【飛行力學】
附加講義
(A) Effects of the variation of parameters on dynamic
system characteristics
(B) Longitudinal responses to atmospheric gusts
103
(A) Effects of the variation of parameters
A.1: Forewords
○ We know that for the longitudinal equation set.
Lu
sU 0  L
 sU 0


Q( s )  s  Du  Tu
D  g
g
 ( s  sSP )( s  sSP
)(s  s Ph )(s  s Ph
),
Mu
sM   M  sM   s 2
and for the lateral equation set
s  Yv
U0
g
~

Q ( s)    Lv
 Lr
s 2  L p s  ( s  sS )(s  s R )( s  s DR )(s  s DR
)
 Nv
s  Nr
 N ps
⇨ sSP : The short period mode, sPh : The Phugoid mode, sS : The spiral
mode, s R : The Roll mode, s DR : The Dutch roll mode
⇨ The values of these dynamic modes are functions of the aerodynamic
derivatives.
○ It is desirable to investigate the change in these dynamic modes due to
variations of the aerodynamic derivatives
104
A.2: For longitudinal motions
A.2.1 The effect on the variation of M
○ Assume that a variation, M , occurs in M so that the equation set of the
perturbed longitudinal motion becomes as follows:
Lu
sU 0  L
 sU 0   u( s )   0 



   0  ( s )
s

D

T
D

g
g

(
s
)
u
u

e






Mu
sM  M  M sM  s2   ( s )   M 
○ Then,
Lu
QM*  ( s )  s  Du  Tu
Mu
Lu
 s  Du  Tu
Mu
sU 0  L
D  g
sM   M   M 
 sU 0
g
sM   s 2
sU 0  L
 sU 0
D  g
sM   M 
g
sM   s 2
 Q ( s )  M     ( s ),
Lu
 M 
s  Du  Tu
 sU 0
.
g
L
  ( s )  s 2  ( Du  Tu ) s  Uu g
0
105
○ We will compare this equation with a feedback system synthesis.
1. For the feedback system shown,  ( s) and
 ( s)
k
 ( s ) are polynomials, and H ( s)   ( s) /  ( s)
( s)
is the open-loop system transfer function.
k H ( s)
k ( s )

2. The closed -loop transfer function will be
.
1  k H ( s )  ( s )  k ( s )
⇨ The so-called closed-loop poles are in fact the roots of the denominator
polynomial ( s)  k ( s) .
3. We can see that  ( s )  k ( s )   ( s )1  k   ( s ) /  ( s )   ( s )1  k H ( s ),
and we use the root locus analysis of 1 k H ( s) to investigate the change
in the roots of ( s)  k ( s) due to variation in k .
4. Likewise, we can write QM*  ( s)  Q( s)1  M    ( s) / Q( s).
⇨  ( s ) Q( s ) will play the role of a fictitious open-loop transfer function, and
M a fictitious feedback gain. And the change in the roots of QM*  ( s) due
to variation in M can then be analyzed by using roots locus theory.
※ Problems: ● How to deal with the minus sign in  M  ?
● Also, how to distinguish M   0 from M   0 ?
106
【 Digression I: Two types of root locus plots 】
○ Root locus plots of  ( s)  k ( s) as k varies.
⇨ Any point on the locus will therefore satisfy  ( s)  k ( s) or 1  kH ( s)  0
or kH ( s)  1 where H ( s)   ( s) /  ( s) .
⇨ Because k  Re and H ( s)  C for all s  C , the need for kH ( s)  1
then implies that H ( s)  1 / k which in turn implies that ∠H(s)=180∘
where ∠H(s) will denote the phase of H(s).
⇨ The need for ∠H(s)=180∘ for every point on the root locus causes the
name of the 180∘ root locus plot.
○ What if a k  kˆ is used?
⇨ In these cases, any point on the locus must instead satisfies  ( s)  kˆ ( s) or
1  kˆH ( s )  0 or kˆH (s)  1.
⇨ This result will require that ∠H(s)=0∘.
⇨ The need for ∠H(s)=0∘ for these cases results in the so called 0∘ root
locus plot.
107
【 Digression II: Similarities and differences of the two root locus plots 】
○ Similarities of the 180∘ root locus plots and the 0∘ root locus plots.
⇨ The locus are symmetric with respect to the real axis.
⇨ The locus start from the roots of  ( s) and end at the roots of  ( s ) or
extend toward infinity through asymptotes.
⇨ The number of asymptotes of the locus equals the order of  ( s) minus
the order of  ( s ) .. (In general,  ( s) is of higher order than  ( s ) )
⇨ The center of asymptotes of the two types of root locus can be computed
using the same formula.
⇨ The asymptotes of the locus plots equally divide the 360∘.
○ Differences between the 180∘ locus and the 0∘ locus.
⇨ The 180∘ locus will not appear on the real axis to the left of even number
of poles and zeros. On the contrary, the 0∘ locus will not appear on the
real axis to the left of odd number of poles and zeros..
⇨ The departure angle of a 180∘ locus will be complementary to those in a 0∘
locus. The same is true for the arrival angles of the locus.
⇨ The asymptotes of a 180∘ locus do not include the positive real axis; they
do for the 0∘ locus.
108
○ An illustrative example using the longitudinal derivatives of Mohawk:
2
1. Entering values for Mohawk, and we have  ( s )  s  0.3058 s  0.074 .
--- Roots of  ( s ) :  0.153  j 0.225 .
2. We already have Q( s )  s  3.6265s  5.6086 s  1.4781s  0.2333.
.  j 0175
.
.
--- Roots of Q( s ) : p1  014
and p2  1.67  j1348
.
3. Therefore, the fictitious open-loop transfer function has 2 zeros and 4 poles.

