Two Suspended Rods and the Division of a Load
A hung weight splits among its supports in inverse proportion to distance
Two horizontal bars are drawn, each with cords tying weights to lettered points (f g above, a o n m below) and numeric loads set beside them. The note works out how a hung weight distributes among its supports: because g p is a subdouble of f g, the weight 6 splits as 2 pounds at f and 4 at p; because o m is a subsesquialter of a o, the weight 5 gives 2 and 3; and because n m is a subquadruple of a n, a load of 2 at a becomes 8 at m. A caption marks the arrangement as 'tested'.
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How a suspended weight divides among its supports
A weight of 6 hung between supports f and p splits as 2 pounds at f and 4 at p, following the spans g p and f g. The reasoning balances each bar by matching load to the inverse of the distance from the fulcrum.
Subdouble, subsesquialter and subquadruple ratios
Because o m is a subsesquialter of a o, the weight 5 divides as 2 and 3; and because n m is a subquadruple of a n, a load of 2 at a wants to be 8 at m. These are set against the opposing arm so the balance stands even.