RE: [Algorithms] It's all Random Numbers

[Zafar Q. <zaf...@ra...>]

Hi,
  As Peter-Pike Sloan said, this is discussed in "NonUniform Random Point
Sets Via Warping", by   Peter Shirley (Graphics Gems III). It's a bit
mathematical though.
On p.81 is a table showing various transformations.

To get a random point on a unit-sphere use:-

theta = arccos(1-2u)
phi   = 2PI * v

where u and v are uniformly random between 0.0 and 1.0,
theta has domain 0 to PI,
phi has domain 0 to 2PI

Cheers
-Zaf

-----Original Message-----
From: Ludvig Tunel [mailto:lu...@it...]
Sent: 31 October 2000 09:42
To: gda...@li...
Subject: Re: [Algorithms] It's all Random Numbers


Hi,

I came to think of a very similar problem:
How do I generate random points in a sphere, ( or on the sphere surface )
using spherical polar coordinates?

If i randomize azimuth and inclination I will get denser distribution around
the "poles".
I would need to transform the randomized[-pi/2 ... pi/2] inclination in some
way, but how?
Taking the square root doesn't feel like the right thing to do here. It
feels
more like I should feed it to cos() or sin() in some way...
The probability of generating a point at a certain inclination must somehow
be transformed to be proportional to the area of the disc at the given
inclination. Right?
But how?

(The problem of denser distribution close to the centre could as I assume
from having read the 2D solution, be solved by taking the cube root of the
radius. ( radius^(1/3) ) Am  I right?)

cheers
Ludde

Gil Gribb wrote:

> > Hi,
> > Thanks for replying.
> >
> > That works spot-on!
> >
> > I was doing everything except the sqrt.
> > Is it trivial to explain why this method works?
>
> Sure, the area of a disk is proportial to the radius squared. And so the
> sqrt "undoes" this.
> -Gil
>
> >
> > Cheers for your time
> > Zaf
> >
> >
> > -----Original Message-----
> > From: Gil Gribb [mailto:gg...@ma...]
> > Sent: 25 October 2000 15:14
> > To: gda...@li...
> > Subject: Re: [Algorithms] It's all Random Numbers
> >
> >
> >
> >
> > > Hi All,
> > > Does anyone know a fast way of calculating 2D points inside a
> > > unit-disc, in a way that leads to a roughly even distribution?
> >
> > Well, the usual way is to just generate points in [-1,1]x[-1,1] and just
> > throw away points that are outside the disk. But, I think this will work
> > too:
> >
> > angle = 2 pi random()
> > r = sqrt(random())
> >
> > x= r cos(angle)
> > y= r sin(angle)
> >
> > -Gil
> >

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