## 3 Calibration Method for the Angular Positioning Deviation of the Rotary Table
### Calibration of The Angular Positioning Deviation of the Rotary Table  
#### 3.3.1 Solving the Center of the Circle of the Laser Tracer Station Coordinate Fitting Circle  
Let the plane fitted by the \(m\) station coordinates of the laser tracer obtained by the self-calibration algorithm be \(\gamma\). The center of the circle fitted by m station coordinates of laser tracer is \(O(x_{c},y_{c},z_{c})\). The projection equation of the plane formed by the plane \(\gamma\) at plane \(z=z_{c}\) is set.  
\[\left(X_{j}^{\prime}-x_{c}\right)^{2}+\left(Y_{j}^{\prime}-y_{c}\right)^{2}=R^ {2} \tag{22}\]  
where \(x_{c}\) is the coordinate value of the center of the fitting circle in the X-axis direction; \(y_{c}\) is the coordinate value of the center of the fitting circle in the Y-axis direction, and \(R\) is the radius of the fitting circle.  
According to the principle of nonlinear least squares, the objective function can be defined as  
\[f=\sum\limits_{j=1}^{m}\left(\left(X_{j}^{\prime}-x_{c}\right)^{2}+\left(Y_{j }^{\prime}-y_{c}\right)^{2}-R^{2}\right)^{2} \tag{23}\]  
Let \(g\left(X_{j}^{\prime},Y_{j}^{\prime}\right)=\left(X_{j}^{\prime}-x_{c}\right) ^{2}+\left(Y_{j}^{\prime}-y_{c}\right)^{2}-R^{2}\), then  
\[f=\sum\limits_{j=1}^{m}g\left(X_{j}^{\prime},Y_{j}^{\prime}\right)^{2} \tag{24}\]  
In order to minimize the objective function \(f\), Equation (24) should satisfy the following conditions,  
\[\left\{\begin{array}{c}\frac{\partial f}{\partial R}=-4R\sum\limits_{j=1}^{m }g\left(X_{j}^{\prime},Y_{j}^{\prime}\right)=0\\ \frac{\partial f}{\partial x_{c}}=-4\sum\limits_{j=1}^{m}g\left(X_{j}^{\prime },Y_{j}^{\prime}\right)\left(X_{j}^{\prime}-x_{c}\right)=-4\sum\limits_{j=1}^{ m}X_{j}^{\prime}g\left(X_{j}^{\prime},Y_{j}^{\prime}\right)=0\\ \frac{\partial f}{\partial y_{c}}=-4\sum\limits_{j=1}^{m}g\left(X_{j}^{\prime },Y_{j}^{\prime}\right)\left(Y_{j}^{\prime}-y_{c}\right)=-4\sum\limits_{j=1}^{ m}Y_{j}^{\prime}g\left(X_{j}^{\prime},Y_{j}^{\prime}\right)=0\end{array}\right. \tag{25}\]  
Let \(u_{j}=X_{j}^{\prime}-\overline{X}^{\prime},u_{c}=x_{c}-\overline{X}^{\prime}, v_{j}=Y_{j}^{\prime}-\overline{Y}^{\prime}\) and \(v_{c}=y_{c}-\overline{Y}^{\prime}\), where \(\overline{X}^{\prime}=\frac{1}{m}\sum\limits_{j=1}^{m}X_{j}^{\prime}, \overline{Y}^{\prime}=\frac{1}{m}\sum\limits_{j=1}^{m}Y_{j}^{\prime}\), then we can get  
\[\left\{\begin{array}{c}\sum\limits_{j=1}^{m}u_{j}g\left(u_{j},v_{j}\right)=0 \\ \sum\limits_{j=1}^{m}v_{j}g\left(u_{j},v_{j}\right)=0\end{array}\right. \tag{26}\]  
Based on Equation (26), we can get  
\[\left\{\begin{array}{c}\sum\limits_{j=1}^{m}\left(u_{j}^{3}-2u_{j}^{2}u_{c} +u_{j}v_{j}^{2}-2u_{j}v_{j}v_{c}\right)=0\\ \sum\limits_{j=1}^{m}\left(u_{j}^{2}v_{j}-2u_{j}v_{j}u_{c}+v_{j}^{3}-2v_{j}^{2} v_{c}\right)=0\end{array}\right. \tag{27}\]According to Equation (27), we can get  
\[\left\{\begin{array}{l}u_{c}=\frac{\sum\limits_{j=1}^{m}u_{j}^{2}v_{j}\sum \limits_{j=1}^{m}u_{j}v_{j}\sqrt{\sum\limits_{j=1}^{m}u_{j}^{3}\sum\limits_{j=1 }^{m}v_{j}^{2}\sqrt{\sum\limits_{j=1}^{m}u_{j}v_{j}^{2}\sum\limits_{j=1}^{m}v_{ j}^{2}\sum\limits_{j=1}^{m}u_{j}v_{j}\frac{\sum\limits_{j=1}^{m}v_{j}^{3}}{\sum \limits_{j=1}^{m}v_{j}^{3}}}}}{2\left(\left(\sum\limits_{j=1}^{m}u_{j}v_{j} \right)^{2}-\sum\limits_{j=1}^{m}u_{j}^{2}\sum\limits_{j=1}^{m}v_{j}^{2}\right) \\ v_{c}=\frac{-\sum\limits_{j=1}^{m}u_{j}^{2}\sum\limits_{j=1}^{m}u_{j}^{2}v_{j} \sqrt{\sum\limits_{j=1}^{m}u_{j}^{3}\sum\limits_{j=1}^{m}u_{j}v_{j}\sum\limits _{j=1}^{m}u_{j}v_{j}^{2}\sum\limits_{j=1}^{m}u_{j}^{2}\sum\limits_{j=1}^{m}v_{ j}^{3}}}{2\left(\left(\sum\limits_{j=1}^{m}u_{j}v_{j}\right)^{2}-\sum\limits_{j=1}^{m}u_{j}^{ 2}\sum\limits_{j=1}^{m}v_{j}^{2}\right)}\end{array}\right. \tag{28}\]  
Then  
\[x_{c}=u_{c}-\overline{X}^{\prime} \tag{29}\]  
\[y_{c}=v_{c}-\overline{Y}^{\prime} \tag{30}\]  
According to Equation (19), we can get  
\[z_{c}=-\frac{ax_{c}+by_{c}+d}{c} \tag{31}\]  
In summary, the coordinates of the center of the circle fitted by the laser tracer O(\(x_{c}\),\(y_{c}\),\(z_{c}\)) are obtained.