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Velocity diagram for 4 bar linkage

  1. Dec 30, 2015 #1
    1. The problem statement, all variables and given/known data

    http://photouploads.com/images/017db8.png
    http://photouploads.com/images/eebec7.png
    http://photouploads.com/images/f472e6.png

    2. Relevant equations
    v= rω

    3. The attempt at a solution
    I don't really have much of an attempt. I missed this lecture so I am completely confused on how to do these questions. I can't really find something online that explains the method from start to end in layman's terms. In the first image above, I'm assuming O stands for origin? Where is the origin in this image? Why is Vao = Vdo = 0?

    Why is Vcb in the direction indicated? If link AB is being pulled to the ''left'', then surely link BC is also pulled to the ''left'' therefore Vcb should also be in this (left) direction?

    Why are the blue axis drawn at the A end rather than the B end?

    For the second image, why is it Vba + Vcb + Vdc = 0, surely the last term should be Vcd as this is written just 2 lines above it?

    For the third image, I really don't understand the whole velocity space. Why are o, a and d in the same spot?

    Perhaps it would be better if someone could link me to a clear explanation of this topic? I tried to search online but didn't yield anything simple.

    Sorry for the lack of explanation but really confused and have been on this for hours now.

    Cheers
     
  2. jcsd
  3. Jan 2, 2016 #2

    haruspex

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    I don't think it matters where O is, it's just a way of stating that A and D are fixed.
    VCB is the velocity of C relative to B. Consequently it is orthogonal to BC. It doesn't matter which way it is drawn. If it turns out to be the other way then its value will be negative.
    In each link-specific diagram, the axes are drawn at the 'local' origin. E.g. in the second such diagram, the velocity of C relative to B is considered, so for that purpose B is the origin.
    VDC is the velocity of D relative to C. VDC=-VCD.
     
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