# The Coupling Characteristic Investigation of Double-Gimbal Magnetically Suspended Control Moment Gyro Used on Agile Maneuver Spacecraft

Authors: Peiling Cui; Jian Cui; Qian Yang; Shiqiang Zheng
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

Double-gimbal magnetically suspended CMG is a novel attitude control actuator for the agile maneuver spacecraft. Taking the double-gimbal magnetically suspended control moment gyro used on agile maneuver spacecraft as the research object, the dynamic model of the magnetically suspended rotor, the inner gimbal, and the outer gimbal of double-gimbal magnetically suspended control moment gyro is built. The nonlinear coupling characteristic between the rotor, the gimbal, and the spacecraft is given. It can be seen that the motion of magnetically suspended rotor does not only rely on magnetic bearing force but also suffer from the influence of gimbal servo system and spacecraft motion. The coupling torque includes the gyro coupling torque and the inertial coupling torque. The work in this paper provides the foundation for further studies.


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## Body

## 1. Introduction

Double-gimbal magnetically suspended control moment gyro (CMG) is a novel solution to realize high precision, long life, and attitude maneuver control of spacecraft by incorporating the advantages of magnetic bearing with double gimbals [1, 2]. Compared with single-gimbal magnetically suspended CMG, double-gimbal magnetically suspended CMG has more than one degree of freedom, and the singularity is smaller. Therefore, it is believed that double-gimbal magnetically suspended CMG can implement the rapid attitude maneuver with high efficiency [3, 4]. By the rotation of inner gimbal and outer gimbal, the angular momentum direction of magnetically suspended high-speed rotor is changed, and then the gyro torque is produced. For magnetic bearing, there are five more degrees of freedom than mechanical stator, including three translation motions and two radial rotation motions [5, 6].Double-gimbal magnetically suspended CMG is composed of high-speed magnetically suspended rotor system, inner gimbal servo system, and outer gimbal servo system [7]. In the agile maneuver spacecraft using double-gimbal magnetically suspended CMG as actuator, there is strong coupling torque on the magnetic bearing resulting from the motion of spacecraft and gimbal, and then the high-speed rotor displacement is increased. When this phenomenon is serious, the rotor will collide with the protecting bearing, and the stability of magnetically suspended rotor will be reduced and even lost. Moreover, the spacecraft maneuver and the rotor radial motion will bring disturbance to the gimbal motion. The output torque precision of double-gimbal magnetically suspended control moment gyro will be influenced. Finally, the spacecraft attitude control precision will be reduced when the gimbal motion and rotor radial motion are large. It is important to build the dynamic model of magnetically suspended rotor, inner gimbal, outer gimbal, and spacecraft when using double-gimbal magnetically suspended control moment gyro as the spacecraft attitude control actuator.For spacecraft dynamic modeling based on mechanical CMG, there is some research result. In [8], the mechanical CMG model is given, and the vector-form output torque of CMG in spacecraft body frame is provided. In [9], the dynamic model of one mechanical double-gimbal CMG is built using Lagrange equation, and the adaptive feedback control law is put forward. In [10], the inertial torque of inner gimbal and outer gimbal is considered, and the spacecraft attitude dynamic model with a cluster of double-gimbal CMGs is built. For spacecraft dynamic modeling based on magnetically suspended CMG, there is little research result. In [11], the spacecraft dynamic model using single-gimbal magnetically suspended CMG is built. It is aimed at large spacecraft, the spacecraft angular velocity is limited to 10−2 rad/s, and the effect of spacecraft maneuver is not involved. As for double-gimbal magnetically suspended CMG, investigation mainly focuses on single part of the magnetically suspended CMG, such as rotor system [12, 13] and gimbal servo system [14]. Models are built separately, omitting the coupling characteristic between different parts. Besides, researchers do not consider maneuver of the spacecraft.In this paper, taking double-gimbal magnetically suspended CMG used for attitude control of agile maneuver spacecraft as the research object, the dynamic model of magnetically suspended rotor, inner gimbal, and outer gimbal of double-gimbal magnetically suspended CMG is built. The nonlinear coupling characteristic between rotor, gimbal, and spacecraft is investigated.


## 2. The Working Principal of Double-Gimbal Magnetically Suspended CMG

Double-gimbal magnetically suspended CMG consists of three subsystems, including magnetically suspended high-speed rotor system, inner gimbal rate servo system, and outer gimbal rate servo system. The angular momentum is provided by magnetically suspended high-speed rotor. The torque is output by the direction changing of angular momentum, which will adjust the spacecraft attitude.Double-gimbal magnetically suspended CMG prototype is shown in Figure1. The gyro room and its subsidiary control system is called the magnetically suspended high-speed rotor system, including the magnetic bearing system and the motor for the high-speed rotor. The gyro room is equipped with the components, including the motor of rotor, axial magnetic bearing, radial magnetic bearing, the displacement sensor, and the protecting bearing. The rotor is driven by the motor, which will provide constant angle momentum. The rotor is suspended by the magnetic bearing, and it has five degrees of freedom. This kind of magnetic bearing is called five-degree-of-freedom magnetic bearing. The displacement of each degree of freedom is measured by the displacement sensor. If the rotor is not on the given zero position, the error signal is passed through the magnetic bearing controller. The amplifier will output the corresponding control current, which will drive the magnetic bearing to produce bearing force to return the rotor to the given position. The base is the mechanical structure that connects double-gimbal magnetically suspended CMG with spacecraft. The output torque of double-gimbal magnetically suspended CMG is imparted to spacecraft by the base.Figure 1
Prototype of double-gimbal magnetically suspended CMG.The magnetic bearing is an elasticity support with gap. When it is suspended normally, the magnetically suspended rotor has small rotation angle and translation motion relative to gyro room. Meanwhile, the angle motion of magnetically suspended high-speed rotor is interrelated with the rotation of inner gimbal and outer gimbal. The reference frames should be defined firstly.At the initial time, it is defined that the gimbal angle, the gimbal rate, the base angle position, and the base angular velocity are all zero. For spacecraft dynamic modeling, the coordinate frames are defined, which are illustrated in Figure2.Figure 2
The illustration of coordinate frames.O
i
X
i
Y
i
Z
i is the inertial frame. The origin is earth center of mass. OiXi points to vernal equinox; OiZi points to the north pole.O
o
X
o
Y
o
Z
o is the orbit frame. The origin is the spacecraft center of mass. OoXo and OoYo represent the roll axis and the pitch axis, respectively. Both of them are located in the orbit plane. The roll axis points to the go-ahead direction. The pitch axis is vertical to orbit plane. Ozo represents the yaw axis, which points to earth’s core. This frame rotates with speed ωo, and the rotation direction is contrary to the pitch axis.O
b
x
b
y
b
z
b is the spacecraft body frame. The origin is the spacecraft center of mass, Oxb, Oyb, and Ozb are fixed on the spacecraft, and they are the moments of inertial principal axis of the spacecraft. φθψT is the 3-1-2 Euler angle with respect to orbit frame.O
s
x
s
y
s
z
s is the zero position frame. At the zero position, it coincides with the outer gimbal frame. It is fixed with body frame.O
j
x
j
y
j
z
j is the outer gimbal frame. It is fixed with outer gimbal. The outer gimbal angle is θj. It is positive when the rotor rotation speed ωj along yj direction is positive.O
g
x
g
y
g
z
g is the inner gimbal frame. It is fixed with inner gimbal. The inner gimbal angle is θg. It is positive when the rotor rotation speed ωg along xg direction is positive.O
1
x
1
y
1
z
1 is the stator frame. It is fixed with gyro room. This frame keeps still with respect to inner gimbal frame. Ox1 and Oy1 point to central directions of the radial magnetic bearing.O
2
x
2
y
2
z
2 is the inner loop frame. It is fixed with the suppositional inner loop. This frame has three translation motions and two rotation degrees of freedom. O2 is the center of rotor mass. xg1yg1zg1 and xg2yg2zg2 in Figure 3 are the suppositional inner loop and the dummy outer loop frames, respectively. The outer gimbal angle α is the Euler angle of xg1yg1zg1 with respect to Ox1y1z1. The inter gimbal angle β is the Euler angle of xg2yg2zg2 with respect to xg1yg1zg1.Figure 3
The coordinate transformation matrix from the stator frame to the rotor frame.O
r
x
r
y
r
z
r is the rotor frame. This frame is fixed with rotor and has one rotation degree of freedom along zr axis with respect to inner loop frame. The rotor rotation angle is Ωt. Ω is the constant angular velocity, and it is expressed in rotor frame coordinate as [0,0,Ω]T.C
a
b is used to denote the coordinate transformation matrix from frame a to frame b. y/m denotes the vector projected in coordinate m. x× denotes the skew symmetric matrix of x=x1x2x3T. On the condition of small angle transformation, then
(1)Cob=cos⁡θ0-sinθcos⁡ϕ01sinϕsinθ0cos⁡θcos⁡ϕ≈1ψ-θ-ψ1φθ-φ1,Cbs=100001010,Csj=cos⁡θj0-sinθj010sinθj0cos⁡θj,Cjg=1000cos⁡θgsinθg0-sinθgcos⁡θg,Cg1=100010001,C1g1=1000cos⁡αsinα0-sinαcos⁡α,Cg1g2=cos⁡β0-sinβ010sinβ0cos⁡β,C12=C1g1Cg1g2≈10-β01αβ-α1,C2r=cos⁡ΩtsinΩt0-sinΩtcos⁡Ωt0001.


