The Coupling Characteristic Investigation of Double
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Hindawi Publishing Corporation
Mathematical Problems in Engineering
Volume 2015, Article ID 278145, 11 pages
http://dx.doi.org/10.1155/2015/278145
Research Article
The Coupling Characteristic Investigation of Double-Gimbal
Magnetically Suspended Control Moment Gyro Used on Agile
Maneuver Spacecraft
Peiling Cui,1,2 Jian Cui,1,2 Qian Yang,1 and Shiqiang Zheng1,2
1
School of Instrumentation Science and Optoelectronics Engineering, Beihang University, Beijing 100191, China
Science and Technology on Inertial Laboratory, Beijing 100191, China
2
Correspondence should be addressed to Peiling Cui; cuiplhh@126.com
Received 4 November 2014; Revised 27 January 2015; Accepted 10 February 2015
Academic Editor: Paulo Batista Gonçalves
Copyright © 2015 Peiling Cui et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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.
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 doublegimbal 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 doublegimbal 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
2
Mathematical Problems in Engineering
Outer gimbal
Inner gimbal
Figure 1: Prototype of double-gimbal magnetically suspended
CMG.
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 singlegimbal 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 highspeed 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 Figure 1. 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 fivedegree-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.
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
Figure 2.
ππ ππ ππ ππ is the inertial frame. The origin is earth center of
mass. ππ ππ points to vernal equinox; ππ ππ points to the north
pole.
ππ ππ ππ ππ is the orbit frame. The origin is the spacecraft
center of mass. ππ ππ and ππ ππ 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. ππ§π represents the yaw axis,
which points to earth’s core. This frame rotates with speed ππ ,
and the rotation direction is contrary to the pitch axis.
ππ π₯π π¦π π§π is the spacecraft body frame. The origin is the
spacecraft center of mass, ππ₯π , ππ¦π , and ππ§π are fixed on
the spacecraft, and they are the moments of inertial principal
π
axis of the spacecraft. [π π π] is the 3-1-2 Euler angle with
respect to orbit frame.
ππ π₯π π¦π π§π is the zero position frame. At the zero position,
it coincides with the outer gimbal frame. It is fixed with body
frame.
ππ π₯π π¦π π§π is the outer gimbal frame. It is fixed with outer
gimbal. The outer gimbal angle is ππ . It is positive when the
rotor rotation speed ππ along π¦π direction is positive.
ππ π₯π π¦π π§π is the inner gimbal frame. It is fixed with inner
gimbal. The inner gimbal angle is ππ . It is positive when the
rotor rotation speed ππ along π₯π direction is positive.
π1 π₯1 π¦1 π§1 is the stator frame. It is fixed with gyro room.
This frame keeps still with respect to inner gimbal frame. ππ₯1
and ππ¦1 point to central directions of the radial magnetic
bearing.
π2 π₯2 π¦2 π§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. π2 is the center
of rotor mass. π₯π1 π¦π1 π§π1 and π₯π2 π¦π2 π§π2 in Figure 3 are the
suppositional inner loop and the dummy outer loop frames,
respectively. The outer gimbal angle πΌ is the Euler angle of
Mathematical Problems in Engineering
3
yg
yi , yj
Base
πg
Magnetically suspended rotor
Rotor
Inner gimbal
Outer gimbal y2 y1
πj
B
π½
B
zj
yb yo
πj
zg
zs
A
πg
zi
o
xi
πg
πj
xj , xg
x2
πΌ
x1
o
z1
A
O
z2
Satellite
xs
xb
xo
O
Orbit
zo
zb
πo
ys
Figure 2: The illustration of coordinate frames.
π₯π1 π¦π1 π§π1 with respect to ππ₯1 π¦1 π§1 . The inter gimbal angle π½
is the Euler angle of π₯π2 π¦π2 π§π2 with respect to π₯π1 π¦π1 π§π1 .
ππ π₯π π¦π π§π is the rotor frame. This frame is fixed with rotor
and has one rotation degree of freedom along π§π axis with
respect to inner loop frame. The rotor rotation angle is Ωπ‘.
Ω is the constant angular velocity, and it is expressed in rotor
frame coordinate as [0, 0, Ω]π .
Cππ is used to denote the coordinate transformation matrix
from frame π to frame π. y/π denotes the vector projected
in coordinate π. x× denotes the skew symmetric matrix
π
of x = [π₯1 π₯2 π₯3 ] . On the condition of small angle
transformation, then
1 π −π
cos π 0 − sin π cos π
]
[
[ 0 1
π
sin π ] ≈ [−π 1 π ]
Cπ = [
],
[ sin π 0 cos π cos π ]
Cπ π
1 0 0
[0 0 1]
=[
],
Cππ
[0 1 0]
cos ππ 0 − sin ππ
]
[
1
0 ]
=[
],
[ 0
[ sin ππ 0 cos ππ ]
1
0
0
]
[
π
[
Cπ = [0 cos ππ sin ππ ]
],
[0 − sin ππ cos ππ ]
1
[ π −π 1 ]
0
1 0 0
[0 1 0]
1
Cπ = [
],
[0 0 1]
0
zg1
z1
zg2 (zr )
π½
yr
yg1 (yg2 )
πΌ
O
y1
Ωt
x1 (xg1 )
xg2
xr
Figure 3: The coordinate transformation matrix from the stator
frame to the rotor frame.
C21
=
π1 π2
C1 Cπ1
1 0 −π½
[0 1 πΌ ]
≈[
],
[π½ −πΌ 1 ]
Cπ2
cos Ωπ‘ sin Ωπ‘ 0
[− sin Ωπ‘ cos Ωπ‘ 0]
=[
].
