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Colored diff for /texts/archimedes/xml/Attic/monte_mecha_02_it_1581.xml between version 1.3 and 1.4

version 1.3, 2002/06/24 12:14:28

version 1.4, 2002/06/24 18:03:54

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<s id="id.2.1.88.7.0">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.2.1.88.8.0">& quantunque <gap/>il <lb/>B &longs;ia piu alto di qual &longs;i voglia al<lb/>tro punto del cerchio, t<gap/>tauia 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. </s>

<s id="id.2.1.88.9.0">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;er<gap/> 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>

<figure id="fig1"></figure><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;er<gap/> 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.2.1.88.10.0">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.2.1.88.11.0">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>

<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>

<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>

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<pb/><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. </s>

<s id="id.2.1.98.5.0">Per laqual <lb/>co&longs;a co'lcentro D, & <lb/>lo &longs;patio CD de&longs;cri<lb/>ua&longs;i il cerchio DGH. </s>

<s id="id.2.1.98.6.0"><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/>

<figure id="fig2"></figure><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. </s>

<s id="id.2.1.98.7.0">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>

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<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>

<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>

<s id="id.2.1.103.1.0">PROPOSITIONE III.</s>

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<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. </s>

<s id="id.2.1.113.2.0">Mo <lb/>ua&longs;i la bilancia come in <lb/>DE, & iui &longs;ia la&longs;cia<lb/>ta. </s>

<s id="id.2.1.113.3.0">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>

<figure id="fig3"></figure><lb/>te che la bilancia DE non &longs;i mouerà, & rimarrà in quel &longs;ito. </s>

<s id="id.2.1.113.4.0">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. </s>

<s id="id.2.1.113.5.0">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.2.1.113.6.0">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. </s>

<s id="id.2.1.113.7.0"><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.2.1.113.8.0">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. </s>

<s id="id.2.1.113.9.0"><lb/>Rimarrà dunque la bilancia oue &longs;arà la&longs;ciata, che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/></s>

<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>

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<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 egual</s><s id="id.2.1.119.2.0">/>& 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>

<figure id="fig4"></figure><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.2.1.119.3.0">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.2.1.119.4.0">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.2.1.119.5.0">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>

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<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>

<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>

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<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 C</s><s id="id.2.1.123.6.0">/>& &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. </s>

<s id="id.2.1.123.7.0"><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/>

<figure id="fig5"></figure><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.2.1.123.8.0">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. </s>

<s id="id.2.1.123.9.0"><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. </s>

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<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>

<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>

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<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. </s>

<s id="id.2.1.130.2.0">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. </s>

<figure id="fig6"></figure><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. </s>

<s id="id.2.1.130.3.0">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.2.1.130.4.0">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. </s>

<s id="id.2.1.130.5.0">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>

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<s id="id.2.1.130.14.0">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. </s>

<s id="id.2.1.130.15.0">Adunque ha<lb/>uer à minore propor<lb/>tione RDH ad HD <lb/>G di quel che haurà <lb/>ODH ad HDG. </s>

<s id="id.2.1.130.16.0"><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. </s>

<figure id="fig7"></figure><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. </s>

<s id="id.2.1.130.17.0">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.2.1.130.18.0">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>

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<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>

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<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. </s>

<s id="id.2.1.140.4.0">Egli <lb/>è da mostrare l'angolo SE <lb/>G e&longs;&longs;ere minore dell'ango <lb/>lo SDG. </s>

<s id="id.2.1.140.5.0">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>

<figure id="fig8"></figure><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.2.1.140.6.0">Tutto l'angolo dun<lb/>que VEG &longs;arà eguale all'angolo SDM. </s>

<s id="id.2.1.140.7.0">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. </s>

<s id="id.2.1.140.8.0">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>

<pb/>

<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. </s><s id="id.2.1.143.4.0">Con&longs;ide-<emph.end type="italics"/><lb/>

<figure id="fig9"></figure><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. </s><s id="id.2.1.143.4.0">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.2.1.143.5.0">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. </s>

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<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 <expan abbr="Lq.">Lque</expan> 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 pagenum="9"/><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. </s><s id="id.2.1.143.9.0">& 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<gap/> Per la<lb/>qual co&longs;a<gap/> pe&longs;o &longs;a <lb/>rà più graue in D, <lb/>che in L, ilche pa <lb/>rimente auiene ne <lb/>punti NM. </s><s id="id.2.1.143.10.0">& <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"/>

<figure id="fig10"></figure><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. </s>

<s id="id.2.1.143.11.0">Per laqual co&longs;a per la pre&longs;uppo&longs;ta il pe<emph.end type="italics"/>

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<s id="id.2.1.153.1.0"><margin.target id="note27"></margin.target><emph type="italics"/>Giordano nella &longs;econda propo&longs;iti<gap/> ne.<emph.end type="italics"/></s>

<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>

<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>

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<s id="id.2.1.159.3.0">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.2.1.159.4.0">Quello ste&longs;&longs;o accade parimente all'altra ragione, laquale adducono dal <lb/>mouimento piu diritto, & piu torto. </s>

<s id="id.2.1.159.5.0">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 D A.</s><s id="id.2.1.159.6.0">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>

<figure id="fig11"></figure><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>

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<s id="id.2.1.161.6.0">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. </s>

<s id="id.2.1.161.7.0">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.2.1.161.8.0">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>

<figure id="fig12"></figure><lb/><emph type="italics"/>de' &longs;iti &longs;arà piu graue, & piu lieue. </s>

<s id="id.2.1.161.9.0">& 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/><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. </s>

<s id="id.2.1.161.10.0"><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. </s>

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<s id="id.2.1.161.19.0"><lb/>Per laqual co&longs;a piu graue &longs;arà in D, che in <lb/>L. </s>

<s id="id.2.1.161.20.0">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.2.1.161.21.0">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"/>

<figure id="fig13"></figure><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 pagenum="11"/><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. </s><s id="id.2.1.161.22.0">& 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.2.1.161.23.0">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>

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<s id="id.2.1.165.1.0"><margin.target id="note30"></margin.target><emph type="italics"/>Per la deci ma ottau<gap/> del terzo<emph.end type="italics"/></s>

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<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. </s>

<figure id="fig14"></figure><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. </s>

<s id="id.2.1.179.3.0">So&longs;tenterà dunque piu la linea CD, che C</s><s id="id.2.1.179.4.0">/>& 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. </s>

<s id="id.2.1.179.5.0">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. </s>

<s id="id.2.1.179.6.0">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. </s>

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<s id="id.2.1.179.9.0">Adunque il pe&longs;o po&longs;to in K &longs;arà<emph.end type="italics"/>

<pb pagenum="13"/><emph type="italics"/>piu graue, che in D. </s>

<s id="id.2.1.179.10.0">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>

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<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. </s>

<s id="id.2.1.181.3.0"><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.2.1.181.4.0">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. </s>

<figure id="fig15"></figure><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. </s>

<s id="id.2.1.181.5.0">Hor percioche LE DM &longs;i taglia<lb/><gap/>o in&longs;ieme in S, &longs;arà il rettangolo LSE eguale al rettangolo DSM. </s>

<s id="id.2.1.181.6.0">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. </s>

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<s id="id.2.1.181.16.0">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. </s>

<s id="id.2.1.181.17.0">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. </s><s id="id.2.1.181.18.0">& 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. </s>

<s id="id.2.1.181.19.0"><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>

<figure id="fig16"></figure><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.2.1.181.20.0">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>

<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>

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<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 quinte.<emph.end type="italics"/></s>

<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>

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<s id="id.2.1.196.3.0">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.2.1.196.4.0">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.2.1.196.5.0">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="p&utilde;to">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.2.1.196.6.0">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. </s>

<s id="id.2.1.196.7.0">Similmente l'arco DA piglie<lb/>rà piu del diretto, che la circon&longs;erenza EV. </s><s id="id.2.1.196.8.0">onde &longs;arà vera la pre&longs;uppo&longs;ta, & le <lb/>altre dimo&longs;trationi rimarranno nella &longs;ua &longs;ua forza. </s>

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<s id="id.2.1.196.10.0">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>

<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>

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<pb/><emph type="italics"/>mouer&longs;i &longs;olamente fin ad S. </s>

<s id="id.2.1.201.9.0">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-rãno">dimostre<lb/>ranno</expan> giamai che per <lb/>uenga in AB. </s>

<s id="id.2.1.201.10.0">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. </s>

<figure id="fig18"></figure><lb/><emph type="italics"/>il pe&longs;o i&longs;te&longs;&longs;o &longs;arà in D graue egualmente, come in A. </s>

<s id="id.2.1.201.11.0">Ma &longs;e le portioni &longs;olamente <lb/>piglieremo da DA, &longs;arà piu graue in A, che in D. </s>

<s id="id.2.1.201.12.0">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.2.1.201.13.0">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.2.1.201.14.0">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>

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<s id="id.2.1.203.8.0">Per la <lb/>qualco&longs;a OVP &longs;arà <lb/>ottu&longs;o. </s>

<s id="id.2.1.203.9.0">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. </s>

<s id="id.2.1.203.10.0"><lb/>Caderà dun que nella li <lb/>nea OT nellaparte di <lb/>VT, et &longs;ia PT. </s><s id="id.2.1.203.11.0">&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. </s>

<figure id="fig19"></figure><lb/><emph type="italics"/>ritta &longs;ce&longs;a della circonferenza OP. </s>

<s id="id.2.1.203.12.0">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. </s>

<s id="id.2.1.203.13.0">Onde la OT &longs;arà parimente maggiore <lb/>della ON. </s><s id="id.2.1.203.14.0">& 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"/>

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<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>

<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>

<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="l&utilde;ge">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. </s>

<figure id="fig20"></figure><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. </s>

<s id="id.2.1.211.2.0"><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.2.1.211.3.0">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.2.1.211.4.0">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>

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<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>

<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>

<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"/>

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<s id="id.2.1.216.5.0">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.2.1.216.6.0">Allequali co&longs;e ri&longs;pondiamo. </s>

<s id="id.2.1.216.7.0">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>

<figure id="fig21"></figure><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>

<pb pagenum="18"/><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. </s><s id="id.2.1.216.9.0">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. </s>

<s id="id.2.1.216.10.0">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. </s>

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<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>

<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>

<s id="id.2.1.221.1.0">PROPOSITIONE V.</s>

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<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. </s>

<s id="id.2.1.257.8.0">Similmente percioche RP è egual<emph.end type="italics"/>

<pb/>

<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. </s>

<figure id="fig22"></figure><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. </s>

<s id="id.2.1.257.9.0">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. </s>

<s id="id.2.1.257.10.0">Dunque il punto P &longs;arà il centro<emph.end type="italics"/><lb/>

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<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>

<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>

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<pb/><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. </s>

<s id="id.2.1.269.14.0">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. </s>

<figure id="fig23"></figure><lb/><emph type="italics"/>&longs;o E in D. </s>

<s id="id.2.1.269.15.0">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. </s>

<s id="id.2.1.269.16.0">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>

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<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>

<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>

<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. </s><s id="id.2.1.276.2.0">faccia&longs;i AD <lb/>eguale ad AB, & <lb/>dal punto D &longs;ia <lb/>fatto <expan abbr="p&emacr;dente">pendente</expan> il pe <lb/>&longs;o G eguale al pe <lb/>&longs;o F, ilquale <expan abbr="eti&amacr;-">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. </s>

<figure id="fig24"></figure><lb/><emph type="italics"/>dio &longs;arà eguale ad E. </s>

<s id="id.2.1.276.3.0">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.2.1.276.4.0">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>

<s id="id.2.1.278.1.0">Altramente.</s>

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<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. </s>

<s id="id.2.1.283.3.0">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. </s>

<s id="id.2.1.283.4.0">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>

<figure id="fig25"></figure><lb/><emph type="italics"/>pe&longs;eranno egualmente. </s>

<s id="id.2.1.283.5.0">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>

<s id="id.2.1.285.1.0">COROLLARIO.</s>

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<s id="id.2.1.290.2.0">Appicchi&longs;i in A il pe&longs;o D, che pe&longs;i egualmente col marco E appic<lb/>cato in F. </s>

<s id="id.2.1.290.3.0">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. </s><s id="id.2.1.290.4.0">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. </s>

<s id="id.2.1.290.5.0"><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. </s>

<figure id="fig26"></figure><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. </s>

<s id="id.2.1.290.6.0">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.2.1.290.7.0">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>

<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>

<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>

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<s id="id.2.1.295.3.0">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. </s>

<s id="id.2.1.295.4.0">Hor perche <lb/>i pe&longs;i DE pe&longs;ano <expan abbr="egualm&emacr;">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. </s><s id="id.2.1.295.5.0">& 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"/>

<figure id="fig27"></figure><lb/><emph type="italics"/>te, &longs;arà D ad E, come CA à CB. </s><s id="id.2.1.295.5.0">& 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>

<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>

<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>

<s id="id.2.1.299.1.0">PROPOSITIONE VII.</s>

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<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. </s>

<figure id="fig28"></figure><lb/><emph type="italics"/>&longs;o in B. </s>

<s id="id.2.1.321.2.0">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/>

<arrow.to.target n="note99"></arrow.to.target> <emph type="italics"/>ver&longs;o la CB. </s>

<s id="id.2.1.321.3.0">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.2.1.321.4.0">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/>

<figure id="fig29"></figure><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. </s>

<s id="id.2.1.321.5.0">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>

<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 Archi mede delleco &longs;ache egual <expan abbr="m&etilde;te">mente</expan> pe&longs;ano.<emph.end type="italics"/></s>

<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>

<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>

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<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/>

<figure id="fig30"></figure><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. </s><s id="id.2.1.327.3.0">& 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. </s>

<s id="id.2.1.327.4.0">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>

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<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>

<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>

<s id="id.2.1.333.1.0">COROLLARIO.</s>

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<s id="id.2.1.337.5.0">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="d&utilde;">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. </s><s id="id.2.1.337.6.0">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. </s>

<figure id="fig31"></figure><lb/><emph type="italics"/>grauezza del pe&longs;o F, co&longs;i è AB ver&longs;o BD. </s><s id="id.2.1.337.6.0">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. </s>

<s id="id.2.1.337.7.0">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 attaccat<gap/> 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. </s>

<s id="id.2.1.337.8.0">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. </s>

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<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>

<s id="id.2.1.345.1.0"><margin.target id="note109"></margin.target><emph type="italics"/>Per lo Corol lario della<emph.end type="italics"/> 4 <emph type="italics"/>del quinto.<emph.end type="italics"/></s>

<s id="id.2.1.346.1.0">Altramente.</s>

<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. </s><s id="id.2.1.347.2.0">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. </s>

<figure id="fig32"></figure><lb/><emph type="italics"/>anco la po&longs;&longs;anza di A ver&longs;o il pe&longs;o E. </s>

<s id="id.2.1.347.3.0">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. </s><s id="id.2.1.347.4.0">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.2.1.347.5.0">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. </s>

<s id="id.2.1.347.6.0">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.2.1.347.7.0">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.2.1.347.8.0">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>

<s id="id.2.1.349.1.0">COROLLARIO I.</s>

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<s id="id.2.1.359.2.0">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. </s>

<s id="id.2.1.359.3.0">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>

<figure id="fig33"></figure><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>

<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>

<s id="id.2.1.362.1.0">COROLLARIO IIII.</s>

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<s id="id.2.1.368.2.0">Di<lb/>co che come AB à BD, co&longs;i è la po&longs;&longs;anza in D al pe&longs;o C. </s>

<s id="id.2.1.368.3.0">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. </s>

<s id="id.2.1.368.4.0"><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. </s>

<s id="id.2.1.368.5.0">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. </s>

