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In Schlick's 1994 paper, "An Inexpensive Model for Physically-Based Rendering", where they derive the approximation, the formula is: $$F_{\lambda}(u) = f_{\lambda} + (1 - f_{\lambda})(1 - u)^{5}$$ Where So, to answer your first question, $\theta$ refers to the angle between the view vector and the half vector. Consider for a minute that the ...

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The Fresnel coefficient should be evaluated using $H$, not $N$. You wrote, I have trouble seeing why we can still use that formula in a BRDF, which is supposed to approximate the integral over all the hemisphere. It's not. The BRDF in itself does not approximate the integral over all the hemisphere. The rendering equation does that: you integrate over ...

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Sample locations with a uniform pattern will create aliasing in the output, whenever there are geometric features of size comparable to or smaller than the sampling grid. That's the reason why "jaggies" exist: because images are made of a uniform square pixel grid, and when you render (for example) an angled line without antialiasing, it crosses rows/columns ...

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Monte Carlo methods rely on the law of large numbers, which states that the average of a random event repeated a large number of times converges toward the expected value (if you flip a coin a gazillion times, on average you will obtain each side half the time). Monte Carlo integration uses that law to evaluate an integral by averaging a large number of ...

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Triangulate the Voronoi cell then write the integral as a sum over the triangles: $$\int_{\Omega}\|P - Pi\|\,dP = \sum_{k=1}^{N}\int_{\Delta_k}\|P-P_i\|\,dP.$$ Write the integration over the triangle in barycebtric coordinates. Let the Jacobian of the transformation, from barycentric coordinates to physical coordinates, for triangle $k$ be $J_k$, and $|det(... 3$D(\omega)$is defined as the area ($m^2$unit in the numerator) of the microsurface with normals pointing in the direction$\omega$.$\mathcal{M}'$is defined as the portion of the microsurface with normals point in the direction$\omega \in \Omega'$. So it's natural that the integral of$D(\omega)$over$\Omega'$gives the area of$\mathcal{M}'$. ... 3 1)What is the V vector? How did they compute that It's the vector toward the camera (view vector), i.e. direction of the reflected ray. The lookup table they're building is parameterized in terms of NdotV, so they are working backward from NdotV to obtain a corresponding V vector. For purposes of doing this integration it doesn't matter what N and V are ... 2 Think of this way: when integrating uniformly over the hemisphere, it's like you are importance-sampling with a constant pdf of$1/2\pi$. The multiplication by$2\pi$at the end, then, can be seen as the division by the pdf, factored out of the sum because it's constant. The constant pdf is$1/2\pi$(as opposed to some other number) because it needs to be ... 1 I recommend to ignore my post (however I consider papers that I've sent valueable) and read post of Nathan Reed, which is much better in explaining this problem! I'm still a junior when it comes to computer graphics so you may want to take my opinion with a grain of salt: To start with - I used to learn/implement stuff using website that you've sent (OpenGL) ... 1 After some research i think i have final answer. General fog equation looks something like this: $$L(t)=L_p\color{red}{T(t)} + \int_0^t\color{blue}{T(x)}\color{green}{\sigma(x)L(x)}dx$$ Transmittance defined as: $$T(x)=e^{-\int_0^x\sigma(x) dx}$$ With const$\sigma$become $$T(x)=e^{-\sigma x}$$ So color that will be visible is: Visible percent$\color{...

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