# Frustum

In geometry, a frustum[a] (from the Latin for “morsel”; plural: frusta or frustums) is the portion of a solid (normally a pyramid or a cone) that lies between one or two parallel planes cutting it. The base faces are polygonal, the side faces are trapezoidal. A right frustum is a right pyramid or a right cone truncated perpendicularly to its axis.[3]

Set of pyramidal right n-gonal frustums
Examples: right pentagonal and square frustums
(n = 5 and n = 4)
Facesn isosceles trapezoids, 2 regular n-gons
Edges3n
Vertices2n
Symmetry groupCnv, [1,n], (*nn)
Dual polyhedronconvex asymmetric right n-gonal bipyramid
Propertiesconvex
Net
Example: net of right trigonal frustum (n = 3)

If a frustum has all its edges of the same length (equilateral figure), then it is a uniform prism.

In computer graphics, the viewing frustum is the three-dimensional region which is visible on the screen. It is formed by a clipped pyramid; in particular, frustum culling is a method of hidden surface determination.

In the aerospace industry, a frustum is the fairing between two stages of a multistage rocket (such as the Saturn V), which is shaped like a truncated cone.

## Elements, special cases, and related concepts

Square frustum

A regular octahedron can be augmented on 3 faces to create a triangular frustum

A frustum's axis 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.

The height of a frustum is the perpendicular distance between the planes of the two bases.

Cones and pyramids can be viewed as degenerate cases of frusta, 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 prismatoids.

Two frusta with two congruent bases joined at these congruent bases make a bifrustum.

## Formulas

### Volume

The formula for the volume of a pyramidal square frustum was introduced by the ancient Egyptian mathematics in what is called the Moscow Mathematical Papyrus, written in the 13th dynasty (c. 1850 BC):

${\displaystyle V={\frac {h}{3}}\left(a^{2}+ab+b^{2}\right),}$

where a and b are the base and top side lengths, and h is the height.

The Egyptians knew the correct formula for the volume of such a truncated square pyramid, but no proof of this equation is given in the Moscow papyrus.

The volume of a conical or pyramidal frustum is the volume of the solid before slicing its "apex" off, minus the volume of this "apex":

${\displaystyle V={\frac {h_{1}B_{1}-h_{2}B_{2}}{3}},}$

where B1 and B2 are the base and top areas, and h1 and h2 are the perpendicular heights from the apex to the base and top planes.

Considering that

${\displaystyle {\frac {B_{1}}{h_{1}^{2}}}={\frac {B_{2}}{h_{2}^{2}}}={\frac {\sqrt {B_{1}B_{2}}}{h_{1}h_{2}}}=\alpha ,}$

the formula for the volume can be expressed as the third of the product of this proportionality, ${\displaystyle \alpha }$ , and of the difference of the cubes of the heights h1 and h2 only:

${\displaystyle V={\frac {h_{1}\alpha h_{1}^{2}-h_{2}\alpha h_{2}^{2}}{3}}=\alpha {\frac {h_{1}^{3}-h_{2}^{3}}{3}}.}$

By using the identity a3b3 = (ab)(a2 + ab + b2), one gets:

${\displaystyle V=(h_{1}-h_{2})\alpha {\frac {h_{1}^{2}+h_{1}h_{2}+h_{2}^{2}}{3}},}$

where h1h2 = h is the height of the frustum.

Distributing ${\displaystyle \alpha }$  and substituting from its definition, the Heronian mean of areas B1 and B2 is obtained:

${\displaystyle {\frac {B_{1}+{\sqrt {B_{1}B_{2}}}+B_{2}}{3}};}$

the alternative formula is therefore:

${\displaystyle V={\frac {h}{3}}\left(B_{1}+{\sqrt {B_{1}B_{2}}}+B_{2}\right).}$

Heron of Alexandria is noted for deriving this formula, and with it, encountering the imaginary unit: the square root of negative one.[4]

In particular:

• The volume of a circular cone frustum is:
${\displaystyle V={\frac {\pi h}{3}}\left(r_{1}^{2}+r_{1}r_{2}+r_{2}^{2}\right),}$
where r1 and r2 are the base and top radii.
• The volume of a pyramidal frustum whose bases are regular n-gons is:
${\displaystyle V={\frac {nh}{12}}\left(a_{1}^{2}+a_{1}a_{2}+a_{2}^{2}\right)\cot {\frac {\pi }{n}},}$
where a1 and a2 are the base and top side lengths.

### Surface area

Conical frustum

3D model of a conical frustum.

For a right circular conical frustum[5][6]

{\displaystyle {\begin{aligned}{\text{Lateral surface area}}&=\pi \left(r_{1}+r_{2}\right)s\\&=\pi \left(r_{1}+r_{2}\right){\sqrt {\left(r_{1}-r_{2}\right)^{2}+h^{2}}}\end{aligned}}}

and

{\displaystyle {\begin{aligned}{\text{Total surface area}}&=\pi \left(\left(r_{1}+r_{2}\right)s+r_{1}^{2}+r_{2}^{2}\right)\\&=\pi \left(\left(r_{1}+r_{2}\right){\sqrt {\left(r_{1}-r_{2}\right)^{2}+h^{2}}}+r_{1}^{2}+r_{2}^{2}\right)\end{aligned}}}

where r1 and r2 are the base and top radii respectively, and s is the slant height of the frustum.

The surface area of a right frustum whose bases are similar regular n-sided polygons is

${\displaystyle A={\frac {n}{4}}\left[\left(a_{1}^{2}+a_{2}^{2}\right)\cot {\frac {\pi }{n}}+{\sqrt {\left(a_{1}^{2}-a_{2}^{2}\right)^{2}\sec ^{2}{\frac {\pi }{n}}+4h^{2}\left(a_{1}+a_{2}\right)^{2}}}\right]}$

where a1 and a2 are the sides of the two bases.

## Examples

Rolo brand chocolates approximate a right circular conic frustum, although not flat on top.