Spacecraft And Aircraft Dynamics

Transcription

Spacecraft and Aircraft DynamicsMatthew M. PeetIllinois Institute of TechnologyLecture 9: 6DOF Equations of Motion

Aircraft DynamicsLecture 9In this Lecture we will cover:Newton’s LawsP d Mi dtHP d Fi m vdtRotating Frames of Reference Equations of Motion in Body-Fixed Frame Often ConfusingM. PeetLecture 9:2 / 24

Review: Coordinate RotationsPositive DirectionsIf in doubt, use the right-hand rules.Figure: Positive RotationsFigure: Positive DirectionsM. PeetLecture 9:3 / 24

Review: Coordinate RotationsRoll-Pitch-YawThere are 3 basic rotations an aircraft can make: Roll Rotation about x-axis Pitch Rotation about y-axis Yaw Rotation about z-axis Each rotation is a one-dimensional transformation.Any two coordinate systems can be related by a sequence of 3 rotations.M. PeetLecture 9:4 / 24

Review: Forces and MomentsForcesThese forces and moments have standard labels. The Forces are:X Axial ForceNet Force in the positive x-directionYSide ForceNet Force in the positive y-directionZ Normal Force Net Force in the positive z-directionM. PeetLecture 9:5 / 24

Review: Forces and MomentsMomentsThe Moments are called, intuitively:LMNRolling MomentPitching MomentYawing MomentM. PeetNet Moment in the positive p-directionNet Moment in the positive q-directionNet Moment in the positive r-directionLecture 9:6 / 24

6DOF: Newton’s LawsForcesNewton’s Second Law tells us that for a particle F ma. In vector form:Xd F i m VF dtiThat is, if F [Fx Fy Fz ] and V [u v w], thenFx mdudtFx mdvdtFz mdwdtDefinition 1. mV is referred to as Linear Momentum.L are defined in an InertialNewton’s Second Law is only valid if F and Vcoordinate system.Definition 2.A coordinate system is Inertial if it is not accelerating or rotating.M. PeetLecture 9:7 / 24

Newton’s LawsMomentsUsing Calculus, this concept can be extended to rigid bodies by integration overall particles.X i d H MMdtiDefinition 3.R ( rc vc )dm is the angular momentum.Where HAngular momentum of a rigid body can be found as I HωIwhere ωI [p, q, r]T is the angular rotation vector of the body about thecenter of mass. p is rotation about the x-axis. q is rotation about the y-axis. r is rotation about the z-axis. ωI is defined in an Inertial Frame.The matrix I is the Moment of Inertia Matrix.M. PeetLecture 9:8 / 24

Newton’s LawsMoment of InertiaThe moment of inertia matrix is defined as Ixx IxyI Iyx Iyy Izx IzyIxy Iyx Z Z ZIxz Izx Z Z ZIyz Izy Z Z ZSo Ixz Iyz IzzIxx Z Z ZxzdmIyy Z Z ZyzdmIzz Z Z Zxydm HxIxx Hy IyxHz Izx IxyIyy Izy y 2 z 2 dm x2 z 2 dm x2 y 2 dm IxzpI Iyz qI IzzrIwhere pI , qI and rI are the rotation vectors as expressed in the inertial framecorresponding to x-y-z.M. PeetLecture 9:9 / 24

Moment of InertiaExamples:Homogeneous Sphere 1 02Isphere mr2 0 150 0M. PeetRing 00 1 12Iring mr2 00Lecture 9:0120 00 110 / 24

Moment of InertiaExamples:Homogeneous Disk 1 13 rh21 2 Idisk mr040M. Peet01 13 rh20F/A-18 2302.97015.130 kslug f t20 I 012.97016.992Lecture 9:11 / 24

Problem:The Body-Fixed FrameThe moment of inertia matrix, I, is fixed in the body-fixed frame. However,Newton’s law only applies for an inertial frame:X i d H MMdtiIf the body-fixed frame is rotating with rotation vector ω , then for any vector, a,d aintheinertialframeisdtd ad a dt I dtB ω aSpecifically, for Newton’s Second Law dVF mdtand dHMdtM. PeetB m ω VB ω HLecture 9:12 / 24

