4, 15, a — 1 and 1/15, d — 4/15, f. Here the 15 is established in the denomination of fifteenths, of which fifteenths one was taken and placed beside the friction of the pole at d, and multiplied by 4, whereby the weight of its body was found, of which we make the counterweight to the friction of the whole load that sustains the pole.

4; 16 — 1; 1/16, b — 4/16, a.

21. Here I find one more, because in a there is 4/16, and in b, where there ought to be 16/16, there is 17/16. Weight 68/16; friction 17/16. Another way. Multiply the arm that sustains the counterweight by 4 as many times as it contains within itself the arm of the friction, and divide the sum by that arm of the counterweight, and what results will be the true counterweight of the friction of the pole, of which its quarter is the friction that the counterweight sends to the pole. Multiply the arm that sustains the counterweight by its opposite weight as many times as it contains within itself its opposite arm, and divide the sum by the weight, and what results will be counterweight to the said weight.

4, b, 4, a — c, 4/4, 1/4 — 4/4. Re-proof against the opponent. And if I did not take the one from the 4 at a, as the opponent wishes, the first proof would answer and confirm it a good rule; but the second rule re-proves it false, as will be shown. The first rule requires that the whole friction answer to the fourth part of the rubbed weight, which here meets the need, since the whole weight of the pole and of the counterweight is 20/4 and the friction of the pole and counterweight is 5/4, that is a quarter of 20, and in this case it stands well. But in the second proof the counterweight of the friction of the pole and of the friction of itself does not answer to the rule, because that counterweight ought to be equal in quantity to the friction of the pole and of that counterweight, the arms of the counterweight and of the friction being equal among themselves; which here the said rule does not answer, since the counterweight is 4/4 and the friction is 5/4, and 5 was never equal to 4. And thus the opponent shows himself not to have been right.

4, a, b, 3 — c, 1, m; 1/3, n — 4/3, f. If from the 4 into which the arm of the counterweight is divided one is taken, and this one is set for the friction of the counterweight that is sought — which counterweight will arise from that one multiplied by 4 — you will find that this counterweight will always be in power equal to the friction of the weight of the pole and of the weight of itself. And the 4, whence that one was taken, has remained 3, which 3 is the denominator of all the parts of the weight and of its friction, because these parts are all named thirds. Therefore every whole is resolved into thirds, and I have 4 thirds of counterweight against 4/3 of friction, and I have 4/3 of friction for the quarter of 16/3 of weight; these 2 rules are found to work, and whatever is in the arms of the said powers being equal, the powers are equal, above all those of weight and friction.