# Sphericity

Sphericity is a measure of how closely the shape of an 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. Defined by Wadell in 1935, the sphericity, $\Psi$ , of a particle is the ratio of the surface area of a sphere with the same volume as the given particle to the surface area of the particle: Schematic representation of difference in grain shape. Two parameters are shown: sphericity (vertical) and rounding (horizontal).
$\Psi ={\frac {\pi ^{\frac {1}{3}}(6V_{p})^{\frac {2}{3}}}{A_{p}}}$ where $V_{p}$ is volume of the particle and $A_{p}$ is the surface area of the particle. The sphericity of a sphere is unity by definition and, by the isoperimetric inequality, any particle which is not a sphere will have sphericity less than 1.

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.

## Ellipsoidal objects

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

$\Psi ={\frac {\pi ^{\frac {1}{3}}(6V_{p})^{\frac {2}{3}}}{A_{p}}}={\frac {2{\sqrt[{3}]{ab^{2}}}}{a+{\frac {b^{2}}{\sqrt {a^{2}-b^{2}}}}\ln {\left({\frac {a+{\sqrt {a^{2}-b^{2}}}}{b}}\right)}}},$

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 particle divided by the actual surface area of the particle.

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

$A_{s}^{3}=\left(4\pi r^{2}\right)^{3}=4^{3}\pi ^{3}r^{6}=4\pi \left(4^{2}\pi ^{2}r^{6}\right)=4\pi \cdot 3^{2}\left({\frac {4^{2}\pi ^{2}}{3^{2}}}r^{6}\right)=36\pi \left({\frac {4\pi }{3}}r^{3}\right)^{2}=36\,\pi V_{p}^{2}$

therefore

$A_{s}=\left(36\,\pi V_{p}^{2}\right)^{\frac {1}{3}}=36^{\frac {1}{3}}\pi ^{\frac {1}{3}}V_{p}^{\frac {2}{3}}=6^{\frac {2}{3}}\pi ^{\frac {1}{3}}V_{p}^{\frac {2}{3}}=\pi ^{\frac {1}{3}}\left(6V_{p}\right)^{\frac {2}{3}}$

hence we define $\Psi$  as:

$\Psi ={\frac {A_{s}}{A_{p}}}={\frac {\pi ^{\frac {1}{3}}\left(6V_{p}\right)^{\frac {2}{3}}}{A_{p}}}$

## Sphericity of common objects

Name Picture Volume Surface Area Sphericity
Platonic Solids
tetrahedron   ${\frac {\sqrt {2}}{12}}\,s^{3}$  ${\sqrt {3}}\,s^{2}$  $\left({\frac {\pi }{6{\sqrt {3}}}}\right)^{\frac {1}{3}}\approx 0.671$
cube (hexahedron)   $\,s^{3}$  $6\,s^{2}$

$\left({\frac {\pi }{6}}\right)^{\frac {1}{3}}\approx 0.806$

octahedron   ${\frac {1}{3}}{\sqrt {2}}\,s^{3}$  $2{\sqrt {3}}\,s^{2}$

$\left({\frac {\pi }{3{\sqrt {3}}}}\right)^{\frac {1}{3}}\approx 0.846$

dodecahedron   ${\frac {1}{4}}\left(15+7{\sqrt {5}}\right)\,s^{3}$  $3{\sqrt {25+10{\sqrt {5}}}}\,s^{2}$

$\left({\frac {\left(15+7{\sqrt {5}}\right)^{2}\pi }{12\left(25+10{\sqrt {5}}\right)^{\frac {3}{2}}}}\right)^{\frac {1}{3}}\approx 0.910$

icosahedron   ${\frac {5}{12}}\left(3+{\sqrt {5}}\right)\,s^{3}$  $5{\sqrt {3}}\,s^{2}$  $\left({\frac {\left(3+{\sqrt {5}}\right)^{2}\pi }{60{\sqrt {3}}}}\right)^{\frac {1}{3}}\approx 0.939$
Round Shapes
ideal cone
$(h=2{\sqrt {2}}r)$
${\frac {1}{3}}\pi \,r^{2}h$

$={\frac {2{\sqrt {2}}}{3}}\pi \,r^{3}$

$\pi \,r(r+{\sqrt {r^{2}+h^{2}}})$

$=4\pi \,r^{2}$

$\left({\frac {1}{2}}\right)^{\frac {1}{3}}\approx 0.794$
hemisphere
(half sphere)
${\frac {2}{3}}\pi \,r^{3}$  $3\pi \,r^{2}$

$\left({\frac {16}{27}}\right)^{\frac {1}{3}}\approx 0.840$

ideal cylinder
$(h=2\,r)$
$\pi r^{2}h=2\pi \,r^{3}$  $2\pi r(r+h)=6\pi \,r^{2}$

$\left({\frac {2}{3}}\right)^{\frac {1}{3}}\approx 0.874$

ideal torus
$(R=r)$
$2\pi ^{2}Rr^{2}=2\pi ^{2}\,r^{3}$  $4\pi ^{2}Rr=4\pi ^{2}\,r^{2}$

$\left({\frac {9}{4\pi }}\right)^{\frac {1}{3}}\approx 0.894$

sphere   ${\frac {4}{3}}\pi r^{3}$  $4\pi \,r^{2}$

$1\,$

Other Shapes
disdyakis triacontahedron   ${\frac {180}{11}}{\sqrt {179-24{\sqrt {5}}}}$  ${\frac {180}{11}}\left(5+4{\sqrt {5}}\right)$  ${\frac {\left(\left(5+4{\sqrt {5}}\right)^{2}{\frac {11\pi }{5}}\right)^{\frac {1}{3}}}{\sqrt {179-24{\sqrt {5}}}}}\approx 0.986$