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Power loss in different drive systems?

  1. Sep 5, 2011 #1
    Firstly, I am not an engineer or engineering student. I come from an art/creative background. However, in what may be arrogance, I'm convinced I can solve a problem that no one but me thinks is a problem. Further, I'm spitting in the face of a system that has been fine for everyone else for the last 100 years. I truly have no idea of how ignorant I am in the matter. Great combination right, arrogance and ignorance. Moreover, not only am I lost on the subject, I don't know where to look, hence my arrival on this forum

    Anyway, I am hoping some kind person in the know here, will be gracious enough to point me to where I might find the correct information.

    The problem: I'm trying to to determine the numbers/formulas for power loss in different drive systems. e.g.

    1. What would be the loss in a simple chain drive like that of a current diamond frame bicycle.

    2. What would be the loss in a system of a/an (eccentric?) drive like that of the old steam trains where the wheels are connected by connecting rod to the drive wheel?
    3. What would be the loss in a shaft drive system that had two 90 degree turns.

    I have experience with repairing motorcycles, (BMW airhead) and discussions about these shaft drive bikes always included the power loss at the 90 degree turn into the rear drive. However, they never mentioned how this loss is calculated.

    Are there any simple formulas that might help? If so, where might I find them? Am I asking the question correctly? Is more information necessary?

  2. jcsd
  3. Sep 5, 2011 #2
    You take the ratio of power out/power in (or energy out/energy in) and you can tell the efficiency. Efficiency will always decrease over time depending on how well components are initially designed and made, maintained, frictional losses and gradual wear and tear.

    Get friction as close to 0 as you can and have the movement smooth, no noise, heat or vibration and you can approach 100% efficiency.
  4. Sep 6, 2011 #3
    Thanks for the reply. That would mean building each and computing the calculations. I thought that perhaps there were standard formulas.

    e. g. (power in - X%) = power out. I was assuming for instance that the unknown X% would be constant. In other words, I thought the 90 degree turn would have the same power loss no matter the application, and that that figure might be in some engineering book some where.
  5. Sep 8, 2011 #4
    Nope. There's no reason for the energy inputted to just disappear as a %loss.

    Take your 90º gears as an example. What is the actual difference between that and two flat gears when they're working? Not much except direction. The forces you put in may point in different directions as it moves through the system but the sum total of them will be the same. I say this abstractly because you shouldn't think of it in this way, to actually calculate where every little bit of force is unnecessary and complicated.

    All of the energy stays in the gears and doesn't transfer/leave it for no reason. Friction will bleed the energy out of your system because it will use a bit of the input force solely for the process of overcoming it. Anything you do that overcomes friction will generate heat and sound, resulting in a loss at the output.

    There's no inherent reason that a 90º gear set will throw away energy any more than other gears. You can get very nitty gritty with it, find out the exact surface contact area etc but this is negligible because there are so many other factors that will affect it in the real world.
  6. Sep 8, 2011 #5
    I think that you'll probably run into the law of diminishing returns, a quick internet search gives chain efficiencies of about 98% and gear efficiencies of 98% to 99%, to try and squeeze the extra 1% out of a drive chain is probably not worth the effort.
  7. Sep 9, 2011 #6
    Thanks for the replies, it is a little clearer to me. I was under the impression that there was a lot of loss for some reason that I did not understand. I suspect--well in the motorcycle world--that the losses are due mainly to the additional weight. The system of gears, shafts, hubs and lubrication necessary is way heaver than a simple chain & sprocket. It almost always makes for a heavier unit for the power to move.
  8. Sep 10, 2011 #7
    Once a change in speed has occurred, weight of components has no effect on power loss. You use power to change the speed of the drive train (change in kinetic energy of rotating parts occurring in a period of time). Once the rotational speed change has been achieved, the losses are due to friction.

    Take an automobile transmission. It is generally thought the drive train consumes about 16%of engine output. That is why there are transmission coolers on transmissions, even some manual transmissions. Remember that 1 BTU equals 778 ft-lbs of energy. That is huge.
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