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Elevators and Mine Hoists

CHAPTER 3

Power supply and automation of elevators and extractors (mine hoists)

Electrical Control of Industrial Mechanisms

Master 1, Semester 3 · 2025–2026 · Prof. Z. TIR — University of El Oued

Photo: Filip.vidinovski, CC BY-SA 4.0, Wikimedia Commons

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What you will be able to do

CHAPTER 3 · LEARNING OUTCOMES

1

Compute the counterweight and the static, dynamic and peak motor torque

2

Check the rope grip on the sheave; place any trip in its quadrant

3

Specify regenerative or dynamic braking and size a braking resistor

4

Explain the jerk-limited S-curve and its effect on comfort and trip time

5

Analyse a fail-safe safety chain: safe versus dangerous failures

6

Extend all of this to large mine hoists

Lectures, Fig. 3.5

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 2

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Meet the reference elevator of the chapter

3.1 · REFERENCE ELEVATOR

Quantity	Value	Remember it as
Car, rated load	1000 kg + 1000 kg	a 13-person office lift
Counterweight	1500 kg (B = 0.5)	car plus half the load
Machine	gearless, sheave 0.60 m, 1:1	8.33 rad/s at 2.5 m/s
Motion	2.5 m/s, 1.0 m/s², 1.5 m/s³	20 m trip in 11.2 s
Worst imbalance	500 kg → 1472 N·m	1635 N·m lifting, 1324 N·m lowering

!

The same lift is solved by hand, simulated in Simscape and used in the tutorials

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 3

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The counterweight: the motor only sees the imbalance

3.1 · COUNTERWEIGHT

Photo: Simiprof, CC BY-SA 4.0, Wikimedia Commons

Lectures, Fig. 3.1

→

M_cw = M_car + B·M_rated with B ≈ 0.4–0.5: the worst imbalance is about half the rated load, not car + load

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 4

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Roping: trading car speed against shaft torque

3.1 · ROPING

1:1 — rope ends on car and counterweight: v_car = v_rope, full imbalance on the motor
2:1 — ropes pass under idlers to fixed hitches: half the speed, half the torque
Hospital bed lift (2500 kg, 1 m/s): 2:1 allows a standard 15 kW gearless motor
Office express lift (1000 kg, 4 m/s): 1:1 keeps the motor speed reasonable

Lectures, Fig. 3.2

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 5

Compare the two drawings. In 1:1 the rope ends are fixed to the car and the counterweight, so the car moves as fast as the rope and the motor sees the full imbalance. In 2:1 the rope is anchored at dead-end hitches and passes under idler pulleys on the car and counterweight: a moving pulley, which halves the car speed and halves the force reflected to the sheave. The equation shows the two levers that keep a lift drive small: the counterweight shrinks the imbalance and the roping ratio i (and a gearbox k_gear) divides what remains — at a proportionally higher motor speed, so the power is the same.

Ask: for the hospital bed lift, why 2:1? (Heavy, slow: halve the torque, smaller motor and drive.) For the express lift, why 1:1? (Light, fast: 2:1 would double the motor speed, noise and rope wear for nothing.)

Timing: 5 minutes. Reference: Lectures section 3.1.1, Fig. 3.2, eq. reflected torque.

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Traction: the ropes grip by friction only

3.1 · TRACTION

Tensions on the two sides may differ only by what the groove can grip
Inertia counts: a side accelerating up pulls M(g + a), down M(g − a)
Reference lift: limit e^fθ = 1.81; full car accelerating up needs 1.64
Emergency braking of the empty car going up is the critical case
With the counterweight on its buffers, the ropes must slip

Lectures, Fig. 3.3

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 6

A traction lift has no rope fixed to the sheave: the ropes are driven by friction in the grooves, like a rope wound round a capstan. The capstan law says the tensions on the two sides can differ only up to a factor e to the f theta, with theta the wrap angle and f the groove friction. For the reference lift, a half-turn of undercut groove gives f about 0.19 and a limit of 1.81. The full car accelerating upwards at 1 m/s² needs 2000 × 10.81 over 1500 × 8.81 = 1.64: safe. But inertia eats the margin, so traction limits the acceleration and the emergency braking. EN 81-50 checks three cases: 125 % load at a landing, emergency braking, and the stalled case — where the ropes must slip, so the motor cannot pull the car into the shaft roof.

