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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.

A gear contact marking test makes a gear contact pattern visible by using a thin contrasting marking layer and rolling the mating gears under stated conditions. The observed transfer or removal of the layer locates the region where the flanks meet closely enough to mark. A useful test record includes the marked flank, rotation direction, mounting position, backlash and applied torque. Manufacturer guidance discusses tooth-surface contact as an assembly and alignment check.^[ref]

A controlled comparison changes one alignment setting while keeping the same marking method and loading condition. A shift of the mark toward an edge then records a change in the mesh geometry. Changing layer thickness at the same time would confound that comparison: the visible pattern is the response of the surfaces and the marking process together.

An original threshold model illustrates this limitation without claiming a universal paint law. Suppose an unloaded gap over a $20\mathrm{mm}$ face isStipulate that a mark appears wherever $g(z)\le h$, with gear marking-layer gap threshold $h$. At $h=2\mu \mathrm{m}$ the marked width is $8\mathrm{mm}$; at $h=4\mu \mathrm{m}$ it is $16\mathrm{mm}$. The indicated coverage doubles even though the underlying gap is unchanged. Real transfer also depends on deformation and the marking material.

Nor does a broad mark uniquely determine the gear face load distribution. Operating torque deforms teeth, shafts and supports, so a lightly loaded pattern can differ from a loaded one. The historical gear-rating standard distinguishes comprehensive loaded assessment from the limited information in a bearing pattern.^[ref]

A meaningful validation pairs the pattern with an independent load-sensitive observation, such as tooth-root strain, or a checked deformation model. The question is whether the setup places contact where intended under the specified condition, rather than whether one visually attractive mark proves uniform pressure at every operating load.

A gear contact pattern is the observed region of tooth-flank contact during a stated mesh test. Its position and extent reveal how the installed tooth surfaces meet. A gear contact marking test uses a thin marking layer and controlled rotation to make that region visible; the mounting position, transmission direction and applied torque belong in the test description. Manufacturer guidance treats surface-contact checks as evidence about the assembled mesh, including shaft alignment.^[ref]

A useful bench comparison holds the gear pair and test conditions fixed, records the initial mark, then changes one alignment setting and repeats the test. A mark moving toward one face edge indicates a change in where the flanks meet. This is an observation of contact location, rather than an automatic measurement of operating pressure or a universal diagnosis of the alignment error.

For example, a mark occupying the full20mm gear effective face width is compatible with a hypothetical uniform line load $w=50N/mm$, but also with the tilted positive distributionBoth integrate to1000N, yet their peak line loads are50 and80N/mm. Merely observing full-width contact cannot choose between those gear face load distributions. The gear face load factor or local stress needs additional load/deformation information.

Loaded and lightly loaded patterns can also differ because shaft, housing and tooth deformation change the gap. A quantitative loaded assessment may use strain measurements or a validated contact model alongside the pattern; the2006 gear-rating standard explicitly distinguishes comprehensive load assessment from the limited information in a bearing pattern.^[ref]

Gear effective face width is the usable axial width participating in a stated cylindrical gear mesh. It must be distinguished from each component's nominal gear face width, and from a claim that every part of that width carries equal load. The normal/transverse tooth geometry and gear overlap ratio use this engaged width, while gear face load distribution describes how the load actually occupies it.^[ref]

An elementary geometric model treats the two toothed widths as intervals $I_1=[a_1,b_1]$ and $I_2=[a_2,b_2]$ on their common axial coordinate. Their available overlap isFor one20mm-wide gear occupying $[-10,10]\mathrm{mm}$ and a16mm-wide mate centered at $6\mathrm{mm}$, the second interval is $[-2,14]\mathrm{mm}$. Their overlap is only12mm. The mean line load for a hypothetical uniformly shared normal force of1200N is then100N/mm, rather than the60N/mm obtained by using the first gear's entire nominal width.

With gear normal module $m_n=2\mathrm{mm}$ and gear helix angle $|\beta |=30{}^{\circ }$, gear axial pitch is $p_x=4\pi \mathrm{mm}$. The12mm overlap gives geometric gear overlap ratio $b_{\mathrm{eff}}/p_x=3/\pi \approx 0.9549$, compared with $5/\pi \approx 1.5915$ for20mm. Axial positioning can therefore alter overlap without changing the ideal gear ratio.

The interval calculation assumes the flanks remain usable throughout their common extent. Relief, damaged regions or partial contact can require a more specific working-width convention. Uneven loading is not generally corrected by shrinking width until a uniform-load formula looks plausible; a stated load-distribution model is needed to predict local loading.

