Macha Posted September 17, 2012 Share Posted September 17, 2012 If I have a volume and calculate the gradient of it, how would I go about finding the voxels where the gradient flips direction, ie the inflection points? Would the sign of the curvature field give me that? 1 Quote Link to comment Share on other sites More sharing options...
mightcouldb1 Posted September 22, 2012 Share Posted September 22, 2012 By gradient did you mean SDF? If so, I think you are looking for the Laplacian which is the divergence of the gradient. That will give you the edge voxels. You can use an exponential function to constrain the voxels to the edge. I did this quickly with the volume calculate node. Not sure if that is what you're looking for! Quote Link to comment Share on other sites More sharing options...
Macha Posted September 23, 2012 Author Share Posted September 23, 2012 (edited) No, unfortunately not because the gradient does not flip at the edges, it merely reduces in magnitude. I am after the center of objects. Not the center of mass, or the geometric center, but those areas where opposing gradient vectors would meet. And if that makes you think of shrinking an isosurface; no that would not give those areas. Somebody suggested it's just the divergence but I havent tested it yet fully. In a sense its the place of maximum divergence. I think it's very close though. Perhaps it is the first derivative of the divergence. I should try that if I figure out how to get it. Or maybe, if I was to normalize the gradients, and then take the divergence... But then that would still get me edges I think... I've kind of solved this problem in a rubbish way already (blurring a gradient, then dot it with original, and test if density is at least a certain value) but it seems to me there should be a much more immediate and clearer way to do it, analogous to finding inflection points in simple functions. Edited September 23, 2012 by Macha Quote Link to comment Share on other sites More sharing options...
Macha Posted September 24, 2012 Author Share Posted September 24, 2012 I tried divergence from a gradient. It's very close. Quote Link to comment Share on other sites More sharing options...
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