M. van Berkel DCT

Transcription

1 Explicit solution of the ODEs describing the 3 DOF Control Moment Gyroscope M. van Berkel DCT Traineeship report Coach(es): Supervisor: dr. N. Sakamoto Prof. dr. H. Nijmeijer Technische Universiteit Eindhoven Department Mechanical Engineering Dynamics and Control Group Eindhoven, October, 28

2 Contents Contents 3 1 Introduction 4 2 Theory of Hamiltonian and Integrable systems Lagrangian and Hamiltonian Liouville integrability Cyclic coordinates and conserved quantities Poisson bracket Control Moment Gyroscope Description of the Control Moment Gyroscope Lagrangian Hamiltonian and conjugate momenta CMG s first integrals Solution in Integral form Dynamics of the CMG-system Equilibrium points of the unperturbed system Libration Rotation Transition from libration to rotation Change of motion Special case of the equilibrium point Summary of the different kind of motions Construction of the solution Constant q Libration of q Total rotation of q Special cases of q Result Optimal control of the CMG Optimal Control problem Linear Optimal Control Nonlinear Sub-Optimal Control Conclusions and Recommendations 28 Bibliography 3 List of symbols 31 2

3 Contents A Analytic solution for special cases 32 A.1 Solution for the case p 1 = A.2 Solution for the case p 1 = and p 4 = B Analysis of f(q 2 ) 34 B.1 Denominator > B.2 Extrema of f(q 2 ) B.3 Maximum q2 Max coincides with a zero if 2a > b B.4 Minimum q2 Min coincides with a zero if 2a > b B.5 Extrema q2 Min and q2 Max coincide with a zero if 2a < b C Numerical values used for the different parameters and simulations 37 3

4 Chapter 1 Introduction In the control of many dynamical systems optimal control plays a important role. Many techniques have already been developed to solve these kind of problems. Next to this many extensions have been developed for instance LQR/LQG and H -control. Compared to linear optimal control, nonlinear optimal control is still rarely applied. This is because most of the existing techniques are only successful when using some simple nonlinearities, even then it is still difficult and often time consuming to apply. However for highly nonlinear systems, optimal linear controllers do not work properly and as a consequence nonlinear optimal control needs to be applied. Therefore in [6] a method is proposed to find an approximation for the stabilizing solution of the Hamilton-Jacobi equation. The proposed method is based on the Hamiltonian perturbation theory. The idea behind this theory is to split the system into two parts, the uncontrolled system and the remaining terms (control). The remaining parts can be treated as a perturbation on the original system. The unperturbed system (without control) needs to be integrable to solve it. With this solution the Hamilton equations can be substantially reduced making it possible to build an approximated nonlinear (sub-)optimal controller. This theory has already been shown to work for some simple nonlinear problems. However to show that the theory also works for more complicated problems a more difficult nonlinear setup has been chosen to design a nonlinear optimal controller by using this theory. The eventual goal is not only to show it to work in simulation but also that it is possible to use feedback to stabilize a real setup. This is the Control Moment Gyroscope system (CMG). Its normal function is the attitude control of spacecrafts. However in this case the control problem will be different. The CMG is a nonholonomic system. This will often lead to problems regarding to control. However in our case some of these properties can be exploited like the existence of cyclic coordinates. In this report the focus will lie on finding a solution of the restricted CMG without control. The solution can be used to synthesize the controller. First a small outline of this theory is given in Chapter 2. After this it will be shown that the system fulfills the necessary conditions to integrate it. Then the solution will be represented in integral form. This is done in Chapter 3. With the help of this solution, the different regimes of motion and equilibrium points are explained (Chapter 4). This is also needed to construct a solution for control purposes from the solution in integral form. After this it will be shown how to construct the solution and some of the results will be compared to the solutions found with ODE-solvers. Finally the perturbation theory can be used to derive a sub-optimal nonlinear feedback controller. How to do this is explained in Chapter 6. 4

5 Chapter 2 Theory of Hamiltonian and Integrable systems Before a start can be made with solving the Control Moment Gyroscope system some theory needs to be introduced. Within the theory the system has to be split up in an uncontrolled part and a controlled part. A requirement is that the uncontrolled part is integrable. In this chapter some of the necessary theory about the Hamiltonian and integrability will be treated. This part is mainly based on the theory described in [2] and [3]. However there are many other books that treat these subjects. 2.1 Lagrangian and Hamiltonian The starting point in general in studying and working with dynamics is the Lagrangian L. A different way to describe the dynamics of a system is the Hamiltonian representation. The main difference is that the Lagrangian equations of motion are expressed in second order differential equations, where the Hamiltonian equations of motion are expressed as first order equations. The consequence of this is that the number of equations is doubled with respect to the Lagrangian representation. This also means that the number of independent variables is doubled. In general the Lagrangian is already known therefore it is most common to transform the Lagrangian to the Hamiltonian. The Lagrangian depends on the independent variables q i and q i. L = L (q 1, q 2,..., q n, q 1, q 2,..., q n ) (2.1) Whereas the Hamiltonian is a function of newly introduced independent coordinates p i and of q i. These p i are called conjugate momenta and are a function of q i and q i. These new coordinates p i together with the q i are called canonical variables. This means that the Hamiltonian has following form: H = H (q 1, q 2,..., q n, p 1, p 2,..., p n, ) (2.2) The conjugate momenta are defined as: p i (q, q) = L(q i, q i ) q j (2.3) The Lagrangian L (q, q) can now be transformed to the Hamiltonian H (q, p). Using some algebraic operations on the conjugate momenta q i can be expressed as function of q and p, ( q (q, p)). The systematic transformation from the Lagrangian to the Hamiltonian representation is done with the help of the Legendre transformation: H (q, p) = i q i (q, p) p i L (q, q (q, p)) (2.4) With this knowledge of H (q, p) also Hamiltonian equations of motions can be derived (2.5). 5