Gear face load distribution describes how tooth-contact load is spread across the engaged gear face width. Equal total transmitted force can produce different local loads when shafts tilt, supports deform or tooth traces are imperfect. A geometrically available gear effective face width therefore does not establish uniform loading. The manufacturer's tooth-flank gear face load factor accounts for support arrangement and the expected quality of contact under load.^[ref]

For a simple axial line-load model, let $w(z)$ have units $N/mm$ over $0\le z\le b$. Its total normal force isUniform loading gives $w=F_n/b$. An illustrative tilted distribution isIts integral remains $F_n$, while its largest line load is $(1+\lambda )F_n/b$. With $F_n=1000\mathrm{N}$, $b=20\mathrm{mm}$ and $\lambda =0.6$, the two edges carry20 and80N/mm, compared with the50N/mm mean. The peak-to-mean ratio1.6 characterizes this constructed model; it is not a substitute for the manufacturer's standard factor calculation.

The gear contact pattern can reveal where mating tooth flanks touch, but an unloaded marking test alone does not measure the operating force density. A loaded comparison must specify applied torque, shaft positioning and support deformation. In helical gears, changing tooth-contact lines and multiple simultaneous contacts also require a consistent choice of time and projected width. A wider tooth can reduce mean load per width while still suffering a concentrated edge load; width and alignment must be considered together.

A gear face load factor quantifies the effect of nonuniform load across the engaged gear face width. For the contact-load convention in ISO 6336-1:2006, $K_{H\beta }$ compares maximum line-load intensity with the mean intensity over the stated width:Here $F_m$ is the total load represented by that distribution and $b$ is its evaluation width. A uniform distribution gives1; an edge-concentrated distribution gives a larger value. The numerator and denominator must refer to the same load convention and operating condition.^[ref]

For an original illustrative distribution over $b=20\mathrm{mm}$,total load is1000N, mean line load is50N/mm and maximum line load is80N/mm. Thus $K_{H\beta }=1.6$. A uniform distribution with the same width and total load gives1. The gear contact pattern can cover the whole width in both cases.

For ideal elastic line contact at fixed geometry and material, Hertzian contact pressure scales with the square root of line load. The corresponding gear contact-stress load factor is therefore $\sqrt{1.6}\approx 1.265$ in this simplified comparison. Multiplying the contact pressure itself by1.6 would confuse a load-intensity ratio with a stress ratio. This scaling does not replace a complete gear-rating calculation.

The manufacturer's treatment connects face-load assessment to shaft-support placement, shaft stiffness and loaded contact quality.^[ref] A value inferred from a toy line-load profile is consequently useful for understanding the factor, while a real design must obtain the relevant distribution or factor from a method appropriate to its geometry and load state.

Gear face width $b$ describes the width of the toothed portion. For a cylindrical spur gear or helical gear, it is measured parallel to the rotation axis, rather than along the slanting tooth trace. A projecting hub need not be part of the toothed width. Manufacturer drawings distinguish face width, hub width and total length as separate dimensions.^[ref]

The available width of one gear is not necessarily the gear effective face width of a pair. For two aligned cylindrical gears with equal width $b$ and relative axial offset $e$, a simple interval-overlap model givesAn original $b=20\mathrm{mm}$ pair offset by $4\mathrm{mm}$ has only $16\mathrm{mm}$ of common axial extent. If a hypothetical normal load $1000\mathrm{N}$ were uniformly spread across that extent, mean load per axial width would rise from $50$ to $62.5N/mm$. This is bookkeeping for an ideal uniform load, not a tooth-strength prediction. Actual gear face load distribution also depends on alignment and deformation.

For helical gears, width also enters gear overlap ratio:With gear normal module $m_n=2\mathrm{mm}$ and gear helix angle $|\beta |=30{}^{\circ }$, reducing engaged width from $20$ to $16\mathrm{mm}$ reduces ideal overlap from $1.5915$ to $1.2732$. The overall gear ratio is unchanged. Width therefore affects engagement and loading even when tooth counts and shaft speeds remain the same.^[ref]

The gear line of action is the common normal along which ideal tooth contact transmits force. For meshing involute profiles in a transverse plane, it is a straight common tangent to the two gear base circles. The angle it makes with the common tangent to the operating pitch circles is the operating pressure angle.^[ref]

Its intersection with the line of gear centers is the gear pitch point. The actual contact point moves along only a finite portion, the gear path of contact, bounded by usable tooth profiles. Thus a diagram's infinite straight line is a geometric construction, not a claim that teeth can contact everywhere on it.

The geometry explains constant ratio. Let the tooth-normal force magnitude be $F_n$ and the base radii $r_{b1},r_{b2}$. The perpendicular distance from each gear center to the gear line of action is its base radius, so the ideal transmitted moments have magnitudesThese moment arms do not change as the contact point advances along that line. The no-penetration contact condition correspondingly gives $r_{b1}{\omega }_1+r_{b2}{\omega }_2=0$ for an external pair. Compatible profiles have $r_{b2}/r_{b1}=N_2/N_1$, recovering the tooth-count gear ratio.

