《航空器的稳定与控制》（英文版）Lecture 4 Aircraft Dynamics

16.333 Lecture 4 Aircraft dynamics . Aircraft non linear eom Linearization -dynamics . Linearization- forces moments o Stability derivatives and coefficients

16.333 Lecture 4 Aircraft Dynamics • Aircraft nonlinear EOM • Linearization – dynamics • Linearization – forces & moments • Stability derivatives and coefficients

Fa2004 16.3334-1 Aircraft Dynamics Note can develop good approximation of key aircraft motion( Phugoid using simple balance between kinetic and potential energies Consider an aircraft in steady level flight with speed Uo and height ho. The motion is perturbed slightly so that U0→U=U70+ ho→h=ho+△h perturbation. It then follows that u N mi t before and after the Assume that e=muz+ mgh is constar e From Newton 's laws we know that. in the vertical direction mh=L-w where weight W= mg and lift L= PSclUZ(S is the wing area) We can then derive the equations of motion of the aircraft mh=l-w OPSCL(U2 PSCL((Uo+u)2-U0x psCi(2uUo)(4 PSCL/94h U ( osce)△h(5) Since h= Ah and for the original equilibrium flight condition L W=3(pSCLU=mg, we get that ScL=2 Combine these result to obtain △+Ω2△h=0,≈V These equations describe an oscillation(called the phugoid oscilla tion of the altitude of the aircraft about it nominal value Only approximate natural frequency (Lanchester ), but value close

Fall 2004 16.333 4–1 Aircraft Dynamics • Note can develop good approximation of key aircraft motion (Phugoid) using simple balance between kinetic and potential energies. • Consider an aircraft in steady, level flight with speed U0 and height h0. The motion is perturbed slightly so that U0 → U = U0 + u (1) h0 → h = h0 + Δh (2) • Assume that E = 1mU2 + mgh is constant before and after the 2 perturbation. It then follows that u ≈ −gΔh U0 • From Newton’s laws we know that, in the vertical direction mh¨ = L − W 1 where weight W = mg and lift L = 2ρSCLU2 (S is the wing area). We can then derive the equations of motion of the aircraft: ¨ 1 mh = L − W = ρSCL(U2 − U0 2 ) (3) 2 1 = ρSCL((U0 + u) 2 − U0 2 ) ≈ 1 ρSCL(2uU0)(4) 2 � � 2 gΔh ≈ −ρSCL U0 = −(ρSCLg)Δh (5) U0 ¨ ¨ Since h = Δh and for the original equilibrium flight condition L = W 1 = 2(ρSCL)U2 = mg, we get that 0 � �2 ρSCLg g = 2 m U0 Combine these result to obtain: Δh¨ + Ω2 Δh = 0 , Ω ≈ g √ 2 U0 • These equations describe an oscillation (called the phugoid oscilla￾tion) of the altitude of the aircraft about it nominal value. – Only approximate natural frequency (Lanchester), but value close

Fa2004 16.3334-2 The basic dynamics are F=mi and th 1F=元+Bb× u. Transport thm m B T=互+B×H e Basic assumptions are 1. Earth is an inertial reference frame 2. A/ C is a rigid body 3. Body frame B fixed to the aircraft(i, j Instantaneous mapping of ie and Blw into the body frame: BW=Pi+Qj+Rk 1c=Ui+Vj+Wk B Uc)B R W By symmetry, we can show that Iru= luz =0, but value of Irz depends on specific frame selected. Instantaneous mapping of the angular momentum H=Hri+hui+hzk into the Body Frame given by H Ix0Ix21「P H h 0I2 yy 0 H 1x201

� k) � � � Fall 2004 16.333 4–2 • The basic dynamics are: �˙ I � ˙ I F = mvc and T� = H 1 �˙ B F = vc + ω × �vc Transport Thm. BI ⇒ � m � ˙ B BI ⇒ T � � = H + ω × H� • Basic assumptions are: 1. Earth is an inertial reference frame 2. A/C is a rigid body 3. Body frame B fixed to the aircraft ( �i,�j, BI • Instantaneous mapping of �vc and ω� into the body frame: BIω� = P�i + Q�j + R� k �vc = U�i + V�j + W� k ⎡ ⎤ ⎡ ⎤ P U ⇒ BIω ⎦ B = ⎣ Q ⎦ ⇒ (vc)B = ⎣ V R W • By symmetry, we can show that Ixy = Iyz = 0, but value of Ixz depends on specific frame selected. Instantaneous mapping of the angular momentum H = Hx �i + Hy �j + Hz � k into the Body Frame given by ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ Hx Ixx 0 Ixz P HB = ⎣ Hy ⎦ = ⎣ 0 Iyy 0 ⎦ ⎣ Q ⎦ Hz Ixz 0 Izz R