6 2.2. Liouville integrability q i = H p i, ṗ i = H q i, i = (1,..., n) (2.5) Within this Hamiltonian formulation many extensions can be made in different directions. 2.2 Liouville integrability In the last section the Lagrangian has been transformed to the Hamiltonian formulation. In this section the notion of integrability is introduced. Within the literature [2], [3] there are many forms of definitions given for Liouville integrability. However commonly following definition is used: A Hamiltonian system is completely integrable in the sense of Liouville if all constants of motion are in involution [2]. Another definition in [2] is: Two functions are in involution if their Poisson bracket is zero. The system needs to be integrable to solve it. Therefore the involution of the constants of motion needs to be checked to be sure it is integrable. A constant of motion is a quantity that is conserved throughout the motion. Another term often used for this is first integral. A constant of motion or first integral can be expressed as follows: f (q 1, q 2,.., q n, q 1, q 2,.., q n ) = constant (2.6) To be able to integrate a system all the first integrals (constants of motion) need to be found like the definition says. In reality the number of first integrals that need to be found is n, the degrees of freedom (2.5). In general the Hamiltonian is also a constant of motion because it is the conserved total energy of the system. Another way to find first integrals is with the help of cyclic coordinates. 2.3 Cyclic coordinates and conserved quantities The easiest way to find first integrals is with the help of cyclic coordinates. A cyclic coordinate is defined as a coordinate q j that does not enter the Hamiltonian function (the same holds for the Lagrangian). In other words H or L are not a function of q j, however H can still be a function of q j. The nice property of this is that its belonging conjugate momentum p j is a constant of motion. This can be easily shown as follows: d L L = d L = d dt q i q j dt q i dt p j = (2.7) A different constant of motion is the total angular momentum. If the needed number of first integrals has been found, then it is necessary to check if they are in involution. 2.4 Poisson bracket Two functions u(q, p) and v(q, p) are in involution if their Poisson bracket is zero, as was earlier stated. The Poisson bracket is defined as follows: [u, v] q,p = u q i v p i u p i v q i (2.8) To check if the Hamiltonian system is integrable all combinations of the first integrals with each other and with the Hamiltonian need to vanish. The Poisson bracket of a first integral and the Hamiltonian is always zero, otherwise it would not be a first integral. This is because: [H, f] q,p = H q i f p i H p i f q i = ṗ i f p i + q i f q i = df dt If f is a first integral of H then df dt. This theory will be used to check integrability of the Control Moment Gyroscope. (2.9) 6

7 Chapter 3 Control Moment Gyroscope In the last chapter the theory about integrable systems has been introduced. In this chapter this theory is used to try to solve the Control Moment Gyroscope system. Therefore first the systems model will be introduced. After this the Lagrangian and the Hamiltonian will be derived. Then the constants of motion are derived and they are used to find the solution of the system in integral form. 3.1 Description of the Control Moment Gyroscope In figure 3.1 a schematic representation of the CMG can be seen. Figure 3.1: Schematic representation Control Moment Gyroscope [ 4] The CMG system consists of four gimbals respectively A, B, C and D. Within these gimbals there are four axis fixed. These are a i, b i, c i and d i with unity vectors i = 1, 2, 3. At the base of the CMG a reference frame N is defined with orthogonal unit vectors N i (i = 1, 2, 3). In this system there are four angles that can rotate. These are q 1 in d 2 direction, q 2 in c 1 direction, q 3 in b 2 direction and q 4 in n 3 direction. The angular velocities ω i are defined in the same directions as q i (i = 1, 2, 3, 4). These are also represented in figure 3.1. All centers of mass are situated at the center of disc D. The body s D and C can be directly actuated through T 1 and T 2 (see figure 3.1). Body D can be actuated in d 2 direction this corresponds to the rotation angle q 1. Body C can be actuated in c 1 direction that corresponds to the rotation angle q 2. 7

8 3.2. Lagrangian All angles q i (i = 1, 2, 3, 4) are observable because these angles are measured by encoders. The elements I x, J x, K x (x = A, B, C, D) are the scalar moments of inertia about the k th (k= 1,2,3) direction respectively in bodies A, B, C, and D. Only the moments of inertia are considered and not the products of inertia. The zero positions resemble the positions as seen in figure 3.1 (q i = ). On the real setup there is the possibility to lock b 2 respectively q 3. This opportunity will be used, so q 3 is locked in zero position. This means no motion around this angle is possible anymore and thereby reducing the system to three degrees of freedom. In many articles and papers like [4] and [5], also a description of the Control Moment Gyroscope can be found. 3.2 Lagrangian In case of the CMG the Lagrangian equals the kinetic energy because the potential energy is zero. For instance the gravity can be neglected in the system. Now the Lagrangian of the system without control can be calculated and is given: in which L = 1 2 ( Jd ω (I c + I d ) ω ( J 2 J 1 cos 2 (q 2 ) ) ω4 2 ) + Jd ω 1 ω 4 sin (q 2 ) (3.1) J 1 = J c + J d K c I d and J 2 = K a + J c + J d + K b with J 1 < J d < J 2 (3.2) From the Lagrangian and the non-conservative forces, Lagrange s equations can be derived. For completeness also the non-conservative quantities are added. J d ( ω 1 + ω 4 sin (q 2 ) + ω 2 ω 4 cos (q 2 )) = T 1 (3.3) (I c + I d ) ω J 1ω 2 4 sin (2q 2 ) J d ω 1 ω 4 cos (q 2 ) = T 2 (3.4) ( J2 J 1 cos 2 (q 2 ) ) ω 4 + J 1 ω 2 ω 4 sin (2q 2 ) + J d ( ω 1 sin (q 2 ) + ω 1 ω 2 cos (q 2 )) = ω 4 d 4 (3.5) Because we want to find the explicit solution of the system, we are only interested in the system without control (T 1 =, T 2 = and d 4 = ). The state space model (with the states: q 1, q 2, q 4, ω 1, ω 2, ω 4 ) is given by: q 1 = ω 1 (3.6) q 2 = ω 2 (3.7) q 4 = ω 4 (3.8) ω 1 = ω 2 cos (q 2 ) ( ( ( J d sin (q 2 ) ω 1 ω 4 J2 J 1 2 cos 2 (q 2 ) ))) J 2 J 1 cos 2 (q 2 ) J d sin 2 (q 2 ) (3.9) ω 2 = ω ( 4 Jd ω 1 cos (q 2 ) J 1ω 4 sin (2 q 2 ) ) I c + I d (3.1) ω 4 = ω ( 2 ω4 sin (2 q 2 ) ( 1 2 J ) d J 1 Jd ω 1 cos (q 2 ) ) J 2 J 1 cos 2 (q 2 ) J d sin 2 (q 2 ) (3.11) The state space model has been used to simulate the unperturbed system. 8