<s id="id.2.1.368.6.0">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. </s>

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<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>

<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>

<s id="id.2.1.374.1.0">Altramente.</s>

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<s id="id.2.1.377.1.0"><emph type="italics"/>Sia la mede&longs;ima bilancia AB, il cui mezo C. </s><s id="id.2.1.377.2.0">dapoitutta la FG &longs;ia la trutina, <lb/>laquale &longs;tia immobile, & &longs;o&longs;tenga la bilancia AB nel punto C. </s><s id="id.2.1.377.3.0">& 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.2.1.377.4.0">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.2.1.377.5.0">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>

<figure id="fig35"></figure><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.2.1.377.6.0">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.2.1.377.7.0">Ma &longs;ia &longs;o&longs;tenuta la trutina da due po&longs;&longs;anze in F cioè, & in G,

<pb/>&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. </s>

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<s id="id.2.1.377.12.0">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>

<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>

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<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.2.1.381.8.0">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>

<figure id="fig36"></figure><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.2.1.381.9.0">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.2.1.381.10.0">& quando <lb/>la trutina &longs;tà in CG, mouer&longs;i in FG. </s>

<s id="id.2.1.381.11.0">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>

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<s id="id.2.1.384.1.0"><margin.target id="note117"></margin.target><emph type="italics"/>il Cardano.<emph.end type="italics"/></s>

<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"/>

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<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. </s>

<s id="id.2.1.390.2.0"><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.2.1.390.3.0">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>

<figure id="fig37"></figure><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>

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<s id="id.2.1.392.7.0"><lb/>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.2.1.392.8.0">Ma per lo con<lb/>trario auiene quando il <lb/>centro è &longs;otto la bilan<lb/>cia. </s>

<s id="id.2.1.392.9.0">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>

<figure id="fig38"></figure><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>

<pb/>

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<s id="id.2.1.395.5.0">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.2.1.395.6.0">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. </s><s id="id.2.1.395.7.0">& 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. </s>

<figure id="fig39"></figure><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. </s>

<s id="id.2.1.395.8.0">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.2.1.395.9.0">ma e&longs;&longs;endo EF eguale ad AB, &longs;arà la circonfe<lb/>renza EAF eguale alla circonferenza AFB. </s>

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<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>

<s id="id.2.1.399.1.0"><margin.target id="note121"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> <gap/>. <emph type="italics"/>del printo.<emph.end type="italics"/></s>

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<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. </s><s id="id.2.1.402.6.0">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/><emph type="italics"/>graue che in verun altro &longs;ito del cerchio DHM. </s>

<s id="id.2.1.402.7.0">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. </s>

<s id="id.2.1.402.8.0">Da <lb/>que&longs;to &longs;ito adunque &longs;i mouera piu velocemente che da qualunque altro. </s>

<s id="id.2.1.402.9.0">& &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. </s>

<s id="id.2.1.402.10.0"><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. </s>

<s id="id.2.1.402.11.0"><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>

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<s id="id.2.1.406.5.0">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. </s><s id="id.2.1.406.6.0">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. </s>

<s id="id.2.1.406.7.0">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. </s>

<figure id="fig41"></figure><lb/><emph type="italics"/>&longs;arà l'angolo LFP minore dell' angolo GEO. </s>

<s id="id.2.1.406.8.0">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. </s>

<s id="id.2.1.406.9.0">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. </s>

<s id="id.2.1.406.10.0">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>

<pb/>

<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>

<arrow.to.target n="note124"></arrow.to.target><emph type="italics"/>La ragione di Ari&longs;totele parimente qui è cbiara. </s>

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<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. </s>

<s id="id.2.1.415.4.0">Ma <lb/>e&longs;&longs;endo HE eguale ad <lb/>HF, &longs;arà EN mag<lb/>gior di NF. </s>

<s id="id.2.1.415.5.0">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 <gap/>i FM. </s><s id="id.2.1.415.6.0">& 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. </s>

<figure id="fig42"></figure><lb/><emph type="italics"/>giore <gap/>i FM. </s><s id="id.2.1.415.6.0">& 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. </s>

<s id="id.2.1.415.7.0">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>

<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>

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<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>

<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>

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<s id="id.2.1.425.1.0"><emph type="italics"/>Sia <expan abbr="all&umacr;gata">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>

<figure id="fig43"></figure><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. </s>

<s id="id.2.1.425.3.0">Adunque il pe&longs;o po&longs;to in E haurà grauezza maggiore del pe&longs;o po&longs;to in F. </s>

<s id="id.2.1.425.4.0"><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>

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<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 pagenum="25"/><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/>di cemmo<gap/>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;</s><s id="id.2.1.428.4.0">/>& 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>

<figure id="fig44"></figure><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.2.1.428.5.0">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>

<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>

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<s id="id.2.1.432.2.0"><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.2.1.432.3.0">&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="p&utilde;">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 A</s><s id="id.2.1.432.4.0">/>congiunga&longs;i la CE. </s><s id="id.2.1.432.5.0">& 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. </s>

<figure id="fig45"></figure><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 A</s><s id="id.2.1.432.4.0">/>congiunga&longs;i la CE. </s><s id="id.2.1.432.5.0">& 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. </s>

<s id="id.2.1.432.6.0">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.2.1.432.7.0">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 CD</s><s id="id.2.1.432.8.0">/>& 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>

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<s id="id.2.1.434.4.0">del le co&longs;e cgual <expan abbr="m&etilde;se">mense</expan> <expan abbr="pes&amacr;ti">pesanti</expan>.<emph.end type="italics"/></s>

<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>

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<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="fattam&emacr;">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>

<figure id="fig46"></figure><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. </s>

<s id="id.2.1.438.7.0">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>

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<s id="id.2.1.438.9.0">& 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.2.1.438.10.0">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"/></s>

<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 rim<gap/>rà.<emph.end type="italics"/></s>

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<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="bil&emacr;-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="fig47"></arrow.to.target><lb/><emph type="italics"/>rà piu pre&longs;&longs;o al &longs;ito egualmente di&longs;tante dall'orizonte. </s>

<figure id="fig47"></figure><lb/><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>

<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>

<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="bil&amacr;">bilam</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>

<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>

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<s id="id.2.1.452.5.0">Ritroui&longs;i il centro della <lb/>grauczza di tutta la magnitudine D, <lb/>& &longs;ia congiunta la CD. </s>

<s id="id.2.1.452.6.0">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="fig49"></arrow.to.target><lb/><emph type="italics"/>ie. </s>

<s id="id.2.1.452.7.0">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>

<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>

<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>

<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/>& lalinea 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>

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<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. </s>

<s id="id.2.1.462.3.0">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>

<figure id="fig51"></figure><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>

<s id="id.2.1.464.1.0">In que&longs;to luogo egli conuiene auertire, il che potcua&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. </s>

<s id="id.2.1.464.2.0">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>

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<s id="id.2.1.469.1.0"><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. </s>

<s id="id.2.1.469.2.0">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>

<figure id="fig52"></figure><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.2.1.469.3.0">& 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.2.1.469.4.0">& 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>

<s id="id.2.1.471.1.0">COROLLARIO I.</s>

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<s id="id.2.1.482.7.0">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'ango<gap/> 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/>

<figure id="fig53"></figure><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. </s>

<s id="id.2.1.482.8.0">Come dunque la circonferenza AE<emph.end type="italics"/>

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<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>

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<s id="id.2.1.489.4.0"><lb/>Moua&longs;i dunque A in sù <lb/>fin'in D; & &longs;ia il mouimen <lb/>to della leua BD. </s><s id="id.2.1.489.5.0">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>

<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>

<figure id="fig54"></figure><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>

<s id="id.2.1.492.1.0">COROLLARIO.</s>

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<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. </s><s id="id.2.1.498.2.0">&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. </s>

<s id="id.2.1.498.3.0">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="p&utilde;">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. </s>

<figure id="fig55"></figure><lb/><emph type="italics"/>ta tagli AF in L, & AC in M. </s>

<s id="id.2.1.498.4.0">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. </s>

<s id="id.2.1.498.5.0">Hor percioche DL &longs;tà à piombo dell' orizonte, il pe&longs;o AC venga ap-<emph.end type="italics"/>

<pb pagenum="40"/><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>

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<s id="id.2.1.498.10.0">che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/></s>

<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>

<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;s&amacr;za&imacr;">po&longs;sanzaim</expan> N al pe&longs;o, come BE à BN, et la <expan abbr="po&longs;s&amacr;za">po&longs;sanza</expan> <expan abbr="&imacr;">im</expan> K al pe&longs;o come FL ad FK.<emph.end type="italics"/></s>

<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>

<figure id="fig56"></figure><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>

<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 <expan abbr="Pq.">Pque</expan> &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. </s><s id="id.2.1.504.2.0">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. </s>

<figure id="fig57"></figure><lb/><emph type="italics"/>nella leua BL ne'punti BP; & nella leua BM dal punto B, & PR. </s><s id="id.2.1.504.2.0">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. </s>

<s id="id.2.1.504.3.0">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.2.1.504.4.0">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>

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<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>

<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>

<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>

<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>

<figure id="fig58"></figure><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>

<pb/>

<s id="id.2.1.512.1.0"><emph type="italics"/>Sia vltimamente la leua. </s>

<s id="id.2.1.512.2.0">AB egualmente di&longs;tante dall'orizonte, il cui &longs;ostegno &longs;ia <lb/>C, & il pe&longs;o DE habb<gap/>a 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>

<figure id="fig59"></figure><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.2.1.512.3.0">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 alpe&longs;o, come C<emph.end type="italics"/>F <emph type="italics"/>à CB, & la po&longs; <lb/>&longs;anza in L alpe&longs;o, come M<emph.end type="italics"/>F <emph type="italics"/>ad ML. </s>

<s id="id.2.1.512.4.0">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 <expan abbr="Kq.">Kque</expan> 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 <expan abbr="lã">lam</expan> 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>

<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>

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<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. </s>

<figure id="fig60"></figure><lb/><emph type="italics"/>po&longs;&longs;anze fo&longs;&longs;ero in CAD. </s>

<s id="id.2.1.516.2.0">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>

<s id="id.2.1.518.1.0">PROPOSITIONE VI.</s>

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<arrow.to.target n="note149"></arrow.to.target> <emph type="italics"/>golo &longs;iano eguali à due angoli <lb/>retti. </s>

<s id="id.2.1.521.6.0">Per laqual co&longs;a anche <lb/>le metà di que&longs;ti, cioè NKB <lb/>&longs;arà minore di MHB. </s><s id="id.2.1.521.7.0">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. </s>

<figure id="fig61"></figure><lb/><emph type="italics"/>&longs;ia eguale all'angolo BHF, &longs;arà NKG maggiore di MHF. </s>

<s id="id.2.1.521.8.0">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. </s>

<s id="id.2.1.521.9.0">Adunque GN &longs;arà maggiore di FM. & co&longs;i per e&longs;&longs;ere BG egua-<emph.end type="italics"/>

<pb pagenum="43"/><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>

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<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>

<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>

<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>

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<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. </s>

<figure id="fig62"></figure><lb/><emph type="italics"/>za &longs;o&longs;tenente in A. </s>

<s id="id.2.1.542.2.0">Moua&longs;i dapoi la leua AB in EF, & &longs;ia il pe&longs;o mo&longs;&longs;o <lb/>in FG. </s>

<s id="id.2.1.542.3.0">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. </s><s id="id.2.1.542.4.0">&longs;ia <lb/>il K il centro della grauezza del pe&longs;o FG. </s>

<s id="id.2.1.542.5.0">Dapoi &longs;iano tirate sì da H, come<emph.end type="italics"/>

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<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>

<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>

<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>

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<s id="id.2.1.558.1.0"><emph type="italics"/>Sia poila 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. </s>

<s id="id.2.1.558.2.0">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. </s>

<s id="id.2.1.558.3.0">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.2.1.558.4.0">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/>

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<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>

<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>

<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 pagenum="46"/><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>

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<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"/>

<figure id="fig64"></figure><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/><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. </s>

<s id="id.2.1.573.3.0">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.2.1.573.4.0">che bi&longs;o<lb/>gnaua mo&longs;trare.<emph.end type="italics"/></s>

<s id="id.2.1.576.1.0"><emph type="italics"/>Di quì <expan abbr="parim&emacr;te">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>

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<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"/>

<pb pagenum="47"/><emph type="italics"/>Ma nella leua EF quanto il centro della grauezza &longs;arà più da lunge dalla leua. </s>

<s id="id.2.1.580.2.0"><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 X<36> 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 C<36> 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>. </s>

<figure id="fig65"></figure><lb/><emph type="italics"/>rà la C<36> 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>. </s>

<arrow.to.target n="note170"></arrow.to.target><lb/><emph type="italics"/>cioche C<36> 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. </s>

<s id="id.2.1.580.4.0">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.2.1.580.5.0">peroche maggiore è CT di CI, & perciò CT hauerà proportio<lb/>ne maggiore à CR, che CI à CR. </s><s id="id.2.1.580.6.0">&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>

<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>

<s id="id.2.1.583.1.0">PROPOSITIONE IX.</s>

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<s id="id.2.1.585.1.0"><emph type="italics"/>Sia la leua AB egualmente <expan abbr="di&longs;t&amacr;te">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. </s>

<s id="id.2.1.585.2.0">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. </s><s id="id.2.1.585.3.0">&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/>

<figure id="fig66"></figure><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. </s>

<s id="id.2.1.585.4.0">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/>

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<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>

<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>

<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>

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<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. </s>

<s id="id.2.1.592.2.0">Moua&longs;i dapoi la leua in BE, & BF, & &longs;i tra<lb/>&longs;porti il pe&longs;o in GH KL. </s><s id="id.2.1.592.3.0">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>

<figure id="fig67"></figure><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.2.1.592.4.0">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/>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>

<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>

<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>

<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>

<figure id="fig68"></figure><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>

<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>

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<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-n&emacr;ce">&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. </s><s id="id.2.1.603.2.0">& 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>

<figure id="fig69"></figure><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. </s><s id="id.2.1.603.2.0">& 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.2.1.603.3.0">che dimo <lb/>&longs;trare era me&longs;tieri.<emph.end type="italics"/></s>

<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>

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<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. </s>

<figure id="fig70"></figure><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. </s>

<s id="id.2.1.610.3.0">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.2.1.610.4.0">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. </s><s id="id.2.1.610.5.0">&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. </s>

<s id="id.2.1.610.6.0">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.2.1.610.7.0">peroche minore è C<foreign lang="greek">*u</foreign> de CT. </s><s id="id.2.1.610.8.0">& 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 pagenum="50"/><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.2.1.610.9.0">& 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>

<s id="id.2.1.612.1.0">PROPOSITIO NE X.</s>

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<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. </s>

<figure id="fig71"></figure><lb/><emph type="italics"/>& il centro della grauezza in LM. </s>

<s id="id.2.1.614.3.0">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.2.1.614.4.0">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/><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. </s>

<s id="id.2.1.614.5.0">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.2.1.614.6.0">che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/></s>

<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>

<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>

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<s id="id.2.1.621.2.0">Dico che il pe&longs;o C &longs;a<lb/>rà mo&longs;&longs;o dalla po&longs;&longs;anza in B. </s>

<s id="id.2.1.621.3.0">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="egualm&emacr;te">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. </s><s id="id.2.1.621.4.0">& 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"/>

<figure id="fig73"></figure><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.2.1.621.5.0">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.2.1.621.6.0">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>

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<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>

<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>

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<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. </s>

<s id="id.2.1.625.4.0">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>

<figure id="fig74"></figure><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>