--- If we consider M  0 , then 180 condition applies.
4
3
2

--- Conversely, 0 root locus must be drawn for the case with M  0 .

M  0 ( 180 locus)
M  0 ( 0 locus)
s=+j
j

--- Decrease in M does not change p1 much, but hurt p2 a lot.
109
A.2.2 The variation of M u
○This time, the variation term is M u , and the equation set becomes
Lu
sU 0  L
 sU 0   u( s )   0 



   0  ( s )
s

D

T
D

g
g

(
s
)
u
u

e





 M u  M u sM  M sM  s2   ( s )   M 
gL
○Then, QM* u ( s )  Q( s )  M u  u ( s ), u ( s )  D ( s  U D ) .
0

. ( s  111
. )  One zero at 111
--- For Mohawk, u ( s )  2912
. .
--- The fictitious system has 4 poles and 1 zero, 3 asymptotes will appear.

M u  0 ( 180 locus)
M u  0 ( 0 locus)
s=+j

j

110
A.3: The lateral motions
A.3.1 The effects of the variation on N v
○ The procedure of Section A.2 will be followed here.
s  Yv
U0
~
○ Original lateral characteristic polynomial: Q ( s )    Lv
 Lr
 Nv
s  Nr
g
s 2  Lp s
 N ps
~
--- For MOHAWK: Q ( s )  ( s  0.04)( s  2.75)( s  0.33  j1.97)( s  0.33  j1.97) .
○ Insert N v  N v  N v into the equation set and expand the determinant:
s  Yv
U0
g 

~

 ~
QN* v ( s)   
 Lv
 Lr
s 2  L p s   Q ( s )  N v ( s )
 ( N v  N v ) s  N r
 N p s 
g
where ( s )  s 2  L p s 
Lr .
U0
2
--- For MOHAWK,  ( s )  s  2.692s  0.202  ( s  0.073)( s  2.765) .
~
~
○ In standard root locus format: QN* v ( s )  Q ( s )1  N v  (( s ) / Q( s)) .
111
--- Using data for MOHAWK:
s  0.073s  2.765
(s)
.
~ 
Q ( s ) s  0.0424s  2.75( s  0.334  j1.97)( s  0.334  j1.97)
--- We have 4 poles and 2 zeros, with one pole and one zero are in LHP.
--- The ‘+’sign in front Of N v will imply 180 locus for N v  0 .
s=+j
N  0 ( 180 locus)
v
N v  0 ( 0 locus)
j

