Conical Frustum

$\begin{figure}\BoxedEPSF{Frustum.epsf scaled 500}\end{figure}$

A conical frustum is a Frustum created by slicing the top off a Cone (with the cut made parallel to the base). For a right circular Cone, let be the slant height and and the top and bottom Radii. Then

$\begin{displaymath} s=\sqrt{(R_1-R_2)^2+h^2}. \end{displaymath}$

(1)

The Surface Area, not including the top and bottom Circles, is

$\begin{displaymath} A=\pi(R_1+R_2)s=\pi(R_1+R_2)\sqrt{(R_1-R_2)^2+h^2}\,. \end{displaymath}$

(2)

The Volume of the frustum is given by

$\begin{displaymath} V=\pi\int_0^h [r(z)]^2\,dz. \end{displaymath}$

(3)

But

$\begin{displaymath} r(z)=R_1+(R_2-R_1){z\over h}, \end{displaymath}$

(4)

$\displaystyle V$	$\textstyle =$	$\displaystyle \pi \int_0^h \left[{R_1+(R_2-R_1){z\over h}}\right]^2\,dz$
	$\textstyle =$	$\displaystyle {\textstyle{1\over 3}}\pi h({R_1}^2+R_1R_2+{R_2}^2).$	(5)

This formula can be generalized to any Pyramid by letting

be the base Areas of the top and bottom of the frustum. Then the Volume can be written as

$\begin{displaymath} V={\textstyle{1\over 3}} h(A_1+A_2+\sqrt{A_1A_2}\,). \end{displaymath}$

(6)

The weighted mean of

over the frustum is

$\begin{displaymath} \left\langle{z}\right\rangle{}=\pi\int_0^h z[r(z)]^2\,dz={\textstyle{1\over 12}}h^2({R_1}^2+2R_1R_2+3{R_2}^2). \end{displaymath}$

(7)

The Centroid is then located at a height

$\begin{displaymath} \bar z={\left\langle{z}\right\rangle{}\over V}={h({R_1}^2+2R_1R_2+3{R_2}^2)\over 4({R_1}^2+R_1R_2+{R_2}^2)} \end{displaymath}$

(8)

(Beyer 1987, p. 133). The special case of the Cone is given by taking

, yielding $\bar z=h/4$ .

See also Cone, Frustum, Pyramidal Frustum, Spherical Segment

References

Beyer, W. H. (Ed.) CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 129-130 and 133, 1987.