<p>I'm not particularly educated in the formal theory of rendering / optics, so I may have something wrong, but I think I can identify the problem here:</p>
<blockquote>
<p>why should the luminous intensity be decreasing with the angle if I can still see the whole surface?</p>
</blockquote>
<p>You're mixing up different cases (one of which must not be what was meant) and getting a contradiction.</p>
<p>For concreteness, imagine you're looking at an illuminated Lambertian disk, at some angle, against a black or unlit background. The image your eye or camera receives is of an ellipse (rotated disk) whose area is proportional to the cosine of the angle:</p>
<p><a href="https://i.sstatic.net/C0Vsu.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/C0Vsu.png" alt="White ellipse on black background" /></a></p>
<p>Now let's consider <em>what you're integrating over</em> — that is, what counts as part of the thing you want to measure the luminous intensity of.</p>
<ol>
<li><p>Suppose you integrate over this entire field of view (or any field of view that contains the entire disk). In that case, the luminous intensity, the <em><a href="https://en.wikipedia.org/wiki/Luminous_flux" rel="nofollow noreferrer">luminous flux</a> per solid angle,</em> must be decreasing as the rotation angle increases, because there is less lit surface (of constant luminance) and more black <em>within the solid angle we are considering.</em> The proportion of the integral's value contributed by the lit surface's luminance is lesser.</p>
<p>This is the case that the cosine law is written for, I think.</p>
</li>
<li><p>Suppose you integrate over the projected area of the disk. In that case, you can compute a luminous intensity that is indeed constant, but the <em>area over which it applies</em> is decreasing with the rotation angle, so anything that <em>uses</em> that luminous intensity value to produce an image will be decreasing its contribution to the image.</p>
</li>
<li><p>Suppose you integrate over a solid angle that that is much smaller than the disk — like one image pixel — but still entirely covered by the disk. Then again the luminous intensity is constant, but “you can see the whole surface” is false because the pixel only sees a part of the disk.</p>
</li>
</ol>

CC BY-SA 4.0
2023-12-11T00:26:53.263Z
I think I understand now where my confusion was coming from. Thank you so much for the thorough explanation!
Bartolini
CC BY-SA 4.0
2023-12-07T23:39:26.497Z
2023-12-11T01:11:51.563Z
2023-12-11T01:11:51.563Z
Kevin Reid
<p>I am currently studying the basics of photometry to better understand the rendering equation of Kajiya. One thing I'm currently struggling with is Lambert's cosine law.</p>
<p>Let's go over the premises:</p>
<ul>
<li>A lambertian surface scatters light evenly in all directions.</li>
<li>The projected surface area decreases with the cosine of the viewing angle.</li>
</ul>
<p>For the luminous intensity from a viewing angle <span class="math-container">$I_\theta$</span> I have the following equation:
<span class="math-container">$$I_\theta = I_0 \cdot cos(\theta).$$</span>
My question is: why should the luminous intensity be decreasing with the angle if I can still see the whole surface? If the light is scattering evenly in every direction, as was the premise for a lambertian surface, why should the projected surface area matter at all. Shouldn't I still recieve the same amount of light, but only in a more concentrated beam?</p>
<p>Thank's in advance to anyone who can help me get out of my confision, which I've been stuck in for the whole last week.</p>

CC BY-SA 4.0
2023-12-06T00:30:19.060Z
Kajiya's equation deals with radiometric quantities, not photometric quantities. As far as the cosine law goes, it doesn't apply only to Lambertian surfaces. If you have integrated flux over a surface in vector calculus then you know that you get a term $r \cdot n$, that's where the cosine comes from. I should mention that I can't say anything much about luminous intensity since I haven't studied photometric quantities and how those should interact with the rendering equation.
lightxbulb
<raytracing><rendering><brdf><light>
Lambertian surface and the luminous intensity
Bartolini
# Lambertian surface and the luminous intensity

(Asked by: username_0 on 2023-12-05)

I am currently studying the basics of photometry to better understand the rendering equation of Kajiya. One thing I'm currently struggling with is Lambert's cosine law.

Let's go over the premises:

* A lambertian surface scatters light evenly in all directions.
* The projected surface area decreases with the cosine of the viewing angle.

For the luminous intensity from a viewing angle $I\_\theta$ I have the following equation:
$$I\_\theta = I\_0 \cdot cos(\theta).$$
My question is: why should the luminous intensity be decreasing with the angle if I can still see the whole surface? If the light is scattering evenly in every direction, as was the premise for a lambertian surface, why should the projected surface area matter at all. Shouldn't I still recieve the same amount of light, but only in a more concentrated beam?

Thank's in advance to anyone who can help me get out of my confision, which I've been stuck in for the whole last week.

**Tags:** raytracing, rendering, brdf, light

> **Comments:**
> **username_1**: Kajiya's equation deals with radiometric quantities, not photometric quantities. As far as the cosine law goes, it doesn't apply only to Lambertian surfaces. If you have integrated flux over a surface in vector calculus then you know that you get a term $r \cdot n$, that's where the cosine comes from. I should mention that I can't say anything much about luminous intensity since I haven't studied photometric quantities and how those should interact with the rendering equation.  

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## 1 answers

### Answer by username_2 on 2023-12-07. Score: 1 | Accepted answer

I'm not particularly educated in the formal theory of rendering / optics, so I may have something wrong, but I think I can identify the problem here:

> why should the luminous intensity be decreasing with the angle if I can still see the whole surface?

You're mixing up different cases (one of which must not be what was meant) and getting a contradiction.

For concreteness, imagine you're looking at an illuminated Lambertian disk, at some angle, against a black or unlit background. The image your eye or camera receives is of an ellipse (rotated disk) whose area is proportional to the cosine of the angle:

[![White ellipse on black background](https://i.sstatic.net/C0Vsu.png)](https://i.sstatic.net/C0Vsu.png)

Now let's consider *what you're integrating over* — that is, what counts as part of the thing you want to measure the luminous intensity of.

1. Suppose you integrate over this entire field of view (or any field of view that contains the entire disk). In that case, the luminous intensity, the *[luminous flux](https://en.wikipedia.org/wiki/Luminous_flux) per solid angle,* must be decreasing as the rotation angle increases, because there is less lit surface (of constant luminance) and more black *within the solid angle we are considering.* The proportion of the integral's value contributed by the lit surface's luminance is lesser.

   This is the case that the cosine law is written for, I think.
2. Suppose you integrate over the projected area of the disk. In that case, you can compute a luminous intensity that is indeed constant, but the *area over which it applies* is decreasing with the rotation angle, so anything that *uses* that luminous intensity value to produce an image will be decreasing its contribution to the image.
3. Suppose you integrate over a solid angle that that is much smaller than the disk — like one image pixel — but still entirely covered by the disk. Then again the luminous intensity is constant, but “you can see the whole surface” is false because the pixel only sees a part of the disk.

> **Comments:**
> **username_0**: I think I understand now where my confusion was coming from. Thank you so much for the thorough explanation!  

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