--- Decreasing N v will help in improving the Spiral mode and the Dutch roll mode.
--- The Roll mode may suffer, but it is all right.
112
A.3.2 The effect of the variation on Lv
U0
g 
 s  Yv
~

 ~
○ Here, QL*v ( s )    ( Lv  Lv )  Lr
s 2  L p s   Q ( s)  Lv ( s )
  N v
s  Nr
 N p s 
and ( s )  ( g  N pU 0 )s  gN r .
--- For MOHAWK, ( s)  72.447( s  0.236) .
--- We have 4 poles and 1 zeros; hence, 3 asymptotes will result.
--- The ‘-’sign in front Of Lv will imply 180 locus for Lv  0 .
s= +j

Lv  0 ( 180 locus)
Lv  0 ( 0 locus)
j

--- Decreasing Lv , i.e. increasing dihedral, will improve the spiral.
--- However, Dutch roll mode will suffer.
113
A.3.3 The effect of the variation on N r
U0
g 
 s  Yv
~

 ~
○ This time, QN* r ( s )     Lv
 Lr
s 2  L p s   Q ( s)  N r ( s)
  N v s  ( N r  N r )  N p s 
and ( s)  s3  Yv  L p s 2  Yv L p s  gLv .
--- For MOHAWK: ( s)  ( s  2.756)( s  0.045  j 0.405)( s  0.045  j 0.405)
--- The ‘-’sign in front of N r will imply 180 locus for N r  0 .
s=+j

N r  0 ( 180 locus)
N r  0 ( 0
 locus)
j

--- Decreasing N r , i.e. increasing the yaw damping, will improve both the spiral
mode and the Dutch roll mode.
114
(B) Longitudinal responses to atmospheric gusts
B.1 Preliminary remarks
○ We are going to deal with the appearance of atmospheric gusts
Atomspheric
disturbances
Flight speed U
⇨ The two components of the gust is often referred to as horizontal gust and
the vertical gust.
Atomspheric Gust
ug : Horizontal gust
wg : Vertical gust
○ What will the atmospheric gusts do to the aircraft motion?
When there are gusts
Without the gusts
The relative wind speed equals the
opposite of the A/C flight speed
Flight speed U
U
Relative wind U
ug
wg
g
Flight speed U
Relative wind U
The horizontal gust introduces additional change in relative wind speed
The vertical gust introduces additional change in effective angle of attack
The aerodynamics of the aircraft motion will be changed by the gusts
115
B.2 Re-derive the longitudinal equations
○ The problem can be resolved by re-deriving the equations of motion.
○ Lets go back to the pre-treated longitudinal equations on p.33:
y
 L  W  mU (  ),
T  D  W (   )  mU ,
M aero
 I y
○ To linearize this equation set, we had introduced the perturbations
U  U 0  U ,    0   , and    0  
--- Among these perturbations, U and   w /U 0 represent
the changes in relative wind due to change in aircraft flight speed.
○ With the presence of the gusts, the following modification to the perturbations
of U and  will be necessary:
U  U 0  U  u g and    0    wg /U 0 .
--- Minus signs reflectt that gust wind changes the relative wind, which
is of opposite sign of the flight speed.
⇨ These changes in the perturbations of U and  will necessitate proper
modifications to the aerodynamic terms of the linearized equations.
⇨ The gravitational terms and the inertial terms of the equations, however,
will not be affected by the presence of atmospheric gusts.
116
○ Linearization of the equation set under the presence of gusts:
 For the drag equation:
1. Expansion of T and D that includes the gust terms:
T
T
T
T  T0 
 (U  u g )  T0 
 U 
u .
U E
U E
U E g
and
wg
w 
D
D 
 (U  u g ) 
    g   Ltrim 
U E
 E 
U0
Uo 
wg
wg
D
D
 D u   D


