Distributing a load between the cords of a suspended beam
Two weights of 7 and 5 balanced on a beam hung by three cords, worked out in fractions
The main text works a statics problem: a beam a, f carries two unequal weights (7 and 5) and is suspended by three cords, and Leonardo divides the occupied length into twelve equal parts to locate the common center of the weights before apportioning the total load among the cords, arriving at shares of 2 18/21 and 6 6/21. A brief opening note sets a proportional division ('divide the 12 at n into the 24 of the proportions'). Both facing pages are covered with schematic beam-and-hanging-weight diagrams and columns of numbers and fractions, most of them left untranscribed in the source.
On this page
Balancing two weights on a beam hung by three cords
Two unequal weights, 7 and 5 (making 12), hang from the ends of the beam a, f, which is suspended at three points chosen at random. Leonardo divides the length taken up by the three cords into twelve equal parts and, placing 5 oppositely against 7, locates the common center of the two weights.
Apportioning the load among the cords in twenty-firsts
Taking the proportions of the three cords as twenty-one, he divides so that each part is 12/21, then sums to find that two of the cords carry shares of 2 18/21 and 6 6/21, distributing the weight exactly.
Schematic beam-and-weight diagrams across the opening
Rows of horizontal beams with weights hung from vertical cords, annotated with fractions and whole numbers, are sketched across both pages to illustrate the load-sharing calculations; these figures are captioned but not transcribed.