tag:blogger.com,1999:blog-3417788486204501901.post2095706586173327266..comments2019-08-24T20:01:41.933+02:00Comments on Trust me, it works on my machine: RBF Node: Interpolate all the thingsMarcus Krautwursthttp://www.blogger.com/profile/11933625388076914378noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-3417788486204501901.post-60426838418649406082019-04-24T21:32:49.580+02:002019-04-24T21:32:49.580+02:00Quote:"The distance matrix A can also be filt...Quote:"The distance matrix A can also be filtered through a basis function before we solve the equation and there are plenty of functions to choose from, eg. "linear", "gaussian" or "multiquadric" just to get some interesting interpolation effects for different datatypes."<br />I found this part to be critical to getting this to work properly. It's a two part process. First create the relational matrix of all the distances to each other... this gives you the x. Second multiply the new position matrix by the x to get your new B matrix values. I used "gaussian" and found the epsilon value was significant to getting numbers that made sense. It determines the width of the gaussian bell curve. Most of my distance values were between 1-20. My epsilon value was 5.<br />So... using numpy<br />import numpy as np<br />a = np.array([[0.120, 0.647, 0.647]])# gaussian kernal applied distances<br />x = np.array([[1.02110848, -0.12303091, -0.05937078], [-0.12303091, 1.05562737, -0.19892568], [-0.05937078, -0.19892568, 1.04425569]])# I had solved for x in my system<br /><br />then...<br />b =a.dot(x)<br />b = array([[ 0.00451912, 0.53952228, 0.53980402]])<br />In my case I was driving RGB values. This distance was equidistant between green and blue.<br />York Schuellernoreply@blogger.comtag:blogger.com,1999:blog-3417788486204501901.post-65048322229905208072018-10-18T05:52:02.980+02:002018-10-18T05:52:02.980+02:00In this example, the driven values are of equal di...In this example, the driven values are of equal dimension to the driving values. The driver has 3 attributes and the driven also has 3, allowing the Ax = B equation to work pretty well. How would this work in the case of a set of driving values affecting N number of driven? Would it matter (I imagine it would) to find the weight values?<br /><br />For example, if three translation values were to drive a set of translation, rotation and scale values on a driven object?<br /><br />Thanks for this write up as well!Alan Weiderhttps://www.blogger.com/profile/03227627236851878094noreply@blogger.comtag:blogger.com,1999:blog-3417788486204501901.post-67055438085376242162018-09-02T10:05:36.873+02:002018-09-02T10:05:36.873+02:00This comment has been removed by the author.王传真https://www.blogger.com/profile/17781950681443698412noreply@blogger.comtag:blogger.com,1999:blog-3417788486204501901.post-5478759848580208452018-08-12T17:15:34.322+02:002018-08-12T17:15:34.322+02:00Hi, I used the website "Linear Equations Solv...Hi, I used the website "Linear Equations Solver" to solve the equation in this post, but I get the value of X is [-16.071 16.071 22.500<br /> 16.071 -16.071 22.500<br /> 22.500 22.500 -31.500]. Did I miss somehing?Unknownhttps://www.blogger.com/profile/03861746663224656071noreply@blogger.com