Forum Replies Created

Viewing 1 - 10 of 10,014 posts
  • Dan Ebberts

    August 8, 2022 at 2:08 pm

    Instead of this:

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

    try this:

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

    August 7, 2022 at 1:42 pm

    I don’t know of a way to do that.

  • Dan Ebberts

    August 5, 2022 at 8:37 pm

    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

    August 2, 2022 at 2:52 pm

    I think I was assuming Time Remap.

  • Dan Ebberts

    July 28, 2022 at 5:51 pm

    Try this:

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

    July 28, 2022 at 3:56 pm

    Or like this:

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

    July 27, 2022 at 4:26 pm

    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

    July 26, 2022 at 12:52 pm

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

    Add this expression to the Source Text property:

    style.font
  • Dan Ebberts

    July 23, 2022 at 5:12 pm

    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

    July 23, 2022 at 1:46 pm

    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);
Viewing 1 - 10 of 10,014 posts

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