[
0
0
1]
π1
]
[
C1 = [0 cos πΌ sin πΌ ] ,
[0 − sin πΌ cos πΌ]
π2
Cπ1
cos π½ 0 − sin π½
[ 0 1
0 ]
=[
],
[ sin π½ 0 cos π½ ]
(1)
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
4
Mathematical Problems in Engineering
y
π½
fby
fay
Ω
Z
A
fbz
faz
B
O
fax
lam
las
fbx
πΌ
lbm
lbs
x
Figure 4: The sketch map of magnetic bearing dynamic model.
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 along π₯, π¦, and π§
direction and rotation motion along π₯, π¦, and π§ 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 π₯, π¦, and π§
direction and the radial rotation along π₯, π¦ 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 πππ₯ and πππ₯ can be synthesized into
radial force ππ₯ . The radial bearing forces πππ¦ and πππ¦ can
be synthesized into radial force ππ¦ . The radial torque in π₯
and π¦ direction is ππ¦ and ππ₯ , 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.
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
σ³¨a and Coriolis inertial
is the implicated inertial force πβ
π
β
σ³¨
force π a π that result from the implicated motion of inner
gimbal and outer gimbal. The rotor dynamic equation in
σ³¨a ) + (−πβ
σ³¨a ). π is
σ³¨a = F + (−πβ
noninertial frame is πβ
π
π
π
β
σ³¨
the mass of magnetically suspended rotor. a π is the relative
translation acceleration of magnetically suspended rotor. F
σ³¨a
σ³¨a and β
is the active force, namely, the bearing force. β
π
π
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
σ³¨a =
σ³¨a = F, where β
suspended rotor is simplified as πβ
π
π
π
π
[π₯πΜ π¦πΜ π§Μπ ] , F = [ππ₯ ππ¦ ππ§ ] .
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 angles πΌ, π½, and
πΎ, 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 by ππ .
(3) The rotation of outer gimbal frame relative to inertial
frame is denoted by ππ .
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 frame ππ₯1 π¦1 π§1 , and the movable frame is the inner
loop frame ππ₯2 π¦2 π§2 .
The angle speed vector of magnetically suspended rotor
β
σ³¨
π ππ includes six parts, namely, the rotation speed of rotor
σ³¨
frame relative to stator frame β
π 1π , the rotation speed of stator
σ³¨
frame relative to inner gimbal frame β
π π1 , the rotation speed
σ³¨
of inner gimbal frame relative to outer gimbal frame β
π ππ , the
rotation speed of outer gimbal frame relative to zero position
σ³¨
frameβ
π π π , the rotation speed of zero position frame relative to
σ³¨
body frameβ
π ππ , and the rotation speed of body frame relative
Mathematical Problems in Engineering
5
σ³¨
σ³¨
σ³¨
σ³¨
σ³¨
σ³¨
σ³¨
to inertial frame β
π ππ : β
π ππ = β
π ππ + β
π ππ + β
π π π + β
π ππ + β
π π1 +
β
σ³¨
π 1π . Because the projection gap is small, the maximal radial
rotation angle of the Euler angles πΌ, π½ is less than 3 × 10−3 rad.
It can be considered that cos πΌ ≈ 1, cos π½ ≈ 1, sin πΌ ≈ 0,
sin π½ ≈ 0.
(1) Derive the angular velocity projection of the movable
frame about each axis. The angular velocity of rotor frame
β
σ³¨Μ σ³¨Μ
σ³¨
σ³¨
πΎ . Its projection
πΌΜ + π½ + β
relative to stator frame is β
π =β
1π
in inner loop frame is
πΌΜ
0
0
πΌΜ
] [ ] [
]
2
π1 [
[0]
β
σ³¨
Μ
]
[ Μ
]
π 1π = [ ] + C1 [
[π½] + [0] = [ π½ cos πΌ ] . (2)
Μ
Μ
[0]
[ 0 ] [πΎ]
[πΎΜ − π½ sin πΌ]
Because stator frame is still relative to inner gimbal frame,
β
σ³¨
π π1 = 0. The angular velocity of inner gimbal relative to outer
σ³¨
gimbal is β
π ππ . Its projection in inner loop frame is
2
β
σ³¨
π ππ
σ³¨
π ππ
= C21 C1πβ
ππΜ
[ ]
]
≈[
[ 0 ].
[0]
(3)
The angular velocity of outer gimbal relative to zero
σ³¨
position frame is β
π π π . Its projection in inner loop frame is
0
]
[ Μ
2
πσ³¨
β
σ³¨
]
π ππ = [
π π π = C21 C1π Cπβ
(4)
[ ππ cos ππ ] .
Μ
[−ππ sin ππ ]
The zero position frame is still relative to body frame,
β
σ³¨
π ππ = 0. The rotation speed of spacecraft relative to orbit
Μ which corresponds to roll, pitch, and
frame is (π,Μ π,Μ π),
yaw. The rotation speed of orbit frame is (0, −π0 , 0)π . So,
the rotation speed of spacecraft can be described in frame
ππ₯π π¦π π§π as follows:
ππππ₯
π
[π ]
β
σ³¨
π ππ = [ πππ¦ ]
σ³¨
The absolute angular velocity β
π ππ of magnetically suspended rotor in inner loop frame is denoted by
2
σ³¨
σ³¨
σ³¨
β
σ³¨
σ³¨
π ππ + C2πβ
π π π + C2πβ
π ππ + β
π ππ = C2πβ
π 1π
2
2
2
2
σ³¨
σ³¨
σ³¨
σ³¨
=β
π ππ + β
π π π + β
π ππ + β
π 1π .
