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Gearbox torque

  1. Aug 15, 2016 #1
    I have an overhead crane and we are having some issues with the gearboxes that drive the bridge. The total crane weight is 44250lbs and has a 4 ton capacity. The wheels are 13.5 inches in diameter and have a wheel load of 16500lbs. The gearbox is driven by a 2HP electric motor at 1750rpm. The gearbox itself is 30:1, the output speed is 233rpm and 2232in-lbs of torque.
    I added the weight of the crane and capacity to give me 52250lbs, and divide it by 4 for each of the wheels of the end truck I get 13062.5lbs.
    Using HP = n*T/5252 I can solve for T which should be the torque required to move the crane correct? I then have 5252*2HP/233rpm=T. This gives me a value of T=45.0015ft-lbs which is 540 in-lbs of torque. So our current gearbox has plenty of torque.

    Am I on the right path? I haven't done this in awhile.

    Thanks in advance.
  2. jcsd
  3. Aug 15, 2016 #2
    Shouldn't the RPM's on the gearbox be part of the 30:1 ratio? Is this a homework problem or a real world problem? If it's a homework problem, I would guess, based on the values given, that other things are being asked for in the problem. In any case, power throughout the system in theory, is going to be constant. So, your math is right if that's all you were solving for, power = torque*rotation speed, as long as the 5252 converts it correctly, I think that's right. It put in into more convenient units, HP * 1100 ft lb/s, and 60 RPM = 1 RPS (obviously). Actually, that gives me a different answer unless I did something wrong with my math or units. T = P/omega = 2200 ft lb/s / [233 RPM/(60 RPM/RPS)]

    It's been a long time since I had to use this kind of stuff too, but I think that's correct. I'm still not sure where the 5252 came from though.

    Anyway, hope that helps.
  4. Aug 16, 2016 #3

    jack action

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    The math is good and the equation is used correctly with the proper units.

    How did you get 233 rpm and 2232 in.lb? This makes 8 hp per the formula above.

    The electric motor torque is 6 ft.lb (= 2*5252/1750) or 72 in.lb.
    The gearbox rpm and torque are 58.3 rpm (= 1750/30) and 2160 in.lb (= 72*30) or 180 ft.lb. This still makes 2 hp (= 58.3*180/5252), as it should.
  5. Aug 16, 2016 #4
    jlefevre76, this is a project from work that sort of got dumped on me.

    Jack Action, I got the 233rpm and 2232in.lb from the gearbox manufacturers catalog.

    I think I figured it out and the current gearboxes are large enough to handle the load.

    Thanks for looking into this and appreciate the review.
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