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Design and Attitude Control of a Satellite with Variable
Geometry
Miguel Fragoso Garrido de Matos Lino
Thesis to obtain the Master of Science Degree in
Aerospace Engineering
Examination Committee
Chairperson: Professor Fernando José Parracho Lau
Supervisor: Professor Paulo Jorge Soares Gil
Committee: Professor João Manuel Gonçalves de Sousa Oliveira
November 2013
i
Resumo
Exploramos um método de controlo de atitude de satélites alternativo a mecanismos convencionais como
rodas de inércia ou thrusters – controlo através da mudança da configuração do satélite.
Um satélite de geometria variável é concebido como sendo composto por partes móveis que podem ter
como único objectivo controlar a orientação do satélite ou podem ser componentes de outros sub-sistemas
que, além das suas respectivas funcionalidades, podem servir também como ferramentas de controlo de
atitude.
Reescrevemos as equações da dinâmica de corpo rígido de modo a obter uma forma geral das equações
de Euler nas quais a variação de geometria do corpo é tida em conta. As influências que a mudança de
configuração tem na energia cinética e no momento angular do corpo são discutidas.
Apresentam-se vários modelos simples de geometrias de satélites que podem mudar a sua forma. Os
tensores de inércia de cada uma das geometrias é obtido em função da configuração do satélite em causa.
A capacidade de mudança de configuração de um dos satélites é aplicada a um caso prático. Considera-
se um satélite que requer que a sua taxa de precessão esteja sincronizada com a velocidade orbital, ao
longo de uma órbita elíptica.
Tentamos explorar a capacidade de um dos satélites concebidos variar os três momentos principais de
inércia, bem como avaliar a estabilidade do movimento em torno dos eixos principais de inércia.
Palavras-chave:
Satélite, Atitude, Controlo, Geometria-Variável.
ii
Abstract
An alternative attitude control method to conventional orientation control mechanisms such as momentum
wheels or thrusters is explored - control by varying the configuration of the satellite.
A satellite with variable geometry would be composed by movable parts whose motion is controlled by the
action of actuators. The movable parts can have the single purpose of controlling the attitude of the satellite
or can be sub-systems, with certain primary purposes concerning the operation of the satellite, which could
also be used to control the attitude.
The rigid body dynamics equations are rewritten to obtain a general form of the Euler equations, where
shape changing capability of the body is taken into account. The influence of configuration changing on
kinetic energy and on angular momentum is also discussed.
Secondly, we envision several simple models for satellite geometries that are able to modify its shape. The
inertia tensors of the geometries are determined as function of the satellite's configuration.
The configuration changing capability of one of the satellite geometries is applied for a practical case. We
consider a satellite that requires its precession rate to be synchronized with the non-uniform orbital velocity
during a period of an elliptic orbit.
We explore the possibility of one of the envisioned satellites of changing the three moments of inertia and
assess the stability of rotation around the principal axes.
Keywords
Satellite, Attitude, Control, Variable-Geometry.
iii
TABLE OF CONTENTS
1 Introduction ...................................................................................................................................... 1
1.1 Objective .................................................................................................................................. 1
1.2 Attitude Control of Spacecraft ................................................................................................... 1
1.3 Satellites with Variable-geometry .............................................................................................. 2
1.4 Current Problems of Attitude Dynamics ..................................................................................... 2
1.4.1 Variable-Structure .............................................................................................................3
2 Attitude Dynamics of a Variable Geometry Body ............................................................................... 5
2.1 Angular Momentum .................................................................................................................. 5
2.2 Equations of Rotational Motion for a Variable-Geometry Body................................................... 6
2.3 General Form of Euler Equations .............................................................................................. 7
2.4 Alternative Derivation .............................................................................................................. 10
2.5 General Euler Equations in the Principal Frame ...................................................................... 11
2.6 Non-Rigid Terms .................................................................................................................... 14
2.6.1 Two-plane Symmetry ...................................................................................................... 14
2.7 Kinetic Energy of a Variable-Geometry Body ........................................................................... 16
2.7.1 Non-Rigid Terms ............................................................................................................. 17
2.7.2 Simplification of the Kinetic Energy Expression ............................................................... 19
2.7.3 Time-derivative of the Kinetic Energy .............................................................................. 20
2.8 Conservation of Angular Momentum ....................................................................................... 21
2.9 Geometrical Solutions for the Dynamics of a Variable Geometry Body .................................... 21
2.9.1 Ellipsoid of the Kinetic Energy ......................................................................................... 22
2.9.2 Ellipsoid of the Angular Momentum ................................................................................. 22
2.9.3 Geometrical Solution ....................................................................................................... 23
2.10 Modified Euler Equations in Terms of Euler Angles ................................................................. 26
2.10.1 Intermediate Frame for an Axisymmetric Body ................................................................ 27
2.11 Angular Momentum For a General Variable-Geometry Body ................................................... 29
2.11.1 Axisymmetric Variable-Geometry Body ........................................................................... 29
iv
3 Design of Variable Geometry Satellites ........................................................................................... 31
3.1 Connected Scissor Mechanism ............................................................................................... 31
3.1.1 Geometry ........................................................................................................................ 31
3.1.2 Body Frame .................................................................................................................... 31
3.1.3 Inertia Tensor .................................................................................................................. 32
3.1.4 Triangle Connected Scissor Mechanism .......................................................................... 36
3.1.5 Point Masses .................................................................................................................. 37
3.2 Octahedron............................................................................................................................. 40
3.2.1 Geometry ........................................................................................................................ 40
3.2.2 Body Frame .................................................................................................................... 41
3.2.3 Inertia Tensor .................................................................................................................. 41
3.2.4 Modified Octahedral ........................................................................................................ 42
3.3 Dandelion ............................................................................................................................... 43
3.3.1 Geometry ........................................................................................................................ 43
3.3.2 Body Frame .................................................................................................................... 44
3.3.3 Inertia Tensor .................................................................................................................. 44
3.4 Eight Point Masses ................................................................................................................. 45
3.4.1 Geometry ........................................................................................................................ 45
3.4.2 Body Frame .................................................................................................................... 46
3.4.3 Trajectory of the Masses ................................................................................................. 46
3.4.4 Geometric Relations with Axisymmetry ............................................................................ 48
3.5 Control Possibilities of axisymmetric satellites ......................................................................... 50
3.6 Compass ................................................................................................................................ 50
3.6.1 Geometry ........................................................................................................................ 50
3.6.2 Body Frame .................................................................................................................... 51
3.6.3 Inertia Tensor .................................................................................................................. 51
3.7 Discussion .............................................................................................................................. 52
4 Solutions of the Free Variable-Geometry Satellite Motion ............................................................... 53
4.1 Synchronous Rotation in an Elliptical Orbit with a Variable-Geometry Satellite ........................ 53
v
4.1.1 Analytical Approximation for Kepler’s Equation ................................................................ 53
4.1.2 Precession Rate as Function of Time .............................................................................. 55
4.1.3 Configuration Variation with Time .................................................................................... 55
4.1.4 Possible Solutions for Synchronous Rotation................................................................... 57
4.1.5 Example: Geosynchronous Orbit ..................................................................................... 58
4.2 Moments of Inertia Swap ........................................................................................................ 62
4.2.1 Relations between Moments of Inertia ............................................................................. 62
4.2.2 General Form of the Euler Equations for the Compass .................................................... 63
4.2.3 Approximated Solution .................................................................................................... 65
4.2.4 Stability Analysis ............................................................................................................. 69
5 Conclusions ................................................................................................................................... 73
5.1 Future Work ............................................................................................................................ 74
6 References..................................................................................................................................... 75
vi
List of Figures
Figure 1 – Lateral perspective from the 𝑥 axis of a Connected Scissor Mechanism of 6 bars of same
length, with a configuration 𝛼1 on the left and a configuration 𝛼2 at the right, being 𝛼2 > 𝛼1. .................. 32
Figure 2 – Top view from the 𝑧 axis of a Connected Scissor Mechanism with 𝑛 = 3 where the angle 𝜙 and
distances ℎ and 𝑙𝑐𝑜𝑠𝛼 are represented. ................................................................................................. 33
Figure 3 – Connected Scissor Mechanism with 6 bars, 3 masses 𝑚1 at the center of the bars, displayed in
red, and 6 masses 𝑚2 at the ends of the bars, displayed in blue. ........................................................... 38
Figure 4 – Octahedron at a configuration 𝛼, where the body frame is displayed as well as the angle of 2𝛼
between two opposite bars. .................................................................................................................... 41
Figure 5 – Dandelion with 𝑛 = 4 bars at a configuration of 𝛼 and the body frame centered at the center of
mass for this particular configuration. ..................................................................................................... 43
Figure 6 – Eight Point Masses in space. The trajectories of the masses so that 𝐴 and 𝐵 are kept constant
are visible in dashed curves. .................................................................................................................. 46
Figure 7 – Compass for two different configurations, 𝛼2 and 𝛼1, with 𝛼2 > 𝛼1.The body frame is centered
at the center of mass.............................................................................................................................. 51
Figure 8 - Minimum 𝛼0 as function of the eccentricity of the orbit for different number of bars. ................ 58
Figure 9 – Configuration as function of time 𝛼(𝑡) during two sidereal days in geosynchronous orbit with
𝑒 = 0.1 for different parameters. ............................................................................................................. 59
Figure 10 - Configuration as function of time 𝛼(𝑡) during two sidereal days in geosynchronous orbit with
𝑒 = 0.01 for different parameters. ........................................................................................................... 59
Figure 11 - 𝐴/𝐶 and 𝐵/𝐶 ratios as function of the configuration 𝛼 for the Compass. ................................ 62
Figure 12 – Perspectives from the 𝜔𝑥 axis of the ellipsoids of the kinetic energy, at darker orange, and of
the angular momentum, at brighter orange, where the curve of the intersection of the ellipsoids is
displayed in blue. The configuration at the left is of 𝛼0 and at the right is of 𝛼. ........................................ 69
vii
List of Tables
Table 1- Sign of variables depending on the nutation angle and the body’s oblateness. .......................... 30
Table 2 – First possible values for 𝜙. ...................................................................................................... 39
Table 3 - Configurations of stability for the Compass .............................................................................. 71
viii
1
1 INTRODUCTION
1.1 OBJECTIVE
Control of the rotational motion of a satellite is often necessary to fulfill the mission requirements. In this
work, we explore the possibility of a satellite with variable geometry and moments of inertia that can vary
according to a control law determined by the user. Some configurations are defined and some possibilities
are explored.
1.2 ATTITUDE CONTROL OF SPACECRAFT
Frequently, controlling the satellite’s attitude is of critical importance. For example, it is fundamental for a
communication satellite to have its antennas correctly oriented. Other examples are spacecraft with solar
panels that are required to be facing the sun so that energy absorption can be optimized or a reconnaissance
satellite whose camera should be pointed to the target. Spacecraft attitude control is also necessary
because satellites are subject to space environmental torques. Examples of this are gravity gradient,
aerodynamic, magnetic or solar radiation pressure torques [1]. Even though the space industry did only
appear on the XX century, similar and simpler mechanical problems, such as the dynamics of tops, had
been solved before.
The first exact analytic solutions for rigid body dynamics problems were obtained around the XIX century.
Firstly, both the dynamics and kinematics of a torque-free rigid body, also called Euler-Poinsot problem,
were solved [2,3]. Classical problems concerning the dynamics and kinematics of tops with different
relations between their principal moments of inertia were solved taking into account the gravity torque [4,5].
Methods for obtaining the solution of the rotational motion of rigid bodies have been used in a broad range
of mechanical applications, including spacecraft control.
The control of the attitude of a spacecraft can be accomplished by several conventional ways, using
mechanisms such as momentum wheels or thrusters [6]. Momentum wheels have an electric motor attached
to a flywheel in such a way that the rotation speed of the wheel can be controlled by computer. This type of
control is precise and requires small amounts of energy, but it does not allow for large angle maneuvers or
changing the angular momentum. On the other hand, thrusters are able to vary the angular momentum and
can perform large changes on the satellite’s orientation. Thrusters can perform small attitude changes as
well, being a more versatile control mechanism. However, the facts of being less precise in comparison with
other control methods and of requiring expenditure of fuel, which can’t be replenished during the mission,
are significant drawbacks [7].
Other alternative control methods include the use of solar sails or magnetic torques [7], as well as the
alternative explored in this work, which is the variable-geometry concept.
2
1.3 SATELLITES WITH VARIABLE-GEOMETRY
Configuration change as a technique to achieve physical equilibrium is well-known in sports. For example,
a ballet dancer or an ice skater that extends his arms towards or away from the body is using the non-rigid
properties of his body by changing its configuration. This technique might be executed in order to perform
a certain motion or to reach an equilibrium condition, for example. This concept can be transferred to the
problem of the attitude control of a satellite.
The change in configuration of a satellite could be performed using a controlled motion of its movable parts.
The movable parts, which could be solar panels, hardware storage containers or any other spacecraft
component that could be convenient to use for this purpose, would be articulated with other components of
the satellite that are fixed in the satellite’s frame. The fixed parts only have rigid-body motion, unlike the
movable parts that also experience motion controlled by the actuators of the spacecraft.
1.4 CURRENT PROBLEMS OF ATTITUDE DYNAMICS
There are many interesting problems related to attitude dynamics. The dynamic problems’ solutions are
more difficult to obtain when the body is not simple, when it has flexible appendages or moving fluids for
example, or when the external forces and torques are complicated functions.
Multi-body mechanics addresses the dynamical behavior of the degrees of freedom of a certain number of
interconnected parts that make up a whole system. Examples of multi-body structures are a tree
configuration [8] or a closed-loop structure [9].
The problem of the multi-body’s dynamics in space environment can be solved by the Lagrangian or by the
Eulerian approaches [10]. The Lagrangian approach is conceptually simple, since no compatibility equations
need to be satisfied, provided that the motion parameters selected are consistent with the compatibility
conditions. However, the computational implementation of the Lagrangian approach is prone to
programming errors because it tends to be computationally heavy [10]. Alternatively, in the Eulerian
approach, the dynamics of each individual part are considered and the compatibility conditions have to be
taken into account. An advantage of the Eulerian approach is that the equations have clearer physical
meaning than in the Lagrangian approach [10].
A particular case of connected rigid bodies is the dual-spin satellite. It is composed of a platform and a rotor
connected to each other, the rotor being able to rotate with respect to the platform. The platform and the
rotor spin around a common and fixed axis, in normal operating conditions [11]. The attitude stability of such
configuration has been analyzed [12], while an exact analytical solution is obtained when both the platform
and the rotor are asymmetric [13]. Recently, other similar problems have been addressed, such as the dual-
spin satellite with a spring-mass damper [14].
3
The problem of rigid-body attitude when time-varying torques are acting on the body has been considered
and approximate analytical solutions were obtained [15,16]. Many solutions for rigid-body dynamics
obtained on the literature rely on linearization techniques [17] and are only applicable for small orientation
changes. However, rigid-body dynamics were solved for large angle excursions from an initial orientation
using numerical simulations and the developed large-angle theory [17]. Other addressed problems were the
dynamics of a slightly asymmetric rigid body [18], and a fully asymmetric rigid body [19], when subjected to
constant torques in three axis. More recently, other exact analytical solutions have been found. For example,
the problem of a rigid body with a spherical ellipsoid of inertia subject to a constant torque was solved using
a parameterization of the rotation by complex numbers [20]. The solution for the dynamics of an
axisymmetric rigid body subject to torques in various initial conditions was also obtained [21]. Other
important issue is to find the time-optimal maneuver that takes the satellite from one orientation to another
[22,23].
The introduction of flexibility in the dynamics of the satellite increases the difficulty of solving the motion
equations. However, more accurate results could be expected from taking into account elastic behavior of
satellite parts such as antennas. Examples of such flexibility problems are: (i) the control of orbiting flexible
structure using a general Lagrangian formulation [24]; (ii) the dynamics of a dumbbell shaped spacecraft
that consists in two mass particles connected by an elastic link [25]; and (iii) the motion stability of a force-
free satellite with attached elastic appendages [26].
Flexibility and the variable-geometry explored in this work are different concepts. They differ in the fact that
the former is usually a consequence of the natural properties of the materials that one tries to cope with
while the latter is a technique to control the attitude of the spacecraft.
1.4.1 Variable-Structure
The consequences of the variation of inertia properties have been addressed before. A parametric study
has been performed to evaluate the influence of many properties, including the moments of inertia, and
parameters of the satellite on the angular momentum vector [27]. However, this study does not consider
any time-varying moments of inertia. Another problem with varying inertia properties is the dynamics of a
body that grows due to accretion of mass while maintaining rigid body motion [28]. The motion of a spinning
rocket that is ejecting mass through a cluster of nozzles was also addressed [29].
One of the methods of variable-structure control is sliding-mode control. This consists of multiple control
structures that can slide over the boundaries of other control structures. The control law of such motion is
not continuous in time since the input is modeled as a consequence of a discontinuous signal that takes the
system from a first continuous structure to a final continuous structure by a discontinuous transition [30].
Even though it is a variable-structure method used for attitude control, sliding-mode control is not a shape-
changing method of the spacecraft as studied at the present work. The sliding-mode technique has been
addressed in the context of large-angle maneuvers [31] and as a method to perform multi-axial maneuvers
[32]. Sliding-mode control can be also used to suppress vibrations on spacecraft [33].
4
The chaotic behavior of the motion of bodies with one time-varying principal moment of inertia was
addressed [34,35]. Unlike the present work, these works do not refer the spacecraft design that would allow
for such a variation of the moment of inertia or consider consequences of varying one moment of inertia in
other components of the inertia tensor. Other limitation is that the variation of one of the principal inertia
moments is considered as a mode of representing the satellite as a deformable rigid-body and not as a
method to control the satellite’s attitude. In addition, the chaos phenomena in gyrostats [36] and the chaotic
modes of motion of an unbalanced gyrostat [37] have been addressed.
Other related topic concerns the dynamics of gyrostats with variable inertia properties, namely the
consequences in the phase space of such dynamic, without referring to the issue of chaotic behavior. The
changes in angular motion due to variable mass and variable inertia of gyrostat composed by multiple
coaxial bodies have already been investigated [38]. In addition, the free motion of a dual-spin satellite
composed of a platform and a rotor with variable moments of inertia has been addressed [39]. The present
work differs from these studies in the geometry studied, since they only consider the particular case of non-
variable geometry gyrostats.
The possibility of varying mass, via jets or thrusters, has been considered [29]. However, the interpretation
of additional terms that appear in the Euler equations due to the configuration variation is limited. The kinetic
energy of a variable-geometry body that is addressed in this work is not discussed in [29].
Other works consider two-tethered systems modelled as a variable length thin rod connecting two endpoint
masses. It has been proven that a length function for a tethered system in a Moon-Earth system exists so
that the rod has a constant angle with respect to the Moon-Earth direction [40]. A length function for an
uniform rotation for a two-tethered body in an elliptic orbit was obtained, as well as the motion stability
intervals for parameters such as eccentricity [41]. In contrast with the present work, where a configuration
function is obtained for a three-dimensional satellite made of an arbitrary number of bars to attain
synchronous rotation in an elliptic orbit, the geometry discussed in [41] is two-dimensional, belonging to the
orbital plane during the entire orbit and being composed of a single bar.
5
2 ATTITUDE DYNAMICS OF A VARIABLE GEOMETRY BODY
2.1 ANGULAR MOMENTUM
In this section we will follow Thomson [29] closely. We consider a generic variable-geometry body Ω. The
system of mass 𝑚 consists of 𝑛 particles of mass 𝑑𝑚 with a certain motion within some specified
boundaries. The mass of the body 𝑚 is the sum of all the particle’s respective masses
𝑚 = ∫𝑑𝑚Ω
. (2.1)
Two different frames are defined: 𝑋𝑌𝑍 is an inertial frame and the 𝑥𝑦𝑧 is a body frame with an arbitrary
origin 0.
The time-derivatives depend whether the time-derivative is determined with respect to the inertial frame or
with respect to the body frame. The relationship between the two derivatives is given by Basic Kinematic
Equation or Transport Equation [11] in its general form,
�� ≡𝑑
𝑑𝑡
𝐼
(𝑨) =𝑑
𝑑𝑡
𝐵
(𝑨) +𝝎𝐵/𝐼 × 𝑨, (2.2)
where 𝑨 is an arbitrary vector and 𝝎𝐵/𝐼 , or 𝝎 as it will be named further on, is the angular velocity of the
body-frame with respect to the inertial frame. The notation of 𝑑
𝑑𝑡
𝐼 , which is equivalent to a dot, and
𝑑
𝑑𝑡
𝐵
refer to time-derivatives in the inertial and in the body frame, respectively. The angular velocity is arbitrary,
because the motion of the body frame with respect to the inertial frame is not specified at this point.
