Induction Motor Torque-Speed Curve Calculator
Induction Motor Torque-Speed Curve Calculator
The whole torque-speed curve of a cage or wound-rotor induction machine from its equivalent-circuit parameters: starting torque, the pull-up through the middle, the breakdown torque and the slip it sits at, and the short stable stretch between breakdown and synchronous speed where the machine actually runs. With your load line on the same axes, and what reduced voltage does to all of it.
Induction motor torque-speed curve
a 7.5 kW, 4-pole machine on 400 V, 50 Hz in star, with R₁ = 0.6 Ω, X₁ = 1.2 Ω, Xm = 40 Ω, R₂ = 0.5 Ω and X₂ = 1.2 Ω, driving 49.7 N·m
The Thevenin reduction, and what it gives in closed form
T(s) = (3 ÷ ωs)·Vth²·(R₂/s) ÷ [(Rth + R₂/s)² + (Xth + X₂)²]
smax = R₂ ÷ √(Rth² + X²) Tmax = 3Vth² ÷ [2ωs(Rth + √(Rth² + X²))]
- ωs
- synchronous speed in mechanical rad/s, 2πf ÷ pole pairs
- s
- slip, (ns − n) ÷ ns. The rotor circuit’s resistance appears as R₂/s, which is why the same machine looks completely different at standstill and at 3% slip
- X
- Xth + X₂, the total leakage reactance in the Thevenin-reduced rotor loop. It is what limits the breakdown torque
- Tmax
- note what is absent: R₂. Rotor resistance sets WHERE the peak is, not how high
Worked example
a 7.5 kW, 4-pole machine on 400 V, 50 Hz in star, with R₁ = 0.6 Ω, X₁ = 1.2 Ω, Xm = 40 Ω, R₂ = 0.5 Ω and X₂ = 1.2 Ω, driving 49.7 N·m
Synchronous speed is 120 × 50 ÷ 4 = 1,500 rpm, which is 157.08 rad/s
Looking back from the rotor branch, Vth = 230.9 V × 40 ÷ √(0.6² + 41.2²) = 224.2 V, and Zth = 0.5654 + j1.1733 Ω
Breakdown sits at s = R₂ ÷ √(Rth² + X²) = 0.5 ÷ 2.4397 = 20.49% slip, which is 1,193 rpm, and its height is 159.7 N·m — 3.21× the 49.74 N·m full-load torque
At standstill (s = 1) the same expression gives 70.92 N·m, or 1.43× full load. Notice it is well below breakdown: a cage machine's peak is in the middle of the curve, not at the start
For a 49.7 N·m load, solving the torque expression for R₂/s gives 17.850 Ω, so s = 2.801% and the machine runs at 1,458 rpm delivering 7.588 kW
At 57.7% voltage everything on the torque curve falls by the square: breakdown becomes 53.17 N·m and starting torque 23.61 N·m. That is the whole star-delta argument, and the whole star-delta problem
What each parameter moves
| Change | Breakdown torque | Breakdown slip | Starting torque |
|---|---|---|---|
| Supply voltage down by a factor k | falls as k² | unchanged | falls as k² |
| Rotor resistance R₂ up by a factor m | unchanged | rises as m | rises, up to the point where breakdown reaches standstill, then falls again |
| Leakage reactance X₁ + X₂ up | falls | falls | falls |
| Supply frequency down at constant volts | rises (until saturation stops it) | rises | rises — which is why a VFD starts a machine gently and a star-delta starter does not |
| Supply frequency down at constant volts per hertz | roughly unchanged | rises as 1/f | roughly unchanged |
Reading the curve, and the three points on it that matter
An induction motor’s torque-speed curve is not a straight line and not a hyperbola. It starts at a moderate value at standstill, dips a little, rises to a peak somewhere around 15–25% slip for an ordinary cage machine, and then falls steeply to zero at synchronous speed. Everything an induction machine does follows from that shape, and the shape follows from one fact: the rotor circuit’s resistance appears in the equivalent circuit as R₂/s. At standstill s = 1 and the rotor looks like a low-resistance, highly reactive short circuit, so a large current produces a mediocre torque at a terrible power factor. At 3% slip R₂/s is thirty-three times larger, the rotor current is nearly in phase, and a much smaller current produces a much larger torque.
The stable region is the last few per cent. Between breakdown slip and synchronous speed the curve falls steeply with speed, which is exactly the condition for stable operation: load the machine a little more and it slows a little, which produces more torque. On the other side of the peak the slope has the wrong sign, so a machine pushed past breakdown does not find a new operating point — it stalls. That is why the breakdown torque, and not the rated torque, is what decides whether a machine survives a momentary overload or a voltage dip. For the synchronous speed and slip arithmetic alone, the induction motor slip calculator does that job and this page builds on it rather than repeating it.
