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Train Trigonometry
Hello Fellows,
I’m trying to build a a Stirling engine with expressions
This is what I have so far…
https://www.youtube.com/watch?v=ZpIzXyB3LAc&feature=youtu.beHow could I change the code so the piston stayed in the tube?
r = thisComp.layer("rotary").transform.rotation;
t = Math.floor((time+3)/2)%2
o = 0;
if (o = t) {
Math.sin(degreesToRadians((r+360)/2))*180
} else {
Math.sin(degreesToRadians(r/2))*180
}Prepare for today and you’ll survive, prepare for tomorrow and you will thrive.
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2 Replies
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Assuming that the rotating piece on the right is the parent of the other rod, which is the parent of the piston, I’d use something like this for the rod’s rotation:
a = parent.transform.rotation;
r1 = 100;
r2 = 200;
b = radiansToDegrees(Math.asin(Math.sin(degreesToRadians(a))*r1/r2));
-(a+b)and something like this for the piston’s rotation:
a = parent.parent.transform.rotation;
r1 = 100;
r2 = 200;
radiansToDegrees(Math.asin(Math.sin(degreesToRadians(a))*r1/r2));r1 is the length of the rotating rod and r2 is the length of the other rod.
Dan
[edit: oops–added missing last line of first expression]
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That works nicely Dan, thanks.
Each time I get a response form you I learn something new.
The parent trick is epic for this.Prepare for today and you’ll survive, prepare for tomorrow and you will thrive.
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