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

Marks: 10 Marks

Year: Dec 2016

Bezier Curve :

(1,1), (2,3), (4,4), (6,1)

The equation of the Bezier Curve is given by;

$P(u)=(1-4)^3P_1+3u(1+-4)^2P_2+3u^2(1-)P_3+u^3P_4$

$P(0.5)$ $=(1-0.5)^3P_1+3 \times 0.5(1-0.5)P_2+3(0.5)^2(1-0.5)P_3+(0.5)^3(P_4) \\ =\dfrac18(1,1)+ \dfrac38 (2,3)+ \dfrac38 (4,4)+ \dfrac18 (6,1) \\ =\bigg[\dfrac18 \times1 +\dfrac38 \times2 +\dfrac38 \times4 +\dfrac18 \times6 ; \dfrac18 \times1 +\dfrac38 \times3 +\dfrac38 \times4 +\times18 \times1 \bigg] \\ =\bigg[\dfrac18+\dfrac34+\dfrac32+\dfrac34 ; \dfrac18+\dfrac98+\dfrac{12}{8}+\dfrac18\bigg] \\ =\bigg[\dfrac{1+6+6+6}{8} ; \dfrac{1+9+12+1}{8} \\ =\bigg[\dfrac{25}{8} ; \dfrac{23}{8}\bigg] \\ =2.175,2.875$

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

$P(u)=[u^3 \ \ u^2 \ \ u \ \ 1]\underbrace{-u^3+3u^2+3u+1;}_\text{x} \underbrace{-3u^3-3u^2+6u+1}_\text{y}$

$\dfrac{dx}{dx}=-3u^2+6u+3 \\ \dfrac{dy}{dy}=-9u^2-6u+6 \\ \bigg|\dfrac{dy}{dx}\bigg|_{0.5}=\dfrac{-9u^2-6u+6}{-3u^2+6u+3} \\ \ \ \ \ \ \ \ \ \ \ \ \ = \dfrac17$