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Compound gear train speed numerical

Short answer

Simple trains have one gear per shaft with idlers reversing direction; compound trains have multiple gears on intermediate shafts for larger reduction; reverted trains are compound with input and output coaxial. For train A(20)-B(60)/C(18)-D(54): output speed 160 rpm in same direction as motor, train value 0.1111 (9:1 reduction).

Differentiate between simple, compound and reverted gear trains. In a compound train, a motor shaft running at 1440 rpm carries gear A (20 teeth) which meshes with gear B (60 teeth). Gear C (18 teeth) is on the same shaft as B and meshes with gear D (54 teeth) on the output shaft. Find the speed and direction of the output shaft relative to the motor shaft and the train value.

Types of gear trains

TypeDescriptionTrain value
SimpleOne gear on each shaft; idlers change only the direction±zfirst/zlast\pm z_{first}/z_{last} (idlers cancel)
CompoundTwo or more gears on one shaft, rotating togetherProduct of drivers / product of followers
RevertedCompound train whose first driver and last driven gears are coaxialSame as compound; compact

For a reverted train, the centre distances must be equal: zA+zB=zC+zDz_A + z_B = z_C + z_D when both pairs have the same module (in general, m1(zA+zB)=m2(zC+zD)m_1(z_A + z_B) = m_2(z_C + z_D)). The train here is not reverted, because 20+60≠18+5420 + 60 \ne 18 + 54.

Formula

Train value=NoutNin=product of teeth on driversproduct of teeth on followers=zA×zCzB×zD\text{Train value} = \frac{N_{out}}{N_{in}} = \frac{\text{product of teeth on drivers}}{\text{product of teeth on followers}} = \frac{z_A \times z_C}{z_B \times z_D}

Calculation

Drivers: A (20) and C (18); followers: B (60) and D (54).

NDNA=20×1860×54=3603240=0.1111\frac{N_D}{N_A} = \frac{20 \times 18}{60 \times 54} = \frac{360}{3240} = 0.1111 ND=1440×0.1111=160 rpmN_D = 1440 \times 0.1111 = 160\ \text{rpm}

Intermediate shaft speed: NB=NC=1440×2060=480N_B = N_C = 1440 \times \dfrac{20}{60} = 480 rpm.

Direction

Each external mesh reverses the sense of rotation. There are two external meshes (A-B and C-D), so the output rotates in the same direction as the motor shaft.

   motor --> A(20) o--o B(60)=C(18) o--o D(54) --> output
   1440 rpm            480 rpm            160 rpm
   (cw)    (ccw)            (ccw)   (cw)

Answer: output speed =160= 160 rpm in the same direction as the motor; train value =0.1111= 0.1111 (speed reduction 9:1).

✓ Answer checked by Claude Opus 5.5 (high effort)

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