## 2 Vehicle System Modeling
### Dynamics Equations of Powertrains  
The motion of the entire train, as shown in Figure 1 and described in Equation (3) (or Equation (7)), is determined by the effective driving force \(4F_{t}\) generated by the drive wheels of the four locomotive cars (including both mechanical and regenerative braking forces), the mechanical braking force \(4F_{b}\) from the four trailing cars, and the total running load \(F^{R}\) of the eight cars (which includes both air resistance and gradient resistance). Converting these effective forces to the basis of a single locomotive car yields the following:  
\[\left\{\begin{array}{c}F_{t}=\frac{M\dot{v}}{4}+F_{b}+\frac{F^{R}}{4}\\ r_{wh}F_{t}=\frac{I_{veh}\dot{\omega}_{wh}}{4}+r_{wh}\Big{(}F_{b}+\frac{F^{R}}{4 }\Big{)}\end{array}\right. \tag{9}\]  
However, each locomotive car (M1, M2) in this study is equipped with four drive systems, each consisting of a motor, reduction gear, drive shaft, and left and right drive wheels. Therefore, the equation of motion for a locomotive car, as given in Equation (9), can be expressed on the basis of a single drive system, resulting in Equation (10).  
\[\left\{\begin{array}{c}\frac{F_{t}}{4}=\frac{M\dot{v}}{16}+\frac{F_{b}}{4}+ \frac{F^{R}}{16}\\ \frac{r_{wh}F_{t}}{4}=\frac{I_{veh}\dot{\omega}_{wh}}{16}+r_{wh}\Big{(}\frac{ F_{b}}{4}+\frac{F^{R}}{16}\Big{)}\end{array}\right. \tag{10}\]  
The concept of performance modeling related to the equation of motion for the traction drive shaft, as formulated in Equation (10), is clearly illustrated in Figure 2.  
The traction force \(F_{t}\) from the locomotive cars (M1, M2) acts as a load torque \(T_{l}=r_{wh}F_{t}/4\) on the drive wheel axis, while the drive motor torque \(T_{m}\) in the traction drivetrain drives the wheels through a reduction gear with a ratio of \(\alpha_{G}\left(\frac{\omega_{wh}}{\omega_{wh}}\right)\). At this point, the motor's rotational inertia \(I_{m}\) is converted to the drive wheel basis. Additionally, \(\eta\) represents the transmission efficiency in the drivetrain. The equation of motion for each motor in the locomotive cars is expanded as shown in Equation (11). The structure of the driving axle is shown in Figure 3.  
\[I_{axl}=I_{g}+I_{a}+2\cdot I_{wh}+I_{p}\cdot(\eta\cdot\alpha_{G})^{2} \tag{11}\]  
\[\Big{[}(\eta\cdot\alpha_{G})^{2}\cdot I_{m}+I_{axl}\Big{]}\dot{\omega}_{wh}=( \eta\cdot\alpha_{G})\cdot T_{m}-\frac{r_{wh}F_{t}}{4} \tag{12}\]  
By eliminating \(F_{t}\) from Equations (10) and (12) and simplifying the resulting expression, we obtain the following equation:  
\[\Big{(}(\eta\cdot\alpha_{G})^{2}\cdot I_{m}+\frac{I_{beh}}{16}+I_{axl}\Big{)} \dot{\omega}_{wh}=(\eta\cdot\alpha_{G})\cdot T_{m}-\frac{r_{wh}F_{b}}{4}-\frac{ r_{wh}F^{R}}{16} \tag{13}\]  
The rotational inertias applied in Equation (13) are listed in Table 1[8; 9; 10].  
Figure 4 illustrates the performance model based on the drivetrain system related to the equation of motion, as shown in Equation (13). It represents the concept of rotational inertia loads acting on the drive shafts, as well as the equivalent vehicle inertia and running load, as the torque generated by the motor is transmitted through the reduction gear to the drive wheels. This study aims to analyze the relative fatigue at the positions of the input and output shafts of the reduction gear, as defined in Figure 4.  
\begin{table}
\begin{tabular}{c c c c} \hline \hline
**Component** & **Mass (kg)** & **Radius (m)** & **Inertia (kg\(\cdot\)m\({}^{2}\))** \\ \hline Pinion & \(m_{p}:13.3\) & \(L_{p}:0.038\) & \(I_{p}:0.0096\) \\ Gear wheel & \(m_{g}:124\) & \(L_{g}:0.0306\) & \(I_{g}:5.805\) \\ Axel & \(m_{a}:410.4\) & \(L_{a}:0.0995\) & \(I_{a}:2.0315\) \\ Wheel & \(m_{wh}:349.7\) & \(L_{wh}:0.43\) & \(I_{wh}:32.3298\) \\ \hline \hline \end{tabular}
\end{table}
Table 1: Properties of rotational components.  
Figure 3: Structure of the driving axle.  
Figure 2: Overview of the dynamic characteristics of each motor within the locomotive.