Dividing a beam's weight among added cords
Middle cords take a sub-double share; four cords double the first two (12; 6 6; 4 4 4)
Three pairs of connected beams are drawn, marked 12, then 6 6, then 4 4 4. A beam hung only by its two ends divides its weight equally between them. Adding a single cord in the middle, being sub-double to the two, removes from each a sub-double share; adding four cords, double the first two, doubles the weight taken from each end cord. Cords equidistant from the beam's centre all keep this proportion.
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A beam hung by its ends splits its weight equally
Because the middle beam is suspended by its ends at the ends of the balance, it divides its weight in equal parts to the two cords from which it hangs. This equal split is the starting case before further cords are added.
Added middle cords relieve the end cords in proportion
A single cord added between the two, being sub-double to them, takes from each a sub-double portion of what remains to each. Place four cords, double the first two, and the weight lacking to each of the two end cords is doubled; cords equidistant from the beam's centre all observe this proportion (12; 6 6; 4 4 4).