Thanks for the help! That gives me a bit more confidence in figuring it out. I need to better understand the big 'B' I think. The examples I've seen with the 3 order PN-Triangles seem a lot simpler than that crazy 'B' formula. There must be a way to simplify it. For example, check this out.
I guess that B number also has a i,j,k index, I guess that is what is creating the decreasing exponent on these terms... somehow. How do you know which exponent goes with which point?
The first thing to understand is that B (a bernstein polynomial) determines how much each point contributes to the final p(u, v) point. The value it returns is between 0 and 1, with 0 meaning that the point won't affect the final output and 1 meaning that it completely determines the output.
The "simplified" equation you give is just that... a simplified version of B when n = 3. You can do the math yourself if you want to take the time. Just take the equation for B, make n always 3, and compute the result for every valid i, j, k combination. Then with a bit of algebraic manipulation you could get the same equation. But the way this equation is written is only true for n=3. Breaking down the case of seven control points into a single equation is going to be too unwieldy. Instead, I would suggest trying to understand how to use:
which works for any n, before trying to simplify it.
Let's look at how this will be used:
Let's say that we have a triangle patch with n = 5 and are trying to determine how much point p301 will contribute to the output point at (u,v). We would use the "big B" to determine this. B of 3,1,1 at u,v. The equation would be:
( 5! / (3! * 1! * 1!) ) * (ui) * (vj) * (1 - u - v)k =
(5*4 *3*2*1) / (3 * 2 * 1 * 1 * 1) * u3 * v * (1 - u -v) =
20*(u3) * v * (1 - u - v)
This means that when u = 0.1 and v = 0.5), the point p301 would contribute to the point:
20*(0.001) *0.1 * 1 * (1 - 0.1 - 0.5) =
Then back in the for-loop I had talked about before, p(3, 0, 1) would be multiplied by 0.0008 and added to the output point.
Note that B can be created as a function in code somewhat easily once you have a factorial function (something that can also be created easily) and a power function (should be available in your favorite programming language's standard library). All that it needs to do is take in multiple parameters (n, u, v, i, j, and k) and output a double value.
...well, that was a long explanation. I probably got something in there wrong
Let me know if something needs to be clarified.