Derivation for equation (8)

This equation tell us how to construct a macrosurface BRDF given microsurface ‘s D, G, F.

In this equation,

is incident vector.

is outgoing vector.

is macrosurface normal.

is microsurface normal.

is the macrosurface BRDF.

is the microsurface BRDF.

is microfacet distribution function.

is shadowing-masking function.

is solid angle in the hemisphere .

How can we get this equation? Please see the figure below.

In this figure, the surface is illuminated by a light source and an observer is looking at the surface. The observer has a microscope so that he will see the microfacets s₁, s₂, ... . These microfacets have different colors because of their orientations are different. When the observer look at the surface at the same location, but without the microscope, he will no longer see the microfacets but a uniform color. This time, he knows that it is colors from microfacets that mix together and form the uniform color. Let‘s denote the colors from microfacets as

In this equation, is the projected area of s_i . According to the definition of microfacet distribution function, we have:

In this equation,

is the area of macrosurface.

is i-th microfacet‘s normal.

is a small solid angle aligned with .

is microfacet distribution function.

Combine these equations, we have:

We can eliminate and using the equation (3) in the paper:

Convert sum to integral, then we get:

Now we can see the term . Let‘s investigate the term further:

In this equation,

is a small solid angle of incident light over the hemisphere .

Finally, we have:

According to the rendering equation

We can regard the inner integral as the equivalent BRDF for macrosurface, so that we get:

Confirm equation (9)

According to the definition of radiance:

In this equation,

is the radiance in outgoing direction

is luminous flux

is area of macrofacet

So the outgoing irradiance is:

Put equation (9) in, we have:

According to equation (10),

So According to equation (9), the overall outgoing irradiance equals the incoming irradiance scaled by a factor , which is less than 1.

Derivation for equation (20)

According to equation (8),

Put equation (15) in it, we have:

When , , then according to equation (10), we have:

Derivation for equation (42)

Let us take a careful look at the definition of .

Suppose there is one point on surface whose normal is , we construct a plane perpendicular to the normal and choose two perpendicular axes and . For a small patch on the plane, we denote the direction pointing from to it , and the small solid angle it occupies .

According to equation (4), which is

We can consider as the probability of finding a microfacet whose normal is inside , so that we have

That‘s exactly what is.

Derivation for equation (45)

Break the ray into many short segments, each with projected length . According to the paper, the probability that the ray is first blocked in segment is , so the probability that ray is always unblocked is:

Then we have:

From calculus we know:

So

So that

Derivation for equation (46)

Let‘s consider the situation that a ray intersects with an short and straight surface segment . In order to do that, the surface height should below the ray at and above the ray at . For a given slope , there exist a set of surface segments that fulfill this condition, which are in the shaded areas in the figures below.

It‘s easy to note that the possible surface height at varies from to . So given a surface with slope , the probability that it intersects with a ray is

In this equation,

is the probability density that surface height reaches at point .

Also we know that the probability that a surface segment has slope is

Combine them, then we get the probability for finding a surface segment with slope as well as intersecting with the ray:

Consider all possible surface slope , the probability that they intersect with the ray is:

Among all surface segments with slope , the probability that a surface segment below the ray is:

Consider all possible surface slope , the probability that they below the ray is:

Assume is independent from , we have:

So the probability that a ray first intersects with surface in is:

Let , so

Derivation for equation (49)

From equation (48), we have:

Note that

So

Then

Put it in equation (45):

Derivation for equation (50)

Notes for GGX paper