<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>

<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>

<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. </s>

<s id="id.2.1.629.2.0">Di <lb/>co che il pe&longs;o C <lb/>&longs;arà mo&longs;&longs;o dal <lb/>la <expan abbr="poß&amacr;za">poßanza</expan> che <lb/>è in D. </s>

<s id="id.2.1.629.3.0">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. </s>

<figure id="fig75"></figure><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. </s>

<s id="id.2.1.629.4.0">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.2.1.629.5.0">Per laqual co&longs;a il pe&longs;o E è maggiore del pe&longs;o C. </s>

<s id="id.2.1.629.6.0"><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>

<s id="id.2.1.631.1.0">Altramente.</s>

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<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. </s>

<s id="id.2.1.632.2.0"><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. </s>

<s id="id.2.1.632.3.0">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/>

<figure id="fig76"></figure><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. </s>

<s id="id.2.1.632.4.0">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>

<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>

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<s id="id.2.1.636.3.0">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><lb/>

<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>

<figure id="fig77"></figure><lb/><emph type="italics"/>to in E, la po&longs;&longs;anza in B lo &longs;o&longs;tentarebbe. </s>

<s id="id.2.1.636.4.0">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>

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<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>

<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 que sto.<emph.end type="italics"/></s>

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<s id="id.2.1.641.2.0">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. </s><s id="id.2.1.641.3.0">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"/>

<figure id="fig78"></figure><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. </s>

<s id="id.2.1.641.4.0">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>

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<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>

<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 que sto.<emph.end type="italics"/></s>

<s id="id.2.1.646.1.0">PROPOSITIONE XII.</s>

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<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. </s>

<s id="id.2.1.649.2.0">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. </s>

<s id="id.2.1.649.3.0">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. </s><s id="id.2.1.649.4.0">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.2.1.649.5.0">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/>

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<s id="id.2.1.652.1.0"><margin.target id="note191"></margin.target><emph type="italics"/>Per lo lein ma di questo.<emph.end type="italics"/></s>

<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>

<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. </s>

<s id="id.2.1.654.2.0">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. </s>

<s id="id.2.1.654.3.0"><lb/>Dunque la po&longs;&longs;anza d<gap/> diece po&longs;ta in C mouerà il pe&longs;o A di cento appiccato in E<emph.end type="italics"/><lb/>

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<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>

<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>

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<s id="id.2.1.659.5.0">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 i<gap/> 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;aaza <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. </s>

<figure id="fig81"></figure><lb/><emph type="italics"/>alla po&longs;&longs;anza in E. </s>

<s id="id.2.1.659.6.0">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>

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<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>

<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>

<s id="id.2.1.665.1.0">COROLLARIO.</s>

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<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. </s><s id="id.2.1.672.2.0">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. </s><s id="id.2.1.672.3.0">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/>

<figure id="fig82"></figure><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. </s>

<s id="id.2.1.672.4.0">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/>

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<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>

<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>

<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 pagenum="54"/><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>

<figure id="fig83"></figure><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>

<s id="id.2.1.679.1.0">PROPOSITIONE XIIII.</s>

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<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"/>

<figure id="fig84"></figure><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/>

<arrow.to.target n="note203"></arrow.to.target> <emph type="italics"/>in E &longs;o&longs;terrebbe i pe&longs;i ABC. </s><s id="id.2.1.682.3.0">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>

<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>

<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>

<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>

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<pb pagenum="55"/><emph type="italics"/>quali taglino la linea AF ne' punti KL. </s>

<s id="id.2.1.691.3.0">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. </s>

<s id="id.2.1.691.4.0">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. </s>

<figure id="fig86"></figure><lb/><emph type="italics"/>come &longs;e fo&longs;&longs;ero appiccati in KL. </s>

<s id="id.2.1.691.5.0">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>

<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>

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<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.2.1.722.5.0">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>

<figure id="fig87"></figure><lb/><emph type="italics"/>F ne'punti CF peroche non po&longs;&longs;ono &longs;egare la girella. </s>

<s id="id.2.1.722.6.0">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. </s>

<s id="id.2.1.722.7.0">Ma percioche BC &longs;ta à piombo sì all'orizonte, come ad e&longs;&longs;a CF &longs;arà<emph.end type="italics"/><lb/>

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<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>

<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 egual mem<gap/>:.<emph.end type="italics"/></s>

<s id="id.2.1.729.1.0">COROLLARIO.</s>

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<pb pagenum="59"/><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/>zain G non potrebbe à modo alcuno &longs;o&longs;tenere il pe&longs;o. </s>

<s id="id.2.1.740.5.0">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. </s>

<figure id="fig88"></figure><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. </s>

<s id="id.2.1.740.6.0">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.2.1.740.7.0">Ma ciò più chiaramente apparerà dapoi. </s>

<s id="id.2.1.740.8.0">Hora percioche la po&longs;&longs;anza al<emph.end type="italics"/>

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<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>

<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"/><gap/> questo nella leua.<emph.end type="italics"/></s>

<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>

<figure id="fig89"></figure><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>

<s id="id.2.1.747.1.0">COROLLARIO I.</s>

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<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. </s>

<s id="id.2.1.772.2.0">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. </s>

<s id="id.2.1.772.3.0">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="t&amacr;to">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. </s>

<figure id="fig90"></figure><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. </s>

<s id="id.2.1.772.4.0">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. </s>

<s id="id.2.1.772.5.0">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. </s>

<s id="id.2.1.772.6.0">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>

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<s id="id.2.1.772.8.0">La po&longs;&longs;anza dunque in B &longs;arà il terzo del pe&longs;o C. </s>

<s id="id.2.1.772.9.0">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>

<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. </s>

<figure id="fig91"></figure><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. </s>

<s id="id.2.1.775.2.0">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>

<s id="id.2.1.777.1.0">PROPOSITIONE V.</s>

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<s id="id.2.1.780.6.0">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 loc<gap/> 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>

<figure id="fig92"></figure><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.2.1.780.7.0">Ma le po&longs;&longs;anze che &longs;ono in BD &longs;ono le i&longs;te&longs;&longs;e, come &longs;<gap/><emph.end type="italics"/>

<pb pagenum="63"/><emph type="italics"/>fu&longs;&longs;ero in HM. </s>

<s id="id.2.1.780.8.0">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>

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<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>

<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>

<s id="id.2.1.787.1.0">COROLLARIO.</s>

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<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;ola parte del <lb/>pe&longs;o, che &longs;o&longs;tiene, &longs;ia come <lb/>BE à BA, & la po&longs;&longs;an<lb/>zain 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><lb/><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"/>

<figure id="fig93"></figure><lb/><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 pagenum="65"/><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. </s>

<s id="id.2.1.797.3.0">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>

<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>

<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>

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<s id="id.2.1.805.7.0">Adunque &longs;o&longs;tiene egualmente sì la cor<lb/>da PO, come la KG. </s>

<s id="id.2.1.805.8.0">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.2.1.805.9.0">& 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. </s><s id="id.2.1.805.10.0">tanto dunque &longs;o&longs;terrà la leua ON, quan-<emph.end type="italics"/><lb/>

<figure id="fig94"></figure><lb/><emph type="italics"/>PO LN quanto le due KG EF. </s><s id="id.2.1.805.10.0">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. </s>

<s id="id.2.1.805.11.0">Onde nell'vna, & l'altra leua ON GF pe&longs;erà egualmente il <lb/>pe&longs;o. </s>

<s id="id.2.1.805.12.0">&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"/>

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<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"/>co rollario della<emph.end type="italics"/> 2. <emph type="italics"/>di questo.<emph.end type="italics"/></s>

<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>

<s id="id.2.1.810.1.0">COROLLARIO I.</s>

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<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. </s>

<s id="id.2.1.825.2.0">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, &longs;ra loro eguali, & ogni vna <lb/>&longs;ar à il doppio di quella tal parte, che è &longs;o&longs;tenuta dalla po&longs;&longs;anza di E. </s>

<figure id="fig95"></figure><lb/><emph type="italics"/>&longs;o G, lequali &longs;ono &longs;o&longs;tenute dalle po&longs;&longs;anze di BD, &longs;ra loro eguali, & ogni vna <lb/>&longs;ar à il doppio di quella tal parte, che è &longs;o&longs;tenuta dalla po&longs;&longs;anza di E. </s>

<s id="id.2.1.825.3.0">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. </s>

<s id="id.2.1.825.4.0">Per laqual co&longs;a anche la parte KL &longs;arà egua<lb/>le alla parte MN. </s>

<s id="id.2.1.825.5.0">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. </s>

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<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>

<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>

<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>

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<s id="id.2.1.858.6.0">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.2.1.858.7.0">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. </s>

<s id="id.2.1.858.8.0">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. </s>

<figure id="fig96"></figure><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. </s>

<s id="id.2.1.858.9.0">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.2.1.858.10.0">& 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-

<s id="id.2.1.858.11.0">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>

<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>

<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>

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<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. </s><s id="id.2.1.866.2.0">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.2.1.866.3.0">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. </s>

<s id="id.2.1.866.4.0">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>

<figure id="fig97"></figure><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.2.1.866.5.0">che era da mo&longs;trare.<emph.end type="italics"/></s>

<s id="id.2.1.868.1.0">PROPOSITIONE XI.</s>

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<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.2.1.870.4.0">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. </s>

<s id="id.2.1.870.5.0"><lb/>Per laqual co&longs;a mentre la po&longs;-<emph.end type="italics"/><lb/>

<arrow.to.target n="fig98"></arrow.to.target>

<pb pagenum="73"/><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. </s>

<s id="id.2.1.870.6.0">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. </s>

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<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>

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<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>

<figure id="fig99"></figure><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 X<36>, 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/><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. </s>

<s id="id.2.1.882.3.0">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. </s>

<s id="id.2.1.882.4.0">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 vetocità 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>

<figure id="fig100"></figure><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>

<s id="id.2.1.886.1.0">PROPOSITIONE XII.</s>

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<s id="id.2.1.888.6.0">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.2.1.888.7.0">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.2.1.888.8.0">& 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><lb/><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>

<figure id="fig101"></figure><lb/><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>

<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>

<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>

<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>

<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>

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<s id="id.2.1.903.6.0">Adunque le tre <lb/>QR EO LP fra loro &longs;aranno <lb/>eguali; à cui &longs;ono etiandio eguali <lb/>BS DT. </s>

<s id="id.2.1.903.7.0">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.2.1.903.8.0">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/><emph type="italics"/>EQ, & per con&longs;e<lb/>quenza QR. </s>

<s id="id.2.1.903.9.0">Lo <lb/>&longs;patio dunque MN <lb/>della traportata po&longs; <lb/>&longs;anza è tre volt<17><lb/>tanto, quanto lo &longs;pa <lb/>tio QR del pe&longs;o <lb/>mo&longs;&longs;o. </s>

<s id="id.2.1.903.10.0">che era da <lb/>mo&longs;trar&longs;i.<emph.end type="italics"/></s>

<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>

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<s id="id.2.1.910.2.0">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.2.1.910.3.0">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.2.1.910.4.0">&longs;iano condotte per li centri delle <lb/>girelle le linee FHQTECVXNZ <lb/>egualmente distanti dall orizonte, lequa <lb/>li tocchino le corde ne i puntl FH QT<emph.end type="italics"/><lb/>

<arrow.to.target n="fig103"></arrow.to.target>

<pb/><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. </s>

<s id="id.2.1.910.5.0"><lb/>Et percioche la corda BCDEFGHZ <lb/>MNO è eguale alla corda BT<36>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/>T<36>Q X<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.2.1.910.6.0">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.2.1.910.7.0">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.2.1.910.8.0">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.2.1.910.9.0">che era da mo&longs;trare.<emph.end type="italics"/></s>

<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>

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<s id="id.2.1.966.3.0"><lb/>Ma la po&longs;&longs;anza di C è due volte tan <lb/>to quanto la po&longs;&longs;anza di K, & per con<emph.end type="italics"/><lb/>

<arrow.to.target n="note253"></arrow.to.target> <emph type="italics"/>&longs;equenza del pe&longs;o <expan abbr="q;">que</expan> peroche egli è <lb/>la me de&longs;ima co&longs;a, come &longs;e in K fo&longs;&longs;e <lb/>appiccato vn pe&longs;o eguale al pe&longs;o Q, <lb/>delquale è due volte tanto la po&longs;&longs;anza <lb/>di C. </s>

<s id="id.2.1.966.4.0">Adunque due po&longs;&longs;anze po&longs;te <lb/>in DC &longs;ono quattro volte tanto quan <lb/>to è il pe&longs;o <expan abbr="q.">que</expan> & concio&longs;ia, che la <lb/>po&longs;&longs;anza di R &longs;o&longs;tenga con le girelle <lb/>il pe&longs;o Q, &longs;arà la po&longs;&longs;anza di R co<lb/>me &longs;e fo&longs;&longs;ero due po&longs;&longs;anze l'vna in D<emph.end type="italics"/><lb/>

<arrow.to.target n="fig104"></arrow.to.target>

<pb pagenum="83"/><emph type="italics"/>& l'altra in C: & l'vna, & l'altra in&longs;ieme &longs;o&longs;tene&longs;&longs;e il pe&longs;o <expan abbr="q.">que</expan> La po&longs;&longs;anza <lb/>dunque di R è quattro volte tanto quanto il pe&longs;o <expan abbr="q.">que</expan> che bi&longs;ognaua dimo&longs;trare.<emph.end type="italics"/></s>

<s id="id.2.1.968.1.0"><margin.target id="note252"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 16. <emph type="italics"/>di questo.<emph.end type="italics"/></s>

<s id="id.2.1.969.1.0"><margin.target id="note253"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 15. <emph type="italics"/>di questo.<emph.end type="italics"/></s>

<s id="id.2.1.970.1.0">COROLLARIO</s>

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<s id="id.2.1.1077.4.0">La leua dunque <lb/>LM non &longs;i mouerà in <lb/>niuna delle parti. </s>

<s id="id.2.1.1077.5.0">Per la <lb/>qual co&longs;a no anche la gi<lb/>rella &longs;i girerà intorno. </s>

<s id="id.2.1.1077.6.0"><lb/>Co&longs;i LM &longs;arà come bi<lb/>lancia, il cni centro D, <lb/>& li pe&longs;i appiccati in LM <lb/>&longs;aranno eguali alla quar-<emph.end type="italics"/><lb/>

<arrow.to.target n="fig105"></arrow.to.target><lb/><emph type="italics"/>ta parte del pe&longs;o C; peroche cia&longs;cheduna corda in LN MQ &longs;o&longs;tiene la quar <lb/>ta parte del pe&longs;o C; &longs;i mouerà dunque tuttala girella, il cui centro è D in sù, <lb/>ma non già voltera&longs;&longs;i intorno.<emph.end type="italics"/></s>

<figure id="fig105"></figure><lb/><emph type="italics"/>ta parte del pe&longs;o C; peroche cia&longs;cheduna corda in LN MQ &longs;o&longs;tiene la quar <lb/>ta parte del pe&longs;o C; &longs;i mouerà dunque tuttala girella, il cui centro è D in sù, <lb/>ma non già voltera&longs;&longs;i intorno.<emph.end type="italics"/></s>

<pb/>

<s id="id.2.1.1080.1.0"><emph type="italics"/>Et &longs;e la corda po&longs;ta in F &longs;iriuolgerà <lb/>d'intorno à due altre girelle, i cui <lb/>centri &longs;iano HK l'aquale dapoi &longs;ia <lb/>rilegata in L; &longs;arà la proportione <lb/>del pe&longs;o alla po&longs;&longs;anza vna volta & <lb/>meza.<emph.end type="italics"/></s>

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<s id="id.2.1.1187.2.0">talche la po&longs;&longs;anza,

