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61 replies to this topic

### #41skytiger  Members

Posted 08 May 2013 - 06:53 AM

Yes I understand the diagram perfectly

If you remove the Lambertian term cos(theta) from the numerator of the angled equation

By imagining that the surface is NOT Lambertian but is a special surface with constant intensity (the first line of my first post)

You will see that there is still a 1 / cos(theta) term that means radiance increases with viewing angle (in my thought experiment)

WHEN and ONLY WHEN you combine this phenomenon with Lambertian surface do you get constant radiance.

The radiance equation IN ISOLATION says that radiance increases with viewing angle.

That is my point and I am sure you can see that now ...

### #42skytiger  Members

Posted 08 May 2013 - 07:16 AM

if it wasn't a Lambertian surface, radiance would increase with viewing angle

### #43skytiger  Members

Posted 08 May 2013 - 07:25 AM

Let me point out more mistakes:

Do you understand this diagram? Yes or no? Can you see why radiance is independent of view angle? (in this case, I mean. with a BRDF it is obviously not constant)

The cosine term in the numerator comes from the Lambertian BRDF

in other words the exact OPPOSITE of what you believe is true!

it is the BRDF that results in constant radiance ...

This is the 3rd time I've posted an really good explanation, please read it this time:

Edited by skytiger, 08 May 2013 - 07:29 AM.

### #44Tasty Texel  Members

Posted 08 May 2013 - 07:27 AM

You can not both understand Lambertian reflectance *and* disagree with my point

There are no physical surfaces for which the emitted flux doesn't fall off to zero at grazing angles. A non-lambertian surfaces simply doesn't emit it's flux distributed in the way of the idealised cosinus lobe. Otherwise the radiance would indeed tend to infinity at grazing angles. The sheer physically implausibility of this idea implies the first statement.

### #45Bacterius  Members

Posted 08 May 2013 - 07:31 AM

By imagining that the surface is NOT Lambertian but is a special surface with constant intensity (the first line of my first post)

But such a surface CANNOT EXIST. IT IS NOT PHYSICAL. Any physical surface will have a falloff with view angle eventually at least as large as the Lambertian reflectance cosine term falloff and radiance shall never tend to infinity. A "special surface with constant intensity" has by definition infinite density and hence zero area, and is thus a point light source which cannot exist in real life and for which radiance is not defined anyway (as it has solid angle zero). If the surface had any area, the power would have to be infinite!

For instance, if you try to evaluate irradiance with your "constant intensity" surfaces, the integral will not converge. Period. For instance, just try to measure the radiant exitance from any point on your theoretical surface! The radiance is proportional to 1 / cos(theta), so let's calculate the radiant exitance by summing up radiance in every direction:

And as you can see, this integral does not converge. Yet the result (radiant exitance) has units W/m^2, and hence should be a physically valid amount of intensity.

Your thought experiment is meaningless. It's like saying "consider this rectangle with negative surface area" or "imagine if this ball of infinite mass collides with the ground". It doesn't mean anything. The theory does not even apply to such surfaces which are outside the realm of physics. I am sure you can see that now.

it is the BRDF that results in constant radiance ...

No, it is NOT. The BRDF is exactly the opposite and is a parameter to help you configure your reflectance distribution to depend on view angle (within physically plausible limits) in order to approximate materials better. Check out the rendering equation. There is a cosine term outside the BRDF. Where do you think it comes from? It's Lambert's cosine law, applied backwards (for irradiance). It must be there.

“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”

### #46Tasty Texel  Members

Posted 08 May 2013 - 07:37 AM

Just to make that clear for the case there is some misconception regarding this pitctures: the length of the arrows in the left image illustrate the perceived brightness (and radiance) whereas in the right part they represent the actual flux emitted in the respective directions.

### #47Hodgman  Moderators

Posted 08 May 2013 - 07:53 AM

On the topic of non-lambertian surfaces:
"Lamberts cosine term" is different to "lambertian surfaces". Lambertian surfaces use BRDF(...) = 1 (not BRDF(...) = cos(theta), where theta is the incident angle). The incident cosine term is not a property of the surface at all, but a fact of geometry.

