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Gear base-roll speed $u_b$ is the rate at which base-circle rolling advances an ideal involute contact along the gear line of action. For a fixed-axis external pair at constant speed,The equality follows from the tooth-count speed ratio and compatible base pitches. It describes the rate of change of contact position along a geometric line, rather than the gear sliding velocity between material points on the flanks. Base circles and their common tangent define this geometry.^[ref]

For standard module2 gears with20 and40 teeth at20degree reference angle, base radii are $r_{b1}=20\cos 20{}^{\circ }\mathrm{mm}$ and $r_{b2}=40\cos 20{}^{\circ }\mathrm{mm}$. Signed angular velocities ${\omega }_1=50rad/s$ and ${\omega }_2=-25rad/s$ give the same $u_b\approx 0.93969m/s$. The first gear's reference-circle tangential speed is1m/s, so base-roll and pitch-circle speeds are distinct.

In one tooth-entry interval, the base circle advances one gear base pitch:With $f_m\approx 159.155\mathrm{Hz}$, the example advances5.90426mm per entry. An active contact length9.65457mm therefore gives gear contact duration $L/u_b\approx 10.2742\mathrm{ms}$.

For compatible involute teeth, changing center distance changes the operating pressure angle and operating pitch circles, while leaving the manufactured base circles unchanged. At the same shaft speeds, base-roll speed is therefore unchanged even though the usable contact length may change. This separates a kinematic rate from the active interval needed to calculate contact time. If speed varies, contact displacement follows $\int u_b(t)dt$; a constant-speed quotient is then only an approximation.

Gear contact duration is the time for a specified tooth pair to traverse its active gear path of contact. For a fixed-axis external spur gear pair at constant speed, the contact location advances along the gear line of action at the common gear base-roll speedThis is the speed of the contact location along a geometric line, not the relative tooth-surface gear sliding velocity. The latter vanishes at the gear pitch point, while the contact location continues moving. For active path length $L$, duration is $T=L/u_b$. The geometry relates gear contact ratio to active path length divided by gear base pitch.^[ref]

The time between successive tooth-pair entries is $1/f_m$, with gear mesh frequency $f_m=N_1|{\omega }_1|/(2\pi )$. Since $r_{b1}=r_{w1}\cos {\alpha }_w$ for compatible operating geometry, the base-circle distance advanced per entry isConsequentlyA contact ratio above1 means one pair remains engaged after the next pair enters; it does not mean one pair contacts for an entire shaft revolution.

For the standard module2,20/40-tooth example with20degree reference angle, active length is9.65457mm and base pitch is5.90426mm. At driving angular velocity $50rad/s$, $u_b=0.93969m/s$, $f_m\approx 159.155\mathrm{Hz}$ and $T\approx 10.2742\mathrm{ms}$. The entry interval is6.28319ms, giving $T{f}_m\approx 1.63519$. Doubling both shaft speeds halves duration while preserving geometric gear contact ratio. Variable speed requires integrating the changing base-roll speed; a helical gear also needs a stated transverse slice or a whole-face contact convention.

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.