• Adobe After Effects Expressions

  Animating a property over a specified time

  Posted by Klemens Hertha on April 15, 2013 at 2:54 pm

  Hi!

  I wasn’t able to find anything here or on google, but this should be really basic and easy… Maybe you could help!

  I have a lot of pitures (3d) in a comp which are not animated on their own. A camera moves backwards and the pics with different z-values “fly away”. All I want to accomplish is that when a picture reaches a certain distance to the camera, it should begin scaling down to 0 over a certain amount of time, let’s say 2 sec.

  The expression I’ve got right now is this one:

  startScale = [100, 100, 100];
  endScale = [0, 0, 0];
  distance = thisComp.layer(“Camera 1”).transform.position[2] – transform.position[2];

  if (distance > (-12000))
  {startScale}
  else
  {ease (time, in_point, out_point, startScale, endScale)}

  Works, but 2 problems: Now the scale-value stays at 100% as it should but when the layer reaches the specified distance to the cam it jumps to the value which it has at this point in time! I want it to smoothly scale down from 100 to 0!
  Second is that I don’t want to use the layer in- and out-points as tMin / tMax. tMin should be the frame where the specified distance is reached (how do I convert that in time??) and tMax 2 sec later!

  Thanks in advance!

  Klemens Hertha replied 13 years, 1 month ago 2 Members · 2 Replies
• 2 Replies

• 

  Dan Ebberts

  April 15, 2013 at 3:54 pm

  Distance triggers aren’e simple. You have to go through the timeline, frame by frame, to find out how long it has been since the threshold was reached.

  Something like this:

  
startScale = [100,100,100];
endScale = [0,0,0];
threshold = 12000;
p2 = thisComp.layer("Camera 1").transform.position;
p1 = transform.position;

  fCur = timeToFrames(time);
f0 = timeToFrames(inPoint);
t = inPoint;
for(i = f0; i <= fCur; i++){
t = framesToTime(i);
if (Math.abs(p1.valueAtTime(t)[2] - p2.valueAtTime(t)[2]) > threshold) break;
}
ease(time-t,0,2,startScale,endScale)


  Dan
• 

  Klemens Hertha

  April 15, 2013 at 5:26 pm

  Thanks again, Dan! Works!
  Although I don’t understand this one anymore… 😉

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