In article <387028BC.31573882@tapestry.tucson.az.us>, Kevin Wampler 
<kev### [at] tapestrytucsonazus> wrote:

> For each evaluation point, you have found the normals of n other nearby
> points, which are s units away from the evaluation point.  For each 
> vector,
> the angle between it and the normal vector at the evaluation point is
> theta.  One way of finding a representation of the average curvature 
> would
> then be:
> 
> K=1/n*sigma(theta/s)

Ok, I mostly understood that-but what is sigma, other than a greek 
letter?
If I understand this correctly, I should divide the angle between the 
normal of each sample and the normal at the evaluation point by the 
distance of the sample from the evaluation point, total these up, and 
divide by the number of samples. Um, never mind, easier for me to think 
this way(pseudocode follows):

SPoint = sample point
EPoint = evaluation point
total = 0;
while(taking N samples)
{
   S = distance of SPoint from EPoint 
   theta = angle between normal at SPoint and normal at EPoint
   total += theta/S;
}
result = total/N;

Is this right?

-- 
Chris Huff
e-mail: chr### [at] yahoocom
Web page: http://chrishuff.dhs.org/

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