aircraft control Lecture 4
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Transcript of aircraft control Lecture 4
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7/29/2019 aircraft control Lecture 4
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Fall2004 16.33341AircraftDynamics
Notecandevelopgoodapproximationofkeyaircraftmotion(Phugoid)using
simple
balance
between
kinetic
and
potential
energies.
Consideranaircraft insteady, levelflightwithspeedU0 andheight
h0. ThemotionisperturbedslightlysothatU0 U =U0+u (1)h0 h=h0+h (2)
Assume that E = 1mU2 +mgh is constant before and after the
2perturbation. Itthenfollowsthatu ghU0
FromNewtonslawsweknowthat,intheverticaldirectionmh=L W
1whereweightW =mgandliftL=2SCLU2 (S isthewingarea).
Wecanthenderivetheequationsofmotionoftheaircraft: 1
mh=
L
W
=
SCL(U
2 U0
2
)
(3)21= SCL((U0+u)2 U02)1SCL(2uU0)(4)22
ghU0 =(SCLg)h (5) SCL
U0 Sinceh= hand for theoriginalequilibriumflightconditionL=1W =2(SCL)U2 =mg,wegetthat0 2
SCLg g=2
m U0Combinetheseresulttoobtain:
h+2h=0 , g2U0
These equations describe an oscillation (called the phugoid oscilla-tion)ofthealtitudeoftheaircraftaboutitnominalvalue.Onlyapproximatenaturalfrequency(Lanchester),butvalueclose.
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k)
Fall2004 16.33342 Thebasicdynamicsare:
I I
F =mvc and T =H1
B vc TransportThm.F =vc + BIm
B
BI T =H + H
Basicassumptionsare:
1.Earthisaninertialreferenceframe2.A/Cisarigidbody3.BodyframeBfixedtotheaircraft(i,j,
BI Instantaneousmappingofvc and intothebodyframe:BI
=Pi+Qj+Rk vc =Ui+Vj+Wk P U BIB =Q (vc)B = VR W
By symmetry, we can show that Ixy = Iyz = 0, but value of Ixzdepends on specific frame selected. Instantaneous mapping of theangularmomentum
H=Hxi+Hyj+HzkintotheBodyFramegivenby
Hx Ixx 0 Ixz PHB =Hy= 0 Iyy 0 Q
Hz Ixz 0 Izz R
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Fall2004 16.33343 Theoverallequationsofmotionarethen:
1 B
F = vc +BI
vcm X U 0 R Q U
mY = V + 0 P V RZ W Q P 0 W
U + QWRV = V + RUPWW+ PV QU
B
BIT = H + H
L IxxP + IxzR 0 R Q Ixx 0 Ixz PM = IyyQ + 0 P 0 Iyy 0 QR
P 0 Ixz 0 Izz RN IzzR+ IxzP Q IxxP + IxzR +QR(IzzIyy) + PQIxz
= IyyQ +PR(IxxIzz) + (R2P2)Ixz
IzzR+ IxzP +PQ(IyyIxx) QRIxz Clearly theseequations arevery nonlinearand complicated, and we
havenotevensaidwhereFandTcomefrom. = Needtolinearize!!Assumethattheaircraft isflying inanequilibriumconditionand
wewilllinearizetheequationsaboutthisnominalflightcondition.
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Fall2004 16.33344Axes
ButfirstweneedtobealittlemorespecificaboutwhichBodyFrame
wearegoinguse. Severalstandards:1.BodyAxes - Xalignedwith fuselagenose. Zperpendicular to
Xinplaneofsymmetry(down). YperpendiculartoXZplane,totheright.
2.WindAxes- Xalignedwithvc. ZperpendiculartoX(pointeddown).
Y
perpendicular
to
XZ
plane,
off
to
the
right.
3.StabilityAxes- Xalignedwithprojectionofvc intothefuselage
planeofsymmetry. ZperpendiculartoX(pointeddown). Ysame.
RELATIVEWIND
( )
( )
( )
B ODY
Z-AXIS
B ODY
Y -AXIS
X-AXIS
WIND
X-AXIS
S T AB ILITY
X-AXIS
B ODY
Advantagestoeach,buttypicallyusethestabilityaxes.Indifferentflightequilibriumconditions,theaxeswillbeoriented
differentlywithrespecttotheA/Cprincipalaxesneedtotrans-form(rotate)theprincipalinertiacomponentsbetweentheframes.Whenvehicleundergoesmotionwith respect to theequilibrium,StabilityAxesremainfixedtoairplaneas ifpaintedon.