9 3.3. Hamiltonian and conjugate momenta 3.3 Hamiltonian and conjugate momenta In Section 2.1 it was explained how the Lagrangian has to be transformed to the Hamiltonian. This transformation is done through the conjugate momenta (2.3). The following conjugate momenta are then found: p 1 = J d ω 1 + J d ω 4 sin (q 2 ) (3.12) p 2 = (I c + I d ) ω 2 (3.13) p 4 = ( J 2 J 1 cos 2 (q 2 ) ) ω 4 + J d ω 1 sin (q 2 ) (3.14) Now all the angular velocities are being replaced with the conjugate momenta. In essence L (q 2, ω 1, ω 2, ω 4 ) has been transformed to the Hamiltonian H (q 2, p 1, p 2, p 4 ) The Hamiltonian has now the following form: ( ) H = 1 p (I c + I d ) + p2 1 (p 4 p 1 sin (q 2 )) 2 + J d J 2 J 1 cos 2 (q 2 ) J d sin 2 (3.15) (q 2 ) 3.4 CMG's rst integrals Two first integrals are easily found with the help of cyclic coordinates. The coordinates q 1 and q 4 are cyclic coordinates because they do not appear in the Hamiltonian. In (2.7) it was already shown that its corresponding conjugate momenta are constant. This means that p 1 and p 4 are first integrals. The Hamiltonian H is also a first integral because it is a conserved quantity (the total energy of the system). It has to be checked that all constants of motion are in involution. Therefore for all combinations the Poisson brackets need to be zero. Because H(q 2, p 1, p 2, p 4 ) is not a function of q 1 and q 4 immediately can be concluded that their Poisson brackets are zero. The Poisson bracket of the Hamiltonian function with the Hamiltonian is always zero. The Poisson bracket can also be used to show that p 2 is not a constant because it is different from zero: [p 2, H] q,p = p 2 q 2 H (q, p) p 2 p 2 p 2 H (q, p) q 2 = H(q 2, p 1, p 2, p 4 ) q 2 (3.16) The Hamiltonian depends on q 2 as a consequence the partial derivative to q 2 is different from zero. Now it can be concluded that the system is integrable. The four degrees of freedom CMG, so without axis 3 locked it is also integrable. This is because the first integrals found for the 3 DOF system are also first integrals of the 4 DOF system. Another conserved quantity that has not been used yet is the total angular momentum. As a consequence the number of first integrals needed is sufficient. However it has to be checked that all their Poisson brackets are zero. This could not be found in the literature. 3.5 Solution in Integral form In Section 3.4 it was already concluded that p 1, p 4 and H are constants of motion. This knowledge is used to find an explicit solution of the corresponding set of ODEs describing the uncontrolled 3 DOF CMG. The value of the constants of motion are determined by the initial conditions. The explicit solution can only be found if the Hamiltonian (3.15) only depends on constants and one variable and its derivative. Because p 2 is not constant it needs to be replaced. This is done by replacing p 2 with (I c + I d ) ω 2. This was derived in (3.13). ( ) H = 1 (I c + I d ) ω2 2 + p2 1 (p 4 p 1 sin (q 2 )) J d J 2 J 1 cos 2 (q 2 ) J d sin 2 (3.17) (q 2 ) Now (3.17) can be rewritten as ω 2 = ω 2 (q 2 ) with the constants p 1, p 4 and H. 9

10 3.5. Solution in Integral form 1 ω 2 = ± 2H p2 1 (p 4 p 1 sin (q 2 )) 2 I c + I d J d J 2 J 1 cos 2 (q 2 ) J d sin 2 (q 2 ) (3.18) By separation of variables (3.18) can be solved. For simplicity the initial conditions will be chosen at t =. The integral will also only be evaluated for the positive root of (3.18). This means that in a later stage the sign of ω 2 should be taken into account. t dt = I c + I d q2 q i 2 2H p2 1 J d dq 2 (p 4 p 1 sin(q 2)) 2 J 2 J 1 cos 2 (q 2) J d sin 2 (q 2) After some algebraic operations the integral can be represented in the following form. (3.19) t = q2 q i 2 J d (I c + I d ) ( J 2 J 1 (J d J 1 ) sin 2 (q 2 ) ) (2J d H p 2 1 ) ( J 2 J 1 (J d J 1 ) sin 2 (q 2 ) ) J d (p 4 p 1 sin (q 2 )) 2 dq 2 (3.2) In a mathematical sense this system is solved. After solving the integral and rewriting it to q 2 the solution will have following form: q 2 = q 2 (t; H, p 1, p 4 ) (3.21) In (3.21) q 2 will only depend on the time t and the constants H, p 1 and p 4. With this information the solution for the entire system can be derived. ω 1 (t; H, p 1, p 4 ) = p 1 p 4 sin (q 2 (t)) p 1 sin 2 (q 2 (t)) J d J 2 J 1 cos 2 (q 2 (t)) J d sin 2 (q 2 (t)) ω 4 (t; H, p 1, p 4 ) = p 4 p 1 sin (q 2 (t)) J 2 J 1 cos 2 (q 2 (t)) J d sin 2 (q 2 (t)) (3.22) (3.23) The integral in (3.2) needs to be solved to be able to solve the entire system. The integral looks like it belongs to the class of elliptic integrals, therefore it was tried to solve (3.2) with the help of elliptic integrals. However because of the cross-term of (p 4 p 1 sin (q 2 )) 2 in (3.19) ( 2p 1 p 4 sin (q 2 )) it was impossible to do this. As a consequence of this it was not possible to find an analytic expression for the solution of this integral. However the solution of this integral is still closely related to the elliptic functions. Therefore the underlying theory of elliptic integrals is used to help solve and test the methods that are used to eventually solve this system. It can also be shown that (3.19) is closely related to elliptic functions. If a simplification is applied for instance p 1 = ( 2p 1 p 4 sin (q 2 ) will vanish). Then this integral reduces to a form where it is solvable with elliptic integrals. The analytic solution for p 1 = and p 1 = p 4 = are shown in Appendix A. Although an analytical solution can be derived in such a way for many cases it is still difficult to work with as can be seen in Appendix A.1. Therefore this system needs to be solved in a different way. However, before this can be done, it is necessary to determine what the essential changes are within the dynamics. 1