Equations of MotionThus we have Fxu̇x̂ Fy m v̇ m det pFzẇuŷqv ẑu̇ qw rvr m v̇ ru pw wẇ pv quand LṗpIxx Ixy IxzIxx Ixy Ixz M Iyx Iyy Iyz q̇ ω Iyx Iyy Iyz q Nṙ Izx IzyIzzr Izx IzyIzz Ixx ṗ Ixy q̇ Ixz ṙpIxx qIxy rIxz Ixy ṗ Iyy q̇ Iyz ṙ ω pIxy qIyy rIyz Ixz ṗ Iyz q̇ Izz ṙ pIxz qIyz rIzz Ixx ṗ Ixy q̇ Ixz ṙ q( pIxz qIyz rIzz ) r( pIxy qIyy rIyz ) Ixy ṗ Iyy q̇ Iyz ṙ p( pIxz qIyz rIzz ) r(pIxx qIxy rIxz ) Ixz ṗ Iyz q̇ Izz ṙ p( pIxy qIyy rIyz ) q(pIxx qIxy rIxz )Which is too much for any mortal. For aircraft, we have symmetry about thex-z plane. Thus Iyz Ixy 0. Spacecraft?M. PeetLecture 9:13 / 24

Equations of MotionReduced EquationsWith Ixy Iyz 0, we have, in summary: Fxu̇ qw rv Fy m v̇ ru pw Fzẇ pv quand Ixx ṗ Ixz ṙ qpIxz qrIzz rqIyyL M Iyy q̇ p2 Ixz prIzz rpIxx r2 Ixz Ixz ṗ Izz ṙ pqIyy qpIxx qrIxzN Right now, Translational variables (u,v,w) depend on rotational variables (p,q,r). Rotational variables (p,q,r) do not depend on translational variables(u,v,w).IFor aircraft, however, Moment forces (L,M,N) depend on rotational andtranslational variables.M. PeetLecture 9:14 / 24

EOMs in Rotating FrameExample: SnipersQuestion: Consider a sniper firing a rifle due east at the equator. Ignoringgravity and drag, what are the equations of motion of the bullet? Use theNorth-East-Up local coordinate system. Muzzle velocity: 1000m/s. Range:4km.rad, orAnswer: The earth is rotating about its axis at angular velocity 2π dayrad.0000727 s . The rotation is positive about the local North-axis. Thus p.0000727 0ω q r0Since the bullet is in free-flight, there are no forces. Thus the Equations ofmotion are 0u̇ qw rvu̇ 0 m v̇ ru pw m v̇ pw 0ẇ pv quẇ pvM. PeetLecture 9:15 / 24

EOMs in Rotating FrameExample: SnipersSimplified EOMs: Using q r 0, we simplify tou̇ 0v̇ pwẇ pv.Solution: For initial condition u(0) 0, v(0) V and w(0) 0 has solutionu(t) 0v(t) v(0) cos(pt)w(t) v(0) sin(pt)Since p is very small compared to flight time, we can approximateu(t) 0v(t) v(0)w(t) v(0)ptWhich yields displacement1U (t) v(0)pt22Conclusion: For a target at range E(ti ) 4km, we have ti 4s and hence theerror at target is:N (t) 0E(t) v(0)t1U (ti ) 2000 .0000727 16 1.1635m2Of course, if we were firing west, the error would be 1.1635m.N (ti ) 0M. PeetLecture 9:16 / 24

