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> > If is't not
> > possible, I have another question (which might be a FAQ).. is it a good
> > way of calculating vertex normals, by adding all the normals of polygons
> > sharing that vertex, and then dividing by the number of normals?
>
> this is good way but not the best
> look at thread:
...
Thanks, I found the thread and your images. Just to make sure I understood
everything, 'weighted' means this: Instead of using normals with length 1,
adding them and dividing by their number, I do this: I take the normals as
returned by the cross product (meaning they have their area multiplied in),
add the length of all those weighted normals together, then add those
normals, and divide by the total length ?
I wouldn't have thought of it, but it sounds like a better way. I must go
try it now..
--
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2)C(.8,1)C(.8,-1)C(1.1,1)C(1.9,-1)pigment{rgb 10}rotate x*90translate<-1,0,4>}
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