## 2 Structural Design
### Kinematic Analysis of Four-Bar Mechanism  
When transplanting, the shape of the sweet potato seedlings planted in the soil is consistent with the movement trajectory of the transplanting manipulator, and the trajectory is composed of horizontal forward and rotation movement. In order to realize the trajectory, we decided to adopt a four-bar mechanism as the prototype of the transplanting manipulator. The gripping levers of the transplanting manipulator are part of the connecting rod in the four-bar mechanism, as depicted in Figure 4. The four-bar mechanism's kinematic analysis is shown below. According to Figure 4, we established the vector equation of the four-bar mechanism [24]:  
\[\left\{ \begin{array}{c}\overset{\rightarrow}{OA}+\overset{\rightarrow}{ AB}=\overset{\rightarrow}{OC}+\overset{\rightarrow}{CB}\\ \overset{\rightarrow}{OE}=\overset{\rightarrow}{OA}+\overset{\rightarrow}{ AE}\end{array} \right. \tag{2}\]  
We established the displacement equation for point E:  
\[\left\{ \begin{array}{c}x_{A}+L_{2}.cos\theta_{2}=x_{C}+L_{3}.cos\theta_{3}\\ y_{A}+L_{2}.sin\theta_{2}=y_{C}+L_{3}.sin\theta_{3}\end{array} \right. \tag{3}\]  
\[\left\{ \begin{array}{c}x_{E}=x_{A}+L_{5}.cos(\theta_{2}+\delta)\\ y_{E}=y_{A}+L_{5}.sin(\theta_{2}+\delta)\end{array} \right. \tag{4}\]  
where \(x_{A}=L_{1}.cos\theta_{1}\); \(y_{A}=L_{1}.sin\theta_{1}\). \(\theta_{2}\) and \(\theta_{3}\) are unknown quantities in Equation (3), and \(x_{E}\) and \(y_{E}\) are unknown quantities in Equation (4). We can solve the displacement equation as follows. \[\text{Let }tg\beta=\frac{y_{A}-y_{C}}{x_{A}-y_{C}}\]  
In triangle ABC, \(\angle\)ACB = \(\beta-\theta_{3}\), so we can know:  
Figure 4: Analysis of the four-bar mechanism. \(V_{T}\) is the forward speed of the transplanter during field work; \(OA\) is the crank; \(O\) is the crank rotation center; \(BAE\) is the link; \(AE\) is the gripping lever; \(BC\) is the rocker; \(OC\) is the frame; the length of OA is \(L_{1}\), mm; the length of \(AB\) is \(L_{2}\), mm; the length of \(BC\) is \(L_{3}\), mm; the length of \(OC\) is \(L_{4}\), mm; the length of \(AE\) is \(L_{5}\), mm; the angle formed by \(OA\) and \(OC\) is denoted by \(\theta_{1}\)\({}^{\circ}\); the angle formed by \(AB\) and the horizontal plane is denoted by \(\theta_{2}\), \({}^{\circ}\); the angle formed by \(BC\) and the horizontal plane is denoted by \(\theta_{3}\), \({}^{\circ}\); the angle of \(\angle\)\(BAE\) is \(\delta\),\({}^{\circ}\); the angle of \(\angle\)\(BCA\) is \(\beta\), \({}^{\circ}\). \[\cos(\beta-\theta_{3})=\frac{\left(x_{A}-x_{\text{C}}\right)^{2}+\left(y_{A}-y_{ \text{C}}\right)^{2}+{L_{3}}^{2}-{L_{2}}^{2}}{2{L_{3}}\sqrt{\left(x_{A}-x_{ \text{C}}\right)^{2}+\left(y_{A}-y_{\text{C}}\right)^{2}}} \tag{5}\]  
We can thus obtain the value of \(\beta-\theta_{3}\), and the value of \(\theta_{3}\). Moreover, we can determine that \(sin\theta_{2}=\frac{y_{\text{C}}+L_{3}sin\theta_{3}-y_{A}}{L_{2}}\), so we obtain the value of \(\theta_{2}\). According to Equations (2) and (3), the equation of velocity is established. Equation (6) is obtained by deriving Equations (3) and (4):  
\[\left\{ \begin{array}{l}\dot{x_{A}}-L_{2}.\dot{\theta_{2}}.sin\theta_{2}=0-L_ {3}.\dot{\theta_{3}}.sin\theta_{3}\\ y_{A}+L_{2}.\dot{\theta_{2}}.cos\theta_{2}=0+L_{3}.\dot{\theta_{3}}.cos\theta _{3}\end{array} \right. \tag{6}\]  
\[\left\{ \begin{array}{l}\dot{x_{E}}=\dot{x_{A}}-L_{5}.\dot{\theta_{2}}.sin( \theta_{2}+\delta)\\ \dot{y_{E}}=\dot{y_{A}}-L_{5}.\dot{\theta_{2}}.cos(\theta_{2}+\delta)\end{array} \right. \tag{7}\]  
where \(\dot{x_{A}}=-L_{1}.\dot{\theta_{1}}.sin\theta_{2}\), and \(\dot{y_{A}}=L_{1}.\dot{\theta_{1}}.cos\theta_{1}\).