3d - calculating distance between a point and a rectangular box (nearest point)

Question

is there a easy formula to calculate this? i've been working on some math but i can only find a way to calculate the distance directed to the center of the box, not directed to the nearest point.. are there some resources on this problem?

question from:https://stackoverflow.com/questions/5254838/calculating-distance-between-a-point-and-a-rectangular-box-nearest-point

Answer 1

Here is a single formula that avoids all the case logic. (I happen to be working in JS right now, so here's a JS implementation). Let rect = {max:{x:_, y:_}, min:{x:_, y:_}} and p={x:_, y:_}

function distance(rect, p) {
  var dx = Math.max(rect.min.x - p.x, 0, p.x - rect.max.x);
  var dy = Math.max(rect.min.y - p.y, 0, p.y - rect.max.y);
  return Math.sqrt(dx*dx + dy*dy);
}

Explanation: This breaks down the problem into calculating the x distance dx and the y distance dy. It then uses distance formula.

For calculating dx, here is how that works. (dy is analogous)

Look at the tuple being provided to the max function: (min-p, 0, p-max). Let's designate this tuple (a,b,c).

If p is left of min, then we have p < min < max, which means the tuple will evaluate to (+,0,-), and so the max function will correctly return a = min - p.

If p is between min and max, then we have min < p < max, which means the tuple will evaluate to (-,0,-). So again, the max function will correctly return b = 0.

Lastly, if p is to the right of max, then we have, min < max < p, and the tuple evaluates to (-,0,+). Once again, Math.max correctly returns c = p - max.

So it turns out all the case logic is taken care of by Math.max, which leads to a nice 3-line, control-flow-free function.