Ask self-check 3.3 now. Tutorials Exercise 3.2 finds an emergency stop at which the ropes slip.

Timing: 5 minutes. Reference: Lectures section 3.1.1, eq. traction, Fig. 3.3.

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Motoring and generating torque are not equal

3.1 · LOAD ANALYSIS

An active load: magnitude and sign depend on payload and direction
Motoring: losses must be supplied — η divides
Generating: friction helps the brake — η multiplies
Add the dynamic torque J·dω/dt to accelerate the whole system

Lectures, Fig. 3.4

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 7

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Torque and power of the reference elevator

3.1 · WORKED EXAMPLE

GIVEN

Reference elevator, full car: ΔM = 500 kg, D = 0.60 m, 1:1, gearless, η = 0.90, 2.5 m/s, 1.0 m/s²

FIND

Static torque lifting and lowering, peak torque accelerating up, power at rated speed, effect of 2:1

SOLUTION

Lossless: 500 × 9.81 × 0.30 = 1472 N·m; lifting 1635 N·m, lowering 1324 N·m
J = 3500 kg × 0.30² = 315 kg·m²; dω/dt = 1.0/0.30 = 3.33 rad/s² → +1050 N·m
Peak while accelerating up ≈ 2685 N·m
8.33 rad/s: 13.6 kW lifting, 11.0 kW returned lowering
2:1 roping: 818 N·m at 159 rpm — same power

✓

Check. P = ΔM g v / η = 500 × 9.81 × 2.5 / 0.9 = 13.6 kW, whatever the roping.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 8

Compute line by line with the class. The lossless torque, then the efficiency: divide when lifting, multiply when lowering — 1635 against 1324 N·m. The dynamic part surprises students: the whole moving mass, car, load and counterweight, 3500 kg, referred to the sheave radius, is 315 kg·m²; at 1 m/s² it needs another 1050 N·m, so the peak is about 2700 N·m, two thirds more than the static torque. The motor's own rotor adds to it. Power at rated speed: 13.6 kW lifting. With 2:1 roping the torque halves and the speed doubles; the power does not change — roping chooses the motor's torque and speed, not its power.

Ask: which number sizes the braking resistor, and which the motor? (Generating for the resistor, the motoring peak for the motor.)

Timing: 6 minutes. Reference: Lectures section 3.1.2, worked example; Tutorials Exercise 3.1.

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Four-quadrant operation of an elevator drive

3.1 · FOUR QUADRANTS

Lectures, Fig. 3.5

→

The quadrant is set by the sign of torque relative to speed — not by the direction of travel

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 9

Conventions first: speed positive when the car goes up; torque positive when the motor lifts. Power P = T omega: positive is motoring, negative is generating. Quadrant I: car heavy, going up — motoring. Quadrant II: counterweight heavy, going up, for example an empty car — the counterweight pulls the car up and the motor restrains it: generating. Quadrant III: counterweight heavy, going down — the motor must pull the light car down: motoring. Quadrant IV: car heavy, going down — the motor holds it back: generating.

Ask self-check 3.2: an empty car travels down — motoring or generating? Most students say generating. It is motoring, Quadrant III, because torque and speed are both negative.

Timing: 6 minutes. Reference: Lectures section 3.1.3, Fig. 3.5, Table 3.2; Tutorials Exercise 3.3.

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Four real trips on the torque–speed plane

3.1 · FOUR QUADRANTS

Lectures, Fig. 3.6

→

Empty car descending: motoring all the way (Q III). Peak regenerated power 16.4 kW, above the 11.0 kW at constant speed

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 10

Simscape simulation of the reference elevator: 1000 kg car, 1000 kg rated load, 50 % counterweight, 0.60 m sheave, 1:1, gearless, efficiency 0.90, 2.5 m/s. Each trip is a closed loop in one quadrant: the vertical sides are the constant acceleration and deceleration, the curved ends the jerk-limited transitions, and the whole constant-speed run collapses to the tip of the loop. Two results. The empty car descending stays in Quadrant III and draws as much power as the full car ascending. And efficiency acts in opposite directions: with a 1472 N·m imbalance the motor gives 1472/0.9 = 1635 N·m motoring but only 1472 × 0.9 = 1324 N·m generating. Peak regenerated power, 16.4 kW during deceleration of an overhauling trip, sizes the resistor.

