Hi I wonder if it exist an easy way to check if a line and a rectangle collide and get the collisionpoint(s). Or do I have to make lines of the rectangle? (I hope not ) Thanks in forward /pb

http://www.gametutorials.com/

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Leyder Dylan (dylan.leyder@slug-production.be.tf
http://users.skynet.be/fa550206/Slug-Production/Index.htm/

Are the lines straight?

Excellent. This is exactly what I was going to ask!

Ilici: I tried your method, but, unless I am doing something wrong, it seems HIGHLY inaccurate. I created a fast and short C# demo (1.0 Framework). It's pretty much the code you posted but instead of testing between a line and a polygon, I modified the code to test between a line and a rectangle. I would appreciate if you or anyone else could take a look and verify it.

C# test

gametutorials.com doesn't have very much on it and nothing as far as this is concerned. Or at least I couldn't find anything.

johnnyBravo: I'll try your method next. It seems the most promising if Ilici's method doesn't work that well.

[edited by - JasonA on November 16, 2003 7:04:14 PM]

its a nice first step, but you need the point of intersection to check if this point is even part of the rectangle side.

your life will be a lot easier if the rectangles are axis aligned, maybe even so much easier that i would think about transforming line and rectangle to make it axis aligned.

after that for the left side:
if (side.y-p0.y)*(side.y-p1.y) is positive both points (start and end point of your line) are on the same side of the side (the same as above with the dot product, just simpler). in that case, go to the next side.

if its negative or 0 you line equation looks like this:
p0 + t*(p1-p0). you know it will intersect, so:
p0.y + t*(p1.y-p0.y) = side.y

solving this shouldnt be hard and you get a t.

short:
t = (side.y - p0.y) / (p1.y - p0.y)

if this t is smaller than 0 or larger than 1 they would intersect, but not in the interval between p0 and p1 (though here they will, we already did a quick and dirty test to figure that out). but the equation for our simplified 2d case is easy enough so forget about the quick and dirty test ,-)

next, plug this t into the equation to get x at the intersection.
x = p0.x + t * (p1.x-p0.x)

this x has to be between the two points of the rectangle side:
if xside.maxx forget it and go to the next side.

also, you might get up to 4 intersections but the one with the smallest t will happen first.

doing the same without the rectangle being axis aligned isnt much different except you would have line equations on both sides and solving this will be a little more work.

p0 + t*(p1-p0) = side0 + s*(side1-side0)

basically twice, for x and y coord. solve one for s and then plug the whole ugly thing into the other instead of s, then solve for t. chances are that a lot of sites will already present you with the formula ,-)

in the end both, s and t will have to be between 0 and 1 for the segments to intersect, if you need the point of intersection just plug t or s into its equation.

[edited by - Trienco on November 18, 2003 12:52:30 PM]