Re: Volume in SQL
From: Jan Hidders <jan.hidders_at_REMOVETHIS.pandora.be>
Date: Sat, 17 Jul 2004 06:46:34 GMT
Message-Id: <pan.2004.07.17.06.47.07.975308_at_REMOVETHIS.pandora.be>
>
> 2 cents. The triangulation center doesn't have to be located in the center
> of mass. It can be anywhere, even outside of the polygon.
Date: Sat, 17 Jul 2004 06:46:34 GMT
Message-Id: <pan.2004.07.17.06.47.07.975308_at_REMOVETHIS.pandora.be>
On Fri, 16 Jul 2004 16:31:58 -0700, Mikito Harakiri wrote:
> "Jan Hidders" <jan.hidders_at_REMOVETHIS.pandora.be> wrote in message
> news:pan.2004.07.16.07.37.25.81889_at_REMOVETHIS.pandora.be...
>> Well, if you look at the 2D case then what you could probably do is take >> the average of the points and triangulate with that point as one corner >> and the other two corners being the end pionts of an edge of >> the convex hull.
>
> 2 cents. The triangulation center doesn't have to be located in the center
> of mass. It can be anywhere, even outside of the polygon.
Hm. You are right. I'm actually only just beginning to see why that is. But in the 3D case I must choose, next to the two points in the CH edge and this aribtrary other point, a point inside the face I'm considering.
> Isn't Stokes' Theorem wonderful?
I dunno yet, I'm seeing it for the first time. :-/
- Jan Hidders