## 2 Robot System Components and Fabrication
### Robot Kinematics  
Figure 6 shows the simplified diagram used for the inverse kinematic analysis of the robot. The actuator is treated as a rigid robotic arm. Symbols are defined as follows: a right subscript \(i\) indicates the arm or the cable number, and a left superscript denotes the coordinate system where the value is defined. \(\Sigma_{0}\) is the base coordinate system, \(\Sigma_{p}\) is the local coordinate system attached on the tip of the over-tube, \(\Sigma_{m_{i}}\) is the coordinate system fixed at the rotation center of an arm (the arm coordinate system), \(\mathbf{r}_{i}\) is the position of the cable attachment point on the end-effector in the local coordinate system, \(\mathbf{s}_{i}\) is a vector along the rotating part of the arm, \(\mathbf{l}_{i}\) is a vector along a cable, \(\mathbf{h}_{i}\) is a vector from the rotation center of an arm to the cable attachment point at the end-effector, \(\mathbf{u}_{i}\) is a unit vector along a cable, \(\mathbf{A}_{i}\) is the position of the center of rotation of an arm, \(\mathbf{F}_{i}\) is the position of the connecting point between the arm and the cable, \(\phi_{m_{i}}\) is the angle between the rotating arm and the \(x\)-axis of the arm coordinate system, \(l_{c_{i}}\) is the cable length, and \(l_{a}\) is the rotating arm length. Figure 4: A prototype of the 3-DOF completely constrained CDPR. The effector was driven to (**a**) translate right, (**b**) turn right, (**c**) turn left, (**d**) translate left, (**e**) translate forward, and (**f**) translate backward. Figure 5: A 3D model of the inflatable CDPR. (**a**) A side view and (**b**) cross-section view of the robot. However, one solution should be discarded as the folded actuator can only generate torque in one direction, and the correct solution should satisfy the following condition:  
\[\left[{{}^{m_{i}}\mathbf{h}_{i}^{p}\times{}^{m_{i}}\mathbf{s}_{i}}\right]_{z}<0 \tag{8}\]  
where \({}^{m_{i}}\mathbf{h}_{i}^{p}\) is the projection vector of \({}^{m_{i}}\mathbf{h}_{i}\) in the \(x\)-\(y\) plane of the arm coordinate system \(\Sigma_{m_{i}}\), and \({}^{m_{i}}\mathbf{s}_{i}=[l_{a}\text{cos}\,\phi_{m_{i}},l_{a}\text{sin}\,\phi_{m_{ i}};]^{T}\). In this case, Equation (6) is the inverse kinematic solution and a condition should be satisfied: \(\phi_{m_{i}}\in[-\pi/2,\pi/2]\). Another condition that should be satisfied is the force and torque equilibrium at the end-effector:  
\[\mathbf{J}^{T}\mathbf{t}+\mathbf{f}=\mathbf{0} \tag{9}\]  
where \(\mathbf{t}=\{t_{1},t_{2},t_{3},t_{4},t_{5},t_{6}\}^{T}\) is the cable tension vector, \(\mathbf{f}\) denotes the external wrench, and the structure matrix depending on the pose of the effector is \(\mathbf{J}^{T}\):  
\[\mathbf{J}^{T}= \begin{bmatrix}\mathbf{u}_{1}\cdots\mathbf{u}_{6}\\ \mathbf{v}_{1}\cdots\mathbf{v}_{6}\end{bmatrix}  \tag{10}\]  
where the unit vector \(\mathbf{u}_{i}={I_{i}}/\|{I_{i}}\|\) and \(\mathbf{v}_{i}=\mathbf{u}_{i}\times\mathbf{P}_{i}\). The pose of the end-effector should satisfy Equation (9) with \(t_{i}>0\), and \(\phi_{m_{i}}\in[-\pi/2,\pi/2]\). The corresponding feasible solutions for vector \(\mathbf{t}\) can be found in [30], and the procedure is as follows: (1) The external wrench\(\mathbf{f}\) can usually be assumed to be \(\mathbf{0}\). (2) An initial value greater than 0 can be assigned to the tension of one cable (e.g., \(t_{1}=1\)) to solve for the tension of all other cables according to Equation (9). If the tension values of other cables are all greater than 0, the current pose of the effector is considered to satisfy the force and torque equilibrium, and the position of the robot tip can be included in the robot's feasible workspace.