Four-quadrant operation

A conveyor lives in quadrant I: positive speed, positive torque. A lift uses the whole torque-speed plane, and it changes quadrant with the load and the direction, sometimes within a single trip. Lesson A8 introduced the plane; this lesson puts a real lift on it (course section 3.1.3).

Signs first

Take positive speed as the car going up, and positive torque as the motor pulling the car up. Then the mechanical power is

P = T \cdot \omega

$P > 0$ : energy flows from the supply into the load (motoring). $P < 0$ : the load drives the motor, which returns energy (generating, or braking).

Quadrant	Speed	Torque	Condition	Energy
I	+ (up)	+	car side heavy, rising: $M_\text{car} + M_\text{load} > M_{cw}$	motoring
II	+ (up)	−	counterweight heavy, rising: $M_{cw} > M_\text{car} + M_\text{load}$	generating
III	− (down)	−	counterweight heavy, descending	motoring
IV	− (down)	+	car side heavy, descending	generating

Quadrants II and IV return energy to the drive. It must be burned in a resistor or sent back to the grid (lesson 4).

FoundationStart here if this is new to you

Ask who is doing the pushing. A heavy car going up: the motor pushes, gravity resists, the motor works (I). A heavy car going down: gravity pushes, the motor holds it back like brakes on a hill, and it makes electricity (IV). An empty car going down: the heavier counterweight wants to pull the car up, so the motor has to pull it down: it works again (III).

Tutorial 3.1: a two-leg journey

The Tutorial 3.1 lift: $M_\text{car} = 1000$ kg, $M_\text{rated} = 1000$ kg, $M_{cw} = 1000 + 0.5 \times 1000 = 1500$ kg, friction neglected.

Leg 1, up with a full load. Car side $1000 + 1000 = 2000$ kg against 1500 kg: the car side is 500 kg heavier. Speed positive; to lift it the motor applies positive torque. Quadrant I, forward motoring.
Leg 2, down empty. Car side 1000 kg against 1500 kg: the counterweight is 500 kg heavier and pulls the car up. Speed negative; to move the car down the motor must pull against it, with negative torque. Quadrant III, reverse motoring.

Both legs draw energy from the grid, although the car goes up once and down once. The opposite journey, the lunchtime rush of the course (down full, then up empty), runs in quadrants IV and II and generates on both legs. That is why a lift drive is sized on its worst motoring torque, not on an average load.

One trip, instant by instant

The widget runs a complete Lab III trip: 20 m, 2.5 m/s, 1 m/s², 1.5 m/s³, sheave 0.60 m, 1:1 gearless, $\eta = 0.9$ . It adds the dynamic torque of lesson 1 to the static torque at every instant.

Try it: one Lab III trip, instant by instant

Mechanical efficiency η0.90

0.9 in the lab; about 0.6 for a worm-gear lift (question 4).

Jerk limit j_max1.50 m/s³

Speed (m/s)
Acceleration (m/s²)
Motor power (kW)

At constant speed: quadrant IV. The drive regenerates for 100 % of the trip.

Trip time: 11.17 s
Peak torque: 2,269 N·m
Peak regenerated power: 16.4 kW
Resistor at 760 V (V²/P): 35.3 Ω

Predict first

Choose 'Full load down'. When is the regenerated power largest?

As the car starts movingDuring the constant-speed runWhen the car starts to slow down from full speed

Three things to read off the widget, which are also Lab III questions 1 to 3:

The dynamic torque is largest during the constant-acceleration phases and exactly zero at constant speed, where $a = 0$ .
"Full load up" and "empty car down" motor throughout the constant-speed run, in quadrants I and III.
"Full load down" and "empty car up" generate on both legs, in quadrants IV and II.

ExplorerGo deeper: derivations and open questions

Energy per trip. Integrate $P(t)$ over each of the four trips in the widget. Which trips return energy, and how much in kJ? Over a busy lunchtime hour with a trip every 40 s, how many kWh could a regenerative drive give back?

Friction. Add a constant friction force of 2 % of the moving weight to the model. It opposes the motion, so it adds to the torque when motoring and subtracts when generating. How does it change the quadrant boundaries near balance?