla par-<lb/>te di B, fin che CE &longs;i <lb/>&longs;accia à piombo dell' ori-<lb/>zonte; cioè fin che la bilan-<lb/>cia AB peruenga in FG. <lb/>Onde è chiaro, che cia&longs;cun <lb/>pe&longs;o po&longs;to in B, &longs;empre <lb/>de&longs;criue la quarta parte <lb/>d'un cerchio.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.437.0.0" xlink:href="figures-it/070_01.jpg"></figure> <p type="main" id="id.2.1.438.0.0"> <s id="id.2.1.438.1.0"><emph type="italics"/>Ma &longs;ia il centro C &longs;otto la bilancia AB, & &longs;ia DCE il perpendicolo. </s> <s id="id.2.1.438.2.0"> &longs;imilmente <lb/>per e&longs;&longs;er il pe&longs;o posto in B, &longs;arà il centro della grauezza della magnitudine compe <lb/>&longs;ta di AB bilancia, & del pe&longs;o po&longs;to in B nella linea DB, come in F; &longs;i <expan abbr="fattamē">fattamen</expan> <lb/>te che come DF &longs;i ha ver&longs;o FB co&longs;i &longs;ia il pe&longs;o po&longs;to in B al pe&longs;o della bilan-<lb/>cia. </s> <s id="id.2.1.438.3.0"> congiunga&longs;i CF. & <lb/>percioche CD è a piombo <lb/>dell' orizonte, non &longs;arà gia <lb/>la linea CF a piombo del <lb/>l'orizonte. </s> <s id="id.2.1.438.4.0"> Per laqual co&longs;a <lb/>la magnitudine compo&longs;ta <lb/>della bilancia AB, & del <lb/>pe&longs;o po&longs;to in B in que&longs;to <lb/>&longs;ito non &longs;tarà mai ferma; <lb/>ma in giu mouera&longs;&longs;i &longs;e alcu <lb/>na co&longs;a non la impedi&longs;ce, <lb/>finche CF peruenga in <lb/>DCE, nel qual &longs;ito la bi-<lb/>lancia rimarrà in&longs;ieme co'l<emph.end type="italics"/> <lb/> <arrow.to.target n="fig46"></arrow.to.target> <lb/><emph type="italics"/>pe&longs;o. </s> <s id="id.2.1.438.5.0"> & il punto B &longs;arà come in G, & il punto A in H, & la bilancia GH <lb/>non hauerà piu il centro di &longs;otto, ma &longs;opra e&longs;&longs;a. </s> <s id="id.2.1.438.6.0"> La qual co&longs;a hauerà &longs;empre, quan-<lb/>tunque &longs;i ponga vn minimo pe&longs;o in B. Auanti che dunque il B peruenga al G, <lb/>egli è nece&longs;&longs;ario, che la bilancia incontri la trutina po&longs;ta di &longs;otto, ouero alcuna altra <lb/>co&longs;a, che &longs;o&longs;tenti il centro C, & iui s'appoggi. </s> <s id="id.2.1.438.7.0"> Da que&longs;to &longs;egue, che il pe&longs;o B &longs;em <lb/>pre &longs;i moue oltre la linea DK, & de&longs;criue &longs;empre vna circonferenza maggiore del <lb/>la quarta parte del cerchio, per e&longs;&longs;ere l'angolo FCE &longs;empre ottu&longs;o, & l'angolo <lb/>DCF &longs;empre acuto. </s> <s id="id.2.1.438.8.0"> & quanto il pe&longs;o posto in B &longs;arà piu leggiero, de&longs;criuerà tut-<lb/>tauia anche circonferenza maggiore. </s> <s id="id.2.1.438.9.0"> Imperoche quanto il pe&longs;o po&longs;to in G &longs;arà piu <lb/>leggiero, tanto piu il pe&longs;o detto posto in G &longs;i alzerà; & la bilancia GA s'acco&longs;te<emph.end type="italics"/> <pb id="p.26v" xlink:href="pageimg-it/071.jpg"/><emph type="italics"/>rà piu pre&longs;&longs;o al &longs;ito egualmente di&longs;tante dall'orizonte. </s> <s id="id.2.1.444.2.0"> Le quali co&longs;e tutte re&longs;tano ma <lb/>ni&longs;e&longs;te da quelle che di &longs;opra &longs;ono &longs;tate dette.<emph.end type="italics"/> </s> </p> <figure id="fig46" place="text" xlink:href="figures-it/070_02.jpg"></figure> <p type="main" id="id.2.1.447.0.0"> <s id="id.2.1.447.1.0"><emph type="italics"/>Prouate que&longs;te co&longs;e, egli è chia <lb/>ro, che il centro della bilan-<lb/>cia è cagione de gli effetti di <lb/>uer&longs;i della bilancia. </s> <s id="id.2.1.447.2.0"> & &longs;i ve <lb/>de ancora che tutte le pro-<lb/>po&longs;itioni di Archimede del <lb/>le co&longs;e, che egualmente pe&longs;a <lb/>no, a ciò pertinenti, in ogni <lb/>&longs;ito &longs;ono vere. </s> <s id="id.2.1.447.3.0"> cioè, &longs;ia pur <lb/>la bilancia di&longs;tante <expan abbr="egualm&etilde;">egualmen</expan> <lb/>te dall'orizonte, ouero non, <lb/>pur che il centro della bilan <lb/>cia &longs;ia collocato in e&longs;&longs;a <expan abbr="bilā">bilan</expan> <lb/>cia, &longs;i come egli la con&longs;ide-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig48"></arrow.to.target> <lb/><emph type="italics"/>rà. </s> <s id="id.2.1.447.4.0"> & quantunque la bilancia habbia difuguali le braccia, auerrd tutt auia l'i&longs;te&longs;&longs;o, & <lb/>&longs;i dimo&longs;trerà co'l modo i&longs;te&longs;&longs;o in tutto, che il centro della bilancia collocato in diuer <lb/>&longs;e maniere produrrà vari effetti.<emph.end type="italics"/> </s> </p> <figure id="fig47" place="text" xlink:href="figures-it/071_01.jpg"></figure> <p type="main" id="id.2.1.449.0.0"> <s id="id.2.1.449.1.0"><emph type="italics"/>Percioche &longs;ia la bilancia <lb/>AB egualmente di&longs;tan <lb/>te dall'orizonte; & &longs;iano <lb/>in AB pe&longs;i di&longs;uguali, il <lb/>centro della grauezza <lb/>dei quali &longs;ia in C, & <lb/>&longs;ia attacata la bilancia <lb/>nell'i&longs;te&longs;&longs;o punto di C, <lb/>& moua&longs;i la bilancia in<emph.end type="italics"/> <lb/> <arrow.to.target n="note134"></arrow.to.target><emph type="italics"/>DE; egliè manife&longs;to, <lb/>che la bilancia rimarrà <lb/>non &longs;olamente in DE, <lb/>ma in qual &longs;i voglia altre <lb/>&longs;ito.<emph.end type="italics"/> </s> </p> <figure id="fig48" place="text" xlink:href="figures-it/071_02.jpg"></figure> <p type="margin" id="id.2.1.450.0.0"> <s id="id.2.1.450.1.0"> <margin.target id="note134"></margin.target><emph type="italics"/>Per la diffi ni ione del centro della grauezza.<emph.end type="italics"/> </s> </p> <pb id="p.27" xlink:href="pageimg-it/072.jpg"/> <p type="main" id="id.2.1.441.0.0"> <s id="id.2.1.441.1.0"><emph type="italics"/>Ma &longs;ia il centro della bilancia AB &longs;opra il C in F; & &longs;ia FC à piombo di AB, <lb/>& dell' orizonte: & &longs;e <lb/>la bilancia &longs;arà mo&longs;&longs;a in<emph.end type="italics"/> <arrow.to.target n="note132"></arrow.to.target> <lb/><emph type="italics"/>DE, la linea CF &longs;arà <lb/>mo&longs;&longs;a in FG, la quale <lb/>per non e&longs;&longs;ere à piombo <lb/>dell' orizonte, la bilancia <lb/>DE &longs;imouerà in giu dalla <lb/>parte di D, finche FG <lb/>ritorni in FC: & allho <lb/>ra la bilancia DE &longs;arà <lb/>in AB, nel qual &longs;ito an <lb/>che rimarrà.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.442.0.0"> <s id="id.2.1.442.1.0"> <margin.target id="note132"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.443.0.0" xlink:href="figures-it/072_01.jpg"></figure> <p type="main" id="id.2.1.444.0.0"> <s id="id.2.1.444.1.0"><emph type="italics"/>Che &longs;e il centro F della bi-<lb/>lancia &longs;arà &longs;otto la <expan abbr="bilē-<lb> cia">bilen-<lb/>cia</expan>, & &longs;iala bilancia mo&longs;<emph.end type="italics"/> <arrow.to.target n="note133"></arrow.to.target> <lb/><emph type="italics"/>&longs;a in DE primier amen <lb/>te egli è manife&longs;to che la <lb/>bilancia rimarrà in AB: <lb/>& in DE mouera&longs;&longs;i in <lb/>giu dalla parte di E, per <lb/>non e&longs;&longs;ere la linea FG <lb/>à piombo dell' orizonte.<emph.end type="italics"/> <lb/> <arrow.to.target n="fig49"></arrow.to.target> <lb/><emph type="italics"/> </s> </p> <p type="margin" id="id.2.1.446.0.0"> <s id="id.2.1.446.1.0"> <margin.target id="note133"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="fig49" xlink:href="figures-it/072_02.jpg"></figure> <pb id="p.27v" xlink:href="pageimg-it/073.jpg"/> <p type="main" id="id.2.1.452.0.0"> <s id="id.2.1.452.1.0"><emph type="italics"/>Da que&longs;te co&longs;e co&longs;i terminate, &longs;e la bilancia fo&longs;&longs;e inarcata, ouero, che le braccia della bi <lb/>lancia forma&longs;&longs;ero vn'angolo, & &longs;i di&longs;poneße il centro diuer&longs;amente, (ben che que-<lb/>&longs;ta propriamente non &longs;arebbe bilancia,) potremo nondimeno anche dimo&longs;trare di lei <lb/>vary effetti. </s> <s id="id.2.1.452.2.0"> Come &longs;ia la bilancia ACB, il cui centro, d'intorno al quale &longs;i volge, <lb/>&longs;i a C, & tiratala linea AB, &longs;ia <lb/>l'arco ouerò l'angolo ACB &longs;opra <lb/>la linea AB; & pongan&longs;i in AB <lb/>icentri della grauezza de'pe&longs;i, i quali <lb/>rimangano in que&longs;to &longs;ito. </s> <s id="id.2.1.452.3.0"> Moua&longs;i poi <lb/>la <expan abbr="bilācia">bilancia</expan> da que&longs;to &longs;ito, come in ECF.<lb/> Dico che la bilancia ECF ritornerà <lb/>in ACB. Ritroui&longs;i il centro della <lb/>grauczza di tutta la magnitudine D, <lb/>& &longs;ia congiunta la CD. Hor percio<emph.end type="italics"/> <lb/> <arrow.to.target n="note135"></arrow.to.target><emph type="italics"/>che i pe&longs;i AB stanno fermi, lali-<lb/>nea CD &longs;arà à piombo dell' orizon-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig49b"></arrow.to.target> <lb/><emph type="italics"/>te. </s> <s id="id.2.1.452.4.0"> Quando dunque la bilancia &longs;arà in ECF, la linea CD &longs;arà come in CG; <lb/>la quale per non e&longs;&longs;ere à piombo dell' orizonte, la bilancia ECF ritornerà in <lb/>ACB. ilche parimente auenir à, &longs;e il centro C &longs;arà me&longs;&longs;o &longs;opra la bilancia, co-<lb/>me in H.<emph.end type="italics"/> </s> </p> <figure id="fig49b" place="text" xlink:href="figures-it/073_01.jpg"></figure> <p type="margin" id="id.2.1.454.0.0"> <s id="id.2.1.454.1.0"> <margin.target id="note135"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.455.0.0"> <s id="id.2.1.455.1.0"><emph type="italics"/>Che &longs;e l'arco, ouero l'angolo ACB <lb/>&longs;arà &longs;otto la linea AB, nel <lb/>modo i&longs;te&longs;&longs;o mo&longs;treremo, la bi-<lb/>lancia ECF, il cui centro &longs;ia <lb/>ouero in C, ouero in H, do-<lb/>uer&longs;i mouere in giu dalla parte <lb/>di F.<emph.end type="italics"/> <lb/> <arrow.to.target n="fig50"></arrow.to.target> </s> </p> <pb id="p.28" xlink:href="pageimg-it/074.jpg"/> <figure id="fig50" place="text" xlink:href="figures-it/073_02.jpg"></figure> <figure place="text" id="id.2.1.458.0.0" xlink:href="figures-it/074_01.jpg"></figure> <figure place="text" id="id.2.1.458.0.1" xlink:href="figures-it/074_02.jpg"></figure> <figure place="text" id="id.2.1.458.0.2" xlink:href="figures-it/074_03.jpg"></figure> <p type="main" id="id.2.1.459.0.0"> <s id="id.2.1.459.1.0"><emph type="italics"/>Et &longs;e l'angolo ACB fo&longs;&longs;e &longs;oprala linea AB, & il centro della bilancia H; & <lb/>& la linea CH &longs;o&longs;tene&longs;&longs;e la bilancia; & &longs;imoue&longs;&longs;e la bilancia in EKF; la bilan<lb/>cia EKF ritornerà in ACB.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.460.0.0"> <s id="id.2.1.460.1.0"><emph type="italics"/>Ma &longs;e il centro della bilancia &longs;arà D, moua&longs;i in qualunque modo la bilancia, doue &longs;i <lb/>la&longs;cierà, iui rimarrà.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.461.0.0"> <s id="id.2.1.461.1.0"><emph type="italics"/>Se poi il punto H &longs;arà &longs;otto la linea AB; allhora la bilancia EKF &longs;i mouerà in <lb/>giu dalla parte di F.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.462.0.0"> <s id="id.2.1.462.1.0"><emph type="italics"/>Et con &longs;imile ragione in tutto, &longs;e l'ango-<lb/>lo ACB &longs;arà &longs;otto la linea AB; <lb/>& &longs;ia il centro della bilancia H, & <lb/>&longs;ia la bilancia &longs;o&longs;tentata dalla linea <lb/>CH; &longs;e la bilancia moueraßi da que&longs;to <lb/>&longs;ito, &longs;i mouerà in giu dalla parte del pe <lb/>&longs;o più ba&longs;&longs;o. </s> <s id="id.2.1.462.2.0"> & &longs;e il centro della bilan-<lb/>cia &longs;ia D; rimarrà doue &longs;i la&longs;cierà. che <lb/>&longs;e &longs;arà in K; & da cotale &longs;ito &longs;i mo <lb/>uerà, ritornerà ad ogni modo nello i&longs;te&longs; <lb/>&longs;o. </s> <s id="id.2.1.462.4.0"> Le quali co&longs;e tutte da quel che in<emph.end type="italics"/> <lb/> <arrow.to.target n="fig51"></arrow.to.target> <lb/><emph type="italics"/>principio dicemmo &longs;ono mani&longs;este. </s> <s id="id.2.1.462.5.0"> &longs;imilmente &longs;e il centro della bilancia &longs;arà po&longs;to <lb/>in vno della bracia della bilancia, ò dentro, ò &longs;uori, ò in qual &longs;i voglia modo trouere <lb/>mo le co&longs;e i&longs;te&longs;&longs;e.<emph.end type="italics"/> </s> </p> <figure id="fig51" place="text" xlink:href="figures-it/074_04.jpg"></figure> <pb id="p.28v" xlink:href="pageimg-it/075.jpg"/> <p type="main" id="id.2.1.464.0.0"> <s id="id.2.1.464.1.0"> In que&longs;to luogo egli conuiene auertire, il che poteua&longs;i anco fare di &longs;opra à carte cin<lb/>que pre&longs;&longs;o la fine della &longs;econda faccia oue è &longs;critto. oltre à ciò po&longs;siamo con&longs;ide-<lb/>rare le co&longs;e che &longs;eguono in tutto al modo i&longs;te&longs;&longs;o. </s> <s id="id.2.1.464.3.0"> Che que&longs;to autore è &longs;tato il <lb/>primo à con&longs;iderare e&longs;qui&longs;itamente la bilancia, & intenderla dalla natura, & dal <lb/>vero e&longs;&longs;er &longs;uo; pero che egli il primiero di tutti ha manife&longs;tato chia ramente il mo <lb/>do del trattarla, & in&longs;egnarla, con proporre tre centri da e&longs;&longs;ere con&longs;iderati in que <lb/>&longs;ta &longs;peculatione; l'uno è il centro del mondo, l'altro il centro della bilancia, & il <lb/>terzo il centro della grauezza della bilancia, che in e&longs;&longs;a era vn na&longs;co&longs;to &longs;ecreto di <lb/>natura. </s> <s id="id.2.1.464.4.0"> Senza que&longs;ti tre centri, chiara co&longs;a è, che non &longs;i puote ve nire in cono&longs;ci-<lb/>mento per&longs;etto, ne dimo&longs;trare gli effetti varij della bilancia, i quali na&longs;cono dalla <lb/>diuer&longs;ità del collo care il centro della bilancia in tre modi, cioè quando il <lb/>centro della bilancia &longs;ta &longs;opra il centro della grauezza di e&longs;&longs;a, ouero quando è <lb/>di &longs;otto, o pure allhorche il centro della bilancia è nell'i&longs;te&longs;&longs;o centro della gra-<lb/>uezza di lei; &longs;i come l'autore in&longs;egna nella tre precedenti dimo&longs;trationi, cioè <lb/>nella <expan abbr="&longs;ecõda">&longs;econda</expan>, nella terza, & nella quarta propo&longs;itione: peroche nella &longs;econda mo-<lb/>&longs;tra quando la bilancia torna &longs;empre egualmente di&longs;tante dall'orizonte; nella ter-<lb/>za quando non &longs;olo non ritorna, ma &longs;i moue al contrario; nella quarta, che <lb/>e&longs;&longs;endo la bilancia &longs;o&longs;tenuta nel &longs;uo centro dalla grauezza &longs;ta ferma douunque el <lb/>la &longs;i troua, il quale effetto in particolare non è piu &longs;tato tocco, ne veduto, ne man <lb/>co da niuno manife&longs;tato, fuor che dall'autore: anzi fin hora tenuto fal&longs;o, & impo&longs; <lb/>&longs;ibile da tutti gli prede ce&longs;&longs;ori no&longs;tri; i quali con molte ragioni &longs;i &longs;ono sforzati di <lb/>prouare non &longs;olamente il contrario, ma hanno etiandio affermato per certo, che <lb/>la &longs;períenza mo&longs;tra la bilancia non dimorare gia mai ferma &longs;e non quando ella è <lb/>egualmente di&longs;tante dall'orizonte. </s> <s id="id.2.1.464.5.0"> Laqual co&longs;a in tutto è contraria alla ragione <lb/>prima, per e&longs;&longs;ere la dimo&longs;tratione della &longs;udetta quarta propo&longs;itione tanto chiara, <lb/>facile, & vera, che non sò, come &longs;e le po&longs;&longs;a in modo alcuno contradire: & poial-<lb/>l'e&longs;perienza concio&longs;ia che l'autore habbia fatto &longs;ottili&longs;simamente lauorare bilan-<lb/>cie giu&longs;te a po&longs;ta per chiarire que&longs;ta verità, vna delle quali hò io veduto in mano <lb/>dell'Illu&longs;tre Signor Gio. Vicenzo Pinello, mandatagli dall'i&longs;te&longs;&longs;o autore, la quale <lb/>per e&longs;&longs;ere &longs;o&longs;tenuta nel centro della &longs;u a grauezza, mo&longs;&longs;a douunque &longs;i vuole, & poi <lb/>la&longs;ciata, &longs;tà ferma in ogni &longs;ito doue ella vien la&longs;ciata. </s> <s id="id.2.1.464.7.0"> Ben è egli vero, che non bi <lb/>&longs;ogna, nel fare cote&longs;ta e&longs;perienza, correr co&longs;i a furia, per e&longs;&longs;ere co&longs;a oltra modo <lb/>difficile, come dice l'áutore di &longs;opra, il fare vna bilancia, la quale &longs;ia nel mezo del <lb/>le &longs;ue braccia &longs;o&longs;tenuta à punto, & nel centro proprio della &longs;ua grauezza. </s> <s id="id.2.1.464.8.0"> Per la <lb/>qual co&longs;a egli è da por <expan abbr="m&etilde;te">mente</expan>, che qual'hora alcuno &longs;i mette&longs;&longs;e à far cotale e&longs;perien<lb/>ra, & non gli riu&longs;ci&longs;&longs;e, non perciò &longs;i deue &longs;gomentare, anzi dica pur fermamente <lb/>di non hauer bene operato, & vn'altra volta ritorni à farne la &longs;perienza, finche la <lb/>bilancia &longs;ia giu&longs;ta, & eguale, & venga &longs;o&longs;tenuta à punto nel centro della grauez-<lb/>za &longs;ua. </s> <s id="id.2.1.464.9.0"> Et benche da altri &longs;iano &longs;tate to cche le altre due predette &longs;peculationi, cioè <lb/>quando la bilancia ritorna &longs;empre egualmente di&longs;tante dall'orizonte, & quando <lb/>&longs;i moue al contrario di que&longs;to &longs;ito, tuttauia non &longs;i è piu inte&longs;a que&longs;ta verità gia <lb/>mai apertamente, &longs;e non dall'autore no&longs;tro; peroche gli altri non hanno co'l &longs;en-<lb/>no penetrato in ciò tanto auanti, che habbiano &longs;aputo con di&longs;tintione con&longs;idera<lb/>re il centro della bilancia in tre modi, come hò narrato. </s> <s id="id.2.1.464.10.0"> Che &longs;e hanno pur diui&longs;a<lb/>to qualche co&longs;a d'intorno à que&longs;to, l'hanno fatto confu&longs;i&longs;simamente, & con ma <lb/>le dimo&longs;trationi, dalle quali non &longs;i puote cauare ferma <expan abbr="cõchiufione">conchiufione</expan>, & chiara. </s> <s id="id.2.1.464.11.0"> Que<lb/>&longs;ti predece&longs;&longs;ori no&longs;tri han&longs;i da intendere i moderni &longs;crittori di cotal materia alle-<lb/>gati in diuer&longs;i luoghi dall'autore, fra quali Giordano, che &longs;cri&longs;&longs;e de'pe&longs;i fù riputa-<pb id="p.29" xlink:href="pageimg-it/076.jpg"/>to a&longs;&longs;ai, & &longs;in qui è &longs;tato &longs;eguito molto nella &longs;ua dottrina. </s> <s id="id.2.1.464.12.0"> Hor l'autore no&longs;tro hà <lb/>procurato con ogni &longs;tudi o di caminare per la via de' buoni Greci antichi, <lb/>mae&longs;tri delle &longs;cienze, & in particolare di Archimede Sira cu&longs;ano prencipe delle ma <lb/>thematiche famo &longs;i&longs;simo, & di Pappo Ale&longs;&longs;andrino, come egli dice, leggendogli <lb/>nella &longs;ua propria fauella, non tradotti; peroche il piu delle volte &longs;ono co&longs;i mal <lb/>trattati, che à gran pen a'&longs;i puote trarre da loro frutto veruno. </s> <s id="id.2.1.464.13.0"> & affine che que&longs;ta <lb/>noua opinion &longs;ua, dimo&longs;trata à pien o nella predetta quarta propo&longs;itione, re&longs;ti to-<lb/>talmente chiara, non &longs;i è gia <expan abbr="cont&etilde;tato">contentato</expan> egli d'hauerla dimo&longs;trata con viue ragioni, <lb/>& certe &longs;olamente, ma come buon filo&longs;ofo, procedente con via di reale dottrina, <lb/>& di fon data &longs;cienza, (imitando Ari&longs;totele, ilqual ne' principii de &longs;uoi libri, inue-<lb/>&longs;tigando dottrina migliore, hà datto contra la opinione de gli antichi, &longs;oluendo <lb/>le ragioni addotte da loro:) hà ben voluto, e&longs;&longs;endo la verità vna &longs;ola, proporre le <lb/>opinioni de'&longs;uoi predece&longs;&longs;ori, & e&longs;aminare le loro ragioni, lequali &longs;embrano pro <lb/>uar il contrario, & &longs;oluerle, la loro fallenza <expan abbr="dimo&longs;trãdo">dimo&longs;trando</expan> co'l pre&longs;ente di&longs;cor&longs;o, che <lb/>incomincia, come è detto à carte cinque nella faccia &longs;econda, & qui fini&longs;ce ilqua <lb/>le di&longs;cor&longs;o &longs;eruirà in que&longs;ta materia, &longs;econdo che &longs;i &longs;uole dire per la opinione de <lb/>gli antichi. </s> <s id="id.2.1.464.14.0"> Et percio che egli contiene co&longs;e di alti&longs;sima &longs;peculatione, ma&longs;simamen-<lb/>te d'intorno al con&longs;iderare doue &longs;ia piu graue vn pe&longs;o &longs;olo po&longs;to in vno braccio <lb/>della bilancia, bi&longs;ogna in ogni modo, per bene intendere, leggerlo, & i&longs;tudiarlo <lb/>con accurati&longs;sima diligenza. </s> <s id="id.2.1.464.15.0"> Ma per certo l'autore è &longs;tato non &longs;olo il primo à tro <lb/>uare que&longs;ta verità, ma il primo etiandio a dim o&longs;trare in qual maniera &longs;ia me&longs;tieri <lb/>con&longs;iderare, & &longs;peculare interamente la pre&longs;ente materia tutta. </s> <s id="id.2.1.464.16.0"> Con laquale &longs;pecu-<lb/>latione proua di nouo, & confermai varij effetti, & accidenti della bilancia già di <lb/>mo&longs;trati nelle pro&longs;sime tre propo&longs;itioni; mo&longs;trando ancora, come &longs;in qui cote&longs;te <lb/>co&longs;e &longs;iano da gli altri &longs;tate malamente con&longs;iderate, & con principij fal&longs;i. </s> <s id="id.2.1.464.17.0"> Anzi di <lb/>piu per confermatione della verità &longs;oggiunge, che que&longs;ti tali non hanno &longs;aputo fa <lb/>re le loro demo&longs;trationi; poi che co'l proprio modo di &longs;peculare v&longs;ato da loro, <lb/>& con le loro mede&longs;ime ragioni proua la &longs;ua intentione, & &longs;entenza e&longs;&longs;ere veri&longs;si <lb/>ma, appoggiando &longs;i alla dottrina di Ari&longs;totele &longs;empre, & facendo toccar con ma-<lb/>no, che egli con e&longs;&longs;o lui è d'accordo nelle que&longs;tioni mechaniche. </s> <s id="id.2.1.464.18.0"> In trattando <lb/>que&longs;ta materia moue l'autore alcuni dubbi molto belli, & curio&longs;i, & poi chiara-<lb/>mente gli &longs;olue. </s> <s id="id.2.1.464.19.0"> In vltimo, accioche non manca&longs;&longs;e nulla al compiuto cono&longs;cimen <lb/>to di que&longs;to &longs;oggetto, egli hà trattato delle bilancie, che hanno le braccia di&longs;ugua<lb/>li, & di quelle che hanno le dette braccia piegate, & torte. </s> <s id="id.2.1.464.20.0"> In &longs;omma &longs;i può ben <lb/>affermare, che in cote&longs;to di&longs;cor&longs;o &longs;iano compre&longs;e tutte quelle co&longs;e, che po&longs;&longs;ono e&longs; <lb/>&longs;ere diui&longs;ate d'intorno à materia tale. </s> <s id="id.2.1.464.21.0"> Le quali &longs;ono di belli&longs;s ima & &longs;ottili&longs;sima &longs;pe <lb/>culatione, & à chiunque &longs;i diletta, & attende à que&longs;ti no bili &longs;tudi nece&longs;&longs;arij &longs;sime, <lb/>& da e&longs;&longs;ere, come hò ricordato piu d'una volta, con molta attentione vedute, & <lb/>con&longs;iderate. </s> </p> <p type="main" id="id.2.1.465.0.0"> <s id="id.2.1.465.1.0"> Doue &longs;i legge que&longs;to vocabolo latino Equilibrio, intenda&longs;i per eguale contrape&longs;o, <lb/>cio è che pe&longs;a tanto da vna banda, quanto dallaltra in pari lance, ò libra, ò bilancia <lb/>che &longs;i dica. </s> </p> <p type="main" id="id.2.1.466.0.0"> <s id="id.2.1.466.1.0"><emph type="italics"/>Librar congiu&longs;te lance.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.467.0.0"> <s id="id.2.1.467.1.0"> Di&longs;&longs;e il Petrarcha. </s> </p> <pb id="p.29v" xlink:href="pageimg-it/077.jpg"/> <p type="head" id="id.2.1.221.0.0"> <s id="id.2.1.221.1.0"> PROPOSITIONE V. </s> </p> <p type="main" id="id.2.1.222.0.0"> <s id="id.2.1.222.1.0"> Due pe&longs;i attaccati nella bilancia, &longs;e la bilancia &longs;arà tra loro in modo <lb/>diui&longs;a, chele parti ri&longs;pondano &longs;cambieuolmente à pe&longs;i; pe&longs;eranno <lb/>tanto ne'punti doue &longs;ono attaccati, quanto &longs;el'uno & l'altro fo&longs;&longs;e <lb/>pendente dal punto della diui&longs;ione. </s> </p> <figure place="text" id="id.2.1.223.0.0" xlink:href="figures-it/077_01.jpg"></figure> <p type="main" id="id.2.1.224.0.0"> <s id="id.2.1.224.1.0"><emph type="italics"/>Siala bilancia AB, il cui centro &longs;ia C, & &longs;iano due pe&longs;i EF pendenti da' punti <lb/>BG: & diuida&longs;i BG in H, &longs;i fattamente, che BH ad HG habbia la pro-<lb/>portione iste&longs;&longs;a, che hà il pe&longs;o E al pe&longs;o F. Dico ipe&longs;i EF pe&longs;are tanto in BG, <lb/>quanto &longs;e amendue pende&longs;&longs;ero dal punto H. faccia&longs;i AC eguale à CH. & &longs;i <lb/>come AC à CG, co&longs;i &longs;accia&longs;i il pe&longs;o E al pe&longs;o L. &longs;imilmente come AC à <lb/>CB, co&longs;i faccia&longs;i il pe&longs;o F al pe&longs;o M. & &longs;iano attaccati ipe&longs;i LM al punto <lb/>A. Horpercioche AC è eguale à CH, &longs;arà BC ver&longs;o CH come il pe&longs;o <lb/>M alpe&longs;o F. & percioche piu grande è BC di CH; &longs;arà anche il pe&longs;o M<emph.end type="italics"/> <lb/> <arrow.to.target n="note58"></arrow.to.target><emph type="italics"/>maggiore di F. Diuida&longs;i dunque il pe&longs;o M in due parti QR, & &longs;iala parte di<emph.end type="italics"/> <lb/> <arrow.to.target n="note59"></arrow.to.target><emph type="italics"/>Q eguale ad F; &longs;arà BC à CH, come RQ à Q: & diuidendo, come BH <lb/>ad HC, co&longs;i R à Q. Dapoi conuertendo, come CH ad HB, co&longs;i Q ad<emph.end type="italics"/> <lb/> <arrow.to.target n="note60"></arrow.to.target><emph type="italics"/>R. Oltre à ciò perche CH è eguale à CA, &longs;arà HC ver&longs;o CG come il pe&longs;o <lb/>E al pe&longs;o L: maè piu grande HC di CG, però &longs;arà anche il pe&longs;o E maggio-<emph.end type="italics"/> <lb/> <arrow.to.target n="note61"></arrow.to.target><emph type="italics"/>re del pe&longs;o L. Onde diuida&longs;i il pe&longs;o E in due parti NO, &longs;i fattamente, che la <lb/>parte di O &longs;ia eguale ad L, &longs;arà HC à CG come tutto lo NO ad O; & <lb/>diuidendo, come HG à GC, co&longs;i N ad O. & conuertendo, come CG à<emph.end type="italics"/> <lb/> <arrow.to.target n="note62"></arrow.to.target><emph type="italics"/>GH, co&longs;i O ad N. & di nuouo componendo, come CH ad HG, co&longs;i ON<emph.end type="italics"/> <lb/> <arrow.to.target n="note63"></arrow.to.target><emph type="italics"/>ad N. & come GH ad HB, co&longs;i è F ad ON. Per la qual co&longs;a per la pro <lb/>portione vguale come CH ad HB, co&longs;i F ad N. Ma come CH ad HB<emph.end type="italics"/> <lb/> <arrow.to.target n="note64"></arrow.to.target><emph type="italics"/>co&longs;i è Q ad R: &longs;arà dunque Q ad R come F ad N. & permutando co-<lb/>me Q ad F; co&longs;i R ad N. ma la parte di Q è egual ad e&longs;&longs;o F. per la qual<emph.end type="italics"/> <lb/> <arrow.to.target n="note65"></arrow.to.target><emph type="italics"/>co&longs;a la parte di R ancora &longs;arà eguale ad N. e&longs;&longs;endo dunque il pe&longs;o L eguale<emph.end type="italics"/> <pb id="p.30" xlink:href="pageimg-it/078.jpg"/><emph type="italics"/>ad O, & il pe&longs;o F eguale parimente al Q, & la parte di R eguale ad N; &longs;a <lb/>ranno ipe&longs;i LM eguali a i pe&longs;i EF. & percioche &longs;i come AC ver&longs;o CG, co<emph.end type="italics"/> <arrow.to.target n="note66"></arrow.to.target> <lb/><emph type="italics"/>&longs;i è il pe&longs;o E al pe&longs;o L, ipe&longs;i EL pe&longs;eranno egualmente. </s> <s id="id.2.1.224.2.0"> &longs;imilmente percioche <lb/>&longs;i come AC è ver&longs;o CB, co&longs;i il pe&longs;o F è alpe&longs;o M, i pe&longs;i FM pe&longs;eranno <lb/>anco egualmente. </s> <s id="id.2.1.224.3.0"> i pe&longs;i dunque LM pe&longs;eranuo egualmente co'pe&longs;i EF attacca-<lb/>ti in BG. & e&longs;&longs;endo la di&longs;tanza CA eguale alla di&longs;tanza CH, &longs;e dunque am <lb/>bidue i pe&longs;i EF &longs;ar anno attaccati in H, i pe&longs;i LM pe&longs;er anno egualmente co'<emph.end type="italics"/> <arrow.to.target n="note67"></arrow.to.target> <lb/><emph type="italics"/>pe&longs;i EF attaccati in H. Ma LM pe&longs;a ancora egualmente con EF in GB. <lb/>Adunque &longs;ar anno egualmente graui i pe&longs;i EF in GB attaccati come in H. pe<emph.end type="italics"/> <arrow.to.target n="note68"></arrow.to.target> <lb/><emph type="italics"/>&longs;eranno dunque tanto in BG quanto attaccati in H.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.225.0.0"> <s id="id.2.1.225.1.0"> <margin.target id="note58"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 17. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.226.0.0"> <s id="id.2.1.226.1.0"> <margin.target id="note59"></margin.target><emph type="italics"/>Per la con-<expan abbr="&longs;egu&etilde;za">&longs;eguenza</expan> del la<emph.end type="italics"/> 4. <emph type="italics"/>del<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.227.0.0"> <s id="id.2.1.227.1.0"> <margin.target id="note60"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 17. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.228.0.0"> <s id="id.2.1.228.1.0"> <margin.target id="note61"></margin.target><emph type="italics"/>Per la con&longs;e<expan abbr="gu&etilde;za">guenza</expan> della<emph.end type="italics"/> 4. <emph type="italics"/>del<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.229.0.0"> <s id="id.2.1.229.1.0"> <margin.target id="note62"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 18. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.230.0.0"> <s id="id.2.1.230.1.0"> <margin.target id="note63"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 16. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.231.0.0"> <s id="id.2.1.231.1.0"> <margin.target id="note64"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.232.0.0"> <s id="id.2.1.232.1.0"> <margin.target id="note65"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 16. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.233.0.0"> <s id="id.2.1.233.1.0"> <margin.target id="note66"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>del primo di Archimede del le co&longs;e che pe &longs;ano egualmente.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.234.0.0"> <s id="id.2.1.234.1.0"> <margin.target id="note67"></margin.target><emph type="italics"/>Per lo<emph.end type="italics"/> 2. <emph type="italics"/> <expan abbr="cõ">com</expan>. della not di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.235.0.0"> <s id="id.2.1.235.1.0"> <margin.target id="note68"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 3. <emph type="italics"/> <expan abbr="cõ">com</expan>. della not ai questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.236.0.0" xlink:href="figures-it/078_01.jpg"></figure> <p type="main" id="id.2.1.237.0.0"> <s id="id.2.1.237.1.0"><emph type="italics"/>Ma &longs;iano i pe&longs;i EF attaccati in CB; & &longs;ia C il centro della bilancia, & diuida&longs;i <lb/>CB in H, per modo che CH ver&longs;o HB &longs;ia come il pe&longs;o F al pe&longs;o E. Dico <lb/>che i pe&longs;i EF pe&longs;er anno tanto in CB quanto nel punto H. faccia&longs;i CA egua <lb/>le à CH, & come CA ver&longs;o CB; co&longs;i faccia&longs;i il pe&longs;o F ver&longs;o vn'altro, che <lb/>&longs;ia D, ilquale &longs;i appicchi in A. Hor percioche CH è eguale à CA, &longs;arà CH <lb/>ver&longs;o CB, come F à D; & ben è maggiore CB di CH, però il pe&longs;o D &longs;a <lb/>rà maggiore del pe&longs;o F. Diuida&longs;i dunque il D in due parti GK, & &longs;ia il G<emph.end type="italics"/> <arrow.to.target n="note69"></arrow.to.target> <lb/><emph type="italics"/>eguale allo F; &longs;arà BC à CH come GK ver&longs;o il G; et diuidendo, come BH <lb/>ad HC, co&longs;i K ver&longs;o G; & conuertendo come CH ad HB, co&longs;i G ver-<emph.end type="italics"/> <arrow.to.target n="note70"></arrow.to.target> <lb/><emph type="italics"/>&longs;o K. & come CH ad HB, co&longs;i è F ver&longs;o E. Dunque come G ver-<lb/>&longs;o K co&longs;i è F ad E. & permutando come G ad F, co&longs;i K ad E. & per-<emph.end type="italics"/> <arrow.to.target n="note71"></arrow.to.target> <lb/><emph type="italics"/>che GF &longs;ono eguali, &longs;aranno anche KE tra loro eguali. </s> <s id="id.2.1.237.2.0"> Concio&longs;ia dunque che <lb/>la parte G &longs;ia eguale ad F, & il K ad e&longs;&longs;o E; &longs;arà tutto il GK eguale a i pe<emph.end type="italics"/> <arrow.to.target n="note72"></arrow.to.target> <lb/><emph type="italics"/>&longs;i EF. & percioche AC è eguale à CH; &longs;e dunque i pe&longs;i EF &longs;aranno penden <lb/>ti dal punto H, il pe&longs;o D pe&longs;erà egualmente co'pe&longs;i EF attaccati in H. Ma <lb/>pe&longs;a anche egualmente con eßi in CB, cioè F in B, & E in C; per e&longs;&longs;ere <lb/>come AC ver&longs;o CB, co&longs;i F ver&longs;o D: percioche il pe&longs;o E pendente da C. <lb/>centro della bilancia non è cau&longs;a, che la bilancia &longs;i moua in alcuna delle due parti. <lb/>tanto &longs;aranno dunque graui i pe&longs;i EF in CB, quanto in H appicati.<emph.end type="italics"/> </s> </p> <pb id="p.30v" xlink:href="pageimg-it/079.jpg"/> <p type="margin" id="id.2.1.239.0.0"> <s id="id.2.1.239.1.0"> <margin.target id="note69"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 17. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.240.0.0"> <s id="id.2.1.240.1.0"> <margin.target id="note70"></margin.target><emph type="italics"/>Per la con&longs;e <expan abbr="gu&etilde;za">guenza</expan> della<emph.end type="italics"/> 4. <emph type="italics"/>del<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.241.0.0"> <s id="id.2.1.241.1.0"> <margin.target id="note71"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.242.0.0"> <s id="id.2.1.242.1.0"> <margin.target id="note72"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 16. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.243.0.0" xlink:href="figures-it/079_01.jpg"></figure> <p type="main" id="id.2.1.244.0.0"> <s id="id.2.1.244.1.0"><emph type="italics"/>Sia <expan abbr="finalm&etilde;te">finalmente</expan> la <expan abbr="bilãcia">bilancia</expan> AB, & da i <expan abbr="pũti">punti</expan> AB &longs;iano <expan abbr="p&etilde;denti">pendenti</expan> ipe&longs;i EF, & &longs;idil centro <lb/>della bilancia C fra ipe&longs;i, & diuida&longs;i la AB in D, talche AD ver&longs;o DB <lb/>&longs;ia come il pe&longs;o F alpe&longs;o E. Dico che i pe&longs;i EF pe&longs;ano tanto in AB, quan <lb/>to &longs;e ambidue &longs;o&longs;&longs;ero pendenti dal punto D. faccia&longs;i CG eguale à CD; & co-<lb/>me DC à CA, co&longs;i faccia&longs;i il pe&longs;o E ad vn'altro pe&longs;o H, ilquale &longs;ia attac <lb/>cato in D. & come GC ver&longs;o CB, co&longs;i faccia&longs;i il pe&longs;o F ad vn'altro che <lb/>&longs;ia K, & attachi&longs;i K in G. Horpercioche, come il BC è ver&longs;o il CG, cioè <lb/>ver&longs;o il CD, co&longs;i il pe&longs;o K ad F; &longs;arà il K maggiore del pe&longs;o F. Per laqual <lb/>co&longs;a diuida&longs;i il pe&longs;o K in L & in MN, & &longs;accia&longs;i la parte L eguale ad F,<emph.end type="italics"/> <lb/> <arrow.to.target n="note73"></arrow.to.target><emph type="italics"/>&longs;arà come BC à CD, co&longs;i tutto LMN ad L; & diuidendo, come BD<emph.end type="italics"/> <lb/> <arrow.to.target n="note74"></arrow.to.target><emph type="italics"/>ver&longs;o DC, co&longs;i la parte MN alla parte L. come dunque BD à DC, co&longs;i <lb/>la parte MN ad F. & come AD à DB, co&longs;i F ad E. Per laqual co&longs;a<emph.end type="italics"/> <lb/> <arrow.to.target n="note75"></arrow.to.target><emph type="italics"/>per la egual proportione, come AD ver&longs;o DC, co&longs;i MN ad E. & e&longs;&longs;endo AD<emph.end type="italics"/> <lb/> <arrow.to.target n="note76"></arrow.to.target><emph type="italics"/>maggiore di CD; &longs;arà anco la parte MN maggiore del pe&longs;o E. Diuida&longs;i dun <lb/>que MN in due parti MN, & &longs;ia M eguale ad E. &longs;arà come AD à<emph.end type="italics"/> <lb/> <arrow.to.target n="note77"></arrow.to.target><emph type="italics"/>DC, co&longs;i NM ad M; & diuidendo, come AC ver&longs;o CD, co&longs;i N ad M:<emph.end type="italics"/> <lb/> <arrow.to.target n="note78"></arrow.to.target><emph type="italics"/>& conuertendo, come DC ver&longs;o CA, co&longs;i M ad N. & come DC à<emph.end type="italics"/> <lb/> <arrow.to.target n="note79"></arrow.to.target><emph type="italics"/>CA, co&longs;i è E ad H; &longs;arà dunque M ad N come E ad H; & permutan <lb/>do come M ad E, co&longs;i N ad H. Ma per e&longs;&longs;ere ME traloro eguali, &longs;aran-<lb/>no anche NH tra &longs;e eguali. </s> <s id="id.2.1.244.2.0"> & percioche co&longs;i è AC ver&longs;o CD, come H <lb/>ad E: i pe&longs;i HE pe&longs;eranno egualmente. </s> <s id="id.2.1.244.3.0"> &longs;imilmente percioche, come è GC à CB, <lb/>co&longs;i F ver&longs;o K, i pe&longs;i etiandio KF pe&longs;eranno egualmente. </s> <s id="id.2.1.244.4.0"> Adunque i pe&longs;i<emph.end type="italics"/> <lb/> <arrow.to.target n="note80"></arrow.to.target><emph type="italics"/>EK HF nella bilancia AB, il cui centro &longs;ia C pe&longs;eranno egualmente. </s> <s id="id.2.1.244.5.0"> & con <lb/>cio&longs;ia che GC &longs;ia eguale à CD, & il pe&longs;o H &longs;ia pur eguale ad N, ipe&longs;i NH<emph.end type="italics"/> <pb id="p.31" xlink:href="pageimg-it/080.jpg"/><emph type="italics"/>pe&longs;eranno egualmente. </s> <s id="id.2.1.244.6.0"> & percioche tutti pe&longs;ano egualmente, tolti via i pe&longs;i HN, <lb/>iquali pe&longs;ano egualmente, i re&longs;tanti pe&longs;eranno egualmente; cioè i pe&longs;i EF, & il pe<emph.end type="italics"/> <arrow.to.target n="note81"></arrow.to.target> <lb/><emph type="italics"/>&longs;o LM pendenti dal centro C della bilancia. </s> <s id="id.2.1.244.7.0"> Ma percioche la parte L è egua-<lb/>le ad F, & la parte M è eguale alla parte E; &longs;arà tutto LM eguale a i pe&longs;i <lb/>FE in&longs;ieme pre&longs;i. </s> <s id="id.2.1.244.8.0"> & e&longs;&longs;endo CG eguale à CD, &longs;e i pe&longs;i EF &longs;aranno &longs;atti <lb/>pendenti dal punto D, ipe&longs;i EF appiccati in D pe&longs;eranno <expan abbr="egualm&etilde;te">egualmente</expan> con LM. <lb/> Per laqual co&longs;a LM pe&longs;erà <expan abbr="egualmēte">egualmente</expan> <expan abbr="tāto">tanto</expan> ad eßi EF appiccati in AB, <expan abbr="quā-<lb> to">quan-<lb/>to</expan> &longs;e fo&longs;&longs;ero appiccati nel punto D; peroche la bilancia rimane &longs;empre nell'i&longs;te&longs;&longs;o<emph.end type="italics"/> <arrow.to.target n="note82"></arrow.to.target> <lb/><emph type="italics"/>modo. </s> <s id="id.2.1.244.9.0"> Adunque i pe&longs;i EF pe&longs;eranno tanto in AB quanto nel punto D; che <lb/>bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.245.0.0"> <s id="id.2.1.245.1.0"> <margin.target id="note73"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 17. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.246.0.0"> <s id="id.2.1.246.1.0"> <margin.target id="note74"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 23. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.247.0.0"> <s id="id.2.1.247.1.0"> <margin.target id="note75"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 17. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.248.0.0"> <s id="id.2.1.248.1.0"> <margin.target id="note76"></margin.target><emph type="italics"/>Corollario della quarta del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.249.0.0"> <s id="id.2.1.249.1.0"> <margin.target id="note77"></margin.target>11. <emph type="italics"/>del<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.250.0.0"> <s id="id.2.1.250.1.0"> <margin.target id="note78"></margin.target>16. <emph type="italics"/>del<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.251.0.0"> <s id="id.2.1.251.1.0"> <margin.target id="note79"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>del<emph.end type="italics"/> 1. <emph type="italics"/>di Archi mede delle co fa che egual <expan abbr="m&etilde;te">mente</expan> pe&longs;ano.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.252.0.0"> <s id="id.2.1.252.1.0"> <margin.target id="note80"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>notitia commune di que&longs;io.