<p>It's the normalization factor to make sure the BRDF always reflects the same amount of energy regardless of the value of the specular exponent n. Without that factor, changing the specular exponent changes the overall reflectance of the material.</p>

<p>Further reading: <a href="http://www.thetenthplanet.de/archives/255" rel="nofollow noreferrer">The Blinn-Phong Normalization Zoo | The Tenth Planet Blog</a></p>

CC BY-SA 3.0
2017-08-16T09:20:58.933Z
Yes, I thought of this after posting the question. And I'm trying to calculate $k_s$.
chaosink
CC BY-SA 3.0
2017-08-16T09:22:03.830Z
Wow, the link you give is amazing!
chaosink
CC BY-SA 3.0
2017-08-16T09:16:36.053Z
2018-03-06T08:54:54.820Z
2018-03-06T08:54:54.820Z
Stefan Werner
<p><a href="https://i.sstatic.net/SSqzx.png" rel="noreferrer"><img src="https://i.sstatic.net/SSqzx.png" alt="reflection"></a></p>

<p>In the book AGI (Advanced Global Illumination), the Phong BRDF is</p>

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

<p>$k_d=\cfrac{albedo}\pi$</p>

<hr>

<p>In Mitsuba and Tungsten, the Phong BRDF is </p>

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

<hr>

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

<p>What's the relation between the two BRDFs?</p>

<pathtracing><brdf><global-illumination>
Phong BRDF in Mitsuba, Tungsten and AGI
chaosink
# Phong BRDF in Mitsuba, Tungsten and AGI

(Asked by: username_0 on 2017-08-16)

[![reflection](https://i.sstatic.net/SSqzx.png)](https://i.sstatic.net/SSqzx.png)

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

$f\_r(x,\Psi\leftrightarrow\Theta)=k\_s\cfrac{(R\cdot\Theta)^n}{N\cdot\Psi}+k\_d$

$k\_d=\cfrac{albedo}\pi$

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In Mitsuba and Tungsten, the Phong BRDF is

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

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

What's the relation between the two BRDFs?

**Tags:** pathtracing, brdf, global-illumination


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## 1 answers

### Answer by username_1 on 2017-08-16. Score: 3 | Accepted answer

It's the normalization factor to make sure the BRDF always reflects the same amount of energy regardless of the value of the specular exponent n. Without that factor, changing the specular exponent changes the overall reflectance of the material.

Further reading: [The Blinn-Phong Normalization Zoo | The Tenth Planet Blog](http://www.thetenthplanet.de/archives/255)

> **Comments:**
> **username_0**: Yes, I thought of this after posting the question. And I'm trying to calculate $k_s$.  
> **username_0**: Wow, the link you give is amazing!  

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