Book 3.4: Sphere Importance Sampling

[rupsis]

Perhaps I'm just slow, but for Listing 16, the f() function took me a little while to figure out.

double f(const vec3& d) {
    auto cosine_squared = d.z()*d.z();
    return cosine_squared;
}

If I understand correctly, $cos(\theta)^2$ here is $(\cos(0^\circ) *z)^2$ (where Z is from the random_unit_vector)
and since $cos(0^\circ) = 1$ we're left with d.z()*d.z().

The text doesn't make that obvious. Perhaps (and very likely) the issue is me. But if my understanding is correct, maybe some clarification could be useful. A small attempt on my part:

[hollasch]

Good point. I don't know if I would have made the connection myself. I would actually expound on this a bit further. Something like:

If the integrand is $\cos^2(\theta)$, and $\theta$ is the angle with the $z$ axis, then $\theta = 0$, and $\cos^2(0) = 1$.

I'm still missing where the $z$ comes in, though. Why is $\cos^2(\theta) = (\cos(0\deg)\cdot z)^2$?

@trevordblack? @petershirley?

[dimitry-ishenko]

@rupsis @hollasch I think you guys got this wrong. 😺

From scalar projection we know that projection of vector $a$ onto $b$ is equal to the length of $a$ multiplied by the cosine of angle $\theta$ between $a$ and $b$.

In our case we have random vector $d$ (of unit length) and from the chapter we know that " $\theta$ is the angle with the $z$ axis." Therefore we can say that projection of $d$ onto axis $z$ is equal $d_z = \lVert d \rVert \cos \theta = 1 \cdot \cos \theta$.

So, in our function $f(\theta, \phi) = \cos^2 (\theta)$ we can replace $\cos \theta$ with $d_z$ and get $f(\theta, \phi) = f(d) = {d_z}^2$.

[hollasch]

See also #1536. Things are a bit muddy in this region of the text.