Mechatronics Engineering DISCUSSION

How do I size a servo or stepper motor for a ball screw axis moving a 20 kg table?

Started by glow motor sizingball screw torquereflected inertiaacceleration torqueservo selection
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Latest activity · 30 Sep 2026

How do I size a servo or stepper motor for a ball screw axis moving a 20 kg table?

#1

I am designing a horizontal linear axis: a 20 kg table on linear guides, driven by a ball screw with a 5 mm lead, 16 mm diameter and 500 mm length. The table has to reach 0.25 m/s in 0.1 s.

Motor catalogues list holding torque, rated torque, peak torque and rotor inertia, and I do not know which of my numbers to compare with which. How do I work out the speed and torque the motor needs, and does the inertia of the screw itself matter?

Community replies 5

Re: How do I size a servo or stepper motor for a ball screw axis moving a 20 kg table?

#2

Convert the linear requirement to the motor shaft first. With a 5 mm lead, 0.25 m/s is 50 rev/s, which is 3000 rpm or 314 rad/s. Reaching that in 0.1 s is an angular acceleration of about 3140 rad/s².

The thrust needed to accelerate the table is F = m × a = 20 kg × 2.5 m/s² = 50 N. A screw converts thrust to torque with T = F × lead / (2π × η). Taking η = 0.9 for a ball screw, 50 N needs 50 × 0.005 / (2π × 0.9), about 0.044 N·m. That looks tiny, but it is only the part of the torque that moves the table.

Re: How do I size a servo or stepper motor for a ball screw axis moving a 20 kg table?

#3

The screw inertia does matter, and on short-lead screws it is often the largest term. Work everything as inertia at the motor shaft:

  • Table reflected through the screw: J = m × (lead / 2π)² = 20 × (0.005 / 6.283)², about 1.3 × 10^-5 kg·m².
  • Steel screw as a solid cylinder: J = (π / 32) × ρ × L × D⁴ = 0.0982 × 7850 × 0.5 × 0.016⁴, about 2.5 × 10^-5 kg·m².

So the screw has about twice the inertia of the 20 kg table. The coupling and the motor rotor have to be added as well.

Re: How do I size a servo or stepper motor for a ball screw axis moving a 20 kg table?

#4

Putting it together with a motor whose rotor inertia is, say, 3.0 × 10^-5 kg·m²: the total is 1.3 + 2.5 + 3.0 = 6.8 × 10^-5 kg·m², and the acceleration torque is J × α = 6.8 × 10^-5 × 3140, about 0.21 N·m. Add friction: if guide and seal drag is 10 N, that is another 0.009 N·m. Peak demand is therefore roughly 0.22 N·m, and it must be available all the way up to 3000 rpm.

Notice that the rotor accounts for almost half of that torque. Picking a much bigger motor just to be safe raises the rotor inertia and can leave you no better off, so recalculate with the real rotor inertia of each candidate.

Re: How do I size a servo or stepper motor for a ball screw axis moving a 20 kg table?

#5

Which catalogue number to compare depends on the motor type. For a stepper, holding torque is a standstill figure; the usable torque falls with speed, and at 3000 rpm many steppers have only a small fraction of it left. Read the torque-speed curve for your driver voltage and keep the required torque 30 to 50 percent below the curve, because an open-loop stepper that stalls loses position without any error being reported.

For a servo, compare the 0.22 N·m with the peak torque at 3000 rpm, and compare the RMS torque over the whole move cycle with the rated continuous torque.

Re: How do I size a servo or stepper motor for a ball screw axis moving a 20 kg table?

#6

Two design changes are worth checking before choosing the motor. A 10 mm lead screw halves the motor speed to 1500 rpm, which puts a stepper in a much better part of its curve; the reflected table inertia rises four times to about 5 × 10^-5 kg·m², but the required angular acceleration halves. Resolution per step halves too.

If the axis is ever mounted vertically, gravity adds a constant torque of m × g × lead / (2π × η) = 20 × 9.81 × 0.005 / 5.65, about 0.17 N·m, which is almost as much as the whole acceleration torque, and the axis then needs a brake or a counterbalance to hold the load with the power off.

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