• Adobe After Effects

  Changing Mask Path after Expression is Applied

  Posted by Marco Polman on April 18, 2013 at 2:23 pm

  I’ve done some extensive searching for previous posts about this issue, but have yet to find any information.

  I’ve found that when I am using an expression on a mask path that I am unable to grab any of the handles or points on the mask until the expression is removed or disabled. I was wondering if anybody else has experienced this problem, and if you’ve found a solution…? I suppose I can keep disabling and re-enabling mask expressions but it’s kind of a hassle considering how many layers and keyframes I’m working with at the moment.

  The specific expression I am using is created using the Ease & Wizz plugin (from AE Scripts), which if you’d like to re-create the problem, is available for free trial. The plugin, if you’re unfamiliar with it, creates an expression to replace AE’s easy-ease functions with curves that cannot be created using the curve handles alone. (It’s quite nice, actually…). I’ve included my specific expression below:

  _________________________________________________________________________________________
  // Curvaceous – all keyframes
  // 2.0.1 : inOutQuart
  // Ian Haigh (https://ianhaigh.com/easeandwizz/)
  // Last built: 2010-10-09T13:35:39+11:00
  function inOutQuart(t, b, c, d) {
  if ((t/=d/2) < 1) return c/2*t*t*t*t + b;
  return -c/2 * ((t-=2)*t*t*t – 2) + b;
  }
  function curvaceous() {
  var n = 0;
  if (numKeys > 0) {
  n = nearestKey(time).index;
  if (key(n).time > time) { n– }
  }
  // after the first two keys, yet before the last two, just do nothing
  if (n > 1 && n < numKeys -1 ) {
  return null;
  }
  try {
  var key1 = key(n);
  var key2 = key(n+1);
  } catch(e) {
  return null;
  }

  t = time – key1.time;
  d = key2.time – key1.time;
  sX = key1.time;
  eX = key2.time – key1.time;
  if ((time < key1.time) || (time > key2.time)) {
  return null;
  } else {
  return valueAtTime(inOutQuart(t, sX, eX, d));
  }
  }
  (curvaceous() || value);
  _________________________________________________________________________________________

  Thanks for any help!

  -Marcus

  Marco Polman replied 13 years, 1 month ago 1 Member · 0 Replies
• 0 Replies

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