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version 1.4, 2003/11/20 19:14:56

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</front>

<body>

<chap>

<pb/>

<s id="id.0.0.0.0.1.0">Guido Ubaldo Marquis del Monte</s>

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<s id="id.0.0.1.0.2.0">To Francisco Maria II Illustrious Duke of Urbino</s>

<pb/>

<s id="id.0.0.1.1.1.0">There are two qualities, Illustrious Prince, that are usually very effective in adding to men's power, namely, utility and nobility.</s>

<s id="id.0.0.1.1.2.0">It seems to me that these join in making the subject of mechanics attractive and in rendering it desirable in comparison with all others.</s>

<s id="id.0.0.1.1.3.0">For if we measure the nobility of something by its origin (as most people now do), the origin of mechanics is, on one side, geometry and, on the other side, physics.</s>

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<s id="id.0.0.1.6.2.0">Hence, too, by means of two forces pulling in opposite directions we have divided stout trees and great masses of marble.</s>

<s id="id.0.0.1.6.3.0">Hence also in war, in the building of ramparts, in fighting at close quarters, in attacking and defending places, there are almost infinite uses [of mechanics].</s>

<pb/>

<s id="id.0.0.1.7.1.0">With the help of mechanics, too, those who work with wood, stone and marble, wines, oils and unguents, iron, gold, and other metals, as well as surgeons, barbers, bakers, tailors, and all workers in the useful arts, make many important contributions to human life.</s>

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<s id="id.0.0.1.12.3.0">And, lo, just as he had been suddenly thrust from the darkness and prison of the body (as we believe) into the light and liberty of mathematics, so at the most inopportune time he left mathematics bereft of its fine and noble father and left us so prostrate that we scarcely seem able even by a long discourse to console ourselves for his loss.</s>

<s id="id.0.0.1.12.4.0">And yet in his endless concern with the elucidation of other parts of mathematics, he either left mechanics completely untreated or touched on it just casually.</s>

<pb/>

<s id="id.0.0.1.13.1.0">Therefore I began to devote myself more eagerly to this study, and, in making my way through every branch of mathematics, I never lost sight of my course to find whatever could be appropriated and derived from each of these branches, so that I may be better equipped to perfect and embellish mechanics.</s>

<s id="id.0.0.1.13.2.0">But now I think that, while I have not completed the treatment of everything that pertains to mechanics, still I have advanced to such a point that I can bring some help to those who learned from Pappus, Vitruvius, and others, what a lever is, a pulley, wheel and axle, wedge, and screw, and how they should be arranged so that weights may be moved, and the many properties present in those machines by virtue of the lever, properties connecting force and weight, which they are eager to learn about.</s>

<s id="id.0.0.1.13.3.0">For that reason I thought it was the proper time for me to emerge and give some example of the work I did on this subject.</s>

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<s id="id.0.0.2.1.1.0">My Revered Lord:-Inasmuch as the science of mechanics is highly useful to many and important actions in our lives, there is good reason that philosophers and ancient kings gave it more than a little study and that princes favored excellent engineers and enriched them.</s>

<s id="id.0.0.2.1.2.0">Certainly this science is of the highest theoretical value and of subtlest structure, for it deals with that part of philosophy which treats of the elements in general, and of the motion and rest of bodies according to their positions; thus we assign the cause of their natural movements, and thus by machines we force bodies to leave their natural places, carrying them upward and in every direction, contrary to their nature.</s>

<pb/>

<s id="id.0.0.2.2.1.0">Both these goals are carried out by propositions arising from matter itself and artificial structures and instruments.</s>

<s id="id.0.0.2.2.2.0">And thus it is necessary to consider this subject in two manners: one that regards theory and the application of reason to things that must be done, making use of arithmetic and geometry, astrology, and natural philosophy; the other that is carried out in practice and requiring activity and manual labor, utilizing architecture, painting, design, the arts of the builder and carpenter and mason and related crafts, and in such a way that these things become intertwined and are a mixture of natural philosophy, mathematics, and the practical arts.</s>

<s id="id.0.0.2.2.3.0">So that whoever is instructed by clever men and has learned from childhood the previously mentioned sciences and can also design and work with his hands may become a skilled mechanic, inventor, and maker of marvelous works.</s>

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<s id="id.0.0.2.7.7.0">Marcellus, seeing that nothing was to be gained by his attempted assaults and that his men were being exposed to danger simply by the existence of that valorous old man, came to share the opinion of his whole army that the defense of Syracuse was governed by divine power.</s>

<s id="id.0.0.2.7.8.0">Hence he changed the course of his warfare turning it into a siege and strictly preventing any foodstuffs from entering the city.</s>

<pb/>

<s id="id.0.0.2.8.1.0">These, then, were the reasons for which mechanics rose to such glory that the Romans later honored it in their armies.</s>

<s id="id.0.0.2.8.2.0">Caesar took prisoner the chief of the smiths of Pompeii, called Magio Cremona; Vitruvius was made captain of catapults by Augustus Caesar, which would be equivalent to captain of artillery in our armies.</s>

<s id="id.0.0.2.8.3.0">This glory was afterward maintained by many eminent writers and masters of mechanics, such as Ctesibius of Alexandria, Hero of Alexandria, another Hero Athenaeus, Bion, and Pappus of Alexandria (who cites Carpus of Antioch), and by Heliodorus, Oribasius, and other Greeks who flourished at various periods.</s>

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<s id="id.0.0.2.13.1.0">At the age of sixteen, you were sent with twelve horsemen, mostly Turkish, and with sufficient funds to serve in the entire war in Italy from the capture of Francis I of France to the general peace which ensued in the year 1529.</s>

<s id="id.0.0.2.13.2.0">This war involved practically every known military movement, through the large armies which confronted one another, the quality and number of undertakings, and a thousand other important events and stratagems that took place; and above all because in one field or another and at all seasons the foremost soldiers of the world were fighting in great numbers with prudence, astuteness, and bravery, for the honor of conquering and being the victors ---</s>

<pb/>

<s id="id.0.0.2.14.1.0">The Christian princes having returned in peace, you dedicated yourself to the service of your Serene Lords, where in the most important and greatest charges and in two wars you have added fifty years of splendid service to the two hundred of your Savorgnan predecessors; and during this time you have made some fifty great catapults in different provinces of their states, well thought out and masterfully made, with great economy of public funds.</s>

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<s id="id.0.0.3.0.1.0">Preface of Filippo Pigafetta</s>

<pb/>

<s id="id.0.0.3.1.1.0">To the Reader:-The present book contains six treatises, the first on the balance and steelyard, the second on the lever, the third on pulleys, the fourth on the windlass, the fifth on the wedge, and the last on the screw, all which are mechanical instruments.</s>

<s id="id.0.0.3.1.2.0">It is entitled Mechanics.</s>

<s id="id.0.0.3.1.3.0">But this word "mechanics" is perhaps not understood by everyone in its true sense, and some are even found who consider it an insulting word, for in many parts of Italy a man is called a mechanic in scorn and degradation, and in some places people are offended to be called even "engineer".</s>

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<s id="id.0.0.3.6.2.0">For these are basic and fundamental, and there may be compounded in various ways combinations of two, three, or more; thus the windlass may be combined with the pulley, the screw with the windlass or the lever, and so on.</s>

<s id="id.0.0.3.6.3.0">This may be done at will by anyone who can proceed with good judgment in various works, as the author notes at the end of this volume.</s>

<pb/>

<s id="id.0.0.3.7.1.0">Now although the author has reasoned of these machines in good method and admirable order, and nothing inherently obscure must be mastered, yet it requires a man's whole mind, and the demonstrations should be read attentively more than once with concentration.</s>

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<s id="id.1.0.1.0.1.0">DEFINITIONS</s>

<pb/>

<s id="id.1.0.1.1.1.0">The center of gravity of any body is a certain point within it, from which, if it is imagined to be suspended and carried, it remains stable and maintains the position which it had at the beginning, and is not set to rotating by that motion.</s>

<s id="id.1.0.1.1.2.0">This definition of the center of gravity is taught by Pappus of Alexandria in the eighth book of his Collections.</s>

<s id="id.1.0.1.1.3.0">But Federico Commandino in his book On Centers of Gravity of Solid Bodies explains this center as follows:</s>

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<s id="id.1.0.3.1.2.0">2. The center of gravity of any body is always in the same place with respect to that body.</s>

<s id="id.1.0.3.1.3.0">3. A heavy body descends according to its center of gravity.</s>

<pb/>

<s id="id.1.1.0.0.1.0">On the Balance</s>

</figure>

<s id="id.1.1.0.1.1.0">Before a discussion of the [actual] balance, to make matters clear , let the straight line AB be a balance, with its support [trutina] CD, which in accordance with common practice is kept always perpendicular to the horizon.</s>

<s id="id.1.1.0.1.2.0">The stationary point C, about which the balance turns, is (though improperly) called the center of the balance, even if it is above or below the balance, and CA and CB [in the first diagram, or their equivalents in the others] are called the arms or distances of the balance.</s>

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<s id="id.1.1.1.0.1.0">LEMMA</s>

</figure>

<s id="id.1.1.1.1.1.0">Let the line AB be perpendicular to the horizon , and describe the circle AEBD having the diameter AB and the center C.</s>

<s id="id.1.1.1.1.2.0">I say that the point B is the lowest place on the circumference AEBD, and the point A is the highest; and that any other points such as D and E, which are equidistant from A, are situated equally below it; and that those points which are closer to A are higher than those which are more distant from A.</s>

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<s id="id.1.1.2.0.1.0">PROPOSITION I</s>

</figure>

<s id="id.1.1.2.1.1.0">If the weight is supported at its center of gravity by a straight line, it will remain stationary only if that straight line is perpendicular to the horizon.</s>

<pb/>

<s id="id.1.1.2.2.1.0">[The proof, omitted here, rests on Postulate 3.]</s>

<s id="id.1.1.2.2.2.0">From this it may be deduced that a weight supported at any point in any manner will never remain at rest unless the line drawn from the center of gravity to the point of support is perpendicular to the horizon.</s>

</figure>

<s id="id.1.1.2.3.1.0">Thus, let the weight be supported by the lines CG and CH.</s>

<s id="id.1.1.2.3.2.0">I say that the line BC being perpendicular to the horizon, the weight will remain at rest; but the line CF being not perpendicular to the horizon, the point F will move downward to D, where it will rest, and the line CD will be perpendicular to the horizon.</s>

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<s id="id.1.1.3.0.1.0">PROPOSITION II</s>

