# 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  
For the closed-chain single-leg mechanism, we adopt the overall analysis method for the kinematic analysis. As shown in Figure 4, a rectangular coordinate system is established at the \(O\) point. The \(X\) axis is horizontal to the right, and the \(Y\) axis is vertically upward. The corresponding \(\theta\) angles are the angles between the vector and the positive direction of the \(X\) axis. A blue triangle represents a component. There are two closed loops in this mechanism, so two vector loop equations can be listed, which are expressed as follows:  
\[\mathbf{r}_{0}+\mathbf{r}_{3}-\mathbf{r}_{2}-\mathbf{r}_{1}=\mathbf{0}_{2\times 1}, \tag{4}\]  
and  
\[\mathbf{r}_{5}+\mathbf{r}_{6}+\mathbf{r}_{7}-\mathbf{r}_{8}=\mathbf{0}_{2\times 1}. \tag{5}\]  
The vector of loop 1 (Equation (4)) is decomposed along the \(X\) and \(Y\) axes, and two trigonometric function equations are obtained:  
\[r_{0}\cos\theta_{0}+r_{3}\cos\theta_{3}-r_{2}\cos\theta_{2}-r_{1}\cos\theta_{1 }=0, \tag{6}\]  
and  
\[r_{0}\sin\theta_{0}+r_{3}\sin\theta_{3}-r_{2}\sin\theta_{2}-r_{1}\sin\theta_{1 }=0, \tag{7}\]  
where \(r_{0},\theta_{0},r_{3},r_{2}\), and \(r_{1}\) are structural parameters, which are constants; \(\theta_{1}\) is the rotation angle of the prime mover, which is a known parameter; and \(\theta_{2}\) and \(\theta_{3}\) are unknown. Figure 4: Establishing the leg coordinate system. After transforming Formulas (6) and (7) to solve \(\theta_{2}\), we can obtain  
\[r_{0}\cos\theta_{0}-r_{2}\cos\theta_{2}-r_{1}\cos\theta_{1}=-r_{3}\cos\theta_{3}, \tag{8}\]  
and  
\[r_{0}\sin\theta_{0}-r_{2}\sin\theta_{2}-r_{1}\sin\theta_{1}=-r_{3}\sin\theta_{3}. \tag{9}\]  
Adding the squares of the two sides of the above formula, we can obtain  
\[A\cos\theta_{2}+B\sin\theta_{2}+C=0, \tag{10}\]  
where  
\[ \begin{cases}A=2r_{0}r_{2}\cos\theta_{0}-2r_{1}r_{2}\cos\theta_{1}\\ B=2r_{0}r_{2}\sin\theta_{0}-2r_{1}r_{2}\sin\theta_{1}\\ C=-r_{0}^{2}-r_{1}^{2}-r_{2}^{2}+r_{3}^{2}+2r_{0}r_{1}\cos(\theta_{0}-\theta_ {1}).\end{cases} \]  
With order \(t=\tan\dfrac{\theta_{2}}{2}\), there are \(\sin\theta_{2}=\dfrac{2t}{1+t^{2}},\cos\theta_{2}=\dfrac{1-t^{2}}{1+t^{2}}\). Substituting these into Equation (10), we can obtain  
\[(C-A)t^{2}+2 Bt+(A+C)=0, \tag{11}\]  
where  
\[ \begin{cases}t=\dfrac{-B\pm\sqrt{B^{2}-C^{2}+A^{2}}}{C-A}\\ \theta_{2}=2\tan^{-1}(t),\end{cases} \]  
and the above equation must satisfy \(B^{2}-C^{2}+A^{2}\geq 0\). In the same way, Formulas (6) and (7) are deformed and solved for \(\theta_{3}\), and we can obtain  
\[r_{0}\cos\theta_{0}+r_{3}\cos\theta_{3}-r_{1}\cos\theta_{1}=r_{2}\cos\theta_{2}, \tag{12}\]  
and  
\[r_{0}\sin\theta_{0}+r_{3}\sin\theta_{3}-r_{1}\sin\theta_{1}=-r_{2}\sin\theta_ {2}. \tag{13}\]  
Adding the squares of the two sides of the above formula, we can obtain  
\[A\cos\theta_{3}+B\sin\theta_{3}+C=0, \tag{14}\]  
where  
\[ \begin{cases}A=2r_{0}r_{3}\cos\theta_{0}-2r_{1}r_{3}\cos\theta_{1}\\ B=2r_{0}r_{3}\sin\theta_{0}-2r_{1}r_{3}\sin\theta_{1}\\ C=r_{0}^{2}+r_{1}^{2}-r_{2}^{2}+r_{3}^{2}-2r_{0}r_{1}\cos(\theta_{0}-\theta_{1 }).\end{cases} \]  
With order \(t=\tan\dfrac{\theta_{3}}{2}\), there are \(\sin\theta_{3}=\dfrac{2t}{1+t^{2}},\cos\theta_{3}=\dfrac{1-t^{2}}{1+t^{2}}\). Substituting these into Equation (14), we can obtain  
\[(C-A)t^{2}+2 Bt+(A+C)=0. \tag{15}\]  
The solution is  
\[ \begin{cases}t=\dfrac{-B\pm\sqrt{B^{2}-C^{2}+A^{2}}}{C-A}\\ \theta_{3}=2\tan^{-1}(t),\end{cases} \]  
and the above equation must satisfy \(B^{2}-C^{2}+A^{2}\geq 0\). Loop 2 (Formula (5)) is decomposed in the \(X\) and \(Y\) directions.