skytiger 294 Report post Posted May 7, 2013 (edited) 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 May 7, 2013 by skytiger 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 7, 2013 (edited) 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 May 7, 2013 by skytiger 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 7, 2013 Hodgman: your top diagram is missing the point you are showing a smaller emitter area I am talking about a smaller emitter PROJECTED area 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 7, 2013 (edited) This doesn't make sense intuitively because Radiance is not a physical quantity and in real life surfaces are close to Lambertian Edited May 7, 2013 by skytiger 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 7, 2013 (edited) 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 May 7, 2013 by skytiger 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 7, 2013 (edited) 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 ? 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 ? to the solid angle subtended by the source as viewed from that detector Edited May 7, 2013 by skytiger 0 Share this post Link to post Share on other sites
David Neubelt 866 Report post Posted May 8, 2013 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. 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 8, 2013 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 0 Share this post Link to post Share on other sites
David Neubelt 866 Report post Posted May 8, 2013 (edited) 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 May 8, 2013 by David Neubelt 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 8, 2013 Exactly as the emitter area increases the flux increases which means the measured radiance *increases* 0 Share this post Link to post Share on other sites
David Neubelt 866 Report post Posted May 8, 2013 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 0 Share this post Link to post Share on other sites
Bacterius 13165 Report post Posted May 8, 2013 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: -- ? is the angle (with respect to the emitter's normal) at which the radiance is being emitted -- d²? is the radiant flux emitted by the surface (this is proportional to the solid angle being emitted into) -- d? 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: -- ? is the angle (with respect to the detector's normal) at which the radiance is being received -- d²? is the radiant flux received by the detector (again, proportional to the solid angle where the light is being received from) -- d? 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): -- ? is 0 // straightforward -- d²? 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 -- d? 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): -- ? is equal to 45 degrees // angle made with the emitter's surface normal -- d²? is equal to P * ((Ad * cos(?)) / Ae) // power over surface area Ae projected into cross-sectional area Ad * cos(?) -- d? 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(?)) / Ae)) / (Ae * Ad * cos(?)) = P / Ae^2 Top-right diagram (radiance measured as it falls on the detector): -- ? is 0 // straightforward -- d²? is equal to P * (Ad / Ae) // power over cross-sectional area Ae -- d? 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): -- ? is equal to 0 degrees // angle made with the detector's surface normal -- d²? 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(?) -- d? 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(?)) = 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(?) 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(?) 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.. 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 8, 2013 -- dA is equal to Ae // straightforward This is a typo? You meant: Ad is equal to Ae -- d²? is equal to P * ((Ad * cos(?)) / Ae) // power over surface area Ae projected into cross-sectional area Ad * cos(?) 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 -- d? 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(?) 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 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 8, 2013 d²? 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 0 Share this post Link to post Share on other sites
Bacterius 13165 Report post Posted May 8, 2013 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) 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 8, 2013 Yes I understand the diagram perfectlyIf you remove the Lambertian term cos(theta) from the numerator of the angled equationBy 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 ... 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 8, 2013 if it wasn't a Lambertian surface, radiance would increase with viewing angle 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 8, 2013 (edited) 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: http://www.oceanopticsbook.info/view/radiative_transfer_theory/level_2/the_lambertian_brdf Edited May 8, 2013 by skytiger 0 Share this post Link to post Share on other sites
Bummel 1888 Report post Posted May 8, 2013 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. 1 Share this post Link to post Share on other sites
Bacterius 13165 Report post Posted May 8, 2013 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. 0 Share this post Link to post Share on other sites
Bummel 1888 Report post Posted May 8, 2013 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. 0 Share this post Link to post Share on other sites
Hodgman 51234 Report post Posted May 8, 2013 (edited) 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. [edit] 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 May 9, 2013 by Hodgman 1 Share this post Link to post Share on other sites
Bacterius 13165 Report post Posted May 8, 2013 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. 0 Share this post Link to post Share on other sites
skytiger 294 Report post Posted May 8, 2013 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? 0 Share this post Link to post Share on other sites
Hodgman 51234 Report post Posted May 8, 2013 (edited) 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 May 8, 2013 by Hodgman 2 Share this post Link to post Share on other sites