Sharing a load among balance cords by the rule of proportion
Dividing an eight-part beam among its suspension cords to find each cord's share of the weight
These calculations distribute the load of a suspended beam among the cords that hold it. Leonardo divides the beam into eight equal parts and shares the spaces between the cords by that number, working the fractions (8/7, 8/6, 8/5) to assign each attachment its portion — for one case 4 4/7 at a and 3 3/7 at o. Another worked example uses a beam whose short arm a b is a subsextuple of c a, apportioning the load so that a carries 2 2/3 and c one and a third. The margins carry ladder-like beam-and-cord diagrams lettered c n r a b with their fractional weights.
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Divide the beam's eight parts among the cords
Always divide the spaces between the two cords by the beam's 8 partitions: with 7 spaces between the cords, each part is named 8/7, and multiplying by the 8 spaces of the beam gives 56/7. Dividing back by the 7 yields 4 4/7 at a and 3 3/7 at o; the other partitions are named 8/6 and 8/5.
Short arm a b as a subsextuple of c a
The weight the smaller arm takes from the greater arm's end stands to the smaller as the smaller middle stands to its greater: a b is a subsextuple of c a, so the load discharged at c a is likewise subsextuple of a b. Apportioning the parts gives a 2 2/3 and c one whole and one third.
Recovering each cord's ounces of load
Were the arm a b detached, the three arms of beam between the cords would give 18 ounces per cord. Attaching a b (a subsextuple of c a) with its 12 ounces lifts 2 ounces from c, so c now feels only 16 ounces — lacking 2 of the first 18 — and the proposed proportion is sufficiently defined.