(7)
Then
2
+ πΌΜ + ππΜ
ππππ₯
+ π½ Μ cos πΌ + πΜ cos π
]
[ 2
2
β
σ³¨
]
π ππ = [
π
π ]
[ ππππ¦
2
Μ
Μ
[ππππ§ + πΎΜ − π½ sin πΌ − ππ sin ππ ]
π
π
− sin ππ ππππ¦
+ πΌΜ + ππΜ
cos ππ ππππ₯
[
[ sin ππ sin ππ ππ + cos ππ ππ + sin ππ cos ππ ππ
πππ₯
πππ§
πππ¦
[
[
Μ
Μ
[
= [ + π½ cos πΌ + ππ cos ππ
[
[ cos π sin π ππ − sin π ππ + cos π cos π ππ
π
π πππ₯
π πππ§
π
π πππ¦
[
Μ sin πΌ − πΜ sin π
Μ
+
πΎ
−
π½
[
π
π
]
]
]
]
].
]
]
]
]
]
(8)
(2) The moment of inertia of magnetic bearing in ππ₯π π¦π π§π
is Jπ = diag (π½ππ₯ π½ππ¦ π½ππ§ ). Because the rotor is symmetric, its
equator moment of inertia is π½ππ₯ = π½ππ¦ = π½ππ , and its pole
moment of inertia is π½ππ§ = π½π . So the rotor moment of inertia
relative to inner loop frame has the following relationship
with Jπ :
J2 =
π
[Cπ2 ] Jπ Cπ2
π½ππ 0 0
[0 π½ 0]
≈[
].
ππ
[0
(9)
0 π½ππ§ ]
The rotor angular momentum in inner loop frame is
[ππππ§ ]
(5)
πΜ
πΜ − π0 π
0
1 π −π
]
[ ] [
[
]
]
[
Μ]
]
[ Μ
=[
[ π ] + [−π 1 π ] [−π0 ] = [ π − π0 ] .
[πΜ ] [ π π 1 ] [ 0 ] [πΜ + π0 π]
σ³¨
The spacecraft rotation speed β
π in inner loop frame is
2
σ³¨
Ηπ = Jπ ⋅ β
π ππ
σ³¨
σ³¨
σ³¨
σ³¨
π ππ + C2πβ
π π π + C2πβ
π ππ + β
= Jπ ⋅ (C2πβ
π 1π )
2
+ πΌΜ + ππΜ )
π½ππ (ππππ₯
]
[
2
Μ
Μ
]
=[
[ π½ππ (ππππ¦ + π½ cos πΌ + ππ cos ππ ) ] .
2
Μ
Μ
[π½ππ§ (ππππ§ + πΎΜ − π½ sin πΌ − ππ sin ππ )]
ππ
2
ππππ₯
[ 2 ]
2
2 1 π π π β
β
σ³¨
σ³¨π
]
π ππ = [
[ππππ¦ ] = C1 Cπ Cπ Cπ Cπ π ππ
2
[ππππ§ ]
Then
π
π
− sin ππ ππππ¦
cos ππ ππππ₯
]
[
π
π
π ]
=[
[ sin ππ sin ππ ππππ₯ + cos ππ ππππ§ + sin ππ cos ππ ππππ¦ ] .
π
[cos ππ sin ππ ππππ₯
−
π
sin ππ ππππ§
(10)
+
π
cos ππ cos ππ ππππ¦
]
(6)
2
σ³¨
πΜ ππ
ΗΜ π = Jπ ⋅ β
×
× π
× π σ³¨
σ³¨
σ³¨
σ³¨
σ³¨
πΜ ππ
π ππ Cπ − C2πβ
π π π Cπ )β
= Jπ {(−β
π 1π C2π − C2πβ
π ππ + C2πβ
×
× π σ³¨
σ³¨
σ³¨
σ³¨
πΜ π π
π π π Cπ )β
+ (−β
π 1π C2π − C2πβ
π π π + C2πβ
6
Mathematical Problems in Engineering
×
σ³¨
σ³¨
σ³¨
σ³¨
πΜ 1π }
πΜ ππ + β
+ (−β
π 1π C2πβ
π ππ ) + C2πβ
2
+ πΌΜ + ππΜ )
π½ππ (πΜ πππ₯
[
[ π½ (πΜ 2 + π½ Μ cos πΌ − π½πΌ
Μ Μ sin πΌ
[ ππ πππ¦
[
[
+ ππΜ cos ππ − ππΜ ππΜ sin ππ )
=[
[
[
2
Μ Μ cos πΌ
+ πΎΜ − π½ Μ sin πΌ − π½πΌ
[ π½ππ§ (πΜ πππ§
[
− ππΜ sin ππ − ππΜ ππΜ cos ππ )
[
2
Μ Μ cos πΌ − πΜ sin π − πΜ πΜ cos π ]
+ πΎΜ − π½ Μ sin πΌ − π½πΌ
π½ππ§ [πΜ πππ§
π
π
π π
π
2
2
+ πΌΜ + ππΜ ] [ππππ¦
+ π½ Μ cos πΌ + ππΜ cos ππ ]
+ π½ππ [ππππ₯
]
]
]
]
]
].
]
]
]
]
2
2
+ πΌΜ + ππΜ ] [ππππ¦
+ π½ Μ cos πΌ + ππΜ cos ππ ] = 0.
− π½ππ [ππππ₯
(15)
]
(11)
The projection of the angular velocity in inner loop frame
can be simplified as
2
+ πΌΜ + ππΜ
ππππ₯
]
[
2
β
σ³¨
2
Μ
Μ
]
π π2 = [
[ππππ¦ + π½ cos πΌ + ππ cos ππ ] .
