Physics and Math of Shading | SIGGRAPH Courses

[@jwallace6913]

invaluable resource - a lot of people give the equations and expect you to just deduce their practical purpose, or just blindly copy them.. it really helps to give an explanation for what F,G,D are actually doing here, and then we can better intuit the math

[@cyuxi]

He's so nice to give a clear explaination of the F G D factor in BRDF. Thank you Sir.

[@djdeluxe76]

Now this gets me even more motivated to pursue that PBR course at my university! Thanks!

[@andraskmeczo575]

Wait y'all have PBR courses?

[@djdeluxe76]

@@andraskmeczo575 yes when studying computer science master at my university PBR concepts were spread out over multiple courses - we had a math heavy course on global illumination one on computer graphics and interactive CG and one on CGI shading tools

[@andraskmeczo575]

@@djdeluxe76 which country is this? I don't think they have PBR courses in my country, but we'll see

[@djdeluxe76]

@@andraskmeczo575 that was in Germany

[@cosmic_gate476]

@@djdeluxe76 so lucky, I had to go out of my way to do the 2 graphics courses offered (and this is a huge university in Toronto). They were both general purpose (and quite detailed but nothing like this)

[@williamforbes6919]

This presentation is absolutely amazing!

[@ArpitAgarwal1]

This is a really useful talk for starting to learn about physics based rendering and material models. thanks

[@PixelPulse168]

Super cool Material for BRDF. Thanks for sharing

[@antoinedevldn]

A++ content! Thanks for sharing this lecture

[@majesticmack]

This is cool. Now I need to figure out how I can implement this in a shader.....

[@antoniobradiano]

look up shadermap4 is simpler

[@DGVFX]

Why am I seeing this at 3 a.m when I should be studing for my thesis?

[@Krunklehorn]

At 11:10 is the speaker implying that a surface could be polished so finely it would almost no longer exhibit the Fresnel effect?

[@dengdengEjoa]

this answers some of questions I had. Thanks a lot.

[@IntegralMoon]

This was awesome :) Thanks for this!

[@NatyHoffman]

You're very welcome!

[@IantoddusSardus]

Greetings mr. Hoffman and thank you. p.s. rip evolveband battleborn

[@husseinalsheikh1808]

thank you very much kind sir

[@williammauritzen4130]

When he says human vision is lossy. What I wonder is whether you had a side by side comparison we could tell the difference. It might appear to be the same color when we are looking at a picture, but would we be able to tell it apart in a "taste test?"

[@NatyHoffman]

You wouldn't be able to tell the difference, unless you were a tetrachromat (one of a a very small number of people with four cone types - Google "Concetta Antico" for an example). The exact mixes of R, G and B lasers needed to give you the same perception as a given broad-spectrum stimulus will vary slightly from person to person (people do differ slightly from the theoretical "CIE standard observer") but there *is* a combination which would be a perfect match to your eyes (your *unaided* eyes - if you looked through an optical filter which blocks some wavelengths and lets others through then you would see differences).

[@williammauritzen4130]

Until there's a study showing that we definitively can't tell the difference I can't take "you have to be tetrachromatic" as proof. The eye is capable of detecting a single photon. It wouldn't surprise me if it was capable of detecting some nuance. Especially given that any specific cell is going to be of slightly different pigment. We don't just create tests and studies to prove ourselves right. Maybe there is some bit of science I'm missing that makes this laughable. But the fact that it varies from person to person kind of undermines the idea that it's a "perfect" match.

[@NatyHoffman]

There is a huge body of studies showing exactly that - basically the entire corpus of color science over almost a century of research. There are studies both from the perceptual side (color matching studies) and the physiological side, on how the S, M and L cones produce a fundamentally 3D signal (assuming luminance is in the photopic range so rods don't contribute significantly) which feeds into another layer of neurons to produce the opponent signals (luminance, blue/yellow, red/green), and so on. The fact that people differ slightly in their color perception because their cone sensitivity curves are slightly different doesn't undermine the fact that each person has only three cone types (again, excluding anomalies like color-blind people, tetrachromats, etc.). If two different spectra produce exactly the same reaction in your S, M and L cones (which is the definition of metamerism), how *could* you tell the difference? Magic?

[@KnakuanaRka]

Naty Hoffman Or shining it through a prism and seeing unambiguously the mixture of wavelengths; you could tell, for example, yellow light from red and green light by this method, since the first will produce one yellow patch when shone through a prism and the other will produce a red and a green patch. Make it more numerical, and you have a spectrometer.

[@AssassinGrudge]

type the dress on youtube

[@antoniobradiano]

TY

[@RoelVanderVeken]

Very nice informative video, subscribed immediately :-)

[@romualdovillalobos9690]

Thanks!!

[@OZTVjjang]

The legend

[@BboyFTW1]

How do I get the Half Vector?

[@busTedOaS]

BboyFTW1 normalize(l+v)

[@raguaviva]

The light separating the protons from the neutrons @ 5m42s almost gave me a heart attack! Atoms don't get polarized by light, they get ionized, what gets polarized is the light (and not always).

[@NatyHoffman]

I'm not an optics expert, but the slide matches the references I used when working on the talk. From the 3rd edition of "Optical Scattering: Measurement and Analysis" by Stover, section 1.1: "The interaction of light... with matter can be viewed through the classical mechanism of polarization. The charged particles... are stretched to form dipoles under the influence of an EM field". From the 7th edition of "Principles of Optics" by Born & Wolf, beginning of section II: "An electromagnetic field produces at a given volume element certain amounts of polarization... Each volume element then becomes the source of a new secondary or scattered wavelet...". The 2nd edition of "Introduction to Modern Optics" by Fowles also discusses light propagation in dielectrics as polarization of the atoms by the EM field perturbations of the light (section 6.4).

[@clickdraw9762]

@@NatyHoffman Damn, what a reply. I must say, i am very impressed by your presentation. Im currently doing research for a bachelors degree, doing blender vizualizations, so your PBR introduction lecture is quite illuminating and inspiring. Hope to see your work in the future.

[@PrithvirajKhelkar]

here

[@pmrMountaineer]

speaking way tooo fast...

[@alexkzy5210]

uff

[@LooterLoser]

what

[@webgpu]

his "S" phoneme sounds funny