## 3. The Dynamic Model of Double-Gimbal Magnetically Suspended CMG

### 3.1. The Dynamic Model of Magnetically Suspended Rotor

In order to obtain large rotor angular momentum, the rotor is designed as disc shape to improve the moment of inertia. The moment of inertia ratio between the pole and the equator is large. In order to perform the dynamic analysis and control system design, the magnetic bearing dynamic model of magnetically suspended CMG must be built. The main nomenclature used in this paper is listed in the Nomenclature section.The magnetically suspended rotor has six degrees of freedom, including translation motion alongx, y, and z direction and rotation motion along x, y, and z direction. The rotation along z direction is controlled by the motor. The other five degrees of freedom are controlled by magnetic bearing, including the translation motion along x, y, and z direction and the radial rotation along x, y direction. The exclusive force between the magnetically suspended rotor and the gyro room is the magnetic bearing force. In Figure 4, the radial bearing forces fax and fbx can be synthesized into radial force fx. The radial bearing forces fay and fby can be synthesized into radial force fy. The radial torque in x and y direction is py and px, which will influence the radial translation and the rotation of magnetically suspended rotor. When the magnetic bearing rotates about the centroid, the translation and rotation motion about the rotor centroid can be considered separately. Firstly, the translation and rotation motion equations of the rotor must be built. By synthesizing the equations, the dynamic equation of the magnetically suspended rotor can be obtained.Figure 4
The sketch map of magnetic bearing dynamic model.The relative motion of magnetically suspended rotor in noninertial frame suffers from two kinds of force. One is the active control force of magnetic bearing and the other is the implicated inertial forcema⃑e and Coriolis inertial force ma⃑c that result from the implicated motion of inner gimbal and outer gimbal. The rotor dynamic equation in noninertial frame is ma⃑r=F+-ma⃑e+-ma⃑c. m is the mass of magnetically suspended rotor. a⃑r is the relative translation acceleration of magnetically suspended rotor. F is the active force, namely, the bearing force. a⃑e and a⃑c are the implicated acceleration and Coriolis acceleration. The relative translation acceleration is greater than the implicated acceleration and Coriolis acceleration, so the implicated inertial force and Coriolis inertial force can be ignored. The relative translation motion model of the magnetically suspended rotor is simplified as ma⃑r=F, where a⃑r=x¨ry¨rz¨rT, F=fxfyfzT.Similar with the translation motion of magnetically suspended rotor, the rotation of magnetically suspended rotor in inertial frame suffers from two kinds of torque. One is the active control torque of magnetic bearing, and the other is implicated inertial torque and Coriolis inertial torque that are produced from the implicated motion of the noninertial frame. Considering the ground experimental condition, the absolute angle motion of magnetically suspended rotor includes the following three kinds of motion.(1)
The rotation of rotor frame relative to magnetic bearing stator is described with Euler anglesα,  β, and γ, which rotates about the inner loop rotation axis, the outer loop rotation axis, and the rotor spin axis.(2)
The rotation of the magnetic bearing stator frame relative to inner gimbal frame is denoted byθg.(3)
The rotation of outer gimbal frame relative to inertial frame is denoted byθj.In the following, based on Newton-Euler method, Euler dynamic equation is used to derive the dynamic model of magnetically suspended rotor of double-gimbal magnetically suspended CMG. The reference frame is the magnetic bearing stator frameOx1y1z1, and the movable frame is the inner loop frame Ox2y2z2.The angle speed vector of magnetically suspended rotorω⃑ir includes six parts, namely, the rotation speed of rotor frame relative to stator frame ω⃑1r, the rotation speed of stator frame relative to inner gimbal frame ω⃑g1, the rotation speed of inner gimbal frame relative to outer gimbal frame ω⃑jg, the rotation speed of outer gimbal frame relative to zero position frame ω⃑sj, the rotation speed of zero position frame relative to body frame ω⃑bs, and the rotation speed of body frame relative to inertial frame ω⃑ib: ω⃑ir=ω⃑ib+ω⃑bs+ω⃑sj+ω⃑jg+ω⃑g1+ω⃑1r. Because the projection gap is small, the maximal radial rotation angle of the Euler angles  α,  β is less than 3 × 10−3 rad. It can be considered that cos⁡α≈1, cos⁡β≈1, sinα≈0, sinβ≈0.(1) Derive the angular velocity projection of the movable frame about each axis. The angular velocity of rotor frame relative to stator frame isω⃑1r=α⃑˙+β⃑˙+γ⃑˙. Its projection in inner loop frame is
