Aircraft Dynamics

Lecture AC-1 Aircraft Dynamics 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 h0 → h = h0 + ∆h (1) (2) – Assume that E = 12 mU 2 + mgh is constant before and after the perturbation. It then follows that g∆h u≈− U0 – From Newton’s laws we know that, in the vertical direction mḧ = L − W where weight W = mg and lift L = 12 ρSCL U 2 (S is the wing area). can then derive the equations of motion of the aircraft: 1 mḧ = L − W = ρSCL(U 2 − U02) 2 1 1 = ρSCL((U0 + u)2 − U02) ≈ ρSCL(2uU0) 2 2
 g∆h ≈ −ρSCL U0 = −(ρSCLg)∆h U0 We (3) (4) (5) Since ḧ = ∆ḧ and for the original equilibrium flight condition L = W = 1 (ρSCL)U02 = mg, we get that 2
g ρSCL g =2 m U0 Combine these result to obtain: 2 g√ 2 U0 – These equations describe an oscillation (called the phugoid oscillation) of the altitude of the aircraft about it nominal value. 3 Only approximate natural frequency, but value very close. ∆ḧ + Ω2∆h = 0 , Ω≈ 16.61 AC 1–3 • The basic dynamics are the same as we had before: I I ˙ F = mv˙c and T = H B 1  F = v˙c + BI ω × vc m B  ˙ + BI ω × H ⇒ T = H ⇒ Transport Thm. 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 BI BI ω = Pi + Qj + Rk  ⇒ BI  ωB =  P Q R ω into the body frame is given by vc = Ui + V j + W k       ⇒ (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  = Hxi + Hyj + Hzk H into the Body Frame given by   HB =  Hx Hy Hz       =   Ixx 0 Ixz 0 Iyy 0 Ixz 0 Izz      P Q R      16.61 AC 1–4 • The overall equations of motion are then: B 1  F = v˙c + m  ⇒ 1   m Fx Fy Fz       =      =     U̇ V̇ Ẇ ⇒     L M N           =  =     ω × vc       +   0 −R Q R 0 −P −Q P 0 U̇ + QW − RV V̇ + RU − P W Ẇ + P V − QU B ˙ + T = H  BI BI      U V W            ω × H IxxṖ + Ixz Ṙ Iyy Q̇ Izz Ṙ + Ixz Ṗ           + 0 −R Q R 0 −P −Q P 0      Ixx 0 Ixz 0 Iyy 0 Ixz 0 Izz IxxṖ + Ixz Ṙ +QR(Izz − Iyy ) + P QIxz +P R(Ixx − Izz ) + (R2 − P 2 )Ixz Iyy Q̇ Izz Ṙ + Ixz Ṗ +P Q(Iyy − Ixx ) − QRIxz      P Q R           • 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. 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. 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 3 So for equilibrium condition, forces balance on the aircraft L = W and T = D – Also assume that Ṗ = Q̇ = Ṙ = U̇ = V̇ = Ẇ = 0 – Impose additional constraints that depend on the flight condition: 3 Steady wings-level flight → Φ = Φ̇ = Θ̇ = Ψ̇ = 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 Lf0 = Cl α0 , where: 3 Cl = lift coefficient, which is a function of the equilibrium condition 3 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) = Lf0 + ∆Lf • Can extend this idea to all dynamic variables and how they influence all aerodynamic forces and moments 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)  0  g  FB = T1(Φ)T2(Θ)T3(Ψ)  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   FBg = mg  − sin Θ0 0 cos Θ0      • Use Euler angles to specify vehicle rotations with respect to the Earth frame Θ̇ = Q cos Φ − R sin Φ Φ̇ = P + Q sin Φ tan Θ + R cos Φ tan Θ Ψ̇ = (Q sin Φ + R cos Φ) sec Θ – Note that if Φ ≈ 0, then Θ̇ ≈ Q • Recall: Φ ≈ Roll, Θ ≈ Pitch, and Ψ ≈ Heading. 16.61 AC 1–8 Recall:  x y z          x y  z       x y z       =      =      =     cψ sψ 0 −sψ cψ 0 0 0 1                          X Y Z x y z x y  z    X      = T3 (ψ)  Y   Z   x      = T2 (θ)  y    z             x      = T1 (φ)  y    z       16.61 AC 1–9 Linearization • Define the trim angular rates and velocities  BI ωBo =     P Q R      (vc )oB  =     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 Velocity Perturbed Velocity ⇒ ⇒ Perturbed Acceleration Velocities U0 , W0 = 0, V0 = 0, U = U0 + u W =w V =v ⇒ ⇒ ⇒ U̇ = u̇ Ẇ = ẇ V̇ = v̇ Angular Rates P0 = 0, Q0 = 0, R0 = 0, P =p Q=q R=r ⇒ ⇒ ⇒ Ṗ = ṗ Q̇ = q̇ Ṙ = ṙ Angles Θ0 , Φ0 = 0, Ψ0 = 0, Θ = Θ0 + θ Φ=φ Ψ=ψ ⇒ ⇒ ⇒ Θ̇ = θ̇ Φ̇ = φ̇ Ψ̇ = ψ̇ 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 Fx0 + ∆Fx m = U̇ + QW − RV ≈ u̇ + qw − rv ≈ u̇ 1 Fy0 + ∆Fy m = V̇ + RU − P W ≈ v̇ + r(U0 + u) − pw ≈ v̇ + rU0 1 Fz0 + ∆Fz m = Ẇ + P V − QU ≈ ẇ + pv − q(U0 + u) ≈ ẇ − qU0  1  ⇒  m ∆Fx ∆Fy ∆Fz           = • Attitude motion:       ⇒     L M N ∆L ∆M ∆N       =               = u̇ v̇ + rU0 ẇ − qU0  1 2 3     Ixx Ṗ + Ixz Ṙ +QR(Izz − Iyy ) + P QIxz +P R(Ixx − Izz ) + (R2 − P 2 )Ixz Iyy Q̇ Izz Ṙ + Ixz Ṗ +P Q(Iyy − Ixx) − QRIxz Ixx ṗ + Ixz ṙ Iyy q̇ Izz ṙ + Ixz ṗ           4 5 6 Key aerodynamic parameters are also perturbed: Total Velocity Perturbed Sideslip angle Perturbed Angle of Attack VT = ((U0 + u)2 + v 2 + w2 )1/2 ≈ U0 + u β = sin−1(v/VT ) ≈ v/U0 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 . 