## 4 Discussion  
### Calculation of Overturning and Stabilizing Moments  
The maximum overturning \(M_{\rm O}\) and corresponding stabilizing moments \(M_{\rm S}\) shall be calculated for the most unfavorable tipping lines. In this case, tipping line is between supports of the scissor lift. For solid and foam-filled tires, according to ISO 4305, the tipping lines determined may be taken at 1/4 of the tire ground contact width from the outside of the ground contact width. The calculations are made with the scissor lift in the most unfavorable extended (configuration 1) with the maximum allowable inclination of the chassis equal to 0.5\({}^{\circ}\), as shown in Figure 11. We assume that the scissor lift used out-of-doors is being affected by wind at a pressure of 100 N/m\({}^{2}\), equivalent to a wind speed of 12.5 m/s (Beaufort Scale 6) according to EN 280-1:2022. All loads and forces which can act simultaneously are considered in their most unfavorable combinations according to the graphical method for on-slope stationary example presented in Figure 13.  
The platform is stable when the condition is met:  
\[M_{\rm S}>M_{O} \tag{5}\]  
Figure 11: Comparison of the force coefficient \(C_{x}\) and \(C_{y}\) for the experimental and numerical results as function of angle of wind attack \(\beta\) in the case of configuration 2. Figure 12: Comparison of the force coefficient \(C_{x}\) and \(C_{y}\) for the experimental and numerical results as function of angle of wind attack \(\beta\) at the case of configuration 3.  
where: \(M_{S}\) is the sum stabilizing moments, \(M_{O}\) is the sum overturning moments. The stabilizing moments, according to the standard EN 280-1:2022 (and level ground stationary), include:  
\[M_{S}=M_{S1}+M_{S2}. \tag{6}\]  
\(M_{S1}\)--is the moment resulting from the product of the structural load \(S\) = 43,164 N acting on the arm \(r_{1}\) equal to 1.1 m from the tipping line. \(S\)--vertical direction. The self-weight and all technical data of a real object are shown in Table 3. The masses of the components of the scissor lift are taken to be static structural loads (not moving), Table 7--factor equal to 1. \(M_{S2}\) = 181.29 Nm--is the moment resulting from the product of the person load equal to 784.8 N acting on the arm \(r_{3}\) equal to 0.231 m.  
To the overturning moments include:  
\[M_{O}=M_{O1}+M_{O2} \tag{7}\]  
\(M_{O1}\) = 7497.73 Nm--is the moment resulting from the product of the wind loads \(W\) = 956.65 N acting on the arm \(h_{1}\) equal to 7.125 m, \(W\)--horizontal direction. If the scissor lift is being affected by wind at a pressure of 100 N/m\({}^{2}\), equivalent to a wind speed of 12.5 m/s that the wind load acting on the scissor lift at the full area of the real object is \(A\) = 9.996 m\({}^{2}\) is equal to \(W\) = 956.65 N. Wind forces are assumed to act horizontally at the center of the area of the parts of the scissor lift and persons and equipment on the work platform \(h_{1}\). Wind load \(W\) is taken to be dynamic forces with factor 1.1 (Table 7). \begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline \multirow{2}{*}{**Working Condition**} & \multirow{2}{*}{**80 kg Up to Rated Load**} & \multirow{2}{*}{**Structural Loads (S)**} & \multirow{2}{*}{**Manual Force (M)**} & \multicolumn{3}{c}{**Wind Load (W)**} \\ \cline{2-2} \cline{5-8}  & \(\times\) 1.0 & \(\times\) 0.1 & \(\times\) 1.0 & \(\times\) 0.1 & \(\times\) 1.0 & \(\times\) 0.1 & \(\times\) 1.0 & \(\times\) 0.1 \\ \hline level ground stationary & V & - & V & - & A & A & H & H \\ \hline \hline \end{tabular}  
* Key: V = vertical; H = horizontal, A = angular, S = at slope angle.  
\end{table} 
Table 7: The load and force directions for stability calculations for scissor lift [17]. Figure 13: The schema of the load acting on real object scissor lift for level ground stationary example according to [17].