## 3 Analysis of Climbing Performance for the Climbing Robot
### Anti-Skid Analysis in Operating Conditions  
Skid results from the lack of friction force between rolling wheels and cables, while friction is mainly transformed from the force generated by the spring. To study the relationship between spring force and friction force, some assumptions were made as follows: (1) The robot is climbing on a smooth cable without any obstacles in a constant velocity. For the wheel with the smallest support force, the driving torque of the motor equals the torque produced by the tangential friction forces. (2) All the driving motors are the same, therefore, we suppose the robot reached the maximum load when the  
Figure 6: Climbing principle. wheel with smallest support force is in the slipping state. (3) To simply the calculation, the self-weight of the swing plate is ignored as it has less effect on the robot climbing ability. That is to say, the two sets of robot swing plates are considered as two two-force bars. (4) All the inner frictions of the robot mechanism are ignored. According to Figures 6 and 7, which shows the schematic diagram of the movement of the robot, the balance equations of the three groups of rolling wheels are established. In Figure 7, \(\alpha\) and \(\gamma\) indicate the angles of two-force bars with the horizontal direction respectively and \(F_{t}\) represents the spring force. \(N_{a}\), \(N_{b}\) and \(N_{c}\) are the support force, which denote the constraint reactions of cables on rolling wheels, while \(F_{1a}\), \(F_{b}\), \(F_{2a}\) and \(F_{c}\) denote forces acting on the bars of the rolling wheels. According to the nature of two-force bars, it is obvious that \(F_{1a}=F_{b}\), \(F_{2a}=F_{c}\):  
As presented in Figure 7, the forces on rolling wheels A, B and C are presented, so the force balance equation for rolling wheel A is  
\[\left\{ \begin{array}{l}N_{a}+m_{a}g\sin\beta=F_{1a}\cos\alpha+F_{2a}\cos \gamma\\ F_{fa}+F_{1a}\sin\alpha=F_{2a}\sin\gamma+m_{a}g\cos\beta\\ \tau_{a}=F_{fa}r\end{array} \right. \tag{13}\]  
For the rolling wheel B we have  
\[\left\{ \begin{array}{l}N_{b}=F_{b}\cos\alpha+m_{b}g\sin\beta\\ F_{fb}=F_{b}\sin\alpha+F_{t}+m_{b}g\cos\beta\\ \tau_{b}=F_{fb}r\end{array} \right. \tag{14}\]  
Therefore we can get: \(N_{b}=\frac{F_{t}-m_{a}g\sin\beta fg\alpha+m_{b}g\cos\beta}{\mu-f\mathrm{ga}}\), For the rolling wheel C  
\[\left\{ \begin{array}{l}N_{c}=F_{c}\cos\gamma+m_{c}g\sin\beta\\ F_{fc}+F_{t}+F_{c}\sin\gamma=m_{c}g\cos\beta\\ \tau_{c}=F_{fc}r\end{array} \right. \tag{15}\]  
Therefore we can deduce: \(N_{c}=\frac{-F_{t}+m_{a}g\sin\beta fg\gamma+m_{c}g\cos\beta}{\mu+f\mathrm{g}}\), For the whole robot (Figure 6):  
\[\left\{ \begin{array}{l}N_{a}+Mg\sin\beta=N_{b}+N_{c}\\ F_{fa}+F_{fb}+F_{fc}\geq Mg\cos\beta\\ F_{fa}r+N_{b}L_{2}=N_{c}L_{1}+F_{fb}r+F_{fc}r\end{array} \right. \tag{16}\]  
In the above equations, \(F_{fa}\), \(F_{fb}\) and \(F_{fc}\) represents the friction force of three rolling wheels respectively. \(m_{a}\), \(m_{b}\) and \(m_{c}\) represents the mass of three rolling wheels. \(\tau_{a}\), \(\tau_{b}\) and \(\tau_{c}\) represents the torques acted on three rolling wheels. \(\mu\) denotes the friction coefficient between the rolling wheels and the cable surface. Figure 7: Force balance relationships of rolling wheels. Therefore we can get: \(N_{a}=N_{b}+N_{c}-Mg\sin\beta\). Combining Equations (13)-(16), the following constraint equation is obtained:  
\[\left\{ \begin{array}{l}N_{a}=N_{b}+N_{c}-Mg\sin\beta\\ N_{b}=\frac{F_{t}-m_{bg}\sin\beta fqx+m_{bg}\cos\beta}{\mu-gq}\\ N_{c}=\frac{-F_{t}+m_{c}g\sin\beta fq\gamma+m_{c}g\cos\beta}{\mu+g\gamma}\end{array} \right. \tag{17}\]  
For the three rolling wheels, we conclude the supporting force of the wheel C is the smallest one from the results the numerical analysis. Therefore, wheel C slides first. For wheel C climbing on the cable vertical cable (\(\beta\)= 0\({}^{\circ}\)), we suppose it is the critical state when \(F_{fc}=N_{c}\cdot\mu\). The motor whose rated torque \(\tau_{c}=F_{fc}r\) is chosen as the source of power.