1First, 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.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.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.The same is true of their other argument, adduced from movements being straighter or more bent.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.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.
12[Figure 12]Draw FG to the center of the world, S, and from S draw also a line tangent to the circle AFBG.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.Still less can it touch the circle at A in the quadrant AF, for it would cut the circle.Therefore it will touch below, and let this line be SO; then add the lines SD and SL, which cut the circumference AOG at points K and H; and join also CK and CH.And thus the closer the weight is to F, the higher it stands above the center, as the weight at D presses more on and stands higher above the turning point C as center; that is, the weight at D weighs down more on the line CD than it would at A on the line CA, and still more at L on the line CL.For, the three angles of each triangle being equal to two right angles, and the angle DCK of the isosceles triangle DCK being less than the angle LCH of the isosceles triangle LCH, the angles CDK and CKD taken together are greater than CLH plus CHL; and the half of this, that is, CDS, will be greater than CLS.And CLS being lesser, the line CL approaches more closely to the natural movement of the weight placed at L when entirely free, that is to say, to the line LS, than CD to the movement DS.For the weight placed at L would move freely toward the center of the world along LS, and the weight at D along DS.But since the weight at L weighs wholly on LS, and that at D on DS, the weight at L will weigh more on the line CL than that at D on DC.Therefore the line CL will more sustain the weight than the line CD; and in the same way, the closer the weight is to F, it will be shown for this reason to be more sustained by the line CL, since the angle CLS is always less, which is obvious.For if the lines CL and LS should come together, which would happen at FCS, then the line CF would sustain the whole weight that is at F and would render it motionless, nor would it have any tendency to descend [gravezza] along the whole circumference of the circle.Therefore the same weight, by diversity of position, will be heavier or lighter, and this not because by reason of its place it sometimes truly acquires greater heaviness and sometimes loses it, being always of the same heaviness wherever it is, but because it presses [grava] more or less On the circumference, as at D it presses more on the circumference DA than at L on the circumference LD.That is, if the weight shall be sustained [jointly] by the circumferences and the straight lines, the circumference AD will more sustain the weight placed at D than the circumference DL sustains the weight placed at L, for CD helps less than CL.Besides this, if the weight at L were completely free, it would move down along LS were it not prevented by the line CL, which forces the weight at L to move beyond the line LS along the circumference LD and in a certain sense pushes it, and, in pushing it, comes partly to sustain it; for, if it did not sustain it and give it resistance, the weight would move down along the line LS, rather than along the circumference LD.Similarly CD offers resistance to the weight placed at D, forcing it to move along the circumference DA.In the same way, the weight being at A, the line CA will constrain it to move outside the line AS along the circumference AO, for the angle CAS is acute, ACS being a right angle.Therefore the lines CA and CD to some degree, though not equally, offer resistance to the weight, and whenever the angle at the circumference of the circle made by the line coming from the center of the world S and that from the center C shall be acute, we shall prove the same thing to occur.Now since the mixed angle CLD is equal to the angle CDA, being contained by radii and the same circumference, and the angle CLS is less than the angle CDS, the-remainder SLD will be greater than the remainder SDA.Hence the circumference DA, which is the path of descent of the weight at D, is closer to the natural movement of the free weight at D (that is, the line DS) than the circumference LD is to the line LS.Therefore the line CD will offer less resistance to the weight placed at D than the line CL to the weight placed at L.So the line CD will sustain less than CL, and the weight will be more free at D than at L, being moved more naturally along DA than along LD.Whence it will be heavier at D than at L.Similarly we shall demonstrate that CA sustains less than CD, and that the weight at A is more free and heavier than at D.Next, in the lower part, for the same reasons, the closer the weight is to G, the more it will be retained, as at H by the line CH than at K by the line CK; for, the angle CHS being greater than the angle CKS, the lines CH and HS approach closer to the [line of] direction than CK and KS, and hence the weight will be more retained by CH than by CK; for, if CH and HS meet in a line, as happens when the weight is at G, then the line CG would sustain the whole weight at G, so that it would remain motionless.Therefore the smaller the angle contained between the line CH and the line of the weight in free fall (that is, between CH and HS), the less the line CH will retain the weight; and where it is less retained, it will be freer and heavier.Besides which, if the weight were free at K, it would move along the line KS; but it is impeded by the line CK, which forces the weight to move from the line KS along the circumference KH.This restricts it in a certain way and thus comes to sustain it; for, if it were not sustained, it would move along the straight line KS and not along the circumference KH.Similarly CH retains the weight, constraining it to move along the circumference HG.And since the angle CHS is greater than the angle CKS, if we take away the