The position of a certain particle with respect to the inertial frame is given by
𝑹 = 𝑹𝟎 + 𝒓 (2.3)
where 𝒓 is the position of that particle with respect to the body frame and 𝑹𝟎 is the position of the origin of
the body frame.
The position of center of mass with respect to the body frame is
𝒓 =
1
𝑚∫𝒓𝑑𝑚Ω
, (2.4)
while the velocity of the center of mass measured in the inertial frame is
�� =
1
𝑚∫ ��𝑑𝑚Ω
. (2.5)
6
The total angular momentum of the body about the arbitrary body frame origin 0 measured in the inertial
frame, since �� is the velocity in the inertial frame of one particle, is
𝒉𝟎 = ∫𝒓 × ��𝑑𝑚
Ω
. (2.6)
By differentiating (2.6) with respect to the inertial frame and using both the definition of total torque
𝑴𝟎 =
1
𝑚∫𝒓× 𝑭𝑑𝑚Ω
(2.7)
and Newton’s Second Law it is possible to obtain [29]
𝑴𝟎 = ��𝟎 + ��𝟎 ×𝑚��. (2.8)
2.2 EQUATIONS OF ROTATIONAL MOTION FOR A VARIABLE-GEOMETRY BODY
With the objective of obtaining angular momentum equations that include the inertia and mass distribution
properties, we first write the general expression for the acceleration in the inertial frame [29]
�� = ��0 + �� × 𝒓 +𝝎× (𝝎× 𝒓) + 2𝝎×𝑑𝒓
𝑑𝑡
𝐵
+𝑑2𝒓
𝑑𝑡2
𝐵
. (2.9)
Equation (2.7) can be rearranged into
𝑴𝟎 = ∫𝒓 × ��𝑑𝑚
Ω
, (2.10)
using Newton’s Second Law. Then, (2.9) can be substituted into (2.10) yielding
𝑴𝟎 = −��𝟎 ×𝑚�� +∫𝒓× (�� × 𝒓)𝑑𝑚Ω
+∫𝒓× (𝝎× (𝝎 × 𝒓))𝑑𝑚Ω
+2∫ 𝒓× (𝝎 ×𝑑𝒓
𝑑𝑡
𝐵
)𝑑𝑚Ω
+∫ 𝒓×𝑑2𝒓
𝑑𝑡2
𝐵
𝑑𝑚Ω
,
(2.11)
where the first term of the right-hand side, ∫ 𝒓 × ��𝟎𝑑𝑚Ω, was replaced by −��𝟎 ×𝑚�� [29].
The first term of (2.11) involves the acceleration of the origin of the body axes ��𝟎 and the position of the
center of mass relative to the origin of the body frame ��. It will be zero whenever one of three circumstances
occur: (i) the origin of the body axes is coincident with the origin of the inertial frame; (ii) ��𝟎 and �� are
parallel, implying ��𝟎 × �� = 0; (iii) the origin of the body frame coincides with the center of mass of the
variable-geometry body.
7
The rotational inertia term 𝑰 ∘ 𝝎 , written in vectorial notation [29], is given by
𝑰 ∘ 𝝎 = ∫ 𝒓 × (𝝎× 𝒓)𝑑𝑚𝛀
. (2.12)
The inertia tensor 𝑰, written in indicial notation, is
𝐼𝑖𝑗 = ∫ (𝛿𝑖𝑗𝑟2 − 𝑟𝑖𝑟𝑗)𝑑𝑚
𝛀
. (2.13)
The second and third terms of (2.11) can be rearranged to incorporate this definition [29], yielding
𝑴𝟎 = 𝑰 ∘ �� +𝝎 × (𝑰 ∘ 𝝎) + 2∫ 𝒓 × (𝝎×
𝑑𝒓
𝑑𝑡
𝐵
)𝑑𝑚Ω
+∫ 𝒓×𝑑2𝒓
𝑑𝑡2
𝐵
𝑑𝑚Ω
. (2.14)
Expression (2.14) is the general form of the Euler Equations, when written in vectorial notation. The first
two-terms on the right-hand side of (2.14) are present in the conventional Euler Equations because they are
related to the rigid-body motion.
2.3 GENERAL FORM OF EULER EQUATIONS
The third and fourth terms of the right-hand side of (2.14) account for non-rigid part of the body motion
because they contain 𝑑𝒓
𝑑𝑡
𝐵 and
𝑑2𝒓
𝑑𝑡2
𝐵
in their expressions. These terms would be zero if the body was purely
rigid and if the selected body frame would have the same rotational motion of the body. We define
𝑽𝑮𝑻 = 2∫ 𝒓 × (𝝎 ×
𝑑𝒓
𝑑𝑡
𝐵
)𝑑𝑚Ω
+∫ 𝒓 ×𝑑2𝒓
𝑑𝑡2
𝐵
𝑑𝑚Ω
, (2.15)
where 𝑽𝑮𝑻 stands for variable-geometry terms.
So that more physical interpretation can be obtained from 𝑽𝑮𝑻, we add and subtract ∫𝑑𝒓
𝑑𝑡
𝐵× (𝝎 × 𝒓)𝑑𝑚
𝛀
to the right-hand side of (2.15):
𝑽𝑮𝑻 = ∫
𝑑𝒓
𝑑𝑡
𝐵
× (𝝎 × 𝒓)𝑑𝑚𝛀
−∫𝑑𝒓
𝑑𝑡
𝐵
× (𝝎 × 𝒓)𝑑𝑚𝛀
+∫ 𝒓 × (𝝎 ×𝑑𝒓
𝑑𝑡
𝐵
)𝑑𝑚𝛀
+∫ 𝒓× (𝝎 ×𝑑𝒓
𝑑𝑡
𝐵
)𝑑𝑚𝛀
+∫ 𝒓×𝑑2𝒓
𝑑𝑡2
𝐵
𝑑𝑚𝛀
.
(2.16)
The first and third terms of the right-hand side of (2.16) will be rearranged separately. Using the vectorial
identity [42]
8
𝑨 × (𝑩 × 𝑪) = 𝑩(𝑨.𝑪) − 𝑪(𝑨. 𝑩), (2.17)
on the sum of the integrand of those two terms, in indicial notation, and rearranging, yields
𝜔𝑖 (2𝑟𝑗
𝑑𝑟𝑗𝑑𝑡
𝐵
)− 𝜔𝑗 (𝑟𝑖𝑑𝑟𝑗𝑑𝑡
𝐵
+ 𝑟𝑗𝑑𝑟𝑖𝑑𝑡
𝐵
). (2.18)
It is possible to simplify (2.18) to
𝜔𝑗
𝑑
𝑑𝑡
𝐵
(𝛿𝑖𝑗𝑟2 − 𝑟𝑖𝑟𝑗), (2.19)
using
2𝑟𝑗
𝑑𝑟𝑗𝑑𝑡
𝐵
=𝑑(𝑟2)
𝑑𝑡
𝐵
, (2.20)
𝑟𝑖
𝑑𝑟𝑗𝑑𝑡
𝐵
+ 𝑟𝑗𝑑𝑟𝑖𝑑𝑡
𝐵
=𝑑(𝑟𝑖𝑟𝑗)
𝑑𝑡
𝐵
(2.21)
and
𝜔𝑖 = 𝛿𝑖𝑗𝜔𝑗 . (2.22)
After inserting the integrand (2.19) in the body integral and taking the derivative outside of the integral, the
derivative of the inertia tensor in indicial notation (2.13) is obtained.
𝜔𝑗
𝑑
𝑑𝑡
𝐵
(∫ (𝛿𝑖𝑗𝑟2− 𝑟𝑖𝑟𝑗)𝑑𝑚
𝛀
) =𝑑𝐼𝑖𝑗𝑑𝑡
𝐵
𝜔𝑗 . (2.23)
This term does not appear in the usual Euler Equations because, for a rigid body, the mass distribution does
not change with time with respect to the body frame. However, for a variable-geometry body, there is no
reference frame where the mass distribution is constant, due to the changes in configuration.
At this point, after including the result (2.23) in vectorial notation, 𝑑𝑰
𝑑𝑡
𝐵∘ 𝝎, the variable geometry terms are
given by
9
𝑽𝑮𝑻 =
𝑑𝑰
𝑑𝑡
𝐵
∘ 𝝎+ ∫ (𝝎 × 𝒓) ×𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
+∫ 𝒓× (𝝎×𝑑𝒓
𝑑𝑡
𝐵
)𝑑𝑚𝛀
+∫ 𝒓 ×𝑑2𝒓
𝑑𝑡2
𝐵
𝑑𝑚𝛀
,
(2.24)
where the sign of the second term was changed using a property of the external product [42]. The last three
terms of the right-hand side of (2.24) can still be rearranged, which is not performed in Thomson [29]. This
can be done by firstly performing the time-derivative in the inertial-frame of the expression
∫ 𝒓 ×
𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
, (2.25)
which, using the Basic Kinematic Equation or Transport Equation (2.2), yields
∫
𝑑𝒓
𝑑𝑡
𝐵
×𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
+∫ (𝝎× 𝒓) ×𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
+∫ 𝒓 ×𝑑2𝒓
𝑑𝑡2
𝐵
𝑑𝑚𝛀
+∫ 𝒓× (𝝎×𝑑𝒓
𝑑𝑡
𝐵
)𝑑𝑚𝛀
.
(2.26)
Since𝑑𝒓
𝑑𝑡
𝐵×
𝑑𝒓
𝑑𝑡
𝐵= 0 , expression (2.26) becomes the same sum that is present in (2.24). For that reason we
can now replace (2.26) into (2.24), obtaining
𝑽𝑮𝑻 =
𝑑𝑰
𝑑𝑡
𝐵
∘ 𝝎 + ��, (2.27)
where we simplified the notation concerning the second non-rigid term by defining:
𝜸 = ∫ 𝒓 ×
𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
. (2.28)
With the variable-geometry terms 𝑽𝑮𝑻 obtained, we can substitute (2.27) and (2.15) into (2.14) and write
the general form of the Euler equations, using a body frame centered at the center of mass, for the variable-
geometry body:
𝑴𝟎 = 𝑰 ∘ �� +𝝎 × (𝑰 ∘ 𝝎) +
𝑑
𝑑𝑡
𝐵
(𝑰) ∘ 𝝎 + ��. (2.29)
In the case of a rigid body, (2.29) would simplify to the usual Euler Equations and the last two terms of the
right-hand side of (2.29) would vanish from the equation, because for a rigid body 𝑑
𝑑𝑡𝑰
𝐵= 0 and
𝑑𝒓
𝑑𝑡
𝐵= 0.
10
2.4 ALTERNATIVE DERIVATION
The general form of the Euler equations (2.29) can be derived through a different path. The velocity in the
inertial frame �� is [29]
�� = ��𝟎 +𝝎 × 𝒓+
𝑑𝒓
𝑑𝑡
𝐵
, (2.30)
which can be replaced in the expression of angular momentum (2.6), yielding
𝒉𝟎 = ∫𝒓× ��0𝑑𝑚
Ω
+∫𝒓× (𝝎 × 𝒓)𝑑𝑚Ω
+∫ 𝒓×𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚. (2.31)
The first term of the right-hand side of (2.31) can be rearranged using the definition of center of mass (2.1),
while the second term is the rotational term (2.12). The general expression for the angular momentum of a
variable-geometry body is:
𝒉𝟎 = −��𝟎 ×𝑚𝒓 + 𝑰 ∘ 𝝎+ 𝜸 (2.32)
After substituting (2.32) into (2.8), we obtain
𝑴𝟎 =
𝑑
𝑑𝑡
𝐼
(−��𝟎 × 𝑚𝒓 + 𝑰 ∘ 𝝎 +∫ 𝒓×𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
) + ��𝟎 ×𝑚��
= −��𝟎 × 𝑚�� − ��𝟎 ×𝑚�� +𝑑
𝑑𝑡
𝐼
(𝑰 ∘ 𝝎) +𝑑
𝑑𝑡
𝐼
(∫ 𝒓 ×𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
) + ��𝟎 ×𝑚��
= −��𝟎 × 𝑚�� +𝑑
𝑑𝑡
𝐼
(𝑰 ∘ 𝝎) +𝑑
𝑑𝑡
𝐼
(∫ 𝒓 ×𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
).
(2.33)
The term −��𝟎 ×𝑚�� is present again, as it was present in (2.11), and vanishes when the body frame is
centered at the center of mass, which is the case we are considering. The term 𝑑
𝑑𝑡
𝐼(𝑰 ∘ 𝝎) can be developed
using (2.2), leading to
𝑑
𝑑𝑡
𝐵
(𝑰 ∘ 𝝎) +𝝎 × (𝑰 ∘ 𝝎) =𝑑
𝑑𝑡
𝐵
(𝑰) ∘ 𝝎 + 𝑰 ∘𝑑
𝑑𝑡
𝐵
(𝝎) + 𝝎× (𝑰 ∘ 𝝎), (2.34)
which can be replaced in (2.33), having 𝑑
𝑑𝑡
𝐵(𝝎) = ��, because the derivative of the angular velocity is the
same on both the frames. Using the definition (2.28) again on the last term of the right-hand side of (2.33),
the same general form of the Euler equations for a body frame centered at the center of mass is obtained:
11
𝑴𝟎 = 𝑰 ∘ �� +𝝎 × (𝑰 ∘ 𝝎) +
𝑑
𝑑𝑡
𝐵
(𝑰) ∘ 𝝎 + ��. (2.35)
Result (2.35) confirms (2.29).
2.5 GENERAL EULER EQUATIONS IN THE PRINCIPAL FRAME
For a rigid body, the selected moving frame is usually one that has the same motion of the body, so that the
inertia tensor remains constant and the equations simplify. The orientation of the axes of this body frame
can be selected so that the inertia tensor is diagonal, which is called a principal frame. This is always
possible because the inertia tensor is positive-definite and symmetric.
In such conditions, when the body frame has the same motion of the body and this body frame is a principal
frame, the orientation of the principal axes with respect to the body remain constant because the body’s
mass distribution and hence its inertia tensor 𝑰 remains unchanged.
However, for a general variable-geometry body, the behavior of the principal frame with respect to the body
is different: a variable-geometry body with a non-rigid motion, which principal frame has a certain orientation
at a certain point in time 𝑡, does not necessarily have the same orientation, relatively to the body, at time
𝑡 + Δ𝑡, with Δ𝑡 > 0. The reason for this is that the configuration change between 𝑡 and 𝑡 + Δ𝑡 in the body
frame generally changes the mass distribution of the body in such a way that the former principal frame, at
𝑡, is altered.
Since there is no reference frame with respect to which a variable-geometry body mass distribution is
constant during a configuration change, the body frame selected should be as convenient as possible. In
this work, we selected the principal frame to be the body frame, so the non-diagonal terms are zero.
The set of vector-base of the principal frame is defined as (𝒆𝑥 , 𝒆𝑦 , 𝒆𝑧), while the set of coordinates is (𝑥, 𝑦, 𝑧).
In a principal frame of a general variable-geometry, at a certain instant of time 𝑡, the inertia tensor in matrix
form is
𝑰 ≡ [𝐴(𝑡) 0 00 𝐵(𝑡) 00 0 𝐶(𝑡)
], (2.36)
where the principal moments of inertia are defined according to:
𝐴(𝑡) = ∫ (𝑦2(𝑡) + 𝑧2(𝑡))𝑑𝑚
Ω
, (2.37)
12
𝐵(𝑡) = ∫ (𝑥2(𝑡) + 𝑧2(𝑡))𝑑𝑚
Ω
, (2.38)
𝐶(𝑡) = ∫ (𝑥2(𝑡) + 𝑦2(𝑡))𝑑𝑚
Ω
. (2.39)
Following the same notation, the time-derivative of the inertia tensor with respect to the body frame is
𝑑
𝑑𝑡
𝐵
(𝑰) ≡ [�� 0 00 �� 00 0 ��
], (2.40)
whereas its components are defined by:
��(𝑡) =
𝑑
𝑑𝑡
𝐵
∫ (𝑦2(𝑡) + 𝑧2(𝑡))𝑑𝑚Ω
, (2.41)
��(𝑡) =
𝑑
𝑑𝑡
𝐵
∫ (𝑥2(𝑡) + 𝑧2(𝑡))𝑑𝑚Ω
, (2.42)
��(𝑡) =
𝑑
𝑑𝑡
𝐵
∫ (𝑥2(𝑡) + 𝑦2(𝑡))𝑑𝑚Ω
. (2.43)
The total applied torque 𝑴𝟎 is written as
𝑴𝟎 = 𝑀𝑥𝒆𝒙 +𝑀𝑦𝒆𝒚 +𝑀𝑧𝒆𝒛, (2.44)
the angular velocity 𝝎 is
𝝎 = 𝜔𝑥𝒆𝒙 +𝜔𝑦𝒆𝒚 +𝜔𝑧𝒆𝒛 (2.45)
and the angular acceleration �� is defined as
�� = ��𝑥𝒆𝒙 + ��𝑦𝒆𝒚 + ��𝑧𝒆𝒛. (2.46)
The derivative of 𝜸 with respect to the inertial frame is given by
�� =𝑑
𝑑𝑡
𝐼
(∫ 𝒓 ×𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
) = ��𝑥𝒆𝒙 + ��𝑦𝒆𝒚 + ��𝑧𝒆𝒛 + 𝛾𝑥��𝒙 + 𝛾𝑦��𝒚 + 𝛾𝑧��𝒛. (2.47)
13
The time-derivatives in the inertial frame of the base-vectors of the body frame are determined using (2.2),
considering that they are constant in the body-frame, which yields:
��𝑥 = 𝜔𝑧𝒆𝒚 −𝜔𝑦𝒆𝒛, (2.48)
��𝑦 = 𝜔𝑥𝒆𝒛 −𝜔𝑧𝒆𝒙, (2.49)
��𝑧 = 𝜔𝑦𝒆𝒙 −𝜔𝑥𝒆𝒚. (2.50)
After replacing (2.48)-(2.50) in (2.47), we obtain
�� = (��𝑥 + 𝛾𝑧𝜔𝑦 − 𝛾𝑦𝜔𝑧)𝒆𝒙 + (��𝑦 + 𝛾𝑥𝜔𝑧 − 𝛾𝑧𝜔𝑥)𝒆𝒚 + (��𝑧 + 𝛾𝑦𝜔𝑥 − 𝛾𝑥𝜔𝑦)𝒆𝒛. (2.51)
Having every term of the general Euler equations (2.35) written in terms of three body frame coordinates,
we write it in vectorial form:
[
𝑀𝑥
𝑀𝑦
𝑀𝑧
] = [𝐴 0 00 𝐵 00 0 𝐶
] . [
��𝑥��𝑦��𝑧
] + [
𝜔𝑥𝜔𝑦𝜔𝑧
] × ([𝐴 0 00 𝐵 00 0 𝐶
] . [
𝜔𝑥𝜔𝑦𝜔𝑧
]) + [�� 0 00 �� 00 0 ��
] . [
𝜔𝑥𝜔𝑦𝜔𝑧
]
+ [
��𝑥��𝑦��𝑧
] + [
𝛾𝑧𝜔𝑦 − 𝛾𝑦𝜔𝑧𝛾𝑥𝜔𝑧 − 𝛾𝑧𝜔𝑥𝛾𝑦𝜔𝑥 − 𝛾𝑥𝜔𝑦
].
(2.52)
Expression (2.52) is written in the three principal directions as:
𝑀𝑥 = ��𝜔𝑥 +𝐴��𝑥 + 𝜔𝑦𝜔𝑧(𝐶 − 𝐵) + ��𝑥 + 𝛾𝑧𝜔𝑦 − 𝛾𝑦𝜔𝑧 , (2.53)
𝑀𝑦 = ��𝜔𝑦 +𝐵��𝑦 +𝜔𝑥𝜔𝑧(𝐴 − 𝐶) + ��𝑦 + 𝛾𝑥𝜔𝑧 − 𝛾𝑧𝜔𝑥 , (2.54)
𝑀𝑧 = ��𝜔𝑧 + 𝐶��𝑧 + 𝜔𝑥𝜔𝑦(𝐵 − 𝐴) + ��𝑧 + 𝛾𝑦𝜔𝑥 − 𝛾𝑥𝜔𝑦 . (2.55)
Equations (2.53)-(2.55) are valid for a general variable-geometry body when its dynamics are written in a
body frame that is a principal frame and whose origin is coincident with the position of the center of mass.