For example, reference radii $20$ and $40\mathrm{mm}$ at reference gear pressure angle $20{}^{\circ }$ give base radii approximately $18.794$ and $37.588\mathrm{mm}$. A $100\mathrm{N}$ normal force produces moment magnitudes $1.8794$ and $3.7588\mathrm{N}\mathrm{m}$. Their ratio stays two as the point of contact moves.

Ideal normal transmission does not eliminate gear tooth sliding. Relative surface slip generally changes along the gear path of contact and vanishes at the gear pitch point for the ideal external pair. Actual friction associated with that slip adds a tangential contact force; its resultant need not lie exactly on the geometric gear line of action.

A gear marking-layer gap threshold is an effective separation below which a stated marking process produces a visible transfer or removal of material. It is a model of the observation in a gear contact marking test, rather than a universal thickness or a material constant. Layer thickness, pressure, rolling history and the marking material can all matter; a visible gear contact pattern does not directly give local operating pressure.

For an original idealization, let the unloaded separation across a face beand stipulate a visible mark where $g(z)\le h$. Here $h$ is the effective threshold, $g_0$ the minimum gap and $a>0$ its slope magnitude. For a centered pattern on face width $b$, the predicted marked width isChoose $b=20\mathrm{mm}$, $g_0=0.5\mu \mathrm{m}$ and $a=0.4\mu m/mm$. Thresholds $h=1.5$ and $2.5\mu \mathrm{m}$ give marked widths $5$ and $10\mathrm{mm}$. The geometry has not changed; the observation threshold has.

Away from the onset and saturation corners,Here a threshold change $0.2\mu \mathrm{m}$ changes the indicated width by $1\mathrm{mm}$. In the shallow-gap limit, small changes of the marking process can therefore cause conspicuous changes in the visible pattern.

A controlled comparison should hold the coating method, applied torque and number/direction of roll-throughs fixed, then vary one geometric setting. A separate fixture with known separations can test whether a proposed threshold model describes the marking process, without treating nominal coating thickness as an exact gap measurement.

Manufacturer contact guidance describes patterns as evidence about assembled tooth surfaces, and the historical rating standard limits the load information inferred from a bearing pattern.^[ref][ref] The threshold equation is a deliberately specified illustrative model; actual deformation and transfer behavior need validation before converting a mark into a quantitative gap or gear face load distribution.

Gear overlap ratio measures the additional engagement overlap caused by a helical gear's tooth inclination across its gear face width. It is distinct from the transverse gear contact ratio, which measures contact length within one transverse section. For face width $b$, reference gear helix angle $\beta$ and gear normal module $m_n$,where $m_t$ is gear transverse module and $p_x$ is gear axial pitch.^[ref]

Unwrapping the reference cylinder explains the ratio: moving across axial width $b$ shifts the tooth pattern circumferentially by $b\tan |\beta |$. Dividing this shift by gear transverse pitch $\pi m_t$ counts how many tooth spacings the engagement progresses across the face.

For an original example, $b=20\mathrm{mm}$, $m_n=2\mathrm{mm}$ and $|\beta |=30{}^{\circ }$ giveDoubling the face width doubles this geometric overlap if the other parameters stay fixed. At the spur gear limit $\beta =0$, the overlap contribution tends to zero.

In the ideal cylindrical-helical model, total geometric contact ratio is the sum of transverse and axial contributions, ${ϵ}_{\gamma }={ϵ}_{\alpha }+{ϵ}_{\beta }$. This does not specify a uniform division of load: flexibility, alignment and the usable flank region affect actual contact. Increasing helix angle can improve overlap while also increasing gear axial force; increasing width also changes the alignment and support demands. These are related geometric tradeoffs, not a guarantee of a specified noise level or load capacity.

The gear pitch point is where the gear line of action meets the line joining the gear centers. For a fixed-axis external involute pair, it is the tangency point of the operating pitch circles. The point is a geometric location in the mesh, rather than a material mark permanently attached to one tooth.^[ref]

Let the gear centers be separated by $a$, with tooth counts $N_1,N_2$. Their operating radii locate the pitch point atFor external signed angular velocities, $r_{w1}{\omega }_1+r_{w2}{\omega }_2=0$. The two hypothetical material points occupying the pitch point therefore have the same instantaneous velocity. This is the local rolling without slipping property used to construct the operating pitch circles.

