DYNAMICS and CONTROL of AUTONOMOUS UNDERWATER VEHICLES with INTERNAL ACTUATORS by Bo Li

DYNAMICS AND CONTROL OF AUTONOMOUS UNDERWATER VEHICLES WITH INTERNAL ACTUATORS by Bo Li

A Dissertation Submitted to the Faculty of College of Engineering and Computer Science in Partial Fulﬁllment of the Requirements for the Degree of Doctor of Philosophy

ACKNOWLEDGEMENTS

I want to express my deep gratitude to my advisor, Professor Tsung-Chow Su, for his tremendous support for my dissertation research at Florida Atlantic University in the past three years. I am blessed and honored to be his student. I also wish to thank Professors Karl von Ellenrieder, Manhar Dhanak, Yuan Wang, and Palaniswamy Ananthakrishnan for serving in my supervisory commit- tee and sharing their time and ideas. I owe special thanks to Professor Karl von Ellenrieder for supporting my research in Maritime Systems Laboratory at SeaT- ech in the past six months. I also want to thank Professor Lei Wang at Shanghai Jiao Tong University for opening up the opportunity for me to study in the United States. I gratefully thank my colleague Yanjun Li and his wife Guifang Tang for their invaluable friendship. I also thank all my friends in Boca Raton for their com- pany and friendship all these years. Thanks also to the students, faculty and staﬀ members in my department who have helped me in the past three years. Finally, I thank my parents for their enduring love and support. I am so lucky to have them in my life.

iv ABSTRACT

Author: Bo Li Title: Dynamics and Control of Autonomous Underwater Vehicles with Internal Actuators Institution: Florida Atlantic University Dissertation Advisor: Dr. Tsung-Chow Su Degree: Doctor of Philosophy Year: 2016

This dissertation concerns the dynamics and control of an autonomous underwater vehicle (AUV) which uses internal actuators to stabilize its horizontal- plane motion. The demand for high-performance AUVs are growing in the field of ocean engineering due to increasing activities in ocean exploration and research. New generations of AUVs are expected to operate in harsh and complex ocean environments. We propose a hybrid design of an underwater vehicle which uses internal actuators instead of control surfaces to steer. When operating at low speeds or in relatively strong ocean currents, the performances of control surfaces will degrade. Internal actuators work independent of the relative flows, thus improving the maneuvering performance of the vehicle. We develop the mathematical model which describes the motion of an underwater vehicle in ocean currents from first principles. The equations of motion of a body-fluid dynamical system in an ideal fluid are derived using both Newton-Euler and Lagrangian formulations. The viscous effects of a real fluid are considered separately. We use a REMUS 100 AUV as the research model, and conduct CFD simu-

v lations to compute the viscous hydrodynamic coefficients with ANSYS Fluent. The simulation results show that the horizontal-plane motion of the vehicle is inherently unstable. The yaw moment exerted by the relative flow is destabilizing. The open-loop stabilities of the horizontal-plane motion of the vehicle in both ideal and real fluid are analyzed. In particular, the effects of a roll torque and a moving mass on the horizontal-plane motion are studied. The results illustrate that both the position and number of equilibrium points of the dynamical system are prone to the magnitude of the roll torque and the lateral position of the moving mass. We propose the design of using an internal moving mass to stabilize the horizontal-plane motion of the REMUS 100 AUV. A linear quadratic regulator (LQR) is designed to take advantage of both the linear momentum and lateral position of the internal moving mass to stabilize the heading angle of the vehicle. Al- ternatively, we introduce a tunnel thruster to the design, and use backstepping and Lyapunov redesign techniques to derive a nonlinear feedback control law to achieve autopilot. The coupling effects between the closed-loop horizontal-plane and vertical-plane motions are also analyzed.

vi To my parents, for always loving and supporting me. DYNAMICS AND CONTROL OF AUTONOMOUS UNDERWATER VEHICLES WITH INTERNAL ACTUATORS

List of Tables ...... xi

List of Figures...... xii

1 Introduction ...... 1 1.1 Motivation...... 1 1.2 Dissertation Overview...... 6

2 Rigid Body Dynamics ...... 8 2.1 Deﬁnition of Coordinates...... 8 2.2 Transformations between Reference Frames...... 9 2.2.1 Transformation of Rotational Motion...... 9 2.2.2 Transformation of Rigid Motion...... 12 2.3 6-DOF Rigid-Body Equations of Motion...... 14

3 Body-Fluid Dynamical System in an Ideal Fluid...... 18 3.1 Kinetic Energy of a Body-Fluid Dynamical System...... 18 3.1.1 Stationary Flow with a Moving Body...... 18 3.1.2 Unsteady Spatially Uniform Flow with a Moving Body...... 24 3.1.3 Steady Spatially Non-uniform Flow with a Moving Body...... 27 3.2 Equations of Motion in a Stationary Flow...... 27 3.2.1 Boltzmann-Hamel Equations...... 27 3.2.2 Lie-Poisson Description...... 28 3.3 Equations of Motion in a Uniform Flow...... 30

viii 3.4 Equations of Motion in a Non-uniform Flow...... 33 3.5 Hydrostatics...... 37 3.6 Summary...... 41

4 Body-Fluid Dynamical System in a Real Fluid...... 45 4.1 CFD Methods to Predict Viscous Forces on AUVs...... 45 4.1.1 Axial Drag...... 46 4.1.2 Cross-Flow Drag...... 49 4.1.3 Body Lift...... 59 4.1.4 Fin Lift...... 66 4.2 Viscous Drag Terms in the Equations of Motion...... 70 4.3 Summary...... 73

5 Stability of an Underwater Vehicle in Ocean Currents ...... 75 5.1 Stability of an Underwater Vehicle in a Stationary Flow...... 77 5.1.1 Special Case in an Ideal Fluid...... 77 5.1.2 Stability of the Horizontal-Plane Motion...... 88 5.1.3 Effect of Roll Torque on the Horizontal-Plane Motion...... 105 5.1.4 Effect of A Moving Mass on the Horizontal-Plane Motion...... 118 5.2 Stability of an Underwater Vehicle in a Steady and Uniform Flow...... 121 5.2.1 Special Case in an Ideal Fluid...... 121 5.2.2 Stability of a Vehicle in a Real Fluid...... 129 5.3 Effect of Unsteadiness and Non-uniformity...... 136

6 Control of an Underwater Vehicle Using Internal Actuators ...... 139 6.1 Control of an Underactuated Vehicle...... 139 6.2 Kinetics of Underwater Vehicles with an Internal Actuator...... 141 6.2.1 Internal Rotor...... 141 6.2.2 Internal Moving Mass...... 146

ix 6.2.3 Equations of Motion on a Horizontal Plane ...... 150 6.3 Heading Autopilot for an Underwater Vehicle ...... 155 6.3.1 Linear Control Design ...... 155 6.3.2 Nonlinear Control Design ...... 177 6.3.3 Nonlinear Control Design with Tunnel Thrusters ...... 184 6.3.4 Coupling with the Vertical-Plane Motion ...... 212

7 Conclusions ...... 222

Appendices ...... 224 A Derivation of Body-Fluid Dynamical System Based on Newton-Euler Formulation ...... 225 A.1 Mathematical Preliminaries...... 225 A.2 Hydrodynamic Loads in a Stationary Flow ...... 232 A.3 Hydrodynamic Loads in a Spatially Uniform Unsteady Flow .... 240 A.4 Hydrodynamic Loads in a Spatially Non-uniform Flow...... 246

Bibliography ...... 252

x LIST OF TABLES

4.1 Main Geometric Parameters of the REMUS AUV ...... 46 4.2 Principal Parameters in Numerical Modeling...... 47 4.3 Principal Parameters in Numerical Modeling...... 50 4.4 Drag Coefficients from Numerical Simulations ...... 52 4.5 Mesh Size in Mesh Independent Study...... 56 4.6 Principal Parameters in Numerical Modeling...... 56 4.7 Principal Parameters in Numerical Modeling...... 59 4.8 Mesh Size in Mesh Independent Study...... 59 4.9 Body Lift Coefficients from Numerical Simulations ...... 66 4.10 REMUS Fin Parameters ...... 67 4.11 Hydrodynamic Coefficients of the AUV of Bare Hull ...... 69

5.1 Parameters and Hydrodynamic Coefficients of the REMUS AUV ...... 94 5.2 REMUS New Vertical Fin Parameters ...... 100 5.3 Hydrodynamic Coefficients of A Modified REMUS AUV ...... 100 5.4 Parameters and Hydrodynamic Coefficients of the REMUS AUV ...... 109

6.1 Linearized Coeﬃcients of the Depth-Plane Motion ...... 213 6.2 Control Parameters of the Vertical-Plane Motion ...... 214 6.3 Hydrodynamic Coeﬃcients of the REMUS AUV...... 215

xi LIST OF FIGURES

2.1 Reference frames of an AUV. xbybzb represents the body-ﬁxed frame, and xeyeze is the earth-ﬁxed frame...... 8

4.1 REMUS 100 AUV model...... 46 4.2 Surface and volume meshes around the AUV...... 47 4.3 Forward Speed vs. vehicle axial drag...... 48 4.4 Streamlines around the AUV when u =1 m/s...... 48 4.5 Surface and volume meshes around the AUV...... 49 4.6 Axial Drag v.s Normal Velocity ...... 51 4.7 Normal Drag v.s Normal Velocity...... 51 4.8 Yaw Moment v.s Normal Velocity...... 52 4.9 Vortex core region in the wake of the incoming ﬂow when the normal speed is 0.4 m/s...... 53 4.10 Velocity ﬁeld around the vehicle with SST k − ω model when the normal speed is 0.4 m/s...... 54

4.11 Velocity field around the vehicle with k−kl−ω model when the normal speed is 0.4 m/s...... 54 4.12 Surface pressure distribution around the cross section at x = 0...... 55 4.13 Surface and volume meshes around the AUV in Fine Mesh Case...... 57 4.14 Axial Drag v.s Normal Velocity in Different Mesh Cases...... 58 4.15 Normal Drag v.s Normal Velocity in Different Mesh Cases...... 58 4.16 Surface and volume meshes around the AUV...... 60 4.17 Axial Drag Force v.s Lateral Velocity (u = 1.5 m/s)...... 60 4.18 Lateral Drag Force v.s Lateral Velocity (u = 1.5 m/s)...... 61 4.19 Yaw Moment v.s Lateral Velocity (u = 1.5 m/s)...... 61

xii 4.20 Body normal (lift) force and the total lateral drag v.s the angle of attack of the relative flow...... 63 4.21 Vortex core region in the wake of the incoming flow when the angle of attack is 33◦...... 64 4.22 Yaw moments (contributed only by lift forces) v.s angle of attack...... 65 4.23 Modified REMUS AUV model...... 68

5.1 The kinetic energy ellipsoid and phase portrait of the dynamical system when rg = 0 and a26 = 0...... 82 5.2 Level surface of kinetic energy and phase portrait of the dynamical system when rg 6= O or a26 6= 0...... 86 5.3 Level sets of kinetic energy and linear impulse of the dynamical system rg 6= O or a26 6= 0...... 87 5.4 Phase portrait of the nonlinear system based on REMUS AUV...... 96 5.5 Phase portrait of the nonlinear system based on modiﬁed conﬁguration of the REMUS AUV...... 101 5.6 The trajectory of the stable equilibrium points as µ increases...... 103 5.7 The trajectory of the stable equilibrium points as η increases...... 104 5.8 One projection of the phase portrait of the lateral-directional motion onto vrφ space...... 111

5.9 The dependence of x5 of the equilibrium point on the parameter Q...... 113 5.10 The dependence of the equilibrium point on the parameter Q observed in Ox1x2x3 space...... 114 5.11 The dependence of the equilibrium point on the parameter Q observed in Ox1x2x3 frame and x5 axis...... 116

5.12 The dependence of x5 of the equilibrium point on yg...... 119

5.13 The dependence of the equilibrium point on yg in Ox1x2x3 frame...... 120

5.14 Trajectories projected on a three-dimensional phase space when rg = 0 and a26 = 0...... 128

5.15 Projection of the phase portrait onto Ox1x2x3 space for a speciﬁc initial value of x4...... 133

5.16 Projection of the phase portrait onto Ox1x2x4 space...... 134

xiii 6.1 Reference frames of an AUV with an internal rotor...... 142 6.2 Reference frames of an AUV with an internal moving mass...... 147 6.3 Feedback block diagram of the LQR controller...... 158 6.4 Simulation of linear and angular velocities of the vehicle (Case 1)...... 159 6.5 Simulation of roll and yaw angle of the vehicle, and the position and linear momentum of the internal moving mass (Case 1)...... 160 6.6 Simulation of linear and angular velocities of the vehicle (Case 2)...... 161 6.7 Simulation of roll and yaw angle of the vehicle, and the position and linear momentum of the internal moving mass (Case 2)...... 162 6.8 Simulation of linear and angular velocities of the vehicle (Case 3)...... 163 6.9 Simulation of roll and yaw angle of the vehicle, and the position and linear momentum of the internal moving mass (Case 3)...... 164 6.10 Step response of the actuator dynamics...... 169 6.11 Simulation of linear and angular velocities of the vehicle (Case 1)...... 172 6.12 Simulation of roll and yaw angle of the vehicle, and the position and local velocity of the internal moving mass (Case 1)...... 173 6.13 Simulation of linear and angular velocities of the vehicle (Case 2)...... 174 6.14 Simulation of roll and yaw angle of the vehicle, and the position and local velocity of the internal moving mass (Case 2)...... 175 6.15 Simulation of linear and angular velocities of the vehicle (Case 3)...... 176 6.16 Simulation of roll and yaw angle of the vehicle, and the position and local velocity of the internal moving mass (Case 3)...... 177 6.17 Simulation of linear and angular velocities of the vehicle (Case 1)...... 196 6.18 Simulation of roll and yaw angle of the vehicle, the position and local velocity of the internal moving mass, and the force exerted by jet pumps (Case 1)...... 196 6.19 The errors of the dynamics during the simulation (Case 1)...... 197 6.20 Simulation of linear and angular velocities of the vehicle (Case 2)...... 197 6.21 Simulation of roll and yaw angle of the vehicle, the position and local velocity of the internal moving mass, and the force exerted by jet pumps (Case 2)...... 198 6.22 The errors of the dynamics during the simulation (Case 2)...... 198 xiv 6.23 Simulation of linear and angular velocities of the vehicle (Case 3)...... 199 6.24 Simulation of roll and yaw angle of the vehicle, the position and local velocity of the internal moving mass, and the force exerted by jet pumps (Case 3)...... 199 6.25 The errors of the dynamics during the simulation (Case 3)...... 200 6.26 The linear and angular velocities of the vehicle in ocean currents when ψd = 0...... 201 6.27 The roll and yaw angles, and the actuations of the vehicle in ocean currents when ψd = 0...... 202

6.28 The trajectory of the vehicle in ocean currents when ψd = 0...... 202 6.29 The linear and angular velocities of the vehicle in ocean currents when ψd = −β...... 203 6.30 The roll and yaw angles, and the actuations of the vehicle in ocean currents when ψd = −β...... 203

6.31 The trajectory of the vehicle in ocean currents when ψd = −β...... 204 6.32 The linear and angular velocities of the controlled vehicle advancing at a low speed...... 206 6.33 The roll and yaw angles, and the actuations of the vehicle advancing at a low speed...... 206 6.34 The linear and angular velocities of the controlled vehicle at a low speed in ocean currents...... 208 6.35 The roll and yaw angles, and the actuations of the vehicle at a low speed in ocean currents...... 208 6.36 The trajectory of the controlled vehicle advancing at a low speed in ocean currents...... 209 6.37 The linear and angular velocities of the controlled vehicle at a low speed in ocean currents when ψd = −β...... 209 6.38 The roll and yaw angles, and the actuations of the vehicle at a low speed in ocean currents when ψd = −β...... 210 6.39 The trajectory of the controlled vehicle at a low speed in ocean currents when ψd = −β...... 210 6.40 Vertical-Plane Control System Block Diagram...... 213 6.41 The linear and angular velocities of the horizontal-plane motion...... 215

xv 6.42 The roll and yaw angles, and the actuations of the horizontal-plane motion. The actuators chatter because of the coupling between the two planes.216 6.43 The vertical-plane motion and the rudder angle...... 217 6.44 The trajectory of the REMUS vehicle in the earth-ﬁxed frame...... 217 6.45 The comparison between the sway thrust and the non-inertial force...... 218 6.46 The linear and angular velocities of the horizontal-plane motion...... 220 6.47 The roll and yaw angles, and the actuations of the horizontal-plane motion. Note that the internal moving mass is locked at yv = 0, when t > 15 s. However, the jet pump still works...... 220 6.48 The vertical-plane motion and the rudder angle...... 221 6.49 The trajectory of the REMUS vehicle in the earth-ﬁxed frame...... 221

xvi CHAPTER 1 INTRODUCTION

1.1 MOTIVATION

Autonomous underwater vehicles (AUVs) are widely used in ocean engineering for a variety of underwater missions. However, AUVs cannot operate ev- erywhere. Conventional AUVs use a stern mounted thruster and control surfaces to steer their motions. Unfortunately, the control surfaces may become ineffective when AUVs move in environments featuring strong ocean currents. When currents approach or exceed the velocity of AUVs, the drag exerted by the flow tends to push them away from the predefined path, but the control surfaces cannot provide enough force to counteract the drag and steer the vehicles back. Therefore, the vehicles will experience ’crabbing’ and navigational drift, both of which can significantly affect data quality when AUVs are used, for instance, in marine geoscience [1]. Future applications of streamlined AUVs require them to be capable of keeping their path and preventing ’crabbing’ in the presence of strong ocean currents. Adding more control surfaces and increasing their effective areas could be a solution [2], however, accurate measurements of the flow velocity around the control surfaces should be available in order to steer them to obtain desired control forces. For the purpose of improving the maneuverability of AUVs at low speeds or their capability of path-keeping in ocean currents, new designs of AUVs, which are equipped with extra thrusters, have also been proposed in the literature. For instance, AUVs could be internally equipped with through-body tunnel thrusters [3, 4, 5] or jet pumps [6]. Some AUVs are externally mounted with azimuth thrusters, which allow

1 the AUVs more maneuverability [7] or the capability of dynamic station keeping [8]. Using additional thrusters is a straightforward approach to enhance the maneuverability of AUVs, especially in environments featuring ocean currents, since they are able to generate forces almost independent of the flow around the vehicle. How- ever, strong ocean currents exert a large drag force on the vehicle, thus the power consumption of the thrusters might be enormous in that circumstance. In addition, the noise generated by the thrusters may jeopardize the capabilities of the survey devices onboard, and they could increase the drag on the vehicle as well. The external thrusters are also susceptible to corrosion and damage since they are exposed to the ocean environment. As a result, it is imperative to develop alternative means of actuation for AUVs. Internal actuators are promising alternatives to traditional thrusters and control surfaces, whose concept and application developed together with underwater gliders. As a new generation of marine autonomous vehicles, underwater gliders fea- ture low power consumption, low speed and long endurance. Unlike propeller-driven AUVs, typical underwater gliders have fixed wings and tails, and use internal actuators to travel by concatenating a series of upwards and downwards glides. The endurance of underwater gliders is significantly longer than propeller-drive AUVs. However, they typically move at a speed of about 0.3 m/s compared to 1.5–2.0 m/s of AUVs, making them unsuitable to operate in areas of strong ocean currents. A detailed review on the history and development of underwater gliders can be found in [9]. Typical internal actuators used by underwater gliders include electrically or thermally driven buoyancy propulsion system and internal moving mass actuators [10]. Internal actuators are inside the hull of the vehicles, thus isolated from the ocean environment and therefore less prone to corrosion and damage than external actuators. They will also preserve the integrity of the streamlined hull, thus not increase the drag on the vehicle and the difficulty of waterproofing. Furthermore,

2 internal actuators give control authorities independent of the relative flow, therefore allowing the vehicles to operate at low speeds or in harsh ocean environments. For the first generation of underwater gliders including Seaglider [11], Spray [12], and Slocum [13], a buoyancy bladder is mounted in the stern wet section of the hull. The volume of the bladder could be changed, thus the surfacing and diving of the vehicles are achieved by controlling the buoyancy. The pitch motion of underwater gliders is controlled by the moving mass actuators which modulate the center of mass relative to the center of buoyancy. For instance, a Spray glider moves a battery pack longitudinally along the central column to change the vehicle pitch. Recently, designs of hybrid underwater vehicle have been proposed, which combine the characteristics of conventional AUVs and underwater gliders. On the one hand, internal actuators could be used in conventional AUVs. For instance, FòlagaIII [6] and its latest evolution eFòlaga[14] are torpedo-like vehicles driven by jet pumps at the vehicle stern, but unlike conventional AUVs, Fòlagavehicles do not have control surfaces. Their maneuverability is achieved using glider-like actuation for pitch and heave, and using jet pumps at the bow and the stern for yaw. A ballast chamber is controlling the vehicle buoyancy by injecting or eject- ing water, while internal displacement of the battery pack is actuated along the surge axis for pitch control. [15] used backstepping approach to design a trajectory- tracking controller for an AUV moving in the vertical plane using an internal point mass and a rear thruster as actuators. On the other hand, propeller units of AUVs could be combined into underwater gliders to increase their forward speeds, therefore they have the capability to overcome ocean currents [16, 17, 18]. Slocum G2 hybrid glider is an evolution of Slocum glider, which is equipped with a collapsi- ble propeller that folds when not being used [19, 20]. The hybrid capability allows the vehicle to pass through strong ocean currents and shallow waters with great efficiency.

3 Most underwater gliders, such as Seaglider and Slocum, make use of rudders to control their headings like conventional AUVs. On the contrary, Spray controls yaw through roll motions. Besides the battery pack controlling pitch, another battery pack is located in the nose of a Spray glider, which could be rotated around the axial column to control roll. Similar design could be found in [21]. The idea of controlling underwater vehicle’s yaw through roll motion is novel, although it is one of the most fundamental control strategies in flight dynamics. As a matter of fact, the phenomenon that vehicle yaw is affected by its roll is obvious from the standpoint of dynamics. Therefore, conventional AUVs are usually properly trimmed in order to eliminate the yaw-roll coupling as much as possible. Actually some devices have been developed to serve the same purpose. For example, Bluefin survey AUV [22], which has no control fins, is propelled and controlled by a Bluefin tailcone. The tailcone is an articulated ring-wing control surface with a ducted propeller, and it can be actuated up to ±12° in the horizontal and vertical directions, acting as both a rudder and an elevator. The tailcone is specially designed to be torque neutral whose stators are angled to counteract the torque induced by the propeller, therefore no roll will be produced due to the propeller. Moreover, control strategies are also studied in order to suppress the coupling effect between the pitch and yaw motions due to the existence of nonzero roll angles [23]. The stabilization of conventional underwater vehicles using internal rotors has been studied by a few authors. The stabilization of steady long-axis translation using three internal rotors was discussed by [24, 25]. The method of controlled Lagrangian was used to design the feedback law for an underwater vehicle with internal rotors, and the effect of viscous damping on the resulting closed-loop system was also discussed. An internal rolling mass was used in [26], which can shift the center of gravity in the sway axis, to stabilize the roll motion. Internal rotors for underwater vehicles are counterparts of reaction wheels for satellites. Both of them

4 can be classified as gyrostabilizers, which are used for land, marine and spacecrafts. A thorough examination of the current state of gyrostabilizer vehicular technology can be found in [27]. In addition to reaction wheels, momentum wheels and control moment gyros (CMGs) are also commonly used gyrostabilizers, especially in spacecrafts. The angular momentum can be transferred between the gyrostabilizers and the craft’s body without changing its overall inertial angular momentum. Therefore, the resulting torque control devices are also called momentum exchange devices [28]. Gy- rostabilizers were first introduced to marine technology more than one hundred years ago to minimize the undesired roll motion, and are nowadays predominantly utilized in luxury motorboats [29, 30]. Applications of CMGs in AUVs were also studied recently for the purpose of attitude control. [31, 32] first proposed the internal actuation of AUVs using CMGs for unrestricted attitude control in a 3D manner. A CMG-actuated test-bed AUV demonstrated the capability of maintain- ing any attitude on the surface of a sphere with a zero radius turning circle and actively stabilize any attitude while translating in surge. Subsequently, CMGs were also used for energy storage onboard AUVs [33]. A proposed control law is capable of separating the energy and momentum transfer functions of the CMG system to control all three rotational axes while simultaneously transferring energy to supply the entire electrical load of the vehicle. Compared with reaction wheels, CMGs can offer considerably larger torque. However, reaction wheels are less mechanically complex and easier to operate so that they are more widely used in spacecrafts. Nevertheless, CMGs still have their advantage in the applications which demand large and agile torque actions. Although a variety of internal actuators have been successfully applied to diving control for AUVs and underwater gliders, a comprehensive study on its use for heading control is rare in the literature. In this dissertation, we address the

5 topic of controlling an AUV’s heading angle by taking advantage of its roll-yaw coupling eﬀect.

1.2 DISSERTATION OVERVIEW

The outlines of the remaining chapters are summarized as follows. In Chapter 2, the reference frames including the body-fixed and earth-fixed frame are defined. The transformation matrix describing the relation between the two reference frames is presented. The 6-DOF rigid-body equations of motion are derived from first principles, and written in a matrix form. Chapter 3 describes the equations of motion of a rigid body moving in an infinite and unbounded ideal fluid. We use the potential flow theory to derive the equations from first principles, and extend the results in a stationary flow to the cases where spatially uniform and non-uniform flow exist. Apart from the classical Newton-Euler formulation which is demonstrated in the appendix, Lagrangian approach is used to verify the results. To close the chapter, we summarize the simplified forms of the equations of motion used in several special cases. In Chapter 4, numerical simulations are conducted to obtain the viscous hydrodynamic coefficients of a REMUS 100 AUV. The simulation results are compared with the values reported in the literature. The hydrodynamic coefficients will be used in the remaining chapters to analyze the stability of the REMUS AUV, and design feedback control laws to stabilize its horizontal-plane motion. The viscous hydrodynamic forces are summarized in the matrix form and incorporated into the equations of motion obtained in the preceding chapter. In Chapter 5, the stabilities of the REMUS AUV’s horizontal-plane motions in both stationary and steady flows are analyzed as dynamical systems. It shows that the vehicle cannot achieve straight-line stability due to the destabilizing hydrodynamic yaw moment. The operating point in the phase portrait of the dynamical

6 system is an unstable equilibrium point, and there are two stable equilibrium points in the domain of interest. We also analyze how a nonzero roll torque and the lateral position of an internal moving mass affect the flow of the dynamical system. The results illustrate that the positions of the equilibrium points in the phase portrait can be significantly affected by the two factors. A bifurcation occurs if the roll torque is large enough or the internal moving mass is far enough from the longitudi- nal center plane of the vehicle, where the unstable equilibrium point and one of the stable point disappear. Chapter 6 presents the novel heading control design for the REMUS 100 using internal actuators. The equations of motion of the vehicle with an internal rotor and internal moving mass are derived. The moving mass is chosen as the internal actuator to stabilize the heading angle of the vehicle in a horizontal plane. First, a linear quadratic regular (LQR) controller is design to stabilize the yaw motion of the vehicle controlling the lateral position of the internal moving mass. Several nonlinear control methods are also proposed. In order to improve the performance of the control design, a tunnel thruster is introduced to the vehicle, which produces a thrust in sway direction. Backstepping and Lyapunov redesign techniques are used to design the control laws for both the moving mass and tunnel thruster. Numerical simulations show that the controllers can successfully stabilize the heading angle of the vehicle operating at normal or low speeds, and in ocean currents. The coupling effects between the closed-loop horizontal-plane and vertical-plane motions are also analyzed. Chapter 7 recapitulates the major contributions of the dissertation, and includes suggestions for future work.

7 CHAPTER 2 RIGID BODY DYNAMICS

2.1 DEFINITION OF COORDINATES

The body-fixed reference frame is assigned to be fixed to the underwater vehicle with its origin at the center of buoyancy of the vehicle as illustrated in the following figure.

Figure 2.1: Reference frames of an AUV. xbybzb represents the body-ﬁxed frame, and xeyeze is the earth-ﬁxed frame.

The linear and angular velocities of the vehicle are generally expressed in the body-ﬁxed reference frame, which are denoted by the following vectors [34]:

T 3 T 3 u = [u, v, w] ∈ R , ω = [p, q, r] ∈ R

in which u, ω denote the linear and angular velocities, respectively. 8 The position and altitude (Euler angles) of the vehicle are usually expressed in the earth-ﬁxed reference frame. In the dissertation, we adopt the North-East- Down (NED) coordinate system, in which the x axis points towards true North, the y axis points towards East and the z axis points downward normal to the Earth’s surface. The position and Euler angles of the vehicle are denoted by the vector

T T T η = [η1 , η2 ] ,

T 3 η1 = [x, y, z] ∈ R (2.1.1)

T 3 η2 = [φ, θ, ψ] ∈ T (2.1.2)

2.2 TRANSFORMATIONS BETWEEN REFERENCE FRAMES

2.2.1 Transformation of Rotational Motion

A rotation matrix R is generally used to denote the relative orientation between the body-ﬁxed and earth-ﬁxed reference frames. R is an element of the Lie group SO(3), which denotes the special orthogonal group of order 3. We can denote cross product operation c = a × b as

c = S(a)b where the map S : R3 → so(3) is deﬁned as 1   0 −a3 a2     S(a) =  a 0 −a  (2.2.1)  3 1   −a2 a1 0

then S(a) belongs to the Lie algebra so(3) of SO(3). Cross product is actually equivalent to the Lie bracket of so(3) as

S(c) = [S(a),S(b)] = S(a)S(b) − S(b)S(a) (2.2.2)

1 3 This map is often denoted as Hat map b :(R , ×) → (so(3), [·, ·]) in the literature, e.g., [35], [36], and [37]. Here we follow the notation in [34].

9 so(3) is the tangent space of SO(3) at the identity, and thus a vector space of 3 × 3 skew-symmetric matrices. The dimension of so(3) is three. If we consider the one-parameter subgroup of SO(3), then we will have the

so-called principal rotations Rx,φ, Ry,θ, Rz,ψ ∈ SO(3), given by       1 0 0 cos θ 0 sin θ cos ψ − sin ψ 0             Rx,φ = 0 cos φ − sin φ , Ry,θ =  0 1 0  , Rz,ψ = sin ψ cos ψ 0             0 sin φ cos φ − sin θ 0 cos θ 0 0 1 (2.2.3) Their derivatives at the identity, i.e., φ = 0, θ = 0, ψ = 0, give the bases of so(3), which are       0 0 0 0 0 1 0 −1 0       T   T   T   S([1, 0, 0] ) = 0 0 −1 ,S([0, 1, 0] ) =  0 0 0 ,S([0, 0, 1] ) = 1 0 0             0 1 0 −1 0 0 0 0 0 (2.2.4) and they are called inﬁnitesimal generators of the one-parameter subgroup of SO(3). Therefore the Lie algebra so(3) can be written as   0 −r q (   )   so(3) =  r 0 −p p, q, r ∈ R (2.2.5)     −q p 0

We can identify so(3) with R3 carrying the cross product as Lie bracket. For the one-parameter subgroup of SO(3), we have the exponential map exp : so(3) → SO(3) given by

S(λ)2t2 R = exp(S(λ)t) = I + S(λ)t + + ··· (2.2.6) λ,t 3×3 2! where λ is the unit vector denoting the direction. The map can also be given by Rodrigues’ formula [35],

2 exp(S(λ)t) = I3×3 + S(λ) sin t + S(λ) (1 − cos t) (2.2.7) 10 This formula can be used to obtain (2.2.3) from (2.2.4). The total rotation matrix R from the body-fixed reference frame to the earth-fixed reference frame is defined as the product of the one-parameter rota-

tion matrices Rx,φ, Ry,θ, Rz,ψ ∈ SO(3). The order of the matrix product is carried out by following the zyx convention in ocean engineering [34]. Then R =

Rz,ψRy,θRx,φ ∈ SO(3), which can be written as:   cψcθ cψsφsθ − cφsψ sφsψ + cφcψsθ     R =  cθsψ cφcψ + sφsψsθ cφsψsθ − cψsφ  (2.2.8)     −s(θ cθsφ cφcθ where s· = sin(·) and c· = cos(·). Then we can verify that

T −1 RR = RR = I3×3 (2.2.9)

We use rotation matrix R to transform the linear velocity u of the vehicle in the body-ﬁxed reference frame into the earth-ﬁxed reference frame as:

η˙ 1 = Ru (2.2.10)

With the rotation matrix R, the body angular velocity ω can be deﬁned in so(3) as: [35, 37] S(ω) = R−1R˙ ∈ so(3) (2.2.11) which is actually the left translation carried on R˙ to so(3). Then we obtain that   0 θ˙ sin φ − ψ˙ cos θ cos φ θ˙ cos φ + ψ˙ cos θ sin φ     S(ω) = −θ˙ sin φ + ψ˙ cos θ cos φ 0 −φ˙ + ψ˙ sin θ      −θ˙ cos φ − ψ˙ cos θ sin φ φ˙ − ψ˙ sin θ 0 (2.2.12) Thus         p φ˙ − ψ˙ sin θ 1 0 − sin θ φ˙                 q =  θ˙ cos φ + ψ˙ cos θ sin φ  = 0 cos φ cos θ sin φ θ˙ (2.2.13)                 r −θ˙ sin φ + ψ˙ cos θ cos φ 0 − sin φ cos θ cos φ ψ˙ 11 Therefore, we can deﬁne the transformation matrix for angular velocities as

−1 ω = T η˙ 2 (2.2.14) where   1 sin φ tan θ cos φ tan θ     T =  0 cos φ − sin φ  (2.2.15)     0 sin φ/ cos θ cos φ/ cos θ Finally, the transformation matrix from the body-ﬁxed reference frame to the earth-ﬁxed reference frame is given by   RO3×3 η˙ = J(η)ν =   ν (2.2.16) O3×3 T where ν = [uT , ωT ]T ∈ R6 is the velocity vector of the rigid body.

2.2.2 Transformation of Rigid Motion

Special Euclidean group SE(3) can be used to describe the kinematics of an AUV [38], which is represented as:

(   ) R η 1 3 SE(3) =   R ∈ SO(3), η1 ∈ R (2.2.17) O1×3 1 where R is the rotation matrix representing the orientation of the rigid body and

η1 describes its position in the earth-ﬁxed reference frame. SE(3) is thus identiﬁed as the left action of a semidirect-product Lie group [37],

3 SE(3) = SO(3) s R

3 If xb ∈ R is the position of a point on the rigid body expressed with respect to the

3 body-fixed reference frame, its position xe ∈ R in earth-fixed reference frame could be obtained by         xe R η1 xb Rxb + η1   =     =   (2.2.18) 1 O1×3 1 1 1 12 The pair (R, η1) satisfies the kinematic equations from the preceding subsection,   R˙ = RS(ω) (2.2.19)   η˙ 1 = Ru

Then the body velocity [uT , ωT ]T ∈ R6 of an AUV can be described by the Lie algebra se(3) associated to SE(3), which is deﬁned as

(   ) S(ω) u 3 se(3) =   S(ω) ∈ so(3), u ∈ R (2.2.20) O1×3 0

∼ 6 being the tangent space of SE(3) at its identity. Conversely, se(3) = R is identiﬁed by the pair (ω, u). The Lie bracket of se(3) is given by       S(ω1) u1 S(ω2) u2 S(ω1)S(ω2) − S(ω2)S(ω1) S(ω1)u2 − S(ω2)u1   ,   =   O1×3 0 O1×3 0 O1×3 0 (2.2.21) or in a compact notation

[(ω1, u1), (ω2, u2)] = (ω1 × ω2, ω1 × u2 − ω2 × u1)

The kinematic equations of an AUV can thus be written as

X˙ = XY (2.2.22) where X ∈ SE(3), and Y ∈ se(3).

13 2.3 6-DOF RIGID-BODY EQUATIONS OF MOTION

If the total force acting on the rigid-body is denoted as F 0, the dynamic equation of translational motion in body-ﬁxed frame can be derived as dp F = 0 dt ZZZ d = (u + ω × r)dm dt V du ZZZ d = m + (ω × r)dm dt dt V d u ZZZ dr = m( r + ω × u) + (ω˙ × r + ω × )dm dt dt V 0 ZZZ 7 drr = m(u˙ + ω × u) + ω˙ × mrg + ω × ( + ω × r)dm dt V

= m(u˙ + ω × u) + m[ω˙ × rg + ω × (ω × rg)]

Hence,

F 0 = m(u˙ + ω˙ × rg) + m[ω × u + ω × (ω × rg)] (2.3.1)

T 3 T where F 0 = [X,Y,Z] ∈ R , and rg = [xg, yg, zg] denotes the position of mass center in the body-ﬁxed reference frame . The angular momentum of a rigid body with respect to the body-ﬁxed reference frame can be expressed as ZZZ π0 = r × νdm (2.3.2) V The above equation can be expanded as ZZZ π0 = r × [u + (ω × r)]dm V ZZZ ZZZ = (r × u)dm + r × (ω × r)dm

V V

14 Since   (y2 + z2)p − xyq − xzr     r × (ω × r) =  −xyp + (x2 + z2)q − yzr  (2.3.3)     −xzp − yzq + (x2 + y2)r then we have       Ixxp − Ixyq − Ixzr Ixx −Ixy −Ixz p ZZZ             r × (ω × r) dm =  −I p + I q − I r  =  −I I −I   q   xy yy yz   xy yy yz    V       −Ixzp − Iyzq + Izzr −Ixz −Iyz Izz r (2.3.4) or ZZZ r × (ω × r) dm = I0ω V where I0 is the tensor of inertia. Thus, π0 can be written as:

π0 = I0ω + mrg × u (2.3.5)

Diﬀerentiating (2.3.2) with respect to time yields dπ ZZZ dr ZZZ dν 0 = × νdm + r × dm (2.3.6) dt dt dt V V The second term is the moment vector acting on the rigid body, ZZZ dν ZZZ r × dm = r × dF = M dt 0 V V The ﬁrst term can be further expanded as ZZZ dr ZZZ × νdm = (ω × r) × (u + ω × r)dm dt V V ZZZ = −u × (ω × r)dm

V ZZZ 0 = −u × (ω × (rg + r ))dm V 0 ZZZ ¨¨* 0¨ = −u × (ω × mrg) − u × (ω × ¨r dm) ¨¨ ¨V

= −u × (ω × mrg)

15 where r0 is the position vector with respect to the center of mass. Consequently, (2.3.6) can be rewritten as

dπ 0 = M − u × (ω × mr ) (2.3.7) dt 0 g

Diﬀerentiating (2.3.5) with respect to time gives

dπ 0 = I ω˙ + ω × (I ω) + (ω × mr ) × u + mr × (u˙ + ω × u) (2.3.8) dt 0 0 g g

Subtracting (2.3.7) by (2.3.8) yields

I0ω˙ + mrg × u˙ + ω × (I0ω) + mrg × (ω × u) = M 0 (2.3.9)

Given the above results, the total 6-DOF rigid-body kinetic equations in the body- ﬁxed reference frame can be written as:

m(u˙ + ω˙ × rg) + m[ω × u + ω × (ω × rg)] = F 0

I0ω˙ + mrg × u˙ + ω × (I0ω) + mrg × (ω × u) = M 0

The first two terms of two equations can be combined into the form of matrix       m(u˙ + ω˙ × rg) mI3×3 −mS(rg) u˙   =     = M RBν˙ (2.3.10) mrg × u˙ + I0ω˙ mS(rg) I0 ω˙ where ν = [uT , ωT ]T = [u, v, w, p, q, r]T ∈ R6 is the velocity vector of the rigid body. Due to (2.3.10), the inertia matrix of the rigid body defined in the body- fixed reference frame can be obtained as   m 0 0 0 mzg −myg      0 m 0 −mz 0 mx   g g       0 0 m myg −mxg 0  M RB =   (2.3.11)    0 −mzg myg Ixx −Ixy −Ixz       mz 0 −mx −I I −I   g g xy yy yz    −myg mxg 0 −Ixz −Iyz Izz 16 And the Coriolis-centripetal inertia matrix can also be derived as:       mω × u + mω × (ω × rg) mS(ω) −mS(ω)S(rg) u   =     ω × (I0ω) + mrg × (ω × u) mS(rg)S(ω) −S(I0ω) ω

= CRB(ω)ν (2.3.12) in which  0 −mr mq    mr 0 −mp     −mq mp 0 CRB(ω) =    −mqyg − mrzg mpyg mpzg    mqx −mpx − mrz mqz  g g g g  mrxg mryg −mpxg − mqyg

 mqyg + mrzg −mqxg −mrxg   −mpy mpx + mrz −mry  g g g g    −mpzg −mqzg mpxg + mqyg    0 −Ixzp − Iyzq + Izzr Ixyp − Iyyq + Iyzr    I p + I q − I r 0 I p − I q − I r  xz yz zz xx xy xz   −Ixyp + Iyyq − Iyzr −Ixxp + Ixyq + Ixzr 0 (2.3.13) Thus, the 6-DOF rigid-body kinetic equations can be expressed as [34]

M RBν˙ + CRB(ω)ν = τ RB (2.3.14)

T T T 6 where τ RB = [F 0 , M 0 ] ∈ R is the vector of excitation forces acting on the rigid body.

17 CHAPTER 3 BODY-FLUID DYNAMICAL SYSTEM IN AN IDEAL FLUID

This chapter develops the equations of motion of a rigid body moving in an infinite and unbounded ideal fluid based on Lagrangian Formulation. The equations of motion in a stationary or uniform flow are well-known in textbooks on theoretical hydrodynamics [39, 40]. Many details of the derivation based on Newton-Euler formulation can be found in [41]. For the completeness and consistency of the dissertation, we reproduce the work proposed in [41] in the appendix, and expand the classical results in a stationary or uniform flow to the case of spatially non-uniform flows. In this chapter, we adopt the energy-based method to derive the equations from first principles, and compare the results with the Newton-Euler formulation.

