# Structural Design and Analysis of Multi-Directional Foot Mobile Robot
## 2 Design and Analysis of Closed-Chain Leg Structure
### Singularity Analysis of Closed-Chain Single-Leg Mechanism  
The Watt-I type closed-chain single-leg mechanism has a single degree of freedom and can be divided into a planar four-bar mechanism and a class II linkage group. In reference ([PERSON], 2018), it is considered that the singularity analysis of a planar multi-ring mechanism can be divided according to the basic rod group. When a basic rod group reaches the singular position type, the complex mechanism is of the singular position type. Based on this principle, the singularity of the planar four-bar mechanism is analyzed. For the four-bar linkage, the axes of the four rotating pairs are parallel to each other. They can be expressed as four line vectors with a zero pitch of \(\mathbf{\hat{s}}_{1},\mathbf{\hat{s}}_{2},\mathbf{\hat{s}}_{3},\mathbf{\hat{s}}_{4}\). We place the origins of the coordinates at point \(A\), and the four kinematic pairs form a screw system \([J]^{T}\) as follows:  
\[[J]^{T}= \begin{bmatrix}\mathbf{\hat{s}}_{1}\\ \mathbf{\hat{s}}_{2}\\ \mathbf{\hat{s}}_{3}\\ \mathbf{\hat{s}}_{4}\end{bmatrix} = \begin{bmatrix}0&0&1&0&0&0\\ 0&0&1&y_{B}&x_{B}&0\\ 0&0&1&y_{C}&x_{C}&0\\ 0&0&1&y_{D}&x_{D}&0\end{bmatrix} , \tag{37}\]  
where \(y_{i},x_{i}\), \((i=B,C,D)\) can be obtained from the positive solutions. Obviously, in this \(4\times 6\) matrix, only a \(4\times 3\) sub-matrix \([J^{5}]^{T}\) is effective. \[[J^{5}]^{T}= \begin{bmatrix}1&0&0\\ 1&y_{B}&x_{B}\\ 1&y_{C}&x_{C}\\ 1&y_{D}&x_{D}\end{bmatrix} . \tag{38}\]  
The rank of \([J^{5}]^{T}\) is 3 at most, so the four spirals \(\mathbf{\hat{s}}_{1},\mathbf{\hat{s}}_{2},\mathbf{\hat{s}}_{3},\mathbf{\hat{s}}_{4}\) are linearly related and belong to a three-series spiral, which can be expressed as  
\[\sum_{j=1}^{4}\omega_{j}\mathbf{\hat{s}}_{j}=0. \tag{39}\]For the spatial three-phase screw, there are three reciprocals that are inverse to it, i.e., three common constraints are imposed on the plane mechanism. They are the force along the \(Z\) direction \(\mathbf{\xi}_{1}^{r}\) and the force couple vectors along the \(X\) and \(Y\) directions \(\mathbf{\xi}_{2}^{r},\mathbf{\xi}_{3}^{r}\):  
\[ \begin{cases}\mathbf{\xi}_{1}^{r}=(0\quad 0\quad 1;\quad 0\quad 0\quad 0)\\ \mathbf{\xi}_{2}^{r}=(0\quad 0\quad 0;\quad 1\quad 0\quad 0)\\ \mathbf{\xi}_{3}^{r}=(0\quad 0\quad 0;\quad 0\quad 1\quad 0).\end{cases}  \tag{40}\]  
For a four-bar mechanism, when \(\mathbf{\xi}_{1}^{r},\mathbf{\xi}_{2}^{r},\mathbf{\xi}_{3}^{r}\) are coplanar, the mechanism is at the limit displacement singularity; when \(\mathbf{\xi}_{2}^{r},\mathbf{\xi}_{3}^{r},\mathbf{\xi}_{4}^{r}\) are coplanar, the mechanism is at the dead-point singularity; when \(\mathbf{\xi}_{3}^{r},\mathbf{\xi}_{4}^{r}\) are coplanar, the mechanism is at the instantaneous degree of freedom change singularity; when \(\mathbf{\xi}_{1}^{r},\mathbf{\xi}_{2}^{r},\mathbf{\xi}_{3}^{r},\mathbf{\xi}_{4}^{r}\) are coaxial, the mechanism is at the continuous geometric singularity. (1) The limit displacement singularity  
When the limit displacement is singular, \(\mathbf{\xi}_{1}^{r},\mathbf{\xi}_{2}^{r},\mathbf{\xi}_{3}^{r}\) are coplanar, i.e.,  
\[det \begin{pmatrix}\begin{bmatrix}1&0&0\\ 1&y_{B}&x_{B}\\ 1&y_{C}&x_{C}\end{bmatrix}\end{pmatrix} =0. \tag{41}\]  
If and only if \(y_{B}\cdot x_{C}-y_{C}\cdot x_{B}=0\), points \(A\), \(B\), and \(C\) are collinear. We solve the input variable \(\theta\) at this time, and the input variable \(\theta\) is obtained as  
\[\theta=\arccos\biggl{(}\frac{(a+b)^{2}+d^{2}-c^{2}}{2d(a+b)}\biggr{)}, \tag{42}\]  
or  
\[\theta=\arccos\biggl{(}\frac{(a-b)^{2}+d^{2}-c^{2}}{2d(a-b)}\biggr{)}, \tag{43}\]  
where \(a\), \(b\), \(c\), \(d\) represent the rod length \(AB\), \(BF\), \(FG\), \(GA\), respectively. (2) The limit displacement singularity  
When the limit displacement is singular, \(\mathbf{\xi}_{2}^{r},\mathbf{\xi}_{3}^{r},\mathbf{\xi}_{4}^{r}\) are coplanar, i.e.,  
\[det \begin{pmatrix}\begin{bmatrix}1&y_{B}&x_{B}\\ 1&y_{C}&x_{C}\\ 1&y_{D}&x_{D}\end{bmatrix}\end{pmatrix} =0.