(2)ω⃑1r2=α˙00+C1g10β˙0+00γ˙=α˙β˙cos⁡αγ˙-β˙sinα.Because stator frame is still relative to inner gimbal frame,ω⃑g1=0. The angular velocity of inner gimbal relative to outer gimbal is ω⃑jg. Its projection in inner loop frame is
(3)ω⃑jg2=C12Cg1ω⃑jg≈θ˙g00.The angular velocity of outer gimbal relative to zero position frame isω⃑sj. Its projection in inner loop frame is
(4)ω⃑sj2=C12Cg1Cjgω⃑jg=0θ˙jcos⁡θg-θ˙jsinθg.The zero position frame is still relative to body frame,ω⃑bs=0. The rotation speed of spacecraft relative to orbit frame is (φ˙,θ˙,ψ˙), which corresponds to roll, pitch, and yaw. The rotation speed of orbit frame is (0,-ω0,0)T. So, the rotation speed of spacecraft can be described in frame Oxbybzb as follows:
(5)ω⃑ibb=ωibxωibyωibz=φ˙θ˙ψ˙+1ψ-θ-ψ1φθφ10-ω00=φ˙-ω0ψθ˙-ω0ψ˙+ω0φ.The spacecraft rotation speedω⃑ib in inner loop frame is
(6)ω⃑ib2=ωibx2ωiby2ωibz2=C12Cg1CjgCsjCbsω⃑ibb=cos⁡θjωibxb-sinθjωibybsinθgsinθjωibxb+cos⁡θgωibzb+sinθgcos⁡θjωibybcos⁡θgsinθjωibxb-sinθgωibzb+cos⁡θgcos⁡θjωibyb.The absolute angular velocityω⃑ir of magnetically suspended rotor in inner loop frame is denoted by
(7)ω⃑ir2=Cb2ω⃑ib+Cj2ω⃑sj+Cg2ω⃑jg+ω⃑1r=ω⃑ib2+ω⃑sj2+ω⃑jg2+ω⃑1r2.Then(8)ω⃑ir2=ωibx2+α˙+θ˙gωiby2+β˙cos⁡α+θ˙jcos⁡θgωibz2+γ˙-β˙sinα-θ˙jsinθg=cos⁡θjωibxb-sinθjωibyb+α˙+θ˙gsinθgsinθjωibxb+cos⁡θgωibzb+sinθgcos⁡θjωibyb+β˙cos⁡α+θ˙jcos⁡θgcos⁡θgsinθjωibxb-sinθgωibzb+cos⁡θgcos⁡θjωibyb+γ˙-β˙sinα-θ˙jsinθg.(2) The moment of inertia of magnetic bearing inoxryrzr is Jr=diagJrxJryJrz. Because the rotor is symmetric, its equator moment of inertia is Jrx=Jry=Jrr, and its pole moment of inertia is Jrz=Jp. So the rotor moment of inertia relative to inner loop frame has the following relationship with Jr:
(9)J2=C2rTJrC2r≈Jrr000Jrr000Jrz.The rotor angular momentum in inner loop frame is(10)Ηr=Jr·ω⃑ir2=Jr·(Cb2ω⃑ib+Cj2ω⃑sj+Cg2ω⃑jg+ω⃑1r)=Jrrωibx2+α˙+θ˙gJrrωiby2+β˙cos⁡α+θ˙jcos⁡θgJrzωibz2+γ˙-β˙sinα-θ˙jsinθg.Then(11)Η˙r=Jr·ω⃑˙ir2=Jr(-ω⃑1r×Cb2-Cg2ω⃑jg×Cbg-Cj2ω⃑sj×Cbj)ω⃑ib+Cb2ω⃑˙ib+(-ω⃑1r×Cj2-Cg2ω⃑sj×Cjg)ω⃑sj+Cj2ω⃑˙sj+-ω⃑1r×Cg2ω⃑jg+Cg2ω⃑˙jg+ω⃑˙1r=Jrrω˙ibx2+α¨+θ¨gJrrω˙iby2+β¨cos⁡α-β˙α˙sinα+θ¨jcos⁡θg-θ˙jθ˙gsinθgJrzω˙ibz2+γ¨-β¨sinα-β˙α˙cos⁡α-θ¨jsinθg-θ˙jθ˙gcos⁡θg.The projection of the angular velocity in inner loop frame can be simplified as(12)ω⃑i22=ωibx2+α˙+θ˙gωiby2+β˙cos⁡α+θ˙jcos⁡θgωibz2-β˙sinα-θ˙jsinθg.(3) In this paper, five-degree-of-freedom magnetic bearing is used. The axial magnetic bearing only provides translation force, and it cannot output torque to the rotor. Therefore, the total torque of the rotor is(13)Mr12=pxpy0.(4) In magnetic bearing frameOxryrzr, the dynamic equation of the ith double-gimbal magnetically suspended CMG is
(14)H˙rx+Hrzωi2y2-Hryωi2z2=Mr1x2,H˙ry+Hrxωi2z2-Hrzωi2x2=Mr1y2,H˙rz+Hryωi2x2-Hrxωi2y2=Mr1z2,
where Mr1x2, Mr1y2, and Mr1z2 are the elements of Mr12. According to (2)~(14), the rotor dynamic equation is
(15)Jrr[ω˙ibx2+α¨+θ¨g]+Jrzω˙ibz2+γ¨-β¨sinα-β˙α˙cos⁡α-θ¨jsinθg-θ˙jθ˙gcos⁡θg·ωiby2+β˙cos⁡α+θ˙jcos⁡θg-Jrrωiby2+β˙cos⁡α+θ˙jcos⁡θg·[ω˙ibz2-β˙sinα-θ˙jsinθg]=px,Jrr[ω˙iby2+β¨cos⁡α-β˙α˙sinα+θ¨jcos⁡θg-θ˙jθ˙gsinθg]+Jrr[ωibx2+α˙+θ˙g][ω˙ibz2-β˙sinα-θ˙jsinθg]-Jrz[ω˙ibz2+γ˙-β˙sinα-θ˙jsinθg][ωibx2+α˙+θ˙g]=py,Jrz[ω˙ibz2+γ¨-β¨sinα-β˙α˙cos⁡α-θ¨jsinθg-θ˙jθ˙gcos⁡θg]+Jrr[ωibx2+α˙+θ˙g][ωiby2+β˙cos⁡α+θ˙jcos⁡θg]-Jrr[ωibx2+α˙+θ˙g][ωiby2+β˙cos⁡α+θ˙jcos⁡θg]=0.Because the projection gap is small and the permitted maximal radial rotation angle is less than 3′′, the rotor motion is restricted in the projection gap.α and β are small, γ˙≫α˙, γ˙≫β˙, Jrr<Jrz, so the above equation can be simplified as
(16)Jrrω˙ibx2+α¨+θ¨g+Hrzωiby2+β˙+θ˙jcos⁡θg=px,Jrrω˙iby2+β¨+θ¨jcos⁡θg-θ˙jθ˙gsinθg-Hrzωibx2+α˙+θ˙g=py,Jrzω˙ibz2+γ˙=0.When inner gimbal and outer gimbal are still on the ground, the dynamic equation of magnetically suspended rotor can be simplified as(17)Jrrα¨+Hrzβ˙=px,Jrrβ¨-Hrzα˙=py.


