Choosing actuators for a quadruped robot

By Tolerance Labs engineering team · Updated 18 October 2026

A quadruped carries its whole body on twelve leg joints that hit the ground several times a second. The actuators set its payload, its top speed, how hard it can land and how long it can stand before the knees overheat.

Three joints per leg

Most quadrupeds use three joints per leg, twelve in all. Hip abduction swings the leg sideways, hip flexion swings it forward and back, and the knee folds the lower leg. In small robots the knee motor often sits beside the hip flexion motor and drives the knee through a belt or rod.

Many designs use one module at all twelve positions, which simplifies spares and tuning; the knee gets its extra torque from a reduction in its belt or linkage.

Torque per joint: a worked example

This example shows how the leg pose decides which joint carries the weight. Sizing joint torque for a legged robot covers the general method and dynamic factors.

Take a 20 kg quadruped with a 0.25 m thigh and a 0.25 m shank. It trots with its hips 0.40 m above the ground and each foot below its hip.

• Knee position: the hip, knee and foot form a triangle with two 0.25 m sides and a 0.40 m base. The knee sits √(0.25² − 0.20²) = √(0.0625 − 0.04) = √0.0225 = 0.15 m to the side of the line from hip to foot.
• Average load per foot over a gait cycle: 20 × 9.81 ÷ 4 ≈ 49 N.
• In a trot each foot is on the ground about half the time, a duty factor of 0.5, so the average force during stance is 49 ÷ 0.5 = 98 N. If the force rises and falls as a half sine, its peak is π/2 times its average: 1.57 × 98 ≈ 154 N.
• Knee torque at peak force: 154 N × 0.15 m ≈ 23 N·m.
• Hip flexion torque: with the foot directly below the hip the lever is zero. With the foot 0.08 m ahead, as when braking or on a slope: 154 × 0.08 ≈ 12 N·m.
• Hip abduction torque: if the foot sits 0.06 m outboard of the abduction axis, 154 × 0.06 ≈ 9 N·m. A sideways foot force of 30 N from a turn or a push adds 30 × 0.40 = 12 N·m, because it acts over the full hip height, for about 21 N·m.

The knee carries the most, but abduction comes close in turns, so do not size it from straight-line trotting alone. Jumps and landings typically raise all three by a factor of two or more. The joint torque calculator runs the leg case with your own numbers.

Why quasi-direct drive dominates

Most quadrupeds use quasi-direct-drive joints: a large, flat motor behind one planetary stage of about 6:1 to 10:1. Two reasons decide it.

Impacts. Each touchdown must stop the rotor along with the leg, and rotor inertia reflects to the output multiplied by the ratio squared. Assume a rotor of 6 × 10⁻⁵ kg·m². At 9:1 it looks like 6 × 10⁻⁵ × 81 ≈ 0.0049 kg·m² at the knee; at 50:1, 6 × 10⁻⁵ × 2,500 = 0.15 kg·m². At a foot 0.25 m from the knee, that is 0.0049 ÷ 0.25² ≈ 0.08 kg against 0.15 ÷ 0.25² = 2.4 kg of extra mass to stop. The high-ratio leg lands like a much heavier leg, and its gear teeth take the shock.

Proprioception. One low-ratio stage loses little torque to friction, so motor current gives a usable estimate of joint torque, and the controller can sense foot force and touchdown without force sensors. The leg can also act as a programmable spring and damper in impedance (MIT) mode. Heavy quadrupeds that need higher ratios must get contact sensing back from torque sensors or series elasticity.

Moving knee mass toward the hip

A knee motor at the knee swings with the thigh on every step. Mounting it at the hip and driving the knee through a belt, chain or push rod takes that mass out of the swinging leg.

Example: a 0.6 kg actuator at the knee, 0.25 m from the hip axis, adds 0.6 × 0.25² ≈ 0.038 kg·m² of inertia about the hip; near the hip axis it adds almost nothing. If the thigh and shank alone are about 0.03 kg·m², the knee motor more than doubles the leg’s inertia, and the torque hip flexion needs to swing it.

• Belt or chain. Can add a second reduction, commonly about 1.5:1 to 2:1. Belts stretch under load, which lowers stiffness, and need a tensioner.
• Push rod or four-bar linkage. Stiffer, with no tension to keep, but the ratio changes with knee angle and the range is limited.
• Motor at the knee. Simplest and stiffest, at the cost of swing inertia.

Speed for trotting and jumping

Jumping. In the example geometry the hip rises 0.15 m per radian of knee rotation, the same lever that set knee torque. To leave the ground at 2 m/s, enough for a jump of 2² ÷ (2 × 9.81) ≈ 0.20 m, the knee must turn at 2 ÷ 0.15 ≈ 13 rad/s while still delivering high torque. At 9:1 that is 13 × 9 ≈ 120 rad/s, about 1,150 rpm, at the motor.

Trotting. At 1.5 m/s with a 0.40 m leg, the average hip flexion speed is about 1.5 ÷ 0.40 ≈ 3.8 rad/s, and the swing peak is commonly two to three times that.

Check speed at high torque and at the sagged bus voltage, not the no-load speed on a full battery; power supplies and batteries for robot joints shows how far the bus can sag during a jump.

Joint mass budget

In many small quadrupeds the twelve joints make up roughly half of the total mass, so the sizing loops back on itself. Example: twelve 0.6 kg joints weigh 7.2 kg. Moving to 0.8 kg joints adds 12 × 0.2 = 2.4 kg, so the 20 kg robot is 12% heavier and needs 12% more torque before any gain. Compare joints by peak torque per kilogram, which commonly falls between about 20 and 60 N·m/kg for quasi-direct-drive modules, and by continuous torque per kilogram. Repeat the estimate two or three times.

Heat while standing

Standing does no mechanical work, but holding torque costs copper loss, and a robot that stands for minutes between tasks can overheat its knees.

In the example, standing on four feet puts 49 N on each foot, and the knee holds 49 × 0.15 ≈ 7.4 N·m. Assume a 9:1 joint whose motor has a torque constant of 0.1 N·m per amp RMS and a phase resistance of 0.2 Ω.

• Motor torque: 7.4 ÷ 9 ≈ 0.82 N·m, ignoring gear losses.
• Current: 0.82 ÷ 0.1 = 8.2 A RMS.
• Copper loss: 3 × 8.2² × 0.2 = 3 × 67 × 0.2 ≈ 40 W per knee, about 160 W for four.
• Standing taller, with the knee 0.10 m off the line instead of 0.15 m, cuts knee torque by a third and copper loss to (2/3)² ≈ 0.44 of its value, about 18 W per knee.

A 1.5:1 belt stage at the knee cuts copper loss by 1.5² = 2.25. Thermal management of robot joint actuators covers temperature rise. If the robot will roll more than it walks, see actuators for wheel-legged robots.

Checklist

• Sketch each leg in its worst poses and compute abduction, hip flexion and knee torque from foot force and lever.
• Use peak foot force from the gait, not body weight alone, and add a dynamic factor for jumps and landings.
• Choose quasi-direct drive unless the torque cannot be reached at an acceptable mass.
• Decide where the knee motor sits and how it drives the knee: belt, rod or direct.
• Check joint speed at high torque against jump and swing needs, at the sagged bus voltage.
• Add up joint mass and repeat the torque estimate with the new total.
• Check standing copper loss against continuous rating in the pose the robot holds longest.

Browse humanoid and legged robot joints and quasi-direct-drive joints, or see complete legged robots.