16.333: Lecture #4

16.333 Lecture 4 Aircraft Dynamics • Aircraft nonlinear EOM • Linearization – dynamics • Linearization – forces & moments • Stability derivatives and coefficients Fall 2004 Aircraft Dynamics 16.333 4–1 • 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 = U 0 + u h0 → h = h0 + Δh (1) (2) • Assume that E = 21 mU 2 + mgh is constant before and after the perturbation. It then follows that u ≈ − gΔh U0 • From Newton’s laws we know that, in the vertical direction ¨ =L−W mh where weight W = mg and lift L = 21 ρSCLU 2 (S is the wing area). We can then derive the equations of motion of the aircraft: ¨ = L − W = 1 ρSCL(U 2 − U 2) mh (3) 0 2 1 1 = ρSCL((U0 + u)2 − U02) ≈ ρSCL(2uU0)(4) 2 2 � � gΔh ≈ −ρSCL U0 = −(ρSCLg)Δh (5) U0 ¨ = Δh ¨ and for the original equilibrium flight condition L = Since h W = 21 (ρSCL)U02 = mg, we get that � �2 ρSCLg g =2 m U0 Combine these result to obtain: g√ 2 ¨ Δh + Ω Δh = 0 , Ω ≈ 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. Fall 2004 16.333 4–2 • The basic dynamics are: I ˙I ˙ � � � F = mv �c and T = H 1 � B F = �v˙c + BI ω � × �vc m ˙B � � � ⇒ T = H + BI ω � ×H ⇒ Transport Thm. • 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 ω � = P�i + Q�j + R�k ⎡ ⎤ P ⇒ BI ωB = ⎣ Q ⎦ R BI ω � into the body frame: �vc = U�i + V �j + W �k ⎡ ⎤ U ⇒ (vc)B = ⎣ 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 � = Hx�i + Hy�j + Hz�k H into the Body Frame given by ⎡ ⎤ ⎡ ⎤⎡ ⎤ Hx Ixx 0 Ixz P HB = ⎣ Hy ⎦ = ⎣ 0 Iyy 0 ⎦ ⎣ Q ⎦ Hz Ixz 0 Izz R Fall 2004 16.333 4–3 • The overall equations of motion are then: 1 � B F = �v˙c + BI ω � × �vc m ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ U̇ X 0 −R Q U 1 ⎣ ⎦ = ⎣ V̇ ⎦ + ⎣ R ⇒ Y 0 −P ⎦ ⎣ V ⎦ m ˙ Z −Q P 0 W W ⎡ ˙ ⎤ U + QW − RV = ⎣ V˙ + RU − P W ⎦ ˙ + P V − QU W ˙B � � T = H + BI � ω � ×H ⎤ ⎡ ⎤⎡ ⎤⎡ ⎤ ˙ IxxṖ + Ixz R 0 −R Q L Ixx 0 Ixz P ⎦+⎣ R ⇒⎣M ⎦ = ⎣ 0 −P ⎦ ⎣ 0 Iyy 0 ⎦ ⎣ Q ⎦ Iyy Q̇ N −Q P 0 Ixz 0 Izz R Izz R˙ + Ixz Ṗ ⎡ ⎤ ⎡ ⎤ IxxṖ + Ixz R˙ +QR(Izz − Iyy ) + P QIxz = ⎣ Iyy Q̇ +P R(Ixx − Izz ) + (R2 − P 2)Ixz ⎦ Izz R˙ + Ixz Ṗ +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. 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. � RE B ODY Y -AXIS B ODY Z-AXIS X-AXIS ( B ODY ) � LA T IV EW IN D X-AXIS ( S T AB ILIT Y ) X-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 4–5 • Can linearize about various steady state conditions of flight. – For steady state flight conditions must have F� = F�aero + F�gravity + F�thrust = 0 and T� = 0 3 So for equilibrium condition, forces balance on the aircraft L = W and T = D ˙ =0 – Also assume that Ṗ = Q˙ = R˙ = U˙ = V˙ = W – 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 Lf0 = CLα α0 where: 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) = Lf0 + ΔLf • Can extend this idea to all dynamic