### 3.2. The Dynamic Model of Inner Gimbal

The absolute angular velocityω⃑ig of inner gimbal is ω⃑ig=ω⃑ib+ω⃑bs+ω⃑sj+ω⃑jg.(1) Taking inner gimbal frameOxgygzg as moveable frame, the angular velocity of outer gimbal relative to zero position is ω⃑sj. Its projection in inner gimbal frame is
(18)ω⃑sjg=ω⃑jg+Cjgω⃑sj=θ˙gθ˙jcos⁡θg-θ˙jsinθg.The zero position frame is still relative to body frame, soω⃑bs=0. The spacecraft rotation speed ω⃑ib in inner gimbal frame is denoted by
(19)ω⃑ibg=ωibxgωibygωibzg=CjgCsjCbsω⃑ibb=cos⁡θjωibyb-sinθjωibxbsinθgsinθjωibyb+cos⁡θgωibzb+sinθgcos⁡θjωibxbcos⁡θgsinθjωibyb-sinθgωibzb+cos⁡θgcos⁡θjωibxb.So the absolute velocityω⃑ig and the time derivative ω⃑˙ig of inner gimbal can be described in inner gimbal frame as
(20)ω⃑igg=Cbgω⃑ib+Cjgω⃑sj+ω⃑jg=ω⃑ibg+ω⃑sjg+ω⃑jgg=ωibxg+θ˙gωibyg+θ˙jcos⁡θgωibzg-θ˙jsinθg=cos⁡θjωibyb-sinθjωibxb+θ˙gsinθgsinθjωibyb+cos⁡θgωibzb+sinθgcos⁡θjωibxb+θ˙jcos⁡θgcos⁡θgsinθjωibyb-sinθgωibzb+cos⁡θgcos⁡θjωibxb-θ˙jsinθg,(21)ω⃑˙igg=ω˙ibxg+θ¨gω˙ibyg+θ¨jcos⁡θg-θ˙jθ˙gsinθgω˙ibzg-θ¨jsinθg-θ˙jθ˙gcos⁡θg.(2) The torque that acts on inner gimbal includes two parts, namely, the magnetic bearing torque and the outer gimbal torque. Then(22)Mgr1=-px-py0.The projection of this torque in inner gimbal frame is(23)Mgrg=C1gMgr1=-px-py0.The torque acting on the inner gimbal by the outer gimbal is(24)Mgjg=pgxpgypgz,
where pgy and pgz are the constraint torque along yg and zg direction. pgx is along xg direction. So the total torque of inner gimbal resulting from the magnetic bearing and outer gimbal is
(25)Mgg=Mgrg+Mgjg=pgx-pxpgy-pypgz.(3)n double-gimbal magnetically suspended CMGs are installed on spacecraft according to certain configurations, which can avoid singularity state. In inner gimbal frame Oxgygzg, the dynamic model of inner gimbal of the ith (i=1,…,n) double-gimbal magnetically suspended CMG is
(26)ω⃑igg×Hgi+H˙gi=Mgg,
where the angular momentum of inner gimbal of the ith double-gimbal magnetically suspended CMG is Hgi=Jg(Cbgω⃑ib+Cjgω⃑sj+ω⃑jg). Then
(27)H˙gi=Jg(-ω⃑jg×CjgCbj-Cjgω⃑sj×Cbj)ω⃑ib+Cbgω⃑˙ib-ω⃑jg×Cjgω⃑sj+Cjgω⃑˙sj+ω⃑˙jg.The moment of inertia of inner gimbal isJgxJgyJgzT; then
(28)Jgxω˙igxg+Jgz-Jgyωigygωigzg=Mgxg,Jgyω˙igyg+Jgx-Jgzωigzgωigxg=Mgyg,Jgzω˙igzg+Jgy-Jgxωigxgωigyg=Mgzg.Substituting (20), (21), and (25) into the above equation, the dynamic model of inner gimbal is
(29)Jgx(ω˙ibxg+θ¨g)+(Jgz-Jgy)(ωibyg+θ˙jcos⁡θg)·(ωibzg-θ˙jsinθg)=pgx-px,Jgy(ω˙ibyg+θ¨jcos⁡θg-θ˙jθ˙gsinθg)+(Jgx-Jgz)(ωibzg-θ˙jsinθg)(ωibxg+θ˙g)=pgy-py,Jgz(ω˙ibzg-θ¨jsinθg-θ˙jθ˙gcos⁡θg)+(Jgy-Jgx)(ωibxg+θ˙g)(ωibyg+θ˙jcos⁡θg)=pgz.


### 3.3. The Dynamic Model of Outer Gimbal

Supposing thatω⃑sj is the angular velocity of outer gimbal relative to zero position frame, ω⃑bs is the angular velocity of zero position frame relative to spacecraft body frame, and ω⃑ib is the implicated angular velocity of spacecraft body frame relative to inertial frame. The absolute angular velocity ω⃑ij of outer gimbal is ω⃑ij=ω⃑ib+ω⃑bs+ω⃑sj.(1) Taking outer gimbal frameOxjyjzj as movable frame, zero position frame is kept stable relative to body frame: ω⃑bs=0.The spacecraft rotating velocity in inner gimbal frame is described as(30)ω⃑ibj=ωibxjωibyjωibzj=CsjCbsω⃑ibb=cos⁡θjωibyb-sinθjωibxbωibzbsinθjωibyb+cos⁡θjωibxb.So the absolute angular velocityω⃑ijj of outer gimbal and its time derivative ω⃑˙ijj can be described as
(31)ω⃑ijj=Cbjω⃑ib+ω⃑sj=ωibxjωibyj+θ˙jωibzj=cos⁡θjωibyb-sinθjωibxbωibzb+θ˙jsinθjωibyb+cos⁡θjωibxb,(32)ω⃑˙ijj=-sinθjθ˙jωibyb+cos⁡θjω˙ibyb-cos⁡θjθ˙jωibxb-sinθjω˙ibxbω˙ibzb+θ¨jcos⁡θjθ˙jωibyb+sinθjω˙ibyb-sinθjθ˙jωibxb+cos⁡θjω˙ibxb=ω˙ibxjω˙ibyj+θ¨jω˙ibzj.(2) Similar with inner gimbal, the torque acting on outer gimbal is the torque from spacecraft and the torque from inner gimbal. The projection of inner gimbal torque on outer gimbal is(33)Mjgj=CgjMjgg=-pgx-pgycos⁡θj-pgzsinθjpgysinθj-pgzcos⁡θj.The torque that the spacecraft is acting on outer gimbal is(34)Mjij=pjxpjypjz,
where pjy is the outer gimbal motor driving torque along yj direction. pjx and pjz are the constraint torque in which the base acts on outer gimbal along xj and zj direction. So the total torque acting on outer gimbal is
(35)Mjj=Mjgj+Mjij=-pgx+pjx-pgycos⁡θg+pgzsinθg+pjy-pgysinθg-pgzcos⁡θg+pjz.(3) In outer gimbal frameOxjyjzj, the outer gimbal dynamic model of the ith double-gimbal magnetically suspended CMG is
(36)ω⃑ijj×Hji+H˙ji=Mjj,
where the angular momentum of outer gimbal of the ith double-gimbal magnetically suspended CMG is Hji=Jj(Cbjω⃑ib+ω⃑sj). Then H˙ji=Jj(-ω⃑sj×Cbjω⃑ib+Cbjω⃑˙ib+ω⃑˙sj).The moment of inertia of outer gimbal isJjxJjyJjzT; then
(37)Jjxω˙ijxj+Jjz-Jjyωijyjωijzj=Mjxj,Jjyω˙ijyj+Jjx-Jjzωijzjωijxj=Mjyj,Jjzω˙ijzj+Jjy-Jjxωijxjωijyj=Mjzj.By substituting (31), (32), and (34) into the above equation, the dynamic equation of outer gimbal is
(38)pjx=Jjxω˙ibxj+pgx,pjy=Jjyω˙ibyj+θ¨j+pgycos⁡θg-pgzsinθg,pjz=Jjzω˙ibzj+pgysinθg+pgzcos⁡θg.


