# Phong BRDF in Mitsuba, Tungsten and AGI

In the book AGI (Advanced Global Illumination), the Phong BRDF is

$f_r(x,\Psi\leftrightarrow\Theta)=k_s\cfrac{(R\cdot\Theta)^n}{N\cdot\Psi}+k_d$

$k_d=\cfrac{albedo}\pi$

In Mitsuba and Tungsten, the Phong BRDF is

$f_r(x,\Psi\leftrightarrow\Theta)=[\cfrac{ratio_{diffuse}}\pi+\cfrac{(R\cdot\Theta)^n(n+2)(1-ratio_{diffuse})}{2\pi}]\cdot albedo$

Why there is a $\cfrac{n+2}{2\pi}$ in the specular part?

What's the relation between the two BRDFs?

• Yes, I thought of this after posting the question. And I'm trying to calculate $k_s$. – chaosink Aug 16 '17 at 9:20