Lecture AC-1 Aircraft Dynamics Copy right 2003 by Jon at h an H ow 1 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 U 0 and height h 0 . The motion is perturbed slightly so that U 0 → U = U 0 + u (1) h 0 → h = h 0 +?h (2) – Assume that E = 1 2 mU 2 + mgh is constant before and after the pertur- bation. It then follows that u ≈? g?h U 0 – From Newton’s laws we know that, in the vertical direction m ¨ h = L?W where weight W = mg and lift L = 1 2 ρSC L U 2 (S is the wing area). We can then derive the equations of motion of the aircraft: m ¨ h = L?W = 1 2 ρSC L (U 2 ?U 2 0 )(3) = 1 2 ρSC L ((U 0 + u) 2 ?U 2 0 ) ≈ 1 2 ρSC L (2uU 0 )(4) ≈?ρSC L parenleftBigg g?h U 0 U 0 parenrightBigg = ?(ρSC L g)?h (5) Since ¨ h =? ¨ h and for the original equilibrium flight condition L = W = 1 2 (ρSC L )U 2 0 = mg,wegetthat ρSC L g m =2 parenleftBigg g U 0 parenrightBigg 2 Combine these result to obtain: ? ¨ h +? 2 ?h =0 , ? ≈ g U 0 √ 2 – These equations describe an oscillation (called the phugoid oscillation) of the altitude of the aircraft about it nominal value. a51 Only approximate natural frequency, but value very close. Spring 2003 16.61 AC 1–3 ? The basic dynamics are the same as we had before: vector F = m ˙ vectorv c I and vector T = ˙ vector H I ? 1 m vector F = ˙ vectorv c B + BI vectorω ×vectorv c Transport Thm. ? vector T = ˙ vector H B + BI vectorω × vector 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 ( vector i, vector j, vector k) ? Instantaneous mapping of vectorv c and BI vectorω into the body frame is given by BI vectorω = P vector i + Q vector j + R vector kvectorv c = U vector i + V vector j + W vector k ? BI ω B = ? ? ? ? ? P Q R ? ? ? ? ? ? (v c ) B = ? ? ? ? ? U V W ? ? ? ? ? ? By symmetry, we can show that I xy = I yz = 0, but value of I xz depends on specific frame selected. Instantaneous mapping of the angular momentum vector H = H x vector i + H y vector j + H z vector k into the Body Frame given by H B = ? ? ? ? ? H x H y H z ? ? ? ? ? = ? ? ? ? ? I xx 0 I xz 0 I yy 0 I xz 0 I zz ? ? ? ? ? ? ? ? ? ? P Q R ? ? ? ? ? Spring 2003 16.61 AC 1–4 ? The overall equations of motion are then: 1 m vector F = ˙ vectorv c B + BI vectorω ×vectorv c ? 1 m ? ? ? ? ? F x F y F z ? ? ? ? ? = ? ? ? ? ? ˙ U ˙ V ˙ W ? ? ? ? ? + ? ? ? ? ? 0 ?RQ R 0 ?P ?QP 0 ? ? ? ? ? ? ? ? ? ? U V W ? ? ? ? ? = ? ? ? ? ? ˙ U + QW ?RV ˙ V + RU ?PW ˙ W + PV ?QU ? ? ? ? ? vector T = ˙ vector H B + BI vectorω × vector H ? ? ? ? ? ? L M N ? ? ? ? ? = ? ? ? ? ? I xx ˙ P + I xz ˙ R I yy ˙ Q I zz ˙ R + I xz ˙ P ? ? ? ? ? + ? ? ? ? ? 0 ?RQ R 0 ?P ?QP 0 ? ? ? ? ? ? ? ? ? ? I xx 0 I xz 0 I yy 0 I xz 0 I zz ? ? ? ? ? ? ? ? ? ? P Q R ? ? ? ? ? = ? ? ? ? ? I xx ˙ P + I xz ˙ R +QR(I zz ?I yy )+PQI xz I yy ˙ Q +PR(I xx ?I zz )+(R 2 ?P 2 )I xz I zz ˙ R + I xz ˙ P +PQ(I yy ?I xx )?QRI xz ? ? ? ? ? ? Clearly these equations are very nonlinear and complicated, and we have not even said where vector F and vector T come from. =? Need to linearize!! – Assume that the aircraft is flying in an equilibrium condition andwewill linearize the equations about this nominal flight condition. 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 vectorv c . Z perpendicular to X (pointed down). Y perpendicular to XZ plane, o? to the right. 