Rescue Hoist: Torque, Efficiency, and a Comfortable-Looking Number
A rescue hoist is being evaluated for a new helicopter fit. The electric motor produces 40 N·m of torque at its output shaft. It drives a gearbox with a 12:1 reduction, and the gearbox output turns the hoist drum, which has a radius of 0.150 m. The gearbox and drum bearings together waste 15% of the torque delivered to the drum, so only 85% is available for lifting. The requirement is to lift a 350 kg rescue load at a steady rate, with cable mass neglected. A maintainer argues the fit is fine because 'the ideal numbers give 3,200 newtons, which is well over 350 kilograms.' Use g = 9.81 m/s².
1. Compute the torque actually available at the drum and the greatest mass the hoist can lift.
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Ideal drum torque: 40 N·m x 12 = 480 N·m Real drum torque: 480 x 0.85 = 408 N·m
Cable tension available: torque / drum radius = 408 / 0.150 = 2,720 N Equivalent mass: 2,720 / 9.81 = 277 kg
Required for the 350 kg load:
- Weight: 350 x 9.81 = 3,433.5 N
- Torque needed at drum: 3,433.5 x 0.150 = 515 N·m
The hoist is short by 515 - 408 = 107 N·m, about 21% of the requirement, and can lift roughly 277 kg - some 73 kg short. The maintainer's 3,200 N was the ideal figure and, more importantly, was never compared to a weight at all: 3,200 N is about 326 kg, not 'well over 350 kg'.
2. What motor torque would be needed to lift the 350 kg load with this gearbox, and how large is the shortfall?
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Work backwards through the same losses:
- Torque required at the drum = 515 N·m
- Torque required at the gearbox input = 515 / (12 x 0.85) = 515 / 10.2 = 50.5 N·m
The motor delivers 40 N·m, so the shortfall is 10.5 N·m, about 26% of the motor's output.
Note the difference between the two framings: the load is 21% beyond capacity, but the motor is 26% short, because the shortfall is measured against a smaller base. Quoting the motor requirement (50.5 N·m) is the useful number - it tells the engineer what to specify, whereas 'it won't lift it' does not.
3. What is the minimum standard reduction ratio that would satisfy the requirement, and what does choosing it cost?
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Required drum torque is 515 N·m and the motor supplies 40 N·m through an 0.85-efficiency train:
- Required ratio = 515 / (40 x 0.85) = 515 / 34 = 15.15
Ratios come in discrete steps, so the answer is 16:1, not 15.15:1. Choosing 16:1 gives a drum torque of 40 x 16 x 0.85 = 544 N·m, i.e. 3,627 N of tension, about 370 kg - a 5% margin rather than a paper-thin one.
The cost: with motor speed fixed, drum speed scales as 1/ratio. Going from 12:1 to 16:1 makes the drum turn at 12/16 = 0.75 of its previous rate, so the hoist lifts 25% more slowly and takes 33% longer to recover the same cable length. A hoist that is strong enough but too slow is a real operational failure in a rescue, so the ratio decision must be checked against recovery time, not just against the lifting requirement.