## 3.1. The Dynamic Model of Magnetically Suspended Rotor

In order to obtain large rotor angular momentum, the rotor is designed as disc shape to improve the moment of inertia. The moment of inertia ratio between the pole and the equator is large. In order to perform the dynamic analysis and control system design, the magnetic bearing dynamic model of magnetically suspended CMG must be built. The main nomenclature used in this paper is listed in the Nomenclature section.The magnetically suspended rotor has six degrees of freedom, including translation motion alongx, y, and z direction and rotation motion along x, y, and z direction. The rotation along z direction is controlled by the motor. The other five degrees of freedom are controlled by magnetic bearing, including the translation motion along x, y, and z direction and the radial rotation along x, y direction. The exclusive force between the magnetically suspended rotor and the gyro room is the magnetic bearing force. In Figure 4, the radial bearing forces fax and fbx can be synthesized into radial force fx. The radial bearing forces fay and fby can be synthesized into radial force fy. The radial torque in x and y direction is py and px, which will influence the radial translation and the rotation of magnetically suspended rotor. When the magnetic bearing rotates about the centroid, the translation and rotation motion about the rotor centroid can be considered separately. Firstly, the translation and rotation motion equations of the rotor must be built. By synthesizing the equations, the dynamic equation of the magnetically suspended rotor can be obtained.Figure 4
The sketch map of magnetic bearing dynamic model.The relative motion of magnetically suspended rotor in noninertial frame suffers from two kinds of force. One is the active control force of magnetic bearing and the other is the implicated inertial forcema⃑e and Coriolis inertial force ma⃑c that result from the implicated motion of inner gimbal and outer gimbal. The rotor dynamic equation in noninertial frame is ma⃑r=F+-ma⃑e+-ma⃑c. m is the mass of magnetically suspended rotor. a⃑r is the relative translation acceleration of magnetically suspended rotor. F is the active force, namely, the bearing force. a⃑e and a⃑c are the implicated acceleration and Coriolis acceleration. The relative translation acceleration is greater than the implicated acceleration and Coriolis acceleration, so the implicated inertial force and Coriolis inertial force can be ignored. The relative translation motion model of the magnetically suspended rotor is simplified as ma⃑r=F, where a⃑r=x¨ry¨rz¨rT, F=fxfyfzT.Similar with the translation motion of magnetically suspended rotor, the rotation of magnetically suspended rotor in inertial frame suffers from two kinds of torque. One is the active control torque of magnetic bearing, and the other is implicated inertial torque and Coriolis inertial torque that are produced from the implicated motion of the noninertial frame. Considering the ground experimental condition, the absolute angle motion of magnetically suspended rotor includes the following three kinds of motion.(1)
The rotation of rotor frame relative to magnetic bearing stator is described with Euler anglesα,  β, and γ, which rotates about the inner loop rotation axis, the outer loop rotation axis, and the rotor spin axis.(2)
The rotation of the magnetic bearing stator frame relative to inner gimbal frame is denoted byθg.(3)
The rotation of outer gimbal frame relative to inertial frame is denoted byθj.In the following, based on Newton-Euler method, Euler dynamic equation is used to derive the dynamic model of magnetically suspended rotor of double-gimbal magnetically suspended CMG. The reference frame is the magnetic bearing stator frameOx1y1z1, and the movable frame is the inner loop frame Ox2y2z2.The angle speed vector of magnetically suspended rotorω⃑ir includes six parts, namely, the rotation speed of rotor frame relative to stator frame ω⃑1r, the rotation speed of stator frame relative to inner gimbal frame ω⃑g1, the rotation speed of inner gimbal frame relative to outer gimbal frame ω⃑jg, the rotation speed of outer gimbal frame relative to zero position frame ω⃑sj, the rotation speed of zero position frame relative to body frame ω⃑bs, and the rotation speed of body frame relative to inertial frame ω⃑ib: ω⃑ir=ω⃑ib+ω⃑bs+ω⃑sj+ω⃑jg+ω⃑g1+ω⃑1r. Because the projection gap is small, the maximal radial rotation angle of the Euler angles  α,  β is less than 3 × 10−3 rad. It can be considered that cos⁡α≈1, cos⁡β≈1, sinα≈0, sinβ≈0.(1) Derive the angular velocity projection of the movable frame about each axis. The angular velocity of rotor frame relative to stator frame isω⃑1r=α⃑˙+β⃑˙+γ⃑˙. Its projection in inner loop frame is
(2)ω⃑1r2=α˙00+C1g10β˙0+00γ˙=α˙β˙cos⁡αγ˙-β˙sinα.Because stator frame is still relative to inner gimbal frame,ω⃑g1=0. The angular velocity of inner gimbal relative to outer gimbal is ω⃑jg. Its projection in inner loop frame is
(3)ω⃑jg2=C12Cg1ω⃑jg≈θ˙g00.The angular velocity of outer gimbal relative to zero position frame isω⃑sj. Its projection in inner loop frame is
(4)ω⃑sj2=C12Cg1Cjgω⃑jg=0θ˙jcos⁡θg-θ˙jsinθg.The zero position frame is still relative to body frame,ω⃑bs=0. The rotation speed of spacecraft relative to orbit frame is (φ˙,θ˙,ψ˙), which corresponds to roll, pitch, and yaw. The rotation speed of orbit frame is (0,-ω0,0)T. So, the rotation speed of spacecraft can be described in frame Oxbybzb as follows:
(5)ω⃑ibb=ωibxωibyωibz=φ˙θ˙ψ˙+1ψ-θ-ψ1φθφ10-ω00=φ˙-ω0ψθ˙-ω0ψ˙+ω0φ.The spacecraft rotation speedω⃑ib in inner loop frame is
(6)ω⃑ib2=ωibx2ωiby2ωibz2=C12Cg1CjgCsjCbsω⃑ibb=cos⁡θjωibxb-sinθjωibybsinθgsinθjωibxb+cos⁡θgωibzb+sinθgcos⁡θjωibybcos⁡θgsinθjωibxb-sinθgωibzb+cos⁡θgcos⁡θjωibyb.The absolute angular velocityω⃑ir of magnetically suspended rotor in inner loop frame is denoted by
(7)ω⃑ir2=Cb2ω⃑ib+Cj2ω⃑sj+Cg2ω⃑jg+ω⃑1r=ω⃑ib2+ω⃑sj2+ω⃑jg2+ω⃑1r2.Then(8)ω⃑ir2=ωibx2+α˙+θ˙gωiby2+β˙cos⁡α+θ˙jcos⁡θgωibz2+γ˙-β˙sinα-θ˙jsinθg=cos⁡θjωibxb-sinθjωibyb+α˙+θ˙gsinθgsinθjωibxb+cos⁡θgωibzb+sinθgcos⁡θjωibyb+β˙cos⁡α+θ˙jcos⁡θgcos⁡θgsinθjωibxb-sinθgωibzb+cos⁡θgcos⁡θjωibyb+γ˙-β˙sinα-θ˙jsinθg.(2) The moment of inertia of magnetic bearing inoxryrzr is Jr=diagJrxJryJrz. Because the rotor is symmetric, its equator moment of inertia is Jrx=Jry=Jrr, and its pole moment of inertia is Jrz=Jp. So the rotor moment of inertia relative to inner loop frame has the following relationship with Jr:
(9)J2=C2rTJrC2r≈Jrr000Jrr000Jrz.The rotor angular momentum in inner loop frame is(10)Ηr=Jr·ω⃑ir2=Jr·(Cb2ω⃑ib+Cj2ω⃑sj+Cg2ω⃑jg+ω⃑1r)=Jrrωibx2+α˙+θ˙gJrrωiby2+β˙cos⁡α+θ˙jcos⁡θgJrzωibz2+γ˙-β˙sinα-θ˙jsinθg.Then(11)Η˙r=Jr·ω⃑˙ir2=Jr(-ω⃑1r×Cb2-Cg2ω⃑jg×Cbg-Cj2ω⃑sj×Cbj)ω⃑ib+Cb2ω⃑˙ib+(-ω⃑1r×Cj2-Cg2ω⃑sj×Cjg)ω⃑sj+Cj2ω⃑˙sj+-ω⃑1r×Cg2ω⃑jg+Cg2ω⃑˙jg+ω⃑˙1r=Jrrω˙ibx2+α¨+θ¨gJrrω˙iby2+β¨cos⁡α-β˙α˙sinα+θ¨jcos⁡θg-θ˙jθ˙gsinθgJrzω˙ibz2+γ¨-β¨sinα-β˙α˙cos⁡α-θ¨jsinθg-θ˙jθ˙gcos⁡θg.The projection of the angular velocity in inner loop frame can be simplified as(12)ω⃑i22=ωibx2+α˙+θ˙gωiby2+β˙cos⁡α+θ˙jcos⁡θgωibz2-β˙sinα-θ˙jsinθg.(3) In this paper, five-degree-of-freedom magnetic bearing is used. The axial magnetic bearing only provides translation force, and it cannot output torque to the rotor. Therefore, the total torque of the rotor is(13)Mr12=pxpy0.(4) In magnetic bearing frameOxryrzr, the dynamic equation of the ith double-gimbal magnetically suspended CMG is
(14)H˙rx+Hrzωi2y2-Hryωi2z2=Mr1x2,H˙ry+Hrxωi2z2-Hrzωi2x2=Mr1y2,H˙rz+Hryωi2x2-Hrxωi2y2=Mr1z2,
where Mr1x2, Mr1y2, and Mr1z2 are the elements of Mr12. According to (2)~(14), the rotor dynamic equation is
(15)Jrr[ω˙ibx2+α¨+θ¨g]+Jrzω˙ibz2+γ¨-β¨sinα-β˙α˙cos⁡α-θ¨jsinθg-θ˙jθ˙gcos⁡θg·ωiby2+β˙cos⁡α+θ˙jcos⁡θg-Jrrωiby2+β˙cos⁡α+θ˙jcos⁡θg·[ω˙ibz2-β˙sinα-θ˙jsinθg]=px,Jrr[ω˙iby2+β¨cos⁡α-β˙α˙sinα+θ¨jcos⁡θg-θ˙jθ˙gsinθg]+Jrr[ωibx2+α˙+θ˙g][ω˙ibz2-β˙sinα-θ˙jsinθg]-Jrz[ω˙ibz2+γ˙-β˙sinα-θ˙jsinθg][ωibx2+α˙+θ˙g]=py,Jrz[ω˙ibz2+γ¨-β¨sinα-β˙α˙cos⁡α-θ¨jsinθg-θ˙jθ˙gcos⁡θg]+Jrr[ωibx2+α˙+θ˙g][ωiby2+β˙cos⁡α+θ˙jcos⁡θg]-Jrr[ωibx2+α˙+θ˙g][ωiby2+β˙cos⁡α+θ˙jcos⁡θg]=0.Because the projection gap is small and the permitted maximal radial rotation angle is less than 3′′, the rotor motion is restricted in the projection gap.α and β are small, γ˙≫α˙, γ˙≫β˙, Jrr<Jrz, so the above equation can be simplified as
(16)Jrrω˙ibx2+α¨+θ¨g+Hrzωiby2+β˙+θ˙jcos⁡θg=px,Jrrω˙iby2+β¨+θ¨jcos⁡θg-θ˙jθ˙gsinθg-Hrzωibx2+α˙+θ˙g=py,Jrzω˙ibz2+γ˙=0.When inner gimbal and outer gimbal are still on the ground, the dynamic equation of magnetically suspended rotor can be simplified as(17)Jrrα¨+Hrzβ˙=px,Jrrβ¨-Hrzα˙=py.