variables and how they influence all aerodynamic forces and moments 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 FBg = T1(Φ)T2(Θ)T3(Ψ) ⎣ 0 ⎦ = mg ⎣ sin Φ cos Θ ⎦ mg cos Φ cos Θ • For symmetric steady state flight equilibrium, we will typically assume that Θ ≡ Θ0, Φ ≡ Φ0 = 0, so ⎡ ⎤ − sin Θ0 ⎦ FBg = mg ⎣ 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. Fall 2004 16.333 4–7 Linearization • Define the trim angular rates and velocities ⎡ ⎤ ⎡ ⎤ P Uo BI o ωB = ⎣ Q ⎦ (vc)oB = ⎣ 0 ⎦ 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 Velocity Perturbed Velocity ⇒ ⇒ Perturbed Acceleration Velocities U0, W0 = 0, V0 = 0, U = U0 + u W =w V =v ⇒ ⇒ ⇒ U̇ = u̇ ˙ = ẇ W 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 ⇒ ⇒ ⇒ Θ˙ = θ̇ Φ˙ = φ̇ Ψ˙ = ψ̇ Fall 2004 16.333 4–8 X Θ X0 � �0 or �0 XE U = U0 + u VT VT0 = U0 � q � Horizontal W C.G. �0 Z Z0 ZE Down Figure 1: Perturbed Axes. The equilibrium condition was that the aircraft was angled up by Θ0 with velocity VT 0 = U0 . The vehicle’s motion has been perturbed (X0 → X) so that now Θ = Θ0 + θ and � VT 0 . Note that VT is no longer aligned with the X­axis, resulting in a non­zero the velocity is VT = 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. 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] = U̇ + QW − RV ≈ u̇ + qw − rv ≈ u̇ m 1 [Y0 + ΔY ] = V̇ + RU − P W m ≈ v̇ + r(U0 + u) − pw ≈ v̇ + rU0 1 ˙ + P V − QU ≈ ẇ + pv − q(U0 + u) [Z0 + ΔZ] = W m ≈ ẇ − qU0 ⎡ ⎤ ⎡ ⎤ ΔX u̇ 1 ⇒ ⎣ ΔY ⎦ = ⎣ v̇ + rU0 ⎦ m ẇ − qU0 ΔZ 1 2 3 • Attitude motion: ⎡ ⎤ ⎡ ⎤ IxxṖ + Ixz R˙ +QR(Izz − Iyy ) + P QIxz L ⎣M ⎦ = ⎣ Iyy Q̇ +P R(Ixx − Izz ) + (R2 − P 2)Ixz ⎦ N Izz R˙ + Ixz Ṗ +P Q(Iyy − Ixx) − QRIxz ⎡ ⎤ ⎡ ⎤ ΔL Ixxṗ + Ixz ṙ ⎦ ⇒ ⎣ ΔM ⎦ = ⎣ Iyy q̇ ΔN Izz ṙ + Ixz ṗ 4 5 6 Fall 2004 16.333 4–10 • Key aerodynamic parameters are also perturbed: Total Velocity VT = ((U0 + u)2 + v 2 + 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 ΔX . . . ΔN . – We must also address the left­hand side (F� , T� ) – Net forces and moments must be zero in equilibrium condition. – Aerodynamic and Gravity forces are a function of equilibrium con­ dition 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 ∂X ∂X ∂X ∂X ∂X g ΔX = ΔU + ΔW + ΔẆ + ΔΘ + . . . + ΔΘ + ΔX c ∂Θ ∂Θ ∂U ∂W ∂ Ẇ g ∂X ∂X ∂X ∂X ∂X = u+ w+ w θ + ... + θ + ΔX c ˙ + ∂Θ ∂Θ ∂U ∂W ∂ Ẇ • ∂X ∂U called stability derivative – evaluated at eq. condition. • Gives dimensional form; non­dimensional form available in tables. • Clearly approximation since ignores lags in the aerodynamics forces (assumes that forces only function of instantaneous values) Fall 2004 16.333 4–11 Stability Derivatives • First proposed by Bryan (1911) – has proven to be a very effec­ tive way to analyze the aircraft flight mechanics – well supported by numerous flight test comparisons. • The