# Gait Planning Research for an Electrically Driven Large-Load-Ratio Six-Legged Robot
## 3 Kinematics Analysis of the Large-Load-Ratio Six-Legged Robot
### Forward Kinematics Analysis of the Large-Load-Ratio Six-Legged Robot  
The foot position of leg \(i\) can be solved when the body posture and joint angles of each leg have been obtained. The analysis of the forward kinematics can be performed for the large-load-ratio six-legged robot. The transformation matrix of leg \(i\) between the joint \(j-1\) and the joint \(j\) is written as follows,  
\[\overset{j-1}{j}\boldsymbol{T}^{(i)}=\left( \begin{array}{cccc}c_{\theta_{i} }^{(i)}&-s_{\theta_{j}^{(i)}}&0&a_{j-1}^{(i)}\\ \\ s_{\theta_{j}^{(i)}}c_{\alpha_{j-1}^{(i)}}&c_{\theta_{j}^{(i)}}c_{\alpha_{j-1} ^{(i)}}&-s_{\theta_{j-1}^{(i)}}&-d_{j}^{(i)}s_{\alpha_{j-1}^{(i)}}\\ \\ s_{\theta_{j}^{(i)}}s_{\alpha_{j-1}^{(i)}}^{(i)}&c_{\theta_{j}^{(i)}}s_{\alpha _{j-1}^{(i)}}^{(i)}&c_{\alpha_{j-1}^{(i)}}^{(i)}&d_{j}^{(i)}c_{\alpha_{j-1}^{(i )}}^{(i)}\\ 0&0&0&1\end{array} \right) \tag{5}\]  
where \(c_{\theta_{j}^{(i)}}s_{\theta_{j}^{(i)}}\), \(c_{\alpha_{j-1}^{(i)}}\), and \(s_{\alpha_{j-1}^{(i)}}\) are the abbreviations for \(\cos\theta_{j}^{(i)}\), \(\sin\theta_{j}^{(i)}\), \(\cos\alpha_{j-1}^{(i)}\), and \(\sin\alpha_{j-1}^{(i)}\), respectively. The transformation matrices among the joints of leg \(i\) can be obtained when the parameters of the D-H model are substituted into Equation (5). Then,  
\[\overset{0}{1}\boldsymbol{T}^{(i)}=\left( \begin{array}{cccc}\cos\theta_{i }&-\sin\theta_{i}&0&0\\ \sin\theta_{i}&\cos\theta_{i}&0&0\\ 0&0&1&0\\ 0&0&0&1\end{array} \right) \tag{6}\]  
\[\overset{1}{2}\boldsymbol{T}^{(i)}=\left( \begin{array}{cccc}\cos\theta_{i }^{\prime}&-\sin\beta_{i}^{\prime}&0&l_{c}\\ 0&0&-1&0\\ \sin\beta_{i}^{\prime}&\cos\beta_{i}^{\prime}&0&0\\ 0&0&0&1\end{array} \right) \tag{7}\]  
\[\overset{2}{3}\boldsymbol{T}^{(i)}=\left( \begin{array}{cccc}\cos\gamma_{i }&-\sin\gamma_{i}&0&l_{t}\\ \sin\gamma_{i}&\cos\gamma_{i}&0&0\\ 0&0&1&0\\ 0&0&0&1\end{array} \right) \tag{8}\]  
Based on Equations (6)-(8), the transformation matrix \(\overset{0}{3}\boldsymbol{T}^{(i)}\) from the coordinate system \(\sum_{K_{i}}\) to the coordinate system \(\sum_{0_{i}}\) can then be gained for the leg \(i\),  
\[\overset{0}{3}\boldsymbol{T}^{(i)}=\left( \begin{array}{cccc}c_{\theta_{i} }c_{\beta_{i}}&-c_{\theta_{i}}s_{\beta_{i}}&s_{\theta_{i}}&l_{c}c_{\theta_{i} }+l_{c}c_{\theta_{i}}\\ \\ s_{\theta_{i}}c_{\beta_{i}}&-s_{\theta_{i}}s_{\beta_{i}}&-c_{\theta_{i}}&l_{s \theta_{i}}c_{\theta_{i}}+l_{c}s_{\theta_{i}}\\ \\ s_{\theta_{i}}&c_{\beta_{i}}&0&l_{\mathbf{i}}s_{\beta_{i}^{\prime}}\\ 0&0&0&1\end{array} \right) \tag{9}\] \begin{table}
\begin{tabular}{c c c c c} \hline \multirow{2}{*}{**Joint \(j\)**} & **Linkage Length** & **Torsion Angle of Link** & **Linkage Offset** & **Rotation Angle of Link** \\  & \(\boldsymbol{a}_{j-1}^{(i)}\) **(mm)** & \(\boldsymbol{a}_{j-1}^{(i)}\) **(deg)** & \(\boldsymbol{a}_{j}^{(i)}\) **(mm)** & \(\boldsymbol{\theta}_{j}^{(i)}\) **(deg)** \\ \hline
1 & 0 & 0\lx@math@degree\) & 0 & 0 & \(\theta_{i}\) \\
2 & \(l_{c}\) & 90\lx@math@degree\) & 0 & \(\beta_{i}^{\prime}\) \\
3 & \(l_{t}\) & 0\lx@math@degree\) & 0 & \(\gamma_{i}\) \\ \hline \end{tabular}
\end{table} 
Table 4: Parameters of the D-H model of leg \(i\). where \(c_{\theta_{i}}\), \(s_{\theta_{i}}\), \(c_{\beta_{i}}\), \(s_{\beta_{i}}\), \(c_{\theta_{i}^{\prime}}\), and \(s_{\theta_{i}^{\prime}}\) are the abbreviations for \(\cos\theta_{i}\), \(\sin\theta_{i}\), \(\cos\beta_{i}\), \(\sin\beta_{i}\), \(\cos\beta_{i}^{\prime}\), and \(\sin\beta_{i}^{\prime}\), respectively. The coordinate matrix \({}^{3}\mathbf{P}_{\mathrm{F}}^{(i)}\) of the foot of leg \(i\) can be written in the coordinate system \(\sum_{K_{i}}\) of the knee joint. Then,  
\[{}^{3}\mathbf{P}_{\mathrm{F}}^{(i)}=\left( \begin{array}{ccc}l_{\mathrm{s}}& 0&0&1\end{array} \right)^{\mathrm{T}} \tag{10}\]  
The transformation matrix \({}^{\mathrm{B}}_{0}\mathbf{T}^{(i)}\) from the coordinate system \(\sum_{0_{i}}\) to the coordinate system \(\sum_{B}\) of the body can then be obtained as follows,  
\[{}^{\mathrm{B}}_{0}\mathbf{T}^{(i)}=\left(