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Problems for Chapter I
Problem 1.1
Consider the Conservation of Linear Momentum Equations (CLMEs) with respect to the
body axes X , Y , Z :


m V  UR  PW   m g   F  F 
m W  PV  QU   m g   F  F 
m U  QW  RV   m g X  FA X  FT X
Y
Z
AY
TY
AZ
TZ
Rewrite the above equations under the following conditions:
-
constant pitching maneuver (loop in the XZ plane);
-
constant rolling around the X axis;
-
steady-turning at constant altitude (with ailerons and rudder maneuver).
Solution of Problem 1.1
Constant Pitching Maneuver
Figure P1.1 shows a constant pitching maneuver.
Figure P1.1.1 - Constant pitching maneuver
Using the steady state conditions,
U1  V1  W1  0
A constant pitching maneuver implies:
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V1  0 , P1  R1  0 , 1  0
With the above conditions, the GEs lead to:
g X   g sin 
gY  g cos  sin   0
g Z  g cos  cos   g cos 
and the KEs lead to:
 1  1 sin 1 tan 1
  
cos 1
 1   0
  0 sin  sec 
1
1
 1 
Which implies:
cos 1 tan 1   P1  1 0 tan 1   0 
 
 
0  Q1 
 sin 1  Q1   0 1
 R  0 0 sec    0 
cos 1 sec 1  
1
1


1  Q1
Finally, the CLMEs for this case are,




m V  UR  PW   m g   F  F   0   F  F 
m W  PV  QU   m g   F  F   m   U   m g cos    F  F 
m U  QW  RV   m g X  FA X  FT X  m  1W1   m g sin 1  FA X  FT X
Y
AY
TY
AY
Z
AZ
TZ
1
TY
1
Constant Rolling Around the X-Axis
Figure P1.2 shows a constant rolling around the X-axis maneuver.
Figure P1.1.2 - Constant rolling around the X-axis
Using the steady state conditions:
U1  V1  W1  0
a constant rolling around the X-axis implies:
Q1  R1  0
1
AZ
TZ
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Using the above conditions the KEs lead to:
 1  1 sin 1 tan 1 cos 1 tan 1   P1  1 sin 1 tan 1
  
 
cos 1
cos 1
 sin 1  Q1   0
 1   0
  0 sin  sec  cos  sec    R  0 sin  sec 
1
1
1
1 1
1
1

 1 
cos 1 tan 1   P1 
 
 sin 1   0 
0 
cos 1 sec 1  

Thus:
1  P1
In this case the CLMEs are given by:




m V  UR  PW   m g   F  F   m   W   m g   F  F 
m W  PV  QU   m g   F  F   m   V   m g   F  F 
m U  QW  RV   m g X  FA X  FT X  0  m g X  FA X  FT X
Y
AY
TY
1
Z
AZ
TZ
1 1
1
Y
Z
Steady-Turning at Constant Altitude
Figure P1.3 shows a steady-turning at constant altitude maneuver.
Figure P1.1.3 - Steady-turning at constant altitude
Using the steady state conditions:
U1  V1  W1  0
A steady-turning at constant altitude implies:
  k  ,   0,   0
Using the IKEs and the above conditions we obtain:
P    sin    sin 
Q  cos   cos  sin   cos  sin 
R  cos  cos   sin   cos  cos 
Thus, the CLMEs for this case are:
AY
TY
AZ
TZ
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


