Industry Applications #02
How to Choose a Servo Motor: Three Numbers That Decide Whether a Mechanism Moves Accurately
"Is this motor strong enough?" is the question asked most often in selection, and the one most often answered wrong. Because not strong enough actually has three completely different meanings: it cannot turn the load, it turns but cannot stop accurately, or it stops accurately but takes too long. Each corresponds to a different number.
This article breaks servo selection down into three must-check numbers, and works through a ball-screw stage from start to finish. Every number can be checked yourself with the formulas given in the text.
1. First, be clear about what "selection" selects
The performance of a servo axis is decided by four things together: the motor, the drive, the mechanism (screw or belt) and the control parameters. Selection can only decide the first two, but whether the target is met is often decided by the third. So the right order is: settle the mechanism first, then calculate the load, and choose the motor last. Doing it the other way round, buying the motor first and then trying to make it work, is the failure mode we see most often.
We use this example throughout:
- Horizontal stage, 30 kg (including fixture and part)
- Ball screw ø20 mm × 600 mm, lead 10 mm, steel
- Target maximum speed 0.5 m/s, acceleration 5 m/s²
- Direct coupling (no reduction)
2. The first number: inertia ratio
The inertia ratio is the moment of inertia of the load referred to the motor shaft, divided by the motor's own rotor inertia. It decides not "whether it can move", but "whether it stops accurately, and how long until it stops vibrating".
2.1 How to calculate reflected inertia
Linear motion is converted to rotational inertia through the screw by way of the lead. A screw with lead L travels L meters per revolution, so the equivalent radius is:
r = L / 2π = 0.010 / 2π = 1.592 × 10⁻³ m
Reflected inertia of the stage:
Jstage = M · r² = 30 × (1.592 × 10⁻³)² = 7.60 × 10⁻⁵ kg·m²
The screw itself is a solid cylinder (J = ½mr²), with steel density 7850 kg/m³:
Jscrew = 7.40 × 10⁻⁵ kg·m²
Adding the two, total load-side inertia = 1.50 × 10⁻⁴ kg·m².
2.2 The same mechanism with different motors
| Motor | Rotor inertia Jm | Inertia ratio | Suitability |
|---|---|---|---|
| 400 W | 0.277 × 10⁻⁴ kg·m² | 5.4 : 1 | Slightly above the 5:1 recommended for measurement stations; settling time on the long side |
| 750 W | 1.13 × 10⁻⁴ kg·m² | 1.3 : 1 | A sensible choice for a measurement station, with ample response margin |
The usual recommended upper limit for general mechanisms is 10:1. But for a measurement station this number is too loose: the higher the inertia ratio, the less the velocity loop gain can be raised, and the result is a longer wait after arrival before vibration stops. Our practical recommendation: for measurement stages that need high response and short settling time, keep it within 5:1, ideally 3:1.
Why? Think of it as the tail wagging the dog. At an inertia ratio of 1:1, motor and load are evenly matched, and when the motor says stop, it stops; at 10:1 the load has ten times the motor's inertia, so after the motor calls a stop the load keeps going and can only be pulled back gradually by feedback, and that "pulling back" is the settling time. Another source of settling time is the motion profile itself, quantified in the motion profile comparison on the industrial automation page: for the same mechanism moving the same 50 mm, the residual vibration is ±27 µm with a trapezoidal profile versus ±5 µm with an S-curve. Both the inertia ratio and the motion profile show up directly in the cycle time.
3. The second number: the torque curve
The "rated torque" on a datasheet is just one point; what really needs to be checked is the torque-speed curve. The curve has two zones with completely different uses.
3.1 Peak torque: set by acceleration
First convert linear acceleration to angular acceleration:
α = a / r = 5.0 / 1.592 × 10⁻³ = 3142 rad/s²
The torque needed to accelerate has to move "the load + the motor itself". Calculating with the 750 W motor:
Taccel = (Jload + Jm) × α = (1.50 × 10⁻⁴ + 1.13 × 10⁻⁴) × 3142 = 0.83 N·m
Then add friction torque (screw, guides and seals; 0.15 N·m in this example):
Tpeak = 0.83 + 0.15 = 0.98 N·m
The peak torque must be below the upper limit of the intermittent zone (the maximum torque curve). For the 750 W motor in this example, 0.98 N·m is even below its rated torque of about 2.4 N·m, so the entire acceleration actually stays within the continuous zone, with a large margin.