Euler equations written in a principal frame [29] are given by:
𝑀𝑥 = 𝐴��𝑥 + 𝜔𝑦𝜔𝑧(𝐶 − 𝐵), (2.56)
𝑀𝑦 = 𝐵��𝑦 +𝜔𝑥𝜔𝑧(𝐴 − 𝐶), (2.57)
𝑀𝑧 = 𝐶��𝑧 + 𝜔𝑥𝜔𝑦(𝐵 − 𝐴). (2.58)
14
We observe that the Euler equations are a particular case of the modified Euler Equations. This is so
because when the geometry is constant in the body frame that is centered at the center of mass, we have
that 𝑑𝒓
𝑑𝑡
𝐵= 0. This implies that not only does �� vanish, but
𝑑
𝑑𝑡
𝐵(𝑰) ∘ 𝝎 vanishes as well.
In this work, we will consider the form of the general form of Euler equations (2.53)-(2.55) where the variable-
geometry body is not subject of any applied torques – the satellite is considered a free body in space. In
addition, we will consider geometries where we have 𝜸 = 0 at all times, which significantly simplifies the
equations. The circumstances under which this assumption is valid will be discussed in section 2.6. Taking
this into account, the equations simplify to:
0 = ��𝜔𝑥 +𝐴��𝑥 +𝜔𝑦𝜔𝑧(𝐶 − 𝐵), (2.59)
0 = ��𝜔𝑦 + 𝐵��𝑦 +𝜔𝑥𝜔𝑧(𝐴 − 𝐶), (2.60)
0 = ��𝜔𝑧 + 𝐶��𝑧 +𝜔𝑥𝜔𝑦(𝐵 − 𝐴). (2.61)
2.6 NON-RIGID TERMS
Of the two non-rigid terms of the general Euler equations (2.27), 𝑑
𝑑𝑡
𝐵(𝑰) ∘ 𝝎 expresses the influence of
varying the geometry of the body, and thus the mass distribution in space, in the body dynamics.
The other variable-geometry term that is present in the equations is ��, has no direct relationship with the
inertia tensor 𝑰, unlike the term 𝑑
𝑑𝑡
𝐵(𝑰) ∘ 𝝎. The physical explanation for �� is related with the orientation of
the principal frame. The geometry variation with time, by changing the mass distribution in space, generally
implies that the principal frame is no longer the same. For example, consider a body with a well-defined
principal frame, due to three perpendicular symmetry planes. If this body changes its geometry in a way that
all the symmetries are no longer present, the orientation of the principal frame with respect to the inertial
frame is no longer clear and the dynamics of it are affected by the non-zero term ��. Thus, �� accounts for
the consequence in the dynamics of the variable-geometry body having a non-rigid motion that is changing
the orientation of its principal axes.
2.6.1 Two-plane Symmetry
We study a particular variable-geometry body which geometry variation is such that �� = 0 at all times –
the variable geometry body with two perpendicular symmetry planes at all its configurations, whose
intersection contains the origin of the body frame.
15
Suppose the planes of symmetry are 𝑥0𝑧 and 𝑦0𝑧, without loss of generality. We define that the position of
a general point of the body, of infinitesimal mass 𝑑𝑚, with respect to the body frame is
𝒓1 = 𝑥𝒆𝒙 + 𝑦𝒆𝒚 + 𝑧𝒆𝒛, (2.62)
having a non-rigid velocity, with respect to the body frame, of
𝑑𝒓1𝑑𝑡
𝐵
= 𝑣𝑥𝒆𝒙+ 𝑣𝑦𝒆𝒚 + 𝑣𝑧𝒆𝒛. (2.63)
Since the body has symmetry with respect to the 𝑥0𝑧 and 𝑦0𝑧 planes, the mirrored points of the first
infinitesimal mass with respect to them, with indexes 2 to 4 must be also infinitesimal masses with the same
mass 𝑑𝑚 that belong to the body. Their positions are:
𝑟2 = −𝑥𝒆𝒙 + 𝑦𝒆𝒚 + 𝑧𝒆𝒛, (2.64)
𝑟3 = 𝑥𝒆𝒙 − 𝑦𝒆𝒚 + 𝑧𝒆𝒛 (2.65)
𝑟4 = −𝑥𝒆𝒙 − 𝑦𝒆𝒚 + 𝑧𝒆𝒛. (2.66)
If the motion is symmetric with respect to these planes, then their velocities in the body frame are also
symmetric, so that the symmetry remains unchanged after the motion. For this reason, the velocities of
points 2, 3 and 4 are defined with respect to the velocity of point 1:
𝑑𝒓𝟐𝑑𝑡
𝐵
= −𝑣𝑥𝒆𝒙 + 𝑣𝑦𝒆𝒚 + 𝑣𝑧𝒆𝒛, (2.67)
𝑑𝒓𝟑𝑑𝑡
𝐵
= 𝑣𝑥𝒆𝒙− 𝑣𝑦𝒆𝒚 + 𝑣𝑧𝒆𝒛, (2.68)
𝑑𝒓𝟒𝑑𝑡
𝐵
= −𝑣𝑥𝒆𝒙 − 𝑣𝑦𝒆𝒚 + 𝑣𝑧𝒆𝒛. (2.69)
The total contribution of the 4 points to 𝜸 is
∑𝒓𝑖 ×
𝑑𝒓𝑖𝑑𝑡
𝐵4
𝑖=1
= 𝒓1 ×𝑑𝒓𝟏𝑑𝑡
𝐵
+ 𝒓2 ×𝑑𝒓𝟐𝑑𝑡
𝐵
+ 𝒓3 ×𝑑𝒓𝟑𝑑𝑡
𝐵
+ 𝒓4 ×𝑑𝒓𝟒𝑑𝑡
𝐵
= 0, (2.70)
using (2.62)-(2.69).
Since (2.70) is true for any set of 4 infinitesimal masses of the body,
16
𝜸 = ∫ 𝒓 ×
𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚Ω
= 0. (2.71)
This derivation proved that, for a body that permanently has at least two perpendicular symmetry planes
whose intersection contains the body frame’s origin, 𝜸 = 0 and hence �� = 0 are valid for all configurations.
This result will be used throughout this work to justify 𝜸 = 0 for the proposed configurations.
It is important to note that if two planes of symmetry are constant with respect to the selected body frame,
then they are always perpendicular to their respective principal axes. Thus, the third principal axis is already
defined by being orthogonal to the plane where the other two principal axes lie on. Therefore, we see in this
particular case that �� = 0 is true when the principal axes have a constant orientation with respect to the
selected body frame.
2.7 KINETIC ENERGY OF A VARIABLE-GEOMETRY BODY
The total kinetic energy of the variable-geometry body is obtained by integrating the square of the modulus
of the velocity in the inertial frame along the body [29]:
𝑇 = ∫
1
2|��|
2
Ω
𝑑𝑚 = ∫1
2��. ��
Ω
𝑑𝑚 (2.72)
Inserting expression for �� (2.30) in (2.72) yields
𝑇 =
1
2∫ ��𝟎. ��𝟎𝑑𝑚Ω
+1
2∫ (𝝎× 𝒓). (𝝎 × 𝒓)𝑑𝑚Ω
+∫ ��𝟎. (𝝎 × 𝒓)Ω
𝑑𝑚
+1
2∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 +∫ ��𝟎 .𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 +∫ (𝝎 × 𝒓).𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚.
(2.73)
We will interpret each one of the terms of (2.73) separately. The first term is the term of translational kinetic
energy, which can be rearranged considering the fact that ��𝟎 is constant over the body resulting in
1
2∫ ��𝟎. ��𝟎𝑑𝑚Ω
=1
2𝑚��𝟎
𝟐. (2.74)
The second term of (2.73) can be rearranged using a vectorial property [42] to obtain the rotational kinetic
energy of the body,
1
2∫ (𝝎 × 𝒓). (𝝎 × 𝒓)𝑑𝑚Ω
=1
2𝝎. (𝑰 ∘ 𝝎), (2.75)
and, if the right-hand side of (2.75) is written in a body frame that is a principal frame, we have
17
1
2𝝎. (𝑰 ∘ 𝝎) =
1
2(𝐴𝜔𝑥
2 +𝐵𝜔𝑦2 + 𝐶𝜔𝑧
2). (2.76)
The third term of the right hand side of (2.73) can be rearranged taking into account that ��𝟎 and 𝝎 are
constant over the body and inserting the definition of center of mass (2.4):
∫ ��𝟎. (𝝎 × 𝒓)Ω
𝑑𝑚 = ��𝟎. 𝝎 ×𝑚��. (2.77)
The fact that the origin of the body frame is coincident with the center of mass implies that �� = 0, making
(2.77) zero.
The other three terms of the right-hand side of (2.73) are terms that are related with non-rigid part of the
motion, since they would all be zero when 𝑑𝒓
𝑑𝑡
𝐵= 0. The first three terms would be already present for pure
rigid-body motion. We define the rigid body component of the kinetic energy
𝑇𝑟𝑖𝑔𝑖𝑑 =
1
2𝑚��𝟎
𝟐 +1
2𝝎. (𝑰 ∘ 𝝎), (2.78)
and the non-rigid component of the kinetic energy
𝑇𝑛𝑜𝑛−𝑟𝑖𝑔𝑖𝑑 =
1
2∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 +∫ ��𝟎 .𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 +∫ (𝝎 × 𝒓) .𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚. (2.79)
The third term in the right-hand side of (2.79) can be rearranged using the vectorial property [42]
(𝑨 × 𝑩). 𝑪 = (𝑩 × 𝑪). 𝑨, (2.80)
taking into account that ��𝟎 does not depend on the position and replacing the definition of 𝜸 (2.28), yielding
𝑇𝑛𝑜𝑛−𝑟𝑖𝑔𝑖𝑑 =
1
2∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 + ��𝟎 . ∫𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 +𝝎.𝜸. (2.81)
2.7.1 Non-Rigid Terms
The first term of (2.81) is the pure non-rigid motion term. Its value is the same regardless of the rigid motion
of the body, both translation and rotation: it only depends on the configuration change and hence on the
distribution of the velocity with respect to the body frame 𝑑𝒓
𝑑𝑡
𝐵. This term is never negative because the
vectorial self-dot product is never negative.
18
Also, the rigid terms of the kinetic energy (2.78) and the first of the non-rigid terms on the right-hand side of
(2.81) are all non-negative:
1
2𝑚��𝟎
𝟐 +1
2𝝎.(𝑰 ∘ 𝝎) +
1
2∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 ≥ 0. (2.82)
However, the other two other terms, ��𝟎. ∫𝑑𝒓
𝑑𝑡
𝐵
Ω𝑑𝑚 and 𝝎.𝜸, can be negative, because the sign of the result
of the dot products between these vectors is generally unknown. This contrasts with the terms already
referred in (2.82) that are non-negative. Moreover, the term ��𝟎 . ∫𝑑𝒓
𝑑𝑡
𝐵
Ω𝑑𝑚 displays a quadratic dependence
on velocity, as it is natural in translational kinetic energy.
To interpret ��𝟎. ∫𝑑𝒓
𝑑𝑡
𝐵
Ω𝑑𝑚 and 𝝎.𝜸 we firstly consider how the system and the problem have been defined.
The system of particles of the body has been looked upon as a mechanical system and its kinetic energy is
a component of the mechanical energy of the system. Nonetheless, the nature of the energy source of the
actuators of the system is not mechanical, it might be electrical or chemical for example, even though the
actuators belong to the system. This is the reason why the actuators can change the mechanical energy of
the whole system.
Varying the geometry of the body as a response to internal actuators of the body can either lead to an
increase or a decrease of the mechanical energy of the system, depending on whether the actuators are
performing positive or negative work. This depends on the direction of the forces generated by the actuator
with respect to the displacement of the particles they are acting on.
If the geometry of the body is not varying, then the kinetic energy will not vary either, because in that case
the body behaves like a rigid body. The kinetic energy will vary if external forces are applied do work on the
system. The actuators are no exception, because, from the point of view of the mechanical system, they
are external forces.
The term ��𝟎. ∫𝑑𝒓
𝑑𝑡
𝐵
Ω𝑑𝑚 in the kinetic non-rigid energy expression (2.81) depends on the translational state
of the body and the geometry variation. It will be positive if the projection of 𝑑𝒓
𝑑𝑡
𝐵 onto ��𝟎 has the direction of
��𝟎 and negative if it has the opposite direction. This term is positive if the actuating internal forces are
increasing the translational energy and negative otherwise. It will be 0 if they are perpendicular and no
translational work is being performed to the body.
19
The term 𝝎.𝜸 of the kinetic non-rigid energy expression (2.81) has an equivalent interpretation. 𝜸 is related
with changes in symmetries and accounts for changes in the orientation of the principal axes. In addition, 𝜸
can be understood as the angular momentum of the body measured in the body frame. When the body
geometry is varying, 𝜸 has the direction of the part of the total angular momentum that the body frame
rotation does not account for. Concerning the sign of this term, 𝝎.𝜸 is positive if the projection of 𝜸 onto 𝝎
is in the direction of 𝝎 and negative otherwise. The physical meaning of this is similar as for the first term:
𝝎.𝜸 is positive when the actuating forces are such that the positive rotational work is being done to the
system, speeding up the rotation and increasing its rotational kinetic energy. The negative and zero work
have a similar interpretation as for ��𝟎. ∫𝑑𝒓
𝑑𝑡
𝐵
Ω𝑑𝑚.
2.7.2 Simplification of the Kinetic Energy Expression
The total kinetic energy for a body frame centered in the body’s center of mass is
𝑇 =
1
2𝑚��𝟎
𝟐 +1
2𝝎. (𝑰 ∘ 𝝎) +
1
2∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 + ��𝟎. ∫𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 +𝝎.𝜸. (2.83)
The first term of the right-hand side of (2.83) will be not accounted for because the central point of this work
is attitude control regardless of the translation state of the body.
The fourth term of the right-hand side of (2.83) can be simplified using the definition of center of mass (2.4):
��𝟎. ∫
𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 = ��𝟎.𝑑
𝑑𝑡
𝐵
∫ 𝒓Ω
𝑑𝑚 = ��𝟎.𝑑(𝑚��)
𝑑𝑡
𝐵
. (2.84)
The selected body frame is centered at the center of mass of the body to simplify the equations, implying
�� = 0 and making equation (2.84) zero.
Usually 𝝎.𝜸 is different from zero, but for the variable-geometry bodies studied in this work the symmetries
are such that 𝜸 = 0. Under these assumptions we have that the total kinetic energy (2.83) reduces to
𝑇 =
1
2𝝎. (𝑰 ∘ 𝝎) +
1
2∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚. (2.85)
Using the vectorial notation used so far, as well as (2.75), yields:
2𝑇 = 𝐴𝜔𝑥
2 + 𝐵𝜔𝑦2 + 𝐶𝜔𝑧
2 +∫ ((𝑑𝒓
𝑑𝑡
𝐵
)𝑥
2
+ (𝑑𝒓
𝑑𝑡
𝐵
)𝑦
2
+ (𝑑𝒓
𝑑𝑡
𝐵
)𝑧
2
)Ω
𝑑𝑚 (2.86)
20
In this work, the geometry changing mechanisms are intended to be simple. For that reason several
geometries have a single varying geometry degree of freedom. This is defined generally as 𝛼, that is a
function of time 𝑡. In these conditions, the integrated terms in (2.86) are given by
(𝑑𝒓
𝑑𝑡
𝐵
)𝑖
2
= ��2 (𝑑𝒓
𝑑𝛼
𝐵
)𝑖
2
, (2.87)
with 𝑖 = 𝑥, 𝑦, 𝑧.
Additionally, we can assume that the geometry variation is slow. This is reasonable because, for attitude
maintenance maneuvers of spacecraft in orbits, the necessary attitude corrections are small. So, we have
that ��2 ≪ 1 and that
��2∫ ((
𝑑𝒓
𝑑𝛼
𝐵
)𝑥
2
+ (𝑑𝒓
𝑑𝛼
𝐵
)𝑦
2
+ (𝑑𝒓
𝑑𝛼
𝐵
)𝑧
2
)Ω
𝑑𝑚 (2.88)
is negligible when compared to the other terms of (2.86), which can be approximated by
2𝑇 ≈ 𝐴𝜔𝑥2 +𝐵𝜔𝑦
2 + 𝐶𝜔𝑧2. (2.89)
The result (2.89) for the kinetic energy is the same as the expression of the kinetic energy for a rigid body
[29]. This proves that the kinetic energy of a variable-geometry body that is slowly changing its configuration
can be approximated by the kinetic energy of a rigid body. However, 𝑇 (2.89) is not a constant because the
moments of inertia depend on time.
2.7.3 Time-derivative of the Kinetic Energy
If we differentiate and rearrange (2.89), we obtain:
2�� = 𝜔𝑥(��𝜔𝑥 + 2𝐴��𝑥) + 𝜔𝑦(��𝜔𝑦 + 2𝐵��𝑦) + 𝜔𝑧(��𝜔𝑧 + 2𝐶��𝑧). (2.90)
If the terms inside the parenthesis in (2.90) are rearranged to have
2�� = 𝜔𝑥(��𝜔𝑥 +𝐴��𝑥 + 𝐴��𝑥) + 𝜔𝑦(��𝜔𝑦 +𝐵��𝑦 + 𝐵��𝑦) + 𝜔𝑧(��𝜔𝑧 + 𝐶��𝑧 + 𝐶��𝑧), (2.91)
then the sums ��𝜔𝑥 + 𝐴��𝑥, ��𝜔𝑦 +𝐵��𝑦 and ��𝜔𝑧 + 𝐶��𝑧 can be replaced using equations (2.59)-(2.61), for a
free body whose non-rigid motion is such that 𝜸 = 0. Then, by using the time-derivative of the product on
the right-hand side of (2.91), we obtain
2�� =
1
2
𝑑
𝑑𝑡(𝐴𝜔𝑥
2 + 𝐵𝜔𝑦2 + 𝐶𝜔𝑧
2) −1
2(��𝜔𝑥
2 + ��𝜔𝑦2 + ��𝜔𝑧
2). (2.92)
21
After identifying the kinetic energy as in (2.89) in the first term of the right-hand side of (2.92), we simplify
and obtain a simple expression for �� that does not depend on the components of the angular acceleration:
�� = −
1
2(��𝜔𝑥
2 + ��𝜔𝑦2 + ��𝜔𝑧
2). (2.93)
Even though the square of the angular velocity components are always positive, the signal of ��, �� and ��
are not known, so �� could be positive, zero or negative.
2.8 CONSERVATION OF ANGULAR MOMENTUM
We recall the expression of the applied torque (2.8)
𝑴𝟎 = ��𝟎 +𝑹�� ×𝑚��. (2.94)
Having the body frame’s origin be the center of mass implies �� = 0, simplifies (2.94) to
𝑴𝟎 = ��𝟎 (2.95)
If no torques are acting upon the body, then 𝑴𝟎 = 0. The satellites will be also considered as a free-body in
this work. Therefore:
��𝟎 = 0 (2.96)
We conclude from (2.96) that the angular momentum of a variable geometry free body with respect to his
center of mass is constant.
This result contrasts with the results regarding the kinetic energy. While varying the geometry of a certain
body generally changes its energy, it does not change the angular momentum. The reason for that
difference is that the forces responsible for changing the geometry of the body are internal to the mechanical
system considered but the energy sources responsible for that do not belong to the mechanical system –
they are external. Consequently, the actuators are able to change the system’s mechanical energy but not
its angular momentum.
2.9 GEOMETRICAL SOLUTIONS FOR THE DYNAMICS OF A VARIABLE GEOMETRY BODY
The geometrical solution consists in obtaining information about the rigid body dynamics using a geometrical
interpretation of two equations: kinetic energy equation and the angular momentum equation.
22
2.9.1 Ellipsoid of the Kinetic Energy
A general variable-geometry body that is free from torque and that has the body frame centered in the center
of mass is considered. Under the assumption that the body’s non-rigid motion is two-plane symmetric, 𝜸 =
0 is valid, and not considering the translation, the kinetic energy is given by
𝑇 =
1
2(𝐴𝜔𝑥
𝟐 + 𝐵𝜔𝑦𝟐 + 𝐶𝜔𝑧
𝟐) +1
2∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚. (2.97)
Expression (2.97) can be further rearranged in order to obtain an equation that allows for geometrical
interpretation:
1 =
𝜔𝑥2
2𝑇 − ∫𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐴
+𝜔𝑦2
2𝑇 − ∫𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐵
+𝜔𝑧2
2𝑇 − ∫𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐶
(2.98)
If we consider a geometrical space given by the coordinates (𝜔𝑥 , 𝜔𝑦 , 𝜔𝑧), (2.98) defines the surface of an
ellipsoid with semi-axes
√2𝑇−∫𝑑𝒓
𝑑𝑡
𝐵.𝑑𝒓
𝑑𝑡
𝐵𝑑𝑚
Ω
𝐴, √
2𝑇−∫𝑑𝒓
𝑑𝑡
𝐵.𝑑𝒓
𝑑𝑡
𝐵𝑑𝑚
Ω
𝐵 , √
2𝑇−∫𝑑𝒓
𝑑𝑡
𝐵.𝑑𝒓
𝑑𝑡
𝐵𝑑𝑚
Ω
𝐶
(2.99)
along the directions of the 𝑥, 𝑦 and 𝑧 axes, respectively.
