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.