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Gear efficiency is the fraction of mechanical input power delivered as useful mechanical output power over a stated operating condition and system boundary:For steady forward operation with positive input power, $P_{\mathrm{loss}}=P_{\mathrm{in}}-P_{\mathrm{out}}$ 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.^[ref]

If the positive reduction gear ratio is $i=|{\omega }_{\mathrm{in}}|/|{\omega }_{\mathrm{out}}|$, thenAn increase in torque therefore does not itself measure gear efficiency: reduction increases torque while decreasing speed. For an original example, let input torque be $2\mathrm{N}\mathrm{m}$ at $1200$ revolutions per minute, and output torque be $7.6\mathrm{N}\mathrm{m}$ at $300$ revolutions per minute. Then $i=4$ andThe 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 $0.95^3\approx 0.8574$, rather than $0.95$.

Gear friction work per engagement is the mechanical energy dissipated by sliding during one tooth pair's active engagement. It is the time integral of that pair's gear sliding-friction loss:The result has units of joules, whereas instantaneous loss has units of watts. It excludes other gearbox losses such as support friction. Manufacturer efficiency guidance distinguishes meshing and support losses and emphasizes their dependence on operating conditions.^[ref]

For a fixed-axis external spur gear pair, let signed contact coordinate $s$ run from $-L_a$ to $L_r$ through the gear pitch point. Gear sliding velocity has magnitude $(|{\omega }_1|+|{\omega }_2|)|s|$. At constant shaft speeds, gear base-roll speed gives $ds/dt=u_b>0$ in this chosen direction. If $\mu$ and the force carried by this pair are constant,With hypothetical $\mu =0.05$, $F_n=100\mathrm{N}$, angular-speed sum75rad/s, $L_a=5.0586\mathrm{mm}$, $L_r=4.5960\mathrm{mm}$ and $u_b=0.93969m/s$, this gives about9.32mJ. Doubling both shaft speeds doubles instantaneous power but halves gear contact duration, leaving this idealized work per engagement unchanged. Real friction and load can themselves change with speed.

Gear tooth load sharing is essential when several pairs engage simultaneously. A constant total mesh load is not a constant full load on each active pair. If identical pair events recur at gear mesh frequency $f_m$, averaging their correctly modeled losses over many events gives $\overline{P}=f_m{E}_f$. The event's force history must already include load sharing; assigning the whole mesh load independently to every overlapping pair would overcount transmitted load and loss.

Gear sliding-friction loss is the rate at which relative tooth-flank sliding dissipates mechanical energy. In a simple Coulomb friction model with normal contact load $F_n$, coefficient of friction $\mu$ and gear sliding velocity $v_s$, the nonnegative dissipated power isThis is a local contact model, rather than the entire gearbox loss. Manufacturer guidance distinguishes gear-meshing losses from losses in rotating supports and warns that overall gear efficiency depends on load and speed.^[ref]

To derive the sign, let $v_s=v_1-v_2$ along the common tangent. The friction force on surface1 is $f_1=-\mu F_n\mathrm{sgn}(v_s)$ and that on surface2 is $-f_1$. Their combined mechanical power is $f_1{v}_1-f_1{v}_2=-\mu F_n|v_s|$. The corresponding positive quantity is the rate of mechanical energy loss. At zero sliding the ideal Coulomb sliding-loss term vanishes, even though another resistance mechanism can remain.

For a fixed-axis external spur gear pair with $|{\omega }_1|+|{\omega }_2|=75rad/s$, $|s|=4\mathrm{mm}$ gives $|v_s|=0.30m/s$. Hypothetical $\mu =0.05$ and $F_n=100\mathrm{N}$ give $P_f=1.5\mathrm{W}$. Holding these load and friction assumptions fixed while doubling both shaft speeds doubles the instantaneous loss at that geometric point.

Gear friction work per engagement integrates this varying loss through the active gear path of contact. If one pair carries a constant $F_n$, the contact coordinate runs from $-L_a$ to $L_r$ at constant speed $\dot{s}=u_b>0$, givingFor $L_a=5.0586\mathrm{mm}$, $L_r=4.5960\mathrm{mm}$ and $u_b=0.93969m/s$, the preceding loads and speeds give about $9.32\mathrm{mJ}$. Real tooth pairs share varying loads; an overall loss calculation must sum the applicable contacts and integrate their actual loads, sliding and friction model.