 Gear tooth sliding

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.