[Dan Ebberts]

Instead of this:

newV2 = parseInt (thisComp.layer("Text Layer").text.sourceText,10);

try this:

newV2 = parseFloat (thisComp.layer("Text Layer").text.sourceText);

[Dan Ebberts]

I don’t know of a way to do that.

[Dan Ebberts]

Try it this way:

m = thisComp.layer(1).marker;
n = 0;
val = "";
if (m.numKeys > 0){
  n = m.nearestKey(time).index;
  if (m.key(n).time > time) n--;
}
if (n > 0){
  t = time - m.key(n).time;
  if (t < m.key(n).duration){
    val = m.key(n).comment;
  }
}
val

[Dan Ebberts]

I think I was assuming Time Remap.

[Dan Ebberts]

Try this:

c = comp("countdown");
c.layer("00").text.sourceText.valueAtTime(time + c.layer(thisComp.name).startTime)

[Dan Ebberts]

Or like this:

L = thisComp.layer("childNull");
L.parent.toComp(L.position)

[Dan Ebberts]

When you say “empty object” are you talking about a null layer? If so, then I think you just need something like this:

L = thisComp.layer("Null 1");
L.toComp(L.anchorPoint)

[Dan Ebberts]

With the text tool, create a new text layer “test”.

Add this expression to the Source Text property:

style.font

[Dan Ebberts]

The part that defines the in tangents? That part takes advantage of the fact the the tangents are parallel to the vector defined by the two points opposite the current point. It’s then just a matter of sizing that vector, and then inverting it to create the out tangent.

[Dan Ebberts]

I created an equilateral triangle shape and converted it to a Bezier path. After discovering that the path had 6 points rather than the 3 I expected, I was able to craft this path expression to add tangent handles. Variable m sets the size of the tangents relative to the sides of the triangle. I hope it’s helpful:

pts = points();
p = [pts[0],pts[2],pts[4]]; // get rid of extra points
m = 1/5; // multiplier
iT = [];
oT = [];
iT[0] = (p[2]-p[1])*m;
iT[1] = (p[0] - p[2])*m;
iT[2] = (p[1] - p[0])*m;
oT[0] = -iT[0];
oT[1] = -iT[1];
oT[2] = -iT[2];
createPath(p,iT,oT,true);