---
title: "Gear ratios of a four-speed car"
subject: "Machine Design II"
chapter: 5
key: "semester_7/machine_design_ii:automobile-gear-ratio-selection"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/automobile-gear-ratio-selection
site: BCT Engineering
---

# Gear ratios of a four-speed car

Machine Design II · Chapter 5: System Design

- Asked in: year not given, 8 marks
- Video: [Gear Ratios 101: What's better? 3.73 or 4.10? | A Simple Breakdown w/ Driving Examples](https://www.youtube.com/watch?v=a2_zPVC4_Js) — Late Model Restoration (LMR)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/automobile-gear-ratio-selection
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-5

## Question

A car of loaded mass 1500 kg has an engine giving a maximum torque of 120 N m. The final drive ratio is 4.1, the transmission efficiency is 88%, the effective wheel radius is 0.32 m and the rolling resistance coefficient is 0.015. The car must climb a gradient of 30% (tan of slope = 0.3) in first gear; air resistance may be neglected. The top gear is direct (ratio 1) and the gearbox has four forward speeds in geometric progression. Find (a) the first-gear ratio, (b) the second and third gear ratios, and (c) the maximum speed in top gear if the engine can run at 5500 rpm. Also check that the drive wheels can transmit the required tractive force if 55% of the weight is on the driving wheels and the tyre-road friction coefficient is 0.7.

## Short answer

First-gear ratio = 3.28; second = 2.21, third = 1.49 in geometric progression. Maximum speed in top direct gear at 5500 rpm is about 162 km/h. Available tractive force 5426 N exceeds required 4440 N, so wheels do not slip on the 30% slope.

## Answer

Take $g = 9.81\ \text{m/s}^2$, $W = mg = 14715$ N.

#### (a) First-gear ratio
Slope angle $\theta = \tan^{-1}(0.3) = 16.70^\circ$; $\sin\theta = 0.2873$, $\cos\theta = 0.9578$.

Tractive force needed (gradient plus rolling resistance):

$$
\begin{aligned} F &= W(\sin\theta + f\cos\theta) \\ &= 14715\,(0.2873 + 0.015\times0.9578) = 4440\ \text{N} \end{aligned}
$$

Torque at the wheels: $T_w = F r = 4440 \times 0.32 = 1421\ \text{N m}$.

Overall ratio from engine to wheels:

$$
G_o = \frac{T_w}{T_e\,\eta} = \frac{1421}{120\times0.88} = 13.45
$$

First-gear ratio of the gearbox:

$$
G_1 = \frac{G_o}{G_f} = \frac{13.45}{4.1} = 3.28
$$

#### (b) Intermediate ratios (GP)
For four speeds, $G_1, G_2, G_3, G_4 = G_1,\ G_1/k,\ G_1/k^2,\ G_1/k^3$ with $G_4 = 1$:

$$
k = (3.28)^{1/3} = 1.486
$$

| Gear | Gearbox ratio | Overall ratio (x 4.1) |
|---|---|---|
| 1 | 3.28 | 13.45 |
| 2 | 2.21 | 9.05 |
| 3 | 1.49 | 6.09 |
| 4 | 1.00 | 4.10 |

#### (c) Maximum speed in top gear
Engine speed $\omega_e = 5500\times2\pi/60 = 575.96$ rad/s.

$$
v = \frac{\omega_e\, r}{G_4 G_f} = \frac{575.96\times0.32}{1\times4.1} = 44.95\ \text{m/s} = 161.8\ \text{km/h}
$$

#### Traction check
Maximum tractive force available $= \mu\,(0.55\,W\cos\theta) = 0.7\times0.55\times14715\times0.9578 = 5426$ N.

Since $5426 > 4440$ N, the wheels do not slip on the 30% slope.

**Answer:** $G_1 = 3.28$; $G_2 = 2.21$, $G_3 = 1.49$; top speed about 162 km/h; traction is sufficient (5426 N available against 4440 N required).

## Check yourself

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

1. How is the first-gear ratio found from the tractive force needed on the 30% slope?
   A. G1 = 13.45, the overall ratio itself
   B. G1 = 2.89, ignoring efficiency
   C. G1 = 3.28, from wheel torque 1421 N m divided by (Te x efficiency) = 13.45, then divided by the final drive 4.1
   D. G1 = 2.54, multiplying by efficiency
   Answer: C. F = W(sin theta + f cos theta) = 4440 N, Tw = F r = 1421 N m, Go = 1421/(120 x 0.88) = 13.45 and G1 = 13.45/4.1 = 3.28. (Key point: First gear ratio)

2. With G4 = 1 and G1 = 3.28 in GP over four speeds, what is the second-gear ratio?
   A. 2.21, using k = 3.28^(1/3) = 1.486
   B. 1.64, halving G1
   C. 2.64
   D. 1.49
   Answer: A. G2 = G1/k with k = 1.486, so G2 = 2.21 and G3 = 1.49. (Key point: Geometric progression of gears)

3. What is the maximum speed in top gear at 5500 rpm, wheel radius 0.32 m and final drive 4.1?
   A. About 45 km/h
   B. About 98 km/h
   C. About 233 km/h
   D. About 162 km/h, from v = omega_e r/(G4 Gf) = 44.95 m/s
   Answer: D. v = 575.96 x 0.32/4.1 = 44.95 m/s = 161.8 km/h. The value 45 would forget the unit conversion. (Key point: Top speed)

4. Why do the drive wheels not slip on the slope?
   A. The available force mu x 0.55 W cos(theta) = 5426 N exceeds the required 4440 N
   B. The required force is larger than the available force
   C. Rolling resistance is zero
   D. Air resistance is included
   Answer: A. Only the weight on the driving wheels counts for traction, and 5426 N is more than 4440 N. (Key point: Traction check)

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