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Torque question involving gears

  1. Jul 2, 2014 #1
    Lets say you have two sets of gears with a combined 200:1 ratio. There's a worm and worm gear + two spur gears that rotate a welding table via a hand crank going to the worm.

    If the torque load from the weldment on the table was, say, 9375 in-lbs... then to calculate the force needed on the hand crank to rotate the table would be as follows:

    9375/200 = 46.88 in-lbs of torque needed in. The hand crank has a radius of 5", so 46.88/5 = 9.4 lbs of force. Is this correct?

    I am also wondering about the speed at which the hand crank turns. 9.4 lbs of force would be what is necessary to induce slight movement right? However realistically we want the table to rotate at a reasonable rate, if even only 0.5-1 rpm. Wouldn't this mean you would need more than just 9.4 lbs of force?

    I'm only seeing relations between torque, horsepower, and RPM. I don't know horsepower or RPM in the above scenario, just how much torque I would need to turn the table (but not necessarily at what rate). Can someone give me some guidance here. Thank you.
  2. jcsd
  3. Jul 2, 2014 #2


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    If you want to turn the table at say 1 RPM, that means, mechanically, the input handle must be turned at 200 RPM. That's the corollary of what a gear ratio does: The 200:1 gear ratio drops the output speed by a factor of 200 while multiplying the input torque by a factor of 200.
  4. Jul 2, 2014 #3
    The torque load of 9375 in-lbf comes from where?

    Your reduction calc is essentially correct. You're missing an efficiency factor and worm gears are not very efficient.

    It all starts with Torque = (J mass moment of inertia) X (Alpha angular acceleration). Your load on the turntable will have a J. Alpha is approximately deltaVelocity/deltaTime. So your 9.4lbf only is sufficient to keep it rotating. You must accelerate that mass to speed within a specific time (dV/dt), and also slow it down. That requires more torque. You must overcome friction, etc....more torque. Sum all the torques: that's what it takes to turn that crank.
  5. Jul 2, 2014 #4
    It would be the load due to the offset of center of gravity for the object on the table. So if the table is at a 90 degree angle to the floor, the object's COG would be say... 4-6 inches away from the pivot (shaft).
  6. Jul 2, 2014 #5
    Right, but there's also a torque requirement to overcome which was my initial calculation. To induce an input torque of 46.88 in-lbs with 9.4 lbs of force, how do I find how what the subsequent shaft RPM would be? Like I said all the relations I can find involve horsepower, which is unknown.

    It seems either I can decide what the input RPM will be or the input torque, but not both simultaneously, either way I am left missing information. The reason I need to know these is to try and calculate the corresponding forces acting on the gears, and most examples I've seen utilize torque, RPM, and horsepower.
  7. Jul 3, 2014 #6


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    It doesn't matter: you still have to turn the input shaft at 200 RPM to get 1 RPM on the table while applying this torque. If, while the input shaft is turning at 200 RPM and the torque applied to the shaft is 46.88 in-lbs (let's round this up to 48 in-lbs [4 lb-ft] for yucks), the power input required is

    P = TN / 5252 = 4 lb-ft * 200 RPM / 5252 = 0.15 HP

    Because of inefficiencies in the gears, as pointed out by tygerdawg, this input power figure might have to be increased to 0.25 HP.
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