# Optimal Dimensional Synthesis of Ackermann Steering Mechanisms for Three-Axle, Six-Wheeled Vehicles
## 2 Position Analysis of Ackermann Steering Mechanism
### Planar Four-Bar Steering Mechanism  
The Ackermann steering mechanism or trapezoidal steering mechanism, as shown in Figure 3, is a symmetric four-bar linkage with equal lengths of the input link and the output link, \(r_{2}=r_{4}\), while the length of the connecting link \(r_{3}\) is smaller or larger than the length of the fixed link \(r_{1}\). If \(r_{3}\) smaller than \(r_{1}\), it isarranged as a trailing Ackermann steering mechanism, otherwise, if \(r_{3}\) larger than \(r_{1}\), it is arranged as a leading Ackermann steering mechanism [7; 13].  
In Figure 3, solid lines represent the trapezoidal steering mechanism during straight travel, while red dotted lines show it during a left turn. The vector loop equation for the four-bar mechanism is as follows:  
\[\overset{\rightharpoonup}{r}_{3}=\overset{\rightharpoonup}{r}_{1}+\overset{ \rightharpoonup}{r}_{4}-\overset{\rightharpoonup}{r}_{2} \tag{5}\]  
Take the inner product itself on each side of Equation (5), as follows:  
\[\overset{\rightharpoonup}{r}_{3}.\overset{\rightharpoonup}{r}_{3}=\left( \overset{\rightharpoonup}{r}_{1}+\overset{\rightharpoonup}{r}_{4}- \overset{\rightharpoonup}{r}_{2}\right).\left(\overset{\rightharpoonup}{r}_{1} +\overset{\rightharpoonup}{r}_{4}-\overset{\rightharpoonup}{r}_{2}\right) \tag{6}\]  
Expanding the inner product as follows:  
\[\begin{array}{c}(2r_{1}r_{4}cos\theta_{1}-2r_{2}r_{4}cos\theta_{2})cos \theta_{4}+(2r_{1}r_{4}sin\theta_{1}-2r_{2}r_{4}sin\theta_{2})sin\theta_{4}=\\ r_{3}^{2}-r_{1}^{2}-r_{2}^{2}-r_{4}^{2}+2r_{1}r_{2}cos(\theta_{2}-\theta_{1}) \end{array} \tag{7}\]  
Which can be briefly expressed as follows:  
\[Acos\theta_{4}+Bsin\theta_{4}=C \tag{8}\]  
Figure 3: Ackerman trapezoidal steering mechanism.  
Figure 2: Trapezoidal steering mechanism when the vehicle is turning left.  
Since \(\theta_{1}=0^{\circ}\), the coefficients can be expressed as follows:  
\[\begin{array}{l}\mathrm{A}=2r_{4}(r_{1}-r_{2}cos\theta_{2})\\ \mathrm{B}=-2r_{2}r_{4}sin\theta_{2}\\ \mathrm{C}= r_{3}^{2}-r_{1}^{2}-r_{2}^{2}-r_{4}^{2}+2r_{1}r_{2}cos\theta_{2}\\ \end{array} \tag{9}\]  
From the half-angle formula of the trigonometric function, the angular position of the output link of the trapezoidal steering mechanism can be expressed as follows:  
\[\theta_{4}=2\ tan^{-1}(\frac{B\pm\sqrt{A^{2}+B^{2}-C^{2}}}{A+C}) \tag{10}\]  
According to the assembly of the steering mechanism, the symbol before the radical sign in Equation (10) is taken positive sign. The angular position of the coupler link is as follows:  
\[\theta_{3}=tan^{-1}(\frac{r_{4}sin\theta_{4}-r_{2}sin\theta_{2}}{r_{1}+r_{4} cos\theta_{4}-r_{2}cos\theta_{2}}) \tag{11}\]  
where  
\[\theta_{2}=\theta_{2o}-\delta_{in} \tag{12}\]  
Since \(\theta_{4o}=\pi-\theta_{2o}\), the steering angle of the outer wheel \(\delta_{out}\) can be expressed as follows:  
\[\delta_{out}=\theta_{4o}-\theta_{4}=\pi-\theta_{2o}-\theta_{4} \tag{13}\]  
In Equations (12) and (13), \(\theta_{2}\), \(\theta_{4}\) are the angular position of the input and output link, respectively, and \(\theta_{2o}\), \(\theta_{4o}\) are the initial input and output angle, respectively, when the vehicle travels straight (as shown in Figure 1).  
As shown in Figure 3, the transmission angle, \(\beta\), of the steering mechanism can be expressed as follows:  
\[\beta=cos^{-1}(\frac{r_{3}^{2}+r_{4}^{2}-r_{1}^{2}-r_{2}^{2}+2r_{1}r_{2}cos \theta_{2}}{2r_{3}r_{4}}) \tag{14}\]  
The transmission angle of a four-bar linkage is a transmission index for transmission efficiency. If the deviation of the transmission angle from 90 degrees, \(|\beta-90|\), is lower than 50 degrees then the transmission efficiency is good and satisfactory [40].