</figure>

<s id="id.1.1.3.1.1.0">A balance parallel to the horizon, with its center above [and] having equal weights at its extremities which are equidistant from the perpendicular [CD], when moved from this position and released, will return and rest in it.</s>

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<s id="id.1.1.4.0.1.0">PROPOSITION III</s>

</figure>

<pb/>

<s id="id.1.1.4.1.1.0"> A balance parallel to the horizon, with its center below [and] having equal weights at its extremities which are equidistant from the perpendicular [CD], will be at rest- but if moved and left tilted, it will move toward the lower side.</s>

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<s id="id.1.1.5.1.1.0">A balance parallel to the horizon, having its center within the balance and with equal weights at its extremities, equally distant from the center of the balance, will remain stable in any position to which it is moved.</s>

</figure>

<s id="id.1.1.5.2.1.0">Let the balance be the straight line AB, parallel to the horizon, with its center C in the line AB and the distance CA equal to the distance CB; and let the weights A and B be equal, and have their centers of gravity in the points A and B.</s>

<s id="id.1.1.5.2.2.0">Let the balance be moved to DE and left there.</s>

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<s id="id.1.1.5.3.2.0">Likewise if the weights are attached to the ends of the balance, in the usual manner, it will be the same, provided that the lines drawn from where the weights are attached toward the center of heavy things (the balance being moved in any manner) go to meet in the center of the world, since, when the weights are attached in this manner, they bear down as if they had their centers of gravity in those same points.</s>

<s id="id.1.1.5.3.3.0">Whence we may consider the results in just the same way.</s>

<pb/>

<s id="id.1.1.5.4.1.0">But with regard to this last conclusion, many things are said by men who believe otherwise.</s>

<s id="id.1.1.5.4.2.0">Hence it will be well to dwell further on this; and according to my ability I shall endeavor to defend not only my own opinion but Archimedes too, who seems to have been of the same opinion.</s>

</figure>

<s id="id.1.1.5.5.1.0">Things being as before, let there be drawn the line FG plumb to AB and to the horizon; and with the center C at the distance CA describe the circle ADFBEC.</s>

<s id="id.1.1.5.5.2.0">The points ABDE will be on the circumference, because the arms of the balance are equal.</s>

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<s id="id.1.1.5.7.2.0">For, the excess of weight D over weight E having some ratio and quantitative part, we imagined it to be not only minimal but also capable of infinite division.</s>

<s id="id.1.1.5.7.3.0">They seek in the following manner to prove that no such weight can be found, since it is not just minimal, but still less.</s>

</figure>

<s id="id.1.1.5.8.1.0">Things being taken as before, and from the points D and E the lines DH and EK being drawn perpendicular to the horizon, let there be taken another equal circle LDM, with center N, which is tangent to the circle FDG at the point D.</s>

<s id="id.1.1.5.8.2.0">NC will be a straight line, and, since the angle KEC is equal to the angle HDN, and the angle CEG is likewise equal to the angle NDM, being contained within equal radii and circumferences , the remaining mixed angle KEG equals that of HDM.</s>

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<s id="id.1.1.5.9.7.0">Hence the angle RDG will be less than the angle ODG, and likewise the angle RDH less than the angle ODH --- And thus, if infinitely many circumferences are drawn between DO and DG, we shall find the ratio diminishing ad infinitum and it follows thus that the ratio of the weight placed at D to that at E is not so small that one infinitely less cannot be found.</s>

<s id="id.1.1.5.9.8.0">And since the angle MDG can be divided in infinitum so also one may divide in infinitum the excess of weight which D has over E.</s>

<pb/>

<s id="id.1.1.5.10.1.0">Nor should it be omitted that they have assumed in their proof as a thing known that the angle KEG is greater than the angle HDG, which indeed is true if DH and EK are parallel.</s>

<s id="id.1.1.5.10.2.0">But since, as they likewise assume, the lines DH and EK meet at the center of the world, they are not ever parallel, and not only will the angle KEG not be greater than the angle HDG, but it will be smaller.</s>

</figure>

<s id="id.1.1.5.11.1.0">For example, draw the line FG to the center of the world S, and join DS and ES.</s>

<s id="id.1.1.5.11.2.0">The angle SEG is to be demonstrated less than the angle SDG.</s>

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<s id="id.1.1.5.11.6.0">Take away from the angle SDM the curvilinear angle MDG, and from the angle VEG the angle VES; and the angle VES formed by straight lines is greater than the angle MDG formed by curved lines, so the remaining angle SEG is less than SDG.</s>

<s id="id.1.1.5.11.7.0">Hence by their own suppositions not only will the weight placed at D fail to be heavier than that at E, but on the contrary the weight at E will be heavier than that at D.</s>

</figure>

<s id="id.1.1.5.12.1.0">Nevertheless, they adduce reasons by which they attempt to show that the balance DE necessarily returns to AB, parallel to the horizon.</s>

<s id="id.1.1.5.12.2.0">First they show that a given weight is heavier at A than at any other place, and this position they call the "level position," the line AB being parallel to the horizon.</s>

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<s id="id.1.1.5.15.1.0">And by the above arguments they attempt to show that the balance DE returns to AB, which arguments seem to me to be easily answered as follows.</s>

<pb/>

<s id="id.1.1.5.16.1.0">First, when they say that the weight placed at A is heavier than in any other position, they deduce this from its varying distances from the line FG, the swiftest and straightest movement being from point A.</s>

<s id="id.1.1.5.16.2.0">To begin with, they do not truly demonstrate that the weight moves more swiftly from A than from any other place, nor does it follow that, since CA is greater than DO, and DO than LP, the weight placed at A is heavier than that at D, and at D than at L.</s>

<s id="id.1.1.5.16.3.0">Now the intellect is not satisfied unless this can be demonstrated from some other cause, for this appears to be merely a sign rather than a cause.</s>

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<s id="id.1.1.5.16.5.0">Besides, all the things adduced from swifter and slower movement to persuade us that the body at A is heavier than that at D do not show that the weight at A, by its being at A, is heavier than the weight at D, by its being at D, but only by their departing from the points D and A.</s>

<s id="id.1.1.5.16.6.0">So, before going further, I shall first show that the closer a weight gets to the line FG, the less it weighs, both as to its position and as to its departure therefrom; and at the same time I shall show it to be false that the weight is heaviest at A of all places.</s>

</figure>

<s id="id.1.1.5.17.1.0">Draw FG to the center of the world, S, and from S draw also a line tangent to the circle AFBG.</s>

<s id="id.1.1.5.17.2.0">This line cannot be drawn from S to touch the circle at A, inasmuch as, if the line AS were drawn, the triangle ACS would have two right angles, that is, SAC and ACS, which is impossible.</s>

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<s id="id.1.1.5.17.39.0">And the line CO will be parallel to the horizon, though not to the horizon of the point C (as they believe), but rather to that of the weight placed at O; for the horizontal must be taken from the center of gravity of the body.</s>

<s id="id.1.1.5.17.40.0">All of which was to be shown.</s>

</figure>

<s id="id.1.1.5.18.1.0">But if the balance arm were greater than CO, say, by the amount CD, the weight placed at O would likewise be heavier.</s>

<s id="id.1.1.5.18.2.0">Describe the circle OH, with center D and radius DO.</s>

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<s id="id.1.1.5.18.5.0">Hence the weight at O will be freer, and consequently heavier, than at C (the center of the balance being at D).</s>

<s id="id.1.1.5.18.6.0">Similarly it will be shown that the longer the arm DO, the heavier will be the weight placed at O.</s>

</figure>

<s id="id.1.1.5.19.1.0">But if the same circle AFBG with its center R shall be closer to the center of the world S, and if a line ST is drawn from the point S tangent to the circle, the point T (where the weight is heaviest) will be farther from the point A than is the point O.</s>

<s id="id.1.1.5.19.2.0">Draw the lines OM and TN from the points O and T, plumb to CS, and add RT, the center R being in the line CS, and the line ARB being parallel to ACB.</s>

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<s id="id.1.1.5.19.11.0">Likewise it may be shown that, the closer the circle is to the center of the world, the farther T will be from A.</s>

<s id="id.1.1.5.19.12.0">Hence, as before, it may be shown that the weight on the circumference TAF will stand upon the center R, while on the circumference TG it will be held by the line, and it will be found heaviest at the point T.</s>

</figure>

<s id="id.1.1.5.20.1.0">And if the point G were the center of the world, then the closer the weight was to G, the heavier it would be; and hence wherever else the weight is placed than at G, it will always get support from the center C; for example , at K.</s>

<s id="id.1.1.5.20.2.0">Draw the line GK, along which the natural motion of the weight would be made; this will make an acute angle with the arm of the balance KC, because the base angles (at K and G) of the isosceles triangle CKG are always acute.</s>

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<s id="id.1.1.5.21.8.0">Therefore the weight placed at K will be heavier than at D.</s>

<s id="id.1.1.5.21.9.0">Similarly it would be shown that the closer the weight was to F (as at L) the less it would weigh, but the closer it is to G (as at H) the heavier it is.</s>

</figure>

<pb/>

<s id="id.1.1.5.22.1.0">And if the center of the world were at S, between the points C and G, first it will be shown in the same way that the weight, wherever it is (as at H), gets support from the center C.</s>

<s id="id.1.1.5.22.2.0">For the lines HG and HS being drawn, the angle at the base GHC of the isosceles triangle CHG is always acute; whereby also SHC, being less than this, is also acute.</s>

<s id="id.1.1.5.22.3.0">But drawing from the point S the line SK plumb to CS, I say that the weight is heavier at K than at any other place in the circumference FKG, and the closer it shall be to F, or to G, the less it will weigh.</s>

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<s id="id.1.1.5.24.2.0">If we next consider the weight of such an arm, we can find the center of gravity of the magnitude made by the weight and the arm; and circumferences can be described according to the distance from the center of the balance to this center of gravity, as if this contained the weight (which indeed it does).</s>

<s id="id.1.1.5.24.3.0">And the things we have found without considering the weight of the arm of the balance can be found in just the same way by considering this weight also.</s>

</figure>

<s id="id.1.1.5.25.1.0">From the things said, if we consider the balance to be removed from the center of the world as these other men have done (and as it is in fact), then it is clearly false for them to say that the weight is heavier at A than at any other place.</s>

<s id="id.1.1.5.25.2.0">And it is also false that the farther the weight is from the line FG, the heavier it is; for the point O is closer to FG than the point A, the line drawn plumb from O to FC being less than CA.</s>

