Pulleys and the Windlass: Cord Gathered in Lifting
How the number of pulleys multiplies the length of cord a windlass must take up
The note works through a block-and-tackle problem: with two pulleys, raising the weight one braccio makes the windlass gather up two braccia of cord. Leonardo labels points n, m, f and t and reasons that the cord n-m-f, being two braccia long, must all leave its place and be taken up as the weight rises. He concludes that the greater the number of wheels turning in the tackle-blocks, the faster the first stretch of cord runs compared with the last. A small pulley diagram is drawn in the lower-left margin.
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Two pulleys double the cord the windlass gathers
With two pulleys, to raise the weight m by one braccio the windlass must gather up two braccia of cord. Leonardo lets n m equal one braccio and m f another, so the cord n m f measures two braccia; as the weight rises this whole length leaves its place and is taken up by the windlass. He adds that the more wheels turn in the tackle-blocks, the faster the first stretch of cord moves than the last.
Block-and-tackle diagram in the lower-left margin
A small drawing at the lower left shows a vertical block-and-tackle: an upper fixed point marked t, with pulleys strung on a cord down to a lower block. It illustrates the windlass-and-pulley arrangement reasoned about in the note.