How do I find rotational inertia?

Read it from CAD mass properties about the spin axis, or sum the parts: a solid cylinder is ½ m r², a thin ring is m r², and a point mass is m r². Watch the units. 1 lb·ft² is 0.0421 kg·m², and mixing the two is the usual error.

Does the gearbox change the inertia?

It changes what the motor feels. A load inertia seen from the motor is divided by the ratio squared, and the motor rotor seen from the output is multiplied by it. The calculator uses the inertia you enter at the output shaft and does not add rotor or gearbox inertia.

Will the real spin-up take as long as the time shown?

The time assumes constant acceleration. A motor’s torque is lowest at the target speed. If it can make the required torque there, it can make more at every lower speed, so the run finishes no later than shown. Battery sag and rotor inertia work the other way.

Power transmission calculator

Angular Acceleration & Motor Calculator

Turn a rotational inertia and a spin-up target into torque, rpm, power, and revolutions, then check them against a motor and ratio.

Input → stages → output

Edit any value and the drivetrain, target, and motor curves update immediately.

Calculator mode

Motor input

Use any motor’s published endpoint data

Motors

Voltage

Free speed

rpm

Stall torque

Free current

Stall current

Controller current cap

Not applied

Gear stages

14.6667:1 total

Efficiency / mesh

Stage 1

Driver

Driven

Stage 2

Driver

Driven

Motion target and results

Rotational inertia

Angular acceleration

Target shaft speed

Additional resisting torque

Bearing friction, process load, or other opposing shaft torque.

Acceleration torque: 7.38 lbf·ft
Time to speed: 3.14 s
Rotation to speed: 7.85 rev

Derived output target: 300 rpm · 7.38 lbf·ft · 314 W peak at target speed

Entered ratio: 14.6667:1
Predicted speed: 310.28 rpm
Speed vs. target: +3.4%
Efficient ratio: 15.399:1
Motor at target: 4,619.61 rpm · 34.9 A
Required power / motor: 334 W · 70.9% peak

Efficient point for the required power

Selected the higher-speed, lower-current point: 77% of free speed, 23% of stall torque, 79.7% modeled efficiency, and 34.9 A per motor.

Drivetrain and Custom motor performance

RPM and torque through each shaft, followed by the interpreted motor curves.

4,619.61 rpm · 34.9 A

Motor input

4,550.71 rpm

0.53 lbf·ft

12T → 48T

4:1

Shaft 1

1,137.68 rpm

2.07 lbf·ft

18T → 66T

3.667:1

Mechanism output

310.28 rpm

7.38 lbf·ft

Entered stages Efficient required-power solution Efficient continuous band 90%+ peak power · transient Low-speed stall risk Torque · 3 N·m stall Current · 145 A stall Power · 471 W modeled peak Efficiency · 87% modeled max

Most efficient solution for the required power

The movement needs 314 W at the output. With 94.1% drivetrain efficiency, each motor must deliver 334 W at its shaft. The calculator selects the higher-efficiency of the two motor-curve points that produce this power.

Efficient solution

Motor speed: 4,619.61 rpm · 77%
Current per motor: 34.9 A
Motor efficiency: 79.7% · 91.9% of max
Required shaft power: 334 W · 70.9% peak
Ratio for desired speed: 15.399:1

Estimated electrical input at this point is 419 W per motor · 419 W total at nominal voltage. This is the steady-state power for the entered speed and torque; acceleration requires additional power.

The curves are an engineering estimate, not a thermal or controller simulation. A breaker rating is not a motor-current cap; brief current can exceed that rating before a time-dependent trip. Battery or supply sag, configured controller limits, commutation, temperature, friction, and manufacturing variation change real performance.

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Quick answer

How it works

Torque to accelerate is inertia times angular acceleration, plus any friction torque. Peak power is that torque times the target speed in rad/s. Enter the inertia about the output shaft. Seen from the motor side it shrinks by the ratio squared.

How to use the calculator

1Get the inertia about the spin axis from your CAD mass properties, or add up solid cylinders (½ m r²) and rings (m r²).
2Turn the spin-up time into acceleration: α = ω ÷ t. Reaching 4,000 rpm in 1 s is 419 rad/s², or 4,000 rpm/s. The unit selector takes either.
3Enter bearing friction or a process load as resisting torque. It adds directly to the acceleration torque.
4Enter inertia as it is on the output shaft. Do not divide it by the ratio yourself.

Key formulas

Angular acceleration formulas

Acceleration torque: T = I × α + T resist; I in kg·m², α in rad/s². 1 rad/s² = 9.55 rpm/s.
Time and turns to speed: t = ω ÷ α · turns = ω² ÷ (4π × α); ω is the target speed in rad/s: rpm × 2π ÷ 60.
Peak power: P = T × ω
Inertia seen by the motor: I motor side = I load ÷ GR²

Worked example

1.5 kg flywheel to 4,000 rpm in about a second

Inputs: A 1.5 kg solid wheel, 4 in across, has I = ½ × 1.5 × 0.0508² = 0.00194 kg·m². It accelerates at 400 rad/s² to 4,000 rpm with no friction.
Result: Torque 0.774 N·m (0.571 lbf·ft), 1.05 s and 34.9 revolutions to speed, 324 W peak. With the default motor on one 97% mesh, the recommended ratio is 1.15:1 at 35 A.

Time = ω ÷ α = 418.9 ÷ 400 = 1.05 s. Power = 0.774 N·m × 418.9 rad/s = 324 W. A light wheel at a speed near the motor’s own needs almost no reduction.

Common questions

What to know before using the result

How do I find rotational inertia?: Read it from CAD mass properties about the spin axis, or sum the parts: a solid cylinder is ½ m r², a thin ring is m r², and a point mass is m r². Watch the units. 1 lb·ft² is 0.0421 kg·m², and mixing the two is the usual error.
Does the gearbox change the inertia?: It changes what the motor feels. A load inertia seen from the motor is divided by the ratio squared, and the motor rotor seen from the output is multiplied by it. The calculator uses the inertia you enter at the output shaft and does not add rotor or gearbox inertia.
Will the real spin-up take as long as the time shown?: The time assumes constant acceleration. A motor’s torque is lowest at the target speed. If it can make the required torque there, it can make more at every lower speed, so the run finishes no later than shown. Battery sag and rotor inertia work the other way.

Formula

F = ma + Fᵣ · T = Fr · P = Fv · GR = ∏(N driven ÷ N driver)

Output speed is motor speed divided by total ratio. Output torque is motor torque times ratio times efficiency. Linear targets use F = ma plus resistance at the wheel radius. Angular targets use τ = Iα plus resisting torque.

Engineering references

Assumptions and limits

Each row is one external mesh between a driver and a driven gear. Gears fixed to the same shaft turn together and add no mesh. Pitch, pressure angle, and center distance are not checked.
The motor is a straight line from free speed and free current to stall torque and stall current, at constant voltage. Battery sag, heating, and controller limits move the real curve.
Efficiency is one number per mesh, applied to every row. Bearing, windage, and lubricant losses are not included.
Acceleration targets assume constant acceleration from rest and no wheel slip. The motor is checked at the target speed with full acceleration torque, the hardest point of the run.

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