<s id="id.1.1.5.25.3.0">It is likewise false that the weight moves more swiftly from the point A than from any other place, for it will move more swiftly from the point O than from A, since at O it is more free than at any other place, and its descent from O will be closer to its straight natural movement than any other descent.</s>

</figure>

<s id="id.1.1.5.26.1.0">Besides this, when they argue by means of the straighter or more curved descent that the weight is heavier at A than at D, and at D than at L, they are certainly wrong; for if any weight were placed at any point on the circumference, as at D, its true descent would be made along the straight line DR parallel to FC, according to its natural movement, as was first said.</s>

<s id="id.1.1.5.26.2.0">For if a weight is placed anywhere, and we regard its natural movement to that proper place to which it moves straight by nature, taking into account the shape of the whole universe, then the space through which it moves naturally will always be along the line drawn from the circumference to the center.</s>

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<s id="id.1.1.5.26.8.0">We may even concede that the weight will be heavier at A than [it will] anywhere else, and that the straight descent of the weight must be along a straight line parallel to FG, and that any points taken in straight lines parallel to the horizon are equally distant from the center; but it will not follow from this that their demonstration is true when they say that the weight is heavier at A than elsewhere, say, at L.</s>

<s id="id.1.1.5.26.9.0">For if it were true that the straighter a weight descended in this sense, the heavier it would be, then it would also follow that where the same weight would descend along equal arcs partaking equally in the straight, it would have equal weights; but this may be shown to be false in the following manner.</s>

</figure>

<s id="id.1.1.5.27.1.0">Let there be the equal arcs AL and AM, and join L and M, cutting AB at X; let LM be parallel to FG and perpendicular to AB, and XM will be equal to XL.</s>

<s id="id.1.1.5.27.2.0">If therefore the weight shall move from L to A along the circumference LA, its straight movement will be measured by the line LX.</s>

Line 672

<s id="id.1.1.5.27.4.0">Hence the descent from L to A will be equal to that from A to M, by reason of the equality of arcs as well as the equal straight lines perpendicular to AB.</s>

<s id="id.1.1.5.27.5.0">Therefore the weight at L will weigh the same as at A, which is false; for it is far heavier at A than at L.</s>

<pb/>

<s id="id.1.1.5.28.1.0">And although AM and LA partake equally of the straight according to these men, perhaps they would say that because the first part of the descent from L, say LD, partakes less of the straight than the first part of the descent from A, that is, AN, the weight will be heavier at A than at L.</s>

<s id="id.1.1.5.28.2.0">For the arc AN being equal to LD, as was assumed, it partakes (according to them) of the straight CT, but LD partakes of the straight PO; hence the weight will be heavier at A than at L.</s>

<s id="id.1.1.5.28.3.0">But if this were true, it would follow that the same weight at the same place, merely considered in a different way with respect to that place, would be heavier and lighter, which is impossible.</s>

Line 686

<s id="id.1.1.5.28.11.0">Moreover, their assumption does not affirm that the positional weight will be greater when at the same place the commencement of the descent is less oblique.</s>

<s id="id.1.1.5.28.12.0">Hence the postulate [they] adopted above, that is, that the weight is positionally heavier according as the descent from the same place is less oblique, is not to be conceded at all, for the reasons we have given; and not only that, but it is not difficult to show the exact opposite; that is, that the less oblique the descent of the same weight along equal arcs, the less it weighs.</s>

</figure>

<s id="id.1.1.5.29.1.0">Let there be as before the equal arcs AL and AM, and the point L close to F, and join L and M perpendicular to AB, and LX will also I be equal to XM.</s>

<s id="id.1.1.5.29.2.0">Then take the point P close to M, between M and G, and let the arc PO be equal to the arc AM; the point O will then be close to A.</s>

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<s id="id.1.1.5.30.1.0">Besides this, even if we concede their assumption, they are far from the true theory of the balance when they argue from it that the balance DE must return to AB; for they always take one weight separately at D, or E, as if now one and now the other were placed in the balance, but never both of them together.</s>

</figure>

<s id="id.1.1.5.31.1.0">Indeed we must do quite the opposite; nor may we consider directly one weight without the other when we reason about them as placed in the balance.</s>

<s id="id.1.1.5.31.2.0">For when they see that the descent of the weight placed at D is less bent than that of the weight placed at E, the weight at D by their assumption must be heavier than the weight placed at E; and by being heavier (they believe) it necessarily moves downward and the balance DE returns to AB.</s>

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<s id="id.1.1.5.31.11.0">So that neither of the two will weigh more than the other; there being no action that proceeds from equality, the weight placed at D will not move the weight placed at E upward, because, if it did, it would be necessary that the weight placed at D should have stronger force in descending than should the weight placed at E in rising.</s>

<s id="id.1.1.5.31.12.0">But these things are equal; therefore the weights will remain at rest and the weighing down of the weight placed at D will be equal to the weighing down of the weight placed at E.</s>

</figure>

<s id="id.1.1.5.32.1.0">Moreover, they assume that the farther the weight is distant from the line of direction FC, the heavier it is.</s>

<s id="id.1.1.5.32.2.0">Now draw from the points D and E the lines DO and El perpendicular to FC; and, as before, the triangle CDO will be demonstrated to be equal to the triangle CEI, and the line DO to be equal to the line El.</s>

Line 750

<s id="id.1.1.5.33.3.0">Thus the descent is said to be more oblique, the more it departs from that space, and straighter the more it approaches it.</s>

<s id="id.1.1.5.33.4.0">Now in this sense the assumption need not give rise to difficulty on the part of anyone, because this is so clear in its truth and its agreement with reason that it does not appear to need to be made evident in any way.</s>

<pb/>

<s id="id.1.1.5.34.1.0">Therefore if the free weight located at D must move to its proper place, and if S is taken as the center of the world, it will doubtless move along the line DS; similarly the free weight placed at E will move along the line ES.</s>

<s id="id.1.1.5.34.2.0">Now if (as is indeed the case) the descent of the weight is to be called more or less oblique according to its departure from or approach to the routes designated by the lines DS and ES, it is clear that, with regard to their natural movements toward their proper places, the descent of E along EC is less oblique than that of D along DA, it having been demonstrated above that the angle SEC is less than the angle SDA.</s>

<s id="id.1.1.5.34.3.0">Whence the weight at E will weigh more than at D, which is completely contrary to that which they have made such an effort to prove.</s>

</figure>

<s id="id.1.1.5.35.1.0">Now they may rise up against us, arguing as follows: If the weight placed at E is heavier than the weight placed at D, the balance DE will never remain in that position, as we have undertaken to maintain, but it will move to FC.</s>

<s id="id.1.1.5.35.2.0">To which we reply that it makes a great deal of difference whether we consider the weights separately, one at a time, or as joined together; for the theory of the weight placed at E when it is not connected with another weight placed at D is one thing, and it is quite another when the weights are joined in such a way that one cannot move without the other.</s>

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<s id="id.1.1.5.35.21.0">Thus the descent of the weight at D will be equal to the rise of the weight at E, and the weight at D will not raise the weight at E.</s>

<s id="id.1.1.5.35.22.0">From which it follows that the weights at D and E, considered in conjunction, are equally heavy.</s>

</figure>

<s id="id.1.1.5.36.1.0">Now the second reason with which they attempt to show that the balance DE returns to AB is that, when the support of the balance is CF, its goal [meta] is CG, and, since the angle DCG is greater than the angle ECG, the weight placed at D will be heavier than that placed at E; therefore the balance DE will return to AB.</s>

<s id="id.1.1.5.36.2.0">In my opinion this does not follow, and this fiction about the support and the goal should just be left out and passed over in silence; for to say anything about it only confuses the issue, the whole thing being arbitrary, since no necessary reason why the weight placed at D at the larger angle will be heavier, or why the greater angle is the cause of greater weight, is given anywhere.</s>

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<s id="id.1.1.5.36.6.0">For what can it matter whether the support is CF or CG, when the balance DE is always sustained at the same point C?</s>

<s id="id.1.1.5.36.7.0">But let us make their delusion still more obvious.</s>

</figure>

<s id="id.1.1.5.37.1.0">Let there be the balance AB with center C and support FG which remains motionless and sustains the balance AB at the point C.</s>

<s id="id.1.1.5.37.2.0">Now let the balance move to DE; and since the support is both above and below the balance, what angle will be the cause of heaviness, the balance DE being sustained always at the same point?</s>

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<s id="id.1.1.5.37.13.0">But neither Aristotle nor experience favors this opinion of theirs, and indeed quite the contrary is true.</s>

<s id="id.1.1.5.37.14.0">They are deceived with regard to experience, since it is clear from experience that this happens when the center of the balance is above or below the balance, rather than when the support is above or below.</s>

</figure>

<s id="id.1.1.5.38.1.0">If the balance AB has its center C above and its support CD below the balance, then when the balance is moved to EF it will return to AB parallel to the horizon.</s>

<s id="id.1.1.5.38.2.0">Likewise if the balance has its center C under the balance while its support CD is above, and the balance is moved to EF, it is manifest that the balance will move down on the side of F, the support being above the balance.</s>

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<s id="id.1.1.5.39.4.0">For, although the center of the balance is always a single point, nevertheless, when it is on top of the balance, it does not matter much if the balance is not sustained precisely at that point; because so long as the center remains above, the balance will always behave the same.</s>

<s id="id.1.1.5.39.5.0">For a like reason, that which occurs when the center is in the balance never happens when it is below the balance, because there will be a difference if it is not sustained always exactly in that center, and it is a very easy thing for that center to change its position when the balance is moved.</s>

<pb/>

<s id="id.1.1.5.40.1.0">Now, it is certainly true that Aristotle did pose two questions only; that is, why, when the support is above, and the balance is not parallel to the horizon in equilibrium, it returns; whereas, if the support is below, it does not return, and moves farther in the direction of the lower side.</s>

<s id="id.1.1.5.40.2.0">But his proofs are not based on the larger or smaller angle and the position of the support, as they pretend, because in this they do not understand the philosopher's meaning when he examines the reason for various effects.</s>

<s id="id.1.1.5.40.3.0">And Aristotle is far from attributing these different effects to the angles; rather he says that the cause is the excess, and that [when the support is] above, more of the distance from the perpendicular along one arm of the balance is now on one side and now on the other.</s>

</figure>

<s id="id.1.1.5.41.1.0">Now with the support above at CF, the perpendicular will be FCG, which, according to Aristotle, always points toward the center of the world and which unequally divides the [actual] balance when it is moved to DE, the larger part being on the side of D, which tends downward.</s>

<s id="id.1.1.5.41.2.0">Therefore on the side of D the balance will move downward until it returns to AB.</s>