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Fall2004 16.33345 Canlinearizeaboutvarioussteadystateconditionsofflight.Forsteadystateflightconditionsmusthave
F Faero+ Fthrust = 0 and T = 0 = Fgravity+ 3Soforequilibriumcondition,forcesbalanceontheaircraft
L= W andT = D AlsoassumethatP = Q= R= U = V = W = 0
Imposeadditionalconstraintsthatdependonflightcondition:3Steadywings-levelflight = = = = 0
Key Point: While nominal forces and moments balance to zero,motion about the equilibrium condition results in perturbations totheforces/moments.RecallfrombasicflightdynamicsthatliftLf = CL0 where:03CL =liftcurveslopefunctionoftheequilibriumcondition30 =nominalangleofattack(anglethatwingmeetsairflow)But,asthevehiclemovesabouttheequilibriumcondition,would
expectthattheangleofattackwillchange= 0+
ThustheliftforceswillalsobeperturbedLf = CL(0+ ) = Lf + Lf0
Canextendthisideatoalldynamicvariablesandhowtheyinfluence
allaerodynamicforcesandmoments
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Fall2004 16.33346GravityForces
GravityactsthroughtheCoMinverticaldirection(inertialframe+Z)Assumethatwehaveanon-zeropitchangle0NeedtomapthisforceintothebodyframeUsetheEulerangletransformation(215)
0 sinFg =T1()T2()T3()
0
=mg
sincos
Bmg coscos
Forsymmetricsteadystateflightequilibrium,wewilltypicallyassumethat0,0 =0,so sin0
Fg =mg 0 Bcos0
UseEuleranglestospecifyvehiclerotationswithrespecttotheEarthframe
= QcosRsin = P+Qsintan+Rcostan = (Qsin+Rcos)sec
Notethatif0,then Q Recall: Roll,Pitch,andHeading.
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Fall2004 16.33347Linearization
Definethetrimangularratesandvelocities P Uo
BIo = Q (vc)o = 0 B BR 0
whichareassociatedwiththeflightcondition. In fact, thesedefinethetypeofequilibriummotionthatwelinearizeabout.Note:W0 = 0 sinceweareusingthestabilityaxes,andV0 = 0 becauseweareassumingsymmetricflight
ProceedwithlinearizationofthedynamicsforvariousflightconditionsNominal Perturbed PerturbedVelocity Velocity Acceleration
Velocities U0, U = U0+ u U = uW0 = 0, W = w W = w
V0 = 0, V = v
V = vAngular P0 = 0,Rates Q0 = 0,
R0 = 0,P = p P = pQ= q
Q =q
R= r R = rAngles 0, = 0+ =0 = 0, = =0 = 0, = =
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Fall2004 16.33348
W C.G.
0VT
VT0= U0
XE
X0
X
q
Z
Z0ZE
or
0
0 0
Horizontal
U = U + u
Down
Figure1:PerturbedAxes.Theequilibriumconditionwasthattheaircraftwasangledupby0withvelocityVT0 = U0.Thevehiclesmotionhasbeenperturbed(X0 X)sothatnow = 0+ andthevelocityisVT = VT0.NotethatVT isnolongeralignedwiththeX-axis,resultinginanon-zerouandw.Theangle iscalledtheflightpathangle,anditprovidesameasureoftheangleofthevelocityvectortotheinertialhorizontalaxis.
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Fall2004 16.33349 Linearization forsymmetricflight
U = U0+ u,V0 = W0 = 0,P0 = Q0 = R0 = 0.Notethattheforcesandmomentsarealsoperturbed.
