Ray and square/rectangle intersection in 3D

Hei. Are making a game and are looking for a ray intersection onto a square or a rectangle only in 3D space. Have search the web and found many solutions but nothing i can understand have a line and line segment intersection script in 2D but i cant figure out have to make it 3D. It is not important from what side it intersect the square or rectangle but it must be able to retrive the point of intersection vector so that later can be tested for distance to se if it occurred before or after other intersections on the same ray intersection.

Any examples in python or other similar scripting languages will be greatly appreciated

Edit: Dont know have to modify the 2D to show an exaple but made a new and posting both.

//this is the exaple it test a ray onto a plane then look to se if that point is in the rectangle and saves it to test for distanse later
list Faces; //triangle faces
list Points; //

vector FindPoint(){
    //calcute the point of intersection onto the plane and returns it
    //if it can intersect
    //else return ZERO_VECTOR
}

integer point-in-quadrilateral(){
    //return 1 if the point is in the rectangular on the plane
    //else return 0
}

default{

    state_entry(){
        integer n = (Faces != []); //return number of elements
        integer x = 0;
        while(x < n){
            vector intersection = FindPoint( FromList(Faces, x) ); //take     out a element and runs it trough the function
            if(intersection != ZERO_VECTOR){
                integer test = point-in-quadrilateral( FromList(Faces,     x) ); //find out if the point is in rectangular
                if(test == 1){ //if so
                    Points += intersection; //save the point
                }
            }
            ++x;
        }

        float first; //the distanse to the box intersection
        integer l = (Points != []);
        integer d;
        while(d < l){
            if(Dist( FromList(Points, d) ) < first) //if the new distanse     is less then first
                return 0; //then end script
            ++d;
        }
    }

}


//this is the 2D version
vector lineIntersection(vector one, vector two, vector three, vector four){
float bx = two.x - one.x;
float by = two.y - one.y;
float dx = four.x - three.x;
float dy = four.y - three.y; 
float b_dot_d_perp = bx*dy - by*dx;
if(b_dot_d_perp == 0.0) {
    return ZERO_VECTOR;
}
float cx = three.x-one.x; 
float cy = three.y-one.y;
float t = (cx*dy - cy*dx) / b_dot_d_perp; 
if(LineSeg){ //if true tests for line segment
    if((t < 0.0) || (t > 1.0)){
        return ZERO_VECTOR;
    }
    float u = (cx * by - cy * bx) / b_dot_d_perp;
    if((u < 0.0) || (u > 1.0)) {
        return ZERO_VECTOR;
    }
}

return <one.x+t*bx, one.y+t*by, 0.0>;

}

Answer

Create a vector equation for a line in R3, then solve for the intersection of that line in the plane of the rectangle that you are testing it against. After that, it’s simple enough to test if that point of solution lies within the bounds.

the parameter t of the solution can be found with:

t = (a * (x0 - rx) + b * (y0 - ry) + c * (x0 - rz)) / (a * vx + b * vy + c * vz)

where:

a(x - x0) + b(y - y0) + c(z - z0) = 0

is the equation of the plane that your rectangle lies on

and:

<x, y, z> = <rx + vx * t, ry + vy * t, rz + vz * t>

is the vector equation of the line in question.

note that:

<rx, ry, rz>

is the initial point of the vector equation, and

<vx, vy, vz>

is the direction vector of the above equation

After that, plugging the parameter t into your vector equation will give you the point to test for distance.

enter image description here

Advertisement

Answer