# Kinematic Modelling and Experimental Validation of a Foldable Pneumatic Soft Manipulator
## 2 Kinematic Modelling of the Foldable Pneumatic Soft Manipulator
### Kinematic Model of the Foldable Pneumatic Module
\begin{array}{l}A_{0}:\left(\begin{array}{cc}\sqrt{3}a&-\frac{a}{2} &0\\ 0&0&0\\ C_{0}:\left(\begin{array}{cc}\sqrt{3}a&\frac{a}{2}&0\\ \end{array}\right)\end{array}\right)\end{array} \right. \tag{4}\]  
In the uninflated state, the coordinates of the \(A\), \(B\) and \(C\) can be obtained by,  
\[\left\{ \begin{array}{l}A:\left(\begin{array}{cc}\sqrt{3}a&-\frac{a}{2}&L_{ 0}\\ 0&0&L_{0}\\ C:\left(\begin{array}{cc}\sqrt{3}a&\frac{a}{2}&L_{0}\\ \end{array}\right)\end{array}\right)\end{array} \right. \tag{5}\]  
Considering the effect of \(R_{1}\), \(R_{2}\) and \(R_{3}\), the deflections of the upper end of the three bellows are  
\[\left\{ \begin{array}{l}v_{1}=\frac{R_{1}L_{1}^{\ast}}{3E_{1}^{\ast}}=\frac{ R_{1}L_{1}^{\ast}2^{\ast}}{3 kT}\\ v_{2}=\frac{R_{2}L_{2}^{\ast}}{3E_{2}^{\ast}}=\frac{R_{2}L_{2}^{\ast}2^{\ast}}{3 kT}\\ v_{3}=\frac{R_{3}L_{3}^{\ast}}{3E_{3}^{\ast}}=\frac{R_{3}L_{2}^{\ast}2^{\ast}}{3 kT}\end{array} \right. \tag{6}\]  
where, \(E\) is the elastic modulus, \(I\) is the moment of inertia, \(EI\) is the bending stiffness. \(S^{\prime}\) is the cross-sectional area of the wall of the bellows, that is, the effective area under axial tension. For the section of the ring whose outer diameter is \(R\) and the inner diameter is \(r\), its moment of inertia is  
\[I=\int y^{2}\mathrm{d}A=\int\limits_{0}^{\pi}\left(\frac{R+r}{2}\right)^{2} \sin^{2}\theta\frac{R^{2}-r^{2}}{2}\mathrm{d}\theta=\frac{\pi(R+r)^{3}(R-r)}{ 16} \tag{7}\]  
Under the action of \(R_{1}\), \(R_{2}\) and \(R_{3}\), the angles generated at the upper end of the three bellows are  
\[\left\{ \begin{array}{l}\theta_{1}=\frac{R_{1}L_{1}^{\ast}}{2E_{1}}^{2}= \frac{R_{1}L_{1}S^{\ast}}{2 kT}\\ \theta_{2}=\frac{R_{2}L_{2}^{\ast}}{2E_{2}T}=\frac{R_{2}L_{2}S^{\ast}}{2 kT}\\ \theta_{3}=\frac{R_{3}L_{3}^{\ast}}{2E_{3}T}=\frac{R_{3}L_{3}S^{\ast}}{2 kT} \end{array} \right. \tag{8}\]  
Considering that the endpoints on the three bellows \(A\), \(B\) and \(C\) have the same normal direction, it means that,  
\[\left\{ \begin{array}{l}\theta_{1}=\theta_{2}\\ \theta_{1}=\theta_{3}\end{array} \right.\Rightarrow\left\{ \begin{array}{l}R_ {1}L_{1}=R_{2}L_{2}\\ R_{1}L_{1}=R_{3}L_{3}\end{array} \right. \tag{9}\]  
The fact that \(A\), \(B\), and \(C\) have the same normal orientation means that the projection of the three bellows are parallel in the X-Y plane. Supposed that the angle from the axis \(x\) counterclockwise to the projection is \(\alpha\), then the coordinates of the upper endpoints of the three bellows after inflation are respectively as follows:  
\[\left\{ \begin{array}{l}A^{\prime}:\left(\begin{array}{ccc}v_{1}\cos\alpha +\frac{\sqrt{3}}{2}a&v_{1}\sin\alpha-\frac{a}{2}&L_{1}\\ B^{\prime}:\left(\begin{array}{ccc}v_{2}\cos\alpha&v_{2}\sin\alpha&L_{2} \\ \end{array}\right)\end{array}\right)\\ C^{\prime}:\left(\begin{array}{ccc}v_{3}\cos\alpha+\frac{\sqrt{3}}{2}a&v_{3} \sin\alpha+\frac{a}{2}&L_{3}\\ \end{array}\right)\end{array} \right. \tag{10}\]Under the constraint of the upper rigid plates, the distance condition satisfies,  
\[L_{A^{\prime}B^{\prime}}=a,\ \ L_{B^{\prime}C^{\prime}}=a,\ \ L_{C^{\prime}A^{ \prime}}=a \tag{11}\]  
To simplify the equation, it is supposed that  
\[R_{1}L_{1}=R_{2}L_{2}=R_{3}L_{3}=W\geq 0,\ \ \ \frac{S^{\prime}}{3 kl}=M>0 \tag{12}\]  
It is assumed that the elongations of the three bellows are all greater than zero and less than \(\Delta L\), then there is \(0\leq\Delta L_{1},\Delta L_{2},\Delta L_{3}\leq\Delta L\). It can be proved that the established condition of the expression \(3a^{2}-3(\Delta L_{1}-\Delta L_{3})^{2}-(2\Delta L_{2}-\Delta L_{3}-\Delta L_{ 1})^{2}\geq 0\) is \(\Delta L/a\leq\sqrt{3}/2\), which means that the ratio of the bellow elongation to the bellow diameter is less than \(\sqrt{3}/2\). Considering that the elongation of each section of the pneumatic actuator is small, the condition is satisfied in actual motion. When \(\Delta L_{1}\
eq\Delta L_{3}\), it can be gotten,  
\[