finding the position of a linear shape layer path

[Dan Ebberts]

I’d have to see an example of what you’re talking about.

Dan

[Peter Zeet]

oh! sorry for the awful explanation, here what I mean:

I was going to use the technique described on the early posts of the thread, but I wanted to do it getting rid of the copy the patm paste as position, etc, so I was wondering if an expression like the one I used, but tweaking the 423 and 1395 values inside of the linear, so instead if the actual X coords I could use the margin left and right of whatever the width of the path could be, making it work always, so the circle is always following the end trim…

[Dan Ebberts]

Ah, OK. It would probably look something like this:

s = thisComp.layer(“yell”);
r = s.sourceRectAtTime(time,false);
[(s.position + s.anchorPoint+ r.left+r.width)[0] + width/2, value[1]]

Dan

[Peter Zeet]

hi!
mmm it kinda works, but I had to tweak a bit the setup (maybe it has to be like this?)

I had to:
1.change the expression to
s = thisComp.layer("yell");
r = s.sourceRectAtTime(time,false);
[(s.position + s.anchorPoint+ r.left+r.width)[0]] //on X pos, separated dimension

2.change the Ball anchor point to its right
3. change the yell shape layer’s anchor point to its left

and this way, it works.. otherwise, it was waw super offseted from the line
it’s weird, but it works 🙂