Timing: 6 minutes. Reference: Lectures section 3.1.3, Fig. 3.6.

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Drive system requirements and control

3.2

Full torque before the brake opens, somewhere to put the regenerated energy, and a ride passengers do not feel.

Lectures, Fig. 3.8

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 11

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Torque before the brake opens

3.2 · ZERO-SPEED TORQUE

Scalar V/Hz needs time to build flux: the car would sink at brake release
Vector control: magnetise (I_d) before release
Pre-torque (I_q) from a load-weighing device
Brake opens only when the motor torque is proven

Lectures, Fig. 3.7

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 12

If the brake released before the motor could hold the load, the imbalance would make the car roll back — a load dip passengers feel and that can be dangerous at a landing. Closed-loop vector control solves it in three steps: inject the magnetising current I_d, compute the torque current I_q from the load-weighing sensor, and only then command the brake open. The figure is the reference elevator with a full car: 500 kg imbalance, 1472 N·m. Starting the speed loop from zero torque, the car sinks 25 mm before it is caught. With a pre-torque from a weighing device that reads 10 % low, 2.5 mm — below what a passenger notices.

Ask self-check 3.5: what does the pre-torque prevent, and how big is the effect? (Rollback; tenfold smaller.)

Timing: 5 minutes. Reference: Lectures section 3.2.1, Fig. 3.7.

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Where does the regenerated energy go?

3.2 · REGENERATION

Generating for about half the duty cycle (Q II and Q IV)
Energy charges the DC bus: above ~800 V the drive trips and the brake slams
Dynamic braking: a chopper switches a resistor in at ~750 V; the energy becomes heat
Active front end: an IGBT rectifier returns the energy to the grid

Lectures, Fig. 3.8

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 13

Remember the Chapter 2 question: a diode rectifier cannot pass power backwards. When the motor generates, the current returns through the inverter's freewheeling diodes into the DC link capacitors and the bus voltage climbs. If it passes the rating, about 800 V for a 400 V drive, the drive trips, the motor coasts and the brake slams shut. Panel (a): dynamic braking — a chopper switches a resistor across the bus at about 750 V; the energy becomes heat in a ventilated resistor bank. Panel (b): an active front end replaces the diodes by an IGBT bridge synchronised to the grid and feeds the energy back. Hotel example from the booklet: about 20 kW per car in an overhauling run, either heat that needs air-conditioning, or power for the lobby.

Timing: 5 minutes. Reference: Lectures section 3.2.1, Fig. 3.8.

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Sizing a braking resistor

3.2 · WORKED EXAMPLE

GIVEN

Chopper threshold V_DC = 750 V, chopper rating I_max = 50 A; peak regenerated power P_gen = 16.4 kW (Fig. 3.6)

FIND

The allowed range of braking resistance and a suitable value

SOLUTION

Upper bound: R_max = V_DC² / P_gen = 750² / 16 400 ≈ 34 Ω
Choose a standard value 10–20 % below it, e.g. 30 Ω
Lower bound: R_min = 750/50 = 15 Ω; 30 Ω draws 25 A
Ratings: peak 750²/30 = 18.8 kW; continuous ≈ 0.25 × 11.0 = 2.8 kW

✓

Check. A resistor above 34 Ω could not absorb 16.4 kW at 750 V: the bus would keep rising until the drive trips.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 14

This is self-check 3.4 developed. The resistor is bounded on both sides. It must be small enough to absorb the peak regenerated power at the chopper threshold: P = V squared over R, so R_max = 750 squared / 16 400 = 34 ohm. And it must be large enough to keep the chopper current within the switch rating: R_min = V_DC / I_max. Pick a standard value 10 to 20 % below R_max, here 30 ohm, and check the chopper current, 25 A. Two power ratings: the peak, 18.8 kW at 30 ohm whenever the chopper conducts, and the continuous rating: if overhauling runs take a quarter of the time at a mean 11 kW, about 2.8 kW. Keep the energy in proportion: one full car going down 20 m returns only 88 kJ, 0.025 kWh. It pays through hundreds of trips a day and through the heat it keeps out of the machine room. The classic error, from the pitfall box: choosing the largest resistance "for safety". A larger R dissipates less power.

