38. Bezier Curves
Dated: 05-07-2025
In the early 1960s, Peter Bezier began looking for a better way to define curves and surfaces one that would be useful to a design engineer.
A Geometric Construction
Select 3 points \(A, B\) and \(C\) .
Select 2 points \(D\) and \(E\) such that \(\frac {|\overline{AD}|}{|\overline{AB}|} = \frac {|\overline{BE}|}{|\overline{BC}|} = u_i\)
Select \(F\) such that \(\frac {|\overline{DF}|}{|\overline{FE}|} = u_i\) , the curve is traced by \(F\) .
Let \(A = (x_A, y_A)\) and \(B = (x_B, y_B)\) then coordinates of \(D\) and \(E\) are
\[
\begin{aligned}
x_D = x_A + u_i(x_B - x_A) \\
y_D = y_A + u_i(y_B - y_A)
\end{aligned}
\]
\[
\begin{aligned}
x_E = x_B + u_i(x_C - x_B) \\
y_E = y_B + u_i(y_C - y_B)
\end{aligned}
\]
Similarly, for \(F\) , we have
\[
\begin{aligned}
x_F = x_D + u_i(x_E - x_D) \\
y_F = y_D + u_i(y_E - y_D)
\end{aligned}
\]
Substituting coordinates for \(D\) and \(E\) in \(F\) and after simplification, we have
\[
\begin{aligned}
x_F = (1-u_i)^2 x_A + 2u_i(1-u_i) x_B + u_i^2 x_C \\
y_F = (1-u_i)^2 y_A + 2u_i(1-u_i) y_B + u_i^2 y_C
\end{aligned}
\]
To generalize
\[
\begin{aligned}
x(u) = x_F \\
y(u) = y_F \\
\end{aligned}
\]
\[
\begin{array}{ccc}
x_0 = x_A & x_1 = x_B & x_2 = x_C \\
y_0 = y_A & y_1 = y_B & y_2 = y_C \\
\end{array}
\]
\[
\begin{aligned}
x(u)=(1-u)^2 x_0 + 2u(1-u)x_1 + u^2 x_2 \\
y(u)=(1-u)^2 y_0 + 2u(1-u)y_1 + u^2 y_2
\end{aligned}
\]
\[
p(u) = (1-u)^2 p_0 + 2u(1-u)p_1 + u^2 p_2
\]
Where \(A = p_0, B = p_1, C = p_2\) .
This is the setBezier curve based on our construction.
The degree of a Bezier curve is equal to \(n - 1\) , where \(n\) is the number of control points.
Cubic version
\(p(u) = (1-u)^3 p_0 + 3u(1-u)^2 p_1 + 3u^2(1-u)p_2 + u^3 p_3\)
\[p(u) = \sum_{i = 0}^n \binom{n}{i} u^i(1 - u)^{n - i} p_i \quad n = \text{Degree of Bezier curve}\]
An Algebraic Definition
Bezier began with the idea that any point \(p(u)\) on a curve segment should be given by an equation such as the following
\[
p(u) = \sum_{i=0}^n p_i f_i(u)
\]
Source: gamemath.com
The \(f_i(u)\) has following important characteristics.
The curve must start on the first control point, \(p_0\) , and end on the last, \(p_n\) . Mathematically, we say that the functions must interpolate these two points.
The curve must be tangentline\(p_1 - p_0\) at \(p_0\) or \(p_n - p_{n - 1}\) at \(p_n\) .
The functions\(f_i(u)\) must be symmetric with respect to \(u\) and \((1 - u)\) . This lets us reverse the sequence of control points without changing the shape of the curve.
Effect of reversing the order of control points on the curve.
We can rewrite the equation as
\[
p(u) = \sum_{i=0}^n p_i B_{i,n}(u)
\]
\[
B_{i,n}(u) = \binom{n}{i} u^i (1-u)^{n-i}
\]
where
\[
\binom{n}{i} = \frac{n!}{i!(n-i)!}
\]
Expanding this equation for second degree Bezier curve, we have
\[
p(u) = p_0 B_{0,2}(u) + p_1 B_{1,2}(u) + p_2 B_{2,2}(u)
\]
\[
\begin{aligned}
B_{0,2}(u) & = (1-u)^2 \\
B_{1,2}(u) & = 2u(1-u) \\
B_{2,2}(u) & = u^2 \\
\end{aligned}
\]
similarly, for \(n = 3\)
\[
p(u) = p_0 B_{0,3}(u) + p_1 B_{1,3}(u) + p_2 B_{2,3}(u) + p_3 B_{3,3}(u)
\]
\[
\implies
p(u) = (1-u)^3 p_0 + 3u(1-u)^2 p_1 + 3u^2(1-u)p_2 + u^3 p_3
\]
\[
\implies
\begin{aligned}
B_{0,3}(u) & = (1-u)^3 \\
B_{1,3}(u) & = 3u(1-u)^2 \\
B_{2,3}(u) & = 3u^2(1-u) \\
B_{3,3}(u) & = u^3
\end{aligned}
\]
\[
\implies
p(u)
=
\begin{bmatrix}
(1-u)^3 \\
3u(1-u)^2 \\
3u^2(1-u) \\
u^3
\end{bmatrix}^T
\begin{bmatrix}
p_0 \\
p_1 \\
p_2 \\
p_3 \\
\end{bmatrix}
\]
\[
\begin{bmatrix}
(1-3u+3u^2-u^3) \\
(3u-6u^2+3u^3) \\
(3u^2-3u^3) \\
u^3
\end{bmatrix}^T
\begin{bmatrix}
p_0 \\
p_1 \\
p_2 \\
p_3
\end{bmatrix}
\]
\[
\begin{array}{cccc}
p(u) =
&
\begin{bmatrix}
u^3 & u^2 & u & 1
\end{bmatrix}
&
\begin{bmatrix}
-1 & 3 & -3 & 1 \\
3 & -6 & 3 & 0 \\
-3 & 3 & 0 & 0 \\
1 & 0 & 0 & 0
\end{bmatrix}
&
\begin{bmatrix}
p_0 \\ p_1 \\ p_2 \\ p_3
\end{bmatrix} \\
&
\Big \downarrow & \Big \downarrow & \Big \downarrow \\
&
U & M & P
\end{array}
\]
\[p(u) = UMP\]
References
