Thanks for your reply, again.
Ok, so the velocity, v, at point C would be:
vC=rOC.ωOC.cos(180-\angleOCG), so would be at a maximum when \angleOCG)=0° or 180°.
and therefore, angular velocity, ωC, would be:
ωC=vC/lCG
I think.
Sorry to go on, but does the maximum velocity of the connecting rod occur at TDC, & BDC, because at that instant there is no horizontal velocity (piston movement)? Therefore, 100% of the tangential velocity of the crankshaft (OC) is utilised to move the connecting rod (CG), obviously...
Thanks CWatters,
Yeah, I think you are right, if inertia and friction are disregarded, as both the velocity of the piston and the tangetial velocity of the crank will be in the same direction. Would therefore, the appropriate crank angle for a off-set crank be 90°, perpendicular, to the angle...
How do you calculate the Maximum Anglular Velocity (and angle at which this occurs) of a connecting rod which uses a slider-mechanism with an offset crank?
ω = 300 rpm, 10\pi rad s-1
crank shaft, r = 50mm (0 → A)
connection rod, l = 200mm (A → B)
When crank angle = 45 degrees, the...