# Lateral Stability of a Mobile Robot Utilizing an Active Adjustable Suspension
## 3 Active Suspension's Model
### Parameters Design  
In this section, we design the parameters of this active suspension. The aim of this section is to achieve the length ranges of the linear actuator when the structural parameters are given. As shown in Figure 4, the active suspension is symmetric. It comprises quadrangle \(\mathrm{E_{1}E_{2}B_{2}B_{1}}\), symmetric triangle EBD, and symmetric quadrangle ACBD. The parameters are defined as follows: \(\angle\mathrm{E_{2}B_{2}B_{1}}=\beta_{1}\), \(\angle\mathrm{EBD}=\beta_{2}\), \(\angle\mathrm{C_{2}B_{2}B_{1}}=\beta_{3}\), \(\angle\mathrm{ACO}=\beta_{4}\), \(\angle\mathrm{DBC}=\alpha_{1}\), and \(\angle\mathrm{ACB}=\pi-\alpha_{3}\). The length of BD, AD, AC, BC, \(\mathrm{C_{1}C_{2}}\), \(\mathrm{B_{1}B_{2}}\), and EB are denoted by \(l_{1}\), \(l_{2}\), \(l_{3}\), \(l_{4}\), \(l_{5}\), \(l_{6}\), and \(l_{7}\), respectively.  
Figure 3: Mobile robot reconfigurations: (**a**) Low ground clearance; (**b**) High ground clearance; (**c**) Leveled configuration.  
Among the overall design of the mobile robot, ground clearance is an important index that can affect the mobility of this mobile robot. Thus, the input of the active suspension (length of the linear actuator) can be determined with the desired lowest and highest ground clearances. The min and max limited positions of the length of the linear actuator (\(L_{emin}\) and \(L_{emax}\)) correspond to the min and max ground clearances (\(H_{Cmin}\) and \(H_{Cmax}\)), respectively. As shown in Figure 4, the relationship between \(L_{e}\) and \(H_{C}\) is shown below:  
\[H_{C}=L_{0}*\cos(\
icefrac{{\alpha}}{{2}})+R+H_{3} \tag{1}\]  
where \(L_{0}\) is the leg length, \(\alpha\) is the two legs, including the angle, and \(H_{3}\) is the vertical displacement between the ground clearance and the hinged point of the leg.  
1. Solving \(\alpha\): Solving Equation (1) leads to the following equation: \[\alpha(H_{C})=2*\arccos\left(\frac{H_{C}-R-H_{3}}{L_{0}}\right).\] (2)
2. Solving \(\alpha_{3}\): As shown in Figure 4, the relationship between \(\alpha_{3}\) and \(\alpha\) can be written as follows: \[\alpha_{3}(H_{C})=\
icefrac{{\alpha(H_{C})}}{{2}}-\beta_{3}+\beta_{4}+ \
icefrac{{\pi}}{{2}}.\] (3)
3. Solving \(\alpha_{1}\): As shown in Figure 4c, \(ACBD\) is a planar four-bar linkage, \(\alpha_{1}\) and \(\alpha_{3}\) are the input and output rotation angles, thus:  
Figure 4: Parameter model of active suspension: (**a**) Whole; (**b**) Upper part; (**c**) Four-bar linkage.  
\[\mathrm{A^{\prime}}\cos\!\alpha_{1}+B^{\prime}\sin\alpha_{1}+C^{\prime}=0, \tag{4}\]  
where:  
\[\left\{\begin{array}{c}\mathrm{A^{\prime}}=l_{3}\cos\alpha_{3}+l_{4}\\ B^{\prime}=l_{3}\sin\alpha_{3}\\ C^{\prime}=\frac{\beta_{2}^{2}-\left(\mathrm{A^{\prime}}\right)^{2}-\left( \mathrm{B^{\prime}}\right)^{2}-l_{1}^{2}}{2l_{1}}\end{array}\right..\]  
Thus:  
\[\alpha_{1}(H_{C})=2\mathrm{arctg}\frac{B^{\prime}\pm\sqrt{\left(\mathrm{A^{ \prime}}\right)^{2}+\left(\mathrm{B^{\prime}}\right)^{2}-\left(\mathrm{C^{ \prime}}\right)^{2}}}{\mathrm{A^{\prime}}-\mathrm{C^{\prime}}}=u(\alpha_{3}(H _{C})). \tag{5}\]  
4. Solving \(L_{e}\) :  
The upper part of the active suspension is shown in Figure 4b. According to quadrangles \(E_{1}E_{2}B_{2}B_{1}\) and \(B_{1}B_{2}C_{2}C_{1}\), which are all symmetric:  
\[L_{e}(H_{C})=2l_{7}*\sin\!\left(\pi-\beta_{2}-\alpha_{1}(H_{C})-\arcsin\! \left(\frac{l_{5}-l_{6}}{2l_{4}}\right)\right)+l_{6}. \tag{6}\]  
Thus, the length of the linear actuator \(L_{e}\) can be achieved when the ground clearance (\(H_{C}\)) is given with Equations (2)-(6), the relationship is described below:  
\[L_{e}=v(H_{C}), \tag{7}\]  
where \(L_{e}\) is the length of the linear actuator, and \(H_{C}\) is the ground clearance.  
According to Equation (7), the min and max length of the linear actuator (\(L_{emin}\) and \(L_{emax}\)) can be achieved when the lowest and highest ground clearance are given, respectively.