The existence of variable-geometry, represented in the term ∫𝑑𝒓
𝑑𝑡
𝐵 .
𝑑𝒓
𝑑𝑡
𝐵𝑑𝑚
Ω and 𝐴(𝑡), 𝐵(𝑡), 𝐶(𝑡) ,
contributes to a decrease in the length of all three semi-axes of the ellipsoid, relatively to the ellipsoid for a
rigid-body. It also changes its shape because the ratios between the semi-axes are modified when the non-
rigid motion term is not zero.
The equation of the ellipsoid of a variable geometry-body reduces to the rigid-body equation of the ellipsoid,
or Poinsot’s ellipsoid [29], when the variable-geometry term ∫𝑑𝒓
𝑑𝑡
𝐵.
𝑑𝒓
𝑑𝑡
𝐵𝑑𝑚
Ω is zero, since
𝑑𝒓
𝑑𝑡
𝐵= 0.
2.9.2 Ellipsoid of the Angular Momentum
The general expression for the angular momentum (2.31), written in the body frame with 𝜸 = 0 and �� = 0,
is
𝒉𝟎 = 𝐴𝜔𝑥𝒆𝒙 + 𝐵𝜔𝑦𝒆𝒚 + 𝐶𝜔𝑧𝒆𝒛. (2.100)
23
In order to obtain an expression with the angular moment that contains the squared components of the
angular velocity, we determine 𝒉𝟎. 𝒉𝟎:
𝒉𝟎. 𝒉𝟎 = ℎ2 = 𝐴2𝜔𝑥2 + 𝐵2𝜔𝑦
2 + 𝐶2𝜔𝑧2 (2.101)
After rearranging, the expression for the angular momentum ellipsoid is obtained:
1 =
𝜔𝑥2
ℎ𝐴
+𝜔𝑦2
ℎ𝐵
+𝜔𝑧2
ℎ𝐶
. (2.102)
The semi-axes are given by ℎ
𝐴, ℎ
𝐵 and
ℎ
𝐶, in the direction of the 𝑥, 𝑦 and 𝑧 axes respectively. It is important to
notice that, even though ℎ is constant, the shape of the ellipsoid of angular momentum varies, since 𝐴, 𝐵
and 𝐶 vary, changing the length of the semi-axes.
2.9.3 Geometrical Solution
Having both of the ellipsoids defined, the angular velocity of the variable-geometry body must belong to a
curve that is the intersection of both the ellipsoids at each instant of time. This curve is not constant because
𝐴, 𝐵 and 𝐶 vary. Additionally, 𝑇 and ∫𝑑𝒓
𝑑𝑡
𝐵 .
𝑑𝒓
𝑑𝑡
𝐵𝑑𝑚
Ω depend on the geometry variation with time as well.
Only ℎ is constant.
This means that, unlike the geometric solution for a rigid-body, the ellipsoid’s shape depends on time
because the variable-geometry body’s inertia properties vary. Therefore, the angular velocity solution at a
particular instant, that lies on that instant’s ellipsoids intersection, has to lie on the next instant’s ellipsoids
intersection, which is different from the previous one.
We will verify that the intersection of the ellipsoids exists for a variable-geometry body. That intersection
would not exist, geometrically, if either one of two situations occur: (i) the angular momentum ellipsoid is
inside the kinetic energy ellipsoid, which implies that, for each of the directions, the kinetic energy ellipsoid
has larger semi-axes than the angular momentum’s ellipsoid:
√2𝑇 − ∫𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐴>ℎ
𝐴,√2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐵>ℎ
𝐵,√2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐶>ℎ
𝐶;
(2.103)
or (ii) the kinetic energy ellipsoid is inside the angular momentum ellipsoid, hence
24
√2𝑇 − ∫𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐴<ℎ
𝐴 ,√2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐵<ℎ
𝐵 ,√2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐶<ℎ
𝐶 .
(2.104)
These geometric restrictions imply that a certain variable-geometry body can’t have arbitrary 𝑇 and ℎ.
We assume 𝐶 ≥ 𝐵 ≥ 𝐴, which is an assumption that does not imply loss of generality because the derivation
would be equivalent if the relation between the different inertia moments would be different. Angular velocity
components’ ratios are defined:
𝜔𝑥 = 𝑐1𝜔𝑧 , (2.105)
𝜔𝑦 = 𝑐2𝜔𝑧 . (2.106)
Inserting the ratios (2.105)-(2.106) in (2.101) and dividing both sides of the equation by 𝐶2, we obtain
ℎ2
𝐶2= 𝜔𝑧
2 ((𝐴
𝐶)2
𝑐12 + (
𝐵
𝐶)2
𝑐22 + 1). (2.107)
If an equivalent derivation for the kinetic energy is done, we obtain
2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐶= 𝜔𝑧
2 (𝐴
𝐶𝑐12 +
𝐵
𝐶𝑐22 + 1).
(2.108)
Solving (2.107) and (2.108) for 𝜔𝑧2 and taking the square root of both sides to obtain a semi-axes ratio on
the left-hand side yields:
ℎ𝐶
√2𝑇 − ∫𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐶
= √(𝐴𝐶)2
𝑐12 + (
𝐵𝐶)2
𝑐22 + 1
𝐴𝐶 𝑐1
2 +𝐵𝐶 𝑐2
2 + 1
(2.109)
Since 𝐴
𝐶≤ 1 and
𝐵
𝐶≤ 1, the numerator of the fraction of the right-hand side of (2.109) is not larger than the
denominator. So:
ℎ
𝐶≤√2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐶
(2.110)
25
So that an inequality similar to (2.110) is obtained, but including 𝐴 instead of 𝐶, two other angular velocity
components’ ratios are defined:
𝜔𝑦 = 𝑎2𝜔𝑥 , (2.111)
𝜔𝑧 = 𝑎3𝜔𝑥 . (2.112)
Following a similar procedure used on the derivation of (2.110), we obtain
ℎ
𝐴≥√2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐴.
(2.113)
We can also write two inequalities concerning the semi-axes of the ellipsoids, using 𝐶 ≥ 𝐵 ≥ 𝐴:
ℎ
𝐴≥ℎ
𝐶, (2.114)
√2𝑇 − ∫𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐴≥√2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐶.
(2.115)
Finally, combining (2.110) and (2.113)-(2.115) results in
ℎ
𝐶≤√2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐶≤√2𝑇 − ∫
𝑑𝒓𝑑𝑡
𝐵
.𝑑𝒓𝑑𝑡
𝐵
𝑑𝑚Ω
𝐴≤ℎ
𝐴.
(2.116)
We conclude from (2.116) that the larger and smaller ellipsoid axes are axes of the angular momentum
ellipsoid. This means that the ellipsoid of angular momentum is always taper than the ellipsoid of kinetic
energy.
Rearranging (2.110) and (2.113) to obtain two inequalities for 𝑇 and then combining them allows to obtain
the upper and lower limits of 𝑇:
1
2(ℎ2
𝐶+∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚Ω
) ≤ 𝑇 ≤1
2(ℎ2
𝐴+∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚Ω
). (2.117)
When the geometry is varying with time, the kinetic energy is limited by two varying bounds. 𝐴 and 𝐶 are
varying and the non-rigid term, which is generally dependent of time, varies as well. The both upper and
26
lower limits of 𝑇 are increased by 1
2∫
𝑑𝒓
𝑑𝑡
𝐵.
𝑑𝒓
𝑑𝑡
𝐵𝑑𝑚
Ω when compared to the rigid-body limits, that are
1
2
ℎ2
𝐶 and
1
2
ℎ2
𝐴, respectively.
As long as 𝐶 ≥ 𝐴, inequality (2.117) is always be true, because the upper limit is always at least as large as
the lower limit, since the term of configuration change is added on both the left and right-hand sides of the
inequality.
2.10 MODIFIED EULER EQUATIONS IN TERMS OF EULER ANGLES
The motion of the body frame could be defined according to many coordinate transformations with respect
to the inertial frame. In the particular case of satellites, using Euler angles is frequent because that allows
for easy interpretation of the orientation of the body. However, other representations could be used, such
as quaternions or Brian angles [7].
The considered Euler angles are: (i) the precession angle 𝜓; (ii) the nutation angle 𝜃; and (iii) the spin angle
𝜑. The orientation of the body frame is given by a first rotation of 𝜓 with respect to the 𝑍 axis, secondly by
a rotation of 𝜃 with respect to the 𝑥′ axis of the second frame, and finally by a rotation of 𝜑 with respect to
the 𝑧′′ axis of the third frame. The number of apostrophes on the axes is the number of rotations the frame
has experienced. The inertial frame has the orientation the body frame, after performing these three
rotations.
The time-derivatives of the Euler angles correspond to angular velocities: �� is the precession rate; �� is the
nutation rate; �� is the spin rate. If these angular velocities, which are parallel to axes of their respective
frame, are projected into the body frame, the angular velocity 𝝎 components, 𝜔𝑥, 𝜔𝑦 and 𝜔𝑧, are obtained
[29]:
𝜔𝑥 = �� sin𝜃 sin𝜑 + �� cos𝜑 (2.118)
𝜔𝑦 = �� sin 𝜃 cos𝜑 − �� sin𝜑 (2.119)
𝜔𝑧 = �� cos𝜃 + �� (2.120)
The general form of the Euler Equations can be written using Euler angles, in a body frame that is a principal
frame centered in the center of mass. This is done by substituting the components of the angular velocity
and of the angular acceleration into (2.53)-(2.55). The components of the angular acceleration are
determined by differentiating (2.118)-(2.120).
27
The equations obtained after doing this are a set 3 coupled non-linear second order equations. Some
simplifications occur if it is assumed: (i) the body is free from torques, 𝑴 = 0; (ii) two plane-symmetry
simplification with 𝜸 = 0; and (iii) axisymmetry with 𝐴 = 𝐵.
However, the equations still remain a set of three coupled non-linear second order equations.
2.10.1 Intermediate Frame for an Axisymmetric Body
The intermediate frame is obtained after the rotation of 𝜓 and then of 𝜃. Therefore, the rotation of the
intermediate frame with respect to the inertial frame is defined as
𝛀 = 𝝎− ��. (2.121)
In this frame, the equations are simpler because the expressions for the angular velocities in terms of the
Euler angles simplify. The system of axes of the intermediate frame is 𝑥′′𝑦′′𝑧′′. In addition, using the
intermediate frame is particularly useful when the body is axisymmetric. For a non-axisymmetric body, the
time-function of 𝐴 and 𝐵 would depend not only on the variation of the body’s geometry but on the
orientation of the body frame at each instant, being a major inconvenience. For an axisymmetric body, the
spin angle of the body frame does not change its inertia tensor.
The alternative derivation of the general form of the Euler equations allows choosing any auxiliary rotation
with the respect to the inertial frame when the time-derivation is performed. In (2.33), the auxiliary reference
frame used to calculate ��𝟎 in the applied torque equation (2.8) was the body frame. In this case, the auxiliary
reference frame to be used will be the intermediate frame. Thus, the Basic Kinematic Equation becomes
𝑑
𝑑𝑡
𝐼
𝑨 =𝑑
𝑑𝑡
𝑀
𝑨 +𝛀 × 𝑨, (2.122)
where 𝑑
𝑑𝑡
𝑀 is the time-derivative measured in the intermediate frame. Inserting the expression of the angular
momentum (2.32) into the expression of the applied torque (2.8) and rearranging results in
𝑴𝟎 =
𝑑
𝑑𝑡
𝐼
(𝑰𝝎) +𝑑
𝑑𝑡
𝐼
(∫ 𝒓 ×𝑑𝒓
𝑑𝑡
𝐵
𝑑𝑚𝛀
)
=𝑑
𝑑𝑡
𝑀
(𝑰) ∘ 𝝎+ 𝑰 ∘𝑑
𝑑𝑡
𝑀
(𝝎) + 𝛀× (𝑰 ∘ 𝝎) + ��,
(2.123)
(2.124)
where having the body frame centered in the center of mass eliminated the term −��𝟎 × 𝑚��.
28
We have that 𝑑
𝑑𝑡
𝑀(𝝎) = �� because the derivative of the angular velocity is the same regardless of the
reference frame. 𝑑
𝑑𝑡
𝑀(𝑰) =
𝑑
𝑑𝑡
𝐵(𝑰), because the body is axisymmetric. In addition, we assume a simplifying
hypothesis: the body is two-plane symmetric and 𝜸 = 0. Taking this into account, (2.124) simplifies to
𝑴𝟎 = 𝑰 ∘ �� + 𝛀 × (𝑰 ∘ 𝝎) +
𝑑
𝑑𝑡
𝐵
(𝑰) ∘ 𝝎. (2.125)
The components of 𝝎 in the intermediate frame are
𝜔𝑥′′ = ��, (2.126)
𝜔𝑦′′ = �� sin𝜃, (2.127)
𝜔𝑧′′ = �� cos 𝜃 + �� (2.128)
The components of 𝛀 in the intermediate frame components are given by:
Ω𝑥′′ = ��, (2.129)
Ω𝑦′′ = �� sin 𝜃, (2.130)
Ω𝑧′′ = �� cos𝜃. (2.131)
The angular acceleration components written in the intermediate frame are obtained by differentiating
(2.126)-(2.128). The inertia tensor (2.36) and its time-derivative (2.40) are written having 𝐴 = 𝐵 and �� = ��.
Having all the variables of (2.125) defined, we write (2.125) for a free-body in the three directions:
0 = ���� + 𝐴�� + (𝐶 − 𝐴)��2 sin𝜃 cos𝜃 + 𝐶���� sin 𝜃 (2.132)
0 = ���� sin 𝜃 + 𝐴(�� sin 𝜃 + 2���� cos 𝜃) − 𝐶��(�� + �� cos 𝜃) (2.133)
0 = ��(�� cos 𝜃 + ��) + 𝐶(�� cos𝜃 − ���� sin 𝜃 + ��) (2.134)
29
2.11 ANGULAR MOMENTUM FOR A GENERAL VARIABLE-GEOMETRY BODY
We have the inertial frame’s 𝑍 axis aligned with the angular momentum 𝒉. Then, 𝒉 can be projected onto
the body frame using Euler angles, yielding
𝒉 = ℎ sin 𝜃 sin𝜑 𝒆𝒙 + ℎ sin 𝜃 cos𝜑 𝒆𝒚 + ℎ cos 𝜃 𝒆𝒛. (2.135)
The expression of the angular momentum, when the variable-geometry body’s symmetry is such that the
term 𝜸 is zero and ��𝟎 = 0, is
𝒉 = 𝐴𝜔𝑥𝒆𝒙 +𝐵𝜔𝑦𝒆𝒚 + 𝐶𝜔𝑧𝒆𝒛. (2.136)
Replacing the components of the angular velocity in terms of Euler angles (2.118)-(2.120) into (2.136) and
equaling (2.136) to (2.135) and separating by the body frame directions results in:
ℎ sin 𝜃 sin𝜑 = 𝐴(�� sin 𝜃 sin𝜑 + �� cos𝜑), (2.137)
ℎ sin𝜃 cos𝜑 = 𝐵(�� sin𝜃 cos𝜑 − �� sin𝜑), (2.138)
ℎ cos𝜃 = 𝐶(�� cos 𝜃 + ��). (2.139)
If the system (2.137)-(2.139) is rearranged to separate the angular rates in different equations, we obtain:
�� = ℎ (
1
𝐴(𝑡)−
1
𝐵(𝑡)) sin 𝜃 sin𝜑 cos𝜑, (2.140)
�� = ℎ(
sin2 𝜑
𝐴(𝑡)+cos2 𝜑
𝐵(𝑡)), (2.141)
�� = ℎ cos 𝜃 (
1
𝐶(𝑡)−sin2𝜑
𝐴(𝑡)−cos2𝜑
𝐵(𝑡)). (2.142)
2.11.1 Axisymmetric Variable-Geometry Body
A variable geometry-body can still be axisymmetric even though its moments of inertia vary: as long as
𝐴(𝑡) = 𝐵(𝑡). Having 𝐴 = 𝐵 simplifies (2.140)-(2.142) to:
�� = 0, (2.143)
�� =
ℎ
𝐴(𝑡)=
ℎ
𝐵(𝑡), (2.144)
30
�� = ℎ cos𝜃 (
1
𝐶(𝑡)−
1
𝐴(𝑡)). (2.145)
Eq. (2.143) implies that the variable geometry body with 𝜸 = 0 has a constant nutation angle, regardless of
how the principal moments of inertia vary with time. A consequence of this is that an axisymmetric variable-
geometry body’s nutation angle cannot be controlled by configuration changing. The nutation angle is
unchangeable and it only depends on the initial conditions of the body dynamics.
We conclude from (2.144) that the precession rate �� is inversely proportional to the moment of inertia 𝐴:
𝐴(𝑡)�� = ℎ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (2.146)
As a consequence, the precession rate increases when the moment of inertia 𝐴 decreases. So, eq. (2.146)
expresses conservation of angular momentum on the 𝑍 axis during the geometry variation of the body.
Since ℎ is the norm of the vector 𝒉, it is always positive, as well the moment of inertia 𝐴. Consequently, ��
is never negative.
Replacing ℎ (2.146) into (2.145) we obtain:
��
��=
𝐶(𝑡)
𝐴(𝑡) − 𝐶(𝑡)sec 𝜃. (2.147)
Eq. (2.147) implies that the proportion between the precession rate and the spin depends not only on the
nutation angle but on the ratio of the inertia moments 𝐶 and 𝐴. Moreover, the ability of the body to change
the moments of inertia, while maintaining its axisymmetry, can change the ratio ��
�� and even change the
signal of the ratio. For example, if the body changes from prolate, which implies 𝐴 > 𝐶, to oblate, which
means 𝐶 > 𝐴, while having a nutation angle between 0 and 𝜋
2, �� that was positive when the body was
prolate, becomes negative when the body becomes oblate.
𝒉 �� ��
𝐶 < 𝐴 , 0 < 𝜃 <𝜋
2 + + −
𝐶 < 𝐴 , 𝜋
2< 𝜃 < 𝜋 + + +
𝐶 > 𝐴 , 0 < 𝜃 <𝜋
2 + + −
𝐶 > 𝐴 , 𝜋
2< 𝜃 < 𝜋 + + +
Table 1- Sign of variables depending on the nutation angle and the body’s oblateness.
31
3 DESIGN OF VARIABLE GEOMETRY SATELLITES
After the derivation of the equations that express the relation between the geometry variation and the body
attitude dynamics, we envisioned satellites that could change its geometry in order to control its attitude.
The geometries studied are rough approximations of possible real variable-geometry satellites so that
insightful information about their dynamics may be obtained easily. For that reason, satellites are
approximated as being made of very simple components: uniform slender bars, springs or point masses. In
this first approach to the problem, we also avoided considering geometries with numerous degrees of
freedom of the shape-changing motion which would difficult the mathematical treatment – geometries with
one degree of freedom were preferred. All the developed geometries have two-plane symmetry so that the
non-rigid term 𝜸 is zero, guaranteeing a major simplification of the dynamics.
3.1 CONNECTED SCISSOR MECHANISM
3.1.1 Geometry
The first developed concept for a variable-geometry satellite is one that is always axisymmetric from a
dynamical point of view and with constant transversal moments of inertia. The inertia tensor resulting from
this, with 𝐴 = 𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, would be arguably the simplest, apart from the trivial spherical case. The control
would be achieved by a single degree of freedom that only changed the moment of inertia 𝐶.
The Connected Scissor Mechanism was conceived as being composed by a certain even number of bars,
a total of 2𝑛. The bars are slender and are articulated to each other, held together on both ends by negligible
mass joints, connected in such a way that they form a closed loop of bars. The number of bars must be
such that 𝑛 ≥ 3, otherwise there would be no closed loop without superimposition of bars. Figure 1 shows
a Connected Scissor Mechanism with 𝑛 = 3, at two different configurations for clearer understanding.
The single degree of freedom of this geometry is the inclination of the bars: 𝑛 bars are inclined with a +𝛼
angle and the other 𝑛 bars have an inclination of –𝛼 with respect to the transversal symmetry plane of the
Connected Scissor Mechanism. In what concerns the mathematical description of the Connected Scissor
Mechanism, 𝛼 will be considered positive and will vary between the limit values 𝛼 = 0 and 𝛼 =𝜋
2.
