## 3 QZS Analysis
### Modeling of the QZS System  
The stiffness of a vibration isolator system strongly influences its effectiveness, especially at low- or ultra-low-frequency vibrations. As a reasonably designed HSLDS isolator can reduce most vibrations, the torsion stiffness should be analyzed during the first stage. Figure 2 demonstrates that two ends of the linear spring damper (\(l_{5}\)) connect with two \"T\"-frame rocking rods at points \(M\) and \(N\). This spring is horizontal at the initial position, and the rod positions \begin{table}
\begin{tabular}{c c c} \hline \hline
**Geometrical Parameters (mm)** & **Mass Parameters (kg)** & **Moment of Inertia (kg-m2)** \\ \hline \(l_{1}=200\) & \(m_{0}=30.0\) & \(l_{20}=0.3215\) \\ \(l_{2}=200\) & \(m_{1}=0.72\) & \(l_{21}=0.0038\) \\ \(l_{3}=242.5\) & \(m_{2}=2.45\) & \(l_{22}=0.0196\) \\ \(l_{4}=282.02\) & \(m_{3}=0.72\) & \(l_{23}=0.0038\) \\ \(l_{5}=299.17\) & \(m_{4}=1.50\) & \(l_{24}=0.0097\) \\ \(R=362\) & \(m_{5}=0.72\) & \(l_{25}=0.0038\) \\ \(L=200\) & \(m_{6}=2.45\) & \(l_{26}=0.0196\) \\ \(H=110\) & \(m_{7}=0.72\) & \(l_{27}=0.0038\) \\ \(d_{1}=141.01\) & - & - \\ \(d_{2}=159.7\) & - & - \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Mechanistic parameters. within the two parallelograms are symmetric. \(x_{M}\) and \(y_{M}\) represent the coordinates of \(M\), \(x_{N}\) and \(y_{N}\) represent the coordinates of \(N\). From the system structure, the following relationship can be obtained:  
\[\left\{ \begin{array}{l}x_{M}=-R\sin\theta+d_{1}\sin\phi-\frac{l_{4}}{2}\cos \phi-\frac{l_{3}}{20}\cos(\frac{\pi}{4}+\theta)\\ y_{M}=R\cos\theta-d_{1}\cos\phi-\frac{l_{4}}{2}\sin\phi-\frac{l_{3}}{20}\sin( \frac{\pi}{4}+\theta)\\ x_{N}=\frac{l_{4}}{2}\cos\phi-R\sin\theta+d_{1}\sin\phi+\frac{l_{3}}{20}\sin( \frac{\pi}{4}+\theta)\\ y_{N}=R\cos\theta-d_{1}\cos\phi+\frac{l_{4}}{2}\sin\phi-\frac{l_{3}}{20}\sin( \frac{\pi}{4}+\theta)\\ \end{array} \right. \tag{3}\]  
where \(k\) is the linear spring stiffness and \(l_{5}\) is the initial spring length. Assuming the length of the spring will change to \(l_{5}^{\prime}\) under the action of a force, we can obtain:  
\[f_{k}=k\big{(}l_{5}^{\prime}-l_{5}\big{)} \tag{4}\]  
where \(l_{5}^{\prime}=\sqrt{\big{(}x_{M}-x_{N}\big{)}^{2}+\big{(}y_{M}-y_{N}\big{)}^ {2}}\), \(l_{5}=l_{4}+\frac{\sqrt{2}}{20}l_{3}\). When \(\theta=0\), the static system stiffness can be obtained using the virtual work principle. Assuming that load \(m_{0}\) rotates around \(o_{2}\) to an angular \(\phi\) under external torque \(M_{t}\), the displacements of rods and elastic potential energy of the spring change simultaneously, and the influence of gravity of the lightweight spring damper on the system can be ignored. The energy caused by joint friction is also ignored in this paper. According to the virtual work principle, we can determine the following relationship:  
\[M_{t}\phi-(m_{0}g\Delta y_{0}+m_{1}g\Delta y_{1}+...+m_{7}g\Delta y_{7}+\frac {1}{2}k\Delta x^{2})=0 \tag{5}\]  
where \(\Delta y_{0}\), \(\Delta y_{1}...\Delta y_{7}\) represent the displacement of each respective rod's center of mass, \(\Delta x=l_{5}^{\prime}-l_{5}\). According to the torsional motion, the equivalent torsion stiffness of the system \(K_{t}\) can be expressed as:  
\[K_{t}=\frac{dM_{t}}{d\phi} \tag{6}\]  
Equations (5) and (6) are used to determine the equivalent stiffness. Torsional moment comprises two components: the torque exerted by the weight of the load and the isolator itself, and equivalent torque produced by the linear spring at \(o_{2}\). According to the mechanism kinematics, when \(\theta\) is equal to zero, the relationship between the equivalent torque, equivalent torsional stiffness, and rotation angle can be depicted as follows. Figure 3a demonstrates the strong non-linearity of the relationship between the equivalent torque and torsional angle at different gravity centers (\(m_{0}\)). Note that \(h=-36.87\) is a critical position where the equivalent torsion is approximately zero.