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by: Jada Daniel

AircraftFlightDynamics&ControlI MEM453

Marketplace > Drexel University > Mechanical Engineering > MEM453 > AircraftFlightDynamics ControlI
Jada Daniel
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This 49 page Class Notes was uploaded by Jada Daniel on Wednesday September 23, 2015. The Class Notes belongs to MEM453 at Drexel University taught by Staff in Fall. Since its upload, it has received 63 views. For similar materials see /class/212409/mem453-drexel-university in Mechanical Engineering at Drexel University.

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Date Created: 09/23/15
Flight Dynamics amp Control Equations of Motion of 6 dof Rigid AircraftDynamics Harry G Kwatny Department of Mechanical Engineering amp Mechanics Drexel University Outline 0 Angular momentum o Dynamical equations of motion 0 Qualitative introduction to aero dynamic forces 0 General forms of aerodynamic forces and moments Angular Momentum particle De nition Let 0 and P denote two points xed in a rigid body and let r denote location of P relative to 0 If f is a force applied at P then the moment of force about 0 or torque about 0 is de ned as T r x f De nition If a particle of mass m is located at P and the particle has inertial velocity V then its moment of momentum about 0 or angular momentum about 0 is de ned as H m r x V Angular Momentum rigid body Suppose O is the origin of a body xed reference frame in a 1i gid body of volume quot 0 and mass distribution dm Let v0 denote the inertial velocity of O and n the angular velocity of the n gid body Then the angular momentum of the rigid body about 0 is de ned as HLjrgtltv0 oogtltrdm Example Torque in Body Coordinates 0 Z y 1i nyyfz mm 0 x fy 2 fo y x 0 f y xfy Example Rigid Body Angular Momentum in com Coordinates p py2227quirxz wbq fbabrb 7pxyqx2zziryz r 7pxziqyzrx2y2 b r x y z O is the com so J17 rbdm 0 D 2 2 2 7 x incz Py 4y IX IV I p HAJ17 7pxyqx2zziryz dm Iv Iy Iyz q I Iyz 12 r 7pxziqyzrx2y2 IXJvy2zzdm IyJvx2zzdm Izjvx2y2dm IvJvxydm IxzJvxzdm 57sz xizsymmetry lyz0lm0 Equations of Motion Consider the motion of a rigid body Let v denote its inertial velocity at its center of mass and HH its angular momentum about its center of mass Then Newton39s second law states d mv F dlt n 2 d H M dt 2 Suppose that all vector quantities v H F M are speci ed in body coordinates then mv m f v ZFb d H 17H17 7 M17 E n w n 2 Now Hg Ibwl Assume mI are constant drop subsuprscripts Aircraft Dynamics in Euler Angles Force Equations m 0 0 L2 rvi qw 731719 0 m 0 3 m pwi ru mg 0039mm 0 0 m w qui pv cos cosy Moment Equations 0 71H 39 542 qu L 1y 0 My 1271Xrq1 r27p2 MMTT r39 0 12 IX 71ypqlxzqr N Z XT Y Longitudinal Aerodynamic Forces Lw Wing angle of attack aw 0 iW Tail angle ofattack at 0tit 85 ft Vt9 8 is downwash angle 139 i are incidence angles of wing and tail w t Yaw Forces and Moments u v The principle aerodynamic yaw forces are divide into two parts 1 those due to wings and fuselage and 2 those produced by the vertical tail Wings amp Fuselage Lw cl nm Dw Venical Tail CDW mews A CLV av PVZSV Di Cov WWVZSV 0 45 17 sidewash due to Wing in uence Roll Forces Roll moment has four primary sources vertical tail lift produces restoring moment fuselage if wings are high produces restoring moment low produces destabilizing moment wing dihedral F produces restoring moment tail dihedral also but o en negligible ailerons 6a Forsmall dihedral angle Lne change n mdgle ofattack is Vsrn lquot Lhaeby producing a positive roll moment Aerodynamic Forces General Form 1 In general the aerodynamic forces