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Solid angle

From Wikipedia, the free encyclopedia
Measure in 3-dimensional geometry
Not to be confused withspherical angle.
Solid angle
Visual representation of a solid angle
Common symbols
Ω
SI unitsteradian
Other units
Square degree,spat
InSI base unitsm2/m2
Conserved?No
Derivations from
other quantities
Ω=A/r2{\displaystyle \Omega =A/r^{2}}
Dimension1{\displaystyle 1}

Ingeometry, asolid angle (symbol:Ω) is a measure of the amount of thefield of view from some particular point that a given object covers. That is, it is a measure of how large the object appears to an observer looking from that point.The point from which the object is viewed is called theapex of the solid angle, and the object is said tosubtend its solid angle at that point.

In theInternational System of Units (SI), a solid angle is expressed in adimensionlessunit called asteradian (symbol: sr), which is equal to one squareradian, sr = rad2. One steradian corresponds to one unit of area (of any shape) on theunit sphere surrounding the apex, so an object that blocks all rays from the apex would cover a number of steradians equal to the totalsurface area of the unit sphere,4π{\displaystyle 4\pi }. Solid angles can also be measured in squares of angular measures such asdegrees, minutes, and seconds.

A small object nearby may subtend the same solid angle as a larger object farther away. For example, although theMoon is much smaller than theSun, it is also much closer toEarth. Indeed, as viewed from any point on Earth, both objects have approximately the same solid angle (and therefore apparent size). This is evident during asolar eclipse.

Definition and properties

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See also:Spherical polygon area

The magnitude of an object's solid angle insteradians is equal to thearea of the segment of aunit sphere, centered at the apex, that the object covers. Giving the area of a segment of a unit sphere in steradians is analogous to giving the length of an arc of aunit circle in radians. Just as the magnitude of a plane angle in radians at the vertex of a circular sector is the ratio of the length of its arc to its radius, the magnitude of a solid angle in steradians is the ratio of the area covered on a sphere by an object to the square of the radius of the sphere. The formula for the magnitude of the solid angle in steradians is

Ω=Ar2,{\displaystyle \Omega ={\frac {A}{r^{2}}},}

whereA{\displaystyle A} is the area (of any shape) on the surface of the sphere andr{\displaystyle r} is the radius of the sphere.

Solid angles are often used inastronomy,physics, and in particularastrophysics. The solid angle of an object that is very far away is roughly proportional to the ratio of area to squared distance. Here "area" means the area of the object when projected along the viewing direction.

Any area on a sphere which is equal in area to the square of its radius, when observed from its center, subtends precisely onesteradian.

The solid angle of a sphere measured from any point in its interior is 4π sr. The solid angle subtended at the center of a cube by one of its faces is one-sixth of that, or 2π/3  sr.The solid angle subtended at the corner of a cube (anoctant) or spanned by aspherical octant isπ/2  sr, one-eighth of the solid angle of a sphere.

Solid angles can also be measured insquare degrees (1 sr = (180/π)2 square degrees), in squarearc-minutes and squarearc-seconds.[a] It can also be expressed in fractions of the sphere (1 sr =⁠1/4π⁠ fractional area), also known asspat (1 sp = 4π sr).

Inspherical coordinates there is a formula for thedifferential,

dΩ=sin⁡θdθdφ,{\displaystyle d\Omega =\sin \theta \,d\theta \,d\varphi ,}

whereθ is thecolatitude (angle from the North Pole) andφ is the longitude.

