The Triangular Balance: Finding Its Centre
Distributing three weights along a balance and a note on halving a pyramid
The verso develops Leonardo's 'triangular balance', working out where its centre falls when three weights (a greater, a middle and a least) hang from its arms. He combines and divides the weights into fractional parts (fifths, sevenths, and divisions into 8 and 56) so that the loads balance reciprocally with the lengths of the arms. A separate top note asks to cut a pyramid transversely into two equal parts, and columns of figures at the foot record the computations.
On this page
Halving a pyramid across
A short note beside suspended triangles instructs: cut me a pyramid into 2 equal parts across (transversely). It poses the problem of dividing a solid into two equal volumes by a single cut.
Combining greater and middle weight
Add the greater and middle weight, 5 and 2, to make 7; divide by that 7 the space between the middle and least weight, and divide the greater weight of 5 into sevenths, giving 2/7 to b c and 5/7 to a b. Applying the weights reciprocally to the spaces and adding the upper weights 2 and 1, one finds 32/7 against 24/7, so the balance centre falls between 32 and 24 when the interval is divided into 56.
Two cords made to feel equal
The cord a holds 2 pounds and the cord b holds 3. Leonardo asks where along the beam to add a further weight of 7 so that a, with its first 2, feels as much as b with its 3, together with their first weights.
Dividing the middle weight into parts
Divide the middle weight into as many parts as the pounds suspended (8), so each unit is split into 4; likewise the 5 becomes 20 parts and the one becomes 4. Take from the 8 as many parts as the pounds of the greater weight, set them against it, and the remainder against the lesser weight, then find the proportional spaces and drop the perpendiculars of the triangular balance upon them.