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.253.0.0"> <s id="id.2.1.253.1.0"> <margin.target id="note81"></margin.target><emph type="italics"/>Per la commune notitia di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.254.0.0"> <s id="id.2.1.254.1.0"> <margin.target id="note82"></margin.target><emph type="italics"/>Per la commune notitia di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.255.0.0"> <s id="id.2.1.255.1.0"> Ma que&longs;te co&longs;e tutte dimo&longs;treremo in altra maniera, & piu Mechani <lb/>camente. </s> </p> <figure place="text" id="id.2.1.256.0.0" xlink:href="figures-it/080_01.jpg"></figure> <p type="main" id="id.2.1.257.0.0"> <s id="id.2.1.257.1.0"><emph type="italics"/>Sia la bilancia AB, & il &longs;uo centro C, & &longs;iano, come nel primo ca&longs;o, due pe&longs;i EF <lb/>pendenti da i punti BG: & &longs;ia GH ad HB, come il pe&longs;o F al pe&longs;o E. Di-<lb/>co che ipe&longs;i EF pe&longs;eranno tanto in GB, quanto &longs;e ambidue &longs;te&longs;&longs;ero pendenti <lb/>dal punto H della diui&longs;ione. </s> <s id="id.2.1.257.2.0"> Siano di&longs;po&longs;te le mede&longs;ime co&longs;e, cioè faccia&longs;i AC <lb/>eguale à CH, & dal punto A &longs;iano appe&longs;i due pe&longs;i LM, per modo che il pe <lb/>&longs;o E ver&longs;o il pe&longs;o L &longs;ia come CA ver&longs;o CG; & come CB ver&longs;o CA, co <lb/>&longs;i &longs;ia il pe&longs;o M ver&longs;o il pe&longs;o F. Ipe&longs;i LM pe&longs;eranno egualmente (come è detto <lb/>di &longs;opra) conlipe&longs;i EF appiccati in GB. Siano dapoi due punti NO li centri <lb/>della grauezza de' pe&longs;i EF; & &longs;iano congiunte le linee GN BO; & &longs;ia con-<lb/>giunta NO, laquale &longs;arà come bilancia; laquale etiandio faccia sì, che le linee <lb/>GN BO &longs;iano traloro egualmente di&longs;tanti; & dal punto H &longs;ia tirata la HP <lb/>à piombo dell'orizonte, laquale tagli NO nel P, & &longs;ia egualmente distante dal <lb/>le linee GN BO. In fine congiunga&longs;i GO, laquale tagli HP in R. Percio<emph.end type="italics"/> <arrow.to.target n="note83"></arrow.to.target> <lb/><emph type="italics"/>che dunque HR è egualmente di&longs;tante dal lato BO del triangolo GBO; &longs;arà <lb/>la GH ver&longs;ola HB, come GR ad RO. Similmente percioche RP è egual<emph.end type="italics"/> <pb id="p.31v" xlink:href="pageimg-it/081.jpg"/> <arrow.to.target n="fig22"></arrow.to.target> <lb/><emph type="italics"/>mente di&longs;tante dal lato GN del triangolo OGN; &longs;arà GR ver&longs;o RO, come <lb/>NP ver&longs;o PO. Per laqual co&longs;a come GH ad HB, così è NP ver&longs;o PO.<emph.end type="italics"/> <lb/> <arrow.to.target n="note84"></arrow.to.target> <emph type="italics"/>Ma come GH ver&longs;o HB, così è il pe&longs;o F ver&longs;o il pe&longs;o E; adunque come NP <lb/>ver&longs;o PO, così è il pe&longs;o F ver&longs;o il pe&longs;o E. Dunque il punto P &longs;arà il centro<emph.end type="italics"/> <lb/> <arrow.to.target n="note85"></arrow.to.target><emph type="italics"/>della grauezza della magnitudine compo&longs;ta di ambidue i pe&longs;i EF. Intendan&longs;i <lb/>dunque i pe&longs;i EF e&longs;&longs;ere in maniera dalla bilancia NO annodati, come &longs;e fo&longs;&longs;e vna <lb/>grandezza &longs;ola d'ambidue ipe&longs;i EF composta, & attacata ne i punti BG, &longs;e dun-<lb/>que &longs;ar anno &longs;cioltii legamenti BG de' pe&longs;i; rimarr anno i pe&longs;i EF <expan abbr="p&etilde;denti">pendenti</expan> da HP; <lb/>&longs;i come prima &longs;tauane in GB. </s> <s id="id.2.1.257.2.10"> Ma i pe&longs;i EF appiccati in GB pe&longs;ano egualmente <lb/>co'ipe&longs;i LM, & ipe&longs;i EF pendenti dal punto H hanno l'i&longs;te&longs;&longs;a di&longs;po&longs;itione ver<emph.end type="italics"/> <lb/> <arrow.to.target n="note86"></arrow.to.target><emph type="italics"/>&longs;o la bilancia AB, come &longs;e &longs;o&longs;&longs;ero appiccati in BG: Gli isteßi pe&longs;i dunque EF <lb/>pendenti da H pe&longs;aranno egualmente con gli i&longs;te&longs;&longs;i pe&longs;i LM. Sono dunque egual-<lb/>mente graui i pe&longs;i EF attaccati in GB, come attaccati in H.<emph.end type="italics"/> </s> </p> <figure id="fig22" place="text" xlink:href="figures-it/081_01.jpg"></figure> <p type="margin" id="id.2.1.259.0.0"> <s id="id.2.1.259.1.0"> <margin.target id="note83"></margin.target><emph type="italics"/>Per la &longs;ecom da del &longs;esta.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.260.0.0"> <s id="id.2.1.260.1.0"> <margin.target id="note84"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>del quinto<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.261.0.0"> <s id="id.2.1.261.1.0"> <margin.target id="note85"></margin.target><emph type="italics"/>Per la &longs;esta del primo di Archimede d'lle co&longs;e, che pe&longs;ano egual mente.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.262.0.0"> <s id="id.2.1.262.1.0"> <margin.target id="note86"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.263.0.0" xlink:href="figures-it/081_02.jpg"></figure> <figure place="text" id="id.2.1.263.0.1" xlink:href="figures-it/081_03.jpg"></figure> <pb id="p.32" xlink:href="pageimg-it/082.jpg"/> <p type="main" id="id.2.1.265.0.0"> <s id="id.2.1.265.1.0"><emph type="italics"/>Similmente dimo&longs;treraßi, che i pe&longs;i EF pe&longs;eranno tanto appiccati in qual &longs;i voglia al-<lb/>tro punto, quanto &longs;e l'vno, & l'altro &longs;o&longs;&longs;e pendente dal punto H della diui&longs;ione. <lb/> Percioche&longs;e, come di &longs;opra habbiamo in&longs;egnato, &longs;i troueranno ipe&longs;i nella bilancia, à <lb/>i quali i pe&longs;i EF pe&longs;ino egualmente; gliisteßi pe&longs;i EF pendenti da H pe&longs;eranno <lb/>egualmente co' mede&longs;imi pe&longs;i trouati; per e&longs;&longs;ere il punto P &longs;empre il centro della <lb/>grauezzaloro; & la HP a piombo dell'orizonte.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.266.0.0"> <s id="id.2.1.266.1.0"> PROPOSITIONE VI. </s> </p> <p type="main" id="id.2.1.267.0.0"> <s id="id.2.1.267.1.0"> I pe&longs;i eguali nella bilancia appiccati hanno in grauezza quella pro-<lb/>portione, che hanno le di&longs;tanze, dalle quali &longs;tanno pendenti. </s> </p> <figure place="text" id="id.2.1.268.0.0" xlink:href="figures-it/082_01.jpg"></figure> <p type="main" id="id.2.1.269.0.0"> <s id="id.2.1.269.1.0"><emph type="italics"/>Sia la bilancia BAC &longs;o&longs;pe&longs;a nel punto A; & &longs;ia &longs;egata la AC, come pare in D. & <lb/>da i punti DC &longs;iano attaccati EF pe&longs;i eguali. </s> <s id="id.2.1.269.2.0"> Dico, che il pe&longs;o F ver&longs;o il pe&longs;o E ba <lb/>quella proportione in grauezza, che hala di&longs;tanza CA alla di&longs;tanza AD. Per-<lb/>cioche faccia&longs;i come CA ver&longs;o AD, co&longs;i il pe&longs;o F ver&longs;o vn'altro pe&longs;o, che &longs;ia G. <lb/> Dico primai p &longs;i GF pendenti dal punto C tanto pe&longs;are, quanto i pe&longs;i EF penden <lb/>ti da punti DC. Tagli&longs;i DC in due parti eguali in H, & da H &longs;iano fatti pendere <lb/>ambidue i pe&longs;i EF. Pe&longs;eranno EF pre&longs;i in&longs;ieme in quel &longs;ito tanto quanto pe&longs;ano<emph.end type="italics"/> <arrow.to.target n="note87"></arrow.to.target> <lb/><emph type="italics"/>in DC. Ponga&longs;i BA eguale ad AH, & &longs;itagli BA in K, di modo, che KA <lb/>&longs;ia eguale ad AD: dapoi dal punto B &longs;ia &longs;atto pendente il pe&longs;o L, ilquale &longs;ia il dop <lb/>pio del pe&longs;o F, cioè eguale a i due pe&longs;i EF, ilqual pe&longs;erà egualmente co'pe&longs;i EF ap <lb/>piccati in H, cioè appiccati in DC. Percioche dunque, come CA ver&longs;o AD, così è <lb/>ilpe&longs;o F ver&longs;o il pe&longs;o G, &longs;arà componendo come CA AD ver&longs;o AD, cioè come <lb/>CK ver&longs;o AD, così ipe&longs;i FG ver&longs;o il pe&longs;o G. Ma per e&longs;&longs;er come CA ver&longs;o AD,<emph.end type="italics"/> <arrow.to.target n="note88"></arrow.to.target> <lb/><emph type="italics"/>così il pe&longs;o F al pe&longs;o G, &longs;arà anche conuertendo, come DA ver&longs;o AC, così il pe&longs;o <lb/>G ver&longs;o il pe&longs;o F; & i doppi de icon&longs;eguenti, come DA alla doppia di e&longs;&longs;a AC, <lb/>così il pe&longs;o G al doppio del pe&longs;o F, cioè al pe&longs;o L. Per laqual co&longs;a come CK ver&longs;o<emph.end type="italics"/> <arrow.to.target n="note89"></arrow.to.target> <lb/><emph type="italics"/>DA, così ipe&longs;i FG al pe&longs;o G; & come AD alla doppia di AC, così il pe&longs;o G al <lb/>pe&longs;o L, adunque dalla egual proportione come CK alla doppia di AC, così ipe&longs;i FG <lb/>al pe&longs;o L. Ma come CK alla doppia di AC, così la metà di CK, cioè AH, cioè<emph.end type="italics"/> <arrow.to.target n="note90"></arrow.to.target> <lb/><emph type="italics"/>BA ver&longs;o AC. </s> <s id="id.2.1.269.2.10"> Adunque come BA ver&longs;o AC, così FG pe&longs;i al pe&longs;o L. Per laqual<emph.end type="italics"/> <pb id="p.32v" xlink:href="pageimg-it/083.jpg"/><emph type="italics"/>co&longs;a per la &longs;e&longs;ta dell'i&longs;te&longs;&longs;o primo di Archimede, i due pe&longs;i FG pendenti dal punto C<emph.end type="italics"/> <lb/> <arrow.to.target n="note91"></arrow.to.target><emph type="italics"/>pe&longs;eranno tanto, quanto il pe&longs;o L pendente dal B; cioè quanto i pe&longs;i EF pen-<lb/>denti da i punti DC. Così percioche i pe&longs;i FG tanto pe&longs;ano quanto ipe&longs;i EF, <lb/>leuato via il pe&longs;o comune F, tanto pe&longs;er à il pe&longs;o G appicato in C, quanto il pe<emph.end type="italics"/> <lb/> <arrow.to.target n="fig23"></arrow.to.target> <lb/><emph type="italics"/>&longs;o E in D. Et perciò il pe&longs;o F al pe&longs;o E hà quella proportione in grauezza, <lb/>che hà al pe&longs;o G. Ma il pe&longs;o F ver&longs;o il G era come CA ver&longs;o AD. adun <lb/>que il pe&longs;o F ancora ver&longs;o il pe&longs;o E hauerà quella proportione in grauczza, che <lb/>ha CA ver&longs;o AD che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig23" place="text" xlink:href="figures-it/083_01.jpg"></figure> <p type="margin" id="id.2.1.271.0.0"> <s id="id.2.1.271.1.0"> <margin.target id="note87"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 5. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.272.0.0"> <s id="id.2.1.272.1.0"> <margin.target id="note88"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 18. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.273.0.0"> <s id="id.2.1.273.1.0"> <margin.target id="note89"></margin.target><emph type="italics"/>Per la con&longs;eguenza della quarta del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.274.0.0"> <s id="id.2.1.274.1.0"> <margin.target id="note90"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 22. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.275.0.0"> <s id="id.2.1.275.1.0"> <margin.target id="note91"></margin.target><emph type="italics"/>Per la &longs;ettima del<emph.end type="italics"/> 5. </s> </p> <p type="main" id="id.2.1.276.0.0"> <s id="id.2.1.276.1.0"><emph type="italics"/>Ma &longs;e nella bilancia BAC &longs;i faranno pendenti da i punti BC, i pe&longs;i EF eguali; <lb/> Dico &longs;imilmente, che il pe&longs;o E ver&longs;o il pe&longs;o F hà quella proportione in grauezza, <lb/>che ha la di&longs;tanza <lb/>CA alla di&longs;tanza <lb/>AB. faccia&longs;i AD <lb/>eguale ad AB, & <lb/>dal punto D &longs;ia <lb/>fatto <expan abbr="pēdente">pendente</expan> il pe <lb/>&longs;o G eguale al pe <lb/>&longs;o F, ilquale <expan abbr="etiā-">etian-</expan><emph.end type="italics"/> <lb/> <arrow.to.target n="fig24"></arrow.to.target> <lb/><emph type="italics"/>dio &longs;arà eguale ad E. Et percioche AD è eguale ad AB; i pe&longs;i FG pe&longs;eran <lb/>no egualmente, & hauranno la mede&longs;ima grauezza. </s> <s id="id.2.1.276.2.0"> Et concio&longs;ia, che la grauezza <lb/>del pe&longs;o E ver&longs;o la grauezza del pe&longs;o G &longs;ia come CA ad AD; &longs;arà la gra-<lb/>uezza del pe&longs;o E ver&longs;o la grauezza del pe&longs;o F, come CA ad AD, cioè CA <lb/>ad AB, che parimente era da mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig24" place="text" xlink:href="figures-it/083_02.jpg"></figure> <p type="head" id="id.2.1.278.0.0"> <s id="id.2.1.278.1.0"> Altramente. </s> </p> <p type="main" id="id.2.1.279.0.0"> <s id="id.2.1.279.1.0"><emph type="italics"/>Sia la bilancia BAC, col&longs;uo centro A: & ne i punti BC &longs;iano appiccati pe&longs;i <lb/>eguali GF, & &longs;ia prima il centro A, come &longs;i vuole, fra B, & C. Dico, che <lb/>il pe&longs;o F ver&longs;o il pe&longs;o G hà quella proportione in grauezza, che ha la di&longs;tanza <lb/>CA alla di&longs;tanza AB. Faccia&longs;i come BA ver&longs;o AC, co&longs;i il pe&longs;o F ad vn-<emph.end type="italics"/> <pb id="p.33" xlink:href="pageimg-it/084.jpg"/><emph type="italics"/>altro H, ilquale &longs;ia appiccato in B: i pe&longs;i HF pe&longs;eranno egualmente de A.<emph.end type="italics"/> <arrow.to.target n="note92"></arrow.to.target> <lb/> <emph type="italics"/>Ma e&longs;&longs;endo i pe&longs;i FG eguali, haurà il pe&longs;o H ver&longs;o il pe&longs;o G la proportione me <lb/>de&longs;ima, che ha ad F. Come dunque CA ver&longs;o AB, co&longs;i è H ver&longs;o G: & <lb/>come H ver&longs;o G, co&longs;i è lagrauezza di H alla grauezza di G, per e&longs;&longs;ere attac<lb/>cati nell i&longs;te&longs;&longs;o punto B. Per laqual co&longs;a come CA ad AB, co&longs;i la grauezza <lb/>del pe&longs;o H alla grauezza del pe&longs;o G. Et concio&longs;ia che la grauezza del pe&longs;o F <lb/>attacato in G &longs;ia<emph.end type="italics"/> <arrow.to.target n="note93"></arrow.to.target> <lb/><emph type="italics"/>eguale alla grauez-<lb/>za del pe&longs;o H attac<lb/>cato in B, &longs;arà la <lb/>grauezza del pe&longs;o F <lb/>ver&longs;o la grauezza <lb/>del pe&longs;o G, come <lb/>CA ver&longs;o AB, <lb/>cioè come la di&longs;tan-<lb/>za alla di&longs;tanza, che <lb/>bi&longs;ognaua mostrare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.280.0.0"> <s id="id.2.1.280.1.0"> <margin.target id="note92"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>del prime di Ar chimede del le co&longs;e che pe &longs;ano egualmente.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.281.0.0"> <s id="id.2.1.281.1.0"> <margin.target id="note93"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 7. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.282.0.0" xlink:href="figures-it/084_01.jpg"></figure> <p type="main" id="id.2.1.283.0.0"> <s id="id.2.1.283.1.0"><emph type="italics"/>Ma &longs;e la bilancia BAC fo&longs;&longs;e tagliata, come &longs;i vuole in D, & appicchin&longs;i in DC <lb/>i pe&longs;i EF eguali. </s> <s id="id.2.1.283.2.0"> Dico &longs;imilmente co&longs;i e&longs;&longs;ere la grauezza del pe&longs;o F alla gra-<lb/>uezza del pe&longs;o E, come la di&longs;tanza CA alla di&longs;tanza AD. Faccia&longs;i AB <lb/>eguale ad AD <lb/>& &longs;ia appicca-<lb/>to in B il pe&longs;o <lb/>G eguale al pe<lb/>&longs;o E, & alpe <lb/>&longs;o F. Hor <lb/>percioche AB <lb/>è eguale ad A <lb/>D; ipe&longs;i GE<emph.end type="italics"/> <lb/> <arrow.to.target n="fig25"></arrow.to.target> <lb/><emph type="italics"/>pe&longs;eranno egualmente. </s> <s id="id.2.1.283.3.0"> Ma per e&longs;&longs;ere la grauezza del pe&longs;o F ver&longs;o la grauezza <lb/>del pe&longs;o G, come CA ad AB, & la grauezza del pe&longs;o E &longs;ia eguale alla <lb/>grauezza del pe&longs;o G; &longs;arà la grauezza del pe&longs;o F ver&longs;o la grauezza del pe&longs;o E, <lb/>come CA ad AB, cioè CA ad AD, che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig25" place="text" xlink:href="figures-it/084_02.jpg"></figure> <p type="head" id="id.2.1.285.0.0"> <s id="id.2.1.285.1.0"> COROLLARIO. </s> </p> <p type="main" id="id.2.1.286.0.0"> <s id="id.2.1.286.1.0"> Da que&longs;to è manife&longs;to, che quanto il pe&longs;o è piu di&longs;tante dal centro <lb/>della bilancia, tanto egli è anco piu graue, & per con&longs;eguente mo-<lb/>uer&longs;i piu velocemente. </s> </p> <p type="main" id="id.2.1.287.0.0"> <s id="id.2.1.287.1.0"> Quinci oltre à ciò &longs;i mo&longs;trerà facilmente anche la ragione della Sta-<lb/>dera. </s> </p> <pb id="p.33v" xlink:href="pageimg-it/085.jpg"/> <p type="main" id="id.2.1.289.0.0"> <s id="id.2.1.289.1.0"> Corollario vocabolo Latino co&longs;tumato da tutti gli altri Scrittori Italiani in cotal ma <lb/>teria, nè di&longs;piacque à Dante nel 28. cap. del Purgatorio. </s> <s id="id.2.1.289.2.0"> Dirotti vn corollario an-<lb/>co per gratia.” vuol dire, &longs;econdo Varrone nel primo libro della lingua Latina, <lb/>quella giunta, & quel &longs;opra piu, che &longs;i dà oltre al pagamento, quando &longs;i comp era <lb/>qualche co&longs;a. </s> <s id="id.2.1.289.4.0"> Al tempo antico allhor che i recitatori di Tragedie, Comedie, & <lb/>altri Poemi nelle &longs;cene &longs;i portauano bene, & piaceuano à gli vditori, era loro do-<lb/>nato oltra al prezzo a&longs;&longs;egnato, vn corollario per cia&longs;cuno, cioè vna piccola coro <lb/>na per douer&longs;ene ornare le tempie per giunta, & &longs;opra piu delle &longs;ue mercedi. </s> <s id="id.2.1.289.5.0"> Co&longs;i <lb/>nelle &longs;cienze matematiche v&longs;a&longs;i di aggiungere certe co&longs;e, oltra le propo&longs;itioni, <lb/>qua&longs;i giunte & con&longs;equenze, le quali na&longs;cono dalle co&longs;e primieramente dimo&longs;tra-<lb/>te, & &longs;ono loro corri&longs;pondenti, & non &longs;ono però nè propo&longs;itioni, nè problemi, <lb/>nè lemmi, ma alla &longs;embianza predetta chiaman&longs;i corollarij, molti de i quali han-<lb/>no congiunta la &longs;ua dimo&longs;tratione. </s> </p> <p type="main" id="id.2.1.290.0.0"> <s id="id.2.1.290.1.0"><emph type="italics"/>Hor &longs;ia AB il fusto della Stadera, la cui trutina &longs;ia in C; & &longs;ia il marco della &longs;ta<emph.end type="italics"/> <lb/> <arrow.to.target n="note94"></arrow.to.target><emph type="italics"/>dera E. Appicchi&longs;i in A il pe&longs;o D, che pe&longs;i egualmente col marco E appic-<lb/>cato in F. Appicchi&longs;i parimente vn'altro pe&longs;o G in A, ilqual anco pe&longs;i egual-<lb/>mente col marco E appiccato in B. Dico, la grauezza del pe&longs;o D ver&longs;o la gra-<lb/>uezza del <lb/>G e&longs;&longs;ere co <lb/>&longs;i, come CF <lb/>ver&longs;o CB. <lb/> Hor per-<lb/>cioche la <lb/>grauezza <lb/>del pe&longs;o D <lb/>è eguale al <lb/>la grauez-<lb/>za del pe-<lb/>&longs;o E at-<lb/>taccato in <lb/>F, & la<emph.end type="italics"/> <lb/> <arrow.to.target n="fig26"></arrow.to.target> <lb/><emph type="italics"/>grauezza del pe&longs;o G è eguale alla grauezza del pe&longs;o E po&longs;to in B; &longs;arà la grauez-<lb/>za del pe&longs;o D alla grauezza del pe&longs;o E po&longs;to in F, come la grauezza del pe&longs;o G alla <lb/>grauezza del pe&longs;o E po&longs;to in B; & permutando come la grauezza del pe&longs;o D alla <lb/>grauezza del pe&longs;o G, co&longs;i la grauezza di E po&longs;to in F alla grauezza di E po&longs;to in B; <lb/>ma la grauezza del pe&longs;o E in F alla grauezza di E in B po&longs;to è come CF <lb/>ver&longs;o CB; come dunque la grauezza del pe&longs;o D alla grauezza del pe&longs;o G, co&longs;i <lb/>è CF ver&longs;o CB. Se dunque la parte del fu&longs;to CB diuidera&longs;&longs;i in parti eguali, po <lb/>&longs;to &longs;olo il pe&longs;o E & piu da pre&longs;&longs;o, & piu da lontano dal punto C; le grauezze de <lb/>pe&longs;i, lequali&longs;tanno pendenti dal punto A &longs;aranno traloro manife&longs;te & note. </s> <s id="id.2.1.290.2.0"> Co-<lb/>me&longs;e la di&longs;tanza CB &longs;arà tripla della di&longs;tanza CF, &longs;arà parimente la grauezza <lb/>di e&longs;&longs;o G tripla della grauezza di D, che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig26" place="text" xlink:href="figures-it/085_01.jpg"></figure> <p type="margin" id="id.2.1.293.0.0"> <s id="id.2.1.293.1.0"> <margin.target id="note94"></margin.target><emph type="italics"/>Ragione del la stadera.<emph.end type="italics"/> </s> </p> <pb id="p.34" xlink:href="pageimg-it/086.jpg"/> <p type="main" id="id.2.1.294.0.0"> <s id="id.2.1.294.1.0"> In altro modo po&longs;&longs;iamo anco v&longs;are la &longs;tadera, affine che le grauezze <lb/>de ipe&longs;i &longs;i facciano note. </s> </p> <p type="main" id="id.2.1.295.0.0"> <s id="id.2.1.295.1.0"><emph type="italics"/>Sia il fu&longs;to della stadera AB, la cui trutina &longs;ia in C, & &longs;ia il marco della stadera <lb/>E, ilquale &longs;ia appiccato in A; & &longs;iano i pe&longs;i DG di&longs;uguali, le proportioni delle <lb/>grauezze de quali cerchia-<lb/>mo: &longs;ia appiccato il pe&longs;o D <lb/>in B talche pe&longs;i egual-<lb/>mente con E. Similmente <lb/>appicchi&longs;i il pe&longs;o G in F, <lb/>ilquale pe&longs;i egualmente con <lb/>l'i&longs;te&longs;&longs;o pe&longs;o E. Dico D <lb/>ver&longs;o G co&longs;i e&longs;&longs;ere; come<emph.end type="italics"/> <arrow.to.target n="note95"></arrow.to.target> <lb/><emph type="italics"/>CF ver&longs;o CB. Hor perche <lb/>i pe&longs;i DE pe&longs;ano <expan abbr="egualmē">egualmen</expan><emph.end type="italics"/> <lb/> <arrow.to.target n="fig27"></arrow.to.target> <lb/><emph type="italics"/>te, &longs;arà D ad E, come CA à CB. & concio&longs;ia, che anche i pe&longs;i GE pe&longs;i-<lb/>no egualmente, &longs;arà il pe&longs;o E ver&longs;o il pe&longs;o G, come FC à CA; Per laqual <lb/>co&longs;a per la proportion eguale il pe&longs;o D al pe&longs;o G, co&longs;i &longs;arà, come CF à CB. <lb/>che parimente bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> <arrow.to.target n="note96"></arrow.to.target> </s> </p> <figure id="fig27" place="text" xlink:href="figures-it/086_01.jpg"></figure> <p type="margin" id="id.2.1.297.0.0"> <s id="id.2.1.297.1.0"> <margin.target id="note95"></margin.target><emph type="italics"/>Per la &longs;esta del primo di Archimede d'lle co&longs;e, che pe&longs;ano egual mente.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.298.0.0"> <s id="id.2.1.298.1.0"> <margin.target id="note96"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 23. <emph type="italics"/>del quinto<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.299.0.0"> <s id="id.2.1.299.1.0"> PROPOSITIONE VII. </s> </p> <p type="head" id="id.2.1.300.0.0"> <s id="id.2.1.300.1.0"> PROBLEMA. </s> </p> <p type="main" id="id.2.1.301.0.0"> <s id="id.2.1.301.1.0"> Dati quanti &longs;i vogliano pe&longs;i nella bilancia, appiccati in qualluogo &longs;i <lb/>&longs;ia, ritrouare il centro della bilancia, dal quale &longs;e &longs;arà fatta penden-<lb/>te la bilancia, i dati pe&longs;i &longs;taranno fermi. </s> </p> <p type="main" id="id.2.1.302.0.0"> <s id="id.2.1.302.1.0"> PROBLEMA. Sotto il nome di Propo&longs;itione &longs;i contiene il Problema ancora vo-<lb/>cabolo greco; ma il Problema ha dauantaggio della Propo&longs;itione in particolare, <lb/>che ordina, & in&longs;egna ad operare qualche effetto; doue la Propo&longs;itione &longs;uole &longs;ta <lb/>re nella nuda &longs;peculatione &longs;olamente. </s> <s id="id.2.1.302.2.0"> Et que&longs;ta è la differenza tra la Propo&longs;itio-<lb/>ne, & il Problema. </s> </p> <pb id="p.34v" xlink:href="pageimg-it/087.jpg"/> <figure place="text" id="id.2.1.304.0.0" xlink:href="figures-it/087_01.jpg"></figure> <p type="main" id="id.2.1.305.0.0"> <s id="id.2.1.305.1.0"><emph type="italics"/>Sia la bilancia AB, & &longs;iano dati quanti &longs;i vogliano pe&longs;i CDEFG prendan&longs;i nel <lb/>la bilancia, a piacere i punti AHKLB, da quali &longs;ian fatti pendenti i dati pe&longs;i. <lb/> Bi&longs;ogna ritrouar il centro della bilancia, dal quale &longs;e &longs;i far à l'appiccamento, rim anga <lb/>no i dati pe&longs;i. </s> <s id="id.2.1.305.2.0"> Diuida&longs;i AH in M, &longs;i che HM ad MA &longs;ia come la grauezza <lb/>del pe&longs;o C alla grauezza del pe&longs;o D. Dapoi diuida&longs;i anco BL in N, &longs;i che <lb/>LN ad NB &longs;ia come la grauezza del pe&longs;o G alla grauezza del pe&longs;o F. Et di-<lb/>uida&longs;i MN in O, &longs;i che MO ver&longs;o ON &longs;ia come la grauezza de pe&longs;i FG <lb/>alla grauezza de'pe&longs;i CD. Et in fine diuida&longs;i KO in P, &longs;i che KP ver&longs;o PO <lb/>&longs;ia come la grauezza de'pe&longs;i CD FG alla grauezza del pe&longs;o E. Hor percio-<emph.end type="italics"/> <lb/> <arrow.to.target n="note97"></arrow.to.target><emph type="italics"/>che i pe&longs;i CDFG tanto pe&longs;ano in O, quanto CD in M, & FG in N; <lb/>pe&longs;eranno egualmente i pe&longs;i CD in M, & FG in N, & il pe&longs;o E in K, <lb/>&longs;e &longs;aranno &longs;o&longs;pe&longs;i nel punto P. Et concio&longs;ia, che ipe&longs;i CD tanto pe&longs;ino in M, <lb/>quanto in AH, & FG in N quanto in LB; ipe&longs;i CDFG pendenti da' <lb/>punti AHLB, & il pe&longs;o E da K, &longs;e da P &longs;aranno &longs;o&longs;pe&longs;i, pe&longs;eranno egual-<lb/>mente, & rimarranno. </s> <s id="id.2.1.305.3.0"> egli è dunque trouato il P centro della bilancia, dalquale <lb/>rimangono i pe&longs;i dati. </s> <s id="id.2.1.305.4.0"> Che bi&longs;ogna operare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.306.0.0"> <s id="id.2.1.306.1.0"> <margin.target id="note97"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 5. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.307.0.0"> <s id="id.2.1.307.1.0"> COROLLARIO. </s> </p> <p type="main" id="id.2.1.308.0.0"> <s id="id.2.1.308.1.0"> Da que&longs;to è chiaro, che &longs;ei centri della grauezza de' pe&longs;i CDEFG <lb/>fo&longs;&longs;ero ne' punti AHKLB, &longs;arebbe il punto P il centro della <lb/>grauezza della magnitudine compo&longs;ta di tutti i pe&longs;i CDEFG. </s> </p> <p type="main" id="id.2.1.309.0.0"> <s id="id.2.1.309.1.0"><emph type="italics"/>Que&longs;to è manife&longs;to dalla diffinitione del centro della grauezza, concio&longs;ia che i pe&longs;i ri-<lb/>mangano, &longs;e &longs;ono &longs;o&longs;tenuti dal punto P.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.310.0.0"> <s id="id.2.1.310.1.0"><emph type="italics"/>Il fine della Bilancia.<emph.end type="italics"/> </s> </p> </chap> <chap id="id.2.2.0.0.3"> <pb id="p.35" xlink:href="pageimg-it/088.jpg"/> <p type="head" id="id.2.1.312.0.0"> <s id="id.2.1.312.1.0"> DELLA LEVA. </s> </p> <p type="head" id="id.2.1.314.0.0"> <s id="id.2.1.314.1.0"> LEMMA. </s> </p> <figure place="text" id="id.2.1.315.0.1" xlink:href="figures-it/088_01.jpg"></figure> <p type="main" id="id.2.1.315.0.0"> <s id="id.2.1.315.1.0"> Siano quattro grandezze ABCD; & &longs;ia la A <lb/>maggiore della B, & C maggiore della D. Dico, <lb/>che A ver&longs;o D hà proportione maggiore di quello <lb/>che hà B ver&longs;o C. </s> </p> <p type="main" id="id.2.1.316.0.0"> <s id="id.2.1.316.1.0"><emph type="italics"/>Hor percioche A ver&longs;o C hà proportion maggio-<lb/>re, che B ver&longs;o C; & A parimente ver&longs;o D<emph.end type="italics"/> <arrow.to.target n="note98"></arrow.to.target> <lb/><emph type="italics"/>hà proportion maggiore di quel che ha ver&longs;o C: <lb/> Dunque A ver&longs;o D l'hauer à maggiore, che B <lb/>ver&longs;o C, Che bi&longs;ognaua mostrare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.317.0.0"> <s id="id.2.1.317.1.0"> <margin.target id="note98"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.319.0.0"> <s id="id.2.1.319.1.0"> PROPOSITIONE I. </s> </p> <p type="main" id="id.2.1.320.0.0"> <s id="id.2.1.320.1.0"> La po&longs;&longs;anza, che &longs;o&longs;tiene il pe&longs;o attaccato alla Leua, ha la proportio <lb/>ne mede&longs;ima al detto pe&longs;o, che ha la di&longs;tanza della Leua fra il &longs;o&longs;te <lb/>gno po&longs;ta, & lo attaccamento del pe&longs;o, alla di&longs;tanza, che è dal &longs;o&longs;te <lb/>gno alla po&longs;&longs;anza. </s> </p> <p type="main" id="id.2.1.321.0.0"> <s id="id.2.1.321.1.0"><emph type="italics"/>Sia la leua AB, il cui &longs;oftegno &longs;ia C; & &longs;iail pe&longs;o D pendente da A con AH, <lb/>&longs;i che AH &longs;ia &longs;empre à piombo dell'orizonte: & &longs;ia la po&longs;&longs;anza &longs;oftenente il pe-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig28"></arrow.to.target> <lb/><emph type="italics"/>&longs;o in B. Dico che la po&longs;&longs;anza posta in B ver&longs;o il pe&longs;o D &longs;ta co&longs;i, come la CA<emph.end type="italics"/> <pb id="p.35v" xlink:href="pageimg-it/089.jpg"/> <arrow.to.target n="note99"></arrow.to.target><emph type="italics"/>ver&longs;o la CB. Faccia&longs;i come la BC alla CA, co&longs;i il pe&longs;o D ad vn'altro pe&longs;o <lb/>E, talche &longs;e egli in B &longs;arà appiccato, pe&longs;erà <expan abbr="egualm&etilde;te">egualmente</expan> con D, per e&longs;&longs;er il C cen <lb/>tro della grauezza di ambidue. </s> <s id="id.2.1.321.2.0"> Per laqual co&longs;a vna po&longs;&longs;anza eguale ad e&longs;&longs;o E po <lb/>&longs;ta nel <lb/>mede&longs;i<lb/>mo lo<lb/>go pe-<lb/>&longs;erà e-<lb/>gual-<lb/>mente <lb/>con e&longs;-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig29"></arrow.to.target> <lb/><emph type="italics"/>&longs;o D, nella leua AB, collocando il &longs;o&longs;tegno &longs;uo in C, cioè impedirà, che il pe-<emph.end type="italics"/> <lb/> <arrow.to.target n="note100"></arrow.to.target><emph type="italics"/>&longs;o D non inchini in giu&longs;o, &longs;i come impedi&longs;ce il pe&longs;o E. Ma la po&longs;&longs;anza di B al <lb/>pe&longs;o D hàla mede&longs;ima proportione, che il pe&longs;o E ha all'iste&longs;&longs;o D: adunque la <lb/>po&longs;&longs;anza di B ver&longs;o il pe&longs;o D &longs;arà come CA ver&longs;o CB; cioè la di&longs;tanza del-<lb/>la leua dal &longs;ostegno al &longs;o&longs;tenimento del pe&longs;o, alla di&longs;tanza dal &longs;ostegno alla po&longs;&longs;an-<lb/>za, che bi&longs;o gnaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig28" place="text" xlink:href="figures-it/088_02.jpg"></figure> <figure id="fig29" place="text" xlink:href="figures-it/089_01.jpg"></figure> <p type="margin" id="id.2.1.324.0.0"> <s id="id.2.1.324.1.0"> <margin.target id="note99"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>del<emph.end type="italics"/> 1. <emph type="italics"/>di Archimede delleco &longs;ache egual <expan abbr="m&etilde;te">mente</expan> pe&longs;ano.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.325.0.0"> <s id="id.2.1.325.1.0"> <margin.target id="note100"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 7. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.326.0.0"> <s id="id.2.1.326.1.0"> Di quì ageuolmente &longs;i puote mo&longs;trare, che <expan abbr="quãto">quanto</expan> il &longs;o&longs;tegno &longs;arà piu <lb/>vicino al pe&longs;o, tanto minor po&longs;&longs;anza &longs;i ricerca à &longs;o&longs;tenere il detto <lb/>pe&longs;o. </s> </p> <p type="main" id="id.2.1.327.0.0"> <s id="id.2.1.327.1.0"><emph type="italics"/>Poste le co&longs;e mede&longs;ime &longs;ia il &longs;o&longs;tegno in F piu da pre&longs;&longs;o ad A, che C; & faccia&longs;i <lb/>come BF ad FA, co&longs;i il pe&longs;o D ad vn'altro pe&longs;o G, ilquale &longs;e in B &longs;ia ap-<emph.end type="italics"/> <lb/> <arrow.to.target n="note101"></arrow.to.target><emph type="italics"/>piccato; i pe&longs;i DG dal &longs;o&longs;tegno F pe&longs;eranno egualmente. </s> <s id="id.2.1.327.2.0"> Hor percioche BF <lb/>è mag-<lb/>giore di <lb/>BC, & <lb/>CA<emph.end type="italics"/> <lb/> <arrow.to.target n="note102"></arrow.to.target><emph type="italics"/>maggio<lb/>re di AF; <lb/>la <lb/>propor-<lb/>tione di <emph.end type="italics"/> <lb/> <arrow.to.target n="fig30"></arrow.to.target> <lb/><emph type="italics"/>BF ver&longs;o FA &longs;arà maggiore, che di BC ver&longs;o CA: & perciò maggiore anco <lb/>&longs;arà la proportione del pe&longs;o D alpe&longs;o G, che de l'iste&longs;&longs;o D ad E: Dunque il<emph.end type="italics"/> <lb/> <arrow.to.target n="note103"></arrow.to.target><emph type="italics"/>pe&longs;o G &longs;arà minore del pe&longs;o E. & concio&longs;ia che la po&longs;&longs;anza po&longs;ta in B eguale à <lb/>G pe&longs;i egualmente con D, auerrà, che minore po&longs;&longs;anza di quella, laquale è eguale <lb/>al pe&longs;o E &longs;o&longs;tenter à il pe&longs;o D; e&longs;&longs;endo la leua AB, & il &longs;o&longs;tegno &longs;uo doue è F, <lb/>che &longs;e egli &longs;o&longs;&longs;e doue è C. Similmente anche mo&longs;trera&longs;&longs;i, che quanto piu dapre&longs;&longs;o &longs;a <lb/>rà il &longs;o&longs;tegno al pe&longs;o D, &longs;empre vi &longs;i ricercherà anco po&longs;&longs;anza minore per &longs;o&longs;tentare <lb/>il detto pe&longs;o D.<emph.end type="italics"/> </s> </p> <figure id="fig30" place="text" xlink:href="figures-it/089_02.jpg"></figure> <p type="margin" id="id.2.1.330.0.0"> <s id="id.2.1.330.1.0"> <margin.target id="note101"></margin.target><emph type="italics"/>Per la mede&longs;ima &longs;esta.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.331.0.0"> <s id="id.2.1.331.1.0"> <margin.target id="note102"></margin.target><emph type="italics"/>Per lo Lemma.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.332.0.0"> <s id="id.2.1.332.1.0"> <margin.target id="note103"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 10. <emph type="italics"/>del quinto<emph.end type="italics"/> </s> </p> <pb id="p.36" xlink:href="pageimg-it/090.jpg"/> <p type="head" id="id.2.1.333.0.0"> <s id="id.2.1.333.1.0"> COROLLARIO. </s> </p> <p type="main" id="id.2.1.334.0.0"> <s id="id.2.1.334.1.0"> Onde &longs;i puote raccogliere chiaramente, che e&longs;&longs;endo AF minore di <lb/>FB, minor po&longs;&longs;anza anco &longs;i ricerca in B per &longs;o&longs;tenere il pe&longs;o D. <lb/>& e&longs;&longs;endo eguale, eguale: & maggiore, maggiore. </s> </p> <p type="head" id="id.2.1.335.0.0"> <s id="id.2.1.335.1.0"> PROPOSITIONE II. </s> </p> <p type="main" id="id.2.1.336.0.0"> <s id="id.2.1.336.1.0"> In altra maniera po&longs;&longs;iamo v&longs;are la Leua. </s> </p> <p type="main" id="id.2.1.337.0.0"> <s id="id.2.1.337.1.0"><emph type="italics"/>Sia la leua AB, il cui &longs;o&longs;tegno &longs;ia B, & il pe&longs;o C &longs;ia attaccato, come &longs;i vuole, in D<emph.end type="italics"/> <arrow.to.target n="note104"></arrow.to.target> <lb/><emph type="italics"/>fra AB; & &longs;ia la po&longs;&longs;anza in A che &longs;ostiene il pe&longs;o C. Dico, che &longs;i come <lb/>BD à BA; co&longs;i è lapo&longs;&longs;anza di A' al pe&longs;o C. Appicchi&longs;i in A il pe&longs;o E <lb/>eguale al C; & come AB ver&longs;o BD, co&longs;i faccia&longs;i il pe&longs;o E ver&longs;o vn'altro pe&longs;o,<emph.end type="italics"/> <arrow.to.target n="note105"></arrow.to.target> <lb/><emph type="italics"/>come F. Et percioche i pe&longs;i CE &longs;ono tra&longs;e eguali, &longs;arà il pe&longs;o C ver&longs;o il pe&longs;o F <lb/>come AB ver&longs;o BD. Attacchi&longs;i parimente il pe&longs;o F in A. & percioche il<emph.end type="italics"/> <arrow.to.target n="note106"></arrow.to.target> <lb/><emph type="italics"/>pe&longs;o E al pe&longs;o F <lb/>è come la grauez <lb/>za del pe&longs;o di E <lb/>alla grauezza di <lb/>F; & il pe&longs;o E <lb/>ad F è come AB <lb/>à BD; come <expan abbr="dũ">dum</expan> <lb/>que la grauezza <lb/>del pe&longs;o E alla<emph.end type="italics"/> <arrow.to.target n="note107"></arrow.to.target> <lb/> <arrow.to.target n="fig31"></arrow.to.target> <lb/><emph type="italics"/>grauezza del pe&longs;o F, co&longs;i è AB ver&longs;o BD. ma come AB à BD, co&longs;i è la <lb/>grauezza del pe&longs;o E alla grauezza del pe&longs;o C: Per laqual co&longs;a la grauezza del <lb/>pe&longs;o E alla grauezza del pe&longs;o F co&longs;i &longs;arà, come la grauezza del pe&longs;o E alla gra-<lb/>uezza del pe&longs;o C. I pe&longs;i dunque CF hanno la mede&longs;ima grauezza: &longs;i che pon-<lb/>ga&longs;i la po&longs;&longs;anza di A che &longs;o&longs;tenga il pe&longs;o F, &longs;arà la po&longs;&longs;anza di A eguale al pe&longs;o <lb/>F. & percioche il pe&longs;o E attaccato in A è graue egualmente, come il C appicca-<emph.end type="italics"/> <arrow.to.target n="note108"></arrow.to.target> <lb/><emph type="italics"/>to in D; hauer à la proportione i&longs;te&longs;&longs;a la po&longs;&longs;anza di A ver&longs;o la grauezza del pe&longs;o <lb/>F appiccato in A, che ha alla grauezza del pe&longs;o C appiccato in D. Mala po&longs;&longs;an <lb/>za di A eguale ad F &longs;o&longs;tiene il pe&longs;o F; dunque la po&longs;&longs;anza di A &longs;o&longs;tenterà anco <lb/>il pe&longs;o C. Et co&longs;i per e&longs;&longs;ere la po&longs;&longs;anza di A eguale al pe&longs;o F, & il pe&longs;o C ver&longs;o <lb/>il pe&longs;o F &longs;ia come AB à BD; &longs;arà il pe&longs;o C ver&longs;o la po&longs;&longs;anza po&longs;ta in A come <lb/>AB à BD. & conuertendo, come BD à BA, co&longs;i la po&longs;&longs;anza po&longs;ta in A ver <lb/>&longs;o il pe&longs;o C. Dunque la po&longs;&longs;anza ver&longs;o il pe&longs;o co&longs;i &longs;arà, come la di&longs;tanza, che è fra<emph.end type="italics"/> <arrow.to.target n="note109"></arrow.to.target> <lb/><emph type="italics"/>il &longs;o&longs;tegno, & l'appiccamento del pe&longs;o alla di&longs;tanza, che è dal &longs;o&longs;tegno alla po&longs;&longs;an-<lb/>za, che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig31" place="text" xlink:href="figures-it/090_01.jpg"></figure> <p type="margin" id="id.2.1.340.0.0"> <s id="id.2.1.340.1.0"> <margin.target id="note104"></margin.target><emph type="italics"/>Nella &longs;esta di questo de la bilancia.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.341.0.0"> <s id="id.2.1.341.1.0"> <margin.target id="note105"></margin.target><emph type="italics"/>Dalla<emph.end type="italics"/> 11. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.342.0.0"> <s id="id.2.1.342.1.0"> <margin.target id="note106"></margin.target><emph type="italics"/>Per la &longs;esta della <expan abbr="bilācia">bilancia</expan><emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.343.0.0"> <s id="id.2.1.343.1.0"> <margin.target id="note107"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 29. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.344.0.0"> <s id="id.2.1.344.1.0"> <margin.target id="note108"></margin.target><emph type="italics"/>Per la &longs;ettima del<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.345.0.0"> <s id="id.2.1.345.1.0"> <margin.target id="note109"></margin.target><emph type="italics"/>Per lo Corollario della<emph.end type="italics"/> 4 <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <pb id="p.36v" xlink:href="pageimg-it/091.jpg"/> <p type="head" id="id.2.1.346.0.0"> <s id="id.2.1.346.1.0"> Altramente. </s> </p> <p type="main" id="id.2.1.347.0.0"> <s id="id.2.1.347.1.0"><emph type="italics"/>Sia la leua AB, il cui &longs;o&longs;tegno &longs;ia B, & il pe&longs;o E &longs;ia pendente dal punto C, & <lb/>&longs;ia in A la forza, che &longs;ostiene l pe&longs;o E. Dico, che &longs;i come BC à BA, co&longs;i è<emph.end type="italics"/> <lb/> <arrow.to.target n="fig32"></arrow.to.target> <lb/><emph type="italics"/>anco la po&longs;&longs;anza di A ver&longs;o il pe&longs;o E. Allunghi&longs;i AB in D, & faccia&longs;i <lb/>BD eguale à BC; & appicchi&longs;i il pe&longs;o F al punto D, che &longs;ia eguale al pe&longs;o E; <lb/>& parimente dal punto A &longs;i faccia pendere il punto G in modo, che il pe&longs;o F hab <lb/>bia la proportione i&longs;te&longs;&longs;a ver&longs;o il pe&longs;o G, che ha AB à BD. ipe&longs;i FG verranno <lb/>à pe&longs;ar egualmente: & concio&longs;ia che CB &longs;ia eguale à BD, anco i pe&longs;i FE egua <lb/>li pe&longs;eranno egualmente. </s> <s id="id.2.1.347.2.0"> Ma ipe&longs;i FEG nella bilancia, ouero nella leua DBA <lb/>appiccati, il cui &longs;o&longs;tegno è B, non pe&longs;eranno egualmente, ma inchineranno à ba&longs;&longs;o <lb/>dalla parte di A. Per laqual co&longs;a ponga&longs;i in A tanta forza, che ipe&longs;i FEG pe&longs;i-<lb/>no egualmente, &longs;arà la po&longs;&longs;anza in A eguale al pe&longs;o G; peroche i pe&longs;i FE pe&longs;a-<lb/>no egualmente, & la forza in A niente altro deue fare, che &longs;o&longs;tenere il pe&longs;o G, ac-<lb/>cio che non de&longs;cenda. </s> <s id="id.2.1.347.3.0"> Et percio che i pe&longs;i FEG, & la po&longs;&longs;anza in A pe&longs;ano egual <lb/>mente, leuati dunque via i pe&longs;i FG, i quali pe&longs;ano egualmente, i re&longs;tanti pe&longs;eran-<lb/>no pur egualmente, cioè la po&longs;&longs;anza in A co'l pe&longs;o E, cioè la po&longs;&longs;anza in A &longs;o-<lb/>sterra ilpe&longs;o E, &longs;i che la leua AB rimanga, come era prima. </s> <s id="id.2.1.347.4.0"> Et per e&longs;&longs;ere la <lb/>po&longs;&longs;anza in A eguale al pe&longs;o G, & il pe&longs;o E eguale al pe&longs;o F, haurà la po&longs;&longs;anza <lb/>in A la proportione iste&longs;&longs;a al pe&longs;o E, che hà BD, cioè BC à BA, che bi&longs;ogna <lb/>ua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig32" place="text" xlink:href="figures-it/091_01.jpg"></figure> <p type="head" id="id.2.1.349.0.0"> <s id="id.2.1.349.1.0"> COROLLARIO I. </s> </p> <p type="main" id="id.2.1.350.0.0"> <s id="id.2.1.350.1.0"> Da que&longs;to etiandio, come prima, puote e&longs;&longs;ere manife&longs;to, che &longs;eil pe&longs;o <lb/>E &longs;arà po&longs;to piu vicino al &longs;o&longs;tegno B, come in H, minore <lb/>po&longs;&longs;anza po&longs;ta in A puote &longs;o&longs;tener il detto pe&longs;o. </s> </p> <p type="main" id="id.2.1.351.0.0"> <s id="id.2.1.351.1.0"><emph type="italics"/>Percioche minor proportione ha HB à BA, che CB à BA. & quanto piu da<emph.end type="italics"/> <lb/> <arrow.to.target n="note110"></arrow.to.target><emph type="italics"/>vicino il pe&longs;o &longs;arà al &longs;o&longs;tegno, &longs;empre anco &longs;i mo&longs;trerà &longs;imilmente minor po&longs;&longs;anza <lb/>poter &longs;o&longs;tener il pe&longs;o E.