Bézier triangle

Bézier triangle
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A cubic Bézier triangle is a surface with the equation

$(\alpha s+\beta t+\gamma u)^3\ \ \ |\ 0 \le s \le 1,\ \ \ 0 \le t \le 1,\ \ \ 0 \le u \le 1,\ \ \ s+t+u=1$

$=\begin{matrix} & & & \ \ \beta^3\ t^3 & & & \\ & & & & & & \\ & & +\ 3\alpha\beta^2\ st^2 & & +\ 3\beta^2\gamma\ t^2 u & & \\ & & & & & & \\ & +\ 3\alpha^2\beta\ s^2 t & & +\ 6\alpha\beta\gamma\ stu & & +\ 3\beta\gamma^2\ tu^2 & \\ & & & & & & \\ +\ \alpha^3\ s^3\ & & +\ 3\alpha^2\gamma\ s^2 u & & +\ 3\alpha\gamma^2\ su^2 & & +\ \gamma^3\ u^3 \end{matrix}$

where α³, β³, γ³, α²β, αβ², β²γ, βγ², αγ², α²γ and αβγ are the control points of the triangle.

An example Bézier triangle with control points marked

The corners of the triangle are the points α³, β³ and γ³. The edges of the triangle are themselves Bézier curves, with the same control points as the Bézier triangle.

It is also possible to create quadratic or other degrees of Bézier triangles, by changing the exponent in the original equation, in which case there will be more or fewer control points. With the exponent 1, the resulting Bézier triangle is actually a regular flat triangle. In all cases, the edges of the triangle will be Bézier curves of the same degree.

By removing the γu term, a regular Bézier curve results. Also, while not very useful for display on a physical computer screen, by adding extra terms, a Bézier tetrahedron or Bézier polytope results.

Due to the nature of the equation, the entire triangle will be contained within the volume surrounded by the control points, and affine transformations of the control points will correctly transform the whole triangle in the same way.

An advantage of Bézier triangles in computer graphics is, they are smooth, and can easily be approximated by regular triangles, by recursively dividing the Bézier triangle into two separate Bézier triangles, until they are considered sufficiently small, using only addition and division by two, not requiring any floating point arithmetic whatsoever.

The following computes the new control points for the half of the full Bézier triangle with the corner α³, a corner halfway along the Bézier curve between α³ and β³, and the third corner γ³.

$\begin{vmatrix} \boldsymbol{\alpha^3}'\\ \boldsymbol{\alpha^2\beta}'\\ \boldsymbol{\alpha\beta^2}'\\ \boldsymbol{\beta^3}'\\ \boldsymbol{\alpha^2\gamma}'\\ \boldsymbol{\alpha\beta\gamma}'\\ \boldsymbol{\beta^2\gamma}'\\ \boldsymbol{\alpha\gamma^2}'\\ \boldsymbol{\beta\gamma^2}'\\ \boldsymbol{\gamma^3}' \end{vmatrix}=\begin{vmatrix} 1&0&0&0&0&0&0&0&0&0\\ {1\over 2}&{1\over 2}&0&0&0&0&0&0&0&0\\ {1\over 4}&{2\over 4}&{1\over 4}&0&0&0&0&0&0&0\\ {1\over 8}&{3\over 8}&{3\over 8}&{1\over 8}&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0&0&0\\ 0&0&0&0&{1\over 2}&{1\over 2}&0&0&0&0\\ 0&0&0&0&{1\over 4}&{2\over 4}&{1\over 4}&0&0&0\\ 0&0&0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0&{1\over 2}&{1\over 2}&0\\ 0&0&0&0&0&0&0&0&0&1 \end{vmatrix}\cdot\begin{vmatrix} \boldsymbol{\alpha^3}\\ \boldsymbol{\alpha^2\beta}\\ \boldsymbol{\alpha\beta^2}\\ \boldsymbol{\beta^3}\\ \boldsymbol{\alpha^2\gamma}\\ \boldsymbol{\alpha\beta\gamma}\\ \boldsymbol{\beta^2\gamma}\\ \boldsymbol{\alpha\gamma^2}\\ \boldsymbol{\beta\gamma^2}\\ \boldsymbol{\gamma^3} \end{vmatrix}$

equivalently, using addition and division by two only,

		β³ := (αβ² + β³)/2
	αβ² := (α<sup>2β + αβ²)/2	β³ := (αβ² + β³)/2
α²β := (α³ + α²β)/2	αβ² := (α²β + αβ²)/2	β³ := (αβ² + β³)/2

		β²γ := (αβγ + β²γ)/2
αβγ := (α²γ + αβγ)/2		β²γ:=(αβγ+β²γ)/2

βγ² := (αγ² + βγ²)/2

where := means to replace the vector on the left with the vector on the right.

Note that halving a bézier triangle is similar to halving Bézier curves of all orders up to the order of the Bézier triangle.

A general nth-order Bezier triangle has (n + 1)(n + 2)/2 control points αⁱ β^j γ^k where i, j, k are nonnegative integers such that i + j + k = n. The surface is then defined as

$(\alpha s + \beta t + \gamma u)^n = \sum_\begin{smallmatrix} i+j+k=n \\ i,j,k \ge 0\end{smallmatrix} {n \choose i\ j\ k } s^i t^j u^k \alpha^i \beta^j \gamma^k = \sum_\begin{smallmatrix} i+j+k=n \\ i,j,k \ge 0\end{smallmatrix} \frac{n!}{i!j!k!} s^i t^j u^k \alpha^i \beta^j \gamma^k$

for all nonnegative real numbers s + t + u = 1.

See also