and moments are expressed in terms of aerodynamic coefficients 0 Forces X CXQS axial force Y CYQS side force Z CZQS normal force 0 Moments L CLQSI rolling moment M CMQSI pitching moment N CAQSI yawing moment Q 39 pV22 dynamic pressure 8 39 characteristic area wing surface I 39 characteristic length bwing span or crnean chord Aerodynamic Forces General Form 2 cXa6cxa a cy a855 pr CM a Cm 15 cm 15 c ap C 00 Cza75 Cza75 C a85 npr cw a8 cm 15 c 005 CP apc ar CMaqu5 Cmav5 C q0 q 0 mp 5W5 r CW a cm 005 cm 15 ziVcP ap CW 00 Notes 1 In highly maneuverable fighters large sideslip velocities may impose nonlinear 3 dependence and 3 crosscoupling with longitudinal forces and moments Also large pitch rates may induce pitch rate affects 39n longitudinal forces 2 For aircraft with large changes in Mach number such as the aerospace plane all coefficients may require Mach number degendence Flight Dynamics amp Control Dynamics Near Equilibria Harry G Kwatny Department of Mechanical Engineering amp Mechanics Drexel University Outline o Program 0 Straight amp Level Flight the primary steady state Perturbation Equations for Straight amp Level Flight o Longitudinal amp Lateral Dynamics o Coordinated Turn 0 Steady Sideslip Program o Determine trim conditions 0 Define and compute steadystate flight conditions 0 Straight and level flight cruise climb descend o Coordinated turn o Steady sideslip 0 Determine perturbation models 0 Small perturbations from a trim condition 0 Examine dynamical behaviors Reduction in Dimension coordinates xs yszs 6 l velocities u vw p qr gt 12 states Suppose I the earth is at and 2 we ignore variations of air density the dynamics are invariant wrt location XS ys ZS the dynamics are invariant wrt with respect to inertial heading Consequently we could drop xs ys ZS l and study an 8 state system 0 Notes 0 These properties do not require linearity 0 These omitted variables can always be included by adding the appropriate kinematic equations we often include 23 andor 1 Reduced Equations m 0 0 u rv qw sm6 XT 0 m 0 i2 m pw ru mg cos smgz Y 0 0 m w qu pv cos600s Z 1x 0 1xz p Iy Izqr1xzpq L 0 1y 0 q Iz Ixrq1xzr2 p2 METT Ixz 0 12 r Ix 1ypq1xzqr N 1 sin tan 6 cos 15 tan 6 p 0 cos sin q 0 sin sec cos sec6 r Eq u i i bria The equations are organized in rst order standard form involving the state vector x input vector u and output vector y E xx f state equations y h x u output equations A set of values x0 uo yo is called an equilibrium point if they satisfy 0 f x0 uo yo h xO uo We are interested in motions that remain close to the equilibrium point The Primary Steady State Straight amp Level flight The straight and level ight condition requires equilibrium ight along a linear path with constant ight path angle 2 constant velocity V zero sideslip and wings level Thus we impose the following conditions for straight and level ight Equilibrium u0v0w0V0a0 0 p0q0r0 090y20w0 Outputs speed V V ight path angle 7 6 a 7 sideslip 0 roll 0 Straight amp Level Flight Proposition An equillibrium point satisfying the straight and level ight conditions eXists if and only if there exists 0 6 T which satisfy the equations 9 XVa6 mgsina7T 0 ZVa6mgcosa70 MVa 5 013TT 0 e In this case the equilibrium values of the states and controls are longitudinal variables states V Va aiq 209 f 0 controls 66 2 1 T lateral variables states 3 Op Or 0 0 controls 6 060 0 Linearization Define xt x0 6xtut u0 6utyt y0 6yt Ex06amp fx0 6xtu0 6ut yo 6yt hx0 6xtu0 6ut Now construct a Taylor series for f h afxo uo ax VV The equations become fx06xu06ufxou0 6XW6uhot hx0 6xu0 6u hx0u0W6xm61thot Notice thatfx0u0 0 and hx0u0 yo SO 6f