Hi, racecar-ga:
The formula for area of a spherical triangle is known as Girard's
theorem:
area = (a + b + c - pi)*R^2
if the radius of the sphere is R. In your application the radius of
the sphere is 1, but it may be helpful to leave the factor R^2 intact
for the time being so the units of area are apparent in the formulas.
(We will measure angles in radians throughout.)
John C. Polking of Rice University maintains this nice Web site:
[The Geometry of the Sphere]
http://math.rice.edu/~pcmi/sphere/
where a proof of this and other "sphere facts" are illustrated by
interactive Java applets for clarity.
First the area of the whole sphere A = 4pi*R^2 is credited to
Archimedes:
[An ancient extra-geometric proof]
http://www.cut-the-knot.com/pythagoras/Archimedes.shtml
See Note 1 below for further comments on this.
From the area of the whole sphere one deduces the area of a "lune"
(region bounded by segments of two great circles). Because the two
points at which two great circles intersect are "antipodal" (at
opposite poles), one can visualize dividing the entire sphere into
some whole number q of parts, for which the area L of the lune is
related to the angle 2pi/q between the great circles in this manner:
L = (4pi*R^2)/q = 2(2pi/q)*R^2
By combining several such "integral subdivisions" of the whole sphere,
one has for any rational p/q the more general case of a lune with
angle 2pi(p/q):
L = (4pi*R^2)*(p/q) = 2(2pi(p/q))*R^2
One then argues by continuous dependence of area of the lune on this
angle that in general (since rational numbers can approximate any real
number with arbitrary accuracy):
L = 2a*R^2
when a is the angle between the great circles at either "corner" of
the lune.
For reference that argument is given here:
[The area of a lune]
http://math.rice.edu/~pcmi/sphere/gos3.html#1
With these two facts (area of the sphere, area of a lune) we have the
ingredients for proving Girard's Theorem:
[The area of a spherical triangle]
http://math.rice.edu/~pcmi/sphere/gos4.html#1
Let T be a spherical triangle with angles a,b,c at its three corners.
Extend the great circles which form the sides of T around the sphere,
and these meet in a congruent spherical triangle T' on the opposite
side of the sphere. Indeed these great circles bound three triangles
which share a side with T and meet T' at a corner and three other
triangles which share a side with T' and meet T at a corner.
What we will describe next may be difficult to visualize without
illustration, so by all means take a look (with a Java enabled
browser) at the nice 3D diagrams presented at the Web site above.
Just as T may be said to be antipodal to T', each of the first three
triangles is paired with one of the last three. Together T,T' and
such corresponding pair of other triangles form two lunes, each with
an angle a, b, or c taken from the respective corner of T. Altogther
then there are six lunes, and we can add up their areas in two
different ways.
On one hand we can simply add the areas of three pairs of lunes
individually:
2 * [2a*R^2 + 2b*R^2 + 2c*R^2] = 4(a + b + c)*R^2
On the other hand the six lunes collectively cover each part of the
sphere once except for T,T' which are covered three times apiece. So
the combined areas of these six lunes is also the area of the sphere
plus two (extra) copies of T,T':
4pi*R^2 + 2*(area of T) + 2*(area of T')
Now T,T' have equal areas, so by equating the above expressions and
solving for the area of T:
4*(area of T) = 4(a + b + c)*R^2 - 4pi*R^2
area of T = (a + b + c - pi)*R^2
Thus Girard's Theorem is proven. See Note 2 below about who Girard
was.
For the sake of completeness we recap the calculation initially
proposed by foodini in answer to your question.
Consider a regular tetrahedron with one vertex at the center of a unit
sphere and the other three vertices on the sphere's surface. The
three angles of the spherical triangle formed by those latter three
vertices are equal, and we can compute them as dihedral angles (angles
between two sides of the tetrahedron).
Drop a perpendicular from one vertex of the tetrahedron to the median
of the opposite equilateral triangular face. This "tetrahedral"
altitude forms a right triangle with a shorter leg that is 1/3 of the
altitude of that triangular face and a hypotenuse (on an adjacent
face) that is an entire "triangular" altitude. Hence, as foodini-ga
claims, the angle between those two triangular faces has cosine of
1/3. Since now R = 1, we have the solid angle as:
area of spherical triangle = 3 arccos(1/3) - pi
= 0.551285598432530807942144151464459... steradians
regards, mathtalk
Note 1:
[Archimedes on Circles and Spheres (scroll to bottom)]
http://www.maths.uwa.edu.au/~schultz/3M3/L7Archimedes1.html
Archimedes' "patented" Method of Exhaustion is more or less equivalent
to our modern day integral calculus, i.e. sum up a series of frustums
(banded sections) of cones to give the area of a hemisphere with ever
increasing accuracy. Archimedes also pointed out the relationship of
sphere volume V to sphere surface area A:
V = (1/3)*R*A
which today we'd likely present in terms of differentiating V with
respect to R, giving A.
A = dV/dR = 3V/R
Note 2:
Girard's Theorem is named after Albert Girard (1595-1632), a French
born mathematician who moved to the Netherlands for religious refuge:
[Mathematicians: Albert Girard]
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Girard_Albert.html
He was first to use the now familiar abbreviations sin, cos, tan for
elementary trigonometric functions.
Note 3:
A more general formula for the solid angle subtended by a triangle is
recounted here by A. van Oosterom:
[Re: Solid angle subtended by rectangle]
http://mathforum.org/epigone/geometry-puzzles/crurdrauku/jpxw4tv0za99@legacy
Let U,V,W be three vectors emanating from the origin, with respective
lengths u,v,w. The (signed) area corresponding to the solid angle
formed at this corner is:
2 arctan( det(M)/(uvw + u V.W + v U.W + w U.V) )
where M is the matrix formed by stacking U,V,W as rows and the period
"." is used to denote the dot product between vectors. Observe that
the denominator here is not affected by the ordering of the three
vectors, but the sign of the numerator alternates with transpositions
of any two vectors.
To duplicate the earlier numerical result one takes unit vectors:
U = (1,0,0)
V = (1/2,sqrt(3)/2,0)
W = (1/2,sqrt(3)/6,sqrt(6)/3)
det(M) = sqrt(2)/2
uvw + u V.W + v U.W + w U.V = 1 + 1/2 + 1/2 + 1/2
2 arctan((sqrt(2)/2)/(5/2)) = 0.551285598432530807942144151464459...
as previously found. |