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From: kneller@socrates.ucsf.edu (Don Kneller)
Newsgroups: comp.graphics
Subject: Re: random points on surface of sphere
Keywords: uniform spherical distribution, random walk
Message-ID: <13956@cgl.ucsf.EDU>
Date: 9 May 90 00:22:29 GMT
References: <1523@ryn.esg.dec.com+ <40768@apple.Apple.COM> <1220@med.Stanford.EDU> <1990May7.132424.23204@laguna.ccsf.caltech.edu>
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Reply-To: kneller@socrates.ucsf.edu (Don Kneller)
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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).

I believe this is correct. I had previously posted that you should
pick phi such that:
	phi = arccos(random number between -1 and 1)

and since z is R * cos(phi), this becomes:
	z = random number between -R and R,

which is what Richard proposed.

To see this is indeed correct, look at the area of a strip of height "dz"
at angle phi on a sphere of radius R.
	
	area	= circumference * width of strip on sphere
		= 2 pi R sin(phi) * R d(phi)

as: z = R cos(phi), dz = - R sin(phi) d(phi) so

	area	= 2 pi R dz	(don't worry about the sign, dz and
				 d(phi) are in opposite directions)

That is, the area on the sphere of a strip of height "dz" is *independent*
of z. If the linear density of points along z is rho, then N = rho * dz
will fall between z and z + dz. The density of points on the sphere
will be:

	rho2	= N / area = rho * dz / (2 pi R dz)
		= rho / (2 pi R)	, independent of z.



The bottom line?:

		1) pick theta randomly between -pi and pi
		2) pick z randomly between -R and R


-----
	Don Kneller
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