Lecture ac 2 Aircraft Longitudinal dynamics Typical aircraft open-loop motions · Longitudinal modes · Impact of actuators . Linear algebra in action! Roll (Rudder) (Aileron) Pitch (Elevator) d A Longitudinal Axis Lateral axis Vertical Axis Copyright 2003 by Jonathan How
Lecture AC 2 Aircraft Longitudinal Dynamics • Typical aircraft open-loop motions • Longitudinal modes • Impact of actuators • Linear Algebra in Action! Copyright 2003 by Jonathan How 1
Spring 2003 1661AC22 Longitudinal Dynamics For notational simplicity, let X=Fn, Y= Fu, and Z= F aF Longitudinal equations(1-15 )can be rewritten as mi=X+X2- mg cos(0+△X mli-qUo)= Zuu+ Zww+ Zii+Zaq-mg sin 000+AZc Iyyq= Mau+ Mww+ Mii+Ma+AMc There is no roll/yaw motion, so q=0 The control commands△X≡△F,△Z≡△F,and△Me≡△M have not yet been specified Rewrite in state space form as n XX -mg cos eo △X (m-Zia ZuZu za+mUo -mg sin 00w △Z Mhiw+ly9 Mu Mw Mg 0 q △M 00 1 0 0001[a1「XnXn mg cos o △X 0 m-Zi 0ai Za zu z+mU0- mg sin e0△Z 0 -Mi lm o Mn ma 00010 EX AX+c descriptor state space form X=E-(AX+c)=AX+c
Spring 2003 16.61 AC 2–2 Longitudinal Dynamics • For notational simplicity, let X = Fx, Y = Fy, and Z = Fz Xu ≡ ∂Fx ∂u ,... • Longitudinal equations (1–15) can be rewritten as: mu˙ = Xuu + Xww − mg cos Θ0θ + ∆Xc m( ˙w − qU0) = Zuu + Zww + Zw˙ w˙ + Zqq − mg sin Θ0θ + ∆Zc Iyyq˙ = Muu + Mww + Mw˙w˙ + Mqq + ∆Mc – There is no roll/yaw motion, so q = ˙ θ. – The control commands ∆Xc ≡ ∆Fc x, ∆Zc ≡ ∆Fc z , and ∆Mc ≡ ∆Mc have not yet been specified. • Rewrite in state space form as mu˙ (m − Zw˙) ˙w −Mw˙w˙ + Iyyq˙ ˙ θ = Xu Xw 0 −mg cos Θ0 Zu Zw Zq + mU0 −mg sin Θ0 Mu Mw Mq 0 00 1 0 u w q θ + ∆Xc ∆Zc ∆Mc 0 m 0 00 0 m − Zw˙ 0 0 0 −Mw˙ Iyy 0 0 0 01 u˙ w˙ q˙ ˙ θ = Xu Xw 0 −mg cos Θ0 Zu Zw Zq + mU0 −mg sin Θ0 Mu Mw Mq 0 00 1 0 u w q θ + ∆Xc ∆Zc ∆Mc 0 EX˙ = AXˆ + ˆc descriptor state space form X˙ = E−1 (AXˆ + ˆc) = AX + c
ng2003 16.61AC2-3 Write out in state space form X -g cos eo wy! [M+Zur]Iwy IM+ Zur wy! [Mg+(Zg+mUo)r| mg sin or 1 M To figure out the c vector, we have to say a little more about how the control inputs are applied to the system
Spring 2003 16.61 AC 2–3 • Write out in state space form: A = Xu m Xw m 0 −g cos Θ0 Zu m−Zw˙ Zw m−Zw˙ Zq+mU0 m−Zw˙ −mg sin Θ0 m−Zw˙ I−1 yy [Mu + ZuΓ] I−1 yy [Mw + ZwΓ] I−1 yy [Mq + (Zq + mU0)Γ] −I−1 yy mg sin ΘΓ 0 0 1 0 Γ = Mw˙ m − Zw˙ • To figure out the c vector, we have to say a little more about how the control inputs are applied to the system
Spring 2003 1661AC24 Longitudinal Actuators Primary actuators in longitudinal direction are the elevators and the thrust Clearly the thrusters elevators play a key role in defining the stead state/equilibrium fight condition Now interested in determining how they also influence the aircraft mo- tion about this equilibrium condition deflect elevator u(t), w(t), q(t), δ.(+) Rudde Recall that we defined AXc 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 the same perturbation approach △Xc=X60e+X6, de is the deflection of the elevator from trim(down positive) Sp change in thrust Xs and Xs, are the control stability derivatives
Spring 2003 16.61 AC 2–4 Longitudinal Actuators • Primary actuators in longitudinal direction are the elevators and the 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 ∆Xc 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 the same perturbation approach ∆Xc = Xδe δe + Xδp δp – δe is the deflection of the elevator from trim (down positive) – δp change in thrust – Xδe and Xδp are the control stability derivatives