## 3.2. The Dynamic Model of Inner Gimbal

The absolute angular velocityω⃑ig of inner gimbal is ω⃑ig=ω⃑ib+ω⃑bs+ω⃑sj+ω⃑jg.(1) Taking inner gimbal frameOxgygzg as moveable frame, the angular velocity of outer gimbal relative to zero position is ω⃑sj. Its projection in inner gimbal frame is
(18)ω⃑sjg=ω⃑jg+Cjgω⃑sj=θ˙gθ˙jcos⁡θg-θ˙jsinθg.The zero position frame is still relative to body frame, soω⃑bs=0. The spacecraft rotation speed ω⃑ib in inner gimbal frame is denoted by
(19)ω⃑ibg=ωibxgωibygωibzg=CjgCsjCbsω⃑ibb=cos⁡θjωibyb-sinθjωibxbsinθgsinθjωibyb+cos⁡θgωibzb+sinθgcos⁡θjωibxbcos⁡θgsinθjωibyb-sinθgωibzb+cos⁡θgcos⁡θjωibxb.So the absolute velocityω⃑ig and the time derivative ω⃑˙ig of inner gimbal can be described in inner gimbal frame as
(20)ω⃑igg=Cbgω⃑ib+Cjgω⃑sj+ω⃑jg=ω⃑ibg+ω⃑sjg+ω⃑jgg=ωibxg+θ˙gωibyg+θ˙jcos⁡θgωibzg-θ˙jsinθg=cos⁡θjωibyb-sinθjωibxb+θ˙gsinθgsinθjωibyb+cos⁡θgωibzb+sinθgcos⁡θjωibxb+θ˙jcos⁡θgcos⁡θgsinθjωibyb-sinθgωibzb+cos⁡θgcos⁡θjωibxb-θ˙jsinθg,(21)ω⃑˙igg=ω˙ibxg+θ¨gω˙ibyg+θ¨jcos⁡θg-θ˙jθ˙gsinθgω˙ibzg-θ¨jsinθg-θ˙jθ˙gcos⁡θg.(2) The torque that acts on inner gimbal includes two parts, namely, the magnetic bearing torque and the outer gimbal torque. Then(22)Mgr1=-px-py0.The projection of this torque in inner gimbal frame is(23)Mgrg=C1gMgr1=-px-py0.The torque acting on the inner gimbal by the outer gimbal is(24)Mgjg=pgxpgypgz,
where pgy and pgz are the constraint torque along yg and zg direction. pgx is along xg direction. So the total torque of inner gimbal resulting from the magnetic bearing and outer gimbal is
(25)Mgg=Mgrg+Mgjg=pgx-pxpgy-pypgz.(3)n double-gimbal magnetically suspended CMGs are installed on spacecraft according to certain configurations, which can avoid singularity state. In inner gimbal frame Oxgygzg, the dynamic model of inner gimbal of the ith (i=1,…,n) double-gimbal magnetically suspended CMG is
(26)ω⃑igg×Hgi+H˙gi=Mgg,
where the angular momentum of inner gimbal of the ith double-gimbal magnetically suspended CMG is Hgi=Jg(Cbgω⃑ib+Cjgω⃑sj+ω⃑jg). Then
(27)H˙gi=Jg(-ω⃑jg×CjgCbj-Cjgω⃑sj×Cbj)ω⃑ib+Cbgω⃑˙ib-ω⃑jg×Cjgω⃑sj+Cjgω⃑˙sj+ω⃑˙jg.The moment of inertia of inner gimbal isJgxJgyJgzT; then
(28)Jgxω˙igxg+Jgz-Jgyωigygωigzg=Mgxg,Jgyω˙igyg+Jgx-Jgzωigzgωigxg=Mgyg,Jgzω˙igzg+Jgy-Jgxωigxgωigyg=Mgzg.Substituting (20), (21), and (25) into the above equation, the dynamic model of inner gimbal is
(29)Jgx(ω˙ibxg+θ¨g)+(Jgz-Jgy)(ωibyg+θ˙jcos⁡θg)·(ωibzg-θ˙jsinθg)=pgx-px,Jgy(ω˙ibyg+θ¨jcos⁡θg-θ˙jθ˙gsinθg)+(Jgx-Jgz)(ωibzg-θ˙jsinθg)(ωibxg+θ˙g)=pgy-py,Jgz(ω˙ibzg-θ¨jsinθg-θ˙jθ˙gcos⁡θg)+(Jgy-Jgx)(ωibxg+θ˙g)(ωibyg+θ˙jcos⁡θg)=pgz.