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.352.0.0"> <s id="id.2.1.352.1.0"> <margin.target id="note110"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.353.0.0"> <s id="id.2.1.353.1.0"> COROLLARIO II. </s> </p> <p type="main" id="id.2.1.354.0.0"> <s id="id.2.1.354.1.0"> Segue etiandio, che la po&longs;&longs;anza in A &longs;empre è minore del pe&longs;o E: </s> </p> <pb id="p.37" xlink:href="pageimg-it/092.jpg"/> <p type="main" id="id.2.1.356.0.0"> <s id="id.2.1.356.1.0"><emph type="italics"/>Percioche pigli&longs;i tra A & B qual punto &longs;i voglia, come C, &longs;empre BC &longs;arà <lb/>minore di BA.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.357.0.0"> <s id="id.2.1.357.1.0"> COROLLARIO III. </s> </p> <p type="main" id="id.2.1.358.0.0"> <s id="id.2.1.358.1.0"> Da que&longs;to parimente &longs;i puote cauare, che &longs;e due &longs;aranno le po&longs;&longs;anze, <lb/>l'vna in A, & l'altra in B, & ambedue &longs;o&longs;tentino il pe&longs;o E, la po&longs;-<lb/>&longs;anza in A ver&longs;o la po&longs;&longs;anza in B è come BC ver&longs;o CA. </s> </p> <p type="main" id="id.2.1.359.0.0"> <s id="id.2.1.359.1.0"><emph type="italics"/>Percioche la leua BA fal'officio di due leue, & AB &longs;ono come due &longs;o&longs;tegni, cioè <lb/>quando AB è leua, & la forza che &longs;o&longs;tiene è in A, &longs;arà il &longs;uo &longs;o&longs;tegno B. Ma <lb/>quando BA è leua, & la po&longs;&longs;anza &longs;ta in B, il &longs;o&longs;tegno &longs;arà A, & il pe&longs;o <lb/>&longs;empre rimane appicca-<lb/>to in C. Et percio che la <lb/>po&longs;&longs;anza in A ver&longs;o il <lb/>pe&longs;o E è come BC à<emph.end type="italics"/> <arrow.to.target n="note111"></arrow.to.target> <lb/><emph type="italics"/>BA, & come il pe&longs;o <lb/>E alla po&longs;&longs;anza, che è <lb/>in B, co&longs;i è BA ad <lb/>AC, &longs;arà per la propor<emph.end type="italics"/> <lb/> <arrow.to.target n="fig33"></arrow.to.target> <lb/><emph type="italics"/>tion eguale la po&longs;&longs;anza in A alla po&longs;&longs;anza in B come BC à CA, & à que <lb/>&longs;to modo facilmente ancora potremo cono&longs;cere la proportione, laquale è po&longs;ta de <lb/>Ari&longs;totele nelle que&longs;tioni Mecaniche alla que&longs;tione 29.<emph.end type="italics"/> </s> </p> <figure id="fig33" place="text" xlink:href="figures-it/092_01.jpg"></figure> <p type="margin" id="id.2.1.361.0.0"> <s id="id.2.1.361.1.0"> <margin.target id="note111"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 22. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.362.0.0"> <s id="id.2.1.362.1.0"> COROLLARIO IIII. </s> </p> <p type="main" id="id.2.1.363.0.0"> <s id="id.2.1.363.1.0"> E manife&longs;to etiandio, che ambedue le po&longs;&longs;anze in A, & in B <lb/>pre&longs;e in&longs;ieme, &longs;ono eguali al pe&longs;o E. </s> </p> <p type="main" id="id.2.1.364.0.0"> <s id="id.2.1.364.1.0"><emph type="italics"/>Percioche il pe&longs;o E alla po&longs;&longs;anza in A è come BA à BC, & l'i&longs;te&longs;&longs;o pe&longs;o E <lb/>ver&longs;o la po&longs;&longs;anza in B è come BA ad AC; Per laqual co&longs;a il pe&longs;o E ver-<lb/>&longs;o l'vna, & l'altra po&longs;&longs;anzain A, & in B pre&longs;e in&longs;ieme, è come AB ver&longs;o <lb/>BC, & CA in&longs;ieme, cioè ver&longs;o BA. il pe&longs;o dunque E è eguale ad amen-<lb/>due le po&longs;&longs;anze pre&longs;e in&longs;ieme.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.365.0.0"> <s id="id.2.1.365.1.0"> PROPOSITIONE III. </s> </p> <p type="main" id="id.2.1.366.0.0"> <s id="id.2.1.366.1.0"> In altro modo ancora po&longs;&longs;iamo v&longs;are la Leua. </s> </p> <pb id="p.37v" xlink:href="pageimg-it/093.jpg"/> <p type="main" id="id.2.1.368.0.0"> <s id="id.2.1.368.1.0"><emph type="italics"/>Sia la leua AB, il cui &longs;o&longs;tegno &longs;ia B. & &longs;ia il pe&longs;o C appiccato al punto A, <lb/>& &longs;ia la po&longs;&longs;anza in D, comunque &longs;i voglia tra AB, &longs;o&longs;tenente il pe&longs;o C. Di-<lb/>co che come AB à BD, co&longs;i è la po&longs;&longs;anza in D al pe&longs;o C. Appicchi&longs;i al <lb/>punto D il pe&longs;o E eguale à C; & come BD à BA, co&longs;i &longs;accia&longs;i il pe&longs;o <lb/>E ad vn'altro pe&longs;o, come F: & per e&longs;&longs;ere i pe&longs;i CE traloro eguali, &longs;arà an-<lb/>co il pe&longs;o C al <lb/>pe&longs;o F, come <lb/>BD à BA. <lb/> Appicchi&longs;i &longs;imil <lb/>mente il pe&longs;o F <lb/>in D. & per-<lb/>che il pe&longs;o E ad <lb/>F è come la gra <lb/>uezza del pe&longs;o <lb/>E alla grauez-<lb/>za del pe&longs;o F; <lb/>& il pe&longs;o E al<emph.end type="italics"/> <lb/> <arrow.to.target n="fig34"></arrow.to.target> <lb/> <arrow.to.target n="note112"></arrow.to.target><emph type="italics"/>pe&longs;o F è come BD à BA. Come dunque la grauezza del pe&longs;o E alla gra-<lb/>uezza del pe&longs;o F, co&longs;i è BD à BA. Ma come BD à BA, co&longs;i è la gra-<lb/>uezza del pe&longs;o E alla grauezza del pe&longs;o C. Per laqual co&longs;a la grauezza del<emph.end type="italics"/> <lb/> <arrow.to.target n="note113"></arrow.to.target><emph type="italics"/>pe&longs;o E alla grauezza del pe&longs;o F ha la proportione mede&longs;ima, che ha alla gra-<lb/>uezza del pe&longs;o C. i pe&longs;i dunque CF hanno la grauezza mede&longs;ma. </s> <s id="id.2.1.368.2.0"> Sia dunque<emph.end type="italics"/> <lb/> <arrow.to.target n="note114"></arrow.to.target><emph type="italics"/>la po&longs;&longs;anza in D &longs;o&longs;tenente il pe&longs;o F, che verrà ad e&longs;&longs;ere la detta po&longs;&longs;anza in <lb/>D eguals al pe&longs;o F. & percioche il pe&longs;o F posto in D è graue egualmente <lb/>come il pe&longs;o C po&longs;to in A; haur à la po&longs;&longs;anza in D la proportione mede&longs;ima <lb/>ver&longs;o la grauezza del pe&longs;o F, che ha alla grauezza del pe&longs;o C. Ma la po&longs;&longs;anza <lb/>in D &longs;o&longs;tiene il pe&longs;o F, dunque la po&longs;&longs;anza in D &longs;o&longs;tenterà anco il pe&longs;o C; & <lb/>il pe&longs;o C alla po&longs;&longs;anza in D &longs;arà co&longs;i come il pe&longs;o C al pe&longs;o F; & C ad F<emph.end type="italics"/> <lb/> <arrow.to.target n="note115"></arrow.to.target><emph type="italics"/>è come BD à BA, &longs;arà dunque il pe&longs;o C alla po&longs;&longs;anza in D, come BD à <lb/>BA: & conuertendo come AB à BD, co&longs;i la po&longs;&longs;anza in D al pe&longs;o C. La <lb/>po&longs;&longs;anza dunque al pe&longs;o, è come la di&longs;tanza dal &longs;ostegno allo appiccamento del pe-<lb/>&longs;o alla distanza dal &longs;o&longs;tegno alla po&longs;&longs;anza. </s> <s id="id.2.1.368.3.0"> che bi&longs;ognaua mostrare.<emph.end type="italics"/> </s> </p> <figure id="fig34" place="text" xlink:href="figures-it/093_01.jpg"></figure> <p type="margin" id="id.2.1.370.0.0"> <s id="id.2.1.370.1.0"> <margin.target id="note112"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo della bilancia.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.371.0.0"> <s id="id.2.1.371.1.0"> <margin.target id="note113"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo della bilancia.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.372.0.0"> <s id="id.2.1.372.1.0"> <margin.target id="note114"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 9. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.373.0.0"> <s id="id.2.1.373.1.0"> <margin.target id="note115"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 7. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.374.0.0"> <s id="id.2.1.374.1.0"> Altramente. </s> </p> <p type="main" id="id.2.1.375.0.0"> <s id="id.2.1.375.1.0"><emph type="italics"/>Sia la leua AB, il cui &longs;o&longs;tegno &longs;ia B. & dal punto A &longs;ia fatto pendente il pe&longs;o <lb/>C, & &longs;ia la po&longs;&longs;anza in D &longs;o&longs;tenente il pe&longs;o C. Dico, che come AB à BD, <lb/>co&longs;i è la po&longs;&longs;anza in D al pe&longs;o C. allunghi&longs;ila AB in E, & faccia&longs;i BE egua-<lb/>le à BA, & al punto E &longs;ia appiccato il pe&longs;o F eguale al pe&longs;o C; & come BD à <lb/>BE co&longs;i faccia&longs;i il pe&longs;o F ad vn'altro pe&longs;o G, ilquale &longs;ia appiccato al punto D, <lb/>i pe&longs;i FG pe&longs;eranno egualmente. </s> <s id="id.2.1.375.2.0"> & percioche AB è eguale à BE, & i pe&longs;i<emph.end type="italics"/> <pb id="p.38" xlink:href="pageimg-it/093.jpg"/><emph type="italics"/>FC &longs;ono eguali, &longs;imilmente i pe&longs;i FC pe&longs;eranno egualmente, ma i pe&longs;i FGC ap-<lb/>piccati nella leua EBA, il cui &longs;o&longs;tegno è in B non pe&longs;eranno egualmente; ma in-<lb/>chineranno in giu&longs;o dalla parte di A. Ponga&longs;i dunque in D tanta forza, che i <lb/>pe&longs;i FGC pe&longs;ino egualmente; &longs;arà la po&longs;&longs;anza in D eguale al pe&longs;o G; peroche<emph.end type="italics"/> <lb/> <arrow.to.target n="fig52"></arrow.to.target> <lb/><emph type="italics"/>i pe&longs;i FG pe&longs;ano egualmente, & la po&longs;&longs;anza in D niente altro deue fare, che <lb/>&longs;o&longs;tenere il pe&longs;o G che non di&longs;cenda. </s> <s id="id.2.1.469.2.0"> & percioche i pe&longs;i FGC, & la po&longs;&longs;anza <lb/>in D pe&longs;ano egualmente, leuati via dunque i pe&longs;i FG, i quali pe&longs;ano egualmente, <lb/>i re&longs;tanti pe&longs;eranno egualmente, cioè la po&longs;&longs;anza in D co'l pe&longs;o C, cioè la po&longs;&longs;an <lb/>za in D &longs;o&longs;terrà il pe&longs;o C, talche la leua AB stia come prima. </s> <s id="id.2.1.469.3.0"> & per e&longs;&longs;ere la <lb/>po&longs;&longs;anza in D eguale al pe&longs;o G, & il pe&longs;o C eguale al pe&longs;o, hauerà la po&longs;&longs;an <lb/>zaposta in D la proportione mede&longs;ima al pe&longs;o C, che EB, cioè AB à BD. <lb/>che bi&longs;ognaua mostrare.<emph.end type="italics"/> </s> </p> <figure id="fig52" place="text" xlink:href="figures-it/094_01.jpg"></figure> <p type="head" id="id.2.1.471.0.0"> <s id="id.2.1.471.1.0"> COROLLARIO I. </s> </p> <p type="main" id="id.2.1.472.0.0"> <s id="id.2.1.472.1.0"> Da que&longs;to è chiaro ancora, come prima, che &longs;e &longs;arà po&longs;to il pe-<lb/>&longs;o più vicino al &longs;o&longs;tegno B, come in H, il pe&longs;o douer&longs;i &longs;o-<lb/>&longs;tenere da forza minore. </s> </p> <p type="main" id="id.2.1.473.0.0"> <s id="id.2.1.473.1.0"><emph type="italics"/>Percioche HB ha proportione minore à BD, che AB à BD. & quanto più<emph.end type="italics"/> <arrow.to.target n="note136"></arrow.to.target> <lb/><emph type="italics"/>da vicino &longs;arà al &longs;o&longs;tegno, &longs;empre anco minore forza vi &longs;i ricercherà.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.474.0.0"> <s id="id.2.1.474.1.0"> <margin.target id="note136"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.475.0.0"> <s id="id.2.1.475.1.0"> COROLLARIO II. </s> </p> <p type="main" id="id.2.1.476.0.0"> <s id="id.2.1.476.1.0"> Egli è parimente manife&longs;to, che la po&longs;&longs;anza in D è &longs;empre <lb/>maggiore del pe&longs;o C. </s> </p> <p type="main" id="id.2.1.477.0.0"> <s id="id.2.1.477.1.0"><emph type="italics"/>Perche &longs;e tra AB &longs;i piglia qual &longs;i voglia punto, come D, &longs;empre AB &longs;arà mag<lb/>giore di BD.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.478.0.0"> <s id="id.2.1.478.1.0"><emph type="italics"/>El è da auertire, che que&longs;te dimo&longs;trationi lequali habbiamo prodotte in mezo, &longs;i po&longs;&longs;o-<lb/>no à tutte que&longs;te co&longs;e commodamente adattare non &longs;olamente e&longs;&longs;endo le leue egual-<lb/>mente distanti dall'orizonte, ma anche inchinate le dette leue all'orizonte. </s> <s id="id.2.1.478.2.0"> ilche è <lb/>chiaro da quel che nella bilancia &longs;i è diui&longs;ato.<emph.end type="italics"/> </s> </p> <pb id="p.38v" xlink:href="pageimg-it/095.jpg"/> <p type="head" id="id.2.1.480.0.0"> <s id="id.2.1.480.1.0"> PROPOSITIONE IIII. </s> </p> <p type="main" id="id.2.1.481.0.0"> <s id="id.2.1.481.1.0"> Se la po&longs;&longs;anza mouerà il pe&longs;o appiccato nella leua, &longs;arà lo &longs;patio <lb/>della po&longs;&longs;anza mo&longs;&longs;a allo &longs;patio del pe&longs;o mo&longs;&longs;o, come la di&longs;tan <lb/>za dal &longs;o&longs;tegno alla po&longs;&longs;anza, alla di&longs;tanza dall'i&longs;te&longs;&longs;o &longs;o&longs;tegno <lb/>fin allo appiccamento del pe&longs;o. </s> </p> <p type="main" id="id.2.1.482.0.0"> <s id="id.2.1.482.1.0"><emph type="italics"/>Sia la leua AB, il cui &longs;o&longs;tegno C, & &longs;ia il pe&longs;o D attaccato al punto B, & &longs;ia <lb/>la po&longs;&longs;anza in A mouente il pe&longs;o D conlaleua AB. Dico lo &longs;patio della po&longs;-<lb/>&longs;anza in A allo &longs;patio del pe&longs;o e&longs;&longs;ere co&longs;i come CA à CB. Moua&longs;i la leua <lb/>AB, & affine che il pe&longs;o D &longs;i moua in sù, bi&longs;ogna che B &longs;i moua in sù, & A in <lb/>giù. </s> <s id="id.2.1.482.2.0"> & percioche C è punto immobile; però mentre A, & B &longs;i mouono, de-<lb/>&longs;criueranno circonferenze di cerchi. </s> <s id="id.2.1.482.3.0"> Moua&longs;i dunque AB in EF; &longs;aranno AEBF <lb/>circonferenze di cerchi, i me-<lb/>zi diametri de' quali &longs;ono CA <lb/>CB. compi&longs;ca&longs;i tutta la cir-<lb/>conferenza AGE, & tut-<lb/>ta la BHF, & &longs;iano KH <lb/>i punti doue AB, & EF ta-<lb/>gliano il cerchio BHF. Hor <lb/>percioche l'angolo BCF è<emph.end type="italics"/> <lb/> <arrow.to.target n="note137"></arrow.to.target><emph type="italics"/>eguale all'angolo HCK, &longs;a-<lb/>rà la circonferenza KH egua<emph.end type="italics"/> <lb/> <arrow.to.target n="note138"></arrow.to.target><emph type="italics"/>le alla circonferenza BF, & <lb/>concio&longs;ia, che le circonferen-<lb/>ze AEKH &longs;iano &longs;otto l'i-<lb/>&longs;te&longs;&longs;o angolo ACE, & la <lb/>circonferenza AE à tutta <lb/>la circonferenza AGE &longs;ia <lb/>come l'angolo ACE à quat <lb/>tro retti, & come l'i&longs;te&longs;&longs;o an-<lb/>golo HCK à quattro retti, <lb/>co&longs;i anche è la circonferenza <lb/>HK à tutta la circonferentia <lb/>HBK, &longs;arà la circonferentia<emph.end type="italics"/> <lb/> <arrow.to.target n="fig53"></arrow.to.target> <lb/><emph type="italics"/>AE à tutta la circonferentia AGE, come la circonferentia KH à tutta la<emph.end type="italics"/> <lb/> <arrow.to.target n="note139"></arrow.to.target><emph type="italics"/>KFH. & permutando come la circonferentia AE alla circonferenza KH, cioè <lb/>BF, co&longs;i tutta la circonferenza AGE à tutta la circonferenza BHF; ma tut-<emph.end type="italics"/> <lb/> <arrow.to.target n="note140"></arrow.to.target><emph type="italics"/>ta la circonferenza AGE co&longs;i &longs;i ha à tutta la BHF, come il diametro del cer-<lb/>chio AEG al diametro del cerchio BHF. Come dunque la circonferenza AE<emph.end type="italics"/> <pb id="p.39" xlink:href="pageimg-it/096.jpg"/><emph type="italics"/>ver&longs;o la circonferenza BF, co&longs;i è il diametro del cerchio AGE al diametro del<emph.end type="italics"/> <arrow.to.target n="note141"></arrow.to.target> <lb/><emph type="italics"/>cerchio BHF: ma come il diametro al diametro, co&longs;i è anche il mezo diametro al <lb/>mezo diametro, cioè CA à CB. Per laqual co&longs;a come la circonferenza AE <lb/>alla circonferenza BF, co&longs;i CA à CB: ma la circonferenza AE è lo <lb/>&longs;patio della po&longs;&longs;anza mo&longs;&longs;a, & la circonferenza BF è eguale allo &longs;patio di D pe-<lb/>&longs;o mo&longs;&longs;o, peroche lo &longs;patio del mouimento del pe&longs;o D &longs;empre è eguale allo &longs;patio <lb/>del mouimento del punto B, per e&longs;&longs;ere attaccato in B. Lo &longs;patio dunque della po&longs; <lb/>&longs;anza mo&longs;&longs;a allo &longs;patio del pe&longs;o mo&longs;&longs;o è come CA à CB; cioè come la di&longs;tan-<lb/>za dal &longs;o&longs;tegno alla po&longs;&longs;anza, alla distanza dall'i&longs;te&longs;&longs;o all'appiccamento del pe&longs;o. <lb/>che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig53" place="text" xlink:href="figures-it/095_01.jpg"></figure> <p type="margin" id="id.2.1.484.0.0"> <s id="id.2.1.484.1.0"> <margin.target id="note137"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 15. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.485.0.0"> <s id="id.2.1.485.1.0"> <margin.target id="note138"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 26. <emph type="italics"/>del terzo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.486.0.0"> <s id="id.2.1.486.1.0"> <margin.target id="note139"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 16. <emph type="italics"/>del<emph.end type="italics"/> 15. </s> </p> <p type="margin" id="id.2.1.487.0.0"> <s id="id.2.1.487.1.0"> <margin.target id="note140"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 23. <emph type="italics"/>del<emph.end type="italics"/> 8. <emph type="italics"/>di Pappo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.488.0.0"> <s id="id.2.1.488.1.0"> <margin.target id="note141"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.489.0.0"> <s id="id.2.1.489.1.0"><emph type="italics"/>Ma &longs;ia la leua AB, il cui &longs;o&longs;tegno B, & la po&longs;&longs;anza mouente in A, & il pe&longs;o <lb/>in C. Dico lo &longs;patio della po&longs;&longs;anza mo&longs;&longs;a allo &longs;patio del pe&longs;o tra&longs;portato co&longs;i e&longs;-<lb/>&longs;ere, come BA à BC.<lb/> Moua&longs;i la leua, & accioche <lb/>il pe&longs;o &longs;ia alzato in sù, egli <lb/>è nece&longs;&longs;ario, che anche i pun <lb/>ti CA &longs;imouano in sù.<lb/> Moua&longs;i dunque A in sù <lb/>fin'in D; & &longs;ia il mouimen <lb/>to della leua BD. mo&longs;tre-<lb/>remo nel modo i&longs;te&longs;&longs;o, come <lb/>prima è detto, che i punti <lb/>CA de&longs;criuono circonferen <lb/>ze di cerchi, i cui mezi dia-<lb/>metri &longs;ono BA BC. & di-<lb/>mo&longs;treremo &longs;imilmente co&longs;i <lb/>e&longs;&longs;ere AD à CE, come il <lb/>mezo diametro AB al me-<lb/>zo diametro BC.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.490.0.0"> <s id="id.2.1.490.1.0"><emph type="italics"/>Et per la ragione i&longs;te&longs;&longs;a, &longs;e la <lb/>po&longs;&longs;anza fo&longs;&longs;e in C, & il <lb/>pe&longs;o in A &longs;i prouerà co&longs;i <lb/>e&longs;&longs;ere CE ver&longs;o AD, co-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig54"></arrow.to.target> <lb/><emph type="italics"/>me BC à BA, cioè la di&longs;tanza dal &longs;o&longs;tegno alla po&longs;&longs;anza; alla di&longs;tanza dal-<lb/>l'iste&longs;&longs;o allo attaccamento del pe&longs;o. </s> <s id="id.2.1.490.2.0"> che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig54" place="text" xlink:href="figures-it/096_01.jpg"></figure> <p type="head" id="id.2.1.492.0.0"> <s id="id.2.1.492.1.0"> COROLLARIO. </s> </p> <p type="main" id="id.2.1.493.0.0"> <s id="id.2.1.493.1.0"> Da que&longs;te co&longs;e è manife&longs;to, che maggiore proportione ha lo &longs;pa <lb/>tio della po&longs;&longs;anza, che moue allo &longs;patio del pe&longs;o mo&longs;&longs;o, che il <lb/>pe&longs;o alla mede&longs;ima po&longs;&longs;anza. </s> </p> <pb id="p.39v" xlink:href="pageimg-it/097.jpg"/> <p type="main" id="id.2.1.494.0.0"> <s id="id.2.1.494.1.0"><emph type="italics"/>Percioche lo &longs;patio della po&longs;&longs;anza allo &longs;patio del pe&longs;o ha la mede&longs;ima proportione, <lb/>che il pe&longs;o alla po&longs;&longs;anza, che &longs;o&longs;liene il detto pe&longs;o. </s> <s id="id.2.1.494.2.0"> Ma la po&longs;&longs;anza, che &longs;ostie-<lb/>ne è minore della po&longs;&longs;anza che moue, però il pe&longs;o haurà proportione minore alla <lb/>po&longs;&longs;anza che lo moue, che alla po&longs;&longs;anza, che lo &longs;ostiene. </s> <s id="id.2.1.494.3.0"> Lo &longs;patio dunque della<emph.end type="italics"/> <lb/> <arrow.to.target n="note142"></arrow.to.target><emph type="italics"/>po&longs;&longs;anza che moue allo &longs;patio del pe&longs;o haur à proportione maggiore, che il pe&longs;o al-<lb/>l'i&longs;le&longs;&longs;a po&longs;&longs;anza.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.495.0.0"> <s id="id.2.1.495.1.0"> <margin.target id="note142"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.496.0.0"> <s id="id.2.1.496.1.0"> PROPOSITIONE V. </s> </p> <p type="main" id="id.2.1.497.0.0"> <s id="id.2.1.497.1.0"> La po&longs;&longs;anza che in qual &longs;i voglia modo &longs;o&longs;tenga il pe&longs;o con la le-<lb/>ua hauerà la proportione mede&longs;ima ad e&longs;&longs;o pe&longs;o, che la di&longs;tan <lb/>za frapo&longs;ta dal &longs;o&longs;tegno al punto, doue dal centro della gra-<lb/>uezza del pe&longs;o tirata vna linea à piombo all'orizonte tagli la <lb/>leua, alla di&longs;tanza che è fra il &longs;o&longs;tegno, & la po&longs;&longs;anza. </s> </p> <p type="main" id="id.2.1.498.0.0"> <s id="id.2.1.498.1.0"><emph type="italics"/>Sia la leua AB egualmente di&longs;tante dall'orizonte, col &longs;uo &longs;o&longs;tegno N. &longs;ia dopo il pe <lb/>&longs;o AC, il cui centro della grauezza &longs;ia D, ilquale &longs;ia prima &longs;otto la leua: ma <lb/>il pe&longs;o &longs;ia appiccato à i punti AO. & dal punto D &longs;ia tirata la linea DE à <lb/>piomho dell' orizonte, & di AB. Che &longs;e vi &longs;aranno altre leue ancora AF AG, <lb/>i cui &longs;o-<lb/>stegni, <lb/>&longs;iano H <lb/>K, & il <lb/>pe&longs;o A <lb/>C &longs;ia ap<lb/>piccato <lb/>nella le-<lb/>ua AG <lb/>ne i pun <lb/>ti AQ, <lb/>& nella <lb/>leua A <lb/>F ne' <expan abbr="pũ">pum</expan> <lb/>ti AP: <lb/>& la li-<lb/>nea DE <lb/>allunga-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig55"></arrow.to.target> <lb/><emph type="italics"/>ta tagli AF in L, & AC in M. Dico che la po&longs;&longs;anza in F &longs;o&longs;tenente il pe&longs;o AC <lb/>ha quella proportione ad e&longs;&longs;o pe&longs;o, che ha KL à KF; & la po&longs;&longs;anza in D ha quella <lb/>proportione al pe&longs;o, che ha NE ad NB; & la po&longs;&longs;anza in G al pe&longs;o quella, che ha <lb/>HM ad HG. Hor percioche DL &longs;tà à piombo dell' orizonte, il pe&longs;o AC venga ap-<emph.end type="italics"/> <pb id="p.40" xlink:href="pageimg-it/098.jpg"/><emph type="italics"/>piccato doue &longs;i voglia nella linea DL, rimarrà nel modo i&longs;te&longs;&longs;o che &longs;i troua. </s> <s id="id.2.1.498.2.0"> Per la-<lb/>qual co&longs;a &longs;e nella leua AB &longs;i &longs;cioglieranno gli appiccamenti, che &longs;ono ad AO, il <lb/>pe&longs;o AC appiccato in E rimarr à nell'i&longs;te&longs;&longs;o modo, come hora rimane, cioè leuato via <lb/>il punto A, & la linea QO, nell'i&longs;te&longs;&longs;o modo il pe&longs;o appiccato in E rimarrà, come <lb/>era &longs;o&longs;tenuto da punti i&longs;te&longs;&longs;i AO, come &longs;i proua per lo commentario di Federico <lb/>Commandino nella &longs;esta propo&longs;itione di Archimede della quadratura della parabo <lb/>la, & dalla prima di que&longs;to della bilancia. </s> <s id="id.2.1.498.3.0"> Co&longs;i percio che il pe&longs;o AC ha &longs;empre la <lb/>i&longs;te&longs;&longs;a di&longs;po&longs;itione ver&longs;o la bilancia, &longs;ia pur in AO &longs;ostentato, ouero pendente dal <lb/>punto E; la po&longs;&longs;anza mede&longs;ima in B &longs;o&longs;tenterà il pe&longs;o i&longs;te&longs;&longs;o AC pendente, ouero <lb/>da E, ouero da AO. ma la po&longs;&longs;anza in B &longs;o&longs;tenente il pe&longs;o AC appiccato in E co&longs;i <lb/>&longs;i hà ad e&longs;&longs;o pe&longs;o, come NE ad NB; La po&longs;&longs;anza dunque in B &longs;o&longs;tenente il pe&longs;o <lb/>AC da punti AO pendente &longs;arà co&longs;i ad e&longs;&longs;o pe&longs;o, come NE ad NB. Non altra-<emph.end type="italics"/> <arrow.to.target n="note143"></arrow.to.target> <lb/><emph type="italics"/>mente &longs;i mo&longs;trerà, che il pe&longs;o AC pendente dal punto L rimane, come &longs;e &longs;o&longs;&longs;e &longs;oste <lb/>nuto da punti AP; & la po&longs;&longs;anza in F ad e&longs;&longs;o pe&longs;o e&longs;&longs;ere co&longs;i come<emph.end type="italics"/> KL <emph type="italics"/>à KF. Ma <lb/>nella leua AG il pe&longs;o AC appiccato in M co&longs;i rimanere, come egli è &longs;o&longs;tenuto da <lb/>punti AQ; & la po&longs;&longs;anza di G co&longs;i e&longs;&longs;ere al pe&longs;o AC, come HM ad HG, cioè co-<lb/>me la di&longs;tanza dal &longs;o&longs;tegno al punto, doue la linea tirata à piombo dell' orizonte <lb/>dal centro della grauezza del pe&longs;o taglia la leua, alla di&longs;tanza dal &longs;o&longs;tegno alla po&longs;-<lb/>&longs;anza. </s> <s id="id.2.1.498.4.0"> che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig55" place="text" xlink:href="figures-it/097_01.jpg"></figure> <p type="margin" id="id.2.1.500.0.0"> <s id="id.2.1.500.1.0"> <margin.target id="note143"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.501.0.0"> <s id="id.2.1.501.1.0"><emph type="italics"/>Che &longs;e FBG &longs;o&longs;&longs;ero i &longs;o&longs;tegni delle leue, & le po&longs;&longs;anze fo&longs;&longs;ero in KNH &longs;o&longs;tenenti il pe <lb/>&longs;o, con &longs;imile modo &longs;i mo&longs;trerà la po&longs;&longs;anza in H, co&longs;i e&longs;&longs;ere al pe&longs;o, come GM à GH, <lb/>et la <expan abbr="po&longs;sāzaī">po&longs;sanzaim</expan> N al pe&longs;o, come BE à BN, et la <expan abbr="po&longs;sāza">po&longs;sanza</expan> <expan abbr="ī">im</expan> K al pe&longs;o come FL ad FK.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.502.0.0"> <s id="id.2.1.502.1.0"><emph type="italics"/>Et &longs;e le leue AB AF AG haue&longs;&longs;ero i &longs;o&longs;legni in A, & il pe&longs;o fo&longs;&longs;e NO; poi dal <lb/>centro D del <lb/>la &longs;ua gra-<lb/>uezza fo&longs;&longs;e <lb/>tirata la li-<lb/>nea DME <lb/>L à piombo <lb/>di AB, & <lb/>dell' orizon <lb/>te, & fo&longs;&longs;e-<lb/>ro le po&longs;&longs;an <lb/>ze in FB <lb/>G; <expan abbr="&longs;imilm&etilde;">&longs;imilmen</expan> <lb/>re mo&longs;tre-<lb/>ra&longs;&longs;i la po&longs; <lb/>&longs;anza di G <lb/>&longs;o&longs;tenente <lb/>il pe&longs;o N<emph.end type="italics"/> <lb/> <arrow.to.target n="fig56"></arrow.to.target> <lb/><emph type="italics"/>O co&longs;i e&longs;&longs;ere ad e&longs;&longs;o pe&longs;o, come AM ad AG, & la po&longs;&longs;anza in B come AE ad <lb/>AB; & la po&longs;&longs;anza in F come AL ad AF.<emph.end type="italics"/> </s> </p> <figure id="fig56" place="text" xlink:href="figures-it/098_01.jpg"></figure> <pb id="p.40v" xlink:href="pageimg-it/099.jpg"/> <p type="main" id="id.2.1.504.0.0"> <s id="id.2.1.504.1.0"><emph type="italics"/>Sia dapoi la leua AB egualmente di&longs;tante dall'orizonte, il cui &longs;o&longs;tegno &longs;ia D, & &longs;ia <lb/>BE il pe&longs;o, il cui centro della grauezza &longs;ia F &longs;opra la leua; & dal punto F riri&longs;i la <lb/>linea FH à piombo, & dell' orizonte, & di e&longs;&longs;a AB; & &longs;ia &longs;o&longs;tenuto il pe&longs;o dal <lb/>punto B, & da PQ &longs;iano po&longs;cia altre leue BLBM, i cui &longs;o&longs;tegni &longs;iano NO; <lb/>& la linea FH allungatatagli BM in K, & BL in G; & venga &longs;o&longs;tenuto il pe&longs;o<emph.end type="italics"/> <lb/> <arrow.to.target n="fig57"></arrow.to.target> <lb/><emph type="italics"/>nella leua BL ne'punti BP; & nella leua BM dal punto B, & PR. Dico, che <lb/>la po&longs;&longs;anza in L &longs;o&longs;tenente il pe&longs;o BE nella leua BL ha quella proportione ad <lb/>e&longs;&longs;o pe&longs;o, che NG ad NL; & la po&longs;&longs;anza in A al pe&longs;o ha quella proportio-<lb/>ne, che DH à DA; & la po&longs;&longs;anza di M al pe&longs;o ha quella proportione, che <lb/>OK ad OM. Hor percioche la linea KF tirata dal centro della grauezza F è<emph.end type="italics"/> <lb/> <arrow.to.target n="note144"></arrow.to.target><emph type="italics"/>à piombo dell' orizonte, &longs;ia pur &longs;ostenuto il pe&longs;o da qual &longs;i voglia punto della linea <lb/>KF, egli rimarrà, come hora &longs;i troua. </s> <s id="id.2.1.504.2.0"> Se dunque &longs;arà &longs;ostenuto in H, rimarrà co <lb/>me prima, cioè leuato via il punto B, & PQ, i quali &longs;o&longs;tengono il pe&longs;o, rimarrà <lb/>il pe&longs;o BE nel modo che da e&longs;&longs;i era &longs;o&longs;tenuto. </s> <s id="id.2.1.504.3.0"> Per la qual co&longs;a grauerà nella le-<lb/>ua AB in H, & haurà alla leua quella di&longs;po&longs;itione mede&longs;ima, che prima, & per-<lb/>ciò &longs;arà come &longs;e fo&longs;&longs;e appiccato in H. La mede&longs;ma po&longs;&longs;anza dunque &longs;o&longs;terrà il me <lb/>de&longs;imo pe&longs;o BE &longs;o&longs;tentato ouero in H, ouero in B & Q. Ma la po&longs;&longs;anza in A<emph.end type="italics"/> <lb/> <arrow.to.target n="note145"></arrow.to.target><emph type="italics"/>&longs;o&longs;tenente il pe&longs;o BE appiccato in H con la leua AB hal'i&longs;te&longs;&longs;a proportione ad e&longs;-<lb/>&longs;o pe&longs;o, che DH à DA; l'i&longs;te&longs;&longs;a po&longs;&longs;anza dunque in A &longs;o&longs;tenente il pe&longs;o BE ne' <lb/>punti BQ &longs;o&longs;tentato, &longs;arà ad e&longs;&longs;o pe&longs;o come DH à DA. Similmente &longs;i mo&longs;trer à <lb/>il pe&longs;o BE, &longs;e in G &longs;arà &longs;o&longs;tenuto, rimanere come egliera &longs;o&longs;tenuto da punti BP: <lb/>& nel punto K, come dapunti BR. Per la qual co&longs;a la po&longs;&longs;anza in L &longs;o&longs;tenente<emph.end type="italics"/> <pb id="p.41" xlink:href="pageimg-it/100.jpg"/><emph type="italics"/>il pe&longs;o BE ad e&longs;&longs;o pe&longs;o co&longs;i &longs;arà come NG ad NL. ma la po&longs;&longs;anza in M al pe <lb/>&longs;o, come OK ad OM; cioè come la di&longs;tanza dal &longs;o&longs;tegno al punto, doue dal cen<lb/>tro della grauezza del pe&longs;o la linea tirata à piombo dell' orizonte taglia la leua, alla <lb/>di&longs;tanza dal &longs;o&longs;tegno alla po&longs;&longs;anza. </s> <s id="id.2.1.504.4.0"> che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig57" place="text" xlink:href="figures-it/099_01.jpg"></figure> <p type="margin" id="id.2.1.506.0.0"> <s id="id.2.1.506.1.0"> <margin.target id="note144"></margin.target><emph type="italics"/>Per la prima di questo della bilancia.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.507.0.0"> <s id="id.2.1.507.1.0"> <margin.target id="note145"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.508.0.0"> <s id="id.2.1.508.1.0"><emph type="italics"/>Che &longs;e LAM fo&longs;&longs;ero i &longs;o&longs;tegni, & le po&longs;&longs;anze in NDO; &longs;imilmente mo&longs;trera&longs;&longs;i <lb/>la po&longs;&longs;anza in N co&longs;i e&longs;&longs;ere al pe&longs;o, come LG ad LN; & la po&longs;&longs;anza in D, <lb/>come AH ad AD, & la po&longs;&longs;anza in O come MK ad MO.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.509.0.0"> <s id="id.2.1.509.1.0"><emph type="italics"/>Et &longs;e le leue BA BL BM haue&longs;&longs;ero i &longs;o&longs;tegni in B, & il pe&longs;o fo&longs;&longs;e NO &longs;opra <lb/>la leua, & dal centro F della grauezza fo&longs;&longs;e tirata la linea FD EG à piombo <lb/>di AB, & dell' orizonte; & fo&longs;&longs;ero le po&longs;&longs;anze in LAM, &longs;imilmente proue-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig58"></arrow.to.target> <lb/><emph type="italics"/>ra&longs;&longs;i la po&longs;&longs;anza in L &longs;o&longs;tenente il pe&longs;o co&longs;i e&longs;&longs;ere ad e&longs;&longs;o pe&longs;o, come BD à BL; <lb/>& la po&longs;&longs;anza in A al pe&longs;o come BE à BA, & la po&longs;&longs;anza in M come BG <lb/>à BM.<emph.end type="italics"/> </s> </p> <figure id="fig58" place="text" xlink:href="figures-it/100_01.jpg"></figure> <pb id="p.41v" xlink:href="pageimg-it/101.jpg"/> <p type="main" id="id.2.1.512.0.0"> <s id="id.2.1.512.1.0"><emph type="italics"/>Sia vltimamente la leua. AB egualmente di&longs;tante dall'orizonte, il cui &longs;ostegno &longs;ia <lb/>C, & il pe&longs;o DE habbia il centro della graueza F nella leua AB; & &longs;iano <lb/>alla fine altre leue GHKL, co i &longs;o&longs;tegni &longs;uoi MN; & il pe&longs;o nella leua GH <lb/>&longs;ia &longs;o&longs;tentato da i punti GO, & nella leua AB da punti AT, & nella leua <lb/>KL da punti KQ, & il centro F della grauezza &longs;ia parimente in amendue le le-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig59"></arrow.to.target> <lb/><emph type="italics"/>ue GH<emph.end type="italics"/> K<emph type="italics"/>L, & &longs;iano le po&longs;&longs;anze in HBL. </s> <s id="id.2.1.512.2.0"> Dico la po&longs;&longs;anza in H co&longs;i e&longs;&longs;ere al <lb/>pe&longs;o, come N<emph.end type="italics"/>F <emph type="italics"/>ad NH; & la po&longs;&longs;anza in B al pe&longs;o, come C<emph.end type="italics"/>F <emph type="italics"/>à CB, & la po&longs; <lb/>&longs;anza in L al pe&longs;o, come M<emph.end type="italics"/>F <emph type="italics"/>ad ML. Hor percioche F è il centro della grauez-<lb/>za del pe&longs;o DE, &longs;e dunque in<emph.end type="italics"/> F <emph type="italics"/>&longs;arà &longs;o&longs;tenuto, &longs;tarà il pe&longs;o DE come prima, per <lb/>la diffinitione del centro della grauezza; & &longs;arà come &longs;e egli fo&longs;&longs;e appiccato in<emph.end type="italics"/> F; <lb/><emph type="italics"/>& &longs;tarà nella leua in quel modo i&longs;te&longs;&longs;o, &longs;o&longs;tenga&longs;i pure ò da punti AP, ouero dal <lb/>punto<emph.end type="italics"/> F. <emph type="italics"/>ilche parimente auerrà nelle leue GH KL, cioè che il pe&longs;o re&longs;terà nel mo <lb/>do i&longs;te&longs;&longs;o, &longs;o&longs;tenti&longs;i pur ò in<emph.end type="italics"/> F, <emph type="italics"/>ouero in GO ouero in KQ. La mede&longs;ma po&longs;&longs;anza <lb/>dunque in B &longs;o&longs;tenterà il pe&longs;o i&longs;te&longs;&longs;o DE appiccato, ouero in<emph.end type="italics"/> F, <emph type="italics"/>ouero in AP: & <lb/>quando egli è appiccato in<emph.end type="italics"/> F, <emph type="italics"/>è ad e&longs;&longs;o pe&longs;o come CF à CB, dunque la po&longs;&longs;anza &longs;o-<lb/>&longs;tenente il pe&longs;o DE appiccato ad AP &longs;arà ad e&longs;&longs;o pe&longs;o come C<emph.end type="italics"/>F <emph type="italics"/>à CB. & nel mo<lb/>do i&longs;te&longs;&longs;o la po&longs;&longs;anza in H &longs;arà al pe&longs;o appiccato in OG co&longs;i, come N<emph.end type="italics"/>F <emph type="italics"/>ad NH. & <lb/>la po&longs;&longs;anza in L &longs;arà al pe&longs;o appiccato in KQ, come M<emph.end type="italics"/>F <emph type="italics"/>ad ML. ilche anco bi&longs;o-<lb/>gnaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig59" place="text" xlink:href="figures-it/101_01.jpg"></figure> <p type="main" id="id.2.1.514.0.0"> <s id="id.2.1.514.1.0"><emph type="italics"/>Ma &longs;e li &longs;o&longs;tegni fo&longs;&longs;ero HBL, & le po&longs;&longs;anze fo&longs;&longs;ero in NCM; &longs;imilmente prouera&longs;&longs;i <lb/>la po&longs;&longs;anza in N co&longs;i e&longs;&longs;ere al pe&longs;o, come HF ad HN & la po&longs;&longs;anza in C come <lb/>BF à BC; & la po&longs;&longs;anza in M come LF ad LM.<emph.end type="italics"/> </s> </p> <pb id="p.42" xlink:href="pageimg-it/102.jpg"/> <p type="main" id="id.2.1.516.0.0"> <s id="id.2.1.516.1.0"><emph type="italics"/>Et &longs;e le leue BA BC BD haue&longs;&longs;ero i &longs;o&longs;tegni in B, & fo&longs;&longs;ero i pe&longs;i in EF GH <lb/>KL, di modo che i loro centri della grauezza MNO fo&longs;&longs;ero nelle leue, & le<emph.end type="italics"/> <lb/> <arrow.to.target n="fig60"></arrow.to.target> <lb/><emph type="italics"/>po&longs;&longs;anze fo&longs;&longs;ero in CAD. Similmente prouera&longs;&longs;i, che la po&longs;&longs;anza in C co&longs;i è <lb/>al pe&longs;o EF, come BM à BC, & la po&longs;&longs;anza in A al pe&longs;o GH, come <lb/>BN à BA, & la po&longs;&longs;anza in D al pe&longs;o KL, come BO à BD.<emph.end type="italics"/> </s> </p> <figure id="fig60" place="text" xlink:href="figures-it/102_01.jpg"></figure> <p type="head" id="id.2.1.518.0.0"> <s id="id.2.1.518.1.0"> PROPOSITIONE VI. </s> </p> <p type="main" id="id.2.1.519.0.0"> <s id="id.2.1.519.1.0"> Sia AB linea retta, ad angoli retti, dellaquale &longs;tia AD, la-<lb/>quale dalla parte di D &longs;ia allungata come &longs;i vuole fin'al C, <lb/>& &longs;ia congiunta la CB, laquale parimente allunghi&longs;i dalla <lb/>parte di B fin ad E. Dapoi &longs;iano dal punto B tirate altre <lb/>linee, come &longs;i vuole BF BG eguali ad AB tra AB BE; <lb/>& da i punti FG &longs;iano tirate le linee FH GK à piombo <lb/>delle &longs;udette, lequali &longs;i facciano eguali fra loro, & ad e&longs;&longs;a A <lb/>D come &longs;e BA AD fo&longs;&longs;ero mo&longs;&longs;e in BF FH, & in BG <lb/>GH; & congiungan&longs;i CH CK, lequali taglino le linee BF <lb/>BG ne'punti MN. Dico che BN è minore di BM, & <lb/>BM di e&longs;&longs;a BA. </s> </p> <pb id="p.42v" xlink:href="pageimg-it/103.jpg"/> <p type="main" id="id.2.1.521.0.0"> <s id="id.2.1.521.1.0"><emph type="italics"/>Congiungan&longs;i BD BH B<emph.end type="italics"/>K, <emph type="italics"/>& percioche due linee DA AB &longs;ono eguali à due<emph.end type="italics"/> <lb/> <arrow.to.target n="note146"></arrow.to.target><emph type="italics"/>HF FB, & l'angolo DAB retto è anco eguale al retto HFB; &longs;aranno i <lb/>re&longs;tanti angoli eguali à i re&longs;tanti angoli, & HB eguale ad e&longs;&longs;a DB. Similmen-<lb/>te mo&longs;trera&longs;&longs;i il triangolo BKG e&longs;&longs;ere eguale al triangolo BHF. Per laqual co <lb/>&longs;a co'l centro B, & con l'in <lb/>teruallo di vna di e&longs;&longs;e de&longs;cri-<lb/>ua&longs;i il cerchio DH<emph.end type="italics"/> K<emph type="italics"/>E, il <lb/>quale tagli le linee CH CK <lb/>ne' punti OP; & congiun-<lb/>gan&longs;i OB PB. Percioche <lb/>dunque il punto K è più vi-<lb/>cino ad E, che H; &longs;arà la<emph.end type="italics"/> <lb/> <arrow.to.target n="note147"></arrow.to.target><emph type="italics"/>linea CK maggiore di CH, <lb/>& CP minore di CO: dun<lb/>que PK &longs;arà maggiore di <lb/>OH. Ma perche il triangolo <lb/>BKP di due lati eguali ha i <lb/>&longs;uoi lati BK BP eguali à i <lb/>lati BH BO del triangolo <lb/>BHO di due lati eguali, ma <lb/>ben la ba&longs;e KP maggiore <lb/>della ba&longs;e HO, &longs;arà l'ango-<lb/>lo KBP maggiore dell' an<emph.end type="italics"/> <lb/> <arrow.to.target n="note148"></arrow.to.target><emph type="italics"/>golo HBO. dunque i restan<lb/>ti angoli alla ba&longs;e, cioè KPB <lb/>PKB pre&longs;i in&longs;ieme, i quali <lb/>tra loro &longs;ono eguali, &longs;aranno <lb/>minori de i re&longs;tanti angol i al-<lb/>la ba&longs;e po&longs;ti, cioè OHB <lb/>HOB, iquali etiandio tra lo <lb/>ro &longs;ono eguali e&longs;&longs;endo che tut <lb/>ti gli angoli di cia&longs;cuno trian<emph.end type="italics"/> <lb/> <arrow.to.target n="note149"></arrow.to.target><emph type="italics"/>golo &longs;iano eguali à due angoli <lb/>retti. </s> <s id="id.2.1.521.2.0"> Per laqual co&longs;a anche <lb/>le metà di que&longs;ti, cioè NKB <lb/>&longs;arà minore di MHB. Et<emph.end type="italics"/> <lb/> <arrow.to.target n="note150"></arrow.to.target><emph type="italics"/>concio&longs;ia, che l'angolo BKG<emph.end type="italics"/> <lb/> <arrow.to.target n="fig61"></arrow.to.target> <lb/><emph type="italics"/>&longs;ia eguale all'angolo BHF, &longs;arà NKG maggiore di MHF. Se dunque nel <lb/>punto K &longs;i &longs;accia l'angolo GKQ eguale ad FHM &longs;i &longs;arà il triangolo GKQ <lb/>eguale al triangolo FHM; Imperoche due angoli in FH divno &longs;ono eguali à <lb/>due in GK d'vn'altro, & il lato FH è eguale al lato GK, &longs;arà GQ eguale <lb/>ad FM. Adunque GN &longs;arà maggiore di FM. & co&longs;i per e&longs;&longs;ere BG egua-<emph.end type="italics"/> <pb id="p.43" xlink:href="pageimg-it/104.jpg"/><emph type="italics"/>le à BF, &longs;arà BN minore di e&longs;&longs;a BM. ma che BM &longs;ia minore di e&longs;&longs;a BA <lb/>è manife&longs;to, percioche BM, è minore di e&longs;&longs;a BF, laquale è eguale à BA. che <lb/>bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig61" place="text" xlink:href="figures-it/103_01.jpg"></figure> <p type="margin" id="id.2.1.523.0.0"> <s id="id.2.1.523.1.0"> <margin.target id="note146"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 4. