Every motor room job starts with the same question: what will this machine draw the moment you close the contactor, and can the board take it. This chapter builds from slip and torque up to the starting method that actually fits the load in front of you.
Three separate windings in the stator, fed with three-phase current 120° apart in time, set up a magnetic field that rotates around the bore of the machine. The field's speed depends on nothing but the supply frequency and how many magnetic poles the winding is arranged into — double the poles and the field turns at half the speed for the same frequency.
The rotor itself carries no external connection in the common squirrel-cage design — just a cage of solid conducting bars shorted together at each end by end-rings. As the stator field sweeps past, it cuts across these bars exactly as a moving magnet cuts across a stationary conductor, inducing an EMF in them by straightforward transformer action. Because the bars are short-circuited, that EMF drives a current, and a current-carrying conductor sitting in a magnetic field feels a force — which is where the shaft torque comes from.
That force always acts to reduce the relative motion between rotor and field, in other words to drag the rotor round after the field, in line with Lenz's law. It is a completely self-regulating machine: no external excitation, no brushes to wear, no separate field circuit to fail, which is a large part of why the induction motor is the default choice for almost everything that turns on a ship.
Nothing drives current into the rotor from outside — the rotor's own current, and therefore its torque, exists only because it is not quite keeping up with the field. Take away that difference in speed and the induced EMF, the rotor current and the torque all disappear with it.
The gap between the field's speed and the rotor's actual speed is called slip, normally expressed as a fraction or percentage of synchronous speed rather than as a raw rpm figure.
At standstill the rotor is stationary while the field still turns at N_s, so slip is 1 (100 %) and the rotor bars are cut at the full line frequency — this is why locked-rotor current is so high, the rotor circuit behaves almost like a short-circuited transformer secondary. As the rotor speeds up, the relative motion between it and the field shrinks, the induced EMF and current fall, and so does torque, until the motor settles at whatever slip generates just enough rotor current to balance the load torque. For a typical cage motor at rated load that settling point is only two to five per cent slip — the shaft is turning at 95–98 % of synchronous speed.
Rotor frequency is worth carrying in your head separately from slip itself: it starts at full line frequency when stationary and falls to only a Hz or two once the motor is up to speed and lightly loaded, which is also why a running motor sounds and feels different from one being started — the rotor's own field is barely rotating relative to the rotor iron at full speed, so eddy-current and hysteresis effects in the rotor stay small in normal running.
Because torque can only exist where there is slip, synchronous speed itself is a limit the motor approaches but can never reach on load — an unloaded motor can get close to it, spinning almost free, but the moment a shaft load is applied the rotor must fall back just enough to induce the current needed to carry that load.
Plot torque against speed for a standard cage motor and the shape is distinctive: torque starts at a moderate value at standstill, dips slightly as speed rises through the first part of run-up (the pull-up torque, usually the lowest point on the curve), climbs to a peak — the breakdown or pull-out torque, typically somewhere around 70–80 % of synchronous speed — and then falls away steeply to the normal full-load operating point, which sits on that final steep section just short of synchronous speed.
The steepness of that final section is the practical point: because a cage motor's running speed hardly moves across a wide range of load, it behaves close to a constant-speed machine right up until it is overloaded past breakdown torque, at which point speed collapses very quickly rather than gracefully. Rotor design changes this curve — a rotor built with higher-resistance bars gives more starting torque for the same starting current at the cost of a less steep, less efficient running characteristic, which is the trade-off behind the different rotor designs built for different duties.
A motor that starts a load does not just need enough torque at standstill — it needs enough torque everywhere along the run-up curve, and the pull-up point, not the starting point, is usually where a marginal start actually fails.
Direct-on-line starting connects the motor straight to full supply voltage and lets it draw whatever current the locked-rotor impedance demands — typically six to eight times full-load current, for perhaps one to a few seconds while the rotor accelerates. On a motor small relative to the supply this is simply absorbed; the difficulty is not the motor's own tolerance of that current but the voltage dip it pulls across the whole switchboard while it lasts, which can upset every other consumer on the board, including sensitive electronics and the torque output of other running motors.
Star-delta starting reduces that current by starting the motor with its windings connected in star, then switching to delta once it has run up. Each winding then sees only 1/√3 of line voltage during starting, and because starting torque follows the square of applied voltage, both the line starting current and the starting torque fall to one third of their direct-on-line values. That is a genuine constraint, not a footnote: a star-delta start only suits a load that can accelerate on roughly a third of full-load torque — typically something started unloaded or very lightly loaded, such as a pump against a closed or part-open valve. Push it onto a load that needs a third or more of its full-load torque to break away and the motor will simply sit at low speed, current high, torque too low to accelerate further, cooking the windings.
