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## Radiance (radiometry) clarification

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### #21skytiger  Members

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Posted 06 May 2013 - 02:47 PM

Mistakes I perceived:

the emitter is always at normal incidence to the detector's view angle

no it isn't

radiance measured at the emitter is the same as radiance measured at the detector

this doesn't mean the radiance is the same when measured from a DIFFERENT ANGLE
it means the radiance doesn't change over DISTANCE

also it means radiance is the same for both combinations of terms:

- solid angle of emitter and projected area of detector
- solid angle of detector and projected area of emitter

The key point is that the cosine term in the definition of radiance does not use the detector's surface normal but the emitter's surface normal

it can use either, see my point above

apparent size of the source is quite clearly constant due to the nature of your setup

no it isn't, the apparent size of the source is SMALLER for the angled detector

you calculated radiance using the following terms:

- power is constant, since the emitter is obviously still emitting the same amount of light in both cases
- area is constant, as the emitter's surface area hasn't changed
- solid angle is constant, because the emitter is held facing the detector at normal incidence in both cases

but you missed PROJECTED AREA (or the cosine term)

Therefore, measured radiance is the same in both cases

not once you factor in the projected area term you missed

but for all practical purposes ™ it’s a point and a direction

no it isn't
you can't just ignore the differential projected area and solid angles ... otherwise what's the point in defining radiance at all?
you CAN simplify like this inside a video game using the simplified Lambertian dot product lighting we use ... but not in Radiometry and definitely not in the example I gave!
that is actually the WHOLE POINT of my post, to clarify why this is the case!

Assuming the receiver is left and the emitter right in your images
Irradiance at every point of the receiver is smaller for the angled configuration

agreed
but the emitter is on the left
so irradiance at the detector is the same
as the flux is the same and the detector area is the same

Think about what happens when you aim the emitter perpendicular to the receiver, the projected area goes to zero and no light moves towards the receiver so radiance is zero, as expected

your logic is flawed
when the emitter is perpendicular there is nothing to discuss, nothing to measure ...
it is the same as if I removed the emitter completely!
you are making the assumption that because the radiance is zero when perpendicular
that it must change from 1 to 0 as the viewing angle increases
this makes no sense
if instead you considered what happened at a viewing angle of 89.99999999 degrees - then you would have seen my point!

The orientation of the emitter doesn't matter since it's defined to emit equally in all directions

it doesn't matter when considering intensity or power ... it DOES matter when measuring radiance
as the differential projected area is LESS

Edited by skytiger, 06 May 2013 - 05:33 PM.

### #22skytiger  Members

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Posted 07 May 2013 - 12:08 AM

If you carefully read these 2 articles

http://en.wikipedia.org/wiki/Lambert%27s_cosine_law

See the section "Details of equal brightness effect"

http://www.oceanopticsbook.info/view/radiative_transfer_theory/level_2/the_lambertian_brdf

You will see that Lambertian cosine * Radiance's 1/cosine _combine_ to give constant radiance

Obviously a 1/cosine term tends to INFINITY when the angle tends to 90°

Also the wikipedia Talk comments are agreeing with ME

you are misinterpreting them

http://en.wikipedia.org/wiki/Talk:Radiance#cosine_term

read them again carefully!

### #23skytiger  Members

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Posted 07 May 2013 - 02:06 AM

And another explanation:

In the figure on the right, if the source is Lambertian and equally bright in all directions, then the amount of energy that falls each the detectors is not the same for each of the detectors, but depends on the angle θ as cosθ.
[Note: A source that is equally bright in all direction does not emit the same amount of energy in all directions.  When determining the brightness of a small source we measure the energy that falls on a detector subtending a given solid angle at the source and then divide by the apparent size (area) of the source.  A small source emitting the same amount of energy into the given solid angle as a larger source is brighter than the larger source.]

http://electron6.phys.utk.edu/optics421/modules/m4/radiometry.htm

Note:

A small source emitting the same amount of energy into the given solid angle as a larger source is brighter than the larger source

### #24Bacterius  Members

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Posted 07 May 2013 - 02:24 AM

Maybe if you stopped attacking everyone and turning this into a "I'm right and you're wrong" argument, people would actually be able to discuss this to benefit everyone's understanding. And it is entirely possible to misinterpret the definitions in such a way that everything is still consistent, but incorrect with respect to the accepted definitions, so the distinction between "right" and "wrong" is meaningless in itself - the real question should be correct with respect to what? Radiometry is a very confusing field where everyone seems to use different definitions (I think this thread is a living example).

