## 2 Theory
### Single Direction Algorithm
\begin{array}{cccccc}G_{u}(q_{1s},q_{1d})&\cdots&G_{u}(q_{1s},q_{ md})&G_{u}(q_{1s},q_{1b})&\cdots&G_{u}(q_{1s},q_{nb})\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ G_{u}(q_{ms},q_{1d})&\cdots&G_{u}(q_{ms},q_{md})&G_{u}(q_{ms},q_{1b})&\cdots&G_{u }(q_{ms},q_{nb})\\ G_{u}\big{(}q_{(m+1)s},q_{1d}\big{)}&\cdots&G_{u}\big{(}q_{(m+1)s},q_{md}\big{)} &G_{u}\big{(}q_{(m+1)s},q_{1b}\big{)}&\cdots&G_{u}\big{(}q_{(m+1)s},q_{nb}\big{)} \\ \vdots&\cdots&\vdots&\vdots&\cdots&\vdots\\ G_{u}\big{(}q_{(m+n)s},q_{1d}\big{)}&\cdots&G_{u}\Big{(}q_{(m+n)s},q_{md} \Big{)}&G_{u}\Big{(}q_{(m+n)s},q_{1b}\big{)}&\cdots&G_{u}\big{(}q_{(m+n)s},q_{ nb}\big{)}\end{array} \right]_{(m+n)\times(m+n)}.\]  
Equation (5) can be obtained according to Equation (4). \[\left[ \begin{array}{cccccc}\pi\cdot w^{2}\cdot m_{1n}e_{1}\cdot(\sin\alpha_ {1}-i\cdot\cos\alpha_{1})+L\cdot w^{2}\cdot m_{1d}U_{1d}\\ \vdots\\ \pi\cdot w^{2}\cdot m_{mu}e_{m}\cdot(\sin\alpha_{m}-i\cdot\cos\alpha_{m})+L \cdot w^{2}\cdot m_{md}U_{md}\\ -L\cdot k_{1s\cdot yx}\cdot V_{1b}-L\cdot k_{1s\cdot yy}\cdot U_{1b}\\ \vdots\\ -L\cdot k_{ns\cdot yx}\cdot V_{nb}-L\cdot k_{ns\cdot yy}\cdot U_{nb}\end{array}  \right]_{m+n}=\frac{EI}{L^{2}}\cdot{H_{1}}^{-1}\left[ \begin{array}{c}U(q_{1s} )\\ \vdots\\ U(q_{ms})\\ \vdots\\ U\Big{(}q_{(m+n)s}\Big{)}\end{array} \right] \tag{5}\]  
The right side of Equation (5) is known. Define \(H_{2}\) as following. \[H_{2}=\frac{EI}{L^{2}}\cdot{H_{1}}^{-1}\left[ \begin{array}{c}U(q_{1s})\\ \vdots\\ U(q_{ms})\\ U\big{(}q_{(m+1)s}\big{)}\\ \vdots\\ U\big{(}q_{(m+n)s}\big{)}\end{array} \right]\]  
Equation (6) is obtained according to Equation (5). \[\left\{ \begin{array}{cccccc}\pi\cdot w^{2}\cdot m_{1n}e_{1}\cdot(\sin\alpha _{1}-i\cdot\cos\alpha_{1})+L\cdot w^{2}\cdot m_{1d}U_{1d}=H_{2}(1,1)\\ \vdots\\ \pi\cdot w^{2}\cdot m_{mu}e_{m}\cdot(\sin\alpha_{m}-i\cdot\cos\alpha_{m})+L \cdot w^{2}\cdot m_{md}U_{md}=H_{2}(m,1)\end{array} \right. \tag{6}\]  
According to Equations (6) and (7) is obtained. \[\left\{ \begin{array}{cccccc}m_{1n}e_{1}\cdot(\sin\alpha_{1}-i\cdot\cos \alpha_{1})=\frac{\left[H_{2}(1,1)-L\cdot w^{2}\cdot m_{1d}U_{1d}\right]}{\pi \cdot w^{2}}\\ \vdots\\ m_{mu}e_{m}\cdot(\sin\alpha_{m}-i\cdot\cos\alpha_{m})=\frac{\left[H_{2}(m,1)-L \cdot w^{2}\cdot m_{md}U_{md}\right]}{\pi\cdot w^{2}}\end{array} \right. \tag{7}\]  
According to Equations (7)-(9) are obtained. \[\left\{ \begin{array}{cccccc}m_{1n}e_{1}=\left|\frac{\left[H_{2}(1,1)-L\cdot w ^{2}\cdot m_{1d}U_{1d}\right]}{\pi\cdot w^{2}}\right|\\ \vdots\\ m_{mu}e_{m}=\left|\frac{\left[H_{2}(m,1)-L\cdot w^{2}\cdot m_{md}U_{md}\right] }{\pi\cdot w^{2}}\right|\end{array} \right. \tag{8}\]\[\left\{ \begin{array}{c}\sin\alpha_{1}-i\cdot\cos\alpha_{1}=\frac{\left[H_{2}(1,1 )-L\cdot w^{2}\cdot m_{1d}U_{lat}\right]}{\left[H_{2}(1,1)-L\cdot w^{2}\cdot m_{1 d}U_{lat}\right]}\\ \vdots\\ \sin\alpha_{m}-i\cdot\cos\alpha_{m}=\frac{\left[H_{2}(m,1)-L\cdot w^{2}\cdot m _{md}U_{lat}\right]}{\left[H_{2}(m,1)-L\cdot w^{2}\cdot m_{md}U_{lat}\right]} \end{array} \right. \tag{9}\]  
Equation (8) is the amplitude of rotor unbalance for each eccentric disc. Moreover, according to Equations (9) and (10) can be obtained, by means of which the angle of rotor unbalance for each eccentric disc can be calculated. \[\left\{ \begin{array}{c}\alpha_{1}=\text{angle}\bigg{(}-\text{imag}\bigg{(} \frac{\left[H_{2}(1,1)-L\cdot w^{2}\cdot m_{1d}U_{lat}\right]}{\pi\cdot w^{2}} \bigg{)}+i\cdot\text{real}\bigg{(}\frac{\left[H_{2}(1,1)-L\cdot w^{2}\cdot m _{1d}U_{lat}\right]}{\pi\cdot w^{2}}\bigg{)}\bigg{)}\\ \vdots\\ \alpha_{m}=\text{angle}\bigg{(}-\text{imag}\bigg{(}\frac{\left[H_{2}(m,1)-L \cdot w^{2}\cdot m_{md}U_{ind}\right]}{\pi\cdot w^{2}}\bigg{)}+i\cdot\text{real }\bigg{(}\frac{\left[H_{2}(m,1)-L\cdot w^{2}\cdot m_{md}U_{ind}\right]}{\pi \cdot w^{2}}\bigg{)}\bigg{)}\end{array} \right. \tag{10}\]  
where angle ( ) is the function used to obtain the angle of a complex number, real ( ) is the real part of a complex number, and imag ( ) is the imaginary part of a complex number.