3.1 KINETIC ENERGY OF A BODY-FLUID DYNAMICAL SYS- TEM

3.1.1 Stationary Flow with a Moving Body

If the translational and rotational velocities are u and ω of the body-ﬁxed

reference frame, the kinetic energy TB of a rigid body can be expressed as:

1 ZZZ 1 ZZZ T = |ν|2dm = |u + ω × r|2dm (3.1.1) B 2 2 V V

18 Using some results in Newton-Euler formulation, the expression of kinetic energy can be further derived as follows:

1 ZZZ T = |u + ω × r|2dm B 2 V 1 ZZZ = (u + ω × r)T (u + ω × r)dm 2 V 1 ZZZ = (uT u + uT (ω × r) + (ω × r)T u + (ω × r)T (ω × r))dm 2 V The ﬁrst term of the above equation can be integrated as:

1 ZZZ 1 uT u dm = uT mu 2 2 V The second term is

1 ZZZ 1 ZZZ 1 1 uT (ω × r)dm = uT (ω × r)dm = uT (ω × mr ) = uT (−mS(r )ω) 2 2 2 g 2 g V V The third term is

1 ZZZ 1 (ω × r)T u dm = (ω × mr )T u 2 2 g V 1 = (−mS(r )ω)T u 2 g 1 = ωT mS(r )u 2 g

The last term is

1 ZZZ 1 ZZZ (ω × r)T (ω × r)dm = ωT (r × (ω × r))dm 2 2 V V 1 = ωT I ω 2 0

Obtaining the above results, the kinetic energy T can be illustrated as:

1 T = uT mu + uT (−mS(r )ω) + ωT mS(r )u + ωT I ω (3.1.2) B 2 g g 0

19 Write the formula into matrix form, we have     1 h i mI3×3 −mS(rg) u 1 T T T TB = u ω = ν M RBν (3.1.3) 2     2 mS(rg) I0 ω

Hence, we also derive the inertia matrix M RB of the rigid body, which is identical to the result in Newton-Euler formulation. Then we can obtain the relations between the kinetic energy and momentum vector of the rigid body     p ∂T /∂u ∂T = M ν = = (3.1.4)   RB   ∂ν π0 ∂T /∂ω

For an unbounded ideal ﬂuid, its kinetic energy is given by ZZZ 1 2 TF = ρ |v| dV 2 V (3.1.5) 1 ZZZ = ρ ∇φ · ∇φ dV 2 V where φ is the velocity potential function. Using Gauss’ theorem

ZZ ZZZ φ ∇φ n dS = ∇ · (φ ∇φ)dV S V ZZZ = (∇φ · ∇φ + φ∇2φ) dV V ZZZ = ∇φ · ∇φ dV V Hence, 1 ZZ T = ρ φ ∇φ n dS F 2 S (3.1.6) 1 ZZ ∂φ = ρ φ dS 2 ∂n S In the scenario where calm fluid is disturbed by a single rigid body moving in it, the kinetic energy of the fluid can be derived explicitly as what follows. In the far field of the control volume, the velocity of fluid is in the order of |r|−3. Hence, the

20 integral on the exterior surface decays as |r|−1, which can be assumed to be zero when the boundary goes to inﬁnity. Using this boundary condition, we have

1 ZZ ∂φ T = ρ φ dS (3.1.7) F 2 ∂n SB

Using boundary condition(A.48) and the deﬁnition of added mass (A.57), we have

1 ZZ T = ρ φ [u · n + ω · (r0 × n)] dS F 2 SB 1 ZZ 1 ZZ = ρ u · φ n dS + ρ ω · φ (r0 × n) dS 2 2 SB SB 1 ZZ 1 ZZ = ρ u · ν φ n dS + ρ ω · ν φ (r0 × n) dS 2 i i 2 i i SB SB 1 1 1 1 = u · (A u) + u · (A ω) + ω · (A u) + ω · (A ω) 2 11 2 12 2 21 2 22 1 1 1 1 = uT A u + uT A ω + ωT A u + ωT A ω 2 11 2 12 2 21 2 22 writing it into matrix form 1 T = νT M ν (3.1.8) F 2 A For the case that the rigid body is translating without rotation, we have then [40]

1 1 T = uT A u = (4πρu · C0 − m¯ uT u) (3.1.9) F 2 11 2

in which C0 is deﬁned in (A.66).

The expressions of TF show that TF is a homogeneous quadratic function of u and ω, so that the motion of the ﬂuid can be expressed in the generalized coordinates [uT , ωT ]. Since

2 TF (λu, λω) = λ TF (u, ω)

according to Euler’s homogeneous function theorem, we have

∂T ∂T u F + ω F = 2T (3.1.10) ∂u ∂ω F 21 This equation also holds for kinetic energy of the rigid body. As a result, ∂(T + T ) ∂(T + T ) u B F + ω B F = 2(T + T ) (3.1.11) ∂u ∂ω B F where TB,TF are the kinetic energy of the rigid body and fluid respectively. This equation can be used to find out the relations between the kinetic energy of the body-fluid system and the impulse wrench on it. Now we are ready to derive the relations between the kinetic energy of the fluid and the hydrodynamic force and moment on the rigid body. ∂T 1 ZZ ∂φ ∂φ ∂ ∂φ F = ρ + φ dS ∂u 2 ∂u ∂n ∂u ∂n SB 1 ZZ ∂φ = ρ ϕ + φn dS 2 ∂n SB Green’s second identity gives that ZZ ZZZ ∂φ ∂ϕ 2 2 ϕ − φ dS = ϕ∇ φ − φ∇ ϕ dV ∂n ∂n S V

Since φ, ϕ are both harmonic functions, neglecting the integral on SC which is of the order |r|−1 will give ZZ ∂φ ZZ ∂ϕ ZZ ϕ dS = φ dS = φ n dS ∂n ∂n SB SB SB Hence, ∂T ZZ F = ρ φ n dS (3.1.12) ∂u SB Similarly, ∂T ZZ F = ρ φ (r0 × n) dS (3.1.13) ∂ω SB The right side of the above equations have the physical meaning of an impulse wrench on the ﬂuid exerted by the rigid body, which generates the motion of the ﬂuid instantaneously from rest. Using the facts that

∂TF = A11u + A12ω ∂u (3.1.14) ∂T F = A u + A ω ∂ω 21 22 22 which can also be written as:   ∂T pB F = M ν = (3.1.15) ∂ν A   πB

and under the assumption that the impulse wrench on the unbounded fluid only comes from the rigid body, we can write the external forces on the rigid body as [39] d ∂TF ∂TF F 0 = − − ω × dt ∂u ∂u (3.1.16) d ∂T ∂T ∂T M = − F − ω × F − u × F 0 dt ∂ω ∂ω ∂u It is also desirable to explore the expressions of the momentum of the fluid. Recalling (A.31), the linear momentum of the fluid can be expressed as:

ZZ ZZ p = pB + pC = ρ φ n dS + ρ φ n dS (3.1.17)

SB SC The two terms on the right side of the above equation represent the impulse acting on the interior and exterior boundaries of the fluid respectively. The difficulty comes when doing the latter integral. In the far field of the rigid body, using (A.67) we obtain that ZZ ZZ 1 ρ φ n dS = ρ C + C0 · ∇ n dS R SC SC ρ ZZ = − n(C0 · R) dS R3 SC (3.1.18) ρ ZZZ = − ∇(C0 · R) dV R3 VC ρV = − C C0 R3 Apparently, the integral goes to a nonzero constant although the exterior boundary is set to an infinite distance, which also seems to depend on the shape of the boundary itself [40]. As pointed out in [42], the total momentum of the fluid is not equivalent to the impulse, and in the present problem indeterminate. Hence we

23 can only consider the change of the momentum of the fluid caused by the action of the rigid body alone. Since the fluid is unbounded, no impulse on the fluid in the infinite distance needs to be accounted for.

3.1.2 Unsteady Spatially Uniform Flow with a Moving Body

Using (3.1.16) to express the hydrodynamic loads exerted on a rigid body moving in a calm flow is a classic result in potential flow theory. The same energy- based approach might be extended to the case where the undisturbed surrounding fluid has a uniform translation. In this scenario, the velocity potential can be expressed as a sum of incident

∗ velocity potential φo and disturbance velocity potential φ as we did in the derivation of hydrodynamic force and moment in the same scenario. Recalling (A.69),

∗ φ = φo + φ

Hence, the kinetic energy of the fluid is 1 ZZ ∂φ T = ρ φ dS F 2 ∂n S 1 ZZ ∂(φ + φ∗) = ρ (φ + φ∗) o dS 2 o ∂n (3.1.19) S 1 ZZ ∂φ ZZ ∂φ 1 ZZ ∂φ∗ = ρ φ o dS + ρ φ∗ o dS + ρ φ∗ dS 2 o ∂n ∂n 2 ∂n S S S The last integral in the above equation can be expressed using the results of preceding subsection as 1 ZZ ∂φ∗ 1 ρ φ∗ dS = νT M ν (3.1.20) 2 ∂n 2 r A r S where νr = [u − U; ω] is the relative velocity of rigid body with respect to the moving fluid. The first integral is the kinetic energy of the undisturbed fluid of the control

24 volume, so that 1 ZZ ∂φ 1 1 ρ φ o dS = MU T U − m¯ U T U (3.1.21) 2 o ∂n 2 2 S where M is the mass of the fluid volume absent the rigid body, which is infinite in the case of an unbounded fluid. Now the second integral should be taken into consideration,

ZZ ∂φ ZZ ∂φ ZZ ∂φ ρ φ∗ o dS = ρ φ∗ o dS + ρ φ∗ o dS ∂n ∂n ∂n S SB SC We have ZZ ∂φ ZZ ρ φ∗ o dS = ρU φ∗ n dS ∂n SB SB

= U · [A11(u − U) + A12ω]

T = νc M Aνr where νc = [U; O3×1] is the unsteady velocity of undisturbed ﬂuid expressed in body-ﬁxed reference frame.

ZZ ∂φ ZZ ρ φ∗ o dS = ρU · φ∗ n dS (3.1.22) ∂n SC SC Here, we encounter the same difficulty as in the preceding subsection where we have to calculate the impulse on the exterior boundary of the fluid. As a result, it seems impossible to obtain the exact expression of the fluid kinetic energy. But if we make an assumption that (3.1.22) can be ignored, then we have that 1 1 1 T = νT M ν + MνT ν − νT Mν¯ + νT M ν F 2 r A r 2 c c 2 c c c A r 1 1 1 = (ν − ν )T M (ν − ν ) + MνT ν − νT Mν¯ + νT M (ν − ν ) (3.1.23) 2 c A c 2 c c 2 c c c A c 1 1 1 = MνT ν + νT M ν − νT (M¯ + M )ν 2 c c 2 A 2 c A c in which the expression of the inertia matrix of the displaced fluid is the same as (A.79)   m¯ I3×3 −mS¯ (rb) M¯ =   b mS¯ (rb) I0 25 If we add the kinetic energy of the rigid body into it, then we have the total energy of the dynamical system,

1 1 1 T = MνT ν + νT (M + M )ν − νT (M¯ + M )ν (3.1.24) tot 2 c c 2 RB A 2 c A c

Due to the derivation in [43], Ttot = const if the ﬂow is steady, which is apparently

a manifestation of energy conservation. When νc = O6×1,(3.1.24) is degenerated into 1 T = νT (M + M )ν (3.1.25) tot 2 RB A which is exactly the total kinetic energy of the body-fluid system in stationary flow. But a fundamental question arises that whether we can use (3.1.24) to derive the dynamical equations based on energy approach. First of all, we can write the impulse wrench on the original undisturbed fluid given by the rigid body as     p A (u − U) + A ω − m¯ U B ¯ 11 12   = M Aνr − Mνc =   (3.1.26) πB A21(u − U) + A22ω − m¯ rb × U

The ﬁrst term on the right side of the equation represents the impulse due to the relative motion of the rigid body in the ﬂow,

ZZ ∗ M Aνr = ρ φ n dS (3.1.27)

SB while the second term reveals the fact the rigid body displaces a certain part of the

0 fluid in the flow. If we construct a expression of TF which satisfies that   0 ∂T pB F = = (M + M¯ )ν − Mν¯ (3.1.28) ∂ν   A r πB

then we have a new expression of the change of the kinetic energy of the ﬂuid as follows (See also [44, 45])

1 1 1 T 0 = νT M ν + νT Mν¯ − νT Mν¯ (3.1.29) F 2 r A r 2 r r 2 26 If we neglect the inﬁnite term in (3.1.23), and write it as

1 1 T = νT M ν − νT (M¯ + M )ν (3.1.30) F 2 A 2 c A c

We can find that (3.1.29) and (3.1.30) are different. As a result, the total kinetic energy given by (3.1.24) can not be used to derive the dynamical equations using energy approach. The cause of the problem with energy approach is that the dynamics of the body-fluid system cannot be expressed by only using the body-fixed reference frame when the flow is not stationary. If we still need to use Lagrangian formulation to derive the hydrodynamic loads on the rigid body with (3.1.16), then (3.1.29) has to be used. But it is not the real expression of the kinetic energy of the fluid, we may have to call it pseudo kinetic energy.

3.1.3 Steady Spatially Non-uniform Flow with a Moving Body

The change of kinetic energy of spatially non-uniform flow can be derived with an important assumption that the flow velocity νc is effectively constant over the length scale of the rigid body. In this case, (3.1.29) is still valid.

3.2 EQUATIONS OF MOTION IN A STATIONARY FLOW

3.2.1 Boltzmann-Hamel Equations

When deriving the kinetic equations, we can directly use the Lagrangian equations for quasi-coordinates, which is also known as Boltzmann-Hamel equations [46]. The Boltzmann-Hamel equations in our special case can be written as [47]   d ∂T ∂T ∂T F 0 + J T Q − J T = (3.2.1) dt ∂ν ∂ν ∂η   M 0 where J is the transformation matrix given by (2.2.16). Deﬁne matrix S as

T JS = I6×6 (3.2.2) 27 Then the 6×6 matrix Q is expressed as    ∂Si:/∂η1        ∂Si:/∂η2  ∂S  T T    Qi: = ν J  .  −  (3.2.3)  .  ∂ηi        ∂Si:/∂η6 6×6

th th where Qi: denotes the i row of matrix Q and ηi is the i component of η. In the case of an stationary ﬂow in which T doesn’t contain η explicitly, we have ∂T = 0 (3.2.4) ∂η Use (2.2.8) and (2.2.15) to compute J T Q, the result is   S(ω) O T 3×3 J Q =   (3.2.5) S(u) S(ω)

Therefore, Boltzmann-Hamel equations (3.2.1) can be simpliﬁed as     d ∂T S(ω) O3×3 ∂T F 0 + = dt ∂ν   ∂ν   S(u) S(ω) M 0

3.2.2 Lie-Poisson Description

Based on the method from geometric mechanics, the kinetics of the dynamical system can be described using Euler-Poincaréequations. This formulation is very useful in stability analysis. For a body-fluid dynamical system in a stationary flow, the configuration space is SE(3) and the velocity phase space is T SE(3). The

Lagrangian of the system L : T SE(3) → R is given by:

˙ L(R, η1, R, η˙ 1) = L(R, η1, RS(ω), Ru)   1 h i u (3.2.6) T T = u , ω (M RB + M A)   2 ω

28 where the pair (R, η1) ∈ SE(3) is the Lie group describing the kinematics of the rigid body, and the pair (ω, u) ∈ se(3) denotes the velocity. It can be proved that the Lagrangian is a left invariant function on T SE(3). Use the fact that

L ¯ (R, η ) = (RR¯ , Rη¯ + η¯ ) (3.2.7) (R,η¯1) 1 1 1

TL ¯ (R˙ , η˙ ) = (R¯ R˙ , R¯ η˙ ) (3.2.8) (R,η¯1) 1 1 we have ˙ ¯ ¯ ¯ ¯ L(L(R¯ ,η¯ )(R, η1),TL(R¯ ,η¯ )(R, η˙ 1)) = L(RR, Rη1 + η¯1, RRS(ω), RRu) 1 1 (3.2.9) ˙ = L(R, η1, R, η˙ 1) According to the theorem of Euler-Poincar´ereduction [36, Theorem 13.5.3], the

dynamics of the system can be reduced from T SE(3) to se(3). Denote l : se(3) → R as the restriction of L to the identity of SE(3),   1 h i u T T l(ω, u) = u , ω (M RB + M A)   (3.2.10) 2 ω The variational derivatives of l are expressed as δl δl (π , p) = , ∈ se(3)∗ (3.2.11) 0 δω δu which denote the angular and linear impulse on the system with respect to the body-fixed reference frame, respectively. The dynamics of the system is thus described by the Euler-Poincaréequations [37, Definition 7.1.2] dπ dp 0 , = ad∗ (π , p) (3.2.12) dt dt (ω,u) 0 in which the operation ad*: se(3) × se(3)∗ → se(3)∗ is given by:

∗ ad(ω,u)(π0, p) = (−S(ω)π0 − S(u)p, −S(ω)p) (3.2.13)

Furthermore, we wish to express the equations using Lie-Poisson bracket on

se(3)∗. The Hamiltonian H : se(3)∗ → R of the system is given by:   1 h i p T T −1 H(π0, p) = p , π (M RB + M A) (3.2.14) 2 0   π0 29 when M RB + M A is symmetric, hence we have that

∂H ∂H = u, = ω ∂p ∂π0

Consider a smooth function f on se(3)∗, then

df(π , p) ∂f dπ ∂f dp 0 = · 0 + · dt ∂π0 dt ∂p dt ∂f ∂f = · (π0 × ω + p × u) + · (p × ω) ∂π0 ∂p ∂f ∂H ∂f ∂H ∂f ∂H = −π0 · × − p · × + × ∂π0 ∂π0 ∂π0 ∂p ∂p ∂π0 ∂f ∂H = − µ, , ∂µ ∂µ =: {f, H} (µ)

∗ ∗ where µ = (π0, p) ∈ se(3) , the pairing h·, ·i : se(3) × se(3) → R denotes the vector inner product, [·, ·] is the Lie bracket of se(3), and the Lie-Poisson bracket on se(3)∗ is also deﬁned [36]. We obtain a Lie-Poisson equation

f˙ = {f, H} (µ) (3.2.15)

The Lie-Poisson bracket may be written in matrix form as:       ∂f ∂H /∂π0 S(π0) S(p) /∂π0 {f, H} (µ) =   ·     (3.2.16) ∂f ∂H /∂p S(p) O3×3 /∂p

Therefore, the kinetic equation of the dynamical system can be expressed as:

µ˙ i = {µi, H} (µ), i = 1, ··· , 6 (3.2.17)

3.3 EQUATIONS OF MOTION IN A UNIFORM FLOW

When in an unsteady spatially uniform flow, the equations of motion can also be written using the pseudo kinetic energy of the fluid defined given by (3.1.29) and Boltzmann-Hamel equation. The Boltzmann-Hamel equation in the case is

30 written as:     d ∂T S(ω) O3×3 ∂T ∂T f 0 + − J T = (3.3.1) dt ∂ν   ∂ν ∂η   S(u) S(ω) m0 Unlike the former case in calm ﬂow, the third term on the left side of the equation cannot reduce to zero since

∂T ∂ν T = r (M + M¯ )ν ∂η ∂η A r (3.3.2) ∂(ν − ν )T = c (M + M¯ )(ν − ν ) ∂η A c

T eT T and νc = J V c, V c = [U , O1×3] stands for the velocity of ﬂow in the earth- ﬁxed reference frame. The transformation matrix J depends on the Euler angles in η. ∂ν ∂J T c = V (3.3.3) ∂η ∂η c

T Since J contains only the Euler angles η2 = [φ, θ, ψ] , but the position vector

T η1 = [x, y, z] , and νc has nonzero linear velocity elements alone, we have   T ∂νc O3×3 O3×3 =   (3.3.4) ∂η eT U (∂R/∂η2) O3×3

We can prove that

∂R −1 = RS(T ei) (3.3.5) ∂η2i

th th 3 where η2i is the i element of η2 and ei is the i base vector for R . As a result, T ∂R e −1 T e −1 U = −S(T ei)R U = −S(T ei)U (3.3.6) ∂η2i

Then it is valid that ∂R U eT = −T −T S(U) (3.3.7) ∂η2 Hence,   T ∂νc O3×3 O3×3 = −   (3.3.8) ∂η −T T S(U) O3×3

31 Consequently,     RT O O O T ∂T 3×3 3×3 3×3 ¯ −J = −     (M A + M)(ν − νc) ∂η T −T O3×3 T T S(U) O3×3 (3.3.9)   O O 3×3 3×3 ¯ = −   (M A + M)νr S(U) O3×3

Furthermore, it is our obligation to prove that the kinetic equation given by kinetic energy will generate the same result as derived by Newton-Euler formulation. The rigid-body kinetic part has already been proved in the preceding subsection, so the key step is to derive the potential hydrodynamic forces from the kinetic energy expressions.       F 0 d ∂T S(ω) O3×3 ∂T O3×3 O3×3 = − F − F + (M + M¯ )ν   dt ∂ν   ∂ν   A r M 0 S(u) S(ω) S(U) O3×3 (3.3.10) Use the pseudo kinetic energy of the ﬂuid given by (3.1.29)

1 1 1 T = νT M ν + νT Mν¯ − νT Mν¯ F 2 r A r 2 r r 2

then we have

∂T F = (M + M¯ )(ν − ν ) − Mν¯ = M (ν − ν ) − Mν¯ (3.3.11) ∂ν A c A c c

Consequently, d ∂T − F = −M (ν˙ − ν˙ ) + M¯ ν˙ (3.3.12) dt ∂ν A c c

The next step is to write F 0 and M 0 explicitly as:

˙ ˙ F 0 =m ¯ U +m ¯ ω × U − [A11(u˙ − U) + A12ω˙ ] − ω × [A11(u − U) + A12ω] (3.3.13)

32 and

˙ ˙ M 0 = − [A21(u˙ − U) + A22ω˙ ] +m ¯ rb × U

− u × [A11(u − U) + A12ω] + u × m¯ U

− ω × [A21(u − U) + A22ω] + ω × (m ¯ rb × U)

+ U × m¯ (u − U) + U × m¯ (ω × rb) + U × [A11(u − U) + A12ω] (3.3.14) ˙ =m ¯ rb × U +m ¯ rb × (ω × U) ˙ − [A21(u˙ − U) + A22ω˙ ] − (u − U) × [A11(u − U) + A12ω]

− ω × [A21(u − U) + A22ω]

Writing the equations in compact matrix notation, we obtain that   F 0 ¯ ¯   = Mν˙ c + C(ν)νc − M Aν˙ r − CA(νr)νr (3.3.15) M 0

The results illustrate that Newton-Euler and Lagrange Formulations give the same expression of potential hydrodynamic force and moment.

3.4 EQUATIONS OF MOTION IN A NON-UNIFORM FLOW

As for a spatially non-uniform ﬂow, an assumption made here for convenience is that the velocity of the ﬂow can be decomposed into two components as:

νc = νs + νu (3.4.1) where νc is the total flow velocity, and νs, νu are the steady but spatially non- uniform and unsteady but spatially uniform components respectively expressed in body-fixed reference frame. We can define that

−1 νc = R V c

−1 νs = R V s (3.4.2)

−1 νu = R V u 33 where V c, V s, V u are the velocities expressed in earth-ﬁxed reference frame. With the assumption that the ﬂow is non-rotational, the velocities can also be written as       U U s U u νc =   , νs =   , νu =   (3.4.3) O3×1 O3×1 O3×1

 e   e   e  U U s U u V c =   , V s =   , V u =   (3.4.4) O3×1 O3×1 O3×1 As a result,

U = U s + U u (3.4.5)

Since only R of J is actually used in (3.4.2) and RT = R−1, J −1 can be replaced

T by R . Another important assumption is that the varying of νs is negligible over the length of the rigid body. In this case, the kinetic energy of the ﬂuid should be written as:

1 T ¯ 1 T ¯ TF = νr (M A + M)νr − ν Mν 2 2 (3.4.6) 1 1 = (ν − ν − ν )T (M + M¯ )(ν − ν − ν ) − νT Mν¯ 2 s u A s u 2 To begin with,

∂TF ¯ ¯ = (M A + M)(ν − νs − νu) − Mν ∂ν (3.4.7) ¯ = M Aν − (M A + M)(νs + νu)

Diﬀerentiating it with respect to time in body-ﬁxed reference frame, we have

d ∂T F = M ν˙ − (M + M¯ )(ν˙ + ν˙ ) (3.4.8) dt ∂ν A A s u in which d ν˙ + ν˙ = (J T V + J T V ) (3.4.9) s u dt s u The ﬁrst term on the right side is

d(J T V ) dJ T dV s = V + J T s dt dt s dt 34 Use (2.2.11), which can be written as

dR = RS(ω) (3.4.10) dt and e e e dU s ∂U s ∂U s = η˙ 1 = Ru (3.4.11) dt ∂η1 ∂η1 we have   T T e T e d(J V ) −S(ω)R U + R (∂U /∂η1)Ru s = s s dt   O3×1

 T e  −S(ω)U s + R (∂U s/∂η1)Ru =   O3×1

Similarly,   T T e d(J V ) −S(ω)U u + R (∂U /∂t) u = u dt   O3×1

As a result, the time-derivative of ﬂow velocity is     ˙ ˙ T e T e U s + U u −S(ω)(U s + U u) + R (∂U s/∂η1)Ru + R (∂U u/∂t) ν˙ c =   =   O3×1 O3×1

 T T e  −S(ω)(U s + U u) + Φ u + R (∂U u/∂t) =   O3×1 (3.4.12) where we deﬁne that ∂U e T Φ = RT s R (3.4.13) ∂η1 and the term given by ΦT u is the convective acceleration in the body-ﬁxed reference frame. Secondly, recalling (3.3.2),

∂T ∂(ν − ν − ν )T = s u (M + M¯ )(ν − ν − ν ) (3.4.14) ∂η ∂η A s u 35 Since ∂(ν + ν ) ∂J T ∂V s u = (V + V ) + J T s (3.4.15) ∂η ∂η s u ∂η using the results (3.3.8), we obtain that   T !T ∂J O3×3 O3×3 (V s + V u) = −   ∂η −T T S(U s + U u) O3×3 and   T e T ∂V (∂U /∂η1) RO3×3 J T s = s ∂η   O3×3 O3×3 hence, we have

 e T  ∂(νs + νu) −(∂U s/∂η1) RO3×3 = −   (3.4.16) ∂η −T T S(U s + U u) O3×3 Consequently,     RT O −(∂U e/∂η )T RO T ∂T 3×3 s 1 3×3 ¯ −J = −     (M A + M)(ν − νc) ∂η T −T O3×3 T T S(U s + U u) O3×3   −Φ O 3×3 ¯ = −   (M A + M)νr S(U s + U u) O3×3 (3.4.17) Now we are ready to write the potential hydrodynamic forces and moments from the kinetic energy.       F 0 d ∂T S(ω) O3×3 ∂T −Φ O3×3 = − F − F + (M +M¯ )ν   dt ∂ν   ∂ν   A r M 0 S(u) S(ω) S(U s + U u) O3×3 (3.4.18) Substitute the kinetic energy terms with the results we obtain above, the forces and moments can be rewritten as     F S(ω) O 0 ¯ 3×3 ¯   = Mν˙ c − M Aν˙ r −   (M Aνr − Mνc) M 0 S(u) S(ω) (3.4.19)   −Φ O 3×3 ¯ +   (M A + M)νr S(U) O3×3 36 In the above equation, the time derivative of νc can be expressed by (3.4.12) in detail. Compared with the case of unsteady uniform velocity alone, the main diﬀer- ence comes from the third term on the right side of the above equation. Therefore,

the expression of M 0 is still the same. However, there is an additional term in the

expression of F 0, which is

−Φ[(A11 +m ¯ I3×3)(u − U) + (A12 − mS¯ (rb))ω] (3.4.20)

This term presents the eﬀect of a spatially slowly-varying non-uniform ﬂow compo-

e nent U s. Recalling our results (A.87) and (A.84) from Newton-Euler formulation, we need to prove Φ are the same in order to demonstrate the equivalence of the two formulations. From Newton-Euler formulation, the velocity gradients expressed in the body-ﬁxed reference frame is

2 ∂U s ∂ φo Φ = , or Φij = ∂x ∂xi∂xj |r|=0

The velocity gradients expressed in the earth-ﬁxed reference frame from Lagrange formulation is ∂U e T Φ = RT s R ∂η1 From the second expression

∂U e (∂RT U e) ∂U ∂x/∂t ∂U ΦT = RT s R = s R = s R = s ∂η1 ∂η1 ∂x ∂η1/∂t ∂x

Since Φ is symmetric for potential ﬂow, the two of them are identical. It denotes the velocity gradients in the body-ﬁxed reference frame.

3.5 HYDROSTATICS

Newton-Euler Approach

The hydrostatic force is treated separately from the hydrodynamic inertial force just for convenience. In the earth-ﬁxed reference frame, the gravity and buoy-

37 ancy vectors acting on an AUV in an unbounded fluid are written as     0 0     e   e   f =  0  and f = −  0  (3.5.1) g   b       W B where W = mg is the weight of the rigid body and B = mg¯ is the buoyancy force. If m = m¯ , the vehicle is said to be neutrally buoyant. In practice, m¯ will be slightly larger than m such that the vehicle will float to the surface of the ocean in case of emergency. The gravity and buoyancy vectors can be transformed into the body-fixed reference frame. Thus the restoring force and moment vector can be immediately obtained based on Newton-Euler formulation as:   R−1(η)(f e + f e) g b 6 g(η) = −   ∈ R (3.5.2) −1 e −1 e rg × R (η)f g + rb × R (η)f b

−1 e T −1 e T T where R (η)f g = mgR k, R (η)f b = mg¯ R k, and k = [0, 0, 1] is the unit vector aligned with the direction of gravity in the earth-fixed reference frame. If we define the direction of gravity with respect to the body-fixed reference frame as [48]

Γ = RT k

then we have   (m − m¯ )gΓ 6 g(η) = −   ∈ R (3.5.3) rg × mgΓ − rb × mg¯ Γ Since the direction of gravity doesn’t vary with time, the equation of motion of Γ is

Γ˙ = Γ × ω (3.5.4)

Lagrangian Approach

We can also derive (3.5.2) using Lagrangian approach. In fact, there is no

diﬃculty to incorporate a potential energy function V : SE(3) → R into (3.2.1), in 38 which T can be replaced by the Lagrangian L = T − V .

eT eT V = −f g (η1 + Rrg) − f b (η1 + Rrb) (3.5.5)

Then it is easy to see that ∂V g(η) = J T (3.5.6) ∂η where     RT O R−1 O T 3×3 3×3 J =   =   (3.5.7) T T O3×3 T O3×3 T

∂V e e = −(f g + f b) (3.5.8) ∂η1 and

∂V eT −1 eT −1 = −f g RS(T ei)rg − f b RS(T ei)rb ∂η2i

−1 T e −1 T e = −[RS(T ei)rg] f g − [RS(T ei)rb] f b

in which (3.3.5) is used. Therefore, similar to (3.3.7)

∂V −T T e −T T e = −T S(rg)R f g − T S(rb)R f b (3.5.9) ∂η2 Hence,

 −1   e e  R O3×3 −(f g + f b) g(η) =     T −T T e −T T e O3×3 T −T S(rg)R f g − T S(rb)R f b

 −1 e e  R (η)(f g + f b) = −   −1 e −1 e S(rg)R (η)f g + S(rb)R (η)f b   (m − m¯ )gΓ = −   rg × mgΓ − rb × mg¯ Γ

Lie-Poisson Description

The origin of the body-ﬁxed reference frame is usually the center of buoy-

ancy of the underwater vehicle such that we have rb = O3×1. Thus, the potential 39 energy V is expressed as:

V = −(m − m¯ )gk · η1 − mgk · Rrg (3.5.10)

It can be shown that V : SE(3) → R is generally not left invariant under the left action of SE(3). Since

V (L ¯ (R, η ),TL ¯ (R˙ , η˙ )) = V (RR¯ , Rη¯ + η¯ , RR¯ S(ω), RRu¯ ) (R,η¯1) 1 (R,η¯1) 1 1 1 ¯ ¯ = −(m − m¯ )gk · (Rη1 + η¯1) − mgk · RRrg ¯ T ¯ T = −(m − m¯ )gR k · η1 − mgR k · Rrg

− (m − m¯ )gk · η¯1

the potential energy depends smoothly on k ∈ (R3)∗, and is left invariant only under the left action of the following Lie group

n o ¯T T Gk = (R, η1) ∈ SE(3) R k = k, k η¯1 = 0 = SE(2) (3.5.11)

Since k is the unit vector pointing to the direction of the gravity in the earth-ﬁxed reference frame, the Lagrangian is unchanged only when the earth-ﬁxed reference frame translates in a horizontal plane and rotates about k. Therefore, the symme- try of the dynamical system in the coincident-center and neutrally buoyant case no longer exists. However, if we consider the special case of a neutrally buoyant underwater vehicle when m =m ¯ , the potential energy becomes

V = −mgk · Rrg (3.5.12)

and it is left invariant under the left action of

n o ¯T Gk = (R, η1) ∈ SE(3) R k = k (3.5.13)

In this case, [48] used semidirect product reduction theorem [49, 50] to show that the dynamics of the vehicle on T SE(3)* can be reduced to a Lie-Poisson system on 40 a nine-dimensional space s∗, the dual of the Lie algebra s of the semidirect product

3 3 3 S = SE(3) ×ρ R , and ρ : SE(3) → Aut(R ) is a left representation of SE(3) on R . The Lie-Poisson bracket is given by:       ∂f ∂H /∂π0 S(π0) S(p) S(Γ) /∂π0             {f, H} (µ) =  ∂f/∂p  ·  S(p) O O   ∂H/∂p  (3.5.14)    3×3 3×3         ∂f ∂H /∂Γ S(Γ) O3×3 O3×3 /∂Γ

∗ where H = T + V is the Hamiltonian of the system, and µ = (π0, p, Γ) ∈ s . The equations can also be written as:   π˙ 0 = π0 × ω + p × u − Γ × mgrg   p˙ = p × ω    Γ˙ = Γ × ω

When the vehicle is not neutrally buoyant, i.e., m 6= m¯ , the second equation above will become p˙ = p × ω + (m − m¯ )gΓ

3.6 SUMMARY

The vectors used here to express the state of the body-ﬂuid system are η, ν

and νc, the first of which denotes the position and Euler angles of the rigid body in the earth-fixed reference frame, and the rest are the velocity vectors of the rigid body and the flow in the body-fixed reference frame respectively. The kinematic equation (2.2.16) expressing the relations between η and ν is   R(η) O3×3 η˙ = R(η)ν =   ν O3×3 T (η)

The Newtonian and Lagrangian formulations yield the same kinetic equations of motion of a rigid body in ideal ﬂuid, which is

¯ ¯ ∗ M RBν˙ + CRB(ν)ν = Mν˙ c + C(ν)νc − M Aν˙ r − CA(νr)νr − g(η) (3.6.1) 41 ∗ where CA(νr) is deﬁned by (A.88). No viscous hydrodynamic and control forces are included in the above equation. The equation can be rewritten as:

−1 ¯ ¯ ∗ ν˙ = (M RB + M A) (M + M A)ν˙ c + C(ν)νc − CA(νr)νr − CRB(ν)ν − g(η) (3.6.2)

assuming that (M RB + M A) is invertible. If the local ﬂow velocity νc is a known

function of position vector η, velocity vector ν and time t (see also (3.4.12) for ν˙ c), we obtain a set of first-order nonlinear differential equations   η˙ = F (η, ν) (3.6.3)  ν˙ = G(η, ν, t) which gives us a velocity phase space of T SE(3), a twelve-dimensional space. In most cases, simplified equations are used. Some commonly used simplified kinetic equations in the literature are summarized as follows. All of them hold the

same assumption that g(η) = O6×1. Moving in a stationary ﬂow

In this case, νc equals zero. Hence, the kinetic equation becomes

−1 ν˙ = −(M RB + M A) [CRB(ν)ν + CA(ν)ν] (3.6.4) which is the most common equation in the literature. It can also be written on

∗ µ = (ω, π0) ∈ se(3) as a Lie-Poisson equation,

µ˙ i = {µi, H} (µ), i = 1, ··· , 6

Moving in a steady, spatially uniform ﬂow

∗ Since the flow is steady and uniform, CA(νr) is replaced by CA(νr) because of the fact that Φ = 0, and from (3.4.12)   −S(ω)U ν˙ c =   (3.6.5) O3×1 42 where U is the linear flow velocity expressed in the body-fixed reference frame. Hence,

−1 ¯ ¯ ν˙ = (M RB + M A) (M + M A)ν˙ c + C(ν)νc − CA(νr)νr − CRB(ν)ν (3.6.6)

Furthermore, we can obtain that     m¯ I −mS¯ (r ) −S(ω)U ¯ ¯ 3×3 b Mν˙ c + C(ω)νc =     b mS¯ (rb) I0 O3×1     mS¯ (ω) O3×3 U +     mS¯ (rb)S(ω) O3×3 O3×1 (3.6.7)   −m¯ S(ω)U +m ¯ S(ω)U =   −mS¯ (rb)S(ω)U +mS ¯ (rb)S(ω)U

= O6×1

Therefore, (3.6.6) can also be written as:

−1 ν˙ = (M RB + M A) [M Aν˙ c − CA(νr)νr − CRB(ν)ν] (3.6.8)

The above equation illustrates that in the case of steady and uniform ﬂow, the displaced mass of ﬂuid doesn’t appear in the equations. Additionally, we can also prove that

M RBν˙ c + CRB(ω)νc = O6×1

Hence

M RBν˙ + CRB(ω)ν = [M RBν˙ + CRB(ω)ν] − [M RBν˙ c + CRB(ω)νc] (3.6.9)

= M RBν˙ r + CRB(νr)νr

Accordingly, we obtain a more compact form of (3.6.6), which is also common in the literature

−1 ν˙ r = −(M RB + M A) [CRB(νr)νr + CA(νr)νr] (3.6.10) 43 or

−1 ν˙ r = −M C(νr)νr (3.6.11) where M = M RB + M A and C(νr) = CRB(νr) + CA(νr) The rigid body is neutrally buoyant (m = m¯ ), its center of gravity co-

incides with center of buoyancy (rg = rb) and mass homogeneously distributed

b (I0 = I0) In this case, ¯ ¯ M RB = M, CRB = C

Therefore,

−1 ∗ ν˙ = ν˙ c − (M RB + M A) [CRB(νr)νr + CA(νr)νr] (3.6.12)

−1 ν˙ r = −M C(νr)νr (3.6.13)

∗ where M = M RB + M A and C(νr) = CRB(νr) + CA(νr). Note that the equation is similar to the one in the preceding case, although the underlying assumption leading to that equation is totally diﬀerent. In both cases, we obtain a equation

similar to (3.6.4), in which ν is replaced by νr. Therefore, the current velocity can

be neglected in the ﬁrst place, then be superposed to νr to calculate the velocity of the rigid body.

44 CHAPTER 4 BODY-FLUID DYNAMICAL SYSTEM IN A REAL FLUID

The fluid domain around a rigid body is usually divided into two regions, the irrotational flow and the boundary layer [51]. The irrotational flow can be approximately predicted by ignoring the existence of the boundary layer and applying irrotational flow theory introduced in the preceding chapter. Then additional terms due to viscous effects of the fluid are introduced into the equations of motion. Note that when the separation of boundary layer occurs, the approximate method may not be very accurate. However, it is still good enough for the applications in ocean engineering. In this chapter, we use CFD methods to predict the viscous forces on an AUV.

4.1 CFD METHODS TO PREDICT VISCOUS FORCES ON AUVS

Apart from towing tank experiments, CFD methods are increasingly used to predict the viscous hydrodynamic forces acting on marine structures. As for the prediction of hull resistance of an AUV by CFD method, reliable results are reported in the literature [52, 53]. In this section, CFD method is used to predict the drag forces on a REMUS AUV [54]. The parameters of a REMUS 100 AUV is provided in [55]. The extru- sion under the nose of the AUV is a sonar transducer.

45 Table 4.1: Main Geometric Parameters of the REMUS AUV Parameter Value Units Description

l +1.33e+000 m Vehicle Total Length d +1.91e−001 m Maximum Hull Diameter

2 Af +2.85e−002 m Hull Frontal Area

2 Ap +2.26e−001 m Hull Projected Area (xz plane)

2 Sw +7.09e−001 m Hull Wetted Surface Area ∇ +3.15e−002 m3 Estimated Hull Volume

Figure 4.1: REMUS 100 AUV model.

4.1.1 Axial Drag

ANSYS Fluent software is used to generate the mesh and conduct the CFD simulations. The origin of the coordinate frame (see Figure 4.1) is the center of buoyancy of the AUV. The velocity inlet boundary is at x = 3 m, and the pressure outlet boundary is at x = −5 m. The width of the ﬂuid domain is 5 m. The mesh around the AUV is displayed in Figure 4.2. 46 Figure 4.2: Surface and volume meshes around the AUV.

The main parameters chosen for the numerical simulations are listed in the following table.

Table 4.2: Principal Parameters in Numerical Modeling

Parameter Value

Water tank size 8 m × 5 m × 4 m Turbulence model SST k − ω model Solution method SIMPLE No. of faces 5,000 No. of cells 184,000

The experimental results [55] together with the CFD results are plotted in Figure 4.3. The CFD results agrees well with the experiments. The experimental results are larger at high forward speeds because a signiﬁcant amount of wave-making took place in the tank.

47 remxfps7

remdxfps8

remdxfps8b

CFD

4 Axial Drag (N) Drag Axial

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

Forward Velocity (m/s)

Figure 4.3: Forward Speed vs. vehicle axial drag.

The streamlines around the AUV are displayed in the following ﬁgure. No signiﬁcant separation of boundary layer is observed except in the region right behind the sonar transducer.

Figure 4.4: Streamlines around the AUV when u =1 m/s.

48 4.1.2 Cross-Flow Drag

In order to study the dynamics of an AUV in ocean currents, the cross-flow drag forces acting on the AUV under a variety of current velocities should be determined. Assuming a fixed AUV is exposed to the current normal to its forward direction, the numerical model constructed in ANSYS Fluent is shown in the following figure. The current velocity is in the positive y axis direction. It could be seen that cells in the wake region of the AUV is refined.

