Here is a basic thobber: ((cos(time/2)+1)*x)+((cos((time-(3.141*2))/2)+1)*(100-x)) https://maxbittker.github.io/Mojulo/#KChjb3ModGltZS8yKSsxKSp4KSsoKGNvcygo

Here is a basic thobber: ((cos(time/2)+1)*x)+((cos((time-(3.141*2))/2)+1)*(100-x)) https://maxbittker.github.io/Mojulo/#KChjb3ModGltZS8yKSsxKSp4KSsoKGNvcygodGltZS0oMy4xNDEqMikpLzIpKzEpKigxMDAteCkp https://maxbittker.github.io/Mojulo/#KChjb3ModGltZS8yKSsxKSp... Here is using the same trick to switch between two patterns: ((cos(time/50)+1)*(A*1000))+((cos((time-(3.141*50))/50)+1)*((x-50)^2+(y-50)^2)) https://maxbittker.github.io/Mojulo/#KChjb3ModGltZS81MCkrMSkqKEEqMTAwMCkpKygoY29zKCh0aW1lLSgzLjE0MSo1MCkpLzUwKSsxKSooKHgtNTApXjIrKHktNTApXjIpKQ== https://maxbittker.github.io/Mojulo/#KChjb3ModGltZS81MCkrMSk... Unfortunately, this trick doesn't work well with patterns with high values, and thus most animated patterns don't work. It works like this: ((cos(time/p)+1)*q)+((cos((time-(3.141*p))/p)+1)*Q) where: p is the speed of the switch (Higher is slower), q is pattern #1, and Q is pattern #2. The cos(...)+1 part in both sections is in charge of timing. In the second part we subtract pi * p, which is half the period, from time to make it start at the half point of the cycle.