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Homework Help: Maximum Velocity of a Con-Rod with Off-set Crank.

  1. Oct 23, 2012 #1
    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[itex]\pi[/itex] rad s-1
    crank shaft, r = 50mm (0 → A)
    connection rod, l = 200mm (A → B)

    When crank angle = 45 degrees, the connection rod is horizontal and in-line with the piston.

    I have calcualted, at 45 degrees, the velocity of the piston relative to 0 (1.11 m s-1) and the angular velocity of the con-rod, AB about A. (5.55 rad s-1 using a vector diagram for v, and ω=v/l

    Not really looking, for an answer just a point in the right direction. Not came across a problem with an off-set crank before and don't really know where to start.

  2. jcsd
  3. Oct 23, 2012 #2


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    I think on a conventional crank the connecting rod is going at maximum velocity when the connecting rod is tangential to the crank disc (eg 90 degrees to the crank arm). In that position the connecting rod is going in the same direction as the crank pin. At other times the crank pin is going in a slightly different direction. I think I've got that right.
  4. Oct 24, 2012 #3
    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 at which the piston is at full extension/retraction, ie. when the velocity of the piston is 0?

    I calculated this angle using sine triangle theory to 8.13°, making the angle of maximum angular velocity 98.13°.
  5. Oct 24, 2012 #4


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    Best ignore my first post as I got totally mixed up somehow.

    Take a look at the enclosed diagram. It shows a general offset crank not the one in the problem.

    GC is the connecting rod
    CO is the crank

    At any instant the connecting rod is rotating about the Gudgeon pin G.
    At the other end the crank is rotating with velocity V tangential to the crank disc.
    The velocity at 90 degrees to the connecting rod is Vcos(∠OCG). That's a maximium when ∠OCG = 0 or 180. Which is also TDC or BDC

    Attached Files:

  6. Oct 24, 2012 #5
    Thanks, CWatters, that makes sence mathematically, since cos 0,180 = 1,-1.
  7. Oct 25, 2012 #6
    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 disregarding friction, inertia, etc.
  8. Oct 25, 2012 #7


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    No you can forget what the piston end is doing. It's all about the other end..

    Imagine you were sat riding on the gudegeon pin looking towards the crank end. From that viewpoint..

    1) the GP end appears statationary
    2) the crank end appears to waggle left and right as per the attached drawing.

    The angular velocity would be zero when it changes direction and a maximium "somewhere" inbetween. It would be dangerous to assume that the fastest point was exactly in the middle as in some situations it might not be..

    To work out the angular velocity you need to know the velocity with which the far end is moving back and forth by writing an equation for the velocity shown dotted. Note that this is not the same as the crank velocity. It varies with the position of the crank. See previous drawing for the trig used.

    Then when you have the equation you work out when the equation is at a maximium.

    It so happens that this is when the piston is at TDC or BDC but for other types of machine it might not be.

    Attached Files:

    Last edited: Oct 25, 2012
  9. Oct 25, 2012 #8

    Thanks for your reply, again.

    Ok, so the velocity, v, at point C would be:

    vC=rOCOC.cos(180-[itex]\angleOCG[/itex]), so would be at a maximum when [itex]\angleOCG[/itex])=0° or 180°.

    and therefore, angular velocity, ωC, would be:


    I think.
  10. Oct 25, 2012 #9


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    I agree.

    I made the angle = Tan-1(50cos(45)/(50+200) = 8.05 degrees.

    Edit: I mean 8.05 degrees to the horizontal
  11. Oct 26, 2012 #10
    Thanks for your help.

    Much appreciated.
  12. Feb 17, 2014 #11
    CWatters or MorgansRocks do you have the velocity diagrams and acceleration diagrams for this question? Its thrown me a little since its a set up I am unfamiliar with.
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