<p><a href="https://en.wikipedia.org/wiki/Schlick%27s_approximation" rel="nofollow noreferrer">The approximation</a> writes the reflection coefficient as<span class="math-container">$$R(\theta)=R_0+(1-R_0)(1-\cos\theta)^5,\,R_0=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2.$$</span>Why is the exponent 5? <a href="https://www.cs.virginia.edu/~jdl/bib/appearance/analytic%20models/schlick94b.pdf" rel="nofollow noreferrer">Schlick 1994</a> introduces this exponent in Eq. (24), claiming it's the right Fresnel approximation, but with no explanation.</p>

CC BY-SA 4.0
2020-01-26T13:42:52.973Z
Before equation 15, they have several constraints. I guess they extended those a bit further in order to get a higher continuity at $F_{\lambda}^{(k)}(1) = 0$.
lightxbulb
CC BY-SA 4.0
2020-01-26T14:16:51.483Z
@lightxbulb Thanks. An exponent of $3$ or $4$ would meet the stated conditions, but I suspect $5$ was obtained either with an approximation of (13) or empirical trial & error.
J.G.
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2020-01-26T14:48:18.480Z
Just plug in a larger $k$ in what I wrote, and you'll get the higher degree.
lightxbulb
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2020-01-26T14:50:23.907Z
@lightxbulb I understand the calculus, but such conditions have no motive apparent in the paper.
J.G.
<lighting><mathematics>
Why does Schlick's approximation contain a \$(1-\cos\theta)^5\$ term?
J.G.
# Why does Schlick's approximation contain a \$(1-\cos\theta)^5\$ term?

(Asked by: username_0 on 2020-01-26)

[The approximation](https://en.wikipedia.org/wiki/Schlick%27s_approximation) writes the reflection coefficient as$$R(\theta)=R\_0+(1-R\_0)(1-\cos\theta)^5,\,R\_0=\left(\frac{n\_1-n\_2}{n\_1+n\_2}\right)^2.$$Why is the exponent 5? [Schlick 1994](https://www.cs.virginia.edu/~jdl/bib/appearance/analytic%20models/schlick94b.pdf) introduces this exponent in Eq. (24), claiming it's the right Fresnel approximation, but with no explanation.

**Tags:** lighting, mathematics

> **Comments:**
> **username_1**: Before equation 15, they have several constraints. I guess they extended those a bit further in order to get a higher continuity at $F_{\lambda}^{(k)}(1) = 0$.  
> **username_0**: @username_1 Thanks. An exponent of $3$ or $4$ would meet the stated conditions, but I suspect $5$ was obtained either with an approximation of (13) or empirical trial & error.  
> **username_1**: Just plug in a larger $k$ in what I wrote, and you'll get the higher degree.  
> **username_0**: @username_1 I understand the calculus, but such conditions have no motive apparent in the paper.  

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## 0 answers

