Loads on a winch drive

Every clutch and brake calculation starts with the same four numbers: the torque the device must carry, the speed of the shaft it sits on, the inertia it must accelerate or stop, and the energy it must turn into heat. Get those right and the rest of this book is arithmetic. Get one wrong, usually by taking it from the wrong shaft or the wrong direction of travel, and a well-made brake can be too weak by a factor of two.

This chapter sets out those numbers for the SW-25 slipway winch. It works out the rope pull hauling up and letting down, carries it through the gear train to the drum, the clutch shaft and the motor, finds the inertia the drive sees, and estimates the energy of a haul and a lowering. It uses two KEVOS® tools: the torque to accelerate a rotating load and the reflected inertia tool. Its mistake is one that appears in many brake calculations: counting a resistance that may not be there.

The slipway and its forces

The cradle runs on rails up a slope of 1 in 12: 1 m of rise for every 12 m of horizontal run. The angle is arctan(1 ÷ 12) = 4.764°, with sin 0.08305 and cos 0.99655. A 25 t boat and cradle weighs 25,000 × 9.81 = 245,250 N.

Two forces act along the slope. Gravity pulls the load down the slope with 245,250 × 0.08305 = 20,367 N. Rolling resistance opposes whichever way the cradle moves, with 0.02 of the force pressing it onto the rails: 0.02 × 245,250 × 0.99655 = 4,888 N. Hauling up, the rope must overcome both; letting down, the rolling resistance helps hold the load back.

Figure 1Rope pull on a wheeled cradle on a slope

F up = m g (sin θ + Crr cos θ)

F down = m g (sin θ − Crr cos θ)

m g

weight of boat and cradle, N

m g sin θ

component of weight along the slope

Crr m g cos θ

rolling resistance, always opposing motion

This is an illustration. The rope runs in a single part from the drum to the cradle, so the rope pull is the load on the drum.

From the rope to each shaft

The drum is 600 mm in diameter to the rope centre, a radius of 0.3 m. Its torque is the rope pull times 0.3 m: 7,576.5 N·m hauling and 6,110.1 N·m holding with no rolling resistance. Ignoring the change in diameter as rope layers build up is acceptable here because the SW-25 spools a single layer for its 60 m slipway.

The drive train has three shafts that matter. The motor runs at 1,450 rpm. A 20:1 helical gearbox with an efficiency of 94% brings that to 72.5 rpm at its output, which is the shaft carrying the friction clutch. An open spur pinion and gear of 22.5:1 and 95% efficiency drive the drum at 3.222 rpm. The overall ratio is 450:1, and the overall efficiency is 0.94 × 0.95 = 0.893.

Figure 2The SW-25 drive train, from motor to rope

The torque on any shaft follows from the drum torque, the ratio between that shaft and the drum, and the efficiency of the gears in between. The efficiency works in opposite directions in the two cases. Hauling, the motor drives the load and the gears lose power on the way, so the motor needs more torque than the ratio alone suggests: 7,576.5 ÷ (450 × 0.893) = 18.85 N·m. Lowering or holding, the load drives backwards through the gears and the losses help the brake, so the load torque at the motor is less than the ratio suggests: 6,110.1 ÷ 450 × 0.893 = 12.13 N·m.

Figure 3SW-25 torque and speed on each shaft

Shaft	Speed	Torque hauling	Torque holding, no rolling resistance
Motor	1,450 rpm	18.85 N·m	12.13 N·m
Clutch shaft (gearbox output)	72.5 rpm	354.5 N·m	258.0 N·m
Drum	3.222 rpm	7,576.5 N·m	6,110.1 N·m

The rope speed while hauling is π × 0.6 × 3.222 ÷ 60 = 0.1012 m/s, or 6.07 m/min, so a full 60 m haul takes just under ten minutes. The power in the rope is 25,255 × 0.1012 = 2.56 kW, and the power at the motor is 18.85 × 151.84 = 2.86 kW, where 151.84 rad/s is 1,450 rpm. The 5.5 kW motor has a rated torque of 5,500 ÷ 151.84 = 36.2 N·m, so it hauls at about half load, which leaves room for starting.

