## 3 Robot Motion Characteristics Analysis
### Kinematic Model Analysis
#### 3.1.2 Motion Analysis of the Robot Wheel-Legged Mode  
Figure 6 shows the switching process of the transformed wheel from wheel to wheel-legged motion.  
Without considering the slip, the robot achieves three cycles of \(120^{\circ}\) phase motion for each week of the rotation of the transformed wheel, as shown in Figure 7. At the beginning of one cycle, trigger leg 1 and trigger leg 3 directly touch the ground. As the motor rotates, trigger leg 1 is gradually lifted and trigger leg 3 begins to make a circular motion at its contact point with the ground. Until trigger leg 3 completes the change from the front leg to the rear leg of the wheel. At this point, trigger leg 3 makes a circular motion along the ground. Trigger leg 2 makes contact with the ground and becomes the front leg of the wheel [22].  
Figure 8 shows the geometric analysis model of the wheel-legged mode. We define the transformed wheel motion process as \(T_{0}=kT/3\), \(T_{1}=kT/3+t_{1}\), \(T_{2}=(k+1)T/3\), where \(O_{1}\) is the circle center of the transformed wheel, \(O_{2}\) is the circle center of the circular arc segment of the wheel trigger leg, point \(B\) is the outermost endpoint of the wheel trigger leg 1, point \(C\) is the contact point between the wheel trigger leg 1 and the ground, and point \(D\) is the location point where the wheel trigger leg 2 is connected to the hub. The distance between \(O_{1}\) and point \(B\) is the effective radius \(r_{w}\) of the wheel, and the distance between  
Figure 6: Wheel and wheel-legged mode switch.  
Figure 7: Wheel change process in one cycle.  
point \(O_{1}\) and point \(O_{2}\) is \(L_{1}\), angle\(O_{2}BO_{1}=\alpha_{1}\), angle\(O_{1}O_{2}P=\alpha_{2}\), angle\(O_{2}O_{1}B=\alpha_{3}\), and angle\(O_{1}O_{2}E=\gamma\).  
According to the geometric relationship in the figure, when \(t=T_{0}\), the relationship between the transformed wheel centers \(S_{t=0}\) and \(H_{t=0}\) is  
\[\left\{\begin{aligned} S_{t=0}=0\\ H_{t=0}=R+L_{1}\cos\gamma\end{aligned}\right. \tag{6}\]  
where:  
\[\varphi_{t=0}=\arcsin(H_{t=0}/r_{w}) \tag{7}\]  
To solve \(H_{t=0}\), it can be inferred that  
\[\left\{\begin{aligned} L_{1}=\sqrt{R^{2}+{r_{w}}^{2}-2{r_{w}}\times R \times\cos\alpha_{1}}\\ \alpha_{1}=\frac{\pi}{6}-\arccos[\frac{{r_{w}}^{2}+(\sqrt{3}R)^{2 }-O_{1}{D}^{2}}{2\sqrt{3}R\times r_{w}}\\ O_{1}D=\sqrt{{L_{1}}^{2}+R^{2}-2L_{1}}\times R\times\cos\alpha_{2}\\ \alpha_{3}=\arccos(\frac{{r_{w}}^{2}+L_{1}^{2}-R^{2}}{2L_{1}\times r _{w}})\end{aligned}\right. \tag{8}\]  
The relationship between the height of the transformed wheel center \(H_{t=0}\), as a function of the effective radius \(r_{w}\) can be expressed as  
\[H_{t=0}=H(r_{w}) \tag{9}\]  
When \(t=T_{1}\), trigger leg 3's outer arc circle \(O_{2}\)' and the ground contact point \(N\) of the line perpendicular to the ground, at the time \(T_{1}\), can be expressed as  
\[T_{1}=\frac{kT}{3}+t_{1}=\frac{kT}{3}+\frac{\pi/2-\varphi_{t=0}+\angle{O_{2}} ^{\prime}{NO_{1}}^{\prime}}{w} \tag{10}\]  
According to the symmetric distribution of the transformed wheel, the relationship between \(S_{L}\) and \(H_  
effective radius of the trigger leg spread. The different degrees of transformation of the transformed wheel affect the other structural parameters of the wheel, and the \(\varphi\) angle at the beginning of each motion cycle also affects the motion characteristics of the wheel.