## 5 Inverse Kinematics Analysis
### The Closed-Form Solution of Inverse Kinematics  
According to Figure 2, the joint axes represented by 2, 3, and 4 in Figure 2 are parallel. According to Pieper criteria, the closed-form solution is utilized to obtain the sequence of the UR10 Robot's joint angles under the constraint of the target pose.  
The closed-form solution is employed to first establish Equation (8) according to Equation (7) and then establish Equation (9) via Equation (8).  
\[\left(\begin{matrix}0\\ 1\end{matrix}T^{-1}\right)T_{robot}=\prod\limits_{i=2}^{6}i_{i}^{-1}T \tag{8}\]  
\[\left[\begin{matrix}c\theta_{65}\theta_{5}\\ s\theta_{65}\theta_{5}\\ c\theta_{5}\\ d_{2}+d_{3}+d_{4}+d_{6}c\theta_{5}+a_{58}\theta_{5}+a_{6}c\theta_{65}\theta_{5} \end{matrix}\right]^{T}=\left[\begin{matrix}y_{5}c\theta_{1}+n_{x}s\theta_{1} \\ o_{y}c\theta_{1}+o_{x}s\theta_{1}\\ a_{y}c\theta_{1}+a_{x}s\theta_{1}\\ yc\theta_{1}+xs\theta_{1}\end{matrix}\right]^{T} \tag{9}\]  
Because \(d_{2}=d_{3}=a_{5}=a_{6}=0\), Equation (9) is expanded to obtain Equation (10). The radian value of joint 1, \(\theta_{1}\), can be obtained via Equation (10). According to the range of motion of joint 1, four solutions can be obtained in for \(\theta_{1}\) in Equation (10).  
\[\begin{split} cd_{4}+d_{6}a_{y}c\theta_{1}+d_{6}a_{x}s\theta_{1} =yc\theta_{1}+xs\theta_{1}\\ d_{4}=\sqrt{\left(y-d_{6}a_{y}\right)^{2}+\left(x-d_{6}a_{x} \right)^{2}}\sin(\theta_{1}+\ell),\tan\ell=\frac{y-d_{6}a_{y}}{x-d_{6}a_{x}} \end{split} \tag{10}\]  
Through the radian value of the first joint \(\theta_{1}\), the radian value of the joint 5, \(\theta_{5}\), and the radian value of the joint 6 \(\theta_{6}\). After Equation (11) is resolved, according to the range of motion of joint 5, four solutions can be obtained for \(\theta_{5}\), two solutions can be obtained for \(\theta_{6}\).  
\[\left[\begin{matrix}\theta_{5}\\ \theta_{6}\end{matrix}\right]=\left[\begin{matrix}a\cdot\cos(a_{y}c\theta_{1 }+a_{x}s\theta_{1})\\ a\cdot\tan 2\big{(}n_{y}c\theta_{1}+n_{x}s\theta_{1},o_{y}c\theta_{1}+o_{x}s \theta_{1}\big{)}\end{matrix}\right] \tag{11}\]  
To obtain \(\theta_{2}\), \(\theta_{3}\), and \(\theta_{4}\), \(\theta_{1}\), \(\theta_{5}\), and \(\theta_{6}\) are utilized. With Equation (12), an equation for \(\theta_{2}\), \(\theta_{3}\), and \(\theta_{4}\), Equation (12) is expanded to obtain Equation (13), where \(\theta_{i_{1}i_{2}\cdots i_{n}}=\sum\limits_{j=1}^{n}\theta_{ij}\).  
\[L=\left(\begin{matrix}0\\ 1\end{matrix}T^{-1}\right)\theta T\left(\begin{matrix}\sharp T^{-1}\end{matrix} \right)\left(\begin{matrix}\sharp T^{-1}\end{matrix}\right)=\prod\limits_{i=2} ^{4}i^{-1}T=R \tag{12}\]  
\[R=\left[\begin{matrix}c\theta_{234}&0&-s\theta_{234}&a_{3}\cdot c\theta_{23}+a _{2}\cdot c\theta_{2}\\ -s\theta_{234}&0&c\theta_{234}&-a_{3}\cdot s\theta_{23}-a_{2}\cdot c\theta_{2} \\ 0&-1&0&d_{4}\\ 0&0&0&1\end{matrix}\right] \tag{13}\]  
Equation (14) provides the solution for \(\theta_{2}\), \(\theta_{3}\), and \(\theta_{4}\). Therefore, \(\theta_{234}\) has two solutions. Both \(\theta_{23}\) and \(\theta_{2}\) have four solutions. In Equations (9), (11), and (13), c represents the cosine function and s represents the sine function.  
\[\left[\begin{matrix}\theta_{234}\\ \theta_{23}\\ \theta_{2}\\ \theta_{3}\\ \theta_{4}\end{matrix}\right]=\left[\begin{matrix}a\tan 2(-L_{21},L_{11})\\ a\cos\left(\frac{L_{14}+L_{23}}{\sqrt{2}a_{3}}\right)-\frac{\pi}{4}\\ a\cos\left(\frac{L_{14}-a_{3}c\theta_{23}}{a_{2}}\right)\\ a\cos\left(\frac{L_{14}-a_{3}c\theta_{23}}{a_{2}}\right)\\ \theta_{23}-\theta_{2}\\ \theta_{234}-\theta_{23}\end{matrix}\right] \tag{14}\]According to the above-mentioned method, there are 1024 closed-form solutions for each pose and hence the initial radian solution of the Euler solution with the least distance needs to be selected.