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<s id="id.1.1.5.41.8.0">The contrary happens when the center is below the balance.</s>

<s id="id.1.1.5.41.9.0">These things are demonstrated in the following manner, what has been said above being assumed: that is, that the weight will be heavier in that place from which its descent is straighter, and is likewise heavier at the place from which its rise would be straighter.</s>

</figure>

<s id="id.1.1.5.42.1.0">Let the balance AB be parallel to the horizon, with its center C above the balance, and let fall the perpendicular CD.</s>

<s id="id.1.1.5.42.2.0">Let the centers of gravity of two equal weights be placed at A and B.</s>

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<s id="id.1.1.5.43.4.0">So that the part NE of the balance is greater than NF.</s>

<s id="id.1.1.5.43.5.0">And since this must be carried downward, the balance EF will move down on the side of E and return to AB.</s>

</figure>

<s id="id.1.1.5.44.1.0">In addition to the things that have been said up to this point, it may be stated that the balance in the position EF will move most swiftly to AB when the line EF, if extended straight, would pass through the center of the world.</s>

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<s id="id.1.1.5.44.10.0">Also the more distant the point H is from the point C, the more swiftly the balance will moves which is manifest not only from what Aristotle says at the beginning of his Questions of Mechanics and from what has been said above, but also from the things that we are going to say in the sixth proposition below.</s>

<s id="id.1.1.5.44.11.0">Hence the balance EF will move the more swiftly, the farther it is from its center.</s>

</figure>

<s id="id.1.1.5.45.1.0">Let there be the balance AB with its center below and let there be equal weights at A and B; and move the balance to EF.</s>

<s id="id.1.1.5.45.2.0">I say that the weight at F has more heaviness than that at E, and therefore the balance EF will move downward on the side F.</s>

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<s id="id.1.1.5.45.10.0">Therefore the natural power of the weight at F will overcome the resistance to force of the weight at E, and thus the weight at F will have greater heaviness than that at E.</s>

<s id="id.1.1.5.45.11.0">Hence the weight at F will move down and the weight at E will move up.</s>

<pb/>

<s id="id.1.1.5.46.1.0">Aristotle's reasoning is equally clear here.</s>

<s id="id.1.1.5.46.2.0">For let the point N be the intersection of the lines CO and EF; NF will be greater than NE, and since the perpendicular CO, according to him, divides the balance unequally with the larger part toward F (that is, NF) the balance EF will move downward on the side F since the greater is carried downward.</s>

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<s id="id.1.1.5.48.1.0">Besides, we may use their logic and their false assumptions to produce the effects and motions of the balance already explained, so that from this one may see the power of truth and how it forces itself to shine forth even from false things.</s>

</figure>

<s id="id.1.1.5.49.1.0">Assuming the same things, that is, the circle AEBF and the balance AB whose center C is above the balance, move the balance to EF; I say that the weight at E has greater heaviness than the weight at F, and that the balance EF will return to AB.</s>

<s id="id.1.1.5.49.2.0">Draw from the points E and F the lines EL and FM perpendicular to AB, which shall be parallel, and let the point N be the intersection of AB and EF.</s>

Line 931

<s id="id.1.1.5.50.4.0">By a like argument it would be shown that the triangle QES is similar to the triangle RFS and that the line EQ is greater than RF, and thus the weight at E will be farther from the line OP than will the weight at F, whence the weight at E will have greater heaviness than the weight at F.</s>

<s id="id.1.1.5.50.5.0">From this it appears evident that the balance will return from EF to AB.</s>

</figure>

<s id="id.1.1.5.51.1.0">But if the center of the balance is below the balance, then it will be shown by the same argument that the lower weight should have greater heaviness than the raised weight.</s>

<s id="id.1.1.5.51.2.0">Draw from the points E and F the lines EL and FM perpendicular to AB.</s>

<s id="id.1.1.5.51.3.0">As before, it is proved that EL is greater than FM, and therefore the descent of the weight at F will partake less of straightness than the rise of the weight at E; hence the resistance to force of the weight at E will overcome the natural inclination of the weight at F, and therefore the weight at E will be heavier than the weight at F.</s>

<pb/>

<s id="id.1.1.5.52.1.0">Extend also CD to O and P and draw from the points E and F the lines EQ and FR perpendicular to that.</s>

<s id="id.1.1.5.52.2.0">It will be proved in the same way that the line EQ is greater than FR, and, since the weight placed at E will be farther from the line of direction OP than the weight at F, the weight at E will have greater heaviness than the weight at F.</s>

<s id="id.1.1.5.52.3.0">From this it follows that the balance EF moves downward on the side of E.</s>

Line 952

<s id="id.1.1.5.53.6.0">Here he affirms not that the balance moves downward, but that it remains, which he seems to have deduced in the last conclusion.</s>

<s id="id.1.1.5.53.7.0">But not only does this not bear against us; when properly understood, it greatly assists us.</s>

</figure>

<s id="id.1.1.5.54.1.0">For let there be the balance AB, parallel to the horizon, with its center E under the balance.</s>

<s id="id.1.1.5.54.2.0">And since Aristotle considers an actual balance, it is necessary to place the support or something else under the center E; let this be EF, and this will be the support that sustains the center E.</s>

Line 968

<s id="id.1.1.5.55.1.0">But someone might add to this that, if a very small weight were placed at B, it would indeed move the balance downward but not all the way to G, and in this position, according to Aristotle, it should remain if the weight were taken away.</s>

<s id="id.1.1.5.55.2.0">This is evident by experience, since the balance tilts more or less when at one end of the balance only there is placed a larger or smaller weight, and this is true enough so long as the center is placed above the balance, but not when it is below or in the balance, as we shall show by way of example.</s>

</figure>

<s id="id.1.1.5.56.1.0">Let there be the balance AB, parallel to the horizon, whose center C is above the balance; let the perpendicular CD be plumb to the horizon, and let this line be extended through D to H.</s>

<s id="id.1.1.5.56.2.0">Now, since we consider the weight of the balance, the point D will be the center of gravity of the balance.</s>

Line 985

<s id="id.1.1.5.56.12.0"> always acute, nor will the point B ever go all the way to the line CH, because the center of gravity of the weight and the balance together will always be between B and D.</s>

<s id="id.1.1.5.56.13.0">The heavier the weight placed at B, the larger will be the arc described, beginning at E and approaching closer to the line CH.</s>

</figure>

<s id="id.1.1.5.57.1.0">But if the balance AB has its center C in the balance, C will also be the center of gravity of the balance; let the line FCG be drawn perpendicular to AB and to the horizon.</s>

<s id="id.1.1.5.57.2.0">Then put any weight you please at B, and let the center of gravity now be at E, so that CE is to EB as the weight placed at B is to the weight of the balance.</s>

<s id="id.1.1.5.57.3.0">And since CE is not perpendicular to the horizon, the balance AB and the weight at B will not remain in this position but will move downward on the side of B until CE becomes perpendicular to the horizon; that is, until the balance AB comes to FG.</s>

<s id="id.1.1.5.57.4.0">Whence it is clear that the weight placed at B always describes a full quadrant.</s>

</figure>

<pb/>

<s id="id.1.1.5.58.1.0">But if the center C is under the balance AB, and DCE is the perpendicular, the placing of a weight at B will similarly make the center of gravity of the system composed of the balance AB and the weight at B be in the line DB, say, at F, in such a way that, as DF is to FB, so is the weight placed at B to the weight of the balance.</s>

<s id="id.1.1.5.58.2.0">Draw CF, and, since CD is perpendicular to the horizon, the line CF will not be; hence the system composed of the balance AB and the weight at B will never remain fixed in this position but will move downward if nothing impedes it, until CF comes to DCE, in which position the balance together with the weight will come to rest.</s>

<s id="id.1.1.5.58.3.0">Now the point B will be at G and the point A at H, so the balance GH will no longer have its center below, but above.</s>

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<s id="id.1.1.5.59.1.0">These things proved, it is clear that the center of the balance is the cause of the various acts of the balance, and it is also seen that all the propositions of Archimedes in his book On Plane Equilibrium are true in every position, whether the balance is horizontal or not, provided only that the center of the balance is located within it, and this is the way he considers it.</s>

<s id="id.1.1.5.59.2.0">And even if the balance has unequal arms, the same will always happen, and it will be proved in exactly the same way that the center of the balance being situated in different manners will produce different effects.</s>

</figure>

<s id="id.1.1.5.60.1.0">Let there be the balance AB parallel to the horizon, and let there be unequal weights at A and B, so that the center of gravity is at C, and let the balance be hung from this point C.</s>

<s id="id.1.1.5.60.2.0">Now move the balance to DE, and it is evident that the balance will rest not only at DE but at any other point.</s>

</figure>

<s id="id.1.1.5.61.1.0">But now let the center of the balance AB be above C at F and let FC be perpendicular to AB and to the horizon.</s>

<s id="id.1.1.5.61.2.0">If the balance shall be moved to DE, the line CF will move to FG, and since this is not perpendicular to the horizon, the balance DE will move down on the side of D until FG returns to FC, when the balance DE will be at AB, in which position it will rest.</s>

</figure>

<s id="id.1.1.5.62.1.0"> But if the center F of the balance shall be below the balance, and the balance is moved to DE, it is evident in the first place that the balance will rest at AB and that at DE it would move downward on the side of E, since the line FG is not perpendicular to the horizon.</s>

</figure>

<s id="id.1.1.5.63.1.0">From these things just finished, if the balance were curved or if the arms of the balance formed an angle, and the center were variously placed-although strictly speaking this would not be a balance-we might nevertheless demonstrate various effects in it also.</s>

<s id="id.1.1.5.63.2.0">Thus let the balance be ACB, which turns about the center C, and draw the line AB so that the curve or angle A CB is above the line AB, and place the centers of gravity of the weights at A and B, which will rest in this position.</s>

<s id="id.1.1.5.63.3.0">Next move the balance from this position as to ECF.</s>

<s id="id.1.1.5.63.4.0">I say that the balance ECF will return to ACB.</s>

</figure>

<pb/>

<s id="id.1.1.5.64.1.0">Find the center of gravity of the whole system, D, and join C to D.</s>

<s id="id.1.1.5.64.2.0">Now since the weights A and B are at rest, the line CD will be perpendicular to the horizon.</s>

<s id="id.1.1.5.64.3.0">Therefore when the balance is at ECF, the line CD will be at CG, and, since this is not perpendicular to the horizon, the balance ECF will return to ACB.</s>

<s id="id.1.1.5.64.4.0">The same will happen if the center C is placed above the balance as at H.</s>