L

 D0 
 U 
  
trim
U0
U E
 E
U E g   E U0
D  D0 
L  Ltrim
ug
U
Relative wind U
wg
--- The last term on the RHS reflects that vertical gust slants the relative wind,
hence the lift vector, causing an additional drag force to appear.
--- We have also assume that u g  U 0 and wg  U 0 so that L  Ltrim .
2. The resulting the drag equation.
(Tu  Du )U  D   U  g (   )
 (Tu  Du ) u g  U10 ( g  D ) wg
.
⇨ Gust effect appears as two additional terms on the RHS.
⇨ These additional terms act as inputs to the longitudinal dynamics.
117
 For the lift equation:
1. Expansion of L that includes the gust terms:
wg
L
L
L  L0 
 (U  u g ) 
 (  )
U E
 E
U0
wg
L
L
L
L
 L0 
U 
 
u 

U E
 E
U E g   E U0
.
2. The resulting lift equation:
Lu U  L   U 0 (   )  Lu u g  ( L / U 0 )  wg
 For the pitching moment equation:
1. In previous note, we had expanded M into as follows:
M  M0 
M
M
M
M
M
 
    
U 
 e .
 E
  E
U E

 E
e
E
2. Here, the following modifications will be made:
U


 U  u g
   wg / U 0
   w g / U 0

 Three additional terms will appear

 on the pitching moment equation

3. The following phenomenon about the vertical gust has also been observed:
118
 w
 g is a result of dwg / dx  0 :
dwg
x
dwg
dx dwg
w g 

 
U0
dt
dx dt
dx
wg( x )
 In such a vertical gust, the tail feels an additional w g (w.r.t. the wing)
wg   X t  (dwg / dx)
--- Pointing upward.
 An additional change in  t will result:
wg
X t dwg X t w g
 t  




U 0 U 0 dx U 0 U 0
 This change in  t results in the following change in pitching moment:
w g
M
M   t 
 M  
 t
U0
(Because
 M Xt

 M  )
 t U
4. The final pitching moment equation:
M   M   M   M u U  
 M uug  U1 [ M wg  ( M  M )w g ]  M  e
0
⇨ Note that a total of four gust terms appear in the RHS.
119
【Modified longitudinal equations with gust inputs】
○ Time domain equation set:
Luu  L   U 0 (   )  Luug  ( L / U 0 )wg
(Tu  Du )u  D   u  g(   )  (Tu  Du )ug  U1 ( g  D )wg
0
M   M   M   M uu  
 M uug  U1 [ M wg  ( M  M )w g ]  M  e
0
※ We had used: U  u ,    ,    and  e   e .※
○ Laplace domain equation set:
Lu
sU 0  L


D  g
 s  Du  Tu

Mu
sM  M
--- Setting
 sU 0   u( s ) 


g

(
s
)


2
sM  s   ( s ) 
 L
u

  Du  Tu

 Mu

e  0
for simplicity.

  u ( s) 
g
1 (D  g)


 w ( s )
U0
 g 
1 [( M  M ) s  M ] 

 

U0
1 L
U0 
120
B.3 Longitudinal response to vertical gusts, or downdraft
○ Let's consider a constant  .
--- An auto-pilot can achieve this maneuver.
 Remaining equations of motion:
1
Lu

 U10 L 
U 0 ( sU 0  L )   u ( s ) 
s  D  T
  w( s )   1 ( D  g ) wg ( s )
1
(
D

g
)
U0
u
u

 U 0 



⇨ We have substituted w / U 0 for  , so that variation in vertical speed of the
aircraft can be analyzed.
 Using Cramer's rule, we will have:
w( s )  H ( s ) wg ( s ) ,
where
b( s )
H ( s) 
,
a( s )
Lu
b( s ) 
s  Du  Tu
and
Lu
a( s ) 
s  Du  Tu
1
U0
L
,
1
U 0 ( D  g )
( sU 0  L )
.
1
U 0 ( D  g )
1
U0
121
 For a constant vertical gust, i.e. wg ( t )  wg , or wg ( s )  wg / s where w g
is some appropriate constant, the steady state response of w will be
wg
lim w( t )  lim s  w( s )  lim s  H ( s ) 
 wg lim H ( s ) .
s
t 
s 0
s 0
s 0
--- We have quoted Final value theorem for this result.
And with
b ( s ) s 0
Lu

