Lecture ac-1 Aircraft dynamics Aircraft Anatomy Ailerons Rudder Cockpit Elevators pinner Fuselage Undercarriage Wing Copyright 2003 by Jonathan How
Lecture AC-1 Aircraft Dynamics Copyright 2003 by Jonathan How 1
Spring 2003 16.61AC1-2 Aircraft Dynamics First note that it is possible to develop a very good approximation of a key motion of an aircraft(called the Phugoid mode) using a very simple balance between the kinetic and potential energies Consider an aircraft in steady, level fight with speed Uo and height ho The motion is perturbed slightly so that U=U0+ h=ho+△h (2) Assume that E=5mu+ mgh is constant before and after the pertur bation. It then follows that g△h 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= PSCL(U4-U6 2SC(C+02-6)≈20C2(2u)(4) g△h ( PSCL9)△h Since h= Ah and for the original equilibrium fight condition L=w (pSCL)UZ=mg, we get that ptcL 2 Combine these result to obtain △+92△h=0,g≈√2 These equations describe an oscillation(called the phugoid oscillation of the altitude of the aircraft about it nominal value o Only approximate natural frequency, but value very close
Spring 2003 16.61 AC 1–2 Aircraft Dynamics • First note that it is possible to develop a very good approximation of a key motion of an aircraft (called the Phugoid mode) using a very simple balance between the 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 + ∆h (2) – Assume that E = 1 2mU2 + mgh is constant before and after the perturbation. It then follows that u ≈ −g∆h U0 – From Newton’s laws we know that, in the vertical direction mh¨ = L − W where weight W = mg and lift L = 1 2ρSCLU2 (S is the wing area). We can then derive the equations of motion of the aircraft: mh¨ = L − W = 1 2 ρSCL(U2 − U2 0 ) (3) = 1 2 ρSCL((U0 + u) 2 − U2 0 ) ≈ 1 2 ρSCL(2uU0) (4) ≈ −ρSCL g∆h U0 U0 = −(ρSCLg)∆h (5) Since h¨ = ∆h¨ and for the original equilibrium flight condition L = W = 1 2 (ρSCL)U2 0 = mg, we get that ρSCLg m = 2 g U0 2 Combine these result to obtain: ∆h¨ + Ω2 ∆h = 0 , Ω ≈ g U0 √ 2 – These equations describe an oscillation (called the phugoid oscillation) of the altitude of the aircraft about it nominal value. ✸ Only approximate natural frequency, but value very close.
Spring 2003 16.61AC1-3 The basic dynamics are the same as we had before F B× Transport Thm +T=H+BIw x H Note the notation change Basic assumptions are 1. Earth is an inertial reference frame 2. A/C is a rigid bod 3. Body frame B fixed to the aircraft(i, j, k) Instantaneous mapping of vc and BW into the body frame is given by Qi+Rk vc=0i P U O R W By symmetry, we can show that Ixy ly2=0, but value of Izz depends on specific frame selected. Instantaneous mapping of the angular momentum H=Hri+Hui+izh into the Body Frame given by H lr0 1 H H Hx Izz 0 I2R
Spring 2003 16.61 AC 1–3 • The basic dynamics are the same as we had before: F = m ˙vc I and T = ˙ H I ⇒ 1 m F = ˙vc B + BIω × vc Transport Thm. ⇒ T = ˙ H B + BIω × H Note the notation change • 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, k) • Instantaneous mapping of vc and BIω into the body frame is given by BIω = Pi + Qj + R k vc = Ui + Vj + W k ⇒ BIωB = P Q R ⇒ (vc)B = U V 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 HB = Hx Hy Hz = Ixx 0 Ixz 0 Iyy 0 Ixz 0 Izz P Q R
Spring 2003 16.61AC1-4 The overall equations of motion are then ×U U 0R QU Fy R 0-P V F W U+QW-Rl V+ RU-PW w+Pt B T= H L TrP+l,R 0R Q Ixr 0 Ix2 P →M IvyQ+R 0-P 0|Q I2R+12P-QP012012R IP+lR +QR(I2z-Iuy)+ PQlzz IyyQ +PR(Irz-I22)+(R2-P)Ix2 I2 R+IrP +PQ(Iwy-Ixx)-QRIza 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 fight condition
Spring 2003 16.61 AC 1–4 • The overall equations of motion are then: 1 m F = ˙vc B + BIω × vc ⇒ 1 m Fx Fy Fz = U˙ V˙ W˙ + 0 −R Q R 0 −P −Q P 0 U V W = U˙ + QW − RV V˙ + RU − PW W˙ + P V − QU T = ˙ H B + BIω × H ⇒ L M N = IxxP˙ + IxzR˙ IyyQ˙ IzzR˙ + IxzP˙ + 0 −R Q R 0 −P −Q P 0 Ixx 0 Ixz 0 Iyy 0 Ixz 0 Izz P Q R = IxxP˙ + IxzR˙ +QR(Izz − Iyy) + PQIxz IyyQ˙ +P R(Ixx − Izz)+(R2 − P2)Ixz IzzR˙ + IxzP˙ +P Q(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.