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.524.0.0"> <s id="id.2.1.524.1.0"> <margin.target id="note147"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del serzo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.525.0.0"> <s id="id.2.1.525.1.0"> <margin.target id="note148"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 25. <emph type="italics"/>del<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.526.0.0"> <s id="id.2.1.526.1.0"> <margin.target id="note149"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 5. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.527.0.0"> <s id="id.2.1.527.1.0"> <margin.target id="note150"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 26. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.528.0.0"> <s id="id.2.1.528.1.0"><emph type="italics"/>Di più &longs;e tra BG BE &longs;i tiri à piacere vn'altra linea eguale à BG; & faccia&longs;i l'ope <lb/>ratione, come di &longs;opra è stato detto, prouera&longs;&longs;i &longs;imilmente la linea BR e&longs;&longs;er mi-<lb/>nore di BN. & quanto più da vicino &longs;arà ad e&longs;&longs;a BE, &longs;arà anche &longs;empre minore.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.529.0.0"> <s id="id.2.1.529.1.0"> Che &longs;e i triangoli eguali BFH BGK fo&longs;&longs;ero di &longs;otto fra BC <lb/>BA collocati; & fo&longs;&longs;ero congiunte le linee HC KC, le-<lb/>quali taglia&longs;&longs;ero le linee BF BG allungate dalla parte di FG <lb/>ne' punti MN, &longs;arà <lb/>la BN maggiore del <lb/>la BM, & la BM di <lb/>e&longs;&longs;a BA. </s> </p> <p type="main" id="id.2.1.530.0.0"> <s id="id.2.1.530.1.0"><emph type="italics"/>Imperoche allunghi&longs;i CH CK <lb/>fin alla circonferenza in OP, <lb/>& congiungan&longs;i BO BP; <lb/>con &longs;imile modo mo&longs;trera&longs;&longs;i <lb/>la linea PK e&longs;&longs;ere maggiore <lb/>ai OH, & l'angolo PKB e&longs; <lb/>&longs;ere minore dell <expan abbr="ãgolo">angolo</expan> OHB. <lb/>& percioche l'angolo BHF <lb/>è eguale dell' angolo BKG, &longs;a <lb/>rà tutto l'angolo PKG mi-<lb/>nore dell' angolo OHF. Per <lb/>laqual co&longs;a il re&longs;tante GKN <lb/>&longs;arà maggiore del re&longs;tante <lb/>FHM. Se <expan abbr="dũque">dunque</expan> fara&longs;&longs;i l'an <lb/>golo GKQ eguale ad FHM <lb/>la linea KQ taglier à in modo <lb/>la GN, che GQ diuenterà <lb/>eguale ad FM. Per laqual <lb/>co&longs;a maggiore &longs;arà GN, che <lb/>FM; allequali &longs;e &longs;ar anno ag <lb/>giunte le eguali BF BG, &longs;a-<lb/>rà BN maggiore di BM. & <lb/>per e&longs;&longs;ere BM maggiore di <lb/>FB, &longs;arà anco maggiore di <lb/>BA. &longs;imilmente prouera&longs;&longs;i <lb/>che <expan abbr="quãto">quanto</expan> più da vicino &longs;arà <lb/>BG à BC, la linea BN &longs;em <lb/>pre &longs;arà maggiore.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.531.0.0" xlink:href="figures-it/104_01.jpg"></figure> <pb id="p.43v" xlink:href="pageimg-it/105.jpg"/> <p type="head" id="id.2.1.532.0.0"> <s id="id.2.1.532.1.0"> PROPOSITIONE VII. </s> </p> <p type="main" id="id.2.1.533.0.0"> <s id="id.2.1.533.1.0"> Sia la linea retta AB, à cui &longs;tia à piombo AD, laquale allun-<lb/>ghi&longs;i dalla parte di D come pare &longs;in'à C, & congiunga&longs;i C <lb/>B, laquale etiandio &longs;i allunghi fin'ad E; & &longs;imilmente tra <lb/>AB BE &longs;iano, come pare, tirate BF BG eguali ad e&longs;&longs;a AB, <lb/>& da punti FG &longs;iano <lb/>tirate le linee FH GK <lb/>pur eguali ad e&longs;&longs;a AD, <lb/>& à piombo di BF BG, <lb/>come &longs;e BA AD fo&longs;-<lb/>&longs;ero mo&longs;&longs;e in BF FH <lb/>BG GK: & congiun-<lb/>gan&longs;i CH CK, lequali <lb/>taglino le linee allunga <lb/>te BF BG ne' punti <lb/>MN. Dico che BN è <lb/>maggiore di BM, & <lb/>BM di e&longs;&longs;a BA. </s> </p> <figure place="text" id="id.2.1.534.0.0" xlink:href="figures-it/105_01.jpg"></figure> <p type="main" id="id.2.1.535.0.0"> <s id="id.2.1.535.1.0"><emph type="italics"/>Congiungan&longs;i BD BH BK, & <lb/>co'l centro B, & conlo &longs;patio <lb/>BD de&longs;criua&longs;i il cerchio. </s> <s id="id.2.1.535.2.0"> &longs;imil-<lb/>mente come nella precedente, di-<lb/>mo&longs;treremo ipunti KHDOP <lb/>e&longs;&longs;ere nella circonferenza del cer <lb/>chio; & <expan abbr="itriāgoli">itriangoli</expan> ABD FBH <lb/>GBK e&longs;&longs;ere tra loro eguali, & <lb/>la linea PK e&longs;&longs;ere maggiore <lb/>della OH, & l'angolo PKB <lb/>e&longs;&longs;ere minore dell'angolo OHB. <lb/> Percioche <expan abbr="dũque">dunque</expan> l'angolo BHF <lb/>è eguale all'angolo BKG, &longs;arà <lb/>tutto l'angolo PKG minore <lb/>dell'angolo OHF. Per laqual <lb/>co&longs;a il re&longs;tante G<emph.end type="italics"/>K<emph type="italics"/>N &longs;arà <lb/>maggiore del re&longs;tante FHM. <lb/> Se <expan abbr="dũque">dunque</expan> &longs;i &longs;arà l'angolo G<emph.end type="italics"/>K<emph type="italics"/>Q<emph.end type="italics"/> <pb id="p.44" xlink:href="pageimg-it/106.jpg"/><emph type="italics"/>eguale ad e&longs;&longs;o FHM, &longs;arà il triangolo GKQ eguale al triangolo FHM, & <lb/>il lato GQ al lato FM eguale; &longs;arà dunque maggiore GN di e&longs;&longs;a FM; & <lb/>perciò BN maggiore &longs;arà di BM. & BM &longs;arà maggiore di BA; impe-<lb/>roche BM èmaggiore di e&longs;&longs;ai BE. Che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.536.0.0"> <s id="id.2.1.536.1.0"><emph type="italics"/>Et nel modo i&longs;te&longs;&longs;o in tutto, quanto più da pre&longs;&longs;o &longs;arà BG ad e&longs;&longs;a BE, &longs;empre la li-<lb/>nea BN &longs;i dimo&longs;trerà e&longs;&longs;er maggiore.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.537.0.0"> <s id="id.2.1.537.1.0"> Che &longs;e &longs;aranno po&longs;ti di <lb/>&longs;otto i triangoli BF <lb/>HB GK tra AB <lb/>BC, & &longs;iano tiratele <lb/>linee CHO GKP, <lb/>lequali taglino le li-<lb/>nee BF BG ne' pun<lb/>ti MN: &longs;arà la linea <lb/>BN minore di e&longs;&longs;a <lb/>BM, & BM di e&longs;sa <lb/>BA. </s> </p> <figure place="text" id="id.2.1.538.0.0" xlink:href="figures-it/106_01.jpg"></figure> <p type="main" id="id.2.1.539.0.0"> <s id="id.2.1.539.1.0"><emph type="italics"/>Congiungan&longs;i BO BP. &longs;imilmen <lb/>te prouera&longs;&longs;i, che l'angolo P <lb/>KB è minore dell' angolo OH <lb/>B. Hor percioche l'angolo F <lb/>HB è eguale all'angolo GKB; <lb/>&longs;arà l'angolo GKN maggio-<lb/>re dell'angolo FHM: per la <lb/>qual co&longs;a la linea GN &longs;arà <lb/>maggiore di e&longs;&longs;a FM. & per-<lb/>ciò la linea BN &longs;arà minore <lb/>della linea BM. & concio-<lb/>&longs;ia che maggiore &longs;ia BF di <lb/>BM; &longs;arà BM minore di <lb/>BA. & con &longs;imile modo <lb/>proueraßi, che quanto più B <lb/>G &longs;arà dapre&longs;&longs;o ad e&longs;&longs;a BC, <lb/>la linea BN &longs;empre &longs;arà <lb/>minore.<emph.end type="italics"/> </s> </p> <pb id="p.44v" xlink:href="pageimg-it/107.jpg"/> <p type="head" id="id.2.1.540.0.0"> <s id="id.2.1.540.1.0"> PROPOSITIONE VIII. </s> </p> <p type="main" id="id.2.1.541.0.0"> <s id="id.2.1.541.1.0"> La po&longs;&longs;anza &longs;o&longs;tenente il pe&longs;o che habbia il centro della grauez-<lb/>za fopra la leuà egualmente di&longs;tante dall'orizonte, quanto <lb/>più il pe&longs;o &longs;i inalzerà da que&longs;to &longs;ito con la leua &longs;empre haurà <lb/>bi&longs;ogno di po&longs;sanza minore per e&longs;sere &longs;o&longs;tenuto: ma &longs;e &longs;arà <lb/>abba&longs;sato di maggiore. </s> </p> <p type="main" id="id.2.1.542.0.0"> <s id="id.2.1.542.1.0"><emph type="italics"/>Sia la leua AB egualmente di&longs;tante dall'orizonte, il cui &longs;o&longs;tegno &longs;ia C, & il pe&longs;o <lb/>BD il centro della grauezza delquale &longs;ia doue è H &longs;opra la leua; & &longs;ia la po&longs;&longs;an<emph.end type="italics"/> <lb/> <arrow.to.target n="fig62"></arrow.to.target> <lb/><emph type="italics"/>za &longs;o&longs;tenente in A. Moua&longs;i dapoi la leua AB in EF, & &longs;ia il pe&longs;o mo&longs;&longs;o <lb/>in FG. Dico primieramente che minore po&longs;&longs;anza po&longs;ta in E &longs;o&longs;tenir à il pe&longs;o <lb/>FG con la leua EF, che la po&longs;&longs;anza in A il pe&longs;o BD con la leua AB. &longs;ia <lb/>il K il centro della grauezza del pe&longs;o FG. Dapoi &longs;iano tirate sì da H, come<emph.end type="italics"/> <pb id="p.45" xlink:href="pageimg-it/108.jpg"/><emph type="italics"/>da K le linee HL KM à piombo de'loro orizonti, lequali &longs;i andaranno à tro-<lb/>uare nel centro del mondo, & &longs;ia HL à piombo anche di e&longs;&longs;a AB. Dapoi &longs;ia <lb/>tirata la linea KN à piombo di EF, laquale &longs;arà eguile ad HL, & la CN <lb/>eguale ad e&longs;&longs;a CL. Hor percioche HL è à piombo dell'orizonte, la po&longs;&longs;anza <lb/>in A &longs;o&longs;tenente il pe&longs;o BD haurà quella proportione ad e&longs;&longs;o pe&longs;o, che CL à<emph.end type="italics"/> <arrow.to.target n="note151"></arrow.to.target> <lb/><emph type="italics"/>CA. Di nuouo, percioche KM è à piombo dell'orizonte, la po&longs;&longs;anza in E &longs;o-<lb/>&longs;tenente il pe&longs;o FG co&longs;i &longs;arà al pe&longs;o come CM à CE. & per e&longs;&longs;ere CN NK <lb/>eguali ad e&longs;&longs;e CL LH, & contenere angoliretti, &longs;arà CM minore di e&longs;&longs;a CL;<emph.end type="italics"/> <arrow.to.target n="note152"></arrow.to.target> <lb/><emph type="italics"/> Dunque CM à CA haurà proportione minore, che CL à CA; & CA <lb/>è eguale à CE, dunque haurà CM proportione minore à CE, che CL à<emph.end type="italics"/> <arrow.to.target n="note153"></arrow.to.target> <lb/><emph type="italics"/>CA: & per e&longs;&longs;erei pe&longs;i BD FG eguali, però che è il pe&longs;o mede&longs;imo. </s> <s id="id.2.1.542.2.0"> Dun-<lb/>que &longs;arà minore proportione della po&longs;&longs;anza in E &longs;o&longs;tenente il pe&longs;o FG ad e&longs;&longs;o <lb/>pe&longs;o, che della po&longs;&longs;anza in A &longs;o&longs;tenente il pe&longs;o BD ad e&longs;&longs;o pe&longs;o. </s> <s id="id.2.1.542.3.0"> Per laqual <lb/>co&longs;a minore po&longs;&longs;anza po&longs;ta in E &longs;o&longs;tenterà il pe&longs;o FG, che la po&longs;&longs;anza in A<emph.end type="italics"/> <arrow.to.target n="note154"></arrow.to.target> <lb/><emph type="italics"/>il pe&longs;o BD. & quanto più &longs;arà inalzato il pe&longs;o, &longs;empre &longs;i mo&longs;trerà po&longs;&longs;anza<emph.end type="italics"/> <arrow.to.target n="note155"></arrow.to.target> <lb/><emph type="italics"/>anche minore douer &longs;o&longs;tenere il pe&longs;o, per e&longs;&longs;ere la linea PC minore della CM. <lb/> Sia dapoi la leua in QR, & il pe&longs;o in QS, il cui centro della grauezza &longs;ia O. <lb/> Dico che po&longs;&longs;anza maggiore &longs;i richiede in R per &longs;o&longs;tenere il pe&longs;o QS, che in <lb/>A per &longs;ostentare il pe&longs;o BD. Tiri&longs;i dal centro O della grauezza la linea OT <lb/>a piombo dell'orizonte. </s> <s id="id.2.1.542.4.0"> & percioche le linee HL OT &longs;e &longs;aranno allungate dal-<lb/>la parte di L, & di T &longs;i andranno à ritrouare nel centro del mondo, &longs;arà la CT mag <lb/>giore della CL: & è la CA eguale ad e&longs;&longs;a CR, dunque la TC haurà pro-<emph.end type="italics"/> <arrow.to.target n="note156"></arrow.to.target> <lb/><emph type="italics"/>portione maggiore à CR, che LC à CA. Maggiore dunque &longs;arà la po&longs;&longs;an-<emph.end type="italics"/> <arrow.to.target n="note157"></arrow.to.target> <lb/><emph type="italics"/>za in R &longs;o&longs;tenente il pe&longs;o QS, che in A &longs;o&longs;tenente il BD. Similmente mo-<lb/>&longs;trera&longs;&longs;i, che quanto la leua RQ abba&longs;&longs;ando&longs;i, &longs;arà più di&longs;tante dalla leua AB,<emph.end type="italics"/> <arrow.to.target n="note158"></arrow.to.target> <lb/><emph type="italics"/>&longs;empre più &longs;i ricercherà po&longs;&longs;anza maggiore à &longs;o&longs;tenere il pe&longs;o: peroche la di&longs;tanza<emph.end type="italics"/> <arrow.to.target n="note159"></arrow.to.target> <lb/><emph type="italics"/>CV è più lunga di CT. Quanto dunque il pe&longs;o &longs;i alzerà più dal &longs;ito egualmente <lb/>di&longs;tante dall'orizonte, &longs;arà &longs;empre &longs;o&longs;tenuto da po&longs;&longs;anza minore; & quanto più &longs;i <lb/>abba&longs;&longs;erà, di po&longs;&longs;anza maggiore haurà me&longs;tieri per e&longs;&longs;er &longs;o&longs;tentato. </s> <s id="id.2.1.542.5.0"> che bi&longs;ogna-<lb/>ua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig62" place="text" xlink:href="figures-it/107_01.jpg"></figure> <p type="margin" id="id.2.1.544.0.0"> <s id="id.2.1.544.1.0"> <margin.target id="note151"></margin.target><emph type="italics"/>Per la quinta di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.545.0.0"> <s id="id.2.1.545.1.0"> <margin.target id="note152"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.546.0.0"> <s id="id.2.1.546.1.0"> <margin.target id="note153"></margin.target><emph type="italics"/>Per la ottaua del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.547.0.0"> <s id="id.2.1.547.1.0"> <margin.target id="note154"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 10. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.548.0.0"> <s id="id.2.1.548.1.0"> <margin.target id="note155"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.549.0.0"> <s id="id.2.1.549.1.0"> <margin.target id="note156"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.550.0.0"> <s id="id.2.1.550.1.0"> <margin.target id="note157"></margin.target><emph type="italics"/>Per la ottaua del<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.551.0.0"> <s id="id.2.1.551.1.0"> <margin.target id="note158"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 10. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.552.0.0"> <s id="id.2.1.552.1.0"> <margin.target id="note159"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.553.0.0"> <s id="id.2.1.553.1.0"> Quinci facilmente &longs;i caua, che la pos&longs;anza in A alla po&longs;sanza <lb/>in E co&longs;i è, come CL à CM. </s> </p> <p type="main" id="id.2.1.554.0.0"> <s id="id.2.1.554.1.0"><emph type="italics"/>Imperoche co&longs;iè LC à CA, come la po&longs;&longs;anza in A al pe&longs;o; & come CA, <lb/>cioè CE à CM, co&longs;i è il pe&longs;o alla po&longs;&longs;anza in E; Per laqual co&longs;a per la pro-<emph.end type="italics"/> <arrow.to.target n="note160"></arrow.to.target> <lb/><emph type="italics"/>portion eguale, la po&longs;&longs;anza in A alla po&longs;&longs;anza in E &longs;arà come CL à CM.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.555.0.0"> <s id="id.2.1.555.1.0"> <margin.target id="note160"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 22. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.556.0.0"> <s id="id.2.1.556.1.0"><emph type="italics"/>Con &longs;imile ragione mo&longs;treraßi non &longs;olamente che la po&longs;&longs;anza in A co&longs;i è alla po&longs;-<lb/>&longs;anza in R, come CL à CT, ma che la po&longs;&longs;anza in E ancora alla po&longs;&longs;anza <lb/>in R è co&longs;i, come CM à CT, & co&longs;i nel re&longs;to.<emph.end type="italics"/> </s> </p> <pb id="p.45v" xlink:href="pageimg-it/109.jpg"/> <p type="main" id="id.2.1.558.0.0"> <s id="id.2.1.558.1.0"><emph type="italics"/>Sia poi la leua AB egualmente di&longs;tante dall'orizonte, il cui &longs;o&longs;tegno &longs;ia B, & il <lb/>centro H della grauezza del pe&longs;o CD &longs;ia &longs;opra la leua; & moua&longs;i la leua in <lb/>BE, & il pe&longs;o in FG. Dico che minore po&longs;&longs;anza po&longs;ta in E &longs;o&longs;tiene il pe&longs;o FG <lb/>con la leua EB, che la po&longs;&longs;anza in A il pe&longs;o CD con la leua AB. Sia K <lb/>il centro della grauezza del pe&longs;o FG, & da i centri delle grauezze HK &longs;iano<emph.end type="italics"/> <lb/> <arrow.to.target n="fig63"></arrow.to.target> <lb/> <arrow.to.target n="note161"></arrow.to.target><emph type="italics"/>tirate le linee HL<emph.end type="italics"/> K<emph type="italics"/>M à piombo de'loro orizonti. </s> <s id="id.2.1.558.2.0"> Hor percioche dalle co&longs;e <lb/>di &longs;opra mo&longs;trate BM è minore di BL, & BE è eguale à BA, haurà pro-<emph.end type="italics"/> <lb/> <arrow.to.target n="note162"></arrow.to.target><emph type="italics"/>portione minore BM à BE, che BL à BA: ma come BM à BE, co&longs;i<emph.end type="italics"/> <lb/> <arrow.to.target n="note163"></arrow.to.target><emph type="italics"/>è la po&longs;&longs;anza in E &longs;o&longs;tenente il pe&longs;o FG ad e&longs;&longs;o pe&longs;o, & come BL a BA, <lb/>co&longs;i la po&longs;&longs;anza in A al pe&longs;o CD; la po&longs;&longs;anza in E al pe&longs;o FG haur à propor-<emph.end type="italics"/> <lb/> <arrow.to.target n="note164"></arrow.to.target><emph type="italics"/>tione minore, che la po&longs;&longs;anza in A al pe&longs;o CD. Dunque la po&longs;&longs;anza in E &longs;a-<lb/>rà minore della po&longs;&longs;anza in A. Similmente mo&longs;trera&longs;&longs;i quanto più il pe&longs;o &longs;i alze-<lb/>rà, &longs;empre minore po&longs;&longs;anza &longs;o&longs;tenere il pe&longs;o, ma &longs;ia la leua in BO, & il pe&longs;o in <lb/>BQ, il cui centro della grauezza &longs;ia R. Dico, che maggior po&longs;&longs;anza &longs;i ricerca <lb/>in O per &longs;o&longs;tenere il pe&longs;o PQ con la leua BO, che per &longs;o&longs;tenere il pe&longs;o CD con <lb/>la leua BA. Sia tirata dal punto R la linea RS à piombo dell'orizonte. </s> <s id="id.2.1.558.3.0"> &<emph.end type="italics"/> <lb/> <arrow.to.target n="note165"></arrow.to.target><emph type="italics"/>percioche BS è maggiore di BL, haurà BS proportione maggiore à BO, che <lb/>BL à BA; Per laqual co&longs;a la po&longs;&longs;anza in O &longs;o&longs;tenente il pe&longs;o PQ &longs;arà maggio <lb/>re della po&longs;&longs;anza in A &longs;o&longs;tenente il pe&longs;o CD. & à que&longs;to modo &longs;i mo&longs;trerà an-<lb/>cora che quanto la leua BO abba&longs;&longs;an lo&longs;i, &longs;arà più di&longs;tante dalla leua AB &longs;em-<lb/>pre vi vorrà po&longs;&longs;anza maggiore à &longs;o&longs;iener il pe&longs;o.<emph.end type="italics"/> </s> </p> <figure id="fig63" place="text" xlink:href="figures-it/109_01.jpg"></figure> <p type="margin" id="id.2.1.560.0.0"> <s id="id.2.1.560.1.0"> <margin.target id="note161"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.561.0.0"> <s id="id.2.1.561.1.0"> <margin.target id="note162"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.562.0.0"> <s id="id.2.1.562.1.0"> <margin.target id="note163"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 5. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.563.0.0"> <s id="id.2.1.563.1.0"> <margin.target id="note164"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 10. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.564.0.0"> <s id="id.2.1.564.1.0"> <margin.target id="note165"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.565.0.0"> <s id="id.2.1.565.1.0"><emph type="italics"/>Di qui parimente, come di &longs;opra è mani&longs;e&longs;to, che la po&longs;&longs;anza in A è alla po&longs;&longs;anza in<emph.end type="italics"/> <pb id="p.46" xlink:href="pageimg-it/110.jpg"/><emph type="italics"/>B, come BL à BM: & la po&longs;&longs;anza in A alla po&longs;&longs;anza in O, come BL à BS. & <lb/>la po&longs;&longs;anza in E alla po&longs;&longs;anza in O, come BM à BS.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.566.0.0"> <s id="id.2.1.566.1.0"><emph type="italics"/>Oltre à ciò &longs;e &longs;i intenderà vn'altra po&longs;&longs;anza in B, per modo che due &longs;iano le po&longs;&longs;an-<lb/>ze, che &longs;o&longs;tentino il pe&longs;o, minore &longs;arà la po&longs;&longs;anza in B, che &longs;o&longs;tiene il pe&longs;o PQ <lb/>con la leua BO, che il pe&longs;o CD con la leua BA. ma per lo contrario &longs;i ri-<lb/>cerca po&longs;&longs;anza maggiore in B per &longs;o&longs;tenere il pe&longs;o FG con la leua BE, che <lb/>il pe&longs;o CD con la leua AB: percioche tirata la linea KN à piombo di EB, <lb/>&longs;arà EN eguale ad AL: Per laqual co&longs;a EM &longs;arà maggiore di LA. Dun<emph.end type="italics"/> <arrow.to.target n="note166"></arrow.to.target> <lb/><emph type="italics"/>que EM haurà proportione maggiore ad EB, che LA ad AB, & LA <lb/>maggiore ad AB, che SO ad OB, lequali &longs;ono proportioni della po&longs;&longs;anza<emph.end type="italics"/> <arrow.to.target n="note167"></arrow.to.target> <lb/><emph type="italics"/>al pë&longs;o.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.567.0.0"> <s id="id.2.1.567.1.0"> <margin.target id="note166"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.568.0.0"> <s id="id.2.1.568.1.0"> <margin.target id="note167"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 5. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.569.0.0"> <s id="id.2.1.569.1.0"><emph type="italics"/>Similmente prouera&longs;&longs;i, che la po&longs;&longs;anza in B &longs;o&longs;tenente il pe&longs;o con la leua AB è al-<lb/>la po&longs;&longs;anza &longs;o&longs;tenente po&longs;ta nell'i&longs;te&longs;&longs;o punto B con la leua EB, come LA <lb/>ad EM; & co&longs;i e&longs;&longs;ere anche alla po&longs;&longs;anza di B &longs;o&longs;tenente il pe&longs;o con la leua OB, <lb/>come AL ad OS. Ma quelle po&longs;&longs;anze che &longs;o&longs;tengono con le leue EB OB <lb/>&longs;ono co&longs;itraloro come EM ad OS.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.570.0.0"> <s id="id.2.1.570.1.0"><emph type="italics"/>Dapoi mo&longs;treremo come nelle co&longs;e che di &longs;opra &longs;ono &longs;tate dette, che la po&longs;&longs;anza in B <lb/>ha quella proportione alla po&longs;&longs;anza in E, che EM ad MB; & la po&longs;&longs;anza<emph.end type="italics"/> <arrow.to.target n="note168"></arrow.to.target> <lb/><emph type="italics"/>in B co&longs;i e&longs;&longs;ere alla po&longs;&longs;anza in A, come AL ad LB, & la po&longs;&longs;anza in B <lb/>alla po&longs;&longs;anza in O, come OS ad SB.<emph.end type="italics"/> <arrow.to.target n="note169"></arrow.to.target> </s> </p> <p type="margin" id="id.2.1.571.0.0"> <s id="id.2.1.571.1.0"> <margin.target id="note168"></margin.target><emph type="italics"/>Per il<emph.end type="italics"/> 3. <emph type="italics"/>corollario.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.572.0.0"> <s id="id.2.1.572.1.0"> <margin.target id="note169"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.573.0.0"> <s id="id.2.1.573.1.0"><emph type="italics"/>Ma &longs;ia la leua AB e-<lb/>gualmente di&longs;tante <lb/>dall'orizonte, il cui <lb/>&longs;o&longs;tegno &longs;ia B, & <lb/>il centro H della <lb/>grauezza del pe&longs;o <lb/>AC &longs;ia &longs;opra la <lb/>leua: & moua&longs;i la <lb/>leua in BE, & il <lb/>pe&longs;o in EF, & la <lb/>po&longs;&longs;anza in G. di <lb/>mo&longs;trera&longs;&longs;i parimen<lb/>te, come di &longs;opra, che <lb/>la po&longs;&longs;anza in G &longs;o <lb/>&longs;tenente il pe&longs;o EF <lb/>è minore della po&longs;-<lb/>&longs;anza in D &longs;o&longs;te-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig64"></arrow.to.target> <lb/><emph type="italics"/>nente il pe&longs;o AC. percioche e&longs;&longs;endo minore BM di BL haurà minore pro-<lb/>portione MB à BG, che LB à BD. & à que&longs;to modo proueraßi, che quan <lb/>to il pe&longs;o più &longs;i alzerà con la leua, &longs;empre minore po&longs;&longs;anza &longs;i ricerca à &longs;o&longs;tenere<emph.end type="italics"/> <pb id="p.46v" xlink:href="pageimg-it/111.jpg"/><emph type="italics"/>il detto pe&longs;o. </s> <s id="id.2.1.573.2.0"> &longs;imilmente &longs;e la leua &longs;i moue in BO, & la po&longs;&longs;anza &longs;o&longs;tenente &longs;ia <lb/>in N, &longs;i mo&longs;trerà <lb/>la po&longs;&longs;anza in N e&longs;-<lb/>&longs;ere maggiore della <lb/>po&longs;&longs;anza in D. pe-<lb/>roche SB ha pro-<lb/>portione maggiore <lb/>à BN che LB <lb/>à BD. Mo&longs;tre-<lb/>raßi ancora, che <lb/>quanto il pe&longs;o più <lb/>s'abb a&longs;&longs;erà, &longs;empre <lb/>ricercar&longs;i po&longs;&longs;anza <lb/>maggiore à &longs;o&longs;tene-<lb/>re il pe&longs;o. </s> <s id="id.2.1.573.3.0"> che bi&longs;o-<lb/>gnaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig64" place="text" xlink:href="figures-it/110_01.jpg"></figure> <figure place="text" id="id.2.1.575.0.0" xlink:href="figures-it/111_01.jpg"></figure> <p type="main" id="id.2.1.576.0.0"> <s id="id.2.1.576.1.0"><emph type="italics"/>Di quì <expan abbr="parimēte">parimente</expan> è chia <lb/>ro, che le po&longs;&longs;anze <lb/>in GDN co&longs;itraloro &longs;ono, come BM à BL, & come BL à BS, & vlti-<lb/>mamente come BM à BS.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.577.0.0"> <s id="id.2.1.577.1.0"> COROLLARIO </s> </p> <p type="main" id="id.2.1.578.0.0"> <s id="id.2.1.578.1.0"> Da que&longs;te co&longs;e è manife&longs;to, che &longs;e la po&longs;sanza con la leua moue<lb/>rà in sù il pe&longs;o, il cui centro della grauezza &longs;ia &longs;opra la leua, <lb/>quanto più &longs;arà alzato il pe&longs;o, &longs;empre vi vorrà po&longs;sanza mi-<lb/>nore per mouere il pe&longs;o. </s> </p> <p type="main" id="id.2.1.579.0.0"> <s id="id.2.1.579.1.0"><emph type="italics"/>Percioche doue la po&longs;&longs;anza &longs;o&longs;tenente il pe&longs;o è &longs;empre minore, &longs;arà parimente la po&longs;-<lb/>&longs;anza, che lo moue &longs;empre minore.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.580.0.0"> <s id="id.2.1.580.1.0"><emph type="italics"/>Da que&longs;te co&longs;e dimo&longs;treraßi etiandio, &longs;ia pur il centro della grauezza del pe&longs;o mede&longs;i-<lb/>mo ò più da pre&longs;&longs;o, ò più da lunge della leua AB egualmente di&longs;tante dall' ori-<lb/>zonte, che la po&longs;&longs;anza mede&longs;ima in A &longs;o&longs;terrà nondimeno il pe&longs;o: come &longs;e il cen <lb/>tro H della grauezza del pe&longs;o BD &longs;ia più da lunge dalla leua BA, che il cen-<lb/>tro N della grauezza del pe&longs;o PV, pur che la linea HL tirata dal punto H <lb/>à piombo dell'orizonte, & della leua AB paßi per N, & &longs;ia il pe&longs;o PV <lb/>eguale al pe&longs;o BD; &longs;arà sì il pe&longs;o BD, & sì il pe&longs;o PV come &longs;e ambidue &longs;o&longs;-<lb/>&longs;ero appiccati ad L; & &longs;ono eguali per e&longs;&longs;ere pre&longs;i in luogo di vn pe&longs;o &longs;olo, dun-<lb/>que la i&longs;te&longs;&longs;a po&longs;&longs;anza in A &longs;o&longs;ienente il pe&longs;o BD &longs;o&longs;terrà anche il pe&longs;o PV.<emph.end type="italics"/><lb/> <pb id="p.47" xlink:href="pageimg-it/112.jpg"/> <emph type="italics"/>Ma nella leua EF quanto il centro della grauezza &longs;arà più da lunge dalla leua. <lb/>tanto più egualmente la po&longs;&longs;anza &longs;o&longs;tenter à il pe&longs;o mede&longs;imo, come &longs;e il centro K <lb/>della grauezza del pe&longs;o FG &longs;o&longs;&longs;e più da lunge dalla leua EF, che il centro X <lb/>dalla grauezza del pe&longs;o <foreign lang="greek">*u</foreign>Z; in modo però, che la lineatirata dal punto<emph.end type="italics"/> K <emph type="italics"/>à <lb/>piombo della leua FE paßi per X; & &longs;ia il pe&longs;o FG eguale al pe&longs;o <foreign lang="greek">*u</foreign>Z; <lb/>& da punti KX &longs;iano tirate le linee KM X36>< à piombo de loro orizonti; &longs;a-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig65"></arrow.to.target> <lb/><emph type="italics"/>rà la C36>< maggiore di CM% & perciò il pe&longs;o FG &longs;arà nella leua co&longs;i come <lb/>&longs;e fo&longs;&longs;e appiccato in M, & il pe&longs;o <foreign lang="greek">*u</foreign>Z come fo&longs;&longs;e appiccato in 36><. Hor per-<emph.end type="italics"/> <arrow.to.target n="note170"></arrow.to.target> <lb/><emph type="italics"/>cioche C36>< ha proportione maggiore à CE, che CM à CE, maggiore <lb/>&longs;arà la po&longs;&longs;anza po&longs;ta in E, che &longs;o&longs;terrà il pe&longs;o <foreign lang="greek">*u</foreign>Z, che FG. Manella leua <lb/>QR per lo contrario &longs;i dimo&longs;trerà, cioè che quanto il centro della grauezza del pe <lb/>&longs;o mede&longs;imo è più da lunge dalla leua, tanto più anche maggiore è la po&longs;&longs;anza che <lb/>&longs;o&longs;tiene il pe&longs;o. </s> <s id="id.2.1.580.2.0"> peroche maggiore è CT di CI, & perciò CT hauerà proportio-<lb/>ne maggiore à CR, che CI à CR. &longs;imilmente dimo&longs;treraßi, &longs;e il pe&longs;o &longs;arà col <lb/>locato fra la po&longs;&longs;anza, & il &longs;o&longs;tegno, ouero la po&longs;&longs;anza po&longs;ta fra il &longs;o&longs;tegno, & il <lb/>pe&longs;o, il che mede&longs;imamente auuenirà alla po&longs;&longs;anzà che moue peroche doue po&longs;&longs;anza <lb/>minore &longs;o&longs;tiene il pe&longs;o, iui po&longs;&longs;anza minore lo mouerà: & doue &longs;iricerca po&longs;&longs;anza <lb/>maggiore in &longs;o&longs;tenere, iui anche maggiore vi vuole in mouere.<emph.end type="italics"/> </s> </p> <figure id="fig65" place="text" xlink:href="figures-it/112_01.jpg"></figure> <p type="margin" id="id.2.1.582.0.0"> <s id="id.2.1.582.1.0"> <margin.target id="note170"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <pb id="p.47v" xlink:href="pageimg-it/113.jpg"/> <p type="head" id="id.2.1.583.0.0"> <s id="id.2.1.583.1.0"> PROPOSITIONE IX. </s> </p> <p type="main" id="id.2.1.584.0.0"> <s id="id.2.1.584.1.0"> La po&longs;&longs;anza &longs;o&longs;tenente il pe&longs;o, che habbia il centro della &longs;ua gra<lb/>uezza &longs;otto la leua egualmente di&longs;tante dall'orizonte, quanto <lb/>più il pe&longs;o &longs;arà alzato da que&longs;to &longs;ito con la leua, haurà egli &longs;em <lb/>preanco me&longs;tieri di po&longs;&longs;anza maggiore ad e&longs;&longs;ere &longs;o&longs;tenuto; <lb/>Ma &longs;e abba&longs;&longs;ato, di minore. </s> </p> <p type="main" id="id.2.1.585.0.0"> <s id="id.2.1.585.1.0"><emph type="italics"/>Sia la leua AB egualmente <expan abbr="di&longs;tāte">di&longs;tante</expan> dall'orizonte, il cui &longs;o&longs;tegno &longs;ia C, & &longs;ia il pe&longs;o AD, <lb/>il cui centro L della grauezza &longs;ia &longs;otto la leua, & &longs;ia in B la po&longs;&longs;anza &longs;o&longs;tenen-<lb/>te il pe&longs;o AD: moua&longs;i dopo la leua in FG, & il pe&longs;o in FH. Dico prima, <lb/>che po&longs;&longs;anza maggiore &longs;i ricerca in G per &longs;o&longs;tenere il pe&longs;o FH con la leua FG, <lb/>di quel che &longs;iala po&longs;&longs;anza in B e&longs;&longs;endo il pe&longs;o AD, ma con la leua AB. &longs;ia<emph.end type="italics"/> <lb/> <arrow.to.target n="fig66"></arrow.to.target> <lb/><emph type="italics"/>M il centro della grauezza del pe&longs;o FH, & da punti LM &longs;iano tirate le linee<emph.end type="italics"/> <lb/> <arrow.to.target n="note171"></arrow.to.target><emph type="italics"/>LK MN à piombo de'loro orizonti; & &longs;iatirata la linea MS à piombo di FG, <lb/>che &longs;arà eguale ad LK, & CK &longs;arà etiandio eguale ad e&longs;&longs;a CS. Percioche dun<emph.end type="italics"/> <lb/> <arrow.to.target n="note172"></arrow.to.target><emph type="italics"/>que CN è maggiore di CK haurà NC proportione maggiore à CG, che CK <lb/>à CB; & la po&longs;&longs;anza in B al pe&longs;o AD ha la mede&longs;ma proportione, che KC<emph.end type="italics"/> <lb/> <arrow.to.target n="note173"></arrow.to.target><emph type="italics"/>à CB: & come la po&longs;&longs;anza in G al pe&longs;o FH, co&longs;i è NC à CG; dunque la<emph.end type="italics"/> <lb/> <arrow.to.target n="note174"></arrow.to.target><emph type="italics"/>po&longs;&longs;anza in G hauer à maggiore proportione al pe&longs;o FH, che la po&longs;&longs;anza in B <lb/>al pe&longs;o AD. Maggiore dunque è la po&longs;&longs;anza in G della po&longs;&longs;anza in B. che&longs;e<emph.end type="italics"/> <pb id="p.48" xlink:href="pageimg-it/114.jpg"/><emph type="italics"/>la leua &longs;arà in OP, & il pe&longs;o in OQ &longs;arà la po&longs;&longs;anza po&longs;ta in B maggiore, <lb/>che in P: percioche &longs;i dimo&longs;trerà nell'i&longs;te&longs;&longs;o modo CR e&longs;&longs;ere minore di CK, & <lb/>CR hauere proportione minore a CP, che CK a CB; & perciò la po&longs;&longs;anza <lb/>po&longs;ta in B e&longs;&longs;ere maggiore della po&longs;&longs;anza po&longs;ta in P. & a que&longs;to modo mo&longs;tre-<lb/>ra&longs;&longs;i che quanto più il pe&longs;o &longs;i alzerà dal &longs;ito AB, &longs;empre vi vorrà po&longs;&longs;anza mag-<lb/>giore à &longs;o&longs;tenerlo. </s> <s id="id.2.1.585.2.0"> ma per lo contrario accaderà &longs;e egli &longs;arà abba&longs;&longs;ato. </s> <s id="id.2.1.585.3.0"> che bi&longs;o-<lb/>gnaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig66" place="text" xlink:href="figures-it/113_01.jpg"></figure> <p type="margin" id="id.2.1.587.0.0"> <s id="id.2.1.587.1.0"> <margin.target id="note171"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 7. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.588.0.0"> <s id="id.2.1.588.1.0"> <margin.target id="note172"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.589.0.0"> <s id="id.2.1.589.1.0"> <margin.target id="note173"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 5. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.590.0.0"> <s id="id.2.1.590.1.0"> <margin.target id="note174"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 10. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.591.0.0"> <s id="id.2.1.591.1.0"><emph type="italics"/>Di quà ancora &longs;i puote ageuolmente cauare, che le po&longs;&longs;anze po&longs;te in PBG &longs;ono in <lb/>modo di&longs;po&longs;te fraloro, come CR à CK; & come CK à CN, & come CN <lb/>à CR.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.592.0.0"> <s id="id.2.1.592.1.0"><emph type="italics"/>Sia dopo la leua AB egualmente di&longs;tante dall'orizonte, co'l &longs;uo &longs;o&longs;tegno B; & il <lb/>pe&longs;o CD habbia il centro O della grauezza &longs;otto la leua, & &longs;ia in A la po&longs;-<lb/>&longs;anza &longs;o&longs;tenente il pe&longs;o CD. Moua&longs;i dapoi la leua in BE, & BF, & &longs;i tra-<lb/>&longs;porti il pe&longs;o in GH KL. Dico, che maggiore po&longs;&longs;anza per &longs;o&longs;tenere il pe&longs;o &longs;i<emph.end type="italics"/> <lb/> <arrow.to.target n="fig67"></arrow.to.target> <lb/><emph type="italics"/>ricerca in E, che in A; & maggiore in A che in F &longs;iano tirate da i centri <lb/>delle grauezze le linee NM OP QR à piombo de gli orizonti, lequali allun<lb/>gate da la parte di NOQ &longs;i andranno à trouare nel centro del mondo. </s> <s id="id.2.1.592.2.0"> Mo&longs;tre-<lb/>ra&longs;&longs;i parimente come di &longs;opra, che BM è maggiore di BP, & BP maggio-<emph.end type="italics"/> <arrow.to.target n="note175"></arrow.to.target> <lb/><emph type="italics"/>re di BR; & che BM ha proportione maggiore à BE, che BP à BA; & <lb/>BP à BA maggiore che BR à BF: & per que&longs;to la po&longs;&longs;anza in E mag-<lb/>giore è della po&longs;&longs;anza in A; & la po&longs;&longs;anza in A maggiore della po&longs;&longs;anza in<emph.end type="italics"/> <lb/>F. <emph type="italics"/>& quanto la leua &longs;i alzerà più dal &longs;ito AB, mo&longs;trera&longs;&longs;i &longs;empre, che mag-<pb id="p.48v" xlink:href="pageimg-it/115.jpg"/>giore po&longs;&longs;anza vi vuole à &longs;o&longs;tenere il pe&longs;o: ma &longs;e abba&longs;&longs;eraßi, minore.<emph.end type="italics"/> </s> </p> <figure id="fig67" place="text" xlink:href="figures-it/114_01.jpg"></figure> <p type="margin" id="id.2.1.594.0.0"> <s id="id.2.1.594.1.0"> <margin.target id="note175"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 7. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.595.0.0"> <s id="id.2.1.595.1.0"><emph type="italics"/>Di quì è chiaro etiandio che le po&longs;&longs;anze po&longs;te in EAF co&longs;itraloro &longs;ono, come BM <lb/>à BP, & come BP à BR, & come BM à BR.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.596.0.0"> <s id="id.2.1.596.1.0"><emph type="italics"/>Di più &longs;e in B &longs;arà vn'altra po&longs;&longs;anza, per modo, che due po&longs;&longs;anze &longs;iano quelle che <lb/>&longs;o&longs;tengano il pe&longs;o. </s> <s id="id.2.1.596.2.0"> Di maggiore po&longs;&longs;anza è bi&longs;ogno in B per &longs;o&longs;tenere il pe&longs;o KL <lb/>con la leua BF, che per &longs;o&longs;tenere il pe&longs;o CD con la leua AB. & dauan-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig68"></arrow.to.target> <lb/><emph type="italics"/>taggio anco maggiore con la leua AB, che con la leua BE: peroche RF ha <lb/>proportione maggiore ad FB, che PA ad AB; & PA ad AB mag-<lb/>giore, che EM ad EB.<emph.end type="italics"/> </s> </p> <figure id="fig68" place="text" xlink:href="figures-it/115_01.jpg"></figure> <p type="main" id="id.2.1.598.0.0"> <s id="id.2.1.598.1.0"><emph type="italics"/>Similmente mo&longs;treraßi, che le po&longs;&longs;anze in B &longs;o&longs;tenenti il pe&longs;o con le leue traloro co&longs;i <lb/>e&longs;&longs;ere, come EM ad AP, & come AP ad FR, & come EM ad FR.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.599.0.0"> <s id="id.2.1.599.1.0"> <arrow.to.target n="note176"></arrow.to.target><emph type="italics"/>Oltre à ciò la po&longs;&longs;anza in B co&longs;i &longs;arà alla po&longs;&longs;anza in F, come RF ad RB; & <lb/>la po&longs;&longs;anza in B alla po&longs;&longs;anza in A come PA à PB, & la po&longs;&longs;anza in B al-<emph.end type="italics"/> <lb/> <arrow.to.target n="note177"></arrow.to.target><emph type="italics"/>la po&longs;&longs;anza in E come EM ad MB.<emph.end type="italics"/> </s> </p> <pb id="p.49" xlink:href="pageimg-it/116.jpg"/> <p type="margin" id="id.2.1.601.0.0"> <s id="id.2.1.601.1.0"> <margin.target id="note176"></margin.target><emph type="italics"/>Per lo<emph.end type="italics"/> 3. <emph type="italics"/>corollario.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.602.0.0"> <s id="id.2.1.602.1.0"> <margin.target id="note177"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.603.0.0"> <s id="id.2.1.603.1.0"><emph type="italics"/>Ma &longs;ia la leua AB egualmente di&longs;tante <expan abbr="dall'orizõte">dall'orizonte</expan>, col &longs;uo &longs;o&longs;tegno B, & il pe&longs;o AC, <lb/>il cui centro della <lb/>grauezza &longs;ia &longs;ot-<lb/>to la leua, & &longs;ia <lb/>la po&longs;&longs;anza <expan abbr="&longs;o&longs;te-<lb> nēce">&longs;o&longs;te-<lb/>nence</expan> il pe&longs;o in D, <lb/>& moua&longs;i la le-<lb/>ua in BE BF, <lb/>& la po&longs;&longs;anza <lb/>in GH; &longs;imil-<lb/>mente mo&longs;trera&longs; <lb/>&longs;i, che la po&longs;&longs;an-<lb/>za in G è mag<lb/>giore della po&longs;&longs;an <lb/>za in D, & la <lb/>po&longs;&longs;anza in D <lb/>maggiore della<emph.end type="italics"/> <lb/> <arrow.to.target n="fig69"></arrow.to.target> <lb/><emph type="italics"/>po&longs;&longs;anza in H. percioche KB ha proportione maggiore à BG, che BL à BD, <lb/>& BL à BD maggiore che MB à BH. & à questa maniera mo&longs;trera&longs;&longs;i che <lb/>quanto la leua più &longs;i alzerà dal &longs;ito AB, dauantaggio douere &longs;empre e&longs;&longs;ere mag <lb/>gior la po&longs;&longs;anza per &longs;o&longs;tenere il pe&longs;o: & quanto più s'abba&longs;&longs;a, minore. </s> <s id="id.2.1.603.2.0"> che dimo <lb/>&longs;trare era me&longs;tieri.<emph.end type="italics"/> </s> </p> <figure id="fig69" place="text" xlink:href="figures-it/116_01.jpg"></figure> <p type="main" id="id.2.1.605.0.0"> <s id="id.2.1.605.1.0"><emph type="italics"/>Similmente in que&longs;te, le po&longs;&longs;anze poste in GDH co&longs;itraloro &longs;aranno, come BK à <lb/>BL, & come BL à BM, & alla &longs;ine come BK à BM.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.606.0.0"> <s id="id.2.1.606.1.0"> COROLLARIO. </s> </p> <p type="main" id="id.2.1.607.0.0"> <s id="id.2.1.607.1.0"> Da que&longs;te co&longs;e etiandio è pale&longs;e, che &longs;e la po&longs;&longs;anza mouerà con <lb/>la leua in sù vn pe&longs;o, che habbia il centro della grauezza &longs;otto <lb/>la leua; Quanto più il pe&longs;o &longs;arà alzato, &longs;empre vi vorrà po&longs;-<lb/>&longs;anza maggiore per mouere il pe&longs;o. </s> </p> <p type="main" id="id.2.1.608.0.0"> <s id="id.2.1.608.1.0"><emph type="italics"/>Imperoche &longs;e la po&longs;&longs;anza &longs;o&longs;tenente il pe&longs;o è &longs;empre maggiore, &longs;arà parimente la <lb/>po&longs;&longs;anza che moue il pe&longs;o &longs;empre maggiore.