Imagine a surface which radiates light equally in all directions from all points
if we measure the radiance of the surface from two positions
one opposite the surface
and one at a 45 degree angle
would the measured radiance be the same?
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Posted 05 May 2013 - 07:11 AM
I will confess this one left me confused for a while. Yes, the radiance is the same, assuming the emitter is at the same distance (it's not clear from the diagram). The reason for this is because the emitter is always at normal incidence to the detector's view angle.
There's this neat theorem called conservation of radiance which is really helpful in puzzling situations like these, which states that radiance measured at the emitter is the same as radiance measured at the detector, assuming no radiance is lost between emitter and detector. In this case, it becomes obvious the radiance should be the same, as in both cases the emitters are identical up to rotation (which conserves areas, distances, and basically everything you could possibly care about when working with radiance).
There are also some helpful notes under the Wikipedia Talk page for radiance, see the "cosine term" paragraph. 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. Notably, the sentence "The cosine factor in the denominator reflects the fact that the apparent size of the source goes to zero as your angle of view approaches 90°." highlights this. In your case, the apparent size of the source is quite clearly constant due to the nature of your setup. This can get horribly confusing especially when you see computer graphics papers such as the rendering equation which apparently use the cosine term with the detector's surface normal, but they make entirely different assumptions.
Another, perhaps more intuitive way of deducing this result is by looking at the units of radiance: watts per meter squared per steradian. Also known as power per area per solid angle. Just to be clear, "power" is the power of the emitter, "area" is the surface area of the emitter, and "solid angle" is the solid angle subtended by the emitter from the detector's point of view. Now, for both cases considered:
- 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
Therefore, measured radiance is the same in both cases.
That said, I could be wrong on this. Please feel free to correct me if I made a mistake.
EDIT: actually, I don't even know anymore. It's all foggy
EDIT 2: no, I think, radiance should really be interpreted as emitted intensity (power / area) per solid angle (direction). There, that's better still mighty confused though,
Edited by Bacterius, 05 May 2013 - 11:16 AM.
“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”
Posted 05 May 2013 - 01:17 PM
Radiometry is a puzzle, but using this special non-Lambertian surface makes things easier
I believe the _flux_ incident to the detector is equal in both cases
but the measured _radiance_ is 1.0 / 0.707 = 1.414 times higher for the angled detector
because from the angled detector's point of view the emitter looks smaller (it has smaller projected area)
which means you have the same amount of flux in a smaller area which increases flux density .... and therefore "brightness"
this is due to the confusing projected area term in the radiance equation:
radiance = flux / solid_angle / projected_area
and THAT is why Lambertian reflectance needs a cosine law term ... to balance out the increase in radiance due to viewing angle
that is why Lambertian surface gives constant radiance when viewing angle changes
Posted 05 May 2013 - 03:02 PM
I think part of your confusion is that you're not asking the full question radiance answers. Radiance is energy flowing at a certain point in a certain direction. (Technically differential area and solid angle, but for all practical purposes ™ it’s a point and a direction.) Radiance is always L(x,w) and you have to pick a point (which you did) _and_ a direction (which you didn't – at least not explicitly). For incident radiance imagine standing at x and looking into direction w with an extremely small fov, so small that it is just a ray. What you see then is the radiance. It doesn't make sense to ask about the radiance a point receives without specifying a single direction -- you can at most ask for the average radiance (averaged over a finite solid angle). If the receiver is left in your images, you are actually illustrating irradiance, because you are using a finite solid angle. Similarly, it doesn't make sense to ask about the radiance of a finite surface.
I think it's best to always talk about radiance with respect to a imaginary surface perpendicular to the chosen direction. In all settings I ever encountered you were free to choose the surface normal, so why not chose the easiest configuration, in which the confusing cosine term simply disappears.
Assuming the receiver is left and the emitter right in your images:
Incident radiance (say) at the center of the receiver (x) coming from the direction pointing to the center of the emitter (w) is the same in both configurations. It's also the same as the exitant radiance from the center of the emitter toward the center of the receiver.
Irradiance at every point of the receiver is smaller for the angled configuration. The solid angle covered by the emitter is the same, but in the case of irradiance you are not free to choose the normal of the surface (since you asking specifically about irradiance of a concrete surface). So in this case the cosine factor is not 1 and it'll give you a smaller value.
