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Gear axial force is the component of tooth-contact load parallel to a gear's rotation axis. In a standard parallel-axis external helical gear pair, ideal force resolution at the reference mounting giveswhere $F_t$ is gear tangential force and $\beta$ is reference gear helix angle. Its direction depends jointly on helix hand and the transmitted-torque direction. Reversing the torque can reverse axial thrust even though the cut helix remains unchanged.^[ref]

For an original example, transmit $6\mathrm{N}\mathrm{m}$ through reference radius $30\mathrm{mm}$ at standard mounting. Then $|F_t|=200\mathrm{N}$. A30degree helix gives axial load $200\tan 30{}^{\circ }\approx 115.47\mathrm{N}$; a15degree helix at the same tangential load gives about $53.59\mathrm{N}$. The support must sustain the appropriate axial reaction in addition to the mesh's radial and tangential loads.

The axial load does not itself imply energy loss. If an ideal gear has no axial velocity, that force does no work through an axial displacement. Real support friction and deflection can nevertheless affect efficiency and tooth alignment. A useful bench check applies a controlled transmitted torque and measures the axial reaction while recording helix hand and the actual power-flow direction; an unsigned force magnitude alone does not establish the direction rule.

For the ideal spur gear limit, $\beta =0$ and this geometric thrust term vanishes. Opposite-handed helical sections can be arranged to cancel net axial thrust under suitable symmetric loading, but unequal loading prevents assuming exact cancellation. Away from the stated reference mounting, force formulas must use a geometrically consistent operating convention rather than mixing operating pitch radii with an unrelated reference angle.

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 $72.79\mathrm{N}$.

A helical gear is a cylindrical gear whose teeth follow a helix around the reference cylinder. The gear helix angle $\beta$ measures their inclination to the axis. In a conventional external pair on parallel axes, mating gears have opposite hands with matching helix-angle magnitudes. Inclination makes engagement progress across the face rather than beginning everywhere along a straight tooth at once, but also creates a gear axial force that the supports must carry.^[ref]

A helical gear must specify its reference plane. The gear normal module $m_n$ is based on a section normal to the tooth direction, whereas gear transverse module $m_t$ is based on the plane perpendicular to the gear axis. Their geometric relation isSimilarly, reference pressure angles satisfy $\tan {\alpha }_t=\tan {\alpha }_n/\cos \beta$. Equal numbers labeled “module” in different planes therefore need not specify interchangeable gears.^[ref]

For an original example, take $m_n=2\mathrm{mm}$, $N=20$ and $\beta =30{}^{\circ }$. Reference diameter is $46.188\mathrm{mm}$, rather than $40\mathrm{mm}$. With ${\alpha }_n=20{}^{\circ }$, the transverse angle is about $22.796{}^{\circ }$.

At standard mounting, a transmitted torque $2\mathrm{N}\mathrm{m}$ gives gear tangential force $F_t\approx 86.60\mathrm{N}$. Ideal contact-force resolution gives axial magnitudeIts direction depends on helix hand and transmitted-torque direction. The corresponding radial component uses the transverse gear pressure angle. This illustrates the design trade: distributed engagement can improve smoothness, but the same transmission requires axial-load support. The example concerns parallel-axis cylindrical gearing; it should not be applied unchanged to crossed-axis arrangements.

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]

A spur gear is a cylindrical gear with straight teeth parallel to its rotation axis. A conventional external pair transmits motion between parallel axes. Unlike the inclined teeth of a helical gear, the ideal straight-tooth geometry produces no axial thrust from the tooth-normal contact force; there are still tangential and separating loads that the mounting must carry.^[ref]

For compatible spur gears, reference diameter is $d=mN$, where $m$ is gear module and $N$ is tooth count. Standard reference axis separation is $(d_1+d_2)/2$. Two external spur gears have signed angular velocity relation ${\omega }_2/{\omega }_1=-N_1/N_2$: they rotate in opposite directions. Matching gear module alone is insufficient; tooth profiles and gear pressure angles must also be compatible.

For an original example, take gear module $3\mathrm{mm}$ and tooth counts 20 and 60. Reference diameters are $60\mathrm{mm}$ and $180\mathrm{mm}$, so standard axis separation is $120\mathrm{mm}$. Driving the smaller spur gear at $600$ revolutions per minute gives the larger one $200$ revolutions per minute in the opposite direction. In a steady ideal transmission, an input torque of $2\mathrm{N}\mathrm{m}$ becomes $6\mathrm{N}\mathrm{m}$ at the output. The corresponding input gear tangential force is $2/0.030\approx 66.7\mathrm{N}$; multiplying this same force by the output reference radius $0.090\mathrm{m}$ recovers the output torque.

A simple bench demonstration marks one tooth on each spur gear and counts complete turns, testing the three-to-one gear ratio. Measuring transmitted torques and rotational speeds additionally tests gear efficiency. Observing the rotation ratio alone cannot establish a tooth's allowable load or the losses caused by friction and backlash.