# Kinematics Model Optimization Algorithm for Six Degrees of Freedom Parallel Platform
## 3 Results
### MPR-NR Algorithm Simulation  
The following experiments were set up to verify that the new method can operate normally according to theoretical expectations. The six-variate quadratic polynomial regression + Newton iterative method (MPR2-NR) and the six-variate cubic polynomial regression + Newton iterative method (MPR3-NR) were used to solve the kinematics positive solution for the same training set and test set. Moreover, compared with the results obtained by the Newton iterative algorithm, the solution accuracy, the calculation time consumption, and the total number of iterations were compared. The results are shown in Table 3.  
It can be seen from Table 3 that, in the two sets of test sets, the calculation accuracy of MPR3-NR is at the same level as that of Newton iteration. The MPR2-NR algorithm is an order of magnitude weaker than the other two algorithms in Test Set 1. Using the calculation method of Equation (18), we can obtain an average single solution time of about 148 microseconds for the Newton iterative algorithm. The average single solution time of the MPR2-NR algorithm is about 79 microseconds. The average single solution time of the MPR3-NR algorithm is about 78 microseconds. For Test Set 1, it can be seen that the number of Newton iterations for both MPR2-NR and MPR3-NR is 0. This phenomenon shows that, in Test Set 1, the accuracy of the MPR algorithm met the system requirements, and there is no Newton iteration process. At this time, compared with the MPR algorithm, the MPR-NR algorithm has one more inverse kinematics solution and the judgment of the accuracy of the solution. Therefore, compared with the corresponding MPR algorithm in Table 2, the average single calculation time of the MPR-NR algorithm is increased by 12 microseconds. For Test Set 2, MPR3-NR takes more time to solve than MPR2-NR when performing a polynomial regression procedure. However, MPR3-NR gives a better initial solution \(X\_0\) than MPR2-NR, making the average number of Newton iterations of MPR3-NR 0.78 times less than MPR2-NR. Therefore, the total solution time of the algorithm is that MPR3-NR is slightly better than MPR2-NR. The following conclusions can be drawn by comparing the number of Newton iterations of the Newton iteration algorithm and the MPR-NR algorithm. After adding the initial solution calculation of polynomial regression, the method of using the Newton iterative algorithm for kinematics forward solution has a better efficiency than directly using the Newton iterative algorithm, and the accuracy can be maintained at the same level. Conclusions can be drawn from the experimental results. The MPR-NR algorithm can guarantee the same accuracy as the Newton iterative algorithm and greatly reduce the solution time, as is consistent with the theoretical analysis conclusion. In order to further explore the accuracy controllability of the Newton iteration algorithm and the MPR-NR algorithm, the relationship between the algorithm solution accuracy \begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline \multirow{2}{*}{**Algorithm**} & \multirow{2}{*}{**Dataset**} & \multicolumn{4}{c}{**Kinematics Positive Solution Mean Absolute**} & \multirow{2}{*}{**Time**} & \multirow{2}{*}{**Iterations**} \\  & & & & & & & & \\ \hline \multirow{2}{*}{Newton iteration} & Test Set 1 & \multirow{2}{*}{[3.2 ]} & 2.7 & 12.5 & 3.2 & 3.0 & 1.4 & \(\times\) 10\({}^{-5}\) & 1.144 & 26,710 \\  & Test Set 2 & & 0.3 & 0.1 & 0.6 & 0.05 & 0.05 & 0.03 & \(\times\) 10\({}^{-3}\) & 1.815 & 41,664 \\ \hline \multirow{2}{*}{MPR2-NR} & Test Set 1 & \multirow{2}{*}{[3.4 ]} & 3.4 & 1.7 & 2.8 & 0.4 & 0.5 & 0.2 & \(\times\) 10\({}^{-4}\) & 0.296 & 0 \\  & Test Set 2 & & 0.3 & 0.1 & 0.6 & 0.05 & 0.05 & 0.03 & \(\times\) 10\({}^{-3}\) & 1.284 & 26,019 \\ \hline \multirow{2}{*}{MPR3-NR} & Test Set 1 & \multirow{2}{*}{[1.8 ]} & 1.9 & 2.7 & 0.3 & 0.2 & 0.2 & \(\times\) 10\({}^{-5}\) & 0.407 & 0 \\  & Test Set 2 & & 0.4 & 0.3 & 0.7 & 0.07 & 0.07 & 0.04 & \(\times\) 10\({}^{-3}\) & 1.162 & 16,998 \\ \hline \hline \end{tabular}
\end{table}