San José State University

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 Spherical Geometry and Trigonometry

Spherical geometry and trigonometry used to be important topics in a technical education because they were essential for navigation. During that time an important element of their presentation was the matter of making accurate computations. This meant largely learning to use logarithms and the tables of logarithms. The modern computer has relieved users of this burden. The computers, however, not only lifted the computational burden they did away with the field. The formulas of spherical trigonometry were programmed into the computers and allowed users to use the results without knowing anything about the formulas or their derivation. Now there are virtually no books on spherical geometry and trigonometry in the libraries which have been published since the early 1940's. To make matters worse the old books in the libraries tend to get discarded so all of the spherical geometry and trigonometry texts are at risk of being discarded.

This lack of recent texts on spherical geometry and trigonometry is puzzling because the use of computers should shift the emphasis from numerical computation to theory. This page is an attempt to present derivations of important results from spherical geometry and trigonometry.

## Great Circle Distances

The problem is to determine the great circle distance between two points given their latitudes and longitudes. Let φ1 and φ2 be the latitudes of the two points and θ1 and θ2 Their longitudes. The basis for the determination of the angular separation of the two points on the great circle which connects them is the Law of Cosines for plane triangles.

The Law of Cosines states that for the above triangle

#### a2 = b2 + c2 - 2bc*cos(A)

When the points are on a great circle this formula reduces to:

#### a2 = 2r2(1 - cos(A))or cos(A) = (1 - (a/r)2/2)

Therefore if we can find the straight line (Euclidean distance between the two points) we can find their great circle angular separation A. The Euclidean distance between the two points can be found from their Euclidean coordinates, (x1, y1, z1) and (x2, y2, z2), by the formula

#### a2 = (x2 - x1)2 + (y2 - y1))2 + (z2 - z1)2

The Euclidian coordinates are given by the transformation

Thus

#### (a/r)2 = (cos(θ2)cos(φ2) - cos(θ1)cos(φ11))2+ (sin(θ2)cos(φ2) - (sin(θ1)cos(φ>1))2+ (sin(φ2) - sin(φ1))2

When the squared terms in the above expression are expanded we get from the first term, in addition to the cross product term, terms of the form

#### cos2(θi)cos2(φi) for i= 1 and 2.

These combine with terms from the second squared term of the form

#### sin2(θi)cos2(φi)

to give terms of the form,

#### cos2(φi).

These, in turn, combine with the terms from the third squared term to give (1+1). Thus,

#### (a/r)2 = 2 - 2cos(θ2)cos(θ1)cos(φ2)cos(φ1) - 2cos(φ2)cos(φ1) - 2sin(φ2)sin(φ1)

Thus the formula for the cosine of the great circle angular separation reduces to:

#### cos(A) = cos(θ2)cos(θ1)cos(φ2)cos(φ1) + sin(θ2)sin(θ1)cos(φ2)cos(φ1) + sin(φ2)sin(φ1)

With rearrangement this can be written as:

#### cos(A) = [cos(θ2)cos(θ1) + sin(θ2)sin(θ1)]cos(φ2)cos(φ1) + sin(φ2)sin(φ1)

The term within the brackets can also be expressed in terms of the difference of longitudes. Hence the standard formula for great circle angular separation:

#### cos(A) = sin(φ2)sin(φ1) + cos(φ2)cos(φ1)cos(θ2 - θ1)

For values of A close to zero the above formulae is highly sensitive to rounding errors in the computation. A better form (developed by Sinnott) for small values of A is:

## The Area of Great Circle Triangles

The method of computing the area of great circle triangles on a sphere is amazingly simple. The area of such a triangle is proportional to the amount by which the sum of the angles of the triangle in radians is in excess of π. The area is also proportional to the square of the radius of the sphere. The formula is easily illustrated. Consider a right triangle with its base on the equator and its apex at the north pole, at which the angle is π/2. The sum of the angles is 3π/2 so the excess is π/2. Such a triangle takes up one eighth of the surface of its sphere, whose area is 4πr2 where r is the radius. The area of the triangle is thus one eighth of the area of the whole sphere or (π/2)r2. The formula checks out for this case. The proof for the general case is given below.

First consider the area of a lune, the area between two great circles as shown below.

