normal at corner of rectangular

[jemidiah]

Yup, using your implementation on a triangle you'll tend to accumulate points around P1, since you're taking x1 and x2 small. Luckily I seem to have a magical intuition about what you want--I don't really know how, but I did figure you wanted a more-or-less uniform distribution, and noticed your method wasn't going to give it to you.


To get a more uniform distribution, use truly random numbers as I originally suggested. In VB (6.0) just use the Randomize/Rnd() functions to get x1 and x2 each between 0 and 1, and plug those into the formula. Generate as many random pairs (x1, x2) as you wish and plug them each into the formula to get a new random point.

If you're interested in getting a truly uniform distribution of random points on an arbitrary triangle, I could figure out the underlying distribution of x1 and x2 that you'd need, but I'm pretty sure that's not where you're going. I could also give a method of "blanketing" the triangle with a uniform grid, but that'd be complicated to think through.

The above paragraph is for a triangle; for a parallelogram, if you simply used uniformly spaced x1 and x2 from 0...1 you'd get both of the above properties. That is, use x1 = 0, 0.2, 0.4, 0.6, 0.8, 1, x2 = 0, 0.2, 0.4, 0.6, 0.8, 1, instead of what you're using which is x1 = 0.5, 0.33, 0.25, 0.20, 0.17, 0.14, ..., 1/n, x2 = 0.5, 0.33, 0.25, 0.20, 0.17, 0.14, ..., 1/m. Using this type of spacing of x1 and x2 for a triangle would also give an approximately uniform result. At the least, the points wouldn't accumulate around the corners nearly as dramatically.