= Gear sliding velocity
{title2=$v_s=(\omega_1-\omega_2)s$}
<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>.<Elements of Metric Gear Technology by SDP-SI>[§3.1, common base-circle tangent, line of action and pitch-circle geometry]{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 $\widehat{\mathbf z}$ for both signed <angular velocities>. The rigid-body velocity law gives
$$
\mathbf v_i(Q)=\omega_i\widehat{\mathbf z}\times(Q-O_i).
$$
Subtracting these expressions and using equal velocities at $P$ gives
$$
\mathbf v_1(Q)-\mathbf v_2(Q)
=(\omega_1-\omega_2)s\,\widehat{\mathbf z}\times\mathbf n.
$$
The 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 $\omega_1=50\,\mathrm{rad/s}$ and $\omega_2=-25\,\mathrm{rad/s}$, at $s=4\,\mathrm{mm}$ the signed tangential sliding velocity is $0.30\,\mathrm{m/s}$; at $s=-4\,\mathrm{mm}$ it is $-0.30\,\mathrm{m/s}$. 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.
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