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  • Posterize Linear interpolation

  • Tudor Baican

    January 14, 2022 at 11:23 pm

    Hello friends,

    I would like to add some values to my position property.

    position + A + B

    Value A should be default linear.

    Value B should be posterize linear by a custom frame rate.

    If I use the posterize(myFrameRate) function, both values are posterized.

    Maybe somebody here have an idea to realize this behavior.



  • Andrei Popa

    January 15, 2022 at 8:14 am

    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

    January 15, 2022 at 3:19 pm

    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;

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