The solid angle for an arbitraryoriented surfaceS subtended at a pointP is equal to the solid angle of the projection of the surfaceS to the unit sphere with centerP, which can be calculated as thesurface integral:

Ω=∬Sr^⋅n^r2dS =∬Ssin⁡θdθdφ,{\displaystyle \Omega =\iint _{S}{\frac {{\hat {r}}\cdot {\hat {n}}}{r^{2}}}\,dS\ =\iint _{S}\sin \theta \,d\theta \,d\varphi ,}

wherer^=r→/r{\displaystyle {\hat {r}}={\vec {r}}/r} is theunit vector corresponding tor→{\displaystyle {\vec {r}}}, theposition vector of an infinitesimal area of surfacedS with respect to pointP, and wheren^{\displaystyle {\hat {n}}} represents the unitnormal vector todS. Even if the projection on the unit sphere to the surfaceS is notisomorphic, the multiple folds are correctly considered according to the surface orientation described by the sign of thescalar productr^⋅n^{\displaystyle {\hat {r}}\cdot {\hat {n}}}.

Thus one can approximate the solid angle subtended by a smallfacet having flat surface areadS, orientationn^{\displaystyle {\hat {n}}}, and distancer from the viewer as:

dΩ=4π(dSA)(r^⋅n^),{\displaystyle d\Omega =4\pi \left({\frac {dS}{A}}\right)\,({\hat {r}}\cdot {\hat {n}}),}

where thesurface area of a sphere isA = 4πr2.

Practical applications

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Solid angles for common objects

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Cone, spherical cap, hemisphere

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Diagram showing a section through the centre of a cone (1) subtending a solid angle of 1 steradian in a sphere of radiusr, along with the spherical "cap" (2). The external surface area A of the cap equalsr2{\displaystyle r^{2}} only if solid angle of the cone is exactly 1 steradian. Hence, in this figureθ =A/2 andr = 1.

The solid angle of acone with its apex at the apex of the solid angle, and withapex angle 2θ, is the area of aspherical cap on aunit sphere

Ω=2π(1−cos⁡θ) =4πsin2⁡θ2.{\displaystyle \Omega =2\pi \left(1-\cos \theta \right)\ =4\pi \sin ^{2}{\frac {\theta }{2}}.}

For smallθ, such thatcosθ ≈ 1 −⁠θ2/2⁠, this reduces toΩ =πθ2.

The above is found by computing the followingdouble integral using the unitsurface element in spherical coordinates:

∫02π∫0θsin⁡θ′dθ′dϕ=∫02πdϕ∫0θsin⁡θ′dθ′=2π∫0θsin⁡θ′dθ′=2π[−cos⁡θ′]0θ=2π(1−cos⁡θ).{\displaystyle {\begin{aligned}\int _{0}^{2\pi }\int _{0}^{\theta }\sin \theta '\,d\theta '\,d\phi &=\int _{0}^{2\pi }d\phi \int _{0}^{\theta }\sin \theta '\,d\theta '\\&=2\pi \int _{0}^{\theta }\sin \theta '\,d\theta '\\&=2\pi \left[-\cos \theta '\right]_{0}^{\theta }\\&=2\pi \left(1-\cos \theta \right).\end{aligned}}}

This formula can also be derived without the use ofcalculus.

Over 2200 years agoArchimedes proved that the surface area of a spherical cap is always equal to the area of a circle whose radius equals the distance from the rim of the spherical cap to the point where the cap's axis of symmetry intersects the cap.[2]

Archimedes' theorem that surface area of the region of sphere below horizontal plane H in given diagram is equal to area of a circle of radius t.

In the above coloured diagram this radius is given as

2rsin⁡θ2.{\displaystyle 2r\sin {\frac {\theta }{2}}.}In the adjacent black & white diagram this radius is given as "t".

Hence for a unit sphere the solid angle of the spherical cap is given as

Ω=4πsin2⁡θ2=2π(1−cos⁡θ).{\displaystyle \Omega =4\pi \sin ^{2}{\frac {\theta }{2}}=2\pi \left(1-\cos \theta \right).}

Whenθ =⁠π/2⁠, the spherical cap becomes ahemisphere having a solid angle 2π.

The solid angle of the complement of the cone is

4π−Ω=2π(1+cos⁡θ)=4πcos2⁡θ2.{\displaystyle 4\pi -\Omega =2\pi \left(1+\cos \theta \right)=4\pi \cos ^{2}{\frac {\theta }{2}}.}

This is also the solid angle of the part of thecelestial sphere that an astronomical observer positioned at latitudeθ can see as the Earth rotates. At the equator all of the celestial sphere is visible; at either pole, only one half.

