## 4 Results
### Example 1
Similarly, the equation of the green arc can be determined, which serves as the boundary of solutions with right-hand configurations. Since the workspace of the robot is divided into areas where point \(P\) can be reached in two ways (left and right-hand--\(f\)) and those reachable with only one type of solution (\(f_{R}\) and \(f_{L}\)), the configuration space should also be divided in a similar way. The algorithm from Figure 9 and Appendix A are used to find IK solutions. For a point \(P(180[\rm mm],120[\rm mm])\) (Figure 11, right), two solutions of the inverse problem are found: for the left-hand it is obtained \(q_{R}(1.06[\rm rad],-1.07[\rm rad])\) and for the right-hand \(q_{L}(0.17[\rm rad],1.07[\rm rad])\) respectively. The result is presented graphically in Figure 12.  
Using the proposed algorithm, multiple solutions are found for the inverse kinematics from the robot's workspace. Using polar coordinates, a set of points, located in the reachable zone according to Figure 8, is generated. The results are given in Figure 13.  
The algorithm removes the points outside the joint constraints. The solutions that are only possible with the left-hand (L) are depicted with small blue circles; those of type R are magenta; and where both types of solutions are possible, they are represented with green circles. Thus, the solution of the inverse problem yields a region \(q(ABCDEF)\) in the configuration space, inside which both left and right-handed solutions are possible. It can be noted that AB is a vertical segment, BC is a horizontal segment, CD is a curved line, DE is a vertical segment, EF is a horizontal segment and FA is again a curved line. Areas from the workspace are transformed into the configuration space, respectively:  
\[\left| \begin{array}{c}f_{L}\stackrel{{ IK}}{{\rightarrow}}q_{L}\\ f\stackrel{{ IK}}{{\rightarrow}}q\\ f_{R}\stackrel{{ IK}}{{\rightarrow}}q_{R}\end{array} \right. \tag{26}\]In Figure 13 (right) two trajectories \(t_{1}\) and \(t_{2}\) are given, which connect point \(P_{1}\in f_{L}\) with point \(P_{2}\in f_{R}\). Both trajectories fall entirely within the robot's workspace. Using the algorithm, points from \(t_{1}\) and \(t_{2}\) are transformed into the configuration space. Trajectory \(t_{1}\) is an arc of a circle with radius R = 90 [mm] and center (x = 60 [mm], y = 116 [mm]). Each point of this arc corresponds to two solutions in the configuration space, which form two red curves \(q_{t1L}\) and \(q_{t1R}\). Parts of these two curves go outside the joint constraints. The trajectory \(t_{2}\) is a spline that passes through point \(P_{3}\), which lies on the boundary of the workspace. Each point from \(t_{2}\) corresponds to two points in the configuration space except point \(P_{3}\), for which there is only one solution because it is reached with a singularity configuration of type \(S_{0}\). The two curves in which \(t_{2}\) is transformed are given in green--\(q_{t2p}\) and blue--\(q_{t2i}\) color, respectively. After solving the IK, point \(P_{1}\) is transformed into two points \(q_{p1L}\) and \(q_{p1R}\) in the configuration space, where \(q_{p1R}\) is outside the joint constraints (not defined). Point \(P_{2}\) is transformed into two points \(q_{P2L}\) and \(q_{P2R}\) in the configuration space, where \(q_{P2L}\) is outside the joint constraints (not defined). Point \(P_{3}\) is transformed into a single point \(q_{P3}\), which lies on the axis \(\theta_{1}\). At this point, the curves \(q_{t2p}\) and \(q_{t2i}\) intersect. It can be said that each of the two trajectories \(t_{1}\) and \(t_{2}\) from the workspace is transformed into two trajectories in the configuration space, and only one of them \(q_{t2p}\) (in green) falls entirely within the joint constraints of the robot.