11 Chapter 4 Dynamics of the CMG-system In Section 3.5 the solution for the CMG system was found in integral form. For future applications (Optimal control theory) it would be easiest to work with an analytical solution for the CMG, but unfortunately this is not possible. That is why another method is needed to find a useful solution for this system. However before this can be done the system s dynamical properties have to be studied. The natural starting point to do this is to determine the equilibrium points of the unperturbed system. Then the two basic motions, oscillation and rotation are described. After this the special cases like the homoclinic and heteroclinic orbit are described and analyzed. At the end of this chapter a summary is given of the different cases. 4.1 Equilibrium points of the unperturbed system A natural way to determine the equilibrium points is by setting the left hand side of the state space model to zero (Section 3.2). This results in the following equilibrium points: q i ω i = (4.1) R This is not a very useful conclusion. However if ω 1, ω 4 and q 2 are taken as constants the system equation (3.18) is still in equilibrium (ω 2 = ). In this case q 1 and q 4 are not considered as coordinates (cyclic coordinates) and ω 1 and ω 4 therefore do not need to be zero. The combination of initial conditions that correspond to this definition of equilibrium are given in table 4.1. Table 4.1: Equilibrium points q i 1 q i 2 q i 4 ω i 1 ω i 2 ω i 4 R ± π 2 R R R R R R R R R R J1 J d ω i 4 sin(q 2 ) R This corresponds to the zeros of (3.18), so it also could be directly derived from (3.18). However it is more difficult to see it in this equation. This is because the initial conditions are embedded within the conjugate momenta and the total energy H. To derive this directly lets first take a look beneath the square root in (3.18). The numerator under the square root will be defined as f(q 2 ). The denominator (under the square root) is always positive, so there are no singular points. This is explained in Appendix B.1. As a consequence of this f(q 2 ) has to be positive or equal to zero otherwise no real solution for ω 2 exists because of the square root in (3.18). To analyze f(q 2 ) for simplicity a, b and c will be used and are defined as follows: f ( q 2 ; q i 2, ω i 1, ω i 2, ω i 4) = a sin 2 (q 2 ) + b sin (q 2 ) + c with (4.2) 11

12 4.2. Libration a = (2J d (J d J 1 )H + J 1 p 2 1), b = 2J d p 1 p 4, c = (2J d H p 2 1)(J 2 J 1 ) J d p 2 4 The numerator f(q 2 ) is just a second order equation in sin (q 2 ), so it is easy to find the zeros of (3.18): sin (q 2 ) = b ± b 2 4ac with a (4.3) 2a Now Z is the set containing all values for q 2 that satisfy (4.3). In other words all the zeros of f(q 2 ) and ω 2. This means that: = f (Z(i)) = ω 2 (Z(i); p 1, p 4, H) with i = 1, 2,... (4.4) However it must be considered that the zeros of f (q 2 ) are not automatically the equilibrium points of the system because q i 2 is also hidden within the constants of motion. The system is in equilibrium as described in (4.1) if: q i 2 within the constants of motion is also a zero of f (q 2 ). This means if q i 2 Z the system is in equilibrium. The length of Z can differ from none to infinity depending on the constants of motion. However because of the sinus wherein q 2 is contained the number of zeros only needs to be considered within the π, π] interval because of the symmetric character of the sin-function. This means that the number of zeros within this interval will only be ranging from none to four. The zeros of f(q 2 ) have important physical meaning even if q i 2 Z. This will be shown in the next sections of this chapter. During the analyzing process it was found that the CMG system has resembles to a simple pendulum system. Although the CMG system is more complex it will be easier to understand when the system is compared to a pendulum. In essence the system can be separated in two basic motions, libration and rotation just like the pendulum. 4.2 Libration The first basic motion corresponds to a simple oscillating behavior. Another name for this motion is libration. Although the CMG system can oscillate in many different ways, it is similar to the simple oscillating pendulum. The main difference is that the oscillation is not around a certain point (where the potential energy is zero in the pendulum) but can change as a function of the constants of motion. In a pendulum the movement is bounded by the zeros of the kinetic energy. The same holds for the CMG, if the kinetic energy is zero, the upper and lower bound of oscillation have been reached. The zeros of the kinetic energy are the same as the zeros of ω 2 respectively f(q 2 ), with as important consequence that the upper and lower value that the angle q 2 can reach will be determined by these zeros. This can be seen in the integral that has to be solved. J d (I c + I d ) ( J 2 J 1 (J d J 1 ) sin 2 (q 2 ) ) t = q2 q i 2 f(q 2 ) dq 2 (4.5) Herein f(q 2 ) has now become the denominator. If f(q 2 ) becomes zero the equation within the integral becomes singular. If this integral is solved, the solution will be of the form t = t(q 2 ; p 1, p 4, H). However the solution needed should be of the form q 2 = q 2 (t; H, p 1, p 4 ). So by switching time t and the angle q 2, t = t(q 2 ; p 1, p 4, H) is inverted to q 2 = q 2 (t; H, p 1, p 4 ). The zeros of f(q 2 ) that lead to singularity will now become the boundaries of q 2. This means that all zeros of f(q 2 ) (Z) need to be understood as the bounds of integration in (4.5). Only the domain where f(q 2 ) is positive can be integrated. The domain that is enclosed by two zeros of f(q 2 ). The two values that enclose the domain where f(q 2 ) is positive are from the set Z. They are called Z LBI, the lower bound of integration and Z UBI, the upper bound of integration. In other words the upper value and the lower value where in between q 2 can oscillate. The difference between this upper and lower bound of integration is the same as the amplitude in the sense of normal periodic functions. 12

