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Solid Angle of Conical Surfaces, Polyhedral Cones, and Intersecting Spherical Caps

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Solid Angle of Conical Surfaces, Polyhedral Cones, and Intersecting Spherical Caps Solid Angle of Conical Surfaces, Polyhedral Cones, and Intersecting Spherical Caps Oleg Mazonka, 2012 Abstract This paper presents formulae for calculation the solid angle of intersecting spherical caps, conical surfaces and polyhedral cones. 1. Introduction ............................................................................................................ 1 2. Conical surfaces ..................................................................................................... 1 3. Polyhedral cones .................................................................................................... 4 4. Intersecting cones................................................................................................... 7 5. Conclusion ........................................................................................................... 13 1. Introduction The problem of calculating solid angles appears in many areas of science and applied mathematics. While calculation of the solid angle of a right circular cone is a simple exercise, calculation of solid angles of arbitrary conical shapes is not trivial. In this paper I would like to show a few solutions for solid angle of conical shapes and as a special case the intersection of two cones. 2. Conical surfaces 2.1 Solid angle of conical surface Let a conical surface be defined by a closed parametric curve in space with apex positioned at the origin of coordinate system. This parametric curve can be projected onto the unit sphere by renormalising the distance to each point on the curve. Let us parameterise a projected curve by the angle (being a linear distancer on the surface of the unit sphere) l, so that the curve on the sphere is defined as (ls ), and denote the first and the second derivatives with symbols u and τ: r d r r d 2 r τ = s u = s 1 dl dl 2 Since the curve is quite arbitrary it can be formed by a number of vertices (corners) δ connected by smooth (up to second derivative) arcs. Let i be a turn angle of a corner δ i, so that i and the inward angle are supplementary. u and τ are not defined at τ − corners. For each corner i it is possible to define the tangent i of one side of the τ + δ corner and the tangent i of the other side. Having both tangents, the angle i can be calculated via trigonometric functions, such as: ~ 1 ~ τ − ×τ + sin δ i i tan δ = i = 2 i cos δ τ − ⋅τ + i i i As will be proven later in this section, the solid angle of a conical surface can be calculated as r 2 r r 2 Ω = π − δ − − ()⋅ 2 ∑ i ∫ dl u us 3 i δ where the sum is taken over all corners i and the line integral is taken along the closed curve excluding corners (in which u is undefined). r A special case of the closed curve s consisting of segments of great circles gives the known formula (Ref [1]) for a spherical polygon: n n Ω = π − δ = ()()π − δ − − π 2 ∑ i ∑ i n 2 4 i=1 i=1 because the integral along segments of great circles is equal to zero. In the absence of corners on the closed curve the second derivative u is well-defined everywhere and hence the above equation can be written as: r r r Ω = 2π − ∫ dl u 2 − ()⋅us 2 5 2.2 Right circular cone example For example, a circle on the unit sphere with an apex angle θ can be parameterised as: r s(ϕ) = (sin θ cos ϕ sin, θ sin ϕ cos, θ ) 6 Then taking into account that dl = dϕ sin θ r r r 2 2 ϕ 2 = sd = sd d = 1 ()− θ ϕ − θ ϕ = u 2 2 2 sin cos , sin sin 0, dl dϕ dl sin θ 1 7 = ()− cos ϕ,−sin ϕ 0, sin θ Hence r 1 1 u 2 = ()sin 2 ϕ + cos 2 ϕ = sin 2 θ sin 2 θ 8 r r 1 ()s ⋅u = ()− sin θ cos 2 ϕ − sin θ sin 2 ϕ −= 1 sin θ And ~ 2 ~ r r r Ω = π − 2 − ()⋅ 2 = π − ϕ θ 1 − = 2 ∫ dl u s u 2 ∫ d sin 2 1 sin θ 9 = 2π − cos θ ∫ dϕ =2π ()1− cos