# Research on Multi-Mode Braking Energy Recovery Control Strategy for Battery Electric Vehicles
## 2 Brake Energy Recovery Control Strategy
### Conventional Brake Energy Recovery Control Strategy
#### 2.3.3 Control Strategy for Energy Recovery of Coasting Brake  
This study of coasting brake energy recovery control strategy is divided into two scenarios: horizontal road coasting and downhill road coasting. (1) Brake energy recovery control strategy for horizontal road skidding:  
Figure 6 shows the analysis of the vehicle coasting mode. The red line in the figure is the coasting curve of the vehicle, and when energy recovery is not added, it can be seen that the driver gradually releases the accelerator pedal from the initial speed, and when it reaches point A, the vehicle speed is \(v_{A}\), leading it to start coasting. When reaching point B, the vehicle speed is \(v_{B}\). The driver steps on the brake pedal to stop the vehicle. The blue line in the figure is the vehicle's coasting curve after joining the coasting brake's energy recovery, starting from point A to enter the coasting and energy recovery, until point E, where the vehicle stops. \begin{table}
\begin{tabular}{c c c c c} \hline \hline \multirow{2}{*}{**k**} & \multirow{2}{*}{**v**} & \multicolumn{3}{c}{**z**} \\ \cline{3-5}  & & **S** & **M** & **L** \\ \hline \multirow{3}{*}{**SOC = S**} & L & S & S & S \\  & M & VL & L & S \\  & H & L & L & S \\ \hline \multirow{3}{*}{**SOC = M**} & L & S & S & S \\  & M & M & M & S \\ \hline \multirow{3}{*}{**SOC = L**} & L & S & S & VS \\  & M & S & S & VS \\ \cline{1-1}  & H & VS & VS & VS \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Conventional braking mode fuzzy rule table. Figure 6: Vehicle skidding mode analysis diagram. From the above analysis, the deceleration of the vehicle when the vehicle does not incorporate energy recovery is a and the deceleration of the vehicle when it incorporates energy recovery is \(a_{E}\). Based on the forces on the vehicle, the following formula can be obtained:  
\[F=m(a_{E}-a) \tag{27}\]  
The expression for \(a\) can be deduced from the slope as follows:  
\[ \begin{array}{l}a=\frac{v_{0}-v_{A}}{t_{A}}=\frac{v_{A}-v_{B}}{\Delta t}\\ \Delta t=\frac{v_{A}-v_{B}}{v_{0}-v_{A}}t_{A}=DC\end{array}  \tag{28}\]  
The equation can be derived when the glide displacement is the same in both scenarios:  
\[S_{ADE}=\frac{1}{2}AD\cdot DE=\frac{1}{2}(AD+BC)DC=S_{ABCD} \tag{29}\]  
Bringing in \(DC\), one can solve for \(DE\):  
\[DE=\frac{(AD+BC)DC}{AD}=\frac{t_{A}(v_{A}^{2}-v_{B}^{2})}{v_{A}(v_{0}-v_{A})} \tag{30}\]  
The expression for \(a_{E}\) is as follows:  
\[a_{E}=\frac{AD}{DE}=\frac{v_{A}^{2}(v_{0}-v_{A})}{v_{A}(v_{0}-v_{A})} \tag{31}\]  
In summary, the expression for the power of the motor mechanism can be obtained as \(F_{e}\):  
\[F_{e}=\frac{mv_{B}^{2}(v_{0}-v_{A})}{t_{A}(v_{A}^{2}-v_{B}^{2})} \tag{32}\]  
According to the constraints on the brake energy recovery control strategy in Section 2.1 of this paper, the motor will only recover brake energy when the system detects that both the brake pedal and accelerator pedal are not operated for 2 s. Therefore, it is 2 s. In addition, the amount of motor power supplied by the motor is also limited by the restrictions imposed by the ECE regulations and the motor's external characteristic curves. Therefore, it can be deduced that the motor outputs braking power in the horizontal road scenario as follows:  
\[F_{e}=\min(\frac{m(17.94-v_{A})^{2}(v_{0}-v_{A})}{2[v_{A}^{2}-(17.94-v_{A})^{2 }]},F_{e\_{\max}},F_{ECE\_max}) \tag{33}\]  
(2) Brake energy recovery control strategy for downhill road coasting:  
Figure 7 shows the force analysis of the vehicle during the downhill coasting scenario; \(F_{\text{G}}\) is the component force of the car's gravity, and \(F\) is the combined driving resistance force; (air resistance, rolling resistance, etc.) on the car. Set the slope angle as \(\alpha\). The ECE braking regulations front and rear wheel braking force curve equation changes to the formula shown in Equation (34):  
\[\left\{ \begin{array}{l}F_{u1}=\frac{z+0.07}{0.85}\cdot\frac{G}{L}(b\cos \alpha+zh_{g}\sin\alpha)\\ F_{u2}=Gz-F_{u1}\end{array} \right.