# Form-Finding Analysis of a Class 2 Tensegrity Robot
## 2 Kinematic Position Analysis
### Forward Kinematic Position Analysis  
Figure 1 shows the class 2 tensegrity robot, which is composed of two rigid substructures joined at one end by a universal joint, and four flexible elements at the remaining ends: two cables, \(C_{1}\) and \(C_{2}\), and two springs, \(r_{1}\) and \(r_{2}\). To maintain this paper to be self-contained, the solution proposed by [PERSON] et al. [11] is presented. This strategy consists of using the Denavit-Hartenberg parameters, contained in Table 1, as well as the auxiliary reference frames \(\Sigma_{W}\), \(\Sigma_{M_{1}}\) and \(\Sigma_{M_{2}}\), which represent the reference frame of the base, the cable joint \(C_{1}\) and the cable joint \(C_{2}\), respectively, shown in Figure 2. \begin{table}
\begin{tabular}{c c c c c} \hline \hline \multirow{2}{*}{\(i\)} & \(d_{i}\) & \(\theta_{i}\) & \(a_{i}\) & \(\alpha_{i}\) \\  & **mm** & **rad** & **mm** & **rad** \\ \hline
1 & 0 & \(\theta_{1}\) & 0 & \(-\pi/2\) \\
2 & 0 & \(\theta_{2}\) & \(a_{2}\) & \(\pi/2\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Denavit–Hartenberg parameters of the structure. Figure 1: Class 2 tensegrity robot. With the geometric parameters of the robot and the reference frames, the rigid links that make up the robot are analyzed following the Denavit-Hartenberg methodology [25], resulting in the following homogeneous transformation matrices:  
\[{}^{W}T_{0} = \begin{bmatrix}1&0&0&l_{1}\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}  \tag{1}\] \[{}^{0}T_{1} = \begin{bmatrix}\mathrm{c}\theta_{1}&0&-\mathrm{s}\theta_{1}&0\\ \mathrm{s}\theta_{1}&0&\mathrm{c}\theta_{1}&0\\ 0&-1&0&0\\ 0&0&0&1\end{bmatrix} \] (2) \[{}^{1}T_{M} = \begin{bmatrix}\mathrm{c}\theta_{2}&0&\mathrm{s}\theta_{2}& \mathrm{a}_{2}\cdot\mathrm{c}\theta_{2}\\ \mathrm{s}\theta_{2}&0&-\mathrm{c}\theta_{2}&\mathrm{a}_{2}\cdot\mathrm{c} \theta_{2}\\ 0&1&0&0\\ 0&0&0&1\end{bmatrix} \] (3) \[{}^{M}T_{M_{1}} = \begin{bmatrix}1&0&0&0\\ 0&1&0&l_{2}^{s}\\ 0&0&1&-l_{2}^{s}\\ 0&0&0&1\end{bmatrix} \] (4) \[{}^{M}T_{M_{2}} = \begin{bmatrix}1&0&0&0\\ 0&1&0&-l_{2}^{s}\\ 0&0&1&l_{2}^{s}\\ 0&0&0&1\end{bmatrix}  \tag{5}\]  
where \(l_{2}^{s}=\frac{l_{2}}{2},\mathrm{s}\theta_{i}\equiv\sin(\theta_{i})\) and \(\mathrm{c}\theta_{i}\equiv\cos(\theta_{i})\) for \(i=1,2\). By multiplying Equations (1)-(5), the relative position and orientation of \(\Sigma_{M_{1}}\) and \(\Sigma_{M_{2}}\) with respect to \(\Sigma_{W}\) is obtained, that is:  
\[{}^{W}T_{M_{1}} = {}^{W}T_{0}\cdot{}^{0}T_{1}\cdot{}^{1}T_{M}\cdot{}^{M}T_{M_{1}} \tag{6}\] \[{}^{W}T_{M_{2}} = {}^{W}T_{0}\cdot{}^{0}T_{1}\cdot{}^{1}T_{M}\cdot{}^{M}T_{M_{2}} \tag{7}\]  
Figure 2: Reference frames. Then, the position of the origins of \(\Sigma_{M_{1}}\) and \(\Sigma_{M_{2}}\) with respect to \(\Sigma_{W}\) are given by:  
\[{}^{W}P_{M_{1}} = \begin{bmatrix}\text{c}\theta_{1}\cdot(-s\theta_{2}\cdot l_{2}^{* }+a_{2}\cdot\text{c}\theta_{2})-s\theta_{1}\cdot l_{2}^{*}+l_{1}\\ s\theta_{1}\cdot(-s\theta_{2}\cdot l_{2}^{*}+a_{2}\cdot\text{c}\theta_{2})+ \text{c}\theta_{1}\cdot l_{2}^{*}\\ -\text{c}\theta_{2}\cdot l_{2}^{*}-a_{2}\cdot\text{s}\theta_{2}\end{bmatrix}  \tag{8}\] \[{}^{W}P_{M_{2}} = \begin{bmatrix}\text{c}\theta_{1}\cdot(s\theta_{2}\cdot l_{2}^{* }+a_{2}\cdot\text{c}\theta_{2})+s\theta_{1}\cdot l_{2}^{*}+l_{1}\\ s\theta_{1}\cdot(s\theta_{2}\cdot l_{2}^{*}+a_{2}\cdot\text{c}\theta_{2})- \text{c}\theta_{1}\cdot l_{2}^{*}\\ c\theta_{2}\cdot l_{2}^{*}-a_{2}\cdot\text{s}\theta_{2}\end{bmatrix}  \tag{9}\]  
To determine the variations in the length of the cables C\({}_{1}\) and C\({}_{2}\), the base points \(B_{1}\) and \(B_{2}\) must be defined (see Figure 2). The points \(B_{1}\) and \(B_{2}\) are defined with respect to \(\Sigma_{W}\), and they can be expressed as:  
\[{}^{W}P_{B_{1}} = \begin{bmatrix}B_{1_{x}}&B_{1_{y}}&B_{1_{z}}\end{bmatrix} ^{T} \tag{10}\] \[{}^{W}P_{B_{2}} = \begin{bmatrix}B_{2_{x}}&B_{2_{y}}&B_{2_{z}}\end{bmatrix}