13 4.2. Libration In reality it is somewhat more difficult because the zeros of f(q 2 ) change, depending on the constants of motion. In essence there can be said that, if there are zeros, they limit the motion of the system. Already it has been explained what the solutions are. However for better understanding (4.2) is represented in figure f : f(q ; π/4, 2,.5,.5) 1 2 f : f(q ;π/4, 1,1,.1) 2 2 f : f(q ;π/4, 1,2.94,1) 3 2 f : f(q ;π/4, 1,3.5,1) 4 2 f(q 2 ) q 2 Figure 4.1: Graphical representation of the numerator f(q 2 ; q2, i ω1, i ω2, i ω4) i for the dierent motions related to libration (dierent initial conditions) In figure 4.1 four possible f(q 2 ) are represented. Within this representation some special configurations are not shown because they correspond to special cases and will be treated later. Important to remember is that ω 2 is only defined in the domain where f(q 2 ). So all information below zero is not important and is as a consequence not represented. f(q 2 ) is represented for four different combinations of initial conditions or constants of motion. f 1 (q 2 ) and f 4 (q 2 ) have only two zeros so these represent two basic oscillations. Only the form of the oscillation will differ significantly. f 3 (q 2 ) has three zeros; there are two domains only separated by one zero where f(q 2 ) is positive. f 2 (q 2 ) has four zeros this means that there are two domains where f(q 2 ) is positive. The best way to understand this picture is to forget for a moment the denominator in (3.18) (then ω 2 = f(q 2 )). Take for instance f 1 (q 2 ), if the initial condition for q2 i is near to the left zero. Then follow f 1 (q 2 ) to the right side. This means that the angular velocity is increasing till the point q 2 = π 2 after this the angular velocity is decreasing till it is zero (f(q 2) = ). From hereon it cannot move further to the right because ω 2 is not defined there (maximum value of q 2 is reached). What will happen then is that ω 2 will decrease again following f 1 (q 2 ) back to the left side. This will happen till the point q 2 = π 2 after this the angular velocity is increasing till it is zero again (minimum value of q 2 is reached). This is because the negative solution of (3.18) also needs to be considered. This motion is repeating itself again and again, so this is a typical oscillating motion. In ω 2 = ω 2 (q 2, p 1, p 4, H) there is still a denominator. This denominator will change the shape of ω 2, but will not change the bounds. The figure 4.1 contains all information necessary because after 2π the picture repeats itself again. Another important conclusion that can be drawn from the figure is that there is symmetry around q 2 = ± π 2. This comes from the relation sin (q 2) = sin (π q 2 ). A minimum of f(q 2 ) is defined as q2 Min. q2 Min is always equal to q 2 = π 2 or q 2 = π 2 or both in the case there are two q2 Min. In the case that f(q 2 ) has only one minimum (q2 Min ) and one maximum (q2 Max ) then q2 Min = π 2 and qmax 2 = π 2 in the case that b > from (4.2). It automatically follows that if b < then q2 Min = π 2 and qmax 2 = π 2. f(q 2) has only one minimum and one maximum if 2a < b. This can be calculated with the help of the first and second derivative of f(q 2 ). This is worked out and explained in Appendix B.2. This seems not to be important in this case however later on it will be explained that there is an important physical meaning if one of these points is also a zero of f(q 2 ). f 2 (q 2 ) in figure 4.1 has four zeros in the domain π, π], this means there are two positive domains (where ω 2 is defined). The domain where the oscillation takes place, depends on the 13

14 4.3. Rotation initial condition q2. i The zeros that bound the domain wherein q2 i is situated are the lower and upper bound of the oscillation. In the case of four zeros in the domain π, π] the amplitude of the oscillation cannot become bigger than π because it is limited between q 2 = ± π 2. By changing the initial conditions in a way that f 2 (q 2 ) moves upwards f 3 (q 2 ) is found. In this case there are three zeros in the interval π, π]. To the right and left of the zero located at q 2 = π 2 (qmin 2 ), the domain is positive. The choice of the domain depends on the side where q2 i is situated. If q2 i is the point between the two positive domains, this corresponds to the equilibrium point as stated in table 4.1. In this point a change occurs where two smaller (oscillation) domains melt together to form one big (oscillation) domain. f 1 (q 2 ) and f 4 (q 2 ) only have two zeros so there is only one positive domain and the zeros still corresponds to the lower and upper bound of the oscillation, Z LBI and Z UBI. 4.3 Rotation This motion corresponds to the angle q 2 rotating totally around in one direction. This can be compared to a pendulum wherein the kinetic energy is higher than the maximum potential energy that can be stored. As a consequence the pendulum keeps rotating. Simply said if you give it enough rotation momentum it will keep on spinning around. This motion corresponds to a case where there are no zeros (Z is empty) and as a consequence the domain where f (q 2 ) > has become R. There is also a very basic case wherein this motion becomes more clear and can be found in Appendix A Transition from libration to rotation The transfer between rotation and libration is the transformation from one or more zeros to no zeros. This also follows from the last two sections. This happens when all zeros are at the same time a minimum of f(q 2 ) (q2 Min ). This case is illustrated in figure f 5 : f(q 2 ;,,4.85, 7) f 6 : f(q 2 ; π/4,1,4.99, 2) f 7 : f(q 2 ; π/3,7,1.97,2).5 f(q 2 ) q 2 Figure 4.2: Transfer from libration to rotation In figure 4.2 the zeros are the bounds as described in Section 4.2 so the motion takes place between these bounds. However now there is no domain where f(q 2 ) is negative. If the energy is slightly increased so that f(q 2 ) moves upwards there will be now zeros left anymore. This means no boundaries so total rotation. This change from libration to rotation can also be sudden, for instance if ω2 i is slightly increased. Then the whole curve will move upwards and there are no zeros anymore, rotation. Now look at f 6 (q 2 ) and f 7 (q 2 ). They have only one q2 Min and one zero in the interval π, π] and the motion is only bounded by one point within this interval. This means that the amplitude of the motion is 2π. 14

