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//****************** Shading Basics - February 4th, 2019 ********************//
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- "Reading: Lines/Hidden Surfaces (last week); Shading (today)"
    - This will be useful; my line-reading skills have been regressing lately, and I've always had some issues with seeing hidden surfaces
- "Technically on the schedule, we should be doing radiometry right now - but for concision's sake, we're going to skip it for now and jump straight into shading. We might come back to it later on in the course, but if you're interested, the textbook is always open!"
    - In other news, Professor Turk figured out how to disable automatic white balancing on the camera, so we can now read the paper he's writing on! Woohoo!
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- QUICK MATH REVIEW FACT:
    - If we have 2 unit vectors "u" and "v", then the dot product is:

        u*v = v*u = cos(theta)

    - This lets us check what the angle between two vectors is, which is VERY useful
- We can also calculate the SURFACE NORMAL of a polygon (i.e. a vector pointing perpendicularly "out" from the surface)
    - Say we've got a triangle ABC (where each one is a 3D point/vector) - in that case, we can do THIS:

        - vector1 = B - A
        - vector2 = C - A
        - Normal = crossProduct(vector1, vector2) = v1 X v2

- Now, let's switch gears to the topic of surface shading
    - When we look at a surface, it isn't just a single color; instead, it's darker or lighter based off of how it's interacting with the light
    - In particular, there are two BIG factors that affect how bright a surface is:
        - Where the light sources shining on the object are
        - Properties of the surface itself
            - There're a LOT of these, but for now, we'll only care about two of them: if the surface reflects light (and how much), and if it absorbs light (and how much)
    - If an object is DIFFUSE (or 'matte' or 'Lambertian'), it means that the surface reflects light equally in all directions
        - A good example of this is chalk, or paper (from most angles)
        - (cue several students excitedly looking at pieces of paper from various angles - thank you, Fresnel effect!)
- Now, imagine that there's a light source (like the sun) where all the photons are coming from a single direction
    - If we hold the paper flat up against the light, a set amount of photons are hitting it in a 2D-area 'a'
    - If we tilt that paper, then the effective surface area where the same amount of photons hits has increased to 'a/cos(theta)' - i.e., the same amount of light has to cover a larger area, so the light will seem dimmer!

- So, to figure out how much light we should draw on the surface, we need to know how much light is hitting the surface, and how much of that light is reaching our eyes
    - Since a diffuse surface reflects light equally in all directions, the position of our eye won't matter when drawing it - but it WILL matter for reflective surfaces
    - So, for a given point on our surface, let's say we know a) the unit-length normal vector N at that point, and b) the unit vector L pointing TOWARDS the light source we want to worry about
        - We can get a vector pointing towards the light pretty easily, by saying "L = lightPos - surfacePos"
        - Using our dot product, we can calculate the angle between the two vectors as:

                N*L = cos(theta)

            - If theta is small, it means the light is striking the surface almost head-on, so the surface will be bright
            - If theta is large, it means the light is just glancing the surface, so the surface will be dim
        - So, this quantity represents approximately how much light is hitting our surface!
- If we then also know the color "Cs" of our surface, and the color "Cl" of our light, we can multiply it by our brightness factor to get what the color the surface should be! Putting this together, we get an equation:

        final surface color = Cs * Cl * (N*L)

    - ...where Cs is the surface color, Cl is the light color, N*L is how much light is hitting the surface, and 'b' is a scaling factor 
    - Since we have to deal with RGB color, we'll do this equation separately for the red, blue, and green channels; so REALLY, we have a matrix that does this 3 separate times:

        C = (Cr, Cg, Cb) =  [Csr * Clr * (N*L)]
                            [Csg * Clg * (N*L)]
                            [Csb * Clb * (N*L)]

- However, what if the angle between the vectors (i.e. the dot product) is negative? Well, it means that the light is BEHIND the object, so no light should be hitting the surface! Therefore, our equation should actually be:

        C = Cs * Cl * max(0, N*L)

- However, is it realistic for objects in shadow to be COMPLETELY pitch black? No! In the real world, light bounces around and can provide "indirect illumination," even to objects the light source can't directly "see"
    - Calculating all those bounces, however, is pretty expensive - so instead, we'll often just fake this indirect light by adding some constant ambient light color 'Ca' to our objects so shadows aren't just pure black:

            C = Cs * (Ca + Cl*max(0, N*L))

        - For a number of years, calculating indirect illumination at all, in fact, was pretty difficult...but today, there are several (computation intensive) well-understood methods for doing it correctly, including:
            - Photon mapping
            - Metropolis light transport
            - Radiosity

- So, those're the basics of Lambertian diffuse shading - but what should we do for SPECULAR (i.e. glossy) surfaces, like plastics and metals? 
    - Well, there are a number of different ways of doing this, but one of the simplest and most common is the PHONG-BLINN ILLUMINATION model
        - Again, let's say we've got a surface normal N and our light-pointing vector L again
        - But now, let's ALSO suppose we have a vector "R" that's the mirrored version of L (flipped about the normal vector), and a vector "E" for which way our eye/camera is pointed
            - If the eye is pointed towards R (i.e. angle between E and R is small), then it should see the full glare of the reflection
            - If the eye is pointed perpendicularly to R, it shouldn't see any glare at all
            - In-between, it'll drop off by some pretty high amount
        - So, one way of representing the reflected light is like this:

                cos(angle between E and R)^(specular power)
                    = (E*R)^P

        - ...which'd give us a final PHONG ILLUMINATION color of:

                C = Cl * max(0, (E*R)^P)

    - This should give us the SPECULAR HIGHLIGHTS (i.e. bright spots) we see on shiny objects like glossy plastics

- So, using these two techniques (and some trial and error), we can simulate a wide variety of materials - cool! However, there's still plenty we can do to make this look better (early 3D animations had a reputation for everything looking plastic, for instance), so we'll keep pushing forward in the next lecture.