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From: gbrown@tybalt.caltech.edu (Glenn C. Brown)
Newsgroups: comp.graphics
Subject: Re: random points on surface of sphere
Keywords: uniform spherical distribution, random walk
Message-ID: <1990May7.132424.23204@laguna.ccsf.caltech.edu>
Date: 7 May 90 13:24:24 GMT
References: <1523@ryn.esg.dec.com+ <40768@apple.Apple.COM> <1220@med.Stanford.EDU>
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Organization: California Institute of Technology
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rick@hanauma.stanford.edu (Richard Ottolini) writes:

>+1. Pick a random angle, 0 <= theta < 360. This is in
>+the x-y plane.
>+
>+2. Pick a random z-value from -1 to 1 ( "<=" or "<" ?
>+I'm not sure).

>Huh?  Each latitude will have the same fraction of points and things
       ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>will appear more crowded at the poles.

Latitude is the angle of elevation (usually refered to as phi)
in polar coordinates.  Choosing a random z is not equivalent to choosing
a random phi.  Therefore, each latitude will not have the same fraction of
points.

I believe the posted method may be right.  If it is, the proof should be
straightforward (I just completed a course in surface integrals), and
I'll post it in a few days if noone else beats me to it. (That's
assuming the above is true.)

--Glenn