<pb/>laquale è &longs;empre ne i raggi, come in F, mentre ella volge <lb/>intorno la rota, & l'a&longs;&longs;e, moua anco in sù il pe&longs;o K appicca<lb/>to all'a&longs;&longs;e con la corda LM riuolta d'intorno all'a&longs;&longs;e. </s>

<s id="id.2.1.1187.3.0">A noi <lb/>re&longs;ta dunque, di mo&longs;trare, perche i gran pe&longs;i da piccola forza, <lb/>

<arrow.to.target n="fig106"></arrow.to.target><lb/>& in che modo etiandio &longs;i mouano con que&longs;to i&longs;trumento: <lb/>& di più manife&longs;tare la ragione del tempo, & dello &longs;patio del<lb/>la po&longs;&longs;anza mouente, & del pe&longs;o mo&longs;&longs;o fra loro, & ridurre l'v<lb/>&longs;o di cote&longs;to i&longs;trumento alla leua. </s>

<figure id="fig106"></figure><lb/>& in che modo etiandio &longs;i mouano con que&longs;to i&longs;trumento: <lb/>& di più manife&longs;tare la ragione del tempo, & dello &longs;patio del<lb/>la po&longs;&longs;anza mouente, & del pe&longs;o mo&longs;&longs;o fra loro, & ridurre l'v<lb/>&longs;o di cote&longs;to i&longs;trumento alla leua. </s>

<s id="id.2.1.1190.1.0">PROPOSITIONE I.</s>

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<s id="id.2.1.1193.2.0">& la po&longs;&longs;anza po&longs;ta in F <lb/>&longs;o&longs;tenga il pe&longs;o K. </s>

<s id="id.2.1.1193.3.0">Dico che la po&longs;&longs;anza in F co&longs;i &longs;i hà al pe&longs;o K, come CB <lb/>à CF. </s>

<s id="id.2.1.1193.4.0">Faccia&longs;i come CF à CB, co&longs;i il pe&longs;o K ad vn altro pe&longs;o come M, il <lb/>quale &longs;ia appi<gap/>ato in F. </s><s id="id.2.1.1193.5.0">& percioche i pe&longs;i MK &longs;ono appiccati in -FB; &longs;arà <lb/>FB come leua, ouero bilancia; ma percioche il C è punto immobile, d'intorno<emph.end type="italics"/><lb/>

<arrow.to.target n="fig107"></arrow.to.target><lb/>

<arrow.to.target n="note297"></arrow.to.target> <emph type="italics"/>alquale l'a&longs;&longs;e, & la rota &longs;i riuolgono; &longs;arà C il &longs;o&longs;tegno della leua FB, ouero il <lb/>centro della bilancia. </s>

<s id="id.2.1.1193.6.0">& per e&longs;&longs;ere co&longs;i CF à CB come K ad M, i pe&longs;i KM <lb/>pe&longs;eranno egualmente. </s>

<s id="id.2.1.1193.7.0">La po&longs;&longs;anza dunque di F &longs;o&longs;tenente il pe&longs;o K contra<lb/>pe&longs;erà egualmente con e&longs;&longs;o pe&longs;o K accioche egli non chini al ba&longs;&longs;o, & &longs;arà eguale <lb/>ad M. </s>

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<s id="id.2.1.1196.1.0"><margin.target id="note298"></margin.target><emph type="italics"/>Per lo corol lario della<emph.end type="italics"/> 4. <emph type="italics"/>del<emph.end type="italics"/> 5.</s>

<s id="id.2.1.1197.1.0"><margin.target id="note299"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 2. <emph type="italics"/>di questo della leua.<emph.end type="italics"/></s>

<s id="id.2.1.1198.1.0">COROLLARIO.</s>

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<s id="id.2.1.1217.1.0"><emph type="italics"/>Oltre à ciò quanto il cerchio FHN d'intorno à i raggi è più grande, tanto anco &longs;i <lb/>con&longs;umerà più tempo in mouere il pe&longs;o, pur che la po&longs;&longs;anza &longs;i moua con eguale ve-<emph.end type="italics"/><lb/>

<arrow.to.target n="note305"></arrow.to.target> <emph type="italics"/>locità; & il tempo tanto &longs;arà maggiore quanto il diametro dell'vno &longs;arà maggiore <lb/>del diametro dell'altro; percioche le circonferenze de'cerchi &longs;i hanno come i diame <lb/>tri. </s>

<s id="id.2.1.1217.2.0">& concio&longs;ia, che per la trige&longs;ima &longs;e&longs;ta del quarto libro di Pappo delle raccolte<emph.end type="italics"/><lb/>

<arrow.to.target n="fig108"></arrow.to.target><lb/><emph type="italics"/>matematiche po&longs;&longs;iamo ritrouare le circonferenze eguali di due cerchi di&longs;uguali; per<lb/>ciò ritroueremo anche il tempo à que&longs;to modo delle portioni di&longs;uguali de' cerchi. </s>

<figure id="fig108"></figure><lb/><emph type="italics"/>matematiche po&longs;&longs;iamo ritrouare le circonferenze eguali di due cerchi di&longs;uguali; per<lb/>ciò ritroueremo anche il tempo à que&longs;to modo delle portioni di&longs;uguali de' cerchi. </s>

<s id="id.2.1.1217.3.0">Ma <lb/>per lo contrario quanto &longs;arà maggiore la circonferenza dell'a&longs;&longs;e, il pe&longs;o mouera&longs;&longs;i <lb/>più pre&longs;to in sù, percioche maggior parte della corda BL in vno giro compiuto, &longs;i <lb/>riuolge d'intorno al cerchio ABO, che &longs;e fo&longs;&longs;e minore, per e&longs;&longs;ere la corda inuolta <lb/>eguale alla circonferenza del cerchio, d'intorno alquale &longs;i riuolge.<emph.end type="italics"/></s>

<s id="id.2.1.1220.1.0"><margin.target id="note305"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 23. <emph type="italics"/>dell'estano li bre di Pappe.<emph.end type="italics"/></s>

<s id="id.2.1.1221.1.0">COROLLARIO.</s>

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<s id="id.2.1.1226.3.0">& &longs;e CB <lb/>fo&longs;&longs;e il mezo diametro del<lb/>l'a&longs;&longs;e, & CA il mezo dia<emph.end type="italics"/>

<arrow.to.target n="note306"></arrow.to.target><lb/><emph type="italics"/>metro della rota co'raggi; <lb/>egli è chiaro, che la po&longs;&longs;an<lb/>za come dieci po&longs;ta in A<emph.end type="italics"/>

<arrow.to.target n="note307"></arrow.to.target><lb/><emph type="italics"/>pe&longs;erebbe egualmente co'l<emph.end type="italics"/><lb/>

<arrow.to.target n="fig109"></arrow.to.target><lb/><emph type="italics"/>pe&longs;o &longs;e&longs;&longs;anta po&longs;to in B. </s><s id="id.2.1.1226.4.0">ma pigli&longs;i tra BC qual &longs;i voglia punto, & &longs;ia D; & <lb/>faccia&longs;i BD il mezo diametro dell'a&longs;&longs;e, & DA il mezo diametro della rota co'rag <lb/>gi, & ponga&longs;i il pe&longs;o &longs;e&longs;&longs;anta in B con vna corda inuolta d'intorno all'a&longs;&longs;e, & la po&longs;<emph.end type="italics"/>

<figure id="fig109"></figure><lb/><emph type="italics"/>pe&longs;o &longs;e&longs;&longs;anta po&longs;to in B. </s><s id="id.2.1.1226.4.0">ma pigli&longs;i tra BC qual &longs;i voglia punto, & &longs;ia D; & <lb/>faccia&longs;i BD il mezo diametro dell'a&longs;&longs;e, & DA il mezo diametro della rota co'rag <lb/>gi, & ponga&longs;i il pe&longs;o &longs;e&longs;&longs;anta in B con vna corda inuolta d'intorno all'a&longs;&longs;e, & la po&longs;<emph.end type="italics"/>

<arrow.to.target n="note308"></arrow.to.target><lb/><emph type="italics"/>&longs;anza in A. </s>

<s id="id.2.1.1226.5.0">Hor percioche AD ha proportione maggiore à DB, che AC à <lb/>CB: haur à proportione maggiore AD à DB, che il pe&longs;o &longs;e&longs;&longs;anta appiccato in <lb/>B alla po&longs;&longs;anza di dieci posta in A. </s>

<s id="id.2.1.1226.6.0">Per laqual co&longs;a la po&longs;&longs;anza di A mouerà il <lb/>pe&longs;o di &longs;e&longs;&longs;anta con l'a&longs;&longs;e nella rota, il mezo diametro delquale è BD, & DA è <lb/>il mezo diametro della rota co'raggi. </s>

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<s id="id.2.1.1230.1.0"><margin.target id="note307"></margin.target><emph type="italics"/>Per il lemma nella pri ma di questo della leua.<emph.end type="italics"/></s>

<s id="id.2.1.1231.1.0"><margin.target id="note308"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 11. <emph type="italics"/>di questo della leua.,<emph.end type="italics"/></s>

<s id="id.2.1.1232.1.0">Altramente.</s>

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<s id="id.2.1.1234.1.0"><emph type="italics"/>Ponga&longs;i l'a&longs;&longs;e, il cui mezo diametro &longs;ia BD, & il centro &longs;uo C, ilquale a&longs;&longs;e &longs;ta<lb/>tuiremo maggiore, ò minore, come la grandezza, & grauezza del pe&longs;o ricerca. </s>

<s id="id.2.1.1234.2.0"><lb/>Allunghi&longs;i po&longs;cia la li<lb/>nea BD fin ad A; & <lb/>faccia&longs;i BC à CA, co <lb/>me diece à &longs;e&longs;&longs;anta., & <lb/>&longs;e CA fo&longs;&longs;e il mezo dia <lb/>metro della rota co'rag <lb/>gi, la po&longs;&longs;anza di diece<emph.end type="italics"/><lb/>

<arrow.to.target n="fig110"></arrow.to.target><lb/><emph type="italics"/>po&longs;ta in A pe&longs;erebbe egualmente co'l pe&longs;o di &longs;e&longs;&longs;anta po&longs;to in B. </s>

<figure id="fig110"></figure><lb/><emph type="italics"/>po&longs;ta in A pe&longs;erebbe egualmente co'l pe&longs;o di &longs;e&longs;&longs;anta po&longs;to in B. </s>

<s id="id.2.1.1234.3.0">Ma allunghi&longs;i, <lb/>BA dalla parte di A, & in que&longs;ta allungata linea prenda&longs;i qual &longs;i voglia punto <lb/>come E, & faccia&longs;i CE il mezo diametro della rota co'raggi; & pong a&longs;i la po&longs;<lb/>&longs;anza di diece in E; haurà EC a CB proportione maggiore, che il pe&longs;o &longs;e&longs;&longs;anta <lb/>po&longs;to in B alla po&longs;&longs;anza di diece po&longs;ta in E. </s>

<s id="id.2.1.1234.4.0">Dunque la po&longs;&longs;anza di diece po&longs;ta in <lb/>E mouerà il pe&longs;o &longs;e&longs;&longs;anta appiccato in B, con la corda inuolta d'intorno all'a&longs;&longs;e, il <lb/>cui mezo diametro è CB, & CE è il mezo diametro della rota co i raggi. </s>

<s id="id.2.1.1234.5.0">che bi<lb/>&longs;ognaua fare.<emph.end type="italics"/></s>

<s id="id.2.1.1236.1.0">Sotto que&longs;ta &longs;orte d'i&longs;trumento &longs;ono gli argani, i molinelli, le tri <lb/>uelle, i timpani, ò rote co' &longs;uoi a&longs;&longs;i, ò &longs;iano dentate, ò nò, & <lb/>&longs;imili. </s>

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<s id="id.2.1.1245.1.0"><emph type="italics"/>Sia il cuneo ABC, & la &longs;ua cima B, & &longs;ia AB eguale à BC, & quel che s'ha <lb/>da fendere &longs;ia DE <lb/>FG; & &longs;ia la par <lb/>te del cuneo HBK <lb/>fra DE FG, & <lb/>HB &longs;ia eguale ad <lb/>e&longs;&longs;a BK. Percuo<lb/>ta&longs;i, come &longs;uol far&longs;i, <lb/>il cuneo in AC, <lb/>mentre il cuneo viè <lb/>perco&longs;&longs;o in AC, &longs;i <lb/>fà AB leua, il cui <lb/>&longs;o&longs;tegno è in H, & <lb/>il pe&longs;o in B. & nel <lb/>modo i&longs;te&longs;&longs;o CB &longs;i <lb/>fa leua, il cui &longs;o&longs;te<lb/>gno è K, & il pe<lb/>&longs;o &longs;imilmente in B. </s>

<s id="id.2.1.1245.2.0"><lb/>Ma mentre il cu<lb/>neo è perco&longs;&longs;o, egli <lb/>entra in e&longs;&longs;o DE<emph.end type="italics"/><lb/>

<arrow.to.target n="fig111"></arrow.to.target><lb/><emph type="italics"/>FG anco con portione di &longs;e maggiore di quel che fo&longs;&longs;e prima: & &longs;ia questa por<lb/>tione MBL; & &longs;ia MB eguale ad e&longs;&longs;a BL. & per e&longs;&longs;ere MB, & BL mag<lb/>giori di HB BK, &longs;arà anco ML maggiore di HK. </s>

<figure id="fig111"></figure><lb/><emph type="italics"/>FG anco con portione di &longs;e maggiore di quel che fo&longs;&longs;e prima: & &longs;ia questa por<lb/>tione MBL; & &longs;ia MB eguale ad e&longs;&longs;a BL. & per e&longs;&longs;ere MB, & BL mag<lb/>giori di HB BK, &longs;arà anco ML maggiore di HK. </s>

<s id="id.2.1.1245.3.0">Mentre dunque ML &longs;arà <lb/>nel &longs;ito di HK; egli è me&longs;tieri che la fe&longs;&longs;a &longs;i faccia maggiore; & che D &longs;i moua<emph.end type="italics"/>

<pb pagenum="108"/><emph type="italics"/>ver&longs;o O, & G ver&longs;o N; & quanto maggior parte del cuneo entra fra DEFG, <lb/>tanto maggior fe&longs;&longs;a &longs;i &longs;accia; & DG &longs;empre più &longs;aranno cacciati ver&longs;o ON. <lb/>dunque la parte KG che &longs;i fende mouera&longs;&longs;i dalla leua AB, ilcui &longs;o&longs;tegno è in H, <lb/>& il pe&longs;o in B; &longs;iche il punto B di e&longs;&longs;a leua AB cacci la parte KG: & la parte <lb/>HD mouera&longs;&longs;i dalla leua CB, il cui &longs;o&longs;tegno è K, &longs;i che B con la leua CB cacci <lb/>la parte HD.<emph.end type="italics"/></s>

<s id="id.2.1.1247.1.0">Ma trouando&longs;i tre maniere di leue, come è &longs;tato di &longs;opra mo<lb/>&longs;trato. </s>

<s id="id.2.1.1247.2.0">però &longs;arà for&longs;e più conueneuole con&longs;iderare il cuneo <lb/>in que&longs;to modo. </s>

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<s id="id.2.1.1249.1.0"><emph type="italics"/>Percioche &longs;ia il cuneo ABC; & &longs;iano due pe&longs;i &longs;eparati DEFG, & HIKL, fra <lb/>quali &longs;ia la parte DBH del cuneo, la cui cima B tenga il mezo tral'vno, & l'al<lb/>tro &longs;ito. </s>

