Relative velocity method_04
- 1. As shown in the following figure, the crank OA makes 150 rpm. Find for the given configuration, the linear velocity of the piston P. Solution: 1 =2 = 2 (150 ) 150 = 2356.5 (a) Solve for the linear velocity of crank OA; 60 (b) Draw the link using CAD, KS = 10: (c) The vector equation connecting known velocity vA to unknown velocity vB on the same link AB is: = + / (d) Sketch the velocity polygon, KV=100:
- 2. Select suitable position of pole O, draw the first velocity vector VA, considering its magnitude and direction. Sketch velocity vB from O, knowing only its direction, which is perpendicular to crank BQ. = (17.7270)(100) = 1772.70 Finally, sketch velocity VB/A from the tip of VA, the direction of VC/A is perpendicular to link BA = (14.1313)(100) = 1413.13 The intersection of Ob and ab is the magnitude of each velocity: and / . (e) Draw a perpendicular line to BC and AC, you will notice that ABC is the velocity image of abc in our velocity polygon. = + / For link BC relative velocity equations: = + / (f) Now, for link PC, relative velocity equation is: Sketch velocity vP/C from c, knowing only its direction, which is perpendicular to crank PC. = (14.8171)(100) = 1481.71 Sketch the path of slider P, the intersection of cp and qp is the velocity of slider P: . Summary of Answer: = . Reference: Theory of Mechanisms 14th Edition by HG Phakatkar ©2009, HG Phakatkar