Three torques, not one
A joint has to produce torque for three separate reasons, and undersized servos almost always come from budgeting only the first. Gravity demands a holding torque proportional to the load, its distance from the axis, and the cosine of the arm angle. Inertia demands more torque during acceleration, proportional to how fast you want the joint to reach speed. Friction — bearings, seals, cable stiffness, a dragging cover — demands a little more, all the time.
Add those three, multiply by a safety factor, and you have the torque the joint output must deliver. Divide by the gear ratio and its efficiency and you have what the motor must deliver. Every step of that is arithmetic; the judgement is in the safety factor and in the friction allowance, both of which are estimates.
The reason this matters more for servos than for most motors is that a servo’s advertised torque is a stall figure, measured at a specified voltage, with the output not moving. A servo asked to hold near its rating gets hot, buzzes, and eventually strips its gear train. A servo asked to move a load at speed near its rating cannot, because available torque falls as speed rises. Sizing at 2× the computed requirement is not padding; it is what makes the rating usable.
Working through the three terms
Static torque. Torque is force times perpendicular distance. The weight m·g acts vertically, and the perpendicular distance from the axis to that line of action is r·cosθ, where θ is measured from horizontal. So the holding torque is m·g·r·cosθ. It is largest with the arm horizontal and zero with the arm vertical, which is why you always size at the horizontal even if the arm rarely stops there.
Which r to use is the step most often got wrong. For a payload at the end of a light arm, r is the full length. For an arm whose own mass dominates — a uniform bar pivoted at one end — the weight acts at the centre of mass, half way along, so r is L/2. The same distinction changes the inertia: a point mass has J = mL², while a uniform rod about its end has J = mL²/3, a third as much. Choosing the wrong model on a heavy arm produces an error of two to three times in one term and three times in the other.
Inertial torque. Newton’s second law for rotation is T = J·α. The angular acceleration you need follows from the speed you want and the time you will allow: α = 2πn ÷ (60t) for n in rpm. This term is entirely under your control — doubling the acceleration time halves it — which makes it the first place to look when the required torque is uncomfortably high.
Gearing. A reduction multiplies torque and divides speed, so the motor needs T ÷ (N × η) where N is the ratio and η the efficiency. Efficiency is not a rounding detail: a worm drive at 50% doubles the motor torque you need compared with an ideal gearbox, and it is the reason a self-locking worm joint costs so much more motor than the ratio suggests. Note also that the load’s inertia reflected to the motor shaft falls with the square of the ratio, which is why geared joints accelerate so much more easily than direct-drive ones.
Worked example: a 500 g gripper on a 300 mm arm
A robot arm carries a 500 g gripper at the end of a 300 mm link. You want the joint to reach 30 rpm in 0.2 s, you allow 0.05 N·m for bearing and cable drag, the servo drives the joint directly, and you size at 2×.
- Static torque, arm horizontal. 0.5 kg × 9.80665 × 0.30 m × cos 0° = 1.4710 N·m.
- Moment of inertia. A point mass at the end: J = 0.5 × 0.30² = 0.045 kg·m².
- Angular acceleration. α = 2π × 30 ÷ (60 × 0.2) = 188.4956 ÷ 12 = 15.708 rad/s².
- Inertial torque. 0.045 × 15.708 = 0.7069 N·m.
- Sum. 1.4710 + 0.7069 + 0.0500 = 2.2279 N·m.
- Apply the safety factor. 2.2279 × 2 = 4.4557 N·m.
- Convert. 4.4557 ÷ 0.0980665 = 45.44 kgf·cm, or 4.4557 ÷ 0.00706155 = 630.9 oz·in.
- Power. ω = 2π × 30 ÷ 60 = 3.1416 rad/s, so P = 4.4557 × 3.1416 = 14.0 W at the output.
That result points straight at a 45–55 kg·cm servo, which in practice means a high-torque digital servo run at 7.4 V rather than a standard 6 V unit. Now look at what each term contributed: gravity 66%, acceleration 32%, friction 2%. Allowing 0.4 s to reach speed instead of 0.2 s halves the inertial term to 0.3534 N·m and drops the requirement to 3.7488 N·m — 38.23 kg·cm, comfortably inside a cheaper servo. Time is the cheapest thing you can spend here.
Reading the result against a real datasheet
Compare your figure with the servo’s stall torque at the voltage you will actually run. Hobby servo ratings are quoted at 4.8 V, 6.0 V and 7.4 V, and the difference between the ends of that range is often 40%. A 45 kg·cm rating at 7.4 V may be 35 kg·cm at 6.0 V, and if your battery sags under a stalled load it is less again.
