Frustum (Truncated Cone or Pyramid) Calculator

How it works

A frustum is what is left when the top of a cone or pyramid is sliced off parallel to its base: a bucket, a lampshade, a flower pot. Its volume is the big cone minus the small cone that was removed, and that difference simplifies to one third of the height times the sum of the two end areas and the square root of their product. For a round frustum that is ⅓πh(R² + Rr + r²); for a square one it is ⅓h(a² + ab + b²), a formula already recorded in the Moscow Mathematical Papyrus around 1850 BC.

The slant height is the straight distance along the sloping side from the bottom rim to the top rim. The difference between the two ends and the height form a right triangle with the slant height as its hypotenuse; for the round frustum the horizontal leg is R − r, for the square one it is half the difference of the edges, because each face slopes in from its own edge. The sloping surface of a round frustum unrolls into part of a ring with area π(R + r)l; the square one has four trapezoidal faces, each ½(a + b)l.

Enter a top of 0 and you get the full cone or pyramid; enter a top equal to the bottom and you get a cylinder or box. The unit you choose labels the results; the capacity line converts the volume in cubic inches, feet, yards, centimetres or metres to US gallons and litres from the legal definitions of those units.

Formula

cone:     V = ⅓ × π × h × (R² + R×r + r²)     l = √((R − r)² + h²)
          lateral = π × (R + r) × l          total = lateral + πR² + πr²
pyramid:  V = ⅓ × h × (a² + a×b + b²)         l = √(((a − b) ÷ 2)² + h²)
          lateral = 2 × (a + b) × l          total = lateral + a² + b²
capacity: 1 US gal = 231 in³ = 3.785411784 L;  1 ft³ = 7.48052 gal;  1 m³ = 1,000 L

Example

A round frustum 2 ft in radius at the bottom, 1 ft at the top and 3 ft tall has a volume of ⅓ × π × 3 × (4 + 2 + 1) = 7π ≈ 21.9911 ft³, which holds 21.9911 × 7.48052 ≈ 164.51 US gallons or 622.72 litres. Its slant height is √(1 + 9) = √10 ≈ 3.1623 ft, its sloping surface is π × 3 × 3.1623 ≈ 29.8038 ft², and with the two circular ends (4π + π ≈ 15.708 ft²) the total surface area is ≈ 45.5117 ft².

A bucket 12 in across the bottom, 10 in across the top and 14 in tall (radii 6 and 5 in) holds ⅓ × π × 14 × (36 + 30 + 25) ≈ 1,334.13 in³, or 1,334.13 ÷ 231 ≈ 5.78 US gallons (21.86 litres) filled to the brim.

A truncated square pyramid 4 ft on the bottom edge, 2 ft on the top edge and 3 ft tall has a volume of ⅓ × 3 × (16 + 8 + 4) = 28 ft³, a slant height of √(1 + 9) ≈ 3.1623 ft, a sloping area of 2 × 6 × 3.1623 ≈ 37.9473 ft², and a total surface area of 37.9473 + 16 + 4 ≈ 57.9473 ft².

Assumptions and limitations

• The cut is parallel to the base and the solid is right: the centre of the top sits directly above the centre of the bottom. An offset top (still parallel) has the same volume but a different surface area; a tilted cut changes both.
• The pyramid option is for a square base. A rectangular frustum needs its own calculation.
• The capacity assumes the whole volume is usable, filled level to the rim; a real bucket or pot has a handle, a lip and a fill line below that.
• The gallon is the US liquid gallon of 231 cubic inches, not the imperial gallon.
• Results are rounded for display only.

Frequently asked questions