m V  UR  PW   m g   F  F   m  cos  cos U  sin W    m g   F  F 
m W  PV  QU   m g   F  F   m  sin V  cos  sin U    m g   F  F 
m U  QW  RV   m g X  FA X  FT X  m  cos  sin W  cos  cos V    m g X  FA X  FT X
Y
Z
AY
TY
AZ
TZ
Y
AY
Z
TY
AZ
TZ
Problem 1.2
Consider a Vertical Take Off Landing (VTOL) aircraft – such as the AV8 Harrier
aircraft. Assume that the aircraft has a jet engine with a constant angular momentum
h  I R R . Derive expressions for hX , hY , hZ at the following configurations:
-
rotor axis at angle of 90 deg tilted upwards with respect to the X axis;
-
rotor axis at angle of 60 deg tilted upwards with respect to the X axis;
-
rotor axis at angle of 30 deg tilted upwards with respect to the X axis;
-
rotor axis aligned with the X axis.
Solution of Problem 1.2
The aircraft features one engine (N=1); therefore:
h  hX i  hY j  hZ k  I R R
Case #1: rotor axis at angle of 90 deg tilted upwards with respect to the X-axis
In this case the engine angular momentum is along the aircraft Z-axis, then:
h  hZ k  I R R  hZ  I R R , hX  hY  0
Case #2: rotor axis at angle of 60 deg tilted upwards with respect to the X-axis
Figure P2.1 shows this setting. It can be seen that:
hX  h cos 60  0.5h  0.5I R R
hZ  h sin 60  0.866h  0.866I R R
hY  0

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h
hZ
hX
Airplane body axis
Rotor axis
Figure P1.2.1 - Rotor tilted at 60 deg
Case 3: rotor axis at angle of 30 deg tilted upwards with respect to the X-axis
Figure P2.2 shows this setting. It can be seen that:
hX  h cos30  0.866h  0.866I R R
hZ  h sin 30  0.5h  0.5I R R
hY  0
hZ
h
hX
Airplane body axis
Rotor axis
Figure P1.2.2 - Rotor tilted at 30 deg
Case 4: rotor axis aligned with the X-axis
In this case the engine angular momentum is along the aircraft X-axis; therefore:
h  hX i  I R R  hX  I R R , hZ  hY  0
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Problem 1.3
Consider a Vertical Take Off Landing (VTOL) aircraft with a thrust vectoring system.
Assume that the aircraft has a jet engine with a constant angular momentum h  I R R .
Derive expressions for hX , hY , hZ at the following configuration:
-
rotor axis at angle of 30 deg tilted upwards with respect to the X axis and at an
angle of 20 deg. tilted to the right with respect to the X axis.
Solution of Problem 1.3
The aircraft features one engine (N=1); therefore:
h  hX i  hY j  hZ k  I R R
Figure P3.1 shows the rotor angular momentum and its components. From the top
triangle we have:
hX  hXZ cos 30  0.866hXZ
hZ  hXZ sin 30  0.5hXZ
From the triangle at the bottom of the figure we have:
hX  hXY cos 20  0.94hXY
hY  hXY sin 20  0.342hXY
The magnitude of the angular momentum is given by,
h2  hX2  hY2  hZ2   0.866hXZ    0.342hXY    0.5hXZ 
2
2
2
Equating the expressions for hX obtained from each triangle,
0.866hXZ  0.94hXY  hXY  0.921hXZ
Therefore:
2
h2  hX2  hY2  hZ2   0.866hXZ    0.342  0.921hXZ    0.5hXZ   1.099hXZ
2
2
h  I RR  1.099hXZ  hXZ  0.954I RR
hXY  0.921hXZ  0.921  0.954I RR  0.879I RR
Finally, using the expressions obtained from the above triangles:
hX  0.866hXZ  0.866  0.954 I RR  0.826 I RR
hZ  0.5hXZ  0.5  0.954 I RR  0.477 I RR
2
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hY  0.342hXY  0.342  0.879I RR  0.3I RR
hXZ
hXZ
hZ
h
hX
hZ
30°
Y
hXY
hXY
hY
hX
hY
X
20°
hX
Z
Figure P1.3.1 - Rotor angular momentum
Problem 1.4
Demonstrate that the relationship between the components of the aircraft velocity in the
body axes and the earth inertial frame (also known as Flight Path Equations), is given by:
 X '  cos  cos   sin  cos   cos  sin  sin  sin  sin   cos  sin  cos   U 
 ' 
 
 Y    sin  cos  cos  cos   sin  sin  sin   sin  cos   sin  sin  cos   V 
 '
 
cos  sin 
cos  cos 
  Z 
 Z    sin 
Solution of Problem 1.4
Starting from:
 X '  cos   sin  0   cos  0 sin   1
0
0  U 
 ' 
 




