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Sphericity

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(Redirected from Sphericity scale) Measure of how closely a shape resembles a sphere

For sphericity in statistics, see Mauchly's sphericity test.
Schematic representation of difference in grain shape. Two parameters are shown: sphericity (vertical) and rounding (horizontal).

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.

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, the sphericity, Ψ {\displaystyle \Psi } , of an object is the ratio of the surface area of a sphere with the same volume to the object's surface area:

Ψ = π 1 3 ( 6 V p ) 2 3 A p {\displaystyle \Psi ={\frac {\pi ^{\frac {1}{3}}(6V_{p})^{\frac {2}{3}}}{A_{p}}}}

where V p {\displaystyle V_{p}} is volume of the object and A p {\displaystyle 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

See also: Flattening Further information: Sphericity of the Earth

The sphericity, Ψ {\displaystyle \Psi } , of an oblate spheroid (similar to the shape of the planet Earth) is:

Ψ = π 1 3 ( 6 V p ) 2 3 A p = 2 a b 2 3 a + b 2 a 2 b 2 ln ( a + a 2 b 2 b ) , {\displaystyle \Psi ={\frac {\pi ^{\frac {1}{3}}(6V_{p})^{\frac {2}{3}}}{A_{p}}}={\frac {2{\sqrt{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 {\displaystyle A_{s}} in terms of the volume of the object being measured, V p {\displaystyle 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 3 2 ( 4 2 π 2 3 2 r 6 ) = 36 π ( 4 π 3 r 3 ) 2 = 36 π V p 2 {\displaystyle 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 = ( 36 π V p 2 ) 1 3 = 36 1 3 π 1 3 V p 2 3 = 6 2 3 π 1 3 V p 2 3 = π 1 3 ( 6 V p ) 2 3 {\displaystyle 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 Ψ {\displaystyle \Psi } as:

