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.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.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.Take the points D and L toward F, and join LC, LS, DC, and DS, and extend the lines LS, DS, KS, and HS to the circumference of the circle at E, M, N, and O, and join CE, CM, CN, and CO.Now since LE and DM come together at S, the straight angle LSE will be equal to the straight angle DSM.And as is LS to DS, so will SM be to SE; but LS is greater than DS, and SM than SE.Therefore LS and SE taken together will be greater than DS and SM, and for the same reason KN will be shown to be less than DM.Moreover, since the straight angle OSH is equal to KSN, by the same reason HO will be greater than KN.And in the same way KN will be shown to be less than any other line passing through S.And since of the two isosceles triangles CLE and DCM the sides LC and CE are equal to the sides DC and CM, and the base LE is greater than DM, the angle LCE will be greater than the angle DCM.Whence the base angles CLE and CEL taken together will be less than the angles CDM and CMD half the sum, that is, the angle CLS, will be less than the angle CDS.Therefore the weight at L will weigh more on the line LC than that at D will on DC, and will be more supported by the center C at L than at D.Similarly it will be shown that the weight at D will be more supported by the center C than at K.Therefore the weight at K will be heavier than at D, and at D than at L; and for the same reason, since KN is less than HO, the angle CKS will be greater than the angle CHS.Whereby the weight at H will be more supported by the center C than at K, and in this manner it will be shown that, wherever the weight is along the circumference FDC, it will be less supported by the center at K than if placed at any other point, and the closer it is to F or to C, the more it will be supported.Then since the angle CKS is greater than CDS, and CDK is equal to CKH, the remainder SKH will be less than the remainder SDK; whereby the circumference KH will be closer to the straight natural movement of the free weight at K, that is, to the line KS, than the circumference DK to the movement DS.Hence the line CD offers more resistance to the weight at D than CK does to the weight at K, and for this reason the [mixed] angle SHC will be shown to be greater than SKH, and consequently the line CH offers more resistance to the weight at H than CK to the weight at K.Similarly it would be shown that the line CL sustains the weight more than CD, and for the same reasons it will be proved that the weight at K will weigh less on the line CK than at any other place along the circumference FDC; and the closer it is to F or to C, the less it will weigh.Therefore it will be heavier at K than at any other place, and it will be less heavy the closer it is to F or to G.
Finally if the center C is the center of the world, it is manifest that the weight placed anywhere [on the circumference] will remain fixed Thus [if the weight is] placed at D, the line CD will sustain the whole weight, being vertical to that weight at D.Therefore the weight will remain at rest.
Now in the things demonstrated thus far, we have made no mention of the weight of the arms of the balance.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).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.
17[Figure 17]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.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.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.
18[Figure 18]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.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.Therefore the natural straight descents of any free weight cannot be made by parallel lines, since all the lines meet in the center of the world.These men assume next that the motion of a weight from D to A along a straight line toward the center of the world is the same quantity as that from O to C, as if the points A and C were equally distant from the center of the world, which is likewise false---.Thus the assumption from which they demonstrate that the balance DE returns to AB turns out to be false, and all their demonstrations fall.Of course they might say that, because of our great distance from the center of the world, these differences are imperceptible, and by reason of that fact may be assumed void, as all those who have treated these matters have assumed, especially since their being imperceptible does not alter the fact that the descent of the weight from L to D partakes more (to use their own phrase) of the straight than the descent DA.Likewise the arc DA will partake less of the straight than the circumference EV, whence [they may say] the supposition will be true, and the other demonstrations will retain their strength.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.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.
19[Figure 19]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.If therefore the weight shall move from L to A along the circumference LA, its straight movement will be measured by the line LX.But if it moves g from A to M along the circumference AM, its straight movement will be measured by the line XM.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.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.