Spherical Trigonometry

Jupsat Pro Astronomy Software

Secrets of the Deep Sky

Get Instant Access

For the coordinate transformations of spherical astronomy, we need some mathematical tools, which we present now.

If a plane passes through the centre of a sphere, it will split the sphere into two identical hemispheres along a circle called a great circle (Fig. 2.1). A line perpendicular to the plane and passing through the centre of the sphere intersects the sphere at the poles P and P'. If a sphere is intersected by a plane not containing the centre, the intersection curve is a small circle. There is exactly one great circle passing through two given points Q and Q' on a sphere (unless these points are an-

Fig. 2.1. A great circle is the intersection of a sphere and a plane passing through its centre. P and P' are the poles of the great circle. The shortest path from Q to Q' follows the great circle

tipodal, in which case all circles passing through both of them are great circles). The arc QQ of this great circle is the shortest path on the surface of the sphere between these points.

A spherical triangle is not just any three-cornered figure lying on a sphere; its sides must be arcs of great circles. The spherical triangle ABC in Fig. 2.2 has the arcs AB, BC and AC as its sides. If the radius of the sphere is r, the length of the arc AB is

|AB\=rc , [c]=rad, where c is the angle subtended by the arc AB as seen from the centre. This angle is called the central angle of the side AB. Because lengths of sides and central

Spherical Triangle And Great Circle
Fig. 2.2. A spherical triangle is bounded by three arcs of great circles, AB, BC and CA. The corresponding central angles are c, a, and b

Hannu Karttunen et al. (Eds.), Spherical Astronomy.

In: Hannu Karttunen et al. (Eds.), Fundamental Astronomy, 5th Edition. pp. 11-45 (2007) DOI: 11685739_2 © Springer-Verlag Berlin Heidelberg 2007

angles correspond to each other in a unique way, it is customary to give the central angles instead of the sides. In this way, the radius of the sphere does not enter into the equations of spherical trigonometry. An angle of a spherical triangle can be defined as the angle between the tangents of the two sides meeting at a vertex, or as the dihedral angle between the planes intersecting the sphere along these two sides. We denote the angles of a spherical triangle by capital letters (A, B, C) and the opposing sides, or, more correctly, the corresponding central angles, by lowercase letters (a, b, c).

The sum of the angles of a spherical triangle is always greater than 180 degrees; the excess

is called the spherical excess. It is not a constant, but depends on the triangle. Unlike in plane geometry, it is not enough to know two of the angles to determine the third one. The area of a spherical triangle is related to the spherical excess in a very simple way:

This shows that the spherical excess equals the solid angle in steradians (see Appendix A.1), subtended by the triangle as seen from the centre.

Spherical Trigonometry
Fig. 2.3. If the sides of a spherical triangle are extended all the way around the sphere, they form another triangle A', antipodal and equal to the original triangle A. The shaded area is the slice S(A)

To prove (2.2), we extend all sides of the triangle A to great circles (Fig. 2.3). These great circles will form another triangle A', congruent with A but antipodal to it. If the angle A is expressed in radians, the area of the slice S( A) bounded by the two sides of A (the shaded area in Fig. 2.3) is obviously 2A/2n = A/n times the area of the sphere, 4nr2. Similarly, the slices S(B) and S(C) cover fractions B/n and C/n of the whole sphere.

Together, the three slices cover the whole surface of the sphere, the equal triangles A and A' belonging to every slice, and each point outside the triangles, to exactly one slice. Thus the area of the slices S(A), S(B) and S(C) equals the area of the sphere plus four times the area of A, A(A):

As in the case of plane triangles, we can derive relationships between the sides and angles of spherical triangles. The easiest way to do this is by inspecting certain coordinate transformations.

