## 3 Simulation Results and Impact Analysis  
### Analysis of Actuator  
In order to determine the rebound coefficient and verify the correctness of the mathematical model of the actuator, the parameters of the actuator on the experimental platform were taken as an example to simulate the actuator motion, and then compared with the experimental data. Table 2 is the parameters of the actuator, in which the maximum action time is the key parameter of the traditional design. The structure and performance parameters of the actuator, including hydraulic force, spring stiffness, impact velocity and rebound coefficient, are discussed. Figure 4 shows the movement curve of the actuator with different rebound coefficients and the same oil pressure. In order to make the results repeatable, two sets of data are selected for analysis. According to the calculation results, the simulated ejection time of the actuator is faster than the experimental value. \(t_{B1}-t_{A1}=0.0041\) s, \(t_{D1}-t_{C1}=0.0042\) s, The reason is that the hydraulic pipe, valve and other resistance losses during the measurement lead to the reduction of the hydraulic force. As can be seen from Table 3, when \(\gamma=-0.1\sim-0.05\), the first rebound distance is basically consistent with the experimental value. Because the output pressure of the hydraulic system is not stable, there are errors, and the measured value has a lot of fluctuations in the balance stage. Since the ejection and withdrawal impact surfaces of the actuator are similar, the rebound coefficient of the actuator is set at \(-0.075\). \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Parameter** & **Symbol** & **Unit** & **Value** \\ \hline Mass & \(m\) & kg & 1.5 \\ Friction & \(f\) & N & 30 \\ Maximum compression of spring & \(x_{max}\) & m & 0.0075 \\ Stroke & \(L\) & m & 0.003 \\ Spring stiffness & \(x_{max}\) & N/m & 18,425 \\ Maximum action time & \(t_{max}\) & s & 0.012 \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Actuator parameters. Figure 5 shows the movement curve of actuator unable to overcome spring force and gas force when the hydraulic pressure is small (p1 = 1.3 MPa). As can be seen from the figure, the difference between \(t_{OA}\) and \(t_{OB}\), \(d_{A}-d_{C}\) and \(d_{B}-d_{D}\) is small. The accuracy and feasibility of the mathematical model are verified. For the ejection process, the hydraulic system is the driving source and the spring is the resistance element. For the withdrawal process, the hydraulic system has no output force, and the spring force releases the accumulated energy during the ejection process as the driving source. Therefore, in the ejection process, the impact velocity is positively correlated with the hydraulic force, and negatively correlated with the spring stiffness. In the withdrawal process, the impact velocity has nothing to do  
Figure 4: Movement curve of the actuator with different rebound coefficients and the same oil pressure. Figure 5: Movement curve of actuator with small hydraulic pressure. \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Rebound** & \multicolumn{2}{c}{**p1 = 4 MPa**} & \multicolumn{2}{c}{**p1 = 3 MPa**} \\
**Coefficient** & **Experiment Value** & **Simulation Value** & **Experiment Value** & **Simulation Value** \\ \hline \(-\)0.2 & & 0.282 & & 0.194 \\ \(-\)0.1 & 0.05 & 0.085 & 0.05 & 0.052 \\ \(-\)0.05 & & 0.044 & & 0.025 \\ \hline \hline \end{tabular}
\end{table} 
Table 3: The first bounce distance of different rebound coefficients. with the hydraulic force, but has a positive correlation with the spring stiffness. Since the speed after the rebound is small, the speed of the rebound is not considered. Take the derivative of Equation (5) to obtain the function of actuator speed, spring stiffness and hydraulic force:  
\[v(t)=\left\{