2
Μ
Μ
[ ππππ§ − π½ sin πΌ − ππ sin ππ ]
(12)
(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
M2π1
ππ₯
[π ]
= [ π¦] .
Because the projection gap is small and the permitted
maximal radial rotation angle is less than 3σΈ σΈ , the rotor motion
is restricted in the projection gap. πΌ and π½ are small, πΎΜ β« πΌ,Μ
πΎΜ β« π½,Μ π½ππ < π½ππ§ , so the above equation can be simplified as
2
2
π½ππ (πΜ πππ₯
+ πΌΜ + ππΜ ) + π»ππ§ (ππππ¦
+ π½ Μ + ππΜ cos ππ ) = ππ₯ ,
2
+ π½ Μ + ππΜ cos ππ − ππΜ ππΜ sin ππ )
π½ππ (πΜ πππ¦
2
+ πΎ)Μ = 0.
π½ππ§ (πΜ πππ§
When inner gimbal and outer gimbal are still on the
ground, the dynamic equation of magnetically suspended
rotor can be simplified as
π½ππ πΌΜ + π»ππ§ π½ Μ = ππ₯ ,
(13)
[0]
(4) In magnetic bearing frame ππ₯π π¦π π§π , the dynamic
equation of the πth double-gimbal magnetically suspended
CMG is
2
2
2
− π»ππ¦ ππ2π§
= ππ1π₯
,
π»Μ ππ₯ + π»ππ§ ππ2π¦
2
2
2
π»Μ ππ¦ + π»ππ₯ ππ2π§
− π»ππ§ ππ2π₯
= ππ1π¦
,
(14)
2
2
2
π»Μ ππ§ + π»ππ¦ ππ2π₯
− π»ππ₯ ππ2π¦
= ππ1π§
,
2
2
2
, ππ1π¦
, and ππ1π§
are the elements of M2π1 .
where ππ1π₯
According to (2)∼(14), the rotor dynamic equation is
2
π½ππ [πΜ πππ₯
− ππΜ sin ππ − ππΜ ππΜ cos ππ ]
2
+ π½ Μ cos πΌ + ππΜ cos ππ ]
⋅ [ππππ¦
2
+ π½ Μ cos πΌ + ππΜ cos ππ ]
− π½ππ [ππππ¦
2
− π½ Μ sin πΌ − ππΜ sin ππ ] = ππ₯ ,
⋅ [πΜ πππ§
2
Μ Μ sin πΌ + πΜ cos π − πΜ πΜ sin π ]
+ π½ Μ cos πΌ − π½πΌ
π½ππ [πΜ πππ¦
π
π
π π
π
2
2
+ πΌΜ + ππΜ ] [πΜ πππ§
− π½ Μ sin πΌ − ππΜ sin ππ ]
+ π½ππ [ππππ₯
2
2
+ πΎΜ − π½ Μ sin πΌ − ππΜ sin ππ ] [ππππ₯
+ πΌΜ + ππΜ ] = ππ¦ ,
− π½ππ§ [πΜ πππ§
(17)
π½ππ π½ Μ − π»ππ§ πΌΜ = ππ¦ .
3.2. The Dynamic Model of Inner Gimbal. The absolute
σ³¨
σ³¨
σ³¨
σ³¨
π ππ = β
π ππ + β
π ππ +
angular velocity β
π ππ of inner gimbal is β
β
σ³¨
β
σ³¨
π π π + π ππ .
(1) Taking inner gimbal frame ππ₯π π¦π π§π as moveable
frame, the angular velocity of outer gimbal relative to zero
σ³¨
position is β
π π π . Its projection in inner gimbal frame is
π
β
σ³¨
π π π
+ πΌΜ + ππΜ ]
2
Μ Μ cos πΌ
+ πΎΜ − π½ Μ sin πΌ − π½πΌ
+ π½ππ§ [πΜ πππ§
(16)
2
+ πΌΜ + ππΜ ) = ππ¦ ,
− π»ππ§ (ππππ₯
ππΜ
]
[
πσ³¨
σ³¨
Μ
]
π π π = [
=β
π ππ + Cπβ
[ ππ cos ππ ] .
Μ
[−ππ sin ππ ]
(18)
The zero position frame is still relative to body frame, so
σ³¨
β
σ³¨
π ππ in inner gimbal
π ππ = 0. The spacecraft rotation speed β
frame is denoted by
π
ππππ₯
[ π ]
π
π π π β
β
σ³¨
σ³¨π
]
π ππ = [
[ππππ¦ ] = Cπ Cπ Cπ π ππ
π
[ππππ§ ]
π
π
− sin ππ ππππ₯
cos ππ ππππ¦
]
[
π
π
π ]
=[
[ sin ππ sin ππ ππππ¦ + cos ππ ππππ§ + sin ππ cos ππ ππππ₯ ] .
