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    Drives / motor sizing

    Motor sizing calculator

    A motor must do two jobs: carry the load and accelerate the inertia. This calculator adds the acceleration torque to the load torque, converts the result into shaft power, and shows how sensitive the sizing is to acceleration time and reflected inertia.

    Speed1500 rpmInertia0.020 kg·m²Efficiency90%
    Peak torque
    11.28 N·m
    2.26× the load torque
    Acceleration torque
    6.28 N·m
    Inertia 0.020 kg·m² over 0.50 s
    Shaft power at peak
    1.97 kW
    At 90% efficiency
    Continuous power
    0.87 kW
    At the load torque
    Torque build-up
    The motor must supply the load plus whatever is needed to accelerate the inertia.
    11.28 N·m
    Torque build-up
    TermValue
    Load torque5.00 N·m
    Angular speed157.1 rad/s
    Angular acceleration314.2 rad/s²
    Acceleration torque6.28 N·m
    Peak torque11.28 N·m
    Peak shaft power1.97 kW

    Reflected inertia must include the load through the gear ratio squared.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Peak torque against acceleration time
    Accelerating faster costs torque in inverse proportion.
    Peak torque against acceleration time0.0008.1616.3124.4732.6340.790.1s0.1s0.3s0.3s0.5s0.5s0.7s0.7s0.9s0.9s1.1s1.1s1.2s1.2s1.4s1.4s1.6s1.6s1.8s1.8s2.0s2.0sN·m
    Torque components
    Load torque against the acceleration allowance.
    Torque components0.0002.665.337.9910.6513.315.00LoadLoad6.28AccelerationAcceleration11.28PeakPeakN·m
    Canonical equations
    The exact relationships implemented by this engine.

    ω = 2πn/60 · n in rpm.

    T_a = J·α · α = ω / acceleration time.

    T_peak = T_load + T_a · What the drive must deliver during acceleration.

    P = T·ω / η · η accounts for gearbox and motor losses.

    engineering methodology

    A clear path from inputs to results.

    A motor must do two jobs: carry the load and accelerate the inertia. This calculator adds the acceleration torque to the load torque, converts the result into shaft power, and shows how sensitive the sizing is to acceleration time and reflected inertia.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the load torque at the shaft.
    2. Enter the reflected inertia including the gearbox.
    3. Set the target speed and acceleration time.
    4. Read the peak torque and the power it implies.

    The formula

    Tpeak=Tload+Jα,P=TωηT_{peak} = T_{load} + J\alpha, \qquad P = \frac{T\omega}{\eta}

    Reference: Newton’s second law for rotation; standard servo and induction motor sizing practice.

    Worked example

    5 N·m load, 0.02 kg·m², 1,500 rpm in 0.5 s

    Inputs
    • Load torque 5 N·m.
    • Inertia 0.02 kg·m².
    • Speed 1,500 rpm.
    • Acceleration 0.5 s.
    • Efficiency 90%.
    Calculation path
    1. 1

      ω = 2π × 1500/60 = 157.1 rad/s.

    2. 2

      α = 157.1 / 0.5 = 314.2 rad/s².

    3. 3

      Acceleration torque: 0.02 × 314.2 = 6.28 N·m.

    4. 4

      Peak torque: 5 + 6.28 = 11.28 N·m.

    5. 5

      Shaft power at peak: 11.28 × 157.1 = 1,772 W; about 1.97 kW electrical at 90% efficiency.

    Result

    Peak 11.28 N·m · 1.77 kW shaft · 1.97 kW electrical.

    Interpreting the result

    Acceleration usually dominates. Here the inertia needs more torque than the load itself, so halving the acceleration time would double the motor size. That is why gearboxes are used: reflecting inertia through a ratio reduces it by the square of the ratio, which is the cheapest way to shrink a drive.

    Assumptions & limitations

    • Steady-state and constant-acceleration sizing only.
    • Inertia is already reflected to the motor shaft.
    • Duty-cycle RMS torque, thermal limits and regeneration are excluded.

    Frequently asked questions

    What is reflected inertia?

    Load inertia divided by the gear ratio squared, plus the motor and coupling inertia.

    Can I size on continuous torque alone?

    No. The drive must supply the peak during acceleration, and the motor must survive the thermal duty cycle.

    Why does efficiency increase the electrical power?

    Shaft power is mechanical output; the drive must supply more to cover losses.

    Methodology last reviewed 2026-10-21. Results are informational and not financial or engineering advice — see our disclaimer.