<emph.end type="italics"/> </s> </p> <pb id="p.49v" xlink:href="pageimg-it/117.jpg"/> <p type="main" id="id.2.1.610.0.0"> <s id="id.2.1.610.1.0"><emph type="italics"/>Da que&longs;te co&longs;e anco &longs;i cauerà facilmente &longs;e &longs;arà il centro della grauezza dell'i&longs;te&longs;&longs;o pe <lb/>&longs;o ò più da pre&longs;&longs;o, ò più da lunge dalla leua AB egualmente di&longs;tante dall'orizon<lb/>te, che la po&longs;&longs;anza mede&longs;ima po&longs;ta in B &longs;o&longs;terrà il pe&longs;o. </s> <s id="id.2.1.610.2.0"> come &longs;e il centro L della <lb/>grauezza del pe&longs;o AD fo&longs;&longs;e più da lunge dalla leua BA, che il centro N <lb/>della grauezza del pe&longs;o PV, pur che la linea LK tirata dal punto L à piom <lb/>bo dell orizonte, & della leua AB pa&longs;&longs;i per N: &longs;imilmente come nella prece-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig70"></arrow.to.target> <lb/><emph type="italics"/>dente &longs;i mo&longs;trerà, che la po&longs;&longs;anza mede&longs;ima in B &longs;ostiene & il pe&longs;o AD, & <lb/>il pe&longs;o PV. Ma nella leua EF quanto il centro della grauezza &longs;arà più da lun<lb/>ge dalla leua, tanto haur à me&longs;tieri di po&longs;&longs;anza maggiore per &longs;ostenere il pe&longs;o. </s> <s id="id.2.1.610.3.0"> co-<lb/>me il centro M della grauezza del pe&longs;o FH &longs;ia più da lunge dalla leua EF, che <lb/>il centro S della grauezza del pe&longs;o XZ. &longs;iano tirate da i punti MS le linee <lb/>MI SG à piombo de gli orizonti; &longs;arà CI maggiore di CG: & perciò la po&longs;&longs;an <lb/>za di E deue e&longs;&longs;ere maggiore &longs;o&longs;tenendo il pe&longs;o FH, che il pe&longs;o XZ. Maper <lb/>lo contrario &longs;i mo&longs;trerà nella leua OR, cioè che quanto il centro della grauezza <lb/>dell'i&longs;te&longs;&longs;o pe&longs;o è più dalunge dalla leua, il pe&longs;o viene &longs;o&longs;tentato da po&longs;&longs;anza mino<lb/>re. </s> <s id="id.2.1.610.4.0"> peroche minore è C<foreign lang="greek">*u</foreign> de CT. & in modo &longs;imile demo&longs;traraßi ancora &longs;tan<lb/>do il pe&longs;o fra la po&longs;&longs;anza, & il &longs;o&longs;tegno, ouero la po&longs;&longs;anzatra il &longs;ostegno, & il<emph.end type="italics"/> <pb id="p.50" xlink:href="pageimg-it/118.jpg"/><emph type="italics"/>pe&longs;o, ilche parimente auerrà alla po&longs;&longs;anza che moue; peroche doue po&longs;&longs;anza mino-<lb/>re &longs;o&longs;tien il pe&longs;o, iui minore po&longs;&longs;anza lo mouerà. </s> <s id="id.2.1.610.5.0"> & doue vuole po&longs;&longs;anza maggio-<lb/>re in &longs;o&longs;tentare, iui anco ella &longs;arà maggiore in mouere.<emph.end type="italics"/> </s> </p> <figure id="fig70" place="text" xlink:href="figures-it/117_01.jpg"></figure> <p type="head" id="id.2.1.612.0.0"> <s id="id.2.1.612.1.0"> PROPOSITIONE X. </s> </p> <p type="main" id="id.2.1.613.0.0"> <s id="id.2.1.613.1.0"> La po&longs;&longs;anza &longs;o&longs;tenente il pe&longs;o che habbia il centro della grauez-<lb/>za nella i&longs;te&longs;&longs;a leua, &longs;ia pure in qual &longs;i voglia modo tra&longs;porta-<lb/>to il pe&longs;o con la leua; vi &longs;arà &longs;empre me&longs;tieri della po&longs;&longs;anza <lb/>i&longs;te&longs;&longs;a, acciò &longs;ia &longs;o&longs;tenuto. </s> </p> <p type="main" id="id.2.1.614.0.0"> <s id="id.2.1.614.1.0"><emph type="italics"/>Sia la leua AB egualmente di&longs;tante dall'orizonte, co'l luo &longs;o&longs;tegno C, & E cen-<lb/>tro della grauezza del pe&longs;o &longs;ia in e&longs;&longs;a leua. </s> <s id="id.2.1.614.2.0"> Moua&longs;i dapoi la leua in FG, & HK,<emph.end type="italics"/> <lb/> <arrow.to.target n="fig71"></arrow.to.target> <lb/><emph type="italics"/>& il centro della grauezza in LM. Dico che la mede&longs;ima po&longs;&longs;anza di KBG &longs;em-<emph.end type="italics"/> <arrow.to.target n="note178"></arrow.to.target> <lb/><emph type="italics"/>pre &longs;o&longs;terrà l'iste&longs;&longs;o pe&longs;o. </s> <s id="id.2.1.614.3.0"> Hor percioche il pe&longs;o nella leua AB è &longs;i fattamen-<lb/>te di&longs;po&longs;to, come &longs;e egli fo&longs;&longs;e appiccato in E; & nella leua GF come &longs;e eglifo&longs; <lb/>&longs;e appiccato in L; & nella leua HK, come &longs;e egli fo&longs;&longs;e appiccato in M; & le<emph.end type="italics"/> <pb id="p.50v" xlink:href="pageimg-it/119.jpg"/><emph type="italics"/>distanze CL CE CM &longs;ono traloro eguali; & parimente CK CB CG pur <lb/>tra loro eguali; &longs;arà la po&longs;&longs;anza in B al pe&longs;o, come CE à CB; & la po&longs;&longs;an-<lb/>za in K al pe&longs;o, come CM à CK, & la po&longs;&longs;anza in G al pe&longs;o, come CL<emph.end type="italics"/> <lb/> <arrow.to.target n="fig72"></arrow.to.target> <lb/><emph type="italics"/>à CG. La po&longs;&longs;anza mede&longs;ma dunque in KBG &longs;osterrà il pe&longs;o mede&longs;mo tra&longs;por <lb/>tato in vari &longs;iti. </s> <s id="id.2.1.614.4.0"> che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig71" place="text" xlink:href="figures-it/118_01.jpg"></figure> <figure id="fig72" place="text" xlink:href="figures-it/118_02.jpg"></figure> <p type="margin" id="id.2.1.617.0.0"> <s id="id.2.1.617.1.0"> <margin.target id="note178"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 5. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.618.0.0"> <s id="id.2.1.618.1.0"><emph type="italics"/>Similmente prouera&longs;&longs;i, &longs;e il pe&longs;o fo&longs;&longs;e tra la po&longs;&longs;anza, & il &longs;o&longs;tegno; ouero la po&longs;-<lb/>&longs;anza tra il &longs;o&longs;tegno, & il pe&longs;o, che il mede&longs;imo auerrà alla po&longs;&longs;anza, che moue.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.619.0.0"> <s id="id.2.1.619.1.0"> PROPOSITIONE XI. </s> </p> <p type="main" id="id.2.1.620.0.0"> <s id="id.2.1.620.1.0"> Se la di&longs;tanza della leua tra il &longs;o&longs;tegno, & la po&longs;&longs;anza haurà pro-<lb/>portione maggiore alla di&longs;tanza trapo&longs;ta dal &longs;o&longs;tegno al pun-<lb/>to, doue dal centro della grauezza del pe&longs;o tirata vna linea à <lb/>piombo dell'orizonte taglia la leua che non ha il pe&longs;o alla po&longs; <lb/>&longs;anza; il pe&longs;o veramente &longs;arà mo&longs;&longs;o dalla po&longs;&longs;anza. </s> </p> <p type="main" id="id.2.1.621.0.0"> <s id="id.2.1.621.1.0"><emph type="italics"/>Sia la leua AB, & dal punto A appicchi&longs;i il pe&longs;o C; cioè il punto A &longs;empre <lb/>&longs;ia quel punto, doue la linea tirata à piombo dal centro della grauezza del pe&longs;o ta-<lb/>gli la leua; & &longs;ia la po&longs;&longs;anza in B, & il &longs;ostegno D; & DB habbia à DA<emph.end type="italics"/> <pb id="p.51" xlink:href="pageimg-it/120.jpg"/><emph type="italics"/>proportione maggiore, che il pe&longs;o C alla po&longs;&longs;anza in B. Dico che il pe&longs;o C &longs;a-<lb/>rà mo&longs;&longs;o dalla po&longs;&longs;anza in B. Faccia&longs;i come BD à DA, co&longs;i il pe&longs;o E alla<emph.end type="italics"/> <arrow.to.target n="note179"></arrow.to.target> <lb/><emph type="italics"/>po&longs;&longs;anza in B; & appicchi&longs;i parimente il pe&longs;o E in A: egliè chiaro che la po&longs;-<lb/>&longs;anza in B pe-<lb/>&longs;a <expan abbr="egualmēte">egualmente</expan> <expan abbr="cõ">com</expan> <lb/>e&longs;&longs;o E; cioè che <lb/>&longs;o&longs;tiene il detto <lb/>pe&longs;o E. & per-<lb/>cioche BD ha <lb/>proportion mag <lb/>giore à DA che <lb/>C alla po&longs;&longs;anza <lb/>in B. & come <lb/>BD à DA, co&longs;i<emph.end type="italics"/> <lb/> <arrow.to.target n="fig73"></arrow.to.target> <lb/><emph type="italics"/>è il pe&longs;o F. alla po&longs;&longs;anza: adunque E haurà proportione maggiore alla po&longs;&longs;an-<emph.end type="italics"/> <arrow.to.target n="note180"></arrow.to.target> <lb/><emph type="italics"/>za, che il pe&longs;o C alla po&longs;&longs;anza i&longs;te&longs;&longs;a. </s> <s id="id.2.1.621.2.0"> Per laqual co&longs;a il pe&longs;o E &longs;arà maggiore <lb/>del pe&longs;o C. & perche la po&longs;&longs;anza pe&longs;a egualmente cone&longs;&longs;o E; dunque la po&longs;&longs;an<lb/>za non pe&longs;erà egualmente cone&longs;&longs;o C, ma per la forza &longs;ua inchinerà al ba&longs;&longs;o. </s> <s id="id.2.1.621.3.0"> àun <lb/>que il pe&longs;o C &longs;arà mo&longs;&longs;o dalla po&longs;&longs;anza in B con la leua AB, il cui &longs;o&longs;tegno <lb/>è in D.<emph.end type="italics"/> </s> </p> <figure id="fig73" place="text" xlink:href="figures-it/120_01.jpg"></figure> <p type="margin" id="id.2.1.623.0.0"> <s id="id.2.1.623.1.0"> <margin.target id="note179"></margin.target><emph type="italics"/>Per la prima di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.624.0.0"> <s id="id.2.1.624.1.0"> <margin.target id="note180"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 10. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.625.0.0"> <s id="id.2.1.625.1.0"><emph type="italics"/>Ma &longs;e la leua fo&longs;&longs;e AB, & il &longs;o&longs;tegno A, & il pe&longs;o C appiccato in D, & la <lb/>po&longs;&longs;anza in B, & BA haue&longs;&longs;e proportione maggiore ad AD, che il pe&longs;o C <lb/>alla po&longs;&longs;anza in B. Dico che il pe&longs;o C mouera&longs;&longs;i dalla po&longs;&longs;anza in B. faccia&longs;i co<emph.end type="italics"/> <arrow.to.target n="note181"></arrow.to.target> <lb/><emph type="italics"/>me BA ad AD, co&longs;i il pe-<lb/>&longs;o E alla po&longs;&longs;anza in B: & <lb/>&longs;e E &longs;arà appiccato in D, la <lb/>po&longs;&longs;anza in B &longs;o&longs;tenterà il pe-<lb/>&longs;o E. Ma per hauere BA pro-<lb/>portione maggiore ad AD, <lb/>che il pe&longs;o C alla po&longs;&longs;anza in <lb/>B; & come BA ad AD, <lb/>co&longs;i è il pe&longs;o E alla po&longs;&longs;anza in <lb/>B; dunque il pe&longs;o E haurà pro<lb/>portione maggiore alla po&longs;&longs;an<emph.end type="italics"/> <arrow.to.target n="note182"></arrow.to.target> <lb/> <arrow.to.target n="fig74"></arrow.to.target> <lb/><emph type="italics"/>za che è in B, che il pe&longs;o C all'i&longs;te&longs;&longs;a po&longs;&longs;anza: & perciò il pe&longs;o E &longs;arà maggio <lb/>re del pe&longs;o C; & la po&longs;&longs;anza in B &longs;o&longs;tiene il pe&longs;o E; dunque la po&longs;&longs;anza in B <lb/>con la leua AB mouerà il pe&longs;o C minore del pe&longs;o E appiccato in D, il cui &longs;o-<lb/>stegno è A.<emph.end type="italics"/> </s> </p> <figure id="fig74" place="text" xlink:href="figures-it/120_02.jpg"></figure> <p type="margin" id="id.2.1.627.0.0"> <s id="id.2.1.627.1.0"> <margin.target id="note181"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.628.0.0"> <s id="id.2.1.628.1.0"> <margin.target id="note182"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 10. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <pb id="p.51v" xlink:href="pageimg-it/121.jpg"/> <p type="main" id="id.2.1.629.0.0"> <s id="id.2.1.629.1.0"><emph type="italics"/>Sia da capo la leua AB, & il &longs;uo &longs;o&longs;tegno A, & il pe&longs;o C &longs;ia appiccato in B, <lb/>& &longs;ia la po&longs;&longs;anza in D: & DA habbia proportione maggiore ad AB, che <lb/>il pe&longs;o C al-<lb/>la po&longs;&longs;anza, <lb/>che è in D. Di<lb/>co che il pe&longs;o C <lb/>&longs;arà mo&longs;&longs;o dal <lb/>la <expan abbr="poßāza">poßanza</expan> che <lb/>è in D. Fac-<lb/>cia&longs;i come D <lb/>A ad AB, <lb/>co&longs;i il pe&longs;o E <lb/>alla po&longs;&longs;anza,<emph.end type="italics"/> <lb/> <arrow.to.target n="fig75"></arrow.to.target> <lb/><emph type="italics"/>che è in D; & &longs;ia il pe&longs;o E pendente dal punto B: la po&longs;&longs;anza in D &longs;o&longs;ter-<lb/>rà il pe&longs;o E. Ma DA tiene proportione maggiore ad AB, che C alla po&longs;-<lb/>&longs;anza in D. & come DA ad AB, co&longs;i è il pe&longs;o E alla po&longs;&longs;anza in D; <lb/>dunque il pe&longs;o E haurà proportione maggiore alla po&longs;&longs;anza che è in D, che il <lb/>pe&longs;o C alla i&longs;te&longs;&longs;a po&longs;&longs;anza. </s> <s id="id.2.1.629.2.0"> Per laqual co&longs;a il pe&longs;o E è maggiore del pe&longs;o C. <lb/>Et percioche la po&longs;&longs;anza in D &longs;o&longs;tiene il pe&longs;o E, dunque la detta po&longs;&longs;anza in <lb/>D mouerà il pe&longs;o C appiccato in B con la leua AB, il cui &longs;o&longs;tegno è A. che <lb/>bi&longs;ognaua prouare.<emph.end type="italics"/> </s> </p> <figure id="fig75" place="text" xlink:href="figures-it/121_01.jpg"></figure> <p type="head" id="id.2.1.631.0.0"> <s id="id.2.1.631.1.0"> Altramente. </s> </p> <p type="main" id="id.2.1.632.0.0"> <s id="id.2.1.632.1.0"><emph type="italics"/>Sia la leua AB, & il pe&longs;o C appiccato in A, & la po&longs;&longs;anza in B, & &longs;ia il <lb/>&longs;o&longs;tegno D; & DB habbia proportione maggiore à DA, che il pe&longs;o C alla <lb/>po&longs;&longs;anza in B. <lb/> Dico che il pe-<lb/>&longs;o C &longs;arà mo&longs; <lb/>&longs;o dalla po&longs;&longs;an-<lb/>za in B. Fac-<lb/>cia&longs;i BE ad <lb/>EA, come il<emph.end type="italics"/> <lb/> <arrow.to.target n="fig76"></arrow.to.target> <lb/><emph type="italics"/>pe&longs;o C &longs;i ha inuer&longs;o la po&longs;&longs;anza, &longs;arà il punto E tra BD: percioche egli è me-<emph.end type="italics"/> <lb/> <arrow.to.target n="note183"></arrow.to.target><emph type="italics"/>&longs;tieri che BE habbia proportione minore ad EA, che DB à DA; & però <lb/>BE &longs;arà minore di BD. & percioche la po&longs;&longs;anza in B &longs;o&longs;tiene il pe&longs;o C ap-<lb/>piccato in A con la leua AB, che hà il &longs;o&longs;tegno E; dunque minore po&longs;&longs;an-<lb/>za po&longs;ta in B, che la data &longs;o&longs;terrà il pe&longs;o mede&longs;imo nel &longs;o&longs;tegno D. La po&longs;&longs;an-<lb/>za data dunque po&longs;ta in B mouerà il pe&longs;o C con la leua AB, che ha il &longs;o&longs;te-<lb/>gno in D.<emph.end type="italics"/> </s> </p> <figure id="fig76" place="text" xlink:href="figures-it/121_02.jpg"></figure> <p type="margin" id="id.2.1.635.0.0"> <s id="id.2.1.635.1.0"> <margin.target id="note183"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <pb id="p.52" xlink:href="pageimg-it/122.jpg"/> <p type="main" id="id.2.1.636.0.0"> <s id="id.2.1.636.1.0"><emph type="italics"/>Sia dapoi la leua AB, & il &longs;uo &longs;o&longs;tegno in A, & il pe&longs;o C appiccato in D, & <lb/>&longs;iala po&longs;&longs;anza in B; & AB habbia proportione maggiore ad AD, che il <lb/>pe&longs;o C alla po&longs;&longs;anza in B. Di<lb/>co che il pe&longs;o C &longs;i mouerà dalla <lb/>po&longs;&longs;anza in B. Faccia&longs;i AB ad <lb/>AE, come il pe&longs;o C alla po&longs; <lb/>&longs;anza; &longs;arà &longs;imilmente il punto E <lb/>tra BD, percioche egli è nece&longs;&longs;a-<emph.end type="italics"/> <arrow.to.target n="note184"></arrow.to.target> <lb/><emph type="italics"/>rio che AE &longs;ia maggiore di A <lb/>D. & &longs;e il pe&longs;o C fo&longs;&longs;e appicca<emph.end type="italics"/> <arrow.to.target n="note185"></arrow.to.target> <arrow.to.target n="fig77"></arrow.to.target> <lb/><emph type="italics"/>to in E, la po&longs;&longs;anza in B lo &longs;o&longs;tentarebbe. </s> <s id="id.2.1.636.2.0"> ma po&longs;&longs;anza minore po&longs;ta in B, <lb/>che la data &longs;o&longs;tiene il pe&longs;o C appiccato in D; dunque la data po&longs;&longs;anza in B mo-<emph.end type="italics"/> <arrow.to.target n="note186"></arrow.to.target> <lb/><emph type="italics"/>uerà il pe&longs;o C appiccato in D con la leua AB, che ha il &longs;uo &longs;o&longs;tegno A.<emph.end type="italics"/> </s> </p> <figure id="fig77" place="text" xlink:href="figures-it/122_01.jpg"></figure> <p type="margin" id="id.2.1.638.0.0"> <s id="id.2.1.638.1.0"> <margin.target id="note184"></margin.target><emph type="italics"/>Per la ottaua del<emph.end type="italics"/> 5. </s> </p> <p type="margin" id="id.2.1.639.0.0"> <s id="id.2.1.639.1.0"> <margin.target id="note185"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.640.0.0"> <s id="id.2.1.640.1.0"> <margin.target id="note186"></margin.target><emph type="italics"/>Per il<emph.end type="italics"/> 1. <emph type="italics"/>corollario del la<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.641.0.0"> <s id="id.2.1.641.1.0"><emph type="italics"/>Sia da capo la leua AB co'l &longs;o&longs;tegno &longs;uo A; & il pe&longs;o C &longs;ia appiccato in B, & <lb/>&longs;ia la po&longs;&longs;anza in D. & DA habbia proportione maggiore ad AB, che il pe <lb/>&longs;o C alla po&longs;&longs;anza in <lb/>D. Dico che il pe&longs;o C<emph.end type="italics"/> <arrow.to.target n="note187"></arrow.to.target> <lb/><emph type="italics"/>&longs;arà mo&longs;&longs;o dalla po&longs;&longs;an-<lb/>za in D. faccia&longs;i come <lb/>il pe&longs;o C a'la po&longs;&longs;anza, <lb/>co&longs;i DA &longs;ia ad AE;<emph.end type="italics"/> <arrow.to.target n="note188"></arrow.to.target> <lb/><emph type="italics"/>&longs;arà AE maggiore di<emph.end type="italics"/> <lb/> <arrow.to.target n="fig78"></arrow.to.target> <lb/><emph type="italics"/>AB; per e&longs;&longs;ere proportione maggiore da DA ad AB, che da DA ad AE.<emph.end type="italics"/> <arrow.to.target n="note189"></arrow.to.target> <lb/> <emph type="italics"/>Che &longs;e il pe&longs;o C &longs;arà appiccato in E, egli è chiaro, che la po&longs;&longs;anza in D &longs;o&longs;ter-<lb/>rà il pe&longs;o C appiccato in E. Ma po&longs;&longs;anza minore che la data &longs;o&longs;tiene l'i&longs;te&longs;&longs;o pe <lb/>&longs;o C in B; dunque la data po&longs;&longs;anza in D mouerà il pe&longs;o C appiccato in B, con <lb/>la leua AB che hà il &longs;o&longs;tegno &longs;uo A. come bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig78" place="text" xlink:href="figures-it/122_02.jpg"></figure> <p type="margin" id="id.2.1.643.0.0"> <s id="id.2.1.643.1.0"> <margin.target id="note187"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.644.0.0"> <s id="id.2.1.644.1.0"> <margin.target id="note188"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 3. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.645.0.0"> <s id="id.2.1.645.1.0"> <margin.target id="note189"></margin.target><emph type="italics"/>Per il<emph.end type="italics"/> 1. <emph type="italics"/>corollario del la<emph.end type="italics"/> 3. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.646.0.0"> <s id="id.2.1.646.1.0"> PROPOSITIONE XII. </s> </p> <p type="head" id="id.2.1.647.0.0"> <s id="id.2.1.647.1.0"> PROBLEMA. </s> </p> <p type="main" id="id.2.1.648.0.0"> <s id="id.2.1.648.1.0"> Fare che vna data po&longs;&longs;anza, moua vn pe&longs;o dato con vna data le-<lb/>ua. </s> </p> <pb id="p.52v" xlink:href="pageimg-it/123.jpg"/> <p type="main" id="id.2.1.649.0.0"> <s id="id.2.1.649.1.0"><emph type="italics"/>Sia il pe&longs;o A come cento, & la po&longs;&longs;anza che ha da mouere &longs;ia come diece; & &longs;ia <lb/>la data leua BC. Egli è bi&longs;ogno che la po&longs;&longs;anza, che è diece moua il pe&longs;o A, che <lb/>è cento, con la leua BC. Diuida&longs;i BC in D con &longs;i fatta maniera che CD hab <lb/>bia la propor tione mede&longs;ima à DB, che ha cento à diece, cioè diece ad vno; per-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig79"></arrow.to.target> <lb/> <arrow.to.target n="note190"></arrow.to.target><emph type="italics"/>cioche &longs;e D &longs;i face&longs;&longs;e &longs;o&longs;tegno, egli è mani&longs;e&longs;to, che la po&longs;&longs;anza in C come diece <lb/>pe&longs;erà egualmente co'l pe&longs;o A appiccato in B, cioè che &longs;o&longs;terrà il pe&longs;o A. Pren<emph.end type="italics"/> <lb/> <arrow.to.target n="note191"></arrow.to.target><emph type="italics"/>da&longs;i tra BD qual &longs;i voglia punto, come E, & faccia&longs;i E il &longs;o&longs;tegno. </s> <s id="id.2.1.649.2.0"> Hor per-<lb/>cioche maggiore è la proportione di CE ad EB, che di CD à DB; CE haurà <lb/>proportione maggiore ad EB, che il pe&longs;o A alla po&longs;&longs;anza di diece po&longs;ta in C; <lb/>dunque la po&longs;&longs;anza di diece po&longs;ta in C mouer à il pe&longs;o A, che è cento, appiccato<emph.end type="italics"/> <lb/> <arrow.to.target n="note192"></arrow.to.target><emph type="italics"/>in B con la leua BC, che ha il &longs;uo &longs;o&longs;tegno E.<emph.end type="italics"/> </s> </p> <figure id="fig79" place="text" xlink:href="figures-it/123_01.jpg"></figure> <p type="margin" id="id.2.1.651.0.0"> <s id="id.2.1.651.1.0"> <margin.target id="note190"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.652.0.0"> <s id="id.2.1.652.1.0"> <margin.target id="note191"></margin.target><emph type="italics"/>Per lo lemma di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.653.0.0"> <s id="id.2.1.653.1.0"> <margin.target id="note192"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.654.0.0"> <s id="id.2.1.654.1.0"><emph type="italics"/>Ma &longs;e la leua fo&longs;&longs;e BC, & il &longs;o&longs;tegno B. diuida&longs;i CB in D per &longs;i fatta maniera, <lb/>che CB habbia la proportione i&longs;te&longs;&longs;a à BD, che ha cento à diece: & &longs;e il pe&longs;o <lb/>A &longs;arà appic<lb/>cato in D, & <lb/>la po&longs;&longs;anza in <lb/>C, la po&longs;&longs;an-<emph.end type="italics"/> <lb/> <arrow.to.target n="note193"></arrow.to.target><emph type="italics"/>za in C come <lb/>diece &longs;o&longs;terrà <lb/>anco il pe&longs;o <lb/>A appiccato<emph.end type="italics"/> <lb/> <arrow.to.target n="fig80"></arrow.to.target> <lb/><emph type="italics"/>in D. Prenda&longs;i qual &longs;i uoglia punto tra DB, come E, & ponga&longs;i il pe&longs;o A in<emph.end type="italics"/> <lb/> <arrow.to.target n="note194"></arrow.to.target><emph type="italics"/>E; & per e&longs;&longs;ere proportione maggiore da CB à BE, che da BC à BD; CB <lb/>haurà proportione maggiore à BE, che il pe&longs;o A di cento alla po&longs;&longs;anza di diece. <lb/> Dunque la po&longs;&longs;anza di diece po&longs;ta in C mouerà il pe&longs;o A di cento appiccato in E<emph.end type="italics"/> <lb/> <arrow.to.target n="note195"></arrow.to.target><emph type="italics"/>con la leua BC, che ha il &longs;o&longs;tegno &longs;uo B. che bi&longs;ognaua menar ad effetto.<emph.end type="italics"/> </s> </p> <figure id="fig80" place="text" xlink:href="figures-it/2123_02.jpg"></figure> <p type="margin" id="id.2.1.656.0.0"> <s id="id.2.1.656.1.0"> <margin.target id="note193"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.657.0.0"> <s id="id.2.1.657.1.0"> <margin.target id="note194"></margin.target><emph type="italics"/>Per la ottaua del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.658.0.0"> <s id="id.2.1.658.1.0"> <margin.target id="note195"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.659.0.0"> <s id="id.2.1.659.1.0"><emph type="italics"/>Ma ciò non &longs;i puote mandar' ad e&longs;ecutione con la leua BC, che habbia il &longs;o&longs;tegno &longs;uo <lb/>in B, & il pe&longs;o A di cento &longs;ia appiccato in C. Percioche ponga&longs;i la po&longs;&longs;anza <lb/>&longs;o&longs;tenente il pe&longs;o A comunque &longs;i &longs;ia tra BC, come in D; &longs;empre la po&longs;&longs;anza <lb/>&longs;arà maggiore del pe&longs;o A. Per laqual co&longs;a egli è me&longs;tieri che &longs;empre la data po&longs;-<emph.end type="italics"/> <pb id="p.53" xlink:href="pageimg-it/124.jpg"/><emph type="italics"/>&longs;anza &longs;ia maggiore del pe&longs;o A. Sia dunque la po&longs;&longs;anza data, come cento cin-<emph.end type="italics"/> <arrow.to.target n="note196"></arrow.to.target> <lb/><emph type="italics"/>quanta. </s> <s id="id.2.1.659.2.0"> Diuida&longs;i BC in D &longs;i fattamente che CB &longs;ia à BD come cento cin-<lb/>quanta à cento, cioè tre à due: & &longs;e la po&longs;&longs;anza &longs;arà po&longs;ta in D, egli è chiaro, <lb/>che la po&longs;&longs;anza in D &longs;o&longs;ter-<lb/>rà il pe&longs;o A appiccato in C.<emph.end type="italics"/> <arrow.to.target n="note197"></arrow.to.target> <lb/><emph type="italics"/>& co&longs;i prenda&longs;i tra DC <lb/>qual &longs;i voglia punto, come <lb/>E, & ponga&longs;i la po&longs;&longs;anza <lb/>mouente in E, & per e&longs;&longs;ere <lb/>proportion maggiore da EB <lb/>à BC, che da DB à BC; <lb/>haurà EB proportione mag<lb/>giore à BC, che il pe&longs;o A<emph.end type="italics"/> <arrow.to.target n="note198"></arrow.to.target> <lb/> <arrow.to.target n="fig81"></arrow.to.target> <lb/><emph type="italics"/>alla po&longs;&longs;anza in E. Dunque la po&longs;&longs;anza di cento cinquanta po&longs;ta in E mouerà il <lb/>pe&longs;o A di cento appiccato in C con la leua BC che hà il &longs;o&longs;tegno B. come bi-<emph.end type="italics"/> <arrow.to.target n="note199"></arrow.to.target> <lb/><emph type="italics"/>&longs;ognaua oprare.<emph.end type="italics"/> </s> </p> <figure id="fig81" place="text" xlink:href="figures-it/124_01.jpg"></figure> <p type="margin" id="id.2.1.661.0.0"> <s id="id.2.1.661.1.0"> <margin.target id="note196"></margin.target><emph type="italics"/>Per il<emph.end type="italics"/> 2. <emph type="italics"/>corollario della<emph.end type="italics"/> 3. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.662.0.0"> <s id="id.2.1.662.1.0"> <margin.target id="note197"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 3. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.663.0.0"> <s id="id.2.1.663.1.0"> <margin.target id="note198"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>del quinto.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.664.0.0"> <s id="id.2.1.664.1.0"> <margin.target id="note199"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.665.0.0"> <s id="id.2.1.665.1.0"> COROLLARIO. </s> </p> <p type="main" id="id.2.1.666.0.0"> <s id="id.2.1.666.1.0"> Di qui è manife&longs;to, &longs;e la data po&longs;&longs;anza &longs;arà maggiore del dato <lb/>pe&longs;o, que&longs;to poter&longs;i fare, ouero &longs;tando in maniera la leua, <lb/>che il &longs;o&longs;tegno &longs;uo &longs;ia fra il pe&longs;o, & la po&longs;&longs;anza; ouero che el-<lb/>la habbia il pe&longs;o fra il &longs;o&longs;tegno, & la po&longs;&longs;anza; ouero alla fine <lb/>e&longs;&longs;endo po&longs;ta la po&longs;&longs;anza fra il pe&longs;o, & il &longs;o&longs;tegno. </s> </p> <p type="main" id="id.2.1.667.0.0"> <s id="id.2.1.667.1.0"> Ma &longs;e la data po&longs;&longs;anza &longs;arà minore, ouero eguale al dato pe&longs;o, <lb/>egli è parimente chiaro, che il mede&longs;imo &longs;i puote mandare ad <lb/>e&longs;ecutione &longs;olamente &longs;tando la leua in maniera, che il &longs;o&longs;te-<lb/>gno &longs;uo &longs;ia tra il pe&longs;o, & la po&longs;&longs;anza; ouero che ella habbia il <lb/>pe&longs;o fra il &longs;o&longs;tegno, & la po&longs;&longs;anza. </s> </p> <p type="head" id="id.2.1.668.0.0"> <s id="id.2.1.668.1.0"> PROPOSITIONE XIII. </s> </p> <p type="head" id="id.2.1.669.0.0"> <s id="id.2.1.669.1.0"> PROBLEMA. </s> </p> <p type="main" id="id.2.1.670.0.0"> <s id="id.2.1.670.1.0"> Dati quanti &longs;i <expan abbr="vogliã">vogliam</expan> pe&longs;i appiccati douunque &longs;i &longs;iano nella leua <lb/>il cui &longs;o&longs;tegno parimente &longs;ia dato, ritrouare vna po&longs;&longs;anza la <lb/>quale &longs;o&longs;tenga i dati pe&longs;i in vn punto dato. </s> </p> <pb id="p.53v" xlink:href="pageimg-it/125.jpg"/> <p type="main" id="id.2.1.672.0.0"> <s id="id.2.1.672.1.0"><emph type="italics"/>Siano i dati pe&longs;i ABC nella leua DE, & il &longs;o&longs;tegno &longs;uo F, douunque ne' pun-<lb/>ti DGH &longs;iano appiccati, & habbia&longs;i à collocare la po&longs;&longs;anza nel punto E. egli <lb/>è me&longs;tieri trouare la po&longs;&longs;anza, laquale &longs;o&longs;tenga in E i dati pe&longs;i ABC con la le <lb/>ua DE. diuida&longs;i DG in K &longs;i fattamente, che DK &longs;ia à KG come il pe-<lb/>&longs;o B al pe&longs;o A; dapoi diuida&longs;i KH in L &longs;i fattamente, che KL &longs;ia ad LH <lb/>come il pe&longs;o C à i pe&longs;i BA, & come FE ad FL, co&longs;i &longs;accian&longs;i i pe&longs;i ABC<emph.end type="italics"/> <lb/> <arrow.to.target n="fig82"></arrow.to.target> <lb/><emph type="italics"/>tutti in&longs;ieme alla po&longs;&longs;anza, laquale ponga&longs;i in E. dico, che la po&longs;&longs;anza in E. &longs;o-<emph.end type="italics"/> <lb/> <arrow.to.target n="note200"></arrow.to.target><emph type="italics"/>&longs;tenterà i dati pe&longs;i ABC appiccati in DGH con la leua DE che ha il &longs;o&longs;te-<emph.end type="italics"/> <lb/> <arrow.to.target n="note201"></arrow.to.target><emph type="italics"/>gno &longs;uo F. Hor percioche &longs;e i pe&longs;i ABC &longs;o&longs;&longs;ero appiccati in&longs;ieme in L, la po&longs; <lb/>&longs;anza in E &longs;o&longs;terrebbe i dati pe&longs;i appiccati in L; ma i pe&longs;i ABC pe&longs;ano tan-<lb/>to in L, quanto &longs;e C in H, & BA in&longs;ieme &longs;o&longs;&longs;ero appiccati in K; & AB<emph.end type="italics"/> <lb/> <arrow.to.target n="note202"></arrow.to.target><emph type="italics"/>nel K tanto pe&longs;ano, quanto &longs;e A in D, & B in G fo&longs;&longs;ero appiccati; dun-<lb/>què la po&longs;&longs;anza in E &longs;o&longs;tenterà i dati pe&longs;i ABC appiccati in DGH con la <lb/>leua DE che ha il &longs;o&longs;tegno F. Che &longs;e la po&longs;&longs;anza haue&longs;&longs;e ad e&longs;&longs;ere po&longs;ta in qual <lb/>&longs;i voglia altro punto dalla leua DE fuor che in F, come in K; faccia&longs;i come <lb/>FK ad FL, co&longs;ii pe&longs;i ABC &longs;iano alla po&longs;&longs;anza: &longs;imilmente dimo&longs;treremo, <lb/>che la po&longs;&longs;anza in K &longs;o&longs;terrà i pe&longs;i ABC ne' punti DGH appiccati. </s> <s id="id.2.1.672.2.0"> come <lb/>bi&longs;ognaua fare.<emph.end type="italics"/> </s> </p> <figure id="fig82" place="text" xlink:href="figures-it/125_01.jpg"></figure> <p type="margin" id="id.2.1.674.0.0"> <s id="id.2.1.674.1.0"> <margin.target id="note200"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.675.0.0"> <s id="id.2.1.675.1.0"> <margin.target id="note201"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 5. <emph type="italics"/>di questo della bilancia.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.676.0.0"> <s id="id.2.1.676.1.0"> <margin.target id="note202"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.677.0.0"> <s id="id.2.1.677.1.0"><emph type="italics"/>Da que&longs;ta, & dalla quinta di que&longs;to, &longs;e i pe&longs;i ABC &longs;aranno po&longs;ti in qual &longs;i voglia <lb/>modo nella leua DE, & che bi&longs;ogni ritrouare la po&longs;&longs;anza, la quale debba &longs;o&longs;te-<lb/>nere in E i dati pe&longs;i &longs;iano tirate da i centri delle grauezze de i pe&longs;i le linee AB <lb/>C à piombo de gli orizonti, lequali taglino la leua DE ne' punti DGH; &<emph.end type="italics"/> <pb id="p.54" xlink:href="pageimg-it/126.jpg"/><emph type="italics"/>&longs;i operino le altre co&longs;e nell'i&longs;te&longs;&longs;o modo: egli è manife&longs;to, che la po&longs;&longs;anza in E,<emph.end type="italics"/> <lb/> <arrow.to.target n="fig83"></arrow.to.target> <lb/><emph type="italics"/>ouero in K &longs;o&longs;tenterà i dati pe&longs;i, percioche egli è l'i&longs;te&longs;&longs;o come &longs;e i pe&longs;i fo&longs;&longs;ero <lb/>appiccati in DGH.<emph.end type="italics"/> </s> </p> <figure id="fig83" place="text" xlink:href="figures-it/126_01.jpg"></figure> <p type="head" id="id.2.1.679.0.0"> <s id="id.2.1.679.1.0"> PROPOSITIONE XIIII. </s> </p> <p type="head" id="id.2.1.680.0.0"> <s id="id.2.1.680.1.0"> PROBLEMA. </s> </p> <p type="main" id="id.2.1.681.0.0"> <s id="id.2.1.681.1.0"> Fare che vna data po&longs;&longs;anza moua quanti pe&longs;i &longs;i vogliano, po&longs;ti <lb/>douunque, & in qualunque modo &longs;i &longs;ia in vna data leua. </s> </p> <p type="main" id="id.2.1.682.0.0"> <s id="id.2.1.682.1.0"><emph type="italics"/>Sia la data leua DE, & &longs;iano i dati pe&longs;i, come è po&longs;to nel precedente corollario, & <lb/>&longs;ia A come cento, B come cinquanta, & C come trenta; & la data po&longs;&longs;an-<lb/>za &longs;ia come trenta. </s> <s id="id.2.1.682.2.0"> &longs;iano po&longs;te le co&longs;e mede&longs;ime, & ritroui&longs;i il punto L; dapoi<emph.end type="italics"/> <lb/> <arrow.to.target n="fig84"></arrow.to.target> <lb/><emph type="italics"/>diuida&longs;i LE in F, &longs;i &longs;attamente che FE ad FL &longs;ia come cento ottanta à <lb/>trenta, cioè &longs;ei ad vno, & &longs;e F &longs;i face&longs;&longs;e &longs;o&longs;tegno, la po&longs;&longs;anza come trenta<emph.end type="italics"/> <pb id="p.54v" xlink:href="pageimg-it/127.jpg"/> <arrow.to.target n="note203"></arrow.to.target><emph type="italics"/>in E &longs;o&longs;terrebbe i pe&longs;i ABC. pigli&longs;i dunque tra LF qualunque punto come <lb/>M, & faccia&longs;i M il &longs;o&longs;tegno: egli è manife&longs;to, che la po&longs;&longs;anza po&longs;ta in E co-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig85"></arrow.to.target> <lb/> <arrow.to.target n="note204"></arrow.to.target><emph type="italics"/>me trenta mouerài pe&longs;i ABC come cento ottanta con la leua DE. che bi&longs;o-<lb/>gnaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig84" place="text" xlink:href="figures-it/126_02.jpg"></figure> <figure id="fig85" place="text" xlink:href="figures-it/127_01.jpg"></figure> <p type="margin" id="id.2.1.685.0.0"> <s id="id.2.1.685.1.0"> <margin.target id="note203"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 13. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.686.0.0"> <s id="id.2.1.686.1.0"> <margin.target id="note204"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.687.0.0"> <s id="id.2.1.687.1.0"><emph type="italics"/>Ma ciò non potremo già vniuer&longs;almente menare ad ef&longs;etto, &longs;e il &longs;o&longs;tegno &longs;o&longs;&longs;e nelle <lb/>&longs;tremità della leua, come in D; peroche la proportione di DE à DL, cioè la <lb/>proportione de' pe&longs;i ABC alla po&longs;&longs;anza, laquale ha da &longs;o&longs;tenere i pe&longs;i &longs;empre <lb/>è data. </s> <s id="id.2.1.687.2.0"> Laqual co&longs;a molto meno anco &longs;i potrebbe fare, &longs;e la po&longs;&longs;anza &longs;i haue&longs;&longs;e <lb/>à porre tra DL.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.688.0.0"> <s id="id.2.1.688.1.0"> PROPOSITIONE XV. </s> </p> <p type="head" id="id.2.1.689.0.0"> <s id="id.2.1.689.1.0"> PROBLEMA. </s> </p> <p type="main" id="id.2.1.690.0.0"> <s id="id.2.1.690.1.0"> Ma percioche mentre i pe&longs;i &longs;i mouono con la leua, ha la leua an-<lb/>cora grauezza, della quale infin qui non &longs;i è fatto mentione <lb/>alcuna: però dimo&longs;triamo primieramente in che modo &longs;i tro <lb/>ui la po&longs;&longs;anza, la quale &longs;o&longs;tenga nel dato punto la leua data, il <lb/>cui &longs;o&longs;tegno &longs;ia parimente dato. </s> </p> <p type="main" id="id.2.1.691.0.0"> <s id="id.2.1.691.1.0"><emph type="italics"/>Sia la leua data AB, il cui &longs;o&longs;tegno C &longs;ia dato: & &longs;ia il punto D nelquale &longs;i hab <lb/>bia à collocare la po&longs;&longs;anza, che debba &longs;o&longs;tentare la leua AB, &longs;i &longs;attamente che <lb/>re&longs;ti immobile. </s> <s id="id.2.1.691.2.0"> &longs;ia dal punto C tirata la linea CE à piombo dell'orizonte la <lb/>quale diuida la leua AB in due parti AE EF; & della parte AE &longs;ia il <lb/>centro G della grauezza, & della parte EF il centro del'a grauezza &longs;ia H, <lb/>& dai punti GH &longs;iano tirate le linee GK HL à piombo de gli orizonti, le<emph.end type="italics"/> <pb id="p.55" xlink:href="pageimg-it/128.jpg"/><emph type="italics"/>quali taglino la linea AF ne' punti KL. Hor percioche la leua AB è diui-<lb/>&longs;a dalla linea CE in due parti, cioè AE EF; però la leua AB, niente altro <lb/>&longs;arà, che due pe&longs;i AE EF nella leua, ouero bilancia AF po&longs;ti; il cui appicca <lb/>mento, ouero &longs;o&longs;tegno è C. Per laqual co&longs;a i pe&longs;i AE EF &longs;aranno co&longs;i po&longs;ti,<emph.end type="italics"/> <lb/> <arrow.to.target n="fig86"></arrow.to.target> <lb/><emph type="italics"/>come &longs;e fo&longs;&longs;ero appiccati in KL. Diuida&longs;i dunque KL in M, &longs;i fattamente, <lb/>che KM &longs;ia ad ML come la grauezza della parte EF alla grauezza della <lb/>parte AE; & come CA à CM, co&longs;i &longs;accia&longs;i la grauezza di tutta la leua <lb/>AB alla po&longs;&longs;anza, laquale &longs;e in D &longs;arà collocata (pur che DA &longs;ia à piombo<emph.end type="italics"/> <arrow.to.target n="note205"></arrow.to.target> <lb/><emph type="italics"/>dell'orizonte) pe&longs;erà egualmente con la leua; cioè &longs;o&longs;terrà la leua AB premendo <lb/>in giù, che bi&longs;ognaua trouare.<emph.end type="italics"/> </s> </p> <figure id="fig86" place="text" xlink:href="figures-it/128_01.jpg"></figure> <p type="margin" id="id.2.1.693.0.0"> <s id="id.2.1.693.1.0"> <margin.target id="note205"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 13. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.694.0.0"> <s id="id.2.1.694.1.0"><emph type="italics"/>Che &longs;e la po&longs;&longs;anza &longs;i haue&longs;&longs;e à porre nel punto B. Faccia&longs;i come CF à CM, <lb/>co&longs;i il pe&longs;o AB alla po&longs;&longs;anza. </s> <s id="id.2.1.694.2.0"> Con &longs;imile modo prouera&longs;&longs;i che la po&longs;&longs;anza in <lb/>B &longs;o&longs;terrà la leua AB. & l'i&longs;te&longs;&longs;o d mo&longs;trera&longs;&longs;i in qualunque altro &longs;ito s'haue&longs;-<lb/>&longs;e à porre la po&longs;&longs;anza, (fuor che in E) come in N. peroche faccia&longs;i CO à <lb/>CM come AB alla po&longs;&longs;anza, laquale &longs;e &longs;i porrà in N &longs;o&longs;tenterà la leua AB.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.695.0.0"> <s id="id.2.1.695.1.0"> Ma aggiunga &longs;i il pe&longs;o appiccato, ouero po&longs;to nella leua; come, <lb/>po&longs;te le co&longs;e i&longs;te&longs;fe, &longs;ia il pe&longs;o P appiccato in A; & la po&longs;-<lb/>&longs;anza s'habbia à porre in B, &longs;i fattamente che &longs;o&longs;tenghi la le <lb/>ua AB in&longs;ieme col pe&longs;o P. </s> </p> <p type="main" id="id.2.1.696.0.0"> <s id="id.2.1.696.1.0"><emph type="italics"/>Diuida&longs;i AM in Q, &longs;i &longs;attamente, che AQ &longs;ia à QM, come la grauezza<emph.end type="italics"/> <arrow.to.target n="note206"></arrow.to.target> <lb/><emph type="italics"/>della leua AB alla grauezza del pe&longs;o P; dapoi come CF à CQ, co&longs;i fac-<emph.end type="italics"/> <arrow.to.target n="note207"></arrow.to.target> <lb/><emph type="italics"/>cia&longs;i la grauezza AB, & P in&longs;ieme alla po&longs;&longs;anza, la quale ponga&longs;i in B: egli <lb/>è manife&longs;to, che la po&longs;&longs;anza in B &longs;o&longs;terrà la leua AB in&longs;ieme co'l pe&longs;o P. Che <lb/>&longs;e fo&longs;&longs;e CA à CM, come AB à P; &longs;arebbe il punto C il loro centro della <lb/>grauezza, & perciò la leua AB in&longs;ieme co'l pe&longs;o P &longs;enza la po&longs;&longs;anza po&longs;ta in <pb id="p.55v" xlink:href="pageimg-it/129.jpg"/>B &longs;tarà &longs;erma. </s> <s id="id.2.1.696.2.0"> Ma &longs;e il centro della grauezza de' pe&longs;i fo&longs;&longs;e tra CF, come in <lb/>O. Faccia&longs;i come CF à CO, co&longs;i AB & P in&longs;ieme alla po&longs;&longs;anza, laqua-<lb/>le in B &longs;o&longs;tenterà sì la leua AB come il pe&longs;o P.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.697.0.0"> <s id="id.2.1.697.1.0"> <margin.target id="note206"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 13 <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.698.0.0"> <s id="id.2.1.698.1.0"> <margin.target id="note207"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di Archimede delle cose che egualmente pe&longs;ano.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.699.0.0" xlink:href="figures-it/129_01.jpg"></figure> <p type="main" id="id.2.1.700.0.0"> <s id="id.2.1.700.1.0"><emph type="italics"/>Similmente mo&longs;trera&longs;&longs;i il mede&longs;imo &longs;e fo&longs;&longs;ero più pe&longs;i nella leua AB douunque, <lb/>& in qual modo &longs;i &longs;ia di&longs;po&longs;ti.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.701.0.0"> <s id="id.2.1.701.1.0"><emph type="italics"/>Oltre à ciò da que&longs;te co&longs;e &longs;i puote cono&longs;cere, come nella decimaquarta propo&longs;itione di <lb/>que&longs;to habbiamo in&longs;egnato, in che modo cioè po&longs;&longs;iamo mouere i dati pe&longs;i po&longs;ti do<lb/>uunque &longs;i voglia nella leua, con vna data po&longs;&longs;anza, e con vna data leua, ilche po&longs;-<lb/>&longs;iamo fare nell'i&longs;te&longs;&longs;o modo non &longs;olamente con&longs;iderando la grauezza della leua; ma <lb/>anco gli altri accidenti, iquali &longs;ono &longs;tati di &longs;opra mo&longs;trati &longs;enza la grauezza del-<lb/>la leua; con &longs;imile modo con&longs;iderata la grauezza della leua in&longs;ieme co' pe&longs;i, ouero <lb/>&longs;enza pe&longs;i &longs;i mo&longs;treranno.