I have a question about radiance and irradiance in a very simple scene.

Let's have a scene with just 2 infinite planes: ground and sky. The camera is
somewhere between them so that both are visible.

The ground is pure lambertian surface (ie. totally matte surface, like a white
piece of paper).

The sky emits light evenly and let's say the outgoing radiance is 1.0.
(Actually, I'm not sure what is the correct unit to define this light source,
radiant exitance is probably correct?) And the camera registers that value as
pure white pixel.

The question is: what is the irradiance at each point in the ground? I guess it
should also be 1.0. So the sky should look identical to ground: pure white.
Right? This is the most important question as I need a reference scene to verify
my renderer.

But how do I calculate it?

The way to do it is to integrate over hemisphere at that point acording to light
transport equation. In 2D scene it would be I(1.0 * cos x  dx) and we have to do
it from -pi/2 to pi/2. Right?

The value of that is sin (pi/2) - sin(-pi/2) = 2. Hmm. Not right. Or is it? Most
likely I'm integrating incorrectly or the wrong thing.

Yes, I'm lost and my maths suck...but the correct result would help a lot.

Thanks,
Severi

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> The way to do it is to integrate over hemisphere at that point acording to 
> light
> transport equation. In 2D scene it would be I(1.0 * cos x  dx) and we have 
> to do
> it from -pi/2 to pi/2. Right?
>
> The value of that is sin (pi/2) - sin(-pi/2) = 2. Hmm. Not right. Or is 
> it? Most
> likely I'm integrating incorrectly or the wrong thing.

You calculated the total light power collected at that point from all 
direction (and hence the amount emitted assuming 100% reflectance).  The 
correct answer is 2.  You'll get the same answer if you try to work out how 
much light power is emitted from a point on the sky, in fact the integral is 
the same.

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scott wrote:
>> The way to do it is to integrate over hemisphere at that point
>> acording to light
>> transport equation. In 2D scene it would be I(1.0 * cos x  dx) and we
>> have to do
>> it from -pi/2 to pi/2. Right?
>>
>> The value of that is sin (pi/2) - sin(-pi/2) = 2. Hmm. Not right. Or
>> is it? Most
>> likely I'm integrating incorrectly or the wrong thing.
> 
> You calculated the total light power collected at that point from all
> direction (and hence the amount emitted assuming 100% reflectance).  The
> correct answer is 2.  You'll get the same answer if you try to work out
> how much light power is emitted from a point on the sky, in fact the
> integral is the same.

Ok. And in 3D the result is? Basically revolving "y=cos x" around y-axis
and calculating the volume. I have forgotten it but I'll figure it out
later...

If irradiance in above case is 2, what about the radiance than is
reflected from that point to a certain direction? (That is equal no
matter what the direction as it is lambertian).

Thanks for your help!

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> If irradiance in above case is 2, what about the radiance than is
> reflected from that point to a certain direction? (That is equal no
> matter what the direction as it is lambertian).

No it's not equal, it's proportional to cos(x), that is what lambertian 
means.  It then "looks" the same brightness to a camera/eye/surface from any 
angle, because the area you see for each pixel/cone/rod/patch is 
proportional to 1/cos(x), so the cos factor cancels out leaving constant 
"brightness" from any angle.

In the case of irradiance of 2, the radiance is actually exactly cos(x), 
because if you integrate cos(x) over -pi/2 to pi/2, you will get 2, as you 
already found out.  So in fact your sky and ground are identically looking 
planes (assuming the sky is not reflective at all), which you proved 
already.

In 3D your radiance is now in units of power per solid angle per area, 
compared to power per angle per area.  Lambertian still means that the 
radiance is proportional to cos(x) (where x is the angle between the surface 
normal and radiated/reflected light), it's just you need to integrate over a 
hemi-sphere rather than a semi-circle.

For lambertian surfaces, in reality this just means that your units are 
different and you only integrate from 0 to pi/2.

So in 3D, if you have a lambertian surface with radiance function 1*cos(x) 
Watts/steradian/metre2, the total irradiance from a point will be 1 
Watt/metre2.

