# A Multibody Model Calibration Method for Vehicle Misuse Testing
## 4 Nonlinear Parameter Calibration Using the Modified Quarter-Vehicle Model Analysis
### Equivalent Obstacle Parameter Calibration by Obstacle Contact Stage  
As depicted in Figure 6, in comparison to the model from stage 3, the lower point of the tire spring ascends as the vehicle moves forward in stage 1. Considering the potential nonlinearity in the actual tire response, an equivalent method to that presented by [PERSON] in the reference [31] is adopted for the speed bump's geometry. Instead of an equivalent shape to that in the reference [31], it is assumed that the lower end moves upward with an equivalent vertical velocity \(v_{\text{e}}\). Then, the obstacle contact process is described by the following equation (ignoring gravity):  
\[ \begin{bmatrix}m_{\text{u}}&0\\ 0&m_{\text{w}}\end{bmatrix}\begin{bmatrix}\dot{z}_{\text{u}}(t)\\ \dot{z}_{\text{w}}(t)\end{bmatrix} + \begin{bmatrix}k_{\text{e}}+k_{\text{s}}&- k_{\text{s}}\\ -k_{\text{s}}&k_{\text{s}}+k_{\text{t}}\end{bmatrix}\begin{bmatrix}z_{\text{u}}(t) \\ z_{\text{w}}(t)\end{bmatrix} = \begin{bmatrix}0\\ k_{\text{t}}v_{\text{e}}t\end{bmatrix} . \tag{32}\]  
The difference between stage 1 and stage 3 is that the former features a slope on the speed bump. When the tire rolls onto the slope, the ground ascends relative to the wheel's center. Therefore, in addition to the term \(k_{\text{t}}\) on the left side, as seen in Equation (25), there is an additional excitation force \(k_{\text{t}}v_{\text{e}}t\) on the right side due to the \"ascending\" ground. Divide the two sides by \(m_{\text{w}}\), and use the relation \(c_{\text{t}}=\left(k_{\text{t}}+k_{\text{s}}\right)/k_{\text{s}}\), \(c_{\text{e}}=\left(k_{\text{e}}+k_{\text{s}}\right)/k_{\text{s}}\); the above equation is simplified as  
\[ \begin{bmatrix}c_{\text{u}}&0\\ 0&1\end{bmatrix}\begin{bmatrix}\dot{z}_{\text{u}}(t)\\ \dot{z}_{\text{w}}(t)\end{bmatrix} +\frac{k_{\text{s}}}{m_{\text{w}}} \begin{bmatrix}c_ {\text{e}}&-1\\ -1&c_{\text{t}}\end{bmatrix}\begin{bmatrix}z_{\text{u}}(t)\\ z_{\text{w}}(t)\end{bmatrix} =\frac{k_{\text{t}}}{m_{\text{w}}} \begin{bmatrix}0 \\ v_{\text{e}}t\end{bmatrix} . \tag{33}\]When an effort is made to solve the two-DOF equation theoretically using Laplace transformation, the expression of the result becomes somewhat complex, making it difficult to extract the necessary information for calibration. Therefore, a single-DOF vibrational model, as depicted in Figure 8, is considered to extract useful information from its theoretical solution. The equivalence is as follows: If two systems have the same natural frequencies, modes, and boundary conditions, their responses will be identical, and thus they will be equivalent. If the main response components of the original system can be preserved in a system with fewer DOFs, then the latter can be regarded as a simplified version of the original system. In fact, only one of the two modes in the two-DOF mass-spring system has principally been excited by the slope of the speed bump. This can be illustrated by the frequency and mode analyses in Equations (16) and (17), where the natural frequencies and directions of motion are distinct for the two modes. Only the synclastic mode is likely to be excited; the opposite cannot be the main one. Therefore, with the same boundary conditions at the lower end and the same natural frequency, the single-DOF vibrational model can be considered a simplification of the original two-DOF system. For the model in Figure 8 with the mass \(m\), the structural spring stiffness \(k\), and the tire stiffness \(k_{\text{t}}\), the system equation is written as  
\[m\ddot{z}+(k+k_{\text{t}})z=k_{\text{t}}v_{\text{e}}t. \tag{34}\]  
Suppose the initial condition is \(\ddot{z}(0)=\dot{z}(0)=0\); we apply the Laplace transformation [32] to Equation (34) and obtain  
\[(ms^{2}+k+k_{\text{t}})Z=\frac{k_{\text{t}}v_{\text{e}}}{s^{2}}, \tag{35}\]  
wherein \(Z(s)=\mathcal{L}(z(t))\) is the Laplace image of the displacement \(z(t)\).