# An Air Spring Vibration Isolator Based on a Negative-Stiffness Structure for Vehicle Seat
## 3 Stability Analysis of the Air-Spring Vibration Isolator System  
We analyzed the restoring force of the ASVIS; the schematic of the crisscross structure and negative stiffness mechanism without the sleeve air spring with stiffness \(K_{s}\) is displayed in Figure 3. The solid line indicates the initial position of the ASVIS, meaning that at this position, two bellows air springs and a sleeve air spring were uncompressed. The dashed line describes the static equilibrium position of the ASVIS when an isolated object with mass (M) is placed on the upper plate. In this diagram, \(z_{s}\) and \(z_{b}\) are the absolute displacementsof the isolated object and the base excitation, respectively, and \(z\) is the vertical static distance between the static equilibrium position and the highest position, which corresponds to the condition with no isolated object. Moreover, \(L_{h}\) is the length of the two straight bars of the crisscross mechanism, and \(\alpha\) is the angle that each straight bar makes with the horizontal direction. The horizontal distance between two rotation joints at points I and K is denoted by the symbol \(b\). Two bellows-type air springs, which are initially located at an angle \(\theta_{0}\) from the horizontal direction, have a stiffness \(K_{b}\) at points M and N. A force \(F_{b}\) is exerted by the bellows-type spring on point O, which is horizontally distanced \(L_{b}\) from the points M and N, while initially located at a height \(h\) from these points. Based on the geometrical relationships in the crisscross structure, the static distance \(z\) caused by mass M is obtained as  
\[z=2h-(b-L_{h}\cos\alpha)\tan\alpha \tag{17}\]  
The increment of the length of the bellows-type air spring can be written as  
\[\Delta L_{b}=L_{0}-L_{b}=\sqrt{h^{2}+{L_{b}}^{2}}-L_{b} \tag{18}\]  
where \(L_{0}\) is the length of the bellows-type air spring in an uncompressed state, and \(L_{b}\) is its length in the static equilibrium position. The force provided by the bellows-type air spring \(F_{b}\) is derived as  
\[F_{b}=2K_{b}\Delta L_{b}\sin\theta \tag{19}\]  
where \(K_{b}\) is the stiffness of the spring, and \(\sin\theta=h/\sqrt{h^{2}+L_{b}^{2}}\). Figure 3: Schematic diagram of the negative-stiffness structure and the crisscross structure. We can describe Equation (19) as follows  
\[F_{b}=2K_{b}h\left(1-\frac{L_{b}}{\sqrt{h^{2}+L_{b}^{2}}}\right) \tag{20}\]  
In the vertical direction, the total force \(F_{nss}\) supplied by the bellows-type air spring can be expressed by  
\[F_{nss}=2F_{b}\cos\alpha/\cos\alpha_{0} \tag{21}\]  
where \(\cos\alpha_{0}=b/L_{h}\) is the initial angle of a straight bar with the horizontal direction. Substituting Equations (17) and (20) into Equation (21), the general equation for the relationship between force \(F_{nss}\) and distance \(z\) in the vertical direction is as follows:  
\[F_{nss}=\left(\frac{2K_{b}\cos\alpha(z+(b-L_{h}\cos\alpha)\tan\alpha)}{\cos \alpha_{0}}\right)\left(1-\frac{L_{b}}{\sqrt{\left(\frac{z+(b-L_{b}\cos \alpha)\tan\alpha}{2}\right)^{2}+L_{b}^{2}}}\right) \tag{22}\]  
The dimensionless parameters are defined as follows  
\[\hat{F}_{nss}=\frac{F_{nss}}{K_{b}L_{b}};\ \hat{z}=\frac{z}{L_{b}};\ \hat{L}_{ h}=\frac{L_{h}}{L_{b}}\text{ and }\hat{b}=\frac{b}{L_{b}}\]  
where \(\hat{F}_{nss}\), \(\hat{z}\), \(\hat{L}_{h}\), and \(\hat{b}\) are the non-dimensional restoring force, the non-dimensional distance of \(z\), the non-dimensional length of the straight bar \(L_{h}\), and the non-dimensional distance of \(b\), respectively.