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Solid angle
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===Tetrahedron=== Let OABC be the vertices of a [[tetrahedron]] with an origin at O subtended by the triangular face ABC where <math>\vec a\ ,\, \vec b\ ,\, \vec c </math> are the vector positions of the vertices A, B and C. Define the [[vertex angle]] {{mvar|θ<sub>a</sub>}} to be the angle BOC and define {{mvar|θ<sub>b</sub>}}, {{mvar|θ<sub>c</sub>}} correspondingly. Let <math>\phi_{ab}</math> be the [[dihedral angle]] between the planes that contain the tetrahedral faces OAC and OBC and define <math>\phi_{ac}</math>, <math>\phi_{bc}</math> correspondingly. The solid angle {{math|Ω}} subtended by the triangular surface ABC is given by <math display=block> \Omega = \left(\phi_{ab} + \phi_{bc} + \phi_{ac}\right)\ - \pi.</math> This follows from the theory of [[spherical excess]] and it leads to the fact that there is an analogous theorem to the theorem that ''"The sum of internal angles of a planar triangle is equal to {{pi}}"'', for the sum of the four internal solid angles of a tetrahedron as follows: <math display=block> \sum_{i=1}^4 \Omega_i = 2 \sum_{i=1}^6 \phi_i\ - 4 \pi,</math> where <math>\phi_i</math> ranges over all six of the dihedral angles between any two planes that contain the tetrahedral faces OAB, OAC, OBC and ABC.<ref>{{cite journal |last1=Hopf |first1=Heinz |title=Selected Chapters of Geometry |journal=ETH Zurich |date=1940 |pages=1–2 |url=http://pi.math.cornell.edu/~hatcher/Other/hopf-samelson.pdf |archive-url=https://web.archive.org/web/20180921122755/http://pi.math.cornell.edu/~hatcher/Other/hopf-samelson.pdf |archive-date=2018-09-21 |url-status=live}}</ref> A useful formula for calculating the solid angle of the tetrahedron at the origin O that is purely a function of the vertex angles {{mvar|θ<sub>a</sub>}}, {{mvar|θ<sub>b</sub>}}, {{mvar|θ<sub>c</sub>}} is given by [[Simon Antoine Jean L'Huilier|L'Huilier]]'s theorem<ref>{{cite web|url=http://mathworld.wolfram.com/LHuiliersTheorem.html|title=L'Huilier's Theorem – from Wolfram MathWorld |publisher=Mathworld.wolfram.com |date=2015-10-19|access-date=2015-10-19}}</ref><ref>{{cite web|url=http://mathworld.wolfram.com/SphericalExcess.html|title=Spherical Excess – from Wolfram MathWorld |publisher=Mathworld.wolfram.com |date=2015-10-19|access-date=2015-10-19}}</ref> as <math display=block> \tan \left( \frac{1}{4} \Omega \right) = \sqrt{ \tan \left( \frac{\theta_s}{2}\right) \tan \left( \frac{\theta_s - \theta_a}{2}\right) \tan \left( \frac{\theta_s - \theta_b}{2}\right) \tan \left(\frac{\theta_s - \theta_c}{2}\right)}, </math> where <math display=block> \theta_s = \frac {\theta_a + \theta_b + \theta_c}{2}. </math> Another interesting formula involves expressing the vertices as vectors in 3 dimensional space. Let <math>\vec a\ ,\, \vec b\ ,\, \vec c </math> be the vector positions of the vertices A, B and C, and let {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} be the magnitude of each vector (the origin-point distance). The solid angle {{math|Ω}} subtended by the triangular surface ABC is:<ref>{{cite journal| first=Folke| last=Eriksson| title= On the measure of solid angles| journal= Math. Mag.| volume=63|issue=3|pages=184–187|year=1990| doi=10.2307/2691141| jstor=2691141}}</ref><ref>{{cite journal| last = Van Oosterom| first = A|author2=Strackee, J | year = 1983| title = The Solid Angle of a Plane Triangle| journal = IEEE Trans. Biomed. Eng.| volume = BME-30| issue = 2| pages = 125–126| doi = 10.1109/TBME.1983.325207| pmid = 6832789| s2cid = 22669644}}</ref> <math display=block>\tan \left( \frac{1}{2} \Omega \right) = \frac{\left|\vec a\ \vec b\ \vec c\right|}{abc + \left(\vec a \cdot \vec b\right)c + \left(\vec a \cdot \vec c\right)b + \left(\vec b \cdot \vec c\right)a}, </math> where <math display=block>\left|\vec a\ \vec b\ \vec c\right|=\vec a \cdot (\vec b \times \vec c)</math> denotes the [[triple product|scalar triple product]] of the three vectors and <math>\vec a \cdot \vec b</math> denotes the [[scalar product]]. Care must be taken here to avoid negative or incorrect solid angles. One source of potential errors is that the scalar triple product can be negative if {{mvar|a}}, {{mvar|b}}, {{mvar|c}} have the wrong [[determinant|winding]]. Computing the absolute value is a sufficient solution since no other portion of the equation depends on the winding. The other pitfall arises when the scalar triple product is positive but the divisor is negative. In this case returns a negative value that must be increased by {{pi}}.
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