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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.

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.