How a linear induction motor works: travelling field, slip and thrust

From a rotating motor to a flat one

Take an ordinary three-phase induction motor, cut the stator along its axis and unroll it flat. The windings now lie in a straight line, and the rotating magnetic field has become a travelling wave that slides along the surface. The rotor, which was a squirrel cage, becomes a flat conductive sheet, usually aluminium or copper, backed by iron.

The physics is unchanged. The travelling field induces eddy currents in the plate, and the currents interact with the field to produce a force along the direction of travel. Torque becomes thrust in newtons, and rotor speed in rpm becomes vehicle speed in m/s or km/h.

Either part can move. In some systems the stator is fixed along the track and the vehicle carries the plate. In others, such as some maglev trains, the vehicle carries the stator and the plate is laid along the guideway. Because the track is a long row of identical stator sections or a long plate, the motor can accelerate a vehicle over hundreds of metres without a gearbox, wheel or rope.

Synchronous speed: v_s = 2 × τ × f

The speed of the travelling field depends only on the supply frequency and the distance between neighbouring poles, called the pole pitch τ. In one electrical cycle the field moves by two pole pitches, so

v_s = 2 × τ × f

where τ is in metres, f in hertz and v_s in metres per second. The number of poles does not appear in the speed. Compare this with a rotating motor, where the synchronous speed is set by the supply frequency and the pole count.

• τ = 120 mm at 50 Hz: v_s = 2 × 0.12 × 50 = 12 m/s, or 43.2 km/h
• τ = 120 mm at 25 Hz: v_s = 6 m/s, or 21.6 km/h
• τ = 250 mm at 50 Hz: v_s = 25 m/s, or 90 km/h

Two things follow. A longer pole pitch gives a faster machine, which is why high-speed LIMs have wide poles. And a variable-frequency drive controls speed directly: halve the frequency and the field, and the vehicle, goes half as fast.

Slip and thrust

As in a rotating induction motor, the plate can never travel at exactly the field speed while producing thrust. If it did, there would be no relative motion, no induced current and no force. Slip measures the lag:

s = (v_s − v) / v_s

where v is the vehicle speed. At standstill s = 1, and at synchronous speed s = 0.

The thrust equals the power crossing the air gap divided by the synchronous speed:

F = P_airgap / v_s

Part of that power becomes useful mechanical power, F × v, and the fraction s of it heats the plate:

P_plate loss = s × P_airgap

This explains why a LIM is most efficient when it runs at small slip, and why it is inefficient while accelerating from rest, when the slip is large. Thrust is zero at s = 0 and positive for any slip between 0 and 1, so the vehicle accelerates until the thrust equals the opposing force from friction, a slope or air drag. A vehicle with no opposing force approaches the synchronous speed but does not quite reach it.

The reaction plate, the air gap and the end effect

How well the plate turns the field into thrust is captured by a dimensionless goodness factor:

G = μ0 × ω × τ² × σ × d / (π² × g_e)

where ω = 2πf, σ is the plate conductivity, d its thickness and g_e the effective air gap. G is also the ratio of the magnetising reactance to the plate resistance. A high G needs a good conductor (copper rather than aluminium), a high frequency, a long pole pitch and a small gap. A thicker plate raises G but also enlarges the effective gap, so it is not a free improvement.

A rotating motor can have a gap of a fraction of a millimetre. A linear motor cannot: the vehicle must clear the track, so the gap plus the plate is several millimetres. The magnetising current grows with the effective gap, so a LIM has a low power factor and efficiency, often well below those of a rotating machine of the same rating. In a single-sided LIM the back iron is also pulled towards the stator by a large normal force, comparable to or larger than the thrust, and the vehicle must support it with wheels or levitation magnets. A double-sided design with a stator on each side of the plate cancels it.

The end effect is the other difference. A rotating stator is a closed ring, but a linear stator has an entry and an exit edge. The eddy currents in the plate cannot build up instantly as it enters the field or vanish instantly as it leaves. At standstill this does not matter. At speed, the plate crosses the stator faster than the currents settle, so the magnetising reactance falls, an extra loss appears and thrust drops. In Duncan's model the effect depends on a parameter Q that grows with the stator length and the plate resistance and falls with speed. A small Q means a strong end effect, which is why it is worst for short stators and high speeds. Longer stators and compensating windings reduce it.

Where linear induction motors are used

LIMs are chosen when the load already moves in a straight line and contact-free drive is valuable.

• Maglev and urban rail: the Linimo line near Nagoya, Japan, carries passengers at up to 100 km/h on a linear-motor-driven magnetic levitation system, and several metro lines use LIM propulsion to save weight and to climb steeper grades.
• Roller coaster launches: a LIM (or a linear synchronous motor) accelerates the train along a straight section in a few seconds, with no friction wheel to wear out.
• Conveyors, baggage handling and sliding doors, where silent contact-free drive matters.
• Test rigs and material-handling shuttles.

A LIM can also brake, by reversing the phase sequence or by lowering the frequency below the vehicle speed, which makes the slip negative and turns the machine into a generator.

How the simulator models a linear induction motor

Curvo Studio's linear induction motor simulator uses a per-phase equivalent circuit, the same as for a rotating induction motor, with no separate plate leakage reactance because the plate sits in the gap. All circuit values are derived from the geometry and from two design numbers, the rated voltage and frequency. It is a teaching model with simplifications: it has no transverse edge effect, magnetic saturation, skin effect, iron loss, or temperature dependence.

• The synchronous speed is v_s = 2 × τ × f, using the drive frequency.
• The effective gap is 1.1 times the sum of the mechanical gap and the plate thickness.
• The plate resistance follows from the conductivity of aluminium (3.0 × 10⁷ S/m) or copper (5.0 × 10⁷ S/m), the thickness and the geometry.
• The drive keeps V/f constant up to the rated frequency. Above it the voltage stops rising and the flux weakens.
• The end effect uses Duncan's model, applied to one stator section of length p × τ. It can be switched off to show the ideal machine.
• The vehicle accelerates under the net force (thrust minus load) and the launch is integrated along the track.

For the default machine (τ = 120 mm, 6 poles, 8 mm gap, 5 mm aluminium plate, 400 V, 50 Hz, 1000 kg, 600 N opposing force) the simulator gives a synchronous speed of 12 m/s. The vehicle settles at about 11.1 m/s (40 km/h), a slip of 7.3%. The thrust at that speed is 600 N, the standstill thrust is about 1900 N, the goodness factor is 6.0, the power factor 0.31 and the efficiency 53%.

Try it in the linear induction motor simulator

Start the linear induction motor simulator from its default settings.

• Change the Drive frequency (VFD) from 50 Hz to 25 Hz. The synchronous speed halves to 6 m/s and the vehicle settles at about 5.2 m/s.
• Raise the Air gap (mechanical) from 8 mm to 16 mm. The goodness factor G falls from 6.0 to 3.7, the slip rises to about 10% and the power factor drops from 0.31 to 0.26. A big gap costs performance.
• Change the Reaction plate material to copper. G rises to about 10.1 and the slip falls to about 5%. The standstill thrust drops to about 1300 N, but the machine runs at a smaller slip.
• Switch End effect from With end effect to Ideal. The efficiency rises from about 53% to 76%, and the readout Thrust with end effect / ideal shows 0.80 at the default speed with the end effect on.
• Raise the Opposing force until the thrust at standstill is below it. The vehicle cannot move.
• Look at the Normal (attraction) force readout: about 1.8 kN, which a real vehicle must support.

The two charts, Thrust against speed and Launch along the track, let you read the standstill thrust, the balance speed and the time to reach the end of the track.

Try it in the simulator