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Maintain X position of shape relative to rotation of a square
Hi everyone.
I have an expression query that I really hope someone can help me with!
Please see x2 attached screenshots. As you can see, I have a square scaling down and rotating in the centre of my comp. I also have two rectangles either side that are to scale up or move along the X axis while remaining perfectly aligned to opposing corners of the square, even while it rotates.
I’m pretty sure it’s something to do with Math.cos expression, however I can’t seem to get it right. Any suggestions?
Thanks in advance.
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3 Replies
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Hi Elliot, I think the easiest solution that comes to mind now is take maximum/minimum of the cos of all four corners.
so, if the radius of the circle those 4 corners fall on is say 200 pixels:
a = thisComp.layer("square").transform.rotation.value;
r = 200;
rightBoundaryX = r * Math.max(
Math.cos(degreesToRadians(a-45))),
Math.cos(degreesToRadians(a+45)),
Math.cos(degreesToRadians(a+135)),
Math.cos(degreesToRadians(a-135))
);
leftBoundaryX = r * Math.min(
Math.cos(degreesToRadians(a-45)),
Math.cos(degreesToRadians(a+45)),
Math.cos(degreesToRadians(a+135)),
Math.cos(degreesToRadians(a-135))
)I haven’t tested this code, but I hope it sets you in the right direction
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Another possibility would be to use sourceRectAtTime() and toComp() to establish the corners:
L = thisComp.layer("square");
r = L.sourceRectAtTime(time,false);
ul = L.toComp([r.left,r.top]);
ur = L.toComp([r.left+r.width,r.top]);
ll = L.toComp([r.left,r.top+r.height]);
lr = L.toComp([r.left+r.width,r.top+r.height]);
xMin = Math.min(ul[0],ur[0],ll[0],lr[0]);
xMax = Math.max(ul[0],ur[0],ll[0],lr[0]);
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Thank you both very much. This is extremely helpful.
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