1
0  0 cos   sin   V 
 Y    sin  cos  0   0
 '
0
1    sin  0 cos   0 sin  cos    Z 
 Z   0
Define the following:
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 cos 
cos   sin  0

A   sin  cos  0 , B   0
  sin 
 0
0
1 
0 sin  
0
0 
1


1
0  , C  0 cos   sin  
0 sin  cos  
0 cos  
Therefore, we have:
X ' 
U 
 '
 
 Y    AB  C V 
Z' 
Z 
 
 
Performing the first multiplication we have:
cos   sin  0   cos  0 sin   cos  cos   sin  cos  sin  
AB   sin  cos  0  0
1
0    sin  cos  cos  sin  sin  
 0
0
1    sin  0 cos     sin 
0
cos  
leading to:
0
0 
cos  cos   sin  cos  sin   1



 AB  C   sin  cos  cos  sin  sin   0 cos   sin   
  sin 
0
cos   0 sin  cos  
cos  cos   sin  cos   cos  sin  sin  sin  sin   cos  sin  cos  
  sin  cos  cos  cos   sin  sin  sin   cos  sin   sin  sin  cos  
  sin 

cos  sin 
cos  cos 
Then,
 X '  cos  cos   sin  cos   cos  sin  sin  sin  sin   cos  sin  cos   U 
 ' 
 
 Y    sin  cos  cos  cos   sin  sin  sin   sin  cos   sin  sin  cos   V 
 '
 
cos  sin 
cos  cos 
  Z 
 Z    sin 
Problem 1.5
Consider the „general‟ expression of the Conservation of Linear Momentum Equations
(CLMEs) with respect to the body axes X , Y , Z .

m U  QW  RV   mg sin   FA X  FT X


m V  UR  PW   mg cos  sin   FAY  FTY


m W  PV  QU   mg cos  cos   FA Z  FT Z

Demonstrate step-by-step how the introduction of the steady state and perturbed flight
conditions, along with the introduction of the small perturbation assumptions leads to the
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following „small perturbations‟ expression for the CLMEs with respect to the body axes
X ,Y , Z .

m u  Q1w  qW1  R1v  rV1   mg cos 1  f A X  fT X



m  w  Pv  pV  Q u  U q   mg cos  sin   mg sin  cos    f  f 
m  v  U1r  uR1  Pw
1  pW1    mg sin 1 sin 1  mg cos 1 cos 1  f AY  fT Y
1
1
1
1
1
1
1
AZ
1
TZ
Specifically, explain why and how different terms out cancel out leading to the
expressions above.
Solution of Problem 1.5
Using the definitions of steady state and perturbed flight:
U  U1  u , V  V1  v , W  W1  w
P  P1  p , Q  Q1  q , R  R1  r
  1   ,   1   ,   1  
FAX  FAX 1  f AX , FAY  FAY 1  f AY , FAZ  FAZ 1  f AZ
FTX  FTX 1  fTX , FTY  FTY 1  fTY , FTZ  FTZ 1  fTZ
U  U1  u , V  V1  v , W  W1  w
By definition, at steady state conditions we have:
U1  V1  W1  0
Therefore:
U u, V  v, W  w
Inserting the above expressions in the CLMEs:

m u   Q1  q W1  w    R1  r V1  v    mg sin  1     FAX 1  f AX  FTX 1  fTX



m  w   P  p V  v    Q  q U  u    mg cos      cos        F  f  F  f 
m v  U1  u  R1  r    P1  p W1  w    mg cos  1    sin  1     FAY 1  f AY  FTY 1  fTY
1
1
1
1
1
1
Performing the multiplications in the left hand side:
AZ 1
AZ
TZ 1

m u  QW
1 1  wQ1  qW1  qw  RV
1 1  vR1  rV1  rv    mg sin  1     FAX 1  f AX  FTX 1  f TX
TZ



