# Design and Experimentation of Tensegrity Jumping Robots
## 3 Dynamic Modeling and Simulations Analysis
### Dynamic Modeling
#### 3.1.2 Dynamic Modeling of Two Modules' Step-Up Jump  
During the step-up process of the tensegrity jumping robot, we choose the module below the robot to release energy in priority, so in the process of releasing energy in the first module, we can accordingly establish the corresponding planar Cartesian coordinate system the Planar Cartesian Coordinate System \(O\)-\(XY\) and set the mid-point of the rod \(A_{2,1}\) as the origin. The X-axis is horizontally to the right and the Y-axis is vertically up, as shown in Figure 4.  
At this time, \(\theta_{i}\) (\(i\) = 1-4) can be seen in Equation (9), the center of mass position \(Y_{0,3}^{0}\), \(X_{0,3}^{0}\) of \(L_{0,3}\) can be seen in Equation (10) and the velocities \(v_{Y_{0,3}}\) and \(v_{X_{0,3}^{0}}\) and accelerations \(a_{Y_{0,3}^{0}}\) and \(a_{X_{0,3}^{0}}\) of the center of mass position of the rack \(L_{0,3}\) can be seen in (12) and (14). After the time \(\Delta t\) has elapsed, the tensegrity jumping robot makes a parabolic motion in the air. At this time, we can obtain (\(i\) = 5-8) (refer to Equation (9)). At this time, the velocity of the robot can be expressed as  
\[\left\{\begin{array}{l}v_{Y_{0,3}}=v_{Y_{0,3}}^{0}+a_{Y_{0,3}}^{0}\Delta t\\ v_{X_{0,3}}=v_{X_{0,3}}^{0}+a_{X_{0,3}}^{0}\Delta t\end{array}\right. \tag{23}\]  
The displacement of the robot during this time period can be represented as  
\[\left\{\begin{array}{l}x_{Y_{0,3}}=v_{Y_{0,3}}^{0}\Delta t+0.5a_{Y_{0,3}}^{0 }\Delta t^{2}\\ X_{X_{0,3}}=v_{X_{0,3}}^{0}\Delta t+0.5a_{X_{0,3}}^{0}\Delta t^{2}\end{array}\right. \tag{24}\]  
At this point, the center of mass position of the rack \(L_{0,3}\) can be represented as  
\[\left\{\begin{array}{l}Y_{0,3}=Y_{0,3}^{0}+x_{Y_{0,3}}\\ X_{0,3}=X_{0,3}^{0}+x_{X_{0,3}}^{0}\end{array}\right. \tag{25}\]  
At this point, the center of mass position of the rack \(L_{0,1}\) can be represented as  
\[\left\{\begin{array}{l}Y_{0,1}=\frac{x_{0,3}^{2}Y_{0,3}+Y_{0,3}^{0}-\sqrt{Y_ {0,3}^{2}s_{xx}^{2}}(x_{0,3}^{2}+Y_{0,3}^{2})}{X_{0,3}^{2}+Y_{0,3}^{2}}\\ X_{0,1}=\frac{v_{0,3}^{2}x_{0,3}+x_{0,3}^{0}-\sqrt{X_{0,3}^{2}s_{xx}^{2}}(x_{ 0,3}^{2}+Y_{0,3}^{2})}{X_{0,3}^{2}+Y_{0,3}^{2}}\end{array}\right. \tag{26}\]  
where the expressions of \(s_{uv}\) can be found in Appendix A.  
Then, at this point, it is known that the velocity at the center of mass position of rack \(L_{0,1}\) is the same as that of \(L_{0,3}\). The center of mass position of rack \(L_{0,5}\) can refer to Equation (11). The velocity at the center of mass position of \(L_{0,5}\) at this time can be referred to as Equation (13). The kinetic energy and potential energy of the robot are  
\[E=\frac{1}{2}\sum_{i=1}^{3}m_{0,i}\Big{(}v_{Y_{0,i}}^{2}+v_{X_{0,i}}^{2}\Big{)} \tag{27}\]  
Figure 4: Diagram of the step-up jump.  
\[U=\frac{k_{4}\Delta l_{4}^{2}}{2}+\sum_{k=1}^{3}m_{0,2i-1}gY_{0,2i-1} \tag{28}\]  
At this point, it is known that the Lagrangian quantity is  
\[L=\sum_{m=1}^{4}\sum_{n=1}^{4}z_{m,n}\dot{\varphi}_{m}\dot{\varphi}_{n}-\frac{k _{4}{\mu}^{2}}{2}-\sum_{i=1}^{3}m_{0,2i-1}gY_{0,2i-1} \tag{29}\]  
where the expressions of \(z_{ij}\) can be found in Appendix A.  
Thus, the Lagrange equation is  
\[\begin{split}&\frac{d}{dt}\frac{\partial L}{\partial\dot{\varphi}_{i}} -\frac{\partial L}{\partial\dot{\varphi}_{i}}=\sum_{j=1}^{4}\frac{dz_{ij}}{dt} \ddot{\varphi}_{\partial_{i}}+\frac{dz_{ij}}{dt}\ddot{\varphi}_{i}\\ &-\sum_{m=1}^{4}\sum_{n=1}^{4}\frac{\partial z_{m_{n}}}{\partial \dot{\varphi}_{i}}\dot{\varphi}_{m}\dot{\varphi}_{n}+k_{4}\frac{\partial \Delta l_{4}}{\partial\dot{\varphi}_{i}}+\sum_{n=1}^{3}m_{0,2n-1}g\frac{ \partial Y_{0,2n-1}}{\partial\dot{\varphi}_{i}}=0(i=1\sim 4)\end{split} \tag{30}\]  
where \(\varphi_{i}\in[0,\,\varphi_{e}]\) and the expression of \(\varphi_{e}\) are referred to in Equations (21) and (22).