Timing: 6 minutes. Reference: Lectures section 3.2.1, eq. braking-resistor bounds; Tutorials Exercise 3.4.

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S-curve profiles: limit the jerk

3.2 · S-CURVE

Lectures, Fig. 3.9

Passengers feel jerk j = da/dt, not speed
A trapezoid has infinite jerk at its corners: a jolt
The S-curve ramps the acceleration: seven phases
Industry practice: a_max 0.8–1.2 m/s², j_max 1.0–2.0 m/s³
Reference lift, 20 m: 8.0 + 2.5 + 0.67 = 11.2 s

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 15

The human body is a sensitive accelerometer: it tolerates constant speed, and even constant acceleration, but feels every change of acceleration — the jerk. The trapezoidal profile switches the acceleration on and off instantly: infinite jerk at each corner, a kick. The S-curve ramps the acceleration up and down, in seven phases: jerk-in, constant acceleration, jerk-out, constant speed, then the same three for braking. Typical industry practice: maximum acceleration 0.8 to 1.2 m/s², jerk 1.0 to 2.0 m/s³ — the safety standards set no comfort limits. The price is a slightly longer trip: the trip-time formula adds exactly a_max over j to the trapezoid. For the reference lift, 8.0 + 2.5 + 0.67 = 11.2 s for 20 m. The formula holds when the trip is long enough to reach full speed; a one-floor trip is not (Tutorials Exercise 3.5).

Ask: why does a smoother profile also reduce rope wear? (No step changes in force, so no rope oscillation.)

Timing: 4 minutes. Reference: Lectures section 3.2.2, Fig. 3.9.

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Tuning the jerk of an express elevator

3.2 · WORKED EXAMPLE

GIVEN

50-storey express lift, 6.0 m/s, a_max = 1.0 m/s²; jerk limit 4.0 m/s³ (poor tuning) or 1.2 m/s³

FIND

The cost in trip time of the smoother tuning

Photo: Filip.vidinovski, CC BY-SA 4.0, Wikimedia Commons

SOLUTION

The acceleration ramps over a_max/j
High jerk: 1.0/4.0 = 0.25 s; low jerk: 1.0/1.2 = 0.83 s
A jerk-limited trip is slower than the trapezoid by exactly a_max/j
Smoother tuning adds only 0.83 − 0.25 ≈ 0.6 s to the journey

✓

Check. 50 storeys of about 3.5 m take some 35 s at 6 m/s: 0.6 s is under 2 % — invisible to the passenger.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 16

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Stopping level with the floor

3.2 · LEVELLING

Lectures, Fig. 3.10

Motor encoder: speed for the vector control, distance by counting pulses
EN 81-20: stop within ±10 mm; hold ±20 mm while loading
Dead reckoning drifts: 0.1 % slip over 100 m = 10 cm
Hoistway vanes correct the position counter at each floor
Direct-to-floor: brake from the slow-down vane straight to the landing

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 17

Two feedback loops. The encoder gives the speed for vector control and the distance by counting pulses — the factor 4 is quadrature decoding. Resolution is no issue: a 2048-line encoder on the reference sheave gives 0.23 mm per count. But counting drifts: rope slip and thermal expansion accumulate, and 0.1 % over 100 m of travel is 10 cm, an unacceptable step. So fixed vanes or magnets in the shaft, read by a sensor on the car, overwrite the counter at each floor. Direct-to-floor landing computes the braking distance v squared over 2a and starts the braking curve at the slow-down vane, landing directly instead of creeping. Re-levelling example: a 2 t forklift drives on, the ropes stretch, the car sinks 20 mm; with the doors open the drive lifts it back at zero speed before the brake closes. Moving with open doors is allowed only in that narrow zone: EN 81-20's unintended car movement protection (UCMP) stops the car if it ever leaves the landing with the doors open for any other reason.

Lab link: TP 3 does the same on the PLC-200 — a stepper plays the traction machine, the encoder on HSC0 checks the position (96 steps per floor), and a ground-floor levelling switch homes the count, exactly like a hoistway vane.

