Compound gear train

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 reverted gear train returns the output rotation axis to the input axis, usually through a two-stage compound gear train. The input and output gears can rotate independently on coaxial shafts, while the intermediate gears share another rotating body. Coaxial axes are a layout condition in addition to the desired gear ratio.

For standard unshifted spur gears, let gears1 and2 mesh in the first stage and gears3 and4 in the second, with2 and3 rigidly connected. The common intermediate-axis separation requiresIf both stages use the same gear module, this reduces to $N_1+N_2=N_3+N_4$. The reduction ratio independently satisfies $i=(N_2/N_1)(N_4/N_3)$. Both conditions must be satisfied; multiplying arbitrary stage ratios does not ensure a reverted layout.^[ref]

For an original design example, choose counts20,60,16 and64, with module $2\mathrm{mm}$ in both stages. Both sums are80, so both separations are $80\mathrm{mm}$. The reduction is $3\times 4=12$, taking an input of1440 revolutions per minute to an output of120. Two external meshes give the same input/output rotation direction.

The gears occupy different axial planes so that coaxial input and output do not require them to be one rigidly joined gear. A practical design still needs independent shaft support, compatible tooth geometry in each stage and adequate clearances. Operating with shifted tooth geometry or another mounting changes the applicable separation equations; the simple tooth-sum condition belongs to the stated standard-gear model.