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.702.0.0"> <s id="id.2.1.702.1.0"> IL FINE DELLA LEVA. </s> </p> </chap> <chap id="id.2.2.0.0.4"> <pb id="p.56" xlink:href="pageimg-it/130.jpg"/> <p type="head" id="id.2.1.705.0.0"> <s id="id.2.1.705.1.0"> DELLA TAGLIA. </s> </p> <p type="main" id="id.2.1.707.0.0"> <s id="id.2.1.707.1.0"> Con l'in&longs;trumento della Taglia &longs;i può mouere il pe <lb/>&longs;o in molti modi: ma percioche in tutti è la ragio-<lb/>ne mede&longs;ima: però affine che la co&longs;a re&longs;ti più chia-<lb/>ra, intenda&longs;i in quello che &longs;i ha da dire, che il pe-<lb/>&longs;o &longs;empre &longs;i habbia da mouere all'insù ad angoli <lb/>retti al piano dell'orizonte in que&longs;to modo. </s> </p> <pb id="p.56v" xlink:href="pageimg-it/131.jpg"/> <p type="main" id="id.2.1.708.0.0"> <s id="id.2.1.708.1.0"><emph type="italics"/>Sia il pe&longs;o A ilquale &longs;i habbia ad alzare in sù ad angoli retti al piano dell'orizonte: <lb/>& come &longs;i co&longs;tuma di fare: &longs;ia <lb/>aitaccata di &longs;opra vna taglia, <lb/>che habbia due girelle, gli a&longs;&longs;etti <lb/>dellequali &longs;iano in BC: & &longs;ia <lb/>anche legata vn'altra taglia al <lb/>pe&longs;o, laquale &longs;imilmente habbia <lb/>due girelle, gli a&longs;&longs;etti delle qua-<lb/>li &longs;iano in DE: & per tutte <lb/>le girelle d'ambedue le taglie &longs;ia <lb/>condotta intorno la corda, la-<lb/>quale in vno de i capi, come in <lb/>F deue e&longs;&longs;ere legata. </s> <s id="id.2.1.708.2.0"> Ponga&longs;i <lb/>ancorala po&longs;&longs;anza che moue in <lb/>G, laquale mentre di&longs;cende, il <lb/>pe&longs;o A per lo contrario &longs;arà le-<lb/>uato in &longs;u&longs;o, &longs;i come afferma Pa<lb/>po nell'ottauo libro delle rac-<lb/>colte matematiche, & Vitruuio <lb/>nel decimo dell'architettura, & <lb/>altri.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.709.0.0" xlink:href="figures-it/131_01.jpg"></figure> <p type="main" id="id.2.1.710.0.0"> <s id="id.2.1.710.1.0"> Hor in che modo que&longs;to <lb/>in&longs;trumento della ta-<lb/>glia &longs;i riduca alla leua, <lb/>& perche vn pe&longs;o gran-<lb/>de &longs;i moua da piccola <lb/>forza, & in qual modo, <lb/>& in quanto tempo; & <lb/>perche la corda debba <lb/>e&longs;&longs;ere legata da vn ca-<lb/>po: & quale debba e&longs;-<lb/>&longs;ere l'officio della ta-<lb/>glia, che è po&longs;ta di &longs;ot-<lb/>to, & quale di quella, <lb/>che &longs;tà di &longs;opra, & in <lb/>che modo &longs;i po&longs;&longs;a tro-<lb/>uare ogni proportio-<lb/>ne data ne i numeri tra la po&longs;&longs;anza, & il pe&longs;o, diciamo. </s> </p> <pb id="p.57" xlink:href="pageimg-it/132.jpg"/> <p type="head" id="id.2.1.712.0.0"> <s id="id.2.1.712.1.0"> LEMMA. </s> </p> <p type="main" id="id.2.1.713.0.0"> <s id="id.2.1.713.1.0"> Siano due linee rette AB CD egualmente di&longs;tanti, lequali <lb/>tocchino il cerchio ACE ne' punti AC, il centro delqual <lb/>cerchio &longs;ia F, & &longs;i congiunghino FA & FC. dico che la <lb/>linea AFC è retta. </s> </p> <p type="main" id="id.2.1.714.0.0"> <s id="id.2.1.714.1.0"><emph type="italics"/>Tiri&longs;i la linea FE egualmente di&longs;tante dal-<emph.end type="italics"/> <arrow.to.target n="note208"></arrow.to.target> <lb/><emph type="italics"/>le linee AB CD. Et percioche AB <lb/>& FE &longs;ono egualmente di&longs;tanti, &<emph.end type="italics"/> <arrow.to.target n="note209"></arrow.to.target> <lb/><emph type="italics"/>l'angolo BAF èretto: &longs;arà anco A <lb/>FE retto, & all'ifte&longs;&longs;o modo CFE &longs;a<emph.end type="italics"/> <arrow.to.target n="note210"></arrow.to.target> <lb/><emph type="italics"/>rà retto: adunque la linea AFC èret-<lb/>ta, ilche s'hauea à dimo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.715.0.0"> <s id="id.2.1.715.1.0"> <margin.target id="note208"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 18. <emph type="italics"/>del terzo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.716.0.0"> <s id="id.2.1.716.1.0"> <margin.target id="note209"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 29. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.717.0.0"> <s id="id.2.1.717.1.0"> <margin.target id="note210"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 14. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.718.0.0" xlink:href="figures-it/132_01.jpg"></figure> <p type="head" id="id.2.1.719.0.0"> <s id="id.2.1.719.1.0"> PROPOSITIONE I. </s> </p> <p type="main" id="id.2.1.720.0.0"> <s id="id.2.1.720.1.0"> Se la corda &longs;i condurrà intorno alla girella della taglia, che &longs;ia <lb/>attaccata di &longs;opra, & che vno delli &longs;uoi capi &longs;i leghi al pe&longs;o, & <lb/>l'altro tratanto &longs;ia pre&longs;o dalla po&longs;sanza, che &longs;o&longs;tiene il detto <lb/>pe&longs;o; la po&longs;sanza &longs;arà eguale al pe&longs;o. </s> </p> <pb id="p.57v" xlink:href="pageimg-it/133.jpg"/> <p type="main" id="id.2.1.722.0.0"> <s id="id.2.1.722.1.0"><emph type="italics"/>Sia il pe&longs;o A alquale venga legata la corda à B: & la taglia, che habbia la girella <lb/>CEF il cui centro D appicchi&longs;i di &longs;opra: & &longs;ia parimente D il centro dell'a&longs; <lb/>&longs;etto, & d'intorno alla girella volga&longs;i la corda BCEFG: & &longs;ia in G la po&longs;-<lb/>&longs;anza, che &longs;o&longs;tiene il pe&longs;o A. Dico la po&longs;&longs;anza po&longs;tain G e&longs;&longs;ere eguale al pe-<lb/>&longs;o A. &longs;ia FG <lb/>egualmente di-<lb/>&longs;tante da CB. <lb/> Percioche <expan abbr="dū-<lb> que">dun-<lb/>que</expan> il pe&longs;o A <lb/>&longs;ta fermo, &longs;a-<emph.end type="italics"/> <lb/> <arrow.to.target n="note211"></arrow.to.target><emph type="italics"/>rà CB à piom<lb/>bo del piano <lb/>dell'orizonte.<emph.end type="italics"/> <lb/> <arrow.to.target n="note212"></arrow.to.target><emph type="italics"/>onde FG &longs;a-<lb/>rà al piano <lb/>i&longs;te&longs;&longs;o à piom-<lb/>bo. </s> <s id="id.2.1.722.2.0"> Siano i <lb/>punti CF nel-<lb/>la girella, da <lb/>quali le corde <lb/>CB FG &longs;cen <lb/>dano nel pia-<lb/>no dell'orizon <lb/>te ad angoli <lb/>retti, tocche-<lb/>ranno le dette <lb/>corde BC FG <lb/>la girella CE<emph.end type="italics"/> <lb/> <arrow.to.target n="fig87"></arrow.to.target> <lb/><emph type="italics"/>F ne'punti CF peroche non po&longs;&longs;ono &longs;egare la girella. </s> <s id="id.2.1.722.3.0"> Siano congiunte le li-<emph.end type="italics"/> <lb/> <arrow.to.target n="note213"></arrow.to.target><emph type="italics"/>nee DC DF. &longs;arà retta la linea CF & &longs;aranno anche retti gli ang oli DCB <lb/>DFG. Ma percioche BC &longs;ta à piombo sì all'orizonte, come ad e&longs;&longs;a CF &longs;arà<emph.end type="italics"/> <lb/> <arrow.to.target n="note214"></arrow.to.target><emph type="italics"/>la detta CF egualmente di&longs;tante dall'orizonte. </s> <s id="id.2.1.722.4.0"> & concio&longs;ia che il pe&longs;o &longs;ia attac-<lb/>cato in CB & la po&longs;&longs;anza &longs;ia in G ch'è il mede&longs;imo, come &longs;e ella fo&longs;&longs;e in F:<emph.end type="italics"/> <lb/> <arrow.to.target n="note215"></arrow.to.target><emph type="italics"/>&longs;arà CF tanto quanto vna bilancia, ouero vna leua, il cui centro, ouero &longs;o&longs;tegno <lb/>&longs;arà D, imperoche la girella è &longs;o&longs;tenuta nell'a&longs;&longs;etto, & il punto D per e&longs;&longs;ere <lb/>centro dell'a&longs;&longs;etto, & della girellarimane immobile, &longs;eben l'vno, & l'altro &longs;i vol<lb/>gono intorno. </s> <s id="id.2.1.722.5.0"> Per laqual co&longs;a e&longs;&longs;endo la di&longs;tanza DC eguale alla di&longs;tanza DF, <lb/>& la po&longs;&longs;anza che è in F contrape&longs;i egualmente al pe&longs;o A attaccato in C &longs;o-<lb/>&longs;tenendo il pe&longs;o in modo, che non cala al ba&longs;&longs;o, &longs;arà la po&longs;&longs;anza a&longs;&longs;egnata in F oue<lb/>ro in G che è tutt'vno, eguale al pe&longs;o A: percioche po&longs;ta in G fal'i&longs;te&longs;&longs;o effet <lb/>to che &longs;e nel mede&longs;imo G &longs;o&longs;&longs;e appiccato vn'altro pe&longs;o eguale al pe&longs;o A, liquali <lb/>pe&longs;i attaccati in CF contrape&longs;eranno egualmente. </s> <s id="id.2.1.722.6.0"> Oltre à ciò non facendo&longs;i moto<emph.end type="italics"/> <pb id="p.58" xlink:href="pageimg-it/134.jpg"/><emph type="italics"/>in niuna delle parti, &longs;arà l'i&longs;te&longs;&longs;o e&longs;&longs;endo circondata in que&longs;to modo la girella intor-<lb/>no con vna corda &longs;ola BC e FG come &longs;e &longs;u&longs;&longs;ero due corde BC FG legate <lb/>alla leua, ouero alla bilancia CF.<emph.end type="italics"/> </s> </p> <figure id="fig87" place="text" xlink:href="figures-it/133_01.jpg"></figure> <p type="margin" id="id.2.1.724.0.0"> <s id="id.2.1.724.1.0"> <margin.target id="note211"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>di questo della bilancia.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.725.0.0"> <s id="id.2.1.725.1.0"> <margin.target id="note212"></margin.target><emph type="italics"/>Per la ottaua dell'vndecimo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.726.0.0"> <s id="id.2.1.726.1.0"> <margin.target id="note213"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 18. <emph type="italics"/>del terzo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.727.0.0"> <s id="id.2.1.727.1.0"> <margin.target id="note214"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 28. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.728.0.0"> <s id="id.2.1.728.1.0"> <margin.target id="note215"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>del<emph.end type="italics"/> 1. <emph type="italics"/>d'Archimede delle co&longs;e che pe&longs;ano egualmente.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.729.0.0"> <s id="id.2.1.729.1.0"> COROLLARIO. </s> </p> <p type="main" id="id.2.1.730.0.0"> <s id="id.2.1.730.1.0"> Da que&longs;to può e&longs;&longs;ere manife&longs;to, che il mede&longs;imo pe&longs;o dalla i&longs;te&longs;-<lb/>&longs;a po&longs;&longs;anza puote e&longs;&longs;ere tuttauia &longs;o&longs;tenuto &longs;enza anche alcu-<lb/>no aiuto di que&longs;ta taglia. </s> </p> <p type="main" id="id.2.1.731.0.0"> <s id="id.2.1.731.1.0"><emph type="italics"/>Percioche &longs;ia il pe&longs;o H eguale al pe&longs;o A à cui &longs;ia legata la corda KL & &longs;iala <lb/>po&longs;&longs;anza, che &longs;o&longs;tiene il pe&longs;o H in L. Hor concio&longs;ia che volendo &longs;o&longs;tenere al-<lb/>cun pe&longs;o &longs;enza aiuto veruno vi bi&longs;ogni tanta forza, quanta &longs;ia eguale al pe&longs;o; la <lb/>po&longs;&longs;anza che è in L &longs;arà eguale al <lb/>pe&longs;o H, ma il pe&longs;o H è po&longs;to <lb/>eguale al pe&longs;o A, alquale è anco <lb/>eguale la po&longs;&longs;anza G. &longs;arà dun-<lb/>que la po&longs;&longs;anza in G eguale alla <lb/>po&longs;&longs;anza in L che è l'i&longs;te&longs;&longs;o, come &longs;e <lb/>la i&longs;te&longs;&longs;a po&longs;&longs;anza &longs;o&longs;tene&longs;&longs;e il pe&longs;o <lb/>mede&longs;imo. </s> <s id="id.2.1.731.2.0"> Oltre à ciò &longs;e le po&longs;&longs;an<lb/>ze, lequali &longs;ono in G & in L &longs;o&longs; <lb/>&longs;ero eguali fra loro, & poi &longs;epara-<lb/>tamente dai pe&longs;i minori, è co&longs;a chia<lb/>ra, che le dette po&longs;&longs;anze non &longs;areb<lb/>bono &longs;ufficienti à &longs;o&longs;tenere quei pe&longs;i <lb/>che &longs;e que&longs;te po&longs;&longs;anze &longs;aranno mag <lb/>giori, egli è manife&longs;to, che e&longs;&longs;e mo-<lb/>ueranno i pe&longs;i. </s> <s id="id.2.1.731.3.0"> & co&longs;i la po&longs;&longs;anza <lb/>in L col pe&longs;o H venirà ad e&longs;&longs;e-<lb/>re nella proportione mede&longs;ima, co-<lb/>me la po&longs;&longs;anza in G col pe&longs;o A.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.732.0.0" xlink:href="figures-it/134_01.jpg"></figure> <p type="main" id="id.2.1.733.0.0"> <s id="id.2.1.733.1.0"><emph type="italics"/>Ma perche nella dimo&longs;tratione è &longs;tato <lb/>pre&longs;uppo&longs;to che l'a&longs;&longs;etto &longs;i volga in <lb/>torno, ilquale il più delle volte &longs;tà <lb/>immobile, però &longs;tando anche immobile il detto a&longs;&longs;etto dimo&longs;tri&longs;i li&longs;te&longs;&longs;o.<emph.end type="italics"/> </s> </p> <pb id="p.58v" xlink:href="pageimg-it/135.jpg"/> <p type="main" id="id.2.1.735.0.0"> <s id="id.2.1.735.1.0"><emph type="italics"/>Sia la girella della taglia CEF, il cui centro &longs;ia D, & &longs;ia l'a&longs;&longs;etto GHK, il cen-<lb/>tro delquale &longs;ia mede&longs;imamenie D: Tiri&longs;i il diametro CGDKF egualmente <lb/>di&longs;tante dall'orizonte. </s> <s id="id.2.1.735.2.0"> et percioche <expan abbr="mē">men</expan> <lb/>tre la girella &longs;i volge, la circonferenza <lb/>del cerchio CEF &longs;empre va egual-<lb/>mente di&longs;tante alla circonferenza del-<lb/>l'a&longs;&longs;etto GHK: percioche ella &longs;i <lb/>volge intorno à l'a&longs;&longs;etto, & le circonfe<lb/>renze de' cerchi egualmente di&longs;tanti <lb/>hanno il centro mede&longs;imo, &longs;arà il pun-<lb/>to D &longs;empre centro & della girella, <lb/>& dell'a&longs;&longs;etto. </s> <s id="id.2.1.735.3.0"> Per laqual co&longs;a e&longs;&longs;en-<lb/>do DC eguale à DF & DG ad <lb/>e&longs;&longs;o DK, &longs;arà GC ad e&longs;&longs;o KF egua<lb/>le. </s> <s id="id.2.1.735.4.0"> Se dunque nella leua, ouero bilan-<lb/>cia CF &longs;i attaccheranno pe&longs;i eguali, <lb/>contrape&longs;eranno egualmente, peroche <lb/>la di&longs;tanza CG è eguale alla di&longs;tan-<lb/>za KF, & l'a&longs;&longs;etto GHK immobi<lb/>le &longs;erue per centro, ouero per &longs;o&longs;tegno. </s> <s id="id.2.1.735.5.0"> Stando dunque immobile l'a&longs;&longs;etto, &longs;e la <lb/>po&longs;&longs;anza &longs;i metterà in F che &longs;o&longs;tenga il pe&longs;o appiccato in C, &longs;arà la po&longs;&longs;anza <lb/>in F ad e&longs;&longs;o pe&longs;o eguale, ilche era da mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.736.0.0" xlink:href="figures-it/135_01.jpg"></figure> <p type="main" id="id.2.1.737.0.0"> <s id="id.2.1.737.1.0"><emph type="italics"/>Et concio&longs;ia che del tutto &longs;ia il mede&longs;imo, che l'a&longs;&longs;etto ouero &longs;i volga intorno, ò non <lb/>&longs;i volga: però &longs;ialecito nelle co&longs;e, che &longs;i hanno à dire, prendere in loco dello a&longs;&longs;etto <lb/>il centro &longs;olamente.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.738.0.0"> <s id="id.2.1.738.1.0"> PROPOSITIONE II. </s> </p> <p type="main" id="id.2.1.739.0.0"> <s id="id.2.1.739.1.0"> Se la corda &longs;i condurrà intorno alla girella della taglia, che &longs;ia <lb/>legata al pe&longs;o, legando l'vn de' capi &longs;uoi in qualche loco, & <lb/>l'altro &longs;ia pre&longs;o dalla po&longs;&longs;anza, che &longs;o&longs;tiene il pe&longs;o, &longs;arà la po&longs;-<lb/>&longs;anza la metà meno del pe&longs;o. </s> </p> <p type="main" id="id.2.1.740.0.0"> <s id="id.2.1.740.1.0"><emph type="italics"/>Sia il pe&longs;o A. &longs;ia BCD la girella della taglia legata al pe&longs;o, il cui centro &longs;ia E, <lb/>&longs;ia dapoi inuolta d'intorno la girella la corda FBCDG, & legata in F, & &longs;ia <lb/>la po&longs;&longs;anza in G che &longs;o&longs;tiene il pe&longs;o A. Dico che la po&longs;&longs;anza in G è la metà <lb/>meno del pe&longs;o A. Siano le corde FB GD perpendicolari all' orizonte del pun<emph.end type="italics"/> <lb/> <arrow.to.target n="note216"></arrow.to.target><emph type="italics"/>to E, lequali &longs;aranno fra loro egualmente di&longs;tanti: & tocchino le dette corde <lb/>FBGD, il cerchio BCD ne i punti BD: congiunga&longs;i la linea BD ella pa&longs;-<emph.end type="italics"/> <pb id="p.59" xlink:href="pageimg-it/136.jpg"/><emph type="italics"/>&longs;erà per E centro, & &longs;arà egualmente di&longs;tante dall'orìzonte di e&longs;&longs;o centro, &<emph.end type="italics"/> <arrow.to.target n="note217"></arrow.to.target> <lb/><emph type="italics"/>concio&longs;ia che la G po&longs;&longs;anza debba &longs;o&longs;tenere il pe&longs;o A con la taglia; bi&longs;ogna, <lb/>che la corda &longs;ia legata dal'vno de' capi, come in<emph.end type="italics"/> F, <emph type="italics"/>&longs;i fattamente, che F fac-<lb/>cia re&longs;i&longs;tenza egualmente almeno alla po&longs;&longs;anza, ch'è in G, altramente e&longs;&longs;a po&longs;&longs;an<lb/>za in G non potrebbe à modo alcuno &longs;o&longs;tenere il pe&longs;o. </s> <s id="id.2.1.740.2.0"> Et perche la po&longs;&longs;anza<emph.end type="italics"/> <lb/> <arrow.to.target n="fig88"></arrow.to.target> <lb/><emph type="italics"/>&longs;o&longs;tiene la girella mediante la corda, & la girella &longs;o&longs;tiene la parte re&longs;tante della <lb/>taglia mediante l'a&longs;&longs;etto, allaqual taglia il pe&longs;o è appiccato, pe&longs;erà que&longs;ta parte del-<lb/>la taglia nell'a&longs;&longs;etto, cioè nel centro E: onde il pe&longs;o A pe&longs;erà &longs;imilmente nel me <lb/>de&longs;imo centro E, come &longs;e egli fo&longs;&longs;e appiccato in E. Po&longs;ta dunque la po&longs;&longs;auza <lb/>che stà in G doue è D (perche egli è totalmente il mede&longs;imo) &longs;arà BD come <lb/>vna lèua, il cui &longs;o&longs;tegno &longs;arà B, & il pe&longs;o attaccato in E, & la po&longs;&longs;anzain D: <lb/>& e&longs;&longs;endo la corda FB immobile, conueneuolmente il B puote &longs;eruire per &longs;o-<lb/>&longs;tegno. </s> <s id="id.2.1.740.3.0"> Ma ciò più chiaramente apparerà dapoi. </s> <s id="id.2.1.740.4.0"> Hora percioche la po&longs;&longs;anza al<emph.end type="italics"/> <arrow.to.target n="note218"></arrow.to.target> <lb/><emph type="italics"/>pe&longs;o ha la proportione mede&longs;ima, che hà BE à BD, & BE in proportione <lb/>è la metà manco di BD: dunque la po&longs;&longs;anza che è in G &longs;arà la metà meno del <lb/>pe&longs;o A. Che bi&longs;ognaua dimo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig88" place="text" xlink:href="figures-it/136_01.jpg"></figure> <p type="margin" id="id.2.1.742.0.0"> <s id="id.2.1.742.1.0"> <margin.target id="note216"></margin.target><emph type="italics"/>Per la &longs;esta dell'vndecimo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.743.0.0"> <s id="id.2.1.743.1.0"> <margin.target id="note217"></margin.target><emph type="italics"/>Per la procedense.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.744.0.0"> <s id="id.2.1.744.1.0"> <margin.target id="note218"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>di questo nella leua.<emph.end type="italics"/> </s> </p> <pb id="p.59v" xlink:href="pageimg-it/137.jpg"/> <p type="main" id="id.2.1.745.0.0"> <s id="id.2.1.745.1.0"><emph type="italics"/>Que&longs;to dunque &longs;tà nell'i&longs;te&longs;&longs;o modo con vna corda &longs;ola FBCDG condotta intor-<lb/>no alla girella, come &longs;e &longs;o&longs;&longs;ero due corde BF GD legate alla leua BD, il cui<emph.end type="italics"/> <lb/> <arrow.to.target n="fig89"></arrow.to.target> <lb/><emph type="italics"/>&longs;o&longs;tegno &longs;arà B, & il pe&longs;o fo&longs;&longs;e attaccato in E & la po&longs;&longs;anza, che lo &longs;o&longs;tiene <lb/>&longs;o&longs;&longs;e in D, ouero in G che è l'iste&longs;&longs;o.<emph.end type="italics"/> </s> </p> <figure id="fig89" place="text" xlink:href="figures-it/137_01.jpg"></figure> <p type="head" id="id.2.1.747.0.0"> <s id="id.2.1.747.1.0"> COROLLARIO I. </s> </p> <p type="main" id="id.2.1.748.0.0"> <s id="id.2.1.748.1.0"> Da que&longs;to dunque è manife&longs;to, che il pe&longs;o è &longs;o&longs;tenuto à que&longs;to <lb/>modo da po&longs;&longs;anza minore in proportlone della metà meno, <lb/>di quel che &longs;arebbe &longs;enza aiuto veruno di cotale taglia. </s> </p> <pb id="p.60" xlink:href="pageimg-it/138.jpg"/> <p type="main" id="id.2.1.750.0.0"> <s id="id.2.1.750.1.0"><emph type="italics"/>Come &longs;ia il pe&longs;o H eguale al <lb/>pe&longs;o A, alquale &longs;ia lega-<lb/>tala corda KL, & la po&longs;-<lb/>&longs;anza, che è in L &longs;o&longs;ten-<lb/>gail pe&longs;o H, &longs;arà la po&longs;-<lb/>&longs;anza in L &longs;eparatamente <lb/>eguale al pe&longs;o H, & al <lb/>pe&longs;o A; ma la po&longs;&longs;an-<lb/>za, che è in G in propor-<lb/>tione è la metà manco del <lb/>pe&longs;o A. Per laqual co&longs;a <lb/>la po&longs;&longs;anzache è in G &longs;a-<lb/>rà la metà meno in propor-<lb/>tione della po&longs;&longs;anza, che è <lb/>in L, & in que&longs;to modo <lb/>ne gli altri tutti di que&longs;ta <lb/>maniera &longs;i potrà ritrouare <lb/>la proportione.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.751.0.0" xlink:href="figures-it/138_01.jpg"></figure> <p type="head" id="id.2.1.752.0.0"> <s id="id.2.1.752.1.0"> COROLLARIO II. </s> </p> <p type="main" id="id.2.1.753.0.0"> <s id="id.2.1.753.1.0"> Egli è manife&longs;to ancora, &longs;e &longs;aranno due po&longs;&longs;anze l'vna in G & <lb/>l'altra in F, lequali &longs;o&longs;tengano il pe&longs;o A, che l'vna, & l'al-<lb/>tra in&longs;ieme &longs;aranno eguali al pe&longs;o A, & cia&longs;cheduna di loro <lb/>&longs;o&longs;terrà la metà del pe&longs;o A. </s> </p> <p type="main" id="id.2.1.754.0.0"> <s id="id.2.1.754.1.0"><emph type="italics"/>Et que&longs;to è mani&longs;e&longs;to dal terzo & dal quarto corollario del &longs;econdo di que&longs;to nel trat<lb/>tato della leua.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.755.0.0"> <s id="id.2.1.755.1.0"> COROLLARIO III. </s> </p> <p type="main" id="id.2.1.756.0.0"> <s id="id.2.1.756.1.0"> Oltre à ciò que&longs;to parimente &longs;i fa noto, perche cioè la corda <lb/>debba e&longs;&longs;ere legata nell'vno de' capi. </s> </p> <pb id="p.60v" xlink:href="pageimg-it/139.jpg"/> <p type="head" id="id.2.1.757.0.0"> <s id="id.2.1.757.1.0"> PROPOSITIONE III. </s> </p> <p type="main" id="id.2.1.758.0.0"> <s id="id.2.1.758.1.0"> Se à cia&longs;cuna dell'vna, & l'altra girella delle due taglie, l'vna del <lb/>le quali &longs;ia po&longs;ta di &longs;opra, & l'altra di &longs;otto, & que&longs;ta &longs;ia lega-<lb/>ta al pe&longs;o; &longs;arà condotta intorno la corda: legando l'vno de' <lb/>capi in qualche loco, & l'altro &longs;ia tenuto dalla po&longs;&longs;anza, che <lb/>&longs;o&longs;tiene il pe&longs;o, &longs;arà la po&longs;&longs;anza la metà meno del pe&longs;o. </s> </p> <p type="main" id="id.2.1.759.0.0"> <s id="id.2.1.759.1.0"><emph type="italics"/>Sia il pe&longs;o A, &longs;ia BCD la girella della <lb/>taglia, che &longs;ia legata al pe&longs;o A, il cui <lb/>centro &longs;ia K, & EFG &longs;ia la girella <lb/>della taglia appiccata di opra, il cui cen-<lb/>tro &longs;ia H, dapoi &longs;ia condotta intorno <lb/>le girelle la corda LBCDMEFGN <lb/>laquale &longs;ia legata in L, & &longs;ia la po&longs;-<lb/>&longs;anza, che &longs;o&longs;tiene il pe&longs;o A in N. <lb/> Dico la po&longs;&longs;anza, che &longs;ta in N e&longs;&longs;e-<lb/>re la metà meno del pe&longs;o A. Percio-<lb/>che &longs;e la po&longs;&longs;anza, che &longs;o&longs;tiene il pe&longs;o<emph.end type="italics"/> <lb/> <arrow.to.target n="note219"></arrow.to.target><emph type="italics"/>A fo&longs;&longs;e collocata doue &longs;ta M, &longs;areb-<lb/>be per certo la po&longs;&longs;anza in M la metà<emph.end type="italics"/> <lb/> <arrow.to.target n="note220"></arrow.to.target><emph type="italics"/>meno del pe&longs;o A: & alla po&longs;&longs;anza in <lb/>M è eguale la forza di N, percioche <lb/>egli è come &longs;e la po&longs;&longs;anza in M &longs;o&longs;te-<lb/>ne&longs;&longs;e la metà del pe&longs;o A &longs;enza taglia, <lb/>alquale egualmente contrape&longs;a il pe&longs;o che <lb/>è in N per e&longs;&longs;ere eguale alla metà del <lb/>pe&longs;o A. Per laqual co&longs;ala forza in N <lb/>che è alla metà del pe&longs;o A eguale, &longs;o-<lb/>&longs;tenirà e&longs;&longs;o A. La po&longs;&longs;anza dunque in <lb/>N che &longs;o&longs;tiene il pe&longs;o A, è la metà <lb/>meno di e&longs;&longs;o A. che bi&longs;ognaua mo-<lb/>&longs;trare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.760.0.0"> <s id="id.2.1.760.1.0"> <margin.target id="note219"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.761.0.0"> <s id="id.2.1.761.1.0"> <margin.target id="note220"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.762.0.0" xlink:href="figures-it/139_01.jpg"></figure> <pb id="p.61" xlink:href="pageimg-it/140.jpg"/> <p type="main" id="id.2.1.764.0.0"> <s id="id.2.1.764.1.0"><emph type="italics"/>Ma &longs;e, come nella &longs;econda figura, la cor <lb/>da BCDEFGHKL &longs;arà inuolta <lb/>d'intorno à le girelle, & legata in <lb/>B: & la po&longs;&longs;anza in L &longs;o&longs;tenga il <lb/>pe&longs;o A, &longs;arà &longs;imilmente la po&longs;&longs;an<lb/>za in L la metà meno del pe&longs;o: <lb/>Peroche la girella della taglia di &longs;o-<lb/>pra, & la taglia i&longs;te&longs;&longs;a &longs;ono del tut-<lb/>to inutili: & è il mede&longs;imo, come &longs;e <lb/>la corda fo&longs;&longs;e legata in F, & che <lb/>la po&longs;&longs;anza in L &longs;o&longs;tene&longs;&longs;e il pe&longs;o <lb/>con la &longs;ola taglia legata al pe&longs;o, la <lb/>qual po&longs;&longs;anza è &longs;tata dimo&longs;trata e&longs;-<lb/>&longs;ere la metà meno del pe&longs;o A.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.765.0.0" xlink:href="figures-it/140_01.jpg"></figure> <p type="head" id="id.2.1.766.0.0"> <s id="id.2.1.766.1.0"> COROLLARIO. </s> </p> <p type="main" id="id.2.1.767.0.0"> <s id="id.2.1.767.1.0"> Seguita da que&longs;te co&longs;e, che &longs;e &longs;aranno due po&longs;&longs;anze in BL, am-<lb/>bedue tra loro &longs;aranno eguali. </s> </p> <p type="main" id="id.2.1.768.0.0"> <s id="id.2.1.768.1.0"><emph type="italics"/>Perciocheogn'vna di loro da per &longs;e è la metàmeno di e&longs;&longs;o A.<emph.end type="italics"/> </s> </p> <pb id="p.61v" xlink:href="pageimg-it/141.jpg"/> <p type="head" id="id.2.1.770.0.0"> <s id="id.2.1.770.1.0"> PROPOSITIONE IIII. </s> </p> <p type="main" id="id.2.1.771.0.0"> <s id="id.2.1.771.1.0"> Sia la leua AB, il cui &longs;o&longs;tegno &longs;ia A, laqual leua &longs;ia diui&longs;a in <lb/>due parti eguali in D, & &longs;ia il pe&longs;o C appiccato in D, & <lb/>&longs;iano due po&longs;&longs;anze eguali in BD, che &longs;o&longs;tengano il pe&longs;o C. <lb/> Dico, che ogn'vna di que&longs;te po&longs;&longs;anze po&longs;te in BD è vn ter-<lb/>zo del pe&longs;o C. </s> </p> <p type="main" id="id.2.1.772.0.0"> <s id="id.2.1.772.1.0"><emph type="italics"/>Hor percioche vna delle due po&longs;&longs;anze è collocata in D, & il pe&longs;o C &longs;tà appiccato <lb/>all'i&longs;te&longs;&longs;o punto D. La po&longs;&longs;anza in D &longs;o&longs;ienirà la parte del pe&longs;o C, che &longs;arà <lb/>eguale ad e&longs;&longs;a po&longs;&longs;an-<lb/>za D. Per laqual co <lb/>&longs;ala po&longs;&longs;anza in B &longs;o <lb/>&longs;tenirà l'altra parte re <lb/>&longs;tante, laqual parte &longs;a <lb/>rà il doppio <expan abbr="tāto">tanto</expan>, quan <lb/>to è la po&longs;&longs;anza di B, <lb/>e&longs;&longs;endo che il pe&longs;o ver <lb/>&longs;o la po&longs;&longs;anza ha la <lb/>proportione i&longs;te&longs;&longs;a, che<emph.end type="italics"/> <lb/> <arrow.to.target n="fig90"></arrow.to.target> <lb/><emph type="italics"/>ha AB ad AD: & le po&longs;&longs;anze po&longs;te in BD &longs;ono eguali, adunque la po&longs;-<lb/>&longs;anza, che è in B &longs;o&longs;tenirà il doppio più di quello, che &longs;o&longs;tenirà la po&longs;&longs;anza, che è <lb/>in D. Diuida&longs;i dunque il pe&longs;o C in due parti, l'vna delle quali &longs;ia il doppio del-<lb/>l'altra: ilche &longs;i farà, &longs;e lo diuideremo in tre parti eguali EFG, & all'hora FG <lb/>&longs;arà il doppio di E. Co&longs;i la po&longs;&longs;anzain D &longs;o&longs;tenirà la parte E, & la po&longs;&longs;anza <lb/>in B le altre due parti FG. Ambedue dunque le po&longs;&longs;anze po&longs;te in BD tra <lb/>loro eguali <expan abbr="&longs;o&longs;terrãno">&longs;o&longs;terranno</expan> in&longs;ieme tutto il pe&longs;o C. & perche la po&longs;&longs;anza in D &longs;o&longs;tie-<lb/>ne la parte E, laquale è la terza parte del pe&longs;o C, & ad e&longs;&longs;o è eguale, &longs;arà la po&longs;-<lb/>&longs;anza in D vn terzo del pe&longs;o C: & concio&longs;ia che la po&longs;&longs;anza di B &longs;o&longs;tenga le <lb/>parti FG, la po&longs;&longs;anza dellequali po&longs;ta in B è la metà meno: &longs;arà la po&longs;&longs;anza <lb/>in B all'vna delle parti FG, come alla G eguale. </s> <s id="id.2.1.772.2.0"> & il G è la terza parte <lb/>del pe&longs;o C. La po&longs;&longs;anza dunque in B &longs;arà il terzo del pe&longs;o C. Cia&longs;cuna delle <lb/>po&longs;&longs;anze dunque in BD è vnterzo del pe&longs;o C, che bi&longs;ognaua dimo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig90" place="text" xlink:href="figures-it/141_01.jpg"></figure> <pb id="p.62" xlink:href="pageimg-it/142.jpg"/> <p type="main" id="id.2.1.775.0.0"> <s id="id.2.1.775.1.0"><emph type="italics"/>Et &longs;e fo&longs;&longs;ero due leue AB EF diui&longs;e in due parti eguali in GD, i &longs;o&longs;tegni delle-<lb/>quali fo&longs;&longs;ero AF, & il pe&longs;o C fo&longs;&longs;e appiccato all'vna, & l'altra leua in DG<emph.end type="italics"/> <lb/> <arrow.to.target n="fig91"></arrow.to.target> <lb/><emph type="italics"/>&longs;i fattamente, però che pe&longs;a&longs;&longs;e egualmente nell'vna, & l'altra: & &longs;o&longs;&longs;ero due po&longs;-<lb/>&longs;anze eguali in BG. Si dimo&longs;trerà con ragione in tutto mede&longs;ima, che ogn'vna <lb/>delle po&longs;&longs;anze po&longs;tein B & G è vn terzo del pe&longs;o C.<emph.end type="italics"/> </s> </p> <figure id="fig91" place="text" xlink:href="figures-it/142_01.jpg"></figure> <p type="head" id="id.2.1.777.0.0"> <s id="id.2.1.777.1.0"> PROPOSITIONE V. </s> </p> <p type="main" id="id.2.1.778.0.0"> <s id="id.2.1.778.1.0"> Se all'vna & l'altra, di cia&longs;cuna girella di due taglie, l'vna delle <lb/>quali &longs;ia po&longs;ta di &longs;opra, & l'altra di &longs;otto, & legata al pe&longs;o; &longs;a-<lb/>rà condotta intornò la corda, legando vno de'&longs;uoi capi alla <lb/>taglia di &longs;otto, & l'altro &longs;ia tenuto dalla po&longs;&longs;anza, che &longs;o&longs;tie-<lb/>ne il pe&longs;o: &longs;arà la po&longs;&longs;anza vn terzo del pe&longs;o. </s> </p> <pb id="p.62v" xlink:href="pageimg-it/143.jpg"/> <p type="main" id="id.2.1.780.0.0"> <s id="id.2.1.780.1.0"><emph type="italics"/>Sia il pe&longs;o A, &longs;ia BCD la girella della taglia legata al pe&longs;o A, il cui centro &longs;ia <lb/>E, & &longs;ia FGH l'altra girella della taglia appiccata di &longs;opra, il cui centro &longs;ia <lb/>K: &longs;ia condotta intorno alle girelle la corda LFGHBCDM, laquale &longs;ia lega<lb/>ta alla taglia di &longs;otto in L; & la po&longs; <lb/>&longs;anza, che &longs;o&longs;tiene il pe&longs;o A &longs;ia in <lb/>M. Dico che la po&longs;&longs;anza in M è vn <lb/>terzo del pe&longs;o A. Siano tirate le li-<lb/>nee FH BD per li centri KE egual<lb/>mente di&longs;tanti dall'orizonte, &longs;i come <lb/>nelle precedenti è detto. </s> <s id="id.2.1.780.2.0"> Hor percio-<lb/>che la corda FL &longs;o&longs;tiene la taglia di <lb/>&longs;otto, laquale &longs;o&longs;tiene la girella nel &longs;uo <lb/>centro E: &longs;arà la corda di L come <lb/>po&longs;&longs;anza che &longs;o&longs;tiene la girella, tanto <lb/>quanto &longs;e fo&longs;&longs;e in e&longs;&longs;o E centro: & <lb/>la po&longs;&longs;anza di M è come &longs;e &longs;te&longs;&longs;e in<emph.end type="italics"/> <lb/> <arrow.to.target n="note221"></arrow.to.target><emph type="italics"/>D; &longs;i farà dunque DB come leua, il <lb/>cui &longs;o&longs;tegno &longs;arà B: ma il pe&longs;o A, <lb/>come di &longs;opra fù dimo&longs;trato, appicca-<lb/>to in E viene &longs;o&longs;tenuto da due po&longs;-<lb/>&longs;anze, l'vna po&longs;ta in D, & l'altra in <lb/>E. & concio&longs;ia, che nel &longs;o&longs;tenere i <lb/>pe&longs;i &longs;tiano le leue FH BD immobi-<lb/>li, &longs;e li pe&longs;i &longs;aranno appiccati alle cor-<emph.end type="italics"/> <lb/> <arrow.to.target n="note222"></arrow.to.target><emph type="italics"/>de FL HB &longs;aranno que&longs;ti i&longs;te&longs;&longs;i egua<lb/>li, per hauere laleua FH il &longs;o&longs;tegno <lb/>nel mezo; altramente dall'vna delle <lb/>parti &longs;i farebbe il mouimento à ba&longs;&longs;o, <lb/>co&longs;a che tuttauia non accade; Adun-<lb/>que tanto &longs;o&longs;tiene la corda FL, quan<lb/>to la HB. Di più percioche dal me-<lb/>zo della leua BD il pe&longs;o pende at-<lb/>taccato, però &longs;e fo&longs;&longs;ero due po&longs;&longs;anze <lb/>in BD che &longs;o&longs;tene&longs;&longs;ero il pe&longs;o, &longs;areb-<lb/>bon fra loro eguali: & benche la cor-<emph.end type="italics"/> <lb/> <arrow.to.target n="note223"></arrow.to.target><emph type="italics"/>da FL &longs;o&longs;tenga e&longs;&longs;a ancora il pe&longs;o, <lb/>poiche ella &longs;ta in loco de la po&longs;&longs;anza<emph.end type="italics"/> <lb/> <arrow.to.target n="note224"></arrow.to.target><emph type="italics"/>E, nondimeno percioche &longs;o&longs;tiene da <lb/>quel mede&longs;imo punto, doue è appicca-<lb/>to il pe&longs;o, non farà però che le po&longs;-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig92"></arrow.to.target> <lb/><emph type="italics"/>&longs;anze, lequali &longs;ono in BD non &longs;iano traloro eguali, peroche aiuta tanto all'v-<lb/>na, quanto all'altra. </s> <s id="id.2.1.780.3.0"> Ma le po&longs;&longs;anze che &longs;ono in BD &longs;ono le i&longs;te&longs;&longs;e, come &longs;e<emph.end type="italics"/> <pb id="p.63" xlink:href="pageimg-it/144.jpg"/><emph type="italics"/>fu&longs;&longs;ero in HM. Per laqual co&longs;a tanto &longs;o&longs;terrà la corda MD quanto la HB: ma <lb/>co&longs;i &longs;o&longs;tiene HB come FL; adunque la corda MD co&longs;i &longs;o&longs;tenirà, come FL, <lb/>cioè come &longs;e in D & in L fo&longs;&longs;ero appiccati pe&longs;i eguali. </s> <s id="id.2.1.780.4.0"> Concio&longs;ia co&longs;a dunque, <lb/>che pe&longs;i eguali &longs;ian &longs;o&longs;tenuti da po&longs;&longs;anze vguali, le po&longs;&longs;anze in ML &longs;aranno egua <lb/>li, delle quali è in tutto vna ragione iste&longs;&longs;a, come &longs;e ambedue fo&longs;&longs;ero in DE. <lb/> Onde, e&longs;&longs;endo che il pe&longs;o A &longs;tia attaccato nel mezo della leua BD, & che due <lb/>po&longs;&longs;anze po&longs;te in DE &longs;o&longs;tenente il pe&longs;o &longs;iano eguali: &longs;arà B il &longs;o&longs;tegno, & <lb/>cia&longs;cheduna po&longs;&longs;anza po&longs;ta in DE ouero in ML &longs;arà vn terzo del pe&longs;o A. <lb/> Adunque la po&longs;&longs;anza in M &longs;o&longs;tenente il pe&longs;o &longs;arà vn terzo del pe&longs;o A. che<emph.end type="italics"/> <arrow.to.target n="note225"></arrow.to.target> <lb/><emph type="italics"/>bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig92" place="text" xlink:href="figures-it/143_01.jpg"></figure> <p type="margin" id="id.2.1.782.0.0"> <s id="id.2.1.782.1.0"> <margin.target id="note221"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.783.0.0"> <s id="id.2.1.783.1.0"> <margin.target id="note222"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.784.0.0"> <s id="id.2.1.784.1.0"> <margin.target id="note223"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 3. <emph type="italics"/>corollario di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.785.0.0"> <s id="id.2.1.785.1.0"> <margin.target id="note224"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo della leua.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.786.0.0"> <s id="id.2.1.786.1.0"> <margin.target id="note225"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 4. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.787.0.0"> <s id="id.2.1.787.1.0"> COROLLARIO. </s> </p> <p type="main" id="id.2.1.788.0.0"> <s id="id.2.1.788.1.0"> Da que&longs;to è manife&longs;to, che ogn'vna delle corde MD FL HB <lb/>&longs;o&longs;tiene la terza parte del pe&longs;o A. </s> </p> <pb id="p.63v" xlink:href="pageimg-it/145.jpg"/> <p type="main" id="id.2.1.789.0.0"> <s id="id.2.1.789.1.0"><emph type="italics"/>Oltre à ciò &longs;e da M &longs;ar à la corda portata intor-<lb/>no ad vn'altra girella po&longs;ta più &longs;u nella ta-<lb/>glia, che &longs;imilmente &longs;ia attaccata di &longs;opra, il <lb/>cui centro &longs;ia N &longs;i fattamente che peruen<lb/>ga in O, & iui &longs;ia tenuta dalla po&longs;&longs;anza; &longs;a <lb/>rà la po&longs;&longs;anza che in O &longs;o&longs;tiene il pe&longs;o A <lb/>parimente vn terzo del pe&longs;o. </s> <s id="id.2.1.789.2.0"> Percioche la <lb/>corda MD &longs;o&longs;tiene tanto di pe&longs;o, come &longs;e in <lb/>D fo&longs;&longs;e appiccato il pe&longs;o eguale alla terza<emph.end type="italics"/> <lb/> <arrow.to.target n="note226"></arrow.to.target><emph type="italics"/>parte del pe&longs;o A, alla quale è pari la po&longs;-<lb/>&longs;anza in O ad e&longs;&longs;a eguale, cioè vnterzo del <lb/>pe&longs;o A. La po&longs;&longs;anza dunque in O è vn <lb/>terzo del pe&longs;o A.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.790.0.0"> <s id="id.2.1.790.1.0"> <margin.target id="note226"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.791.0.0"> <s id="id.2.1.791.1.0"><emph type="italics"/>Et accioche non &longs;i ritorni à dire &longs;pe&longs;&longs;e volte il <lb/>mede&longs;imo, egli fà me&longs;tiero &longs;apere, che lapo&longs; <lb/>&longs;anza in O è &longs;empre eguale à quella, che <lb/>&longs;ta in M. come &longs;arebbe à dire, &longs;e la po&longs;&longs;an-<lb/>za in M fo&longs;&longs;e vn quarto, ouero vn quinto, <lb/>ò &longs;imile co&longs;a di e&longs;&longs;o pe&longs;o, la po&longs;&longs;anza parimen<lb/>te in O &longs;arà vn quarto, ouero vn quinto, <lb/>& co&longs;i di mano in mano dell'i&longs;te&longs;&longs;o pe&longs;o, nel <lb/>modo che è di&longs;po&longs;ta la po&longs;&longs;anza di M.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.792.0.0" xlink:href="figures-it/145_01.jpg"></figure> <p type="main" id="id.2.1.793.0.0"> <s id="id.2.1.793.1.0"> Potrebbbe for&longs;e alcuno dubitare in alcune dimo&longs;trationi delle taglie come in que&longs;ta <lb/>quinta propo&longs;itione, tolta da me per e&longs;&longs;empio per e&longs;&longs;ere piu &longs;chietta delle altre, <lb/>che in fatto con la e&longs;perientia non riu&longs;ci&longs;&longs;ero in proportione le forze a' pe&longs;i, co-<pb id="p.64" xlink:href="pageimg-it/146.jpg"/>me la ragione dimo&longs;tra; peroche pre&longs;upponendo &longs;i nelle dimo&longs;trationi matemati<lb/>che le linee &longs;enza larghezza, & profondità, & co&longs;i le altre co&longs;e imaginando &longs;i &longs;e-<lb/>parate dalla materia, ageuolmente &longs;i per&longs;uadiamo e&longs;&longs;ere vere come dicono. </s> <s id="id.2.1.793.2.0"> Ma <lb/>la e&longs;perientia poi molte volte mo&longs;tra diuer&longs;ità, & &longs;i trouiamo ingannati, facendo <lb/>la materia gran demente variare le co&longs;e. </s> <s id="id.2.1.793.3.0"> In que&longs;ta propo&longs;itione &longs;i narra, che rauol<lb/>gendo d'intorno à due girelle di due taglie vna corda, & quel che &longs;egue, la forza <lb/>&longs;arà vn terzo del pe&longs;o, cioè &longs;e il pe&longs;o &longs;ara trecento, egli verrà &longs;o&longs;tenuto dalla po&longs;<lb/>&longs;anza di cento. </s> <s id="id.2.1.793.4.0"> Direbbe alcuno ciò e&longs;&longs;ere dubbio&longs;o, peroche le girelle, gli a&longs;&longs;etti <lb/>&longs;uoi, le funi, & il pe&longs;o della taglia di &longs;otto fanno re&longs;i&longs;tenza alla forza, & grauano <lb/>sì, che ella non potrà &longs;o&longs;tenere il pe&longs;o. </s> <s id="id.2.1.793.5.0"> Si ri&longs;ponde che que&longs;te co&longs;e ben farebbo-<lb/>no re&longs;i&longs;tenza nel mouere il pe&longs;o, ma non già nel &longs;o&longs;tentarlo: & bi&longs;ogna notaro <lb/>con diligenza che l'autore in que&longs;te dimo&longs;trationi parla &longs;empre del &longs;o&longs;tenere &longs;o-<lb/>lamente con le forze i pe&longs;i che non calino al ba&longs;&longs;o, non del mouere. </s> <s id="id.2.1.793.6.0"> Però con-<lb/>&longs;ideri&longs;i, che quando li pe&longs;i &longs;i hanno da far mouere con le po&longs;&longs;anze, allhora le gi-<lb/>relle, & gli altri impedimenti faranno re&longs;i&longs;tenza; ma quando &longs;i ha da far &longs;olamen-<lb/>te che il pe&longs;o &longs;tia fermo, & habbia il &longs;uo contrape&longs;o &longs;emplicemente &longs;enza porre <lb/>in con&longs;ideratione altri ri&longs;petti, che è officio della po&longs;&longs;anza &longs;o&longs;tenente; all'horz <lb/>nè le girelle, nè altro danno re&longs;i&longs;tenza veruna, & la proua fondata &longs;u la ragione <lb/>torna &longs;empre per eccellentia, anzi pare che quanto piu re&longs;i&longs;tenza vi &longs;ia, tanto piu <lb/>facilmente la forza &longs;o&longs;tenga. </s> <s id="id.2.1.793.7.0"> Auertendo con tutto ciò, che nel fare la e&longs;perienza <lb/>bi&longs;ogna hauere riguardo alla taglia di &longs;otto, & alla corda, lequali hanno la &longs;ua <lb/>grauezza &longs;i fattamente, che &longs;e il pe&longs;o come nell'e&longs;&longs;empio propo&longs;to, &longs;arà trecento <lb/>libre, & la forza cento, & la taglia di &longs;otto con la &longs;ua fune quattordici, è me&longs;tieri <lb/>che alla po&longs;&longs;anza di M &longs;i aggiungano quattro libre, & due terzi di forza, ac-<lb/>cioche po&longs;&longs;a &longs;o&longs;tenere tutto il pe&longs;o, & co&longs;i verrà ad e&longs;&longs;ere in M po&longs;&longs;anza vn ter-<lb/>zo giu&longs;tamente del pe&longs;o. </s> <s id="id.2.1.793.8.0"> Ma per &longs;apere quanta forza bi&longs;ogni aggiungere alla po&longs; <lb/>&longs;anza, accioche per ri&longs;petto alla taglia di &longs;otto, & alla fune, &longs;o&longs;tenghi il pe&longs;o tut-<lb/>to, faccia&longs;i que&longs;ta ragione. </s> <s id="id.2.1.793.9.0"> La taglia di &longs;otto con parte della fune, per gratia di <lb/>e&longs;&longs;empio, è quattordici libre, il pe&longs;o è trecento, & la po&longs;&longs;anza cento. </s> <s id="id.2.1.793.10.0"> Hor per <lb/>la regola detta del tre. </s> <s id="id.2.1.793.11.0"> Se trecento danno cento, che daranno quattordici? </s> <s id="id.2.1.793.12.0"> Tro-<lb/>ueran&longs;i quattro libre, & due terzi da e&longs;&longs;ere aggiunte alla po&longs;&longs;anza di M, per <lb/>&longs;o&longs;tenere il pe&longs;o A. “Laqual co&longs;a tocca in &longs;o&longs;tanza l'auttore più à ba&longs;&longs;o <lb/>dicendo.