Inertia, and where it lives

A brake that only had to hold a load would need only torque. Stopping a moving load also needs inertia, because the brake must take away the kinetic energy as well as resist the load. For a winch the inertia has two parts: the rotating parts themselves, and the boat and cradle moving along the slope.

A mass moving in a straight line at the rope behaves, seen from the drum, like a rotating inertia of m r². For 25 t at 0.3 m that is 25,000 × 0.3² = 2,250 kg·m². The drum and its gear add an estimated 25 kg·m², for 2,275 kg·m² at the drum. Seen from another shaft, inertia scales with the square of the speed ratio: at the motor the whole load and drum look like 2,275 ÷ 450² = 0.0112 kg·m². The motor’s rotor, brake disc and coupling add about 0.040 kg·m², and the gearbox and clutch parts about 0.0006 kg·m², for 0.0518 kg·m² in total at the motor shaft.

That result surprises people who first meet it. At 450:1, the 25 t boat contributes less inertia at the motor than the motor’s own rotor does. The load’s weight still matters enormously, through the torque it applies, but its mass hardly affects how quickly the motor speeds up or stops.

Starting the haul

The reflected inertia tool works this out for a drive. It takes the load inertia at the load shaft, the motor’s inertia, the reduction ratio, the gearbox efficiency, the load speed to reach, the time to reach it and the load torque at the load shaft. It returns the load inertia reflected to the motor, the inertia ratio, the motor speed, the torque to accelerate the motor and the reflected load, the load torque at the motor, and their sum, the peak motor torque. It divides both the load inertia and the load torque by the efficiency as well as the ratio, which is the conservative choice for a motor driving its load. It assumes constant acceleration from rest.

Figure 4Starting the SW-25 haul in the reflected inertia tool

1. 1Reflected load inertia0.0126 kg·m²
2. 2Inertia ratio0.31 : 1
3. 3Motor speed1,450 rpm
4. 4Acceleration torque7.98 N·m
5. 5Load torque at the motor18.85 N·m
6. 6Peak motor torque26.84 N·m

The peak of 26.8 N·m is well under the starting torque of a direct-on-line induction motor, which is typically around twice its rated torque. Two thirds of it is load torque; one third accelerates the motor and everything behind it. The inertia ratio of 0.31:1 tells the same story as before: the motor is accelerating mostly itself.

The same start can be seen from the drum with the torque to accelerate a rotating load tool. It takes the inertia, the starting and final speeds, the time and a friction or load torque, and returns the angular acceleration, the accelerating torque, the total torque, the revolutions turned, the kinetic energy change and the peak power at the end of the change. With 2,275 kg·m², from rest to 3.222 rpm in 1 s and a load torque of 7,576.5 N·m, it gives an angular acceleration of 0.337 rad/s², an accelerating torque of 767.7 N·m, a total torque of 8,344 N·m and a kinetic energy of only 130 J. The drum turns 0.027 revolutions while it speeds up, which is 51 mm of rope.

The drum figure does not include the motor’s own inertia, which is why the two tools do not agree exactly: the motor’s 0.040 kg·m² at 1,450 rpm holds 0.5 × 0.040 × 151.84² = 461 J, more than three times the energy of the boat and drum together.

Energy in a haul and a lowering

The energies that matter for heat are much larger than the kinetic energies. Hauling the full 60 m raises the load by 60 × 0.08305 = 4.98 m and stores 245,250 × 4.98 = 1,222 kJ as potential energy. The motor delivers more than that, because rolling resistance and gears also take their share: the work in the rope alone is 25,255 × 60 = 1,515 kJ.

Lowering, that potential energy must go somewhere. If the cradle is let down on the band brake with the clutch out, the brake absorbs all of it except what rolling resistance takes: 15,479 × 60 = 929 kJ in an ordinary lowering, and up to the full 1,222 kJ if the rails are slippery and rolling resistance is low. Chapter 8 returns to this: it is the band brake’s main heat load, and it is thousands of times the 0.35 kJ of kinetic energy the boat and drum carry at a lowering speed of 10 m/min.