15 4.5. Change of motion A special case will occur if also q2 i Z. As already explained this is an equilibrium point (Section 4.1). This equilibrium point can be compared to the unstable equilibrium point of an inverted pendulum. In theory if an inverted pendulum is in equilibrium and there is a slight perturbation the pendulum will fall down and come back to the equilibrium position taking infinite time. This orbit is called a homoclinic orbit. The same thing happens for these equilibrium points. A slight perturbation will lead to a full rotation of q 2 taking infinite time. In the case of all zeros are also minimums of f(q 2 ) (f 5 (q 2 )) and q2 i Z. Then if there is a slight perturbation around this equilibrium point the angle q 2 will move until it falls back into the other equilibrium point. This is called a heteroclinic orbit. The homoclinic and heteroclinic orbits are both theoretical terms in our case. This is because to fall out of equilibrium a small perturbation is needed. A small perturbation means a change in energy. As a consequence f(q 2 ) will also change and f(q 2 ) will have different properties. However orbits that start close to these conditions will have some similar properties. A similar property is for instance that the orbits will have long periods. These homoclinic and heteroclinic orbits are therefore important conditions. 4.5 Change of motion Now all cases where a sudden change can take place have been described. The cases as described in Section 4.4 and the case of f 3 (q 2 ) in figure 4.1. These cases have all in common that q2 Min coincides with a zero. Although these cases can be identified with the described relations in the preceding sections. It is still quite time consuming and not straightforward to find out when a transition will occur. A nicer expression can be derived where these cases are expressed in the conjugate momenta equal to twice the total energy. Two relations can be derived one for the q 2 = π 2 position and one for the q 2 = π 2 position. This is because qmin 2 is always situated in q 2 = ± π 2. 2 H = (p 1 + p 4 ) 2 + p 2 1 J 2 J d J d 2 H = (p 1 p 4 ) 2 + p 2 1 J 2 J d J d belonging to q Min 2 = π 2 belonging to q Min 2 = π 2 The derivation is made in Appendix B.4 with the help of first and second derivatives of f(q 2 ). Now by filling in the constants of motion it is immediately possible to calculate wether the motion is on the brink of change. There is only one remark about this. That is in the case there are only one q2 Min and one q2 Max (Figure 4.2: f 6, b < 2a). Then one of (4.6) or (4.7) corresponds to a maximum of f(q 2 ) coinciding with a zero (depending on the sign of b, see Appendix B.5). 4.6 Special case of the equilibrium point In the previous sections already an extensive analysis has been made of what happens if q2 Min Z. Now a logical proceeding is what happens if q2 Max Z. The answer is nothing, q 2 is constant in this case. This is because ω 2 is only defined in q2 i (and sometimes defined in π q2 i because of symmetry). It is much easier understood by just showing it. There are two possibilities one zero or two zeros in the interval π, π] this can be seen in figure 4.3. In figure 4.3 f(q 2 ) is plotted again. The figure clearly shows that f 8 (q 2 ) and f 9 (q 2 ) are always negative except where they are zero. This means only here f(q 2 ) is defined. If f(q 2 ) is slightly changed. It moves in that case upwards and there will be a small motion but no big changes. Similarly as in Section 4.5 an expression can be derived here and is explained in Appendix B.3. (4.6) (4.7) 2 H = p 1 2 J d (4.8) This can also be proven with the help of (4.3). As explained the zeros are the boundaries of oscillation, so the boundaries are in this case in the same point. There is no oscillation anymore which means: 15

16 4.7. Summary of the dierent kind of motions x f(q 2 ) f : f(q ; π/2,.5,,.1) 8 2 f 9 : f(q 2 ;π/4,.43,,1) q 2 Figure 4.3: Constant q 2, maxima that coincides with an equilibrium point sin (q 2 ) = b 2a with b 2 4ac = 2 H p 2 1 = (4.9) J d So exactly the same relation has been found. Also in this case a physical interpretation can be given. This case corresponds to a pendulum (without damping) in stable equilibrium. By giving a very small push (increasing energy slightly) the pendulum just begins to oscillate a little. However because the energy is conserved in the system, it will remain oscillating with a very small amplitude, just like the angle q 2 of the CMG. 4.7 Summary of the dierent kind of motions In this chapter the different motions and special cases have been explained. Because they are spread over the chapter they are summarized. In the table underneath the basic motions are summarized: Table 4.2: Summary of the basic motions Basic motions Condition Explanation Equilibrium q i 2 Z q 2 is in rest Libration Z q i 2 Z q 2 is oscillating between Z LBI and Z UBI Rotation Z = q 2 is rotating without any boundaries The special cases are summarized in table 4.3. Although also (4.6) or (4.7) could also be used as conditions to separate the special cases it has here been explained with Z. Table 4.3: Summary of the special cases Special cases Condition Explanation Stable equilibrium (4.8) Comparable to stable equilibrium point of pendulum Libration to rotation f(q 2 ) q2 Min Z If f(q 2 ) Transfer from libration to rotation - Homoclinic orbit f(q 2 ) Small perturbation will lead to homoclinic orbit, Z π, π] = q2 i comparable to unstable equilibrium point of pendulum - Heteroclinic orbit f(q 2 ) q2 i Z Small perturbation will lead n (Z π, π]) = 2 to heteroclinic orbit Amplitude doubling n (Z π, π]) = 3 If f(q 2 ) The amplitude of the oscillation will double n(a) is number of elements in A 16

17 Chapter 5 Construction of the solution In the last chapter all different motions have been separated and explained. In this chapter the focus will be to really construct the solution. In general one can easily solve this system with numerical ODE-solvers. However for control purposes the system needs to be defined for many initial conditions in a certain interval. This is because it is needed in the canonical transformation. That has to be applied to build the feedback controller. It is virtually impossible to generate all these solutions with ODE-solvers and also have a high accuracy. The construction of the solution is done, similar to the separation of the motion that is used in the last chapter. First the solution of q 2 (t; p 1, p 4, H) for the three basic motions are constructed. It is shown how to construct q 2 when it is constant or is as good as constant. Then it is shown how to construct the solution for q 2 if it is oscillating. After this it is shown how to construct the solution in the case that q 2 is rotating. Then the solutions for the special cases are represented. This means that q 2 (t; p 1, p 4, H) is shown in the cases of an amplitude doubling, a homoclinic orbit and a heteroclinic orbit. 5.1 Constant q 2 The solution for a constant q 2 is easily found. If the initial conditions correspond to those in table 4.1 than q 2 is in equilibrium and constant. Then q 2 (t; p 1, p 2, H) = q i 2 is taken as the solution. In a numerical sense if there is a very small oscillation it can also be approximated by a constant q 2. However when one of the parameters slightly is increased it begins to oscillate with a very small amplitude. This is shown in the next figure (a) (b) Angle q 2 [rad] f(q 2 ) q 2 = f(t) q 2 Ode Numerator f(q 2 ) Z LBI Z UBI q 2 i Z LBI and Z UBI Time t [s] Angle q 2 [rad] Figure 5.1: Representation of a nearly constant q 2 with 2H p2 1 J d = In figure 5.1a the solution for q 2 as function of time for a certain set of initial conditions is represented. The dash-dotted line represents the approximated solution, the dashed line the calculated amplitudes. The solid-line is the solution found when solving the state space model numerically. 17