Fa2004 16.3334-3 The overall equations of motion are then F Bl × 0-RQ1「U V+R 0-PIv W Q P 0w 「U+QW-Rv V+RU-Pn W+PV-QU T=互+B×H Imap+lzr 0-RQ1「Ix0Ix21「P →M +|R0-P 01m0Q 2+I2P QP0」LIx20 IP+I2R+QR(I2z-Iyy)+ pQlcz Q +PR(Iz -I2)+(R2- p2)Izz I22r+Ix P +PQ(Iwy-Ixa-QRlzz Clearly these equations are very nonlinear and complicated and we have not even said where f and t come from Need to linearize!! Assume that the aircraft is flying in an equilibrium condition and we will linearize the equations about this nominal flight condition

� 1 Fall 2004 16.333 4–3 • The overall equations of motion are then: 1 �˙ B F = vc + BIω� × �vc m ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ X U˙ 0 −R Q U ⇒ m ⎣ Y ⎦ = ⎣ V˙ ⎦ + ⎣ R 0 −P ⎦ ⎣ V ⎦ Z W˙ −Q P 0 W ⎡ ⎤ U˙ + QW − RV = ⎣ V˙ + RU − PW ⎦ W˙ + PV − QU � ˙ B BI T � � = H + ω × H� ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ L ˙ IxxP˙ + IxzR 0 −R Q Ixx 0 Ixz P ⇒ ⎣ M ⎦ = ⎣ ˙ IyyQ˙ ⎦ + ⎣ R 0 −P ⎦ ⎣ 0 Iyy 0 ⎦ ⎣ Q ⎦ N IzzR + IxzP˙ −Q P 0 Ixz 0 Izz R ⎡ ⎤ ˙ IxxP˙ + IxzR +QR(Izz − Iyy) + PQIxz = ⎣ I ⎦ yyQ˙ +PR(Ixx − Izz) + (R2 − P2)Ixz ˙ IzzR + IxzP˙ +PQ(Iyy − Ixx) − QRIxz • Clearly these equations are very nonlinear and complicated, and we have not even said where F� and T� come from. =⇒ Need to linearize!! – Assume that the aircraft is flying in an equilibrium condition and we will linearize the equations about this nominal flight condition

Fa2004 16.3334-4 Axes But first we need to be a little more specific about which Body Frame we are going use. Several standards 1. Body Axes-X aligned with fuselage nose. Z perpendicular to X in plane of symmetry(down). Y perpendicular to XZ plane, to le rig 2. Wind Axes-X aligned with vc. Z perpendicular to X(pointed down). Y perpendicular to XZ plane, off to the right 3. Stability Axes-X aligned with projection of vc into the fuselage plane of symmetry. Z perpendicular to X(pointed down). Same S(BODY) X-AXIS (STABILITY) BODY X-AXIS Z-AXIS (WIND Advantages to each, but typically use the stability axes In different flight equilibrium conditions, the axes will be oriented differently with respect to the a/c principal axes = need to trans form (rotate) the principal inertia components between the frames When vehicle undergoes motion with respect to the equilibrium Stability Axes remain fixed to airplane as if painted on

Fall 2004 16.333 4–4 Axes • But first we need to be a little more specific about which Body Frame we are going use. Several standards: 1. Body Axes ­ X aligned with fuselage nose. Z perpendicular to X in plane of symmetry (down). Y perpendicular to XZ plane, to the right. 2. Wind Axes ­ X aligned with �vc. Z perpendicular to X (pointed down). Y perpendicular to XZ plane, off to the right. 3. Stability Axes ­ X aligned with projection of �vc into the fuselage plane of symmetry. Z perpendicular to X (pointed down). Y same. R E LATIV E WIND ( ) ( ) ( ) � � B ODY Z-AXIS B ODY Y -AXIS X-AXIS WIND X-AXIS S T AB ILIT Y X-AXIS B ODY • Advantages to each, but typically use the stability axes. – In different flight equilibrium conditions, the axes will be oriented differently with respect to the A/C principal axes ⇒ need to trans￾form (rotate) the principal inertia components between the frames. – When vehicle undergoes motion with respect to the equilibrium, Stability Axes remain fixed to airplane as if painted on