</figure>

<s id="id.1.1.5.65.1.0">Now if the curve or angle ACB shall be beneath the line AB, we may show in the same way that the balance ECF, whether its center is at C or H, must move downward on the side of F.</s>

<s id="id.1.1.5.66.1.0">And if the angle ACB is above the line AB at the center of the balance H, and if the line CH sustains the balance, when the balance is moved to EKF, it will return to ACB.</s>

</figure>

<s id="id.1.1.5.67.1.0">But if the center of the balance is D, one may move the balance in any way, and wherever it is placed, it will rest.</s>

</figure>

<s id="id.1.1.5.68.1.0">Then if the point H shall be beneath the line AB, the balance EKF will move downward on the side of F.</s>

</figure>

<s id="id.1.1.5.69.1.0">And for similar reasons if the angle ACB shall be beneath the line AB, and the center of the balance is at H, and the balance is sustained by the line CH, then, if the balance is moved from this position, it will move downward on the side of the lower weight.</s>

<s id="id.1.1.5.69.2.0">And if the center of the balance is at D, it will rest wherever it is left.</s>

Line 1070

<s id="id.1.1.5.69.4.0">All which things are manifest from what we said at the beginning.</s>

<s id="id.1.1.5.69.5.0">Similarly if the center of the balance is placed in one of the arms of the balance, either within or without, or any other way, we shall find the same things.</s>

<pb/>

<s id="id.1.1.5.70.1.0">Comment by Pigafetta</s>

Line 1110

<s id="id.1.1.5.75.1.0">Wherever the Latin word equilibrium is read, it means "equally counterpoised," that is, weighing as much on one side as on the other in equal scales or balances.</s>

<s id="id.1.1.5.75.2.0">Librar con giuste lance, Petrarch said.</s>

<pb/>

<s id="id.1.1.6.0.1.0">PROPOSITION V</s>

</figure>

<s id="id.1.1.6.1.1.0">If two weights are attached to a balance and the balance is divided between them in such a way that the parts correspond inversely to the weights, they will weigh as much at the points where they are attached as if each were suspended from the point of that division.</s>

<s id="id.1.1.7.0.1.0">PROPOSITION VI</s>

</figure>

<s id="id.1.1.7.1.1.0">Equal weights suspended from a balance have heaviness in proportion to the distances at which they are suspended.</s>

Line 1133

<s id="id.1.1.7.3.1.0">Comment by Pigafetta</s>

<pb/>

<s id="id.1.1.7.4.1.0">Corollary is a Latin word employed by all Italian writers on this subject, nor did it displease Dante in the Twenty-eighth Canto of his Purgatory.</s>

<s id="id.1.1.7.4.2.0">"Also called a corollary, for example," as Varro says in his first book on the Latin language, "is anything over and above that which would normally be paid in buying something."</s>

<s id="id.1.1.7.4.3.0">In ancient times when the actors in tragedies, comedies, and other poems carried off their scenes well and pleased their audiences, something was given to them in addition to the fixed price-a corollary for each-one-that is, a small crown to be placed on their foreheads and added to their rewards.</s>

Line 1145

<s id="id.1.1.8.0.1.0">PROPOSITION VII</s>

</figure>

<s id="id.1.1.8.1.1.0">Problem: Given an indefinite number of weights on the balance, suspended at any places, to find a center on the balance from which, if the balance were to be hung, the given weights would be in equilibrium.</s>

Line 1156

<s id="id.1.1.8.3.1.0">Under the name of "proposition" is included "problem," also a Greek word; the problem goes beyond the proposition in that it proposes and shows how to achieve some result, whereas the proposition gives the bare theory only; and this is the difference between a proposition and a problem.</s>

</figure>

<pb/>

<s id="id.1.1.8.4.1.0">Let there be the balance AB and let there be given any number of weights C, D, E, F, and G hanging from the balance at the points A, H, K, L, and B.</s>

<s id="id.1.1.8.4.2.0">It is required to find the center of the balance from which, if [the balance were] suspended, the weights would be at rest.</s>

</figure>

<s id="id.1.1.8.5.1.0">Divide AH at M so that HM is to MA as the weight C is to the weight D; then divide BL at N so that LN is to NB as the weight G is to the weight F, and divide NM at O so that MO is to N as the weights F and G are to the weights C and D finally, divide KO at P so that KP is to PO as the weights C D, F, and G are to the weight E.</s>

<s id="id.1.1.8.5.2.0">Now since the weights C, D, F, and G weigh as much at O as C and D at M and F and G at N, the weights C and D at M, F and G at N, and E at K</s>

Line 1184

<s id="id.1.2.1.0.1.0">LEMMA</s>

<pb/>

<s id="id.1.2.1.1.1.0">Let there be four magnitudes A, B, C, and D, and let A be greater than B and C be greater than D.</s>

<s id="id.1.2.1.1.2.0">I say that the ratio of A to D is greater than that of B to C.</s>

Line 1201

<s id="id.1.2.2.2.1.0">From this it can easily be demonstrated that the closer the fulcrum is to the weight, the smaller the power required to sustain the weight.</s>

</figure>

<s id="id.1.2.2.3.1.0">Corollary.-Whence it can be quickly deduced that, AF being less than FB, a smaller power is required at B to sustain the weight D; and if they are equal, it is equal; and if AF is greater, the required power is greater.</s>

<pb/>

<s id="id.1.2.3.0.1.0">PROPOSITION II</s>

<s id="id.1.2.3.1.1.0">The lever may be used in a second mode.</s>

</figure>

<s id="id.1.2.3.2.1.0">Proof: Let there be the lever AB with its fulcrum B, and let the weight C be attached at D between A and B, and let the power at A sustain the weight C.</s>

<s id="id.1.2.3.2.2.0">I say, that as BD is to BA, so is the power at A to the weight C---.</s>

</figure>

<s id="id.1.2.3.3.1.0">Corollary I.-From this also, as before, it may be shown that, if the weight E is placed closer to the fulcrum B, as at H, a smaller power is required at A to sustain the weight.</s>

<s id="id.1.2.3.4.1.0">Corollary II.-It also follows that the power at A is always less than the weight E.</s>

</figure>

<s id="id.1.2.3.5.1.0">Corollary III.-From this likewise it may be deduced that, if there are two powers, one at A and the other at B, and both sustain the weight E, the power at A will be to the power at B as BC is to CA.</s>

<pb/>

<s id="id.1.2.3.6.1.0">Corollary IV.-I t is furthermore evident that the two powers at A</s>

<s id="id.1.2.3.6.2.0"> and B taken together are equal to the weight E.</s>

Line 1241

<s id="id.1.2.4.1.1.0">We may also use the lever in a third mode.</s>

</figure>

<s id="id.1.2.4.2.1.0">Let there be the lever AB with its fulcrum at B, and let the weight C be hung from the point A, and let it be the power at D, somewhere between A and B, that sustains the weight C.</s>

<s id="id.1.2.4.2.2.0">I say that, as AB is to BD, so is the power at D to the weight C---.</s>

Line 1253

<s id="id.1.2.4.4.1.0">Corollary II.-It is likewise evident that the power at D is always greater than the weight C.</s>

<pb/>

<s id="id.1.2.5.0.1.0">PROPOSITION IV</s>

<s id="id.1.2.5.1.1.0">If the power shall move the weight hung from the lever, the space through which the power moves will be to the space through which the weight is moved as the distance from the fulcrum to the power is to the distance from the fulcrum to the point from which the weight is hung.</s>

</figure>

<s id="id.1.2.5.2.1.0">Let there be the lever AB with its fulcrum C, and let the weight D be attached at the point B, and let the power at A move the weight D by means of a leverAB.</s>

<s id="id.1.2.5.2.2.0">Then the space of the power at A is to the space of the weight as CA is to CB---.</s>

Line 1272

<s id="id.1.2.5.4.1.0">Corollary.</s>

<s id="id.1.2.5.4.2.0">-From these things it is evident that the ratio of the space of the power which moves to the space of the weight moved is greater than that of the weight to the same power.</s>

</figure>

<s id="id.1.2.5.5.1.0">For the space of the power has the same ratio to the space of the weight as that of the weight to the power which sustains the same weight.</s>

<s id="id.1.2.5.5.2.0">But the power that sustains is less than the power that moves; therefore the weight will have a lesser ratio to the power that moves it than to the power that sustains it.</s>

<s id="id.1.2.5.5.3.0">Therefore the ratio of the space of the power that moves to the space of the weight will be greater than that of the weight to the power.</s>

<pb/>

<s id="id.1.2.6.0.1.0">PROPOSITION V</s>

</figure>

<s id="id.1.2.6.1.1.0">The power that sustains the weight in any way by means of the lever will have the same proportion to the weight as that of the distance from the fulcrum to the point on the lever, vertical to the center of gravity of the weight, to the distance between fulcrum and the power.</s>

<s id="id.1.2.7.0.1.0">PROPOSITION VI</s>

</figure>

<s id="id.1.2.7.1.1.0">Let there be the straight line AB, and perpendicular to it the line AD, prolonged on the side of D to C.</s>

<s id="id.1.2.7.1.2.0">Join C and B and extend this to E.</s>

Line 1300

<s id="id.1.2.7.1.5.0">Draw CH and CK, which cut the lines BF and BG at the points M and N.</s>

<s id="id.1.2.7.1.6.0">I say that BN is shorter than BM, and BM than BA.</s>

</figure>

<s id="id.1.2.7.2.1.0">If the equal triangles BFH and BCK are between BC and BA below, and if there are added the lines HC and KC which cut the lines BF and BG extended at the points M and N, then BN will be greater than BM and BM than BA .</s>

<s id="id.1.2.8.0.1.0">PROPOSITION VII</s>

<pb/>

<s id="id.1.2.8.1.1.0">Let there be the line AB and the perpendicular AD extended as far as C.</s>

<s id="id.1.2.8.1.2.0">Draw CB and extend this to E.</s>

<s id="id.1.2.8.1.3.0">Between AB and BE draw BF and BC equal to AB, and from the points F and C draw the lines FH and CK also equal to AD and perpendicular to BF and BC, as if BA and AD were moved to BF and FH, or to BG and GK.</s>

Line 1326

<s id="id.1.2.9.1.1.0">If the power sustaining the weight which has its center of gravity on a horizontal lever is given, then, the more the weight is raised from this position by means of the lever, the smaller the power required to sustain it.</s>

<s id="id.1.2.9.1.2.0">But if it shall be lower, the power is greater.</s>

</figure>

<s id="id.1.2.9.2.1.0">From this it is easily deduced that the power at A is to the power at E as CL is to CM.</s>

