Loads on Cords and Beams; Dividing Numbers by Proportion
Statics of two cords carrying a loaded beam, worked through ratios and fractions
Two leaves work out the statics of a beam hung from cords and loaded with weights. On 106v Leonardo reasons about two cords, r and s, carrying weights of 8 and 4 at points a and n, showing how shifting a single unit throws the whole load onto one cord, and states that the cords stay equally loaded when their positions match the beam's spans or the opposing weights. On 105r the problem turns arithmetical: numbers such as 5 and 7 are divided into given proportions and reduced to fractions like 5/7, 3/5 and 7/6. Bracket-and-weight diagrams and small tables of numbers fill both pages.
On this page
Shifting one unit throws the load onto cord r or s
If the 8 at a were reduced to 7 while the 4 stayed at n, cord r would carry nothing and cord s the whole 12. Adding or removing a single unit at a or at n moves the entire load from one cord to the other.
Two cords stay equally loaded when in proportion
When the two cords holding the beam are in proportion with the beam's spans, or with the opposing weights, they remain equally loaded however different the weights and their distances. Break that proportion and one cord must carry both weights.
Loads on cords a and b follow the ratio a-d : d-b
The ratio that the segment a-d bears to d-b is the same ratio held by the opposing weights loaded onto the two cords a and b.
Dividing 5 into the proportions 3 and 4
Five is divided into the two proportions a and n, namely 3 and 4; taken together as 7, each part becomes 5/7. The same sharing-out is applied to each of the two weights in turn.
Division of 5 into fifteen parts of 3/5
A small table (6 1, 4 3, 5 2) sums to 15: five split into fifteen parts makes each part 3/5.
Dividing 7 into six proportions
Seven is divided into six proportions — 2 and 3 make 5, and 1 makes six — so that each part is 7/6, before seven is restored to its whole.