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  • How do I get my negative and positive value amounts to show their actual amounts, not relative?

    Posted by Dave Andrade on July 8, 2012 at 2:00 pm

    Sorry I am not including specifics, but I know there is a simple answer, and I cannot locate one. Right now, I am setting up an expression, and I can adjust the value from zero both in the positive and negative direction.

    The PROBLEM lies where whether or not I adjust the amount to lets say -10….or 10…After effects reads that amount as 10. So if I pickwhip lets say opacity it will be transparent at 0, and fully opaque at 10 and -10.

    what I am trying to do it set an expression where -10 is the LOWEST and 10, for example, would be the highest (not the distance relative to 0). I am sure there is a simple javascript script I can insert, but I cannot find one. Again, ideas?

    Adrian Hargrave replied 13 years, 2 months ago 3 Members · 3 Replies
  • 3 Replies
  • John Cuevas

    July 8, 2012 at 2:29 pm

    Sounds like you want an interpolation expression. I made a solid and added a slider control that went from -10 to 10 to create this expression, but you could delete effect(“Slider Control”)(“Slider”) as the controller and pick whip to whatever you want to use as an input.

    input = effect("Slider Control")("Slider");
    ease(input, -10,10,0,100)

    Here’s a link that explains this expression and the other interpolation expressions. Interpolation Methods: linear(), ease()

    Johnny Cuevas, Editor
    Thinkck.com

    “I have not failed 700 times. I have succeeded in proving that those 700 ways will not work. When I have eliminated the ways that will not work, I will find the way that will work.”
    —THOMAS EDISON on inventing the light bulb.

  • Dave Andrade

    July 8, 2012 at 4:20 pm

    Thanks John! I appreciate the help!

  • Adrian Hargrave

    March 5, 2013 at 4:37 am

    Sorry for reviving an old thread, but I’m having the same issue.

    John suggested changing the minimum expected value. But this just increases the expression’s expected value range.

    linear(t, t_min, t_max, value1, value2)

    The way this expression should work is when t equals or is less than t_min, the method will output value1.

    This is not what is happening for me. When t is less than t_min it treats it as a positive number.

    Can anyone help me out where I’m going wrong in all this?

    Thanks!

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