Line 1335

<s id="id.1.2.9.3.1.0">In addition to this, if there is another power at B, so that there are two powers that sustain the weight, the power at B, which sustains the weight PQ by means of the lever BO, will be less than the weight CD on the lever BA.</s>

<s id="id.1.2.9.3.2.0">On the other hand, a greater power is required at B to sustain the weight FC, by means of the lever BE, than the weight CD on the lever AB.</s>

</figure>

<pb/>

<s id="id.1.2.9.4.1.0">Corollary .</s>

<s id="id.1.2.9.4.2.0">-From these things it is evident that, if a power raises, by means of a lever, a weight whose center of gravity is above the lever, then the more the weight is raised, the smaller becomes the power necessary to move the weight.</s>

<s id="id.1.2.10.0.1.0">PROPOSITION IX</s>

</figure>

<s id="id.1.2.10.1.1.0">If the power sustaining a weight that has its center of gravity under the lever is given when that [lever] is horizontal, then the more the weight is raised from this position by means of the lever, the more power will be required to sustain it; but if it is lowered, the power becomes less.</s>

<s id="id.1.2.11.0.1.0">PROPOSITION X</s>

</figure>

<s id="id.1.2.11.1.1.0">The power sustaining a weight that has its center of gravity in the lever itself will always be the same no matter how the weight is moved by means of a lever.</s>

<s id="id.1.2.12.0.1.0">PROPOSITION XI</s>

<pb/>

<s id="id.1.2.12.1.1.0">If the ratio of the distance along the lever between fulcrum and power to the distance between fulcrum and that point on the lever vertical to the center of gravity of the weight is greater than the ratio of the weight to the power, the weight will be moved by the power.</s>

Line 1369

<s id="id.1.2.13.1.1.0">Problem: To move a given weight by means of a given lever with a given power.</s>

</figure>

<s id="id.1.2.13.2.1.0">Let the weight A be 100 and the power that must move it be 10, and let the given lever be BC.</s>

<s id="id.1.2.13.2.2.0">It is required that the power of 10 shall move the weight of 100 by means of the lever BC.</s>

Line 1385

<s id="id.1.2.14.0.1.0">PROPOSITION XIII</s>

<pb/>

<s id="id.1.2.14.1.1.0">Problem: Given an arbitrary number of weights suspended from arbitrary points of a lever whose fulcrum is also given, to find a power which will sustain these weights at a given point.</s>

</figure>

<s id="id.1.2.14.2.1.0">Let there be given the weights A, B, and C on the lever DE (with its fulcrum at F), suspended from the points D, C, and H, and a point E at which the power must be applied---.</s>

<s id="id.1.2.14.2.2.0">Divide DC at K in such a way that DK is to KC as the weight B is to A; then divide KH at L so that KL is to LH as the weight C is to the weights B and A.</s>

Line 1401

<s id="id.1.2.15.1.1.0">Problem: To make a given power move an arbitrary number of weights at arbitrary places on a given lever.</s>

</figure>

<s id="id.1.2.15.2.1.0">Let the given lever be DE, let the given weights be placed as above, and let A be 100, B, 50, and C, 30; and let the given power be 30.</s>

<s id="id.1.2.15.2.2.0">Find the point L as before; then divide LE at F in such a way that FE is to FL as 180 is to 30 (that is, as six is to one), and if F is the fulcrum, the power of 30 at E will sustain the weights A, B,- and C.</s>

Line 1411

<s id="id.1.2.16.0.1.0">PROPOSITION XV</s>

<pb/>

<s id="id.1.2.16.1.1.0">Problem: But since in moving weights with a lever, the lever also has weight, which has not been mentioned up to this point, we shall demonstrate how to find the power which will sustain the lever in a given point, the fulcrum being likewise given.</s>

</figure>

<s id="id.1.2.16.2.1.0">Let there be the lever BA with its fulcrum at C, and let there be the point D at which the power must be applied which must sustain the lever AB so that it remains at rest.</s>

<s id="id.1.2.16.2.2.0">From the point C draw the vertical line CE, dividing the lever AB into AE and EF.</s>

Line 1427

<s id="id.1.2.16.3.1.0">Next, a weight hung from the lever is to be added, such as the weight P hung from A, and the power is to be applied at B in such a way that it sustains the lever AB together with the weight P.</s>

</figure>

<s id="id.1.2.16.4.1.0">Divide AM at Q in such a way that AQ is to QM as the weight of the lever AB is to the weight P.</s>

<s id="id.1.2.16.4.2.0">In whatever ratio CF is to CQ, make the combined weight of AB and P be to the power placed at B.</s>

Line 1446

<s id="id.1.3.0.0.1.0">On the Pulley</s>

<pb/>

<s id="id.1.3.0.1.1.0">By means of the pulley things may be moved in many ways, but, since the theory is the same for all and in order to present the thing most clearly, it is to be understood in that which is about to be said that the weight is always to be moved upward at right angles to the horizontal plane.</s>

Line 1455

<s id="id.1.3.0.2.3.0">Let another block be attached to the weight, also with two pulleys at D and E; and around all the pulleys lead the rope tied at one end, say, at F.</s>

<s id="id.1.3.0.2.4.0">Apply the power at C, so that, when this descends, A is raised, as Pappus shows in the eighth book of his Collections, Vitruvius in the tenth book of his Architecture, and others.</s>

</figure>

<s id="id.1.3.0.3.1.0">Now let us show how the pulley may be reduced to the lever, why a great weight is moved by a small force, in what way, in what time, why the rope must be secured at one end, what is the function of the pulley that is placed below and what that of the one above, and how one may find any given proportion between the power and the weight .</s>

Line 1466

<s id="id.1.3.1.1.1.0">If the rope is led around the pulley that is fastened from above, and one end of the rope is tied to the weight while the power that sustains the said weight is applied to the other end, then the power will be equal to the weight.</s>

</figure>

<s id="id.1.3.1.2.1.0">Corollary.</s>

<s id="id.1.3.1.2.2.0">-From this it is evident that the same weight can always be sustained by the same power without any assistance from this pulley .</s>

<pb/>

<s id="id.1.3.2.0.1.0">PROPOSITION II</s>

Line 1483

<s id="id.1.3.2.2.2.0">Now if we assume the power applied at C to act at D (because it is entirely the same thing), BD will be a lever with its support at B and its weight attached at E and the power applied at D; the rope BF being motionless, B serves as the fulcrum ---.</s>

<s id="id.1.3.2.2.3.0">Now since the power has the same proportion to the weight that BE has to BD --- the power at C will be one-half the weight A.</s>

</figure>

<s id="id.1.3.2.3.1.0">Corollary I.</s>

<s id="id.1.3.2.3.2.0">- From this it is evident that the weight is sustained in this way by a power of only one-half that which would be required without the aid of such a pulley.</s>

Line 1497

<s id="id.1.3.2.5.1.0">Corollary III.</s>

<s id="id.1.3.2.5.2.0">-It is likewise evident why the rope must be secured at one end.</s>

<pb/>

<s id="id.1.3.3.0.1.0">PROPOSITION III</s>

<s id="id.1.3.3.1.1.0">If two pulleys are given, one attached above and the other below, to the latter of which the weight is attached, and a rope is given, led around both with one of its ends fastened at some place, then the power which sustains the weight applied to the other end will be equal to one-half the weight.</s>

</figure>

<s id="id.1.3.3.2.1.0">Corollary.</s>

<s id="id.1.3.3.2.2.0">-If there are two powers at N and L, they will be equal.</s>

Line 1513

<s id="id.1.3.4.0.1.0">PROPOSITION IV</s>

</figure>

<s id="id.1.3.4.1.1.0">Let there be a lever AB with its fulcrum at A, divided into two equal parts at D; let the weight C be suspended from D, and let there be two equal powers applied at B and D which are to sustain the weight C.</s>

<s id="id.1.3.4.1.2.0">I say that each of these powers is one-third of the weight C .</s>

Line 1522

<s id="id.1.3.4.2.1.0">And if two levers AB and EF, bisected at C and D, have their fulcrums at A and F, while the weight C is equally supported by both levers, and if there are two equal powers at B and C, it may be shown that each of those powers is one-third of the weight C.</s>

<pb/>

<s id="id.1.3.5.0.1.0">PROPOSITION V</s>

</figure>

<s id="id.1.3.5.1.1.0">If two pulleys are given, one supported from above and the other attached to the weight, with a rope led around them, one end of which is attached to the lower pulley, then a power equal to one-third of the weight applied to the other end will sustain the weight .</s>

Line 1535

<s id="id.1.3.5.2.2.0">Inasmuch as equal weights are sustained by equal powers, the powers at M and L will be equal, as if they were applied at D and E.</s>

<s id="id.1.3.5.2.3.0">Therefore, since the weight A is centrally attached to the lever BD and the two powers placed at D and E are equal and sustain the weight, then B will be the fulcrum, and each power, whether applied at D and E or M and L, will be one-third the weight A---.</s>

</figure>

<s id="id.1.3.5.3.1.0">Corollary.</s>

<s id="id.1.3.5.3.2.0">-From this it is evident that each of the rope segments MD, FL, and HB sustains one-third of the weight A.</s>

Line 1547

<s id="id.1.3.5.5.1.0">Comment by Pigafetta</s>

<pb/>

<s id="id.1.3.5.6.1.0">Now some people might question these demonstrations about pulleys, for instance this fifth proposition which I select as the best example, asking whether in fact experiment is in agreement with theory as to the ratios of forces and weights.</s>

<s id="id.1.3.5.6.2.0">For in mathematical demonstrations, all lines are assumed to be without breadth or thickness, and all things are abstracted from actual matter, so that it is easy to persuade ourselves of the mathematical truth.</s>

<s id="id.1.3.5.6.3.0">But experience very often shows something different, and we find ourselves deceived, for actual matter changes things quite a bit.</s>

Line 1560

<s id="id.1.3.6.0.1.0">PROPOSITION VI</s>

</figure>

<s id="id.1.3.6.1.1.0">Let there be two levers AB and CD, bisected at E and F, with their fulcrums at B and D; and let there be the weight G, suspended in such a way that it weighs equally on E and F, and two equal powers at A and C which sustain the weight.</s>

<s id="id.1.3.6.1.2.0">I say that each of the powers is one-fourth of the weight G.</s>

</figure>

<s id="id.1.3.6.2.1.0">But if there shall be three levers AB, CD and EF, bisected at C, H, and K, with their fulcrums B, D, and F, and if the weight is similarly suspended from G, H, and K, while three equal powers A, C, and E sustain the weight, it will be likewise seen that each is one-sixth of the weight L.</s>