3.1.2 Body Frame
The body frame used for the Connected Scissor Mechanism is oriented so that, from both the top
perspective and bottom perspective of the 𝑧 axis, the bar structure is seen as a regular polygon of 𝑛 sides.
The 𝑧 axis intersects the geometrical center of the top polygon formed by the slender bars upper ends and
intersects the geometrical center of the bottom polygon formed by the slender bars bottom ends.
32
Figure 1 – Lateral perspective from the 𝑥 axis of a Connected Scissor Mechanism of 6 bars of same length, with a
configuration 𝛼1 on the left and a configuration 𝛼2 at the right, being 𝛼2 > 𝛼1.
Both the 𝑥 and 𝑦 axes are freely oriented within the transversal symmetry plane of the geometry, as long
they are perpendicular to each other and make the body frame a right reference frame.
The selected body frame is centered at the center of mass to simplify terms in the equations, because �� =
0. The geometric center and the center of mass coincide as a result of the bars have uniform density.
The limit value of 𝛼 = 0 corresponds to the situation when all the bars lie horizontally on the 𝑥0𝑦 plane
while 𝛼 =𝜋
2 implies all the bars are vertically oriented within in the 𝑧 axis. These extreme configurations for
the Connected Scissor Mechanism are not feasible in practice because even though the bars are modeled
as slender, having no thickness, the components of the satellite do have thickness and hence it is impossible
that they occupy the same points in space. Because of this, the configurations of the Connected Scissor
Mechanism should be distant of both the lower and upper limits of 𝛼.
3.1.3 Inertia Tensor
We will determine the moments of inertia as a function of: (i) mass of each bar 𝑚; (ii) length of each bar 𝑙;
(iii) the number of bars 2𝑛; and (iv) the Connected Scissor Mechanism’s degree of freedom 𝛼.
The top and the bottom views over the geometry are regular polygons. For that reason, 𝑛 symmetry planes
intersecting the sides of the polygon that are perpendicular to the 𝑥0𝑦 plane exist. This adds up to a total of
𝑛 + 1 symmetry planes. Thus, at least two perpendicular symmetry planes exist permanently in the body
frame, containing the center of mass at their intersection. This implies 𝜸 = 0 for every configuration, as
intended.
33
The 𝐶 moment of inertia is the same regardless of the orientation of the 𝑥 and 𝑦 axes in the 𝑥0𝑦 plane, as
a result of the distance of every point of the Connected Scissor Mechanism to the 𝑧 axis being the same
independently of where the 𝑥 and 𝑦 planes are oriented to. Thus, the evaluation of how 𝐴 and 𝐵 depend on
the orientation of these axes can be regarded as a 2𝐷 problem.
In the 𝑥0𝑦 plane, a total of 𝑛 orientations of the 𝑥 and 𝑦 axes exist so that 𝐴 and 𝐵 don’t change between
the 𝑛 frames, because of the top/bottom polygon’s regularity. According to the Mohr Circle [43], for the two
dimensional inertia tensor, two orientations of the 𝑥 and 𝑦 axes allow for a maximum 𝐴 and minimum 𝐵 and
vice-versa, separated by an angle of amplitude of 𝜋
2. In this case, we have 𝑛 ≥ 3 equivalent inertia tensors
in the plane. Therefore, if 𝑛 orientations have the same inertia tensor then all possible frames in the 𝑥0𝑦
plane have the same inertia tensor. This implies the axisymmetry 𝐴 = 𝐵 , because the maximum and
minimum moments of inertia have to be equal for this to occur.
We will obtain the moments of inertia 𝐴 = 𝐵 and 𝐶. If the top/bottom polygon is divided into 𝑛 geometrically
similar triangles, which are isosceles, we can obtain useful relations for angles and distances from them.
The angle 𝜙 is defined so that it is the double angle of the triangle. Using that the sum of the interior angles
of the triangle is 𝜋 and rearranging yields
𝜙 = 𝜋
𝑛 − 2
2𝑛, (3.1)
being 𝜙 a function of 𝑛. A visualization of an example of 𝜙 for 𝑛 = 3 is available in Figure 2.
The distance between a midpoint of any of the sides of the 2D regular polygon and its geometrical center is
defined as
ℎ =
𝑙
2tan𝜙 cos𝛼. (3.2)
Figure 2 – Top view from the 𝑧 axis of a Connected Scissor Mechanism with 𝑛 = 3 where the angle 𝜙 and distances ℎ
and 𝑙 𝑐𝑜𝑠 𝛼 are represented.
34
The Parallel axes theorem [43] is given by
𝐼 = 𝐼′ +𝑚𝑑2, (3.3)
where 𝐼′ is the moment of inertia with respect to an axis that intersects the center of mass of the body and
𝐼 is the moment of inertia with respect to any parallel axis. 𝑑 is the distance between the two axes.
At this point, we determine the expression for the moment of inertia of an uniform slender bar Ω of mass 𝑚
and length 𝑙 around an arbitrary axis that belongs to a plane that contains both the axis and the bar. Let the
distance between a point of the bar and the axis be named 𝑝 and the infinitesimal bar length be 𝑑𝑠. Thus,
the moment of inertia around the axis is
𝐼 = ∫𝑝2𝑑𝑚
Ω
. (3.4)
The bar has an inclination of 𝛽 with respect to the axis and so 𝑝 can be written in function of 𝑙, 𝑠 and 𝛽, being
the point 𝑠 = 0 corresponds to one of the two ends of the bar:
𝑝 = (𝑠 −
𝑙
2) sin𝛽. (3.5)
Since the mass distribution along the length of the bar is uniform, we have 𝑑𝑚 = 𝑚𝑑𝑠
𝑙. Using this and 𝑝 from
(3.5) in (3.4) and integrating from 0 to 𝑙 along the bar gives the moment of inertia 𝐼
𝐼 =
𝑚𝑙2
12sin2 𝛽 (3.6)
After the derivation of 𝐼 and the definitions concerning the polygon, (3.1) and (3.2), we are able to determine
the 𝐶 moment of inertia of the Connected Scissor Mechanism.
The symmetry of the geometry of the Connected Scissor Mechanism implies that the contribution of each
bar to 𝐶 is the same: they are all similar bars, equally inclined, with +𝛼 or −𝛼 with respect to the 𝑥0𝑦 plane,
having their midpoints all situated in the 𝑥0𝑦 plane. Moreover, each of the bars and a virtual axis, being the
latter an axis that intersects the midpoint of the bars and is parallel to the 𝑧 axis, belong to a plane containing
the bar and the axis.
We gather conditions to use expression (3.6) to obtain the individual contribution of each of the bars to the
total moment of inertia, 𝐶𝑖′, which is determined with respect to the respective virtual axis. Then, we use the
Parallel Axes Theorem (3.3), using ℎ (3.2) as 𝑑 in the expression, since it is the distance between the virtual
35
axis and the 𝑧 axis. The inclination angle of the bars with respect to 𝑧 is not 𝛼 but 𝜋
2− 𝛼 , reason why
𝑚𝑙2
12sin2 𝛼 becomes
𝑚𝑙2
12cos2 𝛼. Performing this yields:
𝐶 = 2𝑛(𝐶𝑖′ + 𝑚ℎ
2) =𝑛𝑚𝑙2
6cos2 𝛼(1 + 3 tan2𝜙). (3.7)
To obtain 𝐴, we use a property of the polar moment of inertia [43]:
𝐴 + 𝐵 + 𝐶 = 2𝐼0. (3.8)
Rearranging (3.8) using 𝐴 = 𝐵 yields
𝐴 = 𝐼0 −
𝐶
2, (3.9)
which allows to obtain 𝐴 if 𝐼0 is known.
The polar inertia moment of a single bar, defined as 𝐼0𝑖, is firstly determined to then obtain the total polar
inertia moment. All the bars have the same 𝐼0𝑖 because for every point in each bar we have a point on any
of the other bars that is equally distant to the origin, which happens for all the points of the bar. Thus:
𝐼0 = 𝑛𝐼0𝑖 . (3.10)
We consider a bar with an orientation that simplifies the calculation to obtain 𝐼0: a bar that lies on a plane
that is perpendicular to the 𝑥 axis. So we write
𝐼0𝑖 = ∫𝑥2 + 𝑦2 + 𝑧2𝑑𝑚
Ω
= ∫𝑥2𝑑𝑚Ω
+∫𝑦2 + 𝑧2𝑑𝑚Ω
. (3.11)
𝑥 is constant along this bar, being 𝑥 = ℎ. Furthermore, ∫ 𝑦2 + 𝑧2𝑑𝑚Ω
is the moment of inertia of this bar, 𝐴𝑖,
which is 𝑚𝑙2
12 regardless of the inclination 𝛼 because, for all 𝛼, the bar belongs to a perpendicular plane to
the 𝑥 axis and is intersected by the 𝑥 axis at its midpoint. Taking this into account simplifies (3.11) into
(𝐼0)𝑖 = 𝑚ℎ2 +
𝑚𝑙2
12=𝑚𝑙2
12(3 tan2𝜙 cos2 𝛼 + 1). (3.12)
Then, the total polar inertia moment is obtained by multiplying (3.12) by the total number of bars 2𝑛:
𝐼0 =
𝑛𝑚𝑙2
6(3 tan2𝜙 cos2 𝛼 + 1). (3.13)
After replacing 𝐼0 (3.13) in (3.9) and simplifying the expression yields
36
𝐴 =
𝑛𝑚𝑙2
12(cos2 𝛼 (3 tan2𝜙 − 1) + 2). (3.14)
3.1.4 Triangle Connected Scissor Mechanism
The dependence of 𝛼 in 𝐴 vanishes when
tan2𝜙 =
1
3, (3.15)
obtaining a constant value for 𝐴, as intended:
𝐴 =
𝑛𝑚𝑙2
6. (3.16)
Since 𝜙 is the double angle in the partial triangle of the regular top/bottom polygon, we have that 𝜙 is limited:
0 < 𝜙 <𝜋
2. 𝑛 is also limited: it is an integer such that 𝑛 ≥ 3. Taking these constraints into account and using
(3.1), (3.15) is rewritten:
tan(𝜋
𝑛 − 2
2𝑛) = ±
√3
3. (3.17)
Firstly, the positive branch of (3.17) is obtained:
tan(𝜋
𝑛 − 2
2𝑛) = tan(
𝜋
6)
⟺ 𝜋𝑛− 2
2𝑛=𝜋
6+ 𝑘𝜋, 𝑘 ∈ ℤ
⟺ 𝑛 =2
23 + 2𝑘
, 𝑘 ∈ ℤ.
(3.18)
The only 𝑘 that satisfies 𝑛 being an integer number larger than 2 is 𝑘 = 0, making 𝑛 = 3 and 𝜙 = 𝜋/6.
Concerning the negative solution, we have:
tan(𝜋
𝑛 − 2
2𝑛) = tan(−
𝜋
6)
⟺ 𝜋𝑛 − 2
2𝑛= −
𝜋
6+ 𝑘𝜋, 𝑘 ∈ ℤ
⟺ 𝑛 =2
−23 + 2𝑘
, 𝑘 ∈ ℤ
(3.19)
37
Eq. (3.19) has no solution, given the restrictions of 𝑛. So we conclude the only solution for having 𝐴 = 𝐵 =
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, or equivalently 𝑑𝐴
𝑑𝛼= 0, is 𝑛 = 3.
This implies that if the polygon is a regular triangle, then the Connected Scissor Mechanism has a constant
𝐴, depending only on 𝑛, 𝑚 and 𝑙, regardless of the degree of freedom of the Connected Scissor Mechanism
𝛼. Thus, when 𝑛 = 3, the moments of inertia are
and
𝐶 = 𝑚𝑙2 cos2 𝛼 (3.21)
The Triangle-Connected Scissor Mechanism is a simple geometry that allows, with a total of 6 bars, to have
an axisymmetric satellite that only varies its 𝐶 moment of inertia according to a very simple expression
(3.21). In addition to that, if a number of bars larger than 6 is selected, 𝐴 also varies with the degree of
freedom 𝛼, allowing for more options in what concerns attitude control.
3.1.5 Point Masses
It would be more unrestrictive if the Connected Scissor Mechanism geometry would allow to vary the 𝐶
moment of inertia, leaving 𝐴 = 𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, without the restraint of having 6 bars. In order to achieve this,
including masses on the joints of the bars was thought of. The masses are modeled as point, representing
structural reinforcements in the bars’ joints.
We define 𝑚1 as the mass of each of the 𝑛 point masses placed at the center of the bars, connecting every
pair of bars. 𝑚2 is the mass of the 2𝑛 point masses placed at the ends of the bars, connecting every pair of
bars.
𝐴 =
𝑚𝑙2
2. (3.20)
38
Figure 3 – Connected Scissor Mechanism with 6 bars, 3 masses 𝑚1 at the center of the bars, displayed in red, and 6
masses 𝑚2 at the ends of the bars, displayed in blue.
The 𝑧 axis and the 𝑛 symmetry axes in the 𝑥0𝑦 plane all remain symmetry axis with the addition of the point
masses. Thus, the body frame is still centered at the center of mass and is a principal frame, where 𝐴 = 𝐵
remains valid, according to the explanation given in sub-section 3.1.3.
In order to simplify the calculations, two non-dimensional quantities with respect to the mass of the slender
bar are introduced:
𝜇1 =
𝑚1
𝑚; (3.22)
𝜇2 =𝑚2
𝑚. (3.23)
With the point masses, expression (3.14) is modified to:
𝐴 =
𝑛𝑚𝑙2
12(cos2 𝛼 ((3+ 3𝜇2 +
3
2𝜇1) tan
2𝜙 − 1 − 3𝜇2) + 2). (3.24)
The condition of having 𝐴 constant regardless of the configuration of the Connected Scissor Mechanism is
having
tan2𝜙 =
1 + 3𝜇2
3 + 3𝜇2 +32𝜇1
(3.25)
If we consider extreme cases for the values of the masses 𝑚, 𝑚1 and 𝑚2 to satisfy (3.25), we have
39
0 ≤ 𝜇1 < ∞, (3.26)
0 ≤ 𝜇2 < ∞. (3.27)
Playing with all possible values of 𝜇1 and 𝜇2 in (3.25) yields
0 ≤ tan2𝜙 ≤ 1. (3.28)
However, 𝜙 cannot have continuous values, which results from (3.1), and 𝑛 is an integer larger than 2. The
first possibilities for 𝜙 are:
𝒏 𝝓 𝐭𝐚𝐧𝝓
3 𝜋
6
1
√3
4 𝜋
4 1
5 3𝜋
10 1.37638192047
6 𝜋
3 √3
… … …
Table 2 – First possible values for 𝜙.
The limit case of infinite bars,
lim𝑛→∞
(𝜋𝑛 − 2
2𝑛) =
𝜋
2, (3.29)
corresponds to a Connected Scissor Mechanism which top/bottom polygon tends to a circle. We also have
lim𝑛→∞
(tan𝜙) = ∞, (3.30)
being tan𝜙 a monotonously crescent function with 𝑛. Therefore, the only possible solutions are 𝑛 = 3 and
𝑛 = 4 because for 𝑛 > 4 the values of tan𝜙 fall out of the interval of (3.28). Only one additional alternative,
𝑛 = 4, becomes possible. Equation (3.25) can be solved to obtain the (𝜇1, 𝜇2) pairs to satisfy 𝑛 = 3 and 𝑛 =
4:
𝑛 = 3 → tan2𝜙 =1
3→ {
𝜇1𝜇2= 4
or𝜇1 = 𝜇2 = 0
, (3.31)
40
𝑛 = 4 → tan2𝜙 = 1 → 𝜇1 = −
4
3 (3.32)
Since the mass ratios cannot be negative, the alternative of 𝑛 = 4 is not possible.
We conclude that adding point masses does not allow for additional options of the number of bars. So, the
two possibilities of having 𝑑𝐴
𝑑𝛼= 0 are: (i) 6 bars with no point masses; and (ii) 6 bars with masses such that
𝑚1
𝑚2= 4. The second alternative is not interesting because the first solution allows
𝑑𝐴
𝑑𝛼= 0, without the point
masses.
3.2 OCTAHEDRON
3.2.1 Geometry
The objective of the concept behind this geometry is to have a variable-geometry axisymmetric body,
preferably with the possibility of selecting particular values for its properties so that 𝐴 = 𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
Consequently, the variability of the inertia tensor would be achieved by varying 𝐶.
This configuration consists in eight slender rigid and uniform bars of mass 𝑚 and length 𝑙 that form an
octahedron with variable shape. The four top bars form the top quadrangular pyramid and the four bottom
bars form the bottom quadrangular pyramid, the pyramids being articulated at the four connections between
the four top bars with the four bottom bars.
The octahedron can change its shape by varying its single degree of freedom 𝛼, which is half of the opening
2𝛼 between opposite bars connected at the top or bottom vertex. The degree of freedom 𝛼 is limited
between 0 and 𝜋/2.
41
Figure 4 – Octahedron at a configuration 𝛼, where the body frame is displayed as well as the angle of 2𝛼 between two
opposite bars.
3.2.2 Body Frame
The 𝑧 axis is defined as the vertical axis that intersects the top vertex of the top pyramid and the bottom
vertex of the bottom pyramid. The 𝑥0𝑦 plane is the transversal symmetry plane, which contains the base of
both pyramids, the orientation of the 𝑥 and 𝑦 axes being free as long as the body frame remains a right
reference frame and 𝑥 and 𝑦 are orthogonal axes. Since the symmetry of the body implies there are four
symmetry axes in the 𝑥0𝑦, the principal moments of inertia in the plane, 𝐴 and 𝐵, are equal, as explained in
sub-section 3.1.3 for the Connected Scissor Mechanism.
The symmetry of the geometry also implies 𝜸 = 0, since the 𝑥0𝑦 plane is perpendicular to each of the other
four symmetry planes of the octahedron. The body frame is centered at the center of mass of the origin, so
that �� = 0.
3.2.3 Inertia Tensor
Performing a similar derivation to what was done in sub-section 3.1.3 for the Connected Scissor Mechanism
and in 3.6.3 for the Compass, we obtain the elements of the inertia tensor:
𝐴 = 𝐵 =
28
3𝑚𝑙2 cos2 𝛼, (3.33)
42
𝐶 =
8
3𝑚𝑙2 sin𝛼2.
(3.34)
This geometry does not allow to select particular constants of the body such that 𝐴 does not depend on 𝛼.
3.2.4 Modified Octahedral
In order to modify the Octahedral so that 𝑑𝐴/𝑑𝛼 could be zero, we considered the possibility of adding an
uniform spring of mass 𝑀 and length 𝑑 that connected the top and bottom vertexes of the top and bottom.
The parameter 𝑑 is an additional degree of freedom to be controlled by actuators, totalizing two degrees of
freedom for this satellite. The objective of this is that 𝑑 could be controlled by a function of 𝛼 such that the
𝐴 would no longer depend on 𝛼.
With this modification, the four top bars are no longer connected to the four bottom bar, even though they
remain symmetric to each other with respect to the 𝑥0𝑦 plane.
The addition of the central spring does not modify the symmetries of the geometry described on sub-section
3.2.2, because 𝑧 axis is still a symmetry axis and the four symmetry axes in the 𝑥0𝑦 plane remain symmetry
axes.
So that there is no intersection between bars, for any 𝛼, we have
𝑑
𝑙≥ 2. (3.35)
Expression (3.36) is modified to
𝐴 = 𝐵 =
2𝑚𝑙2
3(2 cos2 𝛼 +
𝑀
2𝑚(𝑑
𝑙)2
+ 12(𝑑
𝑙−cos𝛼
2)2
), (3.36)
while the addition of the spring does not change (3.34) because the spring lies on the 𝑧 axis:
𝐶 =
8
3𝑚𝑙2 sin𝛼2. (3.37)
The dependence of 𝛼 in (3.36) cannot vanish by selecting of particular 𝑑/𝑙 ratios as a result of all the three
terms in the right-hand side being non-negative.
43
Other attempts to obtain 𝑑𝐴
𝑑𝛼= 0 on octahedral configurations failed, because the addition of other elements
in the geometry such as springs and movable masses added a positive term to the right-hand side of (3.36)
that does not depend on 𝛼.
3.3 DANDELION
3.3.1 Geometry
The concept of the Dandelion was to envision a satellite that, being axisymmetric, would allow a broader
range of geometry possibilities, similarly as the concept of the Connected Scissor Mechanism, being made
of an arbitrary number of bars. An additional objective was to explore the possibility of selecting particular
values for its properties so that 𝐴 = 𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. Consequently, the attitude control would be achieved by
varying 𝐶.
