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Deriving formula for MA of differential windlass

  1. Oct 31, 2014 #1
    Hi,

    I am trying to find an equation that would give the mechanical advantage of this system:

    L-differentialwinde.png

    I am fairly new to this kind of analysis, but my understanding is that to determine the MA I need to consider that:

    [tex]W_{in} = W_{out}[/tex]

    [tex]F_{in} d_{in} = F_{out} d_{out}[/tex]

    [tex]F_{out} / F_{in} = d_{in} / d_{out} = MA[/tex]

    So all I need to do to determine the mechanical advantage is to find the ratio of the distances of the applied and effort forces.

    d_in:

    The crank and large drum act as a wheel and axle, so the distance the large drum travels per full revolution of the crank shaft is:

    [tex]\frac{r_{large \, drum}}{r_{crank}}[/tex]

    d_out

    The movable pulley is going to rise a distance that equals one half the difference in circumferences between the large and small drum:

    [tex]\pi(r_{large \, drum} - r_{small \, drum})[/tex]

    Which give a rough mechanical advantage (ignoring the levers created by the crank) of:

    [tex] 2 \times \frac{r_{large \, drum}}{\pi r_{crank}(r_{large \, drum} - r_{small \, drum})}[/tex]

    I have some concerns though, the main one being that this is different from the only equation I could find online, which is:

    [tex]\frac{2 \, r_{crank}}{r_{large \, drum} - r_{small \, drum}}[/tex]

    Is my analysis correct or have I missed something? Also, if anyone could tell me why the pulley moves only one half the difference between the two radii and not the full difference between the two I would really appreciate it.

    Thanks!
     
  2. jcsd
  3. Oct 31, 2014 #2
    I've thought about it some more and I think I've got it. This complex machine can be broken up into three simple machines: the movable pulley, the wheel and axle, and the rope wrapped around the two drums.

    The movable pulley has a MA of 2.

    The wheel and axle has an MA of [tex]\frac{d_1}{d_2} = \frac{r_{crankshaft}}{r_{large \, drum}}[/tex]

    The rope wrapped around the two drums has an MA of [tex]\frac{d_1}{d_2} = \frac{r_{large \, drum}}{r_{large \, drum} - r_{small \, drum}}[/tex]

    All these mechanical advantage are multiplied to give the MA listed at the end of my post.

    It that it?
     
  4. Oct 31, 2014 #3
    Looks like you got two different results; which are you going with?

    I suggest that you look at this by the principle of virtual work (if you are familiar with that approach).
     
  5. Oct 31, 2014 #4
    Well, I found that equation in a mechanical engineering reference book, and it makes sense, so I'm inclined in that direction. Also, I'm not sure if my d_in equation makes sense, it's a unitless ratio of distances and the MA if the force is applied to the drum, and not an actual distance, isn't it? It seems to be simpler to just break it down into simple machines and take the product. I'll look into virtual work, thanks for that.
     
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