## 5 Example and Verification Experiment
### Verification of Multisolution Selection Method Based on the Key Degrees of the Joints  
In order to verify the correctness of the multisolution selection method based on the key degrees of the joints, the workspace stratification model generated in Section 5.1 was used to select a layer boundary for analysis. As shown in Figure 12, two path points P0 = (258.238, 12.483,200) and P1 = (255.829, 37.333, 200) were taken from the printing path, and Q0 = (2.7675, 99.0989, -117.2417) was known. Verification of the multisolution selection method was performed according to the following steps. 1. According to the inverse kinematic formula in Section 2.2, four groups of solutions of joint angles at path point P1 were solved, and then all possible groups of joint angles were obtained according to Equation (12), as shown in Table 4. They were seen as the possible solutions for the 0 th key joint. 2. According to Section 3.1, the Jacobian matrix J0 was obtainedat the path point P0. The sensitivity Sq of each joint was calculated by the Jacobian matrix J0. \[\text{J}_{0}=\left[ \begin{array}{ccc}98.7823&-34.0912&108.8027\\ -251.6241&-13.3835&42.7137\\ 0&158.3195&136.8850\end{array} \right]\] \[\left\{ \begin{array}{ccc}\text{S}_{\alpha}=270.3195&\\ \text{S}_{\beta}=162.5004&\\ \text{S}_{\gamma}=180.0000&\end{array} \right.\]  
The key degrees of all joints were obtained and all joints were sorted according to their sensitivity. For Sx > Sy > Sg, the A-joint was the first key joint and its key order was one; the C-joint was the second key joint and its key order was two; the B-joint was the third key joint and its key order was three. The nominal increments and the nominal joint angles of all joints were calculated based on Equations (16) and (17) with the Jacobian matrix. The displacement components of the end position from the path point P0 to P1 were calculated as \(\Delta\text{P}_{\text{x}}=-2.409\) mm, \(\Delta\text{P}_{\text{y}}=24.850\) mm, and \(\Delta\text{P}_{\text{z}}=0\) mm. According to the Jacobian matrix J0 and Equation (16), the nominal increments of all joint angles from P0 to P1 were calculated as  
\[\Delta\text{Q}_{1}=\left[\Delta\alpha_{1},\Delta\beta_{1},\Delta\gamma_{1} \right]^{\text{T}}=\left(\text{J}_{0}\text{T}_{0}\right)^{-1}\text{J}_{0} ^{\text{T},\left(\text{P}_{1}-\text{P}_{0}\right)}=\left[0.0965,0.0083,-0.0071 \right]^{\text{T}}\text{rad}=\left[5.5264^{\circ},0.4766^{\circ},-0.4083^{ \circ}\right]^{\text{T}}\]  
Based on Equation (17), the nominal joint angles Q1 of point P1 were obtained. \[\text{Q}_{1}^{\text{n}}=\text{Q}_{0}+\Delta\text{Q}_{1}=\left[\Delta\alpha_{1} {}^{\text{n}},\Delta\beta_{1}{}^{\text{n}},\Delta\gamma_{1}{}^{\text{n}} \right]^{\text{T}}=\left[8.2939^{\circ},99.5755^{\circ},-117.6500^{\circ}\right] ^{\text{T}}\]  
1. For j = 1, the joint with the key order of 1 was found to be the A-joint. Its nominal joint angles \(\theta_{1,1}^{\text{n}}\) were obtained as \(8.2939^{\circ}\) based on Q1. The differences \(\left|\text{q}_{1,1}-\theta_{1,1}^{\text{n}}\right|\) between \(\theta_{1,1}^{\text{n}}\) and \(\text{q}_{1,1}\) in Table 4 were calculated. The smallest difference of \(\alpha\)-joint angles was \(8.3025^{\circ}-8.2939^{\circ}=0.0086^{\circ}\). The possible solutions for the first key order were \(\text{1Q}_{1}\) and \(\text{2Q}_{1}\), and then, j = j + 1 = 2. The second key order was the C-joint, and its nominal joint angles \(\theta_{3,1}^{\text{n}}\) were obtained as \(-117.65^{\circ}\). The differences \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Num.** & \(\alpha\) **(\text{q}\text{Q}_{1}\text{,1}) (\({}^{\circ}\))** & \(\beta\) **(\text{q}\text{Q}_{2}\text{,1}) (\({}^{\circ}\))** & \(\gamma\) **(\text{q}\text{Q}_{2}\text{,1}) (\({}^{\circ}\))** \\ \hline \(\text{1Q}_{1}\) & \(8.3025\) & \(99.0992\) & \(-117.2420\) \\ \(\text{2Q}_{1}\) & \(8.3025\) & \(-32.0948\) & \(117.2420\) \\ \(\text{3Q}_{1}\) & \(-171.6975\) & \(99.0992\) & \(-117.2420\) \\ \(\text{4Q}_{1}\) & \(-171.6975\) & \(-32.0948\) & \(117.2420\) \\ \hline \hline \end{tabular}
\end{table}