Figure 5Energies in one SW-25 cycle, 25 t on the 60 m slipway

Work in the rope, full haul1515

Potential energy gained1222

Brake heat, lowering with rolling resistance929

Kinetic energy at the motor shaft0.6

The mistake: counting a resistance that may not be there

The yard’s first brake calculations used the lowering pull of 15,479 N, which includes the help of rolling resistance. They were checked by a second engineer, who asked a question every brake calculation should answer: which resistances can be relied on at the moment the brake is needed?

Rolling resistance on a slipway varies widely. Wheels with seized bearings drag; greased rails and new bearings run freely; a cradle on wet, weed-covered rails late in a tide may have very little resistance left. A brake that relies on rolling resistance to hold its load is relying on the condition of a part nobody inspects as a brake component. The yard’s rule became: brakes are sized to hold and stop the load with rolling resistance taken as zero. The holding pull rises from 15,479 N to 20,367 N, and the drum torque from 4,643.7 N·m to 6,110.1 N·m: 31.6% more.

The same reasoning applies to any brake calculation. Friction in bearings and seals, wind resistance and the drag of a process helps in normal running, but a brake is a safety device, and the cases it exists for are often those in which the help is missing. Chapters 7 to 10 use the zero-rolling-resistance figures for holding and stopping, and use the lowering figure only to estimate heat in normal use.

Figure 6Before using a load in a clutch or brake calculation

• Avoid: Each force identified with its direction for each motion: hauling, lowering, holding
• Avoid: Resistances that cannot be relied on removed from brake cases
• Avoid: Torque taken to the shaft where the device sits, with efficiency in the right direction
• Avoid: Inertia of linear masses converted to the shaft by m r² and the ratio squared
• Avoid: Potential energy of a lowering estimated, as well as kinetic energy
• Ask: Rope stretch and shock from snatch loads (taken up with the clutch in Chapter 4)

What the numbers mean for the devices

Seen together, the numbers show why the SW-25 has such different devices on its different shafts. The motor brake works at 1,450 rpm and about 12 N·m, so it can be small, but it sees every stop of the drive at full motor speed. The clutch works at 72.5 rpm and a few hundred newton metres. The drum brakes work at a few revolutions a minute and several thousand newton metres, so they are large, but their speeds are so low that a stop releases little kinetic energy; their heat comes from controlled lowering.

Two further points apply to every later chapter. First, a device placed on a slow shaft needs more torque but slips through a smaller angle in a given time; a device on a fast shaft needs less torque but slips through more angle, and the heat it generates is the same for the same energy absorbed. Second, the regulatory frame matters: winches are plant under the model WHS laws, and their design is covered in Australia by AS 1418.2 for serial hoists and winches. This book uses the yard’s working figures for brake factors and tests, labelled as illustrations, not values from that standard.

Exercise: a lighter boat on a steeper slip

Problem. A second slipway at the same yard has a slope of 1 in 10. Its winch has the same 600 mm drum and the same 450:1 drive at 89.3% efficiency. The largest boat and cradle weigh 15 t together, and the rolling resistance is 0.02. Find (a) the hauling rope pull and the hauling torque at the motor, (b) the holding torque at the drum and at the motor with rolling resistance ignored, and (c) the load inertia at the motor from the boat and cradle alone.

Answer. The slope angle is arctan(0.1) = 5.711°, with sin 0.09950 and cos 0.99504. The weight is 15,000 × 9.81 = 147,150 N.

(a) Hauling pull = 147,150 × (0.09950 + 0.02 × 0.99504) = 147,150 × 0.11940 = 17,570 N. Drum torque = 17,570 × 0.3 = 5,271 N·m. Motor torque = 5,271 ÷ (450 × 0.893) = 13.12 N·m.

(b) Holding pull = 147,150 × 0.09950 = 14,642 N. Drum torque = 14,642 × 0.3 = 4,393 N·m. Motor torque = 4,393 ÷ 450 × 0.893 = 8.72 N·m.

(c) Inertia at the drum = 15,000 × 0.3² = 1,350 kg·m²; at the motor = 1,350 ÷ 450² = 0.00667 kg·m².

The error to avoid. Using the hauling efficiency the wrong way round for holding gives 4,393 ÷ (450 × 0.893) = 10.93 N·m at the motor, 25% more than the true load torque. That error is on the safe side for the brake, but the same slip in a calculation of the motor’s lowering duty, or of how much the gear losses help, would not be.