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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.

The gear path of approach is the part of the active gear path of contact followed before the contact reaches the gear pitch point. For a conventional external involute pair in the stated driving direction, contact starts where the driven gear's tip circle intersects the usable gear line of action, and then moves toward the pitch point. This is a straight-line distance along the common normal, not an arc around a pitch circle.^[ref]

Let gear1 drive gear2. Write $r_{a2}$ for gear2's tip radius, $r_{b2}$ for its base radius, $r_{w2}$ for its operating radius and ${\alpha }_w$ for operating pressure angle. A right triangle from the base-circle tangency gives tip-to-tangency distance $\sqrt{r_{a2}^2-r_{b2}^2}$. Subtracting the pitch-point-to-tangency distance givesThis formula assumes the full tip-limited interval is usable without interference or profile truncation.

For an original standard module2,20/40-tooth pair at60mm center distance and20degree reference angle, gear2 has tip radius42mm, operating radius40mm and base radius $40\cos 20{}^{\circ }\mathrm{mm}$. Consequently $L_a\approx 5.0586\mathrm{mm}$. If gear1 rotates at $50rad/s$, contact advances along the line at $u_b=|{\omega }_1|r_{b1}=0.93969m/s$, giving an approach contribution to gear contact duration of about $5.383\mathrm{ms}$. This contact-location speed differs from gear sliding velocity, which tends to zero on arriving at the pitch point.

The total active length adds gear path of recess. Reversing the followed engagement direction exchanges which segment is called approach and which is recess. If altered usable flanks exclude the nominal pitch point, an assumed positive tip-limited formula must not replace an inspection of the actual active interval.

The gear pitch point is where the gear line of action meets the line joining the gear centers. For a fixed-axis external involute pair, it is the tangency point of the operating pitch circles. The point is a geometric location in the mesh, rather than a material mark permanently attached to one tooth.^[ref]

Let the gear centers be separated by $a$, with tooth counts $N_1,N_2$. Their operating radii locate the pitch point atFor external signed angular velocities, $r_{w1}{\omega }_1+r_{w2}{\omega }_2=0$. The two hypothetical material points occupying the pitch point therefore have the same instantaneous velocity. This is the local rolling without slipping property used to construct the operating pitch circles.

For an original example, use20 and40 teeth separated by $60\mathrm{mm}$. The pitch point is20mm from the first center and40mm from the second. With ${\omega }_1=50rad/s$ and ${\omega }_2=-25rad/s$, both surface velocities at that location have magnitude $1m/s$ and the same direction. Ideal relative gear sliding velocity is zero there.

Elsewhere on the gear path of contact, the two tooth surfaces can slide even while maintaining the required normal contact condition. The pitch point remains fixed in this fixed-axis geometry, while successive tooth-contact points advance through the mesh. Whether the pitch point itself belongs to a particular usable contact interval depends on the available flanks; the geometric definition does not assert that every modified tooth profile physically contacts at that point.

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.