# Research on the Use of Hydro-Pneumatic Shock Absorbers for the Rear Suspension of a Vehicle Cabin
## 2 Materials and Methods
### Schematic Diagram of the Loading on an Elastic Element  
To find the force applied on the rear elastic element, we used a simplified diagram of the loading on the cabin suspension, as presented in Figure 3, including the cabin pivot hinge where we have the force R, the front axle axis between a and b and the simplified elastic element of the cabin suspension. We focused on finding the parameters given by the mass of the cabin, leading to finding the working parameters of the shock absorber that will influence not only the driver's position, but also have an impact on the passenger, as their seat usually has less damping than the driver's. We considered two elemental cases, one in which we calculated the force on the elastic rear element considering the loaded cabin, and the other one without any loading, taking into account only the force given by its own weight.  
The force (\(F\)) on the rear elastic element can be computed using the following equation:  
\[F=\frac{Q}{2 cos\alpha} \tag{1}\]  
where \(\alpha\) represents the shock absorber inclination angle.  
The calculation of the stiffness characteristic (\(c_{p}\)) is performed using the volumetric variations in the air cushion of the pneumatic element. For the numerical determination, the relationship below was used, taking into account the additional air volume (\(V_{s}\)) from the elastic element, as well as the dead volume of air (\(V_{m}\)) formed by the fixed internal volume of the cylinder in position 1. From this, the following volumes are subtracted: the cup 2, the elastic buffer 3, the rod 4, and the stroke limiter 5. Determinations will be conducted for two values of the polytropic coefficient (\(n\)), in initial conditions of pressure (\(p_{0}\)) for the empty cabin and a variable displacement (\(f\)). Another influencing factor is the static height of the variable volume (\(H_{0}\)) and the static height of the damping volume (\(h_{0}\)).  
\begin{table}
\begin{tabular}{c c} \hline
**Element** & **Value [mm]** \\ \hline H\({}_{0}\) & 84 \\ h\({}_{0}\) & 65 \\ d & 51 \\ D & 84 \\ t & 15.5 \\ \hline \end{tabular}
\end{table}
Table 1: Dimensions of the pneumatic shock absorber.  
Figure 3: Simplified diagram of the loading on the vehicle cabin suspension. G\({}_{0}\)—force due to the mass of the unloaded cabin; G\({}_{p}\)—force due to the mass of the loaded cabin; R—reaction force from the cabin’s pivot bearing (R\({}_{0}\) is the reaction when the cabin is empty, and \(\Delta\)R is the reaction caused by the load); Q—load in the damping element (Q\({}_{0}\) is the load when the cabin is empty, and \(\Delta\)Q is the load after it has been loaded); d—distance to the center of mass of the loaded cabin; c—distance to the center of mass of the empty cabin; a, b—distances to the wheel axis.  
As for the section, we considered the maximum area of the volume variation (\(A\)) and the area of the damper (\(A_{\text{d}}\)).  
\[c_{p}=\frac{n\cdot p_{0}(V_{s}+V_{m}+A\cdot H_{0}-A_{d}\cdot h_{0})^{n}\cdot(A+A _{d})\cdot A}{2\left[V_{s}+V_{m}+A\cdot H_{0}-A_{d}\cdot h_{0}-\frac{A+A_{d}}{2} \cdot f\right]^{n+1}} \tag{2}\]  
The load capacity of the elastic element (\(F_{p}\)) can be defined using Equation (3):  
\[F_{p}=\frac{p_{0}(V_{s}+V_{m}+A\cdot H_{0}-A_{d}\cdot h_{0})^{n}\cdot A}{\left[ V_{s}+V_{m}+A\cdot H_{0}-A_{d}\cdot h_{0}-\frac{A+A_{d}}{2}\cdot f\right]^{n}} \tag{3}\]  
As a safety measure when using elastic elements, one must check the frequency of the natural frequencies. In order to do so, we must know the required pressure in the elastic element to create the necessary lift to support the cabin. We will consider both cases, loaded cabin and own weight, to obtain the range of the stiffness. The stiffness (\(c\)) of the elastic element corresponding to these pressures (\(p\)) will be:  
\[c=(A+A_{d})\cdot A\frac{p}{2(V_{m}+A\cdot H_{0}-A_{d}\cdot h_{0})} \tag{4}\]  
The natural frequencies (\(n_{f}\)) can be computed using Equation (5), considering the mass of the cabin (\(m\)):  
\[n_{f}=\frac{30}{\pi}\sqrt{\frac{c}{m}} \tag{5}\]