Bezier curves



              
The Bernstein polynomials that form a basis for Bezier curves, from kapittel 4.4.1 in Blending techniques in Curve and Surface constructions

A Bezier curve is expressed by the following formula:  \( C(t) = \displaystyle\sum_{i=0}^{d} b_{d,i}(t)\ c_i \), where \(d\) is the degree of the polynomial , and where  \(c_i\) are \(d+1\) control points and \(b_{d,i}(t)\) er \(d+1\) scalar Bernstein polynomials/basis functions. In the figure above, a set of Bernstein polynomials/basis functions is plotted. Note that the parameter domain is \([0,\ 1]\). In the figure below, we have a set of \(d+1\) control points \(p_i,\ i=0,1,...,d\) which are marked as gray pentagons. Thin black line segments are plotted between the points. Together, these line segments form a control polygon.

The plot is programmed according to chapter 4.4 in the book.

A 3rd degree Bezier curve will have the following formula in matrix form:

\( C(t) = \begin{bmatrix}1- t  & t \end{bmatrix} \begin{bmatrix} 1- t & t  & 0  \\ 0 & 1- t & t  \end{bmatrix} \begin{bmatrix} 1- t & t  & 0 & 0  \\ 0 & 1- t & t  & 0 \\ 0 & 0 & 1- t & t  \end{bmatrix} \begin{bmatrix} c_{0} \\ c_{1} \\ c_{2} \\ c_{3} \end{bmatrix}\),   when   \(t \in[0, 1]\),  se kapittel 4.4.2 i boka. If we want to increase the degree, we just add a new matrix to the right in front of the control points.

The matrix formula above can be written more compactly as \( C(t) = T^3(t)\ \mathbf{c} \), and we can calculate the 1st derivative in the usual way, i.e. with the kernel rule: \( C'(t) = 3\ T^2(t)\ T' \mathbf{c} \).  In matrix form it becomes:

\( C'(t) = 3 \begin{bmatrix}1- t  & t \end{bmatrix} \begin{bmatrix} 1- t & t  & 0  \\ 0 & 1- t & t \end{bmatrix} \begin{bmatrix} -1 & 1  & 0 & 0  \\ 0 & -1 & 1  & 0 \\ 0 & 0 & -1 & 1  \end{bmatrix} \begin{bmatrix} c_{0} \\ c_{1} \\ c_{2} \\ c_{3} \end{bmatrix}\),   when   \(t \in[0, 1]\),  see expression 4.36 i boka, which also shows the formulas for all derivatives.

If we multiply the 3 matrices in the matrix formula of the Bezier curve we get 4 Bernstein polynomials of degree 3: \((1-t)^3\),   \( 3t(1-t)^2\),   \(3t^2(1-t)\)   and   \(t^3\). Set up in a different matrix form, the Bernstein polynomials are:  \( \begin{bmatrix} 1 & -3 & 3 & -1\\ 0 & 3 & -6 & 3\\ 0 & 0 & 3 & 3\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\  t\\ t^2 \\ t^3 \end{bmatrix} \).  The matrix in the expression is then the basis change matrix from a monomial 3rd-degree basis to a Bezier basis, i.e. Bernstein polynomials of degree 3. See  table 4.1 in the book.