• Introduction
• 2D
  • Uniform disk
  • Directions
    • Uniform Unit Hemisphere
    • Power Cosine Weighted Unit Hemisphere Sector
    • Power Cosine Weighted Unit Hemisphere Cap
    • Cosine Weighted Unit Hemisphere
  • Normals
    • Beckmann Distribution
    • Phong Distribution
    • GGX Distribution

Uniform Unit Hemisphere

Sampling of points uniformly on the hemisphere. As with the disk, just randomly choosing two angles won’t give the desired result

function uniform_angular_hemisphere(){


    const u_1 = Math.random();
    const u_2 = Math.random();

    const theta = u_1 * Math.PI / 2.0;
    const phi = u_2 * 2.0 * Math.PI;

    return [theta,phi]
}

Now the result with a uniform area sampling of the upper hemisphere

function uniform_hemisphere(){

    const u_1 = Math.random();
    const u_2 = Math.random();

    // Math.acos(u_1) is also valid
    const theta = Math.acos(1.0 - u_1);
    const phi = u_2 * 2.0 * Math.PI;

    return [theta,phi]
}

Algorithm

Generate uniform points on the hemisphere

1. Choose uniform numbers $u_1, u_2$ in $[0,1]$
2. Calculate $\theta = \cos^{-1}(1-u_1)$ or $\theta = \cos^{-1}(u_1)$
3. Calculate $\phi = 2\pi u_2$

Optionally convert to cartesian coordinates with:

$$\begin{aligned} x &= \sin\theta\cos\phi\\ y &= \sin\theta\sin\phi \\ z &= \cos\theta \end{aligned}$$

PDF Used In Monte Carlo Integration

$$p(\omega) = \frac{1}{2\pi}$$

$$p(\theta, \phi) = \frac{1}{2\pi}\sin\theta$$

Derivation

This is a special case of the general form discussed in the next section, but is shown as an example in more detail.

We want the pdf to be uniform over the unit hemisphere, so it will just be some constant for each solid angle $\Rightarrow p(\omega) = c$

$$\int_\Omega p(\omega)d\omega = \int_\Omega c d\omega = c \int_\Omega d\omega = 1$$

The easy way:

We know that the solid angle of the hemisphere is $2\pi$ . In general, for uniform sampling the pdf is just $\frac{1}{\operatorname{Vol}(\Omega)}$ , where $\operatorname{Vol}$ is the volume of the sample space.

$$c\int_\Omega d\omega \Rightarrow c2\pi = 1 \Rightarrow c = \frac{1}{2\pi}$$

The complicated way:

$$\begin{aligned} \int_\Omega p(\omega)d\omega &= \int_0^{\frac{\pi}{2}}\int_0^{2\pi}c\sin\theta d\theta d\phi \\ &= c 2\pi \int_0^{\frac{\pi}{2}}\sin\theta d\theta \\ &= c2\pi [-\cos\theta]_0^{\frac{\pi}{2}} \\ &= c2\pi (\cos 0 - \cos\frac{\pi}{2}) \\ &= c2\pi \end{aligned}$$

Therefore $c2\pi = 1 \Rightarrow c = \frac{1}{2\pi}$

To convert to the angular representation, we have to add (The factor can be found similarly to the $r$ factor of the disk)

$$p(\omega) = p(\theta,\phi)\sin\theta$$

Marginal And Conditional Densities:

$$p_\theta(\theta) = \int_0^{2\pi} p(\theta,\phi)\sin\theta d\phi = \int_0^{2\pi}\frac{1}{2\pi}\sin\theta d\phi = \frac{1}{2\pi}\sin\theta * 2\pi = \sin\theta$$

$$p(\phi \vert \theta) = \frac{p(\theta,\phi)}{p_\theta(\theta)} = \frac{\sin\theta}{2\pi\sin\theta} = \frac{1}{2\pi}$$

Compute CDFs

$$P_\theta(\theta) = \int_0^\theta \sin\theta' d\theta' = [-\cos(\theta')]_0^\theta = \cos 0 - \cos\theta = 1 - \cos\theta$$

$$P(\phi \vert \theta) = \int_0^\phi p(\phi' \vert \theta)d\phi' = \int_0^\phi \frac{1}{2\pi}d\phi' = \frac{\phi}{2\pi}$$

Invert CDFs

$$u_1 = P_\theta(\theta) = 1 - \cos\theta \Rightarrow \cos\theta = 1 - u_1 \Rightarrow \theta = \cos^{-1}(1- u_1)$$

$$u_2 = P(\phi \vert \theta) = \frac{\phi}{\theta} \Rightarrow 2\pi u_2 = \phi$$