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Torsion spring calcualtion for 12 ounces of weight

  1. Nov 4, 2016 #1
    I am NOT a physics major and clueless to figure out the strength of a custom spring I need to be made. I was hoping someone could help me, either with figuring it out, or turning me to someone/site that may be able to help.
    Here's the sitch':
    The spring is for a center kick-stand on a bicycle. The kick-stand/frame is custom built.
    The spring must be a torsion spring.
    the total weight of the kick-stand (placed on a scale) is 12 ounces.
    The spring should have enough torsional strength to keep the kick-stand in its non-deployed position (up, riding the bike) and keep it there with extra forces of being thrust on it by bumpy roads, etc.
    Make sense?
    Please help, and thank you in advance!
     

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  3. Nov 4, 2016 #2

    Bystander

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  4. Nov 5, 2016 #3

    CWatters

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    This type of stand can be quite dangerous. Spring fails, stand falls, catches on something like a drain cover and...

    To answer the question.. You need to know what g forces act on the bike/stand due to bumps. Suppose the max is 3g (pure guess). Lets also assume that the mass of the arm is evenly distributed so the centre of gravity is half way along it's length. Then I believe you need a spring that can deliver a torque of at least...

    T (in foot pounds) = 3 * (12/16) * 0.5(L)

    where L is the length of the arm (in feet).

    Allow extra for safety.
     
  5. Nov 5, 2016 #4
    CWatters,

    Thanx!
    The center stand's pivot is forward of the legs/feet, so in the event of spring failure the feet would be dragged, as opposed to being pushed. Not an indefensible position take in regards to safety but the same could be said for a single leg, side-hanging kickstand. Failure of either is not the favored experience.

    Okay, onward:
    The big question, indeed, is "what's the unknown?" I haven't a clue what g forces would act upon the bike/stand due to bumps, etc. I would guess that the highly educated in physics would have experience with "guesstimating" what the forces would reasonably be. I'll take your suggestion; seems reasonable.
    I looked at this "problem" from a layman's POV and used "pounds" of weight (over-all, meaning how much weight would hold it up on a balance scale while the extraneous forces are present). I came up with five times its weight of 12ounces, rounded off would be four pounds. So, if I could attach a four-pound weight to the opposite end of the legs, the kick-stand would remain up no matter how bumpy the road.
    Alas, the Spring Maker wouldn't accept this logic. Even if I promised to pay him in cash.

    So, T= 3 x .75 x .333 (length of legs is 8" measured straight-line from pivot to base of feet***).

    3/4 foot pounds?

    *** Wouldn't the center of gravity be on the opposite side of the pivot? Therefore 2/3 for L, not 1/3?


    Again, Many Thanx!

    PS- I gave thought to your cautiousness, and decided I could do even more for safety: I slid the feet "up" (or, "back" on the legs) so if the kick-stand were to drag while being ridden, there would be nothing to "grab" on the ground--the legs would merely "slide" over any obstructions.... (see picture)
     

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    Last edited: Nov 5, 2016
  6. Nov 5, 2016 #5

    CWatters

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    Google found a paper that looks like it might help .

    https://ir.library.oregonstate.edu/xmlui/handle/1957/32759

    Suggests much higher g forces might occur.
     
  7. Nov 6, 2016 #6

    CWatters

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    Not sure I follow that. The drawing suggests to me the centre of mass is in the middle of the leg so at 1/2 L.
     
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