<s id="id.2.1.1249.2.0">Percota&longs;i il cu <lb/>neo in modo, che anche <lb/>dauantaggio più &longs;ia cac <lb/>ciato fra i pe&longs;i, come pri <lb/>maè stato detto; per<lb/>cioche &longs;ono que&longs;ti pe&longs;i <lb/>come &longs;e &longs;o&longs;&longs;ero vno con<lb/>tinuo &longs;olamente GF <lb/>KL, che bi&longs;ogna&longs;&longs;e fen <lb/>dere: percioche nel mo<lb/>do iste&longs;&longs;o la parte DG <lb/>mentre il cuneo è più ol <lb/>tre cacciato, &longs;i mouerà <lb/>ver&longs;o M, & la parte <lb/>HL ver&longs;o N. </s>

<s id="id.2.1.1249.3.0">Moua&longs;i <lb/>dunque la parte DG <lb/>ver&longs;o M, & la parte <lb/>HL ver&longs;o N; & il B<emph.end type="italics"/><lb/>

<arrow.to.target n="fig112"></arrow.to.target><lb/><emph type="italics"/>mentre trapa&longs;&longs;a più oltre, &longs;empre rimanga nel mezo tra l'vn pe&longs;o, & l'altro. </s>

<figure id="fig112"></figure><lb/><emph type="italics"/>mentre trapa&longs;&longs;a più oltre, &longs;empre rimanga nel mezo tra l'vn pe&longs;o, & l'altro. </s>

<s id="id.2.1.1249.4.0">Hor <lb/>mentre D G è mo&longs;&longs;o dal cuneo in uer&longs;o M; egli è manife&longs;to, che B non moue la <lb/>parte DG inuer&longs;o M con la leua CB, il cui &longs;o&longs;tegno è H, perche il punto B non <lb/>tocca il pe&longs;o; ma DG moueraßi dal punto D della leua con e&longs;&longs;a leua AB, che ha <lb/>per &longs;o&longs;tegno B; pereche il punto D tocca il pe&longs;o. </s>

<s id="id.2.1.1249.5.0">& gli i&longs;trumenti mouono per <lb/>toccamento. </s>

<s id="id.2.1.1249.6.0">&longs;imilmente HL moueraßi da H con la leua CB, che ha per &longs;o&longs;te<lb/>gno B; & ambedue le leue &longs;i fanno re&longs;i&longs;tenza l'vna all'altra fra loro in B, talche <lb/>B faccia più tosto officio di &longs;o&longs;tegno, che di mouere il pe&longs;o. </s>

<s id="id.2.1.1249.7.0">laqual co&longs;a anco ma<lb/>nife&longs;teraßi in que&longs;ta maniera.<emph.end type="italics"/></s>

<s id="id.2.1.1252.1.0"><emph type="italics"/>Sia quel che s'ha da fendere vnparallelo grammo rettangolo ABCD; & &longs;iano due <lb/>leue eguali EF GF, & le parti delle leue HF KF &longs;iano tra AB CD; & &longs;ia <lb/>HF eguale ad FK, & &longs;ia HA <lb/>eguale à KB. & faccia me&longs;tieri <lb/>con le leue EF FG fendere AB <lb/>CD &longs;enza perco&longs;&longs;a, cioè &longs;iano le <lb/>po&longs;&longs;anze mouenti in EG eguali. </s>

<s id="id.2.1.1252.2.0"><lb/>Ma per e&longs;&longs;ere fe&longs;&longs;a AB CD, egli <lb/>è me&longs;tieri che la parte HA &longs;i mo <lb/>ua ver&longs;o M; & KB ver&longs;o N: <lb/>ma mentre le leue &longs;i mouono, co<lb/>me per e&longs;&longs;empio l'vna in M, & <lb/>l'altrain N; egli è nece&longs;&longs;ario, <lb/>che il punto F rimanga immobi <lb/>le, perche in e&longs;&longs;o &longs;i fa l'incontro del <lb/>le leue. </s>

<s id="id.2.1.1252.3.0">Per laqual co&longs;a F &longs;arà il <lb/>&longs;o&longs;tegno dell'vna, & l'altra leua; <lb/>& FG mouerà la parte KB, il<emph.end type="italics"/><lb/>

<arrow.to.target n="fig113"></arrow.to.target><lb/><emph type="italics"/>cui &longs;o&longs;tegno &longs;arà F, & la po&longs;&longs;anza mouente in G; & il pe&longs;o in K. &longs;imilmente <lb/>la parte HA moueraßi dalla leua EF, il cui &longs;o&longs;tegno è F, & la po&longs;&longs;anza in E, <lb/>& il pe&longs;o in H.<emph.end type="italics"/></s>

<figure id="fig113"></figure><lb/><emph type="italics"/>cui &longs;o&longs;tegno &longs;arà F, & la po&longs;&longs;anza mouente in G; & il pe&longs;o in K. &longs;imilmente <lb/>la parte HA moueraßi dalla leua EF, il cui &longs;o&longs;tegno è F, & la po&longs;&longs;anza in E, <lb/>& il pe&longs;o in H.<emph.end type="italics"/></s>

<s id="id.2.1.1254.1.0"><emph type="italics"/>Che &longs;e KH fo&longs;&longs;ero i &longs;o&longs;tegni immobili, & i pe&longs;i in F; mentre la leua FG &longs;i sforza di <lb/>mouere il pe&longs;o po&longs;to in F, all'hora le fa re&longs;i&longs;tenza la leua EF, laquale parimente <lb/>&longs;i sforza di mouere il pe&longs;o po&longs;to in F in uer&longs;o la parte oppo&longs;ta; ma percioche le po&longs;<lb/>&longs;anze &longs;ono eguali, & le altre co&longs;e eguali: dunque non &longs;i farà mouimento in F; <lb/>percioche l'eguale non moue l'eguale. </s>

<s id="id.2.1.1254.2.0">Egli è dunque pale&longs;e, che in F &longs;i fà grandißima <lb/>re&longs;i&longs;tenza dalle leue, che iui fra loro &longs;i incontrano, talche F viene ad e&longs;&longs;ere vn cer<lb/>to che immobile. </s>

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<s id="id.2.1.1257.1.0"><emph type="italics"/>Sia il piano egualmente di&longs;tante dall'orizonte, che paßi per AB; &longs;ia anco il cuneo <lb/>CDB; & &longs;ia CD eguale ad e&longs;&longs;a DB: & il lato del cuneo DB &longs;ia &longs;empre nel <lb/>&longs;ottopo&longs;to piano. </s>

<s id="id.2.1.1257.2.0">&longs;ia dopo il pe&longs;o AEFG immobile in A; & &longs;ia la parte del <lb/>cuneo EDH &longs;otto AEFG. </s>

<s id="id.2.1.1257.3.0">Hor percioche mentre il cuneo è perco&longs;&longs;o in CB, <lb/>maggior parte del detto cuneo entra &longs;otto AEFG, di quel che &longs;ia EDH; &longs;ia<emph.end type="italics"/><lb/>

<arrow.to.target n="fig114"></arrow.to.target><lb/><emph type="italics"/>que&longs;ta parte IDH. & perche illato del cuneo DB è &longs;empre nel piano &longs;ottopo<lb/>&longs;to tirato per AB egualmente di&longs;tante dall'orizonte, allhora quando la parte del <lb/>cuneo KDI &longs;arà &longs;otto AEFG; &longs;arà il punto K in H, & I &longs;otto E, ma IK <lb/>è maggiore di HE: dunque il punto E &longs;arà mo&longs;&longs;o in sù. </s>

<figure id="fig114"></figure><lb/><emph type="italics"/>que&longs;ta parte IDH. & perche illato del cuneo DB è &longs;empre nel piano &longs;ottopo<lb/>&longs;to tirato per AB egualmente di&longs;tante dall'orizonte, allhora quando la parte del <lb/>cuneo KDI &longs;arà &longs;otto AEFG; &longs;arà il punto K in H, & I &longs;otto E, ma IK <lb/>è maggiore di HE: dunque il punto E &longs;arà mo&longs;&longs;o in sù. </s>

<s id="id.2.1.1257.4.0">& mentre il cuneo entra <lb/>&longs;otto AEFG, il punto E &longs;i mouerà in sù &longs;opra il lato EI del cuneo; & nel mo <lb/>do i&longs;te&longs;&longs;o, &longs;e il cuneo trapa&longs;&longs;erà più oltre, il punto E moueraßi &longs;empre &longs;opra il la<lb/>to DC del cuneo; dunque il punto E del pe&longs;o &longs;i mouerà &longs;opra il piano DC in<lb/>chinato all orizonte, la cui inclinatione è l'angolo BDC. che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/></s>

<s id="id.2.1.1259.1.0"><emph type="italics"/>In que&longs;to e&longs;&longs;empio con&longs;iderand<gap/> il cuneo, che moue à &longs;embianza dileua, egli è manife&longs;to <lb/>che il cuneo BCD moue il pe&longs;o AEFG conla leua CD: &longs;i che D &longs;ia il &longs;o&longs;te<lb/>gno, & il pe&longs;o po&longs;to in E: ma non già con la leua BD, il cui &longs;o&longs;tegno &longs;ia H, & <lb/>il pe&longs;o po&longs;to in D.<emph.end type="italics"/></s>

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<s id="id.2.1.1262.1.0"><emph type="italics"/>Sia vn piano egualmente di&longs;tante dall'orizonte, che paßi per AB: &longs;ia ilcuneo CAB, <lb/>il cui lato AB &longs;ia &longs;empre nel &longs;ottopo&longs;to piano; & &longs;ia il pe&longs;o AEFG, che non <lb/>habbia verun'altro moto &longs;e non in sù, & in giù ad ang oli retti all'orizonte: talcbe<emph.end type="italics"/><lb/>

<arrow.to.target n="fig115"></arrow.to.target><lb/><emph type="italics"/>tirata la linea IGK à piombo del piano &longs;ottopo&longs;to, & di e&longs;&longs;a AB, il punto G <lb/>venga ad e&longs;&longs;ere &longs;empre nella linea IGK. </s><s id="id.2.1.1262.2.0">& percio che mentre il cuneo è perco&longs;&longs;o <lb/>in CB, egli trapa&longs;&longs;a tutto più oltre &longs;opra AB; il pe&longs;o AEFG &longs;i leuerà, come <lb/>per le co&longs;e predette &longs;i è mo&longs;trato. </s>

<figure id="fig115"></figure><lb/><emph type="italics"/>tirata la linea IGK à piombo del piano &longs;ottopo&longs;to, & di e&longs;&longs;a AB, il punto G <lb/>venga ad e&longs;&longs;ere &longs;empre nella linea IGK. </s><s id="id.2.1.1262.2.0">& percio che mentre il cuneo è perco&longs;&longs;o <lb/>in CB, egli trapa&longs;&longs;a tutto più oltre &longs;opra AB; il pe&longs;o AEFG &longs;i leuerà, come <lb/>per le co&longs;e predette &longs;i è mo&longs;trato. </s>

<s id="id.2.1.1262.3.0">Moua&longs;i il cuneo in modo, che E alla fine peruen <lb/>ga in C, & la giacitura del cuneo ABC venga ad e&longs;&longs;ere MNO, & la giaci<lb/>tura del pe&longs;o AEFG &longs;ia PMQI, & G &longs;ia in I. </s><s id="id.2.1.1262.4.0">co&longs;i perche mentre il cuneo <lb/>&longs;i moue &longs;opra la linea BO, il pe&longs;o AEFG &longs;i moue in sù dalla linea AC. & <lb/>mentre il cuneo ABC trapa&longs;&longs;a più oltre, il pe&longs;o AEFG &longs;empre più dal lato del <lb/>cuneo AC &longs;i leua: dunque il pe&longs;o AEFG &longs;i mouerà &longs;opra il piano del cuneo <lb/>AC; ilche veramente altro non è, &longs;e non vn piano inchinato all'orizonte, la cui <lb/>inclinatione è l'angolo BAC.<emph.end type="italics"/></s>

<s id="id.2.1.1264.1.0"><emph type="italics"/>Que&longs;to mouimento &longs;i riduce ageuolmente alla bilancia, & alla leua; percioche quel <lb/>che &longs;i moue &longs;opra il piano inchinato all'orizonte, &longs;i riduce alla bilancia per la nona <lb/>propo&longs;itione di Pappo dell'ottauo libro delle raccolte matematiche. </s>

<s id="id.2.1.1264.2.0">percioche è vna <lb/>i&longs;te&longs;&longs;a ragione, che ouero &longs;tando fermo il cuneo, il pe&longs;o &longs;i moua &longs;opra il lato del cu<lb/>neo; ouero che e&longs;&longs;endo egli mo&longs;&longs;o, &longs;i moua anco il pe&longs;o &longs;opra il &longs;uo lato, come &longs;o<lb/>pra vn piano inchinato all'orizonte.<emph.end type="italics"/></s>

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<s id="id.2.1.1268.1.0"><emph type="italics"/>Sia il cuneo ABC, & AB &longs;ia eguale ad e&longs;&longs;a BC. </s>

<s id="id.2.1.1268.2.0">Diuida&longs;i AC in due partii<gap/><lb/>D, & &longs;ia congiunta BD. &longs;ia dopo la linea EF, per laquale paßi il piano egual<lb/>mente di&longs;tante dall'orizonte, & &longs;ia BD nella mede&longs;ima linea EF; & mentre il<emph.end type="italics"/><lb/>

<arrow.to.target n="fig116"></arrow.to.target><lb/><emph type="italics"/>cuneo è perco&longs;&longs;o, & mentre &longs;i moue in ver&longs;o E, &longs;empre BD &longs;ia nella linea EF. <lb/>& quel che &longs;i hada fendere &longs;ia GHLM, dentro alquale &longs;ia la parte del cuneo <lb/>KBI: egli è mani&longs;e&longs;to, che mentre il cuneo &longs;i moue in ver&longs;o E, la parte KG mo<lb/>uer&longs;i in ver&longs;o N; & la parte HI in ver&longs;o O. </s><s id="id.2.1.1268.3.0">percota&longs;i il cuneo per modo che la <lb/>linea AC &longs;ia nella linea NO; allhora K &longs;arà in A, & I in C: & K perle <lb/>co&longs;e &longs;udette &longs;arà mo&longs;&longs;o &longs;opra KA, & I &longs;opra IC. </s>

<figure id="fig116"></figure><lb/><emph type="italics"/>cuneo è perco&longs;&longs;o, & mentre &longs;i moue in ver&longs;o E, &longs;empre BD &longs;ia nella linea EF. <lb/>& quel che &longs;i hada fendere &longs;ia GHLM, dentro alquale &longs;ia la parte del cuneo <lb/>KBI: egli è mani&longs;e&longs;to, che mentre il cuneo &longs;i moue in ver&longs;o E, la parte KG mo<lb/>uer&longs;i in ver&longs;o N; & la parte HI in ver&longs;o O. </s><s id="id.2.1.1268.3.0">percota&longs;i il cuneo per modo che la <lb/>linea AC &longs;ia nella linea NO; allhora K &longs;arà in A, & I in C: & K perle <lb/>co&longs;e &longs;udette &longs;arà mo&longs;&longs;o &longs;opra KA, & I &longs;opra IC. </s>

<s id="id.2.1.1268.4.0">Per laqual co&longs;a mentre &longs;i mo <lb/>ue il cuneo, la parte KG &longs;i mouerà &longs;opra il lato BA del cuneo, & la parte IH <lb/>&longs;opra il lato BC. </s>

<s id="id.2.1.1268.5.0">La parte dunque KG &longs;i mouerà &longs;opra il piano inchinato all'o<lb/>rizonte, la cui inclinatione è l'angolo FBA. &longs;imilmente IH &longs;i moue &longs;opra il <lb/>piano BC nell'angolo FBC. </s><s id="id.2.1.1268.6.0">le parti dunque di quel che &longs;i &longs;ende moueran&longs;i &longs;o<lb/>pra piani inchinati all'orizonte. </s>

