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EVSM Theory

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2 comments, last by __MONTE2020 4 years ago

Can someone explain this paragraph from here http://www.punkuser.net/lvsm/lvsm_web.pdf​ ( 7 Other Warps)

“For instance, it is interesting to consider the exponential warping function e^(cx) (where c is a constant) as suggested by [11], but still using a second moment and Chebyshev’s Inequality. This warping has the effect of relatively moving object B toward object A which reduces the above ratio. The resulting exponential variance shadow map (EVSM) has greatly reduced VSM-like light bleeding while still avoiding any bleeding near occluders.”

I can see why the ratio reduces using exponentials, but I don't see how this happens - “This warping has the effect of relatively moving object B toward object A". The paragraph right after that one isn't the clearest either but I would like to understand this first. Is there some shadows expert here that could explain this?

Thanks.

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That sentence is saying the same thing that you're saying: the exponential warp reduces the relative distance between shadow caster A and caster/receiver B compared to the distance between B and receiver C. Let's say A is at a depth of 0.25, B is at a depth 0.5, and C is at a depth of 0.75. In this case B-A is equal to C-B, so deltaX is the same as deltaY. Now let's apply the exponential warp with an exponent of 10. The warped depth of A is now ~12, B is ~148, and C is ~1808. So now deltaX is ~136, and deltaY is ~ 1660, meaning deltaY is now much larger relative to deltaX. This gives us the same equivalent result as moving receiver B up towards A, which is what that quoted sentence is saying.

@MJP Thanks MJP. I was confused by the fact that object C is not mentioned in that sentence (of course now I am aware it was hidden inside “relative distance”), but it's all clear now.

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