Counterweight, roping and reflected torque

The pumps and fans of Chapter 2 resist motion. An elevator is different: a loaded car on a rope stores potential energy, and gravity will release it whether the drive agrees or not. This chapter follows the three requirements that come with that: four-quadrant operation with the energy handled safely, full torque at zero speed before the brake opens, and protection that fails into the stopped state. This lesson is course section 3.1.

The counterweight

In a traction lift the car hangs on ropes that pass over a driven sheave to a counterweight on the other side. The counterweight balances the car and part of the load, so the motor only has to lift the difference:

M_{cw} = M_\text{car} + B \cdot M_\text{rated} \tag{3.1}

The balancing factor $B$ is typically 0.4 to 0.5. With $B = 0.5$ the lift is perfectly balanced when the car is half full, and the largest static imbalance it ever meets is $0.5\,M_\text{rated}$ , whether the car is full or empty. The Lab III and Tutorial 3.1 lift has a 1000 kg car and 1000 kg rated load: $M_{cw} = 1000 + 0.5 \times 1000 = 1500$ kg.

FoundationStart here if this is new to you

Think of a see-saw with the car on one end and the counterweight on the other. If both sides weigh the same, a small push moves it either way. Put a heavy passenger in the car and the car side wants to go down; leave the car empty and the counterweight side wins. The motor only ever fights the difference between the two sides.

Roping: 1:1 or 2:1

The roping ratio $i = v_\text{rope} / v_\text{car}$ says how the ropes are reeved.

1:1: the rope ends are fixed to the car and to the counterweight. The car moves as fast as the rope, and the motor sees the whole imbalance. Preferred for fast passenger lifts, above about 2.5 m/s, where motor speed must stay reasonable.
2:1 (under-slung): the ropes are anchored to the building and pass under pulleys on the car and the counterweight. The car moves at half the rope speed, so the motor turns twice as fast, but the torque it needs is halved. Standard for heavy freight lifts and machine-room-less lifts with small, fast motors.

The course compares two lifts. A hospital bed lift (2500 kg load) uses 2:1 roping, so a standard 15 kW gearless motor is enough; at 1:1 it would need twice the torque, a bigger frame and a costlier drive. An office express lift (1000 kg at 4 m/s) uses 1:1, because 2:1 would push the sheave and motor to noisy, vibrating speeds.

The torque the motor must supply

Let $\Delta M = (M_\text{car} + M_\text{load}) - M_{cw}$ be the mass imbalance: positive when the car side is heavier. Through a sheave of diameter $D$ , roping $i$ , gear ratio $k_\text{gear}$ and efficiency $\eta$ , the static torque at the motor is

T_\text{motoring} = \frac{\Delta M\, g\, D}{2\, i\, k_\text{gear}\, \eta}, \qquad T_\text{generating} = \frac{\Delta M\, g\, D\, \eta}{2\, i\, k_\text{gear}} \tag{3.3, 3.4}

The efficiency divides when the motor drives the load, and multiplies when the load drives the motor: then friction helps to brake. That asymmetry makes the motoring case the sizing case.

Try it: counterweight, roping and the quadrant

Load in the car1,000 kg

Balancing factor B0.50

Quadrant I: motoring, energy from the grid

The car side is heavier by 500 kg.

Counterweight Mcw: 1,500 kg
Motor torque: 1,635 N·m
Motor speed: 80 rpm
Power T·ω: 13.63 kW

T = ΔM · g · D / (2 i) ÷ η

Divide by η when motoring, multiply by η when generating: friction helps to brake.

Predict first

In the widget, the full car goes up at 1:1 and the motor needs 1635 N·m. You switch to 2:1 roping. What happens?

Torque and speed both halveTorque halves and motor speed doubles: the power is the sameNothing changes: the load is the same

Dynamic torque

At constant speed only the imbalance counts. While the car speeds up or slows down, the motor must also accelerate every moving mass:

T_\text{dynamic} = J_\text{total}\,\alpha = J_\text{total}\, \frac{a_\text{linear}\, i\, k_\text{gear}}{D/2}, \qquad J_\text{reflected} = m_\text{total}\left(\frac{D}{2\, i\, k_\text{gear}}\right)^2 \tag{3.5}

$J_\text{total}$ adds the motor rotor, the brake drum and gearbox, and the car, load, counterweight and ropes reflected to the motor shaft. The torque the motor must deliver is $T_\text{total} = T_\text{static} + T_\text{dynamic}$ .

For the Lab III lift going up with a full car, the moving masses total $2000 + 1500 = 3500$ kg. At 1 m/s², the ideal dynamic torque is $3500 \times 1 \times 0.3 = 1050$ N·m, on top of the static $500 \times 9.81 \times 0.3 = 1472$ N·m: $(1472 + 1050) / 0.9 = 2802$ N·m at the motor, the peak of that trip.

ExplorerGo deeper: derivations and open questions

Choosing B. An office lift carries full cars up in the morning and empty cars back down. Compute the energy per round trip for $B = 0.4$ and $B = 0.5$ , with and without regeneration. Which $B$ gives the smaller motor, and which the smaller energy bill?

Rope mass. In a 100 m shaft the ropes themselves weigh hundreds of kilograms, and their weight moves from one side to the other during the trip. Write $\Delta M(x)$ including the rope mass. How do compensation chains or ropes hung under the car fix it? (Mine hoists, lesson 6, meet the same problem at a much larger scale.)