<s id="id.2.1.1268.7.0">& quantun que il piano BC &longs;ia &longs;otto l'orizonte; <lb/>tutta via la parte IH &longs;i moue &longs;opra IC, come &longs;e BC &longs;o&longs;&longs;e &longs;opra l'orizonte nel<lb/>l'angolo DBC: percioche le parti di quel che &longs;i fende &longs;i mouono nel tempo me<lb/>de&longs;imo dall'i&longs;te&longs;&longs;a po&longs;&longs;anza. </s>

<s id="id.2.1.1268.8.0">&longs;arà dunque la mede&longs;ima ragione del mouimento della <lb/>parte IH, & della parte KF. </s><s id="id.2.1.1268.9.0">&longs;imilmente è l'i&longs;te&longs;&longs;a ragione &longs;e EF è egualmente <lb/>di&longs;tante dall'orizonte, ouero &longs;e è à piombo dell'orizonte, ouero in altro modo: pe <lb/>roche egli è nece&longs;&longs;ario, che la po&longs;&longs;anza, laquale moue il cuneo, &longs;ia la mede&longs;ima, re<lb/>&longs;tando le altre co&longs;e le mede&longs;ime. </s>

<s id="id.2.1.1268.10.0">&longs;arà dunque la &longs;te&longs;&longs;a ragione.<emph.end type="italics"/></s>

<s id="id.2.1.1271.1.0"><emph type="italics"/>Dopo que&longs;te co&longs;e egli è da con&longs;iderare, quali &longs;iano quelle co&longs;e, lequali fanno sì, che più <lb/>ageuolmente alcuna co&longs;a &longs;i moua, ouero &longs;i &longs;enda, lequali &longs;ono due.<emph.end type="italics"/></s>

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<s id="id.2.1.1273.3.0">Diuidan&longs;i AC DF in due parti eguali ne'punti GH; <lb/>& &longs;iano congiunte BG & EH. </s>

<s id="id.2.1.1273.4.0">Hor <lb/>percioche le parti di quello, che &longs;i fen <lb/>de dal cuneo ABC &longs;i mouono &longs;opra <lb/>il piano inchinato all'orizonte, la cui <lb/>inclinatione è GBA; & quelle che <lb/>dal cuneo DEF &longs;i mouono &longs;opra il <lb/>piano inchinato all'orizonte, la cui <lb/>inclinatione è HED, & l'angolo <lb/>GBA è minore dell'angolo HED; <lb/>per e&longs;&longs;ere GBA minore di DEF: <lb/>& per la nona di Pappo dell'ottauo <lb/>libro delle raccolte matematiche, quel <lb/>che &longs;i moue &longs;opra il piano AB, &longs;i mo <lb/>uerà più facilmente, & da po&longs;&longs;anza <lb/>minore, che &longs;opra ED. </s>

<s id="id.2.1.1273.5.0">Quel che &longs;i <lb/>&longs;ende dunque dal cuneo ABC più<emph.end type="italics"/><lb/>

<arrow.to.target n="fig117"></arrow.to.target><lb/><emph type="italics"/>ageuolmente, & da po&longs;&longs;anza minore &longs;i fende, che dal cuneo DEF. </s><s id="id.2.1.1273.6.0">&longs;imilmente <lb/>mo&longs;trera&longs;&longs;i, che quanto più acuto &longs;arà l'angolo po&longs;to alla cima del cuneo, tanto più <lb/>ageuolmente mouera&longs;&longs;i, & fendera&longs;&longs;i alcuna co&longs;a. </s>

<figure id="fig117"></figure><lb/><emph type="italics"/>ageuolmente, & da po&longs;&longs;anza minore &longs;i fende, che dal cuneo DEF. </s><s id="id.2.1.1273.6.0">&longs;imilmente <lb/>mo&longs;trera&longs;&longs;i, che quanto più acuto &longs;arà l'angolo po&longs;to alla cima del cuneo, tanto più <lb/>ageuolmente mouera&longs;&longs;i, & fendera&longs;&longs;i alcuna co&longs;a. </s>

<s id="id.2.1.1273.7.0">che bi&longs;ognaua mo&longs;trare.<emph.end type="italics"/></s>

<s id="id.2.1.1275.1.0">Pos&longs;iamo dimo&longs;trare que&longs;to etiandio con altra ragione, con&longs;i<lb/>derando il cuneo come egli moue con le leue contrarie l'vna <lb/>all'altra fra loro, fi come nel &longs;econdo modo fù detto. </s>

<s id="id.2.1.1275.2.0">ma bi&longs;o <lb/>gna prima dimo&longs;trare que&longs;to. </s>

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<s id="id.2.1.1276.1.0"><emph type="italics"/>Sia la leua AB, che habbia il &longs;uo &longs;o&longs;tegno B immobile, & quel che s'ha da mouere <lb/>&longs;ia CD EF rettangolo, co&longs;i di&longs;po&longs;to, che non po&longs;&longs;a mouer&longs;i in giù dalla parte di <lb/>FE; & il punto E &longs;ia immobile, & come centro; &longs;iche il punto D &longs;i moua per <lb/>la circonferenza del cerchio <lb/>DH, il cui centro &longs;ia E. & <lb/>C per la circonferenza CL, <lb/>&longs;i che la linea congiunta CE <lb/>&longs;ia il &longs;uo mezo diametro. </s>

<s id="id.2.1.1276.2.0"><lb/>di più CDEF tocchi la le <lb/>ua AB in C, & la leua <lb/>AB moua il pe&longs;o CDEF, <lb/>& la po&longs;&longs;anza mouente &longs;ia <lb/>in A, il &longs;o&longs;tegno in B, & <lb/>il pe&longs;o in C. &longs;ia dapoi vn'al <lb/>traleua MCN, laquale <lb/>etiandio moua CD EF, il <lb/>cui &longs;o&longs;tegno immobile &longs;ia <lb/>N; la po&longs;&longs;anza mouente in <lb/>M, & il pe&longs;o &longs;imilmente in <lb/>C; & &longs;ia CN eguale ad<emph.end type="italics"/><lb/>

<arrow.to.target n="fig118"></arrow.to.target><lb/><emph type="italics"/>e&longs;&longs;a CB, & CM ad e&longs;&longs;a CA; & moua&longs;i alternamente il pe&longs;o CDEF con le <lb/>leue AB MN. </s>

<figure id="fig118"></figure><lb/><emph type="italics"/>e&longs;&longs;a CB, & CM ad e&longs;&longs;a CA; & moua&longs;i alternamente il pe&longs;o CDEF con le <lb/>leue AB MN. </s>

<s id="id.2.1.1276.3.0">Dico che CDEF più ageuolmente &longs;i mouerà dall'i&longs;te&longs;&longs;a po&longs;&longs;an <lb/>za con la leua AB, che con la leua MN.<emph.end type="italics"/></s>

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<s id="id.2.1.1283.1.0"><emph type="italics"/>Diuidan&longs;i AC DF in due parti eguali in TV, & congiungan&longs;i TBVE, &longs;aranno <lb/>gli angoli po&longs;ti al T, & V retti. </s>

<s id="id.2.1.1283.2.0">congiunga&longs;i IG, laquale tagli BT in X. </s><s id="id.2.1.1283.3.0">Hor<emph.end type="italics"/><lb/>

<arrow.to.target n="fig119"></arrow.to.target><lb/>

<arrow.to.target n="note310"></arrow.to.target> <emph type="italics"/>percioche IB è eguale à BG, & BA eguale à BC: &longs;arà IA eguale ad e&longs;&longs;a<emph.end type="italics"/><lb/>

<arrow.to.target n="note311"></arrow.to.target> <emph type="italics"/>GC. </s>

<s id="id.2.1.1283.4.0">Per laqual co&longs;a BI ad IA è co&longs;i, come BG à GC; dunque IG è egualmen <lb/>te di&longs;tante ad e&longs;&longs;a AC: & perciò gli angoli ad X &longs;ono retti; ma gli angoli XGK<emph.end type="italics"/><lb/>

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<s id="id.2.1.1286.1.0"><margin.target id="note311"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 9. <emph type="italics"/>del primo.<emph.end type="italics"/></s>

<s id="id.2.1.1287.1.0"><margin.target id="note312"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 28. <emph type="italics"/>del primo.<emph.end type="italics"/></s>

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<s id="id.2.1.1293.1.0"><emph type="italics"/>Che &longs;e C &longs;arà mo&longs;&longs;o da qualche po&longs;&longs;anza, <lb/>come per lo manico DE &longs;ia mo&longs;&longs;o. </s>

<s id="id.2.1.1293.2.0">Pri <lb/>ma quanto C &longs;arà più graue; dapoi <lb/>quanto &longs;arà più lungo DE, tanto la <lb/>perco&longs;&longs;a faraßimaggiore: percioche &longs;e <lb/>la po&longs;&longs;anza mouente &longs;arà posta in E, <lb/>&longs;arà il C più di&longs;tante dal centro, & pe <lb/>rò moueraßi più tosto, come Ari&longs;to<lb/>tele dimostra nelle questioni mecani<lb/>che; & puote e&longs;&longs;ere anco chiaro da <lb/>quelle co&longs;e, che furono dette nel trat<lb/>tato della bilancia, che quanto più il<emph.end type="italics"/><lb/>

<arrow.to.target n="fig120"></arrow.to.target>

<pb pagenum="114"/><emph type="italics"/>pe&longs;o C è di&longs;tante dal centro, tanto più far&longs;i graue, & vrterà etiandio con più ga<lb/>gliard'empito, e&longs;&longs;endo la forza in E più po&longs;&longs;ente.<emph.end type="italics"/></s>

<s id="id.2.1.1295.1.0"><emph type="italics"/>Ma que&longs;ta è la <expan abbr="&longs;ecõda">&longs;econda</expan> co&longs;a, laqual è cagione che con que&longs;to i&longs;trumento &longs;imouano gran <lb/>pe&longs;i, & &longs;i fendano. </s>

<s id="id.2.1.1295.2.0">Percioche la perco&longs;&longs;a è vna forza gagliardißima, come è ma<lb/>nife&longs;to da la decimanona delle questioni <lb/>mec aniche di Ari&longs;totele: peroche &longs;e &longs;o<lb/>pra il cuneo &longs;i imporrà vn pe&longs;o grandißi<lb/>mo, allhora il cuneo non farà nulla à pa<lb/>ragone &longs;petialmente della perco&longs;&longs;a. </s>

<s id="id.2.1.1295.3.0">che &longs;e <lb/>anco &longs;i adatta&longs;&longs;e al cuneo vna leua, ouero <lb/>vna vite, ò qualche altro tale &longs;tromento <lb/>per cacciare il cuneo più à dentro nel pe&longs;o, <lb/>non auenir à effetto qua&longs;i di momento niu <lb/>no, ri&longs;petto alla perco&longs;&longs;a. </s>

<s id="id.2.1.1295.4.0">della qual co&longs;a <lb/>puote e&longs;&longs;ere inditio, che &longs;e fo&longs;&longs;e il corpo A <lb/>di pietra, da cui alcuno vole&longs;&longs;e leuar via<emph.end type="italics"/><lb/>

<arrow.to.target n="fig121"></arrow.to.target><lb/><emph type="italics"/>qualche parte, come vn pezzo dell'angolo B, allhora potrebbe rompere ageuolmen <lb/>te con vno martello di ferro, &longs;enza altro &longs;tromento, percotendo in B, qualche pezzo <lb/>dell'angolo B: ilche non potrà fare con ne&longs;&longs;uno altro &longs;tromento, che &longs;ia priuo di per<lb/>co&longs;&longs;a, &longs;e non con difficultà grandißima, &longs;ia ò leua, ò vite, ò qual &longs;i voglia altra co&longs;a <lb/>tale. </s>

<figure id="fig121"></figure><lb/><emph type="italics"/>qualche parte, come vn pezzo dell'angolo B, allhora potrebbe rompere ageuolmen <lb/>te con vno martello di ferro, &longs;enza altro &longs;tromento, percotendo in B, qualche pezzo <lb/>dell'angolo B: ilche non potrà fare con ne&longs;&longs;uno altro &longs;tromento, che &longs;ia priuo di per<lb/>co&longs;&longs;a, &longs;e non con difficultà grandißima, &longs;ia ò leua, ò vite, ò qual &longs;i voglia altra co&longs;a <lb/>tale. </s>

<s id="id.2.1.1295.5.0">La onde la perco&longs;&longs;a è cagione, che &longs;i fendano i gran pe&longs;i. </s>

<s id="id.2.1.1295.6.0">& hauendo la per<lb/>co&longs;&longs;a co&longs;i gran forza, &longs;e le ag giungeremo qualche &longs;tromento accommodato à moue<lb/>re, & fendere, vedremo per certo co&longs;e marauiglio&longs;e. </s>

<s id="id.2.1.1295.7.0">Cote&longs;to &longs;tromento è il cuneo, <lb/>nel quale due co&longs;e, inquanto s'ap <lb/>partiene alla &longs;ua forma, occor<lb/>rono ad e&longs;&longs;ere con&longs;iderate: L'v<lb/>na, che il cuneo è attißimo à ri<lb/>ceuere, & &longs;o&longs;tenere la perco&longs;&longs;a: <lb/>l'altra è, che per la &longs;ua &longs;ottigliez <lb/>za nell vna delle parti <expan abbr="facilm&emacr;te">facilmente</expan> <lb/>entra ne'corpi, come e&longs;pre&longs;&longs;a<lb/>mente &longs;i vede. </s>

<s id="id.2.1.1295.8.0">Il cuneo dunque <lb/>opera&longs;i con la &longs;ua perco&longs;&longs;a, che <lb/>vediamo qua&longs;i miracoli nel fen<lb/>dere i corpi.<emph.end type="italics"/></s>

<pb/>

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<s id="id.2.1.1306.1.0"><emph type="italics"/>Sia il cuneo ABC, ilquale &longs;iriuolga d'intorno al Cilindro DE, & &longs;ia IGH il cu<lb/>neo riuolto d'intorno al cilindro, la cui cima &longs;ia I. &longs;ia dapoi il cilindro in&longs;ieme co'lcu <lb/>neo po&longs;toui d'intorno accomm o dato in modo, che &longs;enza alcuno impedimento &longs;ipo&longs;&longs;a <lb/>volgere intorno co'l manico KF attaccato all'a&longs;&longs;e: & &longs;ia LMNO quel che s'ha <lb/>da fendere, ilquale etiandio dalla parte di MN &longs;ia immobile, &longs;i come &longs;uole far&longs;i <lb/>in quelle co&longs;e, che &longs;i fendono. </s>

<s id="id.2.1.1306.2.0">& &longs;ia la cima I'tra RS. </s>

<s id="id.2.1.1306.3.0">Volga&longs;i intorno KF, &<emph.end type="italics"/><lb/>

<arrow.to.target n="fig122"></arrow.to.target><lb/><emph type="italics"/>peruenga à KP; & mentre che KF &longs;i volge intorno, tutto il cilindro DE anc<gap/><lb/>ra &longs;i volge intorno, & il cuneo IGH. per laqual co&longs;a mentre KF &longs;arà in KP, <lb/>la cima I non &longs;arà più tra RS, ma altr a parte del cuneo, come TV: ma TV è <lb/>maggiore di RS; peroche la parte del cuneo, laquale è più di&longs;tante dalla cima, &longs;em<lb/>pre è mag giore di quella, che è più ad e&longs;&longs;a vicina. </s>

