<p>I see some problems in the quote you posted.</p>

<blockquote>
  <p>In effect, a point rotated around its normal vector will not change
  the way it reflects light.</p>
</blockquote>

<p>This is true, because a Lambertian reflector will <strong>never</strong> change the <em>way</em> it reflects light. The underlying principle stays the same. Also, Lambertian surfaces are isotropic, so the <em>amount</em> of reflected light won't change either (which is probably what this sentence is aiming for).</p>

<blockquote>
  <p>However, the point will change the way it reflects light if it is tilted away from its initial normal vector since the area is illuminated by a smaller fraction of the incident radiation.</p>
</blockquote>

<p>Again not true, because the principle doesn't change. The amount <em>may</em> change, except for the special case that the cosine is &lt;= 0 before and after the tilting. The amount does not necessarily <em>grow</em>, except if we define that the cosine equals 1 before, i.e. that the normal points directly towards the light source.</p>

<p>This whole paragraph should probably be rewritten to be less ambiguous. Including isotropy could make it more complete.</p>

CC BY-SA 3.0
2015-09-04T07:40:47.200Z
2015-09-04T07:40:47.200Z
David Kuri
<p>You are correct, it's badly worded. Illumination falls off with the cosine of the angle between the surface normal and the inverse light direction, so the wording implies the light is shining down the original surface normal, and so any tilting away would be tilting away from the lighting direction.</p>

CC BY-SA 3.0
2015-09-04T17:11:50.277Z
Illumination does not actually fall of the surface area towards the light is just smaller
joojaa
CC BY-SA 3.0
2015-09-05T15:27:52.980Z
You're right I think, "fall-off" to me is anything that makes the surface area smaller from the lights perspective, so distance from and rotation away from the light have the same effect to me, but my definition of "fall-off" is probably not mathematically correct :P
chrisvarnz
CC BY-SA 3.0
2015-09-05T17:38:40.670Z
well yes but that would be hard for a layman to understand. lots of things can fall off.
joojaa
CC BY-SA 3.0
2015-09-04T00:53:07.067Z
2015-09-04T00:53:07.067Z
chrisvarnz
<p>What one has to do in this matter is first define the quantities that are physically at play here, so that everybody speaks about the same thing.</p>

<p>There is:</p>

<ul>
<li>radiance (<a href="https://en.wikipedia.org/wiki/Radiance" rel="nofollow noreferrer">wikipedia</a>)<br>
Flux emitted by a surface per solid angle per projected area.<br>
<a href="https://i.sstatic.net/N2ziI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/N2ziI.png" alt="enter image description here"></a><br>
Unit is <strong>W·sr−1·m−2</strong></li>
<li>radiant intensity (<a href="https://en.wikipedia.org/wiki/Radiant_intensity" rel="nofollow noreferrer">wikipedia</a>)<br>
Origin of radiance, you take the surface area out of the unit.<br>
Unit is <strong>W·sr−1</strong></li>
<li>intensity (<a href="https://en.wikipedia.org/wiki/Luminous_intensity" rel="nofollow noreferrer">wikipedia</a>)<br>
A perception-based unit of power per solid angle.<br>
Unit is <strong>candela</strong></li>
<li>luminance (<a href="https://en.wikipedia.org/wiki/Luminance" rel="nofollow noreferrer">wikipedia</a>)<br>
<em>The luminance is normally obtained by dividing the luminous intensity by the light source area</em> (<a href="http://www.rohm.com/web/eu/faq_search/-/faq_search/FaqId/260" rel="nofollow noreferrer">source</a>)<br>
therefore this is also perception based.<br>
Unit is <strong>cd·m−2</strong></li>
<li>luminuous flux (<a href="https://en.wikipedia.org/wiki/Luminous_flux" rel="nofollow noreferrer">wikipedia</a>)<br>
Same thing but not related to solid angle.<br>
quote:<br>
<em>Luminous flux is a measure of the total amount of light a lamp puts out. The luminous intensity (in candelas) is a measure of how bright the beam in a particular direction is</em><br>
Unit is <strong>lumen</strong></li>
</ul>

<p>You can also talk of irradiance (<a href="https://en.wikipedia.org/wiki/Irradiance" rel="nofollow noreferrer">wiki</a>) when speaking of received radiance.<br>
And one can also talk of total-irradiance when speaking of the irradiance taken for the whole hemisphere.</p>

