《航空器的稳定与控制》（英文版）Lecture 6 Longitudinal Dynamics

16.333: Lecture #6 Aircraft longitudinal dynamics Typical aircraft open-loop motions · Longitudinal modes · Impact of actuators Linear Algebra in Action!

16.333: Lecture # 6 Aircraft Longitudinal Dynamics • Typical aircraft open­loop motions • Longitudinal modes • Impact of actuators • Linear Algebra in Action!

Fa2004 16.3335-1 Longitudinal dynamics Recall: X denotes the force in the X-direction, and similarly for Y and Z, then(as on 4-13) OX Longitudinal equations(see 4-13 )can be rewritten as mu= Xuu+ Xww-mg cos 000+AX mla-qU0= Zuu+ Zww+ Zi+ Zaq-mg sin 000+AZ M2+Mn+ M,iwn+M0q+△Me · There is no roll/yaw motion,soq=θ Control commands△X,△ze,and△ M have not yet been specified

� � Fall 2004 16.333 5–1 Longitudinal Dynamics • Recall: X denotes the force in the X­direction, and similarly for Y and Z, then (as on 4–13) ∂X Xu ≡ , . . . ∂u 0 • Longitudinal equations (see 4–13) can be rewritten as: mu˙ = Xuu + Xww − mg cos Θ0θ + ΔXc m(w˙ − qU0) = Zuu + Zww + Zw˙w˙ + Zqq − mg sin Θ0θ + ΔZc Iyyq˙ = Muu + Mww + Mw˙w˙ + Mqq + ΔMc There is no roll/yaw motion, so q = θ ˙ • . • Control commands ΔXc , ΔZc , and ΔMc have not yet been specified

Fa2004 16.3335-2 Rewrite in state space form as mg cos eo △XC (m-z) ∠n∠zq+mU0- ng sin6o △ZC Maiw+lug Mu Mw Mq △MC 0m-2t00 0 0 mg cos o △XC Lg mo -mg sin e △Zc M M 0 △Mc 0 Ex= Ar+c descriptor state space form →=E-(Ax+c)=A+c

Fall 2004 16.333 5–2 • Rewrite in state space form as ⎡ ⎡⎤ ⎤⎡ ⎡⎤ ⎤ mu˙ Xu Xw 0 −mg cos Θ0 u ΔXc ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎦ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣ ΔZc (m − Zw˙)w˙ Zu Zw Zq + mU0 −mg sin Θ0 Mu Mw Mq 0 w q = + −M ΔMc w˙w˙ + Iyyq˙ θ ˙ 0 0 1 0 θ 0 ⎡ ⎡⎤ ⎤ m 0 0 0 u˙ ⎢ ⎢ ⎢ ⎣ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎦ ⎥ ⎥ ⎥ ⎦ w˙ q˙ 0 m − Zw˙ 0 0 0 −Mw˙ Iyy 0 θ ˙ 0 0 0 1 ⎡ ⎤ ⎡⎤⎡ ⎤ Xu Xw 0 −mg cos Θ0 u ΔXc ⎢ ⎢ ⎢ ⎣ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎦ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎦ ΔZc Zu Zw Zq + mU0 −mg sin Θ0 Mu Mw Mq 0 w q = + ΔMc 0 0 1 0 θ 0 EX˙ = A¯X + ˆc descriptor state space form = E−1 ¯ ⇒ X (AX + ˆc) = AX + c ˙ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣

Fa2004 16.3335-3 Write out in state space form 0 - g cos eo z Zg mlo m-2 -mg sin eo Iy[Mu+Zur Iw [M+ Zur[Ma+(Zg+mUo)r-IwyImg sin Oor Note: slight savings if we defined symbols to embed the mass/inertia X /m, Zu=Zu/m, and M / Iy then a matrix cola X COS Lq 9720 +么[+2[+2+ 1-Z, Check the notation that is being used very carefully To figure out the c vector, we have to say a little more about how the control inputs are applied to the system

� � � � � � Fall 2004 16.333 5–3 • Write out in state space form: ⎡ Xu Xw 0 −g cos Θ0 m m Zu Zw Zq + mU0 −mg sin Θ0 m − Zw˙ m − Zw˙ m − Zw˙ m − Zw˙ I−1 [Mu + ZuΓ] I−1 [Mw + ZwΓ] I−1 [Mq + (Zq + mU0)Γ] −Iyy mg sin Θ0Γ −1 yy yy yy 0 0 1 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ A = Mw˙ Γ = m − Zw˙ • Note: slight savings if we defined symbols to embed the mass/inertia Xˆ ˆ u = Xu/m, Zˆu = Zu/m, and Mq = Mq/Iyy then A matrix collapses to: ⎡ ⎤ ˆ ˆ Xu Xw 0 −g cos Θ0 Aˆ = ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Zˆu Zˆw Z −g sin Θ0 ˆq + U0 1 − Zˆw˙ 1 − Zˆw˙ 1 − Zˆw˙ 1 − Zˆw˙ Mˆ ˆ ˆ ˆ ˆ u + ZˆuΓ Mw + ZˆwΓ Mq + (Zˆq + U0) ˆ Γˆ −g sin Θ0Γ 0 0 1 0 ˆ ˆ Mw˙ Γ = 1 − Zˆw˙ • Check the notation that is being used very carefully • To figure out the c vector, we have to say a little more about how the control inputs are applied to the system