Figure 4.5: Surface and volume meshes around the AUV.

The ﬁrst step to conduct CFD simulations is to choose a proper turbulence model. We choose three candidate models. The main parameters for the numerical simulations are listed in the following table.

49 Table 4.3: Principal Parameters in Numerical Modeling

Parameter Value

Water tank size 8 m × 4 m × 4 m Turbulence model k − model SST k − ω model

k − kl − ω model Solution method SIMPLE No. of surface faces 5,000 No. of volume cells 306,000

Three turbulence methods, including SST k − ω (with low-Re correction enabled), k− model and k−kl−ω model, are used in the numerical simulations. The normal velocity is set to be from 0.1 m/s to 1.0 m/s. Accordingly, the Reynolds number is in the range between 2 × 104 and 2 × 105. For a circular cylinder, the boundary layer remains laminar, and the wake formed behind the cylinder may be completely turbulent [51]. And in two-dimensional flow, the cylinder’s drag coefficient CD stays constant at a value about 1.2 [56]. The results of the numerical simulations are displayed in the following figures.

50 10

SST k-

k-k -

4 Axial Drag (N) Drag Axial

0.0 0.2 0.4 0.6 0.8 1.0

Normal Velocity (m/s)

Figure 4.6: Axial Drag v.s Normal Velocity

100

SST k-

k-k -

40 Normal Drag (N) Drag Normal

0.0 0.2 0.4 0.6 0.8 1.0

Normal Velocity (m/s)

Figure 4.7: Normal Drag v.s Normal Velocity.

51 1

SST k-

-1

k-k -

-2

-3

-4

-5

-6 Yaw Moment (N*m) Moment Yaw

-7

-8

-9

0.0 0.2 0.4 0.6 0.8 1.0

Normal Velocity (m/s)

Figure 4.8: Yaw Moment v.s Normal Velocity.

Table 4.4: Drag Coeﬃcients from Numerical Simulations

Parameter k − SST k − ω k − kl − ω Estimate Units

Xu|u| −3.91e+000 − − −1.62e+000 kg/m

Xvv +7.48e+000 +8.69e+000 +8.71e+000 − kg/m

Yv|v| −6.76e+001 −7.74e+001 −1.00e+002 −1.31e+003 kg/m

Nv|v| +7.57e+000 +6.16e+000 +7.38e+000 −3.18e+000 kg

The experimental result in [55] is actually -3.87e+000 kg/m.

The initial estimate in [55] is -1.31e+002 kg/m based on strip integration methods. Then he multiplied it by an adjustment factor 10 based on the comparisons with experimental data, but no details about the correction and the experimental data were provided.

Assuming the drag coefficients remains constant in a variety of normal velocities listed above, we can determine them using curve fitting method from the 52 numerical results and compare them with the estimated values provided in [55]. From the point of view of fluid dynamics, the correction of multiplying the normal drag coefficients by 10 is not reasonable. Actually, the three-dimensional effects at the ends of the structure will reduce the drag force compared to the strip theory approach due to the eddies developing at those positions [57]. When using strip theory to estimate the drag coefficient, three-dimensional reduction factors should be considered for the region close to the ends of the structures. Conse- quently, since the normal drag coefficient based on strip theory is −1.31 × 102 kg/m,

2 the numerical prediction about −1.00 × 10 kg/m using k − kl − ω turbulence model should be a reasonable value.

Figure 4.9: Vortex core region in the wake of the incoming ﬂow when the normal speed is 0.4 m/s.

Figure 4.9 shows that the vortex system created around the nose of the AUV almost covers one third of the wake behind the midbody of the AUV. The drag coeﬃcients of the cross sections in that region would be certainly aﬀected.

As for the selection of the turbulence model, k − kl − ω model is more capable

53 of handling the cases of sub-critical Reynolds number in which the turbulence is not fully developed, while k − and SST k − ω model (even with low-Re correction) tend to give smaller drag coeﬃcient for this Reynolds number range.

Figure 4.10: Velocity ﬁeld around the vehicle with SST k − ω model when the normal speed is 0.4 m/s.

Figure 4.11: Velocity field around the vehicle with k−kl −ω model when the normal speed is 0.4 m/s. 54 In the yz plane at x = 0 (see Figure 4.10 and 4.11) where the two-dimensional flow is less affected by the three-dimensional effect, the difference of the point of separation and the wake flow could be found using different turbulence models. The distinction is more obvious when the pressure distribution around the cross section is drawn. The shape of pressure distribution and the point of separation based on k − kl − ω turbulence model are more similar to the experimental ones provided in [56].

SST k- 1.0

k-k -

0.5 ) 2

0.0 U

Re = 7.5 X 10

-0.5 p/(0.5

-1.0

-1.5

0 20 40 60 80 100 120 140 160 180

Angle from forward stagnation point (deg.)

Figure 4.12: Surface pressure distribution around the cross section at x = 0.

Since k − kl − ω model proves to be an appropriate turbulence model for our study, we could use it to conduct mesh independent study and to make a more reliable prediction of the cross-flow drag force with two more refined mesh generations. The numbers of faces on the AUV and the cells in the fluid domain of the mesh cases are listed in the following table.

55 Table 4.5: Mesh Size in Mesh Independent Study

Mesh Case No. of surface faces No. of volume cells

Coarse 5,000 300,000 Medium 12,000 500,000 Fine 19,000 900,000

The principal parameters used in the CFD simulations are shown as follows, which are applied to all the three mesh cases.

Table 4.6: Principal Parameters in Numerical Modeling

Parameter Value

Water tank size 8 m × 4 m × 4 m

Turbulence model k − kl − ω model Solution method SIMPLE Discretization schemes Second order upwind

The most refined mesh is demonstrated in the following figure. Much more faces are generated on the surface of the AUV with more cells in the fluid domain around it compared to the coarse mesh case shown in Figure 4.5.

56 Figure 4.13: Surface and volume meshes around the AUV in Fine Mesh Case.

The simulation results are depicted in the following figures, in which we could conclude that the enhancement of the mesh resolution doesn’t alter the simulation results greatly. The results from different mesh cases are very close to each other even for large flow velocities. Consequently, the results in Table 4.4, which are obtained using coarse mesh, can be accepted as reliable predictions of the hydrodynamic coefficients of the AUV.

57 10

Coarse Mesh

Medium Mesh

Fine Mesh

4 Axial Drag (N) Drag Axial

0.0 0.2 0.4 0.6 0.8 1.0

Lateral Velocity (m/s)

Figure 4.14: Axial Drag v.s Normal Velocity in Diﬀerent Mesh Cases.

0 Coarse Mesh

Medium Mesh

Fine Mesh

-20

-40

-60 Normal Drag (N) Drag Normal

-80

-100

-120

0.0 0.2 0.4 0.6 0.8 1.0

Lateral Velocity (m/s)

Figure 4.15: Normal Drag v.s Normal Velocity in Diﬀerent Mesh Cases.

58 4.1.3 Body Lift

When the AUV moves in the flow at an angle of attack, the flow around the hull will be more complicated. The estimates of hydrodynamic coefficients based on empirical methods are not quite reliable. Without any towing tank experimental results, CFD simulation is the best method to predict those hydrodynamic coefficients. In the CFD simulations, the forward velocity of the AUV is fixed at 1.5 m/s and the lateral velocity with respect to the fluid varies from 0.1 m/s to 1 m/s. The main parameters chosen for the numerical simulations are listed in the following table.

Table 4.7: Principal Parameters in Numerical Modeling

Parameter Value

Water tank size 8 m × 5 m × 4 m Turbulence model Realizable k − model Near wall treatment Scalable wall functions Solution method SIMPLE Discretization scheme Second order upwind

Three mesh generations are used to study the mesh independence.

Table 4.8: Mesh Size in Mesh Independent Study

Mesh Case No. of surface faces No. of volume cells

Coarse 5,000 400,000 Medium 12,000 600,000 Fine 19,000 1,100,000

The mesh in the most reﬁned case is displayed in the following ﬁgure.

59 Figure 4.16: Surface and volume meshes around the AUV.

The simulation results are demonstrated in the following ﬁgures, which indicate that the results are not quite sensitive to the mesh size when the mesh is reﬁned well enough.

-6.5

-7.0

-7.5

-8.0

-8.5

-9.0 Axial Drag (N) Drag Axial

-9.5

-10.0

Coarse Mesh

Medium Mesh

-10.5

Fine Mesh

-11.0

0.0 0.2 0.4 0.6 0.8 1.0

Lateral Velocity (m/s)

Figure 4.17: Axial Drag Force v.s Lateral Velocity (u = 1.5 m/s). 60 0

-20

-40

-60

-80 Lateral Drag (N) Drag Lateral

-100

Coarse Mesh

Medium Mesh

-120

Fine Mesh

0.0 0.2 0.4 0.6 0.8 1.0

Lateral Velocity (m/s)

Figure 4.18: Lateral Drag Force v.s Lateral Velocity (u = 1.5 m/s).

-2

-4

-6

-8 Yaw Moment (N*m) Moment Yaw

-10 Coarse Mesh

Medium Mesh

Fine Mesh

-12

0.0 0.2 0.4 0.6 0.8 1.0

Lateral Velocity (m/s)

Figure 4.19: Yaw Moment v.s Lateral Velocity (u = 1.5 m/s).

In an ideal ﬂow, there is an destabilizing moment with no drag or lift forces

61 acting on the rigid body when it is translating at an angle of attack. In viscous ﬂuid, drag and lift forces arise with an additional viscous moment whose direction depends on the shape of the streamline body. For a hull shape with a round and full forebody like the REMUS AUV, the viscous moment is destabilizing [58], as shown in Figure 4.19. In order to determine the viscous hydrodynamic coeﬃcients from the simulations, several assumptions are made based on the observation of the CFD results. To begin with, as shown in Figure 4.17 the axial drag on the AUV doesn’t change much as the lateral velocity varies. Its value is close to the one obtained in the case of a zero lateral velocity. Therefore an assumption regarding the axial drag can be made as follows.

Assumption 4.1. When the AUV translates at a small angle of attack, the axial drag remains unchanged if the forward velocity does so. ♦

As for the lateral force acting on the AUV, it can be assumed to comprise two components, due to circulation and cross flow [58]. The body lift force part is linear with respect to small angle of attack, while the nonlinear part due to cross flow becomes significant at large angle of attack. Hence, we can make the following assumption to derive the body lift coefficients of the AUV.

Assumption 4.2. The total lateral drag is the sum of the body normal (Lift) force which depends on the magnitude and angle of the total velocity, and cross-ﬂow drag which only depends on the magnitude of lateral velocity. ♦

62 10

Body Lift Force

Body Lift Force + Cross-Flow Drag

-10 (kg/m)

-2 (kg/m) 2 2 /U /U

lift -20 total Y Y

-30

-4

-40

-5 0 5 10 15 20 25 30 35

Angle of Attack (deg.)

Figure 4.20: Body normal (lift) force and the total lateral drag v.s the angle of attack of the relative ﬂow.

In Figure 4.20, the curve of body lift coefficient is linear when the angle of attack is smaller than 15◦, which is the stall angle. As for the total lateral drag coefficient, if the angle of attack continues to increase, its value will reach −1.00×102 kg/m at 90◦ (see Table 4.4). Note that the extension is only valid when the normal component of the relative velocity is small enough so that the Reynolds number of the cross-flow is not supercritical, otherwise the cross-flow drag will drop. The physics behind the assumption can be revealed by Figure 4.21. The vortices created around the nose and sonar transducer are washed downstream along the leeward side of the hull, create additional pressure difference between the two sides of the hull. Since the forebody is more affected by the vortices, the viscous moment becomes destabilizing.

63 Figure 4.21: Vortex core region in the wake of the incoming ﬂow when the angle of attack is 33◦.

Assuming the angle of attack α is small, the body lift force can be written as ∂C ∂C v ∂C Y = N αU 2 = N (u2 + v2) ' N uv = Y uv (4.1.1) lift ∂α ∂α u ∂α uv

Then Yuv can be estimated using the slope of the linear part of the body lift curve.

From Figure 4.20, we can find that Yuv = −17.9 kg/m (Note that the force on the vertical rudders are also accounted for). The body lift coefficients is sufficiently

small compared to Yv|v| = −100 kg/m, which is the coefficient due to cross flow. Hence it can probably be ignored in the analysis for the sake of convenience. As for the viscous moment acting on the AUV, it has to be divided into two components, the moment on the bare hull and the moment on the vertical rudders. Although the normal force on the rudders are relatively small compared to the one on the bare hull, its moment has a significant effect on the directional stability of the AUV. Before calculating the moment coefficients, another assumption has to be made.

Assumption 4.3. The total yaw moment (including the contribution from the vertical rudders) is the sum of the body lift moment which depends on the magnitude 64 and angle of the total velocity and the cross-ﬂow moment which only depends on the cross-ﬂow velocity. ♦

0 (kg) 2

-2 /U lift N

-4

-6

Hull (Cross-flow removed)

Vertical Rudders

-8

-5 0 5 10 15 20 25 30 35

Angle of Attack (deg.)

Figure 4.22: Yaw moments (contributed only by lift forces) v.s angle of attack.

It can be seen from Figure 4.22 that the lift moment on the bare hull is destabilizing while the vertical rudders have stabilizing eﬀects as expected (see also the data of an airship in [58]). The hydrodynamic coeﬃcients for lift force and moment can be predicated from the CFD results by following the above assumptions.

The body lift force coeﬃcient Yuv, the moment coeﬃcients for both the bare hull

h r Nuv and vertical rudders Nuv are listed in the following table.

65 Table 4.9: Body Lift Coeﬃcients from Numerical Simulations

Parameter CFD method Estimate Units

Yuv −1.79e+001 −2.86e+001 kg/m

h Nuv −3.90e+001 −3.01e+001 kg

r § Nuv +8.62e+000 +1.32e+001 kg

Only the first four points in Figure 4.22 are used. Note that the Munk moment from ideal flow theory is also included in this term based on CFD simulation. These estimated coefficients are given in [55] based on empirical formula. The Munk

h moment is also included in the estimated Nuv.

§ Both the upper and lower rudders are included.

The bare hull of the REMUS AUV without vertical rudders is severely destabilizing in yaw direction. Therefore when designing an AUV which is expected to work in a relatively strong current environment, a stabilizing hull proﬁle should be selected with a vertical rudder laid at a certain distance from the center of buoyancy in order to provide large stabilizing moment.

4.1.4 Fin Lift

The fins or rudders on the AUV play a significant role in controlling the motion of the vehicle. For an AUV expected to work in relatively strong ocean currents, the vertical rudders should be designed to improve the stability of the AUV’s motion in yaw direction. However, it’s not easy to predict the hydrodynamic characteristics of a rud- 66 der if experimental data is not available. Table 4.9 indicates that the difference

r between the CFD result and empirical estimate of Nuv is considerable. We need to look into this diﬀerence carefully. The parameters of the ﬁns are given as follows.

Table 4.10: REMUS Fin Parameters Parameter Value Units Description

2 Sﬁn +6.65e−003 m Planform Area

bﬁn +8.57e−002 m Span

xﬁn −6.38e−001 m Moment Arm wrt Vehicle Origin at CB

cmean +7.47e−002 m Mean Chord Length

ARe +2.21e+000 n/a Eﬀective Aspect Ratio

First of all, the equation used in [55] to estimate the lift slope of the rudder is [58] ∂C 2πα¯ C = L = (4.1.2) Lα 2¯α ∂α 1 + /ARe where the factor α¯ = 0.9. This equation gives that CLα = 3.12. However, this equation comes from lifting-line theory and is primarily used for high AR wings [59]. In [60], another equation is proposed for marine rudders, which is claimed to be able to give better estimate results than (4.1.2) compared to experimental results. ∂C 1.95π C = L = (4.1.3) Lα 3 ∂α 1 + /ARe

This equation gives that CLα = 2.6. If we use the new CLα, then for the two verti-

r cal rudders together Nuv = −ρCLαSfinxfin = 11 kg. As mentioned before, the motion of an AUV is expected to be stable in yaw direction when moving in relative strong currents without any control force. If the bare hull of the AUV is destabilizing, the vertical rudder or fin should be able to

67 provide enough stabilizing moment to maintain the yaw direction of the vehicle. Consequently, the moment arm of the vertical rudder needs to be so far as possible from the center of buoyancy of the vehicle. As for the REMUS AUV, the vertical rudder cannot provide the very large yaw moment. In order to improve its stability, the vertical rudder can be redeployed to a further position after the AUV from the origin as shown in the following ﬁgure.

Figure 4.23: Modiﬁed REMUS AUV model.

In that case, since the vertical rudders are far from the hull of the AUV, the study on the hydrodynamic coefficients of the hull and rudders can be separated. Without the vertical rudders the flow field around the tail part of the hull will change significantly. Consequently, the hydrodynamic coefficients of the hull should be recalculated. The modelings of the CFD simulations are similar to the ones introduced in the previous subsections. And only the medium mesh generation is adopted in the simulations. The results are listed in the following table. Note that the axial drag force is still assumed to be constant for the angles of attack smaller than 20◦, and the body lift force is too small to consider compared to the one due to cross flow. 68 Table 4.11: Hydrodynamic Coefficients of the AUV of Bare Hull

Parameter CFD simulation results Units

Xu|u| −3.50e+000 kg/m

Yv|v| −8.70e+001 kg/m

Yuv −2.20e+000 kg/m

Nuv −2.00e+001 kg

The relative ﬂow velocity to the ﬁns can be written as

ufin = ur + ω × rfin (4.1.4) where ur is the relative linear velocity of the AUV with respect to the flow, and rfin is the position of the pressure center of the fins in the body-fixed reference frame. As for the vertical rudders, under the assumption that the angle of attach is small, their total lift in y direction can be written as

1 vfin 2 Yrudder ' −2 × ρCLαSfin |ufin| (4.1.5) 2 ufin

Note that the rudder angle is ﬁxed as zero. The components of uﬁn are

ufin = ur + zfinq − yfinr ' ur

vfin = vr + xfinr − zfinp ' vr + xfinr (4.1.6)

zfin = wr + yfinp − xfinq ' wr − xfinq As a result,

vr + xfinr 2 Yrudder ' −ρCLαSfin ur = −ρCLαSfin(urvr + xfinurr) ur r r = Yuvurvr + Yururr where

r Yuv = −ρCLαSﬁn (4.1.7)

r Yur = −ρCLαSﬁnxﬁn (4.1.8) 69 Similarly, we have that

r r Nuv = Yur = −ρCLαSﬁnxﬁn (4.1.9)

r 2 Nur = −ρCLαSﬁnxﬁn (4.1.10)

4.2 VISCOUS DRAG TERMS IN THE EQUATIONS OF MOTION

In practice, the viscous drag forces are usually expanded to the order of two or even three (see [61] and [62] for the details of the expansions). More high-order terms will provide better estimate of drag curves obtained from the experiments or CFD simulations. However, they also increase the diﬃculty for us to study the dynamical system. In our study, only the terms no higher than second order are retained, which means that the angle of attack of the relative ﬂow is always to be assumed as small enough. As a result, the component of the relative velocity in surge direction must be much larger than the ones in the other two directions. In the literature, the viscous drag force is usually written into two parts, linear and quadratic terms [63, 34].

D(νr) = Dl + Dn(νr) (4.2.1) where Dl is the linear drag matrix due to potential damping and possible skin

friction and Dn(νr) is the nonlinear drag matrix due to quadratic damping and higher-order terms. In an unbounded fluid domain, the potential damping vanishes. And for a propeller-driven AUV, its Reynolds number even in sway direction can not be so low that the linear skin friction drag produce a marked effect, especially when an ocean current exists. However, if there is a noticeable difference between the experimental drag curve and the quadratic fitting curve at low speed regime, a linear damping correction can produce more accurate estimate. Some authors point out that in modeling of marine vessels, both linear and nonlinear damping should be included for the reason that exponential convergence 70 is not present without the linear part [64]. We need to take a close look at this statement. Let’s consider a one-dimensional damped oscillator first

x¨ + ax˙ + bx = 0 (4.2.2) where a > 0 is the damping coefficient and b > 0 is the restoring coefficient. The model is a homogeneous second order linear differential equation. Both x and x˙ converge exponentially to zero as t → +∞. If the linear damping term is replaced by a quadratic damping term, the equation becomes

x¨ + ax˙|x˙| + bx = 0 (4.2.3) which can also be written as a planar dynamical system,

x˙ 1 = x2 (4.2.4)

x˙ 2 = −ax2|x2| − bx1

Its ﬁxed point is x0 = (0, 0). And the Lyapunov function is easily found as

1 1 V (x) = x2 + bx2 ≥ 0 (4.2.5) 2 2 2 1

˙ 2 2 We know that V (x0) = 0 and V (x) = −ax2|x2| ≤ 0 for all x ∈ R \{x0}. According to a corollary of LaSalle’s invariance theorem [65, Corollary 4.2], dynamical sys-

tem is globally asymptotically stable. Actually when x is close to x0, the quadratic damping (physically dissipative) term is very small. The phase portrait of the dynamical system looks like the linearized system. However, by applying Bendixson’s Criteria [66], we know that there is no closed orbit lying in the domain. Therefore

the trajectory indeed converges to the ﬁxed point x0, although the rate of conver-

gence is very slow in the nearby neighborhood of x0. By examining the linearized

system, x0 is indeed not an exponentially stable equilibrium point for the nonlinear system unless the linear damping term is included [65, Corollary 4.3]. 71 In conclusion, the linear drag term may have a great effect on the convergence performance of the dynamical system when its state is very close to the equilibrium point, otherwise it can be neglected. Note that the linear damping terms could be important for roll and pitch motion, whose coefficients need to be carefully estimated based on free decay tests. If needed, the linear drag matrix can be written as [34]   Xu 0 0 0 0 0      0 Y 0 Y 0 Y   v p r       0 0 Zw 0 Zq 0  Dl = −   (4.2.6)    0 Kv 0 Kp 0 Kr      0 0 M 0 M 0   w q    0 Nv 0 Np 0 Nr The nonlinear drag matrix may be written in different forms due to the pref- erence of the authors. Based on the results from CFD simulations in the previous section, it can be expressed as follow:   X|u|u|ur| 0 0 0 0 0      0 Y |v | 0 0 0 Y |r|  |v|v r |r|r       0 0 Z|w|w|wr| 0 Z|q|q|q| 0  Dn1(νr) = −   (4.2.7)    0 0 0 K|p|p|p| 0 0       0 0 M |w | 0 M |q| 0   |w|w r |q|q    0 N|v|v|vr| 0 0 0 N|r|r|r|   0 0 0 0 0 0     0 Y u 0 0 0 Y u   uv r ur r      0 0 Zuwur 0 Zuqur 0  Dn2(νr) = −   (4.2.8)   0 0 0 Kupur 0 0      0 0 M u 0 M u 0   uw r uq r    0 Nuvur 0 0 0 Nurur 72 And the total nonlinear drag matrix is

Dn(νr) = Dn1(νr) + Dn2(νr) (4.2.9)

Dn1(νr) deals with the cross-ﬂow drag, which is always dissipative. Dn2(νr) is the expression of the lift force and moment on the hull and ﬁns of an AUV.

4.3 SUMMARY

When considering the viscous effect of real flow, the rule of thumb in ocean engineering is to incorporate the viscous forces into the dynamical equations which are based on ideal flow theory. In reality, decoupling the ideal and viscous flow effects with no consideration of their interactions is not quite accurate except for the case of pure translation of an AUV in calm water. It is assumed here that the angle of attack of the relative flow to the vehicle is small enough that the separation of boundary layer on the leeward side of the hull is not so significant that the hydrodynamic loads based on ideal flow theory could be greatly changed.

Assumption 4.4. The inviscid flow solutions of hydrodynamic loads are not affected by the separation of boundary layer in real flow when the angle of attack of the relative flow is small. ♦

Even though the relative motion of a vehicle in a flow is in steady state, the viscous drag coefficient of the vehicle is not time-independent due to phenomena of vortex shedding. Therefore past motions have impacts on the fluid loads at the present moment [61]. However, the magnitude of the unsteadiness is small compared to the change of the hydrodynamic loads due to the dynamical motion of the vehicle itself. And it is also difficult to capture the oscillations accurately in both model tests and CFD simulations. In order to simplify our analysis of the dynamical system, another assumption could be made,

73 Assumption 4.5. The unsteadiness of the viscous drag coeﬃcients due to vortex shedding can be neglected. ♦

Based on the two assumptions above, we can add the viscous drag terms obtained from experiments or CFD predictions into the kinetic equations given by (3.6.1). Then we have

∗ ¯ ¯ M RBν˙ +CRB(ν)ν +M Aν˙ r +CA(νr)νr +D(νr)νr +g(η) = Mν˙ c +C(ν)νc (4.3.1)

The control term is not included in the equation. In the case of a steady and spatially uniform ﬂow, similar to the derivation of (3.6.10), (4.3.1) can be simpliﬁed as

M RBν˙ r + CRB(νr)νr + M Aν˙ r + CA(νr)νr + D(νr)νr + g(η) = O6×6 (4.3.2)

Generally, the body-ﬂuid dynamical system can still be described in the form of (3.6.3)   η˙ = F (η, ν)

 ν˙ = G(η, ν, t) in which, the two equations present the kinematics and dynamics of the dynamical system, respectively.

74 CHAPTER 5 STABILITY OF AN UNDERWATER VEHICLE IN OCEAN CURRENTS

The governing equations of the 6-DOF motion of an underwater vehicle in ocean currents are usually written in the form of (3.6.3):   η˙ = F (η, ν)

 ν˙ = G(η, ν, t) which is a twelve-dimensional nonlinear dynamical system. In the present research, the unsteadiness of the ﬂow and its loads on the vehicle are neglected. Thus, the dynamical system becomes an autonomous one and the equations can be written as:

    η˙ F (η, ν)   =   (5.0.1) ν˙ G(η, ν) or more generally x˙ = f(x) (5.0.2) where x ∈ R2n, n is the degree of freedom of the dynamical system. The variables of the system consist of the position η ∈ R6 and velocity ν ∈ R6 of the vehicle. Studying the behavior of the above set of equations qualitatively is a formidable task. As a result, the three-dimensional motion of the vehicle is usually decoupled into a horizontal plane and a vertical plane. Then the decoupled motions are analyzed separately in their own planes. Although the assumption of decoupling the motion can be valid in many applications, especially when the roll motion of the vehicle is negligible, it is still desirable to consider the coupled 6-DOF

75 dynamical system, since we may wish to make use of the coupling as a means of controlling the motion of an underwater vehicle in ocean currents. In order to study the coupled 6-DOF model of an underwater vehicle in an ideal fluid, [48] first resorted to the methods from geometric mechanics. The configuration space of the dynamical system is SE(3), whose generalized coordinate

is η ∈ R6. The phase space is T ∗SE(3), the cotangent bundle of SE(3), which is a twelve-dimensional space. In this chapter, the methods used in [48] and [25] are followed to analyze the three-dimensional motion of an underwater vehicle in an ideal fluid. In reality, the dynamical system in an ideal fluid is not practically valid since the system must involve energy dissipation due to the viscous effect in a real fluid. However, the dynamical systems in ideal fluid exhibits invariances of certain quantities as we are going to show shortly. As [25] pointed out that:

Even for those systems which do not exhibit invariance, or symme- try, greater insight into the dynamics often follows from considering an invariant system as a special case.

the study on the stability of an underwater vehicle in a real fluid will certainly benefit from the special case in an ideal fluid. In this chapter, we focus on the open-loop stability of the dynamical system. We will start with the special case of a invariant system, i.e., the dynamics of a rigid body moving in an ideal fluid. Then we will proceed to the real fluid and explore the effect of the viscous hydrodynamic loads on the stability of the horizontal- plane motion of an underwater vehicle in ocean currents. Besides, the effect of roll torque on the horizontal-plane motion will be studied in detail as a preparation for the control design treated in the next chapter.

76 5.1 STABILITY OF AN UNDERWATER VEHICLE IN A STATION- ARY FLOW

5.1.1 Special Case in an Ideal Fluid

Coincident Centers of Buoyancy and Gravity

When a neutrally buoyant underwater vehicle moves in a stationary inviscid ﬂow with no gravitational torque and control forces acting on it, its motion is governed by (3.6.4)

−1 ν˙ = −(M RB + M A) [CRB(ν)ν + CA(ν)ν]

= −M −1C(ν)ν

where M = M RB + M A and C(ν) = CRB(ν) + CA(ν). The equations can also be written in a Lie-Poisson bracket (3.2.17)

µ˙ i = {µi, H} (µ), i = 1, ··· , 6

in which µ = (π0, p) denotes the momentum (strictly speaking, impulse) of the system. We can conclude from physics that the dynamical system is conservative since no dissipative or control forces exist. This fact can be stated in the following theorem:

Theorem 5.1. The total kinetic energy of an unforced body-ﬂuid system in stationary ambient ﬂow remains constant. The total kinetic energy of the dynamical system is given by 1 1 T = νT (M + M )ν = νT Mν (5.1.1) 2 RB A 2 4

Proof. The time derivative of the total kinetic energy is written as

∂T ∂ν T = Mν = −(M −1C(ν)ν)T Mν = −νT CT M −T Mν (5.1.2) ∂t ∂t

77 Accepting that M is symmetric, we have

∂T = −νT CT ν (5.1.3) ∂t

We can actually ﬁnd the equality for inertial forces as   S(ω) O3×3 ∂T C(ν)ν =   = P Mν (5.1.4) S(u) S(ω) ∂ν where     S(ω) O −S(ω) −S(u) 3×3 T P =   , P =   (5.1.5) S(u) S(ω) O3×3 −S(ω) Therefore, ∂T = −νT M T P T ν (5.1.6) ∂t Since       −S(ω) −S(u) u O T 3×1 P ν =     =   (5.1.7) O3×3 −S(ω) ω O3×1 we obtain that ∂T = 0 (5.1.8) ∂t From the standpoint of a Lie-Poisson equation (3.2.15), we can directly obtain that T˙ = {T, H} = {H, H} = 0 (5.1.9)

since T = H in this case. Consequently, we can draw the conclusion that the total kinetic energy of the dynamical system is conservative although there is a kinetic energy exchange between the rigid body and the ﬂuid.

Based on the Lie-Poisson equation (3.2.15), we can ﬁnd conserved quantities of the dynamical system. A conserved quantity C on se(3)∗ is also called a Casimir function, which satisﬁes that

∗ {C, h} (µ) = 0, ∀h : se(3) → R (5.1.10) 78 If we consider the Lie-Poisson bracket in matrix form (3.2.16),       ∂f ∂h /∂π0 S(π0) S(p) /∂π0 {f, h} (µ) =   ·     ∂f ∂h /∂p S(p) O3×3 /∂p       ∂h ∂f /∂π0 S(π0) S(p) /∂π0 = −   ·     ∂h ∂f /∂p S(p) O3×3 /∂p

then the null space of the Hamiltonian matrix above is two-dimensional, and it is spanned by the following two bases     O3×1 p e1 =   , e2 =   p π0

Let ∂C i = e , i = 1, 2 (5.1.11) ∂µ i we have two independent Casimir functions given by

1 C = pT p,C = pT π (5.1.12) 1 2 2 0 which are conserved along the ﬂow of the dynamical system.

Horizontal-Plane Motion Now we conduct a stability analysis of the horizontal-

plane motion of a rigid body as a special case. Firstly, assuming xg = yg = 0 and

a26 = a62 = 0, the governing equations for a typical marine vehicle can be written

as:   (m + a )u ˙ − (m + a )vr = 0  11 22    (m + a )v ˙ + (m + a )ur = 0  22 11    (Izz + a66)r ˙ + (a22 − a11)uv = 0 (5.1.13)   x˙ = u cos ψ − v sin ψ    y˙ = u sin ψ + v cos ψ     ψ˙ = r 79 The first three are kinetic equations of motion and the others are kinematic constraints. Since x, y, ψ don’t appear in the kinetic equations, the dynamics of the rigid body is independent of its position and heading angle. Accordingly, the kinetic equations are decoupled from kinematic ones. Thus we can only analyze the first three first-order nonlinear differential equations, which can also be written in a generalized form as:   x˙ 1 = Ax2x3   x˙ 2 = Bx1x3 (5.1.14)    x˙ 3 = Cx1x2 where

m + a22 m + a11 1 a22 − a11 A = > 0,B = − = − < 0,C = − < 0 (a22 > a11) m + a11 m + a22 A Izz + a66

It is a three-dimensional autonomous nonlinear system. If C = 0 (i.e., a11 = a22),

x3 is a constant accordingly. The nonlinear system becomes a second-order linear system as       x˙ 1 0 Ax¯3 x1   =     (5.1.15) x˙ 2 Bx¯3 0 x2

Meanwhile, if the constant x¯3 = 0, then x1, x2 are both constants. If not, then we can ﬁnd out that x1, x2 conduct oscillatory motions around equilibrium point (0, 0). The rigid body moves in a permanent translation and rotates around its

body origin in a constant angular velocity x3. Normally, the system cannot be reduced into a linear system. In that case, we cannot obtain analytic solutions of the system. Fortunately, the implicit analytic relations between the variables can be found in the above nonlinear system. It is not hard to obtain that   Bx2 − Ax2 = const  1 2  2 2 Cx1 − Ax3 = const (5.1.16)   2 2  Cx2 − Bx3 = const 80 where the constants are determined by the initial conditions. Since A > 0, B < 0, C < 0, we can draw the conclusion that the projection of the trajectory of

state vector [x1, x2, x3] on plane x1Ox2 is part of a ellipsoid, the projection on plane

x2Ox3 is part of a hyperbola, and etc. This conclusion can be explained shortly. Generally, we are likely to analyze the system around an equilibrium point using linearization technique. As for the nonlinear system, all three coordinate axes

are lines of fixed points. As for x0 = 0, we can write a Lyapunov function defined in its neighborhood using the expression of the kinetic energy of the dynamical system, 1 1 − A V (x) = (x2 + Ax2 + x2) (5.1.17) 2 1 2 C 3 where (1 − A)/C = (Izz + a66)/(m + a11) > 0. V (x) satisfies that V (x0) = 0 and

V (x) > 0 for x 6= x0. We have

1 − A V˙ (x) = x x˙ + Ax x˙ + x x˙ 1 1 2 2 C 3 3

= x1x2x3(A − 1 + 1 − A)

= 0

Therefore, x0 = 0 is a stable equilibrium point. The solution curves are on the

manifold given by V (x) = c, i.e., the closed orbits of the ﬂow φt(x) in the phase portrait are on the surfaces of those ellipsoids, and the constant c is determined by the initial state of the dynamical system.

81 Figure 5.1: The kinetic energy ellipsoid and phase portrait of the dynamical system when rg = 0 and a26 = 0.

As a matter of fact, (5.1.14) has the same form as the equations describing the torque-free rotation of rigid body, in whose context the curves in Figure 5.1 is referred to as polhode [67, 68].

T Consider x0 = [U0, 0, 0] as one of the non-isolated equilibrium points on the

x1-axis, in which U0 is an arbitrary constant. Linearizating the system at this point, we obtain that       x˙ 1 0 0 0 x1 − U0             x˙  = 0 0 BU   x  (5.1.18)  2  0  2        x˙ 3 0 CU0 0 x3

82 The eigenvalues and corresponding eigenvectors of the state matrix can be found as:

√ √ λ1 = 0, λ2 = U0 BC, λ3 = −U0 BC       1 0 0          √  √  e1 = 0 , e2 =  −B  , e3 =  −B          √  √  0 − −C −C

One of the eigenvalues of the linearized system is zero, thus x0 is a non-

hyperbolic equilibrium point. For the linearized system, x1-axis is a line of ﬁxed

c u s points or a so-called center subspace E , and E = e2 and E = e3 are the unsta-

ble and stable subspace, respectively. Locally, x0 is a saddle point on the manifold deﬁned by V (x) = c, which is an unstable equilibrium point. If the state of the

system is at x0 initially, it will stay there forever. Physically, the rigid body will always translate along a straight line if there is no rotation. But any small disturbance will result in a diﬀerent mode of motion.

Now take a look at the equilibrium point on the x2-axis. The linearized

T system around point [0,V0, 0] is       x˙ 1 0 0 AV0 x1             x˙  =  0 0 0  x − V  (5.1.19)  2    2 0       x˙ 3 CV0 0 0 x3

The corresponding eigenvalues are

√ √ λ1 = iV0 −AC, λ2 = 0, λ3 = −iV0 −AC

All of the eigenvalues have zero real part. Therefore, the stability of the system

T around [0,V0, 0] cannot be determined from the linearized system. However, in

T Figure 5.1 the point [0,V0, 0] is a center on the manifold given by V (x) = c. Hence

it is a stable equilibrium point. Similarly, any point on the x3-axis is a stable equilibrium point as well [68, Theorem 7.3].

83 A more general situation for the dynamical system is when xg 6= 0, yg = 0

and a26 = a62 =6 0. This is the most common case for marine crafts. Thus the governing equations can be written as:   (m + a )u ˙ − (m + a )vr − (mx + a )r2 = 0  11 22 g 26    (m + a )v ˙ + (mx + a )r ˙ + (m + a )ur = 0  22 g 26 11    (Izz + a66)r ˙ + (mxg + a26)(v ˙ + ur) + (a22 − a11)uv = 0 (5.1.20)   x˙ = u cos ψ − v sin ψ    y˙ = u sin ψ + v cos ψ     ψ˙ = r

or in a more generalized form as  2  x˙ 1 = Ax2x3 + Dx  3  x˙ 2 + Ex˙ 3 = Bx1x3 (5.1.21)    x˙ 3 + F x˙ 2 = Cx1x2 − F x1x3 where the coeﬃcients in the equations are

mx + a mx + a mx + a D = g 26 ,E = g 26 ,F = g 26 m + a11 m + a22 Izz + a66

and the signs of D,E,F depend on the sign of mxg + a26. At last, the equations can be written into the standard form of a ﬁrst-order nonlinear system as:  2  x˙ 1 = Ax2x3 + Dx3   B + EF CE x˙ = x x − x x (5.1.22) 2 1 − EF 1 3 1 − EF 1 2   C F (1 + B)  x˙ 3 = x1x2 − x1x3 1 − EF 1 − EF

or  2  x˙ 1 = Ax2x3 + Dx  3  x˙ 2 = Gx1x2 + Hx1x3 (5.1.23)    x˙ 3 = Ix1x2 + Jx1x3 84 where CE B + EF G F (1 + B) G = − ,H = ,I = − ,J = − 1 − EF 1 − EF E 1 − EF

We can see that the closed orbits of φt(x) would be a little more complicated than

the ones in the previous case. x0 = 0 is still an equilibrium point. Using the expres-

sion of kinetic energy, the Lyapunov function in a neighborhood of x0 is given by a quadratic form as:

1 1 − A V (x) = (x2 + Ax2 + x2 + 2Dx x ) (5.1.24) 2 1 2 C 3 2 3

V (x0) = 0 and as for the point x 6= x0,

√ 2 1 2 1 D 2 1 A(1 − A) − CD 2 V (x) = x + ( Ax2 + √ x3) + x (5.1.25) 2 1 2 A 2 AC 3

Since C < 0, A > 1 and D2 1, the fact that V (x) is positive deﬁnite can be accepted as a general case. An alternative perspective to see this is that the Lya- punov function is actually V (x) = xT M 0x/2, where     m + a11 0 0 1 0 0     0 1     M =  0 m + a22 mxg + a26 = 0 AD  (5.1.26) m + a11        1−A  0 mxg + a26 Izz + a66 0 D C

Then the conditions for M 0 being a positive deﬁnite matrix is that

2 A(1 − A) 2 (Izz + a66)(m + a22) − (mxg + a26) − D = 2 > 0 (5.1.27) C (m + a11) which are generally true in practice [34]. It holds that V˙ (x) = 0. Accordingly, the dynamical system is stable in the

sense of Lyapunov in the neighborhood of x0. The trajectories of φt(x) are on the surface of the ellipsoid deﬁned by V (x) = const. However, the principal axes of the

ellipsoid don’t coincide with Ox1x2x3 axes in this case.

85 Figure 5.2: Level surface of kinetic energy and phase portrait of the dynamical system when rg 6= O or a26 6= 0.

The points on the x1-axis are still unstable equilibrium points, and the ones on the x2-axis are stable points. But the points on the x3-axis are not equilibrium points any longer. The geometric interpretation of the dynamical system are clearly illustrated by Figure 5.1 and 5.2. In fact we can prove that the periodic orbits of the dynamical system are the intersections in R3 of the level sets of two conserved quantities: the kinetic energy ellipsoids V (x) = const and the hyperbolic cylinders W (x) = const which is given by:

W (x) = C1 − V (x) (5.1.28)

86 where C1 given by (5.1.12) is a Casimir function of the system. W (x) is a linear combination of two conserved quantities of the dynamical system. In the case that

rg = O and a26 = 0, i.e., D,E,F = 0, we can obtain the formula of the hyperbolic cylinders as 1 Cx2 + x2 = const 2 A 3 which is exactly the third formula in (5.1.16). In other cases, the level sets of the conserved quantities and their intersections would look like the ones displayed in Figure 5.3. We can see that the equilibria of the system occur at the points where the level sets of V (x) and W (x) are tangent, or one of their gradients vanishes.

Figure 5.3: Level sets of kinetic energy and linear impulse of the dynamical system rg 6= O or a26 6= 0.

