A gear train with five gears is shown in the figure below. The number of teeth on each gear is \(N_1\), \(N_2\), \(N_3\), \(N_4\), and \(N_5\). The idler gear in this gear train is
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Write the overall speed ratio from gear 1 to gear 5 and see which gear's tooth count cancels out.
An idler gear sits in a train only to change the direction of rotation or bridge a gap between shafts, without changing the speed ratio between the first and last gear. Its own tooth count always cancels out of the overall train value.
Gear 4: this gear meshes with gear 3 on one side and gear 5 on the other, but shares no shaft with any other gear. The train value $\omega_1/\omega_5 = (N_2/N_1)(N_4/N_3)(N_5/N_4)$ has $N_4$ once on top and once on the bottom, so it cancels.
Gear 2: part of the compound pair 2-3 fixed on one shaft, and $N_2$ stays in the final ratio, so it changes the output speed.
Gear 3: also part of the compound pair, and $N_3$ stays in the denominator of the final ratio, so it is not an idler.
Gear 1: this is the driver gear itself, not an idler by definition.
Only gear 4 drops out of the speed ratio algebra, so gear 4 is the idler gear.
