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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 mesh stiffness describes the force response of the engaged tooth contacts to a specified relative approach along the mesh direction. In the linear parallel-contact model underlying gear tooth load sharing, all active pairs have the same approach $\delta$ andThe sum assumes compatible displacement coordinates. Adding a bending stiffness, a torsional stiffness and a normal contact stiffness without transforming their coordinates would be dimensionally and mechanically inconsistent.

For an original example, two zero-gap pairs with $k_1=80$ and $k_2=120N/\mu \mathrm{m}$ give $k_{\mathrm{mesh}}=200N/\mu \mathrm{m}$. A $1000\mathrm{N}$ load produces $5\mu \mathrm{m}$ normal approach. If only the first pair is engaged, the same ideal stiffness model instead gives $12.5\mu \mathrm{m}$.

With gaps or nonlinear response, specify whether the reported value is secant stiffness $F/\delta$ from a reference or the local gear mesh tangent stiffness $dF/d\delta$. Supposewhere $\delta$ is measured in micrometres and the result in newtons. The second pair starts carrying load at $\delta =3\mu \mathrm{m}$, corresponding to $F=240\mathrm{N}$. At $F=720\mathrm{N}$, approach is $5.4\mu \mathrm{m}$, so the origin-based secant stiffness is about $133.3N/\mu \mathrm{m}$ while tangent stiffness is $200N/\mu \mathrm{m}$. At the ideal corner itself, a unique two-sided derivative does not exist.

A small-perturbation experiment can hold contact position and mean load approximately fixed, apply a controlled force increment and measure the corresponding relative displacement. Shaft and support mechanical compliance must be separated or included explicitly in the reported system response. Rotating the mesh changes contact position, so one scalar does not necessarily describe a whole cycle.

This matters dynamically: changes in gear mesh stiffness can alter oscillatory response and gear tooth load fractions. The primary NASA analysis used an improved tooth-stiffness model coupled to dynamic-load calculations, rather than prescribing stiffness independently of the load solution.^[ref] The elementary parallel-spring sum explains the linear limit, while real contacts require a stated engagement and deformation model.

The gear tooth load fraction ${\eta }_j$ is the fraction of a stated total mesh force carried by tooth pair $j$ at a particular contact position or time:For nonnegative forces resolved in the same direction, $0\le {\eta }_j\le 1$ and ${\sum }_j{\eta }_j=1$. A normal-force fraction must not mix one pair's normal load with another convention's tangential total. At zero total load the ratio is undefined.

An original quasistatic model treats engaged pairs as parallel springs with the same normal approach $\delta$. If $F_j=k_j\delta$ and every pair has zero initial gap, thenTwo pairs with stiffnesses $60$ and $140N/\mu \mathrm{m}$ under total load $1000\mathrm{N}$ approach by $5\mu \mathrm{m}$ and carry $300$ and $700\mathrm{N}$. Their fractions are $0.3$ and $0.7$, rather than one half each. Counting two geometrically overlapping contacts does not establish equal gear tooth load sharing.

An initial gap $e_j$ changes the model toA pair below its gear tooth engagement threshold carries no load in this unilateral model. Changing total load can change both the active pairs and their fractions, even if all the individual stiffnesses remain constant. Gear contact ratio specifies geometric overlap; load fractions require deformation and engagement information.

The fractions also prevent double counting. If total load is $1000\mathrm{N}$, assigning that full load to both of two pairs gives an impossible sum of $2000\mathrm{N}$ within this model. A gear friction work per engagement calculation must use each pair's own $F_j(t)$ before summing its dissipation.

Dynamic motion can alter the quasistatic distribution. A 1986 NASA analysis calculated tooth stiffness and loads as contact position and speed varied, using a coupled stiffness/load iteration.^[ref] Fractions inferred from the simple spring model should therefore be identified as model results, rather than universal operating values.

Gear tooth load sharing is the division of transmitted mesh load among simultaneously contacting tooth pairs. Gear contact ratio determines how engagement intervals overlap, but it does not by itself specify each pair's force. Tooth and support deformation, profile geometry and operating speed affect that distribution. A primary dynamic analysis explicitly coupled tooth stiffness with calculated dynamic loads rather than assigning a fixed full load to every active pair.^[ref]

An illuminating quasistatic model treats active pairs as parallel springs with a common normal approach $\delta$. With pair stiffnesses $k_j$ and no gaps, $F_j=k_j\delta$, soThe stiffness sum is gear mesh stiffness, and ${\eta }_j$ is a gear tooth load fraction. If two active pairs have stiffness100 and150N/µm under total normal load1000N, their common approach is4µm and loads are400 and600N. Equal sharing requires equal effective stiffness in this particular model.

Include a geometric gap $e_j$ by writing $F_j=k_j\max (\delta -e_j,0)$. If the second pair has a2µm gap, the same1000N total load gives $100\delta +150(\delta -2)=1000$ with $\delta$ measured in micrometres. The result is $\delta =5.2\mu \mathrm{m}$, $F_1=520\mathrm{N}$ and $F_2=480\mathrm{N}$. At only100N, the first pair alone compresses1µm, leaving the second pair out of contact.

For an ideal spur gear mesh with transverse contact ratio between1 and2, the fraction of a tooth-entry cycle having two pairs in contact is ${ϵ}_{\alpha }-1$. A ratio1.635 therefore gives about63.5% two-pair contact and36.5% single-pair contact. This is geometric timing, not a guarantee of equal forces in the overlapping part. A gear friction work per engagement calculation must use each pair's own load history; dynamic motion can require more than the quasistatic spring approximation.