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