--- title: "Compound gear train speed numerical" subject: "Theory of Mechanism and Machine I" chapter: 6 key: "semester_6/theory_of_mechanism_and_machine_i:compound-gear-train-speed-numerical" source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/compound-gear-train-speed-numerical site: BCT Engineering --- # Compound gear train speed numerical Theory of Mechanism and Machine I · Chapter 6: Simple and Planetary gear trains - Asked in: year not given, 6 marks - Video: [Compound Gear Trains](https://www.youtube.com/watch?v=Q_toev7usPk) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/compound-gear-train-speed-numerical - Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-6 ## Question 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. ## 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). ## Answer #### Types of gear trains | Type | Description | Train value | |---|---|---| | Simple | One gear on each shaft; idlers change only the direction | $\pm z_{first}/z_{last}$ (idlers cancel) | | Compound | Two or more gears on one shaft, rotating together | Product of drivers / product of followers | | Reverted | Compound train whose first driver and last driven gears are coaxial | Same as compound; compact | For a reverted train, the centre distances must be equal: $z_A + z_B = z_C + z_D$ when both pairs have the same module (in general, $m_1(z_A + z_B) = m_2(z_C + z_D)$). The train here is not reverted, because $20 + 60 \ne 18 + 54$. #### Formula $$ \text{Train value} = \frac{N_{out}}{N_{in}} = \frac{\text{product of teeth on drivers}}{\text{product of teeth on followers}} = \frac{z_A \times z_C}{z_B \times z_D} $$ #### Calculation Drivers: A (20) and C (18); followers: B (60) and D (54). $$ \frac{N_D}{N_A} = \frac{20 \times 18}{60 \times 54} = \frac{360}{3240} = 0.1111 $$ $$ N_D = 1440 \times 0.1111 = 160\ \text{rpm} $$ Intermediate shaft speed: $N_B = N_C = 1440 \times \dfrac{20}{60} = 480$ 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. ```text 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 $= 160$ rpm in the same direction as the motor; train value $= 0.1111$ (speed reduction 9:1).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a compound gear train, the train value N_out/N_in equals: A. Product of teeth on followers / product of teeth on drivers B. Sum of drivers / sum of followers C. Product of teeth on drivers / product of teeth on followers D. Teeth on last gear / teeth on first gear only Answer: C. Train value is the product of driver teeth divided by the product of follower teeth. (Key point: Train value formula) 2. With A=20, B=60, C=18, D=54 and the motor at 1440 rpm, what is the output speed? A. 480 rpm B. 160 rpm C. 4320 rpm D. 720 rpm Answer: B. N_D = 1440 x (20x18)/(60x54) = 1440 x 0.1111 = 160 rpm. (Key point: Output speed) 3. Why does the output shaft rotate in the same direction as the motor? A. There are two external meshes, each reversing the sense B. Idler gears are present C. Gear C is smaller than B D. The train is reverted Answer: A. Each external mesh reverses rotation; A-B and C-D are two meshes, so the net sense is unchanged. (Key point: Direction of rotation) 4. Why is this train NOT a reverted train? A. Gears B and C are on the same shaft B. It has more than two gears C. The speed ratio is below 1 D. The centre distances differ: 20+60 is not equal to 18+54 Answer: D. A reverted train needs equal centre distances (same module, z_A+z_B = z_C+z_D); 80 is not 72. (Key point: Reverted train condition) ## Related in Theory of Mechanism and Machine I - [Compound planet epicyclic train numerical](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Acompound-planet-gear-train-ring-speed) — Theory of Mechanism and Machine I, chapter 6 - [Assembly condition of planetary gear trains](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aplanetary-gear-assembly-condition-applications) — Theory of Mechanism and Machine I, chapter 6 - [Epicyclic gear train tabular method](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aepicyclic-gear-train-tabular-method) — Theory of Mechanism and Machine I, chapter 6 - [Spur gear pair dimensions numerical](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aspur-gear-pair-geometry-numerical) — Theory of Mechanism and Machine I, chapter 4 - [Straight bevel gear geometry and formative teeth](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Astraight-bevel-gear-geometry-formative-teeth) — Theory of Mechanism and Machine I, chapter 5 - [Tooth force components on a bevel pinion](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Abevel-gear-tooth-forces-numerical) — Theory of Mechanism and Machine I, chapter 8 ## Related in other subjects - [Compound gear train speeds, torques and tooth forces](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Acompound-gear-train-forces) — Machine Design I, chapter 9 - [Gearbox shaft design with spur gear](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Agearbox-shaft-design-asme) — Machine Design II, chapter 5 - [Gear manufacturing and simple indexing](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Agear-milling-indexing-numerical) — Workshop Technology, chapter 11 - [Floating car method: flow, journey and running speed](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Afloating-car-25-5km-flow-speed) — Transportation Engineering II, chapter 1