That is to say, that Lambert proved the cosine term was necessary, and then a Lambertian BRDF is one that does nothing (multiplies by 1), relying on just this mandatory cosine term to give a somewhat correct result.

 this is not true... [/edit]
Using BRDF(...) = 1 ("lambertian") is very common, but it violates helmholtz reciprocity, so it isn't physically plausible. The fix isn't to remove the cosine term, but add another one!

To make it obey helmholtz reciprocity, you actually need to use BRDF(N,V) = dot(N,V) = cos(theta) (where theta is the exitance angle), so that your rendering equation has two cos(theta) multiplications -- one for the path where the light enters the lambertian reflector to begin with (the irradiance at the reflector) and one for the path where the light exits the reflector towards the viewer (the radiance in some direction).

i.e. Seeing as the incident cosine term is mandatory, then helmoltz reciprocity makes an exitance cosine term also mandatory (you can also reverse that, depending on which one you've proven to be required first ), so you can't just remove this term and be physically plausible.

Edited by Hodgman, 08 May 2013 - 11:27 PM.

### #48Bacterius  Members

Posted 08 May 2013 - 08:36 AM

Using BRDF(...) = 1 (lambertian) is very common, but it violates helmholtz reciprocity, so it isn't physically plausible.
To make it obey helmholtz reciprocity, you actually need to use BRDF(N,V) = dot(N,V) = cos(theta) (where theta is the exitance angle), so that your rendering equation has two cos(theta) multiplications -- one for the path where the light enters the lambertian reflector to begin with (the irradiance at the reflector) and one for the path where the light exits the reflector towards the viewer (the radiance in some direction).

Are you sure that's right? BRDF is ratio of radiance to irradiance, so the exitance cosine term is already handled in the incidence cosine term of the next "bounce" of the light ray. And BRDF(..) = 1 clearly obeys Hemlholtz reciprocity, since BRDF(incident, exitant) = BRDF(exitant, incident) = 1. No?

That said for a constant BRDF we need the constant (albedo) to be between 0 and 1 / pi otherwise energy conservation is not achieved, I believe. In theory, anyway. This is usually baked into shaders.

I agree about the cosine term, though. Taking it out makes the definitions meaningless as they no longer stand for "radiant flux", "radiance", but "radiant flux with the cosine term missing", "radiance with the cosine term missing", etc.. it doesn't make sense. The theory doesn't make any sense anymore.

“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”

### #49skytiger  Members

Posted 08 May 2013 - 08:52 AM

So the correct answer to my original question is:

radiance measured from angled detector = 1.414

as given by me in my second post

Yes or no?

### #50Hodgman  Moderators

Posted 08 May 2013 - 08:56 AM

Are you sure that's right?

Ah yeah I've gotten mixed up.

Basic Lambertian is reciprocal, yes; you only run into issues when you try to make it energy conserving by incorporating both reflection and refraction according to Fresnel's law.

The common approach is to use N•L (incident theta) to calculate the reflection/refraction ratio, with only the refracted part using the lambertian (constant) BRDF. However, this doesn't obey helmholtz reciprocity. You also need to calculate the above ratios using N•V (exitant theta) and account for the fraction of the "lambertian reflectance" (which is incident light that is refracted and diffused into exit light) that actually reflects off the inside of the surface.

If you only perform the former calculations, then when swapping L and V, you get different results from the BRDF, which is non-physical.

Edited by Hodgman, 08 May 2013 - 09:15 AM.

### #51skytiger  Members

Posted 08 May 2013 - 09:10 AM

(I apologise, I shouldn't have commented on the BRDF, I don't really understand it)

I would appreciate, for my sanity, a simple confirmation that 1.414 is the correct answer, for my "impossible" question :-)

### #52Bacterius  Members

Posted 08 May 2013 - 09:28 AM

Basic Lambertian is reciprocal, yes; you only run into issues when you try to make it energy conserving by incorporating both reflection and refraction according to Fresnel's law.
The common approach is to use N•L (incident theta) to calculate the reflection/refraction ratio, with only the refracted part using the lambertian (constant) BRDF. However, this doesn't obey helmholtz reciprocity. You also need to calculate the above ratios using N•V (exitant theta) and account for the fraction of the "lambertian reflectance" (which is incident light that is refracted and diffused into exit light) that actually reflects off the inside of the surface.
If you only perform the former calculations, then when swapping L and V, you get different results from the BRDF, which is non-physical.