Ψ = A s A p = π 1 3 ( 6 V p ) 2 3 A p {\displaystyle \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
Sphere 4 3 π r 3 {\displaystyle {\frac {4}{3}}\pi r^{3}} 4 π r 2 {\displaystyle 4\pi \,r^{2}} 1
Disdyakis triacontahedron 180 11 ( 5 + 4 5 ) s 3 {\displaystyle {\frac {180}{11}}\left(5+4{\sqrt {5}}\right)\,s^{3}} 180 11 179 24 5 s 2 {\displaystyle {\frac {180}{11}}{\sqrt {179-24{\sqrt {5}}}}\,s^{2}} ( ( 5 + 4 5 ) 2 11 π 5 ) 1 3 179 24 5 0.9857 {\displaystyle {\frac {\left(\left(5+4{\sqrt {5}}\right)^{2}{\frac {11\pi }{5}}\right)^{\frac {1}{3}}}{\sqrt {179-24{\sqrt {5}}}}}\approx 0.9857}
Rhombic triacontahedron 4 5 + 2 5 s 3 {\displaystyle 4{\sqrt {5+2{\sqrt {5}}}}\,s^{3}} 12 5 s 2 {\displaystyle 12{\sqrt {5}}\,s^{2}} π 1 3 ( 24 5 + 2 5 ) 2 3 12 5 0.9609 {\displaystyle {\frac {\pi ^{\frac {1}{3}}\left(24{\sqrt {5+2{\sqrt {5}}}}\right)^{\frac {2}{3}}}{12{\sqrt {5}}}}\approx 0.9609}
Icosahedron 5 12 ( 3 + 5 ) s 3 {\displaystyle {\frac {5}{12}}\left(3+{\sqrt {5}}\right)\,s^{3}} 5 3 s 2 {\displaystyle 5{\sqrt {3}}\,s^{2}} ( ( 3 + 5 ) 2 π 60 3 ) 1 3 0.939 {\displaystyle \left({\frac {\left(3+{\sqrt {5}}\right)^{2}\pi }{60{\sqrt {3}}}}\right)^{\frac {1}{3}}\approx 0.939}
Dodecahedron 1 4 ( 15 + 7 5 ) s 3 {\displaystyle {\frac {1}{4}}\left(15+7{\sqrt {5}}\right)\,s^{3}} 3 25 + 10 5 s 2 {\displaystyle 3{\sqrt {25+10{\sqrt {5}}}}\,s^{2}} ( ( 15 + 7 5 ) 2 π 12 ( 25 + 10 5 ) 3 2 ) 1 3 0.910 {\displaystyle \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}
Ideal torus
( R = r ) {\displaystyle (R=r)}
2 π 2 R r 2 = 2 π 2 r 3 {\displaystyle 2\pi ^{2}Rr^{2}=2\pi ^{2}\,r^{3}} 4 π 2 R r = 4 π 2 r 2 {\displaystyle 4\pi ^{2}Rr=4\pi ^{2}\,r^{2}} ( 9 4 π ) 1 3 0.894 {\displaystyle \left({\frac {9}{4\pi }}\right)^{\frac {1}{3}}\approx 0.894}
Ideal cylinder
( h = 2 r ) {\displaystyle (h=2\,r)}
π r 2 h = 2 π r 3 {\displaystyle \pi \,r^{2}h=2\pi \,r^{3}} 2 π r ( r + h ) = 6 π r 2 {\displaystyle 2\pi \,r(r+h)=6\pi \,r^{2}} ( 2 3 ) 1 3 0.874 {\displaystyle \left({\frac {2}{3}}\right)^{\frac {1}{3}}\approx 0.874}
Octahedron 1 3 2 s 3 {\displaystyle {\frac {1}{3}}{\sqrt {2}}\,s^{3}} 2 3 s 2 {\displaystyle 2{\sqrt {3}}\,s^{2}} ( π 3 3 ) 1 3 0.846 {\displaystyle \left({\frac {\pi }{3{\sqrt {3}}}}\right)^{\frac {1}{3}}\approx 0.846}
Hemisphere
(half sphere)
2 3 π r 3 {\displaystyle {\frac {2}{3}}\pi \,r^{3}} 3 π r 2 {\displaystyle 3\pi \,r^{2}} ( 16 27 ) 1 3 0.840 {\displaystyle \left({\frac {16}{27}}\right)^{\frac {1}{3}}\approx 0.840}
Cube (hexahedron) s 3 {\displaystyle \,s^{3}} 6 s 2 {\displaystyle 6\,s^{2}} ( π 6 ) 1 3 0.806 {\displaystyle \left({\frac {\pi }{6}}\right)^{\frac {1}{3}}\approx 0.806}
Ideal cone
( h = 2 2 r ) {\displaystyle (h=2{\sqrt {2}}r)}
1 3 π r 2 h = 2 2 3 π r 3 {\displaystyle {\frac {1}{3}}\pi \,r^{2}h={\frac {2{\sqrt {2}}}{3}}\pi \,r^{3}} π r ( r + r 2 + h 2 ) = 4 π r 2 {\displaystyle \pi \,r(r+{\sqrt {r^{2}+h^{2}}})=4\pi \,r^{2}} ( 1 2 ) 1 3 0.794 {\displaystyle \left({\frac {1}{2}}\right)^{\frac {1}{3}}\approx 0.794}
Tetrahedron 2 12 s 3 {\displaystyle {\frac {\sqrt {2}}{12}}\,s^{3}} 3 s 2 {\displaystyle {\sqrt {3}}\,s^{2}} ( π 6 3 ) 1 3 0.671 {\displaystyle \left({\frac {\pi }{6{\sqrt {3}}}}\right)^{\frac {1}{3}}\approx 0.671}

See also

References

  1. Wadell, Hakon (1935). "Volume, Shape, and Roundness of Quartz Particles". The Journal of Geology. 43 (3): 250–280. Bibcode:1935JG.....43..250W. doi:10.1086/624298.

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