<p>refer to: <a href="http://www.crompton.com/light/index.html" rel="nofollow noreferrer">http://www.crompton.com/light/index.html</a><br>
and: <a href="https://pathtracing.wordpress.com/" rel="nofollow noreferrer">https://pathtracing.wordpress.com/</a><br>
and why not: <a href="http://www.nvc-lighting.com/showuseInfo.Aspx?typeID=42&amp;ID=94" rel="nofollow noreferrer">http://www.nvc-lighting.com/showuseInfo.Aspx?typeID=42&amp;ID=94</a></p>

<p>As you can see, there are two classes of units, the perception based and the absolute physical units.<br>
the <strong>radiance</strong> is the measure you want to look at to understand Lambert, you can actually see the cosine falloff directly in the formula.  </p>

<p>You can see the intuition of this in this blog: <a href="https://pathtracing.wordpress.com/" rel="nofollow noreferrer">https://pathtracing.wordpress.com/</a> chapter "Lambert's cosine law"</p>

CC BY-SA 3.0
2015-10-08T08:27:22.417Z
2015-10-09T01:49:20.690Z
2015-10-09T01:49:20.690Z
v.oddou
<p>In reading about <a href="https://en.wikipedia.org/wiki/Lambertian_reflectance#Use_in_computer_graphics" rel="nofollow">Lambertian reflectance</a> on Wikipedia I found the following phrase (in bold) which doesn't sound right to me:</p>

<blockquote>
  <p>In computer graphics, Lambertian reflection is often used as a model for diffuse reflection. This technique causes all closed polygons (such as a triangle within a 3D mesh) to reflect light equally in all directions when rendered. In effect, a point rotated around its normal vector will not change the way it reflects light. However, the point will change the way it reflects light if it is tilted away from its initial normal vector <strong>since the area is illuminated by a smaller fraction of the incident radiation.</strong></p>
</blockquote>

<p>The way I picture the situation described in the paragraph, only tilting away from the light source would cause  less light to be incident in a given area. In general, tilting away from the initial normal vector could lead to either an increase <strong><em>or</em></strong> a decrease in incident light per area, as this says nothing about the location of the light source.</p>

<p>Have I misunderstood the context, or is this something that should be rewritten on Wikipedia?</p>

CC BY-SA 3.0
2015-08-27T13:01:45.767Z
The cross sectional area visible dimishes as your incidence angle grows.
joojaa
CC BY-SA 3.0
2015-08-27T13:10:07.617Z
@joojaa I follow that bit, but the bit in bold seems to be talking about tilting the surface away from its initial normal vector, which would only make sense for the specific case that the incident light is perpendicular to the surface, or I'm missing something.
trichoplax is on Codidact now
CC BY-SA 3.0
2015-10-09T07:44:52.237Z
Yes, the wording is quite strange (what is a rotating point for instance ? :-) ). It's not a mistake, it's poor wording. I'm afraid that once a guy quicky made the whole "basic computer graphics" content of wikipedia for some reason, letting a lot of polishing (or more) to be done. Seems like hot topics are well edited and completed (by academics and master/PhD students ?), but not basic topics (I did, for a very few).
Fabrice NEYRET
<shading>
Is a Lambertian reflector illuminated by a smaller fraction of the incident radiation when it's tilted?
trichoplax is on Codidact now
# Is a Lambertian reflector illuminated by a smaller fraction of the incident radiation when it's tilted?

(Asked by: username_0 on 2015-08-27)

In reading about [Lambertian reflectance](https://en.wikipedia.org/wiki/Lambertian_reflectance#Use_in_computer_graphics) on Wikipedia I found the following phrase (in bold) which doesn't sound right to me:

> In computer graphics, Lambertian reflection is often used as a model for diffuse reflection. This technique causes all closed polygons (such as a triangle within a 3D mesh) to reflect light equally in all directions when rendered. In effect, a point rotated around its normal vector will not change the way it reflects light. However, the point will change the way it reflects light if it is tilted away from its initial normal vector **since the area is illuminated by a smaller fraction of the incident radiation.**

The way I picture the situation described in the paragraph, only tilting away from the light source would cause less light to be incident in a given area. In general, tilting away from the initial normal vector could lead to either an increase ***or*** a decrease in incident light per area, as this says nothing about the location of the light source.

Have I misunderstood the context, or is this something that should be rewritten on Wikipedia?

