Naïf Posted January 10, 2018 Share Posted January 10, 2018 hi everybody, it's my first question here and i'm all out of noobish solutions. I'm trying to recreate this (see attached) using vex only, as an excercise. Now i managed to create the circles, and i want to select random points from where i can influence the radius of the neighbours. In order to do so I created an array of all the points, created an attribute that stores the ptnum of a random one and refreshes every 30 frames. The problem is now i want to store these values in a new array over time, but the append function only refreshes the current number. Is there a way to append a new value to an array every "x" frames? Hope I explained my situation clearly enough, and thanks to anyone who wastes a little bit of his/her time answering optical_with_circles.hipnc Quote Link to comment Share on other sites More sharing options...
j00ey Posted January 10, 2018 Share Posted January 10, 2018 I've put a couple of examples of how to do this in the attached file. First way is using a solver SOP - that way is literally appending values to an array every n frames. The other method isn't technically appending values, it's just generating an array that gets longer every n frames. The advantage of doing it like that is that you can scrub the time line without having to cook the whole frame range. optical_with_circles_edit.hipnc Quote Link to comment Share on other sites More sharing options...
Naïf Posted January 10, 2018 Author Share Posted January 10, 2018 I had the feeling that a solver would have to be involved, or the almighty for loop thanks a lot j00ey! Quote Link to comment Share on other sites More sharing options...
yandao Posted May 11, 2018 Share Posted May 11, 2018 great illustration Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.