Translated from the Greek: 1. Calculate the distance from the test position to each of your selected points.
2. If one of the distances is 0, stop. The corresponding coordinate is 1, the others are 0. 3. Otherwise for each of the distances, calculate 1/distance, or 1/distance^2, or 1/distance^3, or ... . 4. Normalize the values above. In other words, get their sum and for each value calculate value/sum. Ta da! gray From: softimage-boun...@listproc.autodesk.com [mailto:softimage-boun...@listproc.autodesk.com] On Behalf Of Paul Sent: Wednesday, July 17, 2013 12:52 PM To: softimage@listproc.autodesk.com Subject: Re: Maths problem (barycentric coordinates) I have been googling all I can find trying to understand the maths notation but its not my forte and I've not found anything that does exactly what I want. I was hoping someone cleverer than i might take pity on me and provide an idiot proof explanation. On 17 Jul 2013, at 17:07, David Barosin <dbaro...@gmail.com<mailto:dbaro...@gmail.com>> wrote: If you can chew through the greek notation this is helpful. http://en.wikipedia.org/wiki/Inverse_distance_weighting On Wed, Jul 17, 2013 at 11:38 AM, <p...@bustykelp.com<mailto:p...@bustykelp.com>> wrote: Hi, I have spent days on this and I cant work it out I have a selection of points (not on a flat plane) and I have a test position. It returns an array that represents the weighting, related to the proximity to the other points. I want to have it so that when the test position is directly at a point, the value for that point in the array = 1 and the rest will be zero as the test point moves around the area it interpolates these values, but they always add up to 1 It sounds really easy, but I’ve been literally* tearing my hair out over this for days. Ive managed to get barycentric interpolation working for a flat plane, and only 3 points, but I need it to accept multiple points in 3d space. Please help. I’m going bonkers over this Paul *(not really literally)
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