Lecture AC 2 Aircraft Longitudinal Dynamics ? Typical aircraft open-loop motions ? Longitudinal modes ? Impact of actuators ? Linear Algebra in Action! Copy right 2003 by Jon at h an H ow 1 Spring 2003 16.61 AC 2–2 Longitudinal Dynamics ? For notational simplicity, let X = F x , Y = F y ,andZ = F z X u ≡ parenleftBigg ?F x ?u parenrightBigg ,... ? Longitudinal equations (1–15) can be rewritten as: m ˙u = X u u + X w w ?mg cos Θ 0 θ +?X c m(˙w ?qU 0 )=Z u u + Z w w + Z ˙w ˙w + Z q q ?mg sin Θ 0 θ +?Z c I yy ˙q = M u u + M w w + M ˙w ˙w + M q q +?M c – There is no roll/yaw motion, so q = ˙ θ. – The control commands ?X c ≡ ?F c x ,?Z c ≡ ?F c z ,and?M c ≡ ?M c have not yet been specified. ? Rewrite in state space form as ? ? ? ? ? ? ? ? m ˙u (m?Z ˙w )˙w ?M ˙w ˙w + I yy ˙q ˙ θ ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? X u X w 0 ?mg cosΘ 0 Z u Z w Z q + mU 0 ?mg sin Θ 0 M u M w M q 0 00 1 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? u w q θ ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ?X c ?Z c ?M c 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? m 000 0 m?Z ˙w 00 0 ?M ˙w I yy 0 00 01 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ˙u ˙w ˙q ˙ θ ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? X u X w 0 ?mg cos Θ 0 Z u Z w Z q + mU 0 ?mg sin Θ 0 M u M w M q 0 00 1 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? u w q θ ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ?X c ?Z c ?M c 0 ? ? ? ? ? ? ? ? E ˙ X = ? AX +?c descriptor state space form ˙ X = E ?1 ( ? AX +?c)=AX + c Spring 2003 16.61 AC 2–3 ? Write out in state space form: A = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? X u m X w m 0 ?g cos Θ 0 Z u m?Z ˙w Z w m?Z ˙w Z q +mU 0 m?Z ˙w ?mg sinΘ 0 m?Z ˙w I ?1 yy [M u + Z u Γ] I ?1 yy [M w + Z w Γ] I ?1 yy [M q +(Z q + mU 0 )Γ] ?I ?1 yy mg sin ΘΓ 0 0 1 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Γ= M ˙w m?Z ˙w ? 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 2–4 Longitudinal Actuators ? Primary actuators in longitudinal direction are the elevators and the thrust. – Clearly the thrusters/elevators play a key role in defining the steady- state/equilibrium flight 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),... ? Recall that we defined ?X c 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 ?X c = X δ e δ e + X δ p δ p – δ e is the deflection of the elevator from trim (down positive) – δ p change in thrust – X δ e and X δ p are the control stability derivatives Spring 2003 16.61 AC 2–5 ? Nowwehavethat c = E ?1 ? ? ? ? ? ? ? ? ?X c ?Z c ?M c 0 ? ? ? ? ? ? ? ? = E ?1 ? ? ? ? ? ? ? ? X δ e X δ p Z δ e Z δ p M δ e M δ p 00 ? ? ? ? ? ? ? ? ? ? δ e δ p ? ? = Bu ? For the longitudinal case B = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? X δ e m X δ p m Z δ e m?Z ˙w Z δ p m?Z ˙w I ?1 yy [M δ e + Z δ e Γ] I ?1 yy bracketleftBig M δ p + Z δ p Γ bracketrightBig 0 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Typical values for the B747 X δ e = ?16.54 X δ p =0.3mg = 849528 Z δ e = ?1.58·10 6 Z δ p ≈ 0 M δ e = ?5.2·10 7 M δ p ≈ 0 ? Aircraft response y = G(s)u ˙ X = AX + Bu → G(s)=C(sI ?A) ?1 B y = CX ? We now have the means to modify the dynamics of the system, but first let’s figure out what δ e and δ p really do. 