# Energy Consumption Analysis of a Rolling Mechanism Based on a Five-Bow-Shaped-Bar Linkage
## 2 Trajectory Planning of the Mechanism's CoM
### Trajectory Planning of the CoM  
The shape of the mechanism is determined by its location of the CoM, while the component of the CoM in the vertical direction affects potential energy. The speed and acceleration are controlled by the offset of the mechanism's CoM with respect to the grounded point in the horizontal direction. Because the position of the mechanism's CoM is closely related to energy consumption, it is necessary to choose an appropriate CoM trajectory to reduce the energy consumption of the mechanism. Whereupon, the previous work of the trajectory planning is to acquire the workspace of the CoM. In Section 2.2, the relationship between the active joint angles and the rolling angle was analyzed. The workspace of the CoM when \(\phi_{k}=k\phi_{\max}/73\) (\(\phi_{\max}=2\pi/5\), k = 0, 1, 2... 73) can be figured out by Equations (1)-(6). Then, the workspace of the CoM when \(\phi=\phi_{k}\) is illustrated in Figure 5. As shown in Figure 5, the feasible region is encompassed by a red boundary which is formed by connecting outer boundary points. The workspaces of the CoM of the rolling mechanism at the initial and end positions, i.e., \(\phi=0\) and \(\phi=2\pi/5\), are, respectively shown in Figure 6a,b. To ensure the continuity of the rolling locomotion during the alternation of grounded modules, the outer profile curve of the adjacent modules should be continuous. It means that the angle between two successive grounded modules should remain \(2\pi/5\). Correspondingly, the mechanism's CoM must be located on the red line in Figure 6.  
In order to avoid spontaneous rolling, the CoM of the mechanism and the grounded point \(P\) should be in the same vertical line at the initial position, while they also should be in the same vertical line to achieve periodic locomotion at the end position. Therefore, the coordinates of the CoM at the initial and end positions can be written as  
\[\left\{\begin{array}{ll}{}^{O}X_{cstart}=0;&{}^{O}Y_{cstart}=R\\ {}^{O}X_{cend}=\frac{2\pi}{5}R;&{}^{O}Y_{cend}=R\right. \tag{11}\]  
Figure 5: The workspace of the CoM when the rolling angle \(\phi\) is in the range of \(0\) to \(2\pi/5\). Figure 6: The workspace of the mechanism’s CoM at the initial and end positions. (**a**) \(\phi=0\). (**b**) \(\phi=2\pi/5\). It can be seen that both the workspace for \(\phi\in[0,\ 2\pi/5]\) and the workspace for \(\phi=0\) and \(\phi=2\pi/5\) are symmetrical about \({}^{O}X_{c}=0.63R\) in Figures 5 and 6. Thus, points on the symmetry centerline of its workspace of the CoM are used as via points to plan CoM trajectories. The start point, via points, and end point are shown in Figure 7, where height of the via point, denoted as \(h\), ranges from \(0.801R\) to \(1.183R\). The routes of the CoM are start-via-end (1-2-3), and the values of each variable are presented in Table 1.  
[PERSON] and [PERSON] [40] proposed a composite polynomial consisting of Bezier curves and cubic polynomials, exhibiting smoother position, velocity, and acceleration curves than the cubic spline method. In this paper, the quadratic Bezier curve is combined with a cubic polynomial to plan the trajectory of the CoM under the condition that the start, via, and end points have been determined. The direction and shape of the trajectory are specified by the via point, whether the via point is on this curve or not. The \(i\)-\(th\) Bernstein polynomials of degree \(m\) are defined by [42]. \[B_{i}^{m}(\chi)=\binom{m}{i}\chi^{i}(1-\chi)^{m-i},i\in[0,1,\ldots,m] \tag{12}\]  
where \(\binom{m}{i}=\frac{mt}{\mathbb{I}(m-i)! }\) is a binomial coefficient. When \(m=2\), the Bernstein polynomials can be expressed as  
\[B_{0}^{2}(\chi)=\left(1-\chi\right)^{2},\ B_{1}^{2}(\chi)=2\chi(1-\chi),B_{2} ^{2}(\chi)=\chi^{2} \tag{13}\] \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Point** & \(\mathbf{OX_{c}}\) & \(\mathbf{OY_{c}}\) & **Explanation** \\ \hline
1 & 0 & \(R\) & start \\
2 & \(\pi R/5\) & \(h\in[0.801R,1.183R]\) & via \\
3 & \(2\pi R/5\) & \(R\) & end \\ \hline \hline \end{tabular}
\end{table}