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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.

Gear axial pitch $p_x$ is the axial repeat distance between corresponding flanks of successive teeth of a cylindrical helical gear. The measurement follows a line parallel to the rotation axis on a concentric cylinder. It is not the axial advance of one tooth helix through an entire revolution; that larger distance is gear lead $L$. For $N$ equally spaced teeth, $L=N{p}_x$.^[ref]

On the developed cylinder, neighboring tooth traces separated transversely by gear transverse pitch $p_t$ intersect a fixed circumferential coordinate at axial positions separated by $p_x$. Their slope givesFor $d=60\mathrm{mm}$, $N=30$ and gear helix angle $|\beta |=30{}^{\circ }$, $p_t=2\pi \mathrm{mm}$ and $p_x=2\pi \sqrt{3}\mathrm{mm}\approx 10.8828\mathrm{mm}$. The corresponding gear normal pitch is only $5.4414\mathrm{mm}$. These two distances differ by a factor of two in this example because $\sin 30{}^{\circ }=1/2$.

For an engaged gear face width $b=20\mathrm{mm}$, the ideal gear overlap ratio is $b/p_x\approx 1.8378$: successive tooth-contact traces span about1.84 axial pitches across the face. Holding $p_t$ fixed while letting $\beta$ tend to zero makes $p_x$ diverge. Straight spur gear tooth traces do not cross an axial generator successively, so a finite axial-pitch interpretation does not survive that limit. Helix hand determines orientation; the positive pitch magnitude alone does not specify hand.

Gear contact duration is the time for a specified tooth pair to traverse its active gear path of contact. For a fixed-axis external spur gear pair at constant speed, the contact location advances along the gear line of action at the common gear base-roll speedThis is the speed of the contact location along a geometric line, not the relative tooth-surface gear sliding velocity. The latter vanishes at the gear pitch point, while the contact location continues moving. For active path length $L$, duration is $T=L/u_b$. The geometry relates gear contact ratio to active path length divided by gear base pitch.^[ref]

The time between successive tooth-pair entries is $1/f_m$, with gear mesh frequency $f_m=N_1|{\omega }_1|/(2\pi )$. Since $r_{b1}=r_{w1}\cos {\alpha }_w$ for compatible operating geometry, the base-circle distance advanced per entry isConsequentlyA contact ratio above1 means one pair remains engaged after the next pair enters; it does not mean one pair contacts for an entire shaft revolution.

For the standard module2,20/40-tooth example with20degree reference angle, active length is9.65457mm and base pitch is5.90426mm. At driving angular velocity $50rad/s$, $u_b=0.93969m/s$, $f_m\approx 159.155\mathrm{Hz}$ and $T\approx 10.2742\mathrm{ms}$. The entry interval is6.28319ms, giving $T{f}_m\approx 1.63519$. Doubling both shaft speeds halves duration while preserving geometric gear contact ratio. Variable speed requires integrating the changing base-roll speed; a helical gear also needs a stated transverse slice or a whole-face contact convention.

Gear contact ratio measures how much successive tooth engagements overlap. For an external spur gear pair, its transverse value iswhere $L$ is usable gear path of contact length along the gear line of action and $p_b$ is gear base pitch. One tooth pair remains engaged while contact advances through length $L$; successive pairs begin after an advance of $p_b$. Thus ${ϵ}_{\alpha }$ is also the ideal mean number of tooth pairs simultaneously engaged.^[ref]

For a tip-limited, noninterfering external involute pair with tip radii $r_{a1},r_{a2}$, base radii $r_{b1},r_{b2}$, axis separation $a$ and operating pressure angle ${\alpha }_w$, right-triangle geometry givesFor an original example, use module $2\mathrm{mm}$, tooth counts 20 and 40, reference angle $20{}^{\circ }$ and standard separation $60\mathrm{mm}$. With tip radii $22$ and $42\mathrm{mm}$, the formula gives $L\approx 9.6546\mathrm{mm}$ and $p_b\approx 5.9043\mathrm{mm}$, hence ${ϵ}_{\alpha }\approx 1.6352$.

In the ideal periodic timing model, two tooth pairs are engaged for about $63.52\%$ of a tooth-passage period and one pair for about $36.48\%$. This does not mean that force is divided equally between engaged pairs: tooth flexibility and contact geometry affect load sharing. A value below one would leave gaps in ideal engagement, while a value exactly one leaves no overlap margin for tolerances and deformation.

The plot shows contact timing, rather than physical tooth shapes or force sharing. For helical gears, the transverse measure must additionally be distinguished from gear overlap ratio caused by the inclination of teeth across the gear face. Finally, real usable contact can be shortened by modified or unavailable flank portions; the tip-circle formula is conditional on its stated geometric assumptions.

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.

