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A compound gear train has an intermediate rotating body carrying at least two rigidly joined gears. Those gears share angular velocity but can have different tooth counts, allowing successive reductions to multiply without the cancellation found in a simple gear train. For two external stages with the second and third gears rigidly joined,The two external meshes give equal input/output rotation directions.^[ref]

For an original example, use tooth counts $N_1=18$, $N_2=54$, $N_3=12$ and $N_4=60$. The reduction is $3\times 5=15$. Input speed $1800$ revolutions per minute gives intermediate speed $600$ and output speed $120$ revolutions per minute.

This example also permits a reverted gear train, with input and output axes coaxial. If both stages use the same gear module $m$ and standard unshifted spur gears, the two axis separations must agree:Here both tooth-count sums equal72, so $m=2\mathrm{mm}$ gives a common separation $72\mathrm{mm}$. Ratio multiplication alone would not establish this packaging condition.

In steady ideal operation, output torque is fifteen times input torque. With stage gear efficiencies ${\eta }_1,{\eta }_2$, it becomes $15{\eta }_1{\eta }_2$ times input torque, provided the stage power boundaries include the stated losses. A high reduction can therefore trade compact stages against added losses, loads and backlash. Axis separations and support loads remain separate requirements from the desired numerical gear ratio.

A face gear is a gear whose working teeth occupy an annular face and mesh with a mating cylindrical pinion. A common arrangement turns the drive through approximately a right angle; other shaft angles and tapered configurations are possible. Its working surface geometry must maintain compatible tooth contact, rather than simply placing ordinary cylindrical teeth on a flat disk.

Tooth counting still supplies an ideal speed relation. If the pinion has $N_p$ teeth and the face gear $N_f$, equality of tooth-passage rates givesAn original 20-tooth pinion driving a 100-tooth face gear at $1500$ revolutions per minute therefore gives $300$ revolutions per minute at the output. Rotation signs require a stated orientation of the two different shaft axes; the scalar minus sign for an external parallel-axis spur gear pair cannot be copied without that convention.

The annular layout also explains a geometric design issue. With $N$ tooth positions, angular spacing is $2\pi /N$, while arc spacing on a ring of radius $r$ isFor $N=80$, radii $40$ and $60\mathrm{mm}$ give arc spacings about $3.142$ and $4.712\mathrm{mm}$. This radial change does not prescribe two interchangeable gear modules; it shows why a generated face-tooth surface cannot be understood as a single extruded cylindrical tooth profile.

A primary 2002 program built a scaled-power split-torque transmission with two opposed face gears, two input pinions and two idlers.^[ref] Multiple meshes make the geometry useful for distributing and recombining load, but also make alignment and compatibility important.

A bench rotation check can test the tooth-count ratio. A gear contact marking test, dimensional inspection and load-sensitive measurements address different questions about the manufactured surfaces and their operation. Correct output speed alone does not establish an acceptable gear contact pattern, tooth stress or gear efficiency.

Gear efficiency is the fraction of mechanical input power delivered as useful mechanical output power over a stated operating condition and system boundary:For steady forward operation with positive input power, $P_{\mathrm{loss}}=P_{\mathrm{in}}-P_{\mathrm{out}}$ accounts for dissipation within that boundary. Tooth-contact friction is one contribution; a complete gearbox measurement can also include other rotating supports and lubricant losses. Citizen Micro explicitly treats its quoted stage efficiencies as reference values that depend on transmitted torque and speed.^[ref]

If the positive reduction gear ratio is $i=|{\omega }_{\mathrm{in}}|/|{\omega }_{\mathrm{out}}|$, thenAn increase in torque therefore does not itself measure gear efficiency: reduction increases torque while decreasing speed. For an original example, let input torque be $2\mathrm{N}\mathrm{m}$ at $1200$ revolutions per minute, and output torque be $7.6\mathrm{N}\mathrm{m}$ at $300$ revolutions per minute. Then $i=4$ andThe angular-speed conversion factors cancel in the ratio but must be included when computing power in watts.

A bench test places torque and rotational-speed measurements on both sides of the chosen transmission, compares synchronized measurements after reaching steady operation, and reports load, speed and temperature. During acceleration, a difference between input and output power can partly represent changing kinetic energy, rather than loss. Near zero useful load, approximately fixed losses can also produce a much smaller efficiency than at a loaded operating point. For successive stages with consistent boundaries and no extra power path, efficiencies multiply; three hypothetical $0.95$ stages yield $0.95^3\approx 0.8574$, rather than $0.95$.

