## 2 Equivalent Half-Vehicle Model
### Equivalent Half-Vehicle Model Derivation  
When the braking intensity is \(z\), the front wheel load \(F_{sf}\) and rear wheel load \(F_{zr}\) are represented as follows:  
\[\left\{\begin{array}{l}F_{sf}=G(b+zh_{g})/L\\ F_{zr}=G(a-zh_{g})/L\end{array}\right. \tag{1}\]  
where \(G=(m_{b}+m_{ur}+w_{m})g\) is the half-vehicle gravity (g is the acceleration of gravity), \(L\) is the wheelbase, \(L=a+b\), and \(h_{g}\) is the height of the body mass center. Here, \(Gzh_{g}/L\) is the dynamic load, that is, the wheel load transfer caused by the braking strength of the vehicle. According to the effect of braking intensity, ignoring the transfer of the unsprung mass, the dynamic load is equivalent to the front axle equivalent force \(F_{mf}\) and the rear axle equivalent force \(F_{mr}\), the absolute values of which are:  
\[\left\{\begin{array}{l}F_{mf}=abs(m_{b}gzh_{g}/L)\\ F_{mr}=abs(-m_{b}gzh_{g}/L)\end{array}\right. \tag{2}\]  
Here, \(F_{mf}\) and \(F_{mr}\) are a pair of force couples caused by the braking intensity. This force couple will affect the movement of the vehicle body. Therefore, the force couple should be considered when controlling the suspension under braking conditions.  
Figure 1: Equivalent half-vehicle model.  
The following is the process of establishing the dynamic model of the vehicle components without considering the influence of gravity. Because the vertical motion of the vehicle is owing to the existence of the suspension and the unsprung mass stiffness, which cancel the effect of gravity (as gravity is also in the vertical direction), there is no need to consider the influence of gravity. The force couple of \(F_{m\!f}\) and \(F_{m\!r}\) is caused by the longitudinal movement of the vehicle and must be considered.  
A suspension is a component that connects the body and unsprung mass. The forces generated by the front and rear suspensions are:  
\[\left\{\begin{array}{l}F_{f}=K_{sf}(z_{1}-z_{2})+C_{sf}(\dot{z}_{1}-\dot{z}_{ 2})+F_{fc}\\ F_{r}=K_{sr}(z_{3}-z_{4})+C_{sr}(\dot{z}_{3}-\dot{z}_{4})+F_{rc}\end{array}\right. \tag{3}\]  
where \(F_{f}\) and \(F_{r}\) are the forces generated by the front and rear suspensions, respectively: \(\dot{z}_{1}\), \(\dot{z}_{2}\), \(\dot{z}_{3}\), and \(\dot{z}_{4}\) are the vertical velocities of the front unsprung mass, front body, rear unsprung mass, and rear body, respectively.  
For an unsprung mass, applying [PERSON]'s second law, it can be obtained that:  
\[\left\{\begin{array}{l}m_{wf}\ddot{z}_{1}=-K_{tf}z_{1}-F_{f}\\ m_{wf}\ddot{z}_{3}=-K_{tr}z_{3}-F_{r}\end{array}\right. \tag{4}\]  
where \(\ddot{z}_{1}\) and \(\ddot{z}_{3}\) are the vertical accelerations of the front and rear unsprung masses, respectively.  
Applying [PERSON]'s second law for the vehicle body, it can be obtained that:  
\[m_{b}\ddot{z}_{b}=F_{f}+F_{r} \tag{5}\]  
where \(\ddot{z}_{b}\) is the vertical acceleration of the vehicle body mass center.  
The dynamic equation for the pitch angle of the vehicle body is:  
\[\begin{array}{l}I_{p}\ddot{\theta}_{b}=-aF_{f}+bF_{r}-aF_{m\!f}-bF_{m\!r} \\ =-aF_{f}+bF_{r}-m_{b}gz_{b}\end{array} \tag{6}\]  
where \(\ddot{\theta}_{b}\) is the pitch acceleration of the vehicle body.  
When the pitch angle is small, it can be obtained that:  
\[\left\{\begin{array}{l}\ddot{z}_{2}=\ddot{z}_{b}-a\ddot{\theta}_{b}\\ \ddot{z}_{4}=\ddot{z}_{b}+b\ddot{\theta}_{b}\end{array}\right. \tag{7}\]  
where \(\ddot{z}_{2}\) and \(\ddot{z}_{4}\) are the vertical accelerations of the front and the rear bodies, respectively.