I read an interesting article


It is very interesting, but I can not follow the 2nd last paragraph. I don't understand why it is true.

Gearboxes offer a significant benefit in that they affect the inertia ratio by a factor of the gearbox ratio squared.

Please help. Thanks

  • $\begingroup$ Probably related to the fact hat a factor $X$ increase in the gearbox ratio increases the torque by a factor $X$ and also reduces the angular acceleration of the load by a factor $X$. $\endgroup$
    – Johannes
    Aug 8, 2013 at 2:18
  • $\begingroup$ Any equation to support?? $\endgroup$
    – Marco
    Aug 8, 2013 at 2:40
  • $\begingroup$ As @Johannes says: we have a shaft that responds to a torque $\tau$ by $\tau = I \mathrm{d}_t \omega$. Now we bolt a gearbox of ratio $\lambda$ in: so now we find we have to exert torque $\tau^\prime = \lambda^{-1} \tau$ on the gearbox input to get the torque $\tau$ on the output. At the same time, the gearbox's input angular speed $\omega^\prime$ is $\lambda$ times the angular speed of the original shaft, so $\omega^\prime = \lambda \omega$. Plugging these into our original response equation: $\tau = \lambda \tau^\prime = I \mathrm{d}_t \omega = I \mathrm{d}_t \omega^\prime \lambda^{-1}$ ... $\endgroup$ Aug 8, 2013 at 2:40
  • $\begingroup$ ...or $\tau^\prime = I \lambda^{-2} \mathrm{d}_t \omega^\prime$. So the effective inertia is $ I \lambda^{-2}$. This kind of relationship arises whenever a system conserves a product of two variables say $V$ and $I$: if there is a proportional relationship $V = Z I$ between the two variables, and the system scales one of them $V \mapsto \lambda V$ then the system transforms the effective proportion $Z$ by $\lambda^2$. Gearboxes conserve $\tau \omega$ and electrical transformers conserve $V I$ by dint of conservation of energy. $\lambda^2 Z$ is the impedance reflected by a transformer. $\endgroup$ Aug 8, 2013 at 2:46
  • $\begingroup$ @WetSavannaAnimalakaRodVance Would you please put your comment into the answer section? I will accept that as the answer. Actually, it is a good and clear answer. $\endgroup$
    – Marco
    Aug 8, 2013 at 2:55

3 Answers 3


Gearboxes belong to a class of linear system that conserves a product of observable quantities by dint of the principle of conservation of energy. For a gearbox, the product $\tau \omega$, where $\tau$ is the torque exerted on or by a driveshaft and $\omega$ the shaft's angular speed. For a lossless gearbox at steady state, we have:

$\tau_{in} \omega_{in} = \tau_{out} \omega_{out}$

($\tau_{in}$ being the torque exerted on the gearbox's input shaft, $\tau_{out}$ being that exerted by the gearbox's output shaft). So this equation simply says power input = power output. I emphasise steady state because the gearbox can also store energy as rotational kinetic energy in its wheels and other spinning parts. The statement you make assumes that this energy stored by the gearbox is vanishingly small compared to the energy transfers of interest through the gearbox. Therefore we can assume that the above equation, for this argument, holds at all times, whether the system be at steady state or accelerating or decelerating. An "ideal" gearbox is one that is both lossless AND stores no kinetic energy.

Therefore, if the gearing changes the angular speed ratio, so that $\omega_{in} = \lambda \,\omega_{out}$ (here $\lambda > 1$ is "gearing down", so that the input shaft spins faster than the output shaft), the torques are related by the reciprocal ratio $\tau_{in} = \lambda^{-1} \tau_{out}$.

What does this do to the "inertia" of a load linked to the gearbox's output? The mass moment of inertia $I$ of the output load defines how the load responds to torque:

$\tau_{out} = I\, \mathrm{d}_t \omega_{out}$

so that, assuming the gearbox's inertia is small compared to $I$ (this is the same as the small energy storage in gearbox assumption above), then we can plug our reciprocal relationships:

$\lambda \tau_{in} = I\, \mathrm{d}_t \left(\lambda^{-1} \omega_{in}\right)$

or, since we can bring the $\lambda^{-1}$ through our time derivative thanks to the low gearbox inertia assumption:

$\tau_{in} = \lambda^{-2} I\, \mathrm{d}_t \left(\omega_{in}\right)$

So now, if we think of our system as a "black box": as seen from the input, the load responds to torque as though there were no gearbox there, but now with moment of inertia:

$\lambda^{-2} I$

which is the statement you sought to understand. More generally, the effective inertia seen at the input for an output load with inertia $I$ is:

$I_{effective} = I_{gearbox} + \lambda^{-2} I$

where now $I_{gearbox}$ accounts for the gearbox's inertia.