**Tags:** shading

> **Comments:**
> **username_1**: The cross sectional area visible dimishes as your incidence angle grows.  
> **username_0**: @username_1 I follow that bit, but the bit in bold seems to be talking about tilting the surface away from its initial normal vector, which would only make sense for the specific case that the incident light is perpendicular to the surface, or I'm missing something.  
> **username_2**: Yes, the wording is quite strange (what is a rotating point for instance ? :-) ). It's not a mistake, it's poor wording. I'm afraid that once a guy quicky made the whole "basic computer graphics" content of wikipedia for some reason, letting a lot of polishing (or more) to be done. Seems like hot topics are well edited and completed (by academics and master/PhD students ?), but not basic topics (I did, for a very few).  

---

## 3 answers

### Answer by username_3 on 2015-09-04. Score: 5

I see some problems in the quote you posted.

> In effect, a point rotated around its normal vector will not change
> the way it reflects light.

This is true, because a Lambertian reflector will **never** change the *way* it reflects light. The underlying principle stays the same. Also, Lambertian surfaces are isotropic, so the *amount* of reflected light won't change either (which is probably what this sentence is aiming for).

> However, the point will change the way it reflects light if it is tilted away from its initial normal vector since the area is illuminated by a smaller fraction of the incident radiation.

Again not true, because the principle doesn't change. The amount *may* change, except for the special case that the cosine is <= 0 before and after the tilting. The amount does not necessarily *grow*, except if we define that the cosine equals 1 before, i.e. that the normal points directly towards the light source.

This whole paragraph should probably be rewritten to be less ambiguous. Including isotropy could make it more complete.


---

### Answer by username_4 on 2015-09-04. Score: 3

You are correct, it's badly worded. Illumination falls off with the cosine of the angle between the surface normal and the inverse light direction, so the wording implies the light is shining down the original surface normal, and so any tilting away would be tilting away from the lighting direction.

> **Comments:**
> **username_1**: Illumination does not actually fall of the surface area towards the light is just smaller  
> **username_4**: You're right I think, "fall-off" to me is anything that makes the surface area smaller from the lights perspective, so distance from and rotation away from the light have the same effect to me, but my definition of "fall-off" is probably not mathematically correct :P  
> **username_1**: well yes but that would be hard for a layman to understand. lots of things can fall off.  

---

### Answer by username_5 on 2015-10-08. Score: 2

What one has to do in this matter is first define the quantities that are physically at play here, so that everybody speaks about the same thing.

There is:

* radiance ([wikipedia](https://en.wikipedia.org/wiki/Radiance))  
  Flux emitted by a surface per solid angle per projected area.  
  [![enter image description here](https://i.sstatic.net/N2ziI.png)](https://i.sstatic.net/N2ziI.png)  
  Unit is **W·sr−1·m−2**
* radiant intensity ([wikipedia](https://en.wikipedia.org/wiki/Radiant_intensity))  
  Origin of radiance, you take the surface area out of the unit.  
  Unit is **W·sr−1**
* intensity ([wikipedia](https://en.wikipedia.org/wiki/Luminous_intensity))  
  A perception-based unit of power per solid angle.  
  Unit is **candela**
* luminance ([wikipedia](https://en.wikipedia.org/wiki/Luminance))  
  *The luminance is normally obtained by dividing the luminous intensity by the light source area* ([source](http://www.rohm.com/web/eu/faq_search/-/faq_search/FaqId/260))  
  therefore this is also perception based.  
  Unit is **cd·m−2**
* luminuous flux ([wikipedia](https://en.wikipedia.org/wiki/Luminous_flux))  
  Same thing but not related to solid angle.  
  quote:  
  *Luminous flux is a measure of the total amount of light a lamp puts out. The luminous intensity (in candelas) is a measure of how bright the beam in a particular direction is*  
  Unit is **lumen**

You can also talk of irradiance ([wiki](https://en.wikipedia.org/wiki/Irradiance)) when speaking of received radiance.  
And one can also talk of total-irradiance when speaking of the irradiance taken for the whole hemisphere.

refer to: <http://www.crompton.com/light/index.html>  
and: <https://pathtracing.wordpress.com/>  
and why not: <http://www.nvc-lighting.com/showuseInfo.Aspx?typeID=42&ID=94>

As you can see, there are two classes of units, the perception based and the absolute physical units.  
the **radiance** is the measure you want to look at to understand Lambert, you can actually see the cosine falloff directly in the formula.

You can see the intuition of this in this blog: <https://pathtracing.wordpress.com/> chapter "Lambert's cosine law"


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