Question

A simple quick return mechanism is shown in the figure. The forward to return ratio of the quick return mechanism is 2 : 1. If the radius of the crank $O_{1}P$ is 125 mm, then the distance d (in mm) between the crank centre to lever pivot centre point should be

• A. 144.3
• B. 216.5
• C. 240.0
• D. 250.0

Answer 1

The correct option is D 250.0


$O_{1}P=125$ mm
Quick return mechanism:

$\frac{\text{Time of working (forward) stroke}}{\text{Time of return stroke}}=\frac{\beta }{\alpha }$

$=\frac{360-\alpha }{\alpha }$

$\frac{2}{1}=\frac{360-\alpha }{\alpha }$

$\Rightarrow 2\alpha ={360}^o-\alpha$

$\Rightarrow 3\alpha ={360}^o\Rightarrow \alpha ={120}^o$

$\Rightarrow \frac{\alpha }{2}={60}^o$

The extreme position of the crank $\left(O_1\right)$ is shown in figure.
From right angle triangle $O_2{O}_1{P}_1$, we find that
$sin({90}^o-\alpha /2)=\frac{O_1{P}_1}{O_2{P}_2}$

$sin({90}^o-{60}^o)=\frac{125}{O_1{O}_2}=\frac{125}{d}$
$sin{30}^o=\frac{125}{d}$

or $d=\frac{125}{sin{30}^o}=250$ mm