CODEWITHSUNDEEP
mathcollisionbounding
user4428063
user4428063

Reputation:

Checking if any of corners of a rectangle is inside other rectangle

I'm having a mathematical issue.

I have no issue checking if one rectangle is within bounds of the other when they both (or one of them) have no angle and fit the grid perfectly.. but once I change angle of both of them, it all goes downhill.

Here's what I mean - the one to the right works flawlessly, but the ones to the left not so much: enter image description here

Upvotes: 1

Views: 242

Answers (4)

samgak
samgak

Reputation: 24427

It's possible to do this without using either rotations (which means calling trig functions) or normalizing vectors (which requires a sqrt).

Suppose your rectangle is defined by points a, b, c, d in clockwise order. A point is inside the rectangle if it is on the righthand side of each of the vectors ab, bc, cd and da.

You can test point p against vector ab by taking the dot product of the vector ap with a vector perpendicular to ab, which you can get just by swapping the x and y and flipping one of the signs. The sign of the dot product tells you what side of the vector you are on.

If the signs are all positive then the point is inside the rectangle. You don't need to normalize because you are only checking the sign, not the magnitude.

float pointToVector(Vec2 p, Vec2 a, Vec2 b)
{
    Vec2 n = new Vec2(b.y - a.y, a.x - b.x); // no need to normalize

    return n.dot(p - a);
}

bool pointInRect(Vec2 p, Vec2 a, Vec2 b, Vec2 c, Vec2 d)
{
    return (pointToVector(p, a, b) > 0) && // can short-circuit
       (pointToVector(p, b, c) > 0) &&
       (pointToVector(p, c, d) > 0) &&
       (pointToVector(p, d, a) > 0));
}

Note that this is actually an arbitrary point-in-4-sided-convex-polygon test. You could optimize it for rectangles by taking advantage of the fact that ab and cd are parallel, as are bc and da, so you can reuse the perpendicular vectors but reverse the sign of the check. That makes the code a little more complicated however. Also, the above assumes a co-ordinate system with y increasing going up. If not then change the the comparisons to < or define the rectangle in anti-clockwise order.

Upvotes: 1

hansmaad
hansmaad

Reputation: 18915

enter image description here

P = aA + bB

Solve for a and b. P is inside the rect, if a < 1 && b < 1

Upvotes: 0

user3235832
user3235832

Reputation:

I recommend just using Tim's answer and not be so wussy about rotation. However in case you really are too OCD for that, you could try the following vector-based method.

Since the sides of the rectangle are perpendicular, you can use them as a (orthogonal) coordinate system.

enter image description here

New coordinate axes:

enter image description here

Transformed coordinates:

enter image description here

Checks:

enter image description here

Upvotes: 0

Tim Biegeleisen
Tim Biegeleisen

Reputation: 522762

You can rotate both rectangles such that one becomes aligned with the grid. First, calculate the degree of rotation for both rectangles:

angle = arc tan(ength of long side / length of short side)
        arc tan(11 / 7)
      = 57.5 degrees

Now you know that the green triangle is 57.5 degrees counter-clockwise from its long side being parallel to the x axis. So you can rotate both rectangles by 57.5 degrees clockwise to get a configuration which you can handle. Please see here for how to do this.

Upvotes: 0

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