7

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 ...


7

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 surface is a ...


7

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 ...


6

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 ...


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 $\mathcal{M}'$. ...


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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