87 Non-coincident Centers of Buoyancy and Gravity

When the centers of buoyancy and gravity are not coincident, we can ﬁnd conserved quantities of the dynamical system by studying the Lie-Poisson bracket given by (3.5.14). First of all, H is conservative since H˙ = {H, H} = 0. The Hamiltonian matrix of the Lie-Poisson bracket is   S(π0) S(p) S(Γ)     Λ(µ) =  S(p) O O   3×3 3×3   S(Γ) O3×3 O3×3 and its null space is three-dimensional for p, Γ =6 O and p ∦ Γ [48]. Thus the three independent Casimir functions on s∗ are

T T T C1 = p Γ,C2 = p p,C3 = Γ Γ (5.1.29)

5.1.2 Stability of the Horizontal-Plane Motion

Following (4.3.1), the dynamical system of horizontal-plane motion in stationary real ﬂow without any control inputs and linear drag terms can be written as   (m + a )u ˙ − (m + a )vr − (mx + a )r2 = X u|u|  11 22 g 26 u|u|    (m + a22)v ˙ + (mxg + a26)r ˙ + (m + a11)ur = Yv|v|v|v| + Yr|r|r|r| + Yuvuv + Yurur

 (Izz + a66)r ˙ + (mxg + a26)(v ˙ + ur) + (a22 − a11)uv = N v|v| + N r|r| + Nuvuv  v|v| r|r|    + Nurur (5.1.30) Similar to (5.1.21), the equations can be written as  2  x˙ 1 = Ax2x3 + Dx + αx1|x1|  3  x˙ 2 + Ex˙ 3 = Bx1x3 + β1x2|x2| + β2x3|x3| + β3x1x2 + β4x1x3 (5.1.31)    x˙ 3 + F x˙ 2 = Cx1x2 − F x1x3 + γ1x2|x2| + γ2x3|x3| + γ3x1x2 + γ4x1x3

88 where X α = u|u| < 0 m + a11

Yv|v| Yr|r| Yuv Yur β1 = < 0, β2 = > 0, β3 = < 0, β4 = > 0 m + a22 m + a22 m + a22 m + a22

Nv|v| Nr|r| Nuv Nur γ1 = > 0, γ2 = < 0, γ3 = > 0, γ4 = < 0 Izz + a66 Izz + a66 Izz + a66 Izz + a66 The signs of the above coeﬃcients are determined based on the data of the RE- MUS AUV previously introduced. Thus, we can ﬁnd a three-dimensional nonlinear dynamical system in the form of x˙ = f(x) and written explicitly as:  2  x˙ 1 = Ax2x3 + Dx + αx1|x1|  3  0 0 x˙ 2 = G x1x2 + H x1x3 + Γ1x2|x2| + Γ2x3|x3| (5.1.32)   0 0  x˙ 3 = I x1x2 + J x1x3 + Γ3x2|x2| + Γ4x3|x3| where

CE β − Eγ B + EF β − Eγ G0 = − + 3 3 ,H0 = + 4 4 1 − EF 1 − EF 1 − EF 1 − EF C γ − F β F (1 + B) γ − F β I0 = + 3 3 ,J 0 = − + 4 4 1 − EF 1 − EF 1 − EF 1 − EF β − Eγ β − Eγ γ − F β γ − F β Γ = 1 1 ,Γ = 2 2 ,Γ = 1 1 ,Γ = 2 2 1 1 − EF 2 1 − EF 3 1 − EF 4 1 − EF Compared with (5.1.23), some coeﬃcients including G, H, I and J are corrected

due to the existence of viscous eﬀects. In addition, new coeﬃcients Γ1,Γ2,Γ3,Γ4 are introduced into the equations as dissipative terms.

T We can see that x0 = [0, 0, 0] is an equilibrium point of the dynamical

system. According to the linearization in the neighborhood of x0, the point seems like a center. However, considering the nonlinear dissipative terms, it is more likely to be an attractor. Actually, if we add linear drag terms into the dynamical sys-

tem, then x0 is indeed a locally stable equilibrium point as long as the linear drag matrix is negative deﬁnite [65, Theorem 4.7]. Without the linear drag terms, a

Lyapunov function has to be constructed to prove the stability of x0. A natural 89 candidate is given by (5.1.24), which is the Hamiltonian or kinetic energy of the dynamical system. It can be written as

1 V (x) = xT (M + M )x 2 RB A

If we can prove that V˙ (x) is negative for all x in the bounded domain D = {x ∈

3 T R |0 < V (x) < c}, then x0 = [0, 0, 0] is a asymptotically stable equilibrium point in D. From the point of view of physics, V˙ (x) should be negative if the values of parameters in the dynamical equations are in accordance with the physical world. Due to the existence of viscous or dissipative eﬀects, the rigid body will be brought to rest and so does the ﬂuid domain. During this process, the total kinetic energy V (x) of the dynamical system will dwindle from one value to a smaller one, which means that the trajectory of the dynamical system is not able to stay on an isoener- getic surface shown in Figure 5.2. Therefore V˙ (x) is negative in D. As a matter of fact,

˙ T T T V (x) = x (M RB + M A)x˙ = −x (CRB(x) + CA(x))x − x D(x)x (5.1.33)

Since CRB and CA are both skew-symmetric matrices, we have

V˙ (x) = −xT D(x)x (5.1.34) where   X|u|u|u| 0 0     D(x) = −  0 Y |v| + Y u Y |r| + Y u  (5.1.35)  |v|v uv |r|r ur    0 N|v|v|v| + Nuvu N|r|r|r| + Nuru

We can decompose D(x) into symmetric and skew-symmetric matrices. If the symmetric part is positive deﬁnite in D, which must be true in practice, then we

have that x0 is asymptotically stable in D. As a result, without any control input, the velocity x of the rigid body will always converge to the equilibrium point 90 T 3 x0 = [0, 0, 0] . We can extend D to the entire R , x0 is actually a globally asymptotically stable equilibrium point. Now we will deal with a more practical case in which there is a constant

control input in x1 direction. Then (5.1.32) can be written as  2 0  x˙ 1 = Ax2x3 + Dx + αx1|x1| + T  3  0 0 x˙ 2 = G x1x2 + H x1x3 + Γ1x2|x2| + Γ2x3|x3| (5.1.36)   0 0  x˙ 3 = I x1x2 + J x1x3 + Γ3x2|x2| + Γ4x3|x3|

0 where T = T/(m + a11) > 0 is the control input which can be considered as a constant thrust generated by a propulsion system. Then there exists an equilibrium

T point x0 = [U0, 0, 0] which satisﬁes r T 0 U = − > 0 (5.1.37) 0 α

The linearized system in a neighborhood of this equilibrium point is given by       x˙ 1 2αU0 0 0 x1 − U0             x˙  =  0 G0U H0U   x  (5.1.38)  2  0 0  2     0 0    x˙ 3 0 I U0 J U0 x3 where   2αU0 0 0    0 0  Df(x0) =  0 G U H U  (5.1.39)  0 0  0 0  0 I U0 J U0 is the Jacobian matrix of the nonlinear system.

Since U0 > 0, α < 0, the following conditions must be satisﬁed so that x0 is a stable ﬁxed point. G0 + J 0 < 0,G0J 0 − H0I0 > 0 (5.1.40)

Firstly, CE β − Eγ F (1 + B) γ − F β G0 + J 0 = − + 3 3 − + 4 4 1 − EF 1 − EF 1 − EF 1 − EF 91 It is easy to prove that CE F (1 + B) G + J = − − = 0 (5.1.41) 1 − EF 1 − EF therefore we must have that β + γ − Eγ − F β 3 4 3 4 < 0 (5.1.42) 1 − EF Since the fact that (mx + a )2 EF = g 26 < 1 (5.1.43) (m + a22)(Izz + a66) is generally true in practice, then the ﬁrst required condition becomes

β3 + γ4 < Eγ3 + F β4 (5.1.44)

In the case that rg = 0 and a26 = 0, we have E = 0,F = 0. Then the condition is satisﬁed automatically. As for the second condition, if we write

G0 = G + ∆G, H0 = H + ∆H (5.1.45) I0 = I + ∆I,J 0 = J + ∆J

then we have

(GJ − HI) + (G∆J + J∆G + ∆G∆J) − (H∆I + I∆H + ∆H∆I) > 0 (5.1.46)

in which

−BC (m + a11)(a11 − a22) GJ − HI = = 2 < 0 (5.1.47) 1 − EF (m + a22)(Izz + a66) − (mxg + a26) and

(G∆J + J∆G + ∆G∆J) − (H∆I + I∆H + ∆H∆I) β F + β C + γ B + β γ − β γ = − 3 4 3 4 3 3 4 1 − EF β3(m + a22)(mxg + a26) + β4(m + a22)(a11 − a22) − γ3(Izz + a66)(m + a11) = − 2 (m + a22)(Izz + a66) − (mxg + a26)

(β4γ3 − β3γ4)(Izz + a66)(m + a22) − 2 (m + a22)(Izz + a66) − (mxg + a26) (5.1.48) 92 As a result, the condition is

−BC > β3F + β4C + γ3B + β4γ3 − β3γ4 (5.1.49)

T Finally, in order to achieve local stability of the equilibrium point x0 = [U0, 0, 0] , two conditions which are given by (5.1.44) and (5.1.49) have to be satisﬁed simultaneously.

Remark. If we consider linear drag terms in the neighborhood of the equilibrium point, those conditions need not to be strictly satisﬁed for the reason that the stability can be guaranteed by certain selections of linear drag coeﬃcients. If there is

no linear drag and β3 = β4 = γ3 = γ4 = 0, the linearized system is identical to the

one of ideal-ﬂow system, which is unstable in a neighborhood of x0 (see Figure 5.2). In this case, accoriding to Lyapunov’s indirect method [65, Theorem 4.7], the nonlinear system is unstable as well, although its global behavior is not demonstrated yet. N

T In addition to x0 = [U0, 0, 0] , it is very likely that there exits other equi-

3 librium points in the domain D = {x ∈ R |0 < x1 < Uupper, |x2| ≤ Vupper, |x3| ≤

Wupper} that we are interested in. However, finding the equilibrium points is not easy since we have to solve the nonlinear equations. Thus numerical method has to be adopted to search for the equilibrium points. In order to demonstrate the approach to the study of the stability introduced above, several examples are presented based on the parameters and hydrodynamic coefficients of the REMUS 100 AUV [55] studied previously. Some of the hydrodynamic coefficients are adjusted according to the results of CFD simulations.

Example 5.1. The first example is about the AUV without any modification of the layout of vertical rudders. The parameters and hydrodynamic coefficients are listed in the following table.

93 Table 5.1: Parameters and Hydrodynamic Coeﬃcients of the REMUS AUV Parameter Value Units

m +3.05e+001 kg

2 Izz +3.45e+000 kg · m

a11 +9.30e−001 kg

a22 +3.55e+001 kg

a26 −1.93e+000 kg · m/rad

2 a66 +4.88e+000 kg · m /rad

xg +0.00e+000 m

Xu|u| −3.87e+000 kg/m

Yv|v| −1.00e+002 kg/m

2 Yr|r| +6.32e−001 kg · m/rad

Nv|v| +7.38e+000 kg

2 2 Nr|r| −9.40e+001 kg · m /rad

Yuv −1.79e+001 kg/m

Yur +8.62e+000 kg/rad

Nuv +4.57e+000 kg

Nur −6.40e+000 kg · m/rad

Accordingly, the coeﬃcients that are used in the analysis can be obtained as

A = 2.1,C = −4.15, α = −0.123, β1 = −1.515, β2 = 0.0096, β3 = −0.271

β4 = 0.131, γ1 = 0.886, γ2 = −11.285, γ3 = 0.549, γ4 = −0.768

It is impossible to ﬁnd the equilibrium points of the nonlinear system using analytical method. Thus numerical method is utilized to ﬁnd the equilibrium points

94 3 in the domain D = {x ∈ R |0 < x1 < 2, |x2| ≤ 0.4, |x3| ≤ 0.4}, which are

T 0 T 00 T x0 = [1.525, 0, 0] , x0 = [1.282, 0.1678, −0.2291] , x0 = [1.282, −0.1678, 0.2291] (5.1.50)

0 0 As for x0, the linearization in its neighborhood has that G + J < 0 and G0J 0 − H0I0 < 0. Then we know that one of the there eigenvalue is positive, and the other two are negative. The stable subspace of the linearized system is given by Es = span{[1, 0, 0]T , [0, −0.2723, −0.9622]T } and the unstable subspace is Eu = span{[0, −0.3291, 0.9443]T }. According to the stable manifold theorem [66, 69], there exists locally a two-dimensional stable manifold W s tangent to Es and a one

u u s dimensional unstable manifold W tangent to E . Clearly, the x1-axis is on W such that if the initial state of the system is on x1-axis, the trajectory will converge to x0.

0 00 As for x0 and x0, we can see that the nonlinear system is stable at least in a neighborhood of each of them. But what is the global behavior of the nonlinear system in D? From the above analysis, we can imagine that there are three regions in D, including the stable manifold W s and two separate regions of D divided by

s s W . The trajectories starting on W will converge to x0, and others will converge

0 00 to x0 or x0 in their own region. The above description can be illustrated by the following ﬁgure, in which the trajectories are generated using numerical methods.

95 Figure 5.4: Phase portrait of the nonlinear system based on REMUS AUV.

In a nearby neighborhood of x0, the trajectories are topologically equivalent to the ones of the linearized system due to Hartman-Grobman theorem [66]. This fact is clearly displayed in the above ﬁgure. The position of the stable and unstable

s u manifolds of x0, which are W and W respectively, can also be predicted from the

s s ﬁgure. Since x1-axis is on W , we know that W extends along x1-axis in D.

In practice, an AUV can almost never keep its state on x1-axis if without

any control mechanism. As we can see, any small deviation from x1-axis will result in a relatively large sway and yaw velocity, such that the AUV is not able to advance along a straight line. Therefore, it has to adjust the rudder angle constantly to return to the desired state. Using the kinematic equations, we can obtain the 96 steady-state position of the AUV assuming that t → ∞. For example, if the system

converges to x0, then the AUV will translate along a straight line. We can say that the vehicle can achieve straight-line stability. The relation between the horizontal position and vehicle velocity is given by the kinematic equations.   x˙ = u cos ψ − v sin ψ   y˙ = u sin ψ + v cos ψ    ψ˙ = r

0 00 If the system converges to x0 or x0 and we denote the steady-state velocity as

T xss(t) = [uss, vss, rss] , then we can obtain that

 p 2 2 uss + vss  xss(t) = sin(rsst + θ) + C1  rss   p 2 2 uss + vss (5.1.51) yss(t) = − cos(rsst + θ) + C2  rss    ψss(t) = rsst where tan θ = vss/uss, C1,C2 are constants. Hence, the steady-state path of the AUV without control is a circle, whose turning direction depends on the sign of

rss. How should we adjust the parameters of the dynamical system if we expect

T x0 = [U0, 0, 0] to be locally stable? Without considering linear drag terms, the hydrodynamic coeﬃcients of the AUV have to satisfy the conditions given by (5.1.44)

and (5.1.49). If so, then we know that x0 is an asymptotically stable in its neighborhood. But another question arises simultaneously that what the boundary of the neighborhood or the region of attraction is. If we denote     y1 x1 − U0         y = y  =  x  (5.1.52)  2  2      y3 x3 97 and assuming x1 > 0 or y1 > −U0, then (5.1.36) can be written as:         2 y˙1 2αU0 0 0 y1 Ay2y3 + Dy + αy1|y1|        3          y˙  =  0 G0U H0U  y  + G0y y + H0y y + Γ y |y | + Γ y |y |  2  0 0  2  1 2 1 3 1 2 2 2 3 3     0 0     0 0  y˙3 0 I U0 J U0 y3 I y1y2 + J y1y3 + Γ2y2|y2| + Γ4y3|y3|   2 αy − αy1|y1|  1    +  0      0 (5.1.53) Thus, the dynamical system has a new form as follows:

y˙ = Df(x0)y + f(y) + g(y) (5.1.54) where   2 Ay2y3 + Dy + αy1|y1|  3    f(y) = G0y y + H0y y + Γ y |y | + Γ y |y | (5.1.55)  1 2 1 3 1 2 2 2 3 3   0 0  I y1y2 + J y1y3 + Γ3y2|y2| + Γ4y3|y3|   2 αy − αy1|y1|  1    g(y) =  0  (5.1.56)     0

Firstly, let us consider the nonlinear system given by

y˙ = Df(x0)y + g(y) (5.1.57)

Under the conditions of (5.1.44) and (5.1.49), the system converges to zero in y2, y3 directions. As for y1 direction, we have the equation that   2αU0y1, y1 ∈ [0, +∞) 2  y˙1 = 2αU0y1 + αy1 − αy1|y1| = (5.1.58)  2α(U0 + y1)y1, y1 ∈ (−U0, 0)

T It is easy to see that y1 = 0 is a stable ﬁxed point. Therefore, y0 = [0, 0, 0] is a stable equilibrium point for the nonlinear system deﬁned by (5.1.57). 98 As for the term f(y) in (5.1.54), it is exactly the right-hand side of the dif-

ferential equations given by (5.1.32). Thus y0 is a globally asymptotically stable point of the dynamical system given by:

y˙ = f(y) (5.1.59)

We know that y0 is at least a locally asymptotically stable equilibrium point

3 T for (5.1.54) in domain D = {y ∈ R |V (y) < V ([−U0, 0, 0] )}, where V (y) is defined by (5.1.24). From the point of view of vector field, the vector field (5.1.54) is the addition of the vector fields (5.1.57) and (5.1.59). Since D is a positive invariant domain for (5.1.59), if it is still one for (5.1.57), then D is a positive invariant domain for the new vector field. One method to prove D is a positive invariant domain for (5.1.57) is to show that V˙ (y) ≤ 0. If the trajectory generated by the new vector field will always stay inside D. Then there are nonempty, invariant limit sets in D [69, Proposition 8.1.3]. If there are no other limit sets, then all trajectories starting inside D will converge to

T x0 = [x0, 0, 0] . In that case, the entire D is a region of attraction.

Example 5.2. If some modiﬁcations to the design of the AUV are undertaken such

T that the conditions (5.1.44) and (5.1.49) are fulﬁlled, then x0 = [U0, 0, 0] is very likely to be asymptotically stable equilibrium point for all the trajectories in the

domain that we are interested in. The two conditions only concern β3, β4, γ3, γ4, the last three of which are mainly aﬀected by the conﬁguration of the vertical rudders (see (4.1.8),(4.1.9) and (4.1.10)). In order to achieve straight-line stability, new parameters for vertical rudders are adopted (see Figure 4.23).

99 Table 5.2: REMUS New Vertical Fin Parameters Parameter Value Units Description

2 Sﬁn +1.15e−002 m Planform Area

bﬁn +1.31e−002 m Span

xﬁn −1.00e−000 m Moment Arm wrt Vehicle Origin at CB

cmean +8.75e−002 m Mean Chord Length

ARe +1.50e+000 n/a Eﬀective Aspect Ratio

New hydrodynamic coeﬃcients are obtained using CFD methods.

Table 5.3: Hydrodynamic Coeﬃcients of A Modiﬁed REMUS AUV

Parameter Value Units

Yuv −2.52e+001 kg/m

Yur +2.65e+001 kg/rad

Nuv +4.00e+001 kg

Nur −2.65e+001 kg · m/rad

Note that the values of Yuv in Table 5.1 and Table 5.3 are quite different. It is mainly due to the fact that the pressure distribution on the tail part of the AUV is greatly changed after the vertical rudders are deployed away from it (see the coefficients of the bare hull listed in Table 4.11). Thus the new coefficients that are used in the analysis are

β3 = −0.382, β4 = 0.402, γ3 = 4.802, γ4 = −3.181

Using numerical method, the equilibrium points are searched for in the do-

3 main D = {x ∈ R |0 < x1 < 2, |x2| ≤ 0.4, |x3| ≤ 0.4}. Only one equilibrium point

T x0 = [1.525, 0, 0] is found. There probably exist limit cycles inside D, which are nevertheless difficult to detect. 100 Figure 5.5: Phase portrait of the nonlinear system based on modified configuration of the REMUS AUV.

The above phase portrait clearly illustrates the asymptotic stability of the equilibrium point x0. Then we claim that all trajectories starting in the region shown in the ﬁgure will converge to x0 as t → +∞. This result can be proved as follows. As for the system given by (5.1.57), in the domain D = {y ∈ R3|0 <

101 T V (y) < V ([−U0, 0, 0] )} 1 − A V˙ (y) = αy2(2U + y − |y |) + Aβ U y2 + U (−D + γ )y2 1 0 1 1 3 0 2 0 C 4 3 1 − A + U (−A + Aβ + γ )y y 0 4 C 3 2 3 2 A A 1 − A 2 ≤ αy1(2U0 + y1 − |y1|) + U0(Aβ3 − + β4 + γ3)y2 2 2 2C (5.1.60) 1 − A A A 1 − A + U (−D + γ − + β + γ )y2 0 C 4 2 2 4 2C 3 3 2 2 2 = −0.123y1(2U0 + y1 − |y1|) − 0.784U0y2 − 0.774U0y3

< 0

Since V˙ (y) is also a negative deﬁnite function for (5.1.59), then we can say that for the original system (5.1.54),

˙ T T T V (y) = 2y˙ P y = 2[Df(x0))y + g(y)] P y + 2f(y) P y (5.1.61)

in which the inertia tensor is given by   1 0 0   T   P = P = 0 AD  > 0    1−A  0 D C

Thus V˙ (y) in the preceding equation, the sum of the Lyapunov functions of the

T two systems (5.1.57) and (5.1.59), is negative deﬁnite. Therefore, y0 = [0, 0, 0] is an asymptotically stable equilibrium point in D. This coincides with the result

illustrated in the preceding phase portrait. Comparing the phase portrait of the original AUV with the modiﬁed one, we

0 00 can see that the two stable equilibrium points x0 and x0 disappear and x0 turns into a stable equilibrium point. It is likely that as the Jacobian matrix of the dy-

0 00 namical system gradually transforms into a Hurwitz matrix, x0 and x0 move closer

to x0 and eventually become x0.

102 In order to study this phenomenon, we write (5.1.31) as follows,  2  x˙ 1 = Ax2x3 + Dx + αx1|x1|  3  x˙ 2 + Ex˙ 3 = Bx1x3 + β1x2|x2| + β2x3|x3| + β3x1x2 + µβ4x1x3 (5.1.62)   2  x˙ 3 + F x˙ 2 = Cx1x2 − F x1x3 + γ1x2|x2| + γ2x3|x3| + µγ3x1x2 + µ γ4x1x3

Note that a factor µ is introduced to denote the change of the vertical fins’ moment arm xfin. Neglecting the contribution of the main body and the interaction between the vertical fins and the main body of the AUV, we can study the effect of xfin on the stability of the AUV by altering its value. The parameters of the original AUV are used for numerical calculations.

Figure 5.6: The trajectory of the stable equilibrium points as µ increases.

As we can see from the above ﬁgure, the qualitative structure of the ﬂow

0 of the system is dependent on the parameter µ. The stable equilibrium point x0

0 changes continuously as the value of µ increases. When µ ≥ 2.0, x0 moves to the 103 position of x0. Thus, the two stable equilibrium coalesce into one asymptotically

stable equilibrium point on x1-axis (see Figure 5.5). Then the bifurcation value of µ is about 2.0. Similarly, instead of extending the rudders to a further position from the buoyancy center, their planform areas can be enlarged to improve the stability of the horizontal-plane motion. In this case, we write (5.1.31) as:  2  x˙ 1 = Ax2x3 + Dx + αx1|x1|  3  x˙ 2 + Ex˙ 3 = Bx1x3 + β1x2|x2| + β2x3|x3| + ηβ3x1x2 + ηβ4x1x3 (5.1.63)    x˙ 3 + F x˙ 2 = Cx1x2 − F x1x3 + γ1x2|x2| + γ2x3|x3| + ηγ3x1x2 + ηγ4x1x3

The factor η is used to represent the increase of the planform area of the rudders. Then we can plot the trajectory of the stable equilibrium points as η varies, and

ﬁnd out that the dynamical system can achieve stability at x0 when η ≥ 2.0.

Figure 5.7: The trajectory of the stable equilibrium points as η increases. 104 The degree of stability of a vehicle can thus be adjusted by altering the parameters of the rudders. Clearly, a vehicle is easy to maneuver if it is marginally stable or unstable in open loop. It requires more control eﬀort if the vehicle is designed as excessively stable. In applications, the horizontal-plane motion of an uncontrolled AUV is generally unstable for the purpose of good maneuverability, as we have shown in the case of a REMUS AUV. However, straight-line stability can be preferable in some special circumstances, such as in strong ocean currents.

5.1.3 Eﬀect of Roll Torque on the Horizontal-Plane Motion

The eﬀect of rolling on the horizontal-plane motion is usually studied using the lateral-directional equations of motion, which incorporate the motions in sway, roll and yaw directions. In the present research, however, surge motion is also con-

sidered in order to demonstrate the velocity loss in surge. Assuming Ixz,Ixy,Iyz can be ignored, the dynamical system of the motion can be written as:   2  (m + a11)u ˙ − (m + a22)vr − (mxg + a26)r + mzgpr = Xu|u|u|u| + T    (m + a )v ˙ + (mx + a )r ˙ + (m + a )ur − mz p˙ = Y v|v| + Y r|r| + Y uv  22 g 26 11 g v|v| r|r| uv    + Yurur + (W − B) sin φ   (Izz + a66)r ˙ + (mxg + a26)(v ˙ + ur) + (a22 − a11)uv = Nv|v|v|v| + Nr|r|r|r| + Nuvuv     + Nurur + W xg sin φ    (I + a )p ˙ − mz v˙ − mz ur = K p − W z sin φ + Q  xx 44 g g p g    φ˙ = p (5.1.64)

105 where Q is the roll torque acting on the vehicle by an actuator, which can be a rotating propeller. The equations can be written in the standard form as   x˙ = Ax x + Dx2 + Kx x + αx |x | + T 0  1 2 3 3 3 4 1 1    x˙ = G00x x + H00x x + N 0x + Γ 0x |x | + Γ 0x |x | + 0 sin x + Q00  2 1 2 1 3 4 1 2 2 2 3 3 1 5   00 00 0 0 0 0 00  x˙ 3 = I x1x2 + J x1x3 − FN x4 + Γ3x2|x2| + Γ4x3|x3| + 2 sin x5 − FQ  00 00 00 0 0 0  x˙ 4 = −MG x1x2 − M(H + 1)x1x3 + N x4 − MΓ1x2|x2| − MΓ2x3|x3| + 3 sin x5    + Q000     x˙ 5 = x4 (5.1.65)

Denoting R1 = L/(1 − EF ),R2 = (1 − 2E)/(1 − EF ), the coeﬃcients are mz mz mz K K = − g < 0,L = − g < 0,M = − g < 0,N = p < 0 m + a11 m + a22 Ixx + a44 Ixx + a44

W − B W xg W zg 0 Q 1 = < 0, 2 = ≈ 0, 3 = − < 0,Q = m + a22 Izz + a66 Ixx + a44 Ixx + a44 G0 H0 + MR I0 − MR (C + γ ) J 0 − MR γ G00 = ,H00 = 1 ,I00 = 1 3 ,J 00 = 1 4 1 − MR1 1 − MR1 1 − MR1 1 − MR1 NR N R Q0 Q0 N 0 = − 1 ,N 00 = ,Q00 = − 1 ,Q000 = 1 − MR1 1 − MR1 1 − MR1 1 − MR1

0 Γ1 0 Γ2 0 Γ3 − MR1γ1 0 Γ4 − MR1γ2 Γ1 = ,Γ2 = ,Γ3 = ,Γ4 = 1 − MR1 1 − MR1 1 − MR1 1 − MR1

0 R2 − 3R1 0 0 0 3 − MR2 1 = , 2 = 2 − F 1, 3 = 1 − MR1 1 − MR1

Coincident Centers of Buoyancy and Gravity

If the centers of buoyancy and gravity are coincident, we have R1 = 0,K =

0,M = 0, 2 = 0, 3 = 0. Thus, (5.1.65) can be simpliﬁed into   x˙ = Ax x + Dx2 + αx |x | + T 0  1 2 3 3 1 1   0 0 0  x˙ 2 = G x1x2 + H x1x3 + Γ1x2|x2| + Γ2x3|x3| + sin x5  1  0 0 0 x˙ 3 = I x1x2 + J x1x3 + Γ3x2|x2| + Γ4x3|x3| + 2 sin x5 (5.1.66)   0  x˙ 4 = Nx4 + Q     x˙ 5 = x4 106 0 When no actuating roll torque acts on the vehicle, i.e. Q = 0, x4 is exponentially

stable around x4 = 0 since N is negative. Furthermore, if the vehicle is neutrally

0 0 buoyant, we have 1 = 2 = 0. Then the horizontal-plane motion will be decoupled from rolling, and the ﬁrst three equations in (5.1.66) becomes (5.1.36). Generally, the dynamical system given by (5.1.66) has no equilibrium point if Q0 6= 0. In that

case, the vehicle keeps spinning around xb axis, which is unacceptable in applications. Accordingly, the centers of buoyancy and gravity cannot coincide.

Non-coincident Centers of Buoyancy and Gravity

T Case 1: Q=0 If there is no roll torque acting on the vehicle, x0 = [U0, 0, 0, 0, 0]

is an equilibrium point of (5.1.65). The linearized system in a neighborhood of x0 is given by:       x˙ 1 2αU0 0 0 0 0 x1 − U0          00 00 0 0    x˙ 2  0 G U0 H U0 N 1  x2              x˙  =  0 I00U J 00U −FN 0 0   x  (5.1.67)  3  0 0 2  3           00 00 00 0    x˙ 4  0 −MG U0 −M(H + 1)U0 N 3  x4        x˙ 5 0 0 0 1 0 x5 The eigenvalues of the Jacobian matrix determines the stability of the equilibrium

point. One of the eigenvalues is 2αU0, which is negative since α < 0 and U0 > 0. Thus, we need to study the following matrix to determine the stability of the linearized system, which can be partitioned into four 2 × 2 blocks.   00 00 0 0 G U0 H U0 N 1    00 00 0 0     I U0 J U0 −FN  J11 J12  2 J =   =   (5.1.68) −MG00U −M(H00 + 1)U N 00 0  J J  0 0 3 21 22   0 0 1 0

As for J11, due to the fact that MR1 is very small, we assume here that the stabil-

ity of J11 is the same as the Jacobian matrix of (5.1.36). As for J22, we can con-

00 0 clude that it is a Hurwitz matrix since N < 0 and 3 < 0. Therefore, when J11 and 107 J22 are both stable, the coupling terms between the horizontal-plane motion and rolling determine whether J is a stable matrix.

If at least one of the oﬀ-diagonal blocks can be ignored, i.e., J12 = O2×2 or

J21 = O2×2, then we have

det(λI4×4 − J) = det(λI2×2 − J11) × det(λI2×2 − J22)

which implies that the stability of the equilibrium point x0 is totally determined by

the blocks along the diagonal. In matrix J, the elements in J12 are relatively small compared with the ones in the other blocks. Thus J can be considered as a matrix J¯ perturbed by δJ, where     J11 O2×2 O2×2 J12 J¯ =   , δJ =   J21 J22 O2×2 O2×2

The problem here is whether δJ will change the stability of J¯. If the decoupled motions of the horizontal plane and roll direction are both stable, will the perturbation caused by the coupling terms make the system unstable? The answer depends on the sensitivity of the eigenvalues of J¯ to perturbations, and the norm of δJ. The sensitivity of an eigenvalue can be roughly estimated by its condition number. If the condition number is small, the sign of an eigenvalue may not be changed by a small perturbation, so is the stability of J¯.

Example 5.3. Let us study the case of the REMUS AUV described in [55], use the data given by Table 5.1, and the following ones to explore the coupling eﬀect on the lateral-directional motion.

108 Table 5.4: Parameters and Hydrodynamic Coeﬃcients of the REMUS AUV Parameter Value Units

W +2.99e+002 N B +3.06e+002 N

2 Ixx +1.77e−001 kg · m

2 Iyy +3.45e+000 kg · m

2 a44 +7.04e−002 kg · m /rad

2 a66 +4.88e+000 kg · m /rad

zg +1.96e−002 m

2 Kp −1.60e−003 kg · m /(rad · s)

2 2 Kp|p| −1.30e−001 kg · m /rad

The roll damping coeﬃcient considered in

[55] is Kp|p| alone. However, linear damping should be accounted for as well since exponentially convergence in roll motion is expected.

As explained above, the linear damping should be considered in the model

in addition to the quadratic damping. The value of Kp needs to be obtained from experiments. Using the given data, we can have the following matrix J¯ and δJ     −0.5815 −0.5241 0 0 0 0 −0.0001 −0.3301         −5.5369 −0.9264 0 0  0 0 0 −0.0765 ¯     J =   , δJ =   −1.4050 +2.3580 −0.0066 −24.4855 0 0 0 0          0 0 1 0 0 0 0 0

The eigenvalues of J¯ are [0.9584, −2.4662, −0.0033 ± 4.9483i]. The motion on the hor-

izontal plane is unstable around x0, and the roll motion is an underdamped decay, 109 as expected. We can obtain their individual condition numbers as [1.7916, 1.7967, 2.6008]. Apparently, the eigenvalues in roll are more sensitive to perturbations. After adding δJ to J¯, the new eigenvalues are [1.0132, −2.5025, −0.0125 ± 4.9629i]. The perturbation doesn’t change the stability of the linearized system, although the rate of convergence in roll is signiﬁcantly increased. However, the problem is that the non-

T linear system will not converge to x0 = [U0, 0, 0, 0, 0] , so that we need to ﬁnd out

the stable equilibrium points or limit circles.

T If x0 = [U0, 0, 0, 0, 0] is a stable equilibrium point, then it can be proved

T that the nonlinear system is stable. If x0 = [U0, 0, 0, 0, 0] is unstable, as in Exam- ple 5.1, the velocity of the uncontrolled AUV will converge to stable equilibrium

points or limit circles. Since x4 = 0 must hold for a stable equilibrium point, we have to solve the following nonlinear equations,   Ax x + Dx2 + αx |x | + T 0 = 0  2 3 3 1 1   00 00 0 0 0  G x1x2 + H x1x3 + Γ1x2|x2| + Γ2x3|x3| + 1 sin x5 = 0 (5.1.69)  I00x x + J 00x x + Γ 0x |x | + Γ 0x |x | + 0 sin x = 0  1 2 1 3 3 2 2 4 3 3 2 5   0 0  −Mx1x3 + (M1 + 3) sin x5 = 0

Since x3 6= 0, we must have x5 6= 0 as well. Consequently, the roll angle will converge to a nonzero value. This is known as the banking motion during turns. The exact solutions to the nonlinear equations can be found numerically.

Example 5.3 (Continued). We can try to ﬁnd the equilibrium points of the non-

linear system by solving the equations with numerical methods. Except for x0 = [1.525, 0, 0, 0, 0]T , we have two more equilibrium points

0 T 00 T x0 = [1.272, 0.1724, −0.2316, 0, −0.0301] , x0 = [1.272, −0.1724, 0.2316, 0, 0.0301] (5.1.70) Compared with the equilibrium points given by (5.1.50), it shows that the eﬀect of

0 rolling on the horizontal-plane motion is not signiﬁcant. It can be proved that x0 110 00 and x0 are both stable, as depicted in the following ﬁgure.

Figure 5.8: One projection of the phase portrait of the lateral- directional motion onto vrφ space.

The trajectory of the system state is a spiral to the asymptotically stable equilibrium point. The absolute value of the steady-state roll angle is about 2 de- grees, whose signs depend on the turning direction of the horizontal-plane motion. In conclusion, the roll motion of an AUV have little eﬀect on its horizontal-plane motion when the roll torque acting on the vehicle Q = 0.

Case 2: Q =6 0 When there is a constant nonzero roll torque acting on the vehicle. How can we analyze the eﬀect of a nonzero Q on the system with Q = 0? As a matter of fact, Q can be seen as a parameter of the system given by (5.1.65). Then

111 we have a nonlinear dynamical system as x˙ = f(x,Q). The system given by x˙ = f(x,Q 6= 0) can be regarded as a perturbation of the nominal system x˙ = f(x,Q = 0) = f(x), which has been studied above. The perturbed system can be written as the nominal system with a perturbation term,

x˙ = f(x) + g(Q) (5.1.71) where the perturbation therm g(Q) doesn’t depend on the system state. Therefore

T g(Q) is a non-vanishing perturbation [65]. x0 = [U0, 0, 0, 0, 0] is not an equilibrium point of the perturbed system any longer. We need to study the effect of the perturbation on the qualitative structure of the system’s flow, such as the evolution of the position and stability of an equilibrium point. The bifurcation theory [70] might be used to address the problem. However, there are no complete results available for high dimensional system like the one we are dealing with. Consequently, numerical solutions will be implemented to study the effect of Q on the nominal system.

Example 5.4. [55] gives the value of the roll torque exerted by the rotating main

propeller of the REMUS AUV, which is Qprop = −5.43e−1 N · m. We take this value as the base of a series of Q used in the dynamical system, and study the effect of its variation on the qualitative structure of the system. Let us use the data of the original REMUS AUV, whose horizontal-plane motion is unstable around

T x0 = [1.525, 0, 0, 0, 0] , and we can visualize the dependence of the equilibrium

points of the system on Q’s value. The value of x4 of an equilibrium point is con-

stantly zero. Firstly, the values of x5 of the equilibrium points at diﬀerent Q are depicted in Figure 5.9.

112 Stable equilibrium point

Unstable equilibrium point Q = 0

Q = Qprop

Q = 2Qprop

Q = 3Qprop

Q = 4Qprop

-25 -20 -15 -10 -5 0 5 φ (deg.)

Figure 5.9: The dependence of x5 of the equilibrium point on the parameter Q.

The hollow markers represent unstable equilibrium points, while the solid ones are stable. As the absolute value of Q increases, all the equilibrium points gradually move to the left-hand side of the ﬁgure, and the unstable equilibrium point approaches the stable one on the right-hand side. When Q becomes slightly larger than 4Qprop, a bifurcation can be observed, where the unstable equilibrium point coalesces with a stable one, and both of them disappear. Eventually, only the stable equilibrium point on the left-hand side remains.

113 Q = 0 Q = Q 0.3 prop

Q = 2Qprop

Q = 3Qprop Q = 4Q 0.2 Stable equilbrium point prop

0.1 Bifurcation occurs

Unstable equilbrium point

r 0

-0.1

-0.2 Stable equilbrium point

-0.3 -0.2 -0.1 0 1.20 1.15 0.1 1.30 1.25 0.2 1.40 1.35 0.3 1.50 1.45 v 1.55 u

Figure 5.10: The dependence of the equilibrium point on the parameter Q observed in Ox1x2x3 space.

Figure 5.10 demonstrates the projection of the equilibrium points onto the

Ox1x2x3 frame. When Q = 0, we have an unstable equilibrium point on the x1 axis and two stable equilibrium points symmetric about it, as illustrated in Figure 5.4. As we increase the absolute value of Q, the unstable equilibrium points start to

drift away from the x1 axis and move closer to one of the stable equilibrium points who itself comes towards the unstable equilibrium point. At the same time, the

other stable equilibrium point is pushed away from the x1 axis, and eventually becomes the only equilibrium point in the region when the bifurcation occurs. It is impossible to visualize the ﬂow of the dynamical system since it is ﬁve- dimensional. However, we can still estimate the stable and unstable manifolds of the equilibrium points, and use the phase portraits in lower dimensions such as

Figures 5.9 and 5.10 to predict the behavior of the system. 114 Remark. The above figures can also help us to study the possible use of roll torque to control the horizontal-plane motion of an AUV. Firstly, let’s consider how the phase portrait given by Figure 5.4 is affected by a direct turning moment generated by, for instance, a deflected rudder. If the turning moment is denoted as N, then the dynamical system can be written as  2 0  x˙ 1 = Ax2x3 + Dx + αx1|x1| + T  3  0 0 x˙ 2 = G x1x2 + H x1x3 + Γ1x2|x2| + Γ2x3|x3| (5.1.72)   0 0 0  x˙ 3 = I x1x2 + J x1x3 + Γ3x2|x2| + Γ4x3|x3| + N By gradually increasing the value of N, a figure similar to Figure 5.10 can be obtained. It can be found that the equilibrium points of the system are quite sensitive to N, in the sense that the bifurcation occurs even when N is still small. The bifurcation value is important in control. When N is smaller than the bifurcation value, it is likely that the state of the system converges to the undesired equilibrium point if its initial state is in the wrong region. In the case of a roll torque, it is not easy to generate a Q larger than the bifurcation value in applications. However, if the state of the system stays close to

the x1 axis, it is still possible to direct the state to the desired direction by manipu-

lating the position of the unstable equilibrium point correctly. N The preceding example demonstrates the case when the uncontrolled RE-

T MUS AUV is unstable around x0 = [U0, 0, 0, 0, 0] . Let us explore the stable case in the following example when the roll torque Q 6= 0.

Example 5.5. When Q = 0, the stable equilibrium point lies on the x1 axis, and it is the only equilibrium point in the region we are interested in, as displayed in Figure 5.5. When we increase the absolute value of the roll torque Q, the stable

equilibrium point leaves the x1 axis. Figure 5.11 shows that the dynamical system is structural stable under small perturbations of the parameter Q, which means the equilibrium point remains stable and no new equilibrium points emerge in the 115 region being studied. The roll torque Q is negative, so is the steady-state roll angle. The sway and yaw velocity of the vehicle are both positive when reaching their

T steady states. Obviously, x0 = [U0, 0, 0, 0, 0] is no longer a stable equilibrium point when Q 6= 0, which means the uncontrolled vehicle is not able to achieve straight- line stability. Instead, the vehicle will start to turn in circles, although the turning rate is quite small.

Q = 0

Q = Qprop 0.02 Q = 2Qprop

Q = 3Qprop 0.015 Q = 4Qprop

0.01

0.005 Stable equilibrium point

r 0

-0.005

-0.01 -25 -20 -15 -10 -5 0 φ (deg.) -0.015

-0.02 -0.1 -0.05 1.4 0 1.44 1.42 1.48 1.46 0.05 1.52 1.5 1.56 1.54 0.1 1.58 v 1.6 u

Figure 5.11: The dependence of the equilibrium point on the parameter Q observed in Ox1x2x3 frame and x5 axis

Since the roll angle is not zero, the horizontal plane of the earth-ﬁxed reference frame is not exactly aligned with the one of the body-ﬁxed reference frame of the vehicle. However, since the steady-state roll angle is usually quite small, the

diﬀerence is neglected in the qualitative analysis of the dynamical system. Remark. The previous two examples demonstrate the eﬀect of a nonzero roll torque on the horizontal-plane motion of an uncontrolled underwater vehicle. No matter 116 whether the vehicle’s horizontal-plane motion can achieve straight-line stability, the

T torque will always prevent the system’s state from staying at x0 = [U0, 0, 0, 0, 0] . Traditionally, this kind of perturbation is undesirable, and can usually be overcome by a robustly designed control system for the horizontal-plane motion. However, if perturbations are imposed deliberately upon a system, it can be considered as control or actuation forces. For example, surface vessels normally have straight-line stable. When we command them to make turns, turning moment exerted by rudders or tunnel thrusters will manipulate the stable equilibrium point of the system to the desired locations with the aid of a well designed autopilot system. Similarly, the roll torque Q might also be controlled as an actuation force, which can steer the horizontal-plane motion of an underwater vehicle. N

The coupling between the horizontal and roll motion results mainly from the oﬀ-diagonal terms of the inertia matrix, and the fact that the vehicle is not neutrally buoyant. Now let us analyze the case of a vehicle of neutral buoyancy.

