Compound Gear Train That Will Not Lift the Load
A boat-lift winch uses a 2.5 N·m electric motor turning at 1,200 rpm. The motor's 12-tooth pinion drives a 36-tooth gear, which shares a shaft with a 10-tooth sprocket driving a 40-tooth sprocket on the cable drum. Each reduction stage is about 85 percent efficient. The drum is 0.10 m in diameter. The cradle and boat together require a steady 24 N·m at the drum to break free of the water and lift, and they must be raised 1.2 m without stalling. The bearings are freshly lubricated and the gear teeth are intact. The motor draws normal running current right up to the moment the load stops moving, and the drum rocks back and forth a few degrees.
1. Compute the drum speed and the torque actually available at the drum. Will the winch lift the load?
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Speed: two reduction stages, 3:1 then 4:1.
- 1,200 ÷ 3 = 400 rpm on the intermediate shaft
- 400 ÷ 4 = 100 rpm at the drum
Ideal torque multiplication: 2.5 N·m × 3 × 4 = 30 N·m
Real torque: 30 × 0.85 × 0.85 = 30 × 0.7225 = 21.7 N·m
Load required: 24 N·m
Verdict: it will not lift. Available torque is 21.7 N·m against a 24 N·m requirement - a shortfall of about 2.3 N·m, roughly 10 percent. The winch is not broken; it is under-geared for this boat.
Note the ceiling: even with zero losses, this ratio only reaches 30 N·m. So the fix has a hard limit if it relies on friction reduction alone.
2. The owner wants the cheapest possible fix. Evaluate 'just free up the bearings' against 'add one more 2:1 reduction stage,' and recommend one with numbers, including whether the lift speed is still acceptable.
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Option A - eliminate friction losses:
- Torque ceiling = 30 N·m (100% efficiency). Margin over 24 N·m = 6 N·m.
- To make 24 N·m you need total efficiency ≥ 24 ÷ 30 = 80%, i.e. about 89.4% per stage versus the current 85%.
- That is a 4.4-point gain per stage from lubrication alone - possible on paper, but it consumes nearly all the margin. Any binding, cold grease, or a slightly heavier boat puts it back in the stall.
Option B - add a 2:1 stage:
- New total ratio: 3 × 4 × 2 = 24:1 → 1,200 ÷ 24 = 50 rpm at the drum
- Ideal torque 2.5 × 24 = 60 N·m; with three stages at 85%, 60 × 0.614 = 36.8 N·m - a 53% margin over the 24 N·m requirement
- Drum circumference = π × 0.10 = 0.314 m; line speed = 50 × 0.314 = 15.7 m/min, so a 1.2 m lift takes about 4.6 s. Well inside the requirement.
Recommendation: Option B. It roughly doubles the torque margin and the speed penalty is irrelevant at this scale. Option A is a gamble that spends the entire safety margin to save one gear.
3. The operator reports the drum rocks back and forth a few degrees while the motor hums, but never spins free. What does that symptom rule in and rule out, and what would a stripped gear look like instead?
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What the symptom means:
- The motor is energized and producing torque, but the torque is being absorbed and the rotor is stalling near its breakdown point. That matches an under-geared or overloaded train, exactly what the numbers show (21.7 N·m available vs 24 N·m demanded).
- The few degrees of rock is wind-up: the compliant cable and shaft twist elastically, the load slips back, and the cycle repeats.
Ruled out by the symptom:
- Stripped or sheared gear teeth / broken key: the motor would spin freely at no-load speed with the drum dead still, and current would drop, not stay normal. A few degrees of movement is impossible with a fully stripped tooth path.
- Open circuit or dead motor: no hum, no torque, drum silent.
Diagnostic test: briefly add load by hand while it is stalled. If the drum resists in proportion to the hum, the train is intact and the problem is torque. If the drum turns freely while the motor runs, the fault is mechanical downstream of the motor.