18 5.2. Libration of q 2 Also in figure 5.1b f(q 2 ) is represented. Important to notice is that Z LBI the lower bound of integration and Z UBI, the upper bound of integration are both very close to q i 2. In the first case q 2 is nearly constant because the amplitude is very small. This is because (5.1) (Section 4.6). 2 H p 1 2 J d (5.1) Looking at the amplitudes it becomes clear that they are almost the same. Although solving the state space model looks the best, it is time consuming and in longer time frames it results in larger errors. The method described in this section gives immediately a good approximation without solving any differential equation or integral. 5.2 Libration of q 2 This is by far the most difficult case to solve. We first solve the integral on the right hand side of (3.2) by numerical integration. The function that needs to be integrated will be abbreviated with g(q 2, p 1, p 4, H) for simplicity. g(q 2, p 1, p 4, H) = J d (I c + I d ) ( J 2 J 1 (J d J 1 ) sin 2 (q 2 ) ) (2J d H p 2 1 ) ( J 2 J 1 (J d J 1 ) sin 2 (q 2 ) ) J d (p 4 p 1 sin (q 2 )) 2 dq 2 (3.2) and (3.19) can be written in a shorter form. t = q2 q i 2 g(q 2 ; p 1, p 4, H) = q2 q i 2 J d (I c + I d ) ( J 2 J 1 (J d J 1 ) sin 2 (q 2 ) ) f(q 2 ) (5.2) dq 2 (5.3) The integration is done over the domain that is bounded by the two zeros of f(q 2 ) (Z LBI and Z UBI ) that encloses q i 2. The bounds within the programming will be determined from (5.4) instead of (4.3). q 2 = arcsin b ± b 2 4ac with a (5.4) 2a There are maximally two zeros of f(q 2 ) calculated with (5.4). This is because the arcsin is only defined between q 2 = ± π 2. So if necessary the other two zeros of f(q 2) will be constructed with the help of symmetry around q 2 = ± π 2. So that in the end g(q 2; p 1, p 4, H)) is integrated over the correct interval. The zeros of f(q 2 ), Z LBI and Z UBI enclose the interval wherein q2 i is situated. Also the sign of ω2 i has now to be taken care of, because in the form of (5.3) only the positive root has been considered. Time t [s] t = t(q ;p,p,h) Z LBI and Z UBI Angle q 2 [rad] f(q 2 ) Numerator f(q 2 ) Z LBI Z UBI q 2 i Angle q 2 [rad] Figure 5.2: Result of numerically integrating the function between Z LBI and Z UBI 18

19 5.3. Total rotation of q 2 In figure 5.2a this integration is visually represented for a certain set of initial conditions (Appendix C). The dashed lines are the calculated bounds (Z LBI and Z UBI ) and the solid line represents the numerically integrated function t = t(q 2 ; p 1, p 4, H) within this interval. It can be seen that, this line only represents the inverse of q 2 and only for a half period. Within the half period that is calculated all information is already contained. On the time axis the half period T 2 can be directly determined. This can also be expressed as follows: T Z UBI Z LBI Z UBI 2 = g(q 2, p 1, p 4, H)dq 2 g(q 2, p 1, p 4, H)dq 2 = g(q 2, p 1, p 4, H)dq 2 (5.5) q2 i q2 i Z LBI Only a half period is found because in the form of (5.3) only the positive root has been considered. This has as consequence that only half the period has been calculated. If now the negative root of (5.3) is considered then: Z UBI Z LBI g(q 2 ; p 1, p 4, H)dq 2 = g(q 2 ; p 1, p 4, H)dq 2 (5.6) Z LBI Z UBI This is indeed the missing part of the oscillation. However it is not necessary to also calculate this part because it can be constructed from (5.3). This knowledge can now be used to correctly extend t = t(q 2 ; p 1, p 4, H). This is done by mirroring around the horizontal line passing through Z LBI or Z UBI where q 2 has reached its minimal value, respectively maximum value. Then copy this inverse function to form one period T. Now the function can be totally reconstructed by copying and shifting this one period T. This can be seen in figure Time t [s] t = t(q 2 ;p 1,p 4,H) Mirroring 2 Extending Z LBI and Z UBI Mirror axis Angle q 2 [rad] Figure 5.3: Mirroring and extension of the function q 2 In this case the half period T has been mirrored upward to form the entire period T. This whole period can be copied and shifted downward and upward to form the entire t = t(q 2 ; p 1, p 4, H) (here only one downward shift is represented). Now only the time t and q 2 need to be switched to find q 2 = q 2 (t; H, p 1, p 4 ). The solution is shown in figure 5.4. The solution constructed is represented by dots. These dots match the solid line that is found by numerically solving the state space model with ODE Total rotation of q 2 One can find q 2 (t; p 1, p 4, H) by just numerically integrating q 2 from, +. This is possible because there are no zeros of f(q 2 ) that bound the integration. After integration again the form 19