Another, analogous system is the electrical transformer, which conserves the product $V I$, so that $V_{in}(t) I_{in}(t) = V_{out}(t) I_{out}(t)$. If the transformer's voltage "step down" ratio is $\lambda$ so that $V_{out} = V_{in} \lambda^{-1}$ and if it is laden at its output with an impedance $Z$ (in general, the impedance is an integro-differential operator, rather than a simple scalar: $Z$ is analogous to $I\,\mathrm{d}_t$ above), then the impedance "reflected" to the input is:

$Z_{in} = \lambda^{-2} Z$

Analogous comments apply about "ideal" transformers as opposed to real ones that both lose energy and store energy in the magnetic and electric fields inside them:

$Z_{in} = Z_{transformer} + \lambda^{-2} Z$


Consider a motor with rotational inertia $J_m$, spinning a rotational inertia load $J_L$ via a massless, lossless gearbox ($J_g$ = 0). The total system inertia is thus $J_T = J_m + J_L'$ where $J_L'$ is the effective rotational inertia of the load as seen through the gearbox with ratio $G$ (where typically G > 1 for a car differential, and G < 1 for a bicycle chainwheel and sprocket cluster). $J_L'$ can then be derived from power considerations:

The kinetic energy of only the inertia load ($J_L=(1/2)mR^2$ for two examples, a disk on a shaft, or wheels on an axle) is

$$KE_L = \frac{1}{2} J_L \omega_L^2$$

from which the power to the load is

$$\begin{matrix} P_L & = & \frac{d(KE_L)}{dt}\\ & = & J_L \omega_L \alpha_L\\ & = &T_L \omega_L \end{matrix}$$

where $\omega_L = d(\theta_L)/dt$ is angular velocity, $\alpha_L = d(\omega_L)/dt$ is angular acceleration, and $T_L = J_L \alpha_L$ is the torque, all at the load. Using $\alpha_m = \alpha_L G$ and $\omega_m = \omega_L G$, the power “reflected” upstream from the load to the motor is

$$\begin{matrix} P_m &=& J_L (\omega_m/G)(\alpha_m/G)\\ & = & (J_L/G^2) \omega_m \alpha_m\\ & = & J_L' \omega_m \alpha_m\\ & = & T_L' \omega_m \end{matrix}$$

where $J_L' = J_L/G^2$ is the effective inertia load as seen by the motor, and $T_L' = J_L' \alpha_m$ is the torque at the “front” of the gearbox.

So, if G = 10 (say, for a planetary gear), then the effective inertia is 100 times less than the base rotational inertia. Conversely, the effective impedance looking from the load to the motor is amplified by a factor of 100 (which explains why it seems harder to move a reversibly geared machine, when G > 1, from the load side than from the actuator side).

It can be shown [1] that the optimal transfer of power from motor to load occurs when $dP_m/dG = d(N/D)/dG = 0$, where N is a collection of terms that are not a function of G, and $D^{0.5} = [G J_m + J_L/G]$. This condition yields $J_m = J_L'$, which is to say that the effective impedances (upstream to the motor, and downstream to the load) are matched. The optimal gear ratio would then be $G^2 = J_L/J_m$.


[1] D.M. Auslander and C.J. Kempf, 1996, “Mechatronics”, Prentice-Hall, NJ, pg. 122.


  • 1
    $\begingroup$ You should really try to format this a bit better to improve readability. Also, note that we have a system called "MathJax" which allows you to render mathematics as in LaTeX. Please use this to further improve your answer. $\endgroup$
    – Danu
    Jan 13, 2016 at 21:31
  • $\begingroup$ Some general formatting would be nice as well (e.g., paragraph breaks). $\endgroup$
    – Kyle Kanos
    Jan 13, 2016 at 21:47

Moment of inertia is the quotient of torque and angular acceleration:

$$\tau = I \alpha \ \ \implies\ \ I = \frac{\tau}{\alpha}$$

Gear ratio is equivalent to speed ratio, acceleration ratio, and inverse torque ratio: $$N = \frac{driver\ rotations}{driven\ rotations} = \frac{\omega_{input}}{\omega_{output}} = \frac{\alpha_{input}}{\alpha_{output}} = \frac{\tau_{output}}{\tau_{input}}$$

Find the reflected moment of inertia with substitution: $$ \begin{align} I_{input} &= \frac{\tau_{input}}{\alpha_{input}} \\ &= \tau_{input} \div \alpha_{input} \\ &= \left(\frac{\tau_{output}}{N}\right) \div (N\alpha_{output})\\ &= \frac{1}{N^2}\times\frac{\tau_{output}}{\alpha_{output}}\\ \\ I_{input} &= \frac{I_{output}}{N^2} \end{align} $$

Therefore, this is why the reflected moment of inertia is inversely proportional to the square of the gear ratio.


First, do not confuse moment of inertia $I$ with polar moment of inertia $J$. Sometimes $J$ does denote the moment of inertia, but confusion is easy. They have different units so this is a very good way to differentiate.

Second, gear ratio has two different conventions.

  • $N$ = rotations of driver gear : rotations of driven gear. With this convention, $N>1$ increases torque (driver gear is spinning faster) and $N<1$ increases speed (driver gear is spinning slower). This is the convention used above.
  • Gearbox $N$ is always $>1$. This convention recognizes that designer can flip gearbox to either increase torque or to increase speed as needed.

This formula is valid when second convention is used and gearbox is increasing speed:

$$I_{input} = N^2I_{output}$$


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