# Decoupling Control of Multi-DOF Supporting System of MLDSB
## 3 Mathematical Model of the Multi-DOF Supporting System  
### Dynamic Equation of the Multi-DOF Supporting System  
In order to study the supporting characteristics of the multi-DOF system, the following assumptions are made [12; 13; 14; 15]:  
1. The flow condition of the lubrication is laminar and the inertia force is ignored. 2. The viscosity-pressure characteristics of the lubrication are ignored due to the low pressure. 3. The winding magnetic flux leakage is ignored and the flux is evenly distributed in the magnetic circuit. 4. The magnetoresistance in the iron core and rotor is ignored, and the magnetic potential only acts on the air gap. 5. The hysteresis and eddy of magnetic materials are ignored. 6. The supporting surface is assumed to be a rigid body. 7. The gravity of the rotor is ignored. 8. The rigid rotor is geometrically symmetrical uniform, \(J_{x}=J_{y}=J_{r}\). 9. The rotor is in the rotation center, the bearing force is zero, and there is no interference between the DOF supporting system. 10. The installation error of the sensor is not considered. 11. The stiffness of the base is relatively large, and there is no deformation or movement. The rotor's force model of the MLDSB is established as shown in Figure 7.  
The fixed coordinate system \(\text{C}\text{x}\text{y}\text{z}\) is established, and the origin is the mass center of the rotor. According to Figure 7, the expression of the mechanical relationship is assumed as follows. \[\left\{ \begin{array}{l}f_{x,1}=f_{Lx,A2}+f_{Mx,A1}\\ f_{x,2}=f_{Lx,A1}+f_{Mx,A2}\\ f_{x,3}=f_{Lx,B2}+f_{Mx,B1}\\ f_{x,A}=f_{Lx,B1}+f_{Mx,B2}\\ f_{y,1}=f_{Ly,A2}+f_{My,A1}\\ f_{y,2}=f_{Ly,A1}+f_{My,A2}\\ f_{y,3}=f_{Ly,B2}+f_{My,B1}\\ f_{y,4}=f_{Ly,B1}+f_{My,B2}\end{array} \right. \tag{4}\]  
where, the meanings of subscripts \(x\), \(y\), \(L\), \(M\), \(A\), and \(B\) are \(x\) direction, \(y\) direction, hydrostatic force, electromagnetic force, bearing \(A\), and bearing \(B\) respectively. The displacement on the sensors at the left and right ends is converted to the mass center \(C\), and the displacement \(x_{c}\) and \(y_{c}\) are approximately shown as follows. \[\left\{ \begin{array}{l}x_{c}=\frac{l_{b}x_{A}+l_{b}x_{B}}{l}\\ y_{c}=\frac{l_{b}y_{A}+l_{b}y_{B}}{l}\end{array} \right. \tag{5}\]  
where, \(l\) is length of the rotor, the unit is mm. \(l_{a}\), \(l_{b}\) is distance of center and left-unit, distance of center and right-unit respectively, the unit is mm. \(x_{c}\), \(y_{c}\) is displacement of the mass center, the unit is mm. \(x_{A}\), \(y_{A}\) is displacement of the right unit, the unit is mm. \(x_{B}\), \(y_{B}\) is displacement of the left unit, the unit is mm. \[\left\{ \begin{array}{l}\theta_{x}=\frac{x_{B}-x_{A}}{l}\\ \theta_{y}=\frac{y_{B}-y_{A}}{l}\end{array} \right. \tag{6}\]  
where, \(\theta_{x}\) is angle at plane \(yz\), the unit is \({}^{\circ}\). \(\theta_{y}\) is angle at plane \(xz\), the unit is \({}^{\circ}\). The torque of the rotor is shown as follows. \[\left\{ \begin{array}{l}M_{x}=J_{r}\ddot{\theta}_{y}+\omega J_{r}\dot{\theta} _{x}=\big{(}f_{y2}-f_{y1}\big{)}l_{a}+\big{(}f_{y3}-f_{y4}\big{)}l_{b}\\ M_{y}=J_{r}\ddot{\theta}_{x}+\omega J_{r}\ddot{\theta}_{y}=\big{(}f_{x2}-f_{x1 }\big{)}l_{a}+\big{(}f_{x3}-f_{x4}\big{)}l_{b}\end{array} \right. \tag{7}\]  
where, \(J_{r}\) is inertia moment of the rotor around the \(x\) and \(y\)-axis, the unit is kg\(\cdot\)m\({}^{2}\). \(J_{a}\) is inertia moment of the rotor around the \(z\)-axis, the unit is kg\(\cdot\)m\({}^{2}\). Figure 7: Force diagram of the rotor. According to Equations (1)-(7), the force equation of the rotor can be obtained as follows. \[\left\{