1[X0+ X] = U + QWRV u + qwrvu
m 1
[Y0+ Y] = V + RUPWm
v + r(U0+ u)
pw v + rU0
1 [Z0+ Z] = W+ PV QU w +pvq(U0+ u)m
w qU0 X u 1
1Y = v+ rU0 2m
Z w qU0 3
Attitudemotion:
L IxxP + IxzR +QR(IzzIyy) + PQIxz
M = IyyQ +PR(IxxIzz) + (R2P2)IxzN IzzR+ IxzP +PQ(IyyIxx) QRIxz
L Ixxp + Ixzr 4M = Iyyq 5N Izzr+ Ixzp 6
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Fall2004 16.333410 Keyaerodynamicparametersarealsoperturbed:
TotalVelocity2 2)1/2 U0+uVT = ((U0+u)2+v +w
PerturbedSideslipangle = sin1(v/VT)v/U0
PerturbedAngleofAttackx = tan1(w/U)w/U0
Tounderstandtheseequationsindetail,andtheresultingimpactonthevehicledynamics,wemustinvestigatethetermsX ...N.Wemustalsoaddresstheleft-handside(F,T)Netforcesandmomentsmustbezeroinequilibriumcondition.AerodynamicandGravityforcesareafunctionofequilibriumcon-
ditionANDtheperturbationsaboutthisequilibrium. Predictthechangestotheaerodynamicforcesandmomentsusinga
firstorderexpansioninthekeyflightparametersX X X X Xg
X = U+ W+ W+ +...+ +XcU W W X X X X Xg
= u+ w+ w+ +...+ +XcU W W X calledstabilityderivativeevaluatedateq.condition.U Givesdimensionalform;non-dimensionalformavailableintables. Clearly approximation since ignores lags in the aerodynamics forces(assumesthatforcesonlyfunctionofinstantaneousvalues)
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Fall2004 16.333411StabilityDerivatives
First proposed by Bryan (1911) has proven to be a very effec-
tivewaytoanalyzetheaircraftflightmechanicswellsupportedbynumerousflighttestcomparisons.
Theforcesandtorquesactingontheaircraftareverycomplexnonlin-earfunctionsoftheflightequilibriumconditionandtheperturbationsfromequilibrium.Linearizedexpansioncaninvolvemanytermsu, u,...,w, w,...u, w, Typicallyonlyretainafewtermstocapturethedominanteffects.
Dominantbehaviormosteasilydiscussedintermsofthe:Symmetricvariables: U,W,Q&forces/torques: X,Z,andMAsymmetricvariables: V,P,R&forces/torques: Y,L,andN
ObservationfortrulysymmetricflightY,L,andN willbeexactlyzeroforanyvalueofU,W,QDerivativesofasymmetricforces/torqueswithrespecttothesym-
metricmotionvariablesarezero.
Further(convenient)assumptions:
1.Derivativesofsymmetricforces/torqueswithrespecttotheasym-metricmotionvariablesaresmallandcanbeneglected.
2.Wecanneglectderivativeswithrespecttothederivativesofthemotion variables, but keep Z/w and Mw M/w (aero-dynamic lag involved informingnewpressuredistributiononthewinginresponsetotheperturbedangleofattack)
3.X/q isnegligiblysmall.
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Fall2004 16.333412()/() X Y Z L M N
uvwpqr
00
00
000
000
000
000
000
Notethatwemustalsofindtheperturbationgravityandthrustforcesandmoments
Xg
Zg=mgcos0
0 =mgsin00
Aerodynamicsummary:1A X= XU 0u+ XW 0wXu,x w/U02A Y v/U0,p,r3A Zu,x w/U0,x w/U0,q4A Lv/U0,p,r5A M u,x w/U0,x w/U0,q6A N v/U0,p,r
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Fall2004 16.333413 Resultisthat,withtheseforce,torqueapproximations,
equations1,3,5decouplefrom2,4,6
1,3,5arethelongitudinaldynamicsinu,w,andq
X muZ
=
m(wqU0)
M I
yyq
XgX X + Xcu+ w+U W 0 0 0
ZgZ Z Z Zw+ + Zcu+ w+ q+ WU W Q 0 0 00 0M M M w + M q+ Mcu+ w+U 0 W 0 W 0 Q 0
2,4,6arethelateraldynamicsinv,p,andr
Y m(v+ rU0)
L = Ixxp + IxzrN Izzr + Ixzp
Y Y Y v+0p+ r+ +YcV 0 P R 0
L v+ L0p+ L r+ LcV 0 P R 0
N N Nv+0p+ r+ NcV 0 P R 0
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Fall2004 16.333414BasicStabilityDerivativeDerivation
ConsiderchangesinthedragforcewithforwardspeedU
D = 1/2VT2SCD2 2V2 = (u0+ u)2+ v + wT
VT2 VT2= 2(u0+ u) = 2u0u u 0
VT2 VT2Note: = 0 and = 0v 0 w 0
Atreferencecondition: D VT2SCD=Du u 0 u 2 0 S
2 CD V
T2
= u0 + CD02 u 0 u 0S 2 CD= + 2u0CD0u02 u 0
Note D isthestabilityderivative,whichisdimensional.u
Define nondimensional stability coefficient CDu as derivative ofCD withrespecttoanondimensionalvelocityu/u0
D CDand CD0 (CD)0CD = 1VT2S CDu u/u02 0
So( 0 correspondstothevariableatitsequilibriumcondition.
