= Gear tangential force
{title2=$F_t=\tau/r_w$}
<Gear tangential force> is the pitch-circle-equivalent force that transmits <torque> in a <gear> mesh. At operating pitch radius $r_w$,
$$
F_t=\frac{|\tau|}{r_w}=\frac{2|\tau|}{d_w}.
$$
For an external <spur gear> pair without tooth <friction>, it is the component of tooth-normal force along the common tangent to the <operating pitch circles>: $F_t=F_n\cos\alpha_w$. Together with the <gear separating force>, this resolves the mesh load in fixed shaft-loading directions. It is not a claim that the moving contact point always has that same local radial/tangential frame.<Elements of Metric Gear Technology by SDP-SI>[§3.1, common line of action and pressure-angle geometry]{ref}
The relation follows geometrically from the <gear line of action>. Its distance from a gear center is the <gear base circle> radius $r_b$, so the moment of a normal force is $|\tau|=F_nr_b$. Since $r_b=r_w\cos\alpha_w$, substitution yields $|\tau|=F_tr_w$.
For an original example, a 20-tooth input with operating radius $20\,\mathrm{mm}$ transmits $4\,\mathrm{N\,m}$. Its <gear tangential force> is $200\,\mathrm N$. Driving a 60-tooth output at operating radius $60\,\mathrm{mm}$ ideally gives $12\,\mathrm{N\,m}$: the pitch-circle-equivalent tangential load is the same, while its moment arm is three times larger. At input <angular speed> $50\,\mathrm{rad/s}$, pitch speed is $1\,\mathrm{m/s}$ and transmitted <power> is $F_tv=200\,\mathrm W$, agreeing with $\tau\omega$.
The calculation assumes steady ideal force transmission. <Gear efficiency> affects the input/output power balance, while inertia can contribute additional torque during acceleration. A <gear tangential force> is also not the entire contact load: at $\alpha_w=20\,{}^{\circ}$, $F_t=200\,\mathrm N$ corresponds to normal load about $212.84\,\mathrm N$ and separating load about $72.79\,\mathrm N$.
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