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.

A gear ratio states the relation between the rotational motions of connected gears. Because conventions differ, specify the input, output and direction. Here the positive reduction ratio isso $i>1$ means that the output has lower angular speed. For a single external pair on fixed axes, tooth-passage counting givesThe minus sign assumes the same positive rotation convention on both parallel axes. KHK uses the driven-to-driver tooth-count ratio for its speed-reduction examples.^[ref]

Ratios multiply through successive stages of a gear train. Consider a 12-tooth input driving a 36-tooth gear, with a 15-tooth gear rigidly attached to that intermediate rotating body and driving a 45-tooth output. Each stage reduces angular speed by three, so the combined reduction ratio is nine. An input of $1800$ revolutions per minute yields an output of $200$ revolutions per minute. Two external meshes reverse direction twice, so input and output rotate in the same direction.

An intermediate gear used only to pass rotation onwards gives a different result. If a 12-tooth input drives a 36-tooth intermediate gear that itself drives a 48-tooth output, the ratio is $(36/12)(48/36)=4$: the intermediate tooth count cancels. The previous ninefold reduction required two distinct tooth counts rigidly sharing the intermediate rotation.

For steady ideal transmission, conservation of energy gives $|{\tau }_{\mathrm{out}}|=i|{\tau }_{\mathrm{in}}|$. If gear efficiency is $\eta =P_{\mathrm{out}}/P_{\mathrm{in}}$, the corresponding relation is $|{\tau }_{\mathrm{out}}|=\eta i|{\tau }_{\mathrm{in}}|$. With $i=9$, $\eta =0.90$ and input torque $2\mathrm{N}\mathrm{m}$, the output torque is $16.2\mathrm{N}\mathrm{m}$. This power balance assumes steady operation: accelerating rotating components can temporarily store kinetic energy. A measured gear ratio establishes kinematics, while torque capacity and losses require additional mechanical information.

Gear sliding-friction loss is the rate at which relative tooth-flank sliding dissipates mechanical energy. In a simple Coulomb friction model with normal contact load $F_n$, coefficient of friction $\mu$ and gear sliding velocity $v_s$, the nonnegative dissipated power isThis is a local contact model, rather than the entire gearbox loss. Manufacturer guidance distinguishes gear-meshing losses from losses in rotating supports and warns that overall gear efficiency depends on load and speed.^[ref]

To derive the sign, let $v_s=v_1-v_2$ along the common tangent. The friction force on surface1 is $f_1=-\mu F_n\mathrm{sgn}(v_s)$ and that on surface2 is $-f_1$. Their combined mechanical power is $f_1{v}_1-f_1{v}_2=-\mu F_n|v_s|$. The corresponding positive quantity is the rate of mechanical energy loss. At zero sliding the ideal Coulomb sliding-loss term vanishes, even though another resistance mechanism can remain.

For a fixed-axis external spur gear pair with $|{\omega }_1|+|{\omega }_2|=75rad/s$, $|s|=4\mathrm{mm}$ gives $|v_s|=0.30m/s$. Hypothetical $\mu =0.05$ and $F_n=100\mathrm{N}$ give $P_f=1.5\mathrm{W}$. Holding these load and friction assumptions fixed while doubling both shaft speeds doubles the instantaneous loss at that geometric point.

Gear friction work per engagement integrates this varying loss through the active gear path of contact. If one pair carries a constant $F_n$, the contact coordinate runs from $-L_a$ to $L_r$ at constant speed $\dot{s}=u_b>0$, givingFor $L_a=5.0586\mathrm{mm}$, $L_r=4.5960\mathrm{mm}$ and $u_b=0.93969m/s$, the preceding loads and speeds give about $9.32\mathrm{mJ}$. Real tooth pairs share varying loads; an overall loss calculation must sum the applicable contacts and integrate their actual loads, sliding and friction model.