Power collected in total by the receiver is smaller, too (obviously, if irradiance at every point is).
Posted 05 May 2013 - 03:38 PM
Irradiance at every point of the receiver is smaller for the angled configuration. The solid angle covered by the emitter is the same, but in the case of irradiance you are not free to choose the normal
I do not understand exactly what you mean by this. Could you elaborate a bit?
The way I thought about it: imagine a hemisphere around the receiver center point's normal. In the first configuration, the emitting surface would be close to the Zenith of this hemisphere. In the second configuration, the surface would be closer to the Azimuth and thus the cosinus factor would be different.
Is this what you mean?
Posted 05 May 2013 - 05:06 PM
I do not understand exactly what you mean by this. Could you elaborate a bit?
The way I thought about it: imagine a hemisphere around the receiver center point's normal. In the first configuration, the emitting surface would be close to the Zenith of this hemisphere. In the second configuration, the surface would be closer to the Azimuth and thus the cosinus factor would be different.
Is this what you mean?
Yes, that is correct. The incident beam would be smaller due to the cosine factor, hence irradiance decreases by a corresponding amount. That is not radiance, though, but irradiance.
Assuming the receiver is left and the emitter right in your images:
That was my assumption as well. Perhaps this was not the expected interpretation?
Edited by Bacterius, 05 May 2013 - 05:40 PM.
“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”
Posted 05 May 2013 - 06:03 PM
The emitter is on the left
and the receivers are on the right
the blue represents the solid angle subtended by the receivers
the magenta shows the area and the projected area of the emitter
they are meant to be infinitesimally small differential areas - so the direction is from the center of emitter to the center of receiver
Posted 05 May 2013 - 07:13 PM
The emitter is on the left
and the receivers are on the right
the blue represents the solid angle subtended by the receivers
the magenta shows the area and the projected area of the emitter
they are meant to be infinitesimally small differential areas - so the direction is from the center of emitter to the center of receiver
Well in that case, yes, radiance is lower. 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. So a lower emitter projected area for the emitter -> lower radiance, by the factor you stated in your first post.
The blue illustration in the diagram was confusing, though. And it's important to not confuse the cosine term in the emitter's projected area with the cosine term for the irradiance's incidence area, they are not the same!
“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”
Posted 06 May 2013 - 12:33 AM
Irradiance at every point of the receiver is smaller for the angled configuration. The solid angle covered by the emitter is the same, but in the case of irradiance you are not free to choose the normal
I do not understand exactly what you mean by this. Could you elaborate a bit?
The way I thought about it: imagine a hemisphere around the receiver center point's normal. In the first configuration, the emitting surface would be close to the Zenith of this hemisphere. In the second configuration, the surface would be closer to the Azimuth and thus the cosinus factor would be different.
Is this what you mean?
I have to disagree, in my opinion you'll approximately get the same results.
The emitter is on the left
and the receivers are on the right
the blue represents the solid angle subtended by the receivers
the magenta shows the area and the projected area of the emitter
they are meant to be infinitesimally small differential areas - so the direction is from the center of emitter to the center of receiver
Well in that case, yes, radiance is lower. 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. So a lower emitter projected area for the emitter -> lower radiance, by the factor you stated in your first post.
The blue illustration in the diagram was confusing, though. And it's important to not confuse the cosine term in the emitter's projected area with the cosine term for the irradiance's incidence area, they are not the same!
Posted 06 May 2013 - 01:15 AM
I have to disagree, in my opinion you'll approximately get the same results.
- Radiance is the same. You look up from the receiver with in a single direction. Either you see the emitter, then you get the radiance it emits, or you don't see the emitter, then you get zero. The orientation of the emitter doesn't matter since it's defined to emit equally in all directions.
- Irradiance is lower in b than in a. This time not because of the cosine factor(s), but because of the smaller solid angle.
- Power also lower.
Yes, I agree with everything you say. Good point about the orientation of the emitter.
