Finding the center of gravity of a triangular pyramid
Divide the triangle by a perpendicular and weigh the halves on a balance to locate its center S
Leonardo sets out to locate the center of gravity of a triangular pyramid, which he states lies one third of its length toward the base. When the base is not perpendicular he drops a perpendicular over the innermost angle to split the figure into two triangles, each with a perpendicular base, then finds the middle of each weight and sets them against one another on a balance. Because triangle a b d is double b c d in base, it is double in weight, so the balance's center falls over the pyramid's universal center of gravity at S. A large, untranscribed apparatus with a reel and a hatched conical base is drawn down the right side of the sheet.
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Center of gravity lies one third toward the base
The center of gravity of every pyramid is at one third of its length toward the base. This principle governs the constructions that follow.
Splitting a triangle with a non-perpendicular base
When the base of the lying pyramid is not perpendicular, a perpendicular dropped over the innermost angle divides the triangle into two, each of which then has a perpendicular base at their contact.
Weighing the two half-triangles on a balance
To find the center of gravity of triangle a c d, drop the perpendicular b d on angle d and split it into a b d and b d c. Since base a b is double b c, triangle a b d weighs double, so on the balance t f is double t r, and the balance's center sits over the universal center of gravity at S.
Untranscribed standing apparatus
Down the right side is a tall standing device: a vertical shaft carrying a reel or spool, a scoop-like element, and a broad hatched conical base enclosing a ring. No transcription accompanies it.