What is the starting torque?
What is starting torque?
Starting torque describes the turning moment a motor can develop when starting from rest. A machine also has a starting-torque requirement: the resistance and acceleration demand the motor must overcome. Keep available torque separate from required torque. Both are expressed in newton metres, N·m, but they describe opposite sides of the sizing problem. [1, 2]
Getting the shaft moving is only the first test. The drive system must then provide enough torque to bring the load up to its intended operating speed. [5]
On this page: Torque terminology · The correct calculation · Worked example · Starting methods · Troubleshooting
Starting, breakaway and breakdown: read the context
Motor side: For an induction motor, locked-rotor torque specifies standstill torque at rated voltage and frequency. Ts/TN may denote its ratio to rated torque. [1]
Load side: Here, breakaway torque means initial resistance from static friction and process forces. Some catalogues use it for motor starting torque instead, so read the context. It does not mean a component’s fracture limit. [1, 2]
Breakdown torque is the maximum under stated supply conditions, not another name for starting torque. Pull-up torque is the minimum between standstill and breakdown-torque speed. [1]
The distinction explains a common symptom: a motor begins turning but cannot finish accelerating. Compare its torque–speed curve with the load demand, not just the standstill point. Static torque is a broader description of torque without rotation, rather than a universal replacement for these more specific terms. [5]
Why power divided by speed does not give starting torque
At a rotating shaft, P = T × ω and ω = 2πn/60. For n > 0, rearranging gives T(N·m) ≈ 9550 × P(kW) / n(rpm). With power in watts, use approximately 9.55 instead. P must be mechanical shaft power, not electrical input. The guides to torque versus power and rpm explain the quantities.
At zero speed, the rearranged formula is undefined. A stationary shaft can carry finite torque while its mechanical output power is zero. Dividing that power by zero speed cannot recover the torque. Electrical current and heating may still be present. Zero shaft power is not evidence of zero torque.
Hypothetical nameplate example: A 7.5 kW motor rated at 1470 rpm has approximately TN = 9550 × 7.5 / 1470 = 48.7 N·m of rated torque. That calculation does not establish its starting torque. A separately specified starting-to-rated ratio of 1.8 would imply about 87.7 N·m under the corresponding supply conditions, but 1.8 is only an assumption for this example.
Calculate the load and acceleration demands separately
Once the mechanism is moving, a constant-inertia model gives Trequired = Tresisting + Jeq × α. Resisting torque includes friction and process loading; Jeq is the total inertia, including the motor, referred to the same shaft. Inertia is not a fixed breakaway threshold. It determines the extra torque needed for the chosen acceleration. [3, 4]
For a constant-acceleration speed ramp, α = (2π/60) × Δn/Δt. A longer ramp time reduces acceleration torque but does not remove static friction or a jam. With a gearbox, refer torque, speed and inertia consistently to one shaft before combining them.
Worked example: starting the load is not the same as accelerating it
Assume a motor-shaft breakaway resistance of 35 N·m, a constant moving resistance of 25 N·m, and total equivalent inertia of 0.20 kg·m². After breakaway, the intended linear speed ramp runs from zero to 1500 rpm in 2 seconds. These are illustrative inputs, not product specifications.
The acceleration is approximately 78.54 rad/s². Acceleration torque is therefore 0.20 × 78.54 ≈ 15.7 N·m, giving a moving torque demand of 25 + 15.7 = 40.7 N·m. The motor must also overcome the separate 35 N·m breakaway resistance to initiate motion.
Extending the ramp to 4 seconds reduces the moving demand to about 32.9 N·m, but the initial 35 N·m resistance remains. A torque ceiling of 32.9 N·m applied throughout would therefore prevent starting in this model. Check breakaway capability and acceleration capability separately, then assess tolerances, duration and thermal limits for the real installation.
The starting method changes the available torque
Star–delta and soft starting: With star–delta starting, a motor suitable for delta running on that supply has ideally about one-third of its direct-on-line delta starting torque while connected in star. A softstarter also changes torque by reducing voltage. At fixed frequency, the approximate relationship T ∝ V² predicts 64% torque at 80% voltage. This is an approximation, not a rule for every operating condition. [6]
Inverter control: Field-oriented control and flux vector control can manage low-speed torque with suitable motor data and feedback. However, brief starting torque at zero speed and continuous load holding at zero speed are different requirements. Check whether the application needs an encoder; sensorless drives do not all offer the same standstill capability. [7]
Low-speed compensation: Torque boost can compensate winding voltage drop in the low-frequency portion of a V/f characteristic. Excessive boost can raise current and temperature. It is not unlimited extra torque, nor is it equivalent to closed-loop torque control. [5, 11]
Servo, DC and stepper motors need different checks
For a servo system, match the motor, amplifier and position, speed or torque-control architecture. The starting demand must fit within the drive’s peak current and permitted duration, not just the motor’s peak-torque figure. Check RMS torque and thermal duty over the cycle as well. A peak rating is not a continuous standstill rating. [3, 9]
For a permanent-magnet or constant-flux DC motor, electromagnetic torque is approximately T = kt × Ia. Current can therefore produce torque while the shaft is stationary. Actual delivery still depends on losses, current limits and heating; a stall-torque figure is not permission to operate continuously with a blocked shaft. [8]
For a stepper, distinguish holding torque from dynamic capability. Holding torque applies at rest, pull-in describes direct start/stop capability, and pull-out marks the synchronous operating limit at a given speed. Load inertia and pulse rate matter: holding torque alone cannot establish whether the loaded motor will start correctly. [10]
Why starting torque matters in the application
A conveyor, winder or hoist needs motor selection based on torque demand, not kW alone. Starting torque is not automatically higher than rated torque; the requirement follows the load and motion profile. Process conditions can also change the breakaway demand—for example, pressure at the start of a compressor cycle. [2, 4]
Starting successfully once is not proof of continuous suitability. A standard shaft-fan-cooled motor loses cooling as speed falls. Repeated starts and prolonged high current also need assessment. A constant-torque description does not remove those limits. Consider acceleration and deceleration time and S-curve shaping alongside the risk of sudden torque loading. [4]
When the motor will not start under load
Check the mechanical load first. A brake that has not released, a jam, excessive drag or altered process conditions may explain the symptom. Isolate the equipment, release stored energy safely and secure the load before inspection. Do not try to help an energised motor start by turning its shaft by hand.
Then verify supply and configuration. Check nameplate data, winding connection, supply drop, motor parameters, control mode and current/torque limits. Speed mode and torque mode respond differently. Check torque-control settings against the model’s operating instructions. [11]
Interpret measurements carefully. In torque measurement, separate the drive’s estimate from direct shaft data; an appropriate torque transducer may be needed. Starting unloaded but not loaded calls for a comparison between demand and capability, not an immediate diagnosis of one defective part. Avoid repeated start attempts against a blocked mechanism.