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stepper motor speed and torque calculation

Discussion in 'Sensors and Actuators' started by bobdxcool, Apr 23, 2016.

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  1. bobdxcool

    bobdxcool

    98
    0
    Mar 9, 2012
    The bipolar geared stepper motor I am planning to use has a resistance/ phase of 8.5 ohms per phase, inductance of 2.14mH per phase , normal current per phase is 2 amps, holding torque of motor is 0.8Nm, gearbox ratio is 1:8, holding torque of gearbox is 6.4Nm. The maximum permissible RPM of gearbox shaft is 200 rpm. The input voltage rating of the stepper motor is 24-48 volts. I am planning to use a 24 volts power supply and DRV8825 driver to drive this motor. Can anyone help me on how to calculate the maximum RPM of the stepper motor ? And also usually how long can the motor run at this maximum speed ?

    Also, I intend to use this motor for a motorized cord reel application, where I would retract a cable using this motor and a chain mechanism. The total weight of the cable is around 12 kg. Would this torque of 64kgcm (6.4Nm) be sufficient for this application?

    The diameter of the cable drum is 100mm, and the diameter of the cable is 18mm, with the length of the cable being 29 feet

    Attached datasheet of motor here:
     

    Attached Files:

    Last edited: Apr 23, 2016
  2. Minder

    Minder

    2,969
    626
    Apr 24, 2015
    Go to the Gecko stepper site, they have a sizing details.
    M.
     
  3. elektrroda

    elektrroda

    13
    2
    Sep 1, 2015
    do you want to wind cable on cable drum ? diameter is 100mm , it is 10 cm , radius is 5 cm . When radius has 5 cm , max mass how can to put up is about 14kg , for about 100-200RPM . you must calculate how weight of cable which have to bringh up
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    10,070
    2,151
    Nov 17, 2011
  5. Harald Kapp

    Harald Kapp Moderator Moderator

    10,070
    2,151
    Nov 17, 2011
    ... assuming the cable is wound continuously onto the drum. If the width of the drum is insufficient to hold all 29 feet of cable (28 turns*18mm = ~50cm), you will wind the cable onto the drum in multiple layers. The outer layers have a wider radius than the inner layers, therefore the same torque translates into less force, therefore the max. load that can be lifted is reduced proportionally.
    Each layer increases the radius by ~18mm. The max. radius is therefore Rmax=5cm+n*1.8cm (n being the number of layers).
    From Rmax you can calculate Fmax=6.4Nm/Rmax
    which is equivalent to a max. load mass of Mmax = Fmax/(9.81m*kg/s²).
    Don't forget to subtract the mass of the cable from the load.
     
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