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 gear train is an interconnected arrangement of gears that transmits rotation through one or more meshes. Its behavior depends on which gears mesh and which are rigidly attached to the same rotating body. A simple gear train passes motion through successive individual gears; a compound gear train uses at least one intermediate rotating body carrying different gears rigidly together.^[ref]

For a chain of external meshes on fixed parallel axes, each mesh satisfies ${\omega }_{j+1}/{\omega }_j=-N_j/N_{j+1}$. Multiplication gives the overall signed gear ratio; two rigidly joined gears instead share exactly the same angular velocity. These are different connection rules and must not be interchanged.

In a simple gear train with tooth counts 16, 40 and 64, the overall relation isThe intermediate count cancels, and the two reversals give equal input/output rotation directions. That intermediate gear can still change spacing and mounting geometry even though it does not change this reduction ratio.

For a compound gear train, let a 16-tooth input drive a 48-tooth gear, rigidly joined to a 12-tooth gear that drives a 60-tooth output. ThenAn input of $1500$ revolutions per minute produces $100$ revolutions per minute at the output. The two intermediate tooth counts do not cancel because they belong to different gears sharing rotation, rather than one gear engaging both neighbors.

In steady operation, stage gear efficiencies multiply when all intervening losses are included in their stated power boundaries. If the two stages each have efficiency $0.96$, the total is $0.9216$; at input torque $1\mathrm{N}\mathrm{m}$, the output torque is $15(0.9216)=13.824\mathrm{N}\mathrm{m}$. These fixed-axis relations do not automatically describe arrangements whose gear axes move. In all cases, measured speeds test kinematics, while measured power is needed to establish losses.

A simple gear train transmits rotation through individual intermediate gears, each intermediate gear engaging its neighbors without a separate, differently sized gear rigidly sharing that intermediate rotation. For external meshes on fixed parallel axes, consecutive signed ratios multiply and intermediate tooth counts cancel:where $m$ is the number of external meshes. An intermediate idler gear changes placement and potentially output direction, but not this ideal endpoint tooth-count ratio.^[ref]

For example, a chain with tooth counts 18, 30, 42 and 72 has three meshes andAt input speed $800$ revolutions per minute, output speed is $200$ revolutions per minute in the opposite direction. Removing one intermediate gear while maintaining a compatible two-mesh layout preserves the fourfold reduction but reverses the direction conclusion.

The cancellation concerns kinematics, not every engineering consequence. Intermediate tooth counts affect physical sizes, axis positions and angular speeds of intermediate rotating bodies; additional meshes and supports can change gear efficiency and backlash. Moreover, if two different intermediate gears are rigidly coupled, their tooth counts do not cancel by the same argument: that arrangement is a compound gear train.

A split-torque transmission divides transmitted load among two or more mechanical paths and subsequently combines their contributions. Its purpose is to let several meshes carry a total load that would otherwise pass through one path. The gear torque-path load fraction depends on compatibility, stiffness, gaps and mounting; a symmetric layout alone does not prove equal sharing.

For an original idealization, refer both branch torques to the same shaft coordinate and let their common elastic twist be $\theta$. With torsional stiffnesses $K_1,K_2$ and no initial gaps,Values $K_1=40000$ and $K_2=60000\mathrm{N}m/rad$ under total torque $1000\mathrm{N}\mathrm{m}$ give $\theta =0.01\mathrm{rad}$ and branch torques $400$ and $600\mathrm{N}\mathrm{m}$. If each branch were limited to $600\mathrm{N}\mathrm{m}$, this sharing would limit total load to $1000\mathrm{N}\mathrm{m}$; equal sharing would permit $1200\mathrm{N}\mathrm{m}$ within the same simplified capacity assumptions.

Parallel-path gear efficiency also differs from the product rule for serial stages. If input power fractions are $a_j$ with ${\sum }_j{a}_j=1$, and each branch delivers fraction ${\eta }_j$ of its input power to the common output boundary, thenFor fractions $0.4,0.6$ and efficiencies $0.96,0.94$, the result is $0.948$. Multiplying the two branch efficiencies would describe a different arrangement of power boundaries.

The primary face-gear program reported input-pinion torque division about $48\%/52\%$, while its two idlers shared about $57\%/43\%$. These were measured results with stated calibration limitations, not exact consequences of the layout.^[ref]

A meaningful load-sharing test measures both paths over relevant load and position ranges. Strain gauges need an appropriate measurement calibration because different tooth load distributions can produce different strain responses for the same transmitted load. Floating pinions provide one mechanism for reducing imbalance, but the complete assembly and its constraints still determine the operating split.

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.