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Torque calculation needed

  1. Oct 30, 2008 #1
    Can anyone help?

    I need a calculation to find the force required @ the shaft indicated, to acheive torque indicated at the locating faces.
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  2. jcsd
  3. Oct 30, 2008 #2

    brewnog

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    I can't see what that 'torque multiplier" is. Is this a homework question?

    The formulae you need are:

    torque = force x distance
    power = torque x angular speed

    Here, distance is the radius from the axes at which you apply your force.
     
  4. Oct 31, 2008 #3
    Hello Brewnog,

    The torque multiplier has a 125:1 ratio and drives the shaft via the square cut out, this is actualy a work related question, but I'm still unclear as to the solution. This is not really my field of expertise, as I live in a cardboard box just past junction38 on the M1.

    :grumpy:
     
  5. Oct 31, 2008 #4
    Hello Brewnog,

    The torque multiplier has a 125:1 ratio and drives the shaft via the square cut out, this is actualy a work related question, but I am still unclear as to the solution. this is not really my field of expertise, as I live in a cardboard box just past junction38 on the M1
     
  6. Oct 31, 2008 #5
    Well, I can't really understand how your torque multiplier works but if it has a ratio 0f 125:1 then

    T1 = T2 x 125

    or T2 = T1 x 125 depending on what part the torque is acting on.
     
  7. Oct 31, 2008 #6
    Hello Topher925,

    Surely it is not just a direct corelation? doesn't the gearing also multiply the amount of force? I don't know whether I am right but if I apply 19.2 ft/lb x 125 = 2400 x 5" (radius of gear) = 12000 ft/lb.

    Just got to nip out to feed the ferret
     
  8. Oct 31, 2008 #7

    brewnog

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    Well I'll answer, but only since we Yorkshiremen have to stick together.

    If the output gear were twice the size of the input gear, the torque would be doubled (speed would be halved) as per my second equation. Because the gears are the same size, the speed of the output gear is the same as that of the input gear, and torque is not increased. In fact, you lose a bit of torque to mechanical losses.

    I still can't see how the torque multiplier works, but to achieve 12,000ft-lbs at your output shaft, and your torque multiplier is 125:1, then the input torque will be 12,000 / 125, or 12,000 x 125, depending on which way round your torque multiplier multiplies. The input torque would be a little over 96ft-lbs or 1,500,000ft-lbs respectively.

    The first formula I quoted only applies if you are dealing with a force, not a torque.

    I do hope that's not a euphemism.
     
  9. Oct 31, 2008 #8
    The torque multiplier is an enclosed arrangement of gears, with an input socket (to take a torque wrench), and output sprocket,(which fits the drive shaft), designed so you can set your torque wrench to 125th of the final output.

    I'm still a little worried about the 96ft/lb figure, we started off using a lever and chain, knowing exactly how much force to apply, this moved the sleeve from hand-tight to 12000ft/lb approx 1" arouhd the diameter, using this as a guide we have torqued to approx 34ft/lb to acheive the same movement.

    Got to go, the lurcher is pining
     
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