Original Post
Hallo, How can I check if a point is inside a rectangle? The rectangle has been rotated, so I cant simply compare the point to the vertexes (I think). Thanks in advance
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Point inside of rectangle
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Quote:Short answer: transform the point into the local space of the rectangle, and then perform a point-in-axis-aligned-rectangle test (which is straightforward).
Original post by h3ro
Hallo,
How can I check if a point is inside a rectangle? The rectangle has been rotated, so I cant simply compare the point to the vertexes (I think).
Thanks in advance
Sorry for being dump here, but could you please explain a bit more? Math is not really my strongest side. Tried google, but couldnt find anything for "point-in-axis-aligned-rectangle test"
Another method:
If you know where the corners of the rectangle are, you can deduce the equation for the four line-segments that make up the edges, which you can then use to determine the normals of each of the edges.
Get the dot-product of the edge-normal and the vector formed from any point on the line to your test-point, this will tell you which side of the line the point is on.
If the point is on the 'inside' of all of the lines, it's inside the box.
A "point-in-axis-aligned-rectangle test" is just:
If you know where the corners of the rectangle are, you can deduce the equation for the four line-segments that make up the edges, which you can then use to determine the normals of each of the edges.
Get the dot-product of the edge-normal and the vector formed from any point on the line to your test-point, this will tell you which side of the line the point is on.
If the point is on the 'inside' of all of the lines, it's inside the box.
Quote:
Original post by h3ro
Sorry for being dump here, but could you please explain a bit more? Math is not really my strongest side. Tried google, but couldnt find anything for "point-in-axis-aligned-rectangle test"
A "point-in-axis-aligned-rectangle test" is just:
bool Inisde( x, y, l, r, b, t )//x,y are the point, l,r,b,t are the extents of the rectangle{ return x > l && x < r && y > b && y < t;}
Another answer:
* P is the point.
* C is a corner of the rectangle.
* v1 and v2 are the two vectors that define the sides (with C as origin).
* v = P-C
P is in the rectangle if and only if
0<=dot_product(v,v1)<=dot_product(v1,v1) and 0<=dot_product(v,v2)<=dot_product(v2,v2)
* P is the point.
* C is a corner of the rectangle.
* v1 and v2 are the two vectors that define the sides (with C as origin).
* v = P-C
P is in the rectangle if and only if
0<=dot_product(v,v1)<=dot_product(v1,v1) and 0<=dot_product(v,v2)<=dot_product(v2,v2)
Thank you to all of you. I have it working now :)
Quote:
Original post by h3ro
Thank you to all of you. I have it working now :)
What suggestion did you end up using?
I used yours. It was the simplest and the I understood it :P
Thanks
Thanks
Quote:Just for the record, the solution proposed by alvaro is the same as the one I proposed in the first reply to your thread (alvaro, however, was kind enough to spell out the details for you :).
Original post by h3ro
I used yours. It was the simplest and the I understood it :P
Then it's all the same solution on this page!
V is converting to the local space, and uses the dot products for the point-in-rectangle test?
But the dot products are not really necessary if it's AABB, you could use the Hodgman method.
V is converting to the local space, and uses the dot products for the point-in-rectangle test?
But the dot products are not really necessary if it's AABB, you could use the Hodgman method.
For some reason I cant get I to work anymore. Right now it is only drawing half the bounding box. Any ideas on why?
EDIT:
Fooling around for ever fixed it. Not really sure what it was though, but working now :)
[Edited by - h3ro on February 23, 2008 4:47:38 PM]
EDIT:
Fooling around for ever fixed it. Not really sure what it was though, but working now :)
[Edited by - h3ro on February 23, 2008 4:47:38 PM]
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