# A Hybrid Soft Actuator Inspired by Grass-Spike: Design Approach, Dynamic Model, and Applications
## 3 Dynamic Models  
### A Single HSA Model  
Figure 3 depicts the free body diagram of the proposed HSA. To obtain a dynamic model, we start by recapitulating the modeling approach introduced in [21]. Since the obtained model will be utilized for a system identification process to estimate an unknown friction parameter, we substitute the signum function in [21] with a hyperbolic tangent function to lessen the effect of discontinuity during an optimization process. Table 1 summarizes the variables and parameters appearing in the free body diagram shown in Figure 3.  
Figure 3: The free body diagram of a hybrid soft actuator. Figure 2: The switch mechanism that enables the HSA to generate vibration and change the leaning direction by a single motor: The pin will be positioned either in the leftmost or rightmost slot of the switch with respect to the rotating direction of the motor, which in turn determines the leaning direction of the spikes. Blue arrows indicate the HSA’s moving direction according to the pin positions. Suppose that the HSA moves along the \(x\)-axis. The total kinetic energy \(\mathcal{K}\) is equal to the sum of three kinetic energies corresponding to the HSA movement, vibration \(\xi\) with respect to the \(x\)-axis, and vibration \(\eta\) with respect to the \(y\)-axis; therefore:  
\[\mathcal{K}=\frac{m_{0}}{2}(\dot{x}^{2}+\dot{y}^{2})+\frac{m_{1}}{2}(\dot{x}+ \dot{\xi})^{2}+\frac{m_{2}}{2}(\dot{y}+\dot{\eta})^{2}. \tag{1}\]  
The total potential energy \(\mathcal{P}\) is equal to the sum of the potential energies corresponding to the three material points, represented by \(m_{0},m_{1},m_{2}\), and the elasticity of the spikes; thus:  
\[\mathcal{P}=(m_{0}+m_{1})gy+m_{2}g(y+\eta)+n\frac{c}{2}(\phi-\phi_{0})^{2}. \tag{2}\]  
From Figure 3, the position and velocity of \(\mathcal{O}^{\prime}\) with respect to the \(y\)-axis, denoted by \(y\) and \(\dot{y}\), can be represented as:  
\[y = l\cos\phi+a \tag{3}\] \[\dot{y} = -\dot{\phi}l\sin\phi \tag{4}\]  
where \(a\) is the radius of the HSA. Substituting Equations (3) and (4) into Equations (1) and (2) yields:  
\[\mathcal{K} = \frac{1}{2}(m_{0}+m_{1})\dot{x}^{2}+\frac{1}{2}(m_{0}+m_{2})\dot {\phi}^{2}l^{2}(\sin\phi)^{2}+m_{1}\dot{x}\dot{\xi}^{2}-m_{2}|\dot{\phi}\dot{ \eta}\sin\phi+\frac{m_{1}}{2}\dot{\xi}^{2}+\frac{m_{2}}{2}\dot{\eta}^{2} \tag{5}\] \[\mathcal{P} = (m_{0}+m_{1}+m_{2})g(l\cos\phi+a)+m_{2}g\eta+n\frac{c}{2}(\phi- \phi_{0})^{2}. \tag{6}\]  
To apply the Euler-Lagrange equation of (where \(\mathcal{L}=\mathcal{K}-\mathcal{P}\)):  
\[\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial q_{i}}\right)-\frac{ \partial\mathcal{L}}{\partial q_{i}}=Q_{q_{i}}\quad\text{where}\quad q_{i}=x \text{ or }\phi, \tag{7}\]  
an external force \(Q_{i}\) should be defined. By the principle of virtual work, if a point \(x_{i}\) (where the ground and the \(i\)th spike contacts) is moved by \(F_{i}\) as \(\delta x_{i}\), then the sum of \(F_{i}\delta x_{i}\) must be equal to the work done by the external force \(Q_{i}\). Thus, we have:  
Figure 3:0 \begin{table}
\begin{tabular}{c c c c} \hline
**NAME** & **SYMBOL** & **VALUE** & **UNIT** \\ \hline Total mass & \(m_{0}\) & 34 & g \\ Horizontal mass & \(m_{1}\) & 10.5 & g \\ Vertical mass & \(m_{2}\) & 10.5 & g \\ Spike length & \(l\) & 8 & mm \\ Initial spike angle & \(\phi_{0}\) & \(\frac{\pi}{6}\) & rad \\ Number of spikes & \(n\) & 140 & EA \\ Stiffness constant of spikes & \(c\) & \(8.7\times 10^{-4}\) & Nm/rad \\ Coefficient of friction & \(\mu=\mu_{+}\) & Needs to be identified & \\ Sampling time & \(\Delta t\) & 0.0001 & s \\ Motor revolution speed & \(\Omega\) & input variable & rev/s \\ HSA displacement & \(x\) & state variable & mm \\ Spike leaning angle & \(\phi\) & state variable & rad \\ \hline \end{tabular}
\end{table} 
Table 1: The definition of the variables and parameters appearing in Figure 3. rev, revolution.