0
$\begingroup$

I need some maths help for a 2D game I am programming. In this game I have a rectangle, specified by its centers' X and Y coordinates, and its width and height. I then rotate this rectangle via

halfWidth = width / 2;
halfHeight = height / 2;

rad = angle * TO_RADIANS;
cos = FloatMath.cos(rad);
sin = FloatMath.sin(rad);

x1 = -halfWidth * cos - (-halfHeight) * sin;
y1 = -halfWidth * sin + (-halfHeight) * cos;
x2 =  halfWidth * cos - (-halfHeight) * sin;
y2 =  halfWidth * sin + (-halfHeight) * cos;
x3 =  halfWidth * cos - halfHeight * sin;
y3 =  halfWidth * sin + halfHeight * cos;
x4 = -halfWidth * cos - halfHeight * sin;
y4 = -halfWidth * sin + halfHeight * cos;

x1 += x;
y1 += y;
x2 += x;
y2 += y;
x3 += x;
y3 += y;
x4 += x;
y4 += y;

That is working fine.

Additionally I have a point in the same Cartesian coordinate space, specified by pointX and pointY. Now I want check whether or not this point lies within the boudaries of the rotated rectangle -- but unfortunately I do not have a clue how to achieve this :)

Can you help me with this question?
Thanks in advance!

Kind regards, Matthias

  • 0
    Just for the record, I have found two interesting links on that topic: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html [http://www.visibone.com/inpoly/](http://www.visibone.com/inpoly/) Cheers!2011-07-30

1 Answers 1

6

Rotate the point back and test it against the unrotated rectangle.

  • 2
    If you can't bring the rectangle to the point, bring the point to the rectangle. :)2011-07-28
  • 0
    He needs to rotate the test point around the center of the rectangle by -angle, but this is a good plan.2011-07-28
  • 0
    Thanks, that sounds like a plan! But is this also the fastest (in terms of computation time) solution? It seems the rotating all four X/Y pairs of the rect plus the X/Y coords. of the point is a bit much work for the computer (within a loop of a game) -- or isn't there any faster (mathematical) solution?2011-07-28
  • 0
    @Matthias, there is another solution based on half-spaces: the point must be to the left of all edges, when the rotated rectangle is oriented counter-clockwise. Each test reduces to computing a 2x2 determinant.2011-07-28
  • 0
    @lhf, sorry but I do not really understand your second approach. Could you please be so kind and explain that a bit? Thanks a lot in advance for all your help!2011-07-28
  • 0
    @Matthias, it's a point-location algorithm that works for any convex polygon. See also http://mathforum.org/library/drmath/view/54386.html2011-07-28