)
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Fall2004 16.333415Nondimensionalize:
D Su0 CD=
u0 + 2CD0u 0 2 u 0
QS CD= + 2CD0u0 u/u0 0
u0 D= (CDu + 2CD0)QS u 0
Sogivenstabilitycoefficient,cancomputethedragforceincrement. NotethatMachnumberhasasignificanteffectonthedrag:
CDu = CDu/u0 0 =u0a
CDua 0 = M
CDM
where CDM canbeestimatedfromempiricalresults/tables.
Aerodynamic Principles
= Constant
CD
0
Mcr0 1 2
= 0.1CDM
Mach number, M
Thrustforces CTu = CTu/u0 0
Tu 0 = CTu
1u0QS
Foraglider,CTu = 0Forajet,CTu 0Forapropplane,CTu = CD0
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Fall2004 16.333416 Liftforcessimilartodrag
L= 1/2VT
2SCL
L Su0 CL
= + 2CL0 u 0 2 u0 u 0 QS CL
= + 2CL0u0 u/u0 0 u0 L
= (CLu + 2CL0)QS u 0
where CL0 is the lift coefficient for the eq. condition and CLu =MCL
M asbefore. Fromaerodynamictheory,wehavethatCL MCL|M=0CL = = CL
1 M2 M 1 M2M2CLu = 1 M2CL0
Derivatives: Nowconsiderwhathappenswithchangesintheangleofattack. Takederivativesandevaluateatthereferencecondition:Lift: CL
C2L 2CL0Drag: CD = CDmin +eARCD =eARCL
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Fall2004 16.333417 CombineintoX,Z ForcesAtequilibrium,forcesbalance.Usestabilityaxes,so0 =0Include the effect in the force balance of a change in on the
forcerotationssothatwecanseetheperturbations.Assume perturbation is small, so rotations are by cos 1,
sinX = T
D+L
Z = (L+D)C
L
Cx
Cz
z
x
V
Note: = T at t = 0
CD
(t)
(i.e., Static Trim)
So,nowconsiderthederivativesoftheseforces:X T D L
= +L+
TThrustvariationwithverysmall 0 0
Applyatthereferencecondition(=0),i.e. CX Nondimensionalizeandapplyreferencecondition:
CX =
CD +CL02CL0= CL0 CLeAR
CX=
0
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Fall2004 16.333418AndfortheZ direction
Z D L= D
GivingCZ = CD0 CL
RecallthatCM wasalreadyfoundduringthestaticanalysis Canrepeatthisprocessfortheotherderivativeswithrespecttothe
forwardspeed. Forwardspeed:
X T D L= +
u u u uSothat
u0 X u0 T u0 D=
QS u 0 QS u 0 QS u 0 CXu CTu (CDu + 2CD0)
SimilarlyfortheZ direction:Z L D
= u u u
Sothat u0 Z u0 L=
QS u 0 QS u 0(CLu + 2CL0)CZu
M2=
1 M2CL0
2CL0
Manymorederivativestoconsider!
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Fall2004 16.333419Summary
PickedaspecificBodyFrame(stabilityaxes) from the listofalter-
nativesChoice simplifies some of the linearization, but the inertias now
changedependingontheequilibriumflightcondition.
Since the nonlinear behavior is too difficult to analyze, we needed
to consider the linearized dynamic behavior around a specific flightconditionEnablesustolinearizeRHSofequationsofmotion.
Forcesand
moments
also
complicated
nonlinear
functions,
so
we
lin-earizedtheLHSaswell
Enablesustowritetheperturbationsoftheforcesandmomentsintermsofthemotionvariables.Engineering insightallowsustoarguethatmanyofthestability
derivatives that couple the longitudinal (symmetric) and lateral(asymmetric)motionsaresmallandcanbeignored.
Approachrequiresthatyouhavethestabilityderivatives.Thesecanbemeasuredorcalculatedfromtheaircraftplanform
andbasicaerodynamicdata.