Thanks All for the info. I think I best start doing some experimenting
to see how things are looking and then refine the method I use for
picking points from there.
Cheers,
Alan.
On 6/14/05, Andy Jones <andy(at)thefront.com> wrote:
> That's what the quasi-monte-carlo QMC algorithms do for you in Mental
> Ray, right? mi_sample, for instance? I'm not sure what happens to the
> even distribution once you pipe it through the Box-Muller transformation
> or whatever you're doing, though. You could end up with some sort of
> biasing or undesirable jitter patterns.
>
> -Andy
>
> OO's mailbot wrote:
>
> > You are probably after an "evenly distributed" placement rather than
> > truly random. You are likely to get better result faster by jittering
> > an even distribution of sample rather than creating an even distribution.
> >
> > This is the way stocastic sampling works. rather than randomly picking
> > points on a surface, you start by picking uniformaly distributed
> > points and offset them slightly to break up the pattern.
> >
> > Most of what we call "random" in natural patterns are actually caused
> > by cellular or molecular diffusion and are really not random at all
> > but a stable system of attraction/repulsion. ie every point is spaced
> > out more or less evenly across the surface, so that you never have
> > points that are clustered too close or too far away from each other.
> > With a very very high sample a random distribution will approximate
> > that, but you have no controls over overlapping samples with a random
> > number generator...
> >
> > oO
> >
> > On Jun 15, 2005, at 7:11 AM, kim aldis wrote:
> >
> >> 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(at)Softimage.COM
> >>> [mailto:owner-xsi(at)Softimage.COM] On Behalf Of Alan Jones
> >>> Sent: 14 June 2005 21:58
> >>> To: XSI(at)Softimage.COM
> >>> 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(at)aldis.org.uk> wrote:
> >>>
> >>>> Why Gaussian? Wouldn't normalising x/y/z random numbers work?
> >>>>
> >>>>
> >>>>
> >>>>> -----Original Message-----
> >>>>> From: owner-xsi(at)Softimage.com
> >>>>> [mailto:owner-xsi(at)Softimage.com] On Behalf Of Alan Jones
> >>>>> Sent: 14 June 2005 21:42
> >>>>> To: XSI(at)Softimage.com
> >>>>> 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(at)thefront.com> 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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