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Tension in rod of a bicycle wheel skewer

  1. Feb 16, 2015 #1
    The locking mechanism of a bike skewer consists of a rod and slightly out of round disk on one side attached to the skewer. A handle is attached to the out of round disk as well Assuming the skewer is nearly snug with the wheel fork, I think the tension is going to be a function of the distance between the centerline of the disk where it is connected to the skewer and the position of the handle which is related to the disks arc length relative to the unlocked position.
  2. jcsd
  3. Feb 16, 2015 #2


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  4. Feb 17, 2015 #3


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    Could you provide another word for "skewer"? I don't think it's what you mean. (Spoke??)
  5. Feb 17, 2015 #4


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    I believe that he is referring to the "quick release" mechanism on the axle where, for instance, the front wheel is attached to the fork.

    The handle is perhaps 5 cm in length and 1 in breadth. It is attached to the "out of round disk" which is what I would have called a "cam". In my experience, the cam is hidden inside a somewhat larger than usual nut which is threaded onto one end of the axle. The handle is rotated 180 degrees between a released position and a tightened position where it holds by friction. In the released position, the wheel can be removed from the slots at the end of the fork. In the tightened position, the wheel is firmly affixed and the bicycle is ready to ride.

    I would expect the skewer tension to roughly obey Hooke's law with tension directly proportional to the displacement forced by the cam from the position where the parts begin to bind. A slightly out of round cam will have a roughly sinusoidal variation in displacement as a function of handle position. So that will cause one departure from linearity. And I would not expect Hooke's law to be obeyed at all precisely.
  6. Feb 17, 2015 #5
    Thanks for the help
    How's this for a possibility
    perhaps if i can come up with equation that is "close" in terms of its graph. Ie use a cosine based function where the right angular frequency will mimic the curve so that the bottom portion of functions graph is at the point where the tension begins. then integrate that function over an interval where the pressure is applied. My reasoning us that the cosine function is increasing over the interval 3pi/2 to 2pi and hence would cause an increase in the tension within the rod. if i knew the elastic properties of the rod as a fixed value then y=kx could be written with k being the spring constant and "x" being my cosine function in terms of angular change
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