Fa2004 16.3335-4 Longitudinal Actuators Primary actuators in longitudinal direction are the elevators and thrust Clearly the thrusters/elevators play a key role in defining the steady-state/equilibrium flight condition Now interested in determining how they also influence the aircraft motion about this equilibrium condition deflect elevator u(t),w(t), q(t) Canard Rudde Recall that we defined AX as the perturbation in the total force in the x direction as a result of the actuator commands Force change due to an actuator deflection from trim Expand these aerodynamic terms using same perturbation approach △X=X66+X61b de is the deflection of the elevator from trim down positive on change in thrust Xs and Xs, are the control stability derivatives

Fall 2004 16.333 5–4 Longitudinal Actuators • Primary actuators in longitudinal direction are the elevators and thrust. – Clearly the thrusters/elevators play a key role in defining the steady­state/equilibrium flight condition – Now interested in determining how they also influence the aircraft motion about this equilibrium condition deflect elevator → u(t), w(t), q(t), . . . • Recall that we defined ΔXc as the perturbation in the total force in the X direction as a result of the actuator commands – Force change due to an actuator deflection from trim • Expand these aerodynamic terms using same perturbation approach ΔXc = Xδeδe + Xδpδp – δe is the deflection of the elevator from trim (down positive) – δp change in thrust – Xδe and Xδp are the control stability derivatives

Fa2004 16.3335-5 Now we have that △X X6。X △Zc E 26Z6 E △MC Bu Ms M e For the longitudinal case B Iy IMe +Z J Iw [Mo, +Zo T Typical values for the B747 16.54 6n=0.3719=849528 5.2·10 M6n≈0 Aircraft response y=G(sa A=A+ Bu -G(s)=c(sl-a)B y=c We now have the means to modify the dynamics of the system, but first let's figure out what de and dp rea d

� � � � Fall 2004 16.333 5–5 • Now we have that ⎡⎤⎡ ⎤ ΔXc ⎥ ⎥ ⎥ ⎦ = E−1 ⎢ ⎢ ⎢ ⎣ Xδe Xδp Zδe Zδp Mδe Mδp ⎥ ⎥ ⎥ ⎦ c = E−1 ⎢ ⎢ ⎢ ⎣ ΔZc ΔMc δe = Bu δp 0 0 0 • For the longitudinal case ⎡ ⎤ Xδe Xδp m m Zδe Zδp B = m − Zw˙ m − Zw˙ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ I−1 I−1 [Mδe + ZδeΓ] yy yy Mδp + ZδpΓ 0 0 • Typical values for the B747 Xδe = −16.54 Xδp = 0.3mg = 849528 Zδe = −1.58 · 106 Zδp ≈ 0 Mδe = −5.2 · 107 Mδp ≈ 0 • Aircraft response y = G(s)u X˙ = AX + Bu → G(s) = C(sI − A) −1B y = CX • We now have the means to modify the dynamics of the system, but first let’s figure out what δe and δp really do

Fa2004 16.3335-6 Longitudinal Response Final response to a step input u=i s,y=G(su, use the FVt lim y(t)= lims(G(s) S-0 lim y(t)= G(Ou=-(CA B)i ● Initial response to a step input, use theⅣVT )=lim s(G( lim G(s)i For your system, G(s)=C(sI-A-B+D, but D=0,so n S→0 Note: there is No immediate change in the output of the motion variables in response to an elevator input= y(0+)=0 Consider the rate of change of these variables y(o*) i(t)=C=CAR+CBu and normally have that CB#0. Repeat process above to show that (0*)=CBi, and since C=I, v(0+)=Bu Looks good. Now compare with numerical values computed in Mat lab. Plot u, a, and flight path angle y=6-a (since 00=10=0 see picture on 4-8) I Note that C(sI-A)-IB+D=D+CB+CazE+

� � � � Fall 2004 16.333 5–6 Longitudinal Response • Final response to a step input u = u/s ˆ , y = G(s)u, use the FVT uˆ lim y(t) = lim s G(s) t→∞ s→0 s ⇒ lim y(t) = G(0)ˆ u t→∞ u = −(CA−1 B)ˆ • Initial response to a step input, use the IVT uˆ y(0+) = lim s G(s) = lim G(s)uˆ s→∞ s s→∞ – For your system, G(s) = C(sI − A) −1B + D, but D ≡ 0, so lim G(s) → 0 s→∞ – Note: there is NO immediate change in the output of the motion variables in response to an elevator input ⇒ y(0+) = 0 • Consider the rate of change of these variables y˙(0+) y˙(t) = CX˙ = CAX + CBu and normally have that CB =� 0. Repeat process above1 to show that y˙(0+) = CBuˆ, and since C ≡ I, y˙(0+) = Buˆ • Looks good. Now compare with numerical values computed in Mat￾lab. Plot u, α, and flight path angle γ = θ − α (since Θ0 = γ0 = 0 – see picture on 4–8) 1 CB Note that C(sI − A)−1B + D = D + + CA−1B + . . . s s2

