The centre of gravity of weights on a beam, by rule
Dividing loads among cords and the laws of equal loading
Across these two folios Leonardo works out, 'always by rule,' the centre of gravity of weights suspended from cords on a beam, dividing the space between the cords in proportion to the number of weights. He gives worked cases: weights of 3 and 5 sharing eight loads, and a centre b that sustains 4 against a centre a that sustains 5, so that dividing the span a b into 9 parts locates the centre of gravity of the three weights (a 'centre of 9'). He also states the law that an equal number of cords at equal distances from the centre of the weight carry equal loads. Both sheets are crowded with beam-and-weight diagrams, numbers such as 12 9 12, and further mirror-script notes along the margins.
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Find the centre of the weights by rule
Find the centre of the weights always by rule: since a b are 3 equal parts, divide the eight weights into 3, giving 24/8, and do likewise with its eight spaces above.
Centre b sustains 4, centre a sustains 5
b is the centre between 2 and 2, a centre that sustains 4; a is the centre of 3 and 2, that is 5. So a has 5 and b has 4, making 9; dividing the span a b into 9 parts and taking the centre of the spaces 5 against 4 locates the centre of gravity of the three weights, the 'centre of 9.'
Equal cords at equal distances give equal loads
An equal number of cords at equal distance from the centre of the sustained weight causes those cords to be loaded with equal weight; if the centre of the weight falls equally between the cords, all their loads are equal, provided the cords are equal in number as in distance.
Dividing the space between cords by the weights
Working the case s c divided into 5 with weights 3 and 5 (making 8), Leonardo divides the intercord space by the number of weights, draws the line a n, marks the space m a, counts 6 such spaces in m f, and reduces the shares (8/6, then 5 times 6) back to whole numbers.
