Bézier Curve

Curves

Hermite Curve

The parametric equation of curves is:

$$P(t)=a_3t^3+a_2t^2+a_1t+a_0, t∈[0,1]$$

Derivate it:

$$P'(t)=3a_3t^2+2a_2t+a_1$$

Let $t=0$ and $t=1$, we get:

$$P(0)=a_0, P'(0)=a_1, P(1)=a_3+a_2+a_1+a_0, P'(1)=3a_3+2a_2+a_1$$

So, we take $P_0,P_0’,P_1,P_1’$ as:

$$P_0=a_0$$ $$P_0'=a_1$$ $$P_1=a_3+a_2+a_1+a_0$$ $$P_1'=3a_3+2a_2+a_1$$

so we get:

$$a_0=P_0$$ $$a_1=P_0'$$ $$a_2=-3P_0-2P_0'+3P_1-P_1'$$ $$a_3=2P_0+P_0'-2P_1+P_1'$$

Thus,

$$P(t)=(2P_0+P_0'-2P_1+P_1')t^3+(-3P_0-2P_0'+3P_1-P_1')t^2+P_0't+P_0$$

After simplification:

$$P(t)=(2t^3-3t^2+1)P_0+(-2t^3+3t^2)P_1+(t^3-2t^2+t)P_0'+(t^3-t^2)P_1',t∈[0,1]$$

Let $F_0(t)=2t^3-3t^2+1$, $F_1(t)=-2t^3+3t^2$, $G_0(t)=t^3-2t^2+t$, $G_1(t)=t^3-t^2$.

$$P(t)=F_0P_0+F_1P_1+G_0P_0'+G_1P_1'=[F_0,F_1,G_0,G_1] \begin{bmatrix}P_0\\P_1\\P_0'\\P_1'\end{bmatrix}$$

Let $M=\left[\begin{matrix}1&0&0&0\\0&0&1&0\\ -3&3& -2& -1 \\2& -2 &1&1\end{matrix}\right]$ , so we get: $P(t)=[1,t,t^2,t^3]M\left[\begin{matrix}1&0&0&0\\0&0&1&0\\ -3&3&-2&-1\\2&-2&1&1\end{matrix}\right]$

Bézier curve

The curve is defined by four points: the initial position and the terminating position i.e $P_0$ and $P_3$ respectively (which are called “anchors”) and two separate middle points i.e $P_1$ and $P_2$ (which are called “handles”) in our example.

Given distinct points $P_0$ and $P_1$, a linear Bézier curve is simply a straight line between those two points. The curve is given by

$$B(t)=(1-t)P_0+tP_1, t∈[0,1]$$

A quadratic Bézier curve is the path traced by the function $B(t)$, given points $P_0$, $P_1$, and $P_2$,

$$B(t)=(1-t)[(1-t)P_0+tP_1]+t[(1-t)P_1+tP_2],t∈[0,1]$$ $$B(t)=(1-t)^2P_0+2t(1-t)P_1+t^2P_2,t∈[0,1]$$

This can be written in a way that highlights the symmetry with respect to $P_1$:

$$B(t)=P_1+(1-t)^2(P_0-P_1)+t^2(P_2-P_1), t∈[0,1]$$ $$B'(t)=2(1-t)(P_1-P_0)+2t(P_2-P_1)$$ $$B''(t)=2(P_2-2P_1+P_0)$$

Definition

给定空间 $n+1$ 个点的位置矢量 $P_i(i=0,1,2,…,n)$，则 Bézier 曲线可定义为：

$$P(t)=\sum^{n}_{i=0}P_iB_{i,n}(t), t∈[0,1]$$

where $B_{i,n}(t)$ is Bernstein basis polynomials of degree $n$ :

$$B_{i,n}(t)=\mbox{C}_n^it^i(1-t)^{n-i}=\frac{n!}{i!(n-i)!}(1-t)^{n-i},(i=0,1,...,n)$$