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