m  w  PV  vP  pV  pv  QU  uQ  qU  qu   mg cos      cos        F  f  F  f 
m  v  U1R1  rU1  uR1  ur  PW
1 1  wP1  pW1  pw  mg cos  1    sin  1     FAY 1  f AY  FTY 1  f TY
1 1
1
1
1
1
1
1
1
1
AZ 1
Using the following identities:
sin  1     sin 1 cos   cos 1 sin   sin 1   cos 1
cos  1     cos 1 cos  sin 1 sin   cos 1   sin 1
sin  1     sin 1 cos   cos 1 sin   sin 1   cos 1
cos  1     cos 1 cos   sin 1 sin   cos 1   sin 1
AZ
TZ 1
TZ
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The terms that contain „sin‟ and „cos‟ functions can be rewritten as:
cos  1    sin  1      cos 1   sin 1  sin 1   cos 1  
 cos 1 sin 1   cos 1 cos 1   sin 1 sin 1   sin 1 cos 1
cos  1    cos  1      cos 1   sin 1  cos 1   sin 1  
 cos 1 cos 1   cos 1 sin 1   sin 1 cos 1   sin 1 sin 1
Inserting in the CLMEs equations:
m u  QW
1 1  wQ1  qW1  qw  R1V1  vR1  rV1  rv    mg  sin 1   cos 1  


 FAX 1  f AX  FTX 1  fTX
m  v  U1 R1  rU1  uR1  ur  PW
1 1  wP1  pW1  pw 
 mg  cos 1 sin 1   cos 1 cos 1   sin 1 sin 1   sin 1 cos 1  

 FAY 1  f AY  FTY 1  fTY

m  w  PV
1 1  vP1  pV1  pv  QU
1 1  uQ1  qU1  qu  
 mg  cos 1 cos 1   cos 1 sin 1   sin 1 cos 1   sin 1 sin 1  

 FAZ 1  f AZ  FTZ 1  fTZ

Rearranging, we have:
m QW
1 1  R1V1   m  u  wQ1  qW1  qw  vR1  rV1  rv  



 mg sin 1  FAX 1  FTX 1  mg cos 1  f AX  fTX

m U1 R1  PW
1 1   m  v  rU1  uR1  ur  wP1  pW1  pw  


 mg cos 1 sin 1  FAY 1  FTY 1 

 mg  cos 1 cos 1   sin 1 sin 1   sin 1 cos 1   f AY  fTY

m  PV
1 1  QU
1 1   m  w  vP1  pV1  pv  uQ1  qU1  qu  


 mg cos 1 cos 1  FAZ 1  FTZ 1 

 mg   cos 1 sin 1   sin 1 cos 1   sin 1 sin 1   f AZ  fTZ

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The first term in left hand side is equal to the two first terms in the right hand side in each
equation, since they represent the steady state CLMEs. Therfore:

m  u  wQ1  qW1  qw  vR1  rV1  rv   mg cos 1  f AX  fTX
m  v  rU1  uR1  ur  wP1  pW1  pw  


 mg  cos 1 cos 1   sin 1 sin 1   sin 1 cos 1   f AY  fTY
m  w  vP1  pV1  pv  uQ1  qU1  qu  


 mg   cos 1 sin 1   sin 1 cos 1   sin 1 sin 1   f AZ  fTZ

Next, using the small perturbation assumption:
qw  0, rv  0, ur  0, pw  0, pv  0, qu  0,  0
Therefore:
m  u  wQ1  qW1  vR1  rV1   mg cos 1  f AX  fTX



m  v  rU1  uR1  wP1  pW1   mg  cos 1 cos 1   sin 1 sin 1   f AY  fTY


m  w  vP1  pV1  uQ1  qU1   mg   cos 1 sin 1   sin 1 cos 1   f AZ  fTZ
Finally, rearranging, we will have the FINAL “small perturbation” CLMEs:

m u  Q1w  qW1  R1v  rV1   mg cos 1  f A X  fT X



m  w  Pv  pV  Q u  U q   mg cos  sin   mg sin  cos    f  f 
m  v  U1r  uR1  Pw
1  pW1    mg sin 1 sin 1  mg cos 1 cos 1  f AY  fT Y
1
1
1
1
1
1
1
1
AZ
TZ

