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Precision Piston Pump Stepper Motor Calculation and Selection

A stepper driver converts pulse commands into discrete rotation. Pulse count plans displacement and pulse frequency plans speed, but command following still depends on the motor, driver, load and acceleration profile.

1. Start from the motion requirement

Define stroke, cycle time, dwell, moving mass, friction, external force, positioning requirement and transmission. For a piston pump, include hydraulic load and seal friction. Do not select by frame size or holding torque alone.

2. Holding torque, running torque and microstepping

Holding torque describes resistance to external torque while energized and stationary. Running selection must use the torque-speed curve at the intended voltage, current and driver settings.

Microstepping reduces theoretical angle per input pulse and can improve smoothness. Higher command resolution does not produce equal gains in positioning accuracy or repeatability; verify the motor, load and mechanics.

3. Common conversion formulas

For full-step angle θ, full steps per revolution N = 360 ÷ θ. With microstep factor m and input frequency f, motor speed n = f ÷ (N × m) rev/s and rpm = 60 × n.

N = 360 ÷ θ; n = f ÷ (N × m); rpm = 60 × nθ in degrees and f in hertz.

If motor-to-load speed ratio is G and linear travel per load-axis revolution is C, pulse equivalent δ = C ÷ (G × N × m). Use lead for C in a screw drive.

δ = C ÷ (G × N × m)Load and inertia torque must also be reflected to the motor shaft with real efficiency and inertia.

4. Example: 400 mm horizontal reciprocating platform

Assume 400 mm one-way travel, a 4 s round trip with no end dwell, 10 kg moving mass, belt drive, 0.1 s acceleration and 0.1 s deceleration per direction, 1.8 s constant speed and friction coefficient 0.1. This is a calculation example, not a final equipment selection.

The one-way time is 2 s. From the trapezoidal velocity-profile area, 0.4 = vmax × (0.1/2 + 1.8 + 0.1/2), so vmax ≈ 0.2105 m/s and acceleration ≈ 2.105 m/s². Friction is approximately 9.8 N and acceleration force 21.05 N, giving about 30.85 N before pulley inertia and losses.

vmax = 0.4 ÷ (0.1/2 + 1.8 + 0.1/2) ≈ 0.2105 m/sOne-way travel is 2 s: 0.1 s acceleration, 1.8 s constant speed and 0.1 s deceleration.

5. Transmission ratio, microsteps and speed

With a 30 mm load pulley, circumference is about 94.25 mm. Direct drive with a 1.8° motor requires more than 9.42 microsteps for theoretical pulse equivalent below 0.05 mm; this only compares command resolution.

With 3:1 reduction—three motor revolutions per load-axis revolution—and four microsteps, pulse equivalent is about 94.25 ÷ (3 × 200 × 4) = 0.0393 mm/pulse.

Calculated itemExample result
Maximum load-axis speed≈2.234 rev/s
Maximum motor speed at 3:1≈6.70 rev/s or 402 rpm
Input pulse frequency at 4 microsteps≈5,361 pulse/s
Theoretical pulse equivalent≈0.0393 mm/pulse

6. Dynamic-torque check and final acceptance

Ignoring rotational inertia and losses temporarily, required motor-shaft torque is about 30.85 × 0.015 ÷ 3 = 0.154 N·m. A preliminary 2× margin gives 0.309 N·m, but pulley and rotor inertia, efficiency, transmission losses and external force must still be added.

Check the motor-driver dynamic torque near 402 rpm and throughout acceleration. The example 57HS09 motor cannot be accepted from its 0.9 N·m holding torque alone. Finally test missed steps, temperature rise, vibration and reciprocating position at the worst load.

Conclusion

Pulse equivalent proves command resolution only. Final selection requires the torque-speed curve at the actual driver settings and verification of acceleration, temperature, vibration, missed steps and positioning under the most demanding load.

References

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