16.61 AC 1–11 16.61 AC 1–12 • We must also address the left-hand side (F , T ) – Net forces and moments must be zero in the equilibrium condition. – Aerodynamic and Gravity forces are a function of equilibrium condition AND the perturbations about this equilibrium. • Predict the changes to the aerodynamic forces and moments using a first order expansion in the key flight parameters ∆Fx – ∂Fx ∂Fx ∂Fxg ∂Fx ∂Fx ∆U + ∆W + ∆Θ + . . . + ∆Θ + ∆Fxc = ∆Ẇ + ∂U ∂W ∂Θ ∂Θ ∂ Ẇ g ∂Fx ∂Fx ∂Fx ∂F ∂Fx ẇ + u+ w+ θ + . . . + + x θ + ∆Fxc = ∂U ∂W ∂Θ ∂Θ ∂ Ẇ ∂Fx ∂U called a stability derivative. Is a function of the equilibrium condition. Usually tabulated. – Clearly an approximation since there tend to be lags in the aerodynamics forces that this approach ignores (assumes that forces only function of instantaneous values) – First proposed by Bryan (1911), and has proven to be a very effective way to analyze the aircraft flight mechanics – well supported by numerous flight test comparisons. 16.61 AC 1–13 Stability Derivatives • The forces and torques acting on the aircraft are very complex nonlinear functions of the flight equilibrium condition and the perturbations from equilibrium. – Linearized expansion can involve many terms u, u̇, ü, . . . , w, ẇ, ẅ, . . . – Typically only retain a few terms to capture the dominant effects. • Dominant behavior most easily discussed in terms of the: – Symmetric variables: U , W , Q and forces/torques: Fx , Fz , and M – Asymmetric variables: V , P , R and forces/torques: Fy , L, and N • Observation – for truly symmetric flight Y , L, and N will be exactly zero for any value of U , W , Q ⇒ Derivatives of asymmetric forces/torques with respect to the symmetric motion variables are zero. • Further (convenient) assumptions: 1. Derivatives of symmetric forces/torques with respect to the asymmetric motion variables are zero. 2. We can neglect derivatives with respect to the derivatives of the motion variables, but keep ∂Fz /∂ ẇ and Mẇ ≡ ∂M/∂ ẇ (aerodynamic lag involved in forming new pressure distribution on the wing in response to the perturbed angle of attack) 3. ∂Fx/∂q is negligibly small. 16.61 AC 1–14 • Note that we must also find the perturbation gravity and thrust forces and moments ∂Fxg ∂Fzg = −mg cos Θ0 = −mg sin Θ0 ∂Θ 0 ∂Θ 0 • Typical set of stability derivatives. Figure 2: • corresponds to a zero slope - no dependence for small perturbations. No means no dependence for any size perturbation. 16.61 AC 1–15 • Aerodynamic summary: ∂Fx ∂U 0 u 1A ∆Fx = 2A ∆Fy ∼ v, p, r 3A ∆Fz ∼ u, w, ẇ, q 4A ∆L ∼ β, p, r 5A ∆M ∼ u, w, ẇ, q 6A ∆N ∼ β, p, r + ∂Fx ∂W 0 w ⇒ ∆Fx ∼ u, w • Result is that, with these force, torque approximations, equations 1, 3, 5 decouple from 2 4, 6 – 1, 3, 5 are the longitudinal dynamics in u, w, and q     mu̇  ∆Fx             ∆F   m(ẇ − qU )    =   z 0         ∆M Iyy q̇  ≈           ∂Fx u+ ∂U 0 ∂Fz ∂U 0 u + ∂M u+ ∂U 0
g ∂Fx x w + ∂F θ + ∆Fxc ∂W 0 ∂Θ
0

g ∂Fz ∂Fz ∂Fz ∂Fz ∂W 0 w +
∂ Ẇ  0 ẇ +
∂Q  0 q + ∂Θ 0 θ ∂M ∂M ∂M c w + ẇ + q + ∆M ∂W 0 ∂Q 0 ∂ Ẇ 0  + ∆Fzc – 2, 4, 6 are the lateral dynamics in v, p, and r           16.61 AC 1–16 Summary • Picked a specific Body Frame (stability axes) from the list of alternatives ⇒ Choice simplifies some of the linearization, but the inertias now change depending on the equilibrium flight condition. • Since the nonlinear behavior is too difficult to analyze, we needed to consider the linearized dynamic behavior around a specific flight condition ⇒ Enables us to linearize RHS of equations of motion. • Forces and moments also complicated nonlinear functions, so we linearized the LHS as well ⇒ Enables us to write the perturbations of the forces and moments in terms of the motion variables. – Engineering insight allows us to argue that many of the stability derivatives that couple the longitudinal (symmetric) and lateral (asymmetric) motions are small and can be ignored. • Approach requires that you have the stability derivatives. – These can be measured or calculated from the aircraft plan form and basic aerodynamic data.