Mumbai University > Mechanical Engineering > Sem 7 > CAD/CAM/CAE

Marks: 7 M

Year: May 2012

Consider the three control points that define a Bezier curve, i.e., P0 (4, 2), P1 (0, 0) and P2 (2, 8).

Number of points, k = 3

Degree of Bezier curve, n=k-1=3-1=2

Now, to obtain the equation of the Bezier curve in parametric format with parameter ‘u’, we know that

$P(u)= \sum_{i=0}^n P_i×B_{i,n} (u)$

where, n=2 and $P_i$ is the $i^{th}$ control point and B_(i,n) is defined as,

$B_{i,n} = ^nC_r × u^i (1- u)^{n-i}$

Hence,

$P_0 (4,2) and B_{0,2} = (1-u)^2 \\ P_1 (0,0) and B_{1,2} = 2u(1-u) \ and \\ P_2 (2,8) and B_{2,2} =u^2 \\ Thus, P(u)=P_0×B_0,2 (u)+P_1×B_1,2 (u)+P_2×B_{2,2} (u) \\ P(u)=P_0×〖(1-u)〗^2 +P_1×2u(1-u)+P_2×u^2$

which is the equation of the Bezier curve.

It can be further be simplified and written in matrix form as,

$P(u) = \begin{bmatrix} u^2 & u & 1 \\ \end{bmatrix}$ $\begin{bmatrix} 1 & -2 & 1 \\ -2 & 2 & 0 \\ 1 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix} P_0 \\ P_1 \\ P_2 \end{bmatrix}$

$P(u) = \begin{bmatrix} u^2 & u & 1 \\ \end{bmatrix}$ $\begin{bmatrix} 1 & -2 & 1 \\ -2 & 2 & 0 \\ 1 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix} 4 & 2 \\ 0 & 0 \\ 2 & 8 \end{bmatrix}$

$P(u) = \begin{bmatrix} u^2 & u & 1 \\ \end{bmatrix}$ $\begin{bmatrix} 6 & 10 \\ -8 & -4 \\ 4 & 2 \end{bmatrix}$