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Total pressure load on a chain

  1. May 26, 2012 #1
    Scenario- I'm riding a bike up a mountain road, at a constant speed of 5 mph, at a constant grade of 11%. The bike plus me (and all gear) weighs in at 200 pounds.

    Based on this (and excluding frictional losses, wind, etc.) how much TOTAL pressure, in pounds is on the chain? Would it be WxG (200*.11) for a result of 22 pounds? Yea? No?

    I know this is basic.....but I want to be sure I've not overlooked something and thus will take myself on a tangent. Wouldn't be the first time.:yuck:
  2. jcsd
  3. May 26, 2012 #2


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    hi maximiliano! :smile:
    it's not pressure, it's tension, and that's a force (measured in pounds)

    it depends on the gearing

    from the work energy theorem, the rate of work done equals the rate of change of (mechanical) energy

    at constant speed, and excluding friction etc, the energy is just weight (W) times height

    so the rate of change of energy is W times height-per-second, = w times speed times sin11° (∆KE = Wvsinθ)

    work done = force times distance, so rate of work done = force times speed (of the thing doing the force),

    so force = rate of work done / speed of the chain

    = Wvsinθ / speed of the chain

    = Wsinθ times (speed-of-bike)/(speed-of-chain) :wink:

    (similarly, the force you apply to the pedals is Wsinθ times (speed-of-bike)/(speed-of-pedals))
  4. May 26, 2012 #3
    Thanks Tim. Yes, but I'm ONLY looking for the calculation of the tension on the chain. Work done doesn't interest me one bit, and that's why I intentionally excluded any mention of time.

    I was just taking the weight being moved multiplied by the grade/slope of travel. I used a VERY slow speed intentionally as well, because I don't want there to be any additional tension (of any significant amount anyway) added to the equation. Gravity is the only constant force I wanted there.

    So.....it's 200 pounds moving at a slow and perfectly constant rate of speed up a road of 11% slope. The tension on the chain, in pounds, would be 200*.11? Or is the calculation not linear like that?
  5. May 26, 2012 #4


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    yes, and that's why i gave you the calculation for the force (the tension) in the chain :smile:
  6. May 26, 2012 #5
    Thanks Tim! Boy am I a dumb bell! Gee, I can't understand how I made such a basic mistake. Tension on the chain depends on the length of the lever at the rear (drive) point. That is, what final gear I'm in. So....there is a lot more to the formula in order to arrive at pressure on the chain given a constant amount of work.

    Thanks for setting me straight. I was just on a hike and it hit me how dumb my question was. I guess I'll have to keep working on this one using the distance from the center point of the wheel to where the chain is connected to the gear....and use the result as part of the formula for actual chain tension.
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