= 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,
$$
\frac{\omega_4}{\omega_1}=\left(-\frac{N_1}{N_2}\right)\left(-\frac{N_3}{N_4}\right),\qquad
i=\frac{N_2N_4}{N_1N_3}.
$$
The two external meshes give equal input/output rotation directions.<KHK Gear Nomenclature>[Basic Guide B, §1.1, printed p.5, Figure1.3 and equation(1.2)]{ref}
For an original example, use tooth counts $N_1=18$, $N_2=54$, $N_3=12$ and $N_4=60$. The reduction is $3\times5=15$. 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:
$$
\frac m2(N_1+N_2)=\frac m2(N_3+N_4).
$$
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>.
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