This is not so much a question as it is a comment on a previous
discussion.  If that's inappropriate, don't post it!

A few weeks ago, I asked Google Answers how to calculate the length of
a curve on a sphere of radius r, where the curve began at the equator
and ended at the north pole, with each point on the path having its
north latitude and its west longitude equal.  Manuka responded that
the answer would be r. sqrt(2). E(1/2), where E represents a complete
elliptic integral of the second kind.  A decimal approximation of
Manuka's solution matched the decimal approximation I had earlier
computed empirically.

For what it is worth, I found in my CRC Handbook a discussion of
elliptic integrals that allowed me to express this quantity as the sum
of an infinite series.
Thus: 

r. sqrt(2). E(1/2) =

r * pi * sqrt(2) * (1/2 - [(1/2)^2 * (1/2)^2] - [(3/(2*4))^2 * (1/2)^3
* 1/3] - [((3*5)/(2*4*6))^2 * (1/2)^4 * 1/5] - [((3*5*7)/(2*4*6*8))^2
* (1/2)^5 * 1/7] - . . .)

emoll

Makes you glad for computers, doesn't it?

I presume there are more efficient implementations, since this one
doesn't seem to be converging very quickly. Each term is the last one
multiplied by
[(2n-1)/(2n)]^2 . (1/2) . (2n-3)/(2n-1) 
which obviously converges to 1/2 as n grows large. So you need another
3 and a bit terms for every extra digit of accuracy in your answer -
not an appealing prospect!