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I'm writing a Monte Carlo path tracer, and I'm trying to allow any mesh to be an emitter, but I'm not entirely sure about the probabilities to use when I sample them.

Right now, my algorithm for light sampling does the following:

  1. Pick a light $L$ in the scene with some PDF $p_{\mathcal{L}}$.
  2. Pick a random point $y$ on the mesh with uniform probability $p_{\mathcal{S}}(y) = \frac1{S_L}$, where $S_L$ is the total area of the mesh (computed summing over all the areas of all the triangles composing the mesh).
  3. Cast a shadow ray from the current position $x$ to the sampled point $y$.
  4. Check visibility.
  5. Compute the PDF wrt solid angle (I need it wrt to solid angle for my integrator); the PDF wrt area mesure should now be just $p_A(y) = p_{\mathcal{L}}(L) \cdot p_{\mathcal{S}}(y)$ (the choice of light and of a point on the light are independent events), hence the actual PDF I need is EDIT: not $p_\omega(\omega) = \frac{\cos \theta_y}{\lvert y-x \rvert ^2} p_A(y)$ but $p_\omega(\omega) = \frac{\lvert y-x \rvert ^2}{\cos \theta_y} p_A(y)$ :).
  6. Compute the light contribution to the current point as $\frac{E_L \cdot f_r \cdot \cos(\theta_x)}{p_\omega(y)}\cdot \mathrm{thr}$, where $\mathrm{thr}$ is the ray throughput, $E_L$ is the emissive value of the light $L$, $f_r = f_r(-\omega_{\mathrm{incoming}},\omega_{\mathrm{shadow}})$ is the BRDF of the surface containing $x$, $\omega_{\mathrm{incoming}}$ is the unit vector of the current ray direction, and $\omega_{\mathrm{shadow}} = \frac{y-x}{\lvert y-x \rvert ^2}$ is the direction of the shadow ray.
  7. Use that light contribution for next event estimation; use $p_\omega$ as my PDF for the NEE when computing the weights for multiple importance sampling.

I think this should account correctly for the fact that only the projected area of my mesh on the plane of the triangle $x$ lies on should contribute to the illumination of $x$, as the visibility check rules out all the triangles "hidden" by the geometry of the mesh, but I'd like to be sure that I'm not forgetting anything.

EDIT: In particular, the thing that makes me a bit unsure is using the entire surface area of the mesh, even though the part facing the point $x$ might be much smaller.

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  • $\begingroup$ How do you pick a point uniformly on the triangular mesh? Also $p_\omega(\omega)\,d\omega = p_\omega(\omega)\frac{\cos\theta_y}{\|y-x\|^2}\,dA = p_A(y)\,dA$ and then $p_\omega(\omega) = \frac{\|y-x\|^2}{\cos\theta_y}p_A(y)$, so your pdf was wrong unless I'm trippin. $\endgroup$
    – lightxbulb
    Nov 26 '21 at 11:32
  • $\begingroup$ Entire surface is correct since you filter out 0 contribution through the visibility function. It is like having a function $g$ that is nonzero in $[a,b]$ (i.e. $\operatorname{supp}(f) = [a,b]$) and instead of numerically approximating $\int_{a}^{b}g(t)\,dt$ you approximate $\int_{a}^{b+\delta}g(t)\,dt= \int_{a}^{b}g(t)\,dt + \int_{b}^{b+\delta}g(t)\,dt =\int_a^b g(t)\,dt$, with $\delta>0$. But as you can see those are equal (since $g((b,b+\delta]) = 0$) Granted you'll have higher variance when approximating $\int_a^{b+\delta}g(t)\,dt$ because some samples will end up with zero contribution. $\endgroup$
    – lightxbulb
    Nov 26 '21 at 11:48
  • $\begingroup$ @lightxbulb thanks, your second reply was exactly the thing I was missing. If you feel like writing an answer I will accept it. As for your first question 1) yes, I wrote the pdf wrong here (I was the one tripping actually, before editing it was right >.>) but I have implemented it correctly in the code; $\endgroup$
    – uhwo
    Nov 27 '21 at 21:13
  • $\begingroup$ @lightxbulb 2) to randomly sample the mesh, I select a random number $r \in [0,S_L)$, then select a triangle with the inversion method (so that the PDF for picking a triangle is weighted by the surface of each triangle). I have two arrays pre-computed at mesh instantiation: one with the surfaces of each triangle, and one with their cumulative sums (that's the CDF I'm inverting). I do a binary search for $r$ to get the biggest value $\leq r$ in the array with the cumulative sums and pick the corresponding triangle. Then I uniformly sample the triangle. $\endgroup$
    – uhwo
    Nov 27 '21 at 21:13
  • $\begingroup$ so that the PDF for picking a triangle is weighted by the surface of each triangle - I was worried that may not have been the case. It's all good if you're doing this (it's called inverse transform sampling btw). Note that if you non-uniformly rescale your mesh, the CDF becomes invalid. Also I hope you properly uniformly sample the triangle - there's some math in the Global Illumination Compendium, in PBRT, in ray-tracing gems, and in some other places. $\endgroup$
    – lightxbulb
    Nov 27 '21 at 21:35
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Sampling the entire surface would produce a correct result, even if some contributions are zero, as long as you properly account for those being zero. Consider a similar example in 1D, where you are given a function $g$ which is nonzero in $[a,b]$ (i.e. $\operatorname{supp}(g)=[a,b]$ - the support of $g$ is $[a,b]$). Say you want to numerically estimate the integral $\int_a^bg(t)\,dt$. Assume that for some reason you aren't able to generate samples in $[a,b]$ but can instead generate those in $[a,b+\delta],\, \delta>0$. Since $g\big((b,b+\delta]\big) = 0$ then it follows that:

$$\int_a^{b+\delta}g(t)\,dt = \int_a^bg(t)\,dt + \int_b^{b+\delta}g(t)\,dt = \int_a^bg(t)\,dt + \int_b^{b+\delta}0\,dt = \int_a^bg(t)\,dt.$$

Granted, when you numerically estimate $\int_a^{b+\delta}g(t)\,dt$ you will have higher variance since samples in $(b,b+\delta]$ would be used for estimating the zero function.

While the above example is not very realistic, the case in the rendering equation is a realistic one. There the visibility function $V(x,y)$ is zero over some of the domain of integration, but the non-zero part is hard to explicitly integrate over, since the evaluation of $V(x,y)$ generally requires a ray-tracing operation.

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