Spring 2003 16.61AC2-5 ● Now we have that △X X8 X E-1/42 26.26 △M for the longitudinal case B Typical values for the B747 X 16.54 X6n=0.3mg=849528 M6=-5.2·10 M≈0 Aircraft response y=G(su X=AX+ Bu -G(s)=C(sI-A-B y=CX We now have the means to modify the dynamics of the system, but first let's figure out what de and d, really do
Spring 2003 16.61 AC 2–5 • Now we have that c = E−1 ∆Xc ∆Zc ∆Mc 0 = E−1 Xδe Xδp Zδe Zδp Mδe Mδp 0 0 δe δp = Bu • For the longitudinal case B = Xδe m Xδp m Zδe m−Zw˙ Zδp m−Zw˙ I−1 yy [Mδe + ZδeΓ] I−1 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
Spring 2003 16.61AC26 Longitudinal Response Final response to a step input u =i/s, y=G(s)u, use the FVT lim y(t)= lims G(s) =lim y(t)=G(O)u=-CA B)i Initial response to a step input, use the IVT ()=s(G(s)=)=iG(s)i For your system, G(s)=C(sI-A)B+D, but D=0, so linG(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(0) i(t)=CX=CAX+CBu and normally have that CB#0. Repeat process above to show that i(0+)=CBi, and since C=I, (0+)=Bi Looks good. Now compare with numerical values computed in MATLAB Q Plot u, a, and fight path angle ?=8-a(since 00=10=0) See ac 1-10 I Note that C(sI-A)-B+D=D+CB+CArE+
Spring 2003 16.61AC2-7 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 y, and very little a). Settles in >600 secs e Predicted steady state values from code 14.1429 m/s u(speeds up 0.0185 rad a(slight reduction in AOA) -0.0000rad/sq 0.0161radb 0.0024rady Predictions appear to agree well with the numerical results Primary result is a slightly lower angle of attack and a higher sDee Predicted initial rates of the output values from code -0.0001m/s2 0.0233 rad/s a 1.1569rad/s2 0.0000rad/s6 0.0233rad/s All outputs are at zero at t=0, 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 equilibrium condition
Spring 2003 16.61 AC 2–7 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 at 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 equilibrium condition)
Spring 2003 16.61AC2-8 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 0 rad 0 rad/s q 0. 05 rad 0 0.05 rad Predictions appear to agree well with the simulations Primary result is that we are 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/s2 0 0 rad/s 8 0 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
Spring 2003 16.61 AC 2–8 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 is that we are 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)
Spring 2003 16.61AC2-9 o u 8 g 8 a 8 言8 alpha(rad) alpha(rad 8 bb占占 Figure 1: Step Response to 1 deg elevator perturbation-B747 at M=0. 8
Spring 2003 16.61 AC 2–9 0 200 400 600 0 5 10 15 20 25 30 u time 0 200 400 600 −0.03 −0.025 −0.02 −0.015 −0.01 −0.005 0 alpha (rad) time Step response to 1 deg elevator perturbation 0 200 400 600 −0.1 −0.05 0 0.05 0.1 gamma time 0 10 20 30 40 0 5 10 15 20 25 30 u time 0 10 20 30 40 −0.03 −0.025 −0.02 −0.015 −0.01 −0.005 0 alpha (rad) time 0 10 20 30 40 −0.1 −0.08 −0.06 −0.04 −0.02 0 0.02 gamma time Figure 1: Step Response to 1 deg elevator perturbation – B747 at M=0.8
ng2003 16.61AC2-10 O 0 alpha(rad) alpha(rad) 8 gamma gamma 9R88吕吕 88 Figure 2: Step Response to 1/6 thrust perturbation- B747 at M=0. 8
Spring 2003 16.61 AC 2–10 0 200 400 600 −15 −10 −5 0 5 10 15 u time 0 200 400 600 −0.02 −0.015 −0.01 −0.005 0 0.005 0.01 0.015 0.02 alpha (rad) time Step response to 1/6 thrust perturbation 0 200 400 600 0 0.02 0.04 0.06 0.08 0.1 gamma time 0 10 20 30 40 0 1 2 3 4 5 6 7 u time 0 10 20 30 40 −5 0 5 10 15 20 x 10−4 alpha (rad) time 0 10 20 30 40 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 gamma time Figure 2: Step Response to 1/6 thrust perturbation – B747 at M=0.8