How do you calculate the weighting of a point inside of an equilateral triangle compared to its vertices?

Question

In an equilateral triangle that contains a point, how do you calculate 3 weights that sum to 100% and indicate how much influence each vertex has on the point.

When the point is in the center all the weights are 33%:

And if it's on one edge they should be split between the vertices that share that edge:

This is similar to how an HSL color wheel works:

Answer 1

I ended up using another method to solve this. In the diagram below to calculate the weight for a point, find the distance from the control point to the line opposite it and then divide by the triangle's height.

For the weight of A:

weightOfA = lengthOfx / triangleHeight

Answer 2

Let's think of three different points of the plane, $P$, $Q$ and $R$, not all co-linear, each with some coordinate vector in $\mathbb{R}^2$. If we only deal with two at first, $P$ and $Q$, we can write a one parameter interpolation as: $$ X = P(1-t)+Qt, $$ for $t\in[0,1]$. If you like, $100t$ gives you the percentage of $Q$'s weight, and $100(1-t)$ is the percentage of $P$'s weight. Also, if you pick any point $X\in\mathbb{R}^2$ which lies in the segment between $P$ and $Q$, there is only one value of $t$ such that $X = P(1-t)+Qt$, because the equation is linear.

Now, for the third point $R$. Since $P(1-t)+Qt$ already describes all the points in the $PQ$ segment, we interpolate this expression again with the point $R$, obtaining $$ X=[P(1-t)+Qt](1-s)+Rs. $$ for $s\in[0,1]$. Expanding, we get $$ X=P(1-t)(1-s)+Qt(1-s)+Rs. $$ It looks worse than before, but it is the same trick. Any point contained in the triangle $PQR$ can be uniquely identified with two values $t\in[0,1]$ and $s\in[0,1]$. And again, $100(1-t)(1-s)$ is the percentage weight of $P$, $100t(1-s)$ that of $Q$, and $100s$ that of $R$.

I hope this helps!

Answer 3

The mapping between Cartesian coordinates $(x, y)$ and Barycentric coordinates $(u, v, w)$ corresponding to a triangle with vertices $(x_0, y_0)$, $(x_1, y_1)$, and $(x_2, y_2)$ is trivial.

By definition, $u + v + w = 1$, and they directly correspond to the weights of their corresponding vertices. Using the formulae shown in this answer,

• $u$ corresponds to the weight of vertex $(x_1, y_1)$,

• $v$ corresponds to the weight of vertex $(x_2, y_2)$, and

• $w = 1 - u - v$ corresponds to the weight of vertex $(x_0, y_0)$.

Because the mapping is linear, this is also linear interpolation between the three vertices. It can be generalized to any simplex in $K$ dimensions, using $K+1$ barycentric coordinates. In 1D, the simplex is a line segment; in 2D, a triangle; in 3D, a tetrahedron; and so on.

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From Barycentric to Cartesian coordinates we have $$\left\lbrace \begin{aligned} x &= u x_1 + v x_2 + w x_0 \\ y &= u y_1 + v y_2 + w y_0 \\ \end{aligned} \right .$$ by definition; applying $w = 1 - u - v$ we get $$\left\lbrace \begin{aligned} x &= x_0 + u (x_1 - x_0) + v (x_2 - x_0) \\ y &= y_0 + u (y_1 - y_0) + v (y_2 - y_0) \\ \end{aligned} \right .$$ In practice, we often use first vertex and the two edge vectors from it instead, i.e. $$\left\lbrace \begin{aligned} x_u &= x_1 - x_0 \\ y_u &= y_1 - y_0 \\ \end{aligned} \right . , \quad \left\lbrace \begin{aligned} x_v &= x_2 - x_0 \\ y_v &= y_2 - y_0 \\ \end{aligned} \right . \tag{1a}\label{1a}$$ so that the $u$ axis is along the line between vertices $0$ and $1$, $v$ axis along the line between vertices $0$ and $2$, and the $w$ axis between vertices $1$ and $2$. Then, $$\left\lbrace \begin{aligned} x &= x_0 + u x_u + v x_v \\ y &= y_0 + u y_u + v y_v \\ \end{aligned} \right . \tag{1b}\label{1b}$$

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The inverse is $$\left\lbrace \begin{aligned} u &= \frac{(x - x_0) y_v - (y - y_0) x_v}{x_u y_v - x_v y_u} \\ v &= \frac{(y - y_0) x_u - (x - x_0) y_u}{x_u y_v - x_v y_u} \\ \end{aligned} \right.$$ where the divisor, $x_u y_v - x_v y_u$, is twice the area of the triangle, so if the triangle is degenerate (a line or a point), the divisor is zero, and there is no solution.

If you have many points to map to a given triangle, you can save significant computing effort by calculating $$\left\lbrace \begin{aligned} u_0 &= \frac{y_0 x_v - x_0 y_v}{x_u y_v - x_v y_u} \\ u_x &= \frac{y_v}{x_u y_v - x_v y_u} \\ u_y &= \frac{-x_v}{x_u y_v - x_v y_u} \\ \end{aligned} \right., \quad \left\lbrace \begin{aligned} v_0 &= \frac{x_0 y_u - y_0 x_v}{x_u y_v - x_v y_u} \\ v_x &= \frac{-y_u}{x_u y_v - x_v y_u} \\ v_y &= \frac{x_u}{x_u y_v - x_v y_u} \\ \end{aligned} \right. \tag{2a}\label{2a}$$ because then we have $$\left\lbrace \begin{aligned} u &= u_0 + x u_x + y u_y \\ v &= v_0 + x v_x + y v_y \\ \end{aligned} \right. \tag{2b}\label{2b}$$ and we have a very nice symmetry, too, between the two mapping operations.

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In object-oriented programming languages like Python or C++, create a class to represent a triangle. (In C, consider using a structure with room for the twelve precalculated parameters.) Whenever the Cartesian coordinates for the triangle are defined or modified, recalculate or update $x_0$, $y_0$, $x_u$, $y_u$, $x_v$, $y_v$, $u_0$, $u_x$, $u_y$, $v_0$, $v_x$, and $v_y$, and the class can map between Cartesian and Barycentric coordinates with just four multiplications and four additions per operation: extremely lightweight and efficient!

In $K$ dimensions, each mapping operation involves $K^2$ multiplications and additions, but is still quite lightweight and efficient.