Radius of a box in a given direction - math

Question

Let's say I have an axis-aligned box. I also have a location outside that box.
How do I get the point on the boundaries of the box that is along the line from the center of the box to that point? (And thus, know how far the box extends in that direction).
EDIT: preferably, if the point is inside the box, it should just return the point... but that's something I could easily calculate myself.
2D diagram (Though the answer will preferably work in 3D as well):

Answer 1

point3d getIntersection(point3d topfrontleft, point3d backbottomright, point3d externpt)
{
//find the middle
point3d middle(topfrontleft/2+backbottomright/2);
//slide box and source as if the "middle" was at the origin
topfrontleft -= middle;
backbottomright-= middle;
externpt-= middle;
//scale source as if the box is the unit square
externpt/= topfrontleft;
//find the largest magnitude of the source offset
point3d mag = abs(externpt);
auto max = std::max(mag.x,std::max(mag.y,mag.z));
//if it's inside the box, don't scale
if (max < 1)
max = 1;
//scale the source so that it touches the box
externpt/= max;
//then scale and slide that touching point back to original coordinates
externpt= externpt* topfrontleft + middle;
return externpt;
}
http://coliru.stacked-crooked.com/a/9d9504964bc650bb
I'm not experienced with 3d math, so there's probably faster and more effective ways. No idea.
As a bonus, this algorithm doesn't make any reference to the number of dimensions except for finding the largest magnitude, so should work in both 2d, 3d, and any other theoretical number of dimensions.

Answer 2

Here is some code that does what I said in comments modified for rectangular boxes.
#include <stdio.h>
#include <math.h>
// Find intersection p between line A->B and box.
// Point A must be the box center.
// The box is [x_left, y_bottom, x_right, y_top].
void find_box_intersect(double *box, double *a, double *b, double *p)
{
double dx = b[0] - a[0];
double dy = b[1] - a[1];
if (fabs(dx * (box[3] - box[1])) > fabs(dy * (box[2] - box[0]))) {
p[0] = dx > 0 ? box[2] : box[0];
p[1] = a[1] + dy * (p[0] - a[0]) / dx;
} else {
p[1] = dy > 0 ? box[3] : box[1];
p[0] = a[0] + dx * (p[1] - a[1]) / dy;
}
}
int main(void) {
double box[] = { 1, 2,
2, 4 };
double p[2], a[] = { 0.5 * (box[0] + box[2]),
0.5 * (box[1] + box[3]) };
int i, n = 16;
for (i = 0; i < n; i++) {
double theta = 2 * 3.14159 * i / n;
double b[] = { a[0] + cos(theta), a[1] + sin(theta) };
find_box_intersect(box, a, b, p);
printf("%.2f, %.2f\n", p[0], p[1]);
}
return 0;
}
The 3d generalization of this is straightforward. Rather than the if statement that determines only 2 cases, there will be a 3 case if-else chain that determines 3: left-right, top-bottom, front-back.
Output:
2.00, 3.00
2.00, 3.21
2.00, 3.50
1.91, 4.00
1.50, 4.00
1.09, 4.00
1.00, 3.50
1.00, 3.21
1.00, 3.00
1.00, 2.79
1.00, 2.50
1.09, 2.00
1.50, 2.00
1.91, 2.00
2.00, 2.50
2.00, 2.79

Answer 3

Short answer: Intersect the line with the (right) border of the box.
Longer answer:
x is easy: It's at the right border of the box.
for y, solve this: y/line_height = (line_width - box_width/2) / line_width
then add the y of the line lower point
This is assuming the line intersects the right border, as in your picture.
To figure which border the line intersects, compare the ratio of line_height/line_width, and the signs, to the ratio of box_height/box_width.
If the box is square and you're only interested in the radius, and not the intersection point, you can throw away the signs of line_width and line_height, sort them, and just solve one case.
If the box is not square, you can still throw away the signs, but need to compare the ratios to pick one of two remaining cases (in 2D).

Answer 4

Assuming that you know the length and width of the rect, if the angle is pointing left or right, then x is known, so you only need to solve for y (r * sin(angle)); if the angle is pointing up or down, then y is known so you solve for x (r * cos(angle)).
// where the center of the rect = 0, 0 and angles are in degrees 0..360
// determine the 45's first
if (angle in (45, 135, 225, 315)) {
return { x:cos(angle) * (width / 2), y:sin(angle) * (height / 2) };
}
if (angle < 45 || angle > 315) {
// pointing right
return { x:(width / 2), y:sin(angle) * (height / 2) };
}
if (angle > 45 && angle < 135) {
// pointing up
return { x:cos(angle) * (width / 2), y:(height / 2) };
}
if (angle > 135 && angle < 225) {
// pointing left
return { x:(width / -2), y:sin(angle) * (height / 2) };
}
if (angle > 225 && angle < 315) {
// pointing down
return { x:cos( angle ) * (width / 2), y:(height / -2) };
}
There are undoubtedly more elegant mathematical solutions to this problem, but this one works for me--understandable, testable, etc.