# Structural Design and Analysis of Multi-Directional Foot Mobile Robot
## 2 Design and Analysis of Closed-Chain Leg Structure
### Kinematics Analysis of Closed-Chain Single-Leg Mechanism
Since \(\mathbf{r}_{5}=\mathbf{r}_{2}-\mathbf{r}_{4}\), Formula (5) becomes  
\[\mathbf{r}_{2}-\mathbf{r}_{4}+\mathbf{r}_{6}+\mathbf{r}_{8}-\mathbf{r}_{7}=\mathbf{0}_{2\times 1}. \tag{16}\]\(X\) direction decomposition results in  
\[r_{2}\cos\theta_{2}-r_{4}\cos\theta_{4}+r_{6}\cos\theta_{6}+r_{8}\cos\theta_{8}-r _{7}\cos\theta_{7}=0. \tag{17}\]  
\(Y\) direction decomposition results in  
\[r_{2}\sin\theta_{2}-r_{4}\sin\theta_{4}+r_{6}\sin\theta_{6}+r_{8}\sin\theta_{8}- r_{7}\sin\theta_{7}=0, \tag{18}\]  
where \(r_{2},r_{4},\theta_{2},\theta_{4}=\theta_{2}+\theta_{2-4},r_{6},\theta_{6}= \theta_{3}+\theta_{3-6},r_{8},r_{7}\) are all known quantities, and \(\theta_{7}\) and \(\theta_{8}\) are unknown quantities. After transforming Equations (17) and (18) to solve \(\theta_{7}\), we can obtain  
\[r_{2}\cos\theta_{2}-r_{4}\cos\theta_{4}+r_{6}\cos\theta_{6}-r_{7}\cos\theta_{7 }=-r_{8}\cos\theta_{8}, \tag{19}\]  
and  
\[r_{2}\sin\theta_{2}-r_{4}\sin\theta_{4}+r_{6}\sin\theta_{6}-r_{7}\sin\theta_{7 }=-r_{8}\sin\theta_{8}. \tag{20}\]  
After adding the squares of the two sides of the above formula, we simplify it to obtain  
\[A\cos\theta_{7}+B\sin\theta_{7}+C=0, \tag{21}\]  
where  
\[ \begin{cases}A=-2r_{2}r_{7}\cos\theta_{2}+2r_{4}r_{7}\cos\theta_{4}-2r_{6}r_{ 7}\cos\theta_{6}\\ B=-2r_{2}r_{7}\sin\theta_{2}+2r_{4}r_{7}\sin\theta_{4}-2r_{6}r_{7}\sin\theta_{6} \\ C=r_{2}^{2}+r_{4}^{2}+r_{6}^{2}+r_{7}^{2}-r_{8}^{2}-2r_{2}r_{4}\cos(\theta_{2} -\theta_{4})+2r_{2}r_{6}\cos(\theta_{2}-\theta_{6})-2r_{4}r_{6}\cos(\theta_{4 }-\theta_{6}).\end{cases} \]  
With order \(t=\tan\dfrac{\theta_{7}}{2}\), there are \(\sin\theta_{7}=\dfrac{2t}{1+t^{2}},\cos\theta_{7}=\dfrac{1-t^{2}}{1+t^{2}}\). Substituting these into Equation (21), we can obtain  
\[(C-A)t^{2}+2 Bt+(A+C)=0. \tag{22}\]  
The solution is  
\[ \begin{cases}t=\dfrac{-B\pm\sqrt{B^{2}-C^{2}+A^{2}}}{C-A}\\ \theta_{7}=2\tan^{-1}(t),\end{cases} \]  
and the above equation must satisfy \(B^{2}-C^{2}+A^{2}\geq 0\). In the same way, Formulas (17) and (18) are deformed and solved for \(\theta_{8}\), and we can obtain  
\[r_{2}\cos\theta_{2}-r_{4}\cos\theta_{4}+r_{6}\cos\theta_{6}+r_{8}\cos\theta_{8 }=r_{7}\cos\theta_{7}, \tag{23}\]  
and  
\[r_{2}\sin\theta_{2}-r_{4}\sin\theta_{4}+r_{6}\sin\theta_{6}+r_{8}\sin\theta_{8 }=r_{7}\sin\theta_{7}. \tag{24}\]  
Adding the squares of the two sides of the above formula, we can obtain  
\[A\cos\theta_{3}+B\sin\theta_{3}+C=0, \tag{25}\]  
where  
\[ \begin{cases}A=2r_{2}r_{8}\cos\theta_{2}-2r_{4}r_{8}\cos\theta_{4}+2r_{6}r_{8 }\cos\theta_{6}\\ B=2r_{2}r_{8}\sin\theta_{2}-2r_{4}r_{8}\sin\theta_{4}+2r_{6}r_{8}\sin\theta_{6} \\ C=r_{2}^{2}+r_{4}^{2}+r_{6}^{2}+r_{8}^{2}-r_{7}^{2}-2r_{2}r_{4}\cos(\theta_{2} -\theta_{4})+2r_{2}r_{6}\cos(\theta_{2}-\theta_{6})-2r_{4}r_{6}\cos(\theta_{4 }-\theta_{6}).\end{cases} \]With order \(t=\tan\dfrac{\theta_{8}}{2}\), there are \(\sin\theta_{8}=\dfrac{2t}{1+t^{2}},\cos\theta_{8}=\dfrac{1-t^{2}}{1+t^{2}}\). Substituting these into Equation (25), we can obtain  
\[(C-A)t^{2}+2 Bt+(A+C)=0. \tag{26}\]  
The solution is  
\[ \begin{cases}t=\dfrac{-B\pm\sqrt{B^{2}-C^{2}+A^{2}}}{C-A}\\ \theta_{8}=2\tan^{-1}(t),\end{cases} \]  
and the above equation must satisfy \(B^{2}-C^{2}+A^{2}\geq 0\). The leg-end position can be calculated according to the vector \(\mathbf{r}_{1}+\mathbf{r}_{4}+\mathbf{r}_{7}+\mathbf{r}_{9}\), where the angle corresponding to the vector \(\mathbf{r}_{1}\) is \(\theta_{1}\), the angle corresponding to the vector \(\mathbf{r}_{4}\) is \(\theta_{4}=\theta_{2}+\theta_{2-4}\), the angle corresponding to the vector \(\mathbf{r}_{7}\) is \(\theta_{7}\), and the angle corresponding to the vector \(\mathbf{r}_{9}\) is \(\theta_{9}=\theta_{8}+\theta_{8-9}\).