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

Marks: 7 M

Year: May2014

Number of points i.e. k=4,

Hence, we know that the degree of the Bezier curve is n= k-1= 4-1= 3.

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, $n = 3$ 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,

$P_0 (20,20) \ and B_{0,3} = (1-u)^3 , \\ P_1 (60,80) \ and B_{1,3} = 3u(1- u)^2 , \\ P_2 (120,100) \ and B_{2,3} = 3u^2 (1 - u) \ and \\ P_3 (150,30) \ and B_{3,3} = u^3.$

$Thus, P(u)=P_0×B_0,3 (u)+P_1×B_1,3 (u)+P_2×B_2,3 (u)+P_3×B_3,3 (u) P(u)=P_0×(1-u)^3 +P_1×3u(1-u)^2+P_2×3u^2 (1-u)+P_3×u^3$

which is the equation of the Bezier curve.

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

$P(u) = \begin{bmatrix} u^3 & u^2 & u & 1 \end{bmatrix}$ $\begin{bmatrix} 1 & 3 & -3 & 1 \\ 3 & -6 & 3 & 0 \\ -3 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix}P_0 \\ P_1 \\ P_2 \\ P_3\end{bmatrix}$

$P(u) = \begin{bmatrix} u^3 & u^2 & u & 1 \end{bmatrix}$ $\begin{bmatrix} 1 & 3 & -3 & 1 \\ 3 & -6 & 3 & 0 \\ -3 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix} 20 & 20 & 0 \\ 60 & 80 & 0 \\ 120 & 100 & 0 \\ 150 & 30 & 0 \end{bmatrix}$

$P(u) = \begin{bmatrix} u^3 & u^2 & u & 1 \end{bmatrix}$ $\begin{bmatrix} -10 & -10 & 0 \\ 60 & -120 & 0 \\ 120 & 180 & 0 \\ 20 & 20 & 0 \end{bmatrix}$

Now, to obtain the mid-point on the curve we put u=1⁄2

$P(u=1⁄2) = \begin{bmatrix} 1/8 & 1/4 & 1/2 & 1 \end{bmatrix}$ $\begin{bmatrix} -10 & -10 & 0 \\ 60 & -120 & 0 \\ 120 & 180 & 0 \\ 20 & 20 & 0 \end{bmatrix}$ = $\begin{bmatrix} 375/4 & 315/4 & 0 \end{bmatrix}$

Thus, the mid-point of the curve is (375⁄4,315⁄4,0)