Euler AnglesIssue: Equations of motion are expressed in the Body-Fixed frame.Question: How do determine rotation and velocity in the inertial frame. Forintercept, obstacle avoidance, etc.Approach: From Lecture 4, any two coordinate systems can be related througha sequence of three rotations. Recall these transformations are:Roll Rotation (φ) :R1 (φ) 100 0 cos φ sin φ 0 sin φ cos φM. PeetPitch Rotation (θ):R2 (θ) cos θ 0 sin θ10 0 sin θ 0 cos θLecture 9:Yaw Rotation (ψ):R3 (ψ) cos ψ sin ψcos ψ sin ψ00 00 117 / 24

Euler AnglesDefinition 4.The term Euler Angles refers to the angles of rotation (ψ, θ, φ) needed to gofrom one coordinate system to another using the specific sequence of rotationsYaw-Pitch-Roll: BF R1 (φ)R2 (θ)R3 (ψ)V I .VNOTE BENE: Euler angles are often defined differently (e.g. 3-1-3). We usethe book notation.The composite rotation matrix can be writtenR1 (φ)R2 (θ)R3 (ψ) 100cos θ 0 cos φ sin φ 00 sin φ cos φ sin θ010 sin θcos ψ0 sin ψcos θ0 sin ψcos ψ0 00 1This moves a vectorInertial Frame Body-Fixed FrameM. PeetLecture 9:18 / 24

Euler AnglesTo move a vectorBody-Fixed Frame Inertial Framewe need to Invert the Rotations. Rotation matrices are easily inverted, howeverRi (θ) 1 Ri ( θ) I (R1 (φ)R2 (θ)R3 (ψ)) 1 V BF , whereThus V(R1 (φ)R2 (θ)R3 (ψ)) R3 (ψ) 1 R2 (θ) 1 R1 (φ) 1 R3 ( ψ)R2 ( θ)R1 ( φ) cos ψ sin ψ 0cos θ 0 sin θ100cos ψ 0 010 0 cos φ sin φ sin ψ001 sin θ 0 cos θ0 sin φ cos φ cos θ cos ψ sin φ sin θ cos ψ cos ψ sin ψ cos φ sin θ cos ψ sin φ sin ψ cos θ sin ψ sin φ sin θ sin ψ cos φ cos ψ cos φ sin θ sin ψ sin φ cos ψ sin θsin φ cos θcos φ cos θ 1These transformations now describe a Roll-Pitch-Yaw.M. PeetLecture 9:19 / 24

Euler AnglesVelocity vectorThus to find the inertial velocity vector, we must rotate FROM the body-fixedcoordinates to the inertial frame: dx udt dy R3 ( ψ)R2 ( θ)R1 ( φ) v dtdzwdtM. PeetLecture 9:20 / 24

Euler AnglesThe rate of rotation of the Euler Angles can be found by rotating the rotationvector into the inertial frame φ̇p10 sin θ q 0 cos φ cos θ sin φ θ̇ r0 sin θ cos θ cos φψ̇This transformation can also be reversed as φ̇1 sin φ tan θ cos φ tan θp θ̇ 0cos φ sin φ q 0 sin φ sec θ cos φ sec θrψ̇M. PeetLecture 9:21 / 24

SummaryM. PeetLecture 9:22 / 24

ConclusionIn this lecture we have coveredEquations of Motion How to differentiate Vectors in Rotating Frames Derivation of the Nonlinear 6DOF Equations of MotionEuler Angles Definition of Euler Angles Using Rotation Matrices to transform vectors Derivatives of the Euler anglesI Relationship to p-q-r in Body-Fixed FrameM. PeetLecture 9:23 / 24

Next LectureIn the next lecture we will coverLinearized Equations of Motion How to linearize the nonlinear 6DOF EOM How to linearize the force and moment contributionsForce and Moment Contributions The gravity and thrust contributions The full linearized equations of motion including forces and moments How to decouple into Longitudinal and Lateral DynamicsI Reminder on how to create a state-space representation.M. PeetLecture 9:24 / 24

day, or.0000727rad s. The rotation is positive about the local North-axis. Thus ω p q r .0000727 0 0 Since the bullet is in free-flight, there are no forces. Thus the Equations of motion are 0 0 0 m u qw rv v r