This geometry consists in a central bar of mass 𝑀 and length 𝐿 and by 𝑛 other bars of mass 𝑚 and length
𝑙 that have their top end articulated to the top end of the central bar. The 𝑛 bars are equivalently spaced
and form the same angle 𝛼 with the central bar. The attitude control is made by changing 𝛼, the single
degree of freedom of this geometry. 𝛼 is limited between 0 and 𝜋.
Figure 5 – Dandelion with 𝑛 = 4 bars at a configuration of 𝛼 and the body frame centered at the center of mass for this
particular configuration.
44
3.3.2 Body Frame
We select a body frame which 𝑧 axis is aligned with the central bar and is centered at the center of mass of
the geometry. Since there are 𝑛 symmetry axes on the 𝑥0𝑦 plane, because the bars are equally spaced,
the orientation of the 𝑥 and 𝑦 axes can be any and 𝐴 = 𝐵 is valid for all the orientations, taking into account
the discussion in sub-section 3.1.3. In addition, the symmetry of the geometry implies 𝜸 = 0.
For this geometry, the position of the center of mass, with respect to the articulated point at the top end of
the central bar, depends on 𝛼. The distance between the center of mass and the top point of the central bar
is
𝑧𝐶𝑀(𝛼) =
𝑀𝐿 + 𝑛𝑚𝑙 cos 𝛼
2(𝑀 + 𝑛𝑚). (3.38)
The origin of the body frame has the motion along the z axis prescribed by (3.38).
3.3.3 Inertia Tensor
The moment of inertia 𝐶 can be obtained without taking into account the motion of the frame depending on
𝛼, since the distance of the 𝑛 bars to the 𝑧 axis is the same regardless of where the 𝑥0𝑦 plane is with respect
to the body. Consequently:
𝐶 =
𝑛𝑚𝑙2
3sin2 𝛼. (3.39)
The motion of the origin of the body frame has to be accounted for in order to determine 𝐴. Firstly, we
determine 𝐴 for the central bar with respect to an axis that intersects its center of mass, yielding
𝐴𝑂′ =
𝑀𝐿2
12 (3.40)
and the contribution of the 𝑛 bars with respect to an axis that belongs to the plane that intersects their
centers of mass and is parallel to 𝑥0𝑦
𝐴𝑛′ =
𝑛𝑚𝑙2
12cos2 𝛼. (3.41)
The parallel axis theorem can be used on (3.40) and on (3.41) separately, because the distance between
the 𝑥0𝑦 plane and the axis with respect to where (3.40) and on (3.41) were calculated are different and both
depend on 𝑧𝐶𝑀. The former distance is given by
𝑑0 = 𝑧𝐶𝑀 −
𝐿
2, (3.42)
45
while the latter is
𝑑𝑛 = 𝑧𝐶𝑀 −
𝑙
2cos 𝛼. (3.43)
Using the parallel axis theorem (3.3) and then adding the contributions of the central bar and the central
and the set of 𝑛 bars to the total 𝐴 yields:
𝐴 =
𝑀𝐿2
12+ 𝑀(𝑧𝐶𝑀 −
𝐿
2)2
+𝑛𝑚𝑙2
12cos2 𝛼 + 𝑛𝑚 (𝑧𝐶𝑀 −
𝑙
2cos𝛼)
2
. (3.44)
We replace (3.38) in (3.44) and then it is possible to rearrange the expression to obtain
𝐴 =
𝑀𝐿2(𝑀 + 4𝑛𝑚) + 𝑛𝑚𝑙 cos𝛼 (𝑙 cos𝛼 (4𝑀 + 𝑛𝑚) − 6𝑀𝐿)
12(𝑀 +𝑚𝑛). (3.45)
We conclude from (3.45) that it is not possible select constants such that 𝑑𝐴
𝑑𝛼= 0 for all configurations.
3.4 EIGHT POINT MASSES
3.4.1 Geometry
This concept consists in having the attitude control of the satellite be performed by prescribing the variation
of one of the principal moments of inertia with time while keeping the other two constant. We selected 𝐶(𝑡)
as the variable moment of inertia, without loss of generality. Firstly, we consider the general case when 𝐴
and 𝐵 are allowed to be different and then the axisymmetric case.
The variation of 𝐶 is achieved by the controlled motion of eight masses 𝑚 along a specified trajectory within
their respective tubes. The tubes are regarded as having negligible mass comparing with the masses. The
eight masses and the structure required for it are an installation in the satellite to control its attitude rather
than a model for a satellite itself.
The eight-mass system as a whole is conceived as having a single degree of freedom because the position
of the masses are given by the mirrored positions of the other masses, for all configurations.
46
Figure 6 – Eight Point Masses in space. The trajectories of the masses so that 𝐴 and 𝐵 are kept constant are visible in
dashed curves.
3.4.2 Body Frame
The selected body frame for this geometry has the coordinate planes 𝑥0𝑦, 𝑦0𝑧 and 𝑥0𝑧 be the symmetry
planes of the masses motion, which implies the body frame is principal and that the reference frame is
centered at the center of mass, having �� = 0. The symmetries implied by this body frame also ensure 𝜸 =
0 for all configurations.
The eight masses can be regarded as vertexes of a parallelepiped that can change its shape. Each of the
faces of the parallelepiped is intersected at the respective geometric center by a coordinate axis, as can be
observed in Figure 6.
3.4.3 Trajectory of the Masses
The shape of the tubes or, equivalently, the trajectory of the masses, is obtained by imposing that 𝐴 and 𝐵
are constant regardless of their position. At a first instant 𝑡1, the eight mirrored positions of the eight point
masses are given by
𝒓𝒊(𝑡1) = ±𝑥1𝒆𝒙 ± 𝑦1𝒆𝒚 ± 𝑧1𝒆𝒛 with 𝑖 = 1,2, . . . ,8 , (3.46)
where 𝑥1, 𝑦1 and 𝑧1 are the coordinates of the mass whose motion is confined to the first octant, being
therefore positive values for all configurations. The positions of the eight masses at a different instant 𝑡2 are
also defined with respect to the positive coordinates of the mass of the first octant:
𝒓𝒊(𝑡2) = ±𝑥2𝒆𝒙 ± 𝑦2𝒆𝒚 ± 𝑧2𝒆𝒛 with = 1,2, . . . ,8 . (3.47)
47
The total moments of inertia at 𝑡1 are given by
𝐴(𝑡1) = 8𝑚(𝑦12 + 𝑧1
2), (3.48)
𝐵(𝑡1) = 8𝑚(𝑥12 + 𝑧1
2), (3.49)
𝐶(𝑡1) = 8𝑚(𝑥12 + 𝑦1
2), (3.50)
while the moments of inertia at 𝑡2 are
𝐴(𝑡2) = 8𝑚(𝑦22 + 𝑧2
2), (3.51)
𝐵(𝑡2) = 8𝑚(𝑥22 + 𝑧2
2), (3.52)
𝐶(𝑡2) = 8𝑚(𝑥22 + 𝑦2
2). (3.53)
The prescribed change in the moment of inertia 𝐶 is defined:
Δ𝐶 = 𝐶(𝑡2) − 𝐶(𝑡1) (3.54)
After equaling (3.48) to (3.51) and (3.49) to (3.52), so that 𝐴(𝑡1) = 𝐴(𝑡2) and 𝐵(𝑡1) = 𝐵(𝑡2), we can solve a
system of 3 equations to obtain the coordinates at 𝑡2 as function of the coordinates at 𝑡1and Δ𝐶, which is
the known intended change in the variable moment of inertia. Performing this yields:
𝑥2 = ±√𝑥12 +
Δ𝐶
16𝑚, (3.55)
𝑦2 = ±√𝑦12 +
Δ𝐶
16𝑚, (3.56)
𝑧2 = ±√𝑧12 −
Δ𝐶
16𝑚. (3.57)
It can be concluded from (3.57) that the Δ𝐶 is limited by 𝑚 and 𝑧1. Rearranging the expression inside the
square in (3.57) and having, 𝑧1 ≥ 0, as a result of the negative solutions of the coordinates being already
present in the double sign, yields the lower and upper limits of Δ𝐶
−8𝑚(𝑥12 + 𝑦1
2) ≤ Δ𝐶 ≤ 16𝑚𝑧12, (3.58)
48
where the lower limit corresponds to the limit situation Δ𝐶 = −𝐶(𝑡1) when the eight masses are moved to
the 𝑧 axis to have 𝐶 = 0. The upper limit is the situation where the eight point masses are located in the 𝑥0𝑦
plane.
Inequalities (3.58) can be rearranged to obtain the range of possible values of 𝐶, giving
0 ≤ 𝐶 ≤ 8𝑚(𝑥12 + 𝑦1
2 + 2𝑧12). (3.59)
Thus, 𝐶 is limited by the initial coordinates of the point masses and the mass 𝑚.
3.4.4 Geometric Relations with Axisymmetry
Conditions 𝑥1 = 𝑦1 and 𝑥2 = 𝑦2 must be verified in order to have the set of eight masses be axisymmetric,
with 𝐴 = 𝐵.
We define the degree of freedom
𝛼(𝑡) = 𝑥(𝑡) = 𝑦(𝑡), (3.60)
for any instant 𝑡, while the third coordinate is 𝑧(𝑡).
The moments of inertia as function of 𝑧 and 𝛼 are obtained by inserting (3.60) in (3.48)-(3.53), replacing
the instants 𝑡1 and 𝑡2 by a general instant of time 𝑡:
𝐴 = 𝐵 = 8𝑚(𝛼(𝑡)2 + 𝑧(𝑡)2), (3.61)
𝐶 = 16𝑚𝛼(𝑡)2. (3.62)
We now perform an adimensionalisation of the variables, defining
𝑚 =
1
8 (3.63)
and
𝐴 = 𝐵 = 1. (3.64)
Inserting (3.63) and (3.64), for a certain instant 𝑡, into the general expressions (3.61) and (3.62) and solving
them for 𝛼 and 𝑧 as function of 𝐶 yields
𝛼 = √𝐶
2 (3.65)
49
and
𝑧 = √1 −𝐶
2= √1 − 𝛼2. (3.66)
The range of possible values for 𝛼, 𝑧 and 𝐶 is obtained from (3.65) and (3.66):
0 ≤ 𝛼 ≤ 1, (3.67)
0 ≤ 𝑧 ≤ 1, (3.68)
0 ≤ 𝐶 ≤ 2. (3.69)
From (3.69) we conclude that the maximum of 𝐶 is
𝐶𝑚𝑎𝑥 = 2𝐴 = 2𝐵. (3.70)
Expression (3.66) is arranged to obtain a geometrical relation between 𝛼 and 𝑧:
𝛼2 + 𝑧2 = 1, (3.71)
defining a circumference of radius 1 in a plane with 𝛼 and 𝑧 as axis.
It can be concluded from (3.71) that the allowed trajectories for the eight masses belong to the planes 𝑥 = 𝑦
and 𝑥 = −𝑦 form an ellipse in three-dimensional space, one for each of plane. In these planes, the distance
that is perpendicular to the 𝑧 axis is such that coordinate 𝛼, equal in the 𝑥 and 𝑦 axes, projected into the
mentioned planes, becomes √2𝛼. Therefore the geometrical ellipse on the planes 𝑥 = 𝑦 and 𝑥 = −𝑦 is
defined by
(𝛼
√2)2
+ 𝑧2 = 1. (3.72)
Thus, the vertical semi-axis is 1 while the horizontal semi-axis is √2, making the ellipse tapered in the
horizontal direction along 𝑥 = 𝑦 or 𝑥 = −𝑦. This can be observed in Figure 6.
50
3.5 CONTROL POSSIBILITIES OF AXISYMMETRIC SATELLITES
A downside of having variable-geometry axisymmetric satellites as the Connected Scissor Mechanism, the
Octahedron, the Dandelion and the Eight Point Masses with 𝐴 = 𝐵 , is that the nutation angle is not
controllable. In addition, for the Connected Scissor Mechanism geometries, which was the single geometry
where it was possible to obtain 𝑑𝐴
𝑑𝛼= 0 for all 𝛼, the precession rate �� is not controllable as well, which was
concluded from (2.146). Only the spin rate is controllable for this geometry, given by
�� = ℎ cos𝜃
𝐴 − 𝐶(𝛼)
𝐴𝐶(𝛼), (3.73)
which resulted from combining (2.146) and (2.147).
The attitude control possibilities offered by the Connected Scissor Mechanism with constant 𝐴 are limited.
However, for the Connected Scissor Mechanism with variable 𝐴, the Octahedron and the Dandelion, both
the precession rate and the spin rate are modified by a change in 𝛼, since 𝑑𝐴
𝑑𝛼 and
𝑑𝐶
𝑑𝛼 are different from zero.
They are:
�� =
ℎ
𝐴(𝛼), (3.74)
�� = ℎ cos𝜃
𝐴(𝛼) − 𝐶(𝛼)
𝐴(𝛼)𝐶(𝛼).
(3.75)
Consequently, these geometries offer a wider range of attitude control possibilities. Nevertheless, it is
possible to control also the nutation angle of satellite if the satellite is not axisymmetric.
The next envisioned geometry, described in section 3.6, has variable moments of inertia such that 𝐴 ≠ 𝐵 ≠
𝐶, in order to allow for a less limited attitude control than the already discussed geometries.
3.6 COMPASS
3.6.1 Geometry
The Compass consists of two bars articulated at one joint, where one end of each bar coincides. Its degree
of freedom is the opening angle around the joint, 2𝛼. In this work, we limited the range of 𝛼 to 0 ≤ 𝛼 ≤ 𝜋,
but it would be interesting to explore the control possibilities of having an unlimited range of 𝛼. The bars are
slender and have a mass 𝑚 and a length 𝑙, having an uniform mass distribution.
51
3.6.2 Body Frame
The frame selected for the Compass is such that the 𝑦0𝑧 plane contains both the bars. The 𝑧 axis is a
symmetry axis for the compass, having its positive direction pointing towards the joint.
So that �� = 0, the body frame is centered at the center of mass. The position of the center of mass with
respect to the joint depends on the aperture 𝛼 and is fixed with respect to the inertial frame when the
compass has no translational movement.
Figure 7 – Compass for two different configurations, 𝛼2 and 𝛼1, with 𝛼2 > 𝛼1.The body frame is centered at the center
of mass.
The position of the center of mass with respect to the joint is
𝑧𝐶𝑀 = −
𝑙 cos𝛼
2 . (3.76)
The motion of the origin of the body frame with respect to the joint as function of the configuration is defined
by (3.76). This definition of the body frame implies that the 𝑦 axis is intersecting the midpoint of both bars
for every allowed configuration of the Compass.
3.6.3 Inertia Tensor
The selected body frame is a principal frame because the 𝑧 axis is a symmetry axis and the body lies on
the 𝑦0𝑧 plane for all configurations. Moreover, the non-rigid motion term 𝜸 is zero for every 𝛼, because 𝑥0𝑧
and 𝑦0𝑧 are perpendicular symmetry planes for every possible configuration of the Compass.
Expression (3.6) can be used to calculate 𝐵 of the two bars, having the inclination 𝛽 equal to 𝜋
2− 𝛼, yielding
52
𝐵 =
𝑚𝑙2
6cos2 𝛼 (3.77)
We use the parallel axes theorem (3.3), having 𝑑 =𝑙
2sin𝛼, and combine with the expression for the moment
of inertia of the inclined slender bar (3.6) to obtain
𝐶 = 4
𝑚𝑙2
6sin2 𝛼 (3.78)
for the Compass.
Since the compass is contained in the 𝑦0𝑧 plane we use the two-dimensional expression for the polar
moment of inertia 𝐼0, which is in this case, the same as 𝐴, resulting in
𝐴 =
𝑚𝑙2
6(3 sin2 𝛼 + 1). (3.79)
3.7 DISCUSSION
The suggested variable-geometries form a wide range of attitude control possibilities for spacecraft.
If the only objective is to control the spin rate of the satellite, then the axisymmetric solutions that only allow
for 𝐶 to vary can be used. These are the triangular Connected Scissor Mechanism, which could be regarded
as simplified model of a satellite composed of movable parts, and the Eight Point Masses, that could be an
installation to add to the satellite itself.
The other axisymmetric satellites envisioned in the present work are such that varying 𝐶 implies varying the
transverse moments of inertia as well. For these geometries, we verified if it would be possible to select
particular values for the constants, such as length, mass or number of bars, that would allow for the satellite
to vary 𝐶 while keeping 𝐴 constant. This was only achieved on the Connected Scissor Mechanism.
Nevertheless, the satellites where this was not possible, such as the Octahedron and the Dandelion, allow
for a wider range of possibilities of attitude control since they can also change the precession rate, being
interesting possibilities.
The compass is the geometry that allows for more control possibilities because the three Euler angular
velocities can vary.
In what concerns the applicability of these models, both the Connected Scissor Mechanism and the
Dandelion are the most versatile solutions since they can be made of an arbitrary number of bars without
losing dynamic dynamical axisymmetry.
53
4 SOLUTIONS OF THE FREE VARIABLE-GEOMETRY SATELLITE
MOTION
4.1 SYNCHRONOUS ROTATION IN AN ELLIPTICAL ORBIT WITH A VARIABLE-GEOMETRY
SATELLITE
An interesting application of the variable-geometry satellites is the use of the configuration change so that
the satellite could maintain the same orientation with respect to the central body, throughout an elliptic orbit.
Since the velocity of a satellite is not constant during the orbital period in an elliptic orbit, the configuration
of the geometry as function of time would be such that the precession rate is synchronized with the motion
of the satellite in the orbital plane.
This attitude control could be useful to spacecraft that requires its instrumentation to be pointing to the
central body.
We will consider small eccentricity orbits, with 𝑒 ≤ 0.1, which require small geometry variations from the
satellite, because conventional attitude control, by momentum wheels or thrusters for example, would be
more suitable for large-angle maneuver for the synchronous rotation in very eccentric orbits.
We selected the Connected Scissor Mechanism of the set of envisioned satellites to explore this application
of variable-geometry spacecraft. This geometry, being axisymmetric, can change its moment of inertia 𝐴
and therefore is able to control the precession rate as intended. In addition, it is versatile and has simpler
expressions for the moment of inertia 𝐴 (3.14) when compared to the equivalent expression for the
Dandelion (3.45).
The condition for having the rotation of the satellite be synchronous with the movement in the orbital plane
is satisfied by having
�� = �� (4.1)
during all instants of an orbital period, where 𝑓 is the true anomaly. Finding the configuration of the satellite
as function of time 𝛼(𝑡) that satisfies (4.1) is the objective of section 4.1.
4.1.1 Analytical Approximation for Kepler’s Equation
We will consider the solution of the two-body problem in the present section.
The time-derivative of the true anomaly [29] is
54
�� = (1 + 𝑒 cos𝑓)2√𝜇
𝑎3(1 − 𝑒2) , (4.2)
where 𝑒 is the eccentricity of the orbit, 𝜇 is the gravitational constant and 𝑎 is the semi-major axis.
Inserting the definition of mean angular velocity [29]
𝜈 = √
𝜇
𝑎3 (4.3)
into (4.2) so that expression (4.2) becomes function of only two orbital parameters, 𝜈 and 𝑒, and then
substituting cos 𝑓 by the expression that relates the mean anomaly with the eccentric anomaly 𝐸,
cos𝑓 =
cos𝐸 − 𝑒
1 − 𝑒 cos𝐸, (4.4)
yields:
�� = (1 + 𝑒
cos𝐸 − 𝑒
1 − 𝑒 cos 𝐸)2 𝜈
√1 − 𝑒2. (4.5)
In order to determine �� as function of time, 𝐸(𝑡) also needs to be obtained. The relationship between the
mean anomaly 𝑀 and the eccentric anomaly is
𝑀 = 𝐸 − 𝑒 sin𝐸, (4.6)
where the mean anomaly, unlike the eccentric anomaly, can be obtained directly from time:
𝑀 = 𝜈(𝑡 − 𝑡0), (4.7)
being 𝑡0 the instant of time of passing the periapsis point of the orbit.
Equation (4.6) is a transcendental equation whose known analytical solutions are series, which can be
obtained by Lagrange Inversion Theorem or with Bessel functions [44].
Numerical approaches, such as the Newton’s Method, can also be used to obtain 𝐸 from 𝑀 in (4.6).
However, in this work, we use the analytical approximation of the inversion of (4.6).