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 helix angle $\beta$ measures a cylindrical gear tooth's inclination to the rotation axis, normally on the reference cylinder. A straight spur gear has $\beta =0$; a helical gear has a nonzero inclination and a specified hand. The cylinder and reference direction matter because another measurement plane or cylinder radius can give another angle.^[ref]

Let $L$ be the positive gear lead, the axial advance of one tooth helix in a complete turn, and let the reference diameter be $d$. Unwrapping that cylinder produces a right triangle with circumferential side $\pi d$ and axial side $L$:For $d=60\mathrm{mm}$ and $|\beta |=30{}^{\circ }$, $L\approx 326.484\mathrm{mm}$. Doubling the lead on that same cylinder gives angle approximately $16.102{}^{\circ }$, rather than15degrees: the relation involves a tangent, not a linear angle scaling.

With $N$ equally spaced teeth, gear axial pitch is $p_x=L/N=p_t/\tan |\beta |$, where $p_t$ is gear transverse pitch. For nonzero $\beta$, this also gives $p_x=\pi m_n/\sin |\beta |$. The corresponding axial-pitch quantity diverges in the straight-tooth limit; it is not a finite replacement for the transverse tooth spacing of a spur gear.

Larger helix angle changes reference diameter at fixed gear normal module and tooth count, increases axial thrust for a fixed tangential load, and increases face-overlap geometry at fixed face width. These consequences must be evaluated together with gear contact ratio and support loads. A magnitude such as30degrees does not by itself specify right versus left hand, nor the thrust direction under a particular applied torque.

Gear lead $L$ is the axial advance of one tooth helix over a complete turn around a cylindrical helical gear. It differs from gear axial pitch $p_x$, which steps between successive teeth at one circumferential position. With $N$ equally spaced tooth helices, $L=N{p}_x$. The lead magnitude does not specify whether the helix is right-handed or left-handed.^[ref]

Unwrapping a reference cylinder of diameter $d$ gives a right triangle with axial side $L$ and circumferential side $\pi d$. Since gear helix angle $\beta$ is measured from the axis,For $d=60\mathrm{mm}$ and $|\beta |=30{}^{\circ }$, $L=60\pi \sqrt{3}\mathrm{mm}\approx 326.484\mathrm{mm}$. If $N=30$, then gear axial pitch is $10.8828\mathrm{mm}$; confusing the two distances would introduce a factor30.

A useful geometric check keeps the lead fixed while changing the chosen concentric cylinder. The helix's axial advance per revolution remains $L$, but its angle satisfies $\tan |\beta (r)|=2\pi r/L$. For the same lead on diameter $40\mathrm{mm}$, the angle is approximately $21.052{}^{\circ }$, rather than $30{}^{\circ }$. Thus a helix-angle specification needs its cylinder, normally the reference cylinder. Lead can remain finite when the physical tooth is much shorter than one complete turn: it describes the continued helix, not a requirement that the manufactured tooth cover a full revolution.

Gear normal module $m_n$ is the gear normal pitch divided by $\pi$, measured normal to the tooth direction of a helical gear. It describes a different section from gear transverse module $m_t$, which uses the circumferential spacing in a plane perpendicular to the rotation axis.^[ref]

If $\beta$ is reference gear helix angle, geometric projection givesFor an original inference example, reference diameter $d=60\mathrm{mm}$, tooth count $N=30$ and $\beta =30{}^{\circ }$ imply $m_t=2\mathrm{mm}$ but $m_n=\sqrt{3}\mathrm{mm}\approx 1.73205\mathrm{mm}$. The normal pitch is approximately $5.4414\mathrm{mm}$, compared with transverse pitch $6.2832\mathrm{mm}$. This is a geometric example, not a claim that the inferred normal module belongs to a particular manufacturer's preferred series.

Conversely, specifying $m_n=2\mathrm{mm}$ for the same30 teeth and30degree helix angle gives reference diameter $69.282\mathrm{mm}$. Thus two drawings both labeled “module2” can describe different gears when one means normal module and the other transverse module.

Pressure angles need the same care: for reference tooth geometry, $\tan {\alpha }_t=\tan {\alpha }_n/\cos \beta$. Matching only a module number, without its plane, pressure-angle convention and helix geometry, does not establish a compatible pair. In a conventional parallel-axis external mesh, the two helical gears also require appropriate opposite hands; the normal-module definition alone does not specify that connection.

Gear normal pitch $p_n$ is the repeat spacing of corresponding helical gear tooth flanks normal to their tooth direction on the developed reference cylinder. It equals $\pi$ times gear normal module $m_n$. It is a different geometric spacing from gear transverse pitch $p_t$, measured circumferentially, and from gear axial pitch $p_x$, measured parallel to the rotation axis.^[ref]

Unwrap the reference cylinder into a plane. The equally spaced tooth traces become parallel straight lines making gear helix angle $\beta$ with the axis. Projecting the transverse spacing onto the normal to those traces givesFor reference diameter $d=60\mathrm{mm}$, $N=30$ teeth and $\beta =30{}^{\circ }$, $p_t=\pi d/N=2\pi \mathrm{mm}$, whereas $p_n=\sqrt{3}\pi \mathrm{mm}\approx 5.4414\mathrm{mm}$. Multiplying $p_n$ by $N$ gives about $163.24\mathrm{mm}$, not the $188.50\mathrm{mm}$ circumference: normal pitch is not the circumferential step.