## 3.3. The Dynamic Model of Outer Gimbal

Supposing thatω⃑sj is the angular velocity of outer gimbal relative to zero position frame, ω⃑bs is the angular velocity of zero position frame relative to spacecraft body frame, and ω⃑ib is the implicated angular velocity of spacecraft body frame relative to inertial frame. The absolute angular velocity ω⃑ij of outer gimbal is ω⃑ij=ω⃑ib+ω⃑bs+ω⃑sj.(1) Taking outer gimbal frameOxjyjzj as movable frame, zero position frame is kept stable relative to body frame: ω⃑bs=0.The spacecraft rotating velocity in inner gimbal frame is described as(30)ω⃑ibj=ωibxjωibyjωibzj=CsjCbsω⃑ibb=cos⁡θjωibyb-sinθjωibxbωibzbsinθjωibyb+cos⁡θjωibxb.So the absolute angular velocityω⃑ijj of outer gimbal and its time derivative ω⃑˙ijj can be described as
(31)ω⃑ijj=Cbjω⃑ib+ω⃑sj=ωibxjωibyj+θ˙jωibzj=cos⁡θjωibyb-sinθjωibxbωibzb+θ˙jsinθjωibyb+cos⁡θjωibxb,(32)ω⃑˙ijj=-sinθjθ˙jωibyb+cos⁡θjω˙ibyb-cos⁡θjθ˙jωibxb-sinθjω˙ibxbω˙ibzb+θ¨jcos⁡θjθ˙jωibyb+sinθjω˙ibyb-sinθjθ˙jωibxb+cos⁡θjω˙ibxb=ω˙ibxjω˙ibyj+θ¨jω˙ibzj.(2) Similar with inner gimbal, the torque acting on outer gimbal is the torque from spacecraft and the torque from inner gimbal. The projection of inner gimbal torque on outer gimbal is(33)Mjgj=CgjMjgg=-pgx-pgycos⁡θj-pgzsinθjpgysinθj-pgzcos⁡θj.The torque that the spacecraft is acting on outer gimbal is(34)Mjij=pjxpjypjz,
where pjy is the outer gimbal motor driving torque along yj direction. pjx and pjz are the constraint torque in which the base acts on outer gimbal along xj and zj direction. So the total torque acting on outer gimbal is
(35)Mjj=Mjgj+Mjij=-pgx+pjx-pgycos⁡θg+pgzsinθg+pjy-pgysinθg-pgzcos⁡θg+pjz.(3) In outer gimbal frameOxjyjzj, the outer gimbal dynamic model of the ith double-gimbal magnetically suspended CMG is
(36)ω⃑ijj×Hji+H˙ji=Mjj,
where the angular momentum of outer gimbal of the ith double-gimbal magnetically suspended CMG is Hji=Jj(Cbjω⃑ib+ω⃑sj). Then H˙ji=Jj(-ω⃑sj×Cbjω⃑ib+Cbjω⃑˙ib+ω⃑˙sj).The moment of inertia of outer gimbal isJjxJjyJjzT; then
(37)Jjxω˙ijxj+Jjz-Jjyωijyjωijzj=Mjxj,Jjyω˙ijyj+Jjx-Jjzωijzjωijxj=Mjyj,Jjzω˙ijzj+Jjy-Jjxωijxjωijyj=Mjzj.By substituting (31), (32), and (34) into the above equation, the dynamic equation of outer gimbal is
(38)pjx=Jjxω˙ibxj+pgx,pjy=Jjyω˙ibyj+θ¨j+pgycos⁡θg-pgzsinθg,pjz=Jjzω˙ibzj+pgysinθg+pgzcos⁡θg.


## 4. Agile Maneuver Spacecraft Dynamic Modelling Using Double-Gimbal Magnetically Suspended CMG as the Actuator

The torque in spacecraft body frame can be described as(39)H˙b+ω→ibb×Hb=-Mbb+Tdb.Supposing that the moment of inertia of the spacecraft isJbxJbyJbzT, then
(40)Jbxω˙ibxb+Jbz-Jbyωibybωibzb=-Mbxb+Tdxb,Jbyω˙ibyb+Jbx-Jbzωibzbωibxb=-Mbyb+Tdyb,Jbzω˙ibzb+Jby-Jbxωibxbωibyb=-Mbzb+Tdzb,
where Mbb=MbxbMbybMbzbT is the vector form of the total torque that acts on spacecraft. Tdb=TdxbTdybTdzbT is the disturbance, which can be described as
(41)Mbb=∑i=1nMjij=ω⃑ib×∑i=1nCrbh0+∑i=1nCjbω⃑sj×Crjh0+∑i=1nCgbω⃑jg×Crgh0+∑i=1nCrbω⃑1r×h0+∑i=1nTji+∑i=1nCgjTgi+∑i=1nTri,
where h0=00IriγT is the constant rotor angular momentum. The outer disturbance torque can be denoted by
(42)Tgi=ω⃑igg×Hgi+H˙gi,Tji=ω⃑ijj×Hji+H˙ji,Tri=ω⃑ib×Crb+Cjbω⃑sj×Crj+Cgbω⃑jg×Crg+Crbω⃑1r×·JrCbrω⃑ib+Cjrω⃑sj+Cgrω⃑jg+ω⃑1rω⃑ib×Crb+Cjbω⃑sj×Crj+Cgbω⃑jg×Crg+Crbω⃑1r×+CrbJr-ω⃑1r×Cbr-Cgrω⃑jg×Cbg-Cjrω⃑sj×Cbjω⃑ib+Cbrω⃑˙ib+-ω⃑1r×Cjr-Cgrω⃑jg×Cjgω⃑sj+Cjrω⃑˙sj-ω⃑1r×Cgrω⃑jg+Cgrω⃑˙jg+ω⃑˙1r.Because the rotor displacement relative to the gimbal is small,Crg≈E, Crb≈Cgb. Equation (41) can be described as
(43)T=h˙+ω⃑ib×h+Tr,
where the total angular momentum of double-gimbal magnetically suspended CMGs is h=∑i=1nCrbh0. Then
(44)h˙=∑i=1nCjbω⃑sj×Crjh0+∑i=1nCgbω⃑jg×Crgh0≈∑i=1nCjbω⃑sj×Cgjh0+∑i=1nCgbω⃑jg×h0=Jσσ˙,
where σ=σg1⋯σgnσj1⋯σjnT is the inner gimbal angle and the outer gimbal angle of n double-gimbal magnetically suspended CMGs. σgi and σji are the gimbal rate of the ith double-gimbal magnetically suspended CMG. J(σ)=∂h/∂σ is the Jacobin matrix. The outer disturbance torque can be described as
(45)Tr=∑i=1nCrbω⃑1r×h0+∑i=1nTji+∑i=1nCgjTgi+∑i=1nTri.Equations (16), (29), (38), and (40) form the dynamic model of double-gimbal magnetically suspended CMG when the spacecraft maneuvers. It can be seen that the relative motion of any component will result in the relative motion of other components. The motion of all components is coupling, which makes the dynamic model of spacecraft with double-gimbal magnetically suspended CMG complex.


