# Tag Info

15

Using two Fresnel terms is correct in the sense that any given diffuse path will pass through the surface twice. If you're solving diffusion by tracing a path through the medium until it bounces out again then that you will get two (or more) Fresnel terms for that path as it interacts with the surface. However, that's not what you're doing with a diffuse ...

12

DFG pops up in the family of microfacet based BRDFs. It is simply the product of three terms: D : The microfacet distribution. F : The fresnel coefficient. G : The geometric attenuation between microfacets. When someone says Cook-Torrance, they usually mean a microfacet BRDF where the distribution (D) is Beckmann, which I think is what the original Cook-...

11

As you already note, there is no clear cut interpretation/conversion for these values. I think it is even much worse: Depending on your BRDF and internal limitations (like having defined exponents ranging from 2-2048) the interpretation is completely different. Like suggested in the comments, it might be the best to render a series with different values and ...

9

In film production, we almost never use 8-bit textures for color/albedo, because of banding, etc. (JPEG is especially problematic since by spec, it's sRGB rather than linear values.) We either use 'half' (16 bit float) or 16-bit unsigned integer values for color/albedo textures.

9

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 ...

9

It's fine to use photometric units as an overall scale for setting light brightnesses. However, there's a technical subtlety you should be aware of. I'll quote from a blog post I wrote on the subject last year: With RGB images, it’s important to recognize that our display devices behave more radiometrically than photometrically. A red pixel value of 255, ...

8

While browsing to properly write my question, I actually found the answer, which happens to be very simple. Another Fresnel term is also going to weight in as the photons make their way out of the material (so being refracted into the air) and become the diffuse term. Thus the correct factor for the diffuse term would be: $$(1 - F_{in}) * (1 - F_{out})$$

8

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 ...

8

Yes, it's possible in some extreme cases for HDR lighting and tonemapping to expose banding issues in color textures. In those cases, having a higher bit depth for the textures could be useful. However, in my experience the majority of materials and ordinary lighting situations don't exhibit this problem, and most textures in a typical game are fine in 8-bit ...

7

I think I got it! Because $cos(\theta)$ integrates to $\pi$ over the hemisphere (and not $2\pi$). And the incoming light is multiplied by $cos(\theta)$ (and the BRDF).

7

You can combine Oren-Nayar with GGX, if your normalize the result. A BRDF is defined by two properties: Helmholtz reciprocity and energy conservation. $f(l_i, l,_o) = f(l_o, l_i)$ $f(l_i, l_o) \leq 1$ Your Oren-Nayar is the diffuse part $f_d(l_i, l_o)$ and your GGX is your specular part $f_s(l_i, l_o)$. If both are energy conserving, then both are at most ...

7

Using physically based BRDFs only makes sense if your entire pipeline is built for physical units - the extreme range of values can't be displayed properly without some form of tone mapping. You didn't include that part of the code but from the looks of it I'd say you're doing a simple clamp() followed by linear->sRGB conversion, which causes the bad ...

6

I don't know if this is exactly what you're looking for. I work tangentially in the film and TV industry. I don't work for a studio, but I work on the software that studios use for their productions. It's software that any user could buy if they wanted to, but it has been used in feature films, television shows, commercials, music videos, and more. About ...

6

I finally figured out a flaw in my argumentation to use the half vector for the diffuse part. tl;dr version: $\alpha_{hi}$ and $\alpha_{ho}$ are not equal, this assumption only works for the specular part. Therefore the energyconservation is not given. More correct: Per definition $\alpha_{hi} = \alpha_{ho}$, but you are not allowed to use them in the ...

6

You probably know that the BRDF is to calculate the reflected light, from a light source to a camera (In examples a light source and a camera is used, but it does not need to be just that). The property that you are talking about, basically says that when you swap the light source with the camera, it still gives the same result. Lets look at an example I ...

