[Dan Ebberts]

Not directly. You can loop through all the layers though, which might give you what you’re after. What are you trying to do?

Dan

[Dan Ebberts]

You might have to do it this way if your objects are parented to the merry-go-round:

lookAt(thisComp.activeCamera.position,toWorld(anchorPoint));

Dan

[Dan Ebberts]

This should do it:

if (rotation < 0){ linear(rotation,-80,0,[5,7],[0,0]); }else{ linear(rotation,0,140,[0,0],[6,0]); } Dan

[Dan Ebberts]

Recipe for shadow-catching invisible layer:

Use a white solid.
Set mode to Multiply.
Accepts Shadows = on.
Accepts Lights = off.

Dan

[Dan Ebberts]

Something like this maybe:

clockStart = 120; // start time in seconds
clockTime = clockStart – time;
sec = Math.floor(clockTime).toString();
ms = Math.floor((clockTime%1)*100).toString();
if (ms.length < 2) ms = "0" + ms; sec + ":" + ms Dan

[Dan Ebberts]

OK – but it’s a little hard to describe without a diagram.

delta = toWorld(anchorPoint) – thisComp.activeCamera.toWorld([0,0,0]);

This line just calculates the vector (distance and direction) from the camera to the layer. It looks complicated because it’s using the toWorld() layer space transform to convert the layer and camera positions to world coordinates. This is only necessary if the layer or the camera is the child of another layer, but it makes the expression more general to assume that’s the case. If you don’t have to worry about parenting, you could use something like this instead:

delta = position – thisComp.activeCamera.positon;

The next line:

radiansToDegrees(Math.atan2(delta[0],delta[2]))

This is pure trig. Looking from the top, the vector calculated in the first line defines the y rotation that you need to get the layer to turn towards the camera. That y rotation can be calculated from the arc tangent of the x and z components of the vector (that’s what atan2() does). JavaScript represents angles in radians, but AE’s rotation property expects degrees, so we use radiansToDegrees() to do the conversion.

Hope that helps – it’s a lot to bite off if you aren’t comfortable with a little vector math and trig.

Dan

[Dan Ebberts]

Try this on your y rotation property:

delta = toWorld(anchorPoint) – thisComp.activeCamera.toWorld([0,0,0]);
radiansToDegrees(Math.atan2(delta[0],delta[2]))

Dan

[Dan Ebberts]

Yeah, I’d make them all 3D and use an expression like this to sort them in Z space:

below = 0;
for (i = 1; i <= thisComp.numLayers; i++){ if (i == index) continue; if (thisComp.layer(i).position[1] < value[1]) below--; } [value[0],value[1],below] Dan

[Dan Ebberts]

Strangely, noise() returns a value in the range -1 to +1.

dan

[Dan Ebberts]

Possible, but not simple. I think you’d have to write some code that would start at the current comp time and loop backwards through the comp until you found the time when layer1 arrived a y = -102, then calculate where layer2 should be now based on how long it’s been, how fast you want the move to occur, and the distance to be traveled.

Dan