Example 5.6. Assume x0 = [U0, 0, 0, 0, θ0] is one of the stable equilibrium points of the system given by (5.1.65), then we have to solve the following nonlinear equations for θ , 0  0 00  sin θ0 = −Q  1  0 00 2 sin θ0 = FQ (5.1.73)   0 0 0  (M1 + 3) sin θ0 = −Q Generally, there is no solution to the equations. However, if we have W = B and

xg = 0, then the equations turn out to be one single equation

Q sin θ0 = (5.1.74) W zg

and we can always ﬁnd its solution unless Q is extremely large. If the REMUS AUV is inherently stable in the horizontal-plane motion ,then

it can be shown that x0 = [U0, 0, 0, 0, θ0] is still a stable equilibrium point. There- 117 fore, the roll torque Q will not aﬀect the stability of the horizontal-plane motion of

the vehicle. In the previous examples, the roll torque Q is negative. How will the structure of the system’s ﬂow change if Q is positive? As for the system x˙ = f(x) given

by (5.1.65), if we denote fi(x) as the ith dimension of f(x), then it can be proved that

f1(x1, x2, x3, x4, x5,Q) = f1(x1, −x2, −x3, −x4, −x5, −Q)

fi(x1, x2, x3, x4, x5,Q) = −fi(x1, −x2, −x3, −x4, −x5, −Q), i = 2, 3, 4, 5 Therefore, the structure of the system’s ﬂow with Q is symmetric to the case of −Q with respect to the x1 axis.

5.1.4 Eﬀect of A Moving Mass on the Horizontal-Plane Motion

Internal movable masses can be used to control the center of gravity (CG) of the underwater vehicle. We assume that CG is always in the same cross section

as the center of buoyancy (CB), i.e., xg ≡ 0. CG may not be in the central longitu-

dinal section, i.e., yg = 0 is not alway true. In this case, a roll torque W yg cos φ is

introduced into the dynamical system. Since yg should be very small, the coupling

terms involving yg are ignored, and treated as disturbances. Therefore, the constant

roll torque Q in the preceding subsection is replaced by a dynamic one W yg cos φ which depends on yg. Hence, the dynamical system can be written as:

x˙ = f(x) + g(x)u (5.1.75)

in which x˙ = f(x) is given by (5.1.65), u = [yg], and   0     g2 cos x5   g(x) =   (5.1.76) g cos x   3 5   g4 cos x5 118 where g2 = −R1W/((1 − MR1)(Ixx + a44)), g3 = −F g2, and g4 = −g2/R1. Since

it is generally true in applications that x5 ∈ (−π/2, π/2), g(x)u is always nonzero

as long as yg 6= 0. Therefore, as shown in the case when Q 6= 0 and yg = 0, there

are no equilibrium points on the x1 axis. We can ﬁnd the dependence of the equi-

librium points on yg when the vehicle’s horizontal-plane motion is not inherently stable.

Stable equilibrium point

y = 0 Unstable equilibrium point g

yg = -0.4 cm

yg = -0.6 cm

yg = -0.8 cm

yg = -1.0 cm

-30 -25 -20 -15 -10 -5 0 5 φ (deg.)

Figure 5.12: The dependence of x5 of the equilibrium point on yg.

The above ﬁgure is quite similar to Figure 5.9 due to the fact that W yg cos φ

also induced a negative roll torque with yg < 0. When the length of yg is slightly

larger than 0.8 cm, the bifurcation occurs. When yg is beyond the bifurcation value, there is only one roll equilibrium angle, which can be estimated by yg φ0 = arctan (5.1.77) zg

Note that there is a slight diﬀerence between φ0 and its true value given in Figure 5.12 because of the eﬀect of the coupling terms in the equation of roll motion.

119 yg = 0

0.3 yg = -0.4 cm

yg = -0.6 cm

yg = -0.8 cm 0.2 y = -1.0 cm Stable equilbrium point g

0. 1 Bifurcation occurs

Unstable equilbrium point 0 r

-0.1

-0.2 Stable equilbrium point

-0.3 -0.2 -0.1 1.1 0 1.2 1.15 0.1 1.25 1.35 1.3 0.2 1.45 1.4 0.3 1.55 1.5 v u

Figure 5.13: The dependence of the equilibrium point on yg in Ox1x2x3 frame.

Similar to Figure 5.10, one stable equilibrium point and the unstable point move closer to each other as yg decreases. After the bifurcation occurs, only one stable equilibrium point is left in the phase portrait. In the case where the vehicle can achieve straight-line stability in the horizontal plane, we can obtain a similar diagram to Figure 5.11. Since the roll torque W yg cos φ have the same sign as yg, the equilibrium points in the above ﬁgure will be ﬂipped around the x1 axis if we change the sign of yg. In sum, the roll torque which disturbs the dynamical system x˙ = f(x) of the horizontal-plane motion of the vehicle can be induced by a rotating propeller or moving masses in the hull. From the standpoint of actuations, moving a vehicle’s CG is more economical than generating a large roll torque by internal rotors, e.g., 120 reaction wheels.

5.2 STABILITY OF AN UNDERWATER VEHICLE IN A STEADY AND UNIFORM FLOW

5.2.1 Special Case in an Ideal Fluid

For an uncontrolled rigid body in a steady and uniform ideal flow, the total kinetic energy of the system also obeys the principle of conservation of energy. This is not quite obvious at first glance if we are the observers based on the earth-fixed reference frame or body-fixed reference frame. But if we are in the inertial frame of reference which moves with the incident flow at infinity, then we will see that this system is just like the one in a stationary ambient flow.

Theorem 5.2. The total kinetic energy of the body-ﬂuid system in a steady and uniform ﬂow remains constant. The total kinetic energy of the dynamical system is given by (3.1.24),

1 1 1 T = MνT ν + νT (M + M )ν − νT (M¯ + M )ν 4 2 c c 2 RB A 2 c A c

Proof. We denote each term in the above expression as follows

1 T = MνT ν 0 2 c c 1 T = νT (M + M )ν 1 2 RB A 1 T = νT (M¯ + M )ν 2 2 c A c

As for T0, using (3.6.5)   ∂T h i U 0 T T T T = Mν˙ νc = M U S(ω) O = MU S(ω)U = −Mν ν˙ c ∂t c 1×3   c O3×1

T T Since ν˙ c νc = νc ν˙ c ∈ R, we have ∂T 0 = 0 (5.2.1) ∂t 121 Due to (3.6.8)

∂T ∂ν T 1 = (M + M )ν ∂t ∂t RB A

T = [M Aν˙ c − CA(νr)νr − CRB(ν)ν] ν

T T T = ν˙ c M Aν − νr CA(νr)ν

Similar to (A.63), we can write   S(ω) O3×3 CA(νr)νr =   M Aνr (5.2.2) S(u − U) S(ω)

then we have       −S(ω) −S(u − U) u U × ω T T T T νr CA(νr)ν = νr M A     = νr M A   O3×3 −S(ω) ω O3×1

T = νr M Aν˙ c

(3.6.5) is used in the last equality. Hence,

∂T 1 = ν˙ T M ν (5.2.3) ∂t c A c

Moreover,

∂T 2 = ν˙ T (M + M¯ )ν ∂t c A c

Therefore,

∂T ∂T ∂T ∂T = 0 + 1 − 2 = −ν˙ T Mν¯ ∂t ∂t ∂t ∂t c c     h i m¯ I3×3 −mS¯ (rb) U T = U S(ω) O1×3     b mS¯ (rb) I0 O3×1

=m ¯ U T S(ω)U

T T Since U S(ω)U = −(U S(ω)U)T ∈ R, then we obtain that ∂T = 0 ∂t 122 Remark. The kinetic energy of the fluid we use in the theorem is the true one instead of the pseudo one that we use to derive the dynamical equations based on Lagrangian approach. It is not difficult to find that the total kinetic energy of the system with pseudo kinetic energy of the fluid doesn’t obey the principle of energy

conservation at all. N

Horizontal-Plane Motion

Now let us still stick to the horizontal-plane motion. The incident flow velocity is assumed to be steady and uniform. According to (3.6.8), the governing equations can be written as:   (m + a )u ˙ − (m + a )vr − (mx + a )r2 = (a − a )U r sin ψ  11 22 g 26 22 11 x    (m + a )v ˙ + (mx + a )r ˙ + (m + a )ur = −(a − a )U r cos ψ  22 g 26 11 22 11 x    (Izz + a66)r ˙ + (mxg + a26)(v ˙ + ur) + (a22 − a11)(u − Ux cos ψ)(v + Ux sin ψ) = 0   x˙ = u cos ψ − v sin ψ    y˙ = u sin ψ + v cos ψ     ψ˙ = r (5.2.4) where Ux is the flow velocity component in x direction of the earth-fixed reference

frame (Uy is set to zero). The first three are kinetic equations and the last three are kinematic conditions relating velocities expressed in the body-fixed reference frame and the ones expressed in the earth-fixed reference frame. Since the planar position in the earth-fixed reference frame (x, y) does not appear in kinetic equations, they can be excluded from analysis for now. Hence, the nonlinear equations for this

123 dynamical system can be rewritten as follows:   x˙ = Ax x + Dx2 + PU x sin x  1 2 3 3 x 3 4    x˙ 2 + Ex˙ 3 = Bx1x3 + QUxx3 cos x4 (5.2.5)  x˙ + F x˙ = C(x − U cos x )(x + U sin x ) − F x x  3 2 1 x 4 2 x 4 1 3    x˙ 4 = x3 where a − a a − a P = 22 11 = A − 1,Q = − 22 11 = −B − 1 m + a11 m + a22 and in a more general form as:   x˙ = Ax x + Dx2 + PU x sin x  1 2 3 3 x 3 4    x˙ 2 = G(x1 − Ux cos x4)(x2 + Ux sin x4) + Hx1x3 + RUxx3 cos x4 (5.2.6)

 x˙ 3 = I(x1 − Ux cos x4)(x2 + Ux sin x4) + Jx1x3 + SUxx3 cos x4     x˙ 4 = x3 in which Q FQ R = = −(H + 1),S = − = −J 1 − EF 1 − EF and other parameters have the same deﬁnitions as in the preceding subsection. Based on the expression of the kinetic energy T of the dynamical system, we can construct a function like

1 − A V (x) = x2 + Ax2 + x2 + 2Dx x + (A − 1)U 2 cos2 x (5.2.7) 1 2 C 3 2 3 x 4

Since the kinetic energy is constant, we know that the trajectory of the dynamical system lies on the manifold given by V (x) = const, and the constant is determined by the initial state of the system. The ﬁrst step of analyzing the dynamical system is to ﬁnd its equilibrium points and study the linearized system in their neighborhoods.

x1 = Ux cos ψ0, x2 = V0 − Ux sin ψ0 ∈ R, x3 = 0, x4 = ψ0 ∈ R 124 Since ψ0 is an arbitrary constant, we have a continuum of equilibrium points. Hence the rigid body can have arbitrary yaw angle and lateral velocity. Its forward velocity is equal to the ﬂow velocity component in the same direction. In this case, the relative velocity with respect to the ﬂuid is in the transverse direction of the rigid body. Therefore the Munk moment vanishes and there is no lateral force acting on the rigid body due to D’Alembert’s paradox. The linearized system around this point is written as:       x˙ 1 0 0 AV0 + PUx sin ψ0 0 x1 − Ux cos ψ0             x˙ 2 GV0 0 −Ux cos ψ0 GV0Ux sin ψ0 x2 − V0 + Ux sin ψ0         =     x˙   IV 0 0 IV U sin ψ   x   3  0 0 x 0   3        x˙ 4 0 0 1 0 x4 − ψ0 (5.2.8) The eigenvalues of the Jacobian matrix are

2 2 λ1,2 = 0, λ3, λ4 ∈ {λ ∈ C|λ − AIV0 = 0}

Since I = C/(1 − EF ) is generally negative in practice, we have that

√ √ λ3 = i|V0| −AI, λ4 = −i|V0| −AI

The corresponding eigenvectors are     0 −Ux sin ψ0         1  0      e1 =   , e2 =   0  0          0 1

    AV0 − Ux sin ψ0 AV0 − Ux sin ψ0      √   √  −Ux cos ψ0 − iE −AI|V0| −Ux cos ψ0 + iE −AI|V0|     e3 =  √  , e4 =  √   i −AI|V |   −i −AI|V |   0   0      1 1 125 Since all the eigenvalues have zero real parts, the stability of the dynamical system around the equilibrium point can not be determined from the linearized system.

x1 = U0 + Ux cos ψ0 ∈ R, x2 = −Ux sin ψ0, x3 = 0, x4 = ψ0 ∈ R Similar to the preceding case, the rigid body moves with the ﬂow in its transverse direction. Therefore, there are no hydrodynamic force and moment acting on the rigid body. The linearized system around this equilibrium point is written as       x˙ 1 0 0 −Ux sin ψ0 0 x1 − U0 − Ux cos ψ0             x˙ 2 0 GU0 HU0 − Ux cos ψ0 GU0Ux cos ψ0  x2 + Ux sin ψ0          =     x˙  0 IU JU IU U cos ψ   x   3  0 0 0 x 0   3        x˙ 4 0 0 1 0 x4 − ψ0 (5.2.9) The eigenvalues and corresponding eigenvectors of the matrix are

√ √ λ1,2 = 0, λ3 = U0 HI − GJ, λ4 = −U0 HI − GJ     1 0         0 −Ux cos ψ0     e1 =   , e2 =   0  0          0 1     −Ux sin ψ0 −Ux sin ψ0      λ3(JU0 − λ3)   λ4(JU0 − λ4)  −Ux cos ψ0 − /IU0 −Ux cos ψ0 − /IU0     e3 =   , e3 =    λ   λ   3   4      1 1

s s The stable subspace E of the linearized system is given by E = e4, the unstable

u c subspace is E = e3 and the center subspace is E = span{e1, e2}. Thus, this equilibrium point is not a stable one of the nonlinear system.

126 From the above results we can find that the dynamical system with nonzero ambient flow is a generalized case of the one with zero ambient flow. With an addi-

tional dimension in ψ or x4, the trajectories of the dynamical system are extended into a four-dimensional space. It is very diﬃcult to visualize the trajectories in four dimensions and hence to determine the global behavior of the nonlinear system only with the knowledge of the linearized systems in the neighborhood of the ﬁxed points. However, it is possible for us to study the projections of the trajectories on a three-dimensional space. Since the last term in (5.2.7) is bounded, the projection of the hypersurface

of energy onto Ox1x2x3 space is a bounded geometry whose equation is given by

1 − A c − max{(A − 1)U 2 cos2 x } x2 + Ax2 + x2 c − min{(A − 1)U 2 cos2 x } x 4 6 1 2 C 3 6 x 4 where the constant c is determined by the initial state. We know that the trajectories of the dynamical system are inside the geometry. By assigning diﬀerent initial states which have the same kinetic energy, we can draw the trajectories of the dynamical system which are bounded by the same energy surfaces.

127 Figure 5.14: Trajectories projected on a three-dimensional phase space when rg = 0 and a26 = 0.

If Ux is a very small but nonzero number, the trajectories look like (actually

2 2 not) lying on the surface deﬁned by x1 + Ax2 + (1 − A)x3/C = c. However, the trajectories are distorted compared with the ones in Figure 5.1. The intersections between diﬀerent trajectories can occur.

128 5.2.2 Stability of a Vehicle in a Real Fluid

Based on (3.6.8) with viscous terms and constant actuation input T in x1 direction added, the differential equations can be written as:  2  (m + a11)u ˙ − (m + a22)vr − (mxg + a26)r = (a22 − a11)Uxr sin ψ     + Xu|u|(u − Ux cos ψ)|u − Ux cos ψ| + T    (m + a )v ˙ + (mx + a )r ˙ + (m + a )ur = −(a − a )U r cos ψ  22 g 26 11 22 11 x    + Yv|v|(v + Ux sin ψ)|v + Ux sin ψ|   + Yr|r|r|r| + Yuv(u − Ux cos ψ)(v + Ux sin ψ) + Yur(u − Ux cos ψ)r    (Izz + a66)r ˙ + (mxg + a26)(v ˙ + ur) + (a22 − a11)(u − Ux cos ψ)(v + Ux sin ψ)     = Nv|v|(v + Ux sin ψ)|v + Ux sin ψ| + Nr|r|r|r|     + Nuv(u − Ux cos ψ)(v + Ux sin ψ) + Nur(u − Ux cos ψ)r    ψ˙ = r (5.2.10) where Ux is the flow velocity component in x direction of the earth-fixed reference

frame, and Uy is set to zero. We assume that the forward velocity of the vehicle is

sufficiently large so that u − Ux cos ψ > 0 is always true. Thus, the nonlinear equations can be generalized as:   x˙ = Ax x + Dx2 + PU x sin x + α(x − U cos x )|x − U cos x | + T 0  1 2 3 3 x 3 4 1 x 4 1 x 4    x˙ = G0(x − U cos x )(x + U sin x ) + H0x x + Γ (x + U sin x )|x + U sin x |  2 1 x 4 2 x 4 1 3 1 2 x 4 2 x 4   0  + Γ2x3|x3| + R Uxx3 cos x4  0 0  x˙ 3 = I (x1 − Ux cos x4)(x2 + Ux sin x4) + J x1x3 + Γ3(x2 + Ux sin x4)|x2 + Ux sin x4|    + Γ x |x | + S0U x cos x  4 3 3 x 3 4    x˙ 4 = x3 (5.2.11) in which Q − β + Eγ γ + FQ − F β R0 = 4 4 ,S0 = − 4 4 1 − EF 1 − EF 129 The first step to analyze this dynamical system is to find its equilibrium points, which must satisfy that x3 = 0. Now we have the equations that are used to solve

0 0 Since we assume that x1 − Ux cos x4 > 0 and that G L − I Γ1 = 0 is not generally true, a unique equilibrium point can be found as:

T x0 = [Ux cos ψ0 + U0, −Ux sin ψ0, 0, ψ0] (5.2.13)

where ψ0 is an arbitrary constant. Thus the dynamical system has a continuum of equilibrium points. At each equilibrium point, the relative velocity of the AUV with respect to the flow is zero in the lateral direction of the body-fixed frame, such that there is no turning moment acting on the AUV by the fluid. Therefore, the

AUV can advance in a straight line of an arbitrary angle ψ0 with respect to the

direction of the ﬂow. The value of ψ0 depends on the initial state of the system. Now the question is whether the equilibrium point we ﬁnd above is a stable one or not. The linearized system in a neighborhood of the equilibrium point is written as:       x˙ 1 2αU0 0 −Ux sin ψ0 2αU0Ux sin ψ0 x1 − U0 − Ux cos ψ0          0 0 0    x˙ 2  0 G U0 H U0 − Ux cos ψ0 G U0Ux cos ψ0  x2 + Ux sin ψ0          =     x˙   0 I0U J 0U I0U U cos ψ   x   3  0 0 0 x 0   3        x˙ 4 0 0 1 0 x4 − ψ0 (5.2.14) The eigenvalues of the Jacobian matrix are

λ1 = 2αU0, λ2 = 0 130 2 0 0 2 0 0 0 0 λ3, λ4 ∈ {λ ∈ C|λ − U0(G + J )λ + U0 (G J − H I ) = 0}

The corresponding eigenvectors are     1 −Ux sin ψ0         0 −Ux cos ψ0     e1 =   , e2 =   0  0          0 1

    −Ux sin ψ0 −Ux sin ψ0      λ (J0U − λ ) 0   λ (J0U − λ ) 0  −Ux cos ψ0 − 3 0 3 /I U0 −Ux cos ψ0 − 4 0 4 /I U0     e3 =   , e4 =    λ   λ   3   4      1 1 It seems that the criteria for the stability of the dynamical system at least in several directions are the same as the ones in the case with no ﬂow. This can be ex-

plained by (4.3.2), in which the relative velocity νr can be considered as ν in the case of a stationary flow. Therefore, the results in the study of the planar motion in a stationary flow can be directly used. Let us add a steady and spatially uniform flow into Example 5.1 and 5.2 to demonstrate our new result.

Example 5.7. First of all, let us consider the REMUS AUV without modiﬁcations. We know from Example 5.1 the equilibrium points of the system given by (5.1.36).

Since (5.2.11) can be derived from (5.1.36) if ν is substituted with νr, denote an

T equilibrium point of (5.1.36) as [uss, vss, rss] (e.g., see (5.1.50)), then as t → +∞ we have   x = u + U cos x  1 ss x 4    x2 = vss − Ux sin x4 (5.2.15)

 x3 = rss     x4 = rsst + ψ0

When uss = U0, vss = 0, rss = 0, we obtain the unique equilibrium point of

131 the dynamical system (5.2.11), which is given by (5.2.13). This equilibrium point is not stable, and ψ0 depends on the initial and transient state of the system.

When uss 6= U0, vss 6= 0, rss 6= 0, we have   x1 = uss + Ux cos(rsst + ψ0)   x2 = vss − Ux sin(rsst + ψ0) (5.2.16)    x3 = rss

If we project the phase portrait of the dynamical system onto Ox1x2x3 space, the steady-state trajectory is clearly a circle. Moreover, it is a stable limit

T cycle in Ox1x2x3 since [uss, vss, rss] is a stable equilibrium point of (5.1.36) in this case. However, in the four-dimensional space Ox1x2x3x4, the steady-state trajectory is not a periodic orbit. Similar to Figure 5.4, the projection of the phase portrait onto Ox1x2x3 space for one speciﬁc initial value of x4 is illustrated as follows:

132 Figure 5.15: Projection of the phase portrait onto Ox1x2x3 space for a speciﬁc initial value of x4.

The limit cycles in the above ﬁgure are always the same regardless of the

initial state of x4. However, the transient trajectories will be diﬀerent if the initial value of x4 is changed. When a steady state is reached, the yaw rate of the vehicle will be constant and the linear velocity varies periodically. This fact can be clearly indicated in the ﬁgure below. The phase portrait of the system is projected into

Ox1x2x4 space. The spiral in the Figure 5.15 grows at the constant velocity of rss.

In the case of a stationary ﬂow (Ux = 0), x1 and x2 are both constant, the limit cycle in Figure 5.16 will shrink into a single point, and we thus obtain Figure 5.4. Moreover the spiral in Figure 5.16 becomes a straight line.

133 Figure 5.16: Projection of the phase portrait onto Ox1x2x4 space.

Based on the steady-state velocity of the AUV, its trajectory on the horizontal plane can be derived as:

 p 2 2 uss + vss  xss(t) = sin(rsst + θ) + Uxt + C1  rss   p 2 2 uss + vss (5.2.17) yss(t) = − cos(rsst + θ) + C2  rss    ψss(t) = rsst Instead of having a trajectory of a closed circle in calm ﬂow, the AUV moves along a circle which itself is drifting in the same direction as the steady and uniform

flow. Example 5.8. Similar to Example 5.2, the equilibrium point given by (5.2.13) is stable if condition (5.1.44) and (5.1.49) are both satisfied. We can prove as in 134 Example 5.2 that for the projection of the phase portrait in Ox1x2x3 space, the equilibrium point is unique and asymptotically stable given a specific initial state

of the system. However, in Ox1x2x3x4 space the equilibrium points are stable but not asymptotically stable due to the fact that we have a continuum of equilibrium points which are not isolated. The steady state of the dynamical system actually depends on its initial state. From the standpoint of physics, the AUV will travel along a straight line and its lateral velocity will always be equal to the component of the flow velocity in the same direction, as t → +∞. As a matter of fact, when the stability criteria are satisfied, the systems in a both calm flow and steady, uniform flow can be represented by (5.2.11) since they both have the same stable equilibrium points given by (5.2.13). The only difference

is that Ux equals zero in the case of a calm flow. In both cases, we have Z t x4(t) = x3(t)dt + x4(0) 0 therefore, the steady-state yaw angle depends on the transient state of yaw rate. In a calm flow, the direction of the straight line is the same as the yaw angle of the AUV as t → +∞ because its lateral velocity is zero. On the contrary, the vehicle’s heading is not necessarily aligned with the direction of the straight-line path when a steady flow exists, because the AUV drifts with the flow in its lateral direction. In this circumstance, the vehicle experiences ’crabbing’ with a constant sideslip

angle. At this stage, let us consider the case when a roll torque Q is applied on an underwater vehicle in a steady and uniform ﬂow. As the examples in the preceding section presents, the rolling torques acting on the REMUS AUV generally change the positions of the equilibrium points of the dynamical system. Nevertheless, the stability of the equilibrium points are usually preserved. When studying the dynamical system in a steady and uniform ﬂow, the previous results can be directly used due to the fact that the steady states of the 135 system are given by:   x = u + U cos x  1 ss x 4    x2 = vss − Ux sin x4

 x3 = rss     x4 = cos φssrsst + ψ0 where [uss, vss, rss, φss] are the steady states of the system in a stationary ﬂow. For example, the steady state of the horizontal-plane motion of a straight-line stable vehicle can be found in Figure 5.11 when Q 6= 0. Since vss 6= 0 and rss 6= 0, the vehicle will have the same behavior as one which cannot achieve straight-line stability in the ocean current as shown in Figure 5.15. This perturbation caused by the roll torque may be undesirable at ﬁrst glance, since an uncontrolled AUV cannot move along a straight line even when it is stable. However, the roll torque Q may be used on purpose as an actuation to control its horizontal-plane motion. Unlike the turning moment produced by vertical rudders, the roll torque can be generated by internal actuators such as rotors and moving masses. We will address this topic in detail in the following chapters.

5.3 EFFECT OF UNSTEADINESS AND NON-UNIFORMITY

The velocity field of ocean currents is generally not uniform. When considering the non-uniformity of the flow, we always adopt the hypothesis stated by Assumption A.2 that the variation of flow velocity is negligible across the length of the AUV. Then we have proved that there exists additional hydrodynamic forces acting on the AUV. Due to (3.4.19) and (3.6.7), we can write them as:       F ∗ ΦT u −Φ O ¯ 3×3 ¯   = M   +   (M A + M)νr (5.3.1) ∗ M O3×1 O3×3 O3×3

136 If the origin of the body-ﬁxed frame of reference is the buoyancy center of the vehicle, and in ideal ﬂuid velocity gradient tensor is symmetric, we have

 ∗          F m¯ I3×3 O3×3 Φu −Φ O3×3 m¯ I3×3 + A11 A12   =     +     νr ∗ b b M O3×3 I0 O3×1 O3×3 O3×3 A21 I0 + A22   Φ(m ¯ I3×3 + A11)U − Φ(A11u + A12ω) =   O3×1 (5.3.2) We can see that the non-uniformity of the flow field only induces additional force, but not hydrodynamic moment. In a horizontal plane, the velocity gradient tensor can be expressed in the earth-fixed reference frame as:

∂Ux ∂Ux  e ∂U s  ∂x ∂y  =   (5.3.3) ∂η1 ∂Uy ∂Uy  ∂x ∂y

in which [Ux,Uy] is the undisturbed current velocity expressed in the earth-ﬁxed reference frame and (x, y) is the position coordinate. Since velocity gradient tensor is symmetric, using (3.4.13) we have

∂U ∂U ∂U ∂U 1 ∂U ∂U  x cos2 ψ + x sin 2ψ + y sin2 ψ x cos 2ψ + y − x sin 2ψ  ∂x ∂y ∂y ∂y 2 ∂y ∂x  Φ =    ∂U 1 ∂U ∂U ∂U ∂U ∂U   x cos 2ψ + y − x sin 2ψ x cos2 ψ − x sin 2ψ + y sin2 ψ ∂y 2 ∂y ∂x ∂x ∂y ∂y (5.3.4) Thus, Φ is a matrix dependent on the vehicle’s position and yaw angle in a horizontal plane. Hence, the dynamical system has to be extended to be 6-dimensional by incorporating x, y coordinates.

Example 5.9. Assume Uy ≡ 0, ∂Ux/∂x ≡ 0 and ∂Ux/∂y 6= 0, then we have  ∂U ∂U  x sin 2ψ x cos 2ψ    ∂y ∂y  ∂Ux sin 2ψ cos 2ψ Φ =   =   (5.3.5) ∂U ∂U  ∂y x cos 2ψ − x sin 2ψ cos 2ψ − sin 2ψ ∂y ∂y 137 If the origin of the body-ﬁxed frame of reference is the buoyancy center of the vehicle, due to (3.4.20) the additional horizontal force induced by the non-uniformity of the velocity ﬁeld can be written as:   ∂U (m ¯ + a11)(u − Ux cos ψ) sin 2ψ + (m ¯ + a22)(v + Ux sin ψ) cos 2ψ F ∗ = − x ∂y   (m ¯ + a11)(u − Ux cos ψ) cos 2ψ − (m ¯ + a22)(v + Ux sin ψ) sin 2ψ   ∂U a26r cos 2ψ − x ∂y   −a26r sin 2ψ   ∂Ux mu¯ sin 2ψ +mv ¯ cos 2ψ +   ∂y mu¯ cos 2ψ − mv¯ sin 2ψ (5.3.6)

138 CHAPTER 6 CONTROL OF AN UNDERWATER VEHICLE USING INTERNAL ACTUATORS

6.1 CONTROL OF AN UNDERACTUATED VEHICLE

Conventional underwater vehicles are normally steered by control fins. On a horizontal plane, the vertical fins work like rudders of a surface vessel in the sense of providing turning moments to actuate horizontal-plane maneuvers. The dynamic equations of motion of an underwater vehicle in a calm flow is given by:

Mν˙ + C(ν)ν + D(ν)ν + g(η) = Bu (6.1.1)

n where M = M RB + M A, C(ν) = CRB(ν) + CA(ν), ν ∈ R is the velocity vector and u ∈ Rm represents the independent control inputs. Since no actuations are provided in sway, we must have m < n and rank(B) < n, and the underwater vehicle is thus said to be underactuated in the conﬁguration space. Regarding the problems of heading autopilot and trajectory tracking of an underactuated vehicle on a horizontal plane, the dynamical system is described by:  −1 −1  ν˙ = −M (C(ν)ν + D(ν)ν + g(η)) + M Bu

 η˙ = J(η)ν where J(η) is the transformation matrix given by (2.2.16). The system can be written in the form of a nonlinear system aﬃne in control as:

x˙ = f(x) + g(x)u (6.1.2) where x = [ν; η] is the state vector, and f(x), g(x) are both smooth.

139 If the vehicle is fully actuated, many linear and nonlinear control theories can be readily used to design controllers for heading autopilot and trajectory tracking [34]. As for underactuated vehicles, there are two different scenarios. [34] con- siders a vehicle’s workspace as a subspace of its configuration space in which the control objective is defined. In practice, stabilizing and tracking controllers are usually designed for vehicles which are fully actuated in the workspace but underactuated in the configuration space. For example, the workspace of a heading autopilot system is one since only the yaw motion is controlled, while the unactuated sway motion is the internal dynamics which is normally asymptotically stable. Conventional vehicles use rudders to produce independent actuations in yaw, thus the workspace is fully actuated. The other scenario is that the marine vehicle is underactuated in the workspace. Designing a control system that achieves stabilization, path following or trajectory tracking for this case is quite difficult, which is still an active research area for control engineers. The unactuated dynamics in (6.1.1) implies acceleration constraints on the vehicles, which can be written as:

M uν˙ + Cu(ν)ν + Du(ν)ν + gu(η) = 0 (6.1.3)

where M u, Cu(ν), Du(ν), and gu(η) denote the rows of M, C(ν), D(ν), and g(η) that correspond to the dynamics without propulsion forces or moments. [71] derived the sufficient and necessary conditions for the constraints to be partially integrable or totally integrable. It was also shown that underactuated vehicles do not satisfy Brockett’s necessary condition if the elements of the gravitational field g(η) corresponding to the unactuated dynamics are zero and, hence, the vehicles cannot be asymptotically stabilized to a single equilibrium using time-invariant continuous feedback laws. Nevertheless, [72] showed that the system describing the planar motion is strongly accessible and small-time locally controllable at any equilibrium and, hence, it is asymptotically stabilizable to a desired equilibrium using 140 time-invariant discontinuous feedback laws. Besides, a typical result on stabilization of an underactuated surface vessel using smooth time-varying state feedback laws is given in [73]. Regarding the trajectory-tracking control of underactuated vessels, there are two main approaches proposed in the literature [63], including linear time-varying control theory and Lyapunov’s direct method. In [74], the tracking error dynamics of a surface vessel was written as a cascade interconnection of two linear time- varying systems. Linear control theory was used to stabilize the linear subsystems, leading to global exponential stability of the whole system. However, the proposed controller only works under the persistent excitation condition on the reference angular velocity, which excludes a straight-line reference trajectory. [75] wrote the tracking error dynamics in a triangular-like structure through a coordinate transformation, then applied the recursive vectorial backstepping technique to design a state feedback controller. However, the control law is local, and also requires that the yaw reference velocity cannot be zero. [76] achieved global tracking control of an underactuated surface vessel by applying Lyapunov’s direct method and the passivity approach. Nevertheless, the yaw reference velocity still has to satisfy the persistent excitation condition. The condition which requires a nonzero yaw reference velocity was removed in the controller designed by [77], who used a slightly different tracking error dynamics to apply the backstepping technique.

6.2 KINETICS OF UNDERWATER VEHICLES WITH AN INTER- NAL ACTUATOR

6.2.1 Internal Rotor

If an underwater vehicle includes internal rotors as actuators, the dynamical equations governing the body-ﬂuid system should incorporate additional terms. Assume that one internal rotor is installed inside an underwater vehicle. A local

141 body-ﬁxed reference frame is created at the position of the internal rotor, which is

merely a copy of the original body-ﬁxed reference frame translated by a vector rR. The spin axis of the internal rotor coincides with the one of the axes of the local reference frame. The coordinates are illustrated in the following ﬁgure.

Figure 6.1: Reference frames of an AUV with an internal rotor.

The conﬁguration space for the AUV with one internal rotor is SE(3)×R3,

3 It is a direct product of two Lie groups SE(3) and R , denoted as a pair (R, η1, α) where α ∈ R3 indicates the rotation angle of the internal rotor with respect to

the local reference frame. For instance, if its spin axis coincides with the xr axis as shown in Figure 6.1, the rotation angle vector α = [α, 0, 0]T . Now we wish to derive the kinetic energy of an AUV equipped with an internal rotor. Denote the position of a point on the rotor as r0 with respect to its local reference frame. Its absolute velocity expressed in the local reference frame is given by:

0 0 ν = u + ω × rR + ω × r + ΩR × r (6.2.1) 142 3 where u, ω indicate the velocity of the vehicle, ΩR = α˙ ∈ R denotes the angular velocity of the internal rotor with respect to the local body-ﬁxed reference frame.

0 0 Note that r varies with time, and the term ΩR × r represents the time derivative of r0 relative to the local reference frame. Thus the kinetic energy of the internal rotor alone can be expressed as: 1 ZZZ T = |u + ω × r + ω × r0 + Ω × r0|2dm (6.2.2) 2 R R VR Then the kinetic energy of the rotor becomes

T T T T T 0 T T 0 2T = u mRu + u mRS (rR)ω + u mRS (rg)ω + u mRS (rg)ΩR

T T T + ω mRS(rR)u + ω S(rR)mRS (rR)ω

T T 0 T T 0 + ω S(rR)mRS (rg)ω + ω S(rR)mRS (rg)ΩR

T 0 T 0 T T T + ω mRS(rg)u + ω S(rg)mRS (rR)ω + ω IRω + ω IRΩR

T 0 T 0 T T T + ΩRmRS(rg)u + ΩRS(rg)mRS (rR)ω + ΩRIRω + ΩRIRΩR   u h i   T T T   = u ω Ω M R  ω  R     ΩR

and the generalized inertia matrix of the rotor M R is given by:   0 0 mRI3×3 −mRS(rR) − mRS(r ) −mRS(r )  g g  T  0 IR+S(rR)mRS (rR)+ T 0  M R = mRS(rR) + mRS(rg) 0 T T 0 IR + S(rR)mRS (rg)  S(rg)mRS (rR)+S(rR)mRS (rg)   0 0 T  mRS(rg) IR + S(rg)mRS (rR) IR (6.2.3) where mR is the mass of the internal rotor, IR is the moment of inertia about its

0 axis of rotation, and rg is the position vector of the center of gravity of the internal

0 rotor with respect to the local reference frame. Thus IR and rg vary with time while the rotor is spinning. For the sake of simplicity, the rotor is assumed to be homogeneous and ax-

0 isymmetric about its axis of rotation. Then we have that rg = O3×1, and IR is a 143 constant diagonal matrix if the rotor is also symmetric about any two axes of the local reference frame. Therefore, the inertia matrix will be   mRI3×3 −mRS(rR) O3×3    T  M R = m S(r ) I + S(r )m S (r ) I  (6.2.4)  R R R R R R R    O3×3 IR IR in which IR can be denoted as   IR 0 0  xx   R  IR =  0 I 0  (6.2.5)  yy   R  0 0 Izz

The total kinetic energy of the body-ﬂuid system with an internal rotor is given by:

     u 1 h i M RB + M A O6×3   T T T   T = u ω Ω   + M R  ω  (6.2.6) 2 R   O3×6 O3×3   ΩR And the conjugate momenta of the system can thus be expressed as:     p    u       M RB + M A O6×3   π  =   + M R  ω  (6.2.7)  0      O3×6 O3×3   πR ΩR where p denotes the total linear momentum of the dynamical system including

the internal rotor, π0 is the total angular momentum with respect to the origin of

the body-ﬁxed reference frame at the center of buoyancy of the vehicle, and πR represents the angular momentum of the rotor with respect to the origin of the

local body-ﬁxed reference frame. π0, πR could also be written as:

π0 = πpoint + πR (6.2.8)

πR = IR(ω + ΩR) (6.2.9)

144 where πpoint describes the total angular momentum of the system with the internal rotor considered as a mass point at the origin of the local reference frame. When the vehicle is neutrally buoyant in a stationary ideal ﬂuid and free from external torque. The dynamics of the whole vehicle including the internal rotor is given by:       p˙ O3×3 S(p) u   =     (6.2.10) π˙ 0 S(p) S(π0) ω Without external forces acting on the dynamical system, its total momenta are

T conserved, as implied by the Casimir functions (5.1.12), which are C1 = p p/2 and

T C2 = p π0.

Since we have π0 = πpoint + πR, it can be derived that

π˙ point = p × u + πpoint × ω − (π˙ R + ω × πR) (6.2.11)

The last term on the right side of the above equation represents the eﬀect of the rotation of the internal rotor on the underwater vehicle, or equivalently the torque on the rotor exerted by the electric motor rigidly mounted in the vehicle, whose shaft is aligned with the axis of rotation. If the direction of the spin axis is denoted

as i, then π˙ Ri is the control moment developed by the electric motor [78]. Use πR =

IR(ω + ΩR), we have that

(π˙ R + ω × πR) (6.2.12) h ˙ i = [IRω˙ + ω × (IRω)] + IRΩR + ω × (IRΩR)

Therefore, even when the internal rotor is locked in place by the electric motor, i.e.,

ΩR = 0, the eﬀect of the rotor still exists since the rotor has to spin together with the vehicle at the angular velocity ω. As a result, we are more interested in the

terms with the relative angular velocity ΩR.

If we introduce π = πpoint + IRω into the equations, which represents the total angular momentum of the system with the internal rotor locked in place, the 145 equations will be written as

p˙ = p × ω (6.2.13) d(I Ω ) π˙ = p × u + π × ω − R R (6.2.14) dt where d(IRΩR)/dt represents the eﬀect of the relative spin of the internal rotor on the vehicle. As for the internal rotor displayed in Figure 6.1, the relative angular

T velocity ΩR = [α, ˙ 0, 0] . Then we have

d(IRΩR) ˙ =IRΩR + ω × (IRΩR) dt (6.2.15) h iT R R R = Ixxα¨ rIxxα˙ −qIxxα˙ Therefore, the torque generated by the rotor in roll depends on the angular acceleration of α.

Similarly, if there is an internal rotor whose spin axis is aligned with axis zr

R of the local reference frame, then the control moment in yaw direction will be Izzγ¨ where γ is the angle of the relative rotation around zr. However, it is not reason-

able to use an internal rotor along axis zr due to the following reason. The inertia of the vehicle with the added mass is enormous in yaw, which means the control moment must be quite large. Unfortunately, the available reaction wheels which can be ﬁtted inside a small underwater vehicle are normally not able to meet the requirement.

6.2.2 Internal Moving Mass

We have analyzed the effect of a static offset of yg on the dynamics of the system in the preceding chapter. An offset of CG can be achieved by a moving mass inside the vehicle. Since the mass is usually in motion, its dynamic interaction with the vehicle should be studied. Denote the fixed mass of the vehicle as mw, and

the internal moving mass as mv. The total mass of the vehicle is then

m = mw + mv (6.2.16) 146 A local reference frame is fixed to the center of gravity of the movable mass. The different masses and coordinates are illustrated in the following figure.

Figure 6.2: Reference frames of an AUV with an internal moving mass.

We assume that the internal moving mass can only be actuated along yb

T axis. Thus, the position of the ﬁxed mass is denoted as rw = [xw, 0, zg] , and the

T position of the moving mass as rv = [xv, yv, zg] . Then the position of the vehicle’s CG is given by:

mwrw + mvrv rg = (6.2.17) mw + mv Therefore, we have the relation that

y m v = (6.2.18) yg mv

Since the space inside the hull is limited, mv should be as large as possible in order

to minimize yv. However, a heavier mass will be more energetically expensive to move.