<s id="id.1.3.6.2.2.0">And in the same manner, if there were four levers and four weights, each power would be one-eighth of the weight; and so on.</s>

Line 1578

<s id="id.1.3.7.1.1.0">If three pulleys are given, one of which is suspended from above and two from below, and to these latter a weight is attached, and rope is wound around them and one end secured, then a power applied at the other end equal to one-fourth of the weight will sustain the weight.</s>

<pb/>

<s id="id.1.3.7.2.1.0">Corollary I.</s>

<s id="id.1.3.7.2.2.0">-From this it is evident that each of the ropes EF, GK, LN, and OP sustains one-fourth of the weight A.</s>

</figure>

<s id="id.1.3.7.3.1.0">Corollary II.</s>

<s id="id.1.3.7.3.2.0">-It is also clear that the pulley whose center is C sustains no less than that whose center is B .</s>

Line 1598

<s id="id.1.3.9.0.1.0">PROPOSITION IX</s>

</figure>

<s id="id.1.3.9.1.1.0">If there are four pulleys, one of which is attached from above and one of which is attached to the weight, and the rope is led around with one of its ends tied to the lower pulley, while the force is applied to the other end, the force that sustains the weight will be one-fifth of the weight .</s>

<pb/>

<s id="id.1.3.9.2.1.0">Comment by Pigafetta</s>

Line 1621

<s id="id.1.3.10.1.1.0">Let the rope be led around a pulley suspended from above, and let the weight be attached to one end, while the power that moves it is applied to the other.</s>

<s id="id.1.3.10.1.2.0">Then the said power will always move the weight as by a lever always parallel to the horizon.</s>

</figure>

<s id="id.1.3.10.2.1.0">Under the above assumption, the space of the power that moves the weight is equal to the space of the weight that is moved .</s>

<pb/>

<s id="id.1.3.10.3.1.0">Moreover, the power moves the weight through an equal space in an equal time, whether the rope is wound around a pulley supported from above, or [lifts the weight] without any pulley at all, provided that the movements of this power are equal in speed.</s>

<s id="id.1.3.11.0.1.0">PROPOSITION XI</s>

</figure>

<s id="id.1.3.11.1.1.0">Let the rope be led around a pulley to which the weight is attached, and let one end of the rope be sustained at some place while the power that moves the weight is applied at the other.</s>

<s id="id.1.3.11.1.2.0">Then the power will always move the weight as with a lever parallel to the horizon .</s>

Line 1647

<s id="id.1.3.12.0.1.0">PROPOSITION XII</s>

</figure>

<pb/>

<s id="id.1.3.12.1.1.0">If the rope is wound around many pulleys, and one end is fastened at some place, and the force that moves the weight is applied to the other end, the force will always move the weight as by a lever parallel to the horizon.</s>

Line 1661

<s id="id.1.3.14.0.1.0">PROPOSITION XIV</s>

</figure>

<s id="id.1.3.14.1.1.0">If the rope is led around three pulleys in two blocks, with a single pulley attached from above and two below attached to the weight, and one end of the rope is secured at some place while the power that moves the weight is applied to the other, then the space traversed by the power will be four times that through which the weight moves.</s>

Line 1670

<s id="id.1.3.14.2.1.0">Corollary I.</s>

<s id="id.1.3.14.2.2.0">From these things we see why the ratio of the weight to the power that sustains it is the same as the ratio between the space of the moving power and the space of the weight moved.</s>

</figure>

<pb/>

<s id="id.1.3.14.3.1.0">Corollary II.</s>

<s id="id.1.3.14.3.2.0">It is likewise evident from what has been said that the pulleys to which the weight is attached have the function of reducing the space passed through by the weight with respect to the power that moves it, and that the weight goes through the same space in a longer time than [it would if moved] without the pulleys; which function does not belong at at all to the pulley attached from above .</s>

Line 1691

<s id="id.1.3.16.2.1.0">Under these assumptions the space of the weight moved is twice that of the space of the power that moves.</s>

</figure>

<pb/>

<s id="id.1.3.16.3.1.0">Corollary.</s>

<s id="id.1.3.16.3.2.0">From this it is evident that a given weight is drawn with this sytem of pulleys by the same power in an equal time through twice the space traversed without pulleys, provided that the movements of the power are equal in speed .</s>

<s id="id.1.3.17.0.1.0">PROPOSITION XVII</s>

</figure>

<s id="id.1.3.17.1.1.0">If an upper pulley is suspended from above by the power and a lower pulley is securely fixed, while the rope is wound around them with one of its ends tied to the pulley above and the other attached to the weight, then the power will be three times the weight .</s>

Line 1715

<s id="id.1.3.18.2.1.0">Corollary.</s>

<s id="id.1.3.18.2.2.0">ÑFrom this it is evident that if the rope were fastened at C and wound around the pulleys whose centers are B, C, and D, then the power at R would sustain four times the weight Q; for the pulley whose center is at A does nothing---</s>

<pb/>

<s id="id.1.3.18.3.1.0">Corollary.</s>

<s id="id.1.3.18.3.2.0">ÑIn these things it is evident that the pulleys of the upper block constitute the reason for which the weight is moved by a greater power than itself, through a greater space than that of the power or through equal space in less time; but this is in no way caused by the lower pulleys.</s>

Line 1725

<s id="id.1.3.19.1.1.0">If there are two pulleys, one of which is held from above and the other is held by the power [O] that sustains [weight M], while the rope around them has one of its ends secured and the other attached to the weight, then the power will be twice the weight.</s>

</figure>

<s id="id.1.3.19.2.1.0">Corollary.</s>

<s id="id.1.3.19.2.2.0">ÑFrom this it is evident that the pulley below in this case causes the weight to be moved by a greater power than itself and through a greater space than that of the power (or through equal space in less time), which cause does not belong to the pulley attached from above .</s>

Line 1737

<s id="id.1.3.20.1.1.0">If there are two pulleys, the upper one of which is sustained by the power and that below is attached to the weight, while the rope is secured at one end and the other end is attached to the pulley below, the weight will be one and one-half times the power.</s>

<pb/>

<s id="id.1.3.21.0.1.0">PROPOSITION XXI</s>

</figure>

<s id="id.1.3.21.1.1.0">If there are three pulleys, one of which is sustained from above by the power while the other two are below and attached to the weight, and one end of the rope is fastened while the other is attached to the upper pulley, the weight will be one and one-third times the power .</s>

<s id="id.1.3.22.0.1.0">PROPOSITION XXII</s>

</figure>

<s id="id.1.3.22.1.1.0">If there are two pulleys, of which one is sustained from above by the power and the other is attached to the weight, while one end of the rope is fastened and the other is attached to the upper pulley, then the power will be one and one-half times the weight .</s>

Line 1759

<s id="id.1.3.23.1.1.0">If there are two pulleys, one of which is sustained by the power from above while the other is attached to the weight below, and the rope has both ends fastened elsewhere than to the pulleys, then the power will be equal to the weight.</s>

<pb/>

<s id="id.1.3.23.2.1.0">[Propositions 24 and 25 deal similarly with variant systems].</s>

Line 1771

<s id="id.1.3.24.2.1.0">[The author also gives analogous demonstrations for various other problems].</s>

</figure>

<s id="id.1.3.24.3.1.0">Corollary.</s>

<s id="id.1.3.24.3.2.0">ÑFrom these things it is evident that the space of the power that moves has always a greater ratio to the space of the weight moved than that of the weight to the same power.</s>

Line 1781

<s id="id.1.3.26.0.1.0">PROPOSITION XXVII</s>

<pb/>

<s id="id.1.3.26.1.1.0">Problem: To move a given weight with a given power by means of pulleys.</s>

Line 1799

<s id="id.1.3.27.0.1.0">PROPOSITION XXVIII</s>

</figure>

<s id="id.1.3.27.1.1.0">Problem: To provide that the power moving the weight and the weight itself shall move through given spaces which are commensurable.</s>

<pb/>

<s id="id.1.3.27.2.1.0">--- We can achieve the result with a single rope by what was said in the twenty-second and twenty-fifth propositions.</s>

<s id="id.1.3.27.2.2.0">And if we wish to use more ropes, we may do it in an infinite number of ways---</s>

Line 1822

<s id="id.1.4.0.0.1.0">On the Wheel and Axle</s>

</figure>

<s id="id.1.4.0.1.1.0">The construction and nature of this instrument is explained a by Pappus in the eighth book of his Mathematical Collections.</s>

<s id="id.1.4.0.1.2.0">AB is called the axle, CD the drum on the same center (which we will call the wheel), and those rods that are inserted in the holes of the wheel and designated EF, GH, and so on, we shall call handles.</s>

<s id="id.1.4.0.1.3.0">The power or the force is always applied to the handles, as at F, and this turns the wheel, which in turn moves the axle, which draws up the weight K suspended by the rope LM around the axle.</s>

<s id="id.1.4.0.1.4.0">It now remains for us to show how weights may be moved by a small force with this instrument, and in what manner, and, moreover, to show the rule of the times and spaces of the moving power and of the weight moved; and finally, to reduce this instrument to the lever.</s>

<pb/>

<s id="id.1.4.1.0.1.0">PROPOSITION I</s>

<s id="id.1.4.1.1.1.0">The power sustaining the weight by means of the wheel and axle is in the same ratio to the weight as the radius of the axle to the radius of the wheel including the handle.</s>

</figure>

<s id="id.1.4.1.2.1.0">Corollary.</s>

<s id="id.1.4.1.2.2.0">ÑIt is evident that the power is always less than the weight.</s>

Line 1851

<s id="id.1.4.1.5.1.0">Now the power moves the weight by means of the lever FB; that is, when the power at F rotates the wheel, the axle also rotates, and FB serves the function of a lever with its fulcrum at C, its moving power at F, and the weight applied at B---</s>

<pb/>

<s id="id.1.4.1.6.1.0">--- Therefore let the power be where it will, the space of the power will be to the space of the weight moved as CF is to CB; that is, as the radius of the wheel to the radius of the axle.</s>

Line 1872

<s id="id.1.4.2.1.1.0">Problem: To move a given weight by means of the wheel and axle, with a given power.</s>

</figure>

<pb/>

<s id="id.1.4.2.2.1.0">Let there be given a weight of 60 and a power of 10.</s>

<s id="id.1.4.2.2.2.0">Draw the straight line AB, divided at C in such a way that AC is to CB as 60 is to 10.</s>

