= Gear efficiency
{title2=$\eta=P_{\rm out}/P_{\rm in}$}
<Gear efficiency> is the fraction of mechanical input <power> delivered as useful mechanical output <power> over a stated operating condition and system boundary:
$$
\eta=\frac{P_{\rm out}}{P_{\rm in}}
=\frac{|\tau_{\rm out}\omega_{\rm out}|}{|\tau_{\rm in}\omega_{\rm in}|}.
$$
For steady forward operation with positive input <power>, $P_{\rm loss}=P_{\rm in}-P_{\rm out}$ accounts for dissipation within that boundary. Tooth-contact <friction> is one contribution; a complete gearbox measurement can also include other rotating supports and lubricant losses. Citizen Micro explicitly treats its quoted stage efficiencies as reference values that depend on transmitted <torque> and speed.<Technical Data Gearhead by Citizen Micro>[“Gear Efficiency”: mechanical losses and dependence on torque and speed]{ref}
If the positive reduction <gear ratio> is $i=|\omega_{\rm in}|/|\omega_{\rm out}|$, then
$$
|\tau_{\rm out}|=\eta i|\tau_{\rm in}|.
$$
An increase in <torque> therefore does not itself measure <gear efficiency>: reduction increases torque while decreasing speed. For an original example, let input <torque> be $2\,\mathrm{N\,m}$ at $1200$ revolutions per minute, and output <torque> be $7.6\,\mathrm{N\,m}$ at $300$ revolutions per minute. Then $i=4$ and
$$
\eta=\frac{7.6(300)}{2(1200)}=0.95,
\qquad P_{\rm in}=80\pi\,\mathrm W\approx251.33\,\mathrm W,
\qquad P_{\rm loss}=4\pi\,\mathrm W\approx12.57\,\mathrm W.
$$
The angular-speed conversion factors cancel in the ratio but must be included when computing <power> in watts.
A bench test places <torque> and rotational-speed measurements on both sides of the chosen transmission, compares synchronized measurements after reaching steady operation, and reports load, speed and temperature. During acceleration, a difference between input and output <power> can partly represent changing <kinetic energy>, rather than loss. Near zero useful load, approximately fixed losses can also produce a much smaller efficiency than at a loaded operating point. For successive stages with consistent boundaries and no extra power path, efficiencies multiply; three hypothetical $0.95$ stages yield $0.95^3\approx0.8574$, rather than $0.95$.
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