Design a crank$-$rocker quick$-$return mechanism so that the return cycle is $1\mathrm{.}5$ times

faster than the forward cycle for a $90$ degree angle sweep by the rocker. Show

all calculations and verify that it satisfies the Grashof condition $(S+LP+Q).$

Calculate the lengths of all the links. Draw your linkage by hand showing stationary

positions. With a length of the rocker at $20$ cm $.$ Please help, I have attempted this question and I am stuck

Requirements:

return cycle $1\mathrm{.}5$ faster than forward cycle

$90$ deg. angle sweep by rocker

Rocker length $20$ cm

Greshof $(S+LP+Q)$

Find Quick Return Ratio: $\frac{\text{f}\text{o}\text{r}\text{u}\text{e}\text{d}}{\text{O}\text{r}\text{c}\text{t}\text{u}\text{r}\text{n}}=1\mathrm{.}5-ea.$

${}_{\text{f}\text{o}\text{r}\text{w}\text{a}\text{d}}+{}_{\text{r}\text{e}\text{t}\text{u}\text{r}\text{n}}=90-$ eqe $2$

rewrite eq$.1$: A forward $=(1\mathrm{.}5)$ Arcturn $-$ ea$.3$

Sub. eq$.2$ intro eq$.3$: ${}_{\text{r}\text{e}\text{t}\text{u}\text{r}\text{n}}+(1\mathrm{.}5){}_{\text{r}\text{e}\text{t}\text{u}\text{r}\text{n}}=90$

${}_{\text{r}\text{e}\text{t}\text{u}\text{r}\text{n}}=\frac{90}{2\mathrm{.}5}=36,90-36=54$

:$.{}_{\text{r}\text{a}\text{t}\text{u}\text{r}\text{n}}=36,{}_{\text{f}\text{o}\text{r}\text{m}\text{e}\text{d}}=54$

$2to1$

$B_1$

Coupler $+$ crank $=40$

coupler $-$crank$=22\mathrm{.}5$

$2$ couplesr$=62\mathrm{.}5$

coupler $=31\mathrm{.}25$