How to size a robot joint motor

By Tolerance Labs engineering team · Updated 13 October 2026

Pose and load tell you the torque a joint needs. This guide checks whether a motor can deliver it repeatedly without overheating, and fast enough at your bus voltage.

Continuous and peak torque are separate limits

Start from output torque and speed over time for the worst realistic task. The guide on sizing joint torque for a legged robot shows how to get torque from pose and load.

Continuous torque is what the motor can produce indefinitely without the windings exceeding their temperature rating. It depends on mounting, heat sinking, airflow and ambient temperature, so read the conditions next to the number.

Peak torque is limited by iron saturation, magnet demagnetization, drive current and how fast the windings heat. It usually comes with a time limit. Peak torque must cover the highest point in the profile; continuous torque must cover its RMS value.

RMS torque over a duty cycle

Copper loss scales with current squared, and current scales with torque. So heat follows the root mean square of torque, not the average: T_rms = √(Σ T_i² × t_i ÷ Σ t_i).

Example: a knee joint repeats a 2.0 s cycle with three phases.

T_rms = √(361.2 ÷ 2.0) = √180.6 ≈ 13.4 N·m. The average torque is (30 × 0.3 + 12 × 0.5 + 4 × 1.2) ÷ 2.0 = (9 + 6 + 4.8) ÷ 2.0 = 9.9 N·m. Sizing on the average would undersize the joint by about a quarter.

Before margin, this joint needs 13.4 N·m continuous and 30 N·m peak for 0.3 s. RMS averaging holds when the cycle is much shorter than the winding’s thermal time constant. For long holds, check each phase against the continuous rating on its own.

Torque constant, current and heat

The torque constant Kt (N·m/A) links current to motor torque: T_motor = Kt × I. Copper loss is P = I² × R, so loss grows with the square of torque.

Continue the example with a 9:1 gearbox at 90% efficiency, Kt = 0.1 N·m/A and winding resistance R = 0.2 Ω, using the datasheet’s current and resistance definitions throughout.

• RMS motor torque: 13.4 ÷ (9 × 0.9) = 13.4 ÷ 8.1 ≈ 1.65 N·m
• RMS current: 1.65 ÷ 0.1 = 16.5 A
• Copper loss: 16.5² × 0.2 = 272.25 × 0.2 ≈ 54 W
• At the 30 N·m peak: 30 ÷ 8.1 ≈ 3.70 N·m, so 37 A and 37² × 0.2 = 1,369 × 0.2 ≈ 274 W

The peak is about 2.2 times the RMS torque but makes about 5 times the heat. Copper resistance also rises about 0.39% per °C, so a winding 80 °C above room temperature has about 31% more resistance (0.39 × 80). And Kt falls at high current as the iron saturates, so peak current buys less torque than the linear estimate.

KV, back-EMF and bus voltage

KV is the motor’s no-load speed per volt, in rpm/V. A spinning motor generates back-EMF in proportion to speed, and the drive must supply more than the back-EMF plus the I × R drop to push current in.

Kt and KV describe the same magnetic coupling. For a lossless BLDC in SI units, Kt ≈ 9.55 ÷ KV, where 9.55 = 60 ÷ 2π. This is an approximation. The exact factor depends on whether the maker quotes peak or RMS current and phase or line-to-line voltage, so use the datasheet Kt when given.

In the example, Kt = 0.1 N·m/A corresponds to KV ≈ 9.55 ÷ 0.1 = 95.5 rpm/V. On a 48 V bus the no-load motor speed is about 95.5 × 48 ≈ 4,580 rpm, or 4,580 ÷ 9 ≈ 510 rpm at the output.

Now check a loaded point: 30 N·m at 20 rad/s. The motor turns at 20 × 9 = 180 rad/s. Back-EMF is about Kt × ω = 0.1 × 180 = 18 V. The I × R drop at 37 A is 37 × 0.2 = 7.4 V. The total, about 25 V, fits a 48 V bus with room for inductance and drive losses. A 24 V bus would not reach it, since motor drivers cannot use the full bus voltage.

Gear ratio trades torque, speed and inertia

For a fixed motor, ratio N multiplies output torque by N (times efficiency), divides output speed by N and multiplies reflected rotor inertia by N².

Going from 9:1 to 18:1 in the example, with efficiency held at 90%, motor torque drops to 13.4 ÷ (18 × 0.9) ≈ 0.83 N·m, current to 8.3 A, and copper loss to about 8.3² × 0.2 ≈ 14 W, a quarter of before. Top output speed halves to about 4,580 ÷ 18 ≈ 255 rpm, and reflected rotor inertia quadruples.

Higher ratios suit slow, heavily loaded joints. Lower ratios suit joints that need speed, impact tolerance and torque control from current, such as legged-robot hips and knees. See QDD vs high-ratio joints. The same arithmetic applies to a joint built around a frameless torque motor; building with frameless torque motors covers the mechanical side.

Margins

• Continuous torque. Keep RMS torque below the rating with room for heat sinking you have not measured yet. A 20–30% margin is a common starting point. Derate further for enclosed joints and warm environments, since ratings often assume free air or a heat sink at room temperature.
• Peak torque. Cover the highest profile point plus uncertainty in mass, payload and controller overshoot, within the rated peak time.
• Voltage. Leave headroom above back-EMF plus I × R at the fastest loaded point. Battery voltage sags under load and as it discharges.
• Drive current. The drive’s continuous and peak ratings must cover the motor’s.

Checklist

• From the worst realistic profile, compare RMS torque with continuous torque, and the profile peak with peak torque and its time limit, with margin.
• Convert to motor current with Kt and the ratio, then estimate I²R loss with hot resistance.
• Check back-EMF plus I × R at the fastest loaded point against the lowest bus voltage.
• Try the next ratio up and down and compare heat, top speed and reflected inertia.