π
π
π
[cos ππ sin ππ ππππ¦ − sin ππ ππππ§ + cos ππ cos ππ ππππ₯ ]
(19)
Mathematical Problems in Engineering
7
σ³¨
σ³¨
πΜ ππ
So the absolute velocity β
π ππ and the time derivative β
of inner gimbal can be described in inner gimbal frame as
ππ₯π π¦π π§π , the dynamic model of inner gimbal of the πth (π =
1, . . . , π) double-gimbal magnetically suspended CMG is
π
β
σ³¨
π ππ × Hππ + HΜ ππ = Mππ ,
π
π
π
π
πσ³¨
πσ³¨
β
σ³¨
σ³¨
σ³¨
σ³¨
σ³¨
π ππ = Cπβ
π ππ + Cπβ
π π π + β
π ππ = β
π ππ + β
π π π + β
π ππ
where the angular momentum of inner gimbal of the πth
double-gimbal magnetically suspended CMG is Hππ =
πσ³¨
πσ³¨
σ³¨
Jπ (Cπβ
π ππ + Cπβ
π π π + β
π ππ ). Then
π
ππππ₯ + ππΜ
]
[ π
Μ
]
=[
[ππππ¦ + ππ cos ππ ]
π
Μ
[ ππππ§ − ππ sin ππ ]
π
− sin ππ ππππ₯
+ ππΜ
π
π
sin ππ sin ππ ππππ¦
+ cos ππ ππππ§
π
+ sin ππ cos ππ ππππ₯
+ ππΜ cos ππ
π
π
cos ππ sin ππ ππππ¦
− sin ππ ππππ§
π
+ cos ππ cos ππ ππππ₯
− ππΜ sin ππ
π
cos ππ ππππ¦
[
[
[
[
=[
[
[
[
[
[
(20)
]
]
]
]
],
]
]
]
]
× π π
πσ³¨ × π β
π σ³¨Μ
σ³¨
π ππ
π π π Cπ ) σ³¨
π ππ Cπ Cπ − Cπβ
π ππ + Cπβ
HΜ ππ = Jπ ((−β
π
The moment of inertia of inner gimbal is [π½ππ₯ π½ππ¦ π½ππ§ ] ;
then
π
−ππ₯
[−π ]
= [ π¦] .
(21)
(22)
[ 0 ]
The projection of this torque in inner gimbal frame is
(23)
The torque acting on the inner gimbal by the outer gimbal
(24)
where πππ¦ and πππ§ are the constraint torque along π¦π and
π§π direction. πππ₯ is along π₯π direction. So the total torque of
inner gimbal resulting from the magnetic bearing and outer
gimbal is
Mππ
=
Mπππ
+
π
Mππ
πππ₯ − ππ₯
]
[
]
=[
[πππ¦ − ππ¦ ] .
[ πππ§ ]
π
π
π
π
π
(28)
π
π½ππ§ πΜ πππ§ + (π½ππ¦ − π½ππ₯ ) ππππ₯ ππππ¦ = πππ§
.
Substituting (20), (21), and (25) into the above equation,
the dynamic model of inner gimbal is
π
π
π½ππ₯ (πΜ πππ₯ + ππΜ ) + (π½ππ§ − π½ππ¦ ) (ππππ¦ + ππΜ cos ππ )
π
⋅ (ππππ§ − ππΜ sin ππ ) = πππ₯ − ππ₯ ,
π
π½ππ¦ (πΜ πππ¦ + ππΜ cos ππ − ππΜ ππΜ sin ππ )
π
π
+ (π½ππ₯ − π½ππ§ ) (ππππ§ − ππΜ sin ππ ) (ππππ₯ + ππΜ ) = πππ¦ − ππ¦ ,
π
π
+ (π½ππ¦ − π½ππ₯ ) (ππππ₯ + ππΜ ) (ππππ¦ + ππΜ cos ππ ) = πππ§ .
(29)
is
πππ₯
[ ]
]
=[
[πππ¦ ] ,
[πππ§ ]
π
π
[ 0 ]
π
Mππ
π
π½ππ§ (πΜ πππ§ − ππΜ sin ππ − ππΜ ππΜ cos ππ )
−ππ₯
π
[
]
Mπππ = C1 M1ππ = [−ππ¦ ] .
π
π
π½ππ¦ πΜ πππ¦ + (π½ππ₯ − π½ππ§ ) ππππ§ ππππ₯ = πππ¦
,
(2) The torque that acts on inner gimbal includes two
parts, namely, the magnetic bearing torque and the outer
gimbal torque. Then
M1ππ
π
π
,
π½ππ₯ πΜ πππ₯ + (π½ππ§ − π½ππ¦ ) ππππ¦ ππππ§ = πππ₯
]
]
[ π
β
σ³¨
Μ
Μ Μ
]
πΜ ππ = [
[πΜ πππ¦ + ππ cos ππ − ππ ππ sin ππ ] .
π
Μ
Μ Μ
[πΜ πππ§ − ππ sin ππ − ππ ππ cos ππ ]
(27)
× πσ³¨
π σ³¨Μ
σ³¨
σ³¨
πΜ ππ ) .
π π π + β
π π π + Cπβ
−β
π ππ Cπβ
πΜ πππ₯ + ππΜ
π
(26)
(25)
(3) π double-gimbal magnetically suspended CMGs are
installed on spacecraft according to certain configurations,
which can avoid singularity state. In inner gimbal frame
σ³¨
3.3. The Dynamic Model of Outer Gimbal. Supposing that β
π π π
is the angular velocity of outer gimbal relative to zero position
σ³¨
frame, β
π ππ is the angular velocity of zero position frame
σ³¨
relative to spacecraft body frame, and β
π ππ is the implicated
angular velocity of spacecraft body frame relative to inertial
σ³¨
frame. The absolute angular velocity β
π ππ of outer gimbal is
β
σ³¨
β
σ³¨
β
σ³¨
β
σ³¨
π ππ = π ππ + π ππ + π π π .
(1) Taking outer gimbal frame ππ₯π π¦π π§π as movable frame,
zero position frame is kept stable relative to body frame:
β
σ³¨
π ππ = 0.
