---
title: "Epicyclic gear train tabular method"
subject: "Theory of Mechanism and Machine I"
chapter: 6
key: "semester_6/theory_of_mechanism_and_machine_i:epicyclic-gear-train-tabular-method"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/epicyclic-gear-train-tabular-method
site: BCT Engineering
---

# Epicyclic gear train tabular method

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

- Asked in: year not given, 8 marks
- Video: [Analysis of epicyclic gear train: Tabular Method](https://www.youtube.com/watch?v=5ZslMfMKcIo) — Dhananjay Bhoge
- 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/epicyclic-gear-train-tabular-method
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-6

## Question

In an epicyclic gear train, the sun gear has 20 teeth, the planet gear has 30 teeth and the annulus (internal ring) has 80 teeth. The planet is carried on an arm which rotates about the sun axis at 100 rpm clockwise. Using the tabular method, find the speed of (a) the sun gear and the planet when the annulus is fixed, and (b) the annulus and the planet when the sun gear is fixed.

## Short answer

Using the tabular method with arm speed +100 rpm: (a) With annulus fixed, sun = 500 rpm cw, planet = 166.67 rpm ccw. (b) With sun fixed, annulus = 125 rpm cw, planet = 166.67 rpm cw.

## Answer

#### Tabular method
Step 1: Fix the arm and give the sun gear $+x$ revolutions (clockwise is taken as positive in this problem).
Step 2: Find the revolutions of the other gears from the teeth ratios.
Step 3: Give $+y$ revolutions to the whole train (arm included); add the rows to get the actual motions.
Step 4: Use the fixed-gear condition and the arm speed to find $x$ and $y$.

Check geometry: $z_{annulus} = z_{sun} + 2z_{planet} = 20 + 60 = 80$, consistent.

With clockwise positive, the arm speed is $y = +100$ rpm.

| Step | Arm | Sun (20) | Planet (30) | Annulus (80) |
|---|---|---|---|---|
| 1. Arm fixed, sun turns $+x$ | 0 | $+x$ | $-\dfrac{20}{30}x = -0.6667x$ | $-\dfrac{20}{80}x = -0.25x$ |
| 2. Add $+y$ to all | $+y$ | $y + x$ | $y - 0.6667x$ | $y - 0.25x$ |

(The planet turns opposite to the sun in an external mesh. The annulus turns the same way as the planet in the internal mesh, so its sign is opposite to the sun's.)

#### (a) Annulus fixed
$$
y - 0.25x = 0 \Rightarrow x = 4y = 400\ \text{rpm}
$$

$$
N_{sun} = y + x = 100 + 400 = 500\ \text{rpm}
$$

$$
N_{planet} = y - 0.6667x = 100 - 0.6667(400) = -166.67\ \text{rpm}
$$

The sun turns at 500 rpm clockwise (same direction as the arm; a 5:1 speed increase). The planet turns at 166.67 rpm anticlockwise (negative means opposite to the arm).

#### (b) Sun fixed
$$
y + x = 0 \Rightarrow x = -y = -100\ \text{rpm}
$$

$$
N_{annulus} = y - 0.25x = 100 - 0.25(-100) = 125\ \text{rpm}
$$

$$
N_{planet} = y - 0.6667x = 100 - 0.6667(-100) = 166.67\ \text{rpm}
$$

The annulus turns at 125 rpm clockwise and the planet at 166.67 rpm clockwise.

**Answer: (a) sun $= 500$ rpm cw, planet $= 166.67$ rpm ccw; (b) annulus $= 125$ rpm cw, planet $= 166.67$ rpm cw.**

## Check yourself

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

1. In the tabular method, what is done in the first step?
   A. The annulus is fixed and the arm rotates
   B. The arm is fixed and the sun is given +x revolutions, finding the other gears from teeth ratios
   C. All gears are locked and given +y revolutions
   D. The planet is given zero speed
   Answer: B. Step 1 holds the arm fixed and turns the sun by x; the others follow from the teeth ratios. Then +y is added to all. (Key point: Tabular method steps)

2. With the arm fixed and the sun turning +x, the planet (30 teeth) turns:
   A. +x
   B. -0.25x
   C. +0.6667x
   D. -0.6667x
   Answer: D. An external mesh reverses direction: planet = -(20/30)x = -0.6667x. (Key point: Planet motion relative to sun)

3. With the annulus fixed and arm at 100 rpm, what is the sun speed?
   A. 400 rpm
   B. 100 rpm
   C. 500 rpm
   D. 125 rpm
   Answer: C. y - 0.25x = 0 gives x = 400, so N_sun = y + x = 500 rpm. (Key point: Annulus fixed)

4. With the sun fixed and arm at 100 rpm, what is the annulus speed?
   A. 125 rpm
   B. 75 rpm
   C. 166.67 rpm
   D. 100 rpm
   Answer: A. y + x = 0 gives x = -100; N_annulus = 100 - 0.25(-100) = 125 rpm. (Key point: Sun fixed)

## 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
- [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

## 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