I'm in a bit of a hurry, so I'll have to be quick. You probably want to read up on Bidirectional Reflectance Distribution Functions (BRDF's) - a lot of lighting models (such as Phong and Blinn's modification) sacrifice validity to the two laws of a BRDF (bidirectionality and conservation of energy) in favour of simplicity/speed. Several more theoretical models will include normalization functions so that they are more physically plausible w.r.t. these laws.

Pretty sure that should give you something to get started with [smile]

hth
Jack

d is constrained to the range [-p/2, p/2] so that cos(d) is zero at the extremities and unity at d = 0. The resulting curve looks like a hill; one half of a sine wave. Since the image is bound between 0 and 1, any power of it will also be. So here, n acts as a specular power: the higher n is, the narrower the peak at d = 0.
However, because cos < 1, raising the function to a positive power will decrease the area under the graph. Physically, this means that the amount of light being reflected (in all directions) is less than that incident upon the material. This is undesirable. So, to maintain energy conservation, we scale the entire function up so that the enclosed area is independent of n.

Mathematically, the total reflected intensity is the area under the graph, determined by integrating the function over its domain:

I = ∫-p/2p/2 (n+2)/2p . cosn(d) . dd

This value should always be the same, regardless of the value of n.

As for rs, I presume you're right about it being a specular texture value. Maybe somebody who knows more can confirm that.

Edit: On closer inspection, I suspect something is wrong with that formula. Either the rho-factor is rather different to what I thought or I have the whole situation rather wrong. That simple linear factor (n+2)/2p doesn't have a hope in hell of tackling the complexity of a power of cosine under an integral. After a little poking around, I found that the 'correct' scaling factor for even values of n is:

b = p-12n/(n!/((n - n/2)!(n/2)!)

That factorial mess at the end is the binomial coefficient (n, n/2).
For odd n, the situation is slightly different: we lose the factor of pi and introduce some different factorials. I'm not compelled to work it out.

So... any guesses?

Admiral