Spherical Trigonometry

Transcription

1 Rigid Body Mechanics William B. Heard 2006 WILEY-VCH Verlag GmbH & Co. 213 Appendix A Spherical Trigonometry This brief account of spherical trigonometry is included to make the treatment of composition of rotations self-contained. Classical treatments of the subject, such as [86], are primarily analytical and can be a featureless collection of formulas. There is, however a geometric point of view [87, 88] which unifies and motivates the collection of formulas which define the subject. The account given here follows [87,88]. We consider a spherical triangle T defined by unit vectors a, b, andc (Fig. A.1). The lengths of the sides of the triangle, α, β and γ, are the angles between these unit vectors (assumed π) and the vertex angles, A, B and C, are the dihedral angles between the planes formed by pairs of the unit vectors. Associated with this spherical triangle is its polar triangle T, whose vertices are the poles of the great circles completing the sides of T. This relation is fully reciprocal T is the polar of T. Fig. A.1 (a) A triad of unit vectors and defining T and the reciprocal set defining T. (b) Spherical triangle T and its polar T on the unit sphere. Rigid Body Mechanics: Mathematics, Physics and Applications. William B. Heard Copyright 2006 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim ISBN:

2 214 A Spherical Trigonometry The geometrical relations have the following vector expressions: a b = cos γ b c = cos α c a = cos β (A.1) a b = cos C b c = cos A c a = cos B (A.2) a b = sin γ c b c = sin α a c a = sin β b (A.3) a b = sin C c b c = sin A a c a = sin B b (A.4) The minus signs appear in Eqs. (A.2) because the dihedral angle is the supplement of the angle between the normals to the associated planes. Note that here, and in the formulas below, once a relation is obtained others follow by permuting the symbols. The first series of formulas, the so-called law of sines for spherical triangles, follows from the vector identity a b c = b c a = c a b (A.5) so that (A.3) and (A.4) yield a b c = sin γ c c = sin β b b = sin α a a a b c = sin C c c = sin B b b = sin A a a from which the law of sines follows sin γ/sinc = sin β/sinb = sin α/sina. (A.6) (A.7) (A.8) The second series of formulas, the so-called law of cosines for spherical triangles, follows from the vector identity so that (A.3) and (A.4) yield (a b) (c d) =(a c)(b d) (a d)(b c) (A.9) cos α = cos β cos γ + sin β sin γ cos A cos β = cos γ cos α + sin γ sin α cos B cos γ = cos α cos β + sin α sin β cos C cos A = cos B cos C + sin B sin C cos α cos B = cos C cos A + sin C sin A cos β cos C = cos A cos B + sin A sin B cos γ (A.10) (A.11) (A.12) (A.13) (A.14) (A.15) The final series of formulas comes from the following pattern of substitution [86] starting from (A.10): sin β sin γ cos A = cos α cos β cos γ = cos α cos β(cos α cos β + sin α sin β cos C) = cos α sin 2 β cos β sin α sin β cos C

3 A Spherical Trigonometry 215 The complete series is sin γ cos A = cos α sin β sin α cos β cos C (A.16) sin γ cos B = cos β sin α sin β cos α cos C sin β cos C = cos γ sin α sin γ cos α cos B sin β cos A = cos α sin γ sin α cos γ cos B sin α cos B = cos β sin γ sin β cos γ cos A sin α cos C = cos γ sin β sin γ cos β cos A sin C cos α = cos A sin B + sin A cos B cos γ sin C cos β = cos B sin A + sin B cos A cos γ sin B cos γ = cos C sin A + sin C cos A cos β sin B cos α = cos A sin C + sin A cos C cos β sin A cos β = cos B sin C + sin B cos C cos α sin A cos γ = cos C sin B + sin C cos B cos α (A.17) (A.18) (A.19) (A.20) (A.21) (A.22) (A.23) (A.24) (A.25) (A.26) (A.27) The cosine laws (A.10 A.15) are tied together nicely in [88] using matrix notation. First, we obtain (A.10 A.12). Let {v 1, v 2, v 3 } = {a, b, c} Rescale the polar vectors to obtain the reciprocal basis as { a sin α {w 1, w 2, w 3 } = a b c, b sin β a b c, c } sin γ a b c From (A.7) the reciprocal relations are v i w j = δ ij Let matrices V and W consist of columns of these vectors represented in the standard basis V =[v 1 v 2 v 3 ] W =[w 1 w 2 w 3 ] then the reciprocal relation can be expressed W t V = V t W = I Let and let {θ 12, θ 13, θ 23 } = {α, β, γ} {φ 1, φ 2, φ 3 } = {A, B, C} c 12 = cos θ 12 s 12 = sin θ 12 etc.

4 216 A Spherical Trigonometry Now consider the matrices V t V and W t W. 1 c 12 c 13 V t V = c 12 1 c 23 c 13 c 23 1 and W t W =(V t V) 1 because the reciprocal relations yield W t WV t V = V 1 (V t ) 1 V t V = I Thus W t W = 1 det(v t V) s 2 23 c 13 c 23 c 12 c 12 c 23 c 13 c 13 c 23 c 12 s 2 13 c 12 c 13 c 23 c 12 c 23 c 13 c 12 c 13 c 23 s 2 12 or It follows that cos φ 3 (A.12), for example, is That is cos φ 3 = w 1 w 2 w 1 w 2 = (W t W) 12 (W t W) 11 (W t W) 22 = c 13c 23 c 12 s 23 s 13 c 12 = c 13 c 23 + s 23 s 13 cos φ 3 cos γ = cos α cos β + sin α sin β cos C The relations for cos φ 1 and cos φ 2 follow in the same way from the 2 3 and 1 3 entries. To obtain (A.13 A.15) the roles of v 1 and w i are reversed. Let {w 1, w 2, w 3 } = {a, b, c } Rescale the vertex vectors to obtain the reciprocal basis { } a sin α {v 1, v 2, v 3 } = a b c, b sin β a b c, c sin γ a b c The reciprocal relations are again W t V = V t W = I from which again W t W =(V t V) 1

5 A Spherical Trigonometry 217 Now W t W = 1 c 3 c 2 c 3 1 c 1 c 2 c 1 1 and V t V = 1 det(w t W) s 2 1 c 3 + c 1 c 2 c 2 + c 1 c 3 c 3 + c 1 c 2 s 2 2 c 1 + c 2 c 3 c 2 + c 1 c 3 c 1 + c 2 c 3 s 2 3 It follows that cos θ 12 (A.15), for example, is cos θ 12 = v 1 v 2 v 1 v 2 = (V t V) 12 (V t V) 11 (V t V) 22 = c 3 + c 1 c 2 s 1 s 2 or That is c 3 = c 1 c 2 + s 1 s 2 cos θ 12 cos C = cos A cos B + sin A sin B cos γ