How to find the boundary edges

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Thiago Costa

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Dec 11, 2009, 3:57:05 PM12/11/09
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Sorry if this is a dumb question but how do  I find the boundary edges of an arbitrary polymesh? on a friday?

What's the boundary mesh like? What's common and special about those edges that allow me to find them.

in ICE of course.

 

thanks,

 

-Thiago

Hans Payer

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Dec 11, 2009, 4:13:59 PM12/11/09
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Well... doesn't a boundery edge have only one polygon adjacent to it and not two? Could you filter them that way?

Hans

Fabian Schnuer

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Dec 11, 2009, 4:28:47 PM12/11/09
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all the edges that are not shared between two primitives?

Fabian Schnuer

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Dec 11, 2009, 4:30:49 PM12/11/09
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also, if it helps I have some houdini python code that does the
opposite, find all the inner ones. i can mail if off list if you want
it.

On Fri, 2009-12-11 at 15:57 -0500, Thiago Costa wrote:

Grahame Fuller

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Dec 11, 2009, 7:42:35 PM12/11/09
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Using the supplied ICE nodes, it's not easy. There's not a lot of info about connectivity available.

The easiest way is to store the boundary edges in a cluster and check IsElem.

If that's not acceptable, then it may be possible to abuse arrays in the object context and figure out endpoint positions from edge position, length, and ref frame, compare to node positions, look up the previously stored polygon indices, and see if there are one or two matches, but that's guaranteed to fail if there are coincident points so it fails the "arbitrary" condition.

gray

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Thiago Costa

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Dec 12, 2009, 2:06:37 PM12/12/09
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Fabian, please yes I want to see what that's like.

Grahame,
I went the way of clusters for now, but I wish there was this information available in ICE.

-Thiago

2009/12/11 Grahame Fuller <Grahame...@autodesk.com>

Thiago Costa

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Dec 12, 2009, 2:11:30 PM12/12/09
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Hans,
I guess that's something that could work, but how do I get this information in ICE?
if I had edge locations that could return polygon positions, then I could check the size of this array.

I wish this info comes directly from the geometry accessor... that would be much faster I guess.

2009/12/11 Hans Payer <hans...@gmail.com>
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