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IndustryArena Forum > Mechanical Engineering > Linear and Rotary Motion > Stepper motor and Ball Screw torque calculation
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  1. #1
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    Jul 2017
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    25

    Stepper motor and Ball Screw torque calculation

    I am trying to calculate the nema23 motor needed for my Z axis.

    My Z axis with spindle has 22lbs and chinese ball screw SFU1610 (diameter 16mm / pitch (lead) 10mm)

    After much searching for discussions and articles, I found just that
    https://buildyourcnc.com/FAQ/879



    Is this formula correct? Does it apply to the use of ball screw as well (example is with lead screw)?

    Effort = Sf + (Load/(2 x pi x (R/p) x Se))

    Load = 97 N (22 lbs)
    R = 0.314961 inch (8 mm)
    p = 0.393701 inch (10 mm)


    Effort = Sf + (Load/(2 x pi x (R/p) x Se))


    Effort = 5 N + (97 N / (2 x 3.14 x (.314961 / .393701) x .2))


    Effort = 5 N + (97 N / (6.28 x .8 x.2))


    Effort = 5 N + (97 N / 1)


    Effort = 5 N + 97 N


    Effort = 102 N = 22 lbs = 352 oz-in

  2. #2
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    Re: Stepper motor and Ball Screw torque calculation

    Hi Vitali- Heres a simple approach. Plus remember that if you need more acceleration it will require more torque. Peter
    Attached Thumbnails Attached Thumbnails torque.jpg  

  3. #3
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    Re: Stepper motor and Ball Screw torque calculation

    Quote Originally Posted by peteeng View Post
    Hi Vitali- Heres a simple approach. Plus remember that if you need more acceleration it will require more torque. Peter
    Thanks Peter!

    I will study more about that

  4. #4
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    Re: Stepper motor and Ball Screw torque calculation

    Very useful thank you

  5. #5
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    Re: Stepper motor and Ball Screw torque calculation

    Hi,
    the acceleration calculation goes like this:

    dw/dt= T / Jtotal

    where Jtotal is the first moment of inertia of all components, both rotating and linear.
    w is angular velocity in radians/s
    and T is the applied torque in Nm.

    The calculation is really about finding the first moment of inertia.

    Jtotal = Jarmature + Jballscrew + Jlinear axis

    Jarmature is easy, just read the manufacturers spec. For example:
    Object reference not set to an instance of an object.
    This is a typical medium/high torque 23 size stepper and has a J=0.84 x 10-4 kg.m2.

    You don't say how long your ballscrew is, I assume 300mm:

    Jballscrew =m . R2/2
    where m is the mass of the ballscrew, R is the radius of the ballscrew
    m(mass) =l x pi R2 x rho where rho is the density of steel, 8000kg/m3
    =0.3 x 0.0082x PI x 8000
    =0.482 kg
    therefore:
    Jballscrew= 0.482 x 0.0082/2
    =0.15 x 10-4 kg.m2

    Jlinear axis=m . (pitch/2 PI)2
    where m is the axis mass (22lb=10kg), pitch is the ballscrew pitch:
    Jlinear axis=10 x (0.01/2.PI)2
    =10 x 0.001592
    =0.253 x 10-4 kg.m2

    We now combine the individual moments:
    Jtotal= (0.84 + 0.15 + 0.25) x 10-4 kg.m2
    =1.24 x 10-4 kg.m2

    Firstly note that the dominant term in the total momentum is actually the armature of the stepper (0.84) followed by the linear axis (0.25), at about 1/3 that of the
    stepper alone! The ballscrew is the smallest but not insignificacnt term (0.15). Note also that the first moment of inertia of the ballscrew depends on mass, ie proportional
    to the square of the radius and also the radius of gyration, again proportional to the square of ballscrew radius. The net result is that the ballscrew moment is dependent
    on the ballscrew radius to the fourth power. Ballscrew radius ( equivalently diameter) is critical in this calculation.