Ah, I see what you mean, yes. The specular part requires N dot L = N dot V (which is why we use the half-angle vector H = L + V and roll with that, I guess) but the diffuse part does not. So for the exitant theta, you work out how much it would have contributed if it was a diffusely reflected ray, and how much it would have contributed if it was a specularly reflected ray, and combine the two, but this won't work in both directions so we need to take both angles into account. That makes sense!

I would appreciate, for my sanity, a simple confirmation that 1.414 is the correct answer, for my "impossible" question :-)

There is no answer. The radiometric situation presented with the given data cannot occur, the theory cannot predict what would happen (it is outside the range of applicability). But according to your modification of the theory which removes the geometrical condition stated by Lambert's cosine law, yes, 1 / cos(45) = sqrt(2) would be the answer you would expect (for some definition of "expect")

“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”

### #53skytiger  Members

Posted 08 May 2013 - 09:37 AM

Thankyou Radiometry is a pig ... I'm going to think about something else for a few weeks ...

### #54David Neubelt  Members

Posted 08 May 2013 - 10:40 AM

Skytiger,

Let me clarify my post above. I'm not disagreeing with your math that radiance is higher as the emission area grows larger while the solid angle stays constant.

However we can't think of radiance physically in those terms, imagine a scenario where you have a spectroradiometer and you were calibrating a TV. On the TV you have a black screen with a picture of a small red square in the center emitting red light. If you point the spectroradiometer directly at the red square so the red square fills the view of your measuring device then you will get the full radiance. Now, if you start angling the gun closer to the tv so the red square still fills the view then the spectroradiomer will see more of the red square (because the area the gun sees is now larger) and the radiance will go higher. This would be expected (assuming the TV emits photons equivalently in all directions).

However, if you go past 45 degrees to 70 degrees and suddenly not all of the red square and some of the black part of the screen is seen by the spectroradiometer because the angle is so oblique then now the radiance will start to decrease.

If I was able to jam the spectroradiometer into the TV so its parallel to the red square the radiance would fall to 0 because it sees none of the red square.

What happens in our example is the solid angle that is subtended by our red surface element decreases, the flux decreases and the projected area decreases as the gun becomes more oblique to the TV and the radiance decreases to zero.

Edited by David Neubelt, 08 May 2013 - 10:42 AM.

Graphics Programmer - Ready At Dawn Studios

### #55skytiger  Members

Posted 09 May 2013 - 12:50 AM

You are missing the concept of "brightness" which is "flux density"

Same number of photons/sec in a smaller area appears brighter

That is what the projected area term of the radiance equation is all about ...

### #56skytiger  Members

Posted 09 May 2013 - 01:41 AM

Also the radiance equation is only valid for theta [0,90)

between 0 degrees and LESS than 90 degrees

So interpolating between a valid value and an invalid value makes no sense

these are equivalent because at 90 degrees or greater the radiance equation doesn't apply at all

it makes no sense to ask "what is the radiance of a surface I can not see"

Edited by skytiger, 09 May 2013 - 01:42 AM.

### #57skytiger  Members

Posted 09 May 2013 - 01:49 AM

However we can't think of radiance physically in those terms

You can't think of radiance "physically" because it is an abstract concept using differential calculus

So attempting to reason about radiance physically will never make sense ...

Instead you have to first measure the radiance and then convert it to a physical quantity such as flux or intensity

### #58Bacterius  Members

Posted 09 May 2013 - 02:07 AM

Instead you have to first measure the radiance and then convert it to a physical quantity such as flux or intensity

And you believe that with this reasoning, your interpretation of radiance now makes sense and is physically plausible? Or are you just making a separate point here.

“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”

### #59skytiger  Members

Posted 09 May 2013 - 03:03 AM

This is no fun any more

My last friendly word on this topic is this:

You do not understand the concept of radiance

Mentally the concept of radiance has not "clicked" with you yet

Everything you need to understand radiance is in this topic

(The key is understanding "flux density" and the projected area term and isolating radiance from Lambert's cosine law term)

Which I am now enjoying ... as all my radiometric calculations now work ... I hope yours comes soon

Take care

### #60Bacterius  Members

Posted 09 May 2013 - 03:07 AM