Sphericity

Template:Short description Template:Use dmy dates Template:For-multi

Sphericity is a measure of how closely the shape of a physical object resembles that of a perfect sphere. For example, the sphericity of the balls inside a ball bearing determines the quality of the bearing, such as the load it can bear or the speed at which it can turn without failing. Sphericity is a specific example of a compactness measure of a shape.

Sphericity applies in three dimensions; its analogue in two dimensions, such as the cross sectional circles along a cylindrical object such as a shaft, is called roundness.

Definition

Defined by Wadell in 1935,^[1] the sphericity, $Ψ$ , of an object is the ratio of the surface area of a sphere with the same volume to the object's surface area:

Ψ = \frac{π^{\frac{1}{3}} (6 V_{p})^{\frac{2}{3}}}{A_{p}}

where $V_{p}$ is volume of the object and $A_{p}$ is the surface area. The sphericity of a sphere is unity by definition and, by the isoperimetric inequality, any shape which is not a sphere will have sphericity less than 1.

Ellipsoidal objects

Template:See also Template:Further

The sphericity, $Ψ$ , of an oblate spheroid (similar to the shape of the planet Earth) is:

Ψ = \frac{π^{\frac{1}{3}} (6 V_{p})^{\frac{2}{3}}}{A_{p}} = \frac{2 \sqrt[3]{a b^{2}}}{a + \frac{b^{2}}{\sqrt{a^{2} - b^{2}}} \ln (\frac{a + \sqrt{a^{2} - b^{2}}}{b})},

where a and b are the semi-major and semi-minor axes respectively.

Derivation

Hakon Wadell defined sphericity as the surface area of a sphere of the same volume as the object divided by the actual surface area of the object.

First we need to write surface area of the sphere, $A_{s}$ in terms of the volume of the object being measured, $V_{p}$

A_{s}^{3} = {(4 π r^{2})}^{3} = 4^{3} π^{3} r^{6} = 4 π (4^{2} π^{2} r^{6}) = 4 π \cdot 3^{2} (\frac{4^{2} π^{2}}{3^{2}} r^{6}) = 36 π {(\frac{4 π}{3} r^{3})}^{2} = 36 π V_{p}^{2}

therefore

A_{s} = {(36 π V_{p}^{2})}^{\frac{1}{3}} = 3 6^{\frac{1}{3}} π^{\frac{1}{3}} V_{p}^{\frac{2}{3}} = 6^{\frac{2}{3}} π^{\frac{1}{3}} V_{p}^{\frac{2}{3}} = π^{\frac{1}{3}} {(6 V_{p})}^{\frac{2}{3}}

hence we define $Ψ$ as:

Ψ = \frac{A_{s}}{A_{p}} = \frac{π^{\frac{1}{3}} {(6 V_{p})}^{\frac{2}{3}}}{A_{p}}

Sphericity of common objects

Name	Volume	Surface area	Sphericity
Sphere	$\frac{4 π}{3} r^{3}$	$4 π r^{2}$	$1$
Disdyakis triacontahedron	$\frac{900 + 720 \sqrt{5}}{11} s^{3}$	$\frac{180 \sqrt{179 - 24 \sqrt{5}}}{11} s^{2}$	$\frac{{({(5 + 4 \sqrt{5})}^{2} \frac{11 π}{5})}^{\frac{1}{3}}}{\sqrt{179 - 24 \sqrt{5}}} \approx 0.9857$
Tricylinder	$16 - 8 \sqrt{2} r^{3}$	$48 - 24 \sqrt{2} r^{2}$	$\frac{\sqrt[3]{36 π + 18 π \sqrt{2}}}{6} \approx 0.9633$
Rhombic triacontahedron	$4 \sqrt{5 + 2 \sqrt{5}} s^{3}$	$12 \sqrt{5} s^{2}$	$\frac{\sqrt[6]{455625 π^{2} + 202500 π^{2} \sqrt{5}}}{15} \approx 0.9609$
Icosahedron	$\frac{15 + 5 \sqrt{5}}{12} s^{3}$	$5 \sqrt{3} s^{2}$	$\frac{\sqrt[3]{2100 π \sqrt{3} + 900 π \sqrt{15}}}{30} \approx 0.9393$
Bicylinder	$\frac{16}{3} r^{3}$	$16 r^{2}$	$\frac{\sqrt[3]{2 π}}{2} \approx 0.9226$
Ideal bicone $(h = r \sqrt{2})$	$\frac{2 π}{3} r^{2} h = \frac{2 π \sqrt{2}}{3} r^{3}$	$2 π r \sqrt{r^{2} + h^{2}} = 2 π \sqrt{3} r^{2}$	$\frac{\sqrt[6]{432}}{3} \approx 0.9165$
Dodecahedron	$\frac{15 + 7 \sqrt{5}}{4} s^{3}$	$3 \sqrt{25 + 10 \sqrt{5}} s^{2}$	$\frac{\sqrt[6]{2080 + 928 \sqrt{5}} \sqrt[3]{9 π} \sqrt{5}}{30} \approx 0.9105$
Rhombic dodecahedron	$\frac{16 \sqrt{3}}{9} s^{3}$	$8 \sqrt{2} s^{2}$	$\frac{\sqrt[6]{2592 π^{2}}}{6} \approx 0.9047$
Ideal torus $(R = r)$	$2 π^{2} R r^{2} = 2 π^{2} r^{3}$	$4 π^{2} R r = 4 π^{2} r^{2}$	$\frac{\sqrt[3]{18 π^{2}}}{2 π} \approx 0.8947$
Ideal cylinder $(h = 2 r)$	$π r^{2} h = 2 π r^{3}$	$2 π r (r + h) = 6 π r^{2}$	$\frac{\sqrt[3]{18}}{3} \approx 0.8736$
Octahedron	$\frac{\sqrt{2}}{3} s^{3}$	$2 \sqrt{3} s^{2}$	$\frac{\sqrt[3]{3 π \sqrt{3}}}{3} \approx 0.8456$
Hemisphere	$\frac{2 π}{3} r^{3}$	$3 π r^{2}$	$\frac{2 \sqrt[3]{2}}{3} \approx 0.8399$
Cube	$s^{3}$	$6 s^{2}$	$\frac{\sqrt[3]{36 π}}{6} \approx 0.8060$
Ideal cone $(h = 2 r \sqrt{2})$	$\frac{π}{3} r^{2} h = \frac{2 π \sqrt{2}}{3} r^{3}$	$π r (r + \sqrt{r^{2} + h^{2}}) = 4 π r^{2}$	$\frac{\sqrt[3]{4}}{2} \approx 0.7937$
Tetrahedron	$\frac{\sqrt{2}}{12} s^{3}$	$\sqrt{3} s^{2}$	$\frac{\sqrt[3]{12 π \sqrt{3}}}{6} \approx 0.6711$