FLÁIO SILESTRE DYNAMICS & CONTROL OF FLEXIBLE AIRCRAFT LECTURE NOTES LAGRANGIAN MECHANICS APPLIED TO RIGID-BODY DYNAMICS IMAGE CREDITS: BOEING
FLÁIO SILESTRE Introduction Lagrangian Mechanics shall be applied to derive the EoM of the flexible aircraft. However, to get used to the equations, we shall start with the already known EoM of a rigid aircraft (supersonic revision using Newtonian Mechanics). In this part, we will formulate the rigid body aircraft dynamics with Lagrangian Mechanics determining potential and kinetic energy expressions in terms of generalized coordinates. Lagrange s equations shall be enforced - and you will rapidly discover that you will need a symbolic mathematics computational tool for, for example, Mathematica.
FLÁIO SILESTRE Consider the following aircraft as a rigid-body Application of Lagrangian Mechanics! is the angular velocity relative to the inertial frame is the aircraft velocity relative to the inertial frame dm is a mass element p is the position of the mass element relative to the centre of mass R is the position of the mass element relative to the origin of the inertial frame x b roll pitch CM yaw z b y b longitudinal axis
FLÁIO SILESTRE Application of Lagrangian Mechanics ATTENTION! Observe that every vector may be written in any reference frame! x b roll pitch y b I.e. although! is the angular velocity relative to the inertial frame, is does not necessarily BE WRITTEN in the inertial frame. CM z b yaw longitudinal axis Example: in the body frame! b = p q r
FLÁIO SILESTRE Lagrange s Equations Consider a dynamical system of n degrees of freedom. Lagrange s EoM read: d dt @L @L = Q j, j =1,...,n @ q j @q j L is the Lagrangian, defined as L = T Q j are the generalised loads, defined as Q j = f I @ R I @q j U d volume integral distributed force, written in the inertial frame element mass position, written in the inertial frame
FLÁIO SILESTRE Degrees of freedom three translations: components of R 0 in the body frame R 0 b = x CM y CM x b roll pitch CM y b longitudinal axis z CM yaw three rotations: Euler angles z b,, 6
FLÁIO SILESTRE Kinetic Energy Kinetic energy of a mass element dt = 1 dr dt dr dt dm Integrating in the volume of the vehicle T = 1 dr dt dr dt dm d( ) dt is the operator time derivative relative to the inertial frame however, the vector does not need to be written in this frame!!! 7
FLÁIO SILESTRE Kinetic Energy The derivative of R relative to the inertial frame is x b roll pitch y b CM longitudinal axis dr dt = dr 0 dt + dp dt z b yaw = + p t +! p ( ) t is the operator time derivative relative to the body frame 8
FLÁIO SILESTRE Kinetic Energy Thus + pt +! p + pt +! p T = 1 dm or expanding T = 1 dm + 1 (! p) (! p) dm + 1 p t p t dm + T T 1 p t dm + T (! p) dm T Let s analyse each term. T + p t (! p) dm T 6 9
FLÁIO SILESTRE Kinetic Energy Let s analyse each term. T 1= 1 = 1 = 1 m dm dm 10
FLÁIO SILESTRE T = 1 = 1 = 1 = 1 = 1!T (! p) (! p) dm (p!) (p!) dm (X p!) (X p!)dm (X p!) T (X p!)dm = 1!T J! X T p X p dm! J = X p = Kinetic Energy antisymmetric matrix 0 z y z 0 x y x 0 inertia matrix J xx J xy J xz J yx J yy J yz J zx J zy J zz 11
FLÁIO SILESTRE Kinetic Energy T = 1 =0 p t p t dm x b roll pitch y b T = p t dm CM yaw longitudinal axis =0 z b T 6 = =0 p t (! p) dm rigid-body approximation 1
FLÁIO SILESTRE Kinetic Energy T = (! p) dm x b roll pitch CM y b longitudinal axis = apple (! p) dm yaw =! p dm z b =0 1
FLÁIO SILESTRE Final kinetic energy expression (assuming rigid-body) is Kinetic Energy T = 1 m + 1!T J! remember our definition of degrees of freedom q j to apply Lagrange s EoM, energy expressions must be written in terms of these variables and their time-derivatives! Let s now look into the potential energy of the system! 1
FLÁIO SILESTRE Potential Energy Potential (gravitational) energy of a mass element du = R G dm G i = 0 0 g Integrating in the volume of the vehicle U = R G dm = R 0 G dm p G dm U 1 U 1
FLÁIO SILESTRE U 1 = = R 0 G = mgh R 0 G dm dm x b roll pitch CM y b longitudinal axis yaw U = p G dm z b = =0 p dm G 16
FLÁIO SILESTRE Lagrangian Thus L = T U = 1 m + 1!T J! mgh Let s write L in terms of the chosen variables: x CM, y CM, z CM,,, 17
FLÁIO SILESTRE Lagrangian = dr 0 dt = R 0 t +! R 0 R 0 b = x CM y CM z CM R 0 b t = ẋ CM ẏ CM ż CM! b = p q r = 1 0 sin 0 cos sin cos 0 sin cos cos 18
FLÁIO SILESTRE Lagrangian Besides R 0 i = x 0 y 0 = x 0 y 0 H = L ib x CM y CM z 0 z CM Applying the last equations, L may be written as we want and the left side of Lagrange s EoM can be determined. d dt @L @ q j @L @q j = Q j, j =1,...,n don t get scared!!! 19
FLÁIO SILESTRE Generalised Force Let s turn to the right side of Lagrange s EoM: Q j = f I @ R I @q j d volume integral distributed force, written in the inertial frame element mass position, written in the inertial frame Q j = 0 @L ib f x f y f z 1 A @ @q j 0 @L ib x CM y CM z CM + L ib x y z 1 A d aerodynamics thrust 0
FLÁIO SILESTRE Q j = 0 @L ib f x f y f z 1 A @ @q j 0 @L ib x CM y CM z CM Generalised Force 1 x + L ib y A d z Note that, for the first three dof's: Q xcm = f T x f y L T ib L ib Q ycm = Q zcm = f z f x f y f z f x f y f z T T L T ib L ib L T ib L ib 1 0 0 0 1 0 0 0 1 d = F x d = F y d = F z 1
FLÁIO SILESTRE Equations of Motion Applying for all dof s, may simplifications will appear btw. left and right side. You shall get equations in: ẍ CM = ÿ CM = z CM = = = = it shouldn t be too complicate to transform back to u = v = ẇ = ṗ = q = ṙ =