forces and torques acting on the aircraft are very complex nonlin­ ear functions of the flight equilibrium condition and the perturbations from equilibrium. – Linearized expansion can involve many terms u, u, ˙ u, ¨ . . . , w, w, ˙ w, ¨ ... – 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 & forces/torques: X, Z, and M – Asymmetric variables: V , P , R & forces/torques: Y , 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 sym­ metric motion variables are zero. • Further (convenient) assumptions: 1. Derivatives of symmetric forces/torques with respect to the asym­ metric motion variables are small and can be neglected. 2. We can neglect derivatives with respect to the derivatives of the motion variables, but keep ∂Z/∂ẇ and Mẇ ≡ ∂M/∂ẇ (aero­ dynamic lag involved in forming new pressure distribution on the wing in response to the perturbed angle of attack) 3. ∂X/∂q is negligibly small. Fall 2004 16.333 4–12 ∂()/∂() X • u 0 v • w 0 p ≈0 q 0 r Y 0 • 0 • 0 • Z • 0 • 0 • 0 L 0 • 0 • 0 • M • 0 • 0 • 0 N 0 • 0 • 0 • • Note that we must also find the perturbation gravity and thrust forces and moments � � ∂X g �� ∂Z g �� = −mg cos Θ0 = −mg sin Θ0 ∂Θ � ∂Θ � 0 0 • Aerodynamic summary: � ∂X � � ∂X � 1A ΔX = ∂U 0 u + ∂W 0 w ⇒ ΔX ∼ u, αx ≈ w/U0 2A ΔY ∼ β ≈ v/U0, p, r 3A ΔZ ∼ u, αx ≈ w/U0, α̇x ≈ ẇ/U0, q 4A ΔL ∼ β ≈ v/U0, p, r 5A ΔM ∼ u, αx ≈ w/U0, α̇x ≈ ẇ/U0, q 6A ΔN ∼ β ≈ v/U0, p, r Fall 2004 16.333 4–13 • 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 ⎡ ⎤ ⎡ ⎤ ΔX mu̇ ⎣ ΔZ ⎦ = ⎣ m(ẇ − qU0) ⎦ ΔM Iyy q̇ ⎡ � ∂X � � ∂X � � ∂X g � c ∂U ∂W ∂Θ 0 θ + ΔX ⎢ � � � � ⎢� � ⎢ ∂Z u + � ∂Z � w + ∂Z w˙ + ∂Z q + � ∂Z g � θ + ΔZ c ≈ ⎢ ∂U 0 ∂W 0 ∂Q 0 ∂Θ 0 ∂ Ẇ 0 ⎢ � � � � ⎣� � � ∂M � ∂M ∂M ẇ + ∂M q + ΔM c ∂U 0 u + ∂W 0 w + ∂ Ẇ ∂Q u+ 0 w+ 0 0 0 – 2, 4, 6 are the lateral dynamics in v, p, and r ⎡ ⎤ ⎡ ⎤ ΔY m(v̇ + rU0) ⎣ ΔL ⎦ = ⎣ Ixxṗ + Ixz ṙ ⎦ ΔN Izz ṙ + Ixz ṗ ⎡� ∂Y ∂V 0 v � + � ∂Y � ∂P 0 p + � ∂Y � ∂R 0 r + +ΔY ⎢ ⎢ � ∂L � � ∂L � � ∂L � ⎢ ≈ ⎢ ∂V 0 v + ∂P 0 p + ∂R 0 r + ΔLc ⎣� � � ∂N � � ∂N � ∂N c v + p + ∂V 0 ∂P 0 ∂R 0 r + ΔN c ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Fall 2004 16.333 4–14 Basic Stability Derivative Derivation • Consider changes in the drag force with forward speed U D = 1/2ρVT2SCD VT2 = (u0 + u)2 + v 2 + w2 � 2� ∂VT2 ∂VT = 2(u0 + u) ⇒ = 2u0 ∂u ∂u 0 � Note: ∂VT2 ∂v � � = 0 and 0 ∂VT2 ∂w � =0 0 At reference condition: � � � � 2 �� ∂D ∂ ρVT SCD ⇒ Du ≡ = ∂u 0 ∂u 2 � � � 0 � 2� � ρS ∂CD ∂VT = u20 + CD0 2 ∂u 0 ∂u 0 � � � � ∂C ρS D u20 = + 2u0CD0 2 ∂u 0 – Note ∂D ∂u is the stability derivative, which is dimensional. • Define nondimensional stability coefficient CDu as derivative of CD with respect to a nondimensional velocity u/u0 � � D ∂CD and CD0 ≡ (CD)0 CD = 1 2 ⇒ CDu ≡ ∂u/u ρV S 0 0 T 2 – So (•)0 corresponds to the variable