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Following a path without changing position?
Hi dudes,
A lot of thanks for the AE experts who will answering to me.
It’s a problem that I never read a solution on anywhere.
So let me explain, I have a lot of objects going out an optical fiber.
All this object have a mandatory starting and ending position.
Instead of animate all theses objects one by one, I wish set them to follow a predefined curved path without changing starting and ending postion, they just need to follow the curve.How i can do this?
Thanks again.
Val -
5 Replies
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There is no “align to spline” or similar effect in After Effects. One which you could have the layer have its own position while following a spline field. (like those available forces and effectors in 3d apps)
So you’ll have to either animate the layer’s position, or link it to a layer that follows the spline, but since you’re always attributing that motion to the position property, you’ll have to chose one or the other.
Cassius Marques
http://www.zapfilmes.com -
If you have a shape layer with a path, and you want another layer to follow that path, controlled by a slider on the follower layer, you could do something like this:
s = effect(“Slider Control”)(“Slider”).value;
L = thisComp.layer(“Shape Layer 1”);
path = L.content(“Shape 1”).content(“Path 1”).path;
L.toComp(path.pointOnPath(s/100))Dan
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That would absolutely change the start and end position of that layer, would it not?
He expressed the need to keep initial and final position.
Cassius Marques
http://www.zapfilmes.com -
Ah yes, I missed that. That makes it trickier. If it’s really just the start position that needs to be maintained, that’s an easy fix, like this:
s = effect(“Slider Control”)(“Slider”).value;
L = thisComp.layer(“Shape Layer 1”);
path = L.content(“Shape 1”).content(“Path 1”).path;
value + path.pointOnPath(s/100) – path.pointOnPath(0)But if it’s really both the start and end positions, it depends on how those are defined (first and second keyframes?) and if the timing needs to be maintained. Still do-able, I think, but with more math involved.
Dan
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A lot of thanks to all of you !
I knew I will have help here 😀
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