On Fri, 29 Nov 2002 15:01:24 +0100, Mark Weyer
<wey### [at] fregemathematikuni-freiburgde> wrote:

>Why is the lookup array fixed? It might be calculated at program start,
>allowing for varying length.

Probably for the same reason most other jitters use fixed arrays - for
true platform independence (at least that's how I read it).


Peter Popov ICQ : 15002700
Personal e-mail : pet### [at] vipbg
TAG      e-mail : pet### [at] tagpovrayorg

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Hi Christoph,

The cos-theta distribution on the hemisphere is given by creating an 
even distribution on a unit circle and then projecting up or down to a 
unit hemisphere.

You can see this by taking a copy of the radiosity sample array, editing 
it to pov syntax and including it in a small scene which puts a sphere 
at the <X,Y,0> points. If you do an animation which adds 1 or a few 
sample positions each frame you can see how the distribution builds up.

Mike Andrews.

Christoph Hormann wrote:
> Quoting from the source:
> 
> A bit of theory: The goal is to create a set of "random" direction rays
> so that the probability of close-to-normal versus close-to-tangent rolls
> off in a cos-theta curve, where theta is the deviation from normal.
> That is, lots of rays close to normal, and very few close to tangent.
> You also want to have all of the rays be evenly spread, no matter how
> many you want to use.  The lookup array has an array of points carefully
> chosen to meet all of these criteria.
> 
> The problem is similar to the classical "How can I arrange N points evenly
> on a sphere" problem like described in:
> 
> http://www.ogre.nu/sphere.htm
> 
> Just that it is hemispherical, not really uniform and it not only has to
> be good for N points but also for N-X points (X = 1..N-1)
> 
> Christoph
>

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Mark Weyer wrote:
> 
> [...]
> 
> How was the built-in array created? (I would like to know,
> which quality I have to beat). In fact you already mentioned,
> that you want the same quality contraints for initial segments
> of the array, so my earlier suggestion might already do the trick.

You don't need to know how the original distribution was created to
compare quality.  If you follow the link i posted you will find several
measurements of the quality of distributions.  Apart from the maximum
uniformity the distribution should not be regular to be well suited for
radiosity sampling.

Christoph

-- 
POV-Ray tutorials, include files, Sim-POV,
HCR-Edit and more: http://www.tu-bs.de/~y0013390/
Last updated 15 Nov. 2002 _____./\/^>_*_<^\/\.______

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Michael Andrews wrote:
> 
> Hi Christoph,
> 
> The cos-theta distribution on the hemisphere is given by creating an
> even distribution on a unit circle and then projecting up or down to a
> unit hemisphere.
> 
> [...]

It is fairly possible that this is the method how the directions array was
originally created but there is no way how to prove this.

Another method (which probably produces higher quality for a fixed number
of rays if done well) would be a repulsion algorithm with a non uniform
force field (i.e. higher repulsive forces between the points at low angle)

Christoph

-- 
POV-Ray tutorials, include files, Sim-POV,
HCR-Edit and more: http://www.tu-bs.de/~y0013390/
Last updated 15 Nov. 2002 _____./\/^>_*_<^\/\.______

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> If you have a light source almost tangential to the surface it doesn't
> illuminate much and one almost normal to the surface gives high
> illumination. So, if all the samples are weighted the same you want most
> samples from the normal direction and few from tangential to give the
> best estimate.

Ok, that explains it. As a purist I was assuming you take an
even distribution and weight the samples with cos. But it would
be hard to sell that to running-time-theists...

> I suppose it could be, and I've thought about doing this. However you
> would have to calculate an array for each level of radiosity recursion
> because the code reduces the number of samples taken at each subsequent
> recursion level. And the results wouldn't be very different anyway.

The way Christoph was putting it, you reuse an initial segment of the
array. My current approach would be to build in a precomputed array
for everyday use with the possibility to replace it by something larger.

  Mark

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> > The cos-theta distribution on the hemisphere is given by creating an
> > even distribution on a unit circle and then projecting up or down to a
> > unit hemisphere.
> [...]
> Another method (which probably produces higher quality for a fixed number
> of rays if done well) would be a repulsion algorithm with a non uniform
> force field (i.e. higher repulsive forces between the points at low angle)

I do not see why the quality is better than using repulsion
inside the unit circle with a uniform force field before projecting.

  Mark

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Mark Weyer wrote:
> 
> I do not see why the quality is better than using repulsion
> inside the unit circle with a uniform force field before projecting.

To me it seems fairly obvious that projecting a disc on a hemisphere does
not maintain the distance relations and therefore reduces the uniformity
of a distribution.  

To be precise, the points at the rim of the hemisphere will be quite dense
in direction of the circumference but will be moved apart in theta
direction.

Christoph

-- 
POV-Ray tutorials, include files, Sim-POV,
HCR-Edit and more: http://www.tu-bs.de/~y0013390/
Last updated 15 Nov. 2002 _____./\/^>_*_<^\/\.______

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Christoph Hormann wrote:
> 
> Mark Weyer wrote:
> 
>>I do not see why the quality is better than using repulsion
>>inside the unit circle with a uniform force field before projecting.
> 
> 
> To me it seems fairly obvious that projecting a disc on a hemisphere does
> not maintain the distance relations and therefore reduces the uniformity
> of a distribution.  

The point is that you do not want a uniform distribution on the 
hemisphere ...

> 
> To be precise, the points at the rim of the hemisphere will be quite dense
> in direction of the circumference but will be moved apart in theta
> direction.

Ahh! If you used a unit circle to do the repulsion thing then you are 
quite right - you do not get an even distribution at the edge: all the 
outermost points would be pressed against the rim.

This is why I used a 2 radius circle when I was doing the Delaunay 
triangulation, and adding samples until I had enough in the unit circle. 
That way you get an even distribution right across the unit circle - 
right to the rim of the hemisphere.

You could do the same with a repulsion algorithm - fill a circle of 
greater than 1 radius and keep adding a sample and repelling to 
completion untill you have the right number of samples within the 1 unit 
circle.

> 
> Christoph
> 

Mike Andrews.

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Michael Andrews wrote:
> 
> [...]
> 
> Ahh! If you used a unit circle to do the repulsion thing then you are
> quite right - you do not get an even distribution at the edge: all the
> outermost points would be pressed against the rim.
> 
> This is why I used a 2 radius circle when I was doing the Delaunay
> triangulation, and adding samples until I had enough in the unit circle.
> That way you get an even distribution right across the unit circle -
> right to the rim of the hemisphere.

The radius of the circle does not matter, once you project it on a
hemisphere you get an anisotropic distribution (meaning the density of the
points depends on the direction like explained in my last post).  But feel
free to try it out, i'm sure if you check the quality of the results you
will see it.

Christoph

-- 
POV-Ray tutorials, include files, Sim-POV,
HCR-Edit and more: http://www.tu-bs.de/~y0013390/
Last updated 15 Nov. 2002 _____./\/^>_*_<^\/\.______

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Christoph Hormann wrote:
> The radius of the circle does not matter, once you project it on a
> hemisphere you get an anisotropic distribution (meaning the density of the
> points depends on the direction like explained in my last post).  But feel
> free to try it out, i'm sure if you check the quality of the results you
> will see it.

This anisotropy is inevitable - the density of the samples is decreasing 
  with theta but constant in phi. If you plot the radiosity sample 
positions you can see this quite clearly with the official samples.

> 
> Christoph
> 

Mike Andrews.

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