# Energy Consumption Analysis of a Rolling Mechanism Based on a Five-Bow-Shaped-Bar Linkage
## 2 Trajectory Planning of the Mechanism's CoM
### Kinematics Model
\begin{array}{l}{{}^{O}X_{c}=\frac{13}{10}{{}^{O}x_{a}}+\frac{3}{5}l \cos\alpha+\frac{2}{5}l\cos(\alpha+\theta_{2})+\frac{1}{5}l\cos(\alpha+\theta _{2}+\theta_{3})-\frac{1}{5}l\cos(\theta_{1}-\alpha)-\frac{1}{5}R\phi}\\ {{}^{O}Y_{c}=\frac{13}{10}{{}^{O}y_{a}}+\frac{3}{5}l\sin\alpha+\frac{2}{5}l \sin(\alpha+\theta_{2})+\frac{1}{5}l\sin(\alpha+\theta_{2}+\theta_{3})+\frac{ 1}{5}l\sin(\theta_{1}-\alpha)-\frac{1}{5}R}\\ \end{array} \right. \tag{6}\]  
During the rolling process, there is a singular configuration when the five chords of modules form a concave pentagon, which could cause a serious impact on the joint activities. Therefore, the mechanism should maintain a convex pentagon to avoid the generation of singular configurations. For this purpose, [PERSON] et al. [8] proposed one point contact constraint model to limit the dynamic rolling locomotion and acquired the ranges of the joint angles by numerical method. However, the numerical method is not suitable for kinematics analysis. In order to obtain analytical formulas of the range of active joint angles, we establish geometrical models to discuss the two situations for \(\alpha\geq 0\) and \(\alpha<0\), i.e., \(0\leq\phi\leq\pi/5\) and \(\pi/5<\phi\leq 2\pi/5\). The analytical models of the two situations are, respectively illustrated in Figures 3 and 4. (1) \(0\leq\phi\leq\pi/5\). Under this situation of \(\phi\), the range of \(\theta_{1}\) is from \(2\pi/5-2\phi\) to \(\pi-2\arcsin(1/4)\), while \(\theta_{1}\) is divided into two segments for discussion as in Figure 3a,b. According to the range of \(\theta_{1}\) in Figure 3a,b, the corresponding \(\theta_{2}\) can be obtained as Equations (7) and (8). \[\left\{ \begin{array}{l}\theta_{2\min}=\pi-\frac{\theta_{1}}{2}-\arccos \left(\frac{2\cos\theta_{1}-1}{2\sqrt{2+2\cos\theta_{1}}}\right)\\ \theta_{2\max}=\pi-\frac{\theta_{1}}{2}-\arccos\left(\frac{2\cos\theta_{1}+5}{ 2\sqrt{2+2\cos\theta_{1}}}\right)\end{array} \right. \tag{7}\]  
\[\left\{ \begin{array}{l}\theta_{2\min}=0\\ \theta_{2\max}=\pi-\arccos\left(\frac{2\cos\theta_{1}+1}{\sqrt{5+4\cos\theta_{1 }}}\right)-\arccos\left(\frac{\sqrt{5+4\cos\theta_{1}}}{2}\right)\end{array} \right. \tag{8}\]  
(2) \(\pi/5<\phi\leq 2\pi/5\). Under this premise, the range of \(\theta_{2}\) is from \(2\pi/5-2\phi\) to \(\pi-2\arcsin(1/4)\), and \(\theta_{2}\) is divided into two segments for discussion as in Figure 4a,b. According to the range of \(\theta_{2}\) in Figure 4a,b, the corresponding \(\theta_{1}\) can be obtained as Equations (9) and (10). \[\left\{ \begin{array}{l}\theta_{1\min}=\pi-\frac{\theta_{2}}{2}-\arccos \left(\frac{2\cos\theta_{2}-1}{2\sqrt{2+2\cos\theta_{2}}}\right)\\ \theta_{1\max}=\pi-\frac{\theta_{2}}{2}-\arccos\left(\frac{2\cos\theta_{2}+5} {2\sqrt{2+2\cos\theta_{2}}}\right)\end{array} \right. \tag{9}\]  
\[\left\{ \begin{array}{l}\theta_{1\min}=0\\ \theta_{1\max}=\pi-\arccos\left(\frac{2\cos\theta_{2}+1}{\sqrt{5+4\cos\theta_ {2}}}\right)-\arccos\left(\frac{\sqrt{5+4\cos\theta_{2}}}{2}\right)\end{array} \right. \tag{10}\]