On Tue, 22 Aug 2000 12:23:01 -0500, Chris Huff wrote:
>In article <slr### [at] fwicom>, ron### [at] povrayorg 
>wrote:
>
>> Even linear spline prisms?  Whyever for?  There's a simple procedure to
>> determine the distance of a point from an arbitrary polygon[1].  The 
>> proximity is then the minimum over all the edges and the two caps of the 
>> prism.
>
>Because I didn't know this algorithm. :-)
>Hmm, does this take the caps into account? It looks like an infinitely 
>long prism(or proximity to a polygon in a 2D plane).

Yes, it does take the caps into account.  You find the distance to each face,
including the two caps, using the method described, then find the minimum.  
None of what I described is in 2D.

>BTW, what exactly do you mean by "continuously differentiable"? Do you 
>mean that the function may be continuous but the derivative may not be? 
>Where would the function "break"?

Exactly.  It "breaks" in the "corners."  For example, in the interior of a
square, the 2D version of a proximity pattern is undifferentiable along 
the diagonals of the square (as you cross the diagonal, the derivative of
the proximity function abrubtly changes sign and possibly magnitude) and
on the edges (the function is undifferentiable wherever it has a value of
zero, since it is the absolute value of a different function.)

What this means to the isosurface solver is that if it happens to look at
the wrong points, it may miss some intersections.  What really sucks is 
that the function is undifferentiable in exactly the same places that the
object pattern is discontinuous (and thus undifferentiable as well), and 
probably in other places as well.

You can remove some of the areas of concern by making a proximity function
that returns negative values for points on the inside of the shape, but
you can't remove them all.

This problem is not unique to prisms and meshes, either.  The proximity
function is undifferentiable at the center of a sphere, too.

-- 
Ron Parker   http://www2.fwi.com/~parkerr/traces.html
My opinions.  Mine.  Not anyone else's.

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