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Problem 1.6
Consider the „general‟ expression of the Conservation of Angular Momentum Equations
(CAMEs) with respect to the body axes X , Y , Z .
P I XX  R I XZ  PQ I XZ  RQ  I ZZ  IYY   LA  LT
Q IYY  PR  I XX  I ZZ    P 2  R 2  I XZ  M A  M T
R I ZZ  P I XZ  PQ  IYY  I XX   QR I XZ  N A  NT
Demonstrate step-by-step how the introduction of the steady state and perturbed flight
conditions, along with the introduction of the small perturbation assumptions leads to the
following „small perturbations‟ expression for the CAMEs with respect to the body axes
X ,Y , Z .
p I XX  r I XZ   Pq
1  Q1 p  I XZ   R1q  Q1r  I ZZ  I YY    l A  lT 
qIYY   Pr
1  pR1  I XX  I ZZ    2 P1 p  2 R1r  I XZ   mA  mT 
r I ZZ  p I XZ   Pq
1  pQ1  I YY  I XX    Q1r  R1q  I XZ   n A  nT 
Specifically, explain why and how different terms out cancel out leading to the
expressions above.
Solution of Problem 1.6
Using the definitions of steady state and perturbed flight:
P  P1  p , Q  Q1  q , R  R1  r
LA  LA1  l A , M A  M A1  mA , N A  N A1  nA
LT  LT1  lT , M T  M T1  mT , NT  NT1  nT
P  P1  p , Q  Q1  q , R  R1  r
By definition of steady state conditions:
P1  Q1  R1  0
Therefore:
P p, Qq, Rr
Inserting the above expressions in the CAMEs leads to:
p I XX  r I XZ   P1  p  Q1  q  I XZ   R1  r  Q1  q  I ZZ  IYY   LA1  l A  LT1  lT


q IYY   P1  p  R1  r  I XX  I ZZ    P1  p    R1  r  I XZ  M A1  mA  M T1  mT
2
2
r I ZZ  p I XZ   P1  p  Q1  q  IYY  I XX    Q1  q  R1  r  I XZ  N A1  nA  NT1  nT
Performing the multiplications in the left hand side leads to:
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p I XX  r I XZ   PQ
1 1  qP1  pQ1  pq  I XZ   R1Q1  qR1  rQ1  rq  I ZZ  I YY  
 LA1  l A  LT1  lT


2
2
2
2
q IYY   PR
1 1  rP1  pR1  pr  I XX  I ZZ   P1  2 pP1  p   R1  2rR1  r  I XZ 
 M A1  mA  M T1  mT
r I ZZ  p I XZ   PQ
1 1  qP1  pQ1  pq  I YY  I XX    Q1 R1  rQ1  qR1  qr  I XZ 
 N A1  nA  NT1  nT
Rearranging the above equations we have:
 PQ
1 1 I XZ  R1Q1  I ZZ  I YY   p I XX  r I XZ   qP1  pQ1  pq  I XZ 
  qR1  rQ1  rq  I ZZ  IYY   LA1  LT1  l A  lT
2
2
PR
1 1  I XX  I ZZ    P1  R1  I XZ  q I YY   rP1  pR1  pr  I XX  I ZZ  


 2 pP1  p 2   2rR1  r 2  I XZ  M A1  M T1  mA  mT
PQ
1 1  I YY  I XX   Q1 R1 I XZ  r I ZZ  p I XZ   qP1  pQ1  pq  I YY  I XX  
  rQ1  qR1  qr  I XZ  N A1  NT1  nA  nT
The first two terms in left hand side are equal to the two first terms in the right hand side
in each equation, since they represent the steady state CAMEs, then
p I XX  r I XZ   qP1  pQ1  pq  I XZ   qR1  rQ1  rq  I ZZ  IYY   l A  lT


