## 3 Analysis of Suspension Damping Coefficient  
### Calculation of Equivalent Damping Coefficient  
Based on the above introduction to the seat suspension mechanism, the complexity of the seat suspension structure can be simplified. To facilitate the study of this seat type, a simplified air suspension model is developed and presented in Figure 2.  
Figure 1: The air spring suspension. The simplified mathematical structure and the overall force analysis of the simplified model of the seat with scissor suspension springs are shown in Figure 3, where 1 is the seat upper plate, 2 and 3 are the scissor bars, 4 is the seat base plate, 5 is the air spring, and 6 is the damper. The upper plate of the seat and the bottom plate are connected by scissor rods, which are hinged to each other at point O. The air spring force and the damping from the dampers act through the scissor rods. The air spring is arranged between the scissor bar and the base plate, and the air spring force and the damping force of the damping damper act on the seat upper plate through the scissor bars. The scissor bar 2 intersects with the seat upper plate at D, while the scissor bar 3 intersects with the seat upper plate at A. The lower end of the scissor bar B can be slid along the track in the linear groove on the right side of the seat base plate. The air spring 5 is arranged between the scissor bar and the base plate, and vibration damper 6 is arranged above the crossbar in the middle of the base plate and the scissor frame. Firstly, the air spring force \(F_{k}\), damper force \(F_{by}\), and friction force \(F_{fb}\) are calculated, and the calculation results are as follows. \(F_{fb}\) is the friction force caused by the contact between the moving parts, which can cause damping. In the following formula, \(k\) is air  
Figure 3: Simplified mathematical structure of the seat and overall force diagram of the suspension. 1 is the seat upper plate, 2 and 3 are the scissor bars, 4 is the seat base plate, 5 is the air spring, and 6 is the damper. The red boxes refer to rod1 and rod2 respectively. Figure 2: Simplified schematic diagram of the suspension. spring stiffness, \(c\) is the damping coefficient of the damper, \(y_{0}\) is the sinusoidal excitation, and \(y\) is vibration response. \[F_{k}=\frac{k(y-y_{0})l_{2}}{l_{1}+l_{2}} \tag{1}\]  
\[F_{fb}=\mu F_{by} \tag{2}\]  
\[F_{d}=c(\dot{y}-\dot{y}_{0}) \tag{3}\]  
The force equilibrium equations of the scissor frame are established as follows:  
\[ \begin{array}{l}(\mathrm{a}):\sum X=0,F_{cx}+F_{bx}-F_{d}\cos\theta=0\\ (\mathrm{b}):\sum Y=0,F_{cy}+F_{by}-m\ddot{y}+F_{d}\sin\theta-F_{k}=0\\ (\mathrm{c}):\sum M_{c}=0,2F_{by}l_{2}\cos\alpha-m\ddot{y}l_{2}\cos\alpha-F_{k }l_{2}\cos\alpha=0\end{array}  \tag{4}\]  
where (a) is the force balance equation in the x-direction; (b) is the force balance equation in the y-direction; and (c) is the overall moment balance equation of the suspension model. \(F_{bx}\) and \(F_{cx}\) are the horizontal forces at point b and c, and \(F_{by}\) and \(F_{cy}\) are the vertical forces at point b and c. The vertical force at point B can be obtained from Equation (5):  
\[F_{by}=\frac{1}{2}[m\ddot{y}+\frac{k(y-y_{s})}{l_{1}+l_{2}}l_{2}] \tag{5}\]  
The equilibrium equations are established for the local force analysis of the scissor frame, as shown in Figure 4. Figure 4a shows the force diagram for rod1, and Figure 4b shows the force diagram for rod2. The forces acting on rod1 and rod2 are analyzed as follows:  
\[ \begin{array}{l}(\mathrm{a}):\sum X=0,F_{ox}-F_{ax}^{\prime}+F_{bx}=0\\ (\mathrm{b}):\sum Y=0,F_{by}+F_{oy}^{\prime}-F_{ay}=0\\ (\mathrm{c}):\sum M_{O}=0,F_{ax}l_{1}\sin\alpha+F_{ay}l_{1}\cos\alpha+F_{by}l_ {2}\cos\alpha+F_{by}l_{2}\sin\alpha=0\end{array}  \tag{6}\]  
where (a) is the force balance equation in the x-direction; (b) is the force balance equation in the y-direction; and (c) is the force balance equation at point O of rod1 in the suspension model.