Fa2004 16.3335-7 Elevator(1 elevator down-stick forward) See very rapid response that decays quickly(mostly in the first 10 seconds of the a response) Also see a very lightly damped long period response(mostly u, some T, and very little a). Settles in>600 secs Predicted steady state values from code 14. 1429 m/s u(speeds up 0.0185 rad a (slight reduction in AOA) -0.0000rad/sq -0.0161rad6 0.0024rady Predictions appear to agree well with the numerical results Primary result is a slightly lower angle of attack and a higher speed Predicted initial rates of the output values from code 00001m/s32ti -0.0233rad/sa 1.1569 rad s q 0.0000ad/s6 0.0233 rad /s i All outputs are zero at t=of, but see rapid changes in a and q Changes in u and y(also a function of 0)are much more gradual not as easy to see this aspect of the prediction Initial impact Change in a and q(pitches aircraft Long term impact Change in u(determines speed at new equilib rium condition)

Fall 2004 16.333 5–7 Elevator (1◦ elevator down – stick forward) • See very rapid response that decays quickly (mostly in the first 10 seconds of the α response) • Also see a very lightly damped long period response (mostly u, some γ, and very little α). Settles in >600 secs • Predicted steady state values from code: 14.1429 m/s u (speeds up) ­0.0185 rad α (slight reduction in AOA) ­0.0000 rad/s q ­0.0161 rad θ 0.0024 rad γ – Predictions appear to agree well with the numerical results. – Primary result is a slightly lower angle of attack and a higher speed • Predicted initial rates of the output values from code: ­0.0001 m/s2 u˙ ­0.0233 rad/s α˙ ­1.1569 rad/s2 q˙ 0.0000 rad/s θ ˙ 0.0233 rad/s γ˙ – All outputs are zero at t = 0+, but see rapid changes in α and q. – Changes in u and γ (also a function of θ) are much more gradual – not as easy to see this aspect of the prediction • Initial impact Change in α and q (pitches aircraft) • Long term impact Change in u (determines speed at new equilib￾rium condition)

Fa2004 16.3335-8 8 a 88 人 人 导 R pen)n 18 Figure 1: Step Response to 1 deg elevator perturbation -B747 at M=0.8

Fall 2004 16.333 5–8 0 200 400 600 5 0 30 25 20 15 10 u time 0 200 400 600 −0.03 −0.025 −0.02 −0.015 −0.01 −0.005 0 (rad) α time Step response to 1 deg elevator perturbation 0 200 400 600 −0.1 −0.05 0 0.05 0.1 γ time 0 10 20 30 40 5 0 30 25 20 15 10 u time 0 10 20 30 40 −0.03 −0.025 −0.02 −0.015 −0.01 −0.005 0 (rad) α time 0 10 20 30 40 −0.1 −0.02 −0.04 −0.06 −0.08 0 0.02 γ time Figure 1: Step Response to 1 deg elevator perturbation – B747 at M=0.8

Fa2004 16.33359 Thrust(1/6 input) Motion now dominated by the lightly damped long period response Short period motion barely noticeable at beginning o Predicted steady state values from code 0 m/s u 0 rad 0 rad /s q 0.05rad6 0.05 rad Predictions appear to agree well with the simulations Primary result- now climbing with a flight path angle of 0.05 rad at the same speed we were going before. Predicted initial rates of the output values from code 29430m/ 0 rad /s 000 rad/s2 q d/s 8 rad /s Changes to a are very small, and y response initially flat Increase power, and the aircraft initially speeds up Initial impact Change in u(accelerates aircraft Long term impact Change in y( determines climb rate

Fall 2004 16.333 5–9 Thrust (1/6 input) • Motion now dominated by the lightly damped long period response • Short period motion barely noticeable at beginning. • Predicted steady state values from code: 0 m/s u 0 rad α 0 rad/s q 0.05 rad θ 0.05 rad γ – Predictions appear to agree well with the simulations. – Primary result – now climbing with a flight path angle of 0.05 rad at the same speed we were going before. • Predicted initial rates of the output values from code: 2.9430 m/s2 u˙ 0 rad/s α˙ 0 rad/s2 q˙ 0 rad/s θ ˙ 0 rad/s γ˙ – Changes to α are very small, and γ response initially flat. – Increase power, and the aircraft initially speeds up • Initial impact Change in u (accelerates aircraft) • Long term impact Change in γ (determines climb rate)