Haegar!
Thankyou so much for your time explaining that through.
I didn't entirely understand the purpose of calculating the line segments of the triangle edges relative to A, so I didn't think to do the same to the intersection point :)
I was negating distance because I was using (O-A).N not (A-0).N. Guess thats how i saw the formula written to begin with :P
Anyway, here is complete & working, hopefully clearly commented code for anyone else that happens to stumble across this looking for a full implementation of the barycentric intersection technique.
/**Function to return the distance distance along a ray an intersection with the triangle occurs atRay R The ray to compute intersection withfloat maxDist The maximum acceptable distance along the ray an intersection can occur atreturn float +ve the distance along the ray an intersection occurs at or -ve no intersection**/float OBJTriangleProxy::testIntersection (Ray &r, float maxDist){ //Get ray origin and direction to simplify equations later on Vector3d O = r.getOrigin(); Vector3d D = r.getDirection(); //Get triangle vertices Vector3d A = getVert (0); Vector3d B = getVert (1); Vector3d C = getVert (2); //Calculate average normal Vector3d N = (B - A).cross (C-A); N.normalise(); //Calculate distance from ray to the plane the triangle lays in float distance = ((A - O).dot (N)) / (D.dot (N)); //Ray is behind the plane the triangle's plane if (distance < 0) return -1; //Work out dominant axis for projection of triangle points into 2d int dominant = 0; if (abs(N.getX()) > abs(N.getY())) { if (abs(N.getX() > abs(N.getZ()))) dominant = 0; //x dominant else dominant = 2; //z dominant } else { if (abs(N.getY() > abs(N.getZ()))) dominant = 1; //y dominant else dominant = 2; // z dominant } //Find point where the ray intersects the plane Vector3d P = O + D * distance; //Obtain points relative to A B = B - A; C = C - A; P = P - A; //Project vectors onto dominant axis plane switch (dominant) { case 0: //Project onto x B = Vector3d (B.getY(), B.getZ(), 0); C = Vector3d (C.getY(), C.getZ(), 0); P = Vector3d (P.getY(), P.getZ(), 0); break; case 1: //Project onto Y B = Vector3d (B.getX(), B.getZ(), 0); C = Vector3d (C.getX(), C.getZ(), 0); P = Vector3d (P.getX(), P.getZ(), 0); break; case 2: //Project onto Z B = Vector3d (B.getX(), B.getY(), 0); C = Vector3d (C.getX(), C.getY(), 0); P = Vector3d (P.getX(), P.getY(), 0); break; } //Calculate 2d barycentric components float u = ((P.getY() * B.getX()) - (P.getX()*B.getY())) / ((C.getY() * B.getX()) - (C.getX() * B.getY())); float v = ((C.getY() * P.getX()) - (C.getX()*P.getY())) / ((B.getX() * C.getY()) - (B.getY() * C.getX())); //Point in triangle? if (u >= 0 && v >= 0 && u + v <= 1) return distance; //No intersection return -1;}
There is obviously massive potential for precalulation here that i've skipped over, but for the sake of gaining some understanding of the algorithm, it is better to keep it all in one routine to begin with.
[Edited by - skotinator on March 28, 2008 12:53:17 AM]