5

TL;DR: Your $G1$ formula is wrong. Just to avoid confusion, I am assuming the isotropic version of the BRDF, the Smith microfacet model (as opposed to the V-cavity model), and the GGX microfacet distribution. According to Heitz 2014, the masking/shadowing term $G1$ is $$\chi^{+}\left(\omega_{v}\cdot\omega_{m}\right) \frac{2}{1+\sqrt{1+\alpha_{o}^{2}\tan^... 5 I'd like to invite readers to read this article about Quake 2 engine rasterization technology explained in details, if they have the time. If TLDR, please pay attention to this image: What we see is the Albedo channel, that's what you want to encode in 16 bits if I understand your question correctly. I'm not going to say "if it could be encoded in 256 ... 5 Your main idea is more or less correct. The cosine hidden in the projected area measure dA^\perp = dA\cos(θ) compensates the weakening of irradiance due to incident angle (the Lambert's cosine law). This makes radiance independent from the incident angle. My guess is that the main motivation was to make it more practical to work with. The cosine in the ... 5 Yes, I think it's expected that setting roughness = 0, combined with using point lights for illumination, leads to no visible specular highlight. The size of the highlight is infinitesimally small, so the sample points (e.g. pixel centers) almost surely miss it. The math breaks down as well, as the reflectance would become infinite on the zero-sized ... 4 Found the solution, it turns out the lightVec is not the vector of light from the tube but rather the direction the tube will point. Therefore i will need to pass it a light rotation value to be used there. Results: 4 I think you need to split your question into real-time and offline PBR research. For real-time PBR it's mostly about finding fast approximations for path traced equivalent. For example we know how to implement area lighting with contact hardening shadows with a simple MC integrator, but finding fast approximation for this is a complex task requiring strong ... 4 I guess physicists, and before that term was used philosophers, have always been thinking about the nature of light, color and shadow. Knowledge about those matters is not purely rooted in people trying to create the ultimate rendering technique for CGI. In Wikipedia there are whole categories that show formulas and models that were invented to describe ... 3 The most straightforward method of researching physically based lighting is to grab a light and a camera and start taking pictures of an object from all angles when lit from all angles at varying intensities. Then plot out the color values and start fitting a function to that 4+ dimensional data. 3 Just to comment on the 0.16*Specular^2 term that was mentioned in the comments by Karim: Frostbite only remaps the specularity for their internal purpose so that they can pack gemstones specularity into the 8bit channel. They just lose some precision in favor of more variation. The specularity value is the original one when rendering. Check the Disney BRDF ... 3 The problem is in the diffuse term, which can be seen by making the specular portion of IBL not be added into the result. The diffuse only render will not have the darkening, and of course, the specular can't make the diffuse darker. (Note, fresnel vs no fresnel is not the issue here). The core problem is that lambertian diffuse is being used, which is ... 3 I just read notes on moving frostbite to pbr and I found the derivation of the method above. So I will just show the derivation here and quote some of the explanation. One can notice an extra〈n·l〉in the LD term as well as a different weighting 1/(∑Ni〈n·l〉). These empirical terms have been introduce by Karis to allows to improve the reconstructed ... 3 In Monte Carlo integration, the samples x_1, x_2, \ldots x_N are independent, identically-distributed random variables. This implies they all have the same expectation value. The derived quantities f(x_i)/p(x_i) also all have the same expectation value. So the expectation of a sum of N of these is the same as N times the expectation of any one of ... 3 Unfortunately, the iridescence model is not made to be applied to a diffuse term. Pascal and I made it for microfacet models only (that is the specular term). One way to understand how to include it to a game engine might be to look at Unity's HDRP implementation. In the Lit.hlsl to see how to incorporate the iridescence Fresnel into a specular + diffuse ... 2 You need to convert luminous flux (lumens, lm) to luminance (\frac{lm}{m^2sr}) for the rendering equation. For that you need to know the physical properties of the light source. For example luminance L of spherical light source of radius r radiating over 4\pi solid angle and luminous flux \phi has luminance:$$L = \frac{\phi}{4\pi^2r^2} I think ...

2

How should I define the radiance, if I have a directional light, for example the sun? The radiance in physics has different definitions in different books. I will use the definition with irradiance and radiant flux here. Imagine photons having some energy $Q$. Your radiant flux $\Phi$ is the amount of energy $Q$ (and therefore the amount of photons) per ...

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