3.2 RMS torque: set by the whole cycle
Whether the motor overheats depends not on the peak but on the root mean square torque over the whole motion cycle. Taking a single move at this station as an example (acceleration 0.1 s, constant speed 0.05 s, deceleration 0.1 s, dwell 0.2 s, a 0.45 s cycle; during deceleration friction helps braking, so the torque required is 0.15 − 0.83 = −0.68 N·m):
Trms = √( Σ Ti² · ti / Σ ti ) = 0.56 N·m
So this motor's continuous torque must be ≥ 0.56 N·m. A 750 W servo has about 2.4 N·m at its 3000 rpm rated speed (T = P / ω), a very comfortable margin; with a 400 W motor, the same move gives an RMS torque of about 0.39 N·m against roughly 1.3 N·m available at the same speed, so torque is enough there too, but its problem is the inertia ratio, not torque. This is exactly why the three numbers must be checked together: looking at torque alone selects the wrong motor.
4. The third number: encoder resolution
The encoder decides how fine a change in position the controller "can see". Converted for a screw with a 10 mm lead:
| Encoder | counts / rev | Displacement per count |
|---|---|---|
| 17 bit | 131,072 | 0.076 µm |
| 20 bit | 1,048,576 | 0.010 µm |
| 23 bit | 8,388,608 | 0.001 µm |
Resolution is not accuracy. An encoder that can resolve 0.001 µm does not mean the stage stops to within 0.001 µm.
The whole mechanism lies between the two. Actual positioning accuracy is eaten away by:
- Screw lead error: according to THK's specification table, for an effective thread length of 500–630 mm (the 600 mm screw in this example falls in this row), grade C5 has a representative travel error of ±30 µm and a variation of 23 µm; grade C3 has ±16 µm and 12 µm. This is more than two orders of magnitude larger than the 0.076 µm of a 17-bit encoder
- Backlash and preload: lost motion at reversal, usually at the micrometer level
- Thermal drift: steel's coefficient of linear expansion is about 11 × 10⁻⁶/K, so a 600 mm screw grows 6.6 µm for every 1 °C rise. On a continuously running line, this term is often larger than the mechanical errors
- Structural deformation: deflection caused by the stage's weight, and reaction forces during acceleration
These items should be calculated together in the same accuracy budget; the method is in Tolerance Allocation: Where Does Accuracy Leak Away?
Ball screw accuracy grade values follow THK's technical document "Accuracy of the Ball Screw", with grade definitions per the JIS standard.
So pairing a 23-bit encoder with a C7 grade screw will not make positioning better than the screw itself. Encoder resolution should be at least 5 to 10 times finer than the required positioning accuracy, but the lower limit on bit count is usually set by the velocity loop: velocity feedback is derived by differentiating encoder position, and insufficient resolution turns into velocity quantization noise at low speed, limiting velocity loop gain and low-speed smoothness. Mainstream servo motors mostly come with encoders of 20 bits or more as standard, and the bit count itself is rarely where the cost difference lies; to improve real accuracy, the money should go into screw grade, thermal management and measurement feedback.
Beyond bit count there is one more thing to decide: which principle to use. Optical, magnetic and capacitive encoders are vulnerable to completely different things: stray magnetic fields next to a direct-drive motor, oil and condensation on site, and the space left in the mechanism each rule out different options. A comparison of the three and the selection criteria are in section 2.3, Angle feedback, on the industrial automation integration page.
Another practical question is incremental or absolute. Incremental encoders must home at power-up; absolute ones do not. For a line that is switched on and off several times a day, or must resume production immediately after a power cut, the time saved by absolute encoders quickly pays back the price difference.
5. Selection checklist
- Settle the mechanism first: lead, stroke, stage weight, and whether it is horizontal (a vertical axis also needs gravity torque and a brake)
- Calculate reflected inertia: stage + screw + coupling; do not leave out the screw itself
- Calculate the inertia ratio: generally ≤ 10:1, ≤ 5:1 for measurement stations
- Calculate peak torque: (Jload + Jm) × α + friction, and confirm it is below the maximum torque curve
- Calculate RMS torque: including dwell time, and confirm it falls in the continuous zone
- Check torque at high speed: how much torque the curve has left at the target maximum speed
- Choose the encoder: 5–10 times finer than the required accuracy, and confirm velocity feedback resolution is enough at low speed; choose absolute if the machine is switched on and off frequently
- Leave margin: at least 20–30 % on peak torque, because friction increases with service time
Closing: three numbers, three failures
Not enough torque, and the mechanism cannot turn or reach its acceleration; too high an inertia ratio, and it moves but cannot stop accurately, with long settling times; the wrong encoder, and the specification looks impressive while the accuracy is eaten by the mechanism. Each of the three numbers guards against a different failure, and leaving one out causes trouble in a different place.
Worse, all three failures look much alike in early trial runs: "the machine moves a bit oddly". Working out the three numbers before purchasing is far cheaper than tuning parameters afterwards.