The spacecraft rotating velocity in inner gimbal frame is
described as
π
ππππ₯
[ π ]
π
π π β
σ³¨π [
β
σ³¨
]
[
π ππ = [
[ππππ¦ ] = Cπ Cπ π ππ = [
π
[ππππ§ ]
π
π
− sin ππ ππππ₯
cos ππ ππππ¦
π
ππππ§
π
π
]
].
]
[sin ππ ππππ¦ + cos ππ ππππ₯ ]
(30)
8
Mathematical Problems in Engineering
σ³¨
So the absolute angular velocity β
π ππ of outer gimbal and
π
σ³¨
πΜ can be described as
its time derivative β
π
ππ
π
β
σ³¨
π ππ
=
πσ³¨
π ππ
Cπβ
σ³¨
+β
π π π
where the angular momentum of outer gimbal of the πth
double-gimbal magnetically suspended CMG is Hππ =
× πσ³¨
πσ³¨
π σ³¨Μ
σ³¨
σ³¨
σ³¨
πΜ π π ).
π ππ + β
Jπ (Cπβ
π ππ + β
π ππ + Cπβ
π π π ). Then HΜ ππ = Jπ (−β
π π π Cπβ
π
The moment of inertia of outer gimbal is [π½ππ₯ π½ππ¦ π½ππ§ ] ;
then
π
ππππ₯
]
[ π
Μ]
=[
[ππππ¦ + ππ ]
π
[ ππππ§ ]
[
=[
[
(31)
π
π
− sin ππ ππππ₯
cos ππ ππππ¦
π
ππππ§
+ ππΜ
π
π
π
π
π
+ cos ππ πΜ πππ¦
− cos ππ ππΜ ππππ₯
− sin ππ ππΜ ππππ¦
[
π
[ − sin ππ πΜ πππ₯
[
π
[
β
σ³¨
π
Μ
πΜ ππ = [
[ πΜ πππ§ + ππ
[
[ cos π πΜ ππ + sin π πΜ π − sin π πΜ ππ
π π πππ¦
π πππ¦
π π πππ₯
[
π
Μ
[ + cos ππ ππππ₯
]
]
]
]
]
]
]
]
]
π
π
π
π
π
π
π
π
π
π½ππ¦ πΜ πππ¦ + (π½ππ₯ − π½ππ§ ) ππππ§ ππππ₯ = πππ¦ ,
]
π
π
πππ¦ = π½ππ¦ (πΜ πππ¦ + ππΜ ) + πππ¦ cos ππ − πππ§ sin ππ ,
(32)
π
[ πΜ πππ§ ]
(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
−πππ₯
]
[
]
=[
[−πππ¦ cos ππ − πππ§ sin ππ ] .
[ πππ¦ sin ππ − πππ§ cos ππ ]
(33)
The torque that the spacecraft is acting on outer gimbal is
π
Mππ
πππ₯
[ ]
]
=[
[πππ¦ ] ,
[πππ§ ]
π
πππ§ = π½ππ§ πΜ πππ§ + πππ¦ sin ππ + πππ§ cos ππ .
The torque in spacecraft body frame can be described as
HΜ π + πβ πππ × Hπ = −Mππ + Tππ .
Supposing that the moment of inertia of the spacecraft is
π
[π½ππ₯ π½ππ¦ π½ππ§ ] , then
π
π
π
π
π
+ (π½ππ§ − π½ππ¦ ) ππππ¦
ππππ§
= −πππ₯
+ πππ₯
,
π½ππ₯ πΜ πππ₯
(34)
π
π
π
π
πππ¦
πππ§
where Mππ = [πππ₯
] is the vector form of the total
π
π
π
π
πππ¦
πππ§
] is the
torque that acts on spacecraft. Tππ = [πππ₯
disturbance, which can be described as
Mππ = ∑Mππ
(3) In outer gimbal frame ππ₯π π¦π π§π , the outer gimbal
dynamic model of the πth double-gimbal magnetically suspended CMG is
π
π
π
π=1
π=1
π=1
=
+
π
π
β
σ³¨
π ππ × Hππ + HΜ ππ = Mπ ,
(36)
(40)
π
π
π
π
π
+ (π½ππ¦ − π½ππ₯ ) ππππ₯
ππππ¦
= −πππ§
+ πππ§
,
π½ππ§ πΜ πππ§
−πππ₯ + πππ₯
]
[
[
= [−πππ¦ cos ππ + πππ§ sin ππ + πππ¦ ]
] . (35)
[−πππ¦ sin ππ − πππ§ cos ππ + πππ§ ]
π
Mππ
(39)
π
π
π
π
π
+ (π½ππ₯ − π½ππ§ ) ππππ§
ππππ₯
= −πππ¦
+ πππ¦
,
π½ππ¦ πΜ πππ¦
where πππ¦ is the outer gimbal motor driving torque along π¦π
direction. πππ₯ and πππ§ are the constraint torque in which the
base acts on outer gimbal along π₯π and π§π direction. So the
total torque acting on outer gimbal is
π
Mππ
(38)
4. Agile Maneuver Spacecraft
Dynamic Modelling Using
Double-Gimbal Magnetically
Suspended CMG as the Actuator
]
[ π
Μ]
=[
[πΜ πππ¦ + ππ ] .
π
Mπ
(37)
πππ₯ = π½ππ₯ πΜ πππ₯ + πππ₯ ,
π
=
π
By substituting (31), (32), and (34) into the above equation, the dynamic equation of outer gimbal is
πΜ πππ₯
π
Cππ Mππ
π
π½ππ§ πΜ πππ§ + (π½ππ¦ − π½ππ₯ ) ππππ₯ ππππ¦ = πππ§ .
