4, b, a, 3 — 1; 1/3; 4/3. The division of the 4 into which the arm of the friction is divided is made because the weight of the pole too is divided into 4. And I have kept this method so as not to confuse the calculation of the counterweight that is sought for such friction, because here the unit into which the arms of the weights are divided, the unit into which the weights themselves are divided, the unit of the counterweights, and the unit of the frictions are all alike among themselves and of one same nature as regards the number in calculation, which by means of this rule will be easy of investigation. a b is the arm of the counterweight, b i is the arm of the friction, 4 is the number into which the rubbed weight of the pole is divided; and I set this 4 because every heavy body rubbed upon a flat and polished place makes the power of its friction equal to the fourth part of such rubbed weight. When you have established the nature of the number 3 of the arm a h, you will make the one that you took from that a of the nature of the same weights, which will be 1/3, and then you will clothe that one with its due weight, that is with a weight quadruple to it; and this quadruple is the counterweight that is sought, which is the counterweight of the friction of itself and of the friction of the weight of the pole. The arms of the power of the counterweight and of the friction being equal among themselves, equal will be the power of the counterweight and of the friction among themselves. Multiply the arm of the counterweight by 4 as many times as it contains within itself the arm of the friction, and from the sum take one, and with the remainder divide the number of the friction, and what results multiply by 4, and the sum will be the counterweight of the friction of itself and of the pole.

4, b, a, l, 5 — c — 1 and 1/15, d — 4/15, f. Let a b be the arm that sustains the counterweight, which I shall multiply by 4 as many times as it contains within itself its opposite arm of the friction b c. Which arm a b, containing within itself 4 times the arm b e, you will have to multiply by 4 times 4, which means 16. And from the sum take one, which leaves 15, and by that remainder 15 divide the friction of the pole, which is one, at e, and what results is 1/15. And this is the one that I took from the 16, which then remained 15 in a, which one is now named a fifteenth; and this fifteenth gives denomination to all the numbers of the weight, of the counterweight, and of the friction. Then multiply that 1/15 by 4, and the sum of that multiplication will be the counterweight of the friction of itself and of the rubbed pole. The proof that it stands well is that, dividing the friction by the number of the arms of the counterweight, if the result is equal to the counterweight of that friction, the operation has been good. (Turn to the inside, and you will see the proof.)