Fa2004 16.3334-5 Can linearize about various steady state conditions of flight For steady state flight conditions must have aero gravity 0 and T o So for equilibrium condition, forces balance on the aircraft L=wandT=D Also assume that P=Q=R-U=V=W=0 Impose additional constraints that depend on flight condition ◇ Steady wings-level flight→Φ=∮==业=0 Key Point: While nominal forces and moments balance to zero motion about the equilibrium condition results in perturbations to the forces /moments Recall from basic flight dynamics that lift Lo=Clo ao where O Cl= lift curve slope- function of the equilibrium condition o ao= nominal angle of attack(angle that wing meets air flow But, as the vehicle moves about the equilibrium condition, would expect that the angle of attack will change ao+△a Thus the lift forces will also be perturbed =CLn(a0+△a)=L6+△L Can extend this idea to all dynamic variables and how they influence l aerodynamic forces and moments

Fall 2004 16.333 4–5 • Can linearize about various steady state conditions of flight. – For steady state flight conditions must have F Faero + � Fthrust = 0 and T� = 0 � = � Fgravity + � 3 So for equilibrium condition, forces balance on the aircraft L = W and T = D – Also ˙ ˙ ˙ ˙ ˙ assume that P˙ = Q = R = U = V = W = 0 – Impose additional constraints that depend on flight condition: 3 Steady wings­level flight → Φ = Φ =˙ Θ =˙ Ψ = 0 ˙ • Key Point: While nominal forces and moments balance to zero, motion about the equilibrium condition results in perturbations to the forces/moments. – Recall from basic flight dynamics that lift Lf = CLαα0 where: 0 3 CLα = lift curve slope – function of the equilibrium condition 3 α0 = nominal angle of attack (angle that wing meets air flow) – But, as the vehicle moves about the equilibrium condition, would expect that the angle of attack will change α = α0 + Δα – Thus the lift forces will also be perturbed Lf = CLα(α0 + Δα) = Lf + ΔLf 0 • Can extend this idea to all dynamic variables and how they influence all aerodynamic forces and moments

Fa2004 16.33346 gravity Forces Gravity acts through the CoM in vertical direction(inertial frame +Z) Assume that we have a non-zero pitch angle eo Need to map this force into the body frame Use the Euler angle transformation(2-15) sIn B=T()T2On①)0= g sin重cose 9」 COsΦcos日 For symmetric steady state flight equilibrium, we will typically assume that≡60,①≡Φo=0,so SIn b== mg e Use euler angles to specify vehicle rotations with respect to the earth rame =OcosΦ-Rsin Φ=P+Qsin重tan+BcosΦtan (Qin重+Rcos重)sece Note that if更≈0,then≈Q Reca:≈Rol,≈ Pitch,andy≈ Heading

Fall 2004 16.333 4–6 Gravity Forces • Gravity acts through the CoM in vertical direction (inertial frame +Z) – Assume that we have a non­zero pitch angle Θ0 – Need to map this force into the body frame – Use the Euler angle transformation (2–15) ⎡ ⎤ ⎡ ⎤ 0 − sin Θ Fg = T1(Φ)T2(Θ)T3(Ψ) ⎣ 0 ⎦ = mg ⎣ sin Φcos Θ ⎦ B mg cos Φcos Θ • For symmetric steady state flight equilibrium, we will typically assume that Θ ≡ Θ0, Φ ≡ Φ0 = 0, so ⎡ ⎤ − sin Θ0 Fg = mg ⎣ 0 ⎦ B cos Θ0 • Use Euler angles to specify vehicle rotations with respect to the Earth frame Θ˙ = Q cos Φ − R sin Φ Φ˙ = P + Q sinΦtan Θ + R cos Φtan Θ Ψ˙ = (Q sinΦ + R cos Φ)sec Θ – Note that if Φ ≈ 0, then Θ˙ ≈ Q • Recall: Φ ≈ Roll, Θ ≈ Pitch, and Ψ ≈ Heading

Fa2004 16.3334-7 Linearization Define the trim angular rates and velocities P =Q(v) 0 0 which are associated with the flight condition. In fact, these define the type of equilibrium motion that we linearize about. Note Wo=0 since we are using the stability axes, and Vo =0 because we are assuming symmetric flight Proceed with linearization of the dynamics for various flight conditions Noir Perturbed Perturbed Velocity Velocity Acceleration Ⅴ ecocities +u W=0 V=0 A ngular P0=0, P= Rate Q0=0 Q Q=q R0=0, R R=r angle =60+6 业o=0, 0

