Spherical geometry
- From: gregegan@xxxxxxxxxxxxxxxxxxxxx (Greg Egan)
- Date: Sat, 23 Jul 2005 21:19:03 +0800
Questions about spherical geometry seem to come up quite often on this
group, so I thought it might be worth posting proofs of the simplest
results for spherical triangles: the Cosine Laws, the Sine Law, and the
Area formula.
Suppose we have a triangle on the unit sphere, with A, B, C the angles at
the three vertices, and a, b, c the angles subtended at the centre of the
sphere by the sides opposite those vertices.
We can always choose a coordinate system such that the vertices are the
following unit vectors in R^3:
vA = (0, 0, 1)
vB = (sin c, 0, cos c)
vC = (cos A sin b, sin A sin b, cos b)
In terms of spherical polar coordinates, we put vertex A at the north
pole, vertex B on the prime meridian, and vertex C on the meridian with
longitude A. The co-latitudes (angles from the pole) of vertices B and C
are c and b respectively.
Cosine Laws
===========
The angles subtended by the sides of the triangle at the centre of the
sphere are just the angles between the vertices, and hence the cosine of
a, b or c will just be the dot product of the appropriate pair of
vertices. Because we've chosen b and c as parameters for the triangle's
geometry these formulas tell us nothing new:
cos b = vA.vC
cos c = vA.vB
but the formula for cos a gives us more information:
cos a = vB.vC
i.e.
cos a = cos A sin b sin c + cos b cos c
This is the First Cosine Law for spherical triangles.
Now, the sides of our spherical triangle are great circles, i.e. the
intersections of planes through the origin with the unit sphere, and we
can find vectors that are normal to these planes by taking cross products
of pairs of vertices:
sa = vC x vB
sb = vA x vC
sc = vB x vA
We've chosen the orders of the cross products here so that these vectors
point out of the triangle. The angles at the triangle's vertices, which
are just the angles between the planes that define the sides, are related
to the angles between pairs of these side normals:
(angle between sb and sc) = pi-A
(angle between sa and sc) = pi-B
(angle between sa and sb) = pi-C
Now, we can construct a second spherical triangle whose vertices are the
unit vectors found by re-scaling sa, sb and sc to unit length. The angles
subtended at the centre of the sphere by the sides of this second triangle
are thus pi-A, pi-B and pi-C.
If we then go on to construct side normals for this second triangle, they
are (if re-scaled to unit length) just the original vertices vA, vB and
vC. This follows from the fact that, e.g.
sa x sb = (vC x vB) x (vA x vC)
and since sa and sb are both perpendicular to vC, they lie in the plane
normal to vC, and their cross product must be parallel (or anti-parallel)
to vC.
So our second triangle has angles at its vertices of pi-a, pi-b and pi-c.
Substituting the geometry of the second triangle into the First Cosine Law
gives the Second Cosine Law:
cos A = cos a sin B sin C - cos B cos C
Sine Law
========
The scalar triple product of vA, vB and vC is:
(vA x vB).vC =
(0, sin c, 0).(cos A sin b, sin A sin b, cos b) =
sin A sin b sin c
But the scalar triple product of three vectors is unchanged under cyclic
permutations of the vectors, so we must have:
sin A sin b sin c
= sin B sin c sin a
= sin C sin a sin b
If we divide through by (sin a sin b sin c) we get:
sin A / sin a
= sin B / sin b
= sin C / sin c
which is the Sine Law for spherical triangles.
Area Formula
============
Our spherical triangle can be defined as the region on the unit sphere
bounded by three different planes through the origin: the planes normal
to the vectors sa, sb and sc.
The area of the region bounded by just two planes through the origin is
easy to compute. If A is the angle between the planes, then the area is
equal to the fraction A / (2 pi) of the total surface area of the unit
sphere, 4 pi:
4 pi A / (2 pi) = 2A
Where we have three planes, they divide the surface of the sphere into
eight regions, depending on whether the sign of the dot products with sa,
sb and sc are positive or negative. By symmetry, these regions come in
pairs of equal area when we swap all the signs. If we label the areas
according to the eight patterns of signs, and make use of this symmetry,
we get:
Sign of dot product Area of
sa sb sc region
+ + + t
+ + - u
+ - + v
+ - - w
- + + w
- + - v
- - + u
- - - t
Here t is the area of the triangle we're interested in, and u, v and w are
areas of other regions.
But our earlier result about the area of a region bounded by just two
planes also gives us:
t + u
= area of region bounded by planes
normal to sa and sb
= 2C
t + v
= area of region bounded by planes
normal to sa and sc
= 2B
t + w
= area of region bounded by planes
normal to sb and sc
= 2A
Adding these three equations gives:
3t + u + v + w = 2(A+B+C)
We can also sum the areas of four different regions to get half the area
of the unit sphere:
t + u + v + w = 2 pi
>From these two equations, we can solve for t, the area of a spherical
triangle with vertex angles A, B and C:
t = A+B+C-pi
This is the area formula for the unit sphere. For a sphere of radius r,
just multiply by r^2.
--
Greg Egan
Email address (remove name of animal and add standard punctuation):
gregegan netspace zebra net au
.
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