xu 6f xu Exo5x a 05x a 05 E6XA6xB u 6 2C6 D6 Perturbations for Straight amp Level Flight longitudinal equations mcosa mVsina mVsina 0 AV msina chosa chosa 0 d Ad 0 0 1y 0 E q 2 0 0 0 1 A6 QSCX 0552 QSCW 0552 0 mg c056 AV QSCZ 0552 QSCw 0552 Vcosa 0 A0 QSZCM 0562 QSZCW 0562 QSZCmq 0552 0 q 0 1 0 A6 1 QSCMZ 0552 0 QSCZ520 Z AT T QSCM5205E A52 0 0 Perturbations for Straight amp Level Flight lateral equations mV 0 0 0 3 0 1x 1xy 0 i p Z 0 1y 12 0 dt r 0 0 0 1 QSCy a QS Cy a QS C 00 u mgcos 3 QS Cl m QS C a Q p Qszcn a QS Cnp a QS 0 r Longitudinal Dynamics Near straight and level ight the general equations of motion may be divided into two sets which are almost decoupled Longitudinal Dynamics Motion in body XZ plane without yawing or rolling v 0 0l Op Or 0 Longitudinal Variables u w q 9 x5 y the x 2 plane is a plane of symmetry Uncoupled longitudinal motions eXist provided rotor gyroscopic effects are absent the at earth approximation is valid m 0 u qw sin 9 X Force equations m mg 0 m w qu cos 9 Z Moment equations lye M cx cos 9 sin 9 u Kinemat cs 9 6 2 Sin 9 cos 9 v s uuuuuuuu w Lateral Dynamics Lateral Dynamics motion in body Xy plane no pitching w z 0u zu6 6q 00 z 01 Lateral Variables v p r 11 yS motions are smalltrajectory remains close to straight amp level aerodynamic crosscoupling terms are negligible Uncoupled lateral motlons eX1t prov1ded rotor gyroscop1c effects are ab sent the at earth approximation is valid Force equations mx mru mg cos 6sin Y 1 4n 1397 L Moment equatlons XZ Z r N Kinematics 1 COS wan 6 p 11 cos sec 6 r ys cos 6 sin u sinq sin 6 sinwcos cosyv umvmsnv Banked Coordinated Turn The banked turn is de ned by the following conditions 1 equilibrium Va 8 vw pq r are constant rv qw sin9 XT 02m pw ru mg cosHsim Y qu pv cos600s Z Iy Izqr1xzpq L 0 IZ Ixrqlxzr2 p2 METT 1x 1ypq1xzqr N Banked Coordinated Turn2 2 banked turn condition the inertial angular velocity is vertical and constant 0 p sinlt9 as 0 gt q cos6sin 0 0 r cos6cos 3 coordinated turn condition sum of gravity and inertial forces lie in plane of symmetry x 2 plane mpw mru mgcos 9sin 0 gt chos sina eros cosa gcosi9sin 0 4 climb conditions V V 7 7 Banked Coordinated Turn 3 There are 12 equations in 12 unknowns Vaaa ap3q3r363 3T36e36a36y The fact that the velocity is constant in body frame with constant angular velocity about 28 inusures that ground track is circular The coordinated turn condition insures that pilot and passengers will not experience any side forces A pilot achieves coordination by using the rudder in conjunction with an instrument called a turn coordinator which measures the difference between the inertial and gravity forces acting along the y axis The rudder also induces a moment which counteracts the adverse yaw moment resulting from increased decreased drag on the outside inside wing produced by aileron position and which can be significant during the rolling phase ofthe turn Banked Coordinated Turn 4 5 equations explicitly yield 5 unknowns pqrVa or 6 in terms of the others so these can be eliminated leaving chos sina eros cosagcos6sin 0 W qw sim9 X T m 0 mg 0 Y 0 qu pv cos6c0s Z 5 Izqr 1xzpq L Iz Ixrq1 r2 p2 M TT 0 Ix Iypql qr N with p q r known functions of 0 Banked Coordinated Turn 5 Consider the possibility of solutions in which a is small R large Proposition There eXists a solution corresponding to a 0 if and only if there exist 05 5 T satisfying 3 XVot5 mgsinot T 0 ZVa5mgcosot 0 MVa 5 0 TT 0 3 This is the case of straight and level ight in which the equilibrium values of the states and controls are In this case the equilibrium values of the states and controls are longitudinal