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# Subdivision surfaces question

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Hi! I have a really basic question about subdivision surfaces. Basically the idea of subdividing a mesh is that you take each polygon and add some new vertices to it, and then move around the vertices based on some weightings. Well, that''s a rough idea of how it''s done anyways. I''m not too clear on how to implement it though. The thing is, say you''ve subdivided one quad already, and now you''re going to subdivide the quad which is next to the old one. However, the subdivision formulas might need access to the old quad but that doesn''t exist anymore since it''s been subdivided to make 4 smaller ones. I''m not quite sure if that''s clear heh... A sort-of similar example would be this... Suppose you want to blur an image. So you could just say "each pixel is the average of itself plus its surrounding neighbors". But the problem is, once you''ve blurred the first pixel, then when you try to do the second pixel, it will be using the blurred version of the first pixel, when it should actually be using the original value which was there. So, what you might do is create a temporary image, so that now you can safely find the neighbors of each pixel and compute the averages. But I think, with subdivision surfaces, it would be way too efficient to create a temporary mesh or something like that, so there''s got to be some more efficient way... Maybe I''m just missing some important idea or detail. Does anyone have any ideas on how to fix this problem of needing a temporary mesh? (I hope the problem is clear, if not then I can explain more heh) Thanks a lot! I''ll try to keep working on it but so far I''m pretty stumped on how this can be done efficiently. Raj

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Yes, of course you need the old mesh throughout the entire process.

You don't touch it. You do all of your work in a new mesh.

[edited by - jtech on November 4, 2002 2:25:56 PM]

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Oh okay thanks! Yea I spent two days trying to come up with some trick to get it to work without a temporary mesh but i don''t think there''s any easy way to do it. (If it is even possible.) So *phew* thanks for your reply, now I can quit worrying about that hehe

Raj

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