# Prediction of Theoretical Derailments Caused by Cross-Winds with Frequency
## Vehicle Behaviors under a Constant Cross-Wind  
### Formula for a Constant Cross-Wind Acting on a Car Body  
Since railway vehicles run on rails, the lateral and vertical forces applied to the rails can be used to assess derailment. When a cross-wind acts on a car body, the forces acting on suspensions of the vehicle must be known to evaluate its derailment. These forces are closely related to the data of the vertical and horizontal displacements and the roll angle of the body.  
The free body diagram of a railway vehicle is shown in Figure 4[28].  
The free body diagram is divided into a car body, a bogie, and a wheelset when a cross-wind force with an upper angle of \(\varphi\) is applied at a distance of \"\(d\)\" from the center of gravity of the body.  
To take into account the equivalent amount to one wheelset, it is assumed that the mass and rotational inertias of the car body are 1/4 and those of the bogie are 1/2[29].  
For the primary suspension, the vertical and lateral stiffness values of rubber-metal springs installed at an axle box were denoted as k\({}_{\text{II}}\) and k\({}_{\text{1v}}\), respectively, because each axle box is directly connected with the bogie. For the secondary suspension, the vertical and lateral stiffness values of air springs connecting the bogie and the car body were denoted as k\({}_{\text{2l}}\) and k\({}_{\text{2v}}\) considering their half values to each wheelset.  
The static equilibrium equation of a vehicle unit to predict the spring forces transmitted by the cross-wind can be summarized as follows.  
* Static equilibrium equation for the car body: \[\sum F_{\text{x2}}=P_{t}\cos\varnothing+m_{2}\bigg{(}\frac{v^{2}}{r}\cos \varphi-g\sin\varphi\bigg{)}-F_{RH2}-F_{LH2}=0\] (1) \[\sum F_{\text{y2}}=m_{2}\bigg{(}\frac{v^{2}}{r}\sin\varphi+g\cos \varphi\bigg{)}-P_{t}\sin\varnothing-F_{RV2}-F_{LV2}=0\] (2)  
Figure 4: The free body diagram of a vehicle under a cross-wind.  
\[\sum M_{G2}=P_{t}(d\cos\varnothing+l\sin\varnothing)+(F_{RH2}+F_{LH2})h_{2}+(F_{LV2 }-F_{RV2})e_{2}=0 \tag{3}\]
* Static equilibrium equation for the bogie: \[\sum F_{x1}=m_{1}\bigg{(}\frac{v^{2}}{r}\cos\varphi-g\sin\varphi\bigg{)}+F_{RH2 }+F_{LH2}-F_{RH1}-F_{LH1}=0\] (4) \[\sum F_{y1}=m_{1}\bigg{(}\frac{v^{2}}{r}\sin\varphi+g\cos\varphi\bigg{)}+F_{RV2 }+F_{LV2}-F_{RV1}-F_{LV1}=0\] (5)  
\[\sum M_{G1}=(F_{RH2}+F_{LH2})h_{1u}+(F_{RV2}-F_{LV2})e_{2}+(F_{RH1}+F_{LH1})h_ {1l}+(F_{LV1}-F_{RV1})e_{1}=0 \tag{6}\]
* Static equilibrium equation for a wheelset: \[\sum F_{x0}=F_{LH1}+F_{RH1}+m_{0}\bigg{(}\frac{v^{2}}{r}\cos\varphi-g\sin \varphi\bigg{)}-R_{RH}=0\] (7) \[\sum F_{y0}=m_{0}\bigg{(}\frac{v^{2}}{r}\sin\varphi+g\cos\varphi\bigg{)}+F_{LV1 }+F_{RV1}-R_{RV}-R_{LV}=0\] (8) \[\sum M_{G0}=R_{RH}h_{0}+(F_{RV1}-F_{LV1})e_{1}+(R_{LV}-R_{RV})e_{0}=0\] (9)  
The forces of the secondary suspension springs generated by the relative displacements between the body and the bogie are as follows.  
\[F_{RH2}=F_{LH2}=k_{2l}(x_{2}-x_{1}-h_{2}\theta_{2}+h_{1u}\theta_{1}) \tag{10}\]  
\[F_{RV2}=k_{2v}(y_{2}-y_{1}+e_{2}\theta_{2}-e_{2}\theta_{1}) \tag{11}\]  
\[F_{LV2}=k_{2v}(y_{2}-y_{1}-e_{2}\theta_{2}+e_{2}\theta_{1}) \tag{12}\]  
The forces of the primary suspension springs generated by the relative displacements between the bogie and the wheelset are as follows.  
\[F_{RH1}=F_{LH1}=k_{1l}(x_{1}-x_{0}-h_{1l}\theta_{1}) \tag{13}\]  
\[F_{RV1}=k_{1v}(y_{1}-y_{0}+e_{1}\theta_{1}-e_{1}\theta_{0}) \tag{14}\]  
\[F_{LV1}=k_{1v}(y_{1}-y_{0}-e_{1}\theta_{1}+e_{1}\theta_{0}) \tag{15}\]  
The contact forces between the wheel and the rail can be calculated from the primary suspensions deformations as follows.  
\[R_{RH}=k_{0l}(x_{0}-h_{0}\theta_{0}) \tag{16}\]  
\[R_{RV}=k_{0v}(y_{0}+e_{0}\theta_{0}) \tag{17}\]  
\[R_{LV}=k_{0v}(y_{0}-e_{0}\theta_{0}) \tag{18}\]