]
],
]
[sin ππ ππππ¦ + cos ππ ππππ₯ ]
π
Mππ
π
π½ππ₯ πΜ πππ₯ + (π½ππ§ − π½ππ¦ ) ππππ¦ ππππ§ = πππ₯ ,
π
π
π=1
×
×
×
σ³¨
σ³¨
σ³¨
π π π Cππ h0 + ∑Cππβ
π ππ Cππ h0
=β
π ππ ∑Cππ h0 + ∑Cππβ
π
π
π
π
π=1
π=1
π=1
π=1
×
σ³¨
π 1π h0 + ∑Tππ + ∑Cππ Tππ + ∑Tππ ,
+ ∑Cππβ
(41)
Mathematical Problems in Engineering
9
π
where h0 = [0 0 Iππ πΎ] is the constant rotor angular
momentum. The outer disturbance torque can be denoted by
5. The Coupling Characteristic in DoubleGimbal 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
π
σ³¨
π ππ × Hππ + HΜ ππ ,
Tππ = β
σ³¨
Tππ = β
π ππ × Hππ + HΜ ππ ,
π
ππ₯πΜ = ππ₯ ,
×
×
×
×
σ³¨
σ³¨
σ³¨
σ³¨
Tππ = {(β
π π π Cππ + Cππβ
π ππ Cππ + Cππβ
π 1π )
π ππ Cππ + Cππβ
ππ¦πΜ = ππ¦ ,
σ³¨
σ³¨
σ³¨
σ³¨
π ππ + Cππβ
π π π + Cππβ
π ππ + β
π 1π )}
⋅ Jπ (Cππβ
ππ§Μπ = ππ§ ,
2
2
π½ππ (πΜ πππ₯
+ πΌΜ + ππΜ ) + π»ππ§ (ππππ¦
+ π½ Μ + ππΜ cos ππ ) = ππ₯ ,
×
× π
× π σ³¨
σ³¨
σ³¨
σ³¨
π ππ Cπ − Cππβ
π π π Cπ )β
π 1π Cππ − Cππβ
π ππ
+ Cππ Jπ {(−β
+
σ³¨
πΜ ππ
Cππβ
×
× π σ³¨
σ³¨
σ³¨
σ³¨
πΜ π π
π ππ Cπ )β
+ (−β
π 1π Cππ − Cππβ
π π π + Cππβ
×
σ³¨
σ³¨
σ³¨
σ³¨
πΜ 1π } .
πΜ ππ + β
π ππ + Cππβ
−β
π 1π Cππβ
(42)
2
+ π½ Μ + ππΜ cos ππ − ππΜ ππΜ sin ππ )
π½ππ (πΜ πππ¦
2
+ πΌΜ + ππΜ ) = ππ¦ ,
− π»ππ§ (ππππ₯
π
2
+ ππΜ ) + π½ππ πΌΜ
π½ππ₯ (πΜ πππ₯ + ππΜ ) + π½ππ (πΜ πππ₯
2
+ π½ Μ + ππΜ cos ππ ) = πππ₯ ,
+ π»ππ§ (ππππ¦
Because the rotor displacement relative to the gimbal is
small, Cππ ≈ E, Cππ ≈ Cππ . Equation (41) can be described as
σ³¨
T = hΜ + β
π ππ × h + Tπ ,
(43)
where the total angular momentum of double-gimbal magnetically suspended CMGs is h = ∑ππ=1 Cππ h0 . Then
π
π
π
π=1
π=1
π=1
×
×
×
σ³¨
σ³¨
σ³¨
hΜ = ∑Cππβ
π π π Cππ h0 + ∑Cππβ
π ππ Cππ h0 ≈ ∑Cππβ
π π π Cππ h0
+
π
×
σ³¨
π ππ h0
∑Cππβ
π=1
π
π
2
π½ππ¦ (πΜ πππ¦ + ππΜ ) + π½ππ¦ (πΜ πππ¦ + ππΜ ) + π½ππ (πΜ πππ¦
+ ππΜ )
2
+ π½ππ π½ Μ cos ππ − π»ππ§ (ππππ₯
+ πΌΜ + ππΜ ) cos ππ = πππ¦ ,
π
π
π½ππ₯ (πΜ πππ₯ + ππΜ ) + (π½ππ§ − π½ππ¦ ) (ππππ¦ + ππΜ cos ππ )
π
(ππππ§ − ππΜ sin ππ ) = πππ₯ − ππ₯ ,
π
π½ππ¦ (πΜ πππ¦ + ππΜ cos ππ − ππΜ ππΜ sin ππ )
π
π
+ (π½ππ₯ − π½ππ§ ) (ππππ§ − ππΜ sin ππ ) (ππππ₯ + ππΜ ) = πππ¦ − ππ¦ ,
(44)
= J (π) π,Μ
π
π½ππ§ (πΜ πππ§ − ππΜ sin ππ − ππΜ ππΜ cos ππ )
π
π
+ (π½ππ¦ − π½ππ₯ ) (ππππ₯ + ππΜ ) (ππππ¦ + ππΜ cos ππ ) = πππ§ ,
π
where π = [ππ1 ⋅ ⋅ ⋅ πππ ππ1 ⋅ ⋅ ⋅ πππ ] is the inner
gimbal angle and the outer gimbal angle of π double-gimbal
magnetically suspended CMGs. πππ and πππ are the gimbal
rate of the πth double-gimbal magnetically suspended CMG.
J(π) = πh/ππ is the Jacobin matrix. The outer disturbance
torque can be described as
π
π
π
π
π=1
π=1
π=1
π=1
×
σ³¨
π 1π h0 + ∑Tππ + ∑Cππ Tππ + ∑Tππ .