“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”
Posted 06 May 2013 - 02:04 AM
emitter doesn't matter since it's defined to emit equally in all directions
Yes, but it's defined to emit LIGHT equally in all directions
LIGHT <> RADIANCE
If the incident radiance was the same in both cases the surface would be Lambertian ... but it isn't
From wiki:
In optics, Lambert's cosine law says that the radiant intensity or luminous intensity observed from an ideal diffusely reflecting surface or ideal diffuse radiator is directly proportional to the cosine of the angle θ between the observer's line of sight and the surface normal
My surface above is NOT Lambertian as the radiant intensity is not proportional to the cosine ... it is constant
In the radiance equation we are dividing by cosine, as the view angle approaches 90º the radiance approaches infinity ...
Posted 06 May 2013 - 02:14 AM
In the radiance equation we are dividing by cosine, as the view angle approaches 90º the radiance approaches infinity ...
The cosine is counter-balanced by the other terms in the formula (namely radiant flux). Radiance clearly doesn't tend to infinity as the angle approaches grazing incidence. See the Wikipedia talk page for details, someone has asked the same question.
“If I understand the standard right it is legal and safe to do this but the resulting value could be anything.”
Posted 06 May 2013 - 03:25 AM
I strongly disagree
The talk article says "The radiant flux for physical sources falls off at least as fast with angle as cos(?)"
"for physical sources" he is saying that _in real life_ it never happens ...
That is why I created the non-Lambertian geometry above
the surface is not a "physical source" it is constructed deliberately to isolate the cosine / projected area term in the radiance equation
and expose it for what it is - highly confusing!
What happens at 90° is not described in the radiance equation ... it is due to occlusion ... there is no radiance because you can not see the surface
The key to understanding radiance is understanding what happens when the viewing angle _approaches_ 90° ... the radiance approaches infinity
This is due to flux density:
Edited by skytiger, 06 May 2013 - 03:59 AM.
Posted 06 May 2013 - 07:11 AM
Here is another gamedev topic that agrees with me (very messily though ...)
http://www.gamedev.net/topic/533397-questions-about-radiance/
quoting:
That's right. Radiance gives the transmitted radiant power through the given solid angle per unit of projected area. That is why you need to divide by the projected area given by cos(theta)*dA. At a grazing angle the projected area approaches zero, hence the radiant power per unit of projected area approaches infinity.
Posted 06 May 2013 - 10:05 AM
Posted 06 May 2013 - 12:06 PM
I don't think you understand anything I have said
It is an isotropic surface - it radiates intensity equally in all directions
Lambertian surfaces do *not* radiate intensity equally in all directions - they radiate intensity proportional to cosine(viewing_angle)
Radiance depends on viewing angle
As the viewing angle increases the projected area of the emitter decreases and the radiance increases
When you COMBINE Lambertian reflectance's cosine term with radiance's 1/cosine term you get CONSTANT RADIANCE
You seem to believe that both radiance *and* intensity are constant!
your point that "radiance is independent of surface orientation" is simply completely wrong
radiance can ONLY be measured with respect to surface orientation!
Edited by skytiger, 06 May 2013 - 01:26 PM.
Posted 06 May 2013 - 12:08 PM
To reach agreement on this point we need to agree:
(1) that incident power is the same for both detectors
(2) that solid angle subtended by the detector is the same for both detectors
(3) that the area of the emitter does not change
(4) that the projected area of the emitter is smaller for the angled detector (by factor of 0.707)
(5) that radiance = flux / solid_angle / projected_area MUST be greater for the angled detector (by factor of 1.0 / 0.707 = 1.414)
Follow the worked example:
radiance = flux / solid_angle / projected_area
Given flux of 1.0 and solid_angle of 1.0:
For my isotropic surface:
non-angled
radiance = flux / solid_angle / 1.0 = 1.0
angled
radiance = flux / solid_angle / 0.707 = 1.414 // radiance varies
*If* it was Lambertian: (the flux would be reduced due to cosine term)
non-angled
radiance = flux / solid_angle / 1.0 = 1.0
angled
radiance = flux * 0.707 / solid_angle / 0.707 = 1.0 // radiance is constant
Posted 06 May 2013 - 12:49 PM
I think I understood some of it, but some things elude me. And I'm probably not alone.I don't think you understand anything I have said
One last remark, again, mostly for the grinning and/or confused bystanders:your point that "radiance is independent of surface orientation" is simply completely wrong
radiance can ONLY be measured with respect to surface orientation!
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