Load shared between two cords on a suspended rod
Cord m carries 9, cord o carries 3, reasoned from the beam's spaces
A pair of connected, suspended rods marked f, m, n, o is analysed to find how their load divides between two cords. Leonardo reasons that the 4 spaces projecting beyond m must balance 4 inside m n, and that of the 8 spaces between the two cords each cord should carry 4; but the 4 of f m take 2 from o m, so the 8 become 6, of which 3 fall to o and the remaining 9 to cord m. He concludes 'e così sta bene' (and so it stands well). A ruled rod is drawn across the top of the leaf.
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Counting the beam's spaces to balance cords m and o
The 4 spaces projecting beyond m are set to balance 4 inside m n, and of the 8 spaces between the two cords each cord should support 4. But the 4 of f m remove 2 from o m, so the 8 of o m become 6.
Cord m carries 9, cord o carries 3
Of the reduced 6 at o m, 3 fall to o and the rest to cord m, giving 9. By this reckoning the load of the paired rods (9 and 3) is fully accounted for, and Leonardo notes it stands well.