3. Stability Axes - X aligned with projection of vectorv c into the fuselage plane of symmetry. Z perpendicular to X (pointed down). Y same. ? Advantages to each, but typically use the stability axes. – In di?erent flight 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. Spring 2003 16.61 AC 1–6 ? Can linearize about various steady state conditions of flight. – For steady state flight conditions must have vector F = vector F aero + vector F gravity + vector F thrust =0 and vector T =0 a51 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: a51 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 L f 0 = C l α 0 ,where: a51 C l = lift coe?cient, which is a function of the equilibrium condition a51 α 0 = 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 = α 0 +?α – Thus the lift forces will also be perturbed L f = C l (α 0 +?α)=L f 0 +?L f ? Can extend this idea to all dynamic variables and how they influence all aerodynamic forces and moments 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 Θ 0 – Need to map this force into the body frame – Use the Euler angle transformation (2–15) F g B = T 1 (Φ)T 2 (Θ)T 3 (Ψ) ? ? ? ? ? 0 0 mg ? ? ? ? ? = mg ? ? ? ? ? ?sinΘ sinΦ cosΘ cos Φ cosΘ ? ? ? ? ? ? For symmetric steady state flight equilibrium, we will typically assume that Θ ≡ Θ 0 ,Φ≡ Φ 0 =0,so F g B = mg ? ? ? ? ? ?sinΘ 0 0 cos Θ 0 ? ? ? ? ? ? Use Euler angles to specify vehicle rotations with respect to the Earth frame ˙ Θ=Q cosΦ?R sinΦ ˙ Φ=P + Q sinΦ tanΘ + R cosΦ tanΘ ˙ Ψ=(Q sinΦ + R cos Φ) secΘ – Note that if Φ ≈ 0, then ˙ Θ ≈ Q ? Recall: Φ ≈ Roll, Θ ≈ Pitch, and Ψ ≈ Heading. Spring 2003 16.61 AC 1–8 Recall: ? ? ? ? ? x prime y prime z prime ? ? ? ? ? = ? ? ? ? ? cψ sψ 0 ?sψ cψ 0 001 ? ? ? ? ? ? ? ? ? ? X Y Z ? ? ? ? ? = T 3 (ψ) ? ? ? ? ? X Y Z ? ? ? ? ? ? ? ? ? ? x primeprime y primeprime z primeprime ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? x prime y prime z prime ? ? ? ? ? = T 2 (θ) ? ? ? ? ? x prime y prime z prime ? ? ? ? ? ? ? ? ? ? x y z ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? x primeprime y primeprime z primeprime ? ? ? ? ? = T 1 (φ) ? ? ? ? ? x primeprime y primeprime z primeprime ? ? ? ? ? Spring 2003 16.61 AC 1–9 Linearization ? Define the trim angular rates and velocities BI ω o B = ? ? ? ? ? P Q R ? ? ? ? ? (v c ) o B = ? ? ? ? ? U o 0 0 ? ? ? ? ? which are associated with the flight condition. In fact, these define the type of equilibrium motion that we linearize about. Note: – W 0 = 0 since we are using the stability axes, and – V 0 = 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 U 0 ,U= U 0 + u ? ˙ U =˙u W 0 =0,W= w ? ˙ W =˙w V 0 =0,V= v ? ˙ V =˙v Angular Rates P 0 =0,P= p ? ˙ P =˙p Q 0 =0,Q= q ? ˙ Q =˙q R 0 =0,R= r ? ˙ R =˙r Angles Θ 0 , Θ=Θ 0 + θ ? ˙ Θ= ˙ θ Φ 0 =0, Φ=φ ? ˙ Φ= ˙ φ Ψ 0 =0, Ψ=ψ ? ˙ Ψ= ˙ ψ Spring 2003 16.61 AC 1–10 ? Linearization for symmetric flight U = U 0 + u, V 0 = W 0 =0,P 0 = Q 0 = R 0 = 0. Note that the forces and moments are also perturbed. 1 m bracketleftBig F 0 x +?F x bracketrightBig = ˙ U + QW ?RV ≈ ˙u + qw?rv ≈ ˙u 1 m bracketleftBig F 0 y +?F y bracketrightBig = ˙ V + RU ?PW ≈ ˙v + r(U 0 + u)?pw ≈ ˙v + rU 0 1 m bracketleftBig F 0 z +?F z bracketrightBig = ˙ W + PV ?QU ≈ ˙w + pv ?q(U 0 + u) ≈ ˙w ?qU 0 ? 1 m ? ? ? ? ? ?F x ?F y ?F z ? ? ? ? ? = ? ? ? ? ? ˙u ˙v + rU 0 ˙w ?qU 0 ? ? ? ? ? 1 2 3 ? Attitude motion: ? ? ? ? ? L M N ? ? ? ? ? = ? ? ? ? ? I xx ˙ P + I xz ˙ R +QR(I zz ?I yy )+PQI xz I yy ˙ Q +PR(I xx ?I zz )+(R 2 ?P 2 )I xz I zz ˙ R + I xz ˙ P +PQ(I yy ?I xx )?QRI xz ? ? ? ? ? ? ? ? ? ? ? ?L ?M ?N ? ? ? ? ? = ? ? ? ? ? I xx ˙p + I xz ˙r I yy ˙q I zz ˙r + I xz ˙p ? ? ? ? ? 