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# Ray/Triangle Intersection Difficulties

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### #1kmx  Members   -  Reputation: 115

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Posted 21 July 2012 - 07:30 PM

I realize this topic is covered extensively, but I've looked at dozens of academic presentations and I can't seem to make any of them work.

I am attempting to detect a) whether a ray (poiint/direction) intersects with a given triangle and b) where (in world coordinates). I adapted the following code from the lighthouse 3D article:

[source lang="cpp"]int isx3d(Line3D line, Plane3D plane, Number3D *where) { Number3D e1 = vec3d(plane.v0, plane.v1); Number3D e2 = vec3d(plane.v0, plane.v2); Number3D h = cross3d(line.p1, e2); float a = dot3d(e1, h); if (a > -0.00001 && a < 0.00001) //(close_enough(a, 0.0f)) return FALSE; float f = 1.0f/a; Number3D s = vec3d(plane.v0, line.p0); float u = f*dot3d(s,h); if (u < 0.0f || u > 1.0f) return FALSE; Number3D q = cross3d(s, e1); float v = f*dot3d(line.p1,q); if (v < 0.0f || u+v > 1.0f) return FALSE; float t = f*dot3d(e2, q); if (t > 0.00001) { float w = 1.0f-(u+v); Number3D ww; ww.x = w*plane.v0.x+u*plane.v1.x+v*plane.v2.x; ww.y = w*plane.v0.y+u*plane.v1.y+v*plane.v2.y; ww.z = w*plane.v0.z+u*plane.v1.z+v*plane.v2.z; *where = ww; return TRUE; } return FALSE;}[/source]

This code seems very similar to other examples, but it has a couple of problems:

First, it works when the point of origin is on the negative side of the triangle, but when the point of origin is on the positive side, it can trigger a false positive result (as if it is considering the ray being cast negatively from the point of origin as well).

Second, the resulting point is wrong. I have tried transforming this every way I can think of, but it either walks across the face of the triangle in a diagonal line (instead of horizontally if I just moving the point of of origin on the X/Y plane) or it moves along a curve(?).

Edited by kmx, 21 July 2012 - 07:32 PM.

### #2Bacterius  Crossbones+   -  Reputation: 9858

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Posted 21 July 2012 - 09:06 PM

Yeah, ray-triangle intersection is a real pain to get up and running. I'll give you one version that works - it may not be the fastest available but it'll get you back on track until you find better:
/* Let P1, P2, P3 be the three vertices of the triangle, in world space.
Let E1 be the vector P2 - P1, and E2 be the vector P3 - P1 (i.e. the triangle's edge vectors).
Let R be the ray to intersect, where R.Origin is the ray's origin and R.Direction is the ray's normalized direction. */

/* Then you calculate the following quantities: */

s1 = cross(R.Direction, E2);
d = 1 / dot(s1, E1);

/* You then calculate the first barycentric coordinate of the intersection point, u: */

dist = R.Origin - P1;
u = dot(dist, s1) * d;

/* If u is not in [0..1] (account for floating-point inaccuracies here), there is no intersection. */

/* We now calculate the second barycentric coordinate, v, using a quantity s2: */

s2 = cross(dist, E1);
v = dot(R.Direction, s2) * d;

/* If v is less than zero, or u + v is more than 1, there is no intersection. */

/* Otherwise, the intersection distance along the ray is given by: */

Distance = dot(E2, s2) * d;


This will give you the intersection distance along the ray (so R.Origin + R.Direction * Distance will give you the intersection point in world coordinates), and will inform you if the ray does not intersect the triangle (if you are only looking to know whether the ray intersects the triangle, without needing the intersection point, there might be faster ways of doing it!).

This method definitely works as I've used it in my ray tracer, however you need to be careful that all your inputs are in world coordinates, otherwise it obviously won't work. If you need more info/help don't hesitate!

The slowsort algorithm is a perfect illustration of the multiply and surrender paradigm, which is perhaps the single most important paradigm in the development of reluctant algorithms. The basic multiply and surrender strategy consists in replacing the problem at hand by two or more subproblems, each slightly simpler than the original, and continue multiplying subproblems and subsubproblems recursively in this fashion as long as possible. At some point the subproblems will all become so simple that their solution can no longer be postponed, and we will have to surrender. Experience shows that, in most cases, by the time this point is reached the total work will be substantially higher than what could have been wasted by a more direct approach.

- Pessimal Algorithms and Simplexity Analysis

### #3kmx  Members   -  Reputation: 115

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Posted 22 July 2012 - 11:31 AM

Thanks - finally figured out what was going on - my edge vectors were normalized when they should not have been.

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