# A Multibody Model Calibration Method for Vehicle Misuse Testing
## 4 Nonlinear Parameter Calibration Using the Modified Quarter-Vehicle Model Analysis
### Equivalent Stiffness Calibration by Tire Detaching Stage  
It will be demonstrated that the modified model can effectively replicate the obstacle contact process with the equivalent stiffness \(k_{\mathrm{e}}\) and thus yield a practical misuse analysis method. Firstly, the displacement relationship between the sprung mass and the wheel's center is examined. With the wheel's center acceleration recorded during the tests, the wheel center's vertical displacement is computed by integrating twice with respect to time \(t\) as  
\[z_{\mathrm{w}}(t)=\int_{0}^{t}{(\int_{0}^{t}{\dot{z}_{\mathrm{w}}(t)\mathrm{d }t})\mathrm{d}t}. \tag{8}\]  
Comparing the numerical result from Equation (8) with the shock absorber's compression response \(d(t)\) depicted in Figure 7, an approximate quantitative relationship is identified for the current vehicle as  
\[d(t)\approx 0.73\times z_{\mathrm{w}}(t). \tag{9}\]  
Because the shock absorber is nearly vertical, with the compression distance relation  
\[d(t)\approx z_{\mathrm{w}}(t)-z_{\mathrm{u}}(t), \tag{10}\]  
the vertical displacements of the sprung point is obtained as  
\[z_{\mathrm{u}}(t)\approx 0.27\times z_{\mathrm{w}}(t) \tag{11}\]  
or in the vibration mode formulation as  
\[ \begin{bmatrix}z_{\mathrm{u}}(t)\\ z_{\mathrm{w}}(t)\end{bmatrix} \approx \begin{bmatrix}0.27\\ 1\end{bmatrix} z_{\mathrm{w}}(t). \tag{12}\]  
Relation (9) (also Equation (12)) is particularly well satisfied in stage 2, as depicted in Figure 7. In stage 2, the tire detaches from the obstacle, and the wheel's center vibrates vertically under the action of the shock absorber's force. Ignoring the constant gravitational forces, the vibration modes in stage 2 can be computed using the following eigen equations:  
\[ \begin{bmatrix}m_{\mathrm{u}}&0\\ 0&m_{\mathrm{w}}\end{bmatrix}\begin{bmatrix}{\ddot{z}_{\mathrm{u}}(t)}\\ {\ddot{z}_{\mathrm{w}}(t)}\end{bmatrix} + \begin{bmatrix}k_{\mathrm{e}}+k_{ \mathrm{s}}&-k_{\mathrm{s}}\\ -k_{\mathrm{s}}&k_{\mathrm{s}}\end{bmatrix}\begin{bmatrix}z_{\mathrm{u}}(t)\\ z_{\mathrm{w}}(t)\end{bmatrix} =0, \tag{13}\]  
Figure 7: The approximate quantitative relationship between the shock absorber compression \(d(t)\) and 0.73 times the wheel center’s vertical displacement \(0.73\times z_{\mathrm{w}}(t)\), with the same half-period of 0.0675 s.  
wherein \(m_{\rm u}\geq 0\) represents the sprung mass, \(m_{\rm w}\) represents the wheel's center mass, \(k_{\rm e}\geq 0\) represents the equivalent stiffness of the recovery force, and \(k_{\rm s}\) represents the suspension spring's stiffness. Since the tire's mass is neglected in the present paper and the lower end of the tire spring detaches from the ground in this stage, no excitation of the ground is caused by the tire. Consequently, no term related to \(k_{\rm t}\) appears in Equation (13). This stage responds as a free vibration with an initial speed and a position at the lower end. To simplify Equation (13), let \(c_{\rm u}=m_{\rm u}/m_{\rm w}\geq 0\), \(c_{\rm e}=(k_{\rm e}+k_{\rm s})/k_{\rm s}\geq 1\), and divide the equation by \(m_{\rm w}\) to obtain  
\[ \begin{bmatrix}c_{\rm u}&0\\ 0&1\end{bmatrix}\begin{bmatrix}\tilde{z}_{\rm u}(t)\\ \tilde{z}_{\rm w}(t)\end{bmatrix} +\frac{k_{\rm s}}{m_{\rm w}} \begin{bmatrix}c _{\rm e}&-1\\ -1&1\end{bmatrix}\begin{bmatrix}z_{\rm u}(t)\\ z_{\rm w}(t)\end{bmatrix} =0, \tag{14}\]  
Let \(\omega_{0}^{2}=k_{\rm s}/m_{\rm w}\) and hypothesize that the solution is \( \begin{bmatrix}z_{\rm u}(t)\\ z_{\rm w}(t)\end{bmatrix} =z\cdot\sin(\omega t+\varphi)\) with the eigenmode \(z\), the circular frequencies \(\omega\), and the initial phase angle \(\varphi\); the following algebraic equation is obtained:  
\[ \begin{cases}-\omega^{2}\begin{bmatrix}c_{\rm u}&0\\ 0&1\end{bmatrix}+\omega_{0}^{2}\begin{bmatrix}c_{\rm e}&-1\\ -1&1\end{bmatrix}\end{cases} z=0. \tag{15}\]  
Solving Equation (15) yields the eigenvectors as  
\[z_{\pm}= \begin{bmatrix}\frac{\left(c_{\rm u}-c_{\rm e}\right)\pm\sqrt{\left(c _{\rm u}-c_{\rm e}\right)^{2}+4c_{\rm u}}}{2c_{\rm u}}\\ 1\end{bmatrix} .