A soft starter takes a different approach: thyristors ramp the applied voltage up smoothly from a low starting value rather than switching it in two abrupt steps. Torque still falls with the square of the reduced voltage, so the current and torque reduction it offers is broadly similar in size to star-delta, but the ramp removes the mechanical shock of a sudden step change in torque and lets the current profile be tuned rather than fixed at a single ratio. It remains, fundamentally, a voltage-reduction technique with the same underlying torque penalty — it does not get around the star-delta problem, it just makes the transition smoother.
A variable frequency drive sidesteps the starting-current problem altogether rather than managing it: it rectifies the incoming AC to DC, then synthesises a new three-phase output at whatever frequency and voltage it chooses. Starting the motor at a low frequency means synchronous speed itself starts low, so the motor can be brought up to full speed with the rotor almost never far from the field it is chasing — current stays close to full-load current throughout, rather than spiking to six or eight times it.
The drive holds the ratio of output voltage to output frequency constant as it ramps frequency up, which is what keeps the air-gap flux at its rated value at every speed below the motor's rated (base) frequency — and because torque depends on flux and rotor current rather than on frequency directly, the motor can deliver its full rated torque anywhere in that range, not just at one fixed speed. This constant-torque region is the main reason VFDs suit hoists, winches and conveyors as well as pumps and fans.
Above base frequency the drive has run out of headroom: it cannot output more voltage than the supply and the motor's insulation allow, so the V/f ratio — and the flux with it — must fall as frequency rises further. Torque capability falls in step, but because speed is rising at the same time, the power the motor can deliver (torque × speed) stays roughly constant — a constant-power region, useful for loads like some pumps where torque demand naturally falls as speed rises, but a poor match for anything that needs full torque at high speed.
Beyond current and starting torque, a VFD also removes the mechanical shock of starting: instead of a step change in torque slamming through couplings, gearboxes and shafting, the drive's ramp brings torque up gradually, which matters as much to the mechanical drive train as the electrical starting current matters to the switchboard.
A motor's protection relay is built around several distinct failure modes, not one blanket "overcurrent" trip: thermal overload protects the windings against sustained overcurrent that would otherwise cook the insulation over minutes; a locked-rotor or stall trip catches the case where the motor draws starting current but never accelerates away from it, which a thermal element sized for running overload would be too slow to catch; single-phasing (loss of one supply line) protection catches the doubled current in the remaining two phases before it does the same slow thermal damage; and earth-fault protection catches insulation breakdown to frame before it becomes a shock or fire hazard.
Insulation is rated by class, each defined by the maximum hot-spot temperature its materials are designed to tolerate continuously.
Running a motor built to Class F insulation but only ever reaching Class B temperatures in service gives useful thermal margin; running it consistently near its class limit shortens winding life sharply, since insulation ageing roughly doubles for every 8–10 °C above its rated temperature it is allowed to run.
VFDs introduce a failure mode that a direct-on-line motor does not usually see: the fast-switching output waveform induces a common-mode voltage that can find its way across the motor's bearings, discharging as tiny sparks through the oil film and pitting the races over time — a slow, cumulative form of bearing damage with no equivalent on a mains-fed motor. The standard remedies are an insulated bearing at one end of the shaft to break the circuit, or a shaft-earthing brush that gives the common-mode current a lower-resistance path to earth than through the bearing itself, and larger drives often add output filtering to reduce the common-mode voltage at source.
The rating plate current is the motor's full-load current, not its starting current — reading it as the latter when sizing a starter or a protection relay is a straightforward, and surprisingly common, way to get the whole calculation wrong from the first line.
The three problems below sit on top of the sections above rather than testing any one formula in isolation — each one chains two or three of the relationships together the way an actual sizing or fault-finding question on the day usually does.
A 6-pole, 50 Hz induction motor drives a constant-torque load (a positive-displacement pump) and runs at 950 rpm at full load. The VFD supplying it is later set to run the same duty at 40 Hz. Find the synchronous speed and slip at 50 Hz, the slip speed in rpm, and the motor's approximate running speed at 40 Hz.
Poles p = 6 Supply frequency f₁ = 50 Hz Full-load running speed N₁ = 950 rpm Second frequency f₂ = 40 Hz Load torque unchanged (constant-torque duty)
Find the synchronous speed and slip at 50 Hz, the slip speed in rpm, and the motor's approximate running speed at 40 Hz
Find the synchronous speed and slip at 50 Hz.
The rotating field's speed depends only on frequency and pole count, never on load.
Convert slip to slip speed.
Slip speed is the actual rpm the rotor lags the field by — the more useful figure to carry forward, because it stays roughly fixed for a fixed load torque even when frequency changes.
Rotor current.
And therefore torque, is closely proportional to slip frequency for small slip, and slip frequency is itself proportional to slip speed. With the drive's V/f control holding flux constant and the load torque unchanged, the rotor must lag the new field by very nearly the same 50 rpm as before.
