## Thursday, May 31, 2018

### Generating uniform Random rays that hit an axis aligned box

For some tests you want the set of "all" rays that hit a box.   If you want to stratify this is somewhat involved (and I don't know that I have done it nor seen it).   Chris Wyman and I did a jcgt paper on doing this stratified in a square in 2D.   But often stratification isn't needed.   When in doubt I never do it-- the software is always easier to deal with un-stratified, and as soon as dimension gets high most people dont bother because of the Curse of Dimensionality.

We would like to do this with as little math as possible.   First lets consider any side of the box.   (this would apply to any convex polyhedron if we wanted something more general).    If the box in embedded in all possible uniform rays, any ray that hits the box will enter at exactly one point on the surface of the box, and all points are equally likely.   So our first job is to pick a uniform point on the surface.   We can use the technique Greg Turk used to seed points for texture generation on a triangular mesh:

Probability of each side with box of side lengths X, Y, Z is side area of total area.   The side areas are:

XY, YZ, ZX (2 of each)

We can do cumulative area and stuff it in an array of length 6:

c_area = XY
c_area = area + XY
c_area = area + YZ
c_area = area + YZ
c_area = area + ZX
c_area = area + ZX

Now normalize it so it is a cumulative fraction:

for (int i = 0; i < 6; i++)
c_area[i] /= c_area

No take a uniform random real r() in [0,1)

int candidate = 0;
float ra = r();
while (c_area[candidate] < ra) candidate++;

Now candidate is the index to the side.

Let's say the side is in the xy plane and x goes from 0 to X and y goes from 0 to Y.   Now pick a uniform random point on that side:

vec3 ray_entry(X*r(), Y*r(), 0);

Now we have a ray origin.   What is the ray direction?  It is not uniform in all directions.   These are the rays that hit the side.   So the density is proportional to the cosine to the normal-- so they are Lambertian!   This is not obvious.   I will punt on justifying that for now.

So for the xy plane one above, the normal is +z and the ray direction is a uniform random point on a disk projected onto the hemisphere:

float theta = 2*M_PI*r();