## 4 Results  
### Example 1  
Figure 11a shows a 3D printed model of a SCARA type robot. Its structure consists of two links and a base connected by rotational joints, so that the main movements take place in the plane. We are not interested in the vertical translation at this stage. The lengths of the links are respectively \(a_{1}=0.15[\mathrm{m}];\ a_{2}=0.1[\mathrm{m}]\); and the joint constraints are: \(-\frac{\pi}{2}\leq\theta_{1}\leq\frac{\pi}{2};-\pi\leq\theta_{2}\leq\frac{ \pi}{2}\) [rad]. Figure 11: (**a**) 3D printed model of a SCARA type robot; (**b**) Reachability area with left- and right-hand configuration and workspace. Figure 11b shows the left \(U_{L}\) and right \(U_{R}\) hand reachability zones for this robot. The union of these two areas represents the robot's workspace:  
\[w=U_{L}\cup U_{R} \tag{14}\]  
Section \(f\) represents a region where both left-handed and right-handed solutions can exist. \[f=U_{L}\cap U_{R} \tag{15}\]  
There is an area \(f_{L}\), the points from which can only be reached with left-hand configuration:  
\[f_{L}=w-U_{R} \tag{16}\]  
And area \(f_{R}\) which is reachable only with right-hand configuration:  
\[f_{R}=w-U_{L} \tag{17}\]  
Of course, \(w=f_{L}\cup f_{R}\cup f\) Figure 12 (right) illustrates those areas in Cartesian space. Figure 12 (left) shows a graphical representation of the configuration space. What happens at the boundaries of the configuration space, and when \(\theta_{2}=0\)? The solution of the forward problem in this case is simple. The coordinates of point P are determined (see Figure 11 on the right):  
\[ \begin{vmatrix}x_{p}=a_{1}\cos(\theta_{1})+a_{2}\cos(\theta_{1}+\theta_{2}) \\ y_{p}=a_{1}\sin(\theta_{1})+a_{2}\sin(\theta_{1}+\theta_{2})\end{vmatrix}  \tag{18}\]  
When \(\theta_{2}=0\) and \(\theta_{2}=-\pi\) are obtained for points from the configuration space that belong to lines in this space (the red horizontal lines in Figure 12, left), they are transformed into points of circles in the real space (the red arcs on the right of Figure 12). We can verify this from the analytic dependencies as well. Indeed, from (18), for \(\theta_{2}=0\) it follows:  
\[ \begin{vmatrix}x_{p}=(a_{1}+a_{2})\cos(\theta_{1})\\ y_{p}=(a_{1}+a_{2})\sin(\theta_{1})\end{vmatrix}  \tag{19}\]  
After raising to the second degree and summing the two equations from system (19) it is obtained:  
\[ \begin{vmatrix}x_{p}^{2}+y_{p}^{2}=(a_{1}+a_{2})^{2}\Big{(}\sin(\theta_{1})^{ 2}+\cos(\theta_{1})^{2}\Big{)}\end{vmatrix}  \tag{20}\]  
Figure 12: Configuration and workspace of a RR robot. or \[x_{\rm p}^{2}+y_{\rm p}^{2}=\left(a_{1}+a_{2}\right)^{2}\] (21) This is an equation of a circle with radius \(R=\left(a_{1}+a_{2}\right)\) and center (0,0). When \(\theta_{2}=-\pi\), following similar considerations, it is obtained: \[x_{\rm p}^{2}+y_{\rm p}^{2}=\left(a_{1}-a_{2}\right)^{2}\] (22) This is an equation of a circle with radius \(R=\left(a_{1}-a_{2}\right)\). Not all points from these red circles (Figure 12-right) are fully reachable by the robot due to the joint constraints. The points from these arcs can be reached with only one singular configuration of the robot, which means that there exists only one unique solution to both the forward and the inverse kinematics problems. In this case the arm is either fully extended or the links overlap one another. Let us now investigate what happens to the points from the configuration space that lie on the line \(\theta_{1}=\frac{\pi}{2}\). From (19) it follows that:  
\[\left| \begin{array}{c}x_{p}=a_{2}\cos\left(\theta_{2}+\frac{\pi}{2}\right) \\ y_{p}=a_{1}+a_{2}\sin\left(\theta_{2}+\frac{\pi}{2}\right)\end{array} \right. \tag{23}\]  
Analogously, it is obtained an equation of a circle:  
\[x_{p}^{2}+\left(y_{p}-a_{1}\right)^{2}=a_{2}^{2} \tag{24}\]  
Similarly for \(\theta_{1}=-\frac{\pi}{2}\), it follows:  
\[x_{p}^{2}+\left(y_{p}+a_{1}\right)^{2}=a_{2}^{2} \tag{25}\]  
These circles correspond to the blue arcs on Figure 12 (right), and are on the boundaries of the workspace.