variables states V 2 Via 0561 06 f 0f controls 52 5T 2 Ti lateral variables39 states 017 0r 0 0 controls 5r 2 05 0 Banked Coordinated Turn 6 Proposition Suppose 026 Ti constitute a regular solution of the longitudinal equilibrium equations and CW 0 Cys a 0 det Cl a C1539a C15aa 0 Cn a Cn a Cn 0 Then there exists a solution to the coordinated bank turn equations with 6 6a small and 0562 T near a6T To rst order terms the solution is la 0 cm ma 0 0b CW Clara Clara 6 Clo ma AW cm CM CM a CM ma Cm Aa CM 00 0 C152 A e CqlW5m sec 1Clwl Crosswind The steady sideslip is a flight condition that may be used during landing approaches in the presence of a crosswind It is an equilibrium condition in the sense that all accelerations and angular rates are zero In addition the aircraft tracks a linear ground path corresponding to ys0 ie aligned with the X8 axis Flight path angle and speed are also specified Adverse wind conditions were involved in 33 of 76 landing accidents between 1984 and The key issue with the crabbed approach is the post touchdown on the ground dynamics Some combination is generally used crabbed sideslip umvmsnv Crosswind 2 Consider an inertial flat earth fixed frame with x y coordinates on the surface and zx down The intent is to land along the xx aXis so we require y 0 Assume the crosswind velocity is v50 Thus we have the following conditions Regulated outputs M03 y vs0 Vcos cosotcos6sinlIVsin sin sin6sinlcos cosl Vcos sinotcos sin6siny sin cosl O VV y 6 a 7 Equilibrium conditions VQdQ QqQ9QpQfQ QwO Crosswind 3 Proposition crabbing solutions A solution to the ground tracking problem exists with 3 0 0 if and only if there eXist 056 and T which satisfy XVa6 mg sina f T 0 Z Va6mg cosa f 0 MVa 5 013TT 0 e Crosswind 4 Proposition steady sideslip A solution to the ground tracking problem exists with l 0 if and only if there eXist 05 5 and T l which satisfy XVot5 mgsinot 7T 0 ZVa5mgcosot 7 0 MVa50zr 0 and 55 satisfying mgcostingb QSZ Cl 05 C1539Ot5r C15a 0050 0 Cn a Cn539 006 Cn5a 005 0 a Cy a Cy539 006 Cy5a 005 0 0 v 0 cos s1n 0 V Crosswind Example A hypothetical subsonic transport adapted from Etkin has the following data Cy 0168 Cyan 0067 CW 2 0 Cm 0047 C1 0003 CM 004 Cw 203625 CM 2 016 CM 2 0005 Suppose the crosswind is v0 015 V Then the following results are obtained in degrees 3 859437 0121212 6 197409 5 861781 Slip to the left roll to the left hard right rudder left aileron into wind Comments from Airbus Pilots I do not find crosswinds to be anymore challenging in this airplane than any other You have to understand that you cannot quotslipquot this airplane because your are commanding a ROLL RATE with the Side Stick Controller not a BANK ANGLE Here is a suggestion Allow the airplane to do an Auto Land in a crosswind when it is convenient and VFR You will be shocked at the timing of when the airplane leaves the CRAB and applies rudder to align the nose parallel with the runway You think itjust isn39t going to do it and at the very last second it slides it in perfectly I would guess in the last 20 feet or less My technique is just the same as any airplane I39ve ever flown C150 to B767 Crab it down to the flare apply enough rudder to straighten the nose drop the upwind wing to prevent drift land on the upwind main first 5Year Airbus Captain Flight Dynamics amp Control Equations of Motion of 6 dof Rigid AircraftKinematics Harry G Kwatny Department of Mechanical Engineering amp Mechanics Drexel University Outline 0 Rotation Matrix 0 Angular Velocity o Euler Angles o Kinematic equations Consider two reference frames a space frame s and a body frame b with common origin R00 