The heavy body that descends by oblique motion becomes the slower and the lighter, the nearer it comes to the centre of the world.

Let a d be the obliquity of the descent made by the movable; let c be the heavy body that moves along such obliquity; let b be the centre of the world. I say that the more that heavy body moves along the obliquity a d, the slower it becomes, and consequently the lighter, the nearer it approaches the centre of the world b. This proposition is sophistical, because such a line is not obliquely situated in itself; rather it is placed in the site of equality, being with its extremes a d equally distant from the centre b. Now it is manifest that what was proposed is true, by the 3rd of the 2nd, which says that thing to be 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 is 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 a longer path, which descends by a less oblique line.

It is proved. The less oblique line is 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 that descends by the line a r finally stops at the centre of the world in f. The weight o that descends by the oblique a o stops at the right angle that the central line of the weight makes 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. It follows that the third motion is shorter than the second, which is a m, that stops at a right angle and 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 by a line of greater obliquity.

Of percussion in itself and of its mover.

The centre of gravity of every quadrilateral will be at the intersection of its diagonals; and if such a quadrilateral has its opposite sides unequal, that centre of gravity will be shifted, that is, as near to the greater side of the quadrilateral as was the nearness of the said intersection to the lesser side. As if the quadrilateral 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 compass the space r e, and carry it to f t. I say therefore that the centre of gravity of such a body will be found at f.

To manage with whole numbers in these such 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 hang 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, and the arm of the balance s p is 5, and the opposite arm s o is 8; therefore the smaller arm is 5/8 of the larger arm, and the opposite weights must do the same, that the smaller must be 5/8 of the larger. Which here does not go out of the rule, because the 10 of the larger arm is 5/8 of the 16 attached to the smaller 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 smaller arm is 5/16 of the larger arm; for which reason the opposite weights are in the same proportion, because in the smaller arm is 32 and in the larger arm is 10, since one is 5/16 of the other. Then it follows that the balance shifts the centre o to n; and I wish to know how it must be equalized, because the proportion of the arms is 2 against 6, and the weights are 42 against 10. Do it with the rule of three and say: if 2 of the smaller arm wants 6, against the larger 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 is right.