# Design, Experiment, and Improvement of a Quasi-Zero-Stiffness Vibration Isolation System
## 2 The QZS Vibration Isolation System
### Force-Displacement Characteristic of the Two Coupled Magnet Springs  
The configuration of the magnetic negative stiffness spring is shown in Figure 2. The magnetization directions of the inner and outer ring magnets are along the axial direction, as shown in Figure 2b. To understand the force-displacement characteristics, faces 2 and face 3 are used here as an example to calculate the magnetic forces in the mathematical modeling [40]. The magnetic charge of the point \(P\) on face 2 of the inner ring magnet is \(q_{2}\):  
\[q_{2}=\sigma r_{2}dadr_{2}, \tag{1}\]  
where \(r_{2}\) and \(\alpha\) are the polar coordinates of the point on the face 2 and \(\sigma\) is the surface density of magnetic charges of the ring magnets. Figure 2: The negative stiffness spring: (**a**) the negative spring system, (**b**) the cutaway view of the negative spring, and (**c**) the schematic for calculating the magnetic force. According to electromagnetic theory, the magnetic field intensity of point \(Q\) on face 3 of the outer ring magnet produced by point \(P\) is:  
\[\mathbf{H}=\frac{1}{4\pi\mu_{0}}\frac{q_{2}}{r^{2}}\mathbf{r}_{0}, \tag{2}\]  
where \(\mu_{0}\) is the permeability of the vacuum, \(r\) is the distance between point \(P\) on face 2 and point \(Q\) on face 3, and \(\mathbf{r}_{0}=\mathbf{r}/r\) is the unit vector between point \(P\) and point \(Q\). Therefore, the magnetic field force at point \(Q\) is:  
\[d\mathbf{F}=\mathbf{H}q_{3}=\mathbf{H}\sigma r_{3}d\beta dr_{3}=\frac{\sigma^ {2}}{4\pi\mu_{0}}\frac{r_{2}r_{3}dad\beta dr_{2}dr_{3}}{r^{3}}\mathbf{r}. \tag{3}\]  
where \(r_{3}\) and \(\beta\) are the polar coordinates of the point on face 3, and \(q_{3}=\sigma r_{3}d\beta dr_{3}\). According to the static magnetism formula, the surface density of magnetism and the body density of magnetism can be given as:  
\[\left\{ \begin{array}{l}\sigma=\mathbf{j}\mathbf{n}=\mu_{0}\mathbf{M} \mathbf{n}\\ \mathbf{B}=\mu_{0}\Big{(}\overline{\mathbf{H}}+\mathbf{M}\Big{)}\end{array} ,\right. \tag{4}\]  
where \(\mathbf{j}\) is the intensity of the magnetic pole vector, \(\mathbf{n}\) is the unit vector in the normal direction of the end face of the ring magnets, \(\overline{\mathbf{H}}\) is the electromagnetic intension force, and \(\mathbf{M}\) is the intensity of magnetization. Because no electric field is applied, \(\overline{H}=0\) and \(\mathbf{B}=\mu_{0}\mathbf{M}\). Due to the magnetization direction of the ring, the magnetism is parallel with the normal direction of the end face of the ring magnets. Therefore, Equation (4) can be given as:  
\[\sigma=\mu_{0}M=B_{r}, \tag{5}\]  
where \(B_{r}\) is the residual magnetic flux density. Then, Equation (3) can be expressed as:  
\[d\mathbf{F}=\frac{B_{r}^{2}}{4\pi\mu_{0}}\frac{r_{2}r_{3}dad\beta dr_{2}dr_{3 }}{r^{3}}\mathbf{r}=C\frac{r_{2}r_{3}dad\beta dr_{2}dr_{3}}{r^{3}}\mathbf{r}, \tag{6}\]  
where \(C=\frac{B_{r}^{2}}{4\pi\mu_{0}}\). The magnetic field force in the x-direction at point \(Q\) on face 3 of the outer ring magnet produced by point \(P\) is:  
\[dF_{32}=C\frac{r_{2}r_{3}dad\beta dr_{2}dr_{3}}{r^{3}}\mathbf{ri}, \tag{7}\]  
where \(\mathbf{i}\) is the unit vector in the x-direction. The geometric relationship between point \(P\) and point \(Q\) from Figure 2 can be given as:  
\[\mathbf{r}=\mathbf{r}^{{}^{\prime}}-\mathbf{x_{1}}=\mathbf{r_{3}}-\mathbf{r_ {2}}-\mathbf{x_{1}}, \tag{8}\]  
where \(\mathbf{x_{1}}\) is the relative axial displacement of the inner ring magnet relative to the outer ring magnet. Then:  
\[\mathbf{ri}=\mathbf{r_{3}}\mathbf{i}-\mathbf{r_{2}}\mathbf{i}-\mathbf{x_{1}} \mathbf{i}, \tag{9}\]  
and  
\[\left\{ \begin{array}{l}\mathbf{r_{3}}\mathbf{i}=r_{3}\sin\gamma\\ \mathbf{r_{2}}\mathbf{i}=r_{2}\sin\varphi\\ \mathbf{x_{1}}\mathbf{i}=x_{1}\end{array} \right.. \tag{10}\]  
Therefore, Equation (9) can be simplified to:  
\[\mathbf{ri}=r_{3}\sin\gamma-r_{2}\sin\varphi-x_{1}, \tag{11}\]where \(\sin\gamma=\frac{x_{1}}{\sqrt{\gamma_{3}^{2}+x_{1}^{2}}}\) and \(\sin\varphi=\frac{x_{1}}{\sqrt{\gamma_{2}^{2}+x_{1}^{2}}}\).