Timing: 5 minutes. Reference: Lectures section 3.2.3, Fig. 3.10.

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Electrical safety systems

3.3

The brake holds by default; any open contact stops everything.

Lectures, Fig. 3.11

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 18

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The fail-safe brake: springs close it, current opens it

3.3 · BRAKE

Power OFF: springs clamp the disc — holds 125 % of rated load in the car (EN 81)
Power ON: the coil pulls the armature clear
Opened only after the drive has proven its torque
Two independent mechanical sets, each able to slow the rated load
Pick voltage to open fast, lower hold voltage to limit heating
Coil supply in series with the safety chain

Lectures, Fig. 3.11

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 19

Read the cross-section from left to right: shaft, hub keyed to it, the friction disc, the armature plate, the springs in the housing, the coil. Power off: the springs push the armature against the disc and clamp the shaft. Power on: the coil pulls the armature back against the springs and frees it. Losing power can only close the brake. EN 81 requires the brake to stop and hold the car with 125 % of rated load inside — note: that is the imbalance with 1.25 M_rated in the car, not 1.25 times the normal imbalance: for the reference lift 750 kg, 2.2 kN m at the sheave. The brake is built as two independent sets, each able to slow the car with rated load, so no single broken spring disables it. The coil is fed through the safety chain: any open safety contact removes coil power, whatever the software is doing. Power-outage story: the car stops harshly between floors; a technician uses the manual release lever to let it drift to a landing.

Timing: 5 minutes. Reference: Lectures section 3.3.1, Fig. 3.11.

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The safety chain: a hard-wired AND

3.3 · SAFETY CHAIN

Lectures, Fig. 3.12

Normally-closed contacts in series feed the safety relay K1 (110 V or 48 V)
Pit stop, overspeed governor (~115 %), final limits, car door, all landing door locks
Any open contact cuts power to drive and brake, independent of software
More contacts: more nuisance trips, but a safe failure mode

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 20

Trace the chain as a circuit: from the 110 V supply through every normally-closed contact in series to the coil of the safety relay K1 and back to neutral. It is a Boolean AND: every condition must be true for the relay to hold. Name the devices: the pit stop switch, latching, for maintenance staff under the car; the overspeed governor, tripping at about 115 % of rated speed; the final limits beyond the end floors; the car door contact and the landing door locks of every floor. Reliability multiplies, so more contacts mean more nuisance stops — but an open contact is a safe failure. The dangerous failure is a contact that welds or is bridged.

Ask self-check 3.6: why is a bypassed contact worse than a failed-open one?

Timing: 6 minutes. Reference: Lectures section 3.3.2, Fig. 3.12; Tutorials Exercise 3.6.

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The “jumper” practice

3.3 · HAZARD

✗

MYTH

“A bypassed safety contact is just like a failed one.”

✓

WHAT IS TRUE

Opposite classes. An open contact stops the lift: safe. A bridged contact silently removes the protection while the lift runs: dangerous — and a simple series chain cannot see it.

✗

MYTH

“I will remove the test jumper when I have finished.”

✓

WHAT IS TRUE

Forgotten once, the lift runs with a landing door open. Modern controllers detect bypasses or restrict the car to slow inspection speed.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 21

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From relays to microcontrollers

3.3 · CONTROL EVOLUTION

Lectures, Fig. 3.13

→

Group control gives each hall call to the car of lowest cost C = w₁·t_wait + w₂·t_travel + w₃·E_energy

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 22

Walk through the table row by row: where the logic lives, cabinet size, how logic is changed, field wiring, dispatching, drive interface. The modern controller assigns a new hall call by computing a cost for every car: estimated waiting time, travel time and energy, with weights that change with the traffic pattern — up-peak, down-peak, lunch. The call goes to the car with the lowest cost. Modernisation story: a 1920s building replaces a room of relays by a shoebox-sized board, reuses two spare pairs of the old travelling cable as a CAN bus, and gains 20 % handling capacity from the dispatching software alone.

Lab link: TP 3 implements a four-floor elevator in ladder logic on the PLC-200, the 1980s generation of this table.

