Axis-aligned bounding boxes (AABBs) are universally used to bound finite objects in ray-tracing. Ray/AABB intersections are usually faster to calculate than exact ray/object intersections, and allow the construction of bounding volume hierarchies (BVHs) which reduce the number of objects that need to be considered for each ray. (More on BVHs in a later post.) This means that a ray-tracer spends a lot of its time calculating ray/AABB intersections, and therefore this code ought to be highly optimised.

The fastest method for performing ray/AABB intersections is the slab method. The idea is to treat the box as the space inside of three pairs of parallel planes. The ray is clipped by each pair of parallel planes, and if any portion of the ray remains, it intersected the box.

A simple implementation of this algorithm might look like this (in two dimensions for brevity):

bool intersection(box b, ray r) { double tmin = -INFINITY, tmax = INFINITY; if (ray.n.x != 0.0) { double tx1 = (b.min.x - r.x0.x)/r.n.x; double tx2 = (b.max.x - r.x0.x)/r.n.x; tmin = max(tmin, min(tx1, tx2)); tmax = min(tmax, max(tx1, tx2)); } if (ray.n.y != 0.0) { double ty1 = (b.min.y - r.x0.y)/r.n.y; double ty2 = (b.max.y - r.x0.y)/r.n.y; tmin = max(tmin, min(ty1, ty2)); tmax = min(tmax, max(ty1, ty2)); } return tmax >= tmin; } |

However, those divisions take quite a bit of time. Since when ray-tracing, the same ray is tested against many AABBs, it makes sense to pre-calculate the inverses of the direction components of the ray. If we can rely on the IEEE 754 floating-point properties, this also implicitly handles the edge case where a component of the direction is zero - the `tx1`

and `tx2`

values (for example) will be infinities of opposite sign if the ray is within the slabs, thus leaving `tmin`

and `tmax`

unchanged. If the ray is outside the slabs, `tx1`

and `tx2`

will be infinities with the same sign, thus making `tmin == +inf`

or `tmax == -inf`

, and causing the test to fail.

The final implementation would look like this:

bool intersection(box b, ray r) { double tx1 = (b.min.x - r.x0.x)*r.n_inv.x; double tx2 = (b.max.x - r.x0.x)*r.n_inv.x; double tmin = min(tx1, tx2); double tmax = max(tx1, tx2); double ty1 = (b.min.y - r.x0.y)*r.n_inv.y; double ty2 = (b.max.y - r.x0.y)*r.n_inv.y; tmin = max(tmin, min(ty1, ty2)); tmax = min(tmax, max(ty1, ty2)); return tmax >= tmin; } |

Since modern floating-point instruction sets can compute min and max without branches, this gives a ray/AABB intersection test with no branches or divisions.

My implementation of this in my ray-tracer Dimension can be seen here.

This doesn't seem to work out for me for negative direction vectors. Looking forward, I can see a box that's there, but looking backwards, I again see the same box mirrored, even though in this direction it's not "there". Following the algorithm step by step manually for both a positive and a negative z-dir (with x-dir and y-dir set to 0) gives the same near and far planes in both directions:

/*

box = MIN 0,0,0 MAX 256,256,256

ray POS 128,128,-512

case 1: ray DIR 0,0,0.9 -- inverse: (inf,inf,1.1111)

case 2: ray DIR 0,0,-0.9 -- inverse: (inf,inf,-1.1111)

*/

picker.tx1, picker.tx2 = (me.Min.X - ray.Pos.X) * ray.invDir.X, (me.Max.X - ray.Pos.X) * ray.invDir.X // -inf,inf -inf,inf

picker.ty1, picker.ty2 = (me.Min.Y - ray.Pos.Y) * ray.invDir.Y, (me.Max.Y - ray.Pos.Y) * ray.invDir.Y // -inf,inf -inf,inf

picker.tz1, picker.tz2 = (me.Min.Z - ray.Pos.Z) * ray.invDir.Z, (me.Max.Z - ray.Pos.Z) * ray.invDir.Z // -142.22,142.22 142.22,-142.22

picker.txn, picker.txf = math.Min(picker.tx1, picker.tx2), math.Max(picker.tx1, picker.tx2) // -inf,inf -inf,inf

picker.tyn, picker.tyf = math.Min(picker.ty1, picker.ty2), math.Max(picker.ty1, picker.ty2) // -inf,inf -inf,inf

picker.tzn, picker.tzf = math.Min(picker.tz1, picker.tz2), math.Max(picker.tz1, picker.tz2) // -142.22,142.22 -142.22,142.22

picker.tnear = math.Max(picker.txn, math.Max(picker.tyn, picker.tzn)) // -142.22 -142.22

picker.tfar = math.Min(picker.txf, math.Min(picker.tyf, picker.tzf)) // 142.22 142.22

if picker.tfar > picker.tnear {

return true

}

Right, because the test is only for whether the line intersects the box at all. The line extends both forwards and backwards. Just add a tmax >= 0 check. It's tmax, not tmin, since tmin will be < 0 if the ray originates inside the box.

Thanks. Works nicely and fast. I updated it a bit, to use SSE (though Vectormath), floats only.

https://gist.github.com/4412640#file-bbox-cpp-L14

You're welcome! I can't see that gist though (says "OAuth failure"). What kind of performance did the vectorisation give you?