= Gear pressure angle
{title2=$\alpha$}
The <gear pressure angle> specifies the direction of tooth-normal contact relative to tangential motion. For an external <spur gear> pair with <involute> tooth flanks, its operating value is the angle between the <gear line of action> and the common tangent to the <operating pitch circles>. The <gear line of action> is the common normal to the contacting flanks and a common tangent to their <gear base circles>. Reference <gear pressure angle> $\alpha$ and <operating pressure angle> $\alpha_w$ coincide at the corresponding reference mounting, but need not coincide at another axis separation.<Elements of Metric Gear Technology by SDP-SI>[§3.1, definition of pressure angle relative to line of action and line of centers]{ref}
Neglecting <friction>, let the tooth-normal force magnitude be $F_n$. Its components are
$$
F_t=F_n\cos\alpha_w,\qquad F_r=F_n\sin\alpha_w,
\qquad F_r=F_t\tan\alpha_w.
$$
The <gear tangential force> $F_t$ transmits <torque>, $\tau=F_t r_w$, at operating radius $r_w$. The <gear separating force> $F_r$ pushes the gear centers apart in the external mesh and loads their supports. Increasing <gear pressure angle> increases this separating load for the same transmitted tangential load.
For an original example, a <spur gear> transmits $\tau=3\,\mathrm{N\,m}$ at $r_w=30\,\mathrm{mm}$. Then $F_t=100\,\mathrm N$. At $\alpha_w=20\,{}^{\circ}$,
$$
F_r\approx36.40\,\mathrm N,\qquad F_n\approx106.42\,\mathrm N.
$$
If the angle were $25\,{}^{\circ}$ at the same $F_t$, the separating load would rise to approximately $46.63\,\mathrm N$. This comparison concerns load direction, not <gear efficiency>: an ideal normal force can transmit <power> without dissipating it. Actual frictional contact adds another force component and can alter both losses and support loads.
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