Timing: 4 minutes. Reference: Lectures section 3.3.3, Fig. 3.13, eq. dispatch cost.

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Mine hoists (extractors)

3.4

Thirty tonnes from a kilometre deep at 15–18 m/s: the same physics at the largest scale.

Photo: Kj, CC BY-SA 4.0, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 23

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Drum winder or Koepe friction winder

3.4 · CONFIGURATIONS

Drum: rope stored on the drum; suspended rope mass changes over the wind
Koepe: rope driven by friction over one sheave, counterweight on the other side
A balance rope below the conveyances keeps the imbalance constant
Cage for people and materials; skip for bulk ore

Lectures, Fig. 3.14

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 24

Panel (a): the drum winder stores the rope on the drum; the rope leaves at the drum's tangent. As the skip rises, rope is wound on, so the suspended rope mass, and with it the motor torque, changes throughout the wind. Panel (b): the Koepe winder does not store rope. It drives by friction over one sheave, with the skip on one side and a counterweight on the other; a balance rope hangs below both, so as rope moves from one side to the other the imbalance stays the same. The torque equation is the elevator's again, with the gearbox ratio k_gear and the same plus-or-minus efficiency exponent. A good design balances conveyance plus half the payload. A Koepe winder also obeys the traction law of the lift: it caps the acceleration and the emergency braking.

Timing: 4 minutes. Reference: Lectures sections 3.4.1 and 3.4.2, Fig. 3.14, eq. hoist torque.

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Why deep hoists need a balance rope

3.4 · ROPE IMBALANCE

Lectures, Fig. 3.15

→

Over a 1000 m wind, the imbalance falls from 226 kN to −29 kN without a balance rope; with one it stays at 98 kN

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 25

Deep-shaft hoist: 20 t payload, 16 t skip, head ropes of 13 kg/m in total, counterweight equal to skip plus half the payload. At 1000 m the rope itself weighs 13 t — comparable to the payload. Drum winder: carries the full suspended rope, so the force is largest at the bottom. Koepe without a balance rope: as the skip rises the rope weight moves to the other side, and the out-of-balance force falls from 226 kN to minus 29 kN — it even reverses, so the drive goes from motoring to generating within one wind. With a balance rope of the same mass per metre the force is constant at 98 kN, and the drive can be sized for one torque.

Ask: why does the force reverse near the top? (The rope on the counterweight side now outweighs the loaded side.)

Timing: 4 minutes. Reference: Lectures section 3.4.2, Fig. 3.15.

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The winding diagram and the cycle time

3.4 · HOISTING CYCLE

Lectures, Fig. 3.16

Pull-away, acceleration, full-speed run, deceleration, creep, stop
The creep makes stopping accuracy independent of wind length
Production rate is inversely proportional to the cycle time
Raising v_m stops paying when v_m²/a approaches H

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 26

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Capacity of a deep-shaft skip hoist

3.4 · WORKED EXAMPLE

GIVEN

Koepe hoist, one skip against a counterweight: 20 t of ore per skip, H = 1000 m, v_m = 15 m/s, a = 0.8 m/s²; creep and a 15 s dwell

FIND

The trip time and the hourly capacity

Photo: Kj, CC BY-SA 4.0, Wikimedia Commons

SOLUTION

Full speed all the way: H/v_m ≈ 67 s; ramps add only v_m/a ≈ 19 s
Trip ≈ 85 s; with creep and dwell each wind approaches 2 min
The empty skip must also be lowered: one skipload every two winds, ≈ 4 min
Capacity ≈ 15 skips/h × 20 t ≈ 300 t/h; a second skip in place of the counterweight doubles it to 600 t/h

✓

Check. v_m²/a = 281 m is well below H, so full speed is reached; the counterweighted hoist motors on both winds.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 27