20 5.4. Special cases of q q 2 = q 2 (t;p 1,p 4,H) q 2 Ode45 Z LBI and Z UBI Angle q 2 [rad] Time t [s] Figure 5.4: Periodic movement of q 2 t = t(q 2 ; p 1, p 4, H) is found. The points also need to be switched again, so that q 2 = q 2 (t; p1, p4, H) is found. However on account of reducing the number of calculations q 2 is only numerically integrated over the domain π, +π]. Then it is extended over time. This is possible because after one rotation the motion repeats itself. So by just copying the data over this interval and shifting the solution 2π upwards or downwards this interval accordingly, the total solution can be found. The result for a certain set of initial conditions is shown in figure 5.5. Angle q 2 [rad] (a) q 2 = q 2 (t;p 1,p 4,H) q 2 Ode45 f(q 2 ) x (b) Time t [s] 4 f(q 2 ) 2 i q Angle q 2 [rad] Figure 5.5: Total rotation of q 2 In figure 5.5 a constant rotation of q 2 is found. The solid line shows the solution found by solving the state space model numerically and the dots represent the solution found by calculating it by numerical integration as described in this section. Clearly can be seen that they exactly match. 5.4 Special cases of q 2 There are three different special cases the amplitude doubling, the homoclinic orbit and the heteroclinic orbit (they were described in Sections 4.4 and 4.5) For these cases the solution does not have to be constructed differently. The libration and rotation construction as described will suffice. However these special cases need to be tested. Also can be shown that these cases have been analyzed correctly in Chapter 4. Therefore also the solutions have been calculated with the ODEsolver to show that they are indeed correct. In figure 4.1 the amplitude doubling is represented. This happens when there are three zeros and f(q 2 ) is moving upwards a little so that the "middle one" disappears. This can be seen in figure 5.6: 2

21 5.4. Special cases of q 2 Angle q 2 [rad] i q (t) with ω = i q 2 (t) with ω 2 = q 2 Ode45 Z LBI and Z UBI Z (middle) Time t [s] Figure 5.6: Amplitude doubling The same initial conditions are used in the case of f 3 in figure 4.1 (see Appendix C). f 3 represents a f(q 2 ) with three zeros where one zero was also a minimum of f(q 2 ). Only the initial condition of ω2 i has been slightly changed as is shown in the legend. This minimal change will result in a totally different oscillation. The amplitude has doubled (dots in comparison to the stars) and also the period has changed. The zero at q 2 = π 2 (cross-line) that bounded q 2 has disappeared by just changing ω2 i slightly so that the minimum in q 2 = π 2 no longer is a zero. In the following figure the direct transfer from libration to rotation will be considered. This means that q2 i Z. Angle q 2 [rad] i q (t) with ω = i q (t) with ω = Ode45 q 2 (RelTol,AbsTol = 1 5 ) Ode45 q 2 (RelTol,AbsTol = 1 1 ) Z LBI and Z UBI Time t [s] Figure 5.7: Transition from libration to rotation The same initial conditions are used in the case of f 5 in figure 4.2). f 3 represents a f(q 2 ) with two zeros where both zeros are also a minima of f(q 2 ). Only ω i 2 is minimally changed. First the bounds are Z LBI = π 2 and ZUBI = π 2, the system is oscillating (dots). Then the function f(q 2) is lifted up a little by changing ω i 2 so that there are no zeros of f(q 2 ). This will lead to rotation (stars). About the accuracy and calculation time has not really been spoken yet. The accuracy depends on the number of integration points and integration schemes to calculate the half period in (5.6). 21

22 5.5. Result However because of the copying of this half period the error made in the half period will be copied with it, so if the half period is calculated accurate than it will also be accurate over longer time spans. This is not the case if it is solved with ODE-solvers; then it has to be accurate over the entire time domain leading to errors over long time spans. This can be seen in figure 5.7 at the end of the periodic solution. The solid line begins to differ from the dots even if there already is a very high accuracy applied. It is possible that ODE-solvers calculate a totally incorrect solution. Numerical errors can lead to this incorrect solution. Normally the motion is bounded between q 2 = ± π 2. However because of a numerical error the calculated solution can go over the bounds (dashed lines). As a result the period is totally wrong and the new bounds are incorrect. This cannot happen with solutions that are constructed in this report. The bounds of oscillation are calculated analytically. Finally look at the solution of nearly homoclinic and heteroclinic orbits that already have been explained. 1 (a) i q (t) with q = π/ x (b) 2 3 i 5 q (t) with q = π/ Ode45 q 2 Z LBI and Z UBI 8 6 Angle q 2 [rad] 4 5 f(q 2 ) Time t [s] 2 Almost Homoclinic orbit Almost Heteroclinic orbit with f(π/2) = e 12 Z q 2 Figure 5.8: Nearly homoclinic and heteroclinic orbits The f(q 2 ) can be seen again in figure 5.8b where the initial conditions are ω2 i = and q2 i π 2 so it is very close to an equilibrium point (table 4.1). This means that both solutions almost go through a homoclinic orbit (dots) and heteroclinic orbit (stars). The period T will become very big i π 2, T ). The near homoclinic orbit (dots) will go from close to the equilibrium point (q2) i back to the same equilibrium point (q 2 = π 2 2π) taking a long time. In the beginning of the orbit the angular velocity is small. Then just like in an inverted pendulum the angular velocity will increase when it falls "down". Then it will slow down again and approach the equilibrium again very slowly. The same can be observed in figure 5.8a. The near heteroclinic orbit (stars) will go from q 2 = π 2 to q 2 = 3π 2 (q 2 = π 2 ). However instead of the oscillation going up again as you would expect, it will move down after some time. This is because of the perturbation of q2, i as a consequence the bound Z LBI = π 2 is not present anymore (see legend figure 5.8b). It means that it will also go towards q 2 = π 2 2π. There only has to be made one last comment about the solution that has been constructed. When integrating over a domain that is almost flat (nearly homoclinic and heteroclinic orbits) the result will show some overshoot (error) as can be seen in figure 5.8a. However this will have disappeared at the end of T 2 because q 2 will be correct there. This is why it has been decided that this is acceptable. 5.5 Result In this chapter it has been explained how to construct the solution for q 2 = q 2 (t; H, p 1, p 4 ). Also the solutions for q 2 have been shown compared to the solutions found with ODE45, to show that the solutions are indeed correct. All this has been combined into an algorithm wherein only the time and the initial conditions need to be given to find q 2 (q 2 = q 2 (t; q2, i ω1, i ω2, i ω4)). i Remember that the constants of motion are expressed in the initial conditions. Now by substituting q 2 = q 2 (t; q2, i ω1, i ω2, i ω4) i into (3.22), (3.18) and (3.23) also solutions for ω 1, ω 2 and ω 4 are found 22