## 2 Train Motion Constraints and Odometer
### Train Turning Influence on Motion Constraints  
The curved line of a high-speed railway track is generally composed of a straight line, a gentle curve, and a circular curve [22], as shown in Figure 1. There is an installation lever arm between the rotational center of the train body and the MIMU, and the lever arm speed exists at the position of the MIMU during turning [2; 23].  
The train turns under the cooperation of the forward traction, track, bogie, and wheels [24]. The structure of the bogie is shown in Figure 2. Each bogie contains four wheels, and each carriage has two bogies. The carriage is mounted on the bogie through two bogie pins, several springs, and dampers. This structure allows for a certain degree of rotation between the carriage box and the wheels.  
In order to simplify the analysis, select the S section shown in Figure 1, and assume that the train is turning left. The turning motion of the train is shown in Figure 3.  
The bogies of the train sit on a curved track with a radius \(R\). \(P_{1}\) and \(P_{2}\) are the pins of the front and rear bogies; the carriage box lies on the secant between the two pins. The bogies are approximately tangential to the curve, and the carriage lies on the secant between the two pins [24]. There is an misalignment angle \(\beta\) between the axis of the bogie and the carriage [10]. The misalignment angle is given by  
\[\beta\ =\ \arcsin\frac{d}{2R} \tag{4}\]  
Figure 1: Track curve line plan. A, B, C and D are the four points on the train track. Segment S is a circular curve with a turning radius of R.  
Figure 3: Geometric analysis of train turning.  
Figure 2: Physical map of bogie.  
where \(d\) is distance between \(P_{1}\) and \(P_{2}\). The carriage rotates around \(O_{1}\) during turning, and the angular rate of rotation is  
\[\omega=\frac{v_{D1}\sin\beta}{(d/2)}=\frac{v_{D1}}{R} \tag{5}\]  
where \(v_{D1}\) is the velocity of the front bogie's pin. Assuming MIMU is installed at point \(P_{1}\), we have  
\[v_{D1}=\left\|v^{n}_{MIMU}\right\|_{2} \tag{6}\]  
where \(v^{n}_{MIMU}\) is the velocity of the MIMU in the \(n\) frame.  
The assumption is now violated in the \(m\) frame because of the carriage's rotation around \(O_{1}\).We can find the velocity of the carriage in the \(m\) frame as follows:  
\[\begin{array}{lcl}v^{m}_{D}&=&\left[\begin{array}{cc}-v_{D1}\sin\beta&v_{D 1}\cos\beta&0\\ -\frac{v_{D1}d}{2R}&\frac{v_{D1}\sqrt{4R^{2}-d^{2}}}{2R}&0\end{array}\right] ^{T}\\ \end{array} \tag{7}\]  
Assume a velocity of 350 km/h, a distance of 20 m between the bogie pins and a curve radius of 5000 m [22]. The lateral velocity of carriage due to the turning will be  
\[v^{m}_{D,x}=-\frac{v_{D1}d}{2R}\approx-0.19\text{m/s} \tag{8}\]  
This would pose a large risk of corruption of the extended Kalman filter (EKF) state estimates. Substituting Equation (5) into Equation (7) gives  
\[v^{m}_{D}=\left[\begin{array}{cc}-\frac{\omega_{z}d}{2}&\frac{\omega_{z} \sqrt{4R^{2}-d^{2}}}{2}&0\end{array}\right]^{T} \tag{9}\]  
where \(\omega_{z}\) is the output of the \(z\)-axis gyro.  
Of course, turning maneuvers are quite complicated. Both the front and rear bogies have varying angular rates of rotation and the center of rotation is not fixed, which makes it difficult to accurately model the lateral velocity. However, this still provides a theoretical basis for judging whether the hypothesis that the lateral velocity is zero holds.