Fig. 2.4. The location of a point P on the surface of a unit sphere can be expressed by rectangular xyz coordinates or by two angles, ^ and 0. The x'y'z' frame is obtained by rotating the xyz frame around its x axis by an angle x

2.1 Spherical Trigonometry

Sine Terms Coordinate
Fig. 2.5. The coordinates of the point P in the rotated frame are x' = x, y' = y cos x + z sin x, z' = z cos x — y sin x

Suppose we have two rectangular coordinate frames Oxyz and Ox'y'z' (Fig. 2.4), such that the x'y'z' frame is obtained from the xyz frame by rotating it around the x axis by an angle x ■

The position of a point P on a unit sphere is uniquely determined by giving two angles. The angle f is measured counterclockwise from the positive x axis along the xy plane; the other angle 0 tells the angular distance from the xy plane. In an analogous way, we can define the angles f' and 0', which give the position of the point P in the x'y'z' frame. The rectangular coordinates of the point P as functions of these angles are:

x = cos f cos 0 , y = sin f cos 0 , z = sin 0, x' = cos f ' cos 0' , y'= sin f ' cos 0 (2.3)

x = x, y = y cos x + z sin x , z' = -y sin x + z cos x .

Spherical Coordinate Formul Efw
Fig. 2.6. To derive triangulation formulas for the spherical triangle ABC, the spherical coordinates f, 0, f ' and 0' of the vertex C are expressed in terms of the sides and angles of the triangle

In fact, these equations are quite sufficient for all coordinate transformations we may encounter. However, we shall also derive the usual equations for spherical triangles. To do this, we set up the coordinate frames in a suitable way (Fig. 2.6). The z axis points towards the vertex A and the z' axis, towards B. Now the vertex C corresponds to the point P in Fig. 2.4. The angles f, 0, f, 0' and x can be expressed in terms of the angles and sides of the spherical triangle:

We also know that the dashed coordinates are obtained from the undashed ones by a rotation in the yz plane (Fig. 2.5):

By substituting the expressions of the rectangular coordinates (2.3) into (2.4), we have cos f cos 0' = cos f cos 0 , sin f cos 0' = sin f cos 0 cos x + sin 0 sin x , (2.5) sin 0' = — sin f cos 0 sin x + sin 0 cos x ■

f = A — 90°, 0 = 90° — b , f = 90° — B , 0' = 90° — a , x = c ■

Substitution into (2.5) gives cos(90° — B) cos (90° — a) = cos(A — 90°) cos(90° — b), sin(90° — B) cos(90° — a) = sin(A — 90°) cos(90° — b) cos c + sin(90° — b) sin c , sin (90° — a) = — sin(A — 90°) cos(90° — b) sin c + sin (90° — b) cos c ,

or sin B sin a = sin A sin b , cos B sin a = — cos A sin b cos c + cos b sin c , (2.7) cos a = cos A sin b sin c + cos b cos c .

Equations for other sides and angles are obtained by cyclic permutations of the sides a, b, c and the angles A, B, C. For instance, the first equation also yields sin C sin b = sin B sin c , sin A sin c = sin C sin a .

All these variations of the sine formula can be written in an easily remembered form:

sin b sin a sin c

sin A sin B sin C

If we take the limit, letting the sides a, b and c shrink to zero, the spherical triangle becomes a plane triangle. If all angles are expressed in radians, we have approximately sin a cos a :

Substituting these approximations into the sine formula, we get the familiar sine formula of plane geometry:

abc sin A sin B sin C

The second equation in (2.7) is the sine-cosine formula, and the corresponding plane formula is a trivial one:

This is obtained by substituting the approximations of sine and cosine into the sine-cosine formula and ignoring all quadratic and higher-order terms. In the same way we can use the third equation in (2.7), the cosine formula, to derive the planar cosine formula:

Was this article helpful?

0 0
Telescopes Mastery

Telescopes Mastery

Through this ebook, you are going to learn what you will need to know all about the telescopes that can provide a fun and rewarding hobby for you and your family!

Get My Free Ebook

Post a comment