4 1 1/2 1 1/2, 49, n a s, c f b, 1 2 3, 2/3

Rule

4 5/6 4 1/6, f

a b c d e, 1 1/2 1 1/2, 6, 1/2

Say first, in n a against a b n: one times 3 makes 3, and put 3 under a. Then say, in n a: one times one makes one, and put it under b, which represents the one that is between n a, that is the middle of the simple rod n s. Now because the simple rod itself is 2, and the multiplication gave 3 under the a, you will say: if 3 gives me 2 in n s, what will one under b give me? Say: 2 times one makes 2, and 2 into 3 is 2/3, and likewise one leaves 2/3 of one; and because b first sustained 1 1/2 of the 3 braccia that are between b n, before the simple rod n s was added to it, subtract 2/3 (which lightens the simple rod) from 1 and 1/2, and there remains 5/6, and so much does b sustain; and the remainder n sustains, which is 3 and 2/3. Then return to f, which simply sustains the 4 braccia of c b.

Wh… [the text breaks off]

This is drawn from experience. Divide the 4 braccia into 8 equal parts and take half of the braccio that projects outward. One times 6 1/2 makes 6/2, and put it below. Then say: one times 1/2 makes 1/2, and put it under a. Then say: if 6 halves return me one whole, which is between d c, what will that whole return me, which I set down as one half under a? But it is better to say, in exchange for 6 halves, 3 wholes, and to say: if 3 gives me one, which is the braccio of the projection divided in 2, then say: if 3 gives me one, what will one whole give me, which is under a? It will give 1/3; make sixths of it -4- [continues on the left:]

-4- and likewise do with the 1 and 1/2 that is under a, which you will make 9/6. Take away 1/6, and there remain 8/6, which are 1 1/2, and it is correct.

1 1/2 that is 9/6, 1 1/2, p a n, 1 3

Say better: one times 3 makes 3, for a n against a p. Then say for a n: one times one makes one, and put one under p and 3 under a. Then say: if 3 becomes one in a n, what will it become in p 1? It will become 1/3. Subtract a third from 1 1/2, and you then say … /3 […]