Dividing a load among a rod's suspension points by proportion
Summing the section-ratios of a loaded horizontal rod to 20 (doubled to 40)
A dense calculation for distributing a load over a simple horizontal rod ('asta orizzontale') held at several points a, m, n, o, p. Leonardo sums the pairwise ratios of the rod's segments — 3 against 4, 2 against 3, 1 against 4, 1 against 2 — to reach a total of 20, which he doubles to 40 so that the load of 8 can be divided without fractions. The right margin repeats the four ratios (a-p 7, a-o 5, n-o 3, p-n 5), which again total 20. A short ruled rod is drawn across the top of the leaf.
On this page
Load of 8 distributed over a rod held at a, m, n, o, p
The whole sum of the proportions is taken as 20: a m against m p is 3 to 4 (7), a m and m o is 2 to 3 (5, making 12), n m against m p is 1 to 4 (5, making 17), and n m and n o is 1 to 2 (3, making 20). Because dividing 8 by 20 gives fractions, 20 is doubled to 40; dividing gives 5, so each fifth is one whole and each point receives its due share of the load.
Marginal table of the four ratios summing to 20
The right margin lists the four proportions of the weights: a-p 7, a-o 5, n-o 3, p-n 5. These add to 20, from which 40 is made — the number of all the proportions of the said weights.
Ruled horizontal rod at the head of the page
A short horizontal bar divided into equal cells is drawn across the top of the leaf, the diagram to which the segment-ratios a m, m p, m o and n m refer.