Let SW sy 52 be an orthonormal basis for the space frame Let bw by b2 be an orthonormal basis for the body frame The orientation ofthe body frame is specified relative to the space frame if the basis vectors bw by bl are specified in the coordinates ofthe space frame bx bnsx Haysy busz by byxsx IWsy b sz b2 1 s bzysy bzzsz zxx Rotation Matrix 1 b 1 xx Ky xz L byx I by bbb zx zy zz Properties of the Rotation Matrix 1 LT converts body coordinates to space coordinates To see this suppose a vector r has coordinates rbx rw r52 in the body frame and rm rby r172 in the space frame r axbxmyb rtbz rusxmysytrxzsz 39 39 T T T multiply success1ve1y by sX sy sz to prove a 17R 17 bu rm 7 7 T b rxy 7 by bw I72y rb ltgtr iLr 42 by by bu r52 2 Orthonormal unit vectors gt LTL 1 LT L 1 3 Right hand coordinate system gt detL 1 4 Translation plus rotation R R LTrb R R5 LTrb LTib S uccess Ive Rotations Suppose a succession of rotations are made say Ll then L2 then L3 then the total rotation is de ned by LT LzLiLlT Example cos 11 sin11 0 t y l Rotation of angle 11 about z axis Ll isin11 cos 11 0 1 39 0 0 1 1 cos 8 0 7 sin19 2 Rotation of angle 8 about yaxis LZ 0 l 0 sinS 0 cos 8 l 0 0 3 Rotation of angle 11 about xaxis L3 0 cos sin I 0 7s1n cos Euler Angles Consider a reference frame xed in the body With origin located by the position Vector RU and angular orientation denoted by L both relative to a xed inertial space frame L can be parameterized by the Euler angles Ll 1997 Claw pitch roll representing sequential rotations about the axes z y x respectively cosl9cosL1 cosl9sinL1 isinB LLl9 sin sin Bcosyi cos sinL sin sinl9siny1 cos cosL sin cosl9 cos sinl9 cosysin sinL cos sinl9 sing7 sin cosL cos cos l9 2 Standard coordinate frame employs 321 or zyx convention for defining Euler angles Angular Velocity 1 Consider a rotation L gt L AL De nition L limA L Natl A De nition A square matrixA is called antisymmetric ifAT 7A Example A 3gtlt 3 antisymmetric matrix has the general form 0 c b c 0 7a 7b a 0 Notice that there are only 3 independent elements a 175 In this sense every 3gtlt 3 matrix is equal to a 3vector We use the notation 0 c b v and 7 c 0 7a c 7b a 0 De nition The cross product oftWo 3vectors uv is ugtltv N Angular Velocity 2 Proposition LIZ and Z LT and antisymmetric matrices Proof LAL L AL LTLLTALALTLALTAL but L L1LAL LAL1 3 L ALAL LAL AL 0 3 LTZZTL 0 3 H 3LTZT Definition De ne the antisymmetric matrices aquot ZTL 257 LLT I N a 6247 N cob are the angular Velocity in space and body coordinates respectively Note a L QL LQL J SLTZ7cZ47L 3R R i i zrrb i erb RgcZJSL r Lquot R wSxL r L ib Velocity in Body and Space Coordinates Consider a body frame b and a space frame s With common origin and the only relative motion is rotation Ifr is the position vector of any point xed in the body then r I LT Orb r5 constant vs 0 rs r LT W LT w r as W x w ox rs 0 Similarly vb r Low 0 mm w r mm M W t 5 W t 5 ox r 0 Translation Rotation RRnrlt2R Rfr lt31quot R rquot Inertial velocity in space coordinates V5 R LT quot Rg bj quot R agtgtltLbrquot Inertial velocity in body coordinates VquotLRS LRLLT bLR L gLT bLR grbLR cqgtltrb Euler Angle Kinematics Recall the fundamental kinematic relationship L I 7031 ZL I V5 de ne the coordinate vector q 6 11 51Tqwb l sin tan9 cos tan6 l 0 isinq PM 0 cos isinq F 1q 0 cosq cosBsin 0 sin sec9 cos sec6 0 isinq cosBcos Kinematic Equations Summary 5 cosBcosLl 39 a 39 y Luv 39 a Luv Luv u y cosBsinLl y w 39 a 39 39 y my 39 D 39 v i 7515119 00519515197 0051900597 w 1 515197an cosman p 0 cosy ising q 7 9 1 0 singi sec cosq sec r


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