Question 16.6: Design a four-bar mechanism so that θ12=70° and Φ12=50°. Len...

The Freudenstein’s equations becomes

\left[\begin{array}{rrr} -0.736 & -0.190 & 1 \\ -0.926 & 0.259 & 1 \\ -0.901 & 0.656 & 1 \end{array}\right]\left\{\begin{array}{l} k_{1} \\ k_{2} \\ k_{3} \end{array}\right\}=\left\{\begin{array}{l} -0.525 \\ -0.605 \\ -0.263 \end{array}\right\} .

A=\left|\begin{array}{rrr} -0.736 & -0.190 & 1 \\ -0.926 & 0.259 & 1 \\ -0.901 & 0.656 & 1 \end{array}\right|=-0.087 .

A_{1}=\left|\begin{array}{rrr} -0.525 & -0.190 & 1 \\ -0.605 & 0.259 & 1 \\ -0.263 & 0.656 & 1 \end{array}\right|=-0.185, A_{2}=\left|\begin{array}{rrr} -0.736 & -0.525 & 1 \\ -0.926 & -0.605 & 1 \\ -0.901 & -0.263 & 1 \end{array}\right|=-0.063.

A_{3}=\left|\begin{array}{rrr} -0.736 & -0.190 & -0.525 \\ -0.926 & 0.259 & -0.605 \\ -0.901 & 0.656 & -0.263 \end{array}\right|=-0.103 .

k_{1}=\frac{A_{1}}{A}=\frac{-0.185}{-0.087}=0.126=\frac{d}{a}, a=\frac{d}{2.126} .

k_{2}=\frac{A_{2}}{A}=\frac{-0.063}{-0.087}=0.724=-\frac{d}{c} \text { or } \frac{d}{c}=-0.724, c=\frac{-d}{0.724} .

k_{3}=\frac{A_{3}}{A}=\frac{-0.103}{-0.087}=1.184=\frac{a^{2}-b^{2}+c^{2}+d^{2}}{2 a c} .

\left(\frac{d}{2.126}\right)^{2}-b^{2}+\left(\frac{-d^{2}}{0.724}\right)+d^{2}=1.184 \times 2 \times \frac{d}{2.126} \times\left(\frac{-d^{2}}{0.724}\right) .

\frac{d^{2}}{4.52}-b^{2}+\frac{d^{2}}{0.524}+d^{2}=-1.538 d^{2} .

\text { 2. Rotate } O _{2} O _{4} \text { about } O_{4} \text { through } \phi_{12} / 2=25^{\circ} \text { clockwise. The point of intersection of these two lines }\text { locates relative pole } R_{12} .

\text { 3. Construct an angle } \psi_{12}=\frac{1}{2}\left(\theta_{12}-\phi_{12}\right)=\frac{1}{2}\left(70^{\circ}-50^{\circ}\right)=10^{\circ} \text { at } R_{12} . Join any two points on the two arms of this angle to obtain the coupler link AB. Join A with O_{2} \text { and } B \text { with } O_{4} to get the input and output links.

\text { 4. } O_{2} A B O_{4} \text { is the desired mechanism. } .