I think it would be better to write it as 'interpolated surface attributes are affine function of x' rather than 'Barycentric coordinates are affine function of x'

dvernet

I believe we can also find the barycentric coordinates of a point $\vec{p}$ in a triangle by solving a system of homogeneous equations (though as Kayvon mentioned there are better ways to do it) on the vertices of the triangle and $\vec{p}$. For a point $\vec{p} = [p_{x}, p_{y}, p_{z}]^{T}$ in a triangle with vertices $\vec{v_{1}}$, $\vec{v_{2}}$, $\vec{v_{3}}$, to find its barycentric coordinates we could reduce the system to a set of homoegeneous equations as such:

This is due to the fact that nonzero vertices in $\mathbb{R}^{3}$ of a triangle form an affinely independent set in $\mathbb{R}^{3}$, so each point $\vec{p}$ in the triangle is guaranteed to have a unique representation as a convex combination of $\vec{v_{1}}, \vec{v_{2}}$, and $\vec{v_{3}}$. If you solve the above system for a point $\vec{p}$, and the entries given in the system are less than one and sum to 1, then the point is guaranteed to be in the triangle with the barycentric coordinates given by the system of equations.

BryceSummers

@PandaX I think that their is a bijection between barycentric coordinates and "surface attributes". Therefore, saying either statement implies the other.

lucida

Barycentric coordinates as scaled distances also follows directly from showing that Barycentric coordinates are ratios of triangle areas.

For example, we know that $\beta$ is equal to the ratio of the area of the triangle with vertices (a,c,x) to the area of the entire triangle.

Using the formula for area of a triangle (1/2 * height * base), the area of the triangle with vertices (a, c, x) is equal to 1/2 * the distance from x to c-a * the norm of c-a.

The area of the entire triangle is 1/2 * the distance from b to c-a * the norm of c-a.

The ratio of these two areas is thus distance from x to c-a / distance from b to c-a which is the scaled distance described on this slide.

I think it would be better to write it as 'interpolated surface attributes are affine function of x' rather than 'Barycentric coordinates are affine function of x'

I believe we can also find the barycentric coordinates of a point $\vec{p}$ in a triangle by solving a system of homogeneous equations (though as Kayvon mentioned there are better ways to do it) on the vertices of the triangle and $\vec{p}$. For a point $\vec{p} = [p_{x}, p_{y}, p_{z}]^{T}$ in a triangle with vertices $\vec{v_{1}}$, $\vec{v_{2}}$, $\vec{v_{3}}$, to find its barycentric coordinates we could reduce the system to a set of homoegeneous equations as such:

$$ \begin{bmatrix} \vec{v_{1}} & \vec{v_{2}} & \vec{v_{3}} & \vec{p} \\ 1 & 1 & 1 & 1 \end{bmatrix} $$

This is due to the fact that nonzero vertices in $\mathbb{R}^{3}$ of a triangle form an affinely independent set in $\mathbb{R}^{3}$, so each point $\vec{p}$ in the triangle is guaranteed to have a unique representation as a convex combination of $\vec{v_{1}}, \vec{v_{2}}$, and $\vec{v_{3}}$. If you solve the above system for a point $\vec{p}$, and the entries given in the system are less than one and sum to 1, then the point is guaranteed to be in the triangle with the barycentric coordinates given by the system of equations.

@PandaX I think that their is a bijection between barycentric coordinates and "surface attributes". Therefore, saying either statement implies the other.

Barycentric coordinates as scaled distances also follows directly from showing that Barycentric coordinates are ratios of triangle areas.

For example, we know that $\beta$ is equal to the ratio of the area of the triangle with vertices (a,c,x) to the area of the entire triangle.

Using the formula for area of a triangle (1/2 * height * base), the area of the triangle with vertices (a, c, x) is equal to 1/2 * the distance from x to c-a * the norm of c-a.

The area of the entire triangle is 1/2 * the distance from b to c-a * the norm of c-a.

The ratio of these two areas is thus distance from x to c-a / distance from b to c-a which is the scaled distance described on this slide.