# A Worm-Inspired Robot Flexibly Steering on Horizontal and Vertical Surfaces
## 4 Climbing Performance Analysis and Experiments
### Thrust Force and Load Capacity
In a similar way as the algorithm of Segment 1, we can get:  
\[ \begin{array}{l}F_{4}=F+F_{t3}(x)+F_{sl3}(x)+F_{sr3}(x)\\ =F+2 kr\sin(\theta_{3})+2k(D-x-x_{0})\end{array}  \tag{45}\]  
For Segment 2 of the robot, as shown in Figure 17a, we assume that the deflection angle of Segment 1 is \(\theta_{1}\) at the beginning of Phase 2. At some point in Phase 2, when the cylinder of Segment 2 has an elongation of x in the NM direction, the deflection angle of Segment 2 becomes \(\theta_{x}\), as obtained by Equation (32):  
\[\theta_{x}=\sin^{-1}\left(\frac{d\cdot\sin(\theta_{1})}{\sqrt{d^{2}+x^{2}+2 dx \cdot\cos(\theta_{1})}}\right) \tag{46}\]  
Figure 16: The force diagram at a certain point during Phase 1: (**a**) the force on Segment 1; (**b**) the force on Segment 4. In the diagram, \(\theta_{1}\) and \(\theta_{3}\) are the deflection angles of segments 1 and 3, and x is the cylinder elongation. Since the movement of Segment 2 is affected by the simultaneous action of the front and rear cylinders, the combined thrust in the direction of motion is:  
\[F_{cylinder}(x)=F(1+\cos(\theta_{x})) \tag{47}\]  
Since Segment 2 receives the positive tension of the spring at the same time, the following can be obtained in the same way as Equation (41):  
\[\left\{ \begin{array}{l}F_{s11}(x)=k(r\sin(\theta_{x})+MN^{\prime}-x_{0})\\ F_{s71}(x)=k(-r\sin(\theta_{x})+MN^{\prime}-x_{0})\end{array} \right. \tag{48}\]  
Based on the simultaneous use of Equations (47) and (48), the total traction of Segment 2 is:  
\[ \begin{array}{l}F_{2}=F_{cylinder}(x)+F_{s11}(x)+F_{s71}(x)\\ =F(1+\cos(\theta_{x}))+k(D-x-x_{0})\end{array}  \tag{49}\]  
The situation for Segment 3 is similar to that of Segment 2. The difference is that the spring force acting on Segment 3 will hinder the movement of the mechanism. From the geometric relationship, we can get:  
\[ \begin{array}{l}PO^{\prime}=\sqrt{OO^{\prime 2}+OP^{2}+2 OO^{\prime}\cdot OP \cdot\cos(\theta_{2})}\\ =\sqrt{x^{2}+d^{2}+2 xd\cos(\theta_{2})}\end{array}  \tag{50}\]  
\[ \begin{array}{l}OP\\ \sin(\theta_{x})\end{array} =\frac{O^{\prime}P}{\sin(\theta_{2})} \tag{51}\]  
According to simultaneous use of Equations (50) and (51), the following is obtained:  
\[ \begin{array}{l}\theta_{x}=\sin^{-1}\left(\frac{OP\sin(\theta_{2})}{O^{ \prime}P}\right)\\ =\sin^{-1}\left(\frac{d\sin(\theta_{2})}{\sqrt{x^{2}+d^{2}+2 xd\cos(\theta_{2} )}}\right)\end{array}  \tag{52}\]  
Similarly, since the movement of Segment 3 is affected by the simultaneous action of the front and rear cylinders, the combined thrust in the direction of motion (OO\({}^{\prime}\) direction) is:  
\[F_{cylinder}(x)=F(1+\cos(\theta_{x})) \tag{53}\]  
Figure 17: **(a)** A brief diagram of the force in Segment 2 at a certain point during Phase 2; **(b)** a brief diagram of the force in Segment 3 at a certain point during Phase 3. The dotted line in the diagram is the initial position before the movement. In the same way as Equation (41), the tension of the left spring and right spring \(F_{sl3}(x)\) and \(F_{sr3}(x)\) meet:  
\[\left\{ \begin{array}{l}F_{sl3}(x)=k(r\sin(\theta_{3})+PO^{\prime}-x_{0})\\ F_{sr3}(x)=k(-r\sin(\theta_{3})+PO^{\prime}-x_{0})\end{array} \right. \tag{54}\]  
According to simultaneous use of Equations (50), (53), and (54), \(F_{3}\), the total traction of Segment 3 is:  
\[ \begin{array}{l}F_{3}=F_{cylinder}(x)-[F_{sl3}(x)+F_{sr3}(x)]\cos(\theta_{x}) \\ =F(1+\cos(\theta_{x}))-2k\Big{(}\sqrt{x^{2}+d^{2}+2 xd\cos(\theta_{2})}-x_{0} \Big{)}\end{array}  \tag{55}\]  
It can be seen that the thrust output in Segment 3 is mainly affected by the deflection angle of Segment 2 in the initial phase and the cylinder position \(x\). The final calculated results of each of the four segments are shown in Figure 18. If the actual conditions permit \((\theta_{1},\theta_{2},\theta_{3}\leq 25^{\circ}\), \(x\leq 25\) mm), the minimum thrust of segments 1, 2, 3, and 4 are 28.59 N, 36.22 N, 66.96 N, and 62.78 N, respectively.