The solid angle subtended by a segment of a spherical cap cut by a plane at angleγ from the cone's axis and passing through the cone's apex can be calculated by the formula[3]

Ω=2[arccos⁡(sin⁡γsin⁡θ)−cos⁡θarccos⁡(tan⁡γtan⁡θ)].{\displaystyle \Omega =2\left[\arccos \left({\frac {\sin \gamma }{\sin \theta }}\right)-\cos \theta \arccos \left({\frac {\tan \gamma }{\tan \theta }}\right)\right].}

For example, ifγ = −θ, then the formula reduces to the spherical cap formula above: the first term becomesπ, and the secondπ cosθ.

Tetrahedron

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Let OABC be the vertices of atetrahedron with an origin at O subtended by the triangular face ABC wherea→ ,b→ ,c→{\displaystyle {\vec {a}}\ ,\,{\vec {b}}\ ,\,{\vec {c}}} are the vector positions of the vertices A, B and C. Define thevertex angleθa to be the angle BOC and defineθb,θc correspondingly. Letϕab{\displaystyle \phi _{ab}} be thedihedral angle between the planes that contain the tetrahedral faces OAC and OBC and defineϕac{\displaystyle \phi _{ac}},ϕbc{\displaystyle \phi _{bc}} correspondingly. The solid angleΩ subtended by the triangular surface ABC is given by

Ω=(ϕab+ϕbc+ϕac) −π.{\displaystyle \Omega =\left(\phi _{ab}+\phi _{bc}+\phi _{ac}\right)\ -\pi .}

This follows from the theory ofspherical 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 toπ", for the sum of the four internal solid angles of a tetrahedron as follows:

∑i=14Ωi=2∑i=16ϕi −4π,{\displaystyle \sum _{i=1}^{4}\Omega _{i}=2\sum _{i=1}^{6}\phi _{i}\ -4\pi ,}

whereϕi{\displaystyle \phi _{i}} ranges over all six of the dihedral angles between any two planes that contain the tetrahedral faces OAB, OAC, OBC and ABC.[4]

A useful formula for calculating the solid angle of the tetrahedron at the origin O that is purely a function of the vertex anglesθa,θb,θc is given byL'Huilier's theorem[5][6] as

tan⁡(14Ω)=tan⁡(θs2)tan⁡(θs−θa2)tan⁡(θs−θb2)tan⁡(θs−θc2),{\displaystyle \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)}},}

whereθs=θa+θb+θc2.{\displaystyle \theta _{s}={\frac {\theta _{a}+\theta _{b}+\theta _{c}}{2}}.}

Another interesting formula involves expressing the vertices as vectors in 3 dimensional space. Leta→ ,b→ ,c→{\displaystyle {\vec {a}}\ ,\,{\vec {b}}\ ,\,{\vec {c}}} be the vector positions of the vertices A, B and C, and leta,b, andc be the magnitude of each vector (the origin-point distance). The solid angleΩ subtended by the triangular surface ABC is:[7][8]

tan⁡(12Ω)=|a→ b→ c→|abc+(a→⋅b→)c+(a→⋅c→)b+(b→⋅c→)a,{\displaystyle \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}},}

where|a→ b→ c→|=a→⋅(b→×c→){\displaystyle \left|{\vec {a}}\ {\vec {b}}\ {\vec {c}}\right|={\vec {a}}\cdot ({\vec {b}}\times {\vec {c}})}

denotes thescalar triple product of the three vectors anda→⋅b→{\displaystyle {\vec {a}}\cdot {\vec {b}}} denotes thescalar 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 ifa,b,c have the wrongwinding. 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 byπ.