equal angles CHG and CKH, the remainder SHG will be greater than the remainder SKH.Hence the circumference KH (that is, the descent of the weight placed at K) will be closer to the natural movement of the free weight placed at K (that is, to the line KS) than the circumference HG is to the line HS.Hence the line CK retains less than CH, the weight moving more naturally by KH than by HG.With similar reasons it will also be shown that the smaller the angle SKH, the less the line CK will sustain.The weight therefore being at O, since the angle SOC not only is less than the angle CKS but is the least of all angles that come from the points C and S and have their apex on the circumference OKG, the angle SOK will be less than the angle SKH and less than the others so formed.Hence the descent of the weight placed at O will be closer to the natural movement of a free weight at O, than if placed at any other position on the circumference OKG, and the line CO will sustain the weight less than if it were at any other place on the circumference OG.Likewise, since the angle of contact SOK is less than the [mixed] angle SDA, or SAO, or any other such angle, the descent of the weight placed at O will be closer to the natural movement of this weight than if placed at any other site along the circumference ODF.Besides, the line CO cannot push the motion of the weight placed at O when it moves down so that it will move outside the line OS, as the line OS does not cut the circle, but touches it, and the angle SOC is a right angle and not acute; hence the weight placed at O will never weigh against the line CO, nor will it bear upon the center, as would happen at any other point above O.Hence the weight placed at O will for this reason be free, and more completely so at this point than at any other in the circumference FOG; and thus it will be heavier and will bear down more here than elsewhere.And the closer it is to O, the heavier it will be than anywhere farther away.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.All of which was to be shown.
13[Figure 13]But if the balance arm were greater than CO, say, by the amount CD, the weight placed at O would likewise be heavier.Describe the circle OH, with center D and radius DO.The circle OH will touch the circle FOG at the point O and will also touch the line OS at that same point, this being the straight and natural descent of the weight placed at O.And since the angle SOH is less than the angle SOG, the descent of the weight placed at O along the circumference OH will be closer to its natural movement OS than would that along the circumference OG.Hence the weight at O will be freer, and consequently heavier, than at C (the center of the balance being at D).Similarly it will be shown that the longer the arm DO, the heavier will be the weight placed at O.
14[Figure 14]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.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.Then, the triangles COS and RTS being right triangles, SC will be to CO as CO is to CM.Similarly, SR is to RT as RT is to RN.Now, RT being equal to CO, and SC greater than RS, the ratio of SC to CO will be greater than that of SR to RT; whence, likewise, CO has a greater ratio to CM than RT to RN.Thus CM will be less than RN.Then cut RN at P so that RP shall equal CM, and from the point P draw the line PQ parallel to the lines MO and NT, [such that] it will cut the circumference A T at Q; and finally join R and O.Now since CO and CM are equal respectively to RQ and RP, and the angle CMO is equal to the angle RPQ, the angle MCO is equal to the angle PRQ.But the angle MCA is equal to the angle PRA, both being right angles; hence the remainder OCA is equal to the remainder QRA, and the circumference OA is likewise equal to the circumference QA.Thus the point T, being farther from the point A than is Q, will also be farther from the point A than is the point O.Likewise it may be shown that, the closer the circle is to the center of the world, the farther T will be from A.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.
15[Figure 15]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.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.
Now if the weight at K is compared with that at D, the weight at K will be heavier than that at D; for the line DG being drawn, and the three angles of any triangle being equal to two right angles, and the angle DCG of the triangle CDG being greater than the angle KCG of the isosceles triangle CKG, the base angles DGC and GDC taken together will be less than the angles KGC and GKC taken together; and half the sum, that is, the angle CDG, will be less than the angle CKG.Now since the weight at K would move in natural freedom along KG, and the weight at D along DG-these being the lines by which they are brought to the center of the world-the line CD, that is, the balance arm, will approach more nearly to the natural movement of a free weight at D, that is, to the line GD, than CK to the movement made along KG.Therefore the line CD will offer more support than CK, and therefore the weight at K will be heavier, by what has been said, than at D.Besides which, if the weight placed at K were entirely free, it would move down along KG if it were not impeded by the line CK which forces the weight to move beyond the line KG along the circumference KH; the line KG will sustain the weight in part, and will make resistance to it, forcing it to move along the circumference KH.And since the angle CDG is less than the angle CKG, and the angle CDK is equal to the angle CKH, the remaining angle GDK will be greater than the remaining angle GKH.Therefore the circumference KH will be closer to the natural free movement of the weight placed at K, that is, to the line KC, than the circumference DK to the line DG.Whereby the line CD makes more resistance to the weight placed at D than the line CK to the weight placed at K.Therefore the weight placed at K will be heavier than at D.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.
16[Figure 16]