# DX11 Lightmap lumel to world position

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Hey,

I've been spending the last couple of days trying to understand lightmap generation. Here are some of the resources I've used (mostly from flipcode)

http://www.flipcode....Lightmaps.shtml
http://www.flipcode....mentation.shtml

Now, I understand how radiosity is computed (progressive refinement). I understand how you go from hemicube to the value of the lumel. I also get how you use planar mapping to get UV coordinates for a triangle. The part I don't understand no matter how much I try is going from lightmap texture coordinates to world-position. The articles in flipcode only describe it vaguely, without giving actual code.

As far as I understand, you create a quad around a triangle based on its min/max positions (i.e. its bounding box). You then use bilinear interpolation to go to world position based on a specific UV coordinate (i,e. scale the UV texture [0,0]x[1,1] to the quad around the triangle). Is that how its done? Flipcode mentioned something about the equation of a plane but I don't understand how that fits in.

Am I going about this the right way? Can anyone give me any tips or explain how the equation for a plane fits in? I read this article which just uses vertices as sample points instead of lumels. Is that how it's done nowadays?

Thanks

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I haven't tried this myself and I'm pretty much stream of consciousness typing here but I would do something like this: use the triangle's texture coordinates and the lumel's texture space position to find it's barycentric position on the triangle face, then use the triangle's vertex positions to find said barycentric position in object space, then transform said object space position by world matrix and voila.

I apologise if the above is misleading or incorrect

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Hey,

I don't think using barycentric coordinates would work since they are only defined inside the triangle. I would need to sample points outside a specific triangle in order to reduce filtering problems around the edge. But please correct me if I'm wrong.

Anyways, I've figured it out now. The "quad" around a triangle its just the plane that it exists on. We need the plane's equation in order to transform a 2D position on the triangle's (plane's) surface into a 3D position. That 2D position is relatively easy to get if you know the UV coordinates. No interpolation needs to be done!

Thanks!

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