Question

According to Daniel, in his answer, there is no easy way to modify the below function so I bit the bullet and started from scratch. Solution is below (as an answer). Actually, ignore my answer. See Tom Sirgedas’ answer it’s a lot shorter.

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I need to modify the solution found here: Calculate a vector from the center of a square to edge based on radius, that calcs the vector from the center of a rectangle, to work for any point within the rectangle.

Here’s the previous solution, from the link:

    double magnitude;
    double abs_cos_angle= fabs(cos(angle));
    double abs_sin_angle= fabs(sin(angle));
    if (width/2*abs_sin_angle <= height/2*abs_cos_angle)
    {
            magnitude= width/2/abs_cos_angle;
    }
    else
    {
            magnitude= height/2/abs_sin_angle;
    }

    double check_x= x + cos(angle)*magnitude;
    double check_y= y + sin(angle)*magnitude;

check_x and check_y return the point on the edge of the rectangle that a line drawn from the center, at angle, will intersect.

It’s a while since I went to school, so I blindly tried replacing width/2 and height/2 with the point I’m interested in. Unfortunately that didn’t work.

Any ideas?

ETA:

This blind modification always returns the correct result if the line intersects the rectangle at the Top or Left. Depending on the quadrant the originating point is in, it returns a point too far away or too close when the line intersects the Right or Bottom.

Answer 1

Let’s say the rectangle is defined by (x1,y1,x2,y2) and let’s say the ray starts at (px,py).

Let vx = cos(angle)

Let vy = sin(angle)

Traveling a distance of t along the ray will bring you to the point (px+tvx, py+tvy).

Traveling along the ray,

• we hit the left wall when px+t*vx = x1, or t=(x1-px)/vx
• we hit the right wall when px+t*vx = x2, or t=(x2-px)/vx
• we hit the top wall when py+t*vy = y1, or t=(y1-py)/vy
• we hit the bottom wall when py+t*vy = y2, or t=(y2-py)/vy

So, there are four possible solutions for t. The correct value of t (among the four) is the smallest positive one. The actual intersection is at the point (px+tvx, py+tvy). Just be careful not to divide by zero!