Crane motions and the hoist drive

Chapter 3 moved a load along one vertical axis. An overhead crane adds two horizontal ones. Found in steelworks, foundries, power stations, ports and assembly halls, it combines every drive challenge so far: a gravity-driven hoist, high-inertia travel motions, precise positioning, and a suspended load that must never be lost. This lesson is course section 4.1.1.

Three motions, three different loads

An overhead travelling crane has three independent motions:

the hoist raises and lowers the load;
the trolley traverses across the bridge girder;
the bridge travels along the runway rails.

They look alike on a drawing, but they load their drives very differently.

	Hoist	Trolley and bridge
Load character	constant torque, active (overhauling)	high inertia plus rolling friction
Dominant term	gravity, $mg$	acceleration, $d\omega/dt$
Quadrants	all four	mainly I and III
Torque at zero speed	essential, to hold the load	not required
Rated on	continuous overhauling torque	peak accelerating torque and duty cycle
Braking	continuous energy absorption when lowering	at each stop

For the hoist, with drum radius $R$ , gear ratio $i$ and efficiency $\eta$ :

T_\text{hoist} = \frac{m g v}{\omega_m\, \eta^{\pm 1}} = \frac{m g R}{i\, \eta^{\pm 1}} \tag{4.1}

with the exponent $+1$ when raising (motoring) and $-1$ when lowering (generating), exactly as in Chapter 3. For the travel motions the torque is mostly inertial:

T_\text{acc} = \left(J_\text{motor} + \frac{m R^2}{i^2}\right)\frac{d\omega_m}{dt} + T_\text{friction} \tag{4.2}

Worked example: hoist against trolley (course §4.1.1)

A crane hoists 8 t at 0.2 m/s with a 0.35 m drum and $\eta = 0.88$ ; its trolley carries the load plus a 2 t carriage to 0.6 m/s in 3 s, against a rolling resistance of 2 % of the weight.

Hoist. $P = mgv = 8000 \times 9.81 \times 0.2 = 15.7$ kW, so the motor delivers $15.7 / 0.88 = 17.8$ kW, continuously, for the whole lift, in both directions.
Trolley. Rolling resistance $0.02 \times 10\,000 \times 9.81 = 1.96$ kN, only 1.18 kW at full speed. The accelerating force $m\,\Delta v / \Delta t = 10\,000 \times 0.6 / 3 = 2.0$ kN is as large as the whole steady resistance, but only for 3 s in each cycle.

The hoist is rated on continuous power and must hold at zero speed; the trolley on its peak accelerating torque, and thermally on its duty cycle. Sizing either by the other's rule gives the wrong machine.

Tutorial 4.1: sizing a hoist

A hoist raises 5000 kg at 0.25 m/s with a 0.4 m drum and $\eta = 0.85$ ; the DC link is at 620 V.

Raising. $P_\text{load} = 5000 \times 9.81 \times 0.25 = 12.3$ kW; the motor supplies $12.3 / 0.85 = 14.4$ kW: the next standard size is 15 kW.
Speed and torque. The drum turns at $0.25 / 0.4 = 0.625$ rad/s. A 4-pole motor near 1450 rpm (151.8 rad/s) fixes the gear ratio $i \approx 243$ , and the raising torque is $14\,400 / 151.8 \approx 95$ N·m.
Lowering. Gravity now drives the motor and the losses subtract: $P_\text{gen} = 12.3 \times 0.85 = 10.4$ kW returned to the DC link.
Resistor. $R = 620^2 / 10\,400 \approx 37$ Ω: a standard 33 to 39 Ω resistor rated for at least 11 kW, with margin for decelerations. A crane that lowers heavy loads for much of its cycle would return those 10 kW to the grid with an active front end instead.

Try it: the crane hoist of Tutorial 4.1

Load m5,000 kg

Hoisting speed v0.25 m/s

Efficiency η0.85

DC-link voltage U_dc620 V

Raising: quadrant I, motoring. The motor supplies m g v plus the losses.

Load power m g v: 12.26 kW
Gear ratio (motor at 1450 rpm, drum 0.4 m): 243
Motor torque: 95.0 N·m
Motor power T·ω: 14.43 kW

Share of the cycle spent lowering15 %

The resistor's continuous rating follows the duty cycle, not the peak.

Power returned when lowering: 10.42 kW
Braking resistor U_dc²/P: 36.9 Ω
Peak power absorbed: 10.4 kW
Continuous rating needed: 1.56 kW

T = m g R / (i η^±1) ; R = U_dc² / P_gen ; P_cont = P_gen · t_lower / t_cycle

Predict first

In the widget, switch from Raise to Lower with the Tutorial 4.1 load. What happens to the motor torque?

It keeps its sign and its valueIt keeps its sign, but falls from 95 to about 69 N·mIt reverses sign

What hangs on the hook

The load is coupled to the hoist by an end-effector suited to the material, and each adds to the control:

a hook block for slung or palletised loads;
a lifting magnet, a large DC electromagnet for scrap and plate. Its control adds a DC supply, a release sequence for the residual magnetism and, critically, battery back-up, so that a power failure does not drop the load;
a grab (clamshell bucket) for bulk material. It needs two coordinated hoist ropes, one to hold and one to open and close the grab.

The hoist drive requirements underneath stay the same.

FoundationStart here if this is new to you

Lifting a bucket of water up a well is the hoist: gravity pulls all the time, even when you stop. Pushing a heavy cart along a floor is the trolley: hard to get moving and hard to stop, but once it rolls it needs little force. A crane motor has to do the first job and the second job at once, with very different muscles.

ExplorerGo deeper: derivations and open questions

Light hook, fast hoist. With field weakening, the hoist of Tutorial 4.1 can lift an empty hook at two or three times the rated speed. Using the constant-power idea of Chapter 1, what is the largest load it can lift at twice the rated speed with the same 15 kW?

Wheel slip. A bridge accelerating too fast slips its wheels on the rail. With a steel-on-steel adhesion coefficient of about 0.15 on the driven wheels, and half the wheels driven, what acceleration can a 60 t bridge reach?