## 3 Dynamic Model of the Horizontal System  
Based on the thrust analysis in Figure 2, the MRC has two degrees of freedom in the horizontal system: (1) it can move vertically in the Y-direction; (2) it can rotate around the Z-axis with the center of mass of the MRC as the center of rotation, denoted as \(\psi_{z}\). In this paper, [PERSON]'s second law of motion is utilized to describe the vertical motion of the MRC, and the torque equation is employed to describe its rotational motion [2]. \[F_{y}=m\cdot\frac{\partial^{2}y_{y}}{\partial t^{2}} \tag{7}\]  
\[M_{\beta}=J_{\beta}\cdot\frac{\partial^{2}y_{z}}{\partial t^{2}} \tag{8}\]  
In Equation (7), \(F_{y}\) represents the sum of all forces acting on the MRC in the Y-direction, \(m\) is the mass of the MRC, and \(y_{b}\) is the displacement of the center of mass of the MRC in the Y-direction. In Equation (8), \(M_{\beta}\) is the torque of the resultant force \(F_{y}\), \(J_{\beta}\) is the moment of inertia, and \(\psi_{z}\) is the rotation angle. The position of the center of mass, the force-application points of the horizontal thrusts at the left and right ends of the MRC, and the length of the force arm are marked in Figure 2.  
\[J_{\beta}=m_{h}r^{2} \tag{9}\]  
In Equation (9), \(m_{h}\) is the mass from the center of mass of the MRC to the end, and \(r\) is the distance from the center of mass to the force-application point of the horizontal thrust. From this, \(J_{\beta}=1.49\times 10^{-2}\) kg\(\cdot\)m\({}^{2}\) can be calculated. Ignoring other disturbances such as base vibration and suspension force, the sum of all forces \(F_{y}\) on the MRC in the Y-direction is as follows:  
\[F_{y}=F_{lt}+F_{rt} \tag{10}\]  
From the dimensions indicated in Figure 2, the torque equation around the Y-axis is as follows:  
\[M_{\beta}=-0.1275F_{lt}+0.1275F_{rt} \tag{11}\]  
According to Equation (8), the following can be obtained:  
\[-0.1275F_{lt}+0.1275F_{rt}=J_{\beta}\cdot\frac{\partial^{2}y_{z}}{\partial t^ {2}} \tag{12}\]  
From Equations (7) and (10), we obtain the following:  
\[F_{lt}+F_{rt}=m\cdot\frac{\partial^{2}y_{b}}{\partial t^{2}} \tag{13}\]  
Thus, the rigid-body dynamic equation of the MRC horizontal system is derived as follows:  
\[ \begin{bmatrix}1&1\\ -0.1275&0.1275\end{bmatrix} \cdot \begin{bmatrix}F_{lt}\\ F_{rt}\end{bmatrix} = \begin{bmatrix}1.83&0\\ 0&0.0273\end{bmatrix}\begin{bmatrix}\frac{\partial^{2}y_{y}}{\partial t^{2}} \\ \frac{\partial^{2}y_{b}^{2}}{\partial t^{2}}\end{bmatrix}  \tag{14}\]  
The MRC is solely affected by the horizontal thrust in the horizontal direction. According to [PERSON]'s second law, \(F_{lt}\) and \(F_{rt}\) are related to the left-hand side horizontal displacement \(y_{bl}\) and the right-hand side horizontal displacement \(y_{br}\) of the MRC, and they are in the \([y_{bl},y_{br}]\) coordinate system with the horizontal displacement as the research point. However, Equation (14) is in the \([y_{b},\psi_{z}]\) coordinate system with the center of mass as the research point. Therefore, a coordinate transformation is required to convert the horizontal displacement coordinate system to the center-of-mass coordinate system. As can be seen from Figure 2, when the MRC rotates around the Z-axis, there are the following relationships between \(y_{bl},y_{br}\), the displacement of the center of mass \(y_{b}\) and the rotation angle \(\psi_{z}\):  
\[y_{bl}=y_{b}-0.1275\times\tan\psi_{z} \tag{15}\]  
\[y_{br}=y_{b}+0.1275\times\tan\psi_{z} \tag{16}\]  
Since the horizontal displacement is extremely small, the rotation angle \(\psi_{z}\) is also very small, so tan \(\psi_{z}\approx\psi_{z}\). Therefore, Equations (4)-(15) and (4)-(16) can be written as follows:  
\[ \begin{bmatrix}y_{bl}\\ y_{br}\end{bmatrix} = \begin{bmatrix}1&-0.1275\\ 1&0.1275\end{bmatrix}\begin{bmatrix}y_{b}\\ \psi_{z}\end{bmatrix}  \tag{17}\]  
The above formula can be used to transform the horizontal displacement coordinate system into the center-of-mass coordinate system.