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Square <-> hemisphere mapping


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#1 Reitano   Members   -  Reputation: 550

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Posted 25 October 2013 - 04:49 AM

Hi

 

I am looking for a mapping (and its inverse) between a square and a hemisphere. I need it to store samples of a hemispherical function (the sky colour) in a 2D texture, which I can then fetch in shaders. The requirements are:

 

- shader-efficient inverse mapping from the hemisphere to the square (in other words, conversion of a 3D direction to U,V coordinates)

- the mapping should allow some control on the sample distribution. In my case I need more samples near the horizon where the sky colour changes quickly

 

I've done some research these days but could not find anything suitable yet. Most projections use polar coordinates and have costly inverse mappings in terms of ALU (due to atan2, acos functions). I have code for mapping normal vectors to two coordinates for deferred rendering but in that case I cannot control the sample distribution.

 

Perhaps I could use a cylindrical projection with a cheap approximation to atan2 if one exists ?! Any ideas are much appreciated.

 

Thanks,

 

Stefano

 

 



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#2 tonemgub   Members   -  Reputation: 1146

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Posted 25 October 2013 - 05:30 AM

Use a cube map.



#3 Reitano   Members   -  Reputation: 550

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Posted 25 October 2013 - 05:45 AM

That does not answer my question.



#4 tonemgub   Members   -  Reputation: 1146

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Posted 25 October 2013 - 06:56 AM

With a cubemap projection, you won't have to use atan2? :)



#5 Álvaro   Crossbones+   -  Reputation: 13897

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Posted 25 October 2013 - 07:18 AM

You can map the square [-1,1]x[-1,1] to the unit disk like this:

  F(x,y) = (x/d,y/d), where d is the distance from the origin to the edge of the square along the direction (x,y)

  d = sqrt(1 + min(abs(x/y),abs(y/x))^2)

 

Now you can imagine the resulting disk as being in the x-y plane in 3D. Place the hemisphere on top of it and place a point P at (0,0,-w). Use the point P to project from the disk to the hemisphere (meaning, start with a point on the disk, draw the line that joins that point with P and find the intersection of that line with the hemisphere).

 

The number w controls the density of points near the horizon (small w means lots of resolution near the horizon).



#6 Hodgman   Moderators   -  Reputation: 31786

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Posted 25 October 2013 - 07:46 AM

Lots of methods work in either spherical coordinates or Cartesian coordinates, meaning you can implement them with trig, or without.

These all map the sphere to a 2D circle.

http://en.wikipedia.org/wiki/Stereographic_projection

http://en.wikipedia.org/wiki/Sphere_mapping

http://www.opengl.org/archives/resources/code/samples/sig99/advanced99/notes/node177.html (another kind of sphere mapping).

 

If you want to use up the corner space, you could then remap the circle to a square.

There's an article here for mapping a circle to a square, and someone's posted a comment with the inverse:

http://mathproofs.blogspot.com.au/2005/07/mapping-square-to-circle.html

 

I haven't tried it, but you should be able to focus the precision to either the centre or the edges with some code like this:

uv = uv*2-1            // remap from 0-1 to -1 to 1
s = sign(uv)           // pow will destroy the sign, remember it here
uv = pow( abs(uv), x ) // pinch (x>1) or bulge (0<x<1)
uv = uv*sign*0.5+0.5   // remap back to 0-1 range

Edited by Hodgman, 25 October 2013 - 07:46 AM.


#7 Reitano   Members   -  Reputation: 550

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Posted 25 October 2013 - 08:19 AM

Lots of ideas to experiment with. Another option that I should have considered right from the start is the dual paraboloid mapping:

http://www.cs.ubc.ca/~heidrich/Papers/GH.98.pdf

The mapping and its reverse are trivial. I might be able to control the sampling distribution by manipulating the UV space, as suggested by Hodgman. About that, a fixed odd exponent (e.g. 3) would simplify the ALU by removing the sign and abs instructions.

 

@tonemgub: I should have mentioned that I am going to update the map every frame (either on CPU or GPU) and a cubemap likely has a higher cost than a single texture although I haven't tested it. And, apart from this use case, I am interested in this problem from a mathematical point of view.



#8 Reitano   Members   -  Reputation: 550

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Posted 30 October 2013 - 06:19 AM

An update: a paraboloid mapping solves this problem brilliantly. In order to bias the sample distribution near the horizon, I apply an exponent to the sampling direction Z component. For a typical sky, the sampled representation closely matches the original one, and objects far from the camera blend well with the sky. If anyone is interested I can post here some formulas and pseudo-code.






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