Question 16.3: Design a four-bar mechanism to coordinate three positions of...

\cos \theta_{1}=\cos 25^{\circ}=0.9063, \cos \theta_{2}=\cos 35^{\circ}=0.8191, \cos \theta_{3}=\cos 50^{\circ}=0.6428 .

\cos \phi_{1}=\cos 30^{\circ}=0.8660, \cos \phi_{2}=\cos 40^{\circ}=0.7660, \cos \phi_{3}=\cos 60^{\circ}=0.5000 .

\cos \left(\theta_{1}-\phi_{1}\right)=\cos \left(25^{\circ}-30^{\circ}\right)=0.9962 .

\cos \left(\theta_{2}-\phi_{2}\right)=\cos \left(35^{\circ}-40^{\circ}\right)=0.9962 .

\cos \left(\theta_{2}-\phi_{2}\right)=\cos \left(50^{\circ}-60^{\circ}\right)=0.9848 .

A=\left|\begin{array}{lll} 0.8660 & 0.9063 & 1 \\ 0.7660 & 0.8191 & 1 \\ 0.5000 & 0.6428 & 1 \end{array}\right| .

=0.8660(0.8191-0.6428)-0.9063(0.7660-0.5000)+1(0.7666 \times 0.6428 -0.8191 \times 0.5000) .

=5.5652 \times 10^{-3} .

A_{1}=\left|\begin{array}{lll} 0.9962 & 0.9063 & 1 \\ 0.9962 & 0.8191 & 1 \\ 0.9848 & 0.6428 & 1 \end{array}\right| .

=0.9962(0.8191-0.6428)-0.9063(0.9962-0.9848)+1(0.9962 \times 0.6428 -0.8191 \times 0.9848).

=-9.9408 \times 10^{-4} .

A_{2}=\left|\begin{array}{lll} 0.8660 & 0.9063 & 1 \\ 0.7660 & 0.9962 & 1 \\ 0.5000 & 0.9848 & 1 \end{array}\right| .

=0.8660(0.9962-0.9848)-0.9962(0.7660-0.5000)+1(0.7660 \times 0.9848 -0.9962) \times 0.5000).

=-1.14 \times 10^{2} .

A_{3}=\left|\begin{array}{lll} 0.8660 & 0.9962 & 0.9962 \\ 0.7660 & 0.8191 & 0.9962 \\ 0.5000 & 0.6428 & 0.9848 \end{array}\right| .

=0.8660(0.8191 \times 0.9848-0.9962 \times 0.6428)-0.9063(0.7660 \times 0.9848 -0.9962 \times 0.5000) .

=5.746 \times 10^{-3} .

k_{1}=\frac{A_{1}}{A}=\frac{-9.9408 \times 10^{-4}}{-5.5652 \times 10^{-3}}=0.1786=\frac{d}{a} .

Let d=1 unit, then a=5.598 units

k_{2}=\frac{A_{2}}{A}=\frac{1.14 \times 10^{-3}}{-5.5652 \times 10^{-3}}=-0.2048 .

=-\frac{d}{c}, c=4.88 \text { units } .

k_{3}=\frac{A_{3}}{A}=\frac{-5.764 \times 10^{-3}}{-5.5652 \times 10^{-3}}=1.0272 .

=\frac{a^{2}-b^{2}+c^{2}+d^{2}}{2 a c} .

=\frac{(5.598)^{2}=b^{2}+(4.88)^{2}}{2 \times 5.598 \times 4.88}, \quad b=0.176 .

The mechanism is shown in Fig.16.18.