at its equilibrium condition. Fall 2004 16.333 4–15 • Nondimensionalize: � � � � � � ∂D ρSu0 ∂CD = u0 + 2CD0 ∂u 0 2 ∂u 0 �� � � QS ∂CD = + 2CD0 u0 ∂u/u0 0 � �� � ∂D u0 = (CDu + 2CD0 ) QS ∂u 0 So given stability coefficient, can compute the drag force increment. • Note that Mach number has a significant effect on the drag: � � � � ∂CD u0 ∂CD ∂CD �u� CDu = = =M ∂u/u0 0 a∂ a ∂M 0 where ∂CD ∂M can be estimated from empirical results/tables. CD CD = 0.1 M � = Constant 0 Mcr 0 1 Mach number, M 2 Aerodynamic Principles • Thrust forces � CTu = ∂CT ∂u/u0 � � ⇒ 0 – For a glider, CTu = 0 – For a jet, CTu ≈ 0 – For a prop plane, CTu = −CD0 ∂T ∂u � = CTu 0 1 QS u0 Fall 2004 16.333 4–16 • Lift forces similar to drag L = 1/2ρVT2SCL � ⇒ � u0 QS �� ∂L ∂u � ∂L ∂u � 0 � � � � ρSu0 ∂CL = u0 + 2CL0 2 ∂u 0 �� � � QS ∂CL = + 2CL0 u0 ∂u/u0 0 = (CLu + 2CL0 ) 0 where CL0 is the lift coefficient for the eq. condition and CLu = M ∂CL as before. From aerodynamic theory, we have that ∂M CL| ∂CL M CL = √ M=0 = C ⇒ 2 L 2 ∂M 1 − M 1−M ⇒ CLu M2 = CL 1 − M2 0 • α Derivatives: Now consider what happens with changes in the angle of attack. Take derivatives and evaluate at the reference condition: – Lift: ⇒ CLα 2CL0 CL2 CL – Drag: CD = CDmin + ⇒ CDα = πeAR πeAR α Fall 2004 16.333 4–17 • Combine into X, Z Forces – At equilibrium, forces balance. – Use stability axes, so α0 = 0 – Include the effect in the force balance of a change in α on the force rotations so that we can see the perturbations. – Assume perturbation α is small, so rotations are by cos α ≈ 1, sin α ≈ α X = T − D + Lα Z = −(L + Dα) CL ��(t) x Cx V CD Cz z Note: � = �T at t = 0 (i.e., Static Trim) • So, now consider the α derivatives of these forces: ∂X ∂T ∂D ∂L = +L+α − ∂α ∂α ∂α ∂α � � ∂T – Thrust variation with α very small ≈0 ∂α 0 � – Apply at the reference condition (α = 0), i.e. CXα = • Nondimensionalize and apply reference condition: CXα = −CDα + CL0 2CL0 CL = CL0 − πeAR α ∂CX ∂α � 0 Fall 2004 16.333 4–18 • And for the Z direction ∂Z ∂D ∂L = −D − α − ∂α ∂α ∂α Giving CZα = −CD0 − CLα • Recall that CMα was already found during the static analysis • Can repeat this process for the other derivatives with respect to the forward speed. • Forward speed: ∂X ∂T ∂D ∂L = +α − ∂u ∂u ∂u ∂u So that � u0 QS �� ∂X ∂u � � = 0 u0 QS �� ∂T ∂u � � − 0 u0 QS �� ⇒ CXu ≡ CTu − (CDu + 2CD0 ) • Similarly for the Z direction: ∂Z ∂L ∂D =− −α ∂u ∂u ∂u So that � �� � � �� � u0 ∂Z u0 ∂L = − QS ∂u 0 QS ∂u 0 CZu ≡ −(CLu + 2CL0 ) M2 = − CL − 2CL0 1 − M2 0 • Many more derivatives to consider ! ∂D ∂u � 0 Fall 2004 16.333 4–19 Summary • Picked a specific Body Frame (stability axes) from the list of alter­ natives ⇒ 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 lin­ earized 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.