The area of a lune is proportional to the angle defining the lune. The area is a lune is the same proportion of the area of the sphere as the angle is a proportion of 2π radians; i.e.,

#### Area of lune = [(angle of lune)/2π] area of sphere = [(angle of lune)/2π][[(4πr2] = 2(angle of lune)r2 where r is the radius of the sphere

If the unit of area is taken to be r2 then we can say the area of a lune is twice the angle of the lune.

Now consider an arbitrary spherical triangle ABC as shown below.

The great circle passing through the vertices of ABC can be extended where on the other side of the sphere they intersect again. Let A', B' and C' be the antipodal points of A, B and C, respectively.

It is difficult to depict the various triangles and lunes form by the three great circles defining the triangle ABC so in the diagram below the various triangles from the sphere are displayed as though they were peeled off of the sphere and flattened out. The viewer has to make allowances for the fact that the diagram is schematic rather than exact.

In the above diagram:

• ΔABC + ΔBA'C = lune ABA'C
area of lune ABA'C = 2*(angle at A)
• ΔABC + ΔACB' = lune BCB'A
area of lune BCB'A = 2*(angle at B)
• ΔABC + ΔABC' = lune CAC'B
area of lune CAC'B = 2*(angle at C)

If we add up the left and right sides of the above equations we get

#### 2*(area of ΔABC) + (area of (ΔABC + ΔBA'C + ΔACB' + ΔABC') = 2*(angle at A + angle at B + angle at C)

A great circle is the intersection of a sphere and a plane passing through the center of the sphere. Thus a great circle divides the sphere exactly in half and hence the sum of all the areas on one side of any great circle is exactly equal to half of the area of a sphere.

Consider the great circle that the side AB is on. The entire hemisphere containing the vertex C is composed of the following spherical triangles: ABC + BA'C + A'CB' + AB'C. The sum involved in the previous step includes ABC' but not A'CB'.

The area of A'CB' is equal to that of AC'B. These triangles are not congruent but they are mirror images of each other and therefore have equal areas. One can be substituted for the other. Thus the sum of the areas is equal to 2π units of area.

The equation we now have is that:

#### 2*(area of ΔABC) + 2π = 2*(angle A + angle B + angle C). or area of ΔABC = Sum of angles at vertices - π = Spherical Excess.

Although the above proof used the arc AB and the vertex C the same analysis could be done with BC and A or AC and B. The equality of the areas of A'CB' and AC'B applies to any two triangles in which the labels differ only by changing the vertex from no prime to prime or from prime to no prime. Thus the areas of A'BC and AB'C' are equal. Likewise A'B'C' has the same area as ABC.

Thus the area of a spherical triangle on a unit sphere is equal to the spherical exess, which is the sum of vertex angles in excess of π radians. The area of a spherical triangle on a sphere of radius r is equal to the spherical excess times r2.

This relationship for the area of a spherical triangle generalizes to convex spherical polygons with the spherical excess being the sum of the angles - (n-2)π, where n is the number of sides of the polygon.

## The Triangle of the Poles for the Sides of a Spherical Triangle

A side of a spherical triangle is the intersection of a plane passing through the center of a sphere with the surface of the sphere. A line perpendicular to this plane and passing through the center of the sphere would intersect the sphere at what would be the poles of the sphere if the plane were the equatorial plane.

For the spherical triangle ABC let A' be the pole for side BC which is closest to A. Likewise B' and C' are the poles for AC and AB, respectively. The spherical triangle formed by connecting A', B' and C' with great circles is called the polar triangle for the spherical triangle ABC.

Consider what the polar triangle would be for A'B'C'. OC' is by construction perpendicular to OA and OB. A line which is perpendicular to OC' has to be in the plane of OA and OB. Likewise B' is perpendicular to OA and OC. A line which is perpendicular to OB' has to be in the plane of OA and OC. The perpendicular to OC' and OB' which establishes the pole of B'C' therefore has to be in both the two planes, of OA and OB and of OA and OC. The only lines that are in both of those planes are the ones colinear with OA. Since the point A is on the surface of the unit sphere it is the pole for B'C'. Likewise B is the pole for A'C' and C is the pole for A'B'. Thus the polar triangle of A'B'C' is the triangle ABC. So in general the polar triangle of the polar triangle of a spherical triangle is the spherical triangle itself.