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How to increase output speed

  1. Nov 3, 2016 #1
    For a project I am doing I need to increase out put speed to 515 rpm from 1rpm with no more than a 20% decrease in torque, I need help I have been trying a lot of things and can't figure it out.
  2. jcsd
  3. Nov 3, 2016 #2


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    Staff: Mentor

    Welcome to the PF. :smile:

    Can you say more about your project? Is it for schoolwork? What is your power source? What have you tried so far? We can't really help you with so little information to go on...
  4. Nov 3, 2016 #3
    I don't think you can - the output torque will be 515 times less than input torque? You will always loose torque if you are gearing up from 1 rpm to 515rpm.
  5. Nov 3, 2016 #4


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    Welcome to PF.

    Power equals Torque multiplied by RPM.
    If the motor power is fixed then it is impossible to increase RPM without a proportional reduction of torque.
    To increase the RPM, with the same torque, you would need to change to a proportionally more powerfull motor.
  6. Nov 4, 2016 #5
    As long as the torque applied overcomes friction etc., then yes absolutely this can accelerate to 515 RPM. Look up that guy Newton and see what he says about it.

    You need to supply a sketch so your project can be understood better. But the simple estimate is this:

    Torque = (mass moment of inertia) X (angular acceleration) = Jα

    (angular acceleration α ) ≈ (ΔAngularVelocity)/(ΔTimeToAccelerate) = (515 RPM - 1 RPM)/(TimeToReach515RPM - 0)

    If you have a known torque that makes the object rotate at 1 RPM and you increase it +20%, then solve for (TimeToReach515RPM)
  7. Nov 4, 2016 #6
    He would need to increase torque by 80% (~410 times) to achieve 515rpm with a 20% reduction in torque
  8. Nov 4, 2016 #7


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    You can't do that with just a gearbox. As others have said if you step up the rpm by a factor of 515 you will step down the torque by the same factor. This is due to conservation of energy (law of physics).

    There is another potential problem.. Any friction in the bearings of the output shaft is seen at the input shaft but magnified 515 times. This makes frequently makes high ratio step up gearboxes impractical.

    Perhaps if you explain why you need such a step up we can figure out another way. Is this anything to do with hand cranked generators?
  9. Nov 4, 2016 #8
    It's a project dealing with a pto generator so I need
    I am working on a project that is using a pto generator and I need as far as I can figure 240 flbs of torque for continual use under a full load however I am limited with my power source to roughly about 500 flbs and due to the input I could possibly get up to 5 rpms I wish I could tell you more however my hands are tied with other specifics
  10. Nov 4, 2016 #9


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    What you're talking about isn't possible with just a gearbox - you'd need a supplemental power source of some kind, since your output power will be dramatically higher than your input power.
  11. Nov 4, 2016 #10


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    Unfortunately for you, if all you have available is a 500 ft-lb 5rpm source, the best you could do at 240ft-lb is a bit over 10rpm. With just a gearbox, the best possible outcome will result in your torque dropping by the same ratio as your speed increases.
  12. Nov 4, 2016 #11


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    So your power source is..
    500 flbs at 5 rpm which is about 0.5 HP (355W)

    Your load requires...
    240 flbs at 515 rpm which is about 23 HP (17,500W)

    See the impossibility of the problem?
  13. Nov 4, 2016 #12
    Ok yes I do thank you all for the help
  14. Nov 4, 2016 #13
    What is the formula you used to figure this out
  15. Nov 4, 2016 #14

    Randy Beikmann

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    CWatters would have used power = torque x omega, where omega is the shaft speed in radians per second. To use this formula, the values must be in consistent units. You could use power in foot-pounds/sec and torque in foot-pounds; or power in watts and torque in newton-meters.

    Once you have the answer in consistent units, you can convert to horsepower. One horsepower equals 550 foot-pounds/sec, or 745.7 watts.
  16. Nov 5, 2016 #15


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    What Randy said. I generally work in metric/SI units and cheat a lot...

    Power (in Watts) = Torque (in Nm) * angular velocity (in Rads/S).

    I converted 500 flbs to 678Nm using...

    I converted 5 rpm to 0.524 Radians/second using

    678 * 0.524 = 355W

    Convert it back to imperial units..

    1HP = 750W

    355/750 = 0.47 HP (I rounded it to 0.5HP)
  17. Nov 5, 2016 #16


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    If you want to work in Imperial units you have to remember a constant... Google suggests...

    Power (in HP) = T (in foot pounds) * RPM / 5252

    = 500 * 5/5252 = 0.47 HP
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