Rotor resistance is the one parameter that moves the peak without moving its height. Look at the closed forms: the breakdown slip is proportional to R₂ and the breakdown torque does not contain R₂ at all. Triple the rotor resistance and the peak triples its slip and stays the same height — so if you triple it enough, the peak arrives at standstill and the machine develops its maximum torque the instant it is switched on. That is precisely what a wound-rotor machine with external rotor resistors does, and it is why such machines were used for crane hoists and mine winders. A cage machine gets a smaller version of the same effect for free from its own skin effect: at 50 Hz rotor frequency the current crowds into the top of a deep bar and the effective resistance is high, and by 1.5 Hz it has spread out again. NEMA design classes A to D are, in effect, four choices of how much of this to build in.
Reduced-voltage starting throws away torque as the square of voltage. Every torque on the curve scales with V², so a star-delta starter at 57.7% of the winding voltage leaves exactly a third of the torque — and a third of the starting torque is often not enough to break a loaded conveyor away. The chart draws both curves so the gap is visible, and the page says outright when the reduced-voltage starting torque falls below the load. A variable-frequency drive avoids the whole problem by lowering the frequency with the voltage, which keeps the flux and therefore the torque; for what the drive itself costs in losses see the inverter loss calculator, and for the running current at the operating point the full-load current calculator.
What this model leaves out. Core loss is not represented, so the torque is the air-gap torque and the shaft torque is a little lower; friction and windage are not represented either. The parameters are treated as constants, when in a real cage machine R₂ and X₂ both vary strongly with rotor frequency — which is the pull-up dip between starting and breakdown that this model draws as a smooth curve. And saturation is ignored, so the model over-states the torque at large slip where the leakage paths saturate. Get the parameters from a no-load and locked-rotor test done to IEEE Std 112 and the curve will be good to a few per cent in the stable region, which is the part you operate in.
Frequently asked questions
What is breakdown torque and why does it matter more than rated torque?
It is the highest torque the machine can develop, at the peak of the curve. It matters because it is the margin against a stall: a voltage dip reduces every torque on the curve by the square of the dip, so a machine running at full load with a breakdown-to-full-load ratio of 2.5 will stall on a dip to 63% of rated voltage. Standard cage machines are between about 1.8 and 3.
Why is starting torque lower than breakdown torque?
At standstill the rotor frequency equals the supply frequency, so the rotor’s leakage reactance is at its largest and its resistance R₂/s at its smallest. The rotor current is large but badly out of phase with the flux, and torque needs the in-phase part. As the machine speeds up, the rotor frequency falls, the current comes into phase, and the torque rises — until the falling rotor current takes over and the curve turns down.
How do I get the equivalent-circuit parameters?
A no-load test gives Xm and the core and friction losses; a locked-rotor test gives R1 + R2 and X1 + X2; a DC measurement of the stator gives R1 on its own. The split of leakage reactance between stator and rotor is not measurable and is assigned by NEMA design class: 50/50 for A and D, 40/60 for B, 30/70 for C. IEEE Std 112 is the procedure.
Why does extra rotor resistance help a motor start?
It moves the breakdown point to a higher slip without changing its height. Enough of it puts the peak at standstill, so the machine develops maximum torque the moment it is energised. The cost is slip and therefore rotor loss in normal running, which is why a wound-rotor machine shorts its rings out once it is up to speed.
Does star-delta starting really give only a third of the torque?
Yes. Each winding sees the line voltage divided by √3, and torque goes as voltage squared, so it is exactly one third — of the starting torque, of the breakdown torque and of every point in between. The starting current is also a third, which is why the method exists, but you cannot have one without the other.
Does this curve apply under a variable-frequency drive?
Only one slice of it. A drive changes the frequency, which moves synchronous speed and changes every reactance in the circuit in proportion. Holding volts per hertz keeps the flux and gives a family of curves of roughly the same shape sliding along the speed axis — so the breakdown torque stays roughly constant until the drive runs out of voltage, above which it falls off like a field-weakened machine.
Related calculators
References
- Fitzgerald A E, Kingsley C, Umans S D. Electric Machinery, 6th ed. McGraw-Hill. Chapter 6, ‘Polyphase Induction Machines’: the single-phase equivalent circuit, the Thevenin reduction seen from the rotor branch, and the closed-form maximum torque and the slip at which it occurs. Chapter title verified by search; sources give the 6th edition’s date as both 2002 and 2003, so it is not quoted here.
- IEEE Std 112, IEEE Standard Test Procedure for Polyphase Induction Motors and Generators. The no-load, locked-rotor and DC resistance tests that produce the parameters this page consumes, and the conventional split of leakage reactance between stator and rotor by NEMA design class. Copyrighted; cited, not reproduced.
- Chapman S J. Electric Machinery Fundamentals, 5th ed. McGraw-Hill, 2011. The induction-motor chapter, for the torque-speed characteristic, the effect of rotor resistance on the position of the peak, and the NEMA design classes. Edition verified by search; the chapter number is not quoted because it was not.