&longs;rdquo; & &longs;i come habbiamo ciò con&longs;iderato nella decimaquinta, & quel, <lb/>che &longs;egue. </s> <s id="id.2.1.793.13.0"> ilqual loco bi&longs;ogna intendere in que&longs;ta maniera, che le taglie non <lb/>&longs;i deuono pigliare ad vn'i&longs;te&longs;&longs;o modo &longs;empre, ma diuer&longs;amente, come graua-<lb/>no, ilche na&longs;ce dall'e&longs;&longs;ere in vari luoghi, & le po&longs;&longs;anze, & i pe&longs;i collocati, & fer-<lb/>mate le taglie. </s> <s id="id.2.1.793.14.0"> Hor nella &longs;econda propo&longs;itione di que&longs;to trattato has&longs;i da inten-<lb/>dere la po&longs;&longs;anza e&longs;&longs;ere la meta meno del pe&longs;o, prendendo perlo pe&longs;o, & il pe&longs;o, <lb/>& la taglia di &longs;otto in&longs;ieme, à cui &longs;tà attaccato, come &longs;i vede chiaro nella dimo&longs;tra <lb/>tione della detta &longs;econda propo&longs;itione, doue &longs;i proua che la po&longs;&longs;anza &longs;o&longs;tiene la gi<lb/>rella, laquale &longs;o&longs;tiene anche il re&longs;to della taglia nell'a&longs;&longs;etto, alla qual taglia è attac-<lb/>cato il pe&longs;o, oue &longs;i cono&longs;ce e&longs;pre&longs;&longs;o, che la taglia, & il pe&longs;o s'hanno à pigliare <lb/>per tutto il pe&longs;o. </s> <s id="id.2.1.793.15.0"> Per la qual co&longs;a, &longs;e in quel ca&longs;o il pe&longs;o in&longs;ieme con la taglia pe-<lb/>&longs;eranno vinti, la po&longs;&longs;anza che gli &longs;o&longs;tenterà &longs;arà dieci. </s> <s id="id.2.1.793.16.0"> Et per vn'altro e&longs;&longs;empio <lb/>nella nona propo&longs;itione di que&longs;to nel primo ca&longs;o, &longs;e il pe&longs;o con la taglia di &longs;otto <lb/>pe&longs;eranno vinticinque, la po&longs;&longs;anza &longs;o&longs;tenente &longs;arà cinque. </s> <s id="id.2.1.793.17.0"> & co&longs;i egli è me&longs;tieri <lb/>hauer con&longs;ideratione nelle altre, cioè di&longs;tinguere doue è la grauezza della taglia, <lb/> <pb id="p.64v" xlink:href="pageimg-it/147.jpg"/>quando graua di &longs;otto &longs;olamente, come nelle allegate propo&longs;itioni, & &longs;imili: & <lb/>quando &longs;olamente di &longs;opra, come nelle propo&longs;itioni 17. & 18. & &longs;imili: & quan<lb/>do ambedue le taglie grauano di &longs;opra, & di &longs;otto, come nelle propo&longs;itioni 20. <lb/>22. & 23. & &longs;imili: & quando anche nel'vna taglia, ne l'altra grauano, come nella <lb/>prima propo&longs;itione & nella 19. anzi in e&longs;&longs;a 19. la taglia di &longs;otto aiuta la <expan abbr="po&longs;sãza">po&longs;sanza</expan> ad <lb/>e&longs;&longs;ere piu leggiera: & nel &longs;econdo ca&longs;o dopo il corollario della 16. propo&longs;itione, <lb/>& &longs;imili. </s> <s id="id.2.1.793.18.0"> & oltre à ciò deue&longs;i por mente alle corde ancora, la grauezza delle qua-<lb/>li non hà &longs;empre da e&longs;&longs;ere con&longs;iderata, peroche grauano nelle propo&longs;itioni 15. 17. <lb/>ma non grauano già nella 19. </s> </p> <p type="main" id="id.2.1.794.0.0"> <s id="id.2.1.794.1.0"> Ne parmi etiandio che &longs;i habbia ad hauere punto di riguardo alla picciolezza, & <lb/>grandezza delle girelle po&longs;te nelle taglie, & de gli a&longs;&longs;etti &longs;uoi, credendo che per <lb/>neces&longs;ità habbiano da e&longs;&longs;ere lauorati con mi&longs;ura tale, & proportione co&longs;i accu-<lb/>rata, che mancando da quella non rie&longs;cano le dimo&longs;trationi alla e&longs;perientia; per <lb/>roche, &longs;i come nota l'autore poco appre&longs;&longs;o, ba&longs;ta che con certa conueneuole mi&longs;n <lb/>ra, & proportione le girelle nelle taglie &longs;iano maggiori l'vna dell'altra &longs;i fattamen<lb/>te, che le corde non &longs;i to cchino, & freghino fra loro, & co&longs;i vengano ad impedi<lb/>re i mouimenti delle po&longs;&longs;anze, & de' pe&longs;i. </s> </p> <p type="head" id="id.2.1.795.0.0"> <s id="id.2.1.795.1.0"> PROPOSITIONE VI. </s> </p> <p type="main" id="id.2.1.796.0.0"> <s id="id.2.1.796.1.0"> Siano due leue AB CD diui&longs;e in due parti eguali in EF, li <lb/>&longs;o&longs;tegni delle quali &longs;iano in BD; & &longs;ia il pe&longs;o G in EF ap<lb/>piccato all'vna, & l'altra leua &longs;i fattamente, che pe&longs;i dall'vna, <lb/>& dall'altra egualmente: & &longs;iano due po&longs;&longs;anze in AC egua-<lb/>li, che &longs;o&longs;tengano il pe&longs;o. </s> <s id="id.2.1.796.2.0"> Dico, che ogn'vna delle po&longs;&longs;anze <lb/>in AC è vn quarto del pe&longs;o G. </s> </p> <p type="main" id="id.2.1.797.0.0"> <s id="id.2.1.797.1.0"> <arrow.to.target n="note227"></arrow.to.target><emph type="italics"/>Concio&longs;ia che le po&longs;&longs;anze po-<lb/>&longs;te in AC &longs;o&longs;tengano tut<lb/>to il pe&longs;o G, & la po&longs;&longs;an-<lb/>za di A ver&longs;o la parte del <lb/>pe&longs;o, che &longs;o&longs;tiene, &longs;ia come <lb/>BE à BA, & la po&longs;&longs;an-<lb/>za in C alla parte di e&longs;&longs;o <lb/>G pe&longs;o &longs;o&longs;tenuto da lei &longs;ia <lb/>co&longs;i, come DF à DC, & <lb/>come BE à BA, co&longs;i è <lb/>DF à DC: &longs;ar à lapo&longs;&longs;an <lb/>za po&longs;ta in A ver&longs;o la par <lb/>te del pe&longs;o, che &longs;o&longs;tiene, co-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig93"></arrow.to.target><emph type="italics"/>me la po&longs;&longs;anza di C ver&longs;o la parte di e&longs;&longs;o pe&longs;o, che &longs;o&longs;tiene: & le po&longs;&longs;anze po&longs;ie <lb/>in AC &longs;ono eguali; &longs;aranno dunque le parti del pe&longs;o G eguali, lequali &longs;ono &longs;o-<emph.end type="italics"/> <pb id="p.65" xlink:href="pageimg-it/148.jpg"/><emph type="italics"/>&longs;tenute dalle po&longs;&longs;anze. </s> <s id="id.2.1.797.2.0"> Per laqual co&longs;a cia&longs;cuna po&longs;&longs;anza po&longs;ta in AC &longs;o&longs;terrà <lb/>la metà del pe&longs;o G. Mala po&longs;&longs;anza in A è la metà meno del pe&longs;o, che &longs;o&longs;tie-<lb/>ne; adunque la po&longs;&longs;anza in A &longs;arà per lo mezo della metà, cioè eguale alla quar<lb/>ta portione del pe&longs;o G; & però &longs;arà il quarto del pe&longs;o G, nè altramente &longs;i di-<lb/>mo&longs;trerà la po&longs;&longs;anza in C e&longs;&longs;ere vn quarto dell'i&longs;te&longs;&longs;o pe&longs;o G. che bi&longs;ognaua <lb/>mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig93" place="text" xlink:href="figures-it/147_01.jpg"></figure> <p type="margin" id="id.2.1.799.0.0"> <s id="id.2.1.799.1.0"> <margin.target id="note227"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo nella leua.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.800.0.0"> <s id="id.2.1.800.1.0"><emph type="italics"/>Ma &longs;e &longs;aranno tre leue AB <lb/>CD EF diui&longs;e in due <lb/>parti eguali in GHK, li <lb/>&longs;o&longs;tegni delle quali &longs;iano <lb/>BDF, & il pe&longs;o L &longs;ia <lb/>nell'i&longs;te&longs;&longs;o modo appicca-<lb/>to in GHK: & &longs;iano <lb/>tre po&longs;&longs;anze in ACE <lb/>eguali, che &longs;o&longs;tengano il <lb/>pe&longs;o: &longs;i mo&longs;trerà &longs;imil-<lb/>mente cia&longs;cuna po&longs;&longs;anza <lb/>e&longs;&longs;ere vn &longs;e&longs;to del pe&longs;o <lb/>L: & con questo ordi-<lb/>ne &longs;e fo&longs;&longs;ero quattro le-<lb/>ue, & quattro po&longs;&longs;anze, <lb/>cia&longs;cuna po&longs;&longs;anza &longs;arà <lb/>la ottaua parte del pe&longs;o, & co&longs;i di mano in mano in infinito.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.801.0.0" xlink:href="figures-it/148_01.jpg"></figure> <p type="head" id="id.2.1.802.0.0"> <s id="id.2.1.802.1.0"> PROPOSITIONE VII. </s> </p> <p type="main" id="id.2.1.803.0.0"> <s id="id.2.1.803.1.0"> Se à tre girelle di due taglie, l'vna delle quali po&longs;ta di &longs;opra hab <lb/>bia vna &longs;ola girella, & l'altra di &longs;otto ne habbia due, & &longs;ia lega <lb/>ta al pe&longs;o; &longs;ia po&longs;ta d'intorno la corda; legando l'vn de' capi <lb/>&longs;uoi in qualche loco, & l'altro &longs;ia tenuto dalla po&longs;&longs;anza, che <lb/>&longs;o&longs;tiene il pe&longs;o. </s> <s id="id.2.1.803.2.0"> La po&longs;&longs;anza &longs;arà vn quarto del pe&longs;o. </s> </p> <pb id="p.65v" xlink:href="pageimg-it/149.jpg"/> <p type="main" id="id.2.1.805.0.0"> <s id="id.2.1.805.1.0"><emph type="italics"/>Sia il pe&longs;o A: &longs;iano le tre girelle, il centro dellequali &longs;ia BCD: & la girella, il <lb/>cui centro è D, &longs;ia della taglia appiccata di <lb/>&longs;opra: ma quelle girelle, il cui centro è in B <lb/>C &longs;iano della taglia legata al pe&longs;o A: & <lb/>la corda EFGHKLNOP &longs;ia condotta <lb/>intorno à tutte le girelle, & legata in E: & <lb/>&longs;ia la forza che &longs;o&longs;tiene il pe&longs;o A in P. <lb/> Dicola po&longs;&longs;anza in P e&longs;&longs;ere vn quarto del <lb/>pe&longs;o A. Siano tirate le linee KL GF ON <lb/>per li centri delle girelle, &longs;i che &longs;iano egual-<lb/>mente di&longs;tanti dall'orizonte; le quali per le co <lb/>&longs;e, che già &longs;ono dette, &longs;aranno come leue. </s> <s id="id.2.1.805.2.0"> & <lb/>percioche per cagione della leua, ouero bilan-<lb/>cia KL, il cui &longs;o&longs;tegno, ouero centro è nel<emph.end type="italics"/> <lb/> <arrow.to.target n="note228"></arrow.to.target><emph type="italics"/>mezo, tanto &longs;o&longs;tiene la corda KG, quanto <lb/>la NL non &longs;i &longs;acendo mouimento in niu-<lb/>na delle parti: Di più per cau&longs;a della leua <lb/>GF dal cui mezo, come &longs;o&longs;pe&longs;o dipende il <lb/>pe&longs;o; &longs;e fo&longs;&longs;ero due po&longs;&longs;anze in GF, oue-<emph.end type="italics"/> <lb/> <arrow.to.target n="note229"></arrow.to.target><emph type="italics"/>ro in HE, (percioche &longs;i come è &longs;tato più <lb/>volte detto, la ragione dell'vno, & dell'al-<lb/>tro &longs;ito è pari) &longs;arebbono per certo que&longs;te <lb/>tali po&longs;&longs;anze eguali fraloro. </s> <s id="id.2.1.805.3.0"> Onde co&longs;i &longs;o-<lb/>&longs;tiene la corda HG, come EF: &longs;imilmen <lb/>te &longs;imo&longs;trerà tanto &longs;o&longs;tenere la corda PO, <lb/>quanto la NL. Per laqual co&longs;a le corde <lb/>PO KG EF LN &longs;o&longs;tengono egualmen-<lb/>te. </s> <s id="id.2.1.805.4.0"> Adunque &longs;o&longs;tiene egualmente sì la cor-<lb/>da PO, come la KG. Se dunque s'inten-<lb/>de&longs;&longs;ero e&longs;&longs;ere due po&longs;&longs;anze in OG, ouero in <lb/>PH, che è il mede&longs;imo, lequali tuttauia &longs;o-<lb/>&longs;tenghino il pe&longs;o, come &longs;o&longs;tengono le corde, <lb/>&longs;arebbono per certo eguali: & GF ON <lb/>baurebbono le &longs;orze di due leue, il &longs;o&longs;tegno <lb/>delle quali &longs;aranno FN & il pe&longs;o A &longs;a <lb/>rà appiccato in BC, che è il mezo delle le-<lb/>ue. </s> <s id="id.2.1.805.5.0"> & percioche tutte le corde &longs;o&longs;tengo-<lb/>no egualmente, tanto &longs;o&longs;teniranno le due<emph.end type="italics"/> <lb/> <arrow.to.target n="fig94"></arrow.to.target> <lb/><emph type="italics"/>PO LN quanto le due KG EF. tanto dunque &longs;o&longs;terrà la leua ON, quan-<emph.end type="italics"/> <lb/> <arrow.to.target n="note230"></arrow.to.target><emph type="italics"/>to la leua GF. Onde nell'vna, & l'altra leua ON GF pe&longs;erà egualmente il <lb/>pe&longs;o. </s> <s id="id.2.1.805.6.0"> &longs;arà dunque ogni po&longs;&longs;anza che è in PH vn quarto del pe&longs;o A. & e&longs;&longs;en<emph.end type="italics"/> <pb id="p.66" xlink:href="pageimg-it/150.jpg"/><emph type="italics"/>do, che la corda KG &longs;i prenda in loco di po&longs;&longs;anza, come quella, che non &longs;o&longs;tiene <lb/>altramente di quel che faccia PO, &longs;arà la po&longs;&longs;anza di P, che &longs;o&longs;tiene il pe&longs;o A <lb/>vn quarto di e&longs;&longs;o pe&longs;o. </s> <s id="id.2.1.805.7.0"> che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig94" place="text" xlink:href="figures-it/149_01.jpg"></figure> <p type="margin" id="id.2.1.807.0.0"> <s id="id.2.1.807.1.0"> <margin.target id="note228"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.808.0.0"> <s id="id.2.1.808.1.0"> <margin.target id="note229"></margin.target><emph type="italics"/>Per il<emph.end type="italics"/> 2. <emph type="italics"/>corollario della<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.809.0.0"> <s id="id.2.1.809.1.0"> <margin.target id="note230"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.810.0.0"> <s id="id.2.1.810.1.0"> COROLLARIO I. </s> </p> <p type="main" id="id.2.1.811.0.0"> <s id="id.2.1.811.1.0"> Di qui è manife&longs;to, che cia&longs;cuna corda EF GK LN OP &longs;o-<lb/>&longs;tiene la quarta parte del pe&longs;o A. </s> </p> <p type="head" id="id.2.1.812.0.0"> <s id="id.2.1.812.1.0"> COROLLARIO II. </s> </p> <p type="main" id="id.2.1.813.0.0"> <s id="id.2.1.813.1.0"> E chiaro ancora, che non meno &longs;o&longs;tiene la girella il cui centro <lb/>è C, di quello che faccia la girella, il centro dellaquale è B. </s> </p> <pb id="p.66v" xlink:href="pageimg-it/151.jpg"/> <p type="head" id="id.2.1.815.0.0"> <s id="id.2.1.815.1.0"> Altramente. </s> </p> <p type="main" id="id.2.1.816.0.0"> <s id="id.2.1.816.1.0"><emph type="italics"/>Po&longs;te ancora le co&longs;e mede&longs;ime, &longs;e fo&longs;&longs;ero <lb/>due po&longs;&longs;anze eguali, che &longs;o&longs;tene&longs;&longs;ero<emph.end type="italics"/> <lb/> <arrow.to.target n="note231"></arrow.to.target><emph type="italics"/>il pe&longs;o A, l'vna in O, & l'altra in <lb/>C: &longs;arebbe cia&longs;cuna delle dette po&longs;&longs;an-<lb/>ze vn terzo del pe&longs;o A. Ma perche <lb/>la leua GF, il cui &longs;o&longs;tegno è F, è <lb/>diui&longs;a in due parti eguali nel C. &longs;e dun <lb/>que &longs;i porrà la po&longs;&longs;anza in G che &longs;o-<lb/>&longs;tenga l'i&longs;te&longs;&longs;o pe&longs;o, come la po&longs;&longs;anza <lb/>di C, &longs;arà la po&longs;&longs;anza di G la metà <lb/>della po&longs;&longs;anza, che fo&longs;&longs;e in C; per-<lb/>cioche &longs;e la po&longs;&longs;anza di C per &longs;e &longs;te&longs;&longs;a <lb/>&longs;o&longs;tene&longs;&longs;e il pe&longs;o, che è appiccato in C, <lb/>&longs;arebbe per certo eguale ad e&longs;&longs;o pe&longs;o; et <lb/>&longs;e l'i&longs;te&longs;&longs;o pe&longs;o fo&longs;&longs;e &longs;o&longs;tenuto dalla po&longs; <lb/>&longs;anza di G, &longs;arebbe il doppio di e&longs;&longs;a G <lb/>po&longs;&longs;anza, & la po&longs;&longs;anza di C &longs;areb-<emph.end type="italics"/> <lb/> <arrow.to.target n="note232"></arrow.to.target><emph type="italics"/>be vn terzo del pe&longs;o A; dunque la <lb/>po&longs;&longs;anza di G &longs;arebbe vn &longs;e&longs;to della <lb/>po&longs;&longs;anza del pe&longs;o A. Per laqual co <lb/>&longs;a, e&longs;&longs;endo, che la po&longs;&longs;anza di O &longs;ia vn <lb/>terzo del pe&longs;o A, & la po&longs;&longs;anza di <lb/>G vn &longs;e&longs;to: &longs;ara l'vna, & l'altra po&longs;-<lb/>&longs;anza in&longs;ieme po&longs;te in OG la metà <lb/>del pe&longs;o A, percioche la terza par-<lb/>te con la &longs;e&longs;ta &longs;à la metà. </s> <s id="id.2.1.816.2.0"> Ma per-<lb/>cioche la po&longs;&longs;anza di OG, ouero di <lb/>PH, (come prima è detto) &longs;ono fra <lb/>loro eguali, & l'vna, & l'altra in&longs;ie-<lb/>me &longs;ono la metà del pe&longs;o A, &longs;arà <lb/>ogn'vna delle po&longs;&longs;anze po&longs;te in PH <lb/>vn quarto di e&longs;&longs;o A. Adunque la <lb/>po&longs;&longs;anza di P che &longs;o&longs;tiene il pe&longs;o A <lb/>&longs;arà vn quarto di e&longs;&longs;o pe&longs;o A. che era <lb/>da mo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.817.0.0"> <s id="id.2.1.817.1.0"> <margin.target id="note231"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 4. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.818.0.0"> <s id="id.2.1.818.1.0"> <margin.target id="note232"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 3. <emph type="italics"/>di questo della leua.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.819.0.0" xlink:href="figures-it/151_01.jpg"></figure> <pb id="p.67" xlink:href="pageimg-it/152.jpg"/> <p type="main" id="id.2.1.821.0.0"> <s id="id.2.1.821.1.0"><emph type="italics"/>Ma &longs;e'la corda <lb/>&longs;arà legata in <lb/>E, & &longs;ia <lb/>dauantaggio <lb/>inuolta intor <lb/>no à quattro <lb/>girelle, et per <lb/>uenga in P, <lb/>&longs;imo&longs;trerà &longs;i <lb/> <expan abbr="milmēte">milmente</expan>, che <lb/>la po&longs;&longs;anza <lb/>di P &longs;arà <lb/>vn quarto <lb/>del pe&longs;o A; <lb/>peroche egli <lb/>è il mede&longs;i-<lb/>mo, come &longs;e <lb/>la corda fo&longs;-<lb/>&longs;e legata in <lb/>L, & che la <lb/>po&longs;&longs;anza &longs;o-<lb/>&longs;tene&longs;&longs;e il pe <lb/>&longs;o con la cor-<lb/>da inuolta in<lb/>torno à tre gi <lb/>relle &longs;olamen <lb/>te, i centri <lb/>delle quali fo&longs; <lb/>&longs;ero BCQ, <lb/>percioche la <lb/>girella, il cui <lb/>centro è D, <lb/>del tutto è <lb/>inutile.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.822.0.0" xlink:href="figures-it/152_01.jpg"></figure> <pb id="p.67v" xlink:href="pageimg-it/153.jpg"/> <p type="head" id="id.2.1.823.0.0"> <s id="id.2.1.823.1.0"> PROPOSITIONE VIII. </s> </p> <p type="main" id="id.2.1.824.0.0"> <s id="id.2.1.824.1.0"> Siano due leue AB CD diui&longs;e in due parti eguali EF, i &longs;o-<lb/>&longs;tegni delle quali &longs;iano AC, & &longs;ia appiccato il pe&longs;o G ne' <lb/>punti EF all'vna, & l'altra leua, &longs;i fattamente, che dall'vno, <lb/>& l'altro pe&longs;i egualmente: & &longs;iano tre po&longs;&longs;anze eguali in BD <lb/>E che &longs;o&longs;tenghino il pe&longs;o G. Dico, che cia&longs;cuna delle det-<lb/>te po&longs;&longs;anze &longs;eparatamente è vn quinto del pe&longs;o G. </s> </p> <p type="main" id="id.2.1.825.0.0"> <s id="id.2.1.825.1.0"><emph type="italics"/>Percioche il pe&longs;o G &longs;ta appiccato in EF, & &longs;ono le tre po&longs;&longs;anze in EBD egua-<lb/>li: però la po&longs;&longs;anza di E &longs;o&longs;terrà la parte &longs;olamente del pe&longs;o G, che &longs;arà eguale <lb/>ad e&longs;&longs;a po&longs;&longs;anza di E, ma <lb/>le po&longs;&longs;anze di BD &longs;o&longs;terran<lb/>no la parte re&longs;tante, & la<emph.end type="italics"/> <lb/> <arrow.to.target n="note233"></arrow.to.target><emph type="italics"/>parte, che è da B &longs;o&longs;tenu-<lb/>ta, &longs;arà il doppio di e&longs;&longs;o: ma <lb/>la parte &longs;o&longs;tenuta da D &longs;a-<lb/>rà &longs;imilmente il doppio di e&longs; <lb/>&longs;o D per cau&longs;a della pro-<lb/>portione di BA ver&longs;o AE, <lb/>& di DC ver&longs;o CF. Con <lb/>cio&longs;ia dunque, che le po&longs;&longs;an-<emph.end type="italics"/> <lb/> <arrow.to.target n="note234"></arrow.to.target><emph type="italics"/>ze di BD &longs;iano eguali, &longs;a-<lb/>ranno anche (per quel che di <lb/>&longs;opra è detto) le parti del pe<emph.end type="italics"/> <lb/> <arrow.to.target n="fig95"></arrow.to.target> <lb/><emph type="italics"/>&longs;o G, lequali &longs;ono &longs;o&longs;tenute dalle po&longs;&longs;anze di BD, fra loro eguali, & ogni vna <lb/>&longs;arà il doppio di quella tal parte, che è &longs;o&longs;tenuta dalla po&longs;&longs;anza di E. Diuida&longs;i <lb/>dunque il pe&longs;o G in tre parti, delle quali due &longs;iano fra loro eguali, & di più ogni <lb/>vna di loro &longs;eparatamente &longs;ia il doppio dell'altra terza parte, ilche accaderà, &longs;e <lb/>in cinque parti eguali HKLMN &longs;arà diui&longs;o: percioche la parte compo&longs;ta di due <lb/>parti KL è il doppio della parte H, & la parte ancora di MN è &longs;imilmen-<lb/>te il doppio della parte i&longs;te&longs;&longs;a H. Per laqual co&longs;a anche la parte KL &longs;arà egua-<lb/>le alla parte MN. Ma &longs;o&longs;tenga la po&longs;&longs;anza di E la parte di H; & la po&longs;&longs;an <lb/>za di B le parti di KL: & la po&longs;&longs;anza di D le parti MN; adunque le tre <lb/>po&longs;&longs;anze eguali po&longs;te in BDE &longs;o&longs;terranno tutto il pe&longs;o G: & ogn'vna delle <lb/>po&longs;&longs;anze di BD &longs;o&longs;terrà il doppio di quel che &longs;o&longs;tiene la po&longs;&longs;anza di E. Però <lb/>e&longs;&longs;endo che la po&longs;&longs;anza di E &longs;o&longs;tenga la parte di H, laquale è la quinta parte del <lb/>pe&longs;o G, & &longs;ia ad e&longs;&longs;o eguale, &longs;arà la po&longs;&longs;anza di E vn quinto del pe&longs;o G. & <lb/>percioche la po&longs;&longs;anza di B &longs;o&longs;tiene le parti di KL, lequali &longs;ono il doppio & del-<emph.end type="italics"/> <pb id="p.68" xlink:href="pageimg-it/154.jpg"/><emph type="italics"/>la po&longs;&longs;anza di B, & della parte di H, &longs;arà ancora la po&longs;&longs;anza di B ad e&longs;&longs;o H <lb/>eguale. </s> <s id="id.2.1.825.2.0"> Per laqual co&longs;a &longs;arà vn quinto del pe&longs;o G. Ne altrimente &longs;i dimo&longs;tre-<lb/>rà, che la po&longs;&longs;anza di D è vn quinto del pe&longs;o G. cia&longs;cuna po&longs;&longs;anza dunque in <lb/>BDE è vn quinto del pe&longs;o G. che bi&longs;ognaua dimo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig95" place="text" xlink:href="figures-it/153_01.jpg"></figure> <p type="margin" id="id.2.1.827.0.0"> <s id="id.2.1.827.1.0"> <margin.target id="note233"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 4. <emph type="italics"/>di questa nella leua.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.828.0.0"> <s id="id.2.1.828.1.0"> <margin.target id="note234"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.829.0.0" xlink:href="figures-it/154_01.jpg"></figure> <p type="main" id="id.2.1.830.0.0"> <s id="id.2.1.830.1.0"><emph type="italics"/>Che &longs;e &longs;aranno tre leue AB <lb/>CD EF diui&longs;e in due <lb/>parti eguali in GHK, i <lb/>&longs;o&longs;tegni dellequali &longs;iano A <lb/>CE, & il pe&longs;o L nel mo <lb/>do i&longs;te&longs;&longs;o &longs;ia appiccato in <lb/>GHK, & &longs;iano quattro <lb/>po&longs;&longs;anze eguali in BD <lb/>FG che &longs;o&longs;tengano il pe-<lb/>&longs;o L; &longs;i mo&longs;trerà con &longs;imi-<lb/>gliante modo, che cia&longs;cuna <lb/>po&longs;&longs;anza in BD FG &longs;a-<lb/>rà vn &longs;ettimo del pe&longs;o L: <lb/>& &longs;e quattro fo&longs;&longs;ero le le-<lb/>ue, & cinque le po&longs;&longs;anze <lb/>eguali &longs;o&longs;tenenti il pe&longs;o; con l'i&longs;te&longs;&longs;o modo ancora &longs;i mo&longs;trerebbe che ogni vna del-<lb/>le po&longs;&longs;anze &longs;arebbe vn nono del pe&longs;o, & co&longs;i di mano in mano &longs;ucce&longs;&longs;iuamente.<emph.end type="italics"/> </s> </p> <p type="head" id="id.2.1.831.0.0"> <s id="id.2.1.831.1.0"> PROPOSITIONE IX. </s> </p> <p type="main" id="id.2.1.832.0.0"> <s id="id.2.1.832.1.0"> Se à quattro girelle di due taglie, l'vna delle quali &longs;ia po&longs;ta di <lb/>&longs;opra, & l'altra di &longs;otto legata al pe&longs;o, &longs;ia condotta intorno <lb/>la corda, legando l'vno de'&longs;uoi capi alla taglia di &longs;otto, & l'al-<lb/>tro &longs;ia ritenuto dalla po&longs;&longs;anza, che &longs;o&longs;tiene il pe&longs;o. </s> <s id="id.2.1.832.2.0"> &longs;arà la po&longs;-<lb/>&longs;anza vn quinto del pe&longs;o. </s> </p> <pb id="p.68v" xlink:href="pageimg-it/155.jpg"/> <p type="main" id="id.2.1.833.0.0"> <s id="id.2.1.833.1.0"><emph type="italics"/>Sia il pe&longs;o A, alquale &longs;ia legata <lb/>la taglia, che habbia due girel-<lb/>le, i cui centri &longs;iano BC: & <lb/>&longs;ia la taglia appiccata di &longs;opra, <lb/>che habbia due altre girelle, i <lb/>cui centri &longs;iano DE, & la <lb/>corda &longs;ia tirata intorno à tutte <lb/>le girelle, laquale &longs;ia legata al-<lb/>la taglia di &longs;otto in F: & &longs;ia <lb/>la po&longs;&longs;anza in G che &longs;o&longs;tiene <lb/>il pe&longs;o A. Dico che la po&longs;&longs;an <lb/>za di G è vn quinto del pe&longs;o <lb/>A. Siano tirate le linee HK <lb/>LM per li centri BC egual-<lb/>mente di&longs;tanti dall'orizonte, le <lb/>quali nel modo i&longs;te&longs;&longs;o, che di <lb/>&longs;opra è &longs;tato detto, dimo&longs;trere-<lb/>mo e&longs;&longs;ere come leue, i &longs;o&longs;tegni <lb/>delle quali &longs;ono KM, & il pe<lb/>&longs;o A pende attaccato nel me-<lb/>zo BC dell'vna, & l'altra le-<lb/>ua, & le tre po&longs;&longs;anze LHC, che <lb/>&longs;o&longs;tengono il pe&longs;o, lequali con <lb/>&longs;imile modo mo&longs;treremo e&longs;&longs;ere <lb/>eguali: percioche le corde fanno <lb/>l'i&longs;ie&longs;&longs;o officio, come &longs;e fo&longs;&longs;ero <lb/>po&longs;&longs;anze: & percioche il pe&longs;o <lb/>dall'vna, & l'altra leua HK <lb/>LM pe&longs;a egualmente, ilche &longs;i <lb/>dimo&longs;trerà ancora, come nelle <lb/>precedenti è &longs;tato dimo&longs;trato:<emph.end type="italics"/> <lb/> <arrow.to.target n="note235"></arrow.to.target><emph type="italics"/>&longs;arà ogni po&longs;&longs;anza po&longs;ta sì in L <lb/>ouero in G, che è il mede&longs;imo; <lb/>& sì in H & in C, cioè in F <lb/>vn quinto del pe&longs;o A. La po&longs; <lb/>&longs;anza dunque di G, che &longs;o&longs;tie-<lb/>ne il pe&longs;o A. &longs;arà vn quinto <lb/>di e&longs;&longs;o pe&longs;o A. che bi&longs;ognaua <lb/>mo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.834.0.0"> <s id="id.2.1.834.1.0"> <margin.target id="note235"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.835.0.0" xlink:href="figures-it/155_01.jpg"></figure> <pb id="p.69" xlink:href="pageimg-it/156.jpg"/> <p type="main" id="id.2.1.837.0.0"> <s id="id.2.1.837.1.0"><emph type="italics"/>Che &longs;e dauantaggio &longs;i traporterà la cor-<lb/>da in F d'intorno ad vn'altra girella, <lb/>il cui centro &longs;ia N, & &longs;ia legata <lb/>in O, &longs;i prouerà &longs;imilmente per due <lb/>ragioni, come nella &longs;ettima propo&longs;i-<lb/>tione di que&longs;to, che la po&longs;&longs;anza di G <lb/>che &longs;o&longs;tiene il pe&longs;o A, è vn &longs;e&longs;to<emph.end type="italics"/> <arrow.to.target n="note236"></arrow.to.target> <lb/><emph type="italics"/>di e&longs;&longs;o pe&longs;o A. Percioche prima dal <lb/>le treleue LM HK FP licui &longs;o-<lb/>stegni &longs;ono in KP, & il pe&longs;o è ap-<lb/>piccato nel mezo delle leue, & le tre <lb/>po&longs;&longs;anze po&longs;te in LHF che &longs;o&longs;ten-<emph.end type="italics"/> <arrow.to.target n="note237"></arrow.to.target> <lb/><emph type="italics"/>gono il pe&longs;o &longs;ono eguali: poi dalle po&longs; <lb/>&longs;anze di LHN cia&longs;cuna delle quali <lb/>&longs;arebbe vn quinto del pe&longs;o A, per-<lb/>cioche ambedue le po&longs;&longs;anze in&longs;ieme <lb/>po&longs;te in LH &longs;arebbono &longs;otto doppie &longs;e&longs; <lb/>quialtere al pe&longs;o, & la po&longs;&longs;anza di F <lb/>&longs;arebbe vn decimo, e&longs;&longs;endo la metà di <lb/>e&longs;&longs;a N. Ma due quinte parti con <lb/>vna decima parte fanno la metà, la <lb/>qual metà &longs;e &longs;arà diui&longs;a per tre, ri-<lb/>&longs;ponderà la &longs;e&longs;ta parte del pe&longs;o à cia-<lb/>&longs;cuna delle po&longs;&longs;anze po&longs;te in LHF. <lb/> Dalle quali co&longs;e è manife&longs;to la po&longs;&longs;an-<lb/>za di G e&longs;&longs;ere vn &longs;e&longs;to del pe&longs;o A; <lb/>& &longs;i dimo&longs;trerà &longs;imilmente che cia-<lb/>&longs;cuna girella &longs;o&longs;tiene eguale portione <lb/>del pe&longs;o.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.838.0.0"> <s id="id.2.1.838.1.0"> <margin.target id="note236"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 6. <emph type="italics"/>di questo<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.839.0.0"> <s id="id.2.1.839.1.0"> <margin.target id="note237"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.840.0.0" xlink:href="figures-it/156_01.jpg"></figure> <pb id="p.69v" xlink:href="pageimg-it/157.jpg"/> <p type="main" id="id.2.1.842.0.0"> <s id="id.2.1.842.1.0"> In que&longs;to trattato della taglia, &longs;i come in tutti gli altri ancora, l'autore pre&longs;uppone, <lb/>che qualunque per&longs;ona &longs;i mette à leggere il &longs;uo libro delle Mechaniche &longs;ia inten-<lb/>dente di numeri, & di Geometria, & però ha &longs;empre mantenuto quello accurato <lb/>&longs;tile, & dimo &longs;tratiuo co&longs;tumato da buoni Matematici, v&longs;ando i vocaboli proprij <lb/>della &longs;cienza, alcuni de' quali io hò ben potuto volgarizare facilmente, &longs;i che <lb/>ogn'vno gli po&longs;&longs;a intendere, come per e&longs;&longs;empio, nelle proportioni duplum, tri-<lb/>plum, quadruplum, & gli altri &longs;imili, ponendo in vece loro due volte tanto, tre <lb/>volte tanto, & quattro volte tanto: & co&longs;i per 'oppo&longs;ito &longs;ub duplum, &longs;ubtriplu, <lb/>& &longs;ub quadruplum, la metà, vn terzo, & vn quarto: & parimente &longs;e&longs;quialterum, <lb/>&longs;e&longs;quitertium, & &longs;e&longs;quiquartum, & gli altri &longs;imili, che vogliono dire vna volta & <lb/>meza, vna volta, & vn terzo, & vna volta & vn quarto. </s> <s id="id.2.1.842.2.0"> Que&longs;ti dico s'hanno po<lb/>tuto ben dire, & facilmente nella no&longs;tra lingua. </s> <s id="id.2.1.842.3.0"> Ma nell'ampiezza delle propor-<lb/>tioni trouando&longs;i altri vocaboli a&longs;&longs;ai, i quali non è pos&longs;ibile co&longs;i adattare alla no-<lb/>&longs;tra lingua, tra quali alcuni &longs;i trouano po&longs;ti dall'autore in que&longs;to trattato della ta <lb/>glia, & io &longs;ono &longs;tato sforzato à la&longs;ciargli co&longs;i, come erano, per mancamento di pa<lb/>role, che nella no&longs;tra fauella gli po&longs;&longs;ano e&longs;primere; hò giudicato douer e&longs;&longs;ere co<lb/>&longs;a vtile il dichiarare tutti i predetti vocaboli pertinenti alle proportioni, che ha il <lb/>pe&longs;o alla po&longs;&longs;anza, & la po&longs;&longs;anza al pe&longs;o &longs;critti dall'autore in que&longs;to trattato della <lb/>taglia, accio che quelle per&longs;one lequali non po&longs;&longs;edono que&longs;ti termini, non habbia <lb/>no fatica di andare &longs;tudiando iloro &longs;ignificati. </s> </p> <p type="main" id="id.2.1.843.0.0"> <s id="id.2.1.843.1.0"> Dico dunque vna quantità poter&longs;i paragonare, & hauere proportione con vn'altra <lb/>in tre modi principali, la&longs;ciando hora le più &longs;ottili di&longs;tintioni. </s> <s id="id.2.1.843.2.0"> Primieramente <lb/>come maggiore ver&longs;o la minore, dapoi come minore ver&longs;o la maggiore, & in fi-<lb/>ne come eguale ver&longs;o la eguale. </s> <s id="id.2.1.843.3.0"> Tutta la dottrina delle'proportioni, con&longs;i&longs;te in <lb/>que&longs;ti riguardi, cioè dal maggiore al minore, dal min ore al maggiore, & dall'e-<lb/>quale all'equale. </s> <s id="id.2.1.843.4.0"> Hor quando vna quantità, che &longs;ia maggiore è paragonata con <lb/>vn'altra, che &longs;ia minore, che &longs;i dice proportione di maggiore di&longs;uguaglianza, na-<lb/>&longs;cono cinque generi di proportioni, l'vno è il moltiphce &longs;chietto, il &longs;econdo è il <lb/>&longs;opraparticolare, il rerzo il &longs;o prapartiente, il quarto il moltiplice &longs;opraparticola-<lb/>re, & il quinto & vltimo il moltiplice &longs;oprapartiente. </s> <s id="id.2.1.843.5.0"> Ma quando &longs;i fa compara-<lb/>tione della minore quantità ver&longs;o la maggiore, all'hora &longs;i producono cinque altri <lb/>generi oppo&longs;ti apunto à i predetti cinque, & &longs;i dicono di minore di&longs;uguaglian-<lb/>za, à i quali per fargli differenti da loro &longs;i aggiunge da Latini il &longs;ub, cioè &longs;otto, <lb/>&longs;criuendo&longs;t &longs;otto moltiplice, &longs;otto&longs;opra particolare, &longs;otto &longs;oprapartiente, &longs;otto <lb/>moltiplice &longs;opra particolare, & &longs;otto moltiplice &longs;oprapartiente. </s> <s id="id.2.1.843.6.0"> Tutte le propor-<lb/>tioni dunque &longs;ono compre&longs;e in vniuer&longs;ale da que&longs;ti diece generi oppo&longs;ti fra &longs;e <lb/>l'vn l'altro, cia&longs;cheduno de quali poi ha le &longs;ue &longs;petie differenti di proportioni. </s> <s id="id.2.1.843.7.0"> Ma <lb/>io non hò qui intentione di numerarle, nè dichiarare diffu&longs;amente que&longs;ta materia <lb/>delle proportioni, ma &longs;olamente li vocaboli po&longs;ti dall'autore nel pre&longs;ente libro <lb/>della taglia, ba&longs;tando mi hauerne dato in generale vna rozza cognitione. </s> <s id="id.2.1.843.8.0"> Ma chi <lb/>di ciò de&longs;idera hauere intero cono&longs;cimento legga tra i &longs;crittori della lingua Ita-<lb/>liana Fra Luca dal Borgo, il Tartaglia ne i libri della Arithmetica, & il dottis&longs;imo <lb/>Zarlino nella prima parte delle In&longs;titutioni Harmoniche. </s> <s id="id.2.1.843.9.0"> Dice l'autore in que&longs;to <lb/>loco. </s> <s id="id.2.1.843.10.0"> Percio che &longs;arebbono ambedue le po&longs;&longs;anze infieme in LH &longs;otto doppie <lb/>&longs;e&longs;quialtere di e&longs;&longs;o pe&longs;o. </s> <s id="id.2.1.843.11.0"> Cioè le due po&longs;&longs;anze po&longs;te in LH haurebbono quella <lb/>proportione ver&longs;o il pe&longs;o, che ha 2. à 5. cioè &longs;e il pe&longs;o fo&longs;&longs;e come cinque, le po&longs;-<lb/>&longs;anze larebbono come 2. che è la proportione &longs;otto dop pia &longs;e&longs;quialtera. </s> <s id="id.2.1.843.12.0"> Segue <pb id="p.70" xlink:href="pageimg-it/158.jpg"/>poi, Ma due quinte con vna decima fanno la me<lb/>tà, cioè à &longs;ommare in&longs;ieme due quinti, & vn <lb/>decimo fanno la metà di cinque, pero che li <lb/>due quinti &longs;ono due parti del cinque, & la deci <lb/>ma parte è la metà di vn quinto, tanto che met-<lb/>tono in&longs;ieme due, & mezo, che &longs;ono la metà di <lb/>cinque. </s> <s id="id.2.1.843.13.0"> Che &longs;e que&longs;ta metà poi &longs;arà diui&longs;a per <lb/>tre, ne riu&longs;cirà la &longs;e&longs;ta parte da e&longs;&longs;ere attribuita à <lb/>cia&longs;cheduna delle tre po&longs;&longs;anze po&longs;te in LHF. <lb/> Il modo del diuidere la metà per tre è facile, & <lb/>fas&longs;i in que&longs;ta maniera ponendo tre di &longs;opra, & <lb/>vno di &longs;otto; & vno di &longs;opra, & due di &longs;otto <expan abbr="cõ">com</expan> <lb/>la &longs;ua linea nel mezo, come &longs;i co&longs;tuma, & mol-<lb/>tiplicando il tre intero co'l due denominatore <lb/>della metà, ne viene 6, alquale di &longs;opra &longs;i ag-<lb/>giunge vno, & è vn &longs;e&longs;to. </s> </p> <p type="main" id="id.2.1.844.0.0"> <s id="id.2.1.844.1.0"><emph type="italics"/>Che &longs;e come nella terza figura la corda &longs;i allunghe <lb/>rà in O, & &longs;i condarrà intorno ad vn'altra gi-<lb/>rella, il cui centro &longs;ia Q, la qual corda poi &longs;i <lb/>leghi in R alla taglia di &longs;otto; &longs;arà la po&longs;&longs;an-<lb/>za di G vn &longs;ettimo del pe&longs;o. </s> <s id="id.2.1.844.2.0"> & co&longs;i proceden <lb/>do in in&longs;inito, la proportione della po&longs;&longs;anza al pe<emph.end type="italics"/> <arrow.to.target n="note238"></arrow.to.target> <lb/><emph type="italics"/>&longs;o, quanto &longs;i voglia &longs;otto moltiplice ver&longs;o il pe-<lb/>&longs;o &longs;i potrà trouare. </s> <s id="id.2.1.844.3.0"> Dapoi &longs;i mo&longs;trerà &longs;empre, <lb/>come nelle precedenti, che &longs;e la po&longs;&longs;anza, la-<lb/>quale &longs;o&longs;tiene il pe&longs;o &longs;arà vn quarto, ouero vn <lb/>quinto, ouero in qual &longs;i voglia altro modo &longs;arà <lb/>di&longs;po&longs;ta ver&longs;o il pe&longs;o, che &longs;imilmente cia&longs;cuna <lb/>corda &longs;o&longs;terrà la quarta, ò la quinta, ouero qual <lb/>&longs;i voglia altra parte del pe&longs;o, &longs;i come la i&longs;te&longs;&longs;a <lb/>po&longs;&longs;anza: peroche le corde fanno il mede&longs;imo, <lb/>come &longs;e fo&longs;&longs;ero tante po&longs;&longs;anze: & le girelle co-<lb/>me &longs;e fo&longs;&longs;ero tante leue.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.847.0.0" xlink:href="figures-it/158_01.jpg"></figure> <p type="margin" id="id.2.1.845.0.0"> <s id="id.2.1.845.1.0"> <margin.target id="note238"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 8. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.846.0.0"> <s id="id.2.1.846.1.0"> Sotto moltiplice. </s> <s id="id.2.1.846.2.0"> Que&longs;to è il primo genere delle <lb/>proportioni, che &longs;i riguardano dal minore al <lb/>maggiore, detto di minore di&longs;uguaglianza, il <lb/>quale &longs;otto di &longs;e tiene a&longs;&longs;ais&longs;ime &longs;petie, & è op-<lb/>po&longs;to come ho ricordato, al moltiplice. </s> <s id="id.2.1.846.3.0"> Dice <lb/>l'autore: & co&longs;i procedendo in infinito &longs;i potrà <lb/>ritrouare qual &longs;i voglia proportione &longs;otto mol <lb/>tiplice. </s> <s id="id.2.1.846.4.0"> Percio che la po&longs;&longs;anza è minore del pe<lb/>&longs;o, & però ver&longs;o lui ha proportione &longs;otto mol<lb/>tiplice, come di vno ver&longs;o due, & di due ver-<lb/>&longs;o quattro per darne e&longs;&longs;empio, & co&longs;i de gli al-<lb/>tri numeri tali. </s> </p> <pb id="p.70v" xlink:href="pageimg-it/159.jpg"/> <p type="head" id="id.2.1.849.0.0"> <s id="id.2.1.849.1.0"> COROLLARIO </s> </p> <p type="main" id="id.2.1.850.0.0"> <s id="id.2.1.850.1.0"> Di qui è manife&longs;to, che le girelle della taglia, allaquale è legato <lb/>il pe&longs;o, fanno sì, che il pe&longs;o è &longs;o&longs;tenuto da po&longs;&longs;anza minore, <lb/>di quel che &longs;ia e&longs;&longs;o pe&longs;o, co&longs;a che veramente non fanno le gi-<lb/>relle della taglia di &longs;opra. </s> </p> <p type="main" id="id.2.1.851.0.0"> <s id="id.2.1.851.1.0"><emph type="italics"/>Egli nondimeno conuiene &longs;apere, che come &longs;uole &longs;ar&longs;i, la girella della taglia di &longs;otto, <lb/>il cui centro è N, deue e&longs;&longs;ere minore di quella girella, il cui centro è C, & que <lb/>&longs;ta anche minore di quella, che ha il centro in B: & in &longs;omma &longs;e &longs;aranno più gi <lb/>relle nella taglia di &longs;otto legata al pe&longs;o, &longs;empre quella girella deue e&longs;&longs;ere maggiore <lb/>delle altre, che è più vicina al pe&longs;o attaccato: ma al contrario hanno à di&longs;por&longs;i le <lb/>girelle nella taglia di &longs;opra, ilche &longs;i co&longs;tuma di fare, acciò che le corde fra loro non <lb/>&longs;i intrichino; peroche in quanto alle girelle, &longs;iano ò grandi, ò picciole, non importa <lb/>nulla, &longs;eguendone &longs;empre l'i&longs;te&longs;&longs;o.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.852.0.0"> <s id="id.2.1.852.1.0"><emph type="italics"/>Di più è da notare, ilche etiandio dalle co&longs;e dette facilmente appare, che grandi&longs;&longs;ima <lb/>differenza na&longs;ce trala po&longs;&longs;anza, & il pe&longs;o dal legare la corda ouero in R della ta <lb/>glia di &longs;otto, ouero in S, percioche &longs;e &longs;i legherà in S, la po&longs;&longs;anza di G &longs;arà vn <lb/>&longs;e&longs;to del pe&longs;o; ma &longs;e in R vn &longs;ettimo, co&longs;a che non accade alla taglia di &longs;opra: <lb/>percioche leghi&longs;i la corda, come nella precedente figura, ouero in T, ouero in O, <lb/>&longs;empre la po&longs;&longs;anza di G &longs;arà vn &longs;e&longs;to di e&longs;&longs;o pe&longs;o.