<s id="id.1.4.2.2.3.0">And if CB were the radius of the axle and CA the radius of the wheel with its handles, it is clear that the power at A would counterpoise the weight at B.</s>

Line 1893

<s id="id.1.4.2.6.1.0">Here the author has given us five figures, representing five instruments for moving weights which may be reduced under a single property, in order that one may see each to be the same as the wheel and axle already explained.</s>

</figure>

<s id="id.1.4.2.7.1.0">He has put the letters A, B, and C with their lines, in order that one may understand that the weight has the same proportion to the power that sustains it as AC has to CB, and, as the weight shall be moved by a power, the space of the power will likewise be to the space of the weight as AC is to CB.</s>

<s id="id.1.4.2.7.2.0">In each case the power is to be understood as placed at the end of the handles at the distance CA from the center.</s>

Line 1903

<s id="id.1.4.2.7.5.0">Similarly the figure with the drum is to be considered as if the force were at the outside of the drum and the weight were attached to the axle.</s>

<s id="id.1.4.2.7.6.0">As to the bit and brace, or auger, as it is called, since this is an instrument not designed to sustain things but to move them, the power must have a greater proportion to the weight than that of CB to CA, in accordance with the eleventh proposition in the section on the lever.</s>

<pb/>

<s id="id.1.4.2.8.1.0">End of the Wheel and Axle</s>

Line 1917

<s id="id.1.5.0.2.2.0">But when the wedge is struck, it enters into DEFG in a ratio greater than that which was before; let this be the portion MBL.</s>

<s id="id.1.5.0.2.3.0">And since MB and BL are greater than HB and BK, ML will also be greater than HK---</s>

</figure>

<s id="id.1.5.0.3.1.0">But since there are three kinds of levers, as set forth previously, it will be perhaps more convenient to consider the wedge in the following manner .</s>

<s id="id.1.5.0.4.1.0">We may regard AB as a lever with its fulcrum at B and the weight at H --- and similarly the lever CB with its support at B and the weight at K---</s>

<pb/>

<s id="id.1.5.0.5.1.0">Thus let the wedge be ABC, and let there be two separate weights DEFG and HIKL with the part DBH of the wedge between them---</s>

<s id="id.1.5.0.5.2.0">Now while DG is moved by the wedge toward M --- it is by the lever AB with fulcrum B---</s>

<s id="id.1.5.0.5.3.0">Similarly HL is moved from H by the lever CB---</s>

</figure>

<s id="id.1.5.0.6.1.0">If one must split the rectangle ABCD and there are two equal levers EF and GF--- and it is necessary with these levers to split ABCO without striking, then the moving powers at E and G are equal.</s>

</figure>

<s id="id.1.5.0.7.1.0">But since the whole wedge is moved in the splitting, we may consider it also in another manner; that is, when it enters into the thing to be split and is the same as moving a weight upon an inclined plane.</s>

Line 1944

<s id="id.1.5.0.8.1.0">--- In this example, if we consider the wedge as moving like a lever, it is evident that the wedge BCD moves the weight AEFC by means of a lever CD, so that D is the fulcrum and the weight is placed at E, rather than with the lever BD with its support at H and the weight placed at D.</s>

<s id="id.1.5.0.8.2.0">But in order that this may be more clear we shall use another example.</s>

</figure>

<s id="id.1.5.0.9.1.0">--- This movement is easily reduced to the balance and to the lever, since that which is moved on an inclined plane is reduced to the balance by the-ninth proposition of the eighth book of the Mathematical Collections of Pappus.</s>

<s id="id.1.5.0.9.2.0">For it is the same thing whether the wedge stands still-and the weight moves on the side of the wedge, or the wedge is moved and the weight moves along its side, as upon an inclined plane.</s>

Line 1953

<s id="id.1.5.0.10.1.0">Comment by Pigafetta</s>

</figure>

<pb/>

<s id="id.1.5.0.11.1.0">The proposition from Pappus, cited here by our author, I have withheld for a more convenient place in the section on the screw, for it is my opinion that perhaps it is more apropos there and serves more clearly than with regard to the wedge.</s>

<s id="id.1.5.0.11.2.0">This proposition was sent to me by the author, and, though there was nothing wrong with it, I have compared it carefully with the Greek edition of Pappus owned by Signor Pinelli, that it might be most useful and pleasant to those who have never seen anything of Pappus and have never read that marvelous writer on mechanics.</s>

</figure>

<s id="id.1.5.0.12.1.0">Now we shall see how things that are split move as upon inclined planes.</s>

Line 1973

<s id="id.1.5.0.15.1.0">We can also show this by another theory, considering the wedge as moving by two opposed levers, as mentioned previously.</s>

</figure>

<s id="id.1.5.0.16.1.0">Let there be the lever AB that has its fixed support at B and must move the rectangle CDEF, so arranged that it cannot move downward on the side FE.</s>

<s id="id.1.5.0.16.2.0">Let the point E be motionless, and consider it as a center, so that the point D will move along the circumference of the circle DH, whose center is E, while C moves along the circumference CL, and the line CE is its radius---</s>

<s id="id.1.5.0.16.3.0">Now let there be another lever MCN, which also moves CDEF, and this has its fixed support at N---</s>

<s id="id.1.5.0.16.4.0">I say that CDEF is more easily moved by the same power with the lever AB than with the lever MN---</s>

</figure>

<pb/>

<s id="id.1.5.0.17.1.0">Corollary.</s>

<s id="id.1.5.0.17.2.0">ÑFrom this it is clear that the smaller the angle BCF or BCE or BCD, the more easily the weight is moved.</s>

<s id="id.1.5.0.17.3.0">But this may be demonstrated in the same manner.</s>

Line 1996

<s id="id.1.5.0.19.2.0">If by itself, first one must notice that the heavier it is, the greater the stroke will be.</s>

<s id="id.1.5.0.19.3.0">In addition to this, the greater the distance AC, the greater the stroke; for any heavy object when moved takes on more heaviness than when standing still, and the more so, the farther it moves .</s>

</figure>

<s id="id.1.5.0.20.1.0">Now if C is moved by some power, as, for example, the handle DE, then first by the greater weight of C, and second by the greater length of DE, the stroke will be made greater.</s>

<s id="id.1.5.0.20.2.0">For if the moving power is placed at E, C will be more distant from the center and therefore will move more, as Aristotle demonstrates in his Questions of Mechanics.</s>

<s id="id.1.5.0.20.3.0">And it may also be clear from what was said in the section on the balance that the farther C is from the center, the more it will weigh, and it will strike with greater impetus, the force at E being more potent.</s>

</figure>

<s id="id.1.5.0.21.1.0">Now here is the second thing, which is the reason that great weights can be moved and split with this instrument.</s>

<s id="id.1.5.0.21.2.0">Percussion is a very strong force, as is evident from the nineteenth of Aristotle's Questions of Mechanics; for if a very heavy weight shall be placed upon a wedge, the wedge will accomplish nothing compared with its [work by] being struck.</s>

Line 2019

<s id="id.1.5.0.22.1.0">End of the Wedge</s>

<pb/>

<s id="id.1.6.0.0.1.0">On the Screw</s>

</figure>

<s id="id.1.6.0.1.1.0">Pappus in his eighth book deals with many matters of the screw, showing how it should be made and how great weights may be moved with such an instrument; moreover, he gives many useful theories for its understanding.</s>

<s id="id.1.6.0.1.2.0">Now, since among other things he promises to show that the screw is nothing but a wedge used without percussion, which makes its movements by means of a lever, and this is lacking in his book, we shall attempt to show this and, moreover, to reduce the screw to the lever and the balance in order that ultimately we shall understand it completely.</s>

Line 2031

<s id="id.1.6.1.0.1.0">PROPOSITION I</s>

</figure>

<s id="id.1.6.1.1.1.0">If the wedge is adapted in the following manner to the cylinder, it is precisely a screw which has two worms joined together at one point.</s>

Line 2040

<s id="id.1.6.1.2.1.0">Corollary.</s>

<s id="id.1.6.1.2.2.0">ÑFrom this it is evident how one may describe the worms on the screw.</s>

</figure>

<s id="id.1.6.1.3.1.0">We shall now show how weights are moved on the worms of the screw .</s>

</figure>

<pb/>

<s id="id.1.6.1.4.1.0">Now if to the screw in the next diagram below is applied the gear C with twisted teeth, as Pappus shows in the same eighth book, or even with straight teeth made in such a manner as to fit the screw, it is evident that, with the movement of the screw, the gear C will also turn, and the teeth of the gear will move on the worms on the screw; and this is called the perpetual screw, because both the screw and the wheel will go on turning in the same way.</s>

<s id="id.1.6.2.0.1.0">PROPOSITION II</s>

</figure>

<s id="id.1.6.2.1.1.0">Let there be the screw AB with the worm CDEFC; I say that this is nothing but an inclined plane wound around a cylinder .</s>

Line 2069

<s id="id.1.6.2.4.2.0">Therefore he has omitted the proposition of Pappus which he cites here.</s>

<s id="id.1.6.2.4.3.0">But this is so admirably relevant to the explanation of what he says here that I have judged it advisable to add this.</s>

<pb/>

<s id="id.1.6.2.5.1.0">The Problem of Pappus of Alexandria in the Eighth Book of His Mechanical Collections</s>

</figure>

<s id="id.1.6.2.6.1.0">A given force is needed to draw a given weight along a horizontal plane.</s>

<s id="id.1.6.2.6.2.0">It is required to find the force needed to draw the weight up another plane inclined at a given angle to the horizontal plane .</s>

Line 2105

<s id="id.1.6.2.10.1.0">First, what makes the weight easy to move and what especially belongs to the nature of the screw is the worm; for, if around a given screw AB there should be two unequal worms CDA and EFG, I say that the same weight is moved more easily on CDA than on EFG .</s>

</figure>

<pb/>

<s id="id.1.6.2.11.1.0">The other reason for which weights are easily moved consists in the [length of] the radius or handles by which the screw is turned.</s>

</figure>

<s id="id.1.6.2.12.1.0">Corollary.</s>

<s id="id.1.6.2.12.2.0">ÑFrom these things it is evident that the more turns there are to the worm, and the longer the rods or handles, the more easily, but more slowly, the weight is moved.</s>

Line 2139

<s id="id.1.6.2.16.1.0">Anyone will be able then to construct machines and compound several together, such as pulleys and windlasses, or many gears, or in various other ways, and from what we have said one may easily find the relation between the weight and the power.</s>

<pb/>

<s id="id.1.6.2.17.1.0">Comment by Pigafetta</s>

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