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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 line of action is the common normal along which ideal tooth contact transmits force. For meshing involute profiles in a transverse plane, it is a straight common tangent to the two gear base circles. The angle it makes with the common tangent to the operating pitch circles is the operating pressure angle.^[ref]

Its intersection with the line of gear centers is the gear pitch point. The actual contact point moves along only a finite portion, the gear path of contact, bounded by usable tooth profiles. Thus a diagram's infinite straight line is a geometric construction, not a claim that teeth can contact everywhere on it.

The geometry explains constant ratio. Let the tooth-normal force magnitude be $F_n$ and the base radii $r_{b1},r_{b2}$. The perpendicular distance from each gear center to the gear line of action is its base radius, so the ideal transmitted moments have magnitudesThese moment arms do not change as the contact point advances along that line. The no-penetration contact condition correspondingly gives $r_{b1}{\omega }_1+r_{b2}{\omega }_2=0$ for an external pair. Compatible profiles have $r_{b2}/r_{b1}=N_2/N_1$, recovering the tooth-count gear ratio.

For example, reference radii $20$ and $40\mathrm{mm}$ at reference gear pressure angle $20{}^{\circ }$ give base radii approximately $18.794$ and $37.588\mathrm{mm}$. A $100\mathrm{N}$ normal force produces moment magnitudes $1.8794$ and $3.7588\mathrm{N}\mathrm{m}$. Their ratio stays two as the point of contact moves.

Ideal normal transmission does not eliminate gear tooth sliding. Relative surface slip generally changes along the gear path of contact and vanishes at the gear pitch point for the ideal external pair. Actual friction associated with that slip adds a tangential contact force; its resultant need not lie exactly on the geometric gear line of action.

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 path of contact is the locus of actual tooth-contact points during an engagement. For a conventional external involute pair, its active part is a finite straight segment of the gear line of action. Its endpoints are set by the usable tooth flanks, rather than by the infinite geometric line alone. The part before the gear pitch point is the gear path of approach; the part after it is the gear path of recess in the stated direction of transmission.^[ref]

For a standard noninterfering external pair driven by gear1, let $r_{ai}$ be tip radii, $r_{bi}$ base radii, $r_{wi}$ operating radii and ${\alpha }_w$ the operating pressure angle. When tip circles bound the full usable contact,Adding gives $L=L_{\mathrm{app}}+L_{\mathrm{rec}}$. The two lengths need not be equal even when the gears use the same module and reference angle.

For the original module2,20/40-tooth example at20degree reference angle and60mm standard separation, tip radii22 and42mm give $L_{\mathrm{app}}\approx 5.0586\mathrm{mm}$ and $L_{\mathrm{rec}}\approx 4.5960\mathrm{mm}$. Their sum is $9.6546\mathrm{mm}$; dividing by gear base pitch $5.9043\mathrm{mm}$ gives transverse gear contact ratio about1.6352.

These are distances along the line of contact, not arc lengths traveled around a pitch circle. A shortened or altered flank can reduce the active interval, and interference can invalidate the full tip-limited construction. The labels approach and recess also depend on which direction is being followed. A practical contact inspection therefore needs the profile geometry and transmission direction, not just a sketch of two nominal pitch circles.

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.

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 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.