## 5. The Coupling Characteristic in Double-Gimbal Magnetically Suspended CMG

Equations (16), (29), and (38) form the dynamic model of double-gimbal magnetically suspended CMG when spacecraft maneuvers. By writing them together, then
(46)mx¨r=fx,my¨r=fy,mz¨r=fz,Jrrω˙ibx2+α¨+θ¨g+Hrzωiby2+β˙+θ˙jcos⁡θg=px,Jrrω˙iby2+β¨+θ¨jcos⁡θg-θ˙jθ˙gsinθg-Hrz(ωibx2+α˙+θ˙g)=py,Jgx(ω˙ibxg+θ¨g)+Jrr(ω˙ibx2+θ¨g)+Jrrα¨+Hrz(ωiby2+β˙+θ˙jcos⁡θg)=pgx,Jjy(ω˙ibyj+θ¨j)+Jgy(ω˙ibyg+θ¨j)+Jrr(ω˙iby2+θ¨j)+Jrrβ¨cos⁡θg-Hrz(ωibx2+α˙+θ˙g)cos⁡θg=pjy,Jgx(ω˙ibxg+θ¨g)+(Jgz-Jgy)(ωibyg+θ˙jcos⁡θg)(ωibzg-θ˙jsinθg)=pgx-px,Jgy(ω˙ibyg+θ¨jcos⁡θg-θ˙jθ˙gsinθg)+(Jgx-Jgz)(ωibzg-θ˙jsinθg)(ωibxg+θ˙g)=pgy-py,Jgz(ω˙ibzg-θ¨jsinθg-θ˙jθ˙gcos⁡θg)+(Jgy-Jgx)(ωibxg+θ˙g)(ωibyg+θ˙jcos⁡θg)=pgz,pjx=Jjxω˙ibxj+pgx,pjy=Jjyω˙ibyj+θ¨j+pgycos⁡θg-pgzsinθg,pjz=Jjzω˙ibzj+pgysinθg+pgzcos⁡θg.
From the above equations, the following can be seen.(1) Equation (46) is the nonlinear dynamic model of the relative motion between magnetically suspended rotor, inner gimbal, and outer gimbal of double-gimbal magnetically suspended CMG when spacecraft maneuvers. These models are built in inner loop frame, inner gimbal frame, and outer gimbal frame, respectively. The relative motion of the three components must be considered for analyzing the spacecraft stability and precision.(2) The magnetic bearing suspending with gap is used in double-gimbal magnetically suspended CMG. The motion of magnetically suspended rotor does not only rely on magnetic bearing force but also suffer from the influence of gimbal servo system and spacecraft motion.(3) Between the inner gimbal and the outer gimbal, there exists the dynamic coupling that results from the gyro effect. This dynamic coupling not only depends on the gimbal motion but also is related with the radial rotation. When spacecraft maneuvers, the gyro coupling torque of double-gimbal magnetically suspended CMG is proportional to spacecraft angular velocity and the cosine of the radial angle displacement. So this system is a nonlinear system. The disturbance influence of the strong coupling torque on the magnetically suspended rotor can make the magnetically suspended rotor unstable.In (46), px and py can be obtained by using the dynamic equation of magnetically suspended rotor, namely, (16). pgx, pgy, and pgz can be obtained by substituting them into the dynamic model of inner gimbal, namely, (29). And then, they are substituted into the dynamic equation of outer gimbal, namely, (38), to obtain pjx, pjy, and pjy. By ignoring the high-order item, the dynamic model of magnetic bearing along x-direction, y-direction of the inner and outer gimbal can be obtained:
(47)Jrrα¨+Hrzβ˙+Jrrω˙ibx2+θ¨g+Hrzωiby2+θ˙jcos⁡θg=px,Jrrβ¨-Hrzα˙+Jrrω˙iby2+θ¨jcos⁡θg-Hrzωibx2+θ˙g=py,Jrrα¨+Hrzβ˙+Jgx(ω˙ibxg+θ¨g)+Jrr(ω˙ibx2+θ¨g)+Hrz(ωiby2+θ˙jcos⁡θg)=pgx,Jrrβ¨cos⁡θg-Hrz(ωibx2+θ˙g+α˙)cos⁡θg+Jjy(ω˙ibyj+θ¨j)+Jgy(ω˙ibyg+θ¨j)+Jrr(ω˙iby2+θ¨j)=pjy.Based on the dynamic model of double-gimbal magnetically suspended CMG in (46), the coupling torque along x-direction of magnetic bearing, y-direction of magnetic bearing, the inner gimbal, and the outer gimbal can be obtained. The coupling torque can be divided into two portions, including the gyro coupling torque and the inertial coupling torque, where the gyro coupling torque is proportional to the spacecraft angular velocity. The gyro coupling torque is
(48)Mgc2x=Hrzωiby2+θ˙jcos⁡θg,Mgc2y=-Hrzωibx2+θ˙g,Mgcgx=Hrzωiby2+θ˙jcos⁡θg,Mgcjy=-Hrzωibx2+θ˙gcos⁡θg,
where Mgc2x, Mgc2y, Mgcgx, and Mgcjy are the gyro coupling torque along x-direction of magnetic bearing, y-direction of magnetic bearing, the inner gimbal, and the outer gimbal, respectively. The inertial coupling torque is proportional to the angle acceleration of the spacecraft and gimbal:
(49)Mic2x=Jrrω˙ibx2+θ¨g,Mic2y=Jrrω˙iby2+θ¨jcos⁡θg,Micgx=Jgxω˙ibxg+θ¨g+Jrrω˙ibx2+θ¨g,Micjy=Jjyω˙ibyj+θ¨j+Jgyω˙ibyg+θ¨j+Jrrω˙iby2+θ¨j,
where Mic2x, Mic2y, Micgx, and Micjy are the inertial coupling torque along x-direction of magnetic bearing, y-direction of magnetic bearing, the inner gimbal, and the outer gimbal, respectively.


## 6. Conclusion

Double-gimbal magnetically suspended CMG is a novel attitude control actuator for the agile maneuver spacecraft. The rotor has five degrees of freedom besides the rotation degree, including three translation motions and two radial motions. So the dynamic model of the double-gimbal magnetically suspended CMG is complex. Moreover, the maneuver of spacecraft will influence the stability and precision of high speed rotor.In order to verify the interrelation in double-gimbal magnetically suspended CMG, its working principal and basic structure are given firstly. Then, the dynamic model of the magnetically suspended rotor, inter gimbal, outer gimbal, and spacecraft using double-gimbal magnetically suspended control moment gyro as actuator is built. From these models, it can be seen that the model of one component includes the coupling moment produced by other components, and the relative motion of one component will affect the motion of other components. The strong coupling between the components is investigated, which will provide the foundation for the further research.



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*Source: 278145-2015-03-02.xml*