Stockpile Volume (Truncated Pyramid)

V=h3(A1+A2+A1A2)V = \frac{h}{3}\left(A_1 + A_2 + \sqrt{A_1 A_2}\right)

Worked example: 200 m³ pile, 36 m² base and 16 m² top → 7.8947 m high — press Try an example to run it live, then adjust anything.

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Stockpile Volume (Truncated Pyramid) explained

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A stockpile that has been driven over, or a borrow pit dug with a flat floor, is a frustum: a pyramid or cone with the tip cut off. Its volume needs the geometric mean term √(A₁A₂), and leaving it out — just averaging the two areas — always overstates the pile. This is one of the oldest results in mathematics: problem 14 of the Moscow Mathematical Papyrus, written around 1850 BCE, works the volume of a truncated square pyramid with a 4-cubit base, 2-cubit top and 6-cubit height and gets 56 cubic cubits, exactly what this formula gives. Whoever wrote it had no algebra and no proof we know of, and was still right.

A worked example: a sand pile 20 ft square at the base, 10 ft square on top, 9 ft high. A₁ = 400 ft², A₂ = 100 ft², √(A₁A₂) = 200 ft², so V = (9/3)(400 + 100 + 200) = 2,100 ft³ = 77.8 yd³. The naive average of the end areas would have said 2,250 ft³, a 7 % over-count and, at aggregate prices, a real invoice difference. Set A₂ to zero and the formula degenerates correctly to the full pyramid, V = A₁h/3.

Stockpile Volume (Truncated Pyramid) formula

V=h3(A1+A2+A1A2)V = \frac{h}{3}\left(A_1 + A_2 + \sqrt{A_1 A_2}\right)
Where
  • VV= Stockpile volume (yd³)
  • A1A_1= Base area (m²)
  • A2A_2= Top area (m²)
  • hh= Vertical height (m)

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