A four bar mechanism is to be designed, by using three precision points, to generate the function
y=x^{1.5}, \text { for the range } 1 \leq x \leq 4.
Assuming 30^{\circ} starting position and 120^{\circ} finishing position for the input link and 90^{\circ} starting position and 180^{\circ} finishing position for the output link, find the values of x, y, \theta \text { and } \phi corresponding to the three precision points.
Question : A four bar mechanism is to be designed, by using three preci...
Given : x_{ S }=1 ; x_{ F }=4 ; \theta_{ S }=30^{\circ} ; \theta_{ F }=120^{\circ} ; \phi_{ S }=90^{\circ} ; \phi_{ F }=180^{\circ}
Values of x
The three values of x corresponding to three precision points (i.e. for n = 3) according to Chebychev’s spacing are given by
x_{j}=\frac{1}{2}\left(x_{ S }+x_{ F }\right)-\frac{1}{2}\left(x_{ F }-x_{ S }\right) \cos \left[\frac{\pi(2 j-1)}{2 n}\right], where j = 1, 2 and 3
\therefore x_{1}=\frac{1}{2}(1+4)-\frac{1}{2}(4-1) \cos \left[\frac{\pi(2 \times 1-1)}{2 \times 3}\right]=1.2 \ldots(\because j=1)
x_{2}=\frac{1}{2}(1+4)-\frac{1}{2}(4-1) \cos \left[\frac{\pi(2 \times 2-1)}{2 \times 3}\right]=2.5 \ldots(\because j=2)
and x_{3}=\frac{1}{2}(1+4)-\frac{1}{2}(4-1) \cos \left[\frac{\pi(2 \times 3-1)}{2 \times 3}\right]=3.8 \ldots(\because j=3)
Note : The three precision points x_{1}, x_{2} \text { and } x_{3} may be determined graphically.
Values of y
Since y=x^{1.5} , therefore the corresponding values of y are
y_{1}=\left(x_{1}\right)^{1.5}=(1.2)^{1.5}=1.316y_{2}=\left(x_{2}\right)^{1.5}=(2.5)^{1.5}=3.952
y_{3}=\left(x_{3}\right)^{1.5}=(3.8)^{1.5}=7.41
Note : y_{ S }=\left(x_{ S }\right)^{1.5}=(1)^{1.5}=1 \text { and } y_{ F }=\left(x_{ F }\right)^{1.5}=(4)^{1.5}=8
Values of \theta
The three values of \theta corresponding to three precision points are given by
\theta_{j}=\theta_{ S }+\frac{\theta_{ F }-\theta_{ S }}{x_{ F }-x_{ S }}\left(x_{j}-x_{ S }\right) , where j = 1, 2 and 3
\therefore \theta_{1}=30+\frac{120-30}{4-1}(1.2-1)=36^{\circ}
\theta_{2}=30+\frac{120-30}{4-1}(2.5-1)=75^{\circ}
and \theta_{3}=30+\frac{120-30}{4-1}(3.8-1)=114^{\circ}
Values of \phi
The three values of \phi corresponding to three precision points are given by
\phi_{j}=\phi_{ S }+\frac{\phi_{ F }-\phi_{ S }}{y_{ F }-y_{ S }}\left(y_{j}-y_{ S }\right)
\therefore \phi_{1}=90+\frac{180-90}{8-1}(1.316-1)=94.06^{\circ}
\phi_{2}=90+\frac{180-90}{8-1}(3.952-1)=127.95^{\circ}
and \phi_{3}=90+\frac{180-90}{8-1}(7.41-1)=172.41^{\circ}