For an original example, use20 and40 teeth separated by $60\mathrm{mm}$. The pitch point is20mm from the first center and40mm from the second. With ${\omega }_1=50rad/s$ and ${\omega }_2=-25rad/s$, both surface velocities at that location have magnitude $1m/s$ and the same direction. Ideal relative gear sliding velocity is zero there.

Elsewhere on the gear path of contact, the two tooth surfaces can slide even while maintaining the required normal contact condition. The pitch point remains fixed in this fixed-axis geometry, while successive tooth-contact points advance through the mesh. Whether the pitch point itself belongs to a particular usable contact interval depends on the available flanks; the geometric definition does not assert that every modified tooth profile physically contacts at that point.

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 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.^[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 $\hat{\mathbf{z}}$ for both signed angular velocities. The rigid-body velocity law givesSubtracting these expressions and using equal velocities at $P$ givesThe 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 ${\omega }_1=50rad/s$ and ${\omega }_2=-25rad/s$, at $s=4\mathrm{mm}$ the signed tangential sliding velocity is $0.30m/s$; at $s=-4\mathrm{mm}$ it is $-0.30m/s$. 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.

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 $72.79\mathrm{N}$.

Gear tooth sliding is relative tangential motion of mating tooth surfaces at their contact point. Ideal contact prevents relative normal penetration but does not generally require the tangential surface velocities to agree. For a fixed-axis external involute pair, tangential slip vanishes at the gear pitch point and changes as the contact moves along the gear path of contact.^[ref]

Let $P$ be the pitch point, let $\mathbf{n}$ be a unit vector along the gear line of action, and write a contact location as $Q=P+s\mathbf{n}$. With a common positive axis $\hat{\mathbf{z}}$ and signed angular velocities ${\omega }_1,{\omega }_2$, rigid-body velocity differences giveThe velocities at $P$ cancel, leaving this term proportional to distance along the contact line. Since ${\omega }_2$ has opposite sign in an external pair, the sliding-speed magnitude is $(|{\omega }_1|+|{\omega }_2|)|s|$.

For an original example, use ${\omega }_1=50rad/s$ and ${\omega }_2=-25rad/s$. At $s=4\mathrm{mm}$, gear sliding velocity along the common tooth tangent is $0.30m/s$; at $s=-4\mathrm{mm}$ it has the opposite sign. At $s=0$ it is zero. Equality of operating pitch-circle speeds must therefore not be used to conclude that the tooth flanks roll without slipping throughout engagement.

In a simplified Coulomb friction model with coefficient of friction $\mu$ and normal load $F_n$, the dissipated gear sliding-friction loss is $P_f=\mu F_n|v_s|$. Hypothetical values $\mu =0.05$, $F_n=100\mathrm{N}$ and $|v_s|=0.30m/s$ give $1.5\mathrm{W}$. This is a local illustrative loss model, not a lubricant specification. Zero ideal sliding at the pitch point also does not eliminate every source of gearbox loss, such as support friction or deformation-related dissipation. Real load and friction can vary through the engagement, so total loss requires integrating the applicable contact model.

An idler support reaction is the force exerted by the shaft and bearings that balances the mesh loads and other external loads on an idler gear. A vanishing net transmitted torque does not imply a vanishing resultant force: moments can cancel while forces add. A free-body diagram must retain each tooth-contact force and its point of application. The tangential load in each ideal mesh follows from $F_t=|\tau |/r_w$.^[ref]

Consider a steady collinear train with input, idler and output reference radii20,30 and50mm. An input torque $2\mathrm{N}\mathrm{m}$ gives a100N tangential mesh component. On the idler, choose contact positions at $x=-30$ and $x=30\mathrm{mm}$ relative to its center, with both tangential forces upward. Their moments are $-3$ and $+3\mathrm{N}\mathrm{m}$, but their vector sum is200N upward. For equal20degree pressure angles, the two radial mesh components have magnitude $100\tan 20{}^{\circ }\approx 36.4\mathrm{N}$ and opposite directions, so they cancel. Neglecting other loads, the net support reaction is200N downward.

This is the combined support resultant, not automatically the load on each bearing. For a hypothetical shaft supported at axial positions $z=0$ and $z=100\mathrm{mm}$, with the200N upward gear resultant acting at $z=40\mathrm{mm}$, static equilibrium givesThus $R_B=-80\mathrm{N}$ and $R_A=-120\mathrm{N}$. Idler bearing load sharing depends on placement even when the total reaction is fixed.

Noncollinear shaft layouts rotate the mesh-force directions, and helical gears add gear axial force. Acceleration also changes the balance: the sum of torques must supply the idler's angular acceleration, and unequal mesh loads need not cancel radially. The zero-net-torque example therefore establishes a steady ideal force balance, rather than a universal bearing-load formula.

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]