[Dan Ebberts]

I think this is actually a little (half a pixel) more precise:

h = height/2;
w = width/2;
for (i = 0.5; i < width; i++){
  if (sampleImage([i,h],[.5,h])[3]>0) break;
}
left = i;
for (i = width-0.5; i > 0; i--){
  if (sampleImage([i,h],[.5,h])[3]>0) break;
}
right = i;
for (i = 0.5; i < height; i++){
  if (sampleImage([w,i],[w,.5])[3]>0) break;
}
top = i;
for (i = height-0.5; i > 0; i--){
  if (sampleImage([w,i],[w,.5])[3]>0) break;
}
bottom = i;
[(left+right)/2,(top+bottom)/2]

[Dan Ebberts]

The only thing I can think of is the old brute-force sampleImage() method:

h = height/2;
w = width/2;
for (i = 0; i < width; i++){
  if (sampleImage([i,h],[.5,h])[3]>0) break;
}
left = i;
for (i = width-1; i >= 0; i--){
  if (sampleImage([i,h],[.5,h])[3]>0) break;
}
right = i;
for (i = 0; i < height; i++){
  if (sampleImage([w,i],[w,.5])[3]>0) break;
}
top = i;
for (i = height-1; i >= 0; i--){
  if (sampleImage([w,i],[w,.5])[3]>0) break;
}
bottom = i;
[(left+right)/2,(top+bottom)/2]

[Dan Ebberts]

I think if you convert the rectangle to bezier and use this path expression, you won’t need the position or size expressions:

radius = 20;
p = content("Rectangle 1").transform.position;
L = thisComp.layer("more info");
r = L.sourceRectAtTime(time,false);
ul = fromComp(L.toComp([r.left,r.top]))-p;
ur = fromComp(L.toComp([r.left+r.width,r.top]))-p;
lr = fromComp(L.toComp([r.left+r.width,r.top+r.height]))-p;
ll = fromComp(L.toComp([r.left,r.top+r.height]))-p;
ur1 = [ur[0]-radius,ur[1]];
ur2 = [ur[0],ur[1]+radius];
t = radius*0.5525;
it = [[0,0],[-t,0],[0,-t],[0,0],[0,0]];
ot = [[0,0],[t,0],[0,t],[0,0],[0,0]];
createPath([ul,ur1,ur2,lr,ll],it,ot,true)

[Dan Ebberts]

I’m not 100% sure I understand what you’re asking for, but it might be something like this:

L=thisComp.layer("3dNull");
R=L.sourceRectAtTime(time,false);
UL = fromCompToSurface(L.toComp([0,R.top,R.left]));
UR = fromCompToSurface(L.toComp([0,R.top,R.left+R.width]));
LR = fromCompToSurface(L.toComp([0,R.top+R.height,R.left+R.width]));
LL = fromCompToSurface(L.toComp([0,R.top+R.height,R.left]));
createPath([[UR[0],UR[1]],[LR[0],LR[1]],[LL[0],LL[1]],[UL[0],UL[1]]], [], [], true);

[Dan Ebberts]

You probably have File > Project Settings > Expressions set to Legacy ExtendScript. Change it to JavaScript and the error should go away.

[Dan Ebberts]

Unfortunately, any time you apply a source text expression, all the characters are going to get the same styling.

[Dan Ebberts]

I’m guessing this isn’t quite right, but it will probably be something like this:

red = [.4,.0824,.1765,1];
orange = [.851,.4784,.2,1];
yellow = [.9255,.7176,.1922,1];
s = scale[0];
if (s <= 20)
  linear(s,0,20,red,orange)
else if (s <= 50)
  linear(s,20,50,orange,yellow)
else
  yellow

[Dan Ebberts]

Yes, ternary operator. Yours should work if you balance a couple of mismatched parens I think.

[Dan Ebberts]

I can’t imagine that the source code would be accessible. You might be able to reverse engineer something based on information that you do have access to, like the KeyframeEase object and cubic bezier formulas. Sounds like a bit of work though…

[Dan Ebberts]

That looks like a fairly complex physical simulation. It looks like something you might be able to pull of with Newton or Pastiche (or a combination of both) from aescripts. I don’t have those tools, so I’m not sure, but you might want to check them out.