# A Multibody Model Calibration Method for Vehicle Misuse Testing
## 4 Nonlinear Parameter Calibration Using the Modified Quarter-Vehicle Model Analysis
### Tire Stiffness Calibration by Ground Recontact Stage  
After stage 2, the tire makes contact with the road again, causing vibrations in the suspension to generate stage 3. By adding the tire stiffness \(k_{\text{t}}\) depicted in Figure 6c, the equation for the modified quarter-vehicle model of stage 3 is  
\[\begin{bmatrix}m_{\text{u}}&0\\ 0&m_{\text{w}}\end{bmatrix}\begin{bmatrix}\ddot{z}_{\text{u}}(t)\\ \ddot{z}_{\text{w}}(t)\end{bmatrix}+\begin{bmatrix}k_{\text{e}}+k_{\text{s}}&- k_{\text{s}}\\ -k_{\text{s}}&k_{\text{s}}+k_{\text{t}}\end{bmatrix}\begin{bmatrix}z_{\text{u}} (t)\\ z_{\text{w}}(t)\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}. \tag{25}\]  
In stage 3, the excitation of the ground is transmitted through the tire stiffness \(k_{\text{t}}\) to the system. To simplify this expression, the corresponding term is moved to the left side and incorporated into the stiffness matrix in Equation (25). Let \(c_{\text{t}}=\left(k_{\text{t}}+k_{\text{s}}\right)/k_{\text{s}}\geq 1\), and because \(c_{\text{u}}=m_{\text{u}}/m_{\text{w}}\geq 0\), Equation (25) can be rewritten as  
\[\begin{bmatrix}c_{\text{u}}&0\\ 0&1\end{bmatrix}\begin{bmatrix}\ddot{z}_{\text{u}}(t)\\ \ddot{z}_{\text{w}}(t)\end{bmatrix}+\frac{k_{\text{s}}}{m_{\text{w}}}\begin{bmatrix} c_{\text{e}}&-1\\ -1&c_{\text{t}}\end{bmatrix}\begin{bmatrix}z_{\text{u}}(t)\\ z_{\text{w}}(t)\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}. \tag{26}\]Similarly to the vibration solution in Equation (16), in stage 1, the following synclastic eigenmode is generated as  
\[\mathbf{z}_{+,\textbackslash}\textbackslash}=\left[\begin{array}{c}\frac{\left(c _{\text{t}}\,c_{\text{u}}-c_{\text{e}}\right)+\sqrt{\left(c_{\text{t}}\,c_{ \text{u}}-c_{\text{e}}\right)^{2}+4\,c_{\text{u}}}}{2c_{\text{u}}}\\ 1\end{array}\right], \tag{27}\]  
and the synclastic circular frequency for stage 3  
\[\omega_{+,\textbackslash}^{2}=\left(\frac{\left(c_{\text{t}}\,c_{\text{u}}+c_ {\text{e}}\right)-\sqrt{\left(c_{\text{t}}\,c_{\text{u}}-c_{\text{e}}\right)^{2 }+4c_{\text{u}}}}{2c_{\text{u}}}\right)\frac{k_{\text{s}}}{m_{\text{w}}}. \tag{28}\]  
Comparing Equation (28) with Equation (21) yields  
\[\frac{\omega_{+,\textbackslash}^{2}}{\omega_{+,\textbackslash}^{2}}=\frac{ \left(c_{\text{u}}+c_{\text{e}}\right)-\sqrt{\left(c_{\text{u}}-c_{\text{e}} \right)^{2}+4c_{\text{u}}}}{\left(c_{\text{t}}\,c_{\text{u}}+c_{\text{e}} \right)-\sqrt{\left(c_{\text{t}}\,c_{\text{u}}-c_{\text{e}}\right)^{2}+4c_{ \text{u}}}}=\frac{2c_{\text{u}}c_{\textbackslash}}{\left(c_{\text{t}}\,c_{ \text{u}}+c_{\text{e}}\right)-\sqrt{\left(c_{\text{t}}\,c_{\text{u}}-c_{\text {e}}\right)^{2}+4c_{\text{u}}}}, \tag{29}\]  
or rewritten as  
\[\left(c_{\text{t}}\,c_{\text{u}}+c_{\text{e}}\right)-\sqrt{\left(c_{\text{t}} \,c_{\text{u}}-c_{\text{e}}\right)^{2}+4c_{\text{u}}}=\frac{2c_{\text{u}}c_{ \textbackslash}}{\omega_{+,\textbackslash}^{2}}. \tag{30}\]  
We solve the upper algebraic equation to obtain  
\[c_{\text{t}}=\frac{-c_{\text{u}}\,c_{\textbackslash}^{2}}{c_{\textbackslash }^{2}}\,\omega_{+,\textbackslash}^{4}+c_{\text{e}}\,c_{\textbackslash}^{2} \,\omega_{+,\textbackslash}^{2}\,\omega_{+,\textbackslash}^{2}+\omega_{+, \textbackslash}^{4}}{c_{\text{e}}\,\omega_{+,\textbackslash}^{4}}=\frac{2\pi} {2\times 0.0340\,\text{s}}. \tag{31}\]  
Upon examining Figure 7, it is evident that for the tested vehicle, \(\omega_{+,\textbackslash}^{2}=\frac{2\pi}{2\times 0.0675\,\textbackslash}=46.54\text{ rad}/\text{s}\) and \(\omega_{+,\textbackslash}^{3}=\frac{2\pi}{7\times 0.0340\,\textbackslash}=92.40\text{ rad}/\text{s}\), and thus \(c_{\text{t}}=3.86\).