Pyramid

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The solid angle of a four-sided right rectangularpyramid withapex anglesa andb (dihedral angles measured to the opposite side faces of the pyramid) isΩ=4arcsin⁡(sin⁡(a2)sin⁡(b2)).{\displaystyle \Omega =4\arcsin \left(\sin \left({a \over 2}\right)\sin \left({b \over 2}\right)\right).}

If both the side lengths (α andβ) of the base of the pyramid and the distance (d) from the center of the base rectangle to the apex of the pyramid (the center of the sphere) are known, then the above equation can be manipulated to give

Ω=4arctan⁡αβ2d4d2+α2+β2.{\displaystyle \Omega =4\arctan {\frac {\alpha \beta }{2d{\sqrt {4d^{2}+\alpha ^{2}+\beta ^{2}}}}}.}

The solid angle of a rightn-gonal pyramid, where the pyramid base is a regularn-sided polygon of circumradiusr, with a pyramid heighth is

Ω=2π−2narctan⁡(tan⁡(πn)1+r2h2).{\displaystyle \Omega =2\pi -2n\arctan \left({\frac {\tan \left({\pi \over n}\right)}{\sqrt {1+{r^{2} \over h^{2}}}}}\right).}

The solid angle of an arbitrary pyramid with ann-sided base defined by the sequence of unit vectors representing edges{s1,s2}, ...sn can be efficiently computed by:[3]

Ω=2π−arg⁡∏j=1n((sj−1sj)(sjsj+1)−(sj−1sj+1)+i[sj−1sjsj+1]).{\displaystyle \Omega =2\pi -\arg \prod _{j=1}^{n}\left(\left(s_{j-1}s_{j}\right)\left(s_{j}s_{j+1}\right)-\left(s_{j-1}s_{j+1}\right)+i\left[s_{j-1}s_{j}s_{j+1}\right]\right).}

where parentheses (* *) is ascalar product and square brackets [* * *] is ascalar triple product, andi is animaginary unit. Indices are cycled:s0 =sn ands1 =sn + 1. The complex products add the phase associated with each vertex angle of the polygon. However, a multiple of2π{\displaystyle 2\pi } is lost in the branch cut ofarg{\displaystyle \arg } and must be kept track of separately. Also, the running product of complex phases must be scaled occasionally to avoid underflow in the limit of nearly parallel segments.

Latitude-longitude rectangle

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The solid angle of a latitude-longitude rectangle on aglobe is(sin⁡ϕN−sin⁡ϕS)(θE−θW)sr,{\displaystyle \left(\sin \phi _{\mathrm {N} }-\sin \phi _{\mathrm {S} }\right)\left(\theta _{\mathrm {E} }-\theta _{\mathrm {W} }\,\!\right)\;\mathrm {sr} ,}whereφN andφS are north and south lines oflatitude (measured from theequator inradians with angle increasing northward), andθE andθW are east and west lines oflongitude (where the angle in radians increases eastward).[9] Mathematically, this represents an arc of angleϕN −ϕS swept around a sphere byθE −θW radians. When longitude spans 2π radians and latitude spansπ radians, the solid angle is that of a sphere.

A latitude-longitude rectangle should not be confused with the solid angle of a rectangular pyramid. All four sides of a rectangular pyramid intersect the sphere's surface ingreat circle arcs. With a latitude-longitude rectangle, only lines of longitude are great circle arcs; lines of latitude are not.

Celestial objects

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By using the definition ofangular diameter, the formula for the solid angle of a celestial object can be defined in terms of the radius of the object,R{\textstyle R}, and the distance from the observer to the object,d{\displaystyle d}:

Ω=2π(1−d2−R2d):d≥R.{\displaystyle \Omega =2\pi \left(1-{\frac {\sqrt {d^{2}-R^{2}}}{d}}\right):d\geq R.}

By inputting the appropriate average values for theSun and theMoon (in relation to Earth), the average solid angle of the Sun is6.794×10−5 steradians and the average solid angle of theMoon is6.418×10−5 steradians. In terms of the total celestial sphere, theSun and theMoon subtend averagefractional areas of0.0005406% (5.406 ppm) and0.0005107% (5.107 ppm), respectively. As these solid angles are about the same size, the Moon can cause both total and annular solareclipses depending on the distance between the Earth and the Moon during the eclipse.

