sin^-1 available in expressions?

[Mark Walczak]

Thanks everyone!

Dan, that site looks amazing.

I’m still, however having some issues getting it all to work. Perhaps I’ll post the scene if I can.

Thanks again!

What makes you explode?
http://www.explosivegraffix.com

[Dan Ebberts]

Math.asin(), Math.acos(), math.atan(), and math.atan2() are what you’re looking for. This is a good reference:

https://www.w3schools.com/jsref/jsref_obj_math.asp

Dan

[Koby Goldberg]

Yes, they exist. Use:

Math.asin(x)
Math.acos(x)
Math.atan(x)

Koby.

[Koby Goldberg]

Mark,
Note that asin, acos,… return the angle in Radians!
If you want to convert it to Degrees, you should multiply the result by 180 and divide by pi (Math.PI).
However, if you are using sin(x) or cos(x), x should be in Radians.

Koby.

[Mark Walczak]

Thanks for the tip! Fortunately, I was able to find the radians-to-degrees command, which seemed to do the trick. Thanks for all of your help!

What makes you explode?
http://www.explosivegraffix.com

[Jim Lefevre]

Mark (and everyone),

THANK YOU! I’m doing exactly the same thing and have been hitting my head against a wall for a day or so.

You guys are all so great.

Jim

[Jim Lefevre]

Ah… spoke too soon!

As Mark was doing, I’m trying to figure out the angle of a right angled triangle from the lengths of the sides using inverse sin.

I’ve followed all of the replies here and am getting somewhere all up until I try and call the inverse sin using …

Math.asin().

…where the expression gives me an error. Here is the code…

//half the base length
m = thisComp.layer("DRIVER_all_sides").effect("length - X")("Slider")/2;
//pythagorus to find the length of the hypotenuse
a = thisComp.layer("DRIVER_all_sides").effect("length - X")("Slider")/2;
b = thisComp.layer("DRIVER_all_sides").effect("HEIGHT_TRIANGLE")("Slider");
x = Math.sqrt(((a*a) + (b*b)));
//angle via inverse sin
q = (x/m);
p = Math.asin(q);
p

The error is the “Error: couldn’t turn the result into a numeric value” one

I’ve yet to get it to the stage of turning it the Radians into degrees as it’s falling over as soon as Math.asin() is in there.

Does anyone have any suggestions? (and I’ve had a trawl through that wonderful Math resource you posted Dan)

[Jim Lefevre]

HA!!

Done it I think.

For some reason my equation to get the angle via inverse sin had the values dividing the wrong way – I was dividing the base length by the hypotenuse or the other way round – which meant that I was coming out with a value greater than 1 which, it appears, an inverse sine can’t do (I have no idea how this works btw).

I swapped them round and came to a result which worked and then had to take that value away from 90 degrees so that it was rotating in the direction I needed.

Works perfectly now…

//I have two slide values that control the Base length and the height of the triangle (the altitude)

//half the base length to get me a value that creates a right angle triangle essentially
m = thisComp.layer("DRIVER_all_sides").effect("length - X")("Slider")/2;

//pythagorus to find the length of the hypotenuse
a = thisComp.layer("DRIVER_all_sides").effect("length - X")("Slider")/2;
b = thisComp.layer("DRIVER_all_sides").effect("HEIGHT_TRIANGLE")("Slider");
x = Math.sqrt(((a*a) + (b*b)));

//get the angle via inverse sin

q = Math.asin(m/x);

//convert this to degrees

p = (radiansToDegrees(q));

//final result needs offsetting so it rotates the correct way when I change the base triangle value

90-p

[Koby Goldberg]

Hi Jim,

You should be able to do the calculation slightly simpler by using Math.atan2 instead of Math.asin, and thus you will not need the calculation of x (you won’t need the sqrt calculation). Simpler expressions could make your render faster, because expressions are being calculated every frame.

Try this, I think it should give you the same result as above:

a = thisComp.layer(“DRIVER_all_sides”).effect(“length – X”)(“Slider”)/2;
b = thisComp.layer(“DRIVER_all_sides”).effect(“HEIGHT_TRIANGLE”)(“Slider”);

Math.atan2(b,a)

If you still need to do the (90-p) conversion, then just swap a and b in the Math.atan2, i.e.: Math.atan2(a,b), but I think atan2(b,a) should give you the final angle you wanted.

Cheers,

Koby

[Jim Lefevre]

Oh wow. I think this is all fresh enough in my mind that I can restitch it with that and see if it works. I absolutely agree that making things more efficient will help in the long run.

Thank you