The inversion of (4.6) is performed by using both sides of the equation as arguments of sine and using the
properties of the sine function on the right-hand side:
sin(𝑀) = sin(𝐸) cos(𝑒 sin𝐸) − sin(𝑒 sin𝐸) cos(𝐸). (4.8)
55
Writing the functions cos(𝑒 sin𝐸) and sin(𝑒 sin𝐸) as Taylor series and neglecting equal or larger than
second order on the eccentricity, since the eccentricities considered in this work are small, we obtain
sin(𝑀) ≈ sin(𝐸) − 𝑒 sin(𝐸) cos(𝐸) (4.9)
We multiply equation (4.9) by 𝑒 and substitute 𝑒 sin(𝐸) by 𝐸 −𝑀, as given by (4.6). Neglecting the second
order term 𝑒2 sin(𝐸) cos(𝐸) yields the approximated expression for 𝐸(𝑀):
𝐸 ≈ 𝑀 + 𝑒 sin𝑀 (4.10)
For 𝑒 < 0.1, the maximum difference between the solutions for 𝐸(𝑡) throughout an orbit obtained with
Newton’s Method and by (4.10) is smaller than 3.7°.
We replace (4.10) into (4.5), using (4.7) to have time explicitly in the equation, obtaining
�� = (1 + 𝑒
cos(𝜈(𝑡 − 𝑡0) + 𝑒 sin 𝜈(𝑡 − 𝑡0)) − 𝑒
1 − 𝑒 cos(𝜈(𝑡 − 𝑡0) + 𝑒 sin 𝜈(𝑡 − 𝑡0)))
2𝜈
√1 − 𝑒2. (4.11)
4.1.2 Precession Rate as Function of Time
The precession rate for an axisymmetric body is given by (2.146), which becomes
�� =
ℎ
𝑛𝑚𝑙2
12[cos2 𝛼 (3 tan2 (𝜋
𝑛 − 22𝑛
) − 1) + 2] ., (4.12)
when 𝐴(𝛼) of the Connected Scissor Mechanism (3.14) and 𝜙 as function of 𝑛 (3.1) are substituted.
4.1.3 Configuration Variation with Time
In order to obtain 𝛼(𝑡), we equal (4.12) to (4.11), which yields
ℎ𝑛𝑚𝑙2
12
cos2 𝛼 (3 tan2 (𝜋𝑛 − 22𝑛
) − 1) + 2.= (1 + 𝑒
cos(𝜈(𝑡 − 𝑡0) + 𝑒 sin𝜈(𝑡 − 𝑡0)) − 𝑒
1 − 𝑒 cos(𝜈(𝑡 − 𝑡0) + 𝑒 sin 𝜈(𝑡 − 𝑡0)))
2𝜈
√1 − 𝑒2.
(4.13)
By rearranging (4.13), we obtain the analytical expression of 𝛼:
𝛼 = arccos
√ ℎ
𝑛𝑚𝑙2
12(1 + 𝑒
cos(𝜈𝑡 + 𝑒 sin𝜈𝑡) − 𝑒1 − 𝑒 cos(𝜈𝑡 + 𝑒 sin 𝜈𝑡)
)2 𝜈
√1− 𝑒2.
− 2
3 tan2 (𝜋𝑛 − 22𝑛
) − 1.
(4.14)
56
We observe that (4.14) is impossible for 𝑛 = 3, which corresponds to the Triangular Connected Scissor
Mechanism. The reason for this is that the 𝑑𝐴
𝑑𝛼= 0 for 𝑛 = 3, being 𝐴 constant and �� constant as well,
making impossible to have �� = �� for an elliptic orbit.
From (4.14), we conclude also that the existence of the square root on the right-hand side may imply that
combinations of parameters exist such as they will not allow to have �� = ��. For an increasing ratio 𝑛𝑚𝑙2
12/ℎ,
the positive term on the numerator on the left-hand side of (4.14) becomes smaller and, consequently, in
the limit situation when this ratio tends to zero, (4.14) becomes impossible, implying that �� = �� is not
possible.
Having that 𝑛𝑚𝑙2
12~𝐼 and ℎ~𝐼𝜔, we have that the ratio
𝑛𝑚𝑙2
12/ℎ is of order 𝜔. Thus, the smaller the angular
velocity, the closest is the satellite to be near condition where the precession rate is not able to follow orbit
motion. This is reasonable because in a limit situation where 𝜔 = 0, no configuration change could make
the satellite have angular momentum or have a prescribed precession rate.
Equation (4.14) can be rewritten taking into account that ℎ can be expressed in terms of a reference
precession rate ��0 and reference configuration 𝛼0, using (2.144)
ℎ = 𝐴(𝛼0)��0, (4.15)
being ℎ constant.
We arbitrate that the reference conditions in (3.62) are the conditions of the satellite at the periapsis when
𝑡 = 𝑡0, this being the first instant when the �� = �� begins to be satisfied. Nevertheless, we could select any
other point in the orbit to use its conditions as reference in (3.62).
Having the reference conditions defined in the periapsis implies
��0 = ��(𝑡0) = ��(𝑡0), (4.16)
which, by replacing (4.11), becomes
��0 = 𝜈
(1 + 𝑒)2
√1 − 𝑒2. (4.17)
We insert (3.62) into (4.15) and use 𝐴(𝛼) of the Connected Scissor Mechanism as given in (3.14):
ℎ =
𝑛𝑚𝑙2
12[cos2 𝛼0 (3 tan
2 (𝜋𝑛 − 2
2𝑛) − 1) + 2] 𝜈
(1 + 𝑒)2
√1 − 𝑒2. (4.18)
57
With the angular momentum given, we replace (3.62) into (4.14) and rearrange to obtain
𝛼 = arccos
√ [cos2 𝛼0 (3 tan
2 (𝜋𝑛 − 22𝑛
) − 1) + 2] (1 + 𝑒)2
(1 + 𝑒cos(𝜈𝑡 + 𝑒 sin 𝜈𝑡) − 𝑒1 − 𝑒 cos(𝜈𝑡 + 𝑒 sin 𝜈𝑡)
)2 − 2
3 tan2 (𝜋𝑛 − 22𝑛
) − 1,
(4.19)
where 𝑡0 was omitted, because the instant at the periapsis was considered as 𝑡0 = 0.
4.1.4 Possible Solutions for Synchronous Rotation
We will verify under which conditions (4.19) can be satisfied.
Defining
𝑘 = 3 tan2 (𝜋
𝑛 − 2
2𝑛) − 1, (4.20)
which is a constant value that depends only on the number of bars selected for the Connected Scissor
Mechanism, we simplify (4.19) to
𝛼 = arccos√−2
𝑘+𝑘 cos2 𝛼0 + 2
𝑘𝑔(𝑡), (4.21)
where the time-dependent function is
𝑔(𝑡) = (
𝑒 cos(𝜈𝑡 + 𝑒 sin(𝜈𝑡)) − 1
𝑒 − 1)
2
. (4.22)
Taking into account that the square root must have a non-negative argument and that the arccosine must
have an argument smaller or equal to |1| so that (4.19) is possible, the condition for possible solution is
0 ≤
2
𝑘(𝑔(𝑡) − 1) + cos2 𝛼0 𝑔(𝑡) ≤ 1. (4.23)
We conclude from (4.22) that 𝑔(𝑡) is bounded by
1 ≤ 𝑔(𝑡) ≤ (
1 + 𝑒
1 − 𝑒)2
, (4.24)
58
which implies that neither of the terms of (4.23) are negative. Thus, the left inequality in (4.23) is always
valid.
In what concerns the right inequality in (4.23), we rearrange (4.23) to obtain 𝛼0 limited by 𝑒 and 𝑘, taking
the maximum value of 𝑔(𝑡) so that (4.23) is satisfied during the orbital period:
cos2 𝛼0 ≤(1 − 𝑒)2 −
8𝑒𝑘
(1 + 𝑒)2. (4.25)
We conclude from (3.62) that the angle 𝛼0 can be allowed to have any value larger than a certain minimum
𝛼0 that depends on 𝑒 and 𝑘.
It is possible to assess how the minimum 𝛼0 varies with 𝑒 and 𝑘 by plotting different curves of 𝛼0, as function
of 𝑒 for different number of bars.
Figure 8 - Minimum 𝛼0 as function of the eccentricity of the orbit for different number of bars.
We observe from Figure 8 that the range of possible 𝛼0 becomes more limited with the increase in
eccentricity. On the other hand, increasing the number of bars decreases the minimum 𝛼0 allowed, since
the curves in Figure 8 get lower with larger 𝑛, until the theoretical limit of 𝑛 = ∞. Consequently, the most
limiting situation is having the minimum number of bars, 𝑛 = 4 and having the highest eccentricity 𝑒 = 0.1,
which corresponds to a minimum 𝛼0 of 54.4°, which still leaves almost half of the total range of 𝛼 available.
4.1.5 Example: Geosynchronous Orbit
We will obtain 𝛼(𝑡) graphically for a geosynchronous orbit, with 𝜈 = 7.292 × 10−5rad/s [11].
59
In order to observe the influence of the parameters 𝛼0, 𝑛 and 𝑒 in the configuration as function of time 𝛼(𝑡),
we selected two different values for each of the parameters: (i) 𝛼0 = 45º and 𝛼 = 60º; (ii) 𝑛 = 6 and 𝑛 = 10;
and (iii) 𝑒 = 0.1 and 𝑒 = 0.01. This selection adds up to a total of eight different 𝛼(𝑡), which will be plotted
during two sidereal days using the selected values according to (4.21). The curves are be displayed
separately for 𝑒 = 0.1 in Figure 9 and for 𝑒 = 0.01 in Figure 10.
Figure 9 – Configuration as function of time 𝛼(𝑡) during two sidereal days in geosynchronous orbit with 𝑒 = 0.1 for
different parameters.
Figure 10 - Configuration as function of time 𝛼(𝑡) during two sidereal days in geosynchronous orbit with 𝑒 = 0.01 for
different parameters.
60
We observe the solution exists for every point in the orbit for the arbitrated values. This was expected
because the most limitative set of parameters, which corresponds to having maximum eccentricity and
minimum number of bars, according to section 4.1.4, requires having 𝛼0 larger than a certain minimum,
which was respected in the selection of 𝛼0 for the curves.
The most limitative set of parameters is 𝑒 = 0.1 and 𝑛 = 6, which imply a minimum 𝛼0 of 37.5°, according to
(3.62). Since 37.5° is below the arbitrated values of 𝛼0 = 45° and 𝛼0 = 60°, there must exist solution for all
points in the orbit.
We observe in Figure 9 and in Figure 10 that 𝛼(𝑡) is a periodic function with a period equal to the orbital
period, which is in this case a sidereal day.
Additionally, it is noticeable that 𝛼(𝑡) has the maximum value at the periapsis and the minimum value at the
apoapsis. This is explained by the fact that at the periapsis the satellite experiences maximum orbital
velocity while at the apoapsis the satellite has the minimum orbital speed. For the synchronous rotation to
occur, the maximum and minimum of the precession rates must coincide with the extreme points of the
orbital speed. Consequently, the precession rate is maximum at the periapsis and minimum at the apoapsis,
which implies minimum 𝐴 at the periapsis and maximum 𝐴 at the apoapsis, according to (2.144). Since 𝐴 of
the Closed Scissor Mechanism increases when 𝛼 decreases (3.14), the minimum 𝛼 has the maximum at
the periapsis and the maximum at the apoapsis, which is confirmed by Figure 9 and Figure 10.
𝛼(𝑡) has the maximum value at the periapsis, which is 𝛼0 at the initial instant because we arbitrated that the
synchronous rotation would start at the periapsis. If any other point in the orbit had been selected, the shape
of function 𝛼(𝑡) would be the same, but would be shifted horizontally.
The influence of the eccentricity is clear: higher eccentricities imply larger configuration changes during an
orbital period, as can be observed by comparing curves of Figure 9 to Figure 10 that are displayed in the
same color. This is explained by the fact that a more eccentric orbit has larger orbital velocity variations
during a period, implying therefore greater changes in the inertia tensor so that the precession rate is
synchronized with the orbital motion.
Increasing the number of bars implies that the required amplitude of the configuration change becomes
smaller, which is visible if curves with the same eccentricity and the same 𝛼0 are compared in both in Figure
9 and Figure 10.
Moreover, changing the number of bars of the Connected Scissor Mechanism has more influence in the
configuration function over time 𝛼(𝑡) for more eccentric orbits. For example, the two top curves of 𝛼0 = 60°
61
have a much smaller difference in Figure 10 where the eccentricity is lower than the difference the analogue
curves in Figure 10 have, where the eccentricity is higher.
The solution does not depend linearly on the initial configuration 𝛼0, which can be realized graphically. For
example, if we compare the two curves in Figure 9 of 𝑛 = 10 and 𝑒 = 0.1, which differ only in the parameter
𝛼0 by 15°, we obtain that the absolute difference between the two curves is a periodic function with minimum
15° at the periapsis and maximum 22.2° at the apoapsis. Consequently, a solution with a different 𝛼0, having
all the other parameters equal to a known solution 𝛼(𝑡), cannot be obtained by simply adding a constant to
𝛼(𝑡).
62
4.2 MOMENTS OF INERTIA SWAP
As referred in section 3.7, the Compass is, of the envisioned models for variable-geometry satellites, the
geometry that offers the widest range of control possibilities by configuration changing because the three
moments of inertia vary with the configuration.
4.2.1 Relations between Moments of Inertia
An interesting feature of this geometry concerns the ratios between the moments of inertia. Using the
expressions (3.77)-(3.79) obtained in sub-section 3.6.3 for the moments of inertia, we obtain
𝐴
𝐶=
1
4 sin2 𝛼+3
4 (4.26)
and
𝐵
𝐶=
1
4 sin2 𝛼−1
4. (4.27)
Using (4.26) and (3.62), we plot 𝐴/𝐶 and 𝐵/𝐶.
Figure 11 - 𝐴/𝐶 and 𝐵/𝐶 ratios as function of the configuration 𝛼 for the Compass.
The obtained plots are symmetric with respect to the vertical line 𝛼 = 𝜋/2 due to symmetry of the geometry
with respect to the selected body frame. We observe in Figure 11 that 𝐴/𝐶 ≥ 1 for any 𝛼. In addition, the
curve of the 𝐴/𝐶 ratio is always on top of 𝐵/𝐶, implying that 𝐴 is the largest of the three moments of inertia
regardless of the configuration, which is reasonable since 𝐴 = 𝐵 + 𝐶 as explained in sub-section 3.6.3.
Since 𝐶 tends to zero for 𝛼 = 0 or 𝛼 = 𝜋, the curves of Figure 11 tend to infinity at the bounds of the plot.
If we subtract (4.27) from (4.26), we obtain a constant value:
63
𝐴 − 𝐵
𝐶= 1 (4.28)
However, the ratio 𝐵/𝐶 can be larger or smaller than 1 depending on 𝛼. We define the critical configuration,
𝛼𝐶, such that
𝐵(𝛼𝐶) = 𝐶(𝛼𝐶). (4.29)
If we solve (4.27) for 𝐵/𝐶 = 1, we obtain
𝛼𝐶 = arcsin(
1
√5), (4.30)
which has the numerical solutions of 26.6° and 153.4°. Consequently:
𝐵(𝛼)
𝐶(𝛼)> 1, for 0 < 𝛼 < 𝛼𝐶 ∨ 𝜋 − 𝛼𝑐 < 𝛼 < 𝜋 (4.31)
and
𝐵(𝛼)
𝐶(𝛼)< 1, for 𝛼𝐶 < 𝛼 < 𝜋 − 𝛼𝐶 . (4.32)
4.2.2 General Form of the Euler Equations for the Compass
We use
𝑑𝐼
𝑑𝑡=𝑑𝐼
𝑑𝛼
𝑑𝛼
𝑑𝑡= ��
𝑑𝐼
𝑑𝛼 (4.33)
in the general form of the Euler Equations (2.59)-(2.61) for a free variable-geometry body with 𝜸 = 0, which
is the case for the selected body frame for the Compass, as explained in sub-section 3.6.3, which yields:
𝑑𝐴
𝑑𝛼(𝛼(𝑡))��(𝑡)𝜔𝑥 +𝐴(𝛼(𝑡))��𝑥 = (𝐵(𝛼(𝑡)) − 𝐶(𝛼(𝑡)))𝜔𝑦𝜔𝑧 , (4.34)
𝑑𝐵
𝑑𝛼(𝛼(𝑡))��(𝑡)𝜔𝑦 +𝐵(𝛼(𝑡))��𝑦 = (𝐶(𝛼(𝑡)) − 𝐴(𝛼(𝑡)))𝜔𝑦𝜔𝑧 , (4.35)
𝑑𝐶
𝑑𝛼(𝛼(𝑡))��(𝑡)𝜔𝑧 + 𝐶(𝛼(𝑡))��𝑧 = (𝐴(𝛼(𝑡)) − 𝐵(𝛼(𝑡)))𝜔𝑥𝜔𝑦 , (4.36)
64
where the dependence of 𝛼 by the moments of inertia and the respective derivatives is displayed explicitly,
as well as the dependence of 𝛼 and �� on time.
We perform the differentiation of (3.77)-(3.79) with respect to 𝛼, obtaining:
𝑑𝐴
𝑑𝛼= 𝑚𝑙2 sin𝛼 cos𝛼, (4.37)
𝑑𝐵
𝑑𝛼= −
1
3𝑚𝑙2 sin𝛼 cos𝛼, (4.38)
𝑑𝐶
𝑑𝛼=4
3𝑚𝑙2 sin𝛼 cos𝛼, (4.39)
and we replace (3.77)-(3.79) and (4.37)-(3.62) into (4.34)-(4.36), which yields
6��𝜔𝑥 sin𝛼 cos𝛼 + ��𝑥(3 sin2 𝛼 + 1) = 𝜔𝑦𝜔𝑧(1 − 5 sin2 𝛼) (4.40)
−2��𝜔𝑦 sin 𝛼 cos 𝛼 + ��𝑦(1 − sin2 𝛼) = 𝜔𝑥𝜔𝑧(sin
2 𝛼 − 1) (4.41)
2��𝜔𝑧 sin𝛼 cos𝛼 + ��𝑧 sin2 𝛼 = 𝜔𝑥𝜔𝑦 sin
2 𝛼 (4.42)
where 𝑚𝑙2 was divided in all equations. It can be observed that for 𝛼 = 𝜋/2 equation (4.41) becomes 0 = 0,
which is justified by the fact that, for that configuration, both the bars of the Compass lie on the 𝑦 axis,
having therefore zero moment of inertia around that axis. Consequently, the Compass is allowed to have
any 𝜔𝑦 , making indeterminate the system of differential equations.
A similar situation happens for (4.42), which becomes 0 = 0 when 𝛼 = 0 or 𝛼 = 𝜋.
Taking this into account, the range of operation for attitude control of the Compass, according to this model,
should be selected as either 0 < 𝛼 < 𝜋/2 or 𝜋/2 < 𝛼 < 𝜋, preferably with configurations significantly distant
from the limit configurations of those intervals.
It is important to notice that the mentioned singularities come from the assumption of the bars being slender,
which are modelled as infinitely thin by definition. With a more sophisticated model for the bars, where the
thickness would be accounted for, no singularities would exist because there would be no configuration for
the Compass such that a moment of inertia is zero.
65
Equations (4.40)-(4.42) can be further simplified, dividing (4.40) and (4.42) by sin2 𝛼 and (4.41) by cos2 𝛼:
6��𝜔𝑥 cotan𝛼 + ��𝑥(4 + cotan2 𝛼) = 𝜔𝑦𝜔𝑧(cotan
2 𝛼 − 4), (4.43)
−2��𝜔𝑦 tan𝛼 + ��𝑦 = −𝜔𝑥𝜔𝑧 , (4.44)
2��𝜔𝑧 cotan𝛼 + ��𝑧 = 𝜔𝑥𝜔𝑦 . (4.45)
4.2.3 Approximated Solution
The difficulty of finding a solution for the system of differential equations given by (4.43)-(4.45) depends on
the nature of the function 𝛼(𝑡), which simplest form, apart from 𝛼(𝑡) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, is a linear function,
𝛼 = 𝑎 + 𝑏𝑡, (4.46)
where 𝑎 and 𝑏 are constants. Even with the configuration as function of time given by (3.62), the system
(4.43)-(4.45) has no obvious analytic solution.
However, we will obtain an approximate solution for the dynamics of the Compass under two assumptions:
(i) 𝛼(𝑡) is close to the critical configuration, which is having 𝛼 ≈ 𝛼𝐶; (ii) 𝜔𝑧~1 rad/s is larger than the other
two components of the angular velocity, so that
𝜔𝑧2 ≫ 𝜔𝑥𝜔𝑦 , (4.47)
being 𝜔𝑥 , 𝜔𝑦~0.1 rad/s.
