The compound gear train shown in the figure below consists of compound gear B-C and D-E. All gears are mounted on parallel shafts. The motor shaft rotating at 800 rpm is connected to the gear A and the output shaft to the gear F. The number of teeth on gears A, B, C, D, E and F are 24, 56, 30, 80, 32 and 72 respectively. What is the speed of the gear F ?

Concept:

Compound gear train:

When there is more than one gear on a shaft it is known as compound gear.

Speed ratio:

It is the ratio of the speed of driving gear to that of the driven gear.

\(\frac{N_A}{N_B}=\frac{T_B}{T_A}\); \(\frac{N_C}{N_D}=\frac{T_D}{T_C}\); \(\frac{N_E}{N_F}=\frac{T_F}{T_E}\)

Combining

\(\frac{N_A}{N_B}\times\frac{N_C}{N_D}\times\frac{N_E}{N_F}=\frac{T_B}{T_A}\times\frac{T_D}{T_C}\times\frac{T_F}{T_E}\)

Since B,C and D,E are on same shaft, therefore N_C = N_B and N_E = N_D

\(\frac{N_A}{N_F}=\frac{T_B}{T_A}\times\frac{T_D}{T_C}\times\frac{T_F}{T_E}\)

Calculation:

Given:

T_A = 24, TB = 56, TC = 30, TD = 80, T_E = 32 and TF = 72.

N_A = 800 rpm.

\(\frac{N_A}{N_F}=\frac{T_B}{T_A}\times\frac{T_D}{T_C}\times\frac{T_F}{T_E}\)

Therefore

\(\frac{N_F}{N_A}=\frac{T_A}{T_B}\times\frac{T_C}{T_D}\times\frac{T_E}{T_F}\)

\(\frac{N_F}{800}=\frac{24}{56}\times\frac{30}{80}\times\frac{32}{72}\)

N_F = 57.14 rpm

Shortcut TrickIn Compound gear, ratio of last gear to first gear is:

 \(\rm \dfrac{N_{last\;gear}}{N_{first}}=\dfrac{Product\; of\; teeth\; of \;driving\; gear}{Product\; of\; teeth\; of \;driven\; gear}\)