Divide Curve by Curvature

I though some of you might be interested in this. A simple def i wrote to divide a curve 'by curvature' --- with more division points where the curvature is greater.

Right now it only works on one curve at a time... could someone help me with the data trees, so i can get it to work on multiple curves at once?

Attachments:

100125_Divide Curve by Curvature.ghx, 141 KB
100125_Divide Curve by Curvature.jpg, 68 KB

Replies to This Discussion

Permalink Reply by christoph daniel alex on January 25, 2010 at 8:42am

nice one!

Attachments:

100125_DivideCurvebyCurvature_fix.ghx, 149 KB

Permalink Reply by Evan C on January 25, 2010 at 8:46am

ah. excellent. thanks.

Permalink Reply by Charles Aweida on August 11, 2010 at 1:45pm

This is really great and useful. If you guys get a sec an explanation would be awesome. I get the general idea but im not quite sure how the Vector Length Amplitude and the partial results of Mass Addition works. Also there is a secondary line that is drawn, it would be nice to get a better understanding of all this...

Permalink Reply by Evan C on August 11, 2010 at 7:58pm

charles, thanks.

let's see if i can explain the logic....

the vector length component outputs a (scalar) value for curvature at each sample point along the input curve. If you graph this, it looks like the top right image above. The partial result from mass addition (lower right) is to ensure that the x- and y- dim are both always increasing - this means that there are no "peaks" (try bypassing the mass addition component - it still works but you will see gaps at the points of maximum curvature)

I then take this curve and divide it evenly along its length, and project those points down to the x axis - you see how the spacing along the x axis is determined by the slope of the curvature graph. these new x values will be my new division pts along the input curve.

i'm sure there's a better way to explain this (mathematically) - but i hope this helps you see how it's done graphically.

Permalink Reply by Charles Aweida on August 16, 2010 at 5:34pm

actually that is a really good explanation, thanks Evan

Permalink Reply by Chad Westendorf on July 4, 2011 at 7:05pm

Do you still have the corrected .ghx you attached to your post for this thread? Or could you at possibly inform me as to what you corrected if you remember? The link doesn't appear to work.

Permalink Reply by Evan C on July 4, 2011 at 7:24pm

the link works for me....

Permalink Reply by Jacek Jaskólski on August 17, 2010 at 12:18am