# A Worm-Inspired Robot Flexibly Steering on Horizontal and Vertical Surfaces
## 4 Climbing Performance Analysis and Experiments
### Thrust Force and Load Capacity  
In order to withstand its own weight or even extra load during the climbing process, the worm-like robot needs a large traction force. As mentioned earlier, the traction force is primarily provided by the thrust of the cylinders mounted on each segment. However, in the process of the steering movement, the thrust directions of the respective cylinders are different, and considering the action of the spring and the tendon, the thrust that can be actually output finally is more affected by factors such as angles \(\theta_{1}\), \(\theta_{2}\), and \(\theta_{3}\), and the position of the cylinder piston. Below, the thrust of the robot output in each stage will be calculated in detail, and the maximum load that each segment can withstand will also be calculated. The prototype of the robot adopts CDJ2B16-25 cylinders; its performance parameters are shown in Table 1:  
According to the experiment, the maximum air pressure generated by the air pump is 57 kPa, under which the maximum thrust/pull force generated by the cylinder of this type is 57 kPa. The maximum thrust \(F_{0}\) that each cylinder can provide is 11.4 N. Since each segment of the robot uses three cylinders arranged side by side, the thrust generated is \(F=3F_{0}=34.2\)N.  
Segment 1 of the robot, during the elongation of the cylinder, is hindered by the tension of the springs on both sides and the tension of the tendon. As shown in Figure 15(a), it is assumed that at a certain moment of the movement at Phase 1, the motion displacement of Segment 1 is x, the length of the first cylinder is \(\text{MN}=d+x\), and the angle between Segment 1 and Segment 2 is \(\theta_{x}\). Since the value of \(\theta_{x}\) is small, in order to simplify the calculation, we regard AC and BD as approximately parallel. Experimental tests have shown that the above approximation has a very small adverse effect on the results. From the geometric relationship, the length of the left spring \(L_{\text{I1}}(x)\) and the length of the right spring \(L_{\tau 1}(x)\) satisfy:  
\[\left\{ \begin{array}{l}L_{\tau 1}(x)=r\sin(\theta_{1})+MN\\ L_{\tau 1}(x)=-r\sin(\theta_{1})+MN\end{array} \right. \tag{40}\] \begin{table}
\begin{tabular}{c c c c} \hline
**Inner Diameter** & **Total Minimal Length** & **Piston Stroke** & **Maximum Working Pressure** \\ \hline
16 mm & 75 mm & 25 mm & \(\pm\)0.7 MPa \\ \hline \end{tabular}
\end{table} 
Table 1: Cylinder performance parameters. Figure 15: **(a)** Trajectory diagram of the climbing motion on a vertical surface in 120 cycles. At the initial phase (\(L_{2}=400\) mm, \(\theta_{2}=20^{\circ}\)), the trajectory can be regarded as a circle. **(b)** The relationship between the diameter of the circle and the length of tendon \(L_{2}\) in different motion patterns at the initial phase: \(\varphi_{0}=0\). Furthermore, the spring forces of the left spring and the right spring \(F_{sl1}\) and \(F_{sr1}\) meet:  
\[\left\{ \begin{array}{l}F_{sl1}(x)=k(r\sin(\theta_{1})+MN-x_{0})\\ F_{sr1}(x)=k(-r\sin(\theta_{1})+MN-x_{0})\end{array} \right. \tag{41}\]  
According to the equation of torque equilibrium centered on point N, the tension in the right tendon \(F_{t1}(x)\) is obtained to satisfy:  
\[F_{sl1}(x)\cdot r=F_{sr1}(x)\cdot r+F_{t1}\cdot r \tag{42}\]  
Based on simultaneous use of Equations (41) and (42), we can get:  
\[F_{t1}(x)=2 kr\sin(\theta_{1}) \tag{43}\]  
The final remaining propulsion force is:  
\[ \begin{array}{l}F_{1}=F-F_{t1}(x)-F_{sl1}(x)-F_{sr1}(x)\\ =F-2 kr\sin(\theta_{1})-2k(d+x-x_{0})\end{array}  \tag{44}\]  
It can be seen that the thrust output from Segment 1 is mainly affected by the deflection angle \(\theta_{1}\) and the cylinder position \(x\). For Segment 4 of the robot, as shown in Figure 16b, the force is similar to that of Segment 1. The only difference is that the tension of the spring and the tendon is the same as the thrust of the cylinder in one direction, and the external forces will enhance the overall traction of the mechanism.