But as far as I am concerned, any physical quantity which goes to infinity like this is implausible and absolutely useless as it cannot be effectively measured. Regarding the Wikipedia comments - a physical source is defined as a source with an area (i.e. not a point light source). Now Srleffler says:

"The radiant flux for physical sources falls off at least as fast with angle as cos(θ)."

And the radiant flux term in the radiance definition is at the top of the fraction, effectively cancelling the cosine term in the denominator. Radiance does not approach infinity as the angle approaches 90 degrees. The 1 / cosine term does, obviously, but the radiant flux term approaches zero at a rate at least inversely proportional to that. Radiant flux is the amount of radiant power crossing some imaginary cross-section (so, energy per unit of time). And guess what - the radiant flux considered here must pass through the projected area of the light source. And guess what again - this projected area is proportional to cos(theta). So radiant flux is proportional to cos(theta). Lambert's Law! The number of photons/sec (power) directed into any wedge is proportional to the area of the wedge!

And for the record, I see your point about intensity per unit solid angle. But your interpretation that radiance tends to infinity seems inconsistent with other equations, such as, say, the rendering equation. With your current interpretation of "infinite radiance", perceived brightness would increase without bounds when looking at anything that emits or reflect light in your direction (not just a light source) at close to 90 degree angles. That is clearly not the case.

apparent size of the source is quite clearly constant due to the nature of your setup

no it isn't, the apparent size of the source is SMALLER for the angled detector

Nice strawman. But that was my first post where I assumed emitter was on the right and receiver was on the left. Since you said later on that you meant something else, this is obviously not relevant anymore, and I acknowledged it. Good job quoting my post out of context, it is really helping this thread move forward.

----

Either way, your tone and attitude are not making me want to reply to you, and I believe I speak for the other people in the thread as well. You created this thread asking a question. Now it seems you know all the answers, and are literally challenging anyone to prove you wrong. Now it is possible I am wrong. It is possible you are wrong. It is possible everyone in this thread is wrong, or everyone is right (and just talking past one another). It could just be because of some conceptual difference between the definition of radiance as used in computer graphics (where "radiance" might in fact mean "radiant exitance", for instance, which would not surprise me) and in physics (which you are using). You can probably come up with a set of definitions which makes any statement correct. That doesn't mean they will be of any practical use for whatever you are trying to do. Different applications sometimes call for different definitions (which should probably have different names, to avoid confusion, but sometimes that doesn't happen)

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

### #25Hodgman  Moderators

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Posted 07 May 2013 - 02:52 AM

If radiance, which is a measure of energy passing through some part of space, approaches infinity as the viewing angle approaches 90º, then we could build infinite energy machines by placing an LED panel and a photovoltaic panel at 89.999 degress relative to each other.

A small source emitting the same amount of energy into the given solid angle as a larger source is brighter than the larger source

That should just be common sense following from geometry though.
e.g. a big and a small emitter on the same plane, viewed by two detectors on a different angled plane. For the left detector in the image to read the same amount of energy as the right detector, then the small/left source will have to be outputting much more energy per m2 than the big/right source, because more area of the right source is visible to the detector.

If radiance is watts per steridian per area, and area increases with angle, then received watts should decrease as angle increases, because you divide by area. So, we should think that these energy measurements approach zero as the angle increases, because the area approaches infinity...

However, it's important to note that as the detector angle increases and the area visible to the detector increases, the solid angle occupied by the emitter also decreases, which balances things out somewhat.

e.g. above, the top detector, which is at a more shallow angle can see a larger area, but emitter is a smaller fraction of it's visible area. If we draw the wedge taken up by the emitter, it shrinks as the visible area grows.

You will see that Lambertian cosine * Radiance's 1/cosine _combine_ to give constant radiance

Obviously a 1/cosine term tends to INFINITY when the angle tends to 90°

If there's a term that approaches infinity, but the final result remains constant... then there must be another term that approaches zero to balance it out? And then it goes without saying that if radiance remains constant, then radiance can't approach infinity? So which is it, constant or increasing with angle?

Edited by Hodgman, 07 May 2013 - 03:05 AM.

