in a crank and slotted lever quick return mechanism the distance between fixed center is 15cm and the driver crank is 10 cm long.find the ratio of time taken to the the cutting and return strokes?
The schematic of quick return mechanics distance between fixed centers-
AB = 150 mm
Length of driving crank,
BC = 75 mm
In triangle ABC,
"cos{\u03b8}=\\frac{ AB}{BC} =\\frac{ 150}{75}=0.5\u27f9\u03b8=60^ \n0"
Since cutting of metal happens to be in forwarding stroke and angle rotated by crank BC during cutting
"\u03b1=360\u22122\u03b8=360\u2212120=240"
The ratio of cutting stroke time to return stroke time,
"=\\frac{\\alpha}{\\beta}=\\frac{240}{120}=2"
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