nobody — Thu, 27 Jan 2005 18:08:18 GMT

Logged In: NO
it is correct that each coordinate is selected independently,
but with a normal distribution.
(warning: the following is a little fast and loose)
for a normal distribution, the probability of landing at x will be
a^(-b*x^2)
if you use the same distrubution for y and z, the probability of
landing at (x,y,z) will thus be:
a^(-b*x^2) * a^(-b*y^2) * a^(-b*z^2) =
a^(-b*(x^2+y^2+z^2))
you can recognize the latter as:
a^(-b*l^2), where l = length(x,y,z)
in other words, the probability of landing at (x,y,z) is only a
function of the *length* of (x,y,z), i.e. it is unrelated to the
*direction* of (x,y,z)
btw, mathworld has a great page on this subject:
http://mathworld.wolfram.com/SpherePointPicking.html
/michael toksvig

status changed

usu_acm — Fri, 28 Jan 2005 04:31:26 GMT

status assigned → closed

Logged In: YES
user_id=1124226
To: nobody.
Perfect! thnx a lot!

Boost C++ Libraries: Ticket #347: boost::uniform_on_sphere in not uniform on sphere

status changed