4 5 6 Key aerodynamic parameters are also perturbed: Total Velocity V T =((U 0 + u) 2 + v 2 + w 2 ) 1/2 ≈ U 0 + u Perturbed Sideslip angle β =sin ?1 (v/V T ) ≈ v/U 0 Perturbed Angle of Attack α x =tan ?1 (w/U) ≈ w/U 0 ? To understand these equations in detail, and the resulting impact on the vehicle dynamics, we must investigate the terms ?F x ...?N. Spring 2003 16.61 AC 1–12 ? We must also address the left-hand side ( vector F, vector T) –Netforces 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 ?F x = ?F x ?U ?U + ?F x ?W ?W + ?F x ? ˙ W ? ˙ W + ?F x ?Θ ?Θ + ...+ ?F g x ?Θ ?Θ + ?F c x = ?F x ?U u + ?F x ?W w + ?F x ? ˙ W ˙w + ?F x ?Θ θ + ...++ ?F g x ?Θ θ +?F c x – ?F x ?U called a stability derivative. Is a function of the equilibrium con- dition. 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 e?ective way to analyze the aircraft flight mechanics – well supported by numer- ous flight test comparisons. Spring 2003 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, ˙u, ¨u,...,w, ˙w, ¨w,... – Typically only retain a few terms to capture the dominant e?ects. ? Dominant behavior most easily discussed in terms of the: – Symmetric variables: U, W, Q and forces/torques: F x , F z ,andM – Asymmetric variables: V , P, R and forces/torques: F y , L,andN ? Observation – for truly symmetric flight Y , L,andN 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 ?F z /? ˙w and M ˙w ≡ ?M/? ˙w (aerodynamic lag in- volved in forming new pressure distribution on the wing in response to the perturbed angle of attack) 3. ?F x /?q is negligibly small. Spring 2003 16.61 AC 1–14 ? Note that we must also find the perturbation gravity and thrust forces and moments ?F g x ?Θ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 0 = ?mg cos Θ 0 ?F g z ?Θ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 0 = ?mg sin 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. Spring 2003 16.61 AC 1–15 ? Aerodynamic summary: 1A ?F x = parenleftbigg ?F x ?U parenrightbigg 0 u + parenleftbigg ?F x ?W parenrightbigg 0 w ? ?F x ～ u, w 2A ?F y ～ v, p, r 3A ?F z ～ u, w,˙w, q 4A ?L ～ β, p, r 5A ?M ～ u, w,˙w, q 6A ?N ～ β, p, r ? Result is that, with these force, torque approximations, equations 1, 3, 5 decouple from 24, 6 – 1, 3, 5 are the longitudinal dynamics in u, w, and q ? ? ? ? ? ? ? ? ?F x ?F z ?M ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? m ˙u m(˙w ?qU 0 ) I yy ˙q ? ? ? ? ? ? ? ? ≈ ? ? ? ? ? ? ? ? ? ? ? parenleftbigg ?F x ?U parenrightbigg 0 u + parenleftbigg ?F x ?W parenrightbigg 0 w + parenleftBigg ?F g x ?Θ parenrightBigg 0 θ +?F c x parenleftbigg ?F z ?U parenrightbigg 0 u + parenleftbigg ?F z ?W parenrightbigg 0 w + parenleftBigg ?F z ? ˙ W parenrightBigg 0 ˙w + parenleftBigg ?F z ?Q parenrightBigg 0 q + parenleftBigg ?F g z ?Θ parenrightBigg 0 θ +?F c z parenleftbigg ?M ?U parenrightbigg 0 u + parenleftbigg ?M ?W parenrightbigg 0 w + parenleftBigg ?M ? ˙ W parenrightBigg 0 ˙w + parenleftBigg ?M ?Q parenrightBigg 0 q +?M c ? ? ? ? ? ? ? ? ? ? ? – 2, 4, 6 are the lateral dynamics in v, p, and r Spring 2003 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 in- ertias now change depending on the equilibrium flight condition. ? Since the nonlinear behavior is too di?cult 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 func- tions, 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.