Progressive, always positive wiggle?

[Xavier Gomez]

This should work:

startVal = 0;
endVal = 100; // max copies
DTmin = 0.2; // DTmin/max : time interval min/max between changes
DTmax = 0.8;
DNmin = 2; // DNmin/max : change amount min/max
DNmax = 5;

s=t=inPoint;
m=n=startVal;
while (n

Xavier

[Navarro Parker]

Thank you!

Is there a way for this to respect the timing of my existing two keyframes (0 & 100)?

[Xavier Gomez]

Here is a possibility, not optimal since it calculates the whole every frame.
If your keys are not far apart and the param averageChangesPerSec isnt too big, it should be fine

startVal = 0;
endVal = 100; // max copies
t1 = numKeys>0 ? key(1).time : inPoint;
t2 = numKeys>1 ? key(2).time : outPoint;
averageChangesPerSec = 1.5;
seed1 = index+1000;
seed2 = index+2000;
a = 3; // controls time variations (close to 1 = small, bigger=more disparity)
b = 10; // controls value variations (same)

if (time<=t1){startVal;}
else if (t2<=time) {endVal;}
else{
numSteps =Math.ceil((t2-t1)*averageChangesPerSec);
t=0; times = [0];
v=0; values = [0];
for (n=1; n<=numSteps; n++){seedRandom(t+seed1, true);t+=random(1,a); times[n]=t; seedRandom(t+seed2, true);v+=random(1,b); values[n]=v;};

t = (time-t1)/(t2-t1) * t;
x = (endVal-startVal) / v;

a=0; b=numSteps; while(b-a>1){j=Math.floor((a+b)*0.5); if (times[j]<=t){a=j;} else{b=j;};};

startVal + x*linear(t, times[a], times[b], values[a], values[b]);
};

Xavier