Cone Frustum Volume (Truncated Cone)

Also known as truncated cone volume · bucket volume · tapered tank volume

V=πh3(R2+Rr+r2)V = \frac{\pi h}{3}\left(R^2 + Rr + r^2\right)

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Constant used — built into this formula, no need to enter
π=3.141592653589793\pi = 3.141592653589793Pi · exact

Learning zone

A frustum is a cone with its tip sliced off parallel to the base, and it is the shape of almost every real container that tapers: buckets, plant pots, drinking cups, lampshades, silo discharge cones, and the classic paper snow-cone once you truncate it. The volume is V=πh3(R2+Rr+r2)V = \tfrac{\pi h}{3}(R^2 + Rr + r^2), which is the big cone minus the little cone you removed, tidied up so that the vanished apex height never appears.

This is one of the oldest formulas humanity has written down. The Moscow Mathematical Papyrus, copied around 1850 BC, contains a worked problem for the volume of a truncated square pyramid using exactly the same a2+ab+b2a^2 + ab + b^2 structure, some fourteen centuries before Euclid. Whoever discovered it left no proof, only a confident recipe and the answer.

The pitfall is the tempting shortcut of averaging the radii and treating the shape as a cylinder. That gives (R+r2)2\left(\tfrac{R+r}{2}\right)^2 where the truth is 13(R2+Rr+r2)\tfrac{1}{3}(R^2 + Rr + r^2), and for a bucket 2 m across the top and 1 m across the bottom those are 2.25 and 2.33 in units of r2r^2. The average always underestimates, by roughly four percent here and more as the taper steepens. Two reassuring checks live inside the real formula: set R=rR = r and it collapses to the cylinder πr2h\pi r^2 h, and set r=0r = 0 and it collapses to the cone 13πR2h\tfrac{1}{3}\pi R^2 h.

Cone Frustum Volume (Truncated Cone)
V=πh3(R2+Rr+r2)V = \frac{\pi h}{3}\left(R^2 + Rr + r^2\right)
Where
  • VV= Volume
  • hh= Height
  • RR= Large radius
  • rr= Small radius
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