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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.^[ref]

Neglecting friction, let the tooth-normal force magnitude be $F_n$. Its components areThe 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 $\tau =3\mathrm{N}\mathrm{m}$ at $r_w=30\mathrm{mm}$. Then $F_t=100\mathrm{N}$. At ${\alpha }_w=20{}^{\circ }$,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.

Gear separating force is the component of an external gear mesh load along the line joining the gear centers, tending to push those centers apart. For a frictionless external spur gear pair,where $F_n$ is tooth-normal force, $F_t$ is gear tangential force and ${\alpha }_w$ is operating pressure angle. This component acts on the shaft supports even when the mesh is ideal and lossless.^[ref]

For example, take operating radius $r_w=20\mathrm{mm}$ and transmitted torque $4\mathrm{N}\mathrm{m}$. Then $F_t=200\mathrm{N}$. At ${\alpha }_w=20{}^{\circ }$, $F_r\approx 72.79\mathrm{N}$; at $25{}^{\circ }$ and the same tangential load, $F_r\approx 93.26\mathrm{N}$. The larger separating load is a change in load direction, not by itself an efficiency loss.

If the gear centers are ideally fixed, this radial component produces no work through a center-separation displacement, since that displacement is zero. Flexible supports can nevertheless deflect under it and store elastic potential energy. The resulting change in axis separation can change operating pressure angle, backlash and tooth contact, so the assumed geometry and support stiffness interact.

A support-load experiment can apply a known torque slowly to a mounted external pair and measure the reaction along the line of centers while maintaining the stated axis separation. Agreement with $F_r/F_t=\tan {\alpha }_w$ tests the force resolution under low-friction assumptions. Additional reactions, tooth friction and mounting preload must be accounted for before identifying a measured support force with this mesh component. Internal meshes and nonparallel-axis gearing require their own direction conventions; the external separating-force picture must not be transferred unchanged.

Gear tangential force is the pitch-circle-equivalent force that transmits torque in a gear mesh. At operating pitch radius $r_w$,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.^[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 $|\tau |=F_n{r}_b$. Since $r_b=r_w\cos {\alpha }_w$, substitution yields $|\tau |=F_t{r}_w$.

For an original example, a 20-tooth input with operating radius $20\mathrm{mm}$ transmits $4\mathrm{N}\mathrm{m}$. Its gear tangential force is $200\mathrm{N}$. Driving a 60-tooth output at operating radius $60\mathrm{mm}$ ideally gives $12\mathrm{N}\mathrm{m}$: the pitch-circle-equivalent tangential load is the same, while its moment arm is three times larger. At input angular speed $50rad/s$, pitch speed is $1m/s$ and transmitted power is $F_{t}v=200\mathrm{W}$, 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 $72.79\mathrm{N}$.

The operating pressure angle ${\alpha }_w$ is the gear pressure angle of an actual meshing pair at its mounted axis separation. For external spur gears with involute profiles, it is the angle between the gear line of action and the common tangent to the operating pitch circles. Changing the separation changes this angle even though the individual gear base circles remain fixed.^[ref]

Let the two base radii be $r_{b1},r_{b2}$ and the mounted axis separation be $a$. The right-triangle geometry of the gear line of action givesIf $a_0=r_1+r_2$ is the reference separation and both reference gear pressure angles equal $\alpha$, then $r_{bi}=r_i\cos \alpha$, henceThe gear reference circles and gear ratio remain unchanged by this mounting adjustment; the operating pitch circles expand or contract to fit $a$.

For example, reference radii $20\mathrm{mm}$ and $40\mathrm{mm}$ with $\alpha =20{}^{\circ }$ give $a_0=60\mathrm{mm}$. Mounting a compatible pair at $a=61\mathrm{mm}$ givesAt an imposed gear tangential force $100\mathrm{N}$, the associated gear separating force is about $41.30\mathrm{N}$, compared with $36.40\mathrm{N}$ at the reference angle. Holding torque instead of tangential load would additionally require using the changed operating radius.

The angle formula is only a geometric condition. Usable tooth flanks must still reach the contact region and maintain an adequate gear contact ratio; excessive separation can destroy useful contact. Likewise, the existence of a real inverse cosine does not establish that a proposed mounting has suitable backlash or mechanical strength.^[ref]