<figure id="fig122"></figure><lb/><emph type="italics"/>peruenga à KP; & mentre che KF &longs;i volge intorno, tutto il cilindro DE anc<gap/><lb/>ra &longs;i volge intorno, & il cuneo IGH. per laqual co&longs;a mentre KF &longs;arà in KP, <lb/>la cima I non &longs;arà più tra RS, ma altr a parte del cuneo, come TV: ma TV è <lb/>maggiore di RS; peroche la parte del cuneo, laquale è più di&longs;tante dalla cima, &longs;em<lb/>pre è mag giore di quella, che è più ad e&longs;&longs;a vicina. </s>

<s id="id.2.1.1306.4.0">accioche dunque TV &longs;ia tra RS, <lb/>bi&longs;ogna che R ceda, & &longs;i moua ver&longs;o X, & S in ver&longs;o Z, come fanno le co&longs;e, che <lb/>&longs;i fendono. </s>

<s id="id.2.1.1306.5.0">tutto dunque LMNO &longs;i fenderà. </s>

<s id="id.2.1.1306.6.0">Similmente dimo&longs;treremo, che men <lb/>tre il manico KP &longs;arà in KQ, allhora GH &longs;arà fra RS: & mentre GH &longs;arà <lb/>tra RS, egli è nece&longs;&longs;ario che R &longs;ia in X, & S in Z. talche XZ &longs;ia eguale à GH; <lb/>& &longs;empre LM NO &longs;i fenderà dauantaggio. </s>

<s id="id.2.1.1306.7.0">co&longs;i dunque è manife&longs;to, che mentre <lb/>KF &longs;i volge intorno, &longs;empre R &longs;i moue in ver&longs;o X, & S in ver&longs;o Z: & R mo<lb/><gap/>er&longs;i &longs;empre &longs;opra ITG, & S &longs;opra IVH, cioè &longs;opra i lati del cuneo volti <lb/>d'intorno al cilindro.<emph.end type="italics"/></s>

<s id="id.2.1.1309.1.0">PROPOSITIONE I.</s>

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<s id="id.2.1.1311.1.0"><emph type="italics"/>Sia il cuneo ABC; & AB &longs;ia eguale à BC. </s><s id="id.2.1.1311.2.0">diuida&longs;i AC in due parti in D, <lb/>& congiunga&longs;i BD; &longs;arà BD à piombo di AC: & AD eguale à DC, & il <lb/>triangolo ABD eguale al triangolo CBD. </s>

<s id="id.2.1.1311.3.0">Faccia&longs;i dapoi i triangoli rettangoli <lb/>EFG HIK non &longs;olo tra loro eguali, ma etiandio eguali ad ambidue i triang oli<emph.end type="italics"/><lb/>

<arrow.to.target n="fig123"></arrow.to.target><lb/><emph type="italics"/>ADB, & CDB. & &longs;ia il cilindro LMNO, la cui linea che lo circonda detto <lb/>Perimetro &longs;ia eguale ad ambedue FGKI: & LMNO &longs;ia parallelo gram<lb/>mo per l'a&longs;&longs;e. </s>

<figure id="fig123"></figure><lb/><emph type="italics"/>ADB, & CDB. & &longs;ia il cilindro LMNO, la cui linea che lo circonda detto <lb/>Perimetro &longs;ia eguale ad ambedue FGKI: & LMNO &longs;ia parallelo gram<lb/>mo per l'a&longs;&longs;e. </s>

<s id="id.2.1.1311.4.0">& faccia&longs;i MP eguale ad FE, & PN eguale ad HI. & pon <lb/>ga&longs;i HI in NP, & inuolga&longs;i il triangolo HIK d'intorno al cilindro; & &longs;ia de <lb/>&longs;critta la helice NQR &longs;econdo KH, come in&longs;egna anche Pappo nell'ottauo libro <lb/>alla propo&longs;itione vige&longs;ima quarta. </s>

<s id="id.2.1.1311.5.0">& &longs;imilmente ponga&longs;i EF in MP, & in<lb/>uolga&longs;i il triangolo EFG d'intorno al cilindro, & de&longs;criua&longs;i per EG la helice <lb/>PRM. & co&longs;i per e&longs;&longs;ere PM PN eguali ad EF HI, &longs;arà MN eguale ad <lb/>e&longs;&longs;a AC, & per e&longs;&longs;ere le helici PRM PQN eguali alle linee EG HK; &longs;a-<emph.end type="italics"/></s>

<s id="id.2.1.1313.1.0"><emph type="italics"/>Similm ente &longs;e la vite <lb/>haurà più helici co<lb/>me nella &longs;econda &longs;i<lb/>gura, il pe&longs;o A, men <lb/>tre la vite &longs;i volge <lb/>intorno, &longs;empre &longs;i <lb/>mouerà &longs;opr a le he<lb/>lici BCD EFG; <lb/>pur che il pe&longs;o A <lb/>in modo &longs;i adatti, <lb/>che non po&longs;&longs;a mo<lb/>uer&longs;i &longs;e non &longs;opra la <lb/>retta linea HI e<lb/>gualmente di&longs;tante <lb/>da e&longs;&longs;o cilindro. </s>

<s id="id.2.1.1313.2.0">Per <lb/>cioche nell'i&longs;te&longs;&longs;o mo <lb/>do, che &longs;i moue &longs;o<lb/>pra la prima helice, <lb/>&longs;i moue etiandio &longs;o<lb/>pra la &longs;econda, & &longs;o <lb/>pra la terza, et &longs;opra <lb/>le altre. </s>

<s id="id.2.1.1313.3.0">Percioche <lb/>quante &longs;i <expan abbr="vogli&amacr;heli">voglianheli</expan> <lb/>ci che &longs;iano, non &longs;on <lb/>altro niente, che vn <lb/>lato del cuneo inuol<lb/>to d'intorno all'i&longs;te&longs;-<emph.end type="italics"/><lb/>

<arrow.to.target n="fig124"></arrow.to.target><lb/><emph type="italics"/>&longs;o cilindro vna, & più volte. </s>

<figure id="fig124"></figure><lb/><emph type="italics"/>&longs;o cilindro vna, & più volte. </s>

<s id="id.2.1.1313.4.0">et &longs;ia la vite ouero à piombo dell'orizonte, ouero egual <lb/>mente di&longs;tante dall orizonte, ouero in altro modo collocata, nonimporta nulla; per<lb/>cioche &longs;empre valerà l'i&longs;te&longs;&longs;a ragione.<emph.end type="italics"/></s>

<s id="id.2.1.1316.1.0"><emph type="italics"/>che non po&longs;&longs;ano mouer&longs;i &longs;e non &longs;opra la diritta linea LO, laquale &longs;ia egualmente <lb/>di&longs;tante dail'a&longs;&longs;e del cilindro; & &longs;iano MN pre&longs;&longs;o la cima I del cuneo. </s>

<s id="id.2.1.1316.2.0">Volga&longs;i <lb/>intorno KF, & peruenga in KP: & mentre KF &longs;arà in KP, allhora TV &longs;a <lb/>rà fra i pe&longs;i MN, &longs;i come di &longs;opra habbiamo detto. </s>

<s id="id.2.1.1316.3.0">dunque M &longs;i mouerà ver&longs;o<emph.end type="italics"/><lb/>

<arrow.to.target n="fig125"></arrow.to.target><lb/><emph type="italics"/>L, & N ver&longs;o O. </s>

<figure id="fig125"></figure><lb/><emph type="italics"/>L, & N ver&longs;o O. </s>

<s id="id.2.1.1316.4.0">Similmente mo&longs;trera&longs;&longs;i, che mentre KP &longs;arà in KQ, allho<lb/>ra GH &longs;arà tra i pe&longs;i MN; & M &longs;arà in X, & N in Z; &longs;i che XZ &longs;arà <lb/>eguale <gap/> GH. </s>

<s id="id.2.1.1316.5.0">Per laqual co&longs;a mentre KF &longs;i volge intorno, &longs;empre il pe&longs;o N &longs;i <lb/>moue in ver&longs;o O, & &longs;opra la helice IRS; & M &longs;opra l'altra helice.<emph.end type="italics"/></s>

<pb/>

<s id="id.2.1.1319.1.0"><emph type="italics"/>ranno dunque le dette helici eguali ad e&longs;&longs;e AB BC. dunque il cuneo. </s>

<s id="id.2.1.1319.2.0">ABC &longs;arà <lb/>tutto inuolt<foreign lang="greek">q</foreign> d'intorno al cilindro LMNO. </s>

<s id="id.2.1.1319.3.0">Siano tagliate da poile helici, come <lb/>in&longs;egna Pappo, &longs;econdo la larghezza del cuneo; & à que&longs;lo modo il cuneo in&longs;ieme<emph.end type="italics"/><lb/>

<arrow.to.target n="fig126"></arrow.to.target><lb/><emph type="italics"/>co'l cilindro niente altro &longs;arà, che la vite, laquale habbia due helici PRM PQN <lb/>congiunte fra loro d'intorno al cilindro LN in vno &longs;olo punto. </s>

<figure id="fig126"></figure><lb/><emph type="italics"/>co'l cilindro niente altro &longs;arà, che la vite, laquale habbia due helici PRM PQN <lb/>congiunte fra loro d'intorno al cilindro LN in vno &longs;olo punto. </s>

<s id="id.2.1.1319.4.0">che bi&longs;ognaua <lb/>mo&longs;trare.<emph.end type="italics"/></s>

<s id="id.2.1.1321.1.0">COROLLARIO.</s>

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<s id="id.2.1.1326.1.0"><emph type="italics"/>Che &longs;e come nella terza figura, &longs;i imporrà alcuna co&longs;a &longs;opra la vite, come B, che è no <lb/>mata Tilo di&longs;po&longs;to in modo, che dalla parte di &longs;otto egli habbia le helici concaue <lb/>adattate <gap/>olto acconciamente ad e&longs;&longs;a vite. </s>

<s id="id.2.1.1326.2.0">egli potrà e&longs;&longs;ere a&longs;&longs;ai chiaro, che e&longs;&longs;o B, <lb/>mentre la vite &longs;i volge intorno, mouera&longs;&longs;i à quel modo in tutto &longs;opra le helici della<emph.end type="italics"/><lb/>

<arrow.to.target n="fig127"></arrow.to.target><lb/><emph type="italics"/>vite, come &longs;i moueua il pe&longs;o &longs;econdo la prima figura; purche il tilo &longs;i accommo<lb/>di, come in&longs;egna Pappo nell'ottauo libro, in maniera cioè, che egli &longs;i moua egual<lb/>mente di&longs;tante dall'a&longs;&longs;e del cilindro auanti, ouero indietro &longs;olamente.<emph.end type="italics"/></s>

<figure id="fig127"></figure><lb/><emph type="italics"/>vite, come &longs;i moueua il pe&longs;o &longs;econdo la prima figura; purche il tilo &longs;i accommo<lb/>di, come in&longs;egna Pappo nell'ottauo libro, in maniera cioè, che egli &longs;i moua egual<lb/>mente di&longs;tante dall'a&longs;&longs;e del cilindro auanti, ouero indietro &longs;olamente.<emph.end type="italics"/></s>

<pb/>

<s id="id.2.1.1329.1.0"><emph type="italics"/>Et &longs;e in luogo del tilo, che hà le helici concaue nella parte di &longs;otto, &longs;i ponga, come nel <lb/>la quarta figura il cilindro concauo, come D, & nella &longs;ua concaua &longs;uperficie &longs;i de<lb/>&longs;criuano le helici, & &longs;i taglino in modo, che acconciamente &longs;i adattino alla vite; <lb/>(percioche nel mede&longs;imo modo &longs;i de&longs;criueranno le helici nella &longs;uperficie concaua del <lb/>cilindro, come &longs;i fà nella conue&longs;&longs;a) &longs;e la vite poi &longs;arà &longs;ermata ne' poli &longs;uoi, cioè nel<emph.end type="italics"/><lb/>

<arrow.to.target n="fig128"></arrow.to.target><lb/><emph type="italics"/>&longs;uo a&longs;&longs;e, & volga&longs;i intorno, egli è manife&longs;to, che D &longs;i mouerà al mouimento del <lb/>giro della vite, come &longs;a il tilo. </s>

<figure id="fig128"></figure><lb/><emph type="italics"/>&longs;uo a&longs;&longs;e, & volga&longs;i intorno, egli è manife&longs;to, che D &longs;i mouerà al mouimento del <lb/>giro della vite, come &longs;a il tilo. </s>

<s id="id.2.1.1329.2.0">& di più &longs;e D &longs;i &longs;ermerà in EF, &longs;i che rimanga im <lb/>mobile, mentre la vite &longs;i volge intorno, mouera&longs;&longs;i &longs;opra le helici del cilindro D &longs;e<lb/>condo il mouimento del giro &longs;uo, fatto alla de&longs;tra, ouero alla &longs;ini&longs;tra, sì all'innan<lb/>zi, come all'indtetro, & il cilindro D in que&longs;ta maniera accommedato, &longs;i chiama <lb/>volgarmente la madre, ouero la femina della vite.<emph.end type="italics"/></s>

<s id="id.2.1.1332.1.0"><emph type="italics"/>Che &longs;e alla vite (come nella quinta figura) &longs;arà po&longs;ta la rota C co' dentìtorti, come <lb/>in&longs;egna Pappo nel mede&longs;imo ottauo libro, ouero anche diritti; ma in modo &longs;atti, <lb/>che &longs;i adattino facilmente con la vite. </s>

<s id="id.2.1.1332.2.0">egli è &longs;imilmente manife&longs;to, che al mo uimen<emph.end type="italics"/><lb/>

<arrow.to.target n="fig129"></arrow.to.target><lb/><emph type="italics"/>to della vite moueraßi etiandio intorno la rota C. & nell'i&longs;te&longs;&longs;a maniera &longs;i moue<lb/>rannoi denti della rota C &longs;opra le helici della vite. </s>

<figure id="fig129"></figure><lb/><emph type="italics"/>to della vite moueraßi etiandio intorno la rota C. & nell'i&longs;te&longs;&longs;a maniera &longs;i moue<lb/>rannoi denti della rota C &longs;opra le helici della vite. </s>

<s id="id.2.1.1332.3.0">& que&longs;ta &longs;i dice vite perpetua, <lb/>percioche sì la vite, come la rota mentre &longs;i riuolgono &longs;tanno &longs;empre nel modo <lb/>i&longs;te&longs;&longs;o.<emph.end type="italics"/></s>

<s id="id.2.1.1334.1.0"><emph type="italics"/>Que&longs;te co&longs;e habbiamo detto, accioche &longs;ia pale&longs;e, che la vite nel mouere il pe&longs;o fà l'officio <lb/>del cuneo &longs;enza perco&longs;&longs;a. </s>

<s id="id.2.1.1334.2.0">percio che lo rimoue dal luogo oue era, &longs;i come il cuneo <lb/>rimoue quelle co&longs;e che moue, & &longs;ende. </s>

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<s id="id.2.1.1336.1.0"><emph type="italics"/>Et prima come ella moue conle leue; come nella prima figura. </s>

<s id="id.2.1.1336.2.0">giri&longs;i intorno KF, &<emph.end type="italics"/><lb/>