147 The absolute velocity of a mass element of the movable mass is given by:

0 ν = u + r˙ v + ω × (rv + rv) (6.2.19)

0 where rv denotes the position of the mass element relative to the local reference frame. Thus, the kinetic energy of the internal moving mass can be written as:

1 ZZZ T = νT ν dm v 2 VR       u mvI3×3 −mvS(rv) mvI3×3 u       1       =  ω  · m S(r ) I + m S(r )ST (r ) m S(r )  ω  2    v v v v v v v v          r˙ v mvI3×3 −mvS(rv) mvI3×3 r˙ v where Iv is the moment of inertia of the moving mass about the local reference frame. Therefore, the conjugate momenta of the movable mass is given by:

pv = mv(u + ω × rv + r˙ v) (6.2.20)

πv = rv × pv + Ivω (6.2.21)

The motion of the moving mass is actuated by gravity and the constraint forces provided by the vehicle. Accordingly, except for the gravity, the eﬀect of the inter-

nal moving mass on the vehicle also comes from r˙ v and r¨v. Due to the fact that the motion of the moving mass is generally very slow [14], its dynamic eﬀect might

be ignored as unmodeled disturbances. Nevertheless, if mv takes up a sizable portion of the total mass of the vehicle, the contribution from its dynamics should be carefully modeled. In order to simplify the analysis, the following deﬁnition is used to avoid the

direct introduction of r˙ v and r¨v into the equations [10].

¯ p˙ v = F v (6.2.22)

148 The total conjugate momenta of the vehicle can be written as:

p = pw + pv (6.2.23)

π0 = π + rv × pv (6.2.24)

where pw, πw are the conjugate momenta of the vehicle without the internal mov-

ing mass, and π = πw + Ivω. With the substitution into the kinetic equation of the whole system in a calm ﬂuid, the equations in the directions of translational motions are

p˙ w + ω × pw = F ext + (mw − m¯ )gΓ − Fe v (6.2.25) where F ext is the external force acting on the whole system except for gravity, and

Fe v is the force on the moving mass exerted by the vehicle, which is given by

¯ Fe v = F v + ω × pv − mvgΓ (6.2.26)

As for rotational motions, the kinetic equations become

π˙ + ω × π + u × pw = M ext + rw × mwgΓ + rv × mvgΓ u × pv + r˙ v × pv + rv × p˙ v + ω × (rv × pv)

Using the fact that

pv r˙ v = − u − ω × rv mv the second line of the left-hand side of the equation can be simpliﬁed as:

¯ u × pv + r˙ v × pv + rv × p˙ v + ω × (rv × pv) = rv × (F v + ω × pv)

Therefore, we have

π˙ + ω × π + u × pw = M ext + rw × mwgΓ − rv × Fe v (6.2.27)

Since Iv is incorporated into the vehicle, the internal moving mass can be consid-

ered as a point mass, and Fe v, rv × Fe v represent the interaction between the vehicle and the movable mass. 149 In summary, the equations of motion of the vehicle, which include the dynamics of the internal moving mass, can be written as:  ¯  p˙ = p × ω + p × ω + F ext + (m − m¯ )gΓ − F v  w w v   ¯  π˙ = π × ω + pw × u + M ext + rw × mwgΓ − rv × (F v + ω × pv − mvgΓ)   p r˙ = v − u − ω × r v m v  v   p˙ = F¯  v v    Γ˙ = Γ × ω (6.2.28) ¯ in which F v is regarded as the vector of control inputs.

6.2.3 Equations of Motion on a Horizontal Plane

Since the internal moving mass can take up a substantial portion of the total mass of the vehicle, its linear momentum could be quite large. Therefore, its dynamics may not be neglected, whose eﬀect on the vehicle dynamics is given by:

¯ F imm = −F v + pv × ω

M imm = rv × (F imm + mvgΓ)

Since xv and zg are both constant, we have       u + zgq − yvr u + zgq −yvr             p = mv v + x r − z p +y ˙  = mv v + x r − z p + mv  y˙  (6.2.29) v  v g v  v g   v        w + yvp − xvq w − xvq yvp ¯ Because the ﬁrst part is independent of yv, its contribution to F v and pv × ω can be directly incorporated into the inertia and coriolis-centripetal matrices of the vehicle. As for the second part, we have     −yvr y˙vr − yvpq        2 2  mv  y˙  × ω = mv  (p + r )y   v   v      yvp −y˙vp − yvqr 150 Consequently, the part of F imm contributed by a nonzero yv can be written as:     yvr˙ +y ˙vr y˙vr − yvpq        2 2  F imm(yv, y˙v, y¨v) = mv  −y¨  + mv  (p + r )y  (6.2.30)  v   v      −yvp˙ − y˙vp −y˙vp − yvqr

in which the yv terms generate the variations in M RB and CRB(ω) of the vehicle.

At the same time, rv × F imm(yv, y˙v, y¨v) aﬀects the moment of inertia of the vehicle.

Roughly speaking, a nonzero yv will increase the moment of inertia of the vehicle,

based on the parallel axis theorem. As for Ixy and Izz, if yv is much smaller than xv, which is true when the internal mass is placed near the vehicle’s ends, its eﬀect is

negligible. On the contrary, yv could signiﬁcantly increase Ixx since zg is small. The

0 2 increment is given by Ixx = mvyv . If mv = m/6, and yv can reach as far as 6 cm, we

0 2 have Ixx = 0.0183 kg · m , which is approximately 10 percent of the total moment of inertia in roll including the added mass. As for the motion on a horizontal plane, we assume that q = 0, w = 0, and

pv1 can be simpliﬁed as mvu since mvyvr is not signiﬁcant in surge. Following the ¯ strategy used in [10], we use F imm as a whole, thus adding Fv2 and pv2 into the equations given by (5.1.64). If we neglect second-order inertia terms without u, and third-order terms, then the equations of motion on the horizontal plane can be

151 written as:  2  (m + a11)u ˙ − (mw + a22)vr − (mwxw + a26)r + mwzgpr = Xu|u|u|u| + T + pv2r     (mw + a22)v ˙ + (mwxw + a26)r ˙ + (m + a11)ur − mwzgp˙ = Yv|v|v|v| + Yr|r|r|r|    + Y uv + Y ur + (m − m¯ )g sin φ − F¯  uv ur v2    (Izz + a66)r ˙ + (mwxw + a26)(v ˙ + ur) + (a22 − a11)uv = Nv|v|v|v| + Nr|r|r|r|    ¯  + Nuvuv + Nurur + mgxg sin φ − mvxvur − xvFv2   (Ixx + a44)p ˙ − mwzgv˙ − mzgur = Kpp + Kp|p|p|p| − mgzg sin φ + mvgyv cos φ    + z F¯  g v2    φ˙ = p    ˙  ψ = r cos φ   p  v2  y˙v = − v − xvr + zgp  mv   ¯  p˙v2 = Fv2 (6.2.31)

Note that the quadratic damping in roll Kp|p| is considered here due to the fact that roll rates could be large under the eﬀect of the internal moving mass. Besides,

Ixx and Izz are the moments of inertia of the vehicle plus the ones of the moving

0 mass about its own local frame (See the deﬁnition of π). We leave Ixx as a part of

T the unmodeled dynamics of the system. Denoting x = [u, v, r, p, φ, ψ, yv, pv2] , the

152 equations can be rewritten as  2 0  x˙ 1 = Ax2x3 + Dx3 + Kx3x4 + αx1|x1| + T + η1x3x8    x˙ + Ex˙ + Lx˙ = Bx x + β x |x | + β x |x | + β x x + β x x  2 3 4 1 3 1 2 2 2 3 3 3 1 2 4 1 3    + sin x + η F¯  1 5 2 v2   0  x˙ 3 + F x˙ 2 = Cx1x2 − F x1x3 + γ1x2|x2| + γ2x3|x3| + γ3x1x2 + γ4x1x3    ¯  + 2 sin x5 + η3Fv2  x˙ + M 0x˙ = −Mx x + Nx + N x |x | + sin x + x cos x + η F¯  4 2 1 3 4 q 4 4 3 5 4 7 5 4 v2    x˙ 5 = x4     x˙ 6 = x3 cos x5    1  x˙ 7 = x8 − x2 − xvx3 + zgx4  mv   ¯  x˙ 8 = Fv2 (6.2.32) in which

m + a 1 a − a m x + a A = w 22 > 0,B = − < 0,C = 11 22 < 0,D = w w 26 m + a11 A Izz + a66 m + a11 m x + a m x + a mx + a mz E = w w 26 ,F 0 = w w 26 ,F = g 26 ,K = − g < 0 mw + a22 Izz + a66 Izz + a66 m + a11 m z m z mz L = − w g < 0,M 0 = − w g < 0,M = − g < 0 mw + a22 Ixx + a44 Ixx + a44

Kp Kpp N = < 0,Nq = < 0 Ixx + a44 Ixx + a44

Yv|v| Yr|r| Yuv Yur β1 = < 0, β2 = > 0, β3 = < 0, β4 = > 0 mw + a22 mw + a22 mw + a22 mw + a22

Nv|v| Nr|r| Nuv Nur γ1 = > 0, γ2 = < 0, γ3 = > 0, γ4 = < 0 Izz + a66 Izz + a66 Izz + a66 Izz + a66

(m − m¯ )g mgxg mgzg mvg 1 = < 0, 2 = ≈ 0, 3 = − < 0, 4 = > 0 mw + a22 Izz + a66 Ixx + a44 Ixx + a44

1 1 xv zg η1 = > 0, η2 = − < 0, η3 = − , η4 = > 0 m + a11 mw + a22 Izz + a66 Ixx + a44

153 Similar to (5.1.65), we would like to write the equations into the form of x˙ = f(x) + G(x)u.   x˙ = Ax x + Dx2 + Kx x + αx |x | + T 0 + η x x  1 2 3 3 3 4 1 1 1 3 8   00 00 0 0 0 0  x˙ 2 = G x1x2 + H x1x3 + N x4 + N x4|x4| + Γ x2|x2| + Γ x3|x3|  q 1 2   0 0 0 ¯  + 1 sin x5 + 4x7 cos x5 + η2Fv2    00 00 0 0 0 0 0 0  x˙ 3 = I x1x2 + J x1x3 − F N x4 − F Nqx4|x4| + Γ3x2|x2| + Γ4x3|x3|    + 0 sin x − F 00 x cos x + η0 F¯  2 5 4 7 5 3 v2   0 00 0 00 00 00 0 0 x˙ 4 = −M G x1x2 − M (H + 1)x1x3 + N x4 + Nq x4|x4| − M Γ1x2|x2| (6.2.33)    − M 0Γ 0x |x | + 0 sin x + ( − M 00 )x cos x + (η − M 0η0 )F¯  2 3 3 3 5 4 4 7 5 4 2 v2    x˙ 5 = x4     x˙ 6 = x3 cos x5   1   x˙ 7 = x8 − x2 − xvx3 + zgx4  mv   ¯  x˙ 8 = Fv2

0 0 Denoting R1 = L/(1 − EF ),R2 = (1 − 2E)/(1 − EF ), the coeﬃcients are CE β − Eγ B + EF β − Eγ β − Eγ G0 = − + 3 3 ,H0 = + 4 4 ,Γ = 1 1 1 − EF 0 1 − EF 0 1 − EF 0 1 − EF 0 1 1 − EF 0 0 0 β2 − Eγ2 00 G 00 H + MR1 Γ2 = 0 ,G = 0 ,H = 0 1 − EF 1 − M R1 1 − M R1 00 0 00 00 00 0 I = C − F G + γ3,J = −F + H F + γ4

0 Γ1 0 Γ2 0 0 0 0 0 0 Γ1 = 0 ,Γ2 = 0 ,Γ3 = γ1 − F Γ1,Γ4 = γ2 − F Γ2 1 − M R1 1 − M R1 0 0 R2 − R13 0 0 0 0 3 − M R2 0 R14 1 = 0 , 2 = 2 − F 1, 3 = 0 , 4 = − 0 1 − M R1 1 − M R1 1 − M R1 0 NR1 00 N 0 NqR1 00 Nq N = − 0 ,N = 0 ,Nq = − 0 ,Nq = 0 1 − M R1 1 − M R1 1 − M R1 1 − M R1 0 η2 − Lη4 − Eη3 0 0 0 η2 = 0 0 , η3 = η3 − F η2 (1 − EF )(1 − M R1) The objective is to control the heading angle ψ of the underwater vehicle. ¯ The control input is u = [Fv2], thus the input matrix is given by: h i T 0 0 0 0 g (x) = 0 η2 η3 η4 − M η2 0 0 0 1 (6.2.34) 154 6.3 HEADING AUTOPILOT FOR AN UNDERWATER VEHICLE

Heading autopilots for ships and underwater vehicles are usually achieved by controlled rudders. The yaw angle ψ is measured by a compass, and yaw rate can be obtained using a yaw gyro. A feedback control system will use the measure-

ments to compute the necessary rudder angles to track the desired yaw angle ψd. In this study, an internal moving mass is used instead of active rudders to control the yaw angle of an underwater vehicle. As discussed in Subsection 5.1.3 and 5.1.4, it is possible to control the dynamical behavior of the horizontal-plane motion of a vehicle by introducing an oﬀset of its CG.

6.3.1 Linear Control Design

The dynamical system of an AUV controlled by an internal moving mass is highly nonlinear, but we are going to show that it is still possible to stabilize its heading angle using a linear controller based on (6.2.33) if the initial deviations are small. The dynamical system has many equilibrium points. Our control objective is to make it converge to a speciﬁc equilibrium point, which is h iT xd = U0, 0, 0, 0, 0, ψd, 0, 0 (6.3.1)

ud = 0 (6.3.2)

q T 0 in which we have U0 = − α . At this state, the underwater vehicle moves along a straight line at the designed speed, the roll angle φ is zero, and the movable mass

stays at its initial position where yv = 0. Since the advancing speed u of an underwater vehicle cannot be measured in most applications, we only stabilize its lateral motion using a state feedback controller. Let us deﬁne a new state vector

T z = [v, r, p, φ, ψ, yv, pv2] , then the dynamical system can be linearized about the

equilibrium point (zd, ud). Defining δz = z − zd and δu = u − ud, the linearized system is δz˙ = Aδz + Bδu (6.3.3) 155 where the state matrix A is given by:   00 00 0 0 0 G U0 H U0 N 1 0 4 0    00 00 0 0 0 0 0   I U0 J U0 −F N 0 −F 0   2 4    −M 0G00U −M 0(H00 + 1)U N 00 0 0 − M 00 0   0 0 3 4 4      A =  0 0 1 0 0 0 0  (6.3.4)      0 1 0 0 0 0 0       −1 −x z 0 0 0 1/m   v g v   0 0 0 0 0 0 0 and B is defined as:   0 η2      η0   3    η − M 0η0   4 2     B =  0  (6.3.5)      0       0      1 Table 5.1 and 5.4 provide the parameters and hydrodynamic coefficients of

the REMUS AUV given by [55]. The internal moving mass is chosen to be mv = m/6, thus we have that

yv = 6yg (6.3.6)

According to Figure 5.13, yg should be able to reach as long as 1 cm, so that the bifurcation can occur. Therefore, the movable mass must have a range of motion of at last 12 cm, which is possible since the maximum hull diameter is 19.1 cm (See

Table 4.1). As for the ﬁxed mass, we chose xw as -10 cm. Thus, the internal moving

mass is located at xv = 50 cm. Using the data mentioned above, the A and B matrices can be evaluated at the equilibrium point. Then it can be shown that the matrices satisfy the control- lability rank condition. Consequently, the linearized system is controllable. As for 156 the observability, since the yaw rate r, roll rate p, yaw angle φ, roll angle ψ and the

position of the moving mass yv can usually be measured, we have

y = Cδz (6.3.7)

T where the output vector is y = [r, p, φ, ψ, yv] , and the output matrix is   0 1 0 0 0 0 0     0 0 1 0 0 0 0     C = 0 0 0 1 0 0 0 (6.3.8)       0 0 0 0 1 0 0   0 0 0 0 0 1 0

It is easy to prove that the linearized system is also observable. Therefore, we can use the measurements of y to estimate the state vector z by designing an observer.

Linear Optimal Control and Simulation Results

Since the linearized system is both controllable and observable, we can design a linear quadratic regulator (LQR) to achieve an optimal full-state feedback control of the heading angle. The cost function to be minimized is deﬁned as

Z ∞ T T J = y Qyy + δu Ruδu dt (6.3.9) 0

T T where Qy = Qy > 0 and Ru = Ru > 0 are the weighting matrices. They are chosen to prevent large outputs and actuations. Thus, a linear control law is given by δu = −Kδz where K is computed using MATLAB with the given

A, B, C, Qy, Ru. Then we can simulate the vehicle’s motion by solve the nonlinear equations x˙ = f(x) − g(x)Kδz numerically. First, let us consider the case that a REMUS AUV moving along a straight line is aﬀected by small disturbances on its surge and sway velocities. The initial

T state of the system is given by x0 = [1.50, −0.01, 0, 0, 0, 0, 0] . Since the vehicle

157 cannot achieve the straight-line stability, its state will converge to the stable equilibrium point where v < 0, r > 0. The LQR controller is expected to stabilize

the system back to (xd, ud) with ψd = 0. No signiﬁcant tuning is conducted. The weighting matrices are selected to be

3 2 3 Qy = diag(1, 5e , 1e , 1e , 1)

Ru = 1

In addition, the rate limits and saturation constraints on yv are considered in the

simulations. We assume that |y˙v| ≤ 3 cm/s, and the absolute value of yv cannot ex-

ceed 6 cm. The value of yv will be checked at each time step during the simulations to ensure that the constraints will not be violated. The simulation results are illustrated in the following Figure 6.4 and 6.5. Due to the destabilizing Monk moment, the yaw rate of the vehicle becomes positive initially, thus increasing the heading angle. According to Figure 5.13, the unstable equilibrium point will move closer to

the stable point with r > 0 when yg is negative and decreases. After the bifurcation happens, only the stable equilibrium point with r < 0 will remain. Therefore, the

internal moving mass is expected to move towards the negative side of the yv axis

in order to reverse the system’s ﬂow back to the Ox1 axis. During the simulation, all the system states converge to the equilibrium point within 20 seconds.

zd e u x x˙ = f(x) + g(x)u − Feedback Gain K z Feedback Elements

Figure 6.3: Feedback block diagram of the LQR controller

158 1.54

1.52 (m/s) u 1.5 0 10 20 30 40 50 60 70 80 90 100 0.02

0 (m/s) v -0.02 0 10 20 30 40 50 60 70 80 90 100 5

0 (deg/s)

r -5 0 10 20 30 40 50 60 70 80 90 100 50

0 (deg/s)

p -50 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.4: Simulation of linear and angular velocities of the vehicle (Case 1).

159 20

0 (deg)

-20 0 10 20 30 40 50 60 70 80 90 100 5

0 (deg)

-5 0 10 20 30 40 50 60 70 80 90 100 5

(cm) 0

v y -5 0 10 20 30 40 50 60 70 80 90 100 2 m/s) · 0 (kg

v2 -2 p 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.5: Simulation of roll and yaw angle of the vehicle, and the position and linear momentum of the internal moving mass (Case 1).

160 1.55

1.5 (m/s) u 1.45 0 10 20 30 40 50 60 70 80 90 100 0.05

0 (m/s) v -0.05 0 10 20 30 40 50 60 70 80 90 100 20

0 (deg/s)

r -20 0 10 20 30 40 50 60 70 80 90 100 50

0 (deg/s)

p -50 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.6: Simulation of linear and angular velocities of the vehicle (Case 2).

161 50

0 (deg)

-50 0 10 20 30 40 50 60 70 80 90 100 40

20 (deg)

0 0 10 20 30 40 50 60 70 80 90 100 10

(cm) 0

v y -10 0 10 20 30 40 50 60 70 80 90 100 10 m/s) · 0 (kg

v2 -10 p 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.7: Simulation of roll and yaw angle of the vehicle, and the position and linear momentum of the internal moving mass (Case 2).

162 1.54

1.52 (m/s) u 1.5 0 10 20 30 40 50 60 70 80 90 100 0.02

0 (m/s) v -0.02 0 10 20 30 40 50 60 70 80 90 100 10

0 (deg/s) r -10 0 10 20 30 40 50 60 70 80 90 100 50

0 (deg/s) p -50 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.8: Simulation of linear and angular velocities of the vehicle (Case 3).

163 20

0 (deg)

-20 0 10 20 30 40 50 60 70 80 90 100 40

20 (deg)

0 0 10 20 30 40 50 60 70 80 90 100 5

(cm) 0

v y -5 0 10 20 30 40 50 60 70 80 90 100 2 m/s) · 0 (kg

v2 -2 p 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.9: Simulation of roll and yaw angle of the vehicle, and the position and linear momentum of the internal moving mass (Case 3).

Figures 6.6 and 6.7 show the simulation results of the system with a slightly

T larger initial state x0 = [1.50, −0.05, 0, 0, 0, 0, 0] . Since the initial value of v is large, the yaw rate increases rapidly towards its equilibrium state with r > 0. The

internal mass is actuated towards the negative side of yv axis, and saturated at

yv = −6 cm. Then the yaw rate starts to decrease slowly, and when it crosses zero,

the internal mass is on the positive side. Although constraints are imposed on yv, the LQR controller has the capability of stabilizing the heading angle of the vehicle whose linear velocities are slightly disturbed. However, the convergence is certainly not very fast compared with the cases that vertical ﬁns are used as actuators. Aside from stabilizing the heading angle around a given value, the controller is also expected to regulate the vehicle to a new heading direction determined by

π T the guidance system. Let us choose an initial state x0 = [1.50, −0.01, 0, 0, 6 , 0, 0] ,

and our desired heading angle is ψd = 0. The simulation results are displayed in 164 Figures 6.8 and 6.9. Since the initial value of y is negative, the heading angle in-

creases due to the eﬀect of destabilizing Monk moment. yv is thus made negative in

order to reverse the sign of the yaw rate. When r is negative, yv becomes positive and decreases gradually so that r will converge to zero smoothly. In the preceding analysis, we designed a LQR controller for the REMUS vehicle to achieve heading autopilot using the model given by (6.2.33). However, the equations of motion used above do not include the actuator dynamics. And the controller is not practical, because the input is not the position of the moving mass. Therefore, new equations should be derived to address the problems. Firstly, we need to answer the question whether we can neglect some dynamics of the moving mass such as mvy˙v and mvy¨v such that the system given by (6.2.31) can be simplified? If we can do that, only the static effect of the internal mass on the vehicle will be considered, leaving the analysis and control design much easier. Unfortunately, we are going to show that the simplification is not reasonable. Since we have the linear momentum of the internal mass in y direction as

pv2 = mv(y ˙v + v + xvr − zgp)

its local time derivative will be

p˙v2 = mv(¨yv +v ˙ + xvr˙ − zgp˙) (6.3.10)

In the equation of surge motion, we can probably ignore mvy˙v in pv2, and combine

the others with their corresponding terms, because mvy˙v is quite small compared with the thrust T . In sway, yaw and roll directions, however, mvy¨v may have a

pronounced eﬀect on the dynamics, which is mainly caused by a huge value of y¨v.

In applications, it is yv that we control to achieve heading autopilot. If we model the actuator dynamics using a ﬁrst-order model given by

T y˙v + yv = yd(t) 165 where yd(t) is the command signal generated by a controller, y¨v will be quite large when we choose a time constant T < 1. Even if T > 1, mvy¨v could still be compara-

ble with the other terms, especially in sway. As a result, mvy¨v cannot be neglected when we design the control system. Nevertheless, we can still simplify (6.2.31) a little bit using (6.3.10). Thus, ¯ we can replace Fv2 with y¨v, the relative acceleration of the internal moving mass in y direction.  2  (m + a11)u ˙ − (m + a22)vr − (mxg + a26)r + mzgpr = Xu|u|u|u| + T     (m + a22)v ˙ + (mxg + a26)r ˙ + (m + a11)ur − mzgp˙ = Yv|v|v|v| + Yr|r|r|r|     + Yuvuv + Yurur + (m − m¯ )g sin φ − mvy¨v    (I + a )r ˙ + (mx + a )(v ˙ + ur) + (a − a )uv = N v|v| + N r|r|  zz 66 g 26 22 11 v|v| r|r|    + Nuvuv + Nurur + mgxg sin φ − xvmvy¨v   (Ixx + a44)p ˙ − mzgv˙ − mzgur = Kpp + Kp|p|p|p| − mgzg sin φ + mvgyv cos φ    + zgmvy¨v    ˙  φ = p    ˙  ψ = r cos φ    y˙ = y  v vel    y˙vel =y ¨v (6.3.11) where yvel represents the movement speed of the movable mass with respect to the

body-ﬁxed reference frame. Note that Ixx,Izz contain the contribution of the internal moving mass locked in place. Furthermore, we also need to consider actuator dynamics in the dynamical system, which will be modeled as a second-order system. The transfer function is given by:

Yv(s) kv = 2 (6.3.12) Yd(s) mvs + dvs + kv where dv and kv are the damping and feedback gain of the actuator, respectively. 166 Therefore, the dynamical system can be written as:  2  (m + a11)u ˙ − (m + a22)vr − (mxg + a26)r + mzgpr = Xu|u|u|u| + T     (m + a22)v ˙ + (mxg + a26)r ˙ + (m + a11)ur − mzgp˙ = Yv|v|v|v| + Yr|r|r|r|    + Y uv + Y ur + (m − m¯ )g sin φ + d y + k y − k y  uv ur v vel v v v d    (Izz + a66)r ˙ + (mxg + a26)(v ˙ + ur) + (a22 − a11)uv = Nv|v|v|v| + Nr|r|r|r|     + Nuvuv + Nurur + mgxg sin φ + xv(dvyvel + kvyv − kvyd)   (Ixx + a44)p ˙ − mzgv˙ − mzgur = Kpp + Kp|p|p|p| − mgzg sin φ + mvgyv cos φ    − zg(dvyvel + kvyv − kvyd)    ˙  φ = p    ψ˙ = r cos φ     y˙v = yvel    dv kv kv  y˙vel = − yvel − yv + yd mv mv mv (6.3.13) We neglect the equation in surge when designing the controller, and take

T u = U0 under the assumption that u−U0 is small. Denoting x = [v, r, p, φ, ψ, yv, yvel] , the equations can be written in the following form,

x˙ = f(x) + g(x)u

167 where f(x) =  00 00 0 0 0 0 0  G U0v + H U0r + N p + Nqp|p| + Γ1v|v| + Γ2r|r| + 1 sin φ    0 0  + 4yv cos φ − η2(dvyvel + kvyv)    I00U v + J 00U r − FN 0p − FN 0p|p| + Γ 0v|v| + Γ 0r|r| + 0 sin φ  0 0 q 3 4 2   0 0  − F 4yv cos φ − η3(dvyvel + kvyv)    00 00 00 00 0 0 0  − MG U0v − M(H + 1)U0r + N p + Nq p|p| − MΓ1v|v| − MΓ2r|r| + 3 sin φ

0 0  + (4 − M4)yv cos φ − (η4 − Mη2)(dvyvel + kvyv)     p    r cos φ     y  vel   dv kv  − yvel − yv mv mv (6.3.14) and the control input is given by   0 kvη2    0   kvη   3    k (η − Mη0 )  v 4 2      g(x)u =  0  yd (6.3.15)      0       0      kv/mv

The formulas of the coeﬃcients in the above equations can be found in Subsection

5.1.3, and the previous subsection where mw, xw have to be substituted with m, xg, respectively.

168 Actuator Dynamics

We need to determine the parameters dv and kv in the actuator dynamics.

The response of the actuator to the input yd is modeled as a second-order system, which can be written as:

2 Yv(s) ωn = 2 2 (6.3.16) Yd(s) s + 2ζωns + ωn √ 2 where ωn = kv/mv, and ζ = dv/(2 mvkv). Since mv is known, dv and kv can be

determined once we choose the values of ζ and ωn. The actuator is expected to

respond rapidly to the input yd, and at the same time, not exhibit notable oscillations. Hence, the parameters are chosen as follows:

ζ = 0.9, ωn = 2.4

so that the peak time of the second-order system is approximately 3 seconds, whose step response is shown in the following ﬁgure.

Step Response 1.2

0.8

0.6 Amplitude

0.4

0.2

0 0 1 2 3 4 5 Time (seconds)

Figure 6.10: Step response of the actuator dynamics.

169 Linear Approximation of the System

Although augmented by the actuator dynamics, we can still linearize the

T system around the setpoint xd = [0, 0, 0, 0, ψd, 0, 0] , and design a linear controller similar to the preceding subsection. The state matrix A of the linearized system x˙ = Ax + Byd is given by   00 00 0 0 0 0 0 G U0 H U0 N 1 0 4 − η2kv −η2dv    00 00 0 0 0 0 0   I U0 J U0 −FN 0 −F − η kv −η dv   2 4 3 3   0  00 00 00 0 4−M4 0 −MG U0 −M(H + 1)U0 N 3 0 0 −(η4 − Mη2)dv  −(η4−Mη2)kv      A =  0 0 1 0 0 0 0       0 1 0 0 0 0 0       0 0 0 0 0 0 1      0 0 0 0 0 −kv/mv −dv/mv (6.3.17) and the input matrix B = g(x). The output vector of the system is denoted as

T y = [r, p, φ, ψ, yv, yvel] . Thus, we have y = Cx, and   0 1 0 0 0 0 0     0 0 1 0 0 0 0       0 0 0 1 0 0 0 C =   (6.3.18)   0 0 0 0 1 0 0     0 0 0 0 0 1 0     0 0 0 0 0 0 1

170 Using the data of the REMUS AUV, the linearized system is given by     −0.591 −0.533 0 −0.332 0 2.366 0.366 −0.488         −5.629 −0.695 0 −0.0768 0 2.306 1.403  −1.871         −1.428 2.3973 −0.007 −24.48 0 204.76 −0.855  1.140              x˙ =  0 0 1 0 0 0 0  x +  0  yd          0 1 0 0 0 0 0   0           0 0 0 0 0 0 1   0          0 0 0 0 0 −5.76 −4.32 5.76 (6.3.19) It can be shown that the linearized system is both controllable and observable. As a result, a LQR controller can be designed as in the preceding subsection. In addition, (6.3.19) can give us some helpful insight into the dynamics of the nonlinear system in a small neighborhood of its equilibrium point. First of all, the dynamics of the internal moving mass can indeed greatly aﬀect the sway and yaw motions of the vehicle. Secondly, the roll rate p has little eﬀects on the horizontal motion.

Besides, the roll motion is dominated by the position yv of the internal moving mass.

Linear Optimal Control and Simulation Results

In order to design a LQR controller, we chose the weighting matrices as:

3 2 3 3 6 Qy = diag(1, 5e , 1e , 1e , 1e , 1e )

Ru = 1

The feedback gain K is computed using MATLAB, and constraints are imposed

on yv and yvel during the simulations such that the absolute value of yv cannot

be larger than 6 cm, and yvel must be zero while yv is saturated. Firstly, we be-

gin with a small disturbance, and chose the initial state of the system as x0 =

171 [1.50, −0.01, 0, 0, 0, 0, 0]T . The simulation results are shown in Figure 6.11 and Fig- ure 6.12, which illustrate that the system state converge quickly to the equilibrium point. Figure 6.13 and Figure 6.14 show that results of the case with a larger initial

sway velocity v0 = −0.05 m/s. In this case, the internal mass moves quickly to its saturation position, and the heading angle manages to come back to zero within 20 seconds. The last case starts with a nonzero yaw angle, and the controller is

expected to adjust yv to bring it back to zero. A negative initial sway velocity results in a positive yaw rate. Therefore, the yaw angle has to rise at the beginning of the simulation. Then the moving mass manages to switch the yaw rate to negative values, and bring the system state gradually back to the equilibrium point.

1.54

1.52 (m/s) u 1.5 0 10 20 30 40 50 60 70 80 90 100 0.05

0 (m/s) v -0.05 0 10 20 30 40 50 60 70 80 90 100 5

0 (deg/s)

r -5 0 10 20 30 40 50 60 70 80 90 100 50

0 (deg/s)

p -50 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.11: Simulation of linear and angular velocities of the vehicle (Case 1).

172 50

0 (deg)

-50 0 10 20 30 40 50 60 70 80 90 100 10

5 (deg)

0 0 10 20 30 40 50 60 70 80 90 100 5

(cm) 0

v y -5 0 10 20 30 40 50 60 70 80 90 100 20

(cm/s) 0

vel

y -20 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.12: Simulation of roll and yaw angle of the vehicle, and the position and local velocity of the internal moving mass (Case 1).

173 1.54

1.52 (m/s) u 1.5 0 10 20 30 40 50 60 70 80 90 100 0.1

0 (m/s) v -0.1 0 10 20 30 40 50 60 70 80 90 100 20

0 (deg/s)

r -20 0 10 20 30 40 50 60 70 80 90 100 100

0 (deg/s)

p -100 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.13: Simulation of linear and angular velocities of the vehicle (Case 2).

174 50

0 (deg)

-50 0 10 20 30 40 50 60 70 80 90 100 10

5 (deg)

0 0 10 20 30 40 50 60 70 80 90 100 10

(cm) 0

v y -10 0 10 20 30 40 50 60 70 80 90 100 50

(cm/s) 0

vel

y -50 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.14: Simulation of roll and yaw angle of the vehicle, and the position and local velocity of the internal moving mass (Case 2).

175 1.54

1.52 (m/s) u 1.5 0 10 20 30 40 50 60 70 80 90 100 0.05

0 (m/s) v -0.05 0 10 20 30 40 50 60 70 80 90 100 5

0 (deg/s)

r -5 0 10 20 30 40 50 60 70 80 90 100 50

0 (deg/s)

p -50 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.15: Simulation of linear and angular velocities of the vehicle (Case 3).

176 50

0 (deg)

-50 0 10 20 30 40 50 60 70 80 90 100 40

20 (deg)

0 0 10 20 30 40 50 60 70 80 90 100 5

(cm) 0

v y -5 0 10 20 30 40 50 60 70 80 90 100 20

(cm/s) 0

vel y -20 0 10 20 30 40 50 60 70 80 90 100 Time (s)

Figure 6.16: Simulation of roll and yaw angle of the vehicle, and the position and local velocity of the internal moving mass (Case 3).

6.3.2 Nonlinear Control Design

In this subsection, we will try to design a nonlinear controller for the vehicle using the equations of motion with actuator dynamics.

Nonlinear Cascade System

One of the strategies to design a nonlinear controller is to write the dynamical system into the form of a nonlinear cascade system as:   x˙ = f(x) + h(x)ξ + gyd (6.3.20) ˙  ξ = Aξξ + byd

177 T where x = [v, r, p, φ, ψ − ψd] represents the motion of the underwater vehicle,

T ξ = [yv, yvel] denotes the state of the internal moving mass, and we have   00 00 0 0 0 0 0 G U0v + H U0r + N p + Nqp|p| + Γ1v|v| + Γ2r|r| + 1 sin φ    00 00 0 0 0 0 0  I U0v + J U0r − FN p − FNqp|p| + Γ3v|v| + Γ4r|r| + 2 sin φ   00 00 00 00  −MG U0v−M(H +1)U0r+N p+Nq p|p|  f(x) =  0 0 0  (6.3.21)  −MΓ1v|v|−MΓ2r|r|+3 sin φ       p    r cos φ

    0 0 0 0 4 cos φ − kvη2 −dvη2 kvη2      0 0 0   0   −F 4 cos φ − kvη3 −dvη3   kvη3          h(x) = ( − M0 ) cos φ − k (η − Mη0 ) −d (η − Mη0 ) , g = k (η − Mη0 )  4 4 v 4 2 v 4 2   v 4 2           0 0   0      0 0 0 (6.3.22)

    0 1 0 Aξ =   , b =   (6.3.23) − kv − dv kv mv mv mv The uncontrolled dynamics of the vehicle is determined by x˙ = f(x), which

is analyzed in Subsection 5.1.3, and Aξ is Hurwitz. One of the advantages of using the cascade form is that the dynamics of the vehicle and its internal moving mass can be analyzed separately, and their interactions are represented by the interconnection term h(x)ξ. ˙ The origin of the subsystem ξ = Aξξ is globally exponentially stable with

T a Lyapunov function ξ P ξ, in which P satisﬁes the Lyapunov equation PAξ +

T Aξ P = −Q, and Q is positive deﬁnite. Alternatively, we can directly exploit the physics of the subsystem, and take   1 kv 0 P = (6.3.24) 2   0 mv 178 If the subsystem x˙ = f(x) is stable with a known Lyapunov function W (x) ˙ such that W (x) = Lf W (x) ≤ 0, and there exist constants c and M such that for kxk ≥ M, ∂W k kkxk ≤ cW (x) (6.3.25) ∂x

then the origin [x; ξ] = O7×1 of the cascade system   x˙ = f(x) + h(x)ξ (6.3.26) ˙  ξ = Aξξ

is globally stable [79, Proposition 5.3] under the condition

kh(x)ξk ≤ γ1(kξk)kxk + γ2(kξk) (6.3.27)

where γ1 and γ2 are two class K functions, and diﬀerentiable at ξ = O2×1. Besides, a Lyapunov function V (x, ξ) can be constructed for the system, which is given by:

V (x, ξ) = W (x) + Ψ(x, ξ) + ξT P ξ (6.3.28)

˙ T such that V (x, ξ) = Lf W (x) − ξ Qξ ≤ 0 and Ψ(x, ξ) is continuous [79, Theorem 5.8]. As a result, if x˙ = f(x) is globally stable, we can use the total mechanical energy of the vehicle as W (x), which satisﬁes (6.3.25). Since it can be shown that h(x)ξ also satisﬁes (6.3.27), then the system is globally stable. And furthermore, we can follow the control design procedure given by [80] and [79] to write the control law as ∂V ∂V y (x, ξ) = − (x, ξ)g − (x, ξ)b (6.3.29) d ∂x ∂ξ such that ˙ T 2 V (x, ξ) = Lf W (x) − ξ Qξ − yd(x, ξ) < 0 (6.3.30)

In our case, however, since x˙ = f(x) is not stable around x0 = O5×1 (when the vehicle can achieve straight-line stability, the subsystem is stable but not 179 asymptotically around ψ = 0 if without active actuations), we have to use diﬀerent

strategies to design a controller for yd to achieve the heading autopilot. Besides, the states of the system cannot reach far away from the setpoint in applications. If we can design a control law such that the stable equilibrium points of x˙ = f(x) are in the region of attraction of the controlled system, the heading autopilot can be considered as accomplished.

One possible strategy is to design a control law yd(x, ξ) to stabilize the x subsystem, and treat the ξ subsystem as the internal dynamics. Then we need to

check whether the ξ subsystem is stable with the input yd(x, ξ). However, even if the ξ subsystem is stable, we cannot guarantee that ξ will be bounded in the region we expect for the internal moving mass. Therefore, we will consider the subsystems as a whole to design the feedback control law.

Sliding Mode Controller

In order to stabilize the full system and increase its robustness to modeling errors, we can try to design a sliding mode controller to achieve the heading autopilot. Since the underwater vehicle is underactuated due to the fact that rank(g) < dim(x), the system cannot be written into the regular form to follow the standard design procedure [81, 82]. However, the strategy proposed in [83], and generalized in [34] can be adopted to design a robust nonlinear controller for the vehicle and its internal actuator.

Firstly, we will write the full system x˙ = f(x) + gyd in the following form,

∗ x˙ = Ax + f (x) + gyd (6.3.31) where A given by (6.3.17) represents the state matrix of the linearized system, and the nonlinear term in f(x) is denoted as:

f ∗(x) = f(x) − Ax (6.3.32)

180 where   N 0p|p| + Γ 0v|v| + Γ 0r|r| + 0 y φ2/2  q 1 2 4 v     0 0 0 0 2   − FNqp|p| + Γ3v|v| + Γ4r|r| − F 4yvφ /2       00 0 0 0 2   Nq p|p| − MΓ1v|v| − MΓ2r|r| + (4 − M4)yvφ /2   ∗   f (x) =  0  (6.3.33)      2   rφ /2         0      0

The feedback control law yd(x) is composed of two parts:

T ∗ yd(x) = −k x + yd(x) (6.3.34) where the linear part is used to stabilize x˙ = Ax. Thus, we have

T ∗ ∗ x˙ = (A − gk ) x + f (x) + gyd(x, ξ) (6.3.35) | {z } Ac The sliding surface is deﬁned as:

s = wT x (6.3.36) where w ∈ R5 is a design vector to be chosen such that s → 0. Diﬀerentiating s yields

T T ∗ T ∗ s˙ = w Acx + w f (x) + w gyd(x, ξ) (6.3.37)

Choosing −wT fˆ∗(x) − ηsgn(s) y∗(x, ξ) = (6.3.38) d wT g leads to

T T ∗ s˙ = w Acx − ηsgn(s) + w ∆f (x) (6.3.39) where ∆f ∗(x), ∆h(x) represent the modeling errors of f ∗(x) and h(x), respectively.

181 In order to eliminate the ﬁrst term in s˙, w should be a right eigenvector of

T T Ac , which corresponds to a zero eigenvalue. An eigenvalue of Ac can be made zero by letting

T k = [k1, k2, k3, k4, 0, k6, k7] (6.3.40)

Accordingly, the yaw angle ψ is left uncontrolled in the linear control part. And k is chosen to stabilize all the other variables of x. With V = s2/2 as a Lyapunov function candidate fors ˙, we have

V˙ = ss˙ = −η|s| + swT ∆f ∗(x) (6.3.41)

Selecting η as:

∗ η > η0 + kwk · k∆f (x)k (6.3.42) where η0 > 0, yields ˙ V ≤ −η0|s| (6.3.43)

Therefore, s(t) reaches the sliding manifold s = 0 in ﬁnite time. Once s = 0, the dynamics of the subsystem on the sliding plane wT x = 0 is governed by: gwT x˙ = A x + f ∗(x) − fˆ∗(x) (6.3.44) c wT g in which the dynamical equation of ψ is an integrator, and the remaining system is asymptotically stable. Therefore, as t → ∞, we obtain

v, r, p, φ, yv, yvel → 0

T which implies ψ → ψd due to the fact that w x = 0. Thus, the overall control law can be written as: −wT fˆ∗(x) − ηsgn(s) y (x, ξ) = −kT x + (6.3.45) d wT g The sliding mode controller includes a linear part to stabilize the reduced-order system without the ψ dimension on the sliding plane. The nonlinear part in the control law is used to force the system to reach the sliding plane in a ﬁnite time interval. The control design procedure can be summarized as follows. 182 T 1. Compute k to make Ac have a zero eigenvalue in the ψ dimension, and all the other eigenvalues with desired negative real parts.

