Mumbai University > mechanical engineering > Sem 7 > CAD/CAM/CAE

Marks: 0 Marks

Year: Dec 2015

Number of point 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

$$P(u) =\sum_{i=0}^nPi \times Bi,n(u)$$

Where, n=3 & Pi is the i^th control point & bi,n is define as,

$Bi,n =nCr \times u^i (1-u)^{n-i}$

Hence,

$Po( 1,3,0) and B_{0,1} = (1-u)^3 \\ P1( 5,6,0) and B_{1,3} = 3u(1-u)^2 \\ P2( 6,0,0) and B_{2,3} = 3u^2 (1-u) \\ P3 (7,2,0) and B_{3,3} =u^3 \\ Thus, P_0 \times B_{0,3} (u) + P_1 \times B_{1,3} (u) + P_2 \times B_{2,3} (u) + P_3 \times B_{3,3} (u) \\ P(u) = P_0 \times (1-u)^3 + P_1 \times 3u(1-u)^2 + P_2 \times 3u^2 (1-u) +P_3 \times u^3.$

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

$P(u) = (u^3 u^2 u^1 1) \begin{bmatrix}1&3&-3&1 \\ 3&-6&3&0 \\ -3&3&0&0 \\ 1&0&0&0 \end{bmatrix} \times \begin{bmatrix}0&0&0 \\ 7&6&0 \\ 6&5&0 \\ 4&0&0 \end{bmatrix} \\ P(u) = (u^3 u^2 u^1 1) \begin{bmatrix}7&3&0 \\ -24&-21&0 \\ 21&18&0 \\ 0&0&0 \end{bmatrix}$

Now, to obtain the points on the curve for u = 0.5, & 1.

$P(u = 0.5) = \bigg(\dfrac18 \ \ \ \ \dfrac14 \ \ \ \ \dfrac12 \ \ \ \ 1\bigg) \begin{bmatrix}7&3&0 \\ -24&-21&0 \\ 21&18&0 \\ 0&0&0\end{bmatrix} \hspace{3cm}=\bigg[ \dfrac{43}{8} \ \ \ \ \dfrac{33}{8} \ \ \ \ 0 \bigg] \\ P(u = 1) = (1 \ \ \ \ 1 \ \ \ \ 1 \ \ \ \ 1) \begin{bmatrix}7&3&0 \\ -24&-21&0 \\ 21&18&0 \\ 0&0&0 \end{bmatrix} \hspace{4cm}=[ 4 \ \ \ \ 0 \ \ \ \ 0 ]$