Neglecting the term of the right-hand side of (4.45), according to (3.62), and rearranging yields:
𝜔��𝜔𝑧
= −2�� cot 𝛼. (4.48)
We integrate (4.65) from an instant 𝑡0 to a time 𝑡 , which correspond to a configuration 𝛼0 and to a
configuration 𝛼, respectively:
∫
��𝑧𝜔𝑧𝑑𝑡
𝑡
𝑡0
= ∫ −2�� cot𝛼 𝑑𝑡𝑡
𝑡0
= ∫ −2cot 𝛼 𝑑𝛼𝛼
𝛼0
. (4.49)
The integrations are solved to obtain
𝜔𝑧(𝛼) = 𝜔𝑧(𝛼0)
sin2 𝛼0sin2 𝛼
. (4.50)
66
Taking into account that 𝛼 ≈ 𝛼𝐶 and that, according to (4.30), cot𝛼𝐶 = 4, we can neglect the term of the
right-hand-side of (4.43) because it has a product of two small quantities, 𝜔𝑦 and cotan2 𝛼 − 4, with respect
to the terms on the left-hand side. Rearranging the approximated (4.43) and integrating yields
∫��𝑥𝜔𝑥
𝑑𝑡𝑡
𝑡0
= ∫ −6cot 𝛼
4 + cot2𝛼𝑑𝛼
𝛼
𝛼0
. (4.51)
We perform the integrations to obtain
𝜔𝑥(𝛼) = 𝜔𝑥(𝛼0)
1 + 3 sin2 𝛼01 + 3 sin2 𝛼
. (4.52)
Even though we obtained approximated expressions for 𝜔𝑥(𝛼) and 𝜔𝑧(𝛼) from (4.43) and from (4.45)
respectively, 𝜔𝑦 cannot be obtained easily from (4.44) because the term of the right-hand side of (4.44) is
not negligible.
However, we can use the conservation of angular momentum (2.101) to obtain 𝜔𝑦(𝛼) as function of the
initial conditions of the angular velocity and of the configuration. Having
ℎ2
𝑚𝑙2/6= (1 + 3 sin2(𝛼))2𝜔𝑥
2(𝛼) + cos4(𝛼) 𝜔𝑦2(𝛼) + 16 sin4(𝛼) 𝜔𝑧
2(𝛼) (4.53)
at any configuration 𝛼 and
ℎ2
𝑚𝑙2/6= (1 + 3 sin2(𝛼0))
2𝜔𝑥2(𝛼0) + cos4(𝛼0)𝜔𝑦
2(𝛼0) + 16 sin4(𝛼0)𝜔𝑧2(𝛼0) (4.54)
for the initial configuration, we equal (4.53) to (4.54) and replace the obtained components of the angular
velocity (4.50) and (4.52), which eliminates the contributions of the other angular velocities and can be
simplified to:
𝜔𝑦(𝛼) = 𝜔𝑦(𝛼0)
cos2(𝛼0)
cos2(𝛼). (4.55)
The expressions obtained for the components of angular momentum, (4.50), (4.52) and (4.55), show that,
under the mentioned conditions for the approximation, there is no exchange between the three components
of the angular momentum, since each of the components is conserved separately.
67
4.2.3.1 Kinetic Energy
Once the angular components are determined, we can obtain the kinetic energy as function of the
configuration according to (2.86) after determining
1
2∫
𝑑𝒓
𝑑𝑡
𝐵 .
𝑑𝒓
𝑑𝑡
𝐵
Ω𝑑𝑚 as function of 𝛼 as well.
We define 𝑢 as the variable that runs along the length of both bars, starting at 0 at the joint up to the end
point 𝑢 = 𝑙. The position of each point of the compass with respect to the joint 𝒓𝒋 is given by
𝒓𝒋(𝑢, 𝛼) = ±𝑢 sin𝛼 𝒆𝒚 − 𝑢 cos 𝛼 𝒆𝒛, (4.56)
where the positive branch of the first term of right-hand side concerns the right bar and the negative
concerns the left bar of the compass. In order to obtain 𝑑𝒓
𝑑𝑡
𝐵, we add the motion of the center of mass 𝑧𝐶𝑀
as given in (3.76) to the 𝒆𝒛 component of (4.56), which yields
𝒓(𝑢, 𝛼) = ±𝑢 sin𝛼 𝒆𝒚 + (𝑢 −
𝑙
2) cos𝛼 𝒆𝒛. (4.57)
Differentiating (3.62) results in
𝑑𝒓
𝑑𝑡
𝐵
(𝑢, 𝛼) = ��(±𝑢 cos 𝛼 𝒆𝒚 + (𝑙
2− 𝑢) sin𝛼 𝒆𝒛. (4.58)
With the velocity of each point of the compass obtained, we perform the integration using the symmetry of
the bars and determine the non-rigid term of the kinetic energy:
1
2∫
𝑑𝒓
𝑑𝑡
𝐵
.𝑑𝒓
𝑑𝑡
𝐵
Ω
𝑑𝑚 = ∫ ��2 (𝑢2 + sin2 𝛼 (𝑙2
4− 𝑢𝑙))
𝑚
𝑙
𝑙
0
𝑑𝑢 =𝑚𝑙2
12��2(4 − 3 sin2 𝛼 ). (4.59)
We can write the kinetic energy as function of the configuration using the reference configuration 𝛼0 by
replacing (4.59) and the moments of inertia (3.77)-(3.79) in (2.86):
2𝑇(𝛼)
𝑚𝑙2
6
=(1 + 3 sin2 𝛼0)
2
1 + 3 sin2 𝛼𝜔𝑥2(𝛼0) +
(1 − sin2 𝛼0)2
1 − sin2 𝛼𝜔𝑦2(𝛼0) +
sin4 𝛼0
sin2 𝛼𝜔𝑧2(𝛼0) +
��2(4 − 3 sin2 𝛼 )
2. (4.60)
68
4.2.3.2 Ellipsoids of the Kinetic Energy and of the Angular Momentum
Having the kinetic energy determined, it is possible to obtain the semi-axes of the ellipsoid of the kinetic
energy as function of the configuration and initial conditions. We insert the moments of inertia of the
Compass (3.77)-(3.79) in (2.99):
√
2𝑇(𝛼)𝑚𝑙2
6
−��2(4 − 3 sin2 𝛼 )
2
1 + 3 sin2 𝛼,
√
2𝑇(𝛼)𝑚𝑙2
6
−��2(4 − 3 sin2 𝛼 )
2
1 − sin2 𝛼,
√
2𝑇(𝛼)𝑚𝑙2
6
−��2(4 − 3 sin2 𝛼 )
2
4 sin2 𝛼
(4.61)
The initial angular velocity is used on (4.60) to determine 𝑇(𝛼), which eliminates the contribution of �� to the
length of the semi-axes of the ellipsoid of the kinetic energy in (4.61).
The semi-axes of the ellipsoid of the angular momentum are
ℎ
𝑚𝑙2
6(1 + 3 sin2 𝛼)
,ℎ
𝑚𝑙2
6(1 − sin2 𝛼)
,ℎ
𝑚𝑙2
6 4 sin2 𝛼, (4.62)
as obtained by replacing (3.77)-(3.79) in (2.102). ℎ is obtained according to (4.54) as function of certain
initial components of the angular velocity, 𝜔𝑥(𝛼0), 𝜔𝑦(𝛼0) and 𝜔𝑧(𝛼0).
We obtained the ellipsoids for an example with an initial configuration of 𝛼0 = 0.45, which is near the critical
angle 𝛼𝐶 = 0.46, that changes to a configuration of 𝛼 = 0.47.
The initial components of the angular velocity also respect the assumptions made on this section: 𝜔𝑥(𝛼0) =
𝜔𝑦(𝛼0) = 0.1 rad/s and 𝜔𝑧(𝛼0) = 1 rad/s.
69
Figure 12 – Perspectives from the 𝜔𝑥 axis of the ellipsoids of the kinetic energy, at darker orange, and of the angular
momentum, at brighter orange, where the curve of the intersection of the ellipsoids is displayed in blue. The
configuration at the left is of 𝛼0 and at the right is of 𝛼.
For 𝛼 and for 𝛼0, Figure 12 shows nearly circular shapes, because the semi-axes of the ellipsoids on the
𝜔𝑦 and on the 𝜔𝑧 directions are very close, since 𝛼 and 𝛼0 are near the critical configuration 𝛼𝐶, for which
the Compass has 𝐵 = 𝐶.
4.2.4 Stability Analysis
Since the relation between the moments of inertia depends on the configuration, as discussed in section
4.2.1, it is interesting to explore how this influences the stability of the motion of the Compass in each of the
three axes.
4.2.4.1 Constant Configuration
When the Compass is not changing its configuration, the terms on �� vanish in (4.43)-(4.45). We will check
if the motion of the Compass around the 𝑥 axis is stable and in for which configurations it is stable.
The angular velocities are:
𝜔𝑥 = 𝜔0𝑥 + 𝛿𝜔𝑥, (4.63)
𝜔𝑦 = 𝛿𝜔𝑦, (4.64)
𝜔𝑧 = 𝛿𝜔𝑧, (4.65)
being 𝜔0𝑥 ≫ 𝛿𝜔𝑥 , 𝛿𝜔𝑦 , 𝛿𝜔𝑧 and 𝛿𝜔𝑥2, 𝛿𝜔𝑦
2 and 𝛿𝜔𝑧2 negligible. Replacing (4.63)-(3.62) in (4.43)-(4.45) with
�� = 0 yields:
70
𝛿𝜔𝑥 (4 + cotan2 𝛼(𝑡)) = 0 , (4.66)
𝛿𝜔𝑦 = −𝜔0𝑥𝛿𝜔𝑧, (4.67)
𝛿𝜔𝑧 = 𝜔0𝑥𝛿𝜔𝑦, (4.68)
We differentiate eqs. (4.67) and (4.68) and separate the infinitesimal angular velocities into two equations:
𝛿𝜔𝑦 = −𝜔0𝑥2 𝛿𝜔𝑦. (4.69)
𝛿𝜔𝑧 = −𝜔0𝑥2 𝛿𝜔𝑧, (4.70)
The solutions of (4.69) for 𝛿𝜔𝑦 and of (4.70) for 𝛿𝜔𝑧 are stable regardless of the configuration because
−𝜔0𝑥2 is negative and, therefore, induced perturbations on the angular velocity are never amplified. Thus, if
the Compass is spinning around the 𝑥 axis, the motion is stable.
If we perform a similar procedure concerning a situation where the Compass is spinning around the 𝑦 axis,
defining
𝜔𝑥 = 𝛿𝜔𝑥, (4.71)
𝜔𝑦 = 𝜔0𝑦 + 𝛿𝜔𝑦, (4.72)
𝜔𝑧 = 𝛿𝜔𝑧, (4.73)
being 𝜔0𝑦 ≫ 𝛿𝜔𝑥 , 𝛿𝜔𝑦 , 𝛿𝜔𝑧 and neglecting second order infinitesimal angular speeds, we obtain:
𝛿𝜔𝑥 = 𝜔0𝑦
2cotan2 𝛼 − 4
cotan2 𝛼 + 4𝛿𝜔𝑥, (4.74)
𝛿𝜔𝑧 = 𝜔0𝑦
2cotan2 𝛼 − 4
cotan2 𝛼 + 4𝛿𝜔𝑧. (4.75)
In contrast with the 𝑥 axis, the stability for the 𝑧 axis depends on the configuration. The condition for stability
is having
cotan2 𝛼 − 4
cotan2 𝛼 + 4< 0, (4.76)
which is satisfied by having
𝛼𝐶 < 𝛼 < 𝜋 − 𝛼𝐶, (4.77)
according to the definition of (4.30), while the instability occurs for 0 < 𝛼 ≤ 𝛼𝐶 and 𝜋 − 𝛼𝐶 ≤ 𝛼 < 𝜋.
71
The derivation to verify the condition stability around the 𝑧 axis leads to condition (4.76) with an opposite
sign on the left-hand side. Consequently, the configurations that allow for a stable spin around 𝑧 are
0 < 𝛼 < 𝛼𝐶 and 𝜋 − 𝛼𝐶 < 𝛼 < 𝜋. (4.78)
The stability for the Compass without configuration change is resumed on Table 3. We verified that the axis
around which the Compass can rotate stably are the axes that correspond to either the maximum moment
of inertia or to the minimum moment of inertia for the current configuration, as it happens for a rigid body
[29].
Spin around 𝒙 axis Spin around 𝒚 axis Spin around 𝒛 axis
0 < 𝛼 < 𝛼𝐶 𝐴 > 𝐵 > 𝐶 stable unstable stable
𝛼𝐶 < 𝛼 < 𝜋 − 𝛼𝐶 𝐴 > 𝐶 > 𝐵 stable stable unstable
𝜋 − 𝛼𝐶 < 𝛼 < 𝜋 𝐴 > 𝐵 > 𝐶 stable unstable stable
Table 3 - Configurations of stability for the Compass
4.2.4.2 Varying Configuration
If the configuration is changing then all terms of (4.43)-(4.45) must be considered, because �� is no longer
zero. The stability of rotation around each of the three principal axes depends on the function 𝛼(𝑡), which
particular case of 𝛼(𝑡) = 0 was verified in sub-sub-section 4.2.4.1. Consequently, we will be able to obtain
three systems of differential equations for a general prescribed function 𝛼(𝑡), one for the rotation around
each of the principal axes.
We will firstly assess the stability around the 𝑥 axis. Using the definitions of (4.73)-(4.75) on (4.43)-(4.45),
and differentiating the equations on the 𝑦 and 𝑧 directions yields:
𝛿𝜔𝑦 − 2𝛿𝜔𝑦
�� tan𝛼 − 2𝛿𝜔𝑦 (�� tan 𝛼 +��2
cos2 𝛼) = −𝜔0𝑥𝛿𝜔𝑧 , (4.79)
𝛿𝜔𝑧 + 2𝛿𝜔𝑧 �� cotan𝛼 + 2𝛿𝜔𝑧 (�� cotan𝛼 −
��2
sin2 𝛼) = 𝜔0𝑥𝛿𝜔𝑦 . (4.80)
In contrast with what occurs for the case of constant configuration, the equations on the two directions are
coupled and the system of equations has non-linear factors.
The system can be written in State-System Representation. We define
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{𝑋} = [𝛿𝜔𝑧 𝛿𝜔𝑦 𝛿𝜔𝑧 𝛿𝜔𝑦 ]𝑇, (4.81)
add the equations 𝛿𝜔𝑦 = 𝛿𝜔𝑦 and 𝛿𝜔𝑧 = 𝛿𝜔𝑧 to the set of equations (4.79)-(4.80) and rearrange to have
the system written in the form
{𝑋} = −[𝑆]{𝑋}, (4.82)
which does not allow for trivial solution of exponential of [𝑆] for {𝑋}(𝑡) because [𝑆] is not constant.
For the particular case of the rotation around the 𝑥 axis we obtain
[𝑆]𝑥 =
[
0 0 1 00 0 0 1
2(��2
sin2 𝛼− �� cotan𝛼) 0 −2�� cotan𝛼 𝜔0𝑥
0 2(�� tan𝛼 +��2
cos2 𝛼) −𝜔0𝑥 2�� tan𝛼
]
. (4.83)
We determine the dynamics matrix [𝑆] for the other two directions, defining the state-vector for rotation
around 𝑦 , [𝛿𝜔𝑥 𝛿𝜔𝑧 𝛿𝜔𝑥 𝛿𝜔𝑧 ]𝑇
and the state-vector for the rotation around 𝑧 , [𝛿𝜔𝑥 𝛿𝜔𝑦 𝛿𝜔𝑥 𝛿𝜔𝑦 ]𝑇
,
obtaining:
[𝑆]𝑦 =
[
0 0 1 00 0 0 1
6��2
sin2 𝛼− 6�� cotan𝛼
4 + cotan2𝛼−
2𝜔0𝑦�� cotan𝛼
sin2 𝛼4 + cotan2𝛼
2�� cotan𝛼sin2 𝛼
− 6�� cotan𝛼
4 + cotan2 𝛼
𝜔0𝑦(4 − cotan2 𝛼)
4 + cotan2 𝛼
0 2(��2
sin2𝛼− �� cotan𝛼) 𝜔0𝑦 −2�� cotan𝛼
]
, (4.84)
[𝑆]𝑧 =
[
0 0 1 00 0 0 1
6��2
sin2 𝛼− 6�� cotan𝛼
4 + cotan2𝛼−
2𝜔0𝑧�� cotan𝛼sin2 𝛼
4 + cotan2𝛼
2�� cotan𝛼sin2 𝛼
− 6�� cotan𝛼
4 + cotan2 𝛼
𝜔0𝑧(4 − cotan2 𝛼)
4 + cotan2𝛼
0 2(�� tan𝛼 +��2
cos2 𝛼) −𝜔0𝑧 2�� tan𝛼
]
.
(4.85)
We conclude from (4.83)-(4.85) that assessing the stability of the rotation of the Compass around each of
the principal axes is difficult, given that the terms in the dynamics matrixes are not only variable but contain
non-linear functions of the configuration as well.
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5 CONCLUSIONS
The attitude dynamics of a variable-geometry body has been described. We concluded that the added
complexity of the configuration changing is in general large, even though we could obtain some simplifying
assumptions to obtain either approximations or to make the terms related with configuration changing
vanish.
The physical meaning of the terms that are not present in rigid-body motion is also discussed in this work.
We addressed the determination of the kinetic energy of a variable geometry body and concluded that the
kinetic energy varies depending on the relationship between the actuating forces and both the translation
and rotational state of the body.
We obtained a variety of simple geometries that eliminate the extra-terms related to variable geometry to
simplify the problem, but still change their inertia tensors in different ways.
It was concluded that it is possible to obtain an axisymmetric geometry made of connected slender bars
where only one moment of inertia varies, even though we realized that such an inertia tensor does not allow
for either precession rate or nutation rate controls. Other conceived geometries are axisymmetric satellites
with variable transversal moments of inertia, which allow for precession rate control. The most unrestrictive
geometry from the attitude control point of view is a geometry that has three different and variable principal
moments of inertia, in a general configuration.
We developed an application for attitude control using the geometry variation method suggested in this work
– a satellite in an elliptic orbit that requires synchronous rotation. This situation requires attitude control so
that the satellite’s precession rate varies accordingly with the non-uniform orbital velocity, which is more
critical in orbits of higher eccentricities where the velocity variations are larger.
Having selected one of the envisioned geometries to exemplify the variable-geometry control method, we
determined the configuration function as function of time required for the synchronous rotation for small
eccentricities, depending on several parameters: the mean angular velocity and the eccentricity of the orbit,
the number of bars of the satellite, and the initial configuration.
We concluded that the solution for the synchronous rotation is possible in most situations, being not
achievable only in unrestrictive conditions. Interesting physical insight on how the configuration function
depends with various parameters has been discussed by comparing graphics in different situations.
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In the case of the satellite with three variable principal moments of inertia, it was observed that an interesting
feature of the satellites with such a capability is to be able to swap the order of the values of the moments
of inertia, being therefore possible to alter the rotation around principal axes from stable to unstable or
unstable to stable. Having this feature in mind, we obtained the equations of the dynamics of this satellite,
for which we could find an approximate solution around the configuration where the smaller and intermediate
moments of inertia of the satellite swap.
We determined the stable configurations for the rotation around each of the principal axes, when the
geometry is not varying. For a geometry variation situation, a state-space formulation was obtained for the
rotation around the principal axes, even though we concluded the solution is very difficult to determine due
to the non-linearities of the system.
5.1 FUTURE WORK
We used simple models for the envisioned satellites as a first approach. However, for a precise description
of the attitude dynamics, more detailed spacecraft models would be required.
It would be interesting to assess which components of the satellite sub-systems could be used more
efficiently to perform the secondary objective of controlling the attitude. Also, a feasibility study could be
conducted in order to determine how the variable geometry attitude control method compares with
conventional control methods, in terms of precision and energy expense, depending on how large the
required attitude maneuvers are.
In this work, the suggested geometries had symmetries so that a variable-geometry term vanishes, in order
to simplify the problem. A different approach to this would be to study the dynamics of bodies that wouldn’t
have such symmetries and evaluate the consequences on the attitude solutions.
We have faced in the present work the difficulties related to obtaining analytical solutions for variable
geometry satellites, even for simple configuration functions and simple satellite models. Consequently, it
would be interesting to assess the applicability of numerical methods to obtain approximate accurate
solutions for the motion of the satellite.
Swapping the order of the values of the principal moments of inertia, as a consequence of configuration
changing, can be studied in the future as a method to induce controlled instability in the rotation of an axis,
which could result in an intended attitude maneuver to achieve large angle changes on the orientation of
spacecraft.
75
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