## 2 Modeling of Dynamic Rotational Error for Spindle-Rolling Bearing System
### Nonlinear Bearing Model
#### 2.1.1 Modeling of Dynamic Supporting Stiffness
\begin{split}\mathrm{l_{ik}}&=(\mathrm{f_{i}}-0.5) \mathrm{D}+\delta_{\mathrm{ik}}\\ \mathrm{l_{ok}}&=(\mathrm{f_{o}}-0.5)\mathrm{D}+ \delta_{\mathrm{ok}}\end{split} \, \tag{7}\]  
where \(\delta_{\mathrm{ik}}/\delta_{\mathrm{ok}}\) is the contact deformation between the inner/outer bearing ring and the kth rolling element. In accordance with the displacement coordinate relations, the axial distance between the inner and outer raceway groove curvature centers is  
\[ \begin{split}\mathrm{A_{1k}}&=(\mathrm{L_{ok}}+ \mathrm{L_{ik}})\mathrm{Dsin}\alpha^{0}+\delta_{\mathrm{z}}+\eta_{\mathrm{i}} \theta_{\mathrm{x}}\mathrm{sin}\mathrm{f_{bk}}-\eta_{\mathrm{i}}\theta_{ \mathrm{y}}\mathrm{cos}\mathrm{f_{bk}}\\ \mathrm{A_{2k}}&=(\mathrm{L_{ok}}+\mathrm{L_{ik}}) \mathrm{Dcos}\alpha^{0}+\delta_{\mathrm{x}}\mathrm{cos}\mathrm{f_{bk}}+ \delta_{\mathrm{y}}\mathrm{sin}\mathrm{f_{bk}}\end{split} \, \tag{8}\]  
where \(\eta_{\mathrm{i}}\) is the distance between the rotating axis \(Z\) and inner raceway groove curvature center and can be calculated by  
\[\eta_{\mathrm{i}}=\frac{\mathrm{d_{m}}}{2}+(\mathrm{f_{i}}-0.5)\mathrm{D_{b} }\mathrm{cos}\alpha_{0}, \tag{9}\]  
It convenient to introduce new variables \(\mathrm{X_{1k}}\) and \(\mathrm{X_{2k}}\), as shown in Figure 2. It can be seen from Figure 2 that at any rolling element location  
\[ \begin{split}\mathrm{sin}\alpha_{\mathrm{ik}}&= \frac{\mathrm{x_{1k}}}{\mathrm{l_{ik}}}\\ \mathrm{cos}\alpha_{\mathrm{ik}}&=\frac{\mathrm{x_{2k }}}{\mathrm{l_{ik}}}\\ \mathrm{sin}\alpha_{\mathrm{ik}}&=\frac{\mathrm{A_{1k }}-\mathrm{x_{1k}}}{\mathrm{l_{ik}}}\\ \mathrm{cos}\alpha_{\mathrm{oik}}&=\frac{\mathrm{A_{2k }}-\mathrm{x_{2k}}}{\mathrm{l_{ik}}}\end{split}  \tag{10}\]  
Using the Pythagorean Theorem, it can be seen from Figure 2 that [31]. \[ \begin{split}&\left(A_{1k}-X_{1k}\right)^{2}+\left(A_{2k}-X_{2k} \right)^{2}-\left[(f_{\mathrm{i}}-0.5)D+\delta_{\mathrm{ik}}\right]^{2}=0\\ &\mathrm{X_{1k}^{2}}+\mathrm{X_{2k}^{2}}-\left[(f_{\mathrm{o}}-0. 5)D+\delta_{\mathrm{ok}}\right]^{2}=0\end{split}  \tag{11}\]  
As shown in Figure 3, the load equilibrium equations for the kth rolling element can be expressed as [31]. \[ \begin{split}& Q_{ik}\cos\alpha_{ik}-Q_{ok}\cos\alpha_{ok}+\frac{ \lambda_{ik}M_{gk}}{D}\sin\alpha_{ik}-\frac{\lambda_{ok}M_{gk}}{D}\sin\alpha_{ ok}+F_{ck}=0\\ & Q_{ik}\sin\alpha_{ik}-Q_{ok}\sin\alpha_{ok}-\frac{\lambda_{ik}M_ {gk}}{D}\cos\alpha_{ik}+\frac{\lambda_{ok}M_{gk}}{D}\cos\alpha_{ok}=0\end{split}  \tag{12}\]  
where \(Q_{ik}/Q_{ok}\) is the contact force between the inner/outer ring and the rolling element. \(Q_{ik}\) and \(Q_{ok}\) can be expressed as  
\[ \begin{split}& Q_{ik}=K_{ik}\delta_{ik}{}^{1.5}\\ & Q_{ok}=K_{ok}\delta_{ok}{}^{1.5}\end{split}  \tag{13}\]\(\lambda_{\delta k}/\lambda_{\delta k}\) represent the friction coefficients between the inner/outer raceway and rolling element. As for the outer raceway control, \(\lambda_{\delta k}\) = 0 and \(\lambda_{\delta k}\) = 2 The gyroscopic moment M\({}_{\delta k}\) and centrifugal force F\({}_{\mathrm{ck}}\) acting on the kth rolling element can be described as follows:  
\[ \begin{split}\mathrm{M}_{\delta k}&=\mathrm{I}_{ \mathrm{b}}\omega_{\mathrm{bk}}\omega_{\mathrm{sk}}\mathrm{sin}\beta_{\mathrm{ k}}\\ \mathrm{F}_{\mathrm{ck}}&=0.5\mathrm{m}_{\mathrm{b}} \mathrm{d}_{\mathrm{m}}\omega_{\mathrm{bk}}^{2}\end{split}  \tag{14}\]  
where \(\mathrm{m}_{\mathrm{b}}\), \(\mathrm{I}_{\mathrm{b}}\) is the mass and mass moment inertia of the kth rolling element, respectively. Equations (11) and (12) are highly coupled and nonlinear and can be resolved by Newton-Raphson method. Then the contact forces Q\({}_{\mathrm{i}}\), Q\({}_{\alpha}\), contact angles \(\alpha_{\mathrm{i}}\), \(\alpha_{\mathrm{o}}\), and gyroscopic moments M\({}_{\mathrm{i}}\): M\({}_{\mathrm{o}}\), between each rolling element and inner/outer rings can be obtained.