[Proj] Area of a spherical polygon

[Charles Karney]

The area of a polygon on a (unit) sphere is given by the spherical
excess

  A = 2*pi - sum(exterior angles)

However this is badly conditioned if the polygon is small.  In this
case, use

  A = sum S_{i,i+1} over the edges of the polygon

where S12 is the area of the quadrilateral bounded by an edge of the
polygon, two meridians, and the equator, i.e., with vertices

  (phi1,lambda1); (phi2,lambda2);
  (0,   lambda1); (0,   lambda2); 

S12 is given by

  tan(S12/2) = tan(lambda12/2) * ( tan(phi1/2) + tan(phi2/2) )
                           / ( 1 + tan(phi1/2) * tan(phi2/2) )
             = tan(lambda12/2) * tanh( (lam(phi1) + lam(phi2))/2 )

where lambda12 = lambda2 - lambda1 and lam(x) is the lambertian (or
inverse gudermannian) function

  lam(x) = asinh(tan(x)) = atanh(sin(x)) = 2*atanh(tan(x/2))

Notes:

* the formula for S12 is exact, except that...

* it is indeterminate if an edge is a semi-circle;

* the formula for A applies only if the polygon does not include a pole
  (if it does, then add +/- pi to the result);

* in the limit of small phi and lambda, S12 reduces to the trapezoidal
  formula, S12 = (lambda2 - lambda1) * (phi1 + phi2)/2;

* I derived this result from the equation for the area of a spherical
  triangle in terms of two edges and the included angle given by, e.g.,
  Todhunter, Spherical Trig. (1871), Sec. 103, Eq. (2).  See
    http://books.google.com/books?id=3uBHAAAAIAAJ&pg=PA71

* I would be interested to know if this formula for S12 is already
  known.

  --Charles

-- 
Charles Karney <ckarney at sarnoff.com>
Sarnoff Corporation, Princeton, NJ 08543-5300

Tel: +1 609 734 2312
Fax: +1 609 734 2662