The Coordinates of Four control points to a current WCS are given by $– P_0 = [2 \ 2 \ 0]^T , P_1 = [2 3 0]^T , P_2 = [3 3 0]^T , \ and \ P_3 = [3 2 0]^T.$ Find the equation of the resulting Bezier curve. Also find points on the curve for u = 0, ¼, ½, ¾ and 1.

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

Marks: 8 M

Year: Dec 2012

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,

$P_0 (2,2,0) \ and \ B_{0,3} = (1-u)^3 ,\\ P_1 (2,3,0) \ and \ B_{1,3} = 3u(1-u)^2 , \\ P_2 (3,3,0) \ and \ B_{2,3} = 3u^2 (1-u) \ and \\ P_2 (3,2,0) \ and \ B_{3,3} = u^3.$

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,

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

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

$P(u) = \begin{bmatrix} u^3 & u^2 & u & 1 \end {bmatrix}$ $\begin{bmatrix} 2 & 4 & 0 \\ 3 & -3 & 0 \\ 3 & 3 & 0 \\ 3 & 2 & 0 \end{bmatrix}$

Now, to obtain the points on the curve for u=0, 1⁄4, 1⁄2, 3⁄4 and 1,

$P(u = 0) = \begin{bmatrix} 0 & 0 & 0 & 1 \end {bmatrix}$ $\begin{bmatrix} 2 & 4 & 0 \\ 3 & -3 & 0 \\ 3 & 3 & 0 \\ 2 & 2 & 0 \end{bmatrix}$ = $\begin{bmatrix} 2 & 2 & 0 \end {bmatrix}$

$P(u = 1/4) = \begin{bmatrix} 1/64 & 1/16 & 1/4 & 1 \end {bmatrix}$ $\begin{bmatrix} 2 & 4 & 0 \\ 3 & -3 & 0 \\ 3 & 3 & 0 \\ 2 & 2 & 0 \end{bmatrix}$ = $\begin{bmatrix} 71/32 & 21/8 & 0 \end {bmatrix}$

$P(u = 1/2) = \begin{bmatrix} 1/8 & 1/4 & 1/2 & 1 \end {bmatrix}$ $\begin{bmatrix} 2 & 4 & 0 \\ 3 & -3 & 0 \\ 3 & 3 & 0 \\ 2 & 2 & 0 \end{bmatrix}$ = $\begin{bmatrix} 3 & 13/4 & 0 \end {bmatrix}$

$P(u = 3/4) = \begin{bmatrix} 27/64 & 9/16 & 3/4 & 1 \end {bmatrix}$ $\begin{bmatrix} 2 & 4 & 0 \\ 3 & -3 & 0 \\ 3 & 3 & 0 \\ 2 & 2 & 0 \end{bmatrix}$ = $\begin{bmatrix} 145/32 & 17/4 & 0 \end {bmatrix}$

$P(u = 1) = \begin{bmatrix} 1 & 1 & 1 & 1 \end {bmatrix}$ $\begin{bmatrix} 2 & 4 & 0 \\ 3 & -3 & 0 \\ 3 & 3 & 0 \\ 2 & 2 & 0 \end{bmatrix}$ = $\begin{bmatrix} 7 & 6 & 0 \end {bmatrix}$