In geometry, a frustum (plural: frusta or frustums) is the portion of a solid (normally a cone or pyramid) that lies between two parallel planes cutting it. A right frustum is a parallel truncation of a right pyramid.^{[1]}
The term is commonly used in computer graphics to describe the threedimensional region which is visible on the screen, the "viewing frustum", which is formed by a clipped pyramid; in particular, frustum culling is a method of hidden surface determination.
In the aerospace industry, frustum is the common term for the fairing between two stages of a multistage rocket (such as the Saturn V), which is shaped like a truncated cone. It also applies to the essential drive element of the sofar unproven Emdrive.
Contents

Elements, special cases, and related concepts 1

Formulas 2

Volume 2.1

Surface area 2.2

Examples 3

Notes 4

References 5

External links 6
Elements, special cases, and related concepts
Square frustum
A regular
octahedron can be augmented on 3 faces to create a triangular frustum
Each plane section is a floor or base of the frustum. Its axis 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.
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 the prismatoids.
Two frusta joined at their bases make a bifrustum.
Formulas
Volume
The volume formula of a frustum of a square pyramid was introduced by the ancient Egyptian mathematics in what is called the Moscow Mathematical Papyrus, written ca. 1850 BC.:

V = \frac{1}{3} h(a^2 + a b +b^2).
where a and b are the base and top side lengths of the truncated pyramid, and h is the height. The Egyptians knew the correct formula for obtaining the volume of 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 the apex off, minus the volume of the apex:

V = \frac{h_1 B_1  h_2 B_2}{3}
where B_{1} is the area of one base, B_{2} is the area of the other base, and h_{1}, h_{2} are the perpendicular heights from the apex to the planes of the two bases.
Considering that

\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 a product of this proportionality α/3 and a difference of cubes of heights h_{1} and h_{2} only.

V = \frac{h_1 a h_1^2  h_2 a h_2^2}{3} = \frac{a}{3}(h_1^3  h_2^3)
By factoring the difference of two cubes ( a^{3}  b^{3} = (ab)(a^{2} + ab + b^{2}) ) we get h_{1}−h_{2} = h, the height of the frustum, and α(h_{1}^{2} + h_{1}h_{2} + h_{2}^{2})/3.
Distributing α and substituting from its definition, the Heronian mean of areas B_{1} and B_{2} is obtained. The alternative formula is therefore

V = \frac{h}{3}(B_1+\sqrt{B_1 B_2}+B_2)
Heron of Alexandria is noted for deriving this formula and with it encountering the imaginary number, the square root of negative one.^{[2]}
In particular, the volume of a circular cone frustum is

V = \frac{\pi h}{3}(R_1^2+R_1 R_2+R_2^2)
where π is 3.14159265..., and R_{1}, R_{2} are the radii of the two bases.
The volume of a pyramidal frustum whose bases are nsided regular polygons is

V= \frac{n h}{12} (a_1^2+a_1a_2+a_2^2)\cot \frac{\pi}{n}
where a_{1} and a_{2} are the sides of the two bases.
Surface area
For a right circular conical frustum^{[3]}

\begin{align}\text{Lateral Surface Area}&=\pi(R_1+R_2)s\\ &=\pi(R_1+R_2)\sqrt{(R_1R_2)^2+h^2}\end{align}
and

\begin{align}\text{Total Surface Area}&=\pi((R_1+R_2)s+R_1^2+R_2^2)\\ &=\pi((R_1+R_2)\sqrt{(R_1R_2)^2+h^2}+R_1^2+R_2^2)\end{align}
where R_{1} and R_{2} 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 nsided polygons is

A= \frac{n}{4}\left[(a_1^2+a_2^2)\cot \frac{\pi}{n} + \sqrt{(a_1^2a_2^2)^2\sec^2 \frac{\pi}{n}+4 h^2(a_1+a_2)^2} \right]
where a_{1} and a_{2} are the sides of the two bases.
Examples
Notes

1.^ The term "frustum" comes from Latin frustum meaning "piece" or "crumb". The English word is often misspelled as frustrum, a different Latin word cognate to the English word "frustrate".^{[4]} The confusion between these two words is very old: a warning about them can be found in the Appendix Probi, and the works of Plautus include a pun on them.^{[5]}
References

^ William F. Kern, James R Bland,Solid Mensuration with proofs, 1938, p.67

^ Nahin, Paul. "An Imaginary Tale: The story of [the square root of minus one]." Princeton University Press. 1998

^ "Mathwords.com: Frustum". Retrieved 17 July 2011.

^ Clark, John Spencer (1895), Teachers' Manual: Books IVIII.. For Prang's complete course in formstudy and drawing, Books 78, Prang Educational Company, p. 49 .

^ Fontaine, Michael (2010), Funny Words in Plautine Comedy, Oxford University Press, pp. 117, 154, .
External links

Derivation of formula for the volume of frustums of pyramid and cone (Mathalino.com)

Weisstein, Eric W., "Pyramidal frustum", MathWorld.

Weisstein, Eric W., "Conical frustum", MathWorld.

Paper models of frustums (truncated pyramids)

Paper model of frustum (truncated cone)

Design paper models of conical frustum (truncated cones)


Platonic solids (regular)









Dihedral regular



Dihedral uniform



Dihedral others



Degenerate polyhedra are in italics.


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