• Adobe After Effects Expressions

  Wiggle… amplify on Z axis

  Posted by Robert Morris on October 12, 2007 at 9:23 pm

  Hello all expression gurus. I am working on an animation where I have a camera with a wiggle expression applied to it, and the wiggle’s value tied to a slider. What I want to do is have the camera position wiggle on all axis (which it does now), but have the Z axis more pronounced as if the camera is being jerked forward and backward. I supposed I could tie it to a separate wiggle slider and just have the value higher. But is there a way to tie it all to the one slider that controls the amount of wiggle? This is what I tried, but it doesn’t really work:

  x = wiggle(3,thisComp.layer(“null”).effect(“Slider Control”)(“Slider”));
  y = wiggle(3,thisComp.layer(“null”).effect(“Slider Control”)(“Slider”));
  z = wiggle(3,thisComp.layer(“null”).effect(“Slider Control”)(“Slider”))*5;
  [x[0], y[1], z[2]];

  Help?
  -Robert

  Darby Edelen replied 18 years, 7 months ago 3 Members · 3 Replies
• 3 Replies

• 

  Dan Ebberts

  October 12, 2007 at 9:40 pm

  This should work:

  zMult = 5;
  w = wiggle(3,thisComp.layer(“null”).effect(“Slider Control”)(“Slider”)) – value;
  value + [w[0],w[1],w[2]*zMult]

  Dan
• 

  Robert Morris

  October 12, 2007 at 9:55 pm

  Awesome! That seemed to work. I’m a bit fried to figure out why, but it does. Thank you. I’m new to expressions, but I hope to slowly learn more and more.

  Thanks, Dan!
• 

  Darby Edelen

  October 16, 2007 at 11:28 pm

  The wiggle() function returns a modified version of what it is supplied. What this means is that if you have a layer at [360, 240, 200] and apply an expression that reads wiggle(1, 5) you will get results between [355, 235, 195] and [365, 245, 205].

  In your earlier example you were multiplying the results of wiggle() directly. With our previous example and your expression you would get results between [355, 235, 975] and [365, 245, 1025], not only is that a difference of 50 on the Z-axis (5 * 5 should be a difference of 25) but it’s much much farther away on the Z-axis than it should be! What gives?

  What you want to do is multiply the offset that the wiggle() is generating, not the result itself. In Dan’s expression wiggle(3,thisComp.layer("null").effect("Slider Control")("Slider")) - value; gives you the difference between the layer’s ‘normal’ position and it’s wiggled position (in the example we’ve been using this would be values between [5,5,5] and [0,0,0]).

  You don’t want to use this offset directly in your result, instead you want to multiply the offset’s Z value by 5 (giving us [5,5,25]) and then add the original position back into this vector, again from Dan’s expression:

  value + [w[0],w[1],w[2]*zMult]

  The other thing to note about Dan’s expression is that he only used wiggle() once and assigned that value to a variable. This cleaner for the position property because wiggle() will give you a vector in the form of [x,y,z] when applied to the position property. Calling wiggle() three times will give you 3 different [x,y,z] vectors (9 total values of which you’re only using 3) and take (theoretically) three times as long to compute.

  Darby Edelen
  DVD Menu Artist
  Left Coast Digital
  Aptos, CA

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