Solid angles in arbitrary dimensions

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The solid angle subtended by the complete (d − 1)-dimensional spherical surface of the unit sphere ind-dimensional Euclidean space can be defined in any number of dimensionsd. One often needs this solid angle factor in calculations with spherical symmetry. It is given by the formulaΩd=2πd2Γ(d2),{\displaystyle \Omega _{d}={\frac {2\pi ^{\frac {d}{2}}}{\Gamma \left({\frac {d}{2}}\right)}},}whereΓ is thegamma function. Whend is an integer, the gamma function can be computed explicitly.[10] It follows thatΩd={1(d2−1)!2πd2 d even(12(d−1))!(d−1)!2dπ12(d−1) d odd.{\displaystyle \Omega _{d}={\begin{cases}{\frac {1}{\left({\frac {d}{2}}-1\right)!}}2\pi ^{\frac {d}{2}}\ &d{\text{ even}}\\{\frac {\left({\frac {1}{2}}\left(d-1\right)\right)!}{(d-1)!}}2^{d}\pi ^{{\frac {1}{2}}(d-1)}\ &d{\text{ odd}}.\end{cases}}}

This gives the expected results of 4π steradians for the 3D sphere bounded by a surface of area4πr2 and 2π radians for the 2D circle bounded by a circumference of length2πr. It also gives the slightly less obvious 2 for the 1D case, in which the origin-centered 1D "sphere" is the interval[−r,r] and this is bounded by two limiting points.

The counterpart to the vector formula in arbitrary dimension was derived by Aomoto[11][12]and independently by Ribando.[13] It expresses them as an infinite multivariateTaylor series:Ω=Ωd|det(V)|(4π)d/2∑a→∈N0(d2)[(−2)∑i<jaij∏i<jaij!∏iΓ(1+∑m≠iaim2)]α→a→.{\displaystyle \Omega =\Omega _{d}{\frac {\left|\det(V)\right|}{(4\pi )^{d/2}}}\sum _{{\vec {a}}\in \mathbb {N} _{0}^{\binom {d}{2}}}\left[{\frac {(-2)^{\sum _{i<j}a_{ij}}}{\prod _{i<j}a_{ij}!}}\prod _{i}\Gamma \left({\frac {1+\sum _{m\neq i}a_{im}}{2}}\right)\right]{\vec {\alpha }}^{\vec {a}}.}Givend unit vectorsv→i{\displaystyle {\vec {v}}_{i}} defining the angle, letV denote the matrix formed by combining them so theith column isv→i{\displaystyle {\vec {v}}_{i}}, andαij=v→i⋅v→j=αji,αii=1{\displaystyle \alpha _{ij}={\vec {v}}_{i}\cdot {\vec {v}}_{j}=\alpha _{ji},\alpha _{ii}=1}. The variablesαij,1≤i<j≤d{\displaystyle \alpha _{ij},1\leq i<j\leq d} form a multivariableα→=(α12,…,α1d,α23,…,αd−1,d)∈R(d2){\displaystyle {\vec {\alpha }}=(\alpha _{12},\dotsc ,\alpha _{1d},\alpha _{23},\dotsc ,\alpha _{d-1,d})\in \mathbb {R} ^{\binom {d}{2}}}. For a "congruent" integer multiexponenta→=(a12,…,a1d,a23,…,ad−1,d)∈N0(d2),{\displaystyle {\vec {a}}=(a_{12},\dotsc ,a_{1d},a_{23},\dotsc ,a_{d-1,d})\in \mathbb {N} _{0}^{\binom {d}{2}},} defineα→a→=∏αijaij{\textstyle {\vec {\alpha }}^{\vec {a}}=\prod \alpha _{ij}^{a_{ij}}}. Note that hereN0{\displaystyle \mathbb {N} _{0}} = non-negative integers, or natural numbers beginning with 0. The notationαji{\displaystyle \alpha _{ji}} forj>i{\displaystyle j>i} means the variableαij{\displaystyle \alpha _{ij}}, similarly for the exponentsaji{\displaystyle a_{ji}}. Hence, the term∑m≠lalm{\textstyle \sum _{m\neq l}a_{lm}} means the sum over all terms ina→{\displaystyle {\vec {a}}} in which l appears as either the first or second index.Where this series converges, it converges to the solid angle defined by the vectors.