### #26skytiger  Members

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Posted 07 May 2013 - 03:11 AM

I just want everybody to find the right answer

I don't care if I'm right or not

But right now I strongly believe my view is the correct one

and that radiance measured from the angled detector will be 1.414 times greater

All the points that have been made giving a radiance of 1.0 or less than 1.0 seem flawed to me

I don't mean to attack anybody, but let me point out where I think you are going wrong:

And the radiant flux term in the radiance definition is at the top of the fraction

yes it is, but the cosine term at the top comes from the Lambertian reflectance NOT from the radiance equation

If you consider the radiance equation alone (which I am doing by using a non-Lambertian surface)

you can see that radiance increases as the viewing angle increases

The "confusing" part of the radiance equation is that brightness increases as the projected area of the emitter decreases

this is because the power density increases (same number of photons/sec in a smaller area = brighter)

Only when you combine this with a Lambertian surface, where the radiant intensity decreases as viewing angle increases

do you get constant radiance when measured from different viewing angles

I hope I managed to make my point clearly this time

Edited by skytiger, 07 May 2013 - 03:37 AM.

### #27skytiger  Members

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Posted 07 May 2013 - 03:24 AM

If radiance, which is a measure of energy passing through some part of space

you are going wrong right there

radiance is not a measure of energy passing through some part of space - that is flux

radiance is an abstract concept that ALLOWS you to measure energy passing through some part of space if you combine radiance with the viewing geometry

radiance = flux / solid_angle / projected_area

so

flux = radiance * solid_angle * projected_area

because you divide by area

the mistake you are making here is that of "dividing by area" means it will be smaller

but the cosine term is between 0 and 1

dividing something by 0.5 results in an increase, not a decrease

So which is it, constant or increasing with angle

in my diagram the radiance increases as viewing angle increases

IF the surface in my diagram was Lambertian the radiance would be constant

Edited by skytiger, 07 May 2013 - 03:40 AM.

### #28skytiger  Members

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Posted 07 May 2013 - 03:30 AM

Hodgman: your top diagram is missing the point

you are showing a smaller emitter area

I am talking about a smaller emitter PROJECTED area

### #29skytiger  Members

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Posted 07 May 2013 - 04:05 AM

This doesn't make sense intuitively because Radiance is not a physical quantity and in real life surfaces are close to Lambertian

Edited by skytiger, 07 May 2013 - 04:16 AM.

### #30skytiger  Members

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Posted 07 May 2013 - 09:56 AM

Hodgman

very interesting point about the area of the emitter that is visible being greater

however it just confirms the radiance of 1.414

here I show the radiance calculation from both points of view, and the result is 1.414 for both:

Your idea that the solid angle changes is wrong ...

Edited by skytiger, 07 May 2013 - 02:12 PM.

### #31skytiger  Members

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Posted 07 May 2013 - 11:01 AM

However, it's important to note that as the detector angle increases and the area visible to the detector increases, the solid angle occupied by the emitter also decreases, which balances things out somewhat

the problem with this is that the solid angle and projected area terms in the radiance equation have to be one of these combinations:

- solid angle subtended by emitter and projected area of detector

or

- solid angle subtended by detector and projected area of emitter

you are describing an invalid combination:

- solid angle subtended by emitter and projected area of emitter

which is meaningless

quoting wikipedia radiance article:

When calculating the radiance emitted by a source, A refers to an area on the surface of the source, and Ω to the solid angle into which the light is emitted. When calculating radiance at a detector, A refers to an area on the surface of the detector and Ω to the solid angle subtended by the source as viewed from that detector

Edited by skytiger, 07 May 2013 - 11:13 AM.

### #32David Neubelt  Members

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Posted 07 May 2013 - 11:43 PM

Hodgman

very interesting point about the area of the emitter that is visible being greater

however it just confirms the radiance of 1.414

here I show the radiance calculation from both points of view, and the result is 1.414 for both:

Your idea that the solid angle changes is wrong ...

As the angle becomes more oblique to the wall on the left the solid angle the detector subtends will decrease. The solid angle will reach it's maximum when the detector and emitter are parallel.