<arrow.to.target n="fig130"></arrow.to.target><lb/><emph type="italics"/>peruenga in KP, allhora, &longs;i come è detto, TV &longs;arà fra pe&longs;i MN. & &longs;i come <lb/>con&longs;ideriamo le leue nel cuneo, co&longs;i le po&longs;siamo parimente con&longs;ider are nella vite in <lb/>que&longs;ta maniera, cioè &longs;arà IVH la leua co'l &longs;o&longs;tegno &longs;uo I, & il pe&longs;o po&longs;to in <lb/>V. &longs;imilmente ITG la leua co'l &longs;o&longs;tegno &longs;uo I, & il pe&longs;o in T. & le po&longs;&longs;an<lb/>ze mouenti dourebbono e&longs;&longs;ere in GH; ma &longs;i come nel cuneo la po&longs;&longs;anza mouen <lb/>te è la perco&longs;&longs;a, laquale moue il cuneo; però &longs;arà doue la po&longs;&longs;anza moue la vite, co<lb/>me in P colmanico KP; peroche la vite &longs;i moue &longs;enza perco&longs;&longs;a. </s>

<figure id="fig130"></figure><lb/><emph type="italics"/>peruenga in KP, allhora, &longs;i come è detto, TV &longs;arà fra pe&longs;i MN. & &longs;i come <lb/>con&longs;ideriamo le leue nel cuneo, co&longs;i le po&longs;siamo parimente con&longs;ider are nella vite in <lb/>que&longs;ta maniera, cioè &longs;arà IVH la leua co'l &longs;o&longs;tegno &longs;uo I, & il pe&longs;o po&longs;to in <lb/>V. &longs;imilmente ITG la leua co'l &longs;o&longs;tegno &longs;uo I, & il pe&longs;o in T. & le po&longs;&longs;an<lb/>ze mouenti dourebbono e&longs;&longs;ere in GH; ma &longs;i come nel cuneo la po&longs;&longs;anza mouen <lb/>te è la perco&longs;&longs;a, laquale moue il cuneo; però &longs;arà doue la po&longs;&longs;anza moue la vite, co<lb/>me in P colmanico KP; peroche la vite &longs;i moue &longs;enza perco&longs;&longs;a. </s>

<s id="id.2.1.1336.3.0">Ma que&longs;ta con <lb/>&longs;ideratione parerà for&longs;e impropria per cau&longs;a delle leue piegate. </s>

<s id="id.2.1.1336.4.0">Onde &longs;e &longs;i inten<lb/>derà, quello che è mo&longs;&longs;o dalla vite, e&longs;&longs;ere mo&longs;&longs;o &longs;opra vn piano inchinato all' <expan abbr="orizõte">orizonte</expan>; <lb/>per certo cotale con&longs;ideratione &longs;arà più conforme alla figura di e&longs;&longs;a vite, ma&longs;sima<lb/>mente conuenendo anche al cuneo.<emph.end type="italics"/></s>

<s id="id.2.1.1339.1.0">PROPOSITIONE II.</s>

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<s id="id.2.1.1341.1.0"><emph type="italics"/>Sia la vite AB à piombo dell'orizonte, che habbia due helici CDEFG. </s>

<s id="id.2.1.1341.2.0">Ponga&longs;i <lb/>HI eguale à GC, laquale diuida&longs;i in due parti in K. &longs;aranno HK KI non &longs;o<lb/>lamente fra loro, ma etiandio ad e&longs;&longs;e GEEC eguali, & tiri&longs;i ad e&longs;&longs;a HI la li-<emph.end type="italics"/><lb/>

<arrow.to.target n="fig131"></arrow.to.target><lb/><emph type="italics"/>nea LI ad angoliretti; & intenda&longs;i per LI vn piano egualmente di&longs;tante dall'o<lb/>rizonte: & &longs;ia LI due volte tanto quanto la linea che gira intorno al cilindro <lb/>AB che dice&longs;i Perimetro, laquale diuida&longs;i in due parti eguali in M; &longs;ar anno IM <lb/>ML eguali al Perimetro del cilindro. </s>

<figure id="fig131"></figure><lb/><emph type="italics"/>nea LI ad angoliretti; & intenda&longs;i per LI vn piano egualmente di&longs;tante dall'o<lb/>rizonte: & &longs;ia LI due volte tanto quanto la linea che gira intorno al cilindro <lb/>AB che dice&longs;i Perimetro, laquale diuida&longs;i in due parti eguali in M; &longs;ar anno IM <lb/>ML eguali al Perimetro del cilindro. </s>

<s id="id.2.1.1341.3.0">Congiunga&longs;i HL, & da punto M &longs;ia ti-

<pb/>rata la linea MN egualmente di&longs;tante da HI, & congiunga&longs;i KN. </s>

<s id="id.2.1.1341.4.0">Hor per<lb/>cioche i triangoli HIL NML &longs;ono &longs;imili fra loro, per e&longs;&longs;ere NM egualmen-<emph.end type="italics"/><lb/>

<arrow.to.target n="note313"></arrow.to.target> <emph type="italics"/>te di&longs;tante da HI; &longs;arà LI ad IH, come LM ad MN: & permutando co<lb/>me IL ad LM, co&longs;i HI ad NM. </s>

<s id="id.2.1.1341.5.0">Ma IL è due volte tanto quanto LM; dun <lb/>que anco HI &longs;arà il doppio di MN. ma ella è il doppio anche di KI; per laqual<emph.end type="italics"/><lb/>

<arrow.to.target n="fig132"></arrow.to.target><lb/><emph type="italics"/>co&longs;a KI NM &longs;ono tra &longs;e eguali. </s>

<figure id="fig132"></figure><lb/><emph type="italics"/>co&longs;a KI NM &longs;ono tra &longs;e eguali. </s>

<s id="id.2.1.1341.6.0">& percioche gli angoli po&longs;ti ad MI &longs;ono retti, <lb/>&longs;ar à KM vn parallelò grammo rettangolo, & KN &longs;arà eguale ad IM. </s>

<s id="id.2.1.1341.7.0">Per la<lb/>q<gap/> al co&longs;a KN &longs;arà eguale al Perimetro del cilindro AB. </s>

<s id="id.2.1.1341.8.0">Co&longs;i ponga&longs;i HI in <lb/>GC <gap/> HK in GE. </s>

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<s id="id.2.1.1345.1.0"><margin.target id="note313"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 4. <emph type="italics"/><gap/> questo.<emph.end type="italics"/></s>

<s id="id.2.1.1346.1.0"><emph type="italics"/>Ma in che maniera ciò &longs;i riduca alla bilancia è manife&longs;to per la nona dell ottauo libre <lb/>dell'i&longs;te&longs;&longs;o Pappo.<emph.end type="italics"/></s>

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<s id="id.2.1.1351.1.0"><emph type="italics"/>Pa&longs;&longs;i il &longs;ottoposto piano egualmente di&longs;tante dall'orizonte per la linea MN. ma per <lb/>KM pa&longs;si il piano inchinato à que&longs;to nel dato angolo KMN. & &longs;ia il pe&longs;o A <lb/>mo&longs;&longs;o dalla po&longs;&longs;anza C nel &longs;ottopo&longs;to piano. </s>

<s id="id.2.1.1351.2.0">& in vece di A intenda&longs;i vna s&longs;e-<emph.end type="italics"/><lb/>

<arrow.to.target n="fig133"></arrow.to.target><lb/><emph type="italics"/>ra egualmente graue intorno al centro E; laqual &longs;i collochi nel piano por MK, & <lb/>lo tocchi in L. </s><s id="id.2.1.1351.3.0">la linea dunque tirata EL è à piombo al piano, &longs;i come è &longs;tato di<lb/>mo&longs;trato nel quarto teorema de i Sferici. </s>

<figure id="fig133"></figure><lb/><emph type="italics"/>ra egualmente graue intorno al centro E; laqual &longs;i collochi nel piano por MK, & <lb/>lo tocchi in L. </s><s id="id.2.1.1351.3.0">la linea dunque tirata EL è à piombo al piano, &longs;i come è &longs;tato di<lb/>mo&longs;trato nel quarto teorema de i Sferici. </s>

<s id="id.2.1.1351.4.0">et però ella è perpendicolare alla linea KM. </s>

<s id="id.2.1.1351.5.0"><lb/>Tiri&longs;i EH equidi&longs;iante alla MN. & dal punto L &longs;i tiri ad EH la perpendico<lb/>lare LF. </s>

<s id="id.2.1.1351.6.0">Hor percioche l'angolo EHL è dato per e&longs;&longs;er eguale al dato angolo acu <lb/>to KMN; &longs;arà ancora l'angolo ELF dato, cioè eguale all'angolo EHL e&longs;&longs;en<emph.end type="italics"/>

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<s id="id.2.1.1351.8.0">per <lb/>laqual co&longs;a, & la proportion della restante FG ad EF &longs;arà data. </s>

<s id="id.2.1.1351.9.0">Faccia&longs;i come <lb/>GF ad FE, co&longs;i il pe&longs;o A al pe&longs;o B; & la po&longs;&longs;anza C alla po&longs;&longs;anza D. </s>

<s id="id.2.1.1351.10.0">Ma <lb/>la po&longs;&longs;anza del pe&longs;o A è C; adunque la po&longs;&longs;anza del pe&longs;o B nel mede&longs;imo piano <lb/>&longs;arà D. </s><s id="id.2.1.1351.11.0">& perche co&longs;i è laretta linea GF ad FE, come il pe&longs;o A al pe&longs;o B:<emph.end type="italics"/><lb/>

<arrow.to.target n="fig134"></arrow.to.target><lb/><emph type="italics"/>&longs;e li pe&longs;i AB &longs;aranno po&longs;ti ne i centri EG appiccati nel punto F, pe&longs;eranno egual <lb/>mente; come &longs;ostentati dalla ba&longs;e LF, laquale è à piombo all'orizonte. </s>

<figure id="fig134"></figure><lb/><emph type="italics"/>&longs;e li pe&longs;i AB &longs;aranno po&longs;ti ne i centri EG appiccati nel punto F, pe&longs;eranno egual <lb/>mente; come &longs;ostentati dalla ba&longs;e LF, laquale è à piombo all'orizonte. </s>

<s id="id.2.1.1351.12.0">Ma è po <lb/>&longs;to il pe&longs;o A intorno al centro E. percioche in &longs;uo luogo è la s&longs;era. </s>

<s id="id.2.1.1351.13.0">dunque il pe<lb/>&longs;o B posto intorn'al G, pe&longs;erà egualmente; di modo che la s&longs;era per la inclinatio<lb/>ne del piano non de&longs;cender à al ba&longs;&longs;o; ma&longs;tarà &longs;erma, come &longs;e ella fo&longs;&longs;e nel &longs;ottopo<lb/>&longs;to piano. </s>

<s id="id.2.1.1351.14.0">& perche nel &longs;ottopo&longs;to piano ella &longs;arebbe mo&longs;&longs;a dalla po&longs;&longs;anza C; adun<lb/>que nel piano inclinato &longs;arà mo&longs;&longs;a dall'vna el'altra, cioè dalla po&longs;&longs;anza C, & dal <lb/>la po&longs;&longs;anza del pe&longs;o B, cioè dalla po&longs;&longs;anza D. & la po&longs;&longs;anza D è data.<emph.end type="italics"/></s>

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<s id="id.2.1.1358.1.0"><emph type="italics"/>Compia&longs;i il cuneo AD CHI, cioè de&longs;criua&longs;i la helice CHI eguale à CDA, & &longs;ia <lb/>la cima del cuneo C. &longs;imilmente compia&longs;i il cuneo GFEKL, la cui cima &longs;ia E. </s><s id="id.2.1.1358.2.0">pon<emph.end type="italics"/><lb/>

<arrow.to.target n="fig135"></arrow.to.target><lb/><emph type="italics"/>ga&longs;i dapoi la linea retta MN, laquale &longs;ia eguale ad AC, à piombo dellaquale &longs;ia<emph.end type="italics"/><lb/>

<figure id="fig135"></figure><lb/><emph type="italics"/>ga&longs;i dapoi la linea retta MN, laquale &longs;ia eguale ad AC, à piombo dellaquale &longs;ia<emph.end type="italics"/><lb/>

<arrow.to.target n="note314"></arrow.to.target> <emph type="italics"/>tirata la liuea NP, che &longs;ia eguale al Perimetro del cilindro AB: & congiun<lb/>ga&longs;i PM; &longs;arà PM perle co&longs;e dette, eguale ad e&longs;&longs;a CDA. </s>

<s id="id.2.1.1358.3.0">Allunghi&longs;i po&longs;cia<emph.end type="italics"/><lb/>

<arrow.to.target n="note315"></arrow.to.target> <emph type="italics"/>MN in O, et &longs;accia&longs;i ON eguale ad MN, et congiunga&longs;i OP; &longs;arà il cuneo <lb/>OPM eguale al cuneo ADCHI. & &longs;imilmente faccia&longs;i il cuneo STQ eguale<emph.end type="italics"/>

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<s id="id.2.1.1362.1.0"><margin.target id="note316"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 4. <emph type="italics"/>del primo.<emph.end type="italics"/></s>

<s id="id.2.1.1363.1.0">Altramente.</s>

<s id="id.2.1.1364.1.0"><emph type="italics"/>Sia data la vite AB, che habbia due helici eguali CDEFG; &longs;ia dapoi vn'altro ci<lb/>lindro <foreign lang="greek">a b</foreign> eguale ad e&longs;&longs;o AB, nel quale prenda&longs;i OP eguale à CG; & diuida&longs;i <lb/>OP in tre parti eguali OR RT TP; & de&longs;criuan&longs;i tre helici OQ RS TV P; <lb/>&longs;arà cia&longs;cuna delle OR RT TP minore di CE, & di EG; percioche la terza<emph.end type="italics"/><lb/>

<arrow.to.target n="fig136"></arrow.to.target><lb/><emph type="italics"/>parte è minore della metà. </s>

<figure id="fig136"></figure><lb/><emph type="italics"/>parte è minore della metà. </s>

<s id="id.2.1.1364.2.0">dico, che il pe&longs;o mede&longs;imo &longs;i mouerà più facilmente &longs;o<lb/>prale helici OQRS TVP, che &longs;opra CDEFG. </s><s id="id.2.1.1364.3.0">faccia&longs;i HIL triangolo di an <lb/>goli retti, in modo che HI &longs;ia eguale à CG, & IL &longs;ia eguale al doppio del Peri<lb/>metro del cilindro AB, & per LI &longs;i intenda vn piano egualmente di&longs;tante dall'o-<emph.end type="italics"/><lb/>

<arrow.to.target n="note317"></arrow.to.target> <emph type="italics"/>rizonte; &longs;arà HL eguale à CDEFG, & HLI &longs;arà l'angolo della inclinatione. </s>

<s id="id.2.1.1364.4.0"><lb/>faccia&longs;i &longs;imilmente il triangolo X<foreign lang="greek">*u</foreign>Z di angoli retti, in modo che XZ &longs;ia egualc<emph.end type="italics"/>

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<s id="id.2.1.1367.1.0"><margin.target id="note318"></margin.target><emph type="italics"/>Per la<emph.end type="italics"/> 21. <emph type="italics"/>del prime.<emph.end type="italics"/></s>

<s id="id.2.1.1368.1.0"><emph type="italics"/>Che &longs;e OP diuidera&longs;&longs;i in quattro parti eguali, & &longs;i de&longs;criueranno d'intorno <foreign lang="greek">a b</foreign> quat<lb/>tro helici, &longs;i mouerà anco più facilmente il pe&longs;o &longs;opra queste quattro, che &longs;opra le <lb/>tre OQRS TVP, & quanto più helici &longs;aranno, tanto più facilmente &longs;i mouerà <lb/>il pe&longs;o. </s>

<s id="id.2.1.1368.2.0">ilche bi&longs;ognaua mostrare.<emph.end type="italics"/></s>

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<s id="id.2.1.1374.1.0"><emph type="italics"/>Sia la vite che habbia le helici ABCD, & habbia anche le &longs;tanghe EF GH po&longs;te <lb/>ne'buchi della vite. </s>

<s id="id.2.1.1374.2.0">&lo