Notes

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  1. ^The whole sphere contains ~148.510 million square arcminutes and ~534.638 billion square arcseconds.[citation needed]

References

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  1. ^Falla, Romain (2023). "Mesh adaption for two-dimensional bounded and free-surface flows with the particle finite element method".Computational Particle Mechanics.10 (5):1049–1076.doi:10.1007/s40571-022-00541-2.hdl:2268/302810.
  2. ^"Archimedes on Spheres and Cylinders".Math Pages. 2015.
  3. ^abMazonka, Oleg (2012). "Solid Angle of Conical Surfaces, Polyhedral Cones, and Intersecting Spherical Caps".arXiv:1205.1396 [math.MG].
  4. ^Hopf, Heinz (1940)."Selected Chapters of Geometry"(PDF).ETH Zurich:1–2.Archived(PDF) from the original on 2018-09-21.
  5. ^"L'Huilier's Theorem – from Wolfram MathWorld". Mathworld.wolfram.com. 2015-10-19. Retrieved2015-10-19.
  6. ^"Spherical Excess – from Wolfram MathWorld". Mathworld.wolfram.com. 2015-10-19. Retrieved2015-10-19.
  7. ^Eriksson, Folke (1990). "On the measure of solid angles".Mathematics Magazine.63 (3):184–187.doi:10.2307/2691141.JSTOR 2691141.
  8. ^Van Oosterom, A; Strackee, J (1983). "The Solid Angle of a Plane Triangle".IEEE Transactions on Biomedical Engineering. BME-30 (2):125–126.doi:10.1109/TBME.1983.325207.PMID 6832789.
  9. ^"Area of a Latitude-Longitude Rectangle".The Math Forum @ Drexel. 2003.
  10. ^Jackson, FM (1993)."Polytopes in Euclidean n-space".Bulletin of the Institute of Mathematics and Its Applications.29 (11/12):172–174.
  11. ^Aomoto, Kazuhiko (1977)."Analytic structure of Schläfli function".Nagoya Math. J.68:1–16.doi:10.1017/s0027763000017839.
  12. ^Beck, M.; Robins, S.; Sam, S. V. (2010). "Positivity theorems for solid-angle polynomials".Contributions to Algebra and Geometry.51 (2):493–507.arXiv:0906.4031.
  13. ^Ribando, Jason M. (2006)."Measuring Solid Angles Beyond Dimension Three".Discrete & Computational Geometry.36 (3):479–487.doi:10.1007/s00454-006-1253-4.