Then check the speed. Servo speed is quoted as seconds per 60° with no load, and available torque falls roughly linearly from stall as speed rises. A servo rated 0.15 s/60° turns at about 67 rpm unloaded; asked for 30 rpm under load it will deliver well under its stall figure. This is the second reason for a safety factor above 1.5, and it is why the calculator reports mechanical power as well as torque — power is what tells you whether the actuator can do the job at speed rather than only hold it.
Use the angle sweep to find your worst case rather than assuming it. The gravity term follows the cosine, so it is largest at the horizontal and vanishes at the vertical; the inertial and friction terms do not vary with angle at all. That means the required torque falls as the arm rises but never below (inertial + friction) × safety factor — a joint that only ever operates near vertical still needs real torque to move, just not to hold.
Finally, watch the balance between the two main terms. When the calculator reports that the inertial term exceeds the gravity term, the joint is being sized by your acceleration demand, and lengthening the acceleration time is nearly always cheaper than buying a larger actuator. When gravity dominates, the levers are a shorter arm, a lighter payload, or a counterbalance — a spring or counterweight that cancels part of m·g·r removes torque the motor never has to produce, and it costs nothing to hold.
Required torque as the example arm rises
| Angle from horizontal | Static torque | Required torque | Required (kgf·cm) |
|---|---|---|---|
| 0° | 1.4710 N·m | 4.4557 N·m | 45.44 |
| 15° | 1.4209 N·m | 4.3555 N·m | 44.41 |
| 30° | 1.2739 N·m | 4.0615 N·m | 41.42 |
| 45° | 1.0401 N·m | 3.5940 N·m | 36.65 |
| 60° | 0.7355 N·m | 2.9847 N·m | 30.44 |
| 75° | 0.3807 N·m | 2.2752 N·m | 23.20 |
| 90° | 0.0000 N·m | 1.5137 N·m | 15.44 |
The static column is 1.4710 × cosθ. The required column adds the constant 0.7069 N·m of inertia and 0.05 N·m of friction before doubling, so at 90° it settles at (0.7069 + 0.05) × 2 = 1.5137 N·m rather than at zero.
Torque unit conversions
| From | To N·m | To kgf·cm | To oz·in |
|---|---|---|---|
| 1 N·m | 1 | 10.1972 | 141.612 |
| 1 kgf·cm | 0.0980665 | 1 | 13.8874 |
| 1 oz·in | 0.00706155 | 0.0720078 | 1 |
| 1 lbf·in | 0.112985 | 1.15212 | 16 |
| 1 lbf·ft | 1.355818 | 13.8255 | 192 |
A useful sanity check: 1 kg hanging on a 1 cm arm is 1 kgf·cm, and 1 kg on a 1 m arm is 100 kgf·cm — exactly the first test case this calculator ships with.
Mistakes that undersize a joint
- Sizing on holding torque alone. Acceleration frequently adds a third or more, and it is invisible until the arm has to move quickly.
- Using the full length for a heavy arm. A uniform rod’s weight acts at its midpoint and its inertia is mL²/3, not mL². Model the arm as a rod and add the payload separately if both matter.
- Forgetting the arm and gripper in the mass. The payload is often the smaller half of what the joint actually carries.
- Believing the stall rating. It is measured at a stated voltage with the output stationary, and available torque falls as speed rises. Size at 1.5–2.5× and check the speed rating too.
- Ignoring gearbox efficiency. A worm drive at 50% doubles the motor torque required for the same output, which is the price of its self-locking behaviour.
- Sizing at the arm’s resting angle. Gravity peaks at the horizontal. If the arm passes through horizontal at any point in its travel, that is the sizing case.
- Confusing joint speed with motor speed. Behind a 10:1 reduction, 30 rpm at the joint means 300 rpm at the motor, which has to be within its rated range.
- Neglecting the shock case. Sudden stops, hard limits and dropped loads produce torques far above anything in this calculation; those are cases for a mechanical stop or a slip clutch, not a bigger safety factor.
What sits either side of this calculation
Once the torque is settled, the electrical side follows. Mechanical power at the output divided by the motor and driver efficiency gives the electrical draw, which is an ordinary P = VI problem and turns into a current with the watts to amps conversion — the number you need for wiring, fusing and battery sizing. For an industrial gearmotor rather than a hobby servo, the nameplate current comes from the full-load amps calculation instead.
The structure needs checking as well as the actuator. The same m·g·r that sets the holding torque also bends the arm, which is a beam deflection problem — an arm stiff enough to point accurately is often a heavier arm, which feeds back into this calculation. And if the joint is going onto a battery-powered mobile platform or an aircraft, the extra mass and the extra draw both come out of the flight time budget.
One design move sits outside all of this arithmetic and beats it: counterbalancing. A gas spring, an extension spring routed over a cam, or a counterweight opposite the joint cancels part of the gravity term permanently. It costs mass and complexity, but it removes torque the motor would otherwise have to hold continuously, and holding torque is what cooks servos.