T 2. Compute the eigenvector w corresponding to the zero eigenvalue of Ac .

3. Compute the feedback control law given by (6.3.45)

The linear feedback gain k can be calculated based on the reduced-order linear system

x¯˙ = (A¯ − g¯k¯T )x¯

y = Cx¯

T T ¯ T where x¯ = [v, r, p, φ, yv, yvel] , y = [r, p, φ, yv, yvel] , and k = [k1, k2, k3, k4, k6, k7] . The state matrix is given by:   00 00 0 0 0 0 0 G U0 H U0 N − η kv −η dv  1 4 2 2     I00U J 00U −FN 0 0 −F 0 − η0 k −η0 d   0 0 2 4 3 v 3 v   0  00 00 00 0 4−M4 0 −MG U0 −M(H + 1)U0 N 0 −(η4 − Mη )dv  3 −(η4−Mη )kv 2  A¯ =  2     0 0 1 0 0 0       0 0 0 0 0 1      0 0 0 0 −kv/mv −dv/mv (6.3.46) and the output matrix is as follows:   0 1 0 0 0 0     0 0 1 0 0 0     C = 0 0 0 1 0 0 (6.3.47)       0 0 0 0 1 0   0 0 0 0 0 1

Since the pair (A¯ , g¯) is controllable and (A¯ , C) observable given the data of the REMUS AUV, we can compute k¯ using the LQR method. 183 However, the problem with the sliding mode controller is that only local asymptotic stability can be guaranteed due to the nonlinear terms in (6.3.44), whose norms could be signiﬁcant. Besides, the rate of convergence of ψ also de-

T pends on its weight in s = w x. If its weight is small, ψ − ψd could still be large even when s is quite close to zero.

6.3.3 Nonlinear Control Design with Tunnel Thrusters

The LQR controller is able to control the internal moving mass to achieve heading autopilot. However, the success comes at a cost. Unlike rudders which provides turning moment on yaw directly, the internal moving mass generates actu-

ations through roll-yaw coupling. The local position yv of the internal mass adjusts the roll angle, and the diﬀerence between buoyancy and gravity leads to a force in sway when the roll angle is nonzero. Then the sway force makes use of the destabilizing Monk moment to control the heading direction of the vehicle. At the same

time, yv also has a direct eﬀect on sway due to roll-yaw coupling. And the inertia

force mvy¨v cannot be neglected either, as shown in the following equations  2  (m + a11)u ˙ − (m + a22)vr − (mxg + a26)r + mzgpr = Xu|u|u|u| + T     (m + a22)v ˙ + (mxg + a26)r ˙ + (m + a11)ur − mzgp˙ = Yv|v|v|v| + Yr|r|r|r|    + Y uv + Y ur + (m − m¯ )g sin φ − m y¨  uv ur v v    (Izz + a66)r ˙ + (mxg + a26)(v ˙ + ur) + (a22 − a11)uv = Nv|v|v|v| + Nr|r|r|r|   + Nuvuv + Nurur + mgxg sin φ − xvmvy¨v    (Ixx + a44)p ˙ − mzgv˙ − mzgur = Kpp + Kp|p|p|p| − mgzg sin φ + mvgyv cos φ     + zgmvy¨v    ˙  φ = p    ψ˙ = r cos φ (6.3.48)

It can be seen that the indirect eﬀect of yv on sway through mzgp˙ actu- 184 ally neutralizes part of the eﬀort of generating a sway force. Therefore, the lin-

ear controllers take advantage of both yv and mvy¨v to produce the proper actua-

tions. Aside from generating prompt force on the vehicle, mvy¨v can also produce

the proper moment if xv is selected to be negative. However, the realization of a fast moving mass in a small underwater vehicle is quite diﬃcult, if not impossible in practice. In order to address the problem, we write the equations of the dynamical system in the order following the principle of actuation discussed above. We neglect

the equation in surge when designing the controller, and take u = U0. Denoting x = [ψ, r, v, φ, p]T , the equations can be written as:

x˙ = f(x) + g(x)u + δ(¨yv) (6.3.49) where   r cos φ    00 00 0 0 0 0 0  I U0v + J U0r − FN p − FNqp|p| + Γ3v|v| + Γ4r|r| + 2 sin φ     f(x) =  G00U v + H00U r + N 0p + N 0p|p| + Γ 0v|v| + Γ 0r|r| + 0 sin φ  (6.3.50)  0 0 q 1 2 1       p   00 00 00 00  −MG U0v−M(H +1)U0r+N p+Nq p|p| 0 0 0 −MΓ1v|v|−MΓ2r|r|+3 sin φ the control input is given by:   0    0   −F 4     0  g(x)u =   cos φyv (6.3.51)  4       0    0 4 − M4

185 and the disturbance due to δ = mvy¨v is   0    0   η3     0  δ(¨yv) =  η  mvy¨v (6.3.52)  2       0    0 η4 − Mη2

We can use backstepping technique to design a nonlinear controller for the system. In order to write the equations in a strict feedback form, we assume that φ is small so that cos φ ≈ 1 and sin φ ≈ φ. Since F is very small, we ignore the

0 corresponding terms in the yaw equation. We also neglect Γ3 in yaw due to the

0 fact that it is much smaller than Γ4. In addition, the terms with p in sway are also

0 0 ignored since N and Nq are too small. As for the disturbances, only the one in sway will be considered, because the dynamics in yaw and roll are dominated by v and yv, respectively. Although the yv term in sway is undesirable, it is of the same order of magnitude as the φ term. As a result, we will treat it as another disturbance. Hence, we can write the equations as:   ψ˙ = r     r˙ = f1(r) + g1v   v˙ = f2(v, r) + g2φ + δ(yv, y¨v) (6.3.53)    φ˙ = p     p˙ = f3(v, r, p, φ) + g3yv

186 where

00 0 f1(r) = J U0r + Γ4r|r|

00 00 0 0 f2(v, r) = G U0v + H U0r + Γ1v|v| + Γ2r|r|

00 00 00 0 0 00 f3(v, r, p, φ) = −MG U0v − M(H + 1)U0r + N p − MΓ1v|v| − MΓ2r|r| + Nq p|p|

0 + 3 sin φ

00 g1 = I U0

0 g2 = 1

0 g3 = 4 − M4

0 0 δ(yv, y¨v) = 4yv + η2mvy¨v

If δ(yv, y¨v) = 0, we obtain the nominal model of the above system. Since it is in a strict feedback form, the recursive design procedure can be followed to obtain the feedback control law.

Backstepping Control Design ˙ (n) The control objective is speciﬁed by ψd and ψd. We assume that ψd = 0 when n ≥ 2. Then, the backstepping recursive design procedure is demonstrated as follows.

Step 1: Let us deﬁne the tracking error ψe = ψ−ψd, then our ﬁrst equation becomes

˙ ˙ ψe = r − ψd (6.3.54) with r viewed as the input. Thus, we design the feedback control

˙ r = α0 = −k0ψe + ψd (6.3.55)

where k0 > 0 is the feedback gain. To backstep, we use the change of variables

˙ z1 = r − α0 = r + k0ψe − ψd (6.3.56) 187 Hence, the dynamics of the tracking error can be written as:

˙ ψe = −k0ψe + z1 (6.3.57)

Consider a Control Lyapunov Function (CLF)

1 2 V0 = ψe 2

Thus, we have ˙ ˙ 2 V0 = ψeψe = −k0ψe + ψze 1

Step 2: The z1 dynamics is given by:

z˙1 =r ˙ − α˙ 0

= f1(r) + g1v − α˙ 0 where v is viewed as the control input. Taking

1 V = V + z2 1 0 2 1 as a CLF for z1 system yields

˙ ˙ V1 = V0 + z1z˙1

2 = −k0ψe + z1[ψe + f1(r) + g1v − α˙ 0]

Hence, we choose the feedback control law as:

−k1z1 − ψe − f1(r) +α ˙ 0 v = α1 = (6.3.58) g1 where k1 > 0 is the feedback gain. Let us deﬁne a new state variable

z2 = v − α1 (6.3.59)

then we have

˙ 2 2 V1 = −k0ψe − k1z1 + g1z1z2 188 Step 3: The z2 dynamics can be written as:

z˙2 =v ˙ − α˙ 1

= f2(v, r) + g2φ − α˙ 1

where φ is viewed as the control input. A CLF for the z2 system is given by

1 V = V + z2 2 1 2 2

Thus, we have

˙ ˙ V2 = V1 + z2z˙2

2 2 = −k0ψe − k1z1 + z2[g1z1 + f2(v, r) + g2φ − α˙ 1]

We design the stabilizing feedback law as:

−k2z2 − g1z1 − f2(v, r) +α ˙ 1 φ = α2 = (6.3.60) g2 where k2 > 0 is the feedback gain. Introducing a new state variable

z3 = φ − α2 yields ˙ 2 2 2 V2 = −k0ψe − k1z1 − k2z2 + g2z2z3 (6.3.61)

Step 4: The z3 dynamics is given by:

˙ z˙3 = φ − α˙ 2

= p − α˙ 2 where p is viewed as the control input. Taking

1 V = V + z2 3 2 2 3

189 yields

˙ ˙ V3 = V2 + z3z˙3

2 2 2 = −k0ψe − k1z1 − k2z2 + z3(g2z2 + p − α˙ 2)

The stabilizing function is chosen as:

p = α3 = −k3z3 − g2z2 +α ˙ 2 (6.3.62) where k3 > 0 is the feedback gain. We choose a change of variables as:

z4 = p − α3

Then we obtain ˙ 2 2 2 2 V3 = −k0ψe − k1z1 − k2z2 − k3z3 + z3z4 (6.3.63)

Step 5: The z4 dynamics is given by:

z˙4 =p ˙ − α˙ 3

= f3(v, r, p, φ) − α˙ 3 + g3yv where yv is the control input. A CLF for the z4 system is 1 V = V + z2 4 3 2 4 Therefore, we have

˙ ˙ V4 = V3 + z4z˙4

2 2 2 2 = −k0ψe − k1z1 − k2z2 − k3z3 + z4[z3 + f3(v, r, p, φ) − α˙ 3 + g3yv]

And choosing the control law as:

−k4z4 − z3 − f3(v, r, p, φ) +α ˙ 3 yv = (6.3.64) g3 with k4 > 0 yields

˙ 2 2 2 2 2 V4 = −k0ψe − k1z1 − k2z2 − k3z3 − k4z4 < 0 (6.3.65)

for all ψe 6= 0, z1 6= 0, z2 6= 0, z3 6= 0, z4 6= 0. 190 Error Dynamics

The resulting error dynamics is written as:

 ˙          ψe k0 ··· 0 ψe 0 1 ψe                     z˙1  k1  z1 −1 0 g1  z1              . .        z˙  = −  . k .  z  +  −g 0 g  z  (6.3.66)  2  2   2  1 2   2                     z˙3  k3  z3  −g2 0 1 z3           z˙4 0 ··· k4 z4 −1 0 z4

T Denoting z = [ψ,e z1, z2, z3, z4] , the equations become z˙ = −Λz + Sz, where the diagonal matrix Λ is positive deﬁnite, and S is a skew-symmetrical matrix.

1 T ˙ T T Choosing a Lyapunov function V = 2 z z yields V = −z Λz since z Sz = 0. Consequently, the equilibrium point z = 0 is globally exponentially stable. As ˙ t → ∞, we have ψ → ψd and r → ψd.

Remark. The backstepping cancels out all the terms in f1(r), f2(v, r), f3(v, r, p, φ) including the good nonlinearities, forcing the nonlinear system to behave like a linear system in a new set of coordinates. As a matter of fact, backstepping technique has the ﬂexibility to avoid cancellations of useful nonlinearities. In that case, the error dynamics becomes nonautonomous and nonlinear, which can also be proved to be

globally asymptotically stable. N

Control Law Implementation

When implementing the control law

−k4z4 − z3 − f3(v, r, p, φ) +α ˙ 3 yv = g3 we have to compute the time derivatives of α0, α1, α2 and α3.

191 Time derivatives of α0

˙ α0 = −k0ψe + ψd ˙ ˙ α˙ 0 = −k0ψe = −k0(r − ψd)

α¨0 = −k0r˙ = −k0(f1(r) + g1v) ∂f (r) α(3) = −k 1 r˙ − k g v˙ 0 0 ∂r 0 1 ∂2f (r) ∂f (r) α(4) = −k 1 r˙2 − k 1 r¨ − k g v¨ 0 0 ∂r2 0 ∂r 0 1

wherer, ˙ v˙ are given by (6.3.53), andr, ¨ v¨ can be derived accordingly.

Time derivatives of α1

−k1z1 − ψe − f1(r) +α ˙ 0 α1 = g1 ˙ ˙ −k1z˙1 − ψe − f1(r) +α ¨0 α˙ 1 = g1 −r − (k + ∂f1(r) )r ˙ + k α˙ +α ¨ + ψ˙ = 1 ∂r 1 0 0 d g1 2 ∂ f1(r) 2 ∂f1(r) ... −r˙ − ∂r2 r˙ − (k1 + ∂r )¨r + k1α¨0 + α 0 α¨1 = g1 2 2 ∂ f1(r) ∂f1(r) (3) ∂ f1(r) (3) −r¨ − 2 ∂r2 r˙r¨ − (k1 + ∂r )r − ∂r2 r˙r¨ α1 = g1 k α(3) + α(4) + 1 0 0 g1

Time derivatives of α2

−k2z2 − g1z1 − f2(v, r) +α ˙ 1 α2 = g2 ˙ −k2(v ˙ − α˙ 1) − g1(r ˙ − α˙ 0) − f2(v, r) +α ¨1 α˙ 2 = g2 −(k + ∂f2(v,r) )v ˙ − (g + ∂f2(v,r) )r ˙ + g α˙ + k α˙ +α ¨ = 2 ∂v 1 ∂r 1 0 2 1 1 g2 ∂f2(v,r) ∂f2(v,r) (3) −(k2 + ∂v )¨v − (g1 + ∂r )¨r + g1α¨0 + k2α¨1 + α1 α¨2 = g2 2 2 2 ∂ f2(v,r) v˙ 2 + 2 ∂ f2(v,r) v˙r˙ + ∂ f2(v,r) r˙2 − ∂v2 ∂vr ∂r2 g2 192 Time derivatives of α3

α3 = −k3z3 − g2z2 +α ˙ 2

α˙ 3 = −k3(p − α˙ 2) − g2(v ˙ − α˙ 1) +α ¨2

The analytical derivatives of the virtual control signals given above are complicated and cumbersome to derive, since the dynamical system is of high order. An alternative method to derive the control law is called command ﬁltered backstepping [84].

Lyapunov Redesign

The nonlinear feedback control law (6.3.64) is derived based on the nominal model of the dynamical system. In order to stabilize the actual system in the

presence of the disturbance δ(yv, y¨v), we can install jet pumps or tunnel thrusters in the vehicle to provide the balance force in sway direction. The control law u˜ can be designed using Lyapunov redesign technique [65]. The time derivative of the Lyapunov function of the overall system can be written as

˙ ˙ V = V4 + z2δ(yv, y¨v) + z2u˜ (6.3.67)

We need to chooseu ˜ so that z2(δ(yv, y¨v) +u ˜) ≤ 0. Since we have

z2δ(yv, y¨v) + z2u˜ ≤ z2u˜ + |z2||δ(yv, y¨v)|

Taking

u˜ = −ηsgn(z2) (6.3.68) with η ≥ |δ(yv, y¨v)|, yields

z2δ(yv, y¨v) + z2u˜ ≤ 0

Hence, with the control (6.3.64) for the internal moving mass, V˙ of the overall

system is negative deﬁnite. The control law u˜ depends on the magnitude of δ(yv, y¨v) 193 and the sign of z2 = v − α1. The state z2 represents the diﬀerence between the actual and desired sway velocities. The magnitude ofu ˜ is determined by:

0 0 η ≥ |δ(yv, y¨v)| = |4yv + η2mvy¨v| (6.3.69)

Since we can actually measure yv andy ¨v, η is thus given by:

0 0 η(yv, y¨v) = η0 + |4yv + η2mvy¨v| (6.3.70)

where η0 is a positive constant. Then the balance force should be

Ft = (m + a22)˜u = −(m + a22)η(yv, y¨v)sgn(z2) (6.3.71)

The force Ft could be quite large because of the considerable magnitude of a22. In

0 order to reduce Ft, we can try to decrease the absolute value of the constant η2 in

η(yv, y¨v), which is given by:

0 η2 − Lη4 − Eη3 η2 = (1 − EF )(1 − MR1)

Since the constants η3, η4 are

−xv zv η3 = , η4 = Izz + a66 Ixx + a44

0 we can make the absolute value of η2 as small as possible by changing the location

of the internal moving mass in the vehicle. Besides, it can be seen that mvy¨v can

usually produce positive eﬀects on the system. Therefore, we omit the y¨v term in

η(yv, y¨v), leading to

0 Ft = −(m + a22)(η0 + |4yv|)sgn(z2) (6.3.72)

The maximum force the jet pumps or tunnel thrusters should produce can be cal-

culated using the above formula. In order to minimize Ft, the underwater vehicle

is expected to have a smaller a22. Besides, the thruster’s position should be carefully chosen such that it produces no moments in both yaw and roll directions. 194 The control law for Ft is discontinuous, which can lead to chattering. [81] suggest smoothing out the discontinuity inside a boundary layer according to   sgn(s) if |s/φ| > 1 sat(s) = (6.3.73)  s/φ otherwise where φ > 0 can be interpreted as the boundary layer thickness.

Simulation Results

The control coeﬃcients in the nonlinear controller are chosen to be

k0 = 0.5, k1 = k2 = k3 = k4 = 1.0

Then yv computed by (6.3.64) is returned to the full nonlinear dynamical system for numerical simulations in MATLAB. Firstly, we run a simulation with a small

T disturbance, choosing the initial state of the system as x0 = [1.50, −0.05, 0, 0, 0, 0] . The simulation results are shown in Figures 6.17, 6.18 and 6.19, which illustrate that the system state converge quickly to the equilibrium point. Figures 6.20, 6.21 and 6.22 display the results of a case with nonzero initial yaw angle and zero heading setpoint. In this case, the jet pumps exert large forces to counteract the nega-

tive eﬀect of the saturated yv, and make the yaw angle back to zero within a short time. In order to test whether the actuators can bring the dynamical system back

T to the equilibrium point with a large initial state, we set x0 = [1.50, −0.1, 0, 0, 0, 0] in the last case. It is shown that the yaw angle rises rapidly because of the destabilizing moment. The internal moving mass and jet pumps manage to bring the sway velocity to positive, and yaw rate negative. Thus, the yaw angle converges to zero within 10 seconds as shown in Figure 6.24. Because both the displacement of the internal moving mass and the side thruster’s maximum force are constrained, global stability of the system cannot be achieved in practice. 195 1.54

1.52 (m/s) u 1.5 0 10 20 30 40 50 60 0.05

0 (m/s) v -0.05 0 10 20 30 40 50 60 10

0 (deg/s)

r -10 0 10 20 30 40 50 60 50

0 (deg/s)

p -50 0 10 20 30 40 50 60 Time (s)

Figure 6.17: Simulation of linear and angular velocities of the vehicle (Case 1).

0 (deg)

-20 0 10 20 30 40 50 60 10

0 (deg)

-10 0 10 20 30 40 50 60 5

(cm) 0

v y -5 0 10 20 30 40 50 60 5 (N)

-5 Jet pump 0 10 20 30 40 50 60 Time (s)

Figure 6.18: Simulation of roll and yaw angle of the vehicle, the position and local velocity of the internal moving mass, and the force exerted by jet pumps (Case 1).

196 10

0 (deg/s)

z -10 0 10 20 30 40 50 60 0.05

(m/s) 0

2 z -0.05 0 10 20 30 40 50 60 200

(deg) 0

3 z -200 0 10 20 30 40 50 60 500

0 (deg/s)

z -500 0 10 20 30 40 50 60 Time (s)

Figure 6.19: The errors of the dynamics during the simulation (Case 1).

1.6

1.5 (m/s) u 1.4 0 10 20 30 40 50 60 0.2

0 (m/s) v -0.2 0 10 20 30 40 50 60 20

0 (deg/s)

r -20 0 10 20 30 40 50 60 100

0 (deg/s)

p -100 0 10 20 30 40 50 60 Time (s)

Figure 6.20: Simulation of linear and angular velocities of the vehicle (Case 2).

197 50

0 (deg)

-50 0 10 20 30 40 50 60 50

0 (deg)

-50 0 10 20 30 40 50 60 10

(cm) 0

v y -10 0 10 20 30 40 50 60 10 (N)

-10 Jet pump 0 10 20 30 40 50 60 Time (s)

Figure 6.21: Simulation of roll and yaw angle of the vehicle, the position and local velocity of the internal moving mass, and the force exerted by jet pumps (Case 2).

0 (deg/s)

z -50 0 10 20 30 40 50 60 0.2

(m/s) 0

2 z -0.2 0 10 20 30 40 50 60 500

(deg) 0

3 z -500 0 10 20 30 40 50 60 1000

0 (deg/s)

z -1000 0 10 20 30 40 50 60 Time (s)

Figure 6.22: The errors of the dynamics during the simulation (Case 2).

198 1.55

1.5 (m/s) u 1.45 0 10 20 30 40 50 60 0.1

0 (m/s) v -0.1 0 10 20 30 40 50 60 10

0 (deg/s)

r -10 0 10 20 30 40 50 60 100

0 (deg/s)

p -100 0 10 20 30 40 50 60 Time (s)

Figure 6.23: Simulation of linear and angular velocities of the vehicle (Case 3).

0 (deg)

-50 0 10 20 30 40 50 60 20

0 (deg)

-20 0 10 20 30 40 50 60 10

(cm) 0

v y -10 0 10 20 30 40 50 60 10 (N)

-10 Jet pump 0 10 20 30 40 50 60 Time (s)

Figure 6.24: Simulation of roll and yaw angle of the vehicle, the position and local velocity of the internal moving mass, and the force exerted by jet pumps (Case 3).

199 20

0 (deg/s)

z -20 0 10 20 30 40 50 60 0.1

(m/s) 0

2 z -0.1 0 10 20 30 40 50 60 500

(deg) 0

3 z -500 0 10 20 30 40 50 60 1000

0 (deg/s)

z -1000 0 10 20 30 40 50 60 Time (s)

Figure 6.25: The errors of the dynamics during the simulation (Case 3).

Simulation Results in a Steady and Uniform Flow

Consider a REMUS AUV is deployed into a steady and uniform current ﬁeld, where the current velocity is perpendicular to the desired heading direction of the vehicle. The disturbance is introduced by the initial relative velocity in sway. The dynamics of the vehicle is described by (4.3.2), therefore the linear velocities in both (6.3.49) and (6.3.64) should be replaced with the relative velocities. Assume the current velocity is aligned with the positive direction of the y axis in the earth-

ﬁxed reference frame, and its magnitude is Uc = 0.2 m/s. The initial heading angle

of the vehicle is ψ0 = 0. Accordingly, the initial relative velocities are ur(0) = 0 and

vr(0) = −0.2. The same control parameters are used in the simulation as in the previous ones. The results are illustrated in the following three ﬁgures. Although the initial disturbance is relatively large, the actuators are still able to stabilize the heading

200 of the vehicle. As discussed in section 5.2.2, the relative velocities will converge to the desired equilibrium point, where the relative sway and yaw rate are both zero. Since the steady-state heading angle is zero, the sway velocity of the vehicle

is v = Uc = 0.2 m/s. As displayed in Figure 6.28, the AUV is subject to a sideslip angle β. Although the nose of the vehicle is pointing to the desired direction, the ocean current drifts the vehicle oﬀ course. In order to align the velocity with the course, the heading can be adjusted into the ocean current to compensate for its eﬀect [85], if the current velocity is measurable. In our case, the desired heading angle is given by: Uc ◦ ψd = −β = − arcsin ≈ −7.55 (6.3.74) U0 where U0 is the designed speed of the vehicle in calm waters. Since the new objec-

tive is little more challenging, the control parameters are changed to k0 = 0.3, k1 =

0.8, k2 = k3 = k4 = 1.0.

1.6

1.5 (m/s) u 1.4 0 10 20 30 40 50 60 0.4

0.2 (m/s) v 0 0 10 20 30 40 50 60 0.5

0 (rad/s)

r -0.5 0 10 20 30 40 50 60 2

0 (rad/s)

p -2 0 10 20 30 40 50 60 Time (s)

Figure 6.26: The linear and angular velocities of the vehicle in ocean currents when ψd = 0.

201 50

0 (deg)

-50 0 10 20 30 40 50 60 50

0 (deg)

-50 0 10 20 30 40 50 60 10

(cm) 0

v y -10 0 10 20 30 40 50 60 10 (N)

-10 Jet pump 0 10 20 30 40 50 60 Time (s)

Figure 6.27: The roll and yaw angles, and the actuations of the vehicle in ocean currents when ψd = 0.

20 x (m) 15

v = 0.2 m/s 10 c

-5 -20 -10 0 10 20 30 y (m)

Figure 6.28: The trajectory of the vehicle in ocean currents when ψd = 0.

202 1.6

1.4 (m/s) u 1.2 0 10 20 30 40 50 60 0.4

0.2 (m/s) v 0 0 10 20 30 40 50 60 0.5

0 (rad/s)

r -0.5 0 10 20 30 40 50 60 2

0 (rad/s)

p -2 0 10 20 30 40 50 60 Time (s)

Figure 6.29: The linear and angular velocities of the vehicle in ocean currents when ψd = −β.

0 (deg)

-50 0 10 20 30 40 50 60 50

0 (deg)

-50 0 10 20 30 40 50 60 10

(cm) 0

v y -10 0 10 20 30 40 50 60 10 (N)

-10 Jet pump 0 10 20 30 40 50 60 Time (s)

Figure 6.30: The roll and yaw angles, and the actuations of the vehicle in ocean currents when ψd = −β. 203 45

20 x (m) 15

v = 0.2 m/s 10 c

-5 -20 -10 0 10 20 30 y (m)

Figure 6.31: The trajectory of the vehicle in ocean currents when ψd = −β.

The initial heading error and large velocity disturbance make it more dif- ﬁcult for the actuators to stabilize the vehicle, as shown in Figure 6.29 and 6.30. After about 20 seconds, the system reaches the steady state, and the vehicle velocity points to the north as expected. Figure 6.31 shows the constant sideslip angle, and the trajectory of the vehicle in the horizontal plane.

Simulation Results of a Vehicle at Low Speeds

A major advantage that internal actuators have over control surfaces is that they can work independent of relative flows. When underwater vehicles operate at low speeds, their control surfaces are much less effective due to the fact that the hydrodynamic force depends on the square of the relative velocity. As a result, internal actuators and extra thrusters can be used to improve the maneuverability of underwater vehicles at low speeds. If the main thrust acting on the REMUS 204 AUV is 4 N instead of 9 N, its forward speed becomes approximately 1 m/s (see Figure 4.3). Accordingly, the lift forces on the fins are less than half of the ones at the normal speed. When left uncontrolled with T = 4, the equilibrium points of the dynamical system are given by

T x0 =[1.017, 0, 0, 0, 0]

0 T x0 =[0.8467, 0.1147, −0.1541, 0, 0]

00 T x0 =[0.8467, −0.1147, 0.1541, 0, 0]

0 00 in which x0 is a unstable equilibrium point, while both x0 and x0 are stable. Then same control strategy as in the previous subsection is adopted to con-

trol the vehicle at a low speed. The parameters are k0 = 0.3, k1 = 0.8, k2 =

k3 = k4 = 1.0. In the ﬁrst simulation, the initial state of the system is x0 = [1.0, −0.1, 0, 0, 0, 0]T . The results are illustrated in Figure 6.32 and 6.33. The internal moving mass together with a tunnel thruster can successfully stabilize the heading angle of the vehicle at a low speed.

205 1.02

1 (m/s) u 0.98 0 10 20 30 40 50 60 0.1

0 (m/s) v -0.1 0 10 20 30 40 50 60 0.2

0 (rad/s)

r -0.2 0 10 20 30 40 50 60 1

0 (rad/s)

p -1 0 10 20 30 40 50 60 Time (s)

Figure 6.32: The linear and angular velocities of the controlled vehicle advancing at a low speed.

0 (deg)

-50 0 10 20 30 40 50 60 20

0 (deg)

-20 0 10 20 30 40 50 60 10

(cm) 0

v y -10 0 10 20 30 40 50 60 10 (N)

-10 Jet pump 0 10 20 30 40 50 60 Time (s)

Figure 6.33: The roll and yaw angles, and the actuations of the vehicle advancing at a low speed. 206 When operating in a steady and uniform ocean current ﬁeld, a low-speed underwater vehicle can still be stabilized by the actuators, as shown in Figure 6.34 and 6.35. The desired heading direction is north, but the vehicle is driven oﬀ course because of the ocean current. Since the forward speed is low, more deviation along y direction is expected if the vehicle travels a equal distance to north, compared with the case in Figure 6.28. In order to control the vehicle to move north, the setpoint of the heading angle should be Uc ◦ ψd = −β = − arcsin ≈ −11.3 (6.3.75) U0 where β is the sideslip angle. The simulation results of the velocities are displayed

in Figure 6.37. Figure 6.38 shows that the heading angle ψ converges to ψd in about 20 seconds. Then the vehicle is able to stay on the path along the x axis of the earth-ﬁxed reference frame, as illustrated in Figure 6.39. The controller for heading autopilot is not able to steer the vehicle back to the desired path. A path-following or trajectory-tracking controller needs to be designed accordingly, depending on the control objective.

207 1.05

1 (m/s) u 0.95 0 10 20 30 40 50 60 0.4

0.2 (m/s) v 0 0 10 20 30 40 50 60 0.2

0 (rad/s) r -0.2 0 10 20 30 40 50 60 2

0 (rad/s)

p -2 0 10 20 30 40 50 60 Time (s)

Figure 6.34: The linear and angular velocities of the controlled vehicle at a low speed in ocean currents.

0 (deg)

-50 0 10 20 30 40 50 60 50

0 (deg)

-50 0 10 20 30 40 50 60 10

(cm) 0

v y -10 0 10 20 30 40 50 60 10 (N)

-10 Jet pump 0 10 20 30 40 50 60 Time (s)

Figure 6.35: The roll and yaw angles, and the actuations of the vehicle at a low speed in ocean currents. 208 40

x (m) 15

10 v = 0.2 m/s c 5

-5 -20 -10 0 10 20 30 y (m)

Figure 6.36: The trajectory of the controlled vehicle advancing at a low speed in ocean currents.

1.2

1 (m/s) u 0.8 0 10 20 30 40 50 60 0.4

0.2 (m/s) v 0 0 10 20 30 40 50 60 0.2

0 (rad/s)

r -0.2 0 10 20 30 40 50 60 1

0 (rad/s)

p -1 0 10 20 30 40 50 60 Time (s)

Figure 6.37: The linear and angular velocities of the controlled vehicle at a low speed in ocean currents when ψd = −β. 209 50

0 (deg)

-50 0 10 20 30 40 50 60 20

0 (deg)

-20 0 10 20 30 40 50 60 10

(cm) 0

v y -10 0 10 20 30 40 50 60 10 (N)

-10 Jet pump 0 10 20 30 40 50 60 Time (s)

Figure 6.38: The roll and yaw angles, and the actuations of the vehicle at a low speed in ocean currents when ψd = −β.

x (m) 15

10 v = 0.2 m/s c 5

-5 -25 -20 -15 -10 -5 0 5 10 15 20 25 y (m)

Figure 6.39: The trajectory of the controlled vehicle at a low speed in ocean currents when ψd = −β. 210 Ocean Current Observer

Both the linear and nonlinear feedback laws are obtained under the assumption that the full state of the system is measured. However, only the angular velocities and orientations of the AUVs are available in most applications. Accordingly, state observers must be designed to estimate the unmeasured states. For instance,

the state feedback law (6.3.64) requires the unmeasured relative sway velocity vr.

A nonlinear observer can be designed based on the vehicle dynamics to estimate vr. Moreover, when ocean currents exist, the estimation of current velocities is necessary if we need to compensate for their effects on the vehicle, as shown in the above simulation results. Onboard sensors such as Doppler Velocity Log (DVL) can measure vehicle speed, but the distance between the vehicle and the seafloor has to be within a certain range. Low-cost pressure sensor array [86, 87] can be useful in estimating current velocity, if the angle of attack of the relative flow is identified based on the measured pressure distribution along the vehicle hull. Another strategy to estimate current velocity is to design a current observer based on the vehicle kinematics or dynamics. [88] derived a simple current observer based on the kinematic equations. However, measurements or estimations of vehicle position and relative velocity must be available. [89] proposed a pair of coworking nonlinear Luenberger observers to estimate the vehicle and current velocities, which also need the measurements of the position and orientation of the vehicle. In the proposed design, the vehicle model is augmented by a three DOF current induced vessel model, which is the basis for obtaining an estimate of the current velocity. The current observer is designed based on the fact that the relative linear velocity converges to the steady

T state [U0, 0, 0] if the stabilized vehicle travels along linear paths. Therefore, the proposed observer may fail to generate reasonable estimation when the vehicle is following circular routes.

211 6.3.4 Coupling with the Vertical-Plane Motion

The roll motion, which is deliberately generated by an internal moving mass, could be problematic for the vertical-plane motion of the underwater vehicle. Under

the assumptions that the products of inertia Ixy,Ixz and Iyz are zero, pitch and yaw motion are entirely decoupled, if there is no roll motion [23]. For nonzero roll motion φ 6= 0 and p 6= 0, their coupling is noticeable if we look at the kinematic relation given by (2.2.15). Assuming a small pitch angle θ, we have       θ˙ cos φ − sin φ q   =     (6.3.76) ψ˙ sin φ cos φ r

As for the vehicle dynamics, a nonzero yg and roll motion enable many cou-

pling terms in M RB and CRB(ω). However, since yg is really small, its direct eﬀect

on those matrices will be neglected. Therefore, M RB and CRB(ω) are assumed

to be constant, regardless of a varying yg. The y˙v, y¨v terms in (6.2.30) are not considered either in the preceding subsection, because their effects are insignificant compared with the tunnel thrust. Since the coupling effects are not considered when we design the autopilot controller, the burden falls on us to show whether the nonzero roll motion will significantly degrade the performances of the horizontal-plane and vertical-plane controllers due to the coupling effects. We use a simple depth-plane controller introduced by [55], which is based on the linearized vehicle equations of motion:  (m + a )w ˙ − a q˙ − Z w − (mU + Z )q = Z δ  33 35 w 0 q δs s     (Iyy + a55)q ˙ − a35w˙ − Mww − Mqq − Mθθ = Mδs δs (6.3.77)   z˙ = w − U0θ    θ˙ = q where δs is the angle of the horizontal fins. The depth-plane controller consists of an inner proportional and derivative (PD) pitch loop, and an outer proportional 212 depth loop. The inner pitch loop responds sufficiently fast compared with the outer depth loop. The parameters in the above equations are given below in Table 6.1.

Table 6.1: Linearized Coeﬃcients of the Depth-Plane Motion

Parameter Value Units

2 Iyy +3.45e+000 kg · m

a33 +3.55e+001 kg

a35 +1.93e+000 kg · m

2 a55 +4.88e+000 kg · m

Zw −6.66e+001 kg/s

Zq −9.67e+000 kg · m/s

Mw +3.07e+001 kg · m/s

2 Mq −6.87e+000 kg · m /s

2 2 Mθ −5.77e+000 kg · m /s

2 Zδs −5.06e+001 kg · m/s

2 2 Mδs −3.46e+001 kg · m /rad

The depth-plane control system can be expressed by the following block diagram.

z e θ e δ d z γ d θ −K (τ s + 1) s − − p d Plant z θ Feedback Elements

Figure 6.40: Vertical-Plane Control System Block Diagram

The control parameters designed by [55] for U0 = 1.525 m/s are given blow in Table 6.2.

213 Table 6.2: Control Parameters of the Vertical-Plane Motion Parameter Value

γ −0.7720

Kp +10.345

τd +0.2100

A 6-DOF simulation is conducted using a C++ program developed within the Robotic Operating System (ROS) framework. The equations of motion of the REMUS AUV are written as:

M RBν˙ r + CRB(νr)νr + M Aν˙ r + CA(νr)νr + D(νr)νr + g(η) = τ (6.3.78) where the control actuation is given by:   T      F   t     Z δ   δs s  τ =   (6.3.79)   mvgyv cos(φ) − Ftzt      M δ   δs s    0 where T = 9 N denotes the main thrust, and Ft, zt represent the thrust and position of the tunnel thruster, respectively. Note that there is no direct actuation in yaw direction. The horizontal-plane motion is controlled by the internal moving mass and tunnel thruster. The parameters in the matrices are given by Tables 5.1, 5.4, 6.1 and 6.3.

214 Table 6.3: Hydrodynamic Coeﬃcients of the REMUS AUV

Parameter Value Units

Zw|w| −1.00e+002 kg/m

2 Zq|q| −6.32e−001 kg · m/rad

Mw|w| −7.38e+000 kg

2 2 Mq|q| −1.88e+002 kg · m /rad

Zuw −1.79e+001 kg/m

Zuq −8.62e+000 kg/rad

Muw −4.57e+000 kg

Muq +6.40e+000 kg · m/rad

In the simulation, the setpoints are ψd = 0, zd = 1, and there are no ocean currents. The results are illustrated in Figure 6.41- 6.44.

1.5

1 (m/s) u 0.5 0 10 20 30 40 50 60 0.2

0 (m/s) v -0.2 0 10 20 30 40 50 60 0.2

0 (rad/s)

r -0.2 0 10 20 30 40 50 60 5

0 (rad/s)

p -5 0 10 20 30 40 50 60 Time (s)

Figure 6.41: The linear and angular velocities of the horizontal- plane motion.

215 50

0 (deg)

φ -50 0 10 20 30 40 50 60 20

0 (deg)

ψ -20 0 10 20 30 40 50 60 10

(cm) 0

v y -10 0 10 20 30 40 50 60 10 (N) 0

-10 0 10 20 30 40 50 60 Jet pump Time (s)

Figure 6.42: The roll and yaw angles, and the actuations of the horizontal-plane motion. The actuators chatter because of the coupling between the two planes.

216 2 (m) 0

depth -2 0 10 20 30 40 50 60 0.5

(m/s) 0 w -0.5 0 10 20 30 40 50 60 50

(deg) 0

-50 0 10 20 30 40 50 60 20

0 (deg)

s δ -20 0 10 20 30 40 50 60 Time (s)

Figure 6.43: The vertical-plane motion and the rudder angle.

60 -5 50

40 0 30

20 down (m) 5 10 -5 north (m) 0 5 east (m)

Figure 6.44: The trajectory of the REMUS vehicle in the earth- ﬁxed frame.

217 Figure 6.43 shows the vertical-plane motion of the REMUS vehicle controlled by the PD controller. Since there is no integral term in the controller, both the depth z and pitch angle θ have steady-state errors due to the positive-buoyancy property of the vehicle. After convergence, we have w ≈ −0.07 m/s and θ ≈

−0.045 rad. Moreover, the average value of δs is about −5 deg, meaning that the horizontal fins generate downward lift forces to offset the positive buoyancy. The horizontal-plane motion is affected dramatically by the coupling effects. Unlike the cases shown in Figure 6.17 and 6.18, the vehicle keeps rolling with the sway actuation oscillating around its steady state. The persistent oscillation is problematic in applications, especially for the actuators and sensors onboard. Notice that the heave velocity w is nonzero at steady state, which makes the coupling term

(m + a33)pw non-negligible in sway direction. This term keeps disturbing the sway velocity v, if it is not explicitly considered in the horizontal-plane control design. The phenomenon can be illustrated by the following ﬁgure.

40 (m+a )pw 33 20 thruster (N)

-20 Sway force

-40 0 10 20 30 40 50 60 Time (s)

Figure 6.45: The comparison between the sway thrust and the non-inertial force.

When t > 20 s, the non-inertial force (m + a33)pw is in the same phase as the tunnel thrust, which generates a much larger restoring force, and causes the system to oscillate. In order to mitigate the coupling eﬀect, we need to reduce p or

w. Since w = U0θ after reaching the steady state, and θ =6 0 due to the positive buoyancy, our only option is to decrease p. However, the roll motion is deliberately 218 induced by the internal moving mass. In order to reduce p, we have to disable the internal moving mass when appropriate. As a matter of fact, when the roll angle φ is small, the sway actuation is mainly contributed by the jet pump or tunnel

thruster. Therefore, as long as Ft is large enough to counteract the negative eﬀect of a disabled internal moving mass, the tunnel thruster can stabilize the vehicle by itself. The strategy can be validated by the following simulation, in which the internal moving mass will be locked in place when t > 15 s. Figure 6.46 and 6.47 depicts the velocities of the vehicle’s horizontal-plane motion. Note that the initial

sway velocity is v0 = 0.1 m/s, and that yv = 0 when t > 15 s. Although the internal moving mass is disabled, the tunnel thruster still manages to stabilize the heading angle of the vehicle. Since there is no actuation in roll, p and φ decay gradually to

zero. Consequently, the coupling term (m + a33)pw is no longer signiﬁcant enough to oscillate the system, and both the horizontal-plane and depth-plane controllers work reasonable well.

219 1.5

1 (m/s) u 0.5 0 10 20 30 40 50 60 0.2

0 (m/s) v -0.2 0 10 20 30 40 50 60 0.2

0 (rad/s)

r -0.2 0 10 20 30 40 50 60 5

0 (rad/s)

p -5 0 10 20 30 40 50 60 Time (s)

Figure 6.46: The linear and angular velocities of the horizontal- plane motion.

