Posterize Linear interpolation

[Andrei Popa]

I don’t know of a posterize function, but you could use Math.floor().

Assuming that the lineart of your B uses time as first parameter, you could define B like this:

myFrameRate = 3;
t = framesToTime(Math.floor(timeToFrames(time)/myFrameRate)*myFrameRate);
B = linear(t, 5, 6, 0, 100);

First I turn time to frames, so I can work with myFrameRate. Then I divide it by the myFrameRate value to create the steps. Then I multimplicate it by the same value to not alter the actual value of time. Then I make it time again, via framesToTime(), so I can use it in the linear function.

If B is dependent on another value, not time, you could do something similar by using valueAtTime.

There may be some simpler solution, but I can’t think of one.

[Tudor Baican]

Hi Andrei,

lot of thanks for your reply.

I also came on a similar result after I post that. But I start first to turn the values in seconds.

I think your calculation hold a frame 3 frames long. Its not really a frame rate Frames/Sec.

I the end I think this is good and short result .

fRate = 6;
fps = 1/thisComp.frameDuration;
fRateSec = framesToTime(fps/fRate, fps);
pt = fRateSec*Math.floor(time/fRateSec);
a = linear(pt, inPoint, outPoint, [0,0], [500,0]);

b = linear(time, inPoint, outPoint, [0,0], [0,200]);
position + a + b;