Stepper Axis Losing Position on the Conveyor Indexer
An Abitur project team has built a conveyor indexer for its solar-boat hull transport. A NEMA 23 stepper (1.8° step angle, holding torque 3.0 N·m) drives a pulley of 40 mm pitch diameter through a belt. The gantry mass is 6 kg and the measured running friction is 12 N. The index move is 300 mm in 0.60 s with a trapezoidal profile at 8 m/s² acceleration. Microstepping is set to 8, the supply is 24 V and the driver is limited to 20 kHz. The datasheet curve for this motor at 24 V gives 0.82 N·m at the required pulse rate. The team's design rule is a 50 % torque margin. In trials the axis positions correctly at low throughput, but after about 40 cycles at full throughput the zero point has drifted 3 mm; the drift disappears when throughput is reduced. Retensioning the belt made the drift appear after only 25 cycles.
1. Show by calculation whether the axis satisfies the torque requirement. Include the pulse frequency and check it against the driver limit.
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Step 1 - kinematics v = 300 mm / 0.60 s = 500 mm/s Pulley circumference = π × 40 mm = 125.7 mm Rotational speed = 500 / 125.7 = 3.98 rev/s Full steps per second = 3.98 × 200 = 796 steps/s Pulse frequency = 796 × 8 (microstepping) = ≈ 6,370 Hz, comfortably below the 20 kHz driver limit - so the driver is not the bottleneck.
Step 2 - required torque at the pulley F = m·a + F_friction = (6 × 8) + 12 = 48 + 12 = 60 N Torque = F × r = 60 N × 0.020 m = 1.20 N·m
Step 3 - apply the design rule Required with 50 % margin = 1.20 × 1.5 = 1.80 N·m
Step 4 - compare with the motor Available at speed: 0.82 N·m
- Without any margin: 0.82 / 1.20 = 68 % of what is needed.
- Against the team's own rule: 0.82 / 1.80 = 46 %.
Verdict The axis fails by a factor of about 2.2 against the design rule and fails even the un-margined requirement. The motor is undersized for this acceleration, or the acceleration is too aggressive for this motor - those are two descriptions of the same deficit. Note that the holding torque of 3.0 N·m is irrelevant here: it is quoted at standstill, and the failure happens during acceleration, where the torque curve has already fallen to 0.82 N·m.
2. Diagnose the cause of the position drift. Explain in particular why retensioning the belt made the symptom appear sooner, and why belt slip is not the best explanation.
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Diagnosis: the axis is losing steps in open loop during acceleration.
Evidence pattern
- Drift accumulates (3 mm after ~40 cycles) instead of appearing as a single event. Open-loop step loss does exactly this: every acceleration phase where demand exceeds available torque loses a few steps, and the error is never corrected because there is no encoder.
- Drift only at full throughput: at lower throughput the acceleration and speed are lower, the operating point moves left on the torque curve, and the margin becomes sufficient.
- Drift disappears when throughput is reduced - consistent with a torque-margin problem, not with a mechanical fault that would persist.
Why retensioning made it worse Higher belt tension raises the normal force on the pulley bearing and the idler, and therefore raises friction. The design calculation used 12 N; retensioning increased the real value, so the required torque rose above the already insufficient 1.20 N·m. The margin did not change - the load did. This is also evidence against belt slip: if slip had been the cause, tensioning should have helped.
Why belt slip is a weak hypothesis Slip is non-systematic, usually audible, and would show as a sudden jump rather than a steady accumulation. It also does not explain the clean dependence on throughput.
Secondary suspects worth ruling out Driver current reduction from thermal derating, resonance in the mid-speed range, and supply voltage sag on the 24 V rail under acceleration.
3. Write the root-cause paragraph for the project log: state the finding, the quantitative basis, one recommended change and one uncertainty. Your recommendation must be justified numerically.
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Log entry
Finding. The indexer loses steps during acceleration. At 8 m/s² the axis needs (6 × 8 + 12) × 0.020 = 1.20 N·m at the pulley, or 1.80 N·m with the team's 50 % margin, while the motor delivers 0.82 N·m at 6,370 Hz. The axis therefore runs at 68 % of its unmargined requirement. Recommendation. Halve the acceleration to 4 m/s², which gives (6 × 4 + 12) × 0.020 = 0.72 N·m; this fits within 0.82 N·m but leaves only a 14 % margin, so it must be combined with a mechanical change - reducing the pulley to 20 mm pitch diameter halves the torque demand to 0.36 N·m. The resulting speed doubles to about 12,700 Hz, which is still inside the 20 kHz driver limit, but the torque available at that frequency must be re-read from the curve before the change is signed off. Uncertainty. The 12 N friction figure is an estimate from a hand-pull test, and the datasheet torque is quoted at 25 °C; both will be worse in a warm enclosure.
Why this is the right shape for the log It separates what is measured from what is assumed, gives the arithmetic so a supervisor can check it, and names the one number that must be re-measured before the fix is accepted - which is what turns a diagnosis into an engineering record.