0 (deg)

-50 0 10 20 30 40 50 60 50

0 (deg)

-50 0 10 20 30 40 50 60 10

(cm) 0

v y -10 0 10 20 30 40 50 60 10 (N)

-10 Jet pump 0 10 20 30 40 50 60 Time (s)

Figure 6.47: The roll and yaw angles, and the actuations of the horizontal-plane motion. Note that the internal moving mass is locked at yv = 0, when t > 15 s. However, the jet pump still works.

220 2 (m) 0

depth -2 0 10 20 30 40 50 60 0.5

(m/s) 0 w -0.5 0 10 20 30 40 50 60 50

0 (deg)

-50 0 10 20 30 40 50 60 20

(deg) 0

s δ -20 0 10 20 30 40 50 60 Time (s)

Figure 6.48: The vertical-plane motion and the rudder angle.

60 -5 50

0 40 30 5

down (m) 20 -5 0 10 north (m) 5 east (m)

Figure 6.49: The trajectory of the REMUS vehicle in the earth- ﬁxed frame.

221 CHAPTER 7 CONCLUSIONS

The primary theme for this work is the comprehensive development of a nonlinear dynamic model of an underwater vehicle, and a heading control system with internal actuators. The purpose of developing internal actuators is to replace the conventional control surfaces and improve the maneuverabilities of underwater vehicles at low speeds or in environments featuring strong ocean currents. Internal moving mass has already been applied to autonomous underwater vehicles and underwater gliders. However, in most applications, the moving mass is only used to stabilize the diving motions. In our work, its application is extended to the stabilization of an AUV’s horizontal-plane motion. This work starts from the development of the dynamic model from first principles. Both Newton-Euler and Lagrangian formulations are used to describe the hydrodynamic loads acting on the vehicle in an ideal fluid. The contributions of the incoming flow’s non-uniformity are revealed explicitly. The viscous effects of a real fluid are studied using CFD methods. We use ANSYS Fluent to compute the hydrodynamic loads on a REMUS 100 AUV, and compare the results with the values predicted in the literature. A nonlinear dynamical system is obtained which describes the horizontal- plane motion of the REMUS AUV in a real fluid. The nonlinear behavior of the dynamical system is illustrated in the phase portrait, which shows clearly that the horizontal-plane motion is unstable. Further analysis of the effects of a nonzero roll torque and the lateral position of an internal moving mass on the horizontal-plane motion demonstrates how the dynamical flow can be shaped by these two factors.

222 We take advantage of the dynamic coupling between the internal moving mass and the horizontal-plane motion of the vehicle to design a heading control system. A LQR controller is first designed to demonstrate the effectiveness of the strategy. However, the linear controller cannot deal with large perturbations. We introduce a tunnel thruster to the control design and derive nonlinear full-state feedback laws using backstepping and Lyapunov redesign techniques. It is shown in the simulation results that the horizontal-plane motion can be successfully stabilized even under large initial perturbations and constraints on the movement speed of the internal moving mass. The internal actuators can also control the vehicle operating at low speeds and in ocean currents, where they have major advantages over conventional control fins. Recommendations for future work include experimental model validation of the control strategy and development of an observer for an output feedback control strategy. Similar to internal moving mass, a nonzero roll torque may also be deliberately generated to stabilize the horizontal-plane motion of an underwater vehicle. The tailcones of bluefin AUVs are specifically designed to produce a roll torque to balance the propeller-generated torque [22]. The idea of generating desired roll torques by adjusting the stators’ orientation is worth further investigations.

223 APPENDICES

224 APPENDIX A DERIVATION OF BODY-FLUID DYNAMICAL SYSTEM BASED ON NEWTON-EULER FORMULATION

In this section, we adopt Newton-Euler formulation to derive the hydrodynamic forces on a rigid body. The ideal ﬂuid is assumed to be irrotational, so that the velocity ﬁeld can be described by a velocity potential φ as

v = ∇φ which satisﬁes ∇2φ = 0 because of the conservation of mass.

A.1 MATHEMATICAL PRELIMINARIES

A.1.1 Uniqueness Theorem

The uniqueness theorem can be expressed that a vector is uniquely speci- ﬁed by giving its divergence and its curl within a simply connected region and its normal component over the boundary.

A.1.2 Helmholtz’s Theorem

Assume a vector u satisﬁes the following conditions

∇ · u = ϑ (A.1)

∇ × u = ω (A.2)

In ﬂuid mechanics, if u is the velocity, then ϑ has the meaning of local rate of ex- pansion, while ω is the local vorticity. Then Helmhotz’s theorem indicates under certain conditions, u may be written as the sum of two parts, on of which is irrotational, the other of which is solenoidal [90].

u = ∇φ + ∇ × ψ (A.3) 225 where φ and ψ are the scalar and vector potentials of vector u. Clearly, φ and ψ must satisfy the following conditions,

∇2φ = ϑ

∇ × (∇ × ψ) = ∇(∇ · ψ) − ∇2ψ = ω

If ∇ · ψ = 0, we obtain the following conditions

∇2φ = ϑ (A.4)

∇2ψ = −ω (A.5)

then, we can construct that

1 ZZZ ϑ(r0) φ(r) = − dτ(r0) (A.6) 4π |r − r0| V 1 ZZZ ω(r0) ψ(r) = dτ(r0) (A.7) 4π |r − r0| V where r denotes the position of the vector u and r0 is the source point. The integrals are operated inside the volume. The next step is to prove that the construc- tions satisfy the conditions. Since the Laplacian operator ∇2 only operates on r, we have

1 ZZZ ϑ(r0) 1 ZZZ 1 ∇2φ(r) = − ∇2 dτ(r0) = − ϑ(r0)∇2 dτ(r0) 4π |r − r0| 4π |r − r0| V V According to Gauss’ law,  ZZZ −4π if r ∈ V 2 1 0  ∇ 0 dτ(r ) = |r − r |  V 0 if r 6∈ V which can also be conveniently expressed by Dirac delta function as

1 ∇2 = −4πδ(r − r0) = −4πδ(r0 − r) (A.8) |r − r0|

226 Then 1 ZZZ 1 ∇2φ(r) = − ϑ(r0)∇2 dτ(r0) 4π |r − r0| V ZZZ = ϑ(r0)δ(r0 − r)dτ(r0)

V = ϑ(r)

As for ψ(r), ∇2ψ = −ω is also valid as soon as we can prove that

∇ · ψ = 0

Using the facts that 1 1 ∇ = −∇0 |r − r0| |r − r0| where ∇0 means a divergence operation only applied to source point, and

∇ · ω(r0) = ∇ · (∇ × u) = 0 we ﬁnd that 1 ZZZ 1 ∇ · ψ(r) = ω(r0) · ∇ dτ(r0) 4π |r − r0| V 1 ZZZ 1 = − ω(r0) · ∇0 dτ(r0) 4π |r − r0| V 1 ZZZ ω(r0) = − ∇0 · dτ(r0) 4π |r − r0| V 1 ZZ ω(r0) · n0 = − dS(r0) 4π |r − r0| S The surface integral being taken over the entire boundary of the volume will vanish when ω has zero normal component at each point of the boundary.

Furthermore, another potential function φo can be separated from φ with

2 ∇ φo = 0. Eventually, u has a triple decomposition as follow [40],

u = ue + uv + v (A.9)

227 which have the following properties

∇ · ue = ϑ, ∇ × ue = 0

∇ · uv = 0, ∇ × uv = ω (A.10)

∇ · v = 0, ∇ × v = 0

ue and uv can be determined using (A.6) and (A.7). v satisﬁes the harmonic func-

2 tion ∇ φo = 0, which can also be solved by using boundary conditions. ∂φ o = (u − u − u ) · n on S ∂n e v

Then using Green’s identity, we can obtain the integral expression of φo as 1 ZZ ∂ 1 1 ∂φ φ = − φ − o dS(r0) (A.11) o 4π o ∂n |r − r0| |r − r0| ∂n S Hence, the velocity potential function φ can be expressed as 1 ZZZ ϑ(r0) 1 ZZ ∂ 1 1 ∂φ φ(r) = − dτ(r0)− φ − o dS(r0) 4π |r − r0| 4π o ∂n |r − r0| |r − r0| ∂n V S (A.12) Due to uniqueness theorem, the above decomposition is unique.

A.1.3 The Behavior of Velocity Potential at Inﬁnity

Due to (A.11), the velocity potential at point P inside the ﬂuid domain is written as 1 ZZ ∂ 1 1 ∂φ φ = − φ − dS P 4π ∂n r r ∂n S

Write the surface integral into two separate parts, on the rigid body surface SB and

on the surface SC of a sphere with center at the point P in the fluid and sufficiently large radius R. 1 ZZ ∂ 1 1 ∂φ 1 ZZ ∂ 1 1 ∂φ φ = − φ − dS − φ − dS P 4π ∂n r r ∂n 4π ∂n R R ∂n SB SC 1 ZZ ∂ 1 1 ∂φ 1 ZZ 1 ZZ ∂φ = − φ − dS + φ dS + dS 4π ∂n r r ∂n 4πR2 4πR ∂n SB SC SC (A.13) 228 Since ZZ ∂φ ZZ ∂φ ZZ ∂φ ZZZ dS + dS = dS = ∇2φ dV = 0 (A.14) ∂n ∂n ∂n SC SB S V On the surface of rigid body, the normal component of fluid velocity at each point is known as ∂φ = (u + ω × r0) · n (A.15) ∂n where u and ω are the linear and angular velocity of the rigid body respectively and r0 is the position vector of the surface in the body-fixed reference frame. Hence,

ZZ ∂φ ZZ dS = (u + ω × r0) · n dS ∂n SB SB 0 (A.16) ZZ > ZZ 0 = u · n dS + ω · n × r dS SB SB In order to proceed, a corollary of Gauss’ theorem has to be obtained. If B = C × A, and C is a constant vector, Gauss’ theorem is expressed as, ZZZ ZZ ∇ · B dV = B · n dS (A.17) V S Since ZZZ ZZZ ∇ · B dV = ∇ · (C × A) dV V V ZZZ = −C ∇ × A dV V ZZ ZZ B · n dS = (C × A) · n dS

S S ZZ = −C n × A dS

As a result, ZZZ ZZ ∇ × A dV = n × A dS (A.18) V S 229 Using (A.18), we have

ZZ ∂φ ZZ dS = ω · n × r0dS ∂n SB SB ZZZ = −ω · ∇ × r0dS

VB = 0

Therefore, ZZ ∂φ ZZ ∂φ dS = − dS = 0 ∂n ∂n SC SB and 1 ZZ ∂ 1 1 ∂φ 1 ZZ φ = − φ − dS + φ dS P 4π ∂n r r ∂n 4πR2 SB SC We can write the last term as

1 ZZ φ dS = φ¯(P,R) (A.19) 4πR2 SC

representing the mean value of φ over the spherical surface SC , of radius R and centered at point P , Then

1 ZZ ∂ 1 1 ∂φ φ = φ¯(P,R) − φ − dS (A.20) P 4π ∂n r r ∂n SB It is important to know more about φ¯,

1 ZZ 1 ZZ dS 1 ZZ φ¯(P,R) = φ dS = φ = φ dΩ (A.21) 4πR2 4π R2 4π SC SC SC where dΩ is the solid angle subtended at P by dS. Taking the derivative of φ¯ with respect to R, we have

∂φ¯(P,R) 1 ZZ ∂φ 1 ZZ ∂φ 1 ZZ ∂φ = dΩ = dΩ = dS = 0 (A.22) ∂R 4π ∂R 4π ∂n 4πR2 ∂n SC SC SC Therefore, we can say φ¯ is independent of R. Then

φ¯(P,R) = C (A.23) 230 where C is a constant. It is also valid that C is independent of the position of point P , assuming the velocity v vanishes at large R. Eventually, we ﬁnd

1 ZZ ∂ 1 1 ∂φ φ = C − φ − dS (A.24) P 4π ∂n r r ∂n SB Then it can be seen that

φP → C as r → ∞ (A.25) r in the above equation denotes the distance between point P and the point on the rigid body surface, 1 1 1 = = 0 p 0 2 r |r − r | (xi − xi) Write r−1 as Taylor series,

2 1 1 0 ∂ 1 1 0 0 ∂ 1 = − xi + xixj + ··· , (A.26) r |r| ∂xi |r| 2! ∂xi∂xj |r|

The Taylor series can be substituted into (A.24), then we ﬁnd [40]

∂ 1 ∂2 1 φP (r) = C + ci + cij + ··· , (A.27) ∂xi |r| ∂xi∂xj |r| where  1 ZZ ∂φ   ci = −xi + niφ dS  4π ∂n  SB   1 ZZ 1 ∂φ cij = xixj − xinjφ dS (A.28)  4π 2! ∂n  SB    ··· From (A.27), we can draw the conclusion that at large distance from the rigid body, the ﬂuid velocity v = ∇φ is of the order of |r|−3.

231 A.2 HYDRODYNAMIC LOADS IN A STATIONARY FLOW

Assuming the density of the fluid ρ is constant over the fluid control volume, the linear momentum of the fluid can be written as follows: ZZZ p = ρ v dV V ZZZ (A.29) = ρ ∇φ dV V Using Gauss’ theorem, ZZZ ZZ ∇fdV = fn dS (A.30) V S The linear momentum can also be expressed as an integral over the surfaces of fluid control volume, ZZ p = ρ φ n dS (A.31)

S The time derivative of the linear momentum is dp d ZZ = ρ φn dS dt dt S d ZZZ = ρ ∇φ dV dt V (A.32) ZZZ ∂φ ZZ = ρ ∇ dV + ρ ∇φ(b · n) dS ∂t V S ZZ ∂φ = ρ n + ∇φ(b · n) dS ∂t S

The interior control surface of the ﬂuid SB is the surface of rigid body, and the

exterior control surface SC is assumed to be ﬁxed. Thus on SC

b · n = 0

On the surface of rigid body SB,

∂φ b · n = ∂n 232 Since SC is ﬁxed, we can obtain that

ZZ ∂φ d ZZ n dS = φn dS ∂t dt SC SC Hence,

d ZZ ZZ ∂φ ρ φndS = ρ n + ∇φ(b · n) dS dt ∂t SB +SC S ZZ ∂φ ZZ ∂φ = ρ n + ∇φ(b · n) dS + ρ n + ∇φ(b · n) dS ∂t ∂t SB SC ZZ ∂φ ∂φ d ZZ = ρ n + ∇φ dS + ρ φ n dS ∂t ∂n dt SB SC

Consequently, d ZZ ZZ ∂φ ∂φ ρ φ n dS = ρ n + ∇φ dS (A.33) dt ∂t ∂n SB SB The hydrodynamic force on the rigid body can be expressed as the integrals of the

pressure over the body surface SB, thus ZZ F = p n dS (A.34)

SB Using Bernoulli’s equation (the hydrostatic contribution will be treated separately), we obtain ZZ ∂φ 1 F = −ρ + ∇φ · ∇φ n dS (A.35) ∂t 2 SB Combining (A.33) and (A.35) gives

d ZZ ZZ ∂φ 1 F = −ρ φ n dS + ρ ∇φ − ∇φ · ∇φ n dS (A.36) dt ∂n 2 SB SB Since

ZZ ∂φ 1 ZZ 1 ∇φ − ∇φ · ∇φ n dS = ∇φ(∇φ · n) − ∇φ · ∇φ n dS ∂n 2 2 SB +SC SB +SC 233 using the facts that ∇φ(∇φ · n) = (∇φ ∇φ) · n

and Gauss’ theorem, we have

ZZ ∂φ 1 ZZZ 1 ∇φ − ∇φ · ∇φ n dS = ∇ · (∇φ∇φ) − ∇(∇φ · ∇φ) dV ∂n 2 2 SB +SC V From vector analysis,

∇(a · b) = ∇a(a · b) + ∇b(a · b)

= b × (∇ × a) + (b · ∇)a + a × (∇ × b) + (a · ∇)b (A.37)

∇ · (ab) = b(∇ · a) + (a · ∇)b we obtain that 1 ∇ · (∇φ∇φ) − ∇(∇φ · ∇φ) 2 0 0 2 ¨* 1 : =∇φ(¨∇¨φ) + (∇φ · ∇)∇φ − [∇φ × (∇ × ∇φ) + (∇φ · ∇)∇φ 2 : 0 (A.38) + ∇φ ×(∇ × ∇φ) + (∇φ · ∇)∇φ]

=(∇φ · ∇)∇φ − (∇φ · ∇)∇φ

Hence, ZZ ∂φ 1 ∇φ − ∇φ · ∇φ n dS = 0 (A.39) ∂n 2 SB +SC As a result,

d ZZ ZZ ∂φ 1 F = −ρ φ n dS − ρ ∇φ − ∇φ · ∇φ n dS (A.40) dt ∂n 2 SB SC It is also convenient to derive (A.40) using the equation of conservation of linear momentum,

d ZZZ ZZ ∂φ ρ ∇φ dV + ρ ∇φ( − b · n) dS = −F − F C (A.41) dt ∂n V S 234 where F C is the pressure force on the control surface SC . Separate the second term

into the integrals on SB and SC , and use the boundary condition, we obtain that d ZZZ ZZ ∂φ F = −ρ ∇φ dV − ρ ∇φ dS − F C dt ∂n V SC d ZZ d ZZ ZZ ∂φ = −ρ φ n dS − ρ φ n dS − ρ ∇φ dS − F dt dt ∂n C SB SC SC d ZZ ZZ ∂φ ZZ ∂φ = −ρ φ n dS − ρ n dS − ρ ∇φ dS − F dt ∂t ∂n C SB SC SC

Identical to (A.35), F C can be written as ZZ ∂φ 1 F = − + ∇φ · ∇φ n dS C ∂t 2 SC Inserting it into the equation, we can obtain the same expression of F as (A.40). The angular momentum of the ﬂuid with respect to the origin of the earth- ﬁxed reference frame is ZZZ π = ρ r × v dV V ZZZ (A.42) = ρ r × ∇φ dV V Since ∇ × (φr) = ∇φ × r + φ(∇ × r) = ∇φ × r

Using (A.18), we have ZZZ π = ρ r × ∇φ dV V ZZZ = −ρ ∇ × (φr) dV V ZZ (A.43) = −ρ n × (φ r) dS

S ZZ = ρ φ(r × n) dS

S Using the equation of conservation of angular momentum, d ZZ ZZ ∂φ ρ φ(r × n) dS + ρ (r × ∇φ)( − b · n) dS = −M − M (A.44) dt ∂n C S S 235 Similar to the derivation of F , we can obtain the expression of M as:

d ZZ ZZ ∂φ 1 M = −ρ φ(r × n) dS − ρ r × ∇φ − ∇φ · ∇φ n dS (A.45) dt ∂n 2 SB SC

Following the approach given in [41], if the control surface SC is in the far ﬁeld, ∇φ is in the order of |r|−3, hence the integrals on it are assumed to vanish as |r|−4 for force and |r|−3 for moment. Therefore the hydrodynamic force and moment on the rigid body can be simpliﬁed as:

d ZZ F = −ρ φ n dS (A.46) dt SB d ZZ M = −ρ φ (r × n) dS (A.47) dt SB in which the velocity potential φ should satisfy the boundary condition on the surface of the rigid body as:

∂φ = u · n + ω · (r0 × n) (A.48) ∂n where u and ω are the linear and angular velocity of the rigid body respectively and r0 is the position vector of the surface in the body-ﬁxed reference frame. The velocity potential φ can be expressed as:

φ = u · ϕ + ω · χ = νiφi (A.49)

where ϕ, χ are vectors whose components φi are along the body-ﬁxed reference frame. They are solutions of Laplace’s equation whose gradients tends to zero at inﬁnity, and also satisfy the boundary conditions:

∂ϕ = n (A.50) ∂n ∂χ = r0 × n (A.51) ∂n so that ϕ, χ only depend on the geometry of the rigid body and not on its motion. 236 Hence, using the new form of velocity potential, the hydrodynamic force on the rigid body can be rewritten as: d ZZ F = −ρ ν φ n dS 0 dt i i SB ZZ ZZ (A.52) = −ρν˙i φi n dS − ρνiω × φi n dS

SB SB The moment can be expressed as: d ZZ M = −ρ ν φ (r × n) dS dt i i SB d ZZ = −ρ ν φ [(r + r0) × n] dS dt i i 0 SB   d ZZ dr ZZ = r × −ρ ν φ n dS − ρν 0 × φ n dS 0  dt i i  i dt i SB SB (A.53) ZZ ZZ 0 0 − ρν˙i φi (r × n) dS − ρνiω × φi (r × n) dS

SB SB ZZ = r0 × F − ρνiu × φi n dS

SB ZZ ZZ 0 0 − ρν˙i φi (r × n) dS − ρνiω × φi (r × n) dS

SB SB We can obtain the hydrodynamic moment on the rigid body with respect to the origin of the body-ﬁxed reference frame as:

ZZ ZZ ZZ 0 0 M 0 = −ρνiu× φi n dS−ρν˙i φi (r ×n) dS−ρνiω× φi (r ×n) dS (A.54)

SB SB SB

Now we are ready to derive the expression of added-mass tensor aji, ZZ ZZ ∂ϕ φ n dS = φ dS i i ∂n S S B B (A.55) ZZ ZZ ∂χ φ (r0 × n) dS = φ dS i i ∂n SB SB

237 Then ZZ ∂ϕ a = ρ φ j dS j = 1, 2, 3 ji i ∂n S B (A.56) ZZ ∂χ a = ρ φ j−3 dS j = 4, 5, 6 ji i ∂n SB Writing them into an uniﬁed form, we have

ZZ ∂φ a = ρ φ j dS (A.57) ji i ∂n SB The added-mass tensor can be denoted by a so-called added mass matrix as:   a11 a12 ··· a16        a21 a22 ··· a26  A11 A12 M =   = (A.58) A  . . . .     . . .. .  A A   21 22   a61 a62 ··· a66

in which the submatrices are all 3 × 3. Here we need to make an assumption about

the properties of the added-mass matrix M A, which are extremely useful in later derivations.

Assumption A.1. The added-mass matrix is positive semi-deﬁnite :

T M A = M A > 0 ♦

The properties of M A will be accepted as true even for the rigid body moving in real ﬂuid. The hydrodynamic force and moments on the rigid body can be expressed as:

F 0 = −(A11u˙ + A12ω˙ ) − ω × (A11u + A12ω) (A.59)

M 0 = −(A21u˙ + A22ω˙ ) − u × (A11u + A12ω) − ω × (A21u + A22ω) (A.60)

238 The above equations can be written in matrix form as:         F 0 u˙ O3×3 −S(A11u + A12ω) u   = −M A   −     M 0 ω˙ −S(A11u + A12ω) −S(A21u + A22ω) ω

= −M Aν˙ − CA(ν)ν (A.61) where CA(ν) is the Coriolis-centripetal matrix of added mass,   O3×3 −S(A11u + A12ω) CA(ν) =   (A.62) −S(A11u + A12ω) −S(A21u + A22ω)

It can also be written as:     S(ω) O3×3 A11 A12 CA(ν) =     (A.63) S(u) S(ω) A21 A22

Furthermore, using the expressions of added mass, φ can also be rewritten in a more compact form. Recalling (A.28) and (A.27), ∂ 1 −3 φ = C + ci + O(|r| ), i = 1, 2, 3 (A.64) ∂xi |r|

ci can be written in the form of vector as 1 ZZ ∂φ C0 = −r0 + nφ dS 4π ∂n SB (A.65) 1 ZZ = [−r0(u + ω × r0) · n + nφ] dS 4π SB in which 1 ZZ 1 nφ dS = [A u + A ω] 4π 4πρ 11 12 SB and 1 ZZ − r0(u + ω × r0) · n dS 4π SB 1 ZZ 1 ZZ = − r0(u · n) dS − r0(ω × r0) · n dS 4π 4π SB SB 239 where 1 ZZ 1 ZZ − r0(u · n) dS = − u nr0 dS 4π 4π SB SB 1 ZZZ = u ∇r0 dV 4π VB m¯ = u 4πρ 1 ZZ 1 ZZ − r0(ω × r0) · n dS = − n · (ω × r0)r0 dS 4π 4π SB SB 1 ZZZ = ∇ · [(ω × r0)r0] dV 4π VB ZZZ : 0 1 0 0 0 0 = [r (∇ ·(ω× r )) + (ω × r ) · ∇r ] dV 4π VB m¯ = ω × r 4πρ b hence, 1 1 C0 = m¯ (u + ω × r ) + [A u + A ω] (A.66) 4πρ b 4πρ 11 12 Thus we have

1 φ = C + C0 · ∇ + O(|r|−3) (A.67) |r|

A.3 HYDRODYNAMIC LOADS IN A SPATIALLY UNIFORM UN- STEADY FLOW

In more complicated scenarios where the unbounded fluid is not stationary, we need to use (A.40) and (A.45) to derive the equations of hydrodynamic force and moment. In this subsection, we deal with the situation that the velocity of flow is time-varying, but spatially uniform. When the rigid body is absent from the fluid, the incident velocity potential and velocity itself can be expressed as:

U(t) = ∇φo (A.68) 240 When a rigid body moves inside the ﬂuid, the velocity of the ﬂuid will be disturbed. In this case, the velocity potential can be written as:

∗ φ = φo + φ (A.69) where φ∗ denotes the disturbance due to the motion of rigid body. The hydrodynamic force and moment can still be written as:

d ZZ ZZ ∂φ 1 F = −ρ φ n dS − ρ ∇φ − ∇φ · ∇φ n dS dt ∂n 2 SB SC d ZZ ZZ ∂φ 1 M = −ρ φ(r × n) dS − ρ r × ∇φ − ∇φ · ∇φ n dS dt ∂n 2 SB SC

Insert (A.69) into the expression of F , we obtain

d ZZ F = − ρ (φ + φ∗) n dS dt o SB ZZ ∂(φ + φ∗) 1 − ρ ∇(φ + φ∗) o − ∇(φ + φ∗) · ∇(φ + φ∗) n dS o ∂n 2 o o SC The second term in the above equation can be divided into three separate contri-

butions, associated respectively with the undisturbed potential φo, the disturbance potential φ∗, and cross-terms involving both.

2 Since ∇ φo = 0 is valid through the whole ﬂuid including the interior of the

rigid body, the integral solely involving φo can be simpliﬁed as

ZZ ∂φ 1 −ρ ∇φ o − ∇φ · ∇φ n dS = 0 o ∂n 2 o o SC The integral only with φ∗ also vanishes due to the assumption that the exterior surface is at inﬁnity. Now the contribution from cross-term need to be investigated.

ZZ ∂φ∗ ∂φ 1 1 −ρ ∇φ + ∇φ∗ o − ∇φ · ∇φ∗ n − ∇φ∗ · ∇φ n dS (A.70) o ∂n ∂n 2 o 2 o SC

241 ∗ −3 Since ∇φo is constant over the exterior surface and ∇φ is of the order |r| , the integral is O(|r|−1). At a distance large enough, the integral can also be assumed to vanish. As a result, d ZZ F = −ρ (φ + φ∗) n dS dt o S B (A.71) d ZZ d ZZ = −ρ φ n dS − ρ φ∗ n dS dt o dt SB SB The second integral alone in the above equation is similar to the condition of stationary ﬂuid. The only diﬀerence is the expression of boundary condition,

∂φ∗ ∂φ ∂φ = − o = (u + ω × r0 − U) · n = (u − U) · n + ω · (r0 × n) (A.72) ∂n ∂n ∂n

Hence

d ZZ − ρ φ∗ n dS dt SB ˙ = − [A11(u˙ − U) + A12ω˙ ] − ω × [A11(u − U) + A12ω]

The ﬁrst integral is

d ZZ d ZZZ −ρ φo n dS = ρ ∇φo dV dt dt SB VB   d ZZZ = U ρ dV dt   VB d = (m ¯ U) dt =m ¯ U˙ +m ¯ ω × U

Accordingly, we have the expression of hydrodynamic force as

˙ ˙ F =m ¯ U +m ¯ ω × U − [A11(u˙ − U) + A12ω˙ ] − ω × [A11(u − U) + A12ω] (A.73)

242 wherem ¯ is the ﬂuid mass displaced by the rigid body. Similarly, d ZZ d ZZ M = − ρ φ (r × n) dS − ρ φ∗(r × n) dS dt o dt SB SB ZZ ∂φ∗ ∂φ 1 1 − ρ r × ∇φ + ∇φ∗ o − ∇φ · ∇φ∗ n − ∇φ∗ · ∇φ n dS o ∂n ∂n 2 o 2 o SC (A.74) The ﬁrst integral is

d ZZ d ZZ −ρ φ (r × n) dS = ρ (n × φ r) dS dt o dt o SB SB d ZZZ = − ρ (∇ × φor) dV dt VB d ZZZ = − ρ (∇φo × r) dV dt VB   ZZZ d 0 = − U × ρ(r0 + r ) dV dt   VB d = − (U × m¯ r + U × m¯ r ) dt 0 b dU =m ¯ r × − m¯ U × u +m ¯ r × U˙ +m ¯ ω × (r × U) 0 dt b b

The second integral indicates the case of stationary ﬂow being disturbed by the relative motion, so we can use the results from proceeding section. d ZZ ρ φ∗(r × n) dS dt SB ˙ (A.75) = − [A21(u˙ − U) + A22ω˙ ] − u × [A11(u − U) + A12ω]

− ω × [A21(u − U) + A22ω]

The cross-term integrand in the third integral is in the order of |r|−2, so that the in-

243 tegral is O(1). In the following, we need to ﬁnd out the value of the third integral.

ZZ ∂φ∗ ∂φ 1 1 − ρ r × ∇φ + ∇φ∗ o − ∇φ · ∇φ∗ n − ∇φ∗ · ∇φ n dS o ∂n ∂n 2 o 2 o SC ZZ = − ρ r × [U(∇φ∗ · n) + ∇φ∗(U · n) − (U · ∇φ∗) n] dS

SC ZZ ZZ ZZ =ρU × r(∇φ∗ · n)dS − ρ (r × ∇φ∗)(U · n)dS + ρ (r × n)(U · ∇φ∗)dS

SC SC SC

Given the assumption that SC is a spherical surface of radius R centered at the origin of the body-ﬁxed reference frame, the above integral can be written as

ρU ZZ ρ ZZ ρ ZZ : 0 × R(∇φ∗ · R)dS − (R × ∇φ∗)(U · R)dS + (R×R)(U · ∇φ∗)dS R R R SC SC SC

Note that r = r0 + R, the terms which involve r0 have been neglected since only the moment with respect to the body-ﬁxed frame is concerned. Neglecting the O(|r|−3) term, the gradient of it is

1 ∇φ∗ = ∇ C0 · ∇ |r| : 0 0 1 0 1 = C × ∇ × ∇ + (C · ∇)∇ (A.76) |r| |r| r C0 = 3(C0 · r) − |r|5 |r|3

Hence, ρU ZZ ρU ZZ 2C0 · R × R(∇φ∗ · R)dS = × R dS R R R3 SC SC 2ρU ZZ = × n(C0 · R)dS R3 SC 2ρU ZZZ = × ∇(C0 · R)dV R3 VC 2ρV = C (U × C0) R3

244 ρ ZZ ρ ZZ C0 × R − (R × ∇φ∗)(U · R)dS = − (U · R)dS R R R3 SC SC ρ ZZ = − C0 × n(U · R)dS R3 SC ρV = C (U × C0) R3 where VC is the volume enclosed by the spherical surface SC . As a result, the cross- term integral gives

3ρV C (U × C0) = U × m¯ (u − U) + U × (ω × m¯ r ) + U × [A (u − U) + A ω] R3 b 11 12

Eventually the hydrodynamic moment acting on the rigid body with respect to the origin of body-ﬁxed reference frame is

˙ M 0 =m ¯ rb × U +m ¯ u × U +m ¯ ω × (rb × U)

+ U × m¯ u + U × m¯ (ω × rb) + U × [A11(u − U) + A12ω] ˙ − [A21(u˙ − U) + A22ω˙ ] − u × [A11(u − U) + A12ω]

− ω × [A21(u − U) + A22ω] (A.77) ˙ =m ¯ rb × U +m ¯ rb × (ω × U) ˙ − [A21(u˙ − U) + A22ω˙ ] − (u − U) × [A11(u − U) + A12ω]

− ω × [A21(u − U) + A22ω] where rb is the position vector of the buoyancy center in body-ﬁxed reference frame,

and M 0 is the hydrodynamic moment with respect to the origin of the body-ﬁxed reference frame. Writing equations of hydrodynamic force and moment into matrix form, we

245 have           ˙ F 0 m¯ I3×3 −mS¯ (rb) U mS¯ (ω) O3×3 U   =     +     M 0 mS¯ (rb) O3×3 O3×1 mS¯ (rb)S(ω) O3×3 O3×1

− M Aν˙ r − CA(νr)νr

¯ =M¯ ν˙ c + C(ν)νc − M Aν˙ r − CA(νr)νr (A.78)

T T T where νr = [(u − U) , ω ] is the relative velocity of rigid body with respect to the moving fluid, and identical to the case of the rigid body, the inertia and Coriolis- centripetal inertia matrices of the displaced fluid can be written as   m¯ I3×3 −mS¯ (rb) M¯ =   (A.79) b mS¯ (rb) I0   mS¯ (ω) −mS¯ (ω)S(rb) C¯ (ν) =   (A.80) b mS¯ (rb)S(ω) −S(I0ω) b where I0 is the moment of inertia of the displaced fluid with respect to the body- fixed reference frame.

A.4 HYDRODYNAMIC LOADS IN A SPATIALLY NON-UNIFORM FLOW

Except for the unsteady spatially uniform component, we also expect a steady but spatially uniform component in the ﬂow velocity. Just like the case in the previous subsection, the hydrodynamic force can still be written as: d ZZ F = − ρ (φ + φ∗) n dS dt o SB ZZ ∂(φ + φ∗) 1 − ρ ∇(φ + φ∗) o − ∇(φ + φ∗) · ∇(φ + φ∗) n dS o ∂n 2 o o SC 246 where

∇φo = U(t, η1) = U s(η1) + U u(t) (A.81)

U s(η1), U u(t) denote the steady, non-uniform and unsteady, uniform parts respectively. The ﬂow velocity is decomposed into these two parts just for convenience. Now we need to make an assumption about the non-uniformity of the incident ﬂow.

Assumption A.2. The varying of ﬂow velocity across the length of the rigid body is negligible. ♦

Remark. The flow under the above assumption is also called weakly non-uniform flow by [43]. However, when writing the boundary conditions of the disturbance velocity potential φ∗, they still considered the spatial variations of the incident flow velocity

around the surface of the rigid body. N

If we neglect the spatial variations of the incident ﬂow velocity across the surface of the rigid body, the ﬁrst integral in the hydrodynamic force expression

∗ still yields the same result as (A.73). The integrals involving φo and φ alone come to zero. Now let us consider the cross-term integral again. From (A.76), we know that in the far ﬁeld of the rigid body

1 ∇φ∗ = (C0 · ∇)∇ |r| where 1 1 C0 = m¯ [(u − U) + ω × r )] + [A (u − U) + A ω] 4πρ b 4πρ 11 12 Hence, 2 ∗ 0 ∂ 1 (∇φ )i = Cj (A.82) ∂xi∂xj |r|

247 Therefore, the integrand in the cross-term integral takes the form

∂φ∗ ∂φ (∇φ + ∇φ∗ o − ∇φ · ∇φ∗ n) o ∂n ∂n o i 2 2 2 0 ∂φo ∂ 1 0 ∂φo ∂ 1 0 ∂φo ∂ 1 =Cjnk + Cjnk − Cjni ∂xi ∂xj∂xk |r| ∂xk ∂xi∂xj |r| ∂xk ∂xj∂xk |r| 2 2 2 0 ∂φo ∂ 1 ∂φo ∂ 1 ∂φo ∂ 1 =Cj nk + nk − ni ∂xi ∂xj∂xk |r| ∂xk ∂xi∂xj |r| ∂xk ∂xj∂xk |r|

Because C0 is constant at each instant time, it can be moved outside the integral. Let

2 2 2 ∂φo ∂ 1 ∂φo ∂ 1 ∂φo ∂ 1 Γij = nk + nk − ni ∂xi ∂xj∂xk |r| ∂xk ∂xi∂xj |r| ∂xk ∂xj∂xk |r| 1 ∂ ∂2φ ∂2φ ∂ 1 − o + o |r| ∂n ∂xi∂xj ∂xi∂xj ∂n |r|

If i = j = 1, we can prove that

Γ11 = n · (∇ × Λ) where   0  2   ∂φo ∂ 1 ∂ 1 ∂φo 1 ∂ φo   + −  Λ =  ∂x ∂x |r| ∂x |r| ∂x |r| ∂x ∂x  (A.83)  3 1 3 j 3 j   ∂φ ∂ 1 ∂ 1 ∂φ 1 ∂2φ  − o − o + o  ∂x2 ∂x1 |r| ∂x2 |r| ∂xj |r| ∂x2∂xj Therefore,

ZZ ZZ ZZ Γ11dS = n · (∇ × Λ)dS = ∇ · (∇ × Λ)dV = 0

SC SC SC

A similar construction of (A.83) was first given in [41]. However, (A.83) is only valid for i = j = 1. We can construct vectors like (A.83) for the cases when i = j. However, if i 6= j, the vectors will be difficult to find. Assuming it can be proved that ZZ ΓijdS = 0, ∀ i, j ∈ {1, 2, 3}

SC 248 then we have

ZZ 2 2 2 ∂φo ∂ 1 ∂φo ∂ 1 ∂φo ∂ 1 − ρ nk + nk − ni dS ∂xi ∂xj∂xk |r| ∂xk ∂xi∂xj |r| ∂xk ∂xj∂xk |r| SC ZZ 2 2 1 ∂ ∂ φo ∂ φo ∂ 1 = − ρ − + Γij dS |r| ∂n ∂xi∂xj ∂xi∂xj ∂n |r| SC 2 ∂ φo = − 4πρ ∂xi∂xj |r|=0

Denoting 2 ∂ φo Φij = (A.84) ∂xi∂xj |r|=0 which is the symmetric velocity gradient or so-called rate of strain-tensor of the

incident ﬂow U(t, η1)[43]. we obtain that ZZ ∂φ∗ ∂φ − ρ ∇φ + ∇φ∗ o − ∇φ · ∇φ∗ n dS o ∂n ∂n o SC (A.85)

= − Φ[m ¯ (u − U) +m ¯ ω × rb + A11(u − U) + A12ω]

The above expression illustrates the contribution to the hydrodynamic forces of the gradients of the undisturbed ﬂow velocity. If gradients are zero, the contribution vanishes. From the point of view of hydrodynamics, additional pressure gradients are induced by the velocity gradient; therefore the corresponding force is sometimes called ”horizontal buoyancy” [91]. As for the hydrodynamic moment, it can still be written as

d ZZ d ZZ M = − ρ φ (r × n) dS − ρ φ∗(r × n) dS dt o dt SB SB ZZ ∂φ∗ ∂φ 1 1 − ρ r × ∇φ + ∇φ∗ o − ∇φ · ∇φ∗ n − ∇φ∗ · ∇φ n dS o ∂n ∂n 2 o 2 o SC Under the assumption that the undisturbed ﬂow velocity is constant across the length of the rigid body. We can still follow the derivation of the expressions of the ﬁrst two integrals to obtain the same results. It is once again the cross-term

249 integral that requires our full attention.

ZZ ∂φ∗ ∂φ −ρ r × ∇φ + ∇φ∗ o − ∇φ∗ · ∇φ n dS o ∂n ∂n o SC In order to calculate the integral in a simple manner, we still adopt the assumption

that SC is a spherical surface. Therefore, the integrals we need to calculate are

ρ ZZ ρ ZZ (U × R)(∇φ∗ · R)dS − (R × ∇φ∗)(U · R)dS R R SC SC The second integral will yield the same result, while the ﬁrst integral is

ρ ZZ ZZ (U × R)(∇φ∗ · R)dS = −ρ (n × U)(∇φ∗ · R)dS R SC SC ZZ = −ρ (n × U)(∇φ∗ · R)dS

SC ZZZ = −ρ [∇ × (∇φ∗ · R)U]dV

VC 2ρ ZZZ = − [∇ × (C0 · R)U]dV R3 VC where

0 ∂ 0 [∇ × (C · R)U]i = ijk (CmxmUk) ∂xj

0 0 ∂Uk = ijkCjUk + ijkCmxm ∂xj

Hence,

0 0 0 : 0 0 ∇ × (C · R)U = C × U + (C · R)(∇ ×U) = C × U (A.86)

Consequently, the integral becomes

ZZ ZZZ ρ 2ρ 2ρVC (U × R)(∇φ∗ · R)dS = − (C0 × U)dV = (U × C0) R R3 R3 SC VC In conclusion, the hydrodynamic moments still keep the same expression as in the speciﬁc case of uniform ﬂow. Hence, the total hydrodynamic forces and 250 moments can be expressed as   F 0 ¯ ¯   =Mν˙ c + C(ν)νc − M Aν˙ r − CA(νr)νr M 0 (A.87)   −Φ O 3×3 ¯ +   (M A + M)νr O3×3 O3×3

∗ The last term can actually be incorporated into the CA term. Then our new CA becomes  −Φ(A +m ¯ I ) ∗ 11 3×3 CA(νr) =  −S(A11(u − U) + A12ω)  −Φ(A12 − mS¯ (rb)) − S(A11(u − U) + A12ω)  (A.88) −S(A21(u − U) + A22ω) A more sophisticated derivations of the hydrodynamic loads on deformable body in non-uniform flow are given in [43]. They decomposed the total loads into four distinct components: (i) loads exerted on a moving rigid body in a quiescent fluid; (ii) loads due to the presence of a uniform incident flow; (iii) loads induced by the non-uniform incident flow; (iv) loads resulting from the body’s deformation. We obtain their first two components in (A.73) and (A.77). But as for the third component, additional terms arise in [43] due to the fact that they considered the velocity gradient Φ in the boundary conditions of the disturbance velocity potential (A.72).

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