---
title: "Compound planet epicyclic train numerical"
subject: "Theory of Mechanism and Machine I"
chapter: 6
key: "semester_6/theory_of_mechanism_and_machine_i:compound-planet-gear-train-ring-speed"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/compound-planet-gear-train-ring-speed
site: BCT Engineering
---

# Compound planet epicyclic train numerical

Theory of Mechanism and Machine I · Chapter 6: Simple and Planetary gear trains

- Asked in: year not given, 8 marks
- Video: [Compound Sun and Planet Epicyclic Gear Train - Problem](https://www.youtube.com/watch?v=2jKdBA6J9dg) — Mohith Dadu
- 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-planet-gear-train-ring-speed
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-6

## Question

In an epicyclic gear train, a sun gear S of 20 teeth is fixed to the frame. It meshes with planet P1 of 40 teeth. Planet P2 of 30 teeth is compounded with (keyed on the same shaft as) P1 and meshes with the internal teeth of a ring gear R having 90 teeth. P1 and P2 are carried on an arm that rotates at 90 rpm clockwise about the axis of S. Find the speed of the ring gear R and the speed of the compound planet. Verify that the gears can be assembled with the same module.

## Short answer

With arm speed +90 rpm and sun fixed, the tabular method gives ring gear speed 105 rpm clockwise and compound planet speed 135 rpm clockwise.

## Answer

#### Geometry check (same module $m$)
Centre distance of S-P1: $\dfrac{m(20 + 40)}{2} = 30m$.

Centre distance of P2-R (internal mesh): $\dfrac{m(90 - 30)}{2} = 30m$.

Both are equal, so the compound planet P1-P2 can mesh with both S and R at the same arm radius.

```text
        R (ring, 90T, internal)
       /  \
   P2(30)=P1(40)  <- carried on arm
        \
         S (sun, 20T, fixed)
```

#### Tabular method (clockwise positive, arm speed $y = +90$ rpm)
With the arm fixed and S turning $+x$:
- P1 (external mesh with S): $-\dfrac{20}{40}x = -0.50x$.
- P2 turns with P1: $-0.50x$.
- R (internal mesh with P2, same sense as P2): $-0.50x \times \dfrac{30}{90} = -0.1667x$.

| Step | Arm | S (20) | P1-P2 (40, 30) | R (90) |
|---|---|---|---|---|
| 1. Arm fixed, S turns $+x$ | 0 | $+x$ | $-0.50x$ | $-0.1667x$ |
| 2. Add $+y$ to all | $+y$ | $y + x$ | $y - 0.5x$ | $y - 0.1667x$ |

#### Sun fixed
$$
y + x = 0 \Rightarrow x = -y = -90\ \text{rpm}
$$

Ring gear:

$$
N_R = y - 0.1667x = 90 - 0.1667(-90) = 105\ \text{rpm}
$$

Compound planet:

$$
N_P = y - 0.5x = 90 - 0.5(-90) = 135\ \text{rpm}
$$

The ring gear turns in the same direction as the arm, faster than the arm by the ratio $N_R/N_{arm} = 1.1667$. The planet spins about its own axis relative to the arm at $N_P - y = 45$ rpm in the same sense.

**Answer: ring gear $= 105$ rpm clockwise; compound planet $= 135$ rpm clockwise.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why can the compound planet mesh with both S and R at the same arm radius?
   A. Both centre distances equal 30m: m(20+40)/2 and m(90-30)/2
   B. Both gears have the same number of teeth
   C. The ring and sun have equal diameters
   D. The planet has no module
   Answer: A. S-P1 external centre distance is 30m and P2-R internal centre distance is also 30m, so the arm radius is the same. (Key point: Geometry (same module))

2. With the arm fixed and S turning +x, how does the ring R turn?
   A. -0.50x
   B. +0.1667x
   C. -0.1667x
   D. +x
   Answer: C. P1-P2 turn at -0.5x; the internal mesh with P2 gives R = -0.5x x 30/90 = -0.1667x. (Key point: Gear ratio with arm fixed)

3. With the sun fixed and the arm at 90 rpm, what value does x take?
   A. x = +90
   B. x = 0
   C. x = -45
   D. x = -90
   Answer: D. The sun column y + x = 0 gives x = -y = -90. (Key point: Sun fixed condition)

4. What are the ring and compound planet speeds?
   A. Ring 135 rpm, planet 105 rpm
   B. Ring 105 rpm, planet 135 rpm
   C. Ring 90 rpm, planet 135 rpm
   D. Ring 45 rpm, planet 105 rpm
   Answer: B. N_R = 90 + 0.1667x90 = 105 rpm; N_P = 90 + 0.5x90 = 135 rpm. (Key point: Ring speed result)

## Related in Theory of Mechanism and Machine I

- [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
- [Compound gear train speed numerical](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Acompound-gear-train-speed-numerical) — Theory of Mechanism and Machine I, chapter 6
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- [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

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