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Torque for rotating table

  1. Jul 27, 2017 #1
    Is there a way to figure out the torque needed to rotate a work table without knowing the center of gravity?
    I have a customer who wants to use a 40:1 ratio gearbox to rotate a work station, the gearbox would have an 11" wheel on the input shaft for the operator to turn. However the information that the customer has provided is rather limited as most of his answers are "it's proprietary". What I do know is the largest diameter of the fixture and the part being worked on is 40inches. The length is 83inches and has a weight of 930 pounds.
    Since I am rather limited on information, I was wondering if I could calculate the torque by treating the rotating portion as a solid cylinder, and calculate the moment of inertia, and acceleration?

    Thanks to everyone who replies for the help.
  2. jcsd
  3. Jul 27, 2017 #2


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    I assume there is no change in height of the mass centre.
    You do not specify how quickly it is to turn.
    If time is not an issue then the key question is the friction. The torque needed is little more than that needed to overcome frictional torque.
  4. Jul 28, 2017 #3


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    Try adding a long lever to the existing table (scaffolding pole?) and measuring the torque required to rotate it?
  5. Jul 28, 2017 #4


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    Lookup the maximum ergonomic torque that an average worker can apply with the 11" diameter wheel and that plus a bit more will determine the practical maximum output torque that can result. If a worker cannot turn the wheel, then the the amount of resisting torque is irrelevant.
  6. Jul 29, 2017 #5
    Could you add a top view photo of the workstation?
  7. Jul 30, 2017 #6
    The calculation method is straightforward. I assume your worktable surface is parallel to the floor and no tilt is involved. If not, you must include gravity effects.

    Torque required is "max torque" or "peak torque" needed to accelerate the rotational mass inertia.
    Fundamental equation for rotational mass is T = Jα. (analogous to F=ma for translational masses)
    T = peak torque, J = mass moment of inertia, α = angular acceleration of the inertia J.
    J = sum of the inertias (J-Table + J-Workpiece + J-AnyThingElse). Estimate the workpiece inertia from primitives found in a dynamics reference or internet (cylinders, cubes, etc.)
    α ≈ Δω / Δt
    Δω = max rotational velocity - starting rotational velocity
    Δt = time required to change rotational velocities
    Add additional torque "fudge factors" for friction, windage, odd-shaped workpiece inertias, anything else you can think of. Torque is cheap, use plenty of it.

    If, after all of that, your customer continues to be uncooperative and say "it's proprietary" then tell them it will require 43.791 Kazillion lbf-ft or N-m of torque to turn his load.
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