The heavy body which descends by oblique motion will become the slower and lighter the nearer it draws to the centre of the world.

Let a d be the obliquity of the descent made by the movable body; let c be the heavy body that moves along this obliquity; let b be the centre of the world. I say that the more this heavy body moves along the obliquity a d, the slower it will become, and consequently the lighter, the nearer it approaches the centre of the world b. This proposition is sophistical, because such a line in itself is not obliquely situated; rather it is set in the situation of equality, since its extremes a and d are equally distant from the centre b. Now it is manifest to be true what was proposed by the 3rd of the 2nd, which says that that thing is higher which is more distant from the centre of the world; and, by the converse of that 3rd, that is lower which is nearer to that centre. Therefore it is concluded to be true what has been said, because a d is high and goes descending toward c and rising again from c to d, as is shown by the lines a b and c b being shorter one than the other, and the third line d b being equal to a b.

That heavy body shows itself greater and moves with more velocity and by a longer path which descends along a less oblique line.

It is proved. The least oblique line will be a e, which passes through the centre of the world f. More oblique, one than another successively, are the lines a d, a c and a b. The weight r, which descends along the line a r, finally stops at the centre of the world in f. The weight o, which descends along the oblique a o, stops at the right angle made by the central line of the weight with the line of its motion. Therefore the motion a o is shorter than a f, by one of the Elements which says that, of the lines drawn in the circle from the end of the diameter, none is longer than the diameter. There follows the third motion, shorter than the second, which is a m, that stops at a right angle, and it has moved the space a m. And by the same rule is made the 4th motion a n by the weight n, shorter than the 3rd, being made along a line of greater obliquity.

On percussion in itself and on its mover.

The centre of gravity of every square will be at the intersection of its diagonals; and if such a square has its opposite sides unequal, that centre of gravity will be shifted, that is, as near to the greater side of the square as the said intersection was near to the lesser side. As if the square were a b c d, and e were the intersection of the diagonals a d and b c, which intersection is nearer to the lesser side a b than to the greater side c d. Now you will take with the compasses the space r e, and carry it over into f t. I say then that the centre of gravity of such a body will be found in f.

To work with whole numbers in these proportions, it is always necessary to take the numbers of the weights equal to the numbers into which the arms of the balance are divided, where you wish to suspend them, and to divide the proportions of the weights and of the arms equally; and then exchange the proportions, and thus you will do for each weight, and it comes out well.

That the proof above is correct, calculate and you will see that the centre of gravity of 16 and the centre of gravity of 10 is at o,1; and the arm of the balance s p is 5, and the opposite arm s o is 8; therefore the lesser arm is 5/8 of the greater arm, and the opposite weights must do the like, so that the lesser must be 5/8 of the greater. Which here does not depart from the rule, because the 10 of the greater arm is 5/8 of the 16 attached to the lesser arm.

In this second demonstration I find the first arms of the balance o f and o r to be 5 degrees against 16 degrees, that is, the lesser arm is 5/16 of the greater arm; for which reason the opposite weights are in the same proportion, because on the lesser arm is 32 and on the greater arm is 10, since one is 5/16 of the other. Then it follows that the balance changes the centre o into n; and I wish to know how it must be equalized, because the proportion of the arms stands 2 against 6, and the weights are 42 against 10. Do it with the rule of three and say: if 2 of the lesser arm wants 6, against the greater arm which holds 6, what will 42 want? Prove it and you will see that it wants 14; so add 4 above the 10 of r, and it stands right.