Tπ = ∑Cππβ
(45)
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 doublegimbal magnetically suspended CMG complex.
π
πππ₯ = π½ππ₯ πΜ πππ₯ + πππ₯ ,
π
πππ¦ = π½ππ¦ (πΜ πππ¦ + ππΜ ) + πππ¦ cos ππ − πππ§ sin ππ ,
π
πππ§ = π½ππ§ πΜ πππ§ + πππ¦ sin ππ + πππ§ cos ππ .
(46)
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
10
Mathematical Problems in Engineering
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), ππ₯ and ππ¦ can be obtained by using the dynamic
equation of magnetically suspended rotor, namely, (16). πππ₯ ,
πππ¦ , and πππ§ 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 πππ₯ , πππ¦ , and πππ¦ . By ignoring the highorder item, the dynamic model of magnetic bearing along π₯direction, π¦-direction of the inner and outer gimbal can be
obtained:
2
2
+ ππΜ ) + π»ππ§ (ππππ¦
+ ππΜ cos ππ ) = ππ₯ ,
π½ππ πΌΜ + π»ππ§ π½ Μ + π½ππ (πΜ πππ₯
2
2
+ ππΜ cos ππ ) − π»ππ§ (ππππ₯
+ ππΜ ) = ππ¦ ,
π½ππ π½ Μ − π»ππ§ πΌΜ + π½ππ (πΜ πππ¦
π
2
+ ππΜ )
π½ππ πΌΜ + π»ππ§ π½ Μ + π½ππ₯ (πΜ πππ₯ + ππΜ ) + π½ππ (πΜ πππ₯
2
+ ππΜ cos ππ ) = πππ₯ ,
+ π»ππ§ (ππππ¦
π
2
π½ππ π½ Μ cos ππ − π»ππ§ (ππππ₯
+ ππΜ + πΌ)Μ cos ππ + π½ππ¦ (πΜ πππ¦ + ππΜ )
π
2
+ ππΜ ) = πππ¦ .
+ π½ππ¦ (πΜ πππ¦ + ππΜ ) + π½ππ (πΜ πππ¦
(47)
Based on the dynamic model of double-gimbal magnetically suspended CMG in (46), the coupling torque along π₯direction of magnetic bearing, π¦-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
2
+ ππΜ cos ππ ) ,
πππ2π₯ = π»ππ§ (ππππ¦
2
+ ππΜ ) ,
πππ2π¦ = −π»ππ§ (ππππ₯
2
+ ππΜ cos ππ ) ,
πππππ₯ = π»ππ§ (ππππ¦
(48)
2
+ ππΜ ) cos ππ ,
πππππ¦ = −π»ππ§ (ππππ₯
where πππ2π₯ , πππ2π¦ , πππππ₯ , and πππππ¦ are the gyro coupling
torque along π₯-direction of magnetic bearing, π¦-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:
2
+ ππΜ ) ,
πππ2π₯ = π½ππ (πΜ πππ₯
2
+ ππΜ cos ππ ) ,
πππ2π¦ = π½ππ (πΜ πππ¦
π
2
+ ππΜ ) ,
πππππ₯ = π½ππ₯ (πΜ πππ₯ + ππΜ ) + π½ππ (πΜ πππ₯
π
π
2
πππππ¦ = π½ππ¦ (πΜ πππ¦ + ππΜ ) + π½ππ¦ (πΜ πππ¦ + ππΜ ) + π½ππ (πΜ πππ¦
+ ππΜ ) ,
(49)
where πππ2π₯ , πππ2π¦ , πππππ₯ , and πππππ¦ are the inertial coupling
torque along π₯-direction of magnetic bearing, π¦-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.
Nomenclature
ππ₯ :
ππ¦ :
ππ₯ :
ππ¦ :
Jπ :
Jπ :
Jπ :
Jπ :
πΌ:
Radial force in π₯ direction
Radial force in π¦ direction
The radial torque in π₯ direction
The radial torque in π¦ direction
The moments of inertia of spacecraft
The moments of inertia of outer gimbal
The moments of inertia of inner gimbal
The moments of inertia of rotor
The rotation of rotor frame relative to
magnetic bearing stator rotating about
the inner loop rotation axis
π½: The rotation of rotor frame relative to
magnetic bearing stator rotating about
outer loop rotation axis
πΎ: The rotation of rotor frame relative to
magnetic bearing stator rotating about
the rotor spin axis
Mathematical Problems in Engineering
ππ :
ππ :
β
σ³¨
π ππ :
β
σ³¨
π 1π :
β
σ³¨
π π1 :
β
σ³¨
π ππ :
β
σ³¨
π π π :
β
σ³¨
π ππ :
β
σ³¨
π ππ :
The rotation of the magnetic bearing stator
frame relative to inner gimbal frame
The rotation of outer gimbal frame relative to
inertial frame
The angle speed vector of magnetically
suspended rotor
The rotation speed of rotor frame relative to
stator frame
The rotation speed of stator frame relative to
inner gimbal frame
The rotation speed of inner gimbal frame
relative to outer gimbal frame
The rotation speed of outer gimbal frame
relative to zero position frame
The rotation speed of zero position frame
relative to body frame
The rotation speed of body frame relative to
inertial frame.
Conflict of Interests
11
[8]
[9]
[10]
[11]
[12]
[13]
The authors declare that there is no conflict of interests
regarding the publication of this paper.
[14]
Acknowledgments
This research has been supported by National Natural Science
Foundation of China under Grant 61121003, the National
Basic Research Program (973 Program) of China under
Grant 2009CB72400101C, and National Civil Aerospace PreResearch Project.
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