    Lets now calculate the angular acceleration of such a system with 1Nm torque applied:
    dw/dt= 1 / 1.24 x 10-4
    =8064 rad/s2

    and given that the axis velocity is related to angular velocity:
    v= pitch x w/2 .PI so
    dv/dt = pitch x dw/dt/2.PI
    = 0.01 x 8064/6.282
    =12m/s2 or 1.2g which is amazing.

    1Nm =192 oz.in. Remember that steppers lose torque bigtime the faster they go so if you choose low inductance steppers then you might expect to retain 60% of the torque
    at 500 rpm then 192/ 0.6=320 oz.in should be enough torque to accelerate your axis at 1.2 g. Note that if you want thrust in addition to acceleration then you need more torque,
    so 500oz.in is looking like your best bet.

    Craig

  6. #6
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    Nov 2019
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    2

    Re: Stepper motor and Ball Screw torque calculation

    Quote Originally Posted by joeavaerage View Post
    Hi,
    the acceleration calculation goes like this:

    dw/dt= T / Jtotal

    where Jtotal is the first moment of inertia of all components, both rotating and linear.
    w is angular velocity in radians/s
    and T is the applied torque in Nm.

    The calculation is really about finding the first moment of inertia.

    Jtotal = Jarmature + Jballscrew + Jlinear axis

    Jarmature is easy, just read the manufacturers spec. For example:
    [url=https://www.leadshineusa.com/productdetail.aspx?type=products&category=stepper-products&producttype=stepper-motors&series=cm&model=d57cm31[/url] Object reference not set to an instance of an object.[/url webpage
    This is a typical medium/high torque 23 size stepper and has a J=0.84 x 10-4 kg.m2.

    You don't say how long your ballscrew is, I assume 300mm:

    Jballscrew =m . R2/2
    where m is the mass of the ballscrew, R is the radius of the ballscrew
    m(mass) =l x pi R2 x rho where rho is the density of steel, 8000kg/m3
    =0.3 x 0.0082x PI x 8000
    =0.482 kg
    therefore:
    Jballscrew= 0.482 x 0.0082/2
    =0.15 x 10-4 kg.m2

    Jlinear axis=m . (pitch/2 PI)2
    where m is the axis mass (22lb=10kg), pitch is the ballscrew pitch:
    Jlinear axis=10 x (0.01/2.PI)2
    =10 x 0.001592
    =0.253 x 10-4 kg.m2

    We now combine the individual moments:
    Jtotal= (0.84 + 0.15 + 0.25) x 10-4 kg.m2
    =1.24 x 10-4 kg.m2

    Firstly note that the dominant term in the total momentum is actually the armature of the stepper (0.84) followed by the linear axis (0.25), at about 1/3 that of the
    stepper alone! The ballscrew is the smallest but not insignificacnt term (0.15). Note also that the first moment of inertia of the ballscrew depends on mass, ie proportional
    to the square of the radius and also the radius of gyration, again proportional to the square of ballscrew radius. The net result is that the ballscrew moment is dependent
    on the ballscrew radius to the fourth power. Ballscrew radius ( equivalently diameter) is critical in this calculation.

    Lets now calculate the angular acceleration of such a system with 1Nm torque applied:
    dw/dt= 1 / 1.24 x 10-4
    =8064 rad/s2

    and given that the axis velocity is related to angular velocity:
    v= pitch x w/2 .PI so
    dv/dt = pitch x dw/dt/2.PI
    = 0.01 x 8064/6.282
    =12m/s2 or 1.2g which is amazing.

    1Nm =192 oz.in. Remember that steppers lose torque bigtime the faster they go so if you choose low inductance steppers then you might expect to retain 60% of the torque
    at 500 rpm then 192/ 0.6=320 oz.in should be enough torque to accelerate your axis at 1.2 g. Note that if you want thrust in addition to acceleration then you need more torque,
    so 500oz.in is looking like your best bet.

    Craig
    Thanks a lot!

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