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15936166
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2013-04-10T20:53:21.683
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2013-04-10T21:03:41.470
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2013-04-10T21:03:41.470
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First of all, instead of thinking of two rectangles moving with speed <code>(x1, y1)</code> and <code>(x2, y2)</code> respectively, you may fix one of them (set its speed to <code>(0, 0)</code> ) and think of another one moving with speed <code>(x2 - x1, y2 - y1)</code>. This way, the situation looks like one rectangle is immovable, and another one is passing by, possibly hitting the first. <ul> <li>Assuming you don't have any angular velocity</li> </ul> <img src="https://i.stack.imgur.com/WlC7D.png" alt="enter image description here"> Not hard to see, that you can then intersect 4 trajectories of a second rectangle (they are rays starting from different corners of a bounding box in <code>(x2 - x1, y2 - y1)</code> direction) with 4 sides of the first rectangle, standing still. Then you'll have to do the same vice versa - find the intersection of a first rectangle moving in reverse direction - <code>(-(x2 - x1), -(y2 - y1))</code> with 4 sides of a second rectangle. Choose the minimum distance between the all intersection points you've found (there might be 0-8 of them) and you're done. Don't forget to consider many special cases - when the sides of both rectangles are parallel, when there's no intersection at all etc. Note, this all is done in <code>O(1)</code> time, though the calculations are quite complex - 32 intersections of a ray and a segment. <ul> <li>If you really wish your rectangles to rotate with some speed, I would suggest considering what @comingstorm has said: this is a problem of finding roots of a non-linear equation, however, even in such case, if you have a limited angular speed of your rectangles, you may split the task into a series of <a href="http://en.wikipedia.org/wiki/Ternary_search" rel="nofollow noreferrer">ternary search</a> subtasks, though I suppose this is just one of possible methods of solving non-linear problems.</li> </ul>
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singulars
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1. This table or related slice is empty.
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1. POFinding time and position of intersection of two moving, rotating bounding boxes
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1. PTAnswer
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1. USIxanezis
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1. USIxanezis
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