Definitions
Nearby Words

Frustum

[fruhs-tuhm]
Set of pyramidal frusta

Examples: Pentagonal and square frustums
Faces n trapezoids,
2 n-gon
Edges 3n
Vertices 2n
Symmetry group Cnv
Dual polyhedron -
Properties convex
For the graphics technique known as Frustum culling, see Hidden surface determination
A frustum (plural: frusta or frustums) is the portion of a solid – normally a cone or pyramid – which lies between two parallel planes cutting the solid.

Elements, special cases, and related concepts

Each plane section is a base of the frustum. The axis of the frustum, if any, is that of the original cone or pyramid. A frustum is circular if it has circular bases; it is right if the axis is perpendicular to both bases, and oblique otherwise.

Cones and pyramids can be viewed as degenerate cases of frustums, where one of the cutting planes passes through the apex (so that the corresponding base reduces to a point). The pyramidal frusta are a subclass of the prismatoids.

Two frusta joined at their bases make a bifrustum.

Formulas

The volume of a frustum is the difference between the volume of the cone (or other figure) before slicing the apex off, minus the volume of the cone (or other figure) that was sliced off:

$V = left | frac\left\{1\right\}\left\{3\right\} h_1 B_1 - frac\left\{1\right\}\left\{3\right\} h_2 B_2 right |.$
where $h_1$ and $h_2$ are the perpendicular heights from the apex to the planes of the smaller and larger base, $B_1$, $B_2$ are the areas of the two bases.

Let $h$ be the height of the frustum, that is, the perpendicular distance between the two planes. Considering that $h = left | h_1 - h_2 right | ,$ and $frac\left\{B_1\right\}\left\{h_1^2\right\}=frac\left\{B_2\right\}\left\{h_2^2\right\}$, one gets the alternative formula for the volume

$V = frac\left\{1\right\}\left\{3\right\} h\left(B_1+sqrt\left\{B_1 B_2\right\}+B_2\right)$
(See Heronian mean.)

In particular, the volume of a circular cone frustum is

$V = frac\left\{1\right\}\left\{3\right\} pi h\left(R_1^2+R_1 R_2+R_2^2\right)$
where $pi$ is 3.14159265..., and $R_1$, $R_2$ are the radii of the two bases.

Circular Frustum

Using the definitions above, in the case of a circular frustum (or truncated cone), the volume function reduces to:
$V = frac\left\{pi\right\}\left\{12\right\} h D_1^2 left\left(1 - left\left(frac\left\{D_2\right\}\left\{D_1\right\}right\right)^2right\right)$ , where 'D' is the diameter of the respective base.
Equivalently:
$V = frac\left\{pi\right\}\left\{12\right\} h left\left(D_1^2 - frac\left\{D_2^2\right\}\left\{D_1/D_2\right\}right\right)$
(Although the former equation can be reduced further, this form is more intuitive.)

Also, the volume ratio can be written as a function of length ratios, or area ratios:

$frac\left\{V_1\right\}\left\{V_2\right\} = left\left(frac\left\{D_1\right\}\left\{D_2\right\}right\right)^3 = left\left(frac\left\{R_1\right\}\left\{R_2\right\}right\right)^3 = left\left(frac\left\{h_1\right\}\left\{h_2\right\}right\right)^3 = left\left(frac\left\{B_1\right\}\left\{B_2\right\}right\right)^frac\left\{3\right\}\left\{2\right\}.$