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Torque Calcuation

  1. Jun 1, 2010 #1
    I've got a torque problem. Let me break it down. i've been researching for hours and don't want to commit to calculations without knowing if what I'm doing is 100% accurate.

    I need to calculate torque, and then based on the torque from a motor, I want to calculate the coefficient of friction for the surface area in contact with a rotating cylinder... roughly .300" in contact with the cylinder.

    The shaft is rotating by a DC motor which turns at 1725rpm @ 90v. We've got it clocked down to 1500 rpm through a DC motor speed controller. It goes to a gear box (5:1 ratio) which clocks down the shaft rotations to 300rpm.

    There is a driveshaft, more or less, that allows a large cylinder to be incorporated into the middle of this... about 4" in diameter. It rotates, and 6 samples with a known load of 2.84lb are forced into the spinning cylinder, - .049" in surface area each.

    How do I find torque of the shaft? I know my RPMs, my motor is 1 HP and operates at 1725rpm natively, with a F/L torque of 586 oz-in.

    I want to get torque on the shaft @ 300rpm after the gearbox, and then apply the load and calculate the extra torque needed to spin the shaft after the loaded samples are applied to it.

    I need:
    Torque before a mass is applied to the rotational system,
    Torque after the mass is applied.
    and the CoF of the material.

    I feel like I can use amperage and voltage through the motor to get power, but I'm not positive if its correct... I'm also assuming a 20% loss in power through the right angle gearbox and friction from seals and bearings, etc.

    Thanks!
     
  2. jcsd
  3. Jun 2, 2010 #2
    78 views and no comments? am i out of line?
     
  4. Jun 2, 2010 #3
    If the other 78 are like me, they were hoping to open this one and explain to someone in less than a paragraph how to calculate torque about a pivot! Your question is rather more involved, and a little too on the engineering side of things for me to venture an answer.

    ,Simon
     
  5. Jun 2, 2010 #4
    oh. whoops.
     
  6. Jun 2, 2010 #5

    jack action

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    If I understand correctly, you want to know if the mechanical power is equal to the electrical power? The answer would be yes: Torque X rpm = Voltage X Amperage (in SI units).

    Although, you can expect some minor losses in the transformation process. Manufacturer of high precision motor should be able to tell you the exact relationship between mechanical output and electrical input of their motors.
     
  7. Jun 2, 2010 #6
    the book only supplies full load specs - 1hp and 585 oz in torque i believe, at 90v and 10 amps. i pulled out the multimeter and found that the motor under its max load which it will experience is running at 75.1v and 1.75A.

    through my calculations with those numbers, i diagnosed it running at .34hp and about 3.4 ft-lbs tq.

    not sure if I trust my math. I may post up the calculator i made in excel. I made the calculator with a motor efficiency through the system of 80%.
     
  8. Jun 3, 2010 #7
    will someone look over this excel file and tell me their opinions? I think i've got it all figured out, but would like a second opinion.
     

    Attached Files:

  9. Jun 3, 2010 #8

    jack action

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    First, your motor is obviously not at full load with this voltage and amperage, because:

    75.1 V X 1.75 A = 131 W (= 0.176 hp)

    This your power (electrical) input, which is a lot less than 1 hp.

    Then, assuming an overall eff of 80% (including motor AND gearbox):

    0.176 hp X 0.8 = 0.141 hp

    That is the power output of your system. It has to be lower than the input.

    Then, the final torque is:

    (0.141 hp / 300 rpm) * 5252 = 2.47 lb.ft

    As for your motor specs, here's how I analyzed them (without knowing more about it):

    585 oz.in @ 1725 rpm gives you 1 hp, which is the motor power output (mechanical)

    10 A @ 90 V gives you 1.2 hp, which would be the motor power input (electrical)

    That would mean that in those conditions, your motor has an 80% eff. Is it the same eff at lower power ? I don't know. Although I doubt it, but I'm not an expert on electric motor design either.

    EDIT: I just saw your Excel sheet. In the line "Torque - lb ft", you actually calculate the torque in N.m.

    I don't quite understand how you get a CoF by dividing a torque by a force. That would give you the radius of the drum actually (in ft).
     
  10. Jun 3, 2010 #9
    @jack action

    thank you for your input, its valued.

    I followed this thread (i thought) to get the CoF, but I may have misunderstood it.

    https://www.physicsforums.com/showthread.php?t=151516

    I corrected my torque calculations, thank you. Please do give more input if you have time to look over this other thread.
     
  11. Jun 3, 2010 #10
    alright, this came too easy. please confirm....

    torque = Radius x force

    which means rotational force = torque/radius

    loaded rotational force = 41.83/2" = 20.92 lbsf
    no load rotational force = 38.25/2" = 19.123 lbsf

    resistive force from load = 20.92 lbsf - 19.123 lbs f = 1.7928 lbs

    coefficient friction is (resistive force divided by rotational force) divided by 2

    CoF = (1.7928/20.92)/2 = 0.0428

    This seems realistic as this is plastic wearing against polished steel, in hot water. What do you guys think?
     
  12. Jun 3, 2010 #11

    jack action

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    From you modified Excel sheet:

    Torque:

    loaded: 2.47 lb.ft = 29.62 lb.in
    unloaded: 2.26 lb.ft = 27.08 lb.in

    Rotational forces (dia = 4"):

    loaded: 14.81 lb
    unloaded: 13.54 lb

    Resistive force = 14.81 - 13.54 = 1.27 lb

    Cof = friction force / normal force

    In your case, the normal force would be 2.84 lb (which you gave in the OP)

    So:

    CoF = 1.27 / 2.84 = 0.447
     
  13. Jun 3, 2010 #12
    thanks jackaction.

    Hmmmm. Wouldn't the normal force (if we go back to the known loaded masses) be the 6x the 2.84 then? there is essentially 17.16 lbs of force into the spinning barrel/shaft...

    I followed everything else. I had made an error one of the formulas in my excel sheet.

    CoF = friction force/normal force = 1.27/17.16 = 0.074?
     
  14. Jun 3, 2010 #13

    jack action

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    It is the total normal force. If it is 6 samples with a force of 2.84 lb on each, then you are right, it is 17 lb. If the 2.84 lb is applied on all of the 6 samples, then it is only 2.84 lb.
     
  15. Jun 4, 2010 #14
    it looks like we figured this one out... thank jackaction.

    now the only problem is datalogging motor voltage and amperage with the DAQ so we can get this calculations done in real time :)
     
  16. Jun 14, 2010 #15
    I'm back boys....

    Why doesn't this need to account for the amount of surface area in contact with the rotating shaft?

    I feel like if I had the same force applied on the shaft through more contact surface area, it would slow down more....

    As I just typed that.... I realized the answer to my question. The pressure on the barrell is the same regardless. the linear force into the shaft is all that matters.... nevermind.
     
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