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,someγ,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/s 2 ˙u -0.0233 rad/s ˙α -1.1569 rad/s 2 ˙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 16.61 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: 0m/su 0radα 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/s 2 ˙u 0 rad/s ˙α 0 rad/s 2 ˙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 16.61 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 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 Spring 2003 16.61 AC 2–11 ? Summary: –To increase equilibrium climb rate, add power. –To increase equilibrium speed, increase δ e (move elevator further down). – Transient (initial) e?ects are the opposite andtendtobemoreconsistentwith what you would intuitively expect to occur Spring 2003 16.61 AC 2–12 Modal Behavior ? Analyze the model of the vehicle dynamics to quantify the responses we saw. – Homogeneous dynamics are of the form ˙ X = AX, so the response is X(t)=e At X(0) – a matrix exponential. ? To simplify the investigation of the system response, find the modes of the system using the eigenvalues and eigenvectors – λ is an eigenvalue of A if det(λI ?A)=0 which is true i? there exists a nonzero v (eigenvector)forwhich (λI ?A)v =0 ? Av = λv – If A (n×n), typically will get n eigenvalues and eigenvectors Av i = λ i v i – Assuming that the eigenvectors are linearly independent,canform A bracketleftBig v 1 ··· v n bracketrightBig = bracketleftBig v 1 ··· v n bracketrightBig ? ? ? ? ? λ 1 0 . . . 0 λ n ? ? ? ? ? AT = TΛ ? T ?1 AT =Λ ,A= TΛT ?1 – Given that e At = I + At + 1 2! (At) 2 + ...,andthatA = TΛT ?1 ,thenitis easy to show that X(t)=e At X(0) = Te Λt T ?1 X(0) = n summationdisplay i=1 v i e λ i t β i – State solution is a linear combination of the system modes v i e λ i t e λ i t – determines the nature of the time response v i – determines the extent to which each state contributes to that mode β i – determines the extent to which the initial condition excites the mode Spring 2003 16.61 AC 2–13 ? Thus the total behavior of the system can be found from the system modes ? Consider numerical example of B747 A = ? ? ? ? ? ? ? ? ?0.0069 0.0139 0 ?9.8100 ?0.0905 ?0.3149 235.8928 0 0.0004 ?0.0034 ?0.4282 0 001.0000 0 ? ? ? ? ? ? ? ? which gives two sets of complex eigenvalues λ = ?0.3717±0.8869i,ω=0.962,ζ=0.387, short period λ = ?0.0033±0.0672i,ω=0.067,ζ=0.049, Phugoid - long period – result is consistent with step response - heavily damped fast re- sponse, and a lightly damped slow one. ? To understand the eigenvectors, we have to do some normalization (scales each element appropriately so that we can compare relative sizes) – ?u = u/U 0 ,?w = w/U 0 ,?q = q/(2U 0 /c) – Then divide through so that θ ≡ 1 Short Period Phugoid ?u 0.0156 + 0.0244i ?0.0254 + 0.6165i ?w 1.0202 + 0.3553i 0.0045 + 0.0356i ?q ?0.0066 + 0.0156i ?0.0001 + 0.0012i θ 1.0000 1.0000 ? Short Period – primarily θ and α =?w in the same phase. The ?u and ?q response is very small. ? Phugoid – primarily θ and ?u,andθ lags by about 90 ? .The?w and ?q response is very small ? consisitent with approximate solution on AC 2–1? ? Dominant behavior agrees with time step responses – note how initial con- ditions were formed. Spring 2003 16.61 AC 2–14 0.96166 0.54017 1.0803 30 210 60 240 90 270 120 300 150 330 180 0 0.067282 0.5 1 30 210 60 240 90 270 120 300 150 330 180 0 0 5 10 15 ?1 ?0.5 0 0.5 Perturbation States u,w,q time (sec) Short Period 0 100 200 300 400 500 600 ?1 ?0.5 0 0.5 1 Perturbation States u,w,q time (sec) Phugoid uwq uwq Figure 3: Mode Response – B747 at M=0.8