Load shared between two cords supporting a beam
A general rule drawn from experience: the end cord bears 1 1/3 and the inner cord 2 2/3 of four braccia
On folio 167r Leonardo gives a 'general rule drawn from experience' for how a hanging load divides between two cords that support a beam, illustrated by two tick-marked beam diagrams. A worked example divides four braccia into eight parts and uses the half-length of the shorter arm as a multiplier to find each support's counterpoise. He concludes that when two cords hold four braccia, one at the extreme end and the other one braccio inward, the end cord carries 1 1/3 braccia of load and the inner cord 2 2/3. A 'short rule' at the upper left restates the calculation procedure compactly.
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Load divided between an end cord and an inner cord
If two cords support four braccia, one at the extreme end and the other one braccio inward, the cord at the extreme bears one braccio and 1/3 of load while the inner cord bears 2 and 2/3. The counterpoise of the overhanging arm is derived from the half-length b c of the shorter arm b d.
Short rule for the counterpoise calculation
The half of the shorter simple rod multiplies each like part of the greater rod and then multiplies itself; the products are set over a fraction bar, and the resulting total is subtracted from the number that the half of the greater rod weighs by itself.
General rule drawn from experience
Dividing four braccia into eight parts, one times six gives six placed under b, and the overhang squared gives one under e; reducing yields 1/6, which taken from 9/6 leaves 8/6, that is one braccio and 1/3, with the rest of the weight resting on a.