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In Fig 87a we see a schematic picture of the mechanism The time is 60/1200 = 1/20 sec/rev; the time for total rise of 11/4 in is 160/360 1/20 = 1/45 sec
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In Fig 87b the displacement diagram divides into two parts, denoted by 1 and 2 We are given q1 + q2 = 160 degrees Since the ratio of acceleration is 3 : 1, we see that q1 = 40 degrees q 2 = 120 degrees The time t1 = t2 = 40 1 1 = sec 45 180 160 120 1 1 = sec 160 45 60
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Also, we know that the displacements 1 y1 + y2 = 1 4 For the parabolic curve from Eq (224) the follower displacement y= where A = follower accceleration, in sec 2 The displacement y1 = y2 = 1 1 5 (20, 200) = in 180 2 16 1 1 15 (6750) = in 60 2 16
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The displacement diagram is shown plotted in Fig 87b The inertia loads from Eq (81) Fi = where w is the equivalent weight: w= weight of spring + weight of linkage 3 = assumed negligible + 2 = 2 lb w y g
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The inertia force which increases the cam surface load Fi = 2 (20, 200) = +104 lb 386
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The critical inertia force tending to remove follower from cam
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CAM MECHANISM FORCES
Displacement y 1 " 14
2 15 " 16
5 " 16
Transition point 40
20 dwell 120 160 160 (b) Displacement diagram Cam angle q, degrees +20,200 in/sec2 +20,200 in/sec2 Dwell
(a) Cam-follower schematic
(c) Acceleration, in/sec2
+104 lb
+104 lb 6750 in/sec2
(d ) Inertia force, lb 35 lb
+