## 2 Materials and Methods
### Analytical Study of Forward Kinematics  
The goal here is to determine the coordinates of point \(C\) (\(x_{c}\), \(y_{c}\)), which represents the position of the end-effector, given the input angles \(\theta_{1}\) and \(\theta_{4}\) of the actuators. This process involves analyzing the manipulator's geometry based on the constraints of its closed kinematic chain, as shown in Figure 1.  
As depicted in Figure 1, the manipulator configuration consists of two fixed points, \(A\) and \(E\):  
* Point \(A\) is located at the origin \((0,0)\). * Point \(E\) is displaced along the \(x\)-axis at position (\(x_{E},0\)). The links \(L_{1}\), \(L_{2}\), \(L_{3}\), and \(L_{4}\) connect the joints at points \(A\), \(B\), \(C\), \(D\), and \(E\), where each joint has a specific angle: \(\theta_{1}\) at \(A\); \(\theta_{2}\) at \(B\); \(\theta_{3}\) at \(D\); and \(\theta_{4}\) at \(E\). The positions of points \(B\) and \(D\) were derived using trigonometric relationships of the link lengths and input angles. The position of points B and D can be found from an analysis of the configuration shown in Figure 1. Given the fixed base at \(A\), the coordinates of point \(B\) and D are as follows:  
\[x_{B}=L_{1}\cos\theta_{1} \tag{1}\]  
\[y_{B}=L_{1}\sin\theta_{1} \tag{2}\]  
\[x_{D}=x_{E}+L_{4}\cos\theta_{4} \tag{3}\]  
\[y_{D}=y_{E}+L_{4}\sin\theta_{4} \tag{4}\]  
To determine the position of point \(C\), which lies at the intersection of links \(L_{2}\) and \(L_{3}\), the following conditions must be satisfied:  
\[L_{2}^{2}=(x_{C}-x_{B})^{2}+(y_{C}-y_{B})^{2} \tag{5}\]  
Substituting Equations (1) and (2) in Equation (5), we obtain the following:  
\[L_{2}^{2}=(x_{C}-L_{1}\cos\theta_{1})^{2}+(y_{C}-L_{1}\sin\theta_{1})^{2} \tag{6}\]  
Figure 1: Two-degree-of-freedom planar parallel manipulator 2-RRR. \[L_{2}^{2}=x_{C}^{2}+y_{C}^{2}-2x_{C}L_{1}\text{cos}\,\theta_{1}-2y_{C}L_{1}\sin \theta_{1} \tag{7}\]  
Figure 1.0  
Figure 1.1  
Figure 1.2  
To determine the position of \(C\), subtracting Equation (8) from Equation (10), to eliminate \(x_{C}^{2}\) and \(y_{C}^{2}\), and solving for \(x_{c}\) in terms of \(y_{c}\), Equation (12) is obtained:  
\[ \begin{array}{c}x_{C}(2x_{E}+2L_{4}\text{cos}\,\theta_{4}-2L_{1}\text{cos}\, \theta_{1})+y_{C}(2L_{4}\text{sin}\,\theta_{4}-2L_{1}\text{cos}\,\theta_{1}) \ldots\\ \ldots+L_{1}^{2}+L_{3}^{2}-L_{2}^{2}-L_{4}^{2}-x_{E}^{2}-2L_{4}x_{E}\text{cos} \,\theta_{4}=0\end{array}  \tag{11}\]  
\[x_{C}=e\times y_{c}+f \tag{12}\]  
where  
\[e=\frac{(L_{1}\text{sin}\,\theta_{1}-L_{4}\text{sin}\,\theta_{4})}{(x_{E}+L_{4 }\text{cos}\,\theta_{4}-L_{1}\text{cos}\,\theta_{1})} \tag{13}\]  
\[f=\frac{\left(L_{1}^{2}+L_{3}^{2}-L_{2}^{2}-L_{4}^{2}-x_{E}^{2}-2L_{4}x_{E} \text{cos}\,\theta_{4}\right)}{(2L_{1}\text{cos}\,\theta_{1}-2L_{4}\text{cos }\,\theta_{4}-2x_{E})} \tag{14}\]  
Substituting Equation (12) into Equation (7),  
\[L_{2}^{2}=(e\times y_{c}+f)^{2}+y_{C}^{2}-2(e\times y_{c}+f)L_{1}\text{cos}\, \theta_{1}-2y_{C}L_{1}\text{sin}\,\theta_{1}+L_{1}^{2} \tag{15}\]  
Simplifying to obtain a quadratic equation,  
\[ \begin{array}{c}y_{C}^{2}\big{(}e^{2}+1\big{)}+y_{C}(2 ef-2 eL_{1}\text{cos} \,\theta_{1}-2L_{1}\text{sin}\,\theta_{1})+f^{2}-2 fL_{1}\text{cos}\,\theta_{ 1}\ldots\\ \ldots+L_{1}^{2}-L_{2}^{2}=0\end{array}  \tag{16}\]  
\[dy_{C}+gy_{c}+h=0 \tag{17}\]  
where  
\[d=\Big{(}e^{2}+1\Big{)} \tag{18}\]  
\[g=(2 ef-2 eL_{1}\text{cos}\,\theta_{1}-2L_{1}\text{sin}\,\theta_{1}) \tag{19}\]  
\[h=\Big{(}f^{2}-2 fL_{1}\text{cos}\,\theta_{1}+L_{1}^{2}-L_{2}^{2}\Big{)} \tag{20}\]  
Solving for \(y_{C}\),  
\[y_{C}=\frac{-g\pm\sigma\sqrt{g^{2}-4 dh}}{2d} \tag{21}\]  
The parameter \(\sigma=1\)\(o\)\((-1)\) yields two possible solutions [11] corresponding to two assembly modes (Figure 2):  
* Upper configuration \(\sigma=1\) (Figure 2a). * Lower configuration \(\sigma=1\) (Figure 2b). In this study, the upper configuration was used.