## 2 Structure and Model
### Mathematical Model
 \tag{8}\]  
Then, the dimensionless pressure and gas film clearance can be expressed as:  
\[ \begin{cases}\ P=P_{0}+P_{X}\cdot\Delta X+P_{X}\cdot\Delta\dot{X}+P_{Y}\cdot \Delta Y+P_{Y}\cdot\Delta\dot{Y}\\ H=H_{0}+H_{X}\cdot\Delta X+H_{X}\cdot\Delta\dot{X}+H_{Y}\cdot\Delta Y+H_{Y}\cdot \Delta\dot{Y}\end{cases}  \tag{9}\]  
On the basis of the distribution of static pressure \(p_{0}\) and static clearance \(H_{0}\), the static performance parameters of the bearing can be obtained through integration as follows:  
\[ \begin{array}{l}\text{Dimensionless gas film force:}\begin{bmatrix}F_{X}\\ F_{Y}\end{bmatrix}=\frac{1}{p_{a}r^{2}}\begin{bmatrix}f_{x}\\ f_{y}\end{bmatrix}=\int_{-l/d}^{+l/d}\int_{0}^{2\pi}\begin{bmatrix}-P_{0}\cos \theta\\ -P_{0}\sin\theta\end{bmatrix}\text{d}\theta\text{d}Z\end{array}  \tag{10}\]  
\[ \begin{array}{l}\text{Bearing load capacity:}\quad\ W=p_{a}r^{2}\sqrt{F_{X}^{2}+ F_{Y}^{2}}\end{array}  \tag{11}\]\[\text{Bearing attitude angle:}\quad\theta_{0}^{k+1}=\tan^{-1}\left[\frac{F_{X} \sin\left(\theta_{0}^{k}\right)-F_{Y}\cos\left(\theta_{0}^{k}\right)}{F_{X}\cos \left(\theta_{0}^{k}\right)+F_{Y}\sin\left(\theta_{0}^{k}\right)}\right] \tag{12}\] \[\text{Friction torque:}\quad\quad f_{t}=p_{a}r^{2}\int_{-1/d}^{+1/d }\int_{0}^{2\pi}\left[\frac{H_{0}}{2}\frac{\partial P_{0}}{\partial\theta}+ \frac{\Lambda}{6H_{0}}\right]\text{d}\theta\text{d}Z \tag{13}\]  
When the rotor has a low displacement disturbance \(\left(\Lambda x,\ \Delta y\right)\) and a velocity disturbance \(\left(\Delta\dot{x},\ \Delta\dot{y}\right)\) close to the balance position, the gas film force can be expressed as:  
\[F=F_{0}+\left(\frac{\partial F}{\partial X}\right)_{0}\Delta X+\left(\frac{ \partial F}{\partial Y}\right)_{0}\Delta Y+\left(\frac{\partial F}{\partial \dot{X}}\right)_{0}\Delta\dot{X}+\left(\frac{\partial F}{\partial\dot{Y}} \right)_{0}\Delta\dot{Y}+O\left(X^{2}+Y^{2}+\dot{X}^{2}+Y^{2}\right) \tag{14}\]  
where the subscript \"0\" represents the static balance position. Then, the dimensionless stiffness coefficient and dimensionless damping coefficient of the bearing can be expressed as:  
\[\text{Dimensionless stiffness coefficient:}\left[ \begin{matrix}K_{xx}&K_{yx} \\ K_{xy}&K_{yy}\end{matrix} \right]=\frac{k}{p_{a}r^{2}}\left[ \begin{matrix}k_{xx} &k_{yx}\\ k_{xy}&k_{yy}\end{matrix} \right]=\left( \begin{matrix}\left[\frac{\partial F}{ \partial\dot{X}}\right]\\ \left[\frac{\partial F}{\partial\dot{Y}}\right]\end{matrix} \right)_{0} \tag{15}\] \[\text{Dimensionless damping coefficient:}\left[ \begin{matrix}D_{xx}&D_{ yx}\\ D_{xy}&D_{yy}\end{matrix} \right]=\frac{c_{0}}{p_{a}r^{2}}\left[ \begin{matrix}d_{xx} &d_{yx}\\ d_{xy}&d_{yy}\end{matrix} \right]=\left( \begin{matrix}\left[\frac{\partial F}{ \partial\dot{X}}\right]\\ \left[\frac{\partial F}{\partial\dot{Y}}\right]\end{matrix} \right)_{0} \tag{16}\]  
The rigid symmetrical rotor model is used to analyze the stability, and the rotor mass is assumed to be 2 M. Then, the motion equation of a small disturbance close to the balance position of the rotor can be obtained as follows:  
\[\left\{ \begin{matrix}M\Delta\ddot{x}+k_{xx}\Delta x+k_{xy}\Delta y+d_{xx} \Delta\dot{x}+d_{xy}\Delta\dot{y}=0\\ M\Delta\ddot{y}+k_{yx}\Delta x+k_{yy}\Delta y+d_{yx}\Delta\dot{x}+d_{yy}\Delta \dot{y}=0\end{matrix} \right. \tag{17}\]  
The solution of \(\Delta x,\ \Delta y\) is assumed to take the following form:  
\[\left\{ \begin{matrix}\Delta x=\Delta x_{0}e^{st}\\ \Delta y=\Delta y_{0}e^{st}\end{matrix} \right. \tag{18}\]  
where \(s=\lambda+iv\), \(\lambda\) is the system damping coefficient, and \(v\) is whirl frequency. In the critical state, the system damping is zero, that is, \(\lambda=0\). Under this condition, Equation (18) can be written as [28]:  
\[\left[ \begin{matrix}Z_{xx}-v^{2}M&Z_{xy}\\ Z_{yx}&Z_{yy}-v^{2}M\end{matrix} \right]\left[ \begin{matrix}\Delta x_{0}\\ \Delta y_{0}\end{matrix} \right]=0 \tag{19}\]  
where \(Z_{jk}=K_{jk}+ivD_{jk}\) is the compound stiffness.