---
title: "Tractive effort and speed from gear ratios"
subject: "Automotive Technology"
chapter: 5
key: "semester_5/automotive_technology:gear-ratio-tractive-effort-numerical"
source: https://www.bctengineer.com/q/semester_5/automotive_technology/gear-ratio-tractive-effort-numerical
site: BCT Engineering
---

# Tractive effort and speed from gear ratios

Automotive Technology · Chapter 5: Transmission Systems

- Asked in: year not given, 8 marks
- Video: [Automobile Design Calculation - Tractive Force](https://www.youtube.com/watch?v=DIkZBjiZ_Gs) — Nelson Winfred
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/automotive_technology/gear-ratio-tractive-effort-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/automotive_technology/chapter-5

## Question

A car of mass 1300 kg has an engine producing a maximum torque of 150 N m. The first gear ratio is 3.8:1, the top gear ratio is 0.8:1, the final drive ratio is 4.1:1, transmission efficiency is 90 %, and the loaded tyre radius is 0.30 m. Calculate (i) the torque at the driving wheels in first gear, (ii) the tractive force, (iii) the initial acceleration in first gear if the rolling resistance coefficient is 0.015 (neglect air drag and rotational inertia), and (iv) the vehicle speed at 3000 rpm in first and in top gear.

## Short answer

Wheel torque = engine torque × gear ratios × efficiency = 2103 N m. Tractive force = wheel torque / tyre radius = 7011 N. Initial acceleration = (tractive force – rolling resistance) / mass = 5.25 m/s². Speed at 3000 rpm: 21.8 km/h in first gear, 103.4 km/h in top gear.

## Answer

**Given:** $m = 1300$ kg, $T_e = 150$ N m, $G_1 = 3.8$, $G_4 = 0.8$, $G_f = 4.1$, $\eta = 0.9$, $r = 0.30$ m, $f = 0.015$.

#### (i) Wheel torque in first gear

$$
T_w = T_e\, G_1 G_f\, \eta = 150 \times 3.8 \times 4.1 \times 0.9 = 2103.3\ \text{N m}
$$

#### (ii) Tractive force

$$
F_t = \frac{T_w}{r} = \frac{2103.3}{0.30} = 7011\ \text{N}
$$

#### (iii) Initial acceleration

Rolling resistance:

$$
F_{rr} = f\, m g = 0.015 \times 1300 \times 9.81 = 191.3\ \text{N}
$$

$$
a = \frac{F_t - F_{rr}}{m} = \frac{7011 - 191.3}{1300} = 5.25\ \text{m/s}^2
$$

This is the engine-limited value. The real value is smaller because of air drag and rotating inertia, and it is also limited by tyre grip: with $\mu \approx 0.8$ and about 60 % of the weight on the driven axle, the grip limit is about $0.8 \times 0.6 \times 9.81 \approx 4.7$ m/s², so the driven wheels would spin at full torque.

#### (iv) Vehicle speed at 3000 rpm

$$
v = \frac{2\pi N r}{60\, G\, G_f}
$$

First gear:

$$
v_1 = \frac{2\pi \times 3000 \times 0.30}{60 \times 3.8 \times 4.1} = 6.05\ \text{m/s} = 21.8\ \text{km/h}
$$

Top gear:

$$
v_4 = \frac{2\pi \times 3000 \times 0.30}{60 \times 0.8 \times 4.1} = 28.7\ \text{m/s} = 103.4\ \text{km/h}
$$

**Answer:** $T_w = 2103$ N m; $F_t = 7011$ N; $a = 5.25$ m/s²; $v = 21.8$ km/h (first) and $103.4$ km/h (top) at 3000 rpm.

## Check yourself

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

1. Which formula gives wheel torque in first gear?
   A. Te x G1 x Gf x efficiency = 2103 N m
   B. Te / (G1 x Gf) = 9.6 N m
   C. Te x G1 = 570 N m
   D. Te x Gf x r
   Answer: A. 150 x 3.8 x 4.1 x 0.9 = 2103.3 N m. (Key point: Wheel torque)

2. What is the tractive force?
   A. T_w / r = 7011 N
   B. T_w x r = 631 N
   C. T_w / m
   D. T_w / g
   Answer: A. Divide wheel torque by tyre radius: 2103.3/0.30 = 7011 N. (Key point: Tractive force)

3. How is acceleration found, ignoring drag and inertia?
   A. a = (Ft - f m g)/m = 5.25 m/s^2
   B. a = Ft/m = 5.39 m/s^2 only
   C. a = Ft x m
   D. a = f g
   Answer: A. Rolling resistance is 0.015 x 1300 x 9.81 = 191.3 N, so a = (7011 - 191.3)/1300 = 5.25 m/s^2. (Key point: Initial acceleration)

4. What is the speed at 3000 rpm in first gear and in top gear?
   A. 21.8 km/h and 103.4 km/h
   B. 103.4 km/h and 21.8 km/h
   C. 6.05 km/h and 28.7 km/h
   D. 21.8 km/h and 28.7 km/h
   Answer: A. v = 2 pi N r/(60 G Gf) gives 6.05 m/s (21.8 km/h) in first and 28.7 m/s (103.4 km/h) in top. (Key point: Vehicle speed)

## Related in Automotive Technology

- [Engine displacement, compression ratio and power](https://www.bctengineer.com/question.md?key=semester_5%2Fautomotive_technology%3Aengine-displacement-bmep-numerical) — Automotive Technology, chapter 1
- [Automatic transmission and torque converter](https://www.bctengineer.com/question.md?key=semester_5%2Fautomotive_technology%3Aautomatic-transmission-torque-converter) — Automotive Technology, chapter 5
- [Air flow and fuel flow of an SI engine](https://www.bctengineer.com/question.md?key=semester_5%2Fautomotive_technology%3Afuel-consumption-air-flow-numerical) — Automotive Technology, chapter 2

## Related in other subjects

- [Gear ratios of a four-speed car](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aautomobile-gear-ratio-selection) — Machine Design II, chapter 5
- [8-pole motor: speed at max torque (50 Kg-m)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-8pole-50kgm-starting-torque) — Electrical Machines, chapter 5, asked 4×