θ which is a known formula for the solid angle of a cone with apex angle θ. 2.3 Derivation Now let us prove Eq 3 by deriving it from Eq 4. Let us break the smooth parts of the curve into infinitesimally small segments of size dl . Let δ * be an angle enumerated r rj r by index j on the sphere between a tangent τ at the point s( l) and a vector τ ′ which r r is a projection of the tangent τ (l + dl ) to the plane normal to s . Here the smooth δ * curve is approximated by a polygon with N edges and turn angles j . As the polygon → ∞ δ * δ tends to the curve ( N ) all angles j go to zero while i are fixed at n vertices. δ * δ The final result can be expressed if the sum of all j and i between segments is known, Eq 4: n N * Ω = π − δ = π − δ + δ 2 ∑ 2 ∑ i lim ∑ j 10 N →∞ i=1 j=1 r r r r r r Given that τ (l + dl ) =τ + u dl , (s ⋅τ ) = 0 , and a projection of any vector x on the r r r r r sphere is x − s( s ⋅ x), τ ′ becomes: r r r r r r r r r r r τ ′ = τ (l + dl )− s( s ⋅τ (l + dl )) = τ + udl − s( s ⋅ (τ + udl)) = r r r r r 11 =τ + ()u − s() s ⋅u dl δ * The angles j can be calculated from the scalar products: r r r r r r r r τ ⋅τ ′ τ ⋅ (τ + (u − s( s⋅ u))dl ) 1 cos δ * = = = ′ r r r r r 2 2 r r r r 2 12 τ τ τ ()τ + ()u − s() s⋅ u dl 1+ dl ()u − s() s⋅ u r r because (u ⋅τ ) = 0 and τ = 1. Here index j is implicit. Since both δ * and dl are small Eq 12 can be rewritten as: δ 2* 1 r r r r 1 r r r 1− = 1− dl 2 ()u − s() s⋅ u 2 = 1− dl 2 (u 2 − ()s ⋅u 2 ) ⇒ 2 2 2 13 r r r ⇒ δ * = dl u 2 − ()s ⋅u 2 δ * In the limit the sum of j must be replaced by line integral: N r r r δ * = 2 − ()⋅ 2 lim ∑ j dl u s u 14 N →∞ ∫ j =1 Combining together Eqs 10 and 14 proves Eq 3. ~ 3 ~ 2.4 Arbitrary smooth generating curve In a similar line of thought a generic formula for an arbitrary parameterised line in space L(t) without corners can be derived: r r L × L Ω = π − 1 2 15 2 ∫ 2 dt L1 where L1 and L2 are sphere tangential projections of the first and the second derivatives of L : r r r r r = ′ − ( ⋅ ′) L1 L s s L r r r r r = ′′ − ()⋅ ′′ L2 L s s L 16 r r s = L L The numerator in the expression under the integral in Eq 15 is the absolute value of the vector product. Eq 15 reduces to Eq 5 when L( t) = (ls ): r r r r r r =τ − ( ⋅τ ) = τ L1 dt dl s s dl dl r r r r r ( )2 = ( )2 − (⋅ )( )2 L2 dt u dl s s u dl r r ()2 = ()()τ 2 = 2 L1 dt dl dl r r 17 L × L r r r r r 1 2 = τ × ()− ()⋅ = 2 dt u s s u dl L1 r r r r r r r r r r r r r r = dl τ 2 ()()u − s() s ⋅u 2 − ()τ ⋅ ()u − s s ⋅u 2 = dl ()u − s() s ⋅u 2 3. Polyhedral cones 3.1 Solid angle of discrete shape Eq 3 is quite general but not always very efficient when many corner angles δ have to be calculated. Often the generating curve is approximated by a sequence of points in which case the line integral of Eq 3 has to be replaced with the sum over corner angles. It might also be important that the sum is calculated in an efficient way. r r r Let a polyhedral cone be defined by a list of n unit vectors {s ,s ,..., s }. Let j be a r r r r r 1 2 n = = circular index in s j so that sn+1 s1 and s0 sn . Let us also define the following values: r r a = s − ⋅ s + j r j 1 r j 1 = ⋅ bj s j −1 s j r r 18 c = s ⋅ s + = b + j rj j r1 rj 1 = ⋅ []× d j s j −1 s j s j+1 ~ 4 ~ r r r Vectors {s − ,s ,s + } form a spherical triangle. a, b, and c are cosines of the triangle j 1 j j 1 {r r r } sides, and d is a volume of the parallelepiped spanned by the vectors s j −1,s j,s j +1 . The angle δ is supplementary to the inward angle (π − δ ) of the triangle at the r j j vertex s j . The spherical law of cosines reads (assuming implicit index j): a = bc + 1− b2 1− c2 cos (π − δ ) = bc − 1− b2 1− c2 cos δ 19 from which sine of the angle δ can be found: d = 1− b2 1− c2 sin δ 20 This can be directly verified by taking into account that d 2 = 1+ 2abc − a2 − b2 − c2 (which in turn can be obtained from the determinant formula for a scalar triple product ).
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