At $\beta =0$, normal and transverse pitches agree, as expected for a spur gear. For nonzero gear helix angle, using one in place of the other introduces a factor $\cos \beta$. Normal pitch is also distinct from gear base pitch, which refers to the base-circle geometry of an involute. Matching a pitch value without its measurement convention, pressure angle and tooth geometry is insufficient to establish a compatible mesh.

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.

Gear transverse module $m_t$ specifies tooth spacing in the plane perpendicular to a cylindrical gear's axis. If $N$ is tooth count and $d$ is reference diameter, $m_t=d/N$, so reference-circle gear transverse pitch is $p_t=\pi m_t$. For a helical gear, this transverse section differs from the section normal to the tooth direction.^[ref]

Unwrapping the reference cylinder makes the tooth direction a slanted line. Projecting its normal spacing into the circumferential direction giveswhere $m_n$ is gear normal module and $\beta$ is reference gear helix angle. At zero helix angle the normal and transverse modules coincide, recovering the spur gear case.

For an original example, take $m_n=2.5\mathrm{mm}$, $\beta =30{}^{\circ }$ and $N=24$. Then $m_t\approx 2.88675\mathrm{mm}$ and $d\approx 69.282\mathrm{mm}$. A compatible 48-tooth gear has reference diameter $138.564\mathrm{mm}$, giving standard separation $103.923\mathrm{mm}$. Treating2.5 as the transverse module instead would incorrectly give diameters60 and120mm and separation90mm.

The transverse specification is convenient for reference diameters and axis separations, while the normal specification is convenient for the tooth-normal geometry. A drawing must identify which module and pressure-angle planes it uses; a numerical module value alone is incomplete for a helical gear. These are reference dimensions: a changed operating separation must additionally be interpreted through the mounted geometry, rather than redefining the gear's original module.

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.

Idler bearing load sharing determines how an idler support reaction is divided among the shaft's supporting bearing elements. Zero net torque on an ideal steady idler gear does not remove its mesh-induced transverse force. The bearing loads additionally depend on axial positions and the supported-shaft arrangement. Manufacturer guidance separates a radial load between two bearings from an overhung load and treats gear loads as applications of those cases.^[ref]

Model the shaft as a simply supported beam with supports A and B at axial coordinates $z=0$ and $z=L$. Let a gear resultant $P$ act upward at $z=a$, and take upward reactions as positive. Static equilibrium requiressoFor a200N mesh resultant at $a=40\mathrm{mm}$ between supports100mm apart, the downward reactions are120N and80N. Equal division occurs only for the centered load in this symmetric ideal support model.

For overhung idler loading at $a=130\mathrm{mm}$ with the same support span and resultant, $R_A=+60\mathrm{N}$ and $R_B=-260\mathrm{N}$. The farther support's force reverses direction, and the nearer bearing carries more than the applied resultant. This does not violate force balance: $60-260+200=0$. If a particular support cannot provide the required reaction direction, the assumed support model must change.

For loads in two transverse directions, solve each component balance before forming the resultant at each bearing. Helical gears add gear axial force; allocating that thrust requires the actual bearing constraints and cannot be inferred from the transverse lever rule. More than two effective supports generally require deformation compatibility as well as equilibrium. These calculations estimate support forces, rather than directly determining bearing life or internal rolling-element contact loads.

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.

A locating bearing support establishes a shaft's axial position relative to its housing while supplying the required mechanical support. Depending on the arrangement, a locating support can consist of one bearing or a coordinated set. Its function must be distinguished from a non-locating bearing support that accommodates axial displacement, including relative thermal expansion.^[ref]

For an ideal two-support example, place the locating support A at $z=0$ and a radially supporting, axially free support B at $z=200\mathrm{mm}$. An upward transverse gear load300N at $z=80\mathrm{mm}$ gives downward radial reactions180N at A and120N at B by static equilibrium. If the gear also exerts an axial force of120N and A is the sole axial restraint, the axial reaction at A is $-120\mathrm{N}$ and that at B is zero.

The magnitude of A's combined external reaction vector is $\sqrt{180^2+120^2}\approx 216.3\mathrm{N}$. This vector magnitude is a force-balance result, not automatically the catalogue's equivalent dynamic bearing load or life-rating input. Those calculations distinguish radial and axial loading according to bearing type and operating conditions. Helical gears provide a practical example requiring both load components.^[ref]

The location of the axial restraint also determines how shaft expansion changes component positions. If A fixes one end and the other support permits extension, a point a distance $z$ from A moves approximately $\alpha z\Delta T$ in a simple uniformly heated bar model. Thus choosing a locating support affects alignment and axial positioning as well as which support carries thrust.

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.