q IYY   rP1  pR1  pr  I XX  I ZZ   2 pP1  p 2   2rR1  r 2  I XZ  mA  mT
r I ZZ  p I XZ   qP1  pQ1  pq  IYY  I XX    rQ1  qR1  qr  I XZ  n A  nT
Next, using the small perturbation assumption, we have:
pq  0, rq  0, pr  0, p 2  0, r 2  0
Therefore:
p I XX  r I XZ   qP1  pQ1  I XZ   qR1  rQ1  I ZZ  IYY   l A  lT
q IYY   rP1  pR1  I XX  I ZZ    2 pP1  2rR1  I XZ  mA  mT
r I ZZ  p I XZ   qP1  pQ1  IYY  I XX    rQ1  qR1  I XZ  nA  nT
Finally, rearranging, we will have the FINAL „small perturbation‟ CAMEs:
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p I XX  r I XZ   Pq
1  Q1 p  I XZ   R1q  Q1r  I ZZ  I YY    l A  lT 
qIYY   Pr
1  pR1  I XX  I ZZ    2 P1 p  2 R1r  I XZ   mA  mT 
r I ZZ  p I XZ   Pq
1  pQ1  I YY  I XX    Q1r  R1q  I XZ   n A  nT 
Problem 1.7
Consider the Inverted Kinematic Equations (IKEs) under the assumption of small
perturbations:
p    1 cos 1  sin 1
q  1 sin 1   cos 1  1 sin 1 sin 1
1 cos 1 cos 1   sin 1 cos 1
r  1 cos 1 sin 1  1 sin 1 cos 1
 cos 1 cos 1  1 cos 1   sin 1
Explain under which conditions the above equations can be reduced to the following
expressions:
p 
q 
r 
Solution of Problem 1.7
Assuming initial steady state flight conditions we have:
1  0 , 1  0 , 1  0
Thus, the given equations become:
p     sin 1
q   cos 1   sin 1 cos 1
r   cos 1 cos 1
Next, assuming initial wing level flight ( 1  0 ) the above equations become:
p     sin 1
q 
r   cos 1
Finally, assuming no initial pitch angle (1  0) , we have:
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p 
q 
r 
Thus the required conditions are:

Initial steady state flight

Initial wing level flight ( 1  0 )

