Question

The Whitworth quick return motion mechanism has the driving crank AC=150 mm long.
The distance between fixed centres CD= 100 mm. The line of stroke of the ram passes
through the centre of rotation of the slotted lever whose free end is connected to the
ram by a connecting link. Find the ratio of time of cutting to time of return.

Answer 1

Quick return ratio = {Time of cutting/time of return}

= $\frac{a}{b}$

= $360-\frac{a}{a}$

AC₂ =150 mm ,CD = 100mm

Cos $\frac{b}{2}$ = $\frac{CD}{AC}$

  = $\frac{100}{150}$

$\frac{b}{2}$ = 48.2°

b = 96.4°

QRR = $\frac{360-96.4}{96.4}$

=2.735