Example 2-14 — a motor connected to a load through gears
Two shafts turning at different speeds, and one equation of motion. Put a gearbox between a motor and its load and the load’s inertia and damping still arrive at the motor shaft — but scaled by the square of the gear ratio. That single factor n² is what lets a small motor move a large load, and what decides the ratio you should choose.
The mechanism
A step torque Tm is applied to the motor at t = 0. Change the tooth counts and watch both the speeds and the equivalent inertia change — the gears turn at whatever ratio you set, but what the motor actually has to accelerate is the number in the Je tile.
In hardware
Every symbol in the model is standing in for a real object. Here is a gearmotor driving a conveyor drum — the commonest place this exact problem turns up — with the six parts numbered, and the same six numbers on the idealised model underneath. Note which side of the gearbox each part lands on: that is what decides whether its inertia gets multiplied by n² or not.
| # | COMPONENT | SYMBOL | WHAT IT PHYSICALLY IS |
|---|---|---|---|
| 1 | Motor rotor | Jm | Everything that spins with the motor shaft: rotor iron, shaft, cooling fan — and the pinion keyed to it. |
| 2 | Motor bearings and windage | bm | Bearing drag, air resistance on the rotor and fan, and the speed-dependent part of the iron losses. |
| 3 | Pinion, 15 teeth | N₁ | Cut on or keyed to the motor shaft, so it turns at ωm. |
| 4 | Gear, 30 teeth | N₂ | On the load shaft. Twice the teeth, so half the speed and twice the torque. |
| 5 | Conveyor drum | JL | The drum and its shaft, the gear bolted to it, and the belt and product being dragged along. |
| 6 | Belt and product drag | bL | Belt tension losses, product sliding friction, air drag — every loss that grows with belt speed. |
The pinion is on the motor’s side. It is easy to think of the whole gear set as “the gearbox” and reflect all of it, but the pinion turns at ωm, so its inertia belongs to Jm and is not multiplied by anything. The gear turns at ωL, so it joins JL and is reflected by n². On a hard reduction that difference matters: the big gear can dominate the load inertia while the pinion is negligible.
Neither damper is a real part. Nothing in the drawing above is a dashpot. bm and bL are what is left after lumping together every loss that happens to grow in proportion to speed, and calling the sum viscous.
The reflection
The motor never sees JL or bL directly. It sees them through the gears, multiplied by n². These two bars are live: move the tooth-count sliders above and watch how much of the equivalent inertia and damping is the load’s contribution.
Je = Jm + n²JL
| Motor, Jm | 0.0200 | 21% |
| Load reflected, n²JL | 0.0750 | 79% |
| Load as it really is, JL | 0.3000 | ×0.250 |
be = bm + n²bL
| Motor, bm | 0.0100 | 33% |
| Load reflected, n²bL | 0.0200 | 67% |
| Load as it really is, bL | 0.0800 | ×0.250 |
Because the reflection has to conserve energy, not torque. The load turns at ωL = nωm, so its kinetic energy is ½JL(nωm)² = ½(n²JL)ωm² — the same energy an inertia of n²JL would have on the motor shaft. The identical argument on dissipated power, bLωL² = (n²bL)ωm², gives the damping. One factor of n comes from the speed and one from the torque; the algebra in the next section produces the same n² without ever mentioning energy.
The reduction
Start on the load side, where the only applied torque is T₂ coming through the gears. Then use the two gear relations to push everything onto the motor shaft.
The gear torque drives it; the load damper opposes it.
The motor torque drives it; the motor damper and the mesh reaction T₁ oppose it.
Rearranging the load equation and substituting θL = (N₁/N₂)θm,
The reaction on the pinion is smaller than the torque delivered to the load, by the same tooth ratio: T₁ = (N₁/N₂)T₂. Substituting into the motor equation brings in the second factor of n:
which rearranges to a single-shaft system:
With Je = Jm + n²JL and be = bm + n²bL, the gearbox has disappeared entirely:
A pole at the origin and one at −be/Je: the angle ramps without bound while the speed is a first order system with time constant τ = Je/be settling at Tm/be. The load transfer function is the motor’s, scaled by n — nothing more.
Step response
Both curves come from 2.30a and 2.30b with the sliders’ current values, so they move as you drag. The marker is where the animation above has got to.
Choosing the ratio
Gearing down multiplies torque but costs speed, so somewhere in between there is an optimum — and the two things you might want to maximise have different optima. Both curves below are live; the dashed line is the optimum, the dot is where your tooth counts currently sit.
αL(0) = nTm/Je rad/s² · how fast it gets moving
ωL(∞) = nTm/be rad/s · how fast it ends up
The acceleration optimum is an inertia match. Maximising nTm/(Jm + n²JL) over n gives
At the best ratio the reflected load inertia is exactly equal to the motor’s own, and the equivalent inertia is twice the motor’s. Gear down further and you keep multiplying torque, but you are now accelerating a rotor that is mostly your own motor. The best acceleration you can ever get is Tm/(2√(JmJL)), and no choice of gearing beats it.
The speed optimum is a damping match — same algebra, other pair of terms: n* = √(bm/bL), giving Tm/(2√(bmbL)). The two optima only coincide when Jm/JL = bm/bL, so in general a drive train has to be designed for one or the other.
Exercises
With the defaults, compute √(Jm/JL), then set the tooth counts to hit it. Check that the reflection bar splits 50/50 and that Je equals exactly 2Jm.
Raise bL to its maximum. The speed optimum slides left while the acceleration optimum stays put. Which one should a positioning stage be designed around, and which one a conveyor?
Set N₁ = N₂ so n = 1. Verify that Je is now just Jm + JL and that the two shafts turn at the same speed — the reduction collapses to two masses bolted together.
Gear down hard (n small) and τ approaches Jm/bm; gear up and it approaches JL/bL. Show that from τ = Je/be by taking the two limits.