Center of Gravity of a Pyramid Suspended from a Balance
The five-ninths toward the base balances the remainder; natural versus accidental center
A balance beam labelled K, x, h and r at the top of the sheet suspends a triangular 'pyramid' whose center of gravity Leonardo locates: the third part, or rather the five-ninths, measured toward the base stands in equilibrium against the weight of the whole remainder. He proves it from an earlier proposition that the middle of any weight falls beneath the middle of its support, so the pyramid's center must hang perpendicular below the beam's middle. A second note relates the natural and 'accidental' centers of gravity, arguing that the nearer the accidental center lies to the base rather than the apex, the nearer the natural center lies to it than to the mid-length.
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Five-ninths toward the base balances the remainder of the pyramid
The third part, or five-ninths, of any pyramid measured toward its base rests at equal height in equilibrium against the weight of the whole joined remainder. The beam is labelled K x h r with the apparatus running to point o and to m, n at the pyramid.
Proof: the middle of a weight falls below the middle of its support
By the fifth proposition, the middle of each weight falls below the center of its support; since the balance is joined to the suspended pyramid, the middle of the pyramid's weight must hang perpendicular below the middle of the sustaining beam.
Natural versus accidental center of gravity along the pyramid
The nearer the accidental center of gravity lies to the base of the transverse pyramid rather than to its apex, the nearer the natural center of gravity lies to that accidental center than does the center of the pyramid's length.