Graphics Programmer - Ready At Dawn Studios

### #33skytiger  Members

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Posted 08 May 2013 - 12:58 AM

The solid angle the detector subtends is constant in all my diagrams

because the distance between detector and emitter (center points) is equal

and the detector is always normal to the light

In my first post the solid angle the emitter subtends does decrease with angle

But in the post above the solid angles subtended by both emitter and detector are constant

In the first post the emitter area is constant, in the last post the emitter *projected* area is constant

Either way the result is the same, radiance increases with viewing angle

because radiance = flux / solid_angle / projected_area

as the solid_angle decreases radiance must increase

as the projected_area decreases radiance must increase

as the flux increases radiance must increase

In my diagram above you can measure the plane angle (depicting the solid angle) with a protractor and see it is constant

### #34David Neubelt  Members

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Posted 08 May 2013 - 01:27 AM

The solid angle the detector subtends is constant in all my diagrams

because the distance between detector and emitter (center points) is equal

and the detector is always normal to the light

In my first post the solid angle the emitter subtends does decrease with angle

But in the post above the solid angles subtended by both emitter and detector are constant

In the first post the emitter area is constant, in the last post the emitter *projected* area is constant

Either way the result is the same, radiance increases with viewing angle

because radiance = flux / solid_angle / projected_area

as the solid_angle decreases radiance must increase

as the projected_area decreases radiance must increase

as the flux increases radiance must increase

In my diagram above you can measure the plane angle (depicting the solid angle) with a protractor and see it is constant

If the solid angle is constant then the area on the left wall will increase as the angle goes further oblique.

Edited by David Neubelt, 08 May 2013 - 01:27 AM.

Graphics Programmer - Ready At Dawn Studios

### #35skytiger  Members

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Posted 08 May 2013 - 01:33 AM

Exactly

as the emitter area increases the flux increases which means the measured radiance *increases*

### #36David Neubelt  Members

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Posted 08 May 2013 - 02:06 AM

That only makes sense if you let the emitter area increase unbounded but physical devices are bounded in area so it doesn't make sense to think in those terms.

-= Dave

Graphics Programmer - Ready At Dawn Studios

### #37Bacterius  Members

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Posted 08 May 2013 - 03:41 AM

Exactly

as the emitter area increases the flux increases which means the measured radiance *increases*

And now you conveniently forget mentioning the cosine term in the denominator which you've been arguing about for the past two days. Funny how that works, isn't it?

Let's go with the diagram you posted and walk through it step by step:

Emitter is on the left, detector is on the right. For the diagrams on the left, we are considering radiance emitted towards the detector, and for the diagrams on the right, we are considering radiance received by the detector (they should, of course, be the same). Assume the detector and emitter have respective areas Ad and Ae and unit distance from each other. Let P be the total power emitted by the emitter (in every direction over its entire surface). Now:

Now, for calculating radiance emitted by a surface into a given solid angle:

-- θ is the angle (with respect to the emitter's normal) at which the radiance is being emitted

-- d²ϕ is the radiant flux emitted by the surface (this is proportional to the solid angle being emitted into)

-- dΩ is the solid angle being emitted into

-- dA is a surface patch on the emitter's surface

And, for calculating radiance received by a detector from a given solid angle:

-- θ is the angle (with respect to the detector's normal) at which the radiance is being received

-- d²ϕ is the radiant flux received by the detector (again, proportional to the solid angle where the light is being received from)

-- dΩ is the solid angle being received from

-- dA is a surface patch on the detector's surface

Now, the top-left diagram (radiance measured as it exits the emitter towards the detector):

-- θ is 0 // straightforward

-- d²ϕ is equal to P * (Ad / Ae) // power over surface Ae projected into cross-sectional area Ad - note projected area = real area since we're at normal incidence

-- dΩ is equal to Ad // fine, we see the detector at normal incidence so its solid angle subtends over its entire area

-- dA is equal to Ae // straightforward

--> L = (P * (Ad / Ae)) / (Ae * Ad) = P / Ae^2

Bottom-left diagram (radiance measured as it exits the emitter towards the detector):

-- θ is equal to 45 degrees // angle made with the emitter's surface normal

-- d²ϕ is equal to P * ((Ad * cos(θ)) / Ae) // power over surface area Ae projected into cross-sectional area Ad * cos(θ)

-- dΩ is equal to Ad // the solid angle is the same as the detector is always facing the emitter at normal incidence

-- dA is equal to Ae // straightforward

--> L = (P * ((Ad * cos(θ)) / Ae)) / (Ae * Ad * cos(θ)) = P / Ae^2

Top-right diagram (radiance measured as it falls on the detector):

-- θ is 0 // straightforward

-- d²ϕ is equal to P * (Ad / Ae) // power over cross-sectional area Ae

-- dΩ is equal to Ae // fine, we see the emitter at normal incidence so its solid angle subtends over its entire area

-- dA is equal to Ad // straightforward

--> L = (P * (Ad / Ae)) / (Ad * Ae) = P / Ae^2

Bottom-right diagram (radiance measured as it falls on the detector):

