<p>I don't really understand what are you doing. I think your first equation shouldn't have a cos factor in it.</p>

<p>We have the relation,</p>

<p><span class="math-container">$BRDF = dL_r / dE_i$</span></p>

<p>That is the brdf is the ratio of reflected radiance to incoming "irradiance". Re-arranging this gives us,</p>

<p><span class="math-container">$ dL_r =  BRDF * dE_i$</span></p>

<p>For diffuse surfaces we know,</p>

<p><span class="math-container">$BRDF = \alpha/\pi$</span></p>

<p>Substituting in above equation we have,</p>

<p><span class="math-container">$dL_r = \alpha/\pi * dE_i$</span></p>

<p>Irrespective of what angle we view, the radiance is the same. Now for a diffuse surface with albedo = 1 we have what you said,</p>

<p><span class="math-container">$dL_r = dE_i/\pi$</span></p>

<p>Now in order to measure Irradiance, we can use the relation,</p>

<p><span class="math-container">$E = \int_{2\pi} L \cos(\theta) d\omega$</span></p>

<p><span class="math-container">$E = \displaystyle\int_{\phi = 0}^{2\pi} \displaystyle\int_{\theta = 0}^{\pi/2} L \cos(\theta) \sin(\theta) d\theta d\phi$</span></p>

<p><span class="math-container">$E = E/\pi * \pi$</span></p>

<p><span class="math-container">$E = E$</span></p>

<p>So you got back the same irradiance.</p>

CC BY-SA 4.0
2019-05-23T03:59:06.423Z
The first cosine is for the direction of incoming light, to get the "true" irradiance or whatever. E.g. $E_0$ coming in at a $90^\circ$ angle doesn't light up the surface at all. I think my discrepancy comes from the fact you integrated $L\cos(\theta)$ instead of just integrating $L$.
HiddenBabel
CC BY-SA 4.0
2019-05-23T06:34:54.730Z
1). I know. But if you are putting in the cosine then you have to do it properly. $dE$ will change into $dL_f$.  Because as shown below $dE = dL \cos(\theta) d\omega$ 2) Why would you just integrate $L$. You get irradiance by integrating $L \cos(\theta)$.
gallickgunner
CC BY-SA 4.0
2019-05-23T08:23:48.977Z
Ah I thought I had already taken care of the $\cos\theta$ when I was calculating solid angle for my actual problem (not a flat plane). But that's just it, I needed $\cos^2\theta$ in total, and now I get $E=E$.
HiddenBabel
CC BY-SA 4.0
2019-05-23T01:57:06.720Z
2019-05-23T01:57:06.720Z
gallickgunner
<p>For a Lambertian surface, the radiance at a point is</p>

<p><span class="math-container">$$L = \frac{\alpha}{\pi}\cos\theta_i E_0$$</span></p>

<p>If we have an infinite flat surface, albedo = 1, illuminated directly normal, then we get </p>

<p><span class="math-container">$$L = \frac{E_0}{\pi}$$</span></p>

<p>But now suppose we are sitting very close to the surface. The surface occupies <span class="math-container">$2\pi$</span> sr of our view, so the irradiance we measure is</p>

<p><span class="math-container">$$E_r = 2\pi L = 2E_0$$</span></p>

<p>Is this correct? I was expecting to get <span class="math-container">$E_0$</span> back again. <span class="math-container">$2E_0$</span> makes it feel like we're violating the conservation of energy or something.</p>

<global-illumination><reflection><diffuse>
Factor of 2 converting radiance to irradiance
HiddenBabel
# Factor of 2 converting radiance to irradiance

(Asked by: username_0 on 2019-05-23)

For a Lambertian surface, the radiance at a point is

$$L = \frac{\alpha}{\pi}\cos\theta\_i E\_0$$

If we have an infinite flat surface, albedo = 1, illuminated directly normal, then we get

$$L = \frac{E\_0}{\pi}$$

But now suppose we are sitting very close to the surface. The surface occupies $2\pi$ sr of our view, so the irradiance we measure is

$$E\_r = 2\pi L = 2E\_0$$

Is this correct? I was expecting to get $E\_0$ back again. $2E\_0$ makes it feel like we're violating the conservation of energy or something.

**Tags:** global-illumination, reflection, diffuse


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## 1 answers

### Answer by username_1 on 2019-05-23. Score: 1 | Accepted answer

I don't really understand what are you doing. I think your first equation shouldn't have a cos factor in it.

We have the relation,

$BRDF = dL\_r / dE\_i$

That is the brdf is the ratio of reflected radiance to incoming "irradiance". Re-arranging this gives us,

$ dL\_r = BRDF \* dE\_i$

For diffuse surfaces we know,

$BRDF = \alpha/\pi$

Substituting in above equation we have,

$dL\_r = \alpha/\pi \* dE\_i$

Irrespective of what angle we view, the radiance is the same. Now for a diffuse surface with albedo = 1 we have what you said,

$dL\_r = dE\_i/\pi$

Now in order to measure Irradiance, we can use the relation,

$E = \int\_{2\pi} L \cos(\theta) d\omega$

$E = \displaystyle\int\_{\phi = 0}^{2\pi} \displaystyle\int\_{\theta = 0}^{\pi/2} L \cos(\theta) \sin(\theta) d\theta d\phi$

$E = E/\pi \* \pi$

$E = E$

So you got back the same irradiance.

> **Comments:**
> **username_0**: The first cosine is for the direction of incoming light, to get the "true" irradiance or whatever. E.g. $E_0$ coming in at a $90^\circ$ angle doesn't light up the surface at all. I think my discrepancy comes from the fact you integrated $L\cos(\theta)$ instead of just integrating $L$.  
> **username_1**: 1). I know. But if you are putting in the cosine then you have to do it properly. $dE$ will change into $dL_f$.  Because as shown below $dE = dL \cos(\theta) d\omega$ 2) Why would you just integrate $L$. You get irradiance by integrating $L \cos(\theta)$.  
> **username_0**: Ah I thought I had already taken care of the $\cos\theta$ when I was calculating solid angle for my actual problem (not a flat plane). But that's just it, I needed $\cos^2\theta$ in total, and now I get $E=E$.  

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