Ackermann steering geometry calculation

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.

 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.