 | | | Random points on a unit sphere without biasing | Random points on a unit sphere without biasing 2005-06-14 - By kim aldis
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True.
A Gaussian distribution would concentrate the generated points towards the
centre so you'd have fewer outside the sphere. Not perfect but a deal
better.
<trivia>
Gaussian distribution just requires averaging a number of random numbers.
The average of an infinite number of random numbers between lying between -1
and 1 is zero. I once knew a Cambridge graduate mathematician who couldn't
get that.
</trivia>
> -- --Original Message-- --
> From: owner-xsi@(protected)
> [mailto:owner-xsi@(protected)] On Behalf Of Alan Jones
> Sent: 14 June 2005 21:58
> To: XSI@(protected)
> Subject: Re: Random points on a unit sphere without biasing
>
> X, Y, Z ones would be the same as doing a cube getting more
> in the corners. Apparently gaussian is special... At least
> that's what I assume from what I read on the mathworld page.
>
> Cheers,
>
> Alan.
>
> On 6/14/05, kim aldis <kim@(protected)> wrote:
> > Why Gaussian? Wouldn't normalising x/y/z random numbers work?
> >
> >
> >
> > > -- --Original Message-- --
> > > From: owner-xsi@(protected)
> > > [mailto:owner-xsi@(protected)] On Behalf Of Alan Jones
> > > Sent: 14 June 2005 21:42
> > > To: XSI@(protected)
> > > Subject: Re: Random points on a unit sphere without biasing
> > >
> > > Thanks Andy. After looking at the options I think I'm going to go
> > > with the one at the bottom of the mathworld page. It suggested 3
> > > gaussian random numbers for X, Y and Z in a vector and
> normalizing
> > > the vector (at least that's what I think it said from my limited
> > > math). I managed to track down an apparently fast way to generate
> > > gaussian random numbers from evenly distributed ones so with any
> > > luck I'm set.
> > >
> > > Thanks again,
> > >
> > > Alan.
> > >
> > > On 6/14/05, Andy Jones <andy@(protected)> wrote:
> > > > The mathworld article answers this perfectly. Basically,
> > > you pick one
> > > > angle (theta) at random and a height (h) on the sphere at
> > > random. You
> > > > can check this by calculating the approximated surface
> area (width
> > > > *
> > > > radius) of the bands around the sphere at different heights
> > > along the
> > > > sphere, and take the limit as dh -> 0. At the poles, the
> > > width of the
> > > > band is larger for a given height, and at the equator, the
> > > radius of
> > > > the band is obviously larger. In the limit, they have
> a perfectly
> > > > inverse relationship such that each band has the same
> area. More
> > > > generally, a good way to randomly sample vectors within a
> > > given angle
> > > > of an average vector is to restrict the height range to
> > > [cos(angle), 1].
> > > >
> > > > I think you can't pick points in a cube and normalize
> > > because you'll
> > > > bias more points at the corners and fewer points in the
> > > middles of the
> > > > faces, since a cube has more and less volume in those
> > > directions. The
> > > > same problem occurs if you sample on a cube's surface area.
> > > The cube
> > > > method is especially problematic because the biasing isn't
> > > determined
> > > > by the original vector direction. Of course, that
> doesn't mean it
> > > > doesn't work okay in practice for many applications.
> > > >
> > > > -Andy
> > > >
> > > > Alan Jones wrote:
> > > >
> > > > >Hi All,
> > > > >
> > > > >This is to the maths geniuses in the room. I want to
> generate X
> > > > >number of points on a unit sphere, but be sure I won't
> have any
> > > > >biasing involved.
> > > > >
> > > > >My first thought was just to use a couple of random numbers
> > > > >(let's assume they don't have any bias) and then use
> those with a
> > > > >few sin and cos function etc to generate the points. Though I
> > > > >have
> > > a feeling
> > > > >that would give me more points around the poles.
> > > > >
> > > > >Anyone have some good suggestions?
> > > > >
> > > > >Cheers,
> > > > >
> > > > >Alan.
> > > > >
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