# Development and Evaluation of the Traction Characteristics of a Crawler EOD Robot
## 4 Simulation Singularity Crossing Obstacles
### Simulation of Crossing the Step-Type Obstacle  
The simulation of crossing a step-type obstacle was performed in two variants (Figure 22): when climbing the obstacle; and suspended on the obstacle. Using the schematic in Figure 22, the following geometric relationships are calculated:  
\[\cos\theta=\frac{R-h}{R}\ [-],\ \sin\theta=\frac{\sqrt{h(2R-h)}}{R}\ [-] \tag{35}\]  
The equilibrium equations for the forces and moments acting on the vehicle are as follows:  
\[\sum F_{X}=0=\mu\gamma_{r}F_{Nr}+\mu\gamma_{f}F_{Nf}\cos\theta-F_{Nf}\sin \theta\ [N], \tag{36}\]  
\[\sum F_{Y}=0=F_{Nr}-mg+F_{Nf}\cos\theta+\mu\gamma_{f}F_{Nf}\sin\theta\ [N], \tag{37}\]  
\[\sum M=0=-F_{Nr}(L_{w}+R\sin\theta)+mg\big{(}L_{cg}+R\sin\theta\big{)}\ [ Nm]. \tag{38}\]  
The following notations are introduced:  
\[ \begin{array}{c}A=L_{\text{CG}}+R\sin\theta\ [m],\ B=L_{w}-L_{cg}\ [m],\ S=\sin \theta\ [-],\\ C=\cos\theta\ [-]\end{array}  \tag{39}\]  
From Equations (35)-(42), the following second-degree equation is finally obtained:  
\[\mu^{2}\gamma_{r}\gamma_{f}AS+\mu C\Big{(}A\gamma_{r}+B\gamma_{f}\Big{)}-SB=0, \tag{40}\] \begin{table}
\begin{tabular}{c c c} \hline \hline
**Characteristics** & **U.M.** & **Amount** \\ \hline Rear wheel diameter, \(D_{rw}\) & mm & 350 \\ Front wheel diameter, \(D_{fw}\) & mm & 350 \\ Wheelbase, \(L_{w}\) & mm & 1200 \\ Distance between front axe and center of gravity, \(L_{cg}\) & mm & 600 \\ Height of the center of gravity relative to the & mm & 650 \\ running track, \(h_{mg}\) & kg & 250 \\ The mass of the vehicle, \(m\) & N & 2450 \\ The weight of the vehicle, \(mg\) & - & 0.8 \\ Coefficient of adhesion, \(\mu\) & - & 0.8 \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Characteristics of the crawler robot used for simulation. Figure 22: Schematic of crossing a step-type obstacle: (**a**) shows the passage over the obstacle with the first cheek, in front of the robot; (**b**) follows the approach of the obstacle with the driving wheel. with the next positive solution:  
\[\mu\geq\frac{-C\Big{(}A\gamma_{r}+B\gamma_{f}\Big{)}+\sqrt{C^{2}\Big{(}A\gamma_{r }+B\gamma_{f}\Big{)}^{2}+4\gamma_{r}\gamma_{f}ABS^{2}}}{2\gamma_{r}\gamma_{f}AS}. \tag{41}\]  
Relationship (41) allows one to determine the coefficient of adhesion required to cross the obstacle. From Equations (36)-(38) result the expressions of the normal forces acting on the drive wheels:  
\[F_{Nr}=\frac{mgA}{L_{w}+RS}\;[N],\;F_{Nf}=\frac{mgB}{(L_{w}+RS)\Big{(}C+\mu \gamma_{f}S\Big{)}}\;[N] \tag{42}\]  
If the obstacle is crossed by a single axis, the following equations are used:  
\[\sum F_{X}=0=\mu\gamma_{r}F_{Nr}+\mu\gamma_{f}F_{Nf}\cos\theta-F_{Nf}\sin \theta/2\;[N]. \tag{43}\]  
\[\mu\geq\frac{-C\Big{(}A\gamma_{r}+B\gamma_{f}\Big{)}+\sqrt{C^{2}\Big{(}A\gamma _{r}+B\gamma_{f}\Big{)}^{2}+2\gamma_{r}\gamma_{f}ABS^{2}}}{2\gamma_{r}\gamma _{f}AS}\;[-]. \tag{44}\]  
\[F_{Nr}=\frac{mgA}{L_{w}+RS}\;[N].,\;F_{Nf}=\frac{mgB}{(L_{w}+RS)\Big{(}C+\mu \gamma_{f}S\Big{)}}\;[N] \tag{45}\]  
The calculations were performed tabularly in Microsoft Excel. The results obtained are presented in Tables 3 and 4. The data obtained allow for an assessment of the demands of the driving wheels that attack the obstacle; the coefficient of adhesion required for the development of the traction force; the requirements imposed on the energy aggregate (engine and transmission) to overcome the obstacle. From the analysis of the obtained data, it can be seen that the most important variation is the coefficient of adhesion necessary to allow for the crossing of the obstacle. For the two situations studied, Figures 23 and 24 show the variation in the minimum adhesion coefficient that ensures the crossing of the obstacle, depending on its height.