Fall 2004 16.333 4–7 Linearization • Define the trim angular rates and velocities ⎡ ⎤ ⎡ ⎤ P Uo BIωo = ⎣ Q ⎦ (vc) o = ⎣ 0 ⎦ B B R 0 which are associated with the flight condition. In fact, these define the type of equilibrium motion that we linearize about. Note: – W0 = 0 since we are using the stability axes, and – V0 = 0 because we are assuming symmetric flight • Proceed with linearization of the dynamics for various flight conditions Nominal Perturbed ⇒ Perturbed Velocity Velocity ⇒ Acceleration Velocities U0, U = U0 + u U ⇒ ˙ = u˙ W ˙ 0 = 0, W = w W = w˙ V0 = 0, V = v ⇒ ⇒ V˙ = v˙ Angular P0 = 0, Rates Q0 = 0, R0 = 0, P = p P˙ = p˙ Q = q ⇒ Q˙ = q˙ R = r ⇒ ⇒ R˙ = r˙ Angles Θ ˙ 0, Θ = Θ0 + θ Θ = θ ⇒ ˙ Φ ˙ 0 = 0, Φ = φ ⇒ Φ = φ˙ Ψ ˙ 0 = 0, Ψ = ψ ⇒ Ψ = ψ˙

Fa2004 16.33348 Oo or yo Horizontal Down Figure 1: Perturbed Axes. The equilibrium condition was that the aircraft was angled up by eo with velocity VTo=Uo. The vehicle's motion has been perturbed(Xo- X)so that now 0=00+0 and the velocity is VT+ vro. Note that Vr is no longer aligned with the X-axis, resulting in a non-zero u and w. The angle y is called the flight path angle, and it provides a measure of the angle of the velocity vector to the inertial horizontal axis

� Fall 2004 16.333 4–8 W C.G. 0 VT VT0 = U0 XE X0 X q � Z Z0 ZE � Θ � or �0 �0 �0 Horizontal U = U + u Down Figure 1: Perturbed Axes. The equilibrium condition was that the aircraft was angled up by Θ0 with velocity VT0 = U0. The vehicle’s motion has been perturbed (X0 → X) so that now Θ = Θ0 +θ and the velocity is VT = VT0. Note that VT is no longer aligned with the X­axis, resulting in a non­zero u and w. The angle γ is called the flight path angle, and it provides a measure of the angle of the velocity vector to the inertial horizontal axis

Fa2004 16.3334-9 Linearization for symmetric fight U=Uo+u, Vo=Wo=0, Po= Qo= Ro=0 Note that the forces and moments are also perturbed AX0+△Ⅺ=U+QW-F 1Y0+△Y]=V+RU-PW +T(U0+)-p≈t+r z0+△Z]=W+PV-QU≈t (U0+ △X △Y 0+rU0 △Z ● Attitude motion L IrrP+IzR +QR(I2z-Iuy)+ PQI, M Iyy Q +PR(Ixz-I2)+(R IzaR+Ix p+PQ(Iwu-Ixx)-QR △L cp+ →△M 5 △N I2+Ix2p」6

Fall 2004 16.333 4–9 • Linearization for symmetric flight U = U0 + u, V0 = W0 = 0, P0 = Q0 = R0 = 0. Note that the forces and moments are also perturbed. 1 [X0 + ΔX] = U˙ + QW − RV u˙ + qw − rv ≈ u˙ m ≈ 1 [Y0 + ΔY ] = V˙ + RU − PW m ≈ v˙ + r(U0 + u) − pw ≈ v˙ + rU0 1 ˙ [Z0 + ΔZ] = W + PV − QU w˙ + pv − q(U0 + u) m ≈ ≈ w˙ − qU0 ⎡ ⎤ ⎡ ⎤ ΔX u˙ 1 1 ⎣ ΔY ⎦ = ⎣ v˙ + rU0 ⇒ ⎦ 2 m ΔZ w˙ − qU0 3 • Attitude motion: ⎡ ⎤ ⎡ ⎤ L ˙ IxxP˙ + IxzR +QR(Izz − Iyy) + PQIxz ⎣ M ⎦ = ⎣ I ⎦ yyQ˙ +PR(Ixx − Izz) + (R2 − P2)Ixz N ˙ IzzR + IxzP˙ +PQ(Iyy − Ixx) − QRIxz ⎡ ⎤ ⎡ ⎤ ΔL Ixxp˙ + Ixzr˙ 4 ⇒ ⎣ ΔM ⎦ = ⎣ Iyyq˙ ⎦ 5 ΔN Izzr˙ + Ixzp˙ 6