I am doing some programming, and I hit a bump. I got a square that I can rotate at will. I got an angle from $0$ to $360$ The square is a fixed size, let's say 3.

Now I need to find the position of the 4 corners. This is "relatively" easy to do when the square is at a fixed angle, as I just need to plus on the $X$ & $Y$ axis, half the width & half the height, to get to a corner. The difficult part for me is to do the same thing while the square is rotated.

Say at $45$degrees I can't figure out the position of the "top left corner" anymore. Is there a simple equation I could use/create that would use say -1.5, 1.5 (on either or both XY axis) that would give me the position $(XY)$ of one of the corners?

Thanks in advance Concerned mathdoer

(Note: I am using this in a script, to spawn objects in a square with a fixed rotation and size, from 1 to 9, from top left to bottom right in orderly fashion)

  • $\begingroup$ Are you familiar at all with matrices? $\endgroup$ – G Tony Jacobs Aug 18 '17 at 15:52
  • $\begingroup$ Nope, not at all. $\endgroup$ – Blueduckraider50 Aug 18 '17 at 15:54
  • $\begingroup$ I can think of possibilities using pythagorean theorem to some extent possibly if you need that ( assuming you know the pythagorean theorem). okay I guess my idea is more of a changing center thing. $\endgroup$ – user451844 Aug 18 '17 at 16:08

If the center of the square is at $(0,0)$, and one of the corners is originally at $(x,y)$, then after rotating through an angle of $\theta$ counterclockwise, that angle's position is given by:

$$\left[\begin{matrix}\cos \theta & -\sin\theta \\ \sin\theta & \cos\theta \end{matrix}\right]\left[\begin{matrix} x \\ y \end{matrix}\right],$$

Unpacking that, since you haven't worked with matrices, you get:

$$(x\cos\theta - y\sin\theta, x\sin\theta+y\cos\theta),$$

as your new coordinates.

If the center of your square is originally at $(h,k)$, and your corner is originally at $(x,y)$, then your new location is:

$$(h+ (x-h)\cos\theta - (y-k)\sin\theta, k+(x-h)\sin\theta + (y-k)\cos\theta).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.