Spring 2003 16.61AC1-5 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 Uc into the fuselage plane of symmetry. Z perpendicular to X(pointed down).Y same X-AXIS e(BODY) BODY Y-AXIS STABILITY) X-AXIS Z-AXIS WIND) Advantages to each, but typically use the stability axes In different fight equilibrium conditions, the axes will be oriented dif- ferently with respect to the A/C principal axes = need to transform (rotate) the principal Inertia components between the frames When vehicle undergoes motion with respect to the equilibrium, the Stability Axes remain fixed to the airplane as if painted on o
Spring 2003 16.61 AC 1–5 • 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. • 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 transform (rotate) the principal Inertia components between the frames. – When vehicle undergoes motion with respect to the equilibrium, the Stability Axes remain fixed to the airplane as if painted on.
Spring 2003 16.61AC1-6 Can linearize about various steady state conditions of fight For steady state fight conditions must have F= Faero+ Gravity+ Thrust=0 and T=0 o So for equilibrium condition, forces balance on the aircraft L=W andt= d Also assume that P=Q=R=U=V=W=0 Impose additional constraints that depend on the fight condition ◇ Steady wings-level flight→==6=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 fight dynamics that lift Lo=Clao, where o CI= lift coefficient, which is a function of the equilibrium condition ao= nominal angle of attack (angle that the wing meets the air fow) But, as the vehicle moves about the equilibrium condition, would expect that the angle of attack will change Q+△ Thus the lift forces will also be perturbed C(a0+△a)=L+△ Can extend this idea to all dynamic variables and how they influence all aerodynamic forces and moments
Spring 2003 16.61 AC 1–6 • Can linearize about various steady state conditions of flight. – For steady state flight conditions must have F = Faero + Fgravity + Fthrust = 0 and T = 0 ✸ 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 the flight condition: ✸ Steady wings-level 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 0 = Clα0 , where: ✸ Cl = lift coefficient, which is a function of the equilibrium condition ✸ α0 = nominal angle of attack (angle that the wing meets the air flow). – But, as the vehicle moves about the equilibrium condition, would expect that the angle of attack will change α = α0 + ∆α – Thus the lift forces will also be perturbed Lf = Cl(α0 + ∆α) = Lf 0 + ∆Lf • Can extend this idea to all dynamic variables and how they influence all aerodynamic forces and moments
Spring 2003 16.61AC1-7 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) F=71()T()1(0)0= mg sin g cos6 cosΦcos日 For symmetric steady state fight equilibrium, we will typically assume that ≡60,Φ≡亚o=0 sin e Fg=mg cos e Use Euler angles to specify vehicle rotations with respect to the Earth frame 6=Qcos中-Rsin ④=P+Qsin重tan+ R cos p tan 亚=(Qsin更+Rcos)sec Note that if g≈0,then≈Q Recall:≈Roll,e≈ Pitch,and业≈ Heading