Compute together: 1000 / 15 = 66.7 s; 15 / 0.8 = 18.75 s; the trip about 85 s. Add the creep and the 15 s dwell and each wind approaches two minutes. Now the trap most students fall into: with a single skip and a counterweight, the skip that has just discharged at the top must be lowered empty before it can be loaded again. One skipload therefore costs two winds, about four minutes: 15 skips an hour, about 300 t of ore per hour. Replace the counterweight by a second skip — one rises full while the other descends empty — and every wind delivers a load: about 600 t per hour from the same machine. The check verifies the formula's assumption: 15 squared over 0.8 is 281 m, well below 1000 m, so a genuine full-speed run exists. End with the energy, and a trap: does this hoist regenerate when the empty skip goes down? No — the counterweight is heavier than the empty skip, so the motor must pull the skip down: Quadrant III, like the empty lift car. Even the braking ramps stay motoring: decelerating about 100 t at 0.8 m/s² needs 81 kN, less than the 98 kN imbalance. A drum winder lowering its skip, or a hoist lowering men and materials, does regenerate — megawatts.

Ask: what is gained by raising v_m to 18 m/s? (H/v falls by 11 s but v/a rises by 3.75 s: about 7 s per trip.)

Timing: 5 minutes. Reference: Lectures section 3.4.5, worked example.

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A century of four-quadrant hoist drives

3.4 · DRIVE GENERATIONS

Lectures, Fig. 3.17

→

Tell them apart by where the regenerated energy goes: into the flywheel, back through the thyristor bridge, or out through the active front end

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 28

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Overspeed supervision that follows the position

3.4 · SAFETY

Lectures, Fig. 3.18

→

Limit = 110 % of the programmed speed: a drive that fails to slow is caught at 874 m and stopped at 919 m, 81 m short of the landing

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 3 · 29

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Common mistakes with vertical drives

WATCH OUT

✗

MYTH

“Going down means the motor generates.”

✓

WHAT IS TRUE

The sign of torque relative to speed decides. An empty car descending against a heavier counterweight is motoring, Quadrant III.

✗

MYTH

“The counterweight cancels the load, so the motor only overcomes friction.”

✓

WHAT IS TRUE

It balances one load only (40–50 % of rated). Size the drive for full load in the least favourable direction.

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Quick check — try before you look

SELF-CHECK 3.1–3.6

3.1

Car 1000 kg, rated load 1000 kg, B = 0.45: counterweight mass and largest imbalance?

3.2

An empty car travels down. Motoring or generating, and in which quadrant?

3.3

Reference lift, e^fθ = 1.81: is traction kept with the full car accelerating up at 1.0 m/s²?

3.4

Chopper at 750 V, peak regenerated power 16.4 kW: largest resistance, and how is the value chosen?

3.5

What does a pre-torque before brake release prevent, and how large is the effect in Fig. 3.7?

3.6

Why is a bypassed safety contact more dangerous than one that has failed open?

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Summary

CHAPTER 3 · SUMMARY

The counterweight reduces the motor's task to the imbalance; roping trades speed for torque; traction limits a
Four quadrants: generating about half the time — resistor or active front end
Full torque at zero speed before the brake opens; S-curves for comfort
Encoder plus hoistway vanes for accurate levelling
Safety is hard-wired: series NC chain and spring-applied brake fail stopped
Mine hoists: the same principles, balance ropes and position-dependent limits

KEY EQUATIONS

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

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Practise this

CHAPTER 3 · NEXT STEPS

TD

Tutorials

Exercises 3.1–3.7: torque and roping, traction, quadrants, braking resistor, S-curve, safety chain, mine hoist.

TP

Lab works

TP 3 — a four-floor elevator on the PLC-200.

?

Before next week

Redo the self-check questions; then read Chapter 4 §4.1 on the three motions of a crane.

Photo: Simiprof, CC BY-SA 4.0, Wikimedia Commons

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Image credits

CHAPTER 3 · SOURCES

Figures and diagrams: Lectures booklet, Electrical Control of Industrial Mechanisms (computed in Python/Simscape or drawn in TikZ from the chapter's own equations).
Elevator machine room — Filip.vidinovski, CC BY-SA 4.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Elevator_machine_room.jpg
ElevatorCounterweight — Simiprof, CC BY-SA 4.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:ElevatorCounterweight.jpg
Headframe of Campi shaft — Kj, CC BY-SA 4.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Headframe_of_Campi_shaft.jpg
Licence texts: creativecommons.org/licenses (CC BY, CC BY-SA); creativecommons.org/publicdomain/zero/1.0 (CC0). Photos are shown unmodified except for cropping and scaling.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

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