This question often pops up, when you need a random direction vector to place things in 3D or you want to do a particle simulation.

We recall that a 3D unit-sphere (and hence a direction) is parametrized only by two variables; elevation \theta \in [0; \pi] and azimuth \varphi \in [0; 2\,\pi] which can be converted to Cartesian coordinates as

\begin{aligned} x &= \sin\theta \, \cos\varphi \\ y &= \sin\theta \, \sin\varphi \\ z &= \cos\theta \end{aligned}If one takes the easy way and uniformly samples this parametrization in numpy like

```
phi = np.random.rand() * 2 * np.pi
theta = np.random.rand() * np.pi
```

One (i.e. you as you are reading this) ends with something like this:

While the 2D surface of polar coordinates uniformly sampled (left), we observe a bias of sampling density towards the poles when projecting to the Cartesian coordinates (right).

The issue is that the cos mapping of the elevation has an uneven step size in *Cartesian* space, as you can easily verify: cos^{'}(x) = sin(x).

The solution is to simply sample the elevation in the Cartesian space instead of the spherical space – i.e. sampling z \in [-1; 1]. From that we can get back to our elevation as \theta = \arccos z:

```
z = 1 - np.random.rand() * 2 # convert rand() range 0..1 to -1..1
theta = np.arccos(z)
```

As desired, this compensates the spherical coordinates such that we end up with uniform sampling in the Cartesian space:

## Custom opening angle

If you want to further restrict the opening angle instead of sampling the full sphere you can also easily extend the above. Here, you must re-map the cos values from [1; -1] to [0; 2] as

```
cart_range = -np.cos(angle) + 1 # maximal range in cartesian coords
z = 1 - np.random.rand() * cart_range
theta = np.arccos(z)
```

## Optimized computation

If you do not actually need the parameters \theta, \varphi, you can spare some trigonometric functions by using \sin \theta = \sqrt { 1 - z^2} as

\begin{aligned} x &= \sqrt { 1 - z^2} \, \cos\varphi \\ y &= \sqrt { 1 - z^2} \, \sin\varphi \end{aligned}