Du  Tu
1
U0
L
 a( s ) s 0 ,
1
(
D

g
)
U0

we will have
lim H ( s ) 
s 0
hence, lim w( t )  wg .
b( s )
 1;
a( s ) s 0
t 
⇨ In the end, aircraft rises or falls with same
velocity as a vertical gust.
122
B.4 Longitudinal response to horizontal gusts
○We will stick with the constant  assumption. But let's go back to the equation
set in terms of  . The remaining equation set will be:
Lu
sU 0  L   u( s )   Lu 


ug ( s ) .
s  D  T




D  g   ( s )  Du  Tu 

u
u
○ For a step horizontal gust, u g (t )   , or u g ( s )   / s , where  is some constant:
 Cramer's rule will give us
 Du  Tu s  UL0  Du  Tu   ULu0  D  g 
u( s ) 
ug ( s )
L
L
Lu
2
s  U  Du  Tu s  U  Du  Tu   U  D  g 


0
0
0
 For this step gust, u g ( s )   / s ; hence,
Du  Tu s  UL Du  Tu   UL D  g 

u ( s)  2 L

L
L






s  U  Du  Tu s  U Du  Tu  U D  g s

u
0
0


u
0
0
0
 At the steady state, we will have
limt  u(t )  lims0 s  u(s)   .
⇨ The A/C is drifted with same velocity as gust..
123
○ But we are more concerned about the aircraft response to a ramp horizontal
gust, namely u g (t )   t where  is some constant:
 Such a gust, called a wind shear, is caused by strong convection of air.
 The phenomenon is called a microburst..
Actual gust
Horizontal gust, ug
The air flow
A typical
microburst
 Inside a microburst, u g (t )     t where  and  are constants.
 We have already treated u g (t )   , the step portion of the gust, now we
will dealt with its ramp portion, u g (t )   t .
124
(A) Change of speed in a wind shear:
-- Using data of Mohawk.
 For a ramp gust, u g ( s )   / s 2 ; hence,
u (s)
0.3058s  0.3138
1 3.197
3.212
0.015
 2 2
 2


.

s ( s  1.309s  0.3138) s
s
s  0.316 s  0.993
 u (t )  (t  3.197  3.212e 0.316 t  0.015e 0.993t )  
 This value of u(t) computes the inertial flight speed of the aircraft.
But we are also interested in the change in relative air speed.
⇨ Change in relative air speed: u rel (t )  u (t )  u g (t )  u (t )  u~g t .
⇨ At steady state: limt 1 urel (t )  3.197   0 .
10
 Observations:
⇨ For a tail wind shear (順風),   0 , the 8
inertial flight speed is increasing, but
the relative air speed is decreasing.
Eventually, the relative air speed will
decrease by an amount  3.197 .
⇨ Decrease in relative air speed reduces
6
ug ( t )
urel
4
2
0
u( t )
0
2
4
6 t
lift, causing the A/C to loss altitude.
⇨ May cause crashes at take off or landing, see discussion below.
8
10
125
(B) Change in altitude in a wind shear:


x -axis
 Rate of change in altitude:

h   U 0  (   ) U 0 .
h   U0
 A   0 is assumed; hence,
Horizon
   .
 Therefore,
h( s)  U 0   ( s) / s and
( Lu / U 0 ) s

 ( s)  2 L
 2
L
Lu






s  U 0  Du  Tu s  U 0 Du  Tu  U0 D  g s
 Using data of Mohawk as an example:
h( s )
 0.3841
5.106 1.224
5.682
0.576
 2 2

 2 

.

s ( s  1.309s  0.3138)
s
s
s  0.316 s  0.993
 h(t )  (5.106  1.224t  5.682e 0.316 t  0.576e 0.992 t )  
 Observations:
⇨ Eventually, limt 1 h(t )  0 , indicating a A/C crash, for   0 .
⇨ This danger of crash may occur near an airport where an aircraft is in
low altitude doing take off or landing.
126