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.853.0.0"> <s id="id.2.1.853.1.0"><emph type="italics"/>Dopo que&longs;te co&longs;e egli è da con&longs;iderare in che modo la forza moua il pe&longs;o, & di più lo <lb/>&longs;patio, & il tempo della po&longs;&longs;anza, che moue, & del pe&longs;o che è mo&longs;&longs;o.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.854.0.0"> <s id="id.2.1.854.1.0"> “Di piu egli è da notare ilche etiandio è manife&longs;to dalle co&longs;e dette &c. </s> <s id="id.2.1.854.2.0"> Qui potreb-<lb/>be for&longs;e ad alcuno parere difficile in che modo po&longs;&longs;a e&longs;&longs;ere, che dal legare la cor-<lb/>da in R, ouero in S, come &longs;i vede in que&longs;ta figura, na&longs;ca tanta differenza. </s> <s id="id.2.1.854.3.0"> Onde <lb/>noti&longs;i che legando la corda in S, la girella Q re&longs;ta del tutto inutile, & è come <lb/>&longs;e ella non vifo&longs;&longs;e; & la corda per non e&longs;&longs;ere attaccata in R alla taglia di &longs;otto, <lb/>ma in S fuori non &longs;o&longs;tiene la taglia, talche la forza di G viene ad e&longs;&longs;ere &longs;olamen<lb/>te vn &longs;e&longs;to del pe&longs;o. </s> <s id="id.2.1.854.4.0"> &longs;oggiunge poi ilche non auiene alla taglia di &longs;opra. Doue <lb/>auerta&longs;i che mentre &longs;i ha tenuto propo&longs;ito delle lettere S & R, ha bi&longs;ognato guar <lb/>dare nella qui &longs;opra&longs;critta figura, ma in parlando di TO, egli è me&longs;tieri per in-<lb/>tendere que&longs;to loco mirare nella figura precedente, che è la &longs;econda della nona <lb/>propo&longs;itione, peroche iui&longs;ono le lettere TO. La ragione per la quale non na&longs;ca <lb/>differenza nella po&longs;&longs;anza à legare la corda in T ouero in O, ma &longs;ia tutto vno, <lb/>è che la taglia di &longs;opra &longs;ta &longs;empre ferma, per modo, che non importa nulla il le-<lb/>gare la corda in O nella taglia di &longs;opra, ouero in T fuori di e&longs;&longs;a, poiche am-<lb/>bidue i luoghi &longs;ono immobili, & iui la corda &longs;ta ferma. </s> <s id="id.2.1.854.5.0"> Lequali tutte co&longs;e l'auto <lb/>re hà toccato breuis&longs;imamente per e&longs;&longs;ere que&longs;to trattato della taglia lungo, la-<lb/>&longs;ciando al lettore ancora qualche co&longs;a da &longs;peculare per &longs;e mede&longs;imo. </s> </p> <pb id="p.71" xlink:href="pageimg-it/160.jpg"/> <p type="head" id="id.2.1.856.0.0"> <s id="id.2.1.856.1.0"> PROPOSITIONE X. </s> </p> <p type="main" id="id.2.1.857.0.0"> <s id="id.2.1.857.1.0"> Se la corda &longs;arà inuolta intorno alla girella della taglia appicca-<lb/>ta di &longs;opra, all'vno de'capi, dellaqual corda &longs;ia attaccato il pe<lb/>&longs;o, & all'altro po&longs;ta la po&longs;&longs;anza, che moue. </s> <s id="id.2.1.857.2.0"> La detta po&longs;&longs;anza mo <lb/>uerà con la leua &longs;empre egualmente di&longs;tante dall'orizonte. </s> </p> <p type="main" id="id.2.1.858.0.0"> <s id="id.2.1.858.1.0"><emph type="italics"/>Sia il pe&longs;o A. &longs;ia la girella della taglia appiccata di &longs;opra, che habbia il centro K. <lb/> Sia dapoi la corda HB CDEF legata al pe&longs;o A in H, & &longs;ia inuolta d'intor <lb/>no alla girella; & &longs;ia la taglia per modo appiccata <lb/>in L, che non habbia alcun altro mouimento &longs;uor <lb/>che il volgimento libero della girella d'intorno al <lb/>&longs;uo a&longs;&longs;etto, & &longs;ia la po&longs;&longs;anza in F che moua il <lb/>pe&longs;o A. Dico, che la po&longs;&longs;anza di F mouerà <lb/>&longs;empre il pe&longs;o A con la leua egualmente di&longs;tan-<lb/>te dall'orizonte. </s> <s id="id.2.1.858.2.0"> &longs;ia tirata la linea BKE egual-<lb/>mente di&longs;tante dall'orizonte, & &longs;iano i punti BE<emph.end type="italics"/> <arrow.to.target n="note239"></arrow.to.target> <lb/><emph type="italics"/>doue le corde BH & EF toccano il cerchio: <lb/>&longs;arà BKE la leua, il &longs;o&longs;tegno dellaquale è nel <lb/>&longs;uo mezo, che è K, come di &longs;opra è detto. </s> <s id="id.2.1.858.3.0"> Men-<lb/>tre che dunque la forza di F inchina al ba&longs;&longs;o ver <lb/>&longs;o M, la leua EB &longs;i mouerà, mouendo&longs;i tut-<lb/>ta la girella, cioè volgendo&longs;i attorno. </s> <s id="id.2.1.858.4.0"> Mentre <lb/>che dunque F &longs;ta in M &longs;ia il punto E della <lb/>leua mo&longs;&longs;o fin ad I, & il B &longs;in'al C, di mo-<lb/>do, che la leua &longs;ia in CI. Dapoi &longs;i faccia la li <lb/>nea NM eguale ad e&longs;&longs;a FE: & quando il <lb/>punto E, &longs;arà in I all'hora il punto della cor <lb/>da, ilquale era in E &longs;arà in N, & quello,<emph.end type="italics"/> <lb/> <arrow.to.target n="fig96"></arrow.to.target> <lb/><emph type="italics"/>che era in B &longs;arà in C di modo, che tirata la linea CI pa&longs;&longs;erà per lo centro <lb/>K. Hormentre il B &longs;ta in C &longs;ia il punto H in G, & &longs;arà BH al CBG <lb/>eguale, e&longs;&longs;endo la mede&longs;ima corda. </s> <s id="id.2.1.858.5.0"> & percioche mentre EF inchina in MN <lb/>rimane pur &longs;empre EFM à piombo dell'orizonte, & tocca il cerchio nel punto <lb/>E di modo, che la linea tirata dal punto E per lo centro K &longs;ia &longs;empre egualmen<lb/>te di&longs;tante dall'orizonte, ilche mede&longs;imamente auiene alla corda BG & al pun-<pb id="p.71v" xlink:href="pageimg-it/161.jpg"/>to B. Mentre dunque il cerchio, ouero la girella &longs;i volge intorno, &longs;empre &longs;i mo-<lb/>ue la leua EB, & &longs;em-<lb/>pre ancora rimane vn'al-<lb/>tra leua in EB, e&longs;&longs;endo <lb/>che per natura di e&longs;&longs;a gi-<lb/>rella, nellaquale &longs;empre, <lb/>mentre &longs;i moue, re&longs;ti il <lb/>diametro da B in E, <lb/>(ilquale è in loco di le-<lb/>ua) auuiene che parten <lb/>do&longs;ene vna, &longs;ucceda <lb/>l'altra &longs;empre, durando <lb/>però cotale aggiramen-<lb/>to; & co&longs;i accade, che <lb/>la po&longs;&longs;anza moua il pe<lb/>&longs;o &longs;empre con la leua <lb/>EB egualmente di&longs;tan <lb/>te dall'orizonte, ilche <lb/>bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig96" place="text" xlink:href="figures-it/160_01.jpg"></figure> <p type="margin" id="id.2.1.860.0.0"> <s id="id.2.1.860.1.0"> <margin.target id="note239"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.861.0.0" xlink:href="figures-it/161_01.jpg"></figure> <p type="main" id="id.2.1.862.0.0"> <s id="id.2.1.862.1.0"> Po&longs;te le co&longs;e i&longs;te&longs;&longs;e, lo &longs;patio della po&longs;&longs;anza, che moue il pe&longs;o, è <lb/>eguale allo &longs;patio dello i&longs;te&longs;&longs;o pe&longs;o, che è mo&longs;&longs;o. </s> </p> <p type="main" id="id.2.1.863.0.0"> <s id="id.2.1.863.1.0"><emph type="italics"/>Percioche egli è &longs;tato dimo&longs;trato, che mentre F &longs;tà in M, il pe&longs;o A, cioè il punto <lb/>H è in G: & concio&longs;ia che la corda HBCDEF &longs;ia eguale alla GBCDEN<lb/>FM per e&longs;&longs;ere la corda i&longs;te&longs;&longs;a: leuata via dunque la commune GBCDENF <lb/>&longs;arà la HG alla FM eguale, & &longs;imilmente &longs;i mo&longs;trerà la di&longs;ce&longs;a di F e&longs;&longs;ere <lb/>&longs;empre eguale alla &longs;alita di H. Adunque lo &longs;patio della po&longs;&longs;anza è eguale allo <lb/>&longs;patio del pe&longs;o. </s> <s id="id.2.1.863.2.0"> che era da dimo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.864.0.0"> <s id="id.2.1.864.1.0"> Oltre à ciò la po&longs;&longs;anza moue il pe&longs;o i&longs;te&longs;&longs;o per i&longs;patio eguale in <lb/>tempo eguale, tanto con la corda inuolta intorno alla girella <lb/>della taglia appiccata di &longs;opra, quanto &longs;enza taglia, pur che li <lb/>mouimenti di e&longs;&longs;a po&longs;&longs;anza in velocità &longs;iano eguali. </s> </p> <pb id="p.72" xlink:href="pageimg-it/162.jpg"/> <p type="main" id="id.2.1.866.0.0"> <s id="id.2.1.866.1.0"><emph type="italics"/>Stando le co&longs;e i&longs;te&longs;&longs;e, &longs;ia vn'altro pe&longs;o P eguale al pe&longs;o A, alquale &longs;ia legata la cor <lb/>da TQ à piombo dell'orizonte: & &longs;ia TQ eguale ad e&longs;&longs;a HB: & muoua <lb/>la po&longs;&longs;anza di Q il <lb/>pe&longs;o P all'insù ad <lb/>angoli retti all'orizon<lb/>te, come &longs;i moue il pe <lb/>&longs;o A. Dico, che per <lb/>eguale &longs;patio, & in <lb/>vno i&longs;te&longs;&longs;o tempo la <lb/>po&longs;&longs;anza di <expan abbr="q.">que</expan> mo-<lb/>ue il pe&longs;o P, & la <lb/>po&longs;&longs;anza di<emph.end type="italics"/> F <emph type="italics"/>il pe-<lb/>&longs;o A: ilche è il me-<lb/>de&longs;imo, come &longs;e l'i-<lb/>&longs;te&longs;&longs;o pe&longs;o fo&longs;&longs;e mo&longs;-<lb/>&longs;o in tempo eguale, <lb/>&longs;econdo che habbia-<lb/>mo propo&longs;to. </s> <s id="id.2.1.866.2.0"> Sia <lb/>allungata la EF in <lb/>S, & la TQ in R, <lb/>& &longs;iano le QRFS <lb/>fatte eguali non &longs;olo <lb/>fra &longs;e, ma etiandio <lb/>ad e&longs;&longs;a BH. Hor <lb/>concio&longs;ia che le TQ <lb/>QR &longs;iano eguali ad <lb/>e&longs;&longs;e HB FS, & <lb/>la &longs;orza di Q mo-<lb/>ua il pe&longs;o P per <lb/>la linea retta TQ <lb/>R: & dall'altro<emph.end type="italics"/> <lb/> <arrow.to.target n="fig97"></arrow.to.target> <lb/><emph type="italics"/> canto la forza di F moua A per la retta HB, & le velocità de i mouimenti <lb/>dell'una, & l'altra po&longs;&longs;anza &longs;iano eguali, all'hor che nell'i&longs;te&longs;&longs;o tempo la po&longs;&longs;anza di <lb/>Q &longs;arà in R, & la po&longs;&longs;anza di F &longs;arà in S, e&longs;&longs;endo gli &longs;patij eguali: & men <lb/>tre la po&longs;&longs;anza di Q è in R, il pe&longs;o P, cioè il punto T &longs;arà in Q, per e&longs;&longs;e-<lb/>rela TQ eguale ad e&longs;&longs;a QR, & mentre che la po&longs;&longs;anza di F &longs;ta in S, il pe-<lb/>&longs;o A, cioè il punto H &longs;arà in B; ma lo &longs;patio TQ è eguale allo &longs;patio HB: <lb/>adunque le po&longs;&longs;anze di FQ mo&longs;&longs;e egualmente moueranno i pe&longs;i PA eguali <lb/>per eguali &longs;patij in tempo eguale. </s> <s id="id.2.1.866.3.0"> che era da mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig97" place="text" xlink:href="figures-it/162_01.jpg"></figure> <pb id="p.72v" xlink:href="pageimg-it/163.jpg"/> <p type="head" id="id.2.1.868.0.0"> <s id="id.2.1.868.1.0"> PROPOSITIONE XI. </s> </p> <p type="main" id="id.2.1.869.0.0"> <s id="id.2.1.869.1.0"> Se la corda &longs;arà inuolta intorno alla girella della taglia legata al <lb/>pe&longs;o, laqual corda con vno de' &longs;uoi capi &longs;ia legata in qualche <lb/>luogo, & con l'altro pre&longs;a dalla po&longs;&longs;anza che moue il pe&longs;o; La <lb/>po&longs;&longs;anza mouerà &longs;empre con la leua egualmente di&longs;tante dal<lb/>l'orizonte. </s> </p> <p type="main" id="id.2.1.870.0.0"> <s id="id.2.1.870.1.0"><emph type="italics"/>Sia il pe&longs;o A: &longs;ia la girella CED <lb/>della taglia legata al pe&longs;o A, <lb/>da KH, & &longs;ia KH ad ango-<lb/>li retti dell'orizonte, di modo <lb/>che il pe&longs;o &longs;egua &longs;empre il mo-<lb/>uimento della taglia, &longs;ia pur fat-<lb/>to all'insù, ouero all'ingiù, & <lb/>&longs;ia il centro della girella K, & <lb/>la corda inuolta intorno alla gi-<lb/>rella &longs;ia BCDEF, la quale <lb/>&longs;ia legata in B, di modo che <lb/>&longs;tia immobile in B: & &longs;ia in F <lb/>la po&longs;&longs;anza, che moue il pe&longs;o A. <lb/> Dico che la po&longs;&longs;anza di F mo-<lb/>ue &longs;empre il pe&longs;o A con la le-<lb/>ua egualmente di&longs;tante dall'ori-<lb/>zonte. </s> <s id="id.2.1.870.2.0"> Siano BC EF egual <lb/>mente di&longs;tanti sì fra loro, come <lb/>ad e&longs;&longs;a KH, & à piombo al-<emph.end type="italics"/> <lb/> <arrow.to.target n="note240"></arrow.to.target><emph type="italics"/>l'orizonte della i&longs;te&longs;&longs;a KH, & <lb/>toccanti il cerchio CED ne i <lb/>punti EC, & &longs;ia congiunta la <lb/>EC laquale pa&longs;&longs;erà per lo cen <lb/>tro K, & &longs;arà egualmente di <lb/>&longs;tante dall'orizonte, &longs;i come pri-<lb/>ma è detto. </s> <s id="id.2.1.870.3.0"> Hor percioche la <lb/>girella CED &longs;i volge d'intor-<lb/>no K &longs;uo centro, però mentre <lb/>la forza di F tira sù il punto E <lb/>dourebbe di&longs;cendere il punto C <lb/>& tirare in giù B: mala cor-<lb/>da po&longs;ta in B è immobile, on-<lb/>de BC non può di&longs;cendere. <lb/> Per laqual co&longs;a mentre la po&longs;-<emph.end type="italics"/> <lb/> <arrow.to.target n="fig98"></arrow.to.target> <pb id="p.73" xlink:href="pageimg-it/164.jpg"/><emph type="italics"/>&longs;anza di F tira sùlo E, tutta la girella &longs;i mouerà in sù, & per con&longs;equenza tut-<lb/>ta la taglia, & il pe&longs;o; & EKC &longs;arà come leua, il cui &longs;o&longs;tegno &longs;arà C: pero-<emph.end type="italics"/> <arrow.to.target n="note241"></arrow.to.target> <lb/><emph type="italics"/>che il punto C per cau&longs;a di BC qua&longs;i è immobile, ma la po&longs;&longs;anza che moue la <lb/>leua è in F con la corda EF, & il pe&longs;o &longs;ta appiccato in K. Che &longs;e il punto <lb/>C fo&longs;&longs;e del tutto immobile, & &longs;i moua la leua EC in NC, & &longs;i diuida NC <lb/>in due parti eguali in L: &longs;aranno CL LN eguali ad e&longs;&longs;e CK KE. Per la <lb/>qual co&longs;a &longs;e la leua EC fo&longs;&longs;e in CN, il punto K &longs;arebbe in L: & &longs;e &longs;i con<lb/>duce&longs;&longs;e la linea LM à piombo dell'orizonte, laquale &longs;ia anche eguale alla KH, <lb/>&longs;arebbe il pe&longs;o A, cioè il punto H in M. Ma percioche la po&longs;&longs;anza di F men<lb/>tre và in &longs;u&longs;o mouendo la girella &longs;empre &longs;i moue &longs;opra la linea retta EFG, laquale <lb/>è anco egualmente di&longs;tante &longs;empre da BC, &longs;arà nece&longs;&longs;ario, che la girella della ta-<lb/>glia &longs;empre &longs;i trouitra le linee EG BC, & il centro K &longs;tando nel mezo, &longs;i mo-<lb/>uerà &longs;empre &longs;opra la linea retta HKT. Sia condotta adunque per L la linea <lb/>PT LQ egualmente di&longs;tante sì dall orizonte, come dalla EC, laquale &longs;eghi la <lb/>HK allungata in T, & co'l centro T, & lo &longs;patio TQ &longs;i formi il cerchio QR <lb/>PS, ilquale &longs;arà eguale al cerchio CED; & li punti PQ toccheranno le cor-<lb/>de FE BC ne i punti PQ. Peroche il rettangolo PECQ & la PT & la <lb/>TQ &longs;ono eguali ad e&longs;&longs;e EK KC. Dapoi per T &longs;ia tirato RTS diametro <lb/>del cerchio PQS egualmente di&longs;tante ad e&longs;&longs;a NC, & &longs;ia fatta TO eguale<emph.end type="italics"/> <arrow.to.target n="note242"></arrow.to.target> <lb/><emph type="italics"/>alla KH. Hor mentre il centro K &longs;arà mo&longs;&longs;o fin alla linea PQ all'hora il cen-<lb/>tro K &longs;arà in T. Ma egli è &longs;tato dimo&longs;trato, che il centro della girella &longs;i moue <lb/>&longs;empre per la linearetta HT. </s> <s id="id.2.1.870.3.10"> Onde accioche il centro K &longs;ia nella linea PQ egual-<lb/>mente di&longs;tante ad e&longs;&longs;a EC, egli è nece&longs;&longs;ario, che e&longs;&longs;o &longs;ia in T: & accioche an-<lb/>chora la leua EC &longs;i alzi nell'angolo ECN egli è nece&longs;&longs;ario, che &longs;ia in RS & <lb/>non in CN percioche l'angolo RSE all'angolo NCE è eguale & co&longs;i il &longs;o-<emph.end type="italics"/> <arrow.to.target n="note243"></arrow.to.target> <lb/><emph type="italics"/>&longs;tegno C non è del tutto immobile, mouendo&longs;i tutta la girella all'insù, & tutta <lb/>mutt'il luogo: nondimeno il C ha ragione di &longs;o&longs;tegno, peroche meno &longs;i moue C <lb/>di quel che fà K & E, percioche &longs;i moue il punto E fin ad R, & il K fin al T, <lb/>ma il punto C fin ad S &longs;olamente. </s> <s id="id.2.1.870.4.0"> Per laqual co&longs;a mentre il centro K &longs;i troua <lb/>in T, il &longs;ito della girella &longs;arà QRPS: & il pe&longs;o A, cioè il punto H &longs;arà <lb/>in O, e&longs;&longs;endo TO eguale à KH; ma il &longs;ito di EC, cioè della leua mo&longs;&longs;a, &longs;arà <lb/>RS: & la po&longs;&longs;anza di F &longs;arà mo&longs;&longs;a in &longs;u&longs;o per la rettalinea EFG: ma nel-<lb/>l'i&longs;te&longs;&longs;o tempo, che K &longs;arà in T, &longs;ia la po&longs;&longs;anza in G; & mentre la leua EC in <lb/>que&longs;to modo &longs;i moue, rimangono pur &longs;empre GPBQ fraloro egua mente di-<lb/>&longs;tanti, & à piombo dell'orizonte, talche doue toccano la girella, come ne' punti <lb/>PQ, &longs;empre la linea PQ &longs;arà il diametro della girella & come leua egualmen-<lb/>te di&longs;tante dall'orizonte. </s> <s id="id.2.1.870.5.0"> Mentre dunque la girella &longs;i moue, & và attorno, &longs;em <lb/>pre anche &longs;i moue la leua EC, & &longs;empre rimane vn'altra leua nellagirella egual<lb/>mente di&longs;tante dall'orizonte, come PQ, per modo, che la po&longs;&longs;anza di F moua <lb/>il pe&longs;o, &longs;tando la leua egualmente di&longs;tante all'orizonte, il cui &longs;o&longs;tegno &longs;acà &longs;empre <lb/>nella linea CB, & il pe&longs;o nel mezo della leua appiccato: & la po&longs;&longs;anza nella li-<lb/>nea EG, che era da mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig98" place="text" xlink:href="figures-it/163_01.jpg"></figure> <p type="margin" id="id.2.1.873.0.0"> <s id="id.2.1.873.1.0"> <margin.target id="note240"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.874.0.0"> <s id="id.2.1.874.1.0"> <margin.target id="note241"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.875.0.0"> <s id="id.2.1.875.1.0"> <margin.target id="note242"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 34. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.876.0.0"> <s id="id.2.1.876.1.0"> <margin.target id="note243"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 29. <emph type="italics"/>del primo.<emph.end type="italics"/> </s> </p> <pb id="p.73v" xlink:href="pageimg-it/165.jpg"/> <p type="main" id="id.2.1.877.0.0"> <s id="id.2.1.877.1.0"> Stando le co&longs;e i&longs;te&longs;&longs;e. </s> <s id="id.2.1.877.2.0"> Lo &longs;patio della po&longs;&longs;anza, che moue il pe-<lb/>&longs;o è il doppio dello &longs;patio dell'i&longs;te&longs;&longs;o pe&longs;o mo&longs;&longs;o. </s> </p> <p type="main" id="id.2.1.878.0.0"> <s id="id.2.1.878.1.0"><emph type="italics"/>E&longs;&longs;endo &longs;tato dimo&longs;trato, che mentre il K &longs;tà nel T, il pe&longs;o A cioè il punto H <lb/>e&longs;&longs;ere in O: & nell'i&longs;te&longs;&longs;o tempo ancora la po&longs;&longs;anza di F e&longs;&longs;ere in G: & per-<lb/>cioche la corda BCDEF eguale è alla corda EQSPG, peroche è la mede&longs;ima <lb/>corda: & la corda che è inuolta intorno al mezo cerchio CDE eguale è alla cor-<lb/>da, che &longs;ta d'intorno al mezo cerchio QSP: tolti via dunque li due pezzi di cor<lb/>da communi BQ, & FP: &longs;arà il re&longs;lante della corda FG eguale ad e&longs;&longs;i due <lb/>pezzi di corda rima&longs;i CQ & EP in&longs;ieme pre&longs;i. </s> <s id="id.2.1.878.2.0"> Ma EP eguale è al TK, <lb/>& il CQ &longs;arà anche eguale ad e&longs;&longs;o TK, peroche &longs;ono PK & TC parallelo-<lb/>grammi rettangoli. </s> <s id="id.2.1.878.3.0"> Per laqual co&longs;a le linee EPCQ in&longs;ieme &longs;ono due volte tan <lb/>to, quanto è TK. Adunque la corda FC &longs;arà due volte tanto quanto la TK. <lb/>& percioche la KH è eguale alla TO, leuando via la corda commune KO &longs;a <lb/>rà la KT eguale ad e&longs;&longs;a KO. Per laqual co&longs;a la corda FG &longs;arà due volte tan-<lb/>to quanto e&longs;&longs;a HO: cioè lo &longs;patio d'lla po&longs;&longs;anza due volte tanto quanto lo &longs;pa-<lb/>tio del pe&longs;o, che era da mo&longs;trare.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.879.0.0"> <s id="id.2.1.879.1.0"> Parallelogrammi rettangoli. </s> <s id="id.2.1.879.2.0"> Vuol dire &longs;igure di linee egualmente diftanti fra loro, <lb/>lequali formino angoli retti à differenza di altre figure, che &longs;e ben &longs;ono di linee <lb/>egualmente di&longs;tanti, non formano tuttauia angoli retti. </s> </p> <p type="main" id="id.2.1.880.0.0"> <s id="id.2.1.880.1.0"> Dapoi la po&longs;&longs;anza mouerà il pe&longs;o i&longs;te&longs;&longs;o in tempo eguale per la <lb/>metà dello &longs;patio, con la corda inuolta d'intorno alla girella <lb/>della taglia legata al pe&longs;o, che &longs;enza taglia; pur che le veloci-<lb/>tà de' mouimenti di e&longs;&longs;a po&longs;&longs;anza &longs;iano eguali. </s> </p> <pb id="p.74" xlink:href="pageimg-it/166.jpg"/> <p type="main" id="id.2.1.882.0.0"> <s id="id.2.1.882.1.0"><emph type="italics"/>Peroche &longs;ia, &longs;tando le co&longs;e i&longs;te&longs;&longs;e, vn'altro pe&longs;o V eguale al pe&longs;o A al quale &longs;ia <lb/>legata la corda 36>X< & &longs;ia in X la po&longs;&longs;anza, che moue il pe&longs;o V. Dico, &longs;e le ve <lb/>locità de' mouimenti dell'vna, & l'altra po&longs;&longs;anza &longs;aranno eguali, che la po&longs;&longs;anza<emph.end type="italics"/> <lb/> <arrow.to.target n="fig99"></arrow.to.target> <lb/><emph type="italics"/>di<emph.end type="italics"/> F <emph type="italics"/>mouerà il pe&longs;o A nell'i&longs;te&longs;&longs;o tempo per la metà dello &longs;patio, per lo quale <lb/>il pe&longs;o V &longs;arà mo&longs;&longs;o dalla po&longs;&longs;anza di X, che è il mede&longs;imo, come &longs;el'i&longs;te&longs;&longs;o pe-<lb/>&longs;o in tempo eguale fo&longs;&longs;e mo&longs;&longs;o. </s> <s id="id.2.1.882.2.0"> Moua la po&longs;&longs;anza di X il pe&longs;o V, & la po&longs;&longs;an-<lb/>za peruenga in <foreign lang="greek">*u</foreign>; & &longs;ia X<foreign lang="greek">*u</foreign> eguale ad e&longs;&longs;a FG: & &longs;i faccia <foreign lang="greek">*u</foreign>Z eguale <lb/>à X36><, talche quando la po&longs;&longs;anza di X &longs;arà in <foreign lang="greek">*u</foreign>, &longs;ia il pe&longs;o V cioè il punto 36><<emph.end type="italics"/> <pb id="p.74v" xlink:href="pageimg-it/167.jpg"/><emph type="italics"/>in Z; ma 36>Z< è eguale ad FG, e&longs;&longs;endo eguale ad X<foreign lang="greek">*u</foreign>: dunque 36>Z< &longs;arà due <lb/>volte tanto, quanto OH. Per laqual co&longs;a mentre le po&longs;&longs;anze &longs;aranno in G<foreign lang="greek">*u</foreign>, i <lb/>pe&longs;i AV &longs;aranno in OZ. Hor nell'i&longs;te&longs;&longs;o tempo &longs;aranuo le po&longs;&longs;anze in G<foreign lang="greek">*u</foreign>, <lb/>peroche le velocità de mouimenti &longs;ono eguali: onde la forza di F mouerà il pe-<lb/>&longs;o A nel mede&longs;imo tempo per la metà di quello &longs;patio, per loquale il pe&longs;o V &longs;a<emph.end type="italics"/> <lb/> <arrow.to.target n="fig100"></arrow.to.target> <lb/><emph type="italics"/>rà mo&longs;&longs;o dalla po&longs;&longs;anza di X: & li pe&longs;i &longs;ono eguali, adunque la po&longs;&longs;anza moue-<lb/>rà il pe&longs;o i&longs;l e&longs;&longs;o in tempo eguale per la metà dello &longs;patio, con la corda, & la taglia <lb/>legata in que&longs;to modo al pe&longs;o, che &longs;enza taglia; purche le velocità della po&longs;&longs;anza <lb/>de'mouimenti &longs;iano eguali, che era da mo&longs;trar&longs;i.<emph.end type="italics"/> </s> </p> <figure id="fig99" place="text" xlink:href="figures-it/166_01.jpg"></figure> <figure id="fig100" place="text" xlink:href="figures-it/167_01.jpg"></figure> <pb id="p.75" xlink:href="pageimg-it/168.jpg"/> <p type="head" id="id.2.1.886.0.0"> <s id="id.2.1.886.1.0"> PROPOSITIONE XII. </s> </p> <p type="main" id="id.2.1.887.0.0"> <s id="id.2.1.887.1.0"> Se la corda &longs;arà riuolta d'intorno à più girelle, legando l'vno de' <lb/>capi &longs;uoi in qualche loco, & l'altro &longs;ia tenuto dalla po&longs;&longs;anza, <lb/>che moue il pe&longs;o: La po&longs;&longs;anza mouerà con le leue &longs;empre <lb/>egualmente di&longs;tanti dall'orizonte. </s> </p> <p type="main" id="id.2.1.888.0.0"> <s id="id.2.1.888.1.0"><emph type="italics"/>Sia il pe&longs;o A. &longs;ia la girella CED della <lb/>taglia legata al pe&longs;o da KS ad angoli ret<lb/>ti all'orizonte; di modo, che il pe&longs;o &longs;egua <lb/>&longs;empre il &longs;uo mouimento ò &longs;u&longs;o, ò giu&longs;o, <lb/>che &longs;ia fatto. </s> <s id="id.2.1.888.2.0"> Sia dapoi la girella intorno <lb/>al centro L della taglia appiccata di &longs;opra; <lb/>& &longs;iala corda BCDEHMNO riuol-<lb/>ta d'intorno alle girelle, laquale &longs;ia legata <lb/>in B; & &longs;ia in O la forza mouente il <lb/>pe&longs;o A, mouendo&longs;i al ba&longs;&longs;o per OP. Di-<lb/>co che la po&longs;&longs;anza di O mouerà &longs;empre il <lb/>pe&longs;o A con le leue &longs;empre egualmente <lb/>di&longs;tanti dall'orizonte. </s> <s id="id.2.1.888.3.0"> &longs;ia tirata la linea <lb/>NH per lo centro L egualmente di&longs;tan-<lb/>te dall'orizonte, che &longs;arà la leua della girel-<emph.end type="italics"/> <arrow.to.target n="note244"></arrow.to.target> <lb/><emph type="italics"/>la, il cui centro è L: &longs;ia tirata da poi la <lb/>EC per lo centro K, &longs;imilmente di&longs;tan-<emph.end type="italics"/> <arrow.to.target n="note245"></arrow.to.target> <lb/><emph type="italics"/>te egualmente dall'orizonte, la quale &longs;arà <lb/>anche la leua della girella, il cui centro è<emph.end type="italics"/> <arrow.to.target n="note246"></arrow.to.target> <lb/><emph type="italics"/>K. Moua&longs;i la po&longs;&longs;anza di O in giu&longs;o, la <lb/>quale mentre in giu&longs;o &longs;i moue, mouerà la <lb/>leua NH, & mentre la leua &longs;i moue, la <lb/>N &longs;i mouerà in giu&longs;o, & la H in &longs;u&longs;o, <lb/>come è detto di &longs;opra. </s> <s id="id.2.1.888.4.0"> Mamentre la H <lb/>&longs;i moue in &longs;u&longs;o, moue etiandio in &longs;u&longs;o la E, <lb/>& la leua EC, il cui &longs;o&longs;tegno è C, ma <lb/>il &longs;o&longs;tegno C non puote mouere in giu&longs;o <lb/>il B; però la girella il cui centro è K mo <lb/>uera&longs;&longs;i in &longs;u&longs;o, & per con&longs;equenza la ta-<lb/>glia, & il pe&longs;o A, come nella preceden-<lb/>te è stato detto. </s> <s id="id.2.1.888.5.0"> & perche per la mede&longs;i-<lb/>ma cau&longs;a, che è stata a&longs;&longs;egnata nelle pre-<lb/>cedenti, rimangono &longs;empre le leue egual-<lb/>mente distanti dall'orizonte in HN, &<emph.end type="italics"/> <lb/> <arrow.to.target n="fig101"></arrow.to.target> <pb id="p.75v" xlink:href="pageimg-it/169.jpg"/><emph type="italics"/>in EC, la po&longs;&longs;anza dun-<lb/>que mouente il pe&longs;o A <lb/>lo mouerà &longs;empre &longs;tando <lb/>le leue egualmente distan-<lb/>ti dall'orizonte; che erada <lb/>mo&longs;trar&longs;i.<emph.end type="italics"/> </s> </p> <figure id="fig101" place="text" xlink:href="figures-it/168_01.jpg"></figure> <p type="margin" id="id.2.1.890.0.0"> <s id="id.2.1.890.1.0"> <margin.target id="note244"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 1. <emph type="italics"/>&<emph.end type="italics"/> 10. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.891.0.0"> <s id="id.2.1.891.1.0"> <margin.target id="note245"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="margin" id="id.2.1.892.0.0"> <s id="id.2.1.892.1.0"> <margin.target id="note246"></margin.target><emph type="italics"/>Par la<emph.end type="italics"/> 10. <emph type="italics"/>di questo.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.893.0.0"> <s id="id.2.1.893.1.0"><emph type="italics"/>Et &longs;e la corda &longs;arà riuolta d'in <lb/>torno à più girelle; &longs;imil-<lb/>mente &longs;i dimo&longs;trerà la po&longs;-<lb/>&longs;anza mouere il pe&longs;o con <lb/>le leue &longs;empre egualmente <lb/>di&longs;tanti dall'orizonte: & <lb/>le leue delle girelle della ta<lb/>glia di &longs;opra &longs;empre e&longs;&longs;e-<lb/>re come HN, i &longs;o&longs;tegni <lb/>delle quali &longs;aranno &longs;empre <lb/>nel mezo: ma le leue delle <lb/>girelle della taglia di &longs;otto <lb/>&longs;empre e&longs;&longs;ere, come EC; <lb/>li cui &longs;o&longs;tegni &longs;aranno nel-<lb/>le &longs;tremità delle leue.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.894.0.0"> <s id="id.2.1.894.1.0"> Stando le co&longs;e i&longs;te&longs;&longs;e, <lb/>lo &longs;patio della po&longs;-<lb/>&longs;anza, è il doppio <lb/>dello &longs;patio del pe-<lb/>&longs;o. </s> </p> <figure place="text" id="id.2.1.895.0.0" xlink:href="figures-it/169_01.jpg"></figure> <p type="main" id="id.2.1.896.0.0"> <s id="id.2.1.896.1.0"><emph type="italics"/>Sia mo&longs;&longs;o il centro K fin al centro R; & &longs;ia la girella FTG: poi &longs;ia per lo cen-<lb/>tro R condotta la linea GF egualmente di&longs;tante da e&longs;&longs;a EC: le corde EH <lb/>CB toccheranno la girella ne i punti GF. Faccia&longs;i alla fine RQ eguale à <lb/>KS. Mentre dunque K &longs;arà in R, il pe&longs;o A, cioè il punto S &longs;arà in Q,<emph.end type="italics"/> <pb id="p.76" xlink:href="pageimg-it/170.jpg"/><emph type="italics"/>& mentre il centro della girella è in R, &longs;ia la po&longs;&longs;anza di O mo&longs;&longs;a in P. & <lb/>percioche la corda BCDEHMNO eguale è alla corda BFTGHMNP <lb/>per e&longs;&longs;er la corda iste&longs;&longs;a, & FTG è eguale à CDE; leuate via dunque le com-<lb/>muni BF & GHMNO, &longs;arà la re&longs;tante OP eguale ad e&longs;&longs;e FC EG pre-<lb/>&longs;e in&longs;ieme: & per con&longs;equenza due volte tanto, quanto è KR, & QS. & e&longs;-<lb/>&longs;endo OP lo &longs;patio della po&longs;&longs;anza mo&longs;&longs;a, & SQ lo &longs;patio del pe&longs;o mo&longs;&longs;o; <lb/>&longs;arà lo &longs;patio della po&longs;&longs;anza due volte tanto quanto lo &longs;patio del pe&longs;o. </s> <s id="id.2.1.896.2.0"> che era <lb/>da mostrar&longs;i.<emph.end type="italics"/> </s> </p> <p type="main" id="id.2.1.897.0.0"> <s id="id.2.1.897.1.0"> Oltre à ciò la po&longs;&longs;anza mouerà il pe&longs;o i&longs;te&longs;&longs;o in tempo eguale <lb/>per la metà dello &longs;patio, con vna corda riuolta d'intorno à <lb/>due girelle, l'una delle quali &longs;ia della taglia di &longs;opra, & l'altra <lb/>&longs;ia della taglia legata al pe&longs;o; che &longs;enza taglie: pur che i mo-<lb/>uimenti di e&longs;&longs;a po&longs;&longs;anza &longs;iano egualmente veloci. </s> </p> <pb id="p.76v" xlink:href="pageimg-it/171.jpg"/> <p type="main" id="id.2.1.898.0.0"> <s id="id.2.1.898.1.0"><emph type="italics"/>Percioche &longs;tando le co-<lb/>&longs;e i&longs;te&longs;&longs;e, &longs;ia il pe&longs;o <lb/>V eguale ad e&longs;&longs;o A, <lb/>alquale &longs;ia legata la <lb/>corda X36>%< & &longs;ia <lb/>la po&longs;&longs;anza in X che <lb/>moue il pe&longs;o V; la <lb/>quale mentre moue <lb/>il pe&longs;o, peruenga in <lb/> <foreign lang="greek">*u</foreign>: & &longs;iano fatte <lb/>X<foreign lang="greek">*u</foreign> Z36>< eguali ad <lb/>e&longs;&longs;a OP; &longs;arà Z36><lb/>due volte tanto qua-<lb/>to QS. & &longs;e le <lb/>velocità de' moui-<lb/>menti dell'vna, & <lb/>l'altra po&longs;&longs;anza &longs;a-<lb/>ranno eguali; egli è <lb/>manife&longs;to, che il pe-<lb/>&longs;o V trapa&longs;&longs;a due <lb/>volte tanto &longs;patio <lb/>nell'i&longs;te&longs;&longs;o tempo, di <lb/>quel che trapa&longs;&longs;i il <lb/>pe&longs;o A: percioche <lb/>nel tempo mede&longs;imo <lb/>la po&longs;&longs;anza di X per-<lb/>uiene ad <foreign lang="greek">*u</foreign>, & la <lb/>po&longs;&longs;anza di O à P; <lb/>& li pe&longs;i &longs;imilmen-<lb/>te in <expan abbr="Zq.">Zque</expan> che era <lb/>da mo&longs;trar&longs;i.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.899.0.0" xlink:href="figures-it/171_01.jpg"></figure> <p type="head" id="id.2.1.900.0.0"> <s id="id.2.1.900.1.0"> PROPOSITIONE XIII. </s> </p> <p type="main" id="id.2.1.901.0.0"> <s id="id.2.1.901.1.0"> Riuolgendo la corda d'intorno à due girelle di due taglie, I'vna <lb/>delle quali &longs;ia di &longs;opra, & l'altra di &longs;otto, & legata al pe&longs;o; e&longs;-<lb/>&longs;endo anche l'vno de' capi di detta corda legato alla taglia di <lb/>&longs;otto, & l'altro tenuto dalla po&longs;&longs;anza che moue; &longs;arà lo &longs;patio <lb/>cor&longs;o della po&longs;&longs;anza, che tira, tre volte tanto quanto lo &longs;pati <lb/>del pe&longs;o mo&longs;&longs;o. </s> </p> <pb id="p.77" xlink:href="pageimg-it/172.jpg"/> <p type="main" id="id.2.1.903.0.0"> <s id="id.2.1.903.1.0"><emph type="italics"/>Sia il pe&longs;o A; &longs;ia BCD la girel-<lb/>la della taglia legata al pe&longs;o A, <lb/>attaccato da EQ, & &longs;ia E il <lb/>centro della girella; &longs;ia dapoi F <lb/>GH la girella della taglia appic-<lb/>cata di &longs;opra, il cui centro K; & <lb/>&longs;ia la corda LFGHDBCM ri-<lb/>uolta intorno à tutte le girelle, & <lb/>legata alla taglia di &longs;otto in L: <lb/>& &longs;ia in M la po&longs;&longs;anza, che <lb/>moue. </s> <s id="id.2.1.903.2.0"> Dico lo &longs;patio cor&longs;o dalla <lb/>po&longs;&longs;anza di M, mentre moue il <lb/>pe&longs;o, e&longs;&longs;ere triplo dello &longs;patio del <lb/>pe&longs;o mo&longs;&longs;o A. Moua&longs;i la po&longs;&longs;an <lb/>za di M fin ad N; & il centro <lb/>E &longs;ia mo&longs;&longs;o fin ad O; & L fin <lb/>à P; & il pe&longs;o A, cioè il pun-<lb/>to Q fin ad R; & la girella <lb/>mo&longs;&longs;a &longs;ia TSV. Siano condot-<lb/>te per EO le linee ST BD <lb/>egualmente di&longs;tanti dall'orizonte, <lb/>lequali &longs;aranno anche tra loro e-<lb/>gualmente di&longs;tanti. </s> <s id="id.2.1.903.3.0"> Ma percio-<lb/>che mentre E &longs;ta in O, il pun-<lb/>to Q &longs;ta in R; &longs;arà EQ egua <lb/>le ad OR, & EO ade&longs;&longs;o QR <lb/>eguale; &longs;imilmente LQ &longs;arà <lb/>eguale à PR, & LP ad e&longs;&longs;o <lb/>QR eguale. </s> <s id="id.2.1.903.4.0"> Adunque le tre <lb/>QR EO LP fra loro &longs;aranno <lb/>eguali; à cui &longs;ono etiandio eguali <lb/>BS DT. Et percioche la corda <lb/>LFGHDCBM è eguale alla <lb/>corda PFGHTVSN e&longs;&longs;en-<lb/>do vna corda i&longs;te&longs;&longs;a, & la corda, <lb/>che è intorno al mezo cerchio <lb/>TVS è eguale alla corda, che è <lb/>intorno al mezo cerchio BCD; <lb/>tolte via dunque le communi PF <lb/>GHT, & SM; &longs;arà la re&longs;tan-<lb/>te MN eguale alle tre BS <lb/>LP DT pre&longs;e in&longs;ieme. </s> <s id="id.2.1.903.5.0"> ma BS LP DT in&longs;ieme &longs;ono tre volte tanto, quanto<emph.end type="italics"/> <lb/> <arrow.to.target n="fig102"></arrow.to.target> <pb id="p.77v" xlink:href="pageimg-it/173.jpg"/><emph type="italics"/>EO, & per con&longs;e-<lb/>quenza QR. Lo <lb/>&longs;patio dunque MN <lb/>della traportata po&longs;<lb/>&longs;anza è tre volte<lb/>tanto, quanto lo &longs;pa<lb/>tio QR del pe&longs;o <lb/>mo&longs;&longs;o. </s> <s id="id.2.1.903.6.0"> che era da <lb/>mo&longs;trar&longs;i.<emph.end type="italics"/> </s> </p> <figure id="fig102" place="text" xlink:href="figures-it/172_01.jpg"></figure> <p type="main" id="id.2.1.905.0.0"> <s id="id.2.1.905.1.0"><emph type="italics"/>Il tempo ancora di que<lb/>sto mouimento è <lb/>mani&longs;e&longs;to, percio-<lb/>che la po&longs;&longs;anza i&longs;te&longs; <lb/>&longs;a in tempo eguale <lb/>mouerà l'i&longs;te&longs;&longs;o pe-<lb/>&longs;o in i&longs;patio tre co-<lb/>tanto maggiore &longs;en-<lb/>za tali taglie, di <lb/>quel che &longs;arebbe <lb/>con e&longs;&longs;e taglie à que <lb/>&longs;to modo commoda <lb/>te. </s> <s id="id.2.1.905.2.0"> Lo &longs;patio del <lb/>pe&longs;o mo&longs;&longs;o &longs;enza le <lb/>taglie è eguale allo <lb/>&longs;patio della po&longs;&longs;an-<lb/>za. </s> <s id="id.2.1.905.3.0"> & in que&longs;to <lb/>modo ritrouaremo <lb/>in tutte il tempo.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.906.0.0" xlink:href="figures-it/173_01.jpg"></figure> <pb id="p.78" xlink:href="pageimg-it/174.jpg"/> <p type="head" id="id.2.1.908.0.0"> <s id="id.2.1.908.1.0"> PROPOSITIONE XIIII. </s> </p> <p type="main" id="id.2.1.909.0.0"> <s id="id.2.1.909.1.0"> Legando la corda d'intorno à tre girelle di due taglie, l'vna del-<lb/>lequali &longs;ia di &longs;opra, & habbia vna &longs;ola girella, & l'altra di &longs;ot-<lb/>to, & ne habbia due, & &longs;ia lega<lb/>ta al pe&longs;o; laqual corda &longs;ia le-<lb/>gata con l'vno de' capi &longs;uoi in <lb/>qualche loco, & l'altro tenu-<lb/>to dalla po&longs;&longs;anza, che moue il <lb/>pe&longs;o: &longs;arà lo &longs;patio cor&longs;o dal-<lb/>la po&longs;&longs;anza, che tira, quattro <lb/>volte tanto, quanto è lo &longs;patio <lb/>del pe&longs;o mo&longs;&longs;o. </s> </p> <p type="main" id="id.2.1.910.0.0"> <s id="id.2.1.910.1.0"><emph type="italics"/>Sia il pe&longs;o A, &longs;iano le due girelle, i cui cen<lb/>tri<emph.end type="italics"/> K <emph type="italics"/>I della taglia legata al pe&longs;o con <lb/>K<foreign lang="greek">a</foreign>; dimodo, che il pe&longs;o &longs;empre &longs;egua il <lb/>mouimento della taglia in &longs;u&longs;o, ouero in <lb/>giu&longs;o: &longs;ia dapoi la girella il cui centro L <lb/>della taglia appe&longs;a di &longs;opra in <foreign lang="greek">d</foreign>; & &longs;ia la <lb/>corda BCDEFGHZMNO riuolta <lb/>intorno à tutte le girelle, & legata in B; <lb/>& &longs;ia in O la po&longs;&longs;anza, che moue il pe-<lb/>&longs;o A. Dico lo &longs;patio, ilquale la po&longs;&longs;an <lb/>za di O mouendo trapa&longs;&longs;a, e&longs;&longs;ere quat-<lb/>tro volte tanto, quanto lo &longs;patio del pe-<lb/>&longs;o A mo&longs;&longs;o. </s> <s id="id.2.1.910.2.0"> Mouan&longs;i le girelle della <lb/>taglia legata al pe&longs;o; & mentre il centro <lb/>K è in R, il centro I &longs;ia in S, & il <lb/>pe&longs;o A, cioè il punto <foreign lang="greek">a</foreign> in <foreign lang="greek">b</foreign>: &longs;aranno <lb/>IS KR <foreign lang="greek">ab</foreign> tra &longs;e eguali, & parimen-<lb/>te KI ad e&longs;&longs;a RS eguale: percioche le <lb/>girelle mantengono fra &longs;e la di&longs;tanza me<lb/>de&longs;ima &longs;empre; & K<foreign lang="greek">a</foreign> &longs;arà eguale ad e&longs; <lb/>&longs;a R<foreign lang="greek">b</foreign>. </s> <s id="id.2.1.910.3.0"> &longs;iano condotte per li centri delle <lb/>girelle le linee FHQTECVXNZ <lb/>egualmente distanti dall orizonte, lequa <lb/>li tocchino le corde nei punti FH QT<emph.end type="italics"/> <lb/> <arrow.to.target n="fig103"></arrow.to.target> <pb id="p.78v" xlink:href="pageimg-it/175.jpg"/><emph type="italics"/>EC VX NZ che parimente &longs;aranno <lb/>fra loro egualmente di&longs;tanti: & EQ CT <lb/>VN XZ non &longs;olamente fra &longs;e, ma <lb/>ancora ad e&longs;&longs;e IS KR <foreign lang="greek">ab</foreign> &longs;aranno e-<lb/>guali: & mentre li centri KI &longs;ono in <lb/>RS, la po&longs;&longs;anza di O &longs;ia mo&longs;&longs;a in P. <lb/> Et percioche la corda BCDEFGHZ<lb/>MNO è eguale alla corda BT36>QF<lb/>GHX<foreign lang="greek">*u</foreign>VP e&longs;&longs;endo vna corda mede-<lb/>&longs;ima, & le corde d'intorno à mezi cerchi <lb/>T36>QX<foreign lang="greek">*u</foreign>V &longs;ono eguali alle corde, che <lb/>&longs;ono d'intorno à CDE ZMN; tolte <lb/>via dunque le communi BT, QFGHX, <lb/>& VO; &longs;arà OP eguale ad e&longs;&longs;e VN <lb/>XZ CT QE pre&longs;e tutte in&longs;ieme. </s> <s id="id.2.1.910.4.0"> male <lb/>quattro VN ZX CT QE &longs;ono tra&longs;e <lb/>eguali, & in&longs;ieme quattro volte tanto <lb/>quanto KR & <foreign lang="greek">ab</foreign>. </s> <s id="id.2.1.910.5.0"> Per laqual co&longs;a OP <lb/>&longs;arà quattro volte tanto quanto è e&longs;&longs;a <lb/> <foreign lang="greek">ab</foreign>. </s> <s id="id.2.1.910.6.0"> Adunque lo &longs;patio della po&longs;&longs;anza <lb/>è quattro volte tanto quanto è lo &longs;palio <lb/>del pe&longs;o. </s> <s id="id.2.1.910.7.0"> che era da mo&longs;trare.<emph.end type="italics"/> </s> </p> <figure id="fig103" place="text" xlink:href="figures-it/175_01.jpg"></figure> <p type="main" id="id.2.1.912.0.0"> <s id="id.2.1.912.1.0"><emph type="italics"/>Et &longs;e la corda in P &longs;arà dauantaggio ri-<lb/>uolta d'intorno ad vn'altra girella ver&longs;o il <lb/> <foreign lang="greek">d</foreign>, & la po&longs;&longs;anza mouendo&longs;i in giù mo<lb/>ua in sù il pe&longs;o: &longs;imilmente &longs;i mo&longs;trerà <lb/>lo &longs;patio della po&longs;&longs;anza e&longs;&longs;ere quattro <lb/>volte tanto quanto lo &longs;patio del pe&longs;o.<emph.end type="italics"/> </s> </p> <figure place="text" id="id.2.1.913.0.0" xlink:href="figures-it/2000.30.0179.jpg"></figure> <p type="main" id="id.2.1.914.0.0"> <s id="id.2.1.914.1.0"><emph type="italics"/>Ma&longs;e la corda in B &longs;i riuolgerà d'intorno ad <lb/>vn'altra girella, laqual corda &longs;ilegbi da<emph.end type="italics"/> <lb/> <arrow.to.target n="note247"></arrow.to.target><emph type="italics"/>poi alla taglia di &longs;otto; &longs;arà la po&longs;&longs;anza <lb/>di O, che &longs;o&longs;tiene il pe&longs;o A vn quinto <lb/>dal pe&longs;o. </s> <s id="id.2.1.914.2.0"> &a