Find the new synchronous speed and running speed at 40 Hz.
Answer≈750 rpm at 40 Hz (new slip ≈ 50 / 800 = 6.25 %).
The trap: assuming slip as a percentage stays constant when frequency changes — it is the slip speed, not the slip percentage, that stays roughly constant for constant load torque, because slip percentage is referenced to a synchronous speed that has itself moved.
A cargo pump motor has a full-load current of 45 A. Its rating plate data gives a locked-rotor starting current of 7 × FLC direct-on-line, with a locked-rotor starting torque of 1.8 × full-load torque. The switchboard can tolerate a maximum starting current of 200 A on this feeder without an unacceptable voltage dip elsewhere. Determine whether DOL starting is acceptable and, if not, whether star-delta starting solves the current problem — and whether the resulting starting torque is enough for a pump that needs about 40 % of full-load torque to break away.
Full-load current FLC = 45 A DOL starting current = 7 × FLC DOL starting torque = 1.8 × full-load torque Board limit = 200 A Load breakaway torque required ≈ 0.4 × full-load torque
Determine whether DOL starting is acceptable and, if not, whether star-delta starting solves the current problem — and whether the resulting starting torque is enough for a pump that needs about 40 % of full-load torque to break away
Check DOL starting current against the board limit.
315 A is well above the 200 A the board can absorb.
So direct-on-line starting is ruled out on this feeder — it is the voltage dip pulled on the rest of the switchboard that fails, not the motor's own tolerance of the current.
Check star-delta starting.
Reconfiguring the windings from delta to star for starting drops the voltage across each winding to 1/√3 of line voltage, which cuts the line starting current — and, because torque follows the square of applied voltage, the starting torque too — both to one third of their direct-on-line values.
Compare with the board limit and the load's needs.
105 A clears the 200 A limit comfortably. The available starting torque, 0.6 × full-load torque, is also well above the 0.4 × full-load torque this pump needs to break away, so the motor will accelerate rather than stall on the flat part of the curve.
AnswerDOL is not acceptable (315 A > 200 A); star-delta is acceptable — 105 A and 0.6 × FLT, both within the board's and the load's limits.
The trap: checking only that star-delta clears the current limit and not the torque — a load needing more than about a third of its full-load torque to break away may simply sit and hum in star without ever accelerating.
A motor is rated 400 V at 50 Hz (its base speed) and is driven by a VFD that holds V/f constant below base speed. Find the voltage the drive applies at 25 Hz and confirm the motor can deliver full rated torque there. Then find the flux ratio and the torque the motor can deliver at 80 Hz, where the drive's output voltage is limited to the 400 V it cannot exceed, and check what this means for the power the motor can deliver at that speed.
Rated voltage = 400 V at rated (base) frequency 50 Hz Below base speed: V/f held constant Above base speed: voltage capped at 400 V (cannot rise further) Test points: 25 Hz (below base) and 80 Hz (above base)
Find the voltage the drive applies at 25 Hz and confirm the motor can deliver full rated torque there
Find the rated V/f ratio and the voltage at 25 Hz.
At 25 Hz the drive has applied exactly 8 V/Hz.
The same ratio as at rated frequency, so the air-gap flux is unchanged from its rated value. Torque depends on flux and rotor current, not on frequency directly, so with flux held at rated value the motor can still deliver its full rated torque at this lower speed.
Find the flux ratio at 80 Hz.
Above base speed the drive cannot exceed the 400 V the motor and supply are rated for, so the V/f ratio — and with it the flux — falls as frequency rises further.
Relate this to torque and power.
Torque capability follows the flux, so at 80 Hz the motor can deliver only about 62.5 % of its rated torque. But speed itself has risen by a factor of 80/50 = 1.6, so multiplying torque capability by speed shows the power capability barely moves.
Answer200 V and full rated torque at 25 Hz; 62.5 % of rated torque at 80 Hz, but power capability stays at roughly 100 % of rated.
The trap: assuming a VFD gives full torque at any frequency you dial in — above base speed the drive runs out of voltage headroom, flux falls, and torque capability falls with it even though the motor is perfectly healthy.
N_s = 120f/pSynchronous speed, rpmSlip s = (N_s − N)/N_sTypically 2–5 % at full loadRotor frequency = s·fHighest at standstill, falls as the motor runs upDOL starting current 6–8×FLCStarting torque ≈ 1.5–2 × full-load torqueStar-delta: I_line and T both ÷3Only suits lightly-loaded startsSoft starter: T ∝ V²Smooth ramp; similar current/torque cut to star-deltaVFD: V/f constant below base speedConstant torque regionVFD above base speed: V capped, flux ∝ 1/fConstant power; torque fallsInsulation class B / F / H130 / 155 / 180 °C hot-spot limitsBearing currents on VFD-fed motorsMitigate with insulated bearing or shaft-earthing brush