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scott wrote:
>> If irradiance in above case is 2, what about the radiance than is
>> reflected from that point to a certain direction? (That is equal no
>> matter what the direction as it is lambertian).
> 
> No it's not equal, it's proportional to cos(x), that is what lambertian
> means.  It then "looks" the same brightness to a camera/eye/surface from
> any angle, because the area you see for each pixel/cone/rod/patch is
> proportional to 1/cos(x), so the cos factor cancels out leaving constant
> "brightness" from any angle.
> 
> In the case of irradiance of 2, the radiance is actually exactly cos(x),
> because if you integrate cos(x) over -pi/2 to pi/2, you will get 2, as
> you already found out.  So in fact your sky and ground are identically
> looking planes (assuming the sky is not reflective at all), which you
> proved already.
> 
> In 3D your radiance is now in units of power per solid angle per area,
> compared to power per angle per area.  Lambertian still means that the
> radiance is proportional to cos(x) (where x is the angle between the
> surface normal and radiated/reflected light), it's just you need to
> integrate over a hemi-sphere rather than a semi-circle.
> 
> For lambertian surfaces, in reality this just means that your units are
> different and you only integrate from 0 to pi/2.
> 
> So in 3D, if you have a lambertian surface with radiance function
> 1*cos(x) Watts/steradian/metre2, the total irradiance from a point will
> be 1 Watt/metre2.


Damn it. I have to chew on this one a bit. I need my renderer to give
accurate results and I want to be able to define light sources using
real units.

Now ssRay gives good looking results which are or are not accurate. I
have no way to verify it. I'll be back later...

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scott wrote:

> So in 3D, if you have a lambertian surface with radiance function 
> 1*cos(x) Watts/steradian/metre2, the total irradiance from a point will 
> be 1 Watt/metre2.

All of human endevour is in here, my friends!

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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scott wrote:

> No it's not equal, it's proportional to cos(x), that is what lambertian
> means.  It then "looks" the same brightness to a camera/eye/surface from
> any angle, because the area you see for each pixel/cone/rod/patch is
> proportional to 1/cos(x), so the cos factor cancels out leaving constant
> "brightness" from any angle.

I have to ask about this. Many sources say that radiance is analogous to
luminance. And they also say that luminance "is most often used to
characterize the brightness". Like here:

http://www.optics.arizona.edu/Palmer/rpfaq/rpfaq.htm

So that means radiance also describes "brightness"?? And lambertian
surface has equal brightness no matter where you look at it from if the
light source remains stationary.

So what is the deal here?

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> I have to ask about this. Many sources say that radiance is analogous to
> luminance.

OK sorry I didn't understand the radiance/irradiance thing (I usually work 
with luminance and luminous intensity etc) - so I looked them up.

Illuminance is the same as irradiance, and describes how much power per unit 
area hits a surface from all directions.  If you were dealing with infra-red 
light, you could use this to work out how hot something got.

Luminance is the same as radiance, it describes how bright something seems 
to be from a particular direction.  It tells you how much power is coming in 
a certain direction in a certain area, it is directly related to how many 
photons would hit a pixel in a camera or a rod in your eye (or anything else 
with finite area).

Luminous intensity is the same as radiant intensity, this describes how much 
power is emitted in a certain direction, but it says nothing about the area 
that the energy is spread over.  So the same luminous intensity could be 
coming from two planes, but because one is tilted differently from the other 
the luminance would appear to be different.

For a lambertian surface, the luminance is constant from all viewing angles, 
but the luminous intensity of each point depends on cos(x).

In your calculations, you can either use luminous intensity and factor in 
for the geometry, or use the luminance directly and not worry about the 
geometry.  The 1st will simpplify to the 2nd in most cases anyway, but I 
don't know, maybe it is easier to understand the 1st case, or better for 
later when you have non-lambertian surfaces, I don't know.

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Severi Salminen nous illumina en ce 2008-09-15 13:40 -->

> 
> Damn it. I have to chew on this one a bit. I need my renderer to give
> accurate results and I want to be able to define light sources using
> real units.
>
Just what do you call "real" units for your light?
Each and every "real" unit we use are totaly arbitrary, and somewhet dependent 
on the method used to mesure them.
You can define the power and intensity of a given light several way, and 
different light that have the same value for any given unit can look very different.
Just try to compare a 1000 LUX incadessent light bulb, sodium light, mercury 
light, fluoressent light, and argon light. They all look different. Some will 
look brighter, some dimmer, not counting the variation in colouration...
In fact, the sodium will probably look the brighter, and the argon one dimmer.
The yellow sodium keep your iris more open, while the bluish argon light will 
cause your iris to contract a lot.

-- 
Alain
-------------------------------------------------
If you are good, you will be assigned all the work. If you are really good, you 
will get out of it.

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Alain wrote:

> Just what do you call "real" units for your light?
> Each and every "real" unit we use are totaly arbitrary, and somewhet
> dependent on the method used to mesure them.

My point was that if I use real meaningful units I can verify that the
integration works properly. For example: if I have a theoretical light
source emitting certain power and a perfect lambertian surface, I can
check if the surface reflects the correct radiance. I didn't mean that I
would necessarily get more photorealistic images with those units. Well,
actually, I will, if using real units helps to solve errors.

Now I don't specify light sources that way and I can't verify the
results (and I do have some very possible errors now in the code...). I
only get pretty images and that is only half the fun :)

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