0 4, 1 1, a b c d e, 1 2 2

0 8, 2 2

0 16, 4 4, 42

1 3/4 12 1/4, 4 4, a b c d f n K, 9 6 24, 54/24, 2 and 6/24, 1/4

1 2 4

Rule of weight.

a c is a piece of rod which is 2 braccia, of which the support a bears one and c the other. Now I wish to add 2 more braccia of rod between c and e, and I want to know how much that rod takes away from the weight at a.

Do it in this way: take half of the simple rod which you added at c e, the half being d e, and see how many times c d goes into a c, which is 2 times; whence you will say: one times 2 makes 2, and put the 2 under d. Then say likewise for c d: one times one makes one, and put one under a. Then, because the added simple rod was 2 braccia, you will say: if I put 2 under d, which gives me 2 at c e, what will it return to me under a? Do it by the rule of three and say: 2 times one makes 2, and you have the 2 under d; from that 2 nothing remains, and nothing will remain at a of the weight it first bore.

You will say, in the half simple rod, in f n against a f: 3 times 8 is 24, and put it under n. Then say: 3 times 3 is 9, and put it under a; then say: if 24 returns to me in f K, what will the 9 under a return to me? Say: six times 9 is 54; divide 54 by 24, which gives 2 and 6/24; reduce it, and it comes to 1/4. Now subtract 2 and 1/4 from the weight that was borne at a, from the weight a f, which was 4; subtract 2 and 1/4 from 4, and 1 and 3/4 remain.