Is squared motor gearbox ratio proportional to inertia ratio? I read an interesting article
http://m.machinedesign.com/news/motor-sizing-made-easy
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
 A: 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:
$Z_{in} = Z_{transformer} + \lambda^{-2} Z$
A: 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 $G^2 = J_L/J_m$.
Refs: 
[1] D.M. Auslander and C.J. Kempf, 1996, “Mechatronics”, Prentice-Hall, NJ, pg. 122.
4/10/2016
