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skytiger    294

(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 :-)

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Bacterius    13165

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")

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skytiger    294

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

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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

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skytiger    294

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 ...

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skytiger    294

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

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skytiger    294

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

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Bacterius    13165

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.

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skytiger    294

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

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Bacterius    13165

skytiger    294

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Bummel    1888

@Skytiger: Are you willing to share your list of links to all those articles you've read? :)