---
title: "Ackermann steering geometry calculation"
subject: "Automotive Technology"
chapter: 6
key: "semester_5/automotive_technology:ackermann-steering-numerical"
source: https://www.bctengineer.com/q/semester_5/automotive_technology/ackermann-steering-numerical
site: BCT Engineering
---

# Ackermann steering geometry calculation

Automotive Technology · Chapter 6: Suspension and Steering

- Asked in: year not given, 5 marks
- Video: [Ackerman Steering Geometry and Anti Ackerman](https://www.youtube.com/watch?v=YVisLuiU-Oo) — XF Motorsports
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/automotive_technology/ackermann-steering-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/automotive_technology/chapter-6

## Question

A car has a wheelbase of 2.7 m and a distance of 1.4 m between the king-pin (pivot) centres. State the Ackermann condition for correct steering. When the inner front wheel is turned through 20 degrees, find the angle of the outer front wheel for true rolling, and the turning radii of the inner and outer front wheels (measured at the wheel centre line to the instantaneous centre).

## Short answer

Ackermann condition: cot(outer) – cot(inner) = c/L. For inner 20°, wheelbase 2.7m, pivot distance 1.4m, outer angle = 17.02°. Turning radii: inner = L/sin(20°) = 7.89 m; outer = L/sin(17.02°) = 9.22 m.

## Answer

#### Condition for correct steering

For pure rolling (no tyre scrub) during a turn, the axes of all four wheels must meet at one point on the extended line of the rear axle (the instantaneous centre). For this the inner wheel turns more than the outer wheel and, with inner angle $\theta$ and outer angle $\phi$,

$$
\cot\phi - \cot\theta = \frac{c}{L}
$$

where $c$ is the distance between the pivots and $L$ the wheelbase. The Ackermann linkage (trapezium) approximates this.

```text
 Instantaneous centre O ---------- rear axle line
        |   \               |
        |    \  R_in        |  L
        |      \ theta      |
     inner wheel         outer wheel (phi)
        |<-------- c ------->|
```

#### Numerical

**Given:** $L = 2.7$ m, $c = 1.4$ m, $\theta = 20^\circ$.

$$
\begin{aligned} \cot 20^\circ &= 2.7475 \\ \frac{c}{L} &= \frac{1.4}{2.7} = 0.5185 \\ \cot\phi &= 2.7475 + 0.5185 = 3.2660 \\ \phi &= \tan^{-1}\left(\frac{1}{3.266}\right) = 17.02^\circ \end{aligned}
$$

Turning radii (wheel centre-line to instantaneous centre, $R = L/\sin$ of wheel angle):

$$
R_{in} = \frac{L}{\sin\theta} = \frac{2.7}{\sin 20^\circ} = 7.89\ \text{m}, \qquad R_{out} = \frac{L}{\sin\phi} = \frac{2.7}{\sin 17.02^\circ} = 9.22\ \text{m}
$$

Check: the difference $R_{out} - R_{in}$ projected along the axle is $9.22\cos 17.02^\circ - 7.89\cos 20^\circ = 8.82 - 7.42 = 1.40$ m $= c$. Correct.

**Answer:** outer wheel angle $= 17.02^\circ$; $R_{in} = 7.89$ m; $R_{out} = 9.22$ m.

## Check yourself

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

1. What must be true of the wheel axes for pure rolling (no scrub) during a turn?
   A. Only the inner wheel axis passes through the rear axle
   B. Both front wheels turn through the same angle
   C. All four wheel axes meet at one point on the extended rear axle line
   D. The front wheel axes stay parallel to each other
   Answer: C. Pure rolling needs all wheel axes to meet at the instantaneous centre on the rear axle line, so the inner wheel turns more than the outer. (Key point: Ackermann condition)

2. Which relation gives the outer angle phi for inner angle theta?
   A. tan(phi) - tan(theta) = c/L
   B. sin(phi) = L/R
   C. cot(phi) - cot(theta) = c/L
   D. cot(theta) - cot(phi) = c/L
   Answer: C. The condition is cot(phi) - cot(theta) = c/L, with c the pivot distance and L the wheelbase. (Key point: Ackermann equation)

3. With L = 2.7 m, c = 1.4 m and theta = 20 degrees, what is the outer wheel angle?
   A. 12.5 degrees
   B. 23.2 degrees
   C. 17.02 degrees
   D. 20.0 degrees
   Answer: C. cot(phi) = 2.7475 + 0.5185 = 3.266, so phi = 17.02 degrees, smaller than the inner angle. (Key point: Outer wheel angle)

4. How is the turning radius of a wheel centre line obtained and what are the results?
   A. R = L sin(angle); 0.92 m and 0.79 m
   B. R = c/sin(angle); 4.09 m and 4.79 m
   C. R = L/sin(angle); R_in = 7.89 m and R_out = 9.22 m
   D. R = L/tan(angle) ; both 7.42 m
   Answer: C. R = L/sin of the wheel angle: 2.7/sin20 = 7.89 m for the inner and 2.7/sin17.02 = 9.22 m for the outer wheel. (Key point: Turning radii)

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