-- θ is equal to 0 degrees // angle made with the detector's surface normal

-- d²ϕ is equal to P * ((Ad * cos(45)) / Ae) // same reasoning as with bottom-left, power over surface area Ae projected into cross-sectional area Ad * cos(θ)

-- dΩ is equal to Ae * cos(45) // the projected area of the emitter, no problem

-- dA is equal to Ad // straightforward

--> L = (P * ((Ad * cos(45)) / Ae)) / (Ad * Ae * cos(45) * cos(θ)) = P / Ae^2

And we see the radiance is constant with view angle, as expected. It is also dependent on the emitter's surface area and power - of course, a higher power leads to larger radiance and a larger surface area leads to a smaller radiance as the emitter subtends a larger solid angle). And it doesn't go to infinity unless the emitter has no area which is not physical (and radiance is then meaningless). And it is of course independent of detector surface area, which only comes into play when you consider irradiance at the detector (where you multiply the radiance by Ad * cos(θ) to obtain the irradiance)

Worth noting P already depends on the emitter's surface area (since a higher surface area leads to higher power) so in reality radiance is proportional to 1 / Ae if you keep the emitter's power per unit area (also known as intensity) constant, which is generally the case.

Your "non-lambertian geometry" concept is meaningless. "Lambertian" doesn't refer to any form of "geometry", at least not to do with aiming detectors and emitters at various angles with respect to each other. Lambert's cosine law simply makes an observation that at grazing angles, the flux tends to zero, and that is true in general:

It may vary for various materials (that is where the BRDF comes in - the Lambertian model is simply a base case for the notion of BRDF) but it is completely unphysical for a surface to be able to emit constant flux (or, in general, flux which does not decrease at least as fast as cos(θ) with view angle) at grazing angles, as that would mean it has infinite power. That is what you have been assuming all along, I believe. It just does not happen for any emitter which has area.

I would nevertheless like to thank you for making this thread as it has really challenged (and improved, in many ways) my understanding of radiance, radiant flux, etc..

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

### #38skytiger  Members

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Posted 08 May 2013 - 06:02 AM

-- dA is equal to Ae // straightforward

This is a typo? You meant: Ad is equal to Ae

-- d²ϕ is equal to P * ((Ad * cos(θ)) / Ae) // power over surface area Ae projected into cross-sectional area Ad * cos(θ)

this is where you are going wrong

if you project the area of the detector onto the plane of the emitter the projected area will be 1.414

(think of your shadow at 5pm) the increased emitter area visible to the detector results in HIGHER flux and HIGHER radiance

if you project the area of the emitter onto the plane of the detector the projected area will be 0.707

(think of your shadow at 1pm) now you have the same flux compressed into a smaller area which results in HIGHER radiance

-- dΩ is equal to Ad // the solid angle is the same as the detector is always facing the emitter at normal incidence

YES you are correct about this ... can't you see how you are contradicting yourself above?

Your "non-lambertian geometry" concept is meaningless. "Lambertian" doesn't refer to any form of "geometry", at least not to do with aiming detectors and emitters at various angles with respect to each other. Lambert's cosine law simply makes an observation that at grazing angles, the flux tends to zero, and that is true in general:

It may vary for various materials (that is where the BRDF comes in - the Lambertian model is simply a base case for the notion of BRDF) but it is completely unphysical for a surface to be able to emit constant flux (or, in general, flux which does not decrease at least as fast as cos(θ) with view angle) at grazing angles, as that would mean it has infinite power. That is what you have been assuming all along, I believe. It just does not happen for any emitter which has area.

I would nevertheless like to thank you for making this thread as it has really challenged (and improved, in many ways) my understanding of radiance, radiant flux, etc..

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

It is logically inconsistent

### #39skytiger  Members

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Posted 08 May 2013 - 06:17 AM

d²ϕ is equal to P * (Ad / Ae)

P / Ae

power / area

W / m²

gives radiant exitance

radiant exitance * solid angle gives ...

W sr / m²

there is no such radiometric unit!

You need to divide by 2 pi to get radiant intensity (pi for a Lambertian surface)

then multiply by area Ae and solid angle Ad to get flux incident to detector

Your flux is proportional to the correct flux however

so this is a minor mistake, not the cause of your misunderstanding

### #40Bacterius  Members

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Posted 08 May 2013 - 06:47 AM

Your interpretation is "radiance goes to infinity as view angle tends to 90 degrees". Yes or no?

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)

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

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