# Barycentric Coordinate for Surface Sampling

To convert a mesh into a point cloud, one has to sample points that can uniformly cover the surface. To do so, one must choose the number of samples proportional to the area of a face (polygon).

First, we iterate through all polygons, splitting them into triangles since measuring the area of a triangle is much easier than computing the area of a polygon. Then, for each triangle, we compute its area and store it in an array. Next, we select a triangle with probability proportional to its area.

Sampling uniformly on a triangle requires a trick: Barycentric coordinate system.

## Barycentric Coordinate System

The Barycentric coordinate is a way to define a point on a $n$ dimensional simplex using $n$ coordinates. You define $n$ weights $[a_1 ,…, a_n]$ to denote a point $p= a_1 \mathbf{x}_1 + \cdots + a_n \mathbf{x}_n$. It is just a convex hull of the simplex.

Then, for each selected triangle (let’s denote the vertices of the triangle as $A, B, C$ ), we sample a point on its surface by generating two random numbers, $r_1$ and $r_2$ between 0 and 1, and compute the coordinate using the equation on the next section.

## Computing the Area of Faces

First, you have to convert all polygons into triangles. You can easily do this in many commercial mesh converters.

Let the vertices of a triangle $A, B, C$ in 3D coorindate as $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$. Then, the area of the triangle formed by the three vectors is

\[\frac12 \| (\mathbf{a} - \mathbf{c}) \times (\mathbf{b} - \mathbf{c}) \|_2\]## Setting the number of samples for each face

Just sum up all the triangle areas and convert them into a probability distribution. If you multiply the number of points you want by the sample to the distribution, you will get the number of samples per face.

## Sampling points using the Barycentric coordinate

For two random variables $r_1, r_2$ uniformly distributed from 0 to 1, we sample a new point $d$ as follows.

\[\mathbf{d} = (1 - \sqrt{r_1})\mathbf{a} + \sqrt{r_1} (1 - r_2) \mathbf{b} + \sqrt{r_1} r_2 \mathbf{c}\]Intuitively, $r_1$ sets the percentage from vertex $a$ to the opposing edge. Taking the square-root of $r_1$ gives a uniform random point with respect to surface area.

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