Timeline for Where does sin(theta) go in estimators of The Rendering Equation?
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Jun 30, 2023 at 11:15 | vote | accept | Ocelot | ||
Jun 30, 2023 at 1:55 | answer | added | lightxbulb | timeline score: 1 | |
Jun 29, 2023 at 23:13 | comment | added | Ocelot | @lightxbulb I would really appreciate a fully-fledged answer with detailed explanations, which I'll upvote and accept. I cannot do so with a comment that mostly puts a blame on me. | |
Jun 29, 2023 at 21:21 | comment | added | lightxbulb | It's because you very sloppily failed to define the measure associated with your probability density function. Since the $|\sin\theta|$ is missing, for this to be correct the density must be defined w.r.t. the solid angle measure. For clarity and correctness I would have written instead: $I\approx \frac{1}{N}\sum_{k=1}^N \frac{albedo}{\pi} \frac{\cos\theta_k \sin\theta_k}{p(\phi_k, \theta_k)}$ where $(\phi_k, \theta_k)$ are distributed according to $p$ (wrt $d\phi d\theta$). In fact $p$ does not need to include a sine term if you pick other mappings, and would thus not cancel in general. | |
Jun 29, 2023 at 18:30 | comment | added | Ocelot | @lightxbulb I'm aware of that, what I'm asking about is why for the MC estimator we don't include the metric tensor of the spherical-to-cartesian mapping. Does this imply we don't use the mapping at all? | |
Jun 29, 2023 at 17:17 | comment | added | lightxbulb | It accounts for the term that allows you to rewrite the integral over the hemisphere as an iterated integral. The iterated integral is over a rectangle $[0,2\pi)\times [0,\pi/2]$, this rectangle gets mapped to the hemisphere using spherical coordinates. Now if you take a small square at a point from this rectangle it gets mapped to a tangent parallelogram at the corresponding point on the sphere. The $|\sin\theta|$ term is the area of that parallelogram. It's basically a sirface integration term. You can derive it as $\sqrt{det[J^TJ]}$ where $J$ is the Jacobian of spherical coords wrt angles. | |
Jun 29, 2023 at 15:50 | answer | added | Hubble | timeline score: 1 | |
Jun 29, 2023 at 13:17 | history | asked | Ocelot | CC BY-SA 4.0 |