No initial pitch angle (1  0)
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Problem 1.8
Consider the drawings below with the dimensions of 4 „extra large‟ aircraft.
Figure P1.8.1 –
(Source: http://upload.wikimedia.org/wikipedia/commons/9/96/Giant_Plane_Comparison.jpg)
Using your best technical judgment, rank the values of all the moments of inertia for the
different aircraft. Document your response with simple calculations using approximate
distances.
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Solution of Problem 1.8
The moment of inertia I XX which represents the inertial resistance of the aircraft to a
change in its angular velocity around the X- axis (rolling velocity). Clearly, it is more
difficult to change the rolling velocity of an aircraft with a larger wingspan; the presence
of engines in the wing can only increase this inertia, since they add mass to the wing.
According to this and assuming that the engines of all the aircrafts have approximately
the same weight (the H4 engines are smaller but they are piston engines while the other
three aircraft have turbofan engines) the I XX can be ranked according to Table P1.8.1.
Similarly, the moment of inertia IYY represents the inertial resistance of the aircraft to a
change in its angular velocity around the Y- axis (pitching velocity), which is located at
the aircraft CG. The CG is located at approximately 25% of the mean aerodynamic
chord; even without this information, it can be stated that the CG is located at a stating
approximately on the wing. Therefore, the weight distributed in the fuselage – not in the
wings - is the main provider to this moment of inertia. Therefore, the value for this
parameter can be ranked according to the fuselage length; this, the longer the fuselage the
larger the value of IYY . The results are presented in Table P1.8.1.
Similarly, the moment of inertia I ZZ represents the inertial resistance of the aircraft to a
change in its angular velocity around the Z- axis (yawing velocity), which is located at
the aircraft CG. In this case, the contribution of the masses located in both the fuselage
and wings are important. The Antonov‟s wings and fuselage are larger than the Airbus‟;
the Airbus‟ wings and fuselage are larger than the Boeing‟s; therefore, it can be stated
that their I ZZ are in the same order of magnitude. For this case it is difficult to say if
Hughes‟ I ZZ is greater than Antonov‟s since, although the Hughes H-4 has a larger
wingspan, it is a lighter aircraft (made of wood) and has a smaller fuselage; therefore, a
more detailed knowledge of the structural density and mass distribution is required for an
accurate analysis.
A comprehensive ranking is provided below (with some approximation for I ZZ ).
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Aircraft
Rank ( I XX )
Rank ( IYY )
Rank ( I ZZ )
nd
st
1st
2nd
3rd
1st
Antonov An-225
2
1
Airbus A380-800
3rd
2nd
Boeing 747-400
4th
3rd
st
Hughes H-4
1
4th
Table P1.8.1 – Ranking of the Moments of Inertia
Problem 1.9
Consider the F111 aircraft shown in the drawing below.
Figure P1.9.1 - 3D View of the General Dynamic F-111 Aircraft
(Source: http://www.aviastar.org/index2.html)
Using your best technical judgment, explain how the values of all the moments of inertia
change when the configuration of the aircraft is changed from the low-subsonic forward
wing position (low sweep angle) to the high-subsonic high-sweep angle position.
Solution of Problem 1.9
The moment of inertia I XX which represents the inertial resistance of the aircraft to a
change in its angular velocity around the X- axis (rolling velocity). I XX will decrease as
the wing sweep angle increases, since it is more difficult to change this velocity of an
aircraft with a larger wingspan ( the mass distribution is now closer to the axis of
rotation). A similar trend is expected for I ZZ . The opposite trend is expected for the
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moment of inertia IYY . This is due to the fact that the masses in the wing will shift further
down toward the tail with respect to the aircraft CG. Therefore, IYY will increase as the
wing sweep angle increases. Table P9.1 summarizes the trends.
Moment of inertia
I XX
IYY
FROM low wing sweep angle Decreases
Increases
(subsonic)
TO high wing
sweep angle (supersonic)
Table P1.9.1 – Trends for moments of inertia
I ZZ
Decreases
Problem 1.10
Consider the Lockheed L-1011 “Tristar” aircraft shown in the drawing below.
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Figure P1.10.1 - 3D View of the Lockheed L-1011 Aircraft
(Source: http://www.aviastar.org/index2.html)
Assuming that at steady state conditions the throttle setting is the same for each of the 3
aircraft engines, first introduce generic distances along the body axes X , Y , Z of the
distances of each of the engines with respect to the aircraft center of gravity. Next,
qualitatively describe how the terms FT X , FTY , FT Z , LT , M T , NT in the CLMEs and CAMEs
are modified under the following engine failure conditions:
-
failure of the tail engine;
-
failure of the wing engine on the right of the pilot.
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Assume that in both cases the failures occur while the aircraft is in steady-state rectilinear
flight conditions.
Solution of Problem 1.10
At steady-state rectilinear flight conditions:
0  mg sin 1  FAX 1  FTX 1




0  mg cos  cos    F  F 
0  mg cos 1 sin 1  FAY 1  FTY 1
1
1
AZ 1
TZ 1
0  LA1  LT1
0  M A1  M T1
0  N A1  NT1
Figure P1.10.2 shows the distances of each of the engines from the aircraft center of
gravity.
Figure P1.10.2 - Distances of the engines with respect to the aircraft center of gravity
Assume:
FTX 1  3FE
FTY 1  FTZ 1  0
This means that each of the engines generates the same thrust , FE , and only in the X
direction. Additionally:
LT1  0
MT1  2 zEW FE  zET FE
0
NT1  yEW FE  yEW FE  0
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Condition 1. Failure of the tail engine
In this case, the above force/moments equations will become:
FTX 1  2FE
 NOMINAL THRUST
FTY 1  FTZ 1  0
LT1  0
M T1  2 zEW FE  0
NT1  yEW FE  yEW FE  0
Condition 2. Failure of the wing engine on the right of the pilot.
In this case, the above force/moments equations will become:
FTX 1  2FE
 NOMINAL THRUST
FTY 1  FTZ 1  0
LT1  0
MT1  zEW FE  zET FE  0
NT1  yEW FE  0
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