Lecture 9 : Radiometry and Cameras

soohark

So, this diagram is in 2D, but in reality we would take the intersection line of the two planes and find vector perpendicular to this line on each plane. Then take the dot product of the two, right?

brianjo

This should work (up to a sign difference), but it's easier to just dot the plane normals, which you presumably have from the plane eqns.

Tianye

This illustration from CS148 might help understanding. We basically just need the angle between the ray and the surface normal.

clemire

Is there an error here? How did sin theta d-theta d-phi turn into d cos theta d-phi?

Tianye

Remember that $d cos\theta = -sin\theta d \theta$.

Therefore, $\int_0^{\pi}sin\theta d\theta = -\int_1^{-1}d cos\theta = \int_{-1}^1 d cos\theta$

The integration in terms of $\phi$ remains unchanged

wmonroe

If I'm understanding this correctly, the claim that radiance is constant along rays is another way of stating the steady-state assumption: if the distribution of light throughout the scene isn't changing, any photon at a particular point traveling in a particular direction is accompanied uniformly by other photons from the same source behind it and in front of it along the line of its direction of travel. If instead, for example, we were to simulate MIT's trillion-FPS camera experiment, which depended fundamentally on changes in light intensity with time, this assumption would have to be discarded.

mmp

That's a good intuition; showing that radiance is constant along rays basically involves this sort of approach to show photon densities are the same at different points along the ray.

Note, though, that steady state is a necessary but not sufficient condition. In a foggy environment, radiance isn't constant along rays, as some is absorbed by the fog, but it can still be in steady-state...

mgao12

I wonder what the relationship here between $L_i$ and $L_o$ is for reflection case. I think it may depend on the property of the surface. For a mirror, the strongest direction of reflection certainly follows the reflection law. However, diffuse reflection surface may result in uniform radiance in all directions. So does the "cos" relation still hold here?

Tianye

There are a lot of models describing the relationship between $L_i$ and $L_o$, which can be put into different categories including BRDF(model reflection), BTDF(model transmission) and BSSRDF(model both reflection and transmission). There are models as simple as the Phong reflection model (where the diffuse reflection property for different material is characterized by a constant, $L_o$ = $k_d$$L_i$cos$\theta_i$) and also ones way more complicated. Seems we will have several lectures to discuss these models...

clemire

I think in this case we would use a cosine measuring the difference between the outgoing ray in the direction we are computing and the ray reflected across the surface from the incident ray.

mmp

The cosine factor is still present regardless of how the surface is reflecting light (mirror, diffuse, etc). It turns out that it's there to account for the reduction in energy arriving at the surface from the light in the first place as the surface's orientation changes w.r.t. the incident light.

Hopefully this will be clear after Tuesday's lecture next week. :-)

soohark

hmm, how hard is it to extend this to recreate other lens effects like lens flare, bokkeh, and chromatic abberations? For example, with chromatic abberations I think I could do it by sending three seperate rays for rgb values and deflecting them at slightly different angles.

wmonroe

It looks like people have done these in previous years' final projects (one example is here). Bokkeh could be added simply by modifying the shape of the aperture (in Assignment 3 we assumed a circular aperture). Sending R, G, and B at slightly different angles would be a reasonable approximation; real chromatic aberration would create smoothly varying fringes for continuous spectra, which would require modeling the entire visible part of the spectrum instead of just R/G/B channels.

The link above goes into detail about lens flare. The most prominent effect can be created by modeling the realistic lens as a Fresnel reflector, with both reflection and transmission (we only allowed transmission). This gives the series of circles/hexagons that one normally notices in an image with lens flare. The student who did this project also added an element that I probably would not have thought to include: if the light is bright enough, you can see the diffraction of the light around the aperture in addition to internal reflections in the lens apparatus. This produces an interesting starburst effect that doesn't appear with reflection alone.

jingpu

The term of cosine theta' here is introduced because of the change of variable of integration. It took me some time to figure it out.

mmp

Zach kindly points out an error in this slide:

I think

d^2 = cos^2(theta) / ||p' - p ||^2

should be:

||p' - p ||^2 = d^2 / cos^2(theta)

or more simply:

||p' - p || = d / cos(theta)

bmild

What is d, particularly for the current assignment? In the paper they indicate that it is the distance to the exit pupil, but I couldn't figure out what that was supposed to be either. (I'm trying to use this to figure out the appropriate weight for GenerateRay to return.)

EDIT: Now I see that d is arbitrary since we are integrating over area, which scales with d^2, so the effect of scaling d will be canceled out by the dA' differential in the integral.