Further reading

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  • Jaffey, A. H. (1954). "Solid angle subtended by a circular aperture at point and spread sources: formulas and some tables".Rev. Sci. Instrum.25 (4):349–354.doi:10.1063/1.1771061.
  • Masket, A. Victor (1957). "Solid angle contour integrals, series, and tables".Rev. Sci. Instrum.28 (3): 191.doi:10.1063/1.1746479.
  • Naito, Minoru (1957). "A method of calculating the solid angle subtended by a circular aperture".J. Phys. Soc. Jpn.12 (10):1122–1129.doi:10.1143/JPSJ.12.1122.
  • Paxton, F. (1959)."Solid angle calculation for a circular disk".Rev. Sci. Instrum.30 (4): 254.doi:10.1063/1.1716590.
  • Khadjavi, A. (1968). "Calculation of solid angle subtended by rectangular apertures".J. Opt. Soc. Am.58 (10):1417–1418.doi:10.1364/JOSA.58.001417.
  • Gardner, R. P.; Carnesale, A. (1969). "The solid angle subtended at a point by a circular disk".Nucl. Instrum. Methods.73 (2):228–230.doi:10.1016/0029-554X(69)90214-6.
  • Gardner, R. P.; Verghese, K. (1971). "On the solid angle subtended by a circular disk".Nucl. Instrum. Methods.93 (1):163–167.doi:10.1016/0029-554X(71)90155-8.
  • Gotoh, H.; Yagi, H. (1971). "Solid angle subtended by a rectangular slit".Nucl. Instrum. Methods.96 (3):485–486.doi:10.1016/0029-554X(71)90624-0.
  • Cook, J. (1980). "Solid angle subtended by a two rectangles".Nucl. Instrum. Methods.178 (2–3):561–564.doi:10.1016/0029-554X(80)90838-1.
  • Asvestas, John S..; Englund, David C. (1994). "Computing the solid angle subtended by a planar figure".Opt. Eng.33 (12):4055–4059.doi:10.1117/12.183402. Erratum ibid. vol 50 (2011) page 059801.
  • Tryka, Stanislaw (1997). "Angular distribution of the solid angle at a point subtended by a circular disk".Opt. Commun.137 (4–6):317–333.doi:10.1016/S0030-4018(96)00789-4.
  • Prata, M. J. (2004). "Analytical calculation of the solid angle subtended by a circular disc detector at a point cosine source".Nucl. Instrum. Methods Phys. Res. A.521 (2–3): 576.arXiv:math-ph/0305034.doi:10.1016/j.nima.2003.10.098.
  • Timus, D. M.; Prata, M. J.; Kalla, S. L.; Abbas, M. I.; Oner, F.; Galiano, E. (2007). "Some further analytical results on the solid angle subtended at a point by a circular disk using elliptic integrals".Nucl. Instrum. Methods Phys. Res. A.580:149–152.doi:10.1016/j.nima.2007.05.055.
Wikimedia Commons has media related toSolid angle.

External links

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  • Arthur P. Norton, A Star Atlas, Gall and Inglis, Edinburgh, 1969.
  • M. G. Kendall, A Course in the Geometry of N Dimensions, No. 8 of Griffin's Statistical Monographs & Courses, ed. M. G. Kendall, Charles Griffin & Co. Ltd, London, 1961
  • Weisstein, Eric W."Solid Angle".MathWorld.
Linear/translational quantitiesAngular/rotational quantities
Dimensions1LL2Dimensions1θθ2
Ttime:t
s		absement:A
m s		Ttime:t
s
1distance:d,position:r,s,x,displacement
m		area:A
m2		1angle:θ,angular displacement:θ
rad		solid angle:Ω
rad2, sr
T−1frequency:f
s−1,Hz		speed:v,velocity:v
m s−1		kinematic viscosity:ν,
specific angular momentum: h
m2 s−1		T−1frequency:f,rotational speed:n,rotational velocity:n
s−1,Hz		angular speed:ω,angular velocity:ω
rad s−1
T−2acceleration:a
m s−2		T−2rotational acceleration
s−2		angular acceleration:α
rad s−2
T−3jerk:j
m s−3		T−3angular jerk:ζ
rad s−3
Mmass:m
kg		weighted position:M ⟨x⟩ = ∑mxmoment of inertia: I
kg m2		ML
MT−1Mass flow rate:m˙{\displaystyle {\dot {m}}}
kg s−1		momentum:p,impulse:J
kg m s−1,N s		action:𝒮,actergy:ℵ
kg m2 s−1,J s		MLT−1angular momentum:L,angular impulse:ΔL
kg m rad s−1
MT−2force:F,weight:Fg
kg m s−2,N		energy:E,work:W,Lagrangian:L
kg m2 s−2,J		MLT−2torque:τ,moment:M
kg m rad s−2,N m
MT−3yank:Y
kg m s−3, N s−1		power:P
kg m2 s−3, W		MLT−3rotatum:P
kg m rad s−3, N m s−1
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