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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.

The gear path of recess is the part of the active gear path of contact followed after passing the gear pitch point, in the stated driving direction. For a standard external involute pair driven by gear1, contact ends at the driving gear's tip circle when that tip bounds the usable profile. The recess length lies on the gear line of action; it is not an arc of the gear pitch circle.^[ref]

Use driving-gear tip radius $r_{a1}$, base radius $r_{b1}$, operating radius $r_{w1}$ and operating pressure angle ${\alpha }_w$. The base-circle tangent triangle givesFor a standard module2,20/40-tooth pair at60mm separation and20degree reference angle, $r_{a1}=22\mathrm{mm}$, $r_{w1}=20\mathrm{mm}$ and $r_{b1}=20\cos 20{}^{\circ }\mathrm{mm}$ yield $L_r\approx 4.5960\mathrm{mm}$. The corresponding gear path of approach is about5.0586mm, so the pitch point does not bisect the active contact interval.

At driving angular velocity $50rad/s$, the contact-location speed is $u_b=|{\omega }_1|r_{b1}\approx 0.93969m/s$. Recess therefore lasts about4.891ms. Adding approach gives gear contact duration $T\approx 10.274\mathrm{ms}$. Tooth-passage frequency is $20(50)/(2\pi )\approx 159.155\mathrm{Hz}$, so $T{f}_m\approx 1.6352$, agreeing with the transverse gear contact ratio calculated as active length divided by gear base pitch.

A shorter usable driving flank reduces recess and can change gear sliding-friction loss, because sliding grows with distance from the pitch point. The tip-circle formula alone cannot describe profile modifications or an interfering contact. The geometry and direction must be specified before applying the approach/recess labels.

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.

Gear sliding velocity is the relative tangential surface velocity at a tooth contact. It must distinguish the common point where the tooth surfaces touch from the two material points instantaneously occupying it. For an ideal fixed-axis external involute pair, relative normal velocity vanishes, but relative tangential velocity generally does not. It is zero at the gear pitch point and grows with distance from it along the gear line of action.^[ref]

Let $P$ be the pitch point, $Q=P+s\mathbf{n}$ a contact point and $\mathbf{n}$ a unit direction along the line of action. Use a common positive rotation axis $\hat{\mathbf{z}}$ for both signed angular velocities. The rigid-body velocity law givesSubtracting these expressions and using equal velocities at $P$ givesThe result is perpendicular to the common normal, as the contact constraint requires. External gears rotate in opposite directions, so the sliding-speed magnitude is $(|{\omega }_1|+|{\omega }_2|)|s|$, rather than their speed difference in magnitude.

For ${\omega }_1=50rad/s$ and ${\omega }_2=-25rad/s$, at $s=4\mathrm{mm}$ the signed tangential sliding velocity is $0.30m/s$; at $s=-4\mathrm{mm}$ it is $-0.30m/s$. Both vanish at $P$. Doubling both shaft speeds doubles gear sliding velocity at the same contact location, while leaving the ideal gear ratio unchanged. This connects geometry and running speed to gear sliding-friction loss. The formula concerns a two-dimensional spur gear contact; general three-dimensional contact requires the complete surface velocities and common normal.

Gear tooth sliding is relative tangential motion of mating tooth surfaces at their contact point. Ideal contact prevents relative normal penetration but does not generally require the tangential surface velocities to agree. For a fixed-axis external involute pair, tangential slip vanishes at the gear pitch point and changes as the contact moves along the gear path of contact.^[ref]

Let $P$ be the pitch point, let $\mathbf{n}$ be a unit vector along the gear line of action, and write a contact location as $Q=P+s\mathbf{n}$. With a common positive axis $\hat{\mathbf{z}}$ and signed angular velocities ${\omega }_1,{\omega }_2$, rigid-body velocity differences giveThe velocities at $P$ cancel, leaving this term proportional to distance along the contact line. Since ${\omega }_2$ has opposite sign in an external pair, the sliding-speed magnitude is $(|{\omega }_1|+|{\omega }_2|)|s|$.

For an original example, use ${\omega }_1=50rad/s$ and ${\omega }_2=-25rad/s$. At $s=4\mathrm{mm}$, gear sliding velocity along the common tooth tangent is $0.30m/s$; at $s=-4\mathrm{mm}$ it has the opposite sign. At $s=0$ it is zero. Equality of operating pitch-circle speeds must therefore not be used to conclude that the tooth flanks roll without slipping throughout engagement.

In a simplified Coulomb friction model with coefficient of friction $\mu$ and normal load $F_n$, the dissipated gear sliding-friction loss is $P_f=\mu F_n|v_s|$. Hypothetical values $\mu =0.05$, $F_n=100\mathrm{N}$ and $|v_s|=0.30m/s$ give $1.5\mathrm{W}$. This is a local illustrative loss model, not a lubricant specification. Zero ideal sliding at the pitch point also does not eliminate every source of gearbox loss, such as support friction or deformation-related dissipation. Real load and friction can vary through the engagement, so total loss requires integrating the applicable contact model.