Spring 2003 16.61 AC 1–7 Gravity Forces • Gravity acts through the CoM in vertical direction (inertial frame +Z) – Assume that we have a non-zero pitch angle Θ0 – Need to map this force into the body frame – Use the Euler angle transformation (2–15) Fg B = T1(Φ)T2(Θ)T3(Ψ) 0 0 mg = mg − sin Θ sin Φ cos Θ cos Φ cos Θ • For symmetric steady state flight equilibrium, we will typically assume that Θ ≡ Θ0, Φ ≡ Φ0 = 0, so Fg B = mg − sin Θ0 0 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 sin Φ tan Θ + R cos Φ tan Θ Ψ=( ˙ Q sin Φ + R cos Φ) sec Θ – Note that if Φ ≈ 0, then Θ˙ ≈ Q • Recall: Φ ≈ Roll, Θ ≈ Pitch, and Ψ ≈ Heading
Spring 2003 16.61AC1-8 Recall: ca six X 0Y=T3)lY 0012 y|=T2() y "|=n(9)y
Spring 2003 16.61 AC 1–8 Recall: x y z = cψ sψ 0 −sψ cψ 0 0 01 X Y Z = T3(ψ) X Y Z x y z = x y z = T2(θ) x y z x y z = x y z = T1(φ) x y z
Spring 2003 16.61AC1-9 Linearization Define the trim angular rates and velocities PQR (v)=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 fight Proceed with the linearization of the dynamics for various fight conditions Nominal Perturbed→ Perturbed Velocit Velocit Acceleration Velocities U=Uo+u Vo=0, Angular rates P=p P=p Q0=0 Q=q Q=q R0=0 R=r R=r noles 60, =60+6 Φ0=0 o=0 业= y
Spring 2003 16.61 AC 1–9 Linearization • Define the trim angular rates and velocities BIωo B = P Q R (vc) o B = Uo 0 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 the linearization of the dynamics for various flight conditions Nominal Perturbed ⇒ Perturbed Velocity Velocity ⇒ Acceleration Velocities U0, U = U0 + u ⇒ U˙ = ˙u W0 = 0, W = w ⇒ W˙ = ˙w V0 = 0, V = v ⇒ V˙ = ˙v Angular Rates P0 = 0, P = p ⇒ P˙ = ˙p Q0 = 0, Q = q ⇒ Q˙ = ˙q R0 = 0, R = r ⇒ R˙ = ˙r Angles Θ0, Θ=Θ0 + θ ⇒ Θ =˙ ˙ θ Φ0 = 0, Φ = φ ⇒ Φ =˙ φ˙ Ψ0 = 0, Ψ = ψ ⇒ Ψ =˙ ψ˙
Spring 2003 16.61AC1-10 Qo= Ro=O. Note that the forces and moments are also perturbed Linearization for symmetric fight U= Uo +u, Vo= Wo=0 P+△F]=U+QW-RV≈i+q-r≈t 0+△F=V+RU-PW≈i+r(U+)-p≈t+r F+△F]=W+PV-QU≈t+p-q(o+) qUo △F 元+rUo △F 3 Attitude motion L IxzP+lxR +QR(I2x-Iwy+ pQl M Q+PR(Ix-12)+(R2-P2)l2 I2zR+lxP +PQ(Iyy-Ixz)-QRIzz △L ap+1, 4 →|△M △N 6 Key aerodynamic parameters are also perturbed Total Velocity VT =((Uo+u)2+v2+w2)1/2NUo+ Perturbed Sideslip angle B= sin(u/vr)av/U Perturbed Angle of Attack ar tan(W/U)Nw/Uo To understand these equations in detail, and the resulting impact on the vehicle dynamics, we must investigate the terms AF N
Spring 2003 16.61 AC 1–10 • Linearization for symmetric flight U = U0 + u, V0 = W0 = 0, P0 = Q0 = R0 = 0. Note that the forces and moments are also perturbed. 1 m F0 x + ∆Fx = U˙ + QW − RV ≈ u˙ + qw − rv ≈ u˙ 1 m F0 y + ∆Fy = V˙ + RU − PW ≈ v˙ + r(U0 + u) − pw ≈ v˙ + rU0 1 m F0 z + ∆Fz = W˙ + P V − QU ≈ w˙ + pv − q(U0 + u) ≈ w˙ − qU0 ⇒ 1 m ∆Fx ∆Fy ∆Fz = u˙ v˙ + rU0 w˙ − qU0 1 2 3 • Attitude motion: L M N = IxxP˙ + IxzR˙ +QR(Izz − Iyy) + PQIxz IyyQ˙ +P R(Ixx − Izz)+(R2 − P2)Ixz IzzR˙ + IxzP˙ +P Q(Iyy − Ixx) − QRIxz ⇒ ∆L ∆M ∆N = Ixxp˙ + Ixzr˙ Iyyq˙ Izzr˙ + Ixzp˙ 4 5 6 Key aerodynamic parameters are also perturbed: Total Velocity VT = ((U0 + u) 2 + v2 + w2 ) 1/2 ≈ U0 + u Perturbed Sideslip angle β = sin−1 (v/VT ) ≈ v/U0 Perturbed Angle of Attack αx = tan−1 (w/U) ≈ w/U0 • To understand these equations in detail, and the resulting impact on the vehicle dynamics, we must investigate the terms ∆Fx ... ∆N
