Light passing through alpha-transparent surface
Author
Hi,
I am adding alpha transparency support to our simple raytracing lightmapper, and I just realized I am not quite sure how exactly the color of a ray of light is affected when it passes through a surface, given a color and alpha value at the point of intersection.
Let's say a ray of bright white light (R: 1, G: 1, B: 1) passes through a red surface (R: 1, G: 0; B: 0),
1) When Alpha is 0, the result should be bright white
2) When Alpha is 1, the result should be black
3) When Alpha in (0, 1), the result should be somewhere in between, with a hue based on the red material color.
I tried things like light * color * ( 1-alpha ), that don't work for obvious reasons. I am now considering simple 2-part linear interpolation from black -> materialColor -> white, based on the alpha. However, I can't shake the feeling that there should be a much more direct solution. Maybe I am looking at this problem the wrong way?
Any input on this issue would be most welcome.
Best regards,
Paul-Jan
p.s. Note how I am not looking for physically correct behaviour of light here, I am not doing refraction or anything similar, I just want the result to look... ok.
Quote:Original post by pauljan
I tried things like light * color * ( 1-alpha ), that don't work for obvious reasons. I am now considering simple 2-part linear interpolation from black -> materialColor -> white, based on the alpha. However, I can't shake the feeling that there should be a much more direct solution. Maybe I am looking at this problem the wrong way?
Depends on what you'd like to achieve. In the real world, a perfect red filter will produce perfectly red monochromatic light, that will never turn white even at high intensities. Now, a typical sheet of red transparent plastic (or similar) is far from being a perfect filter. It will not completely absorb other wavelengths, just attenuate them. This is why at high intensities, this residual light is still strong enough to saturate the receptors on your retina. The net result is a white light.
Now, let's transpose this to computer graphics. If you have a perfectly red filter, ie. (1, 0, 0), then the resulting light must always be perfectly red (assuming white incident light), regardless of the intensity. So this behaviour is correct. But if your filter is not perfect, for example (1, 0.3, 0.3), then the light must saturate to white from a certain intensity on.
The way to simulate this effect is by using HDR, high dynamic range lighting. Essentially, your light sources have a larger dynamic range than your usual 0 to 1. Maybe 0 to 2, or more. Then, use the equation you showed above, a simple colour filter, and clamp the end result to the [0..1] range.
Author
Thanks, that clears things up, a lot. Using HDR (clamping) crossed my mind, but that didn't solve the problem of the light staying red. Following your explanation, I now see that that is how things are supposed to work. Just like a perfectly black filter will never let any light through, no matter what alpha it has. In real life, there is no such thing as a pitch black yet transparent surface.
I am not sure yet wether I'll want to simply extend the dynamic range of my lights and clamp, or put things through an exposure-type exp function, but the basic principles are clear now. Thanks!
I am not sure yet wether I'll want to simply extend the dynamic range of my lights and clamp, or put things through an exposure-type exp function, but the basic principles are clear now. Thanks!
it seems you can approximate this with an exponecial funcition for example
1 - e^(-L*k)
the bigger L the more close the result is to one, this gives high (infinate to be eact)dynamic range and the sharpness of the thing is controled by the constant k.
filters are controled by beers law anyway, and this has a similar form
Tim
1 - e^(-L*k)
the bigger L the more close the result is to one, this gives high (infinate to be eact)dynamic range and the sharpness of the thing is controled by the constant k.
filters are controled by beers law anyway, and this has a similar form
Tim
Author
Yes, that's what I refered to as 'exposure'. It's Huge Elias' photography analogy used to describe exactly this solution to the HDR problem, see http://freespace.virgin.net/hugo.elias/graphics/x_posure.htm for details.
For the moment I've settled with a 4x range for the color filtering with local clipping, simply because it looks good. For the next version I'll probably add exposure type HDR handling for the light equation as a whole.
For the moment I've settled with a 4x range for the color filtering with local clipping, simply because it looks good. For the next version I'll probably add exposure type HDR handling for the light equation as a whole.
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