The tensor flight dynamics tutor

iv
2020 by Peter H. Zipfel.
Published by Modeling and Simulation Technologies, Shalimar, Florida USA.
This publication may be freely copied and shared. If parts are included in other
publications proper citation is requested.
Send inquiries, suggestions, and corrections to mastech.zipfel@cox.net or
mastech. zipfel@gmail.com.
Modeling and Simulation Technologies

v
Modeling and Simulation Technologies
Preface
I have condensed the essentials of my book “Introduction to Tensor
Flight Dynamics” into 87 slides. They should serve the student as a thorough
review, and the instructor as visual aids in the class room by requesting the
complimentary PowerPoint slides from mastech.zipfel@cox.net or
mastech.zipfel@gmail.com.
Tensors  Matrices  Computer is the modus operandi of modern
flight dynamics, taking advantage of today’s immense computer power. I
wish this booklet will be your companion as you keep Tensor Flight
Dynamics close to your heart.
Peter H Zipfel, Shalimar, Florida, 2020

1
2020 Modeling and Simulation Technologies
Tensor Flight DynamicsTutor
The Tensor Flight Dynamics Tutor
Dr. Peter H. Zipfel
mastech.zipfel@cox.net
mastech.zipfel@gmail.com
Modeling and
Simulation
Technologies
Tensors  Matrices  Computer
CHAPTER SECTION PAGE
1 Introduction 1
2
Geometry
2.1 Vectors are Tensors of Rank One 5
2.2 Multiplying Tensors of Rank One 11
2.3 Frames and Coordinate Systems 17
3
Kinematics
3.1 Rotational Time Derivative 25
3.2 Euler Transformation of Frames 31
3.3 Attitude Determination 39
4
Point-Mass
Dynamics
4.1 Newton’s Law 45
4.2 Vehicles Requiring Inertial Frame 51
4.3 Vehicles Using Earth Frame 59
5
Rigid Body
Dynamics
5.1 Newton’s and Euler’s Laws 67
5.2 Missiles and Rockets 77
5.3 Aircraft and UAVs 85

2
Introduction
2MaSTech 2020 Tensor Flight Dynamics Tutor
Welcome to Tensor Flight Dynamics!
• Hi, I am your Tutor. I chisel you with my hammer until you
conform to Tensor Flight Dynamics
• I assume that you have studied my book “Introduction to
Tensor Flight Dynamics”, Amazon 2019, 2020
• Before taking your exam I assist you in reviewing the entire
material
• If you are the instructor you can use these PowerPoint
slides to project them as you teach the course
– Just drop me an e-mail
• mastech.zipfel@cox.net (primary)
• mastech.zipfel@gmail.com
– And I will send you the PowerPoint slides free of charge

3
Introduction
3MaSTech 2020 Tensor Flight Dynamics Tutor
Directions
• The PowerPoint charts follow the book
– Four chapters
• Geometry, Kinematics, Point-Mass Dynamics, Rigid Body Dynamics
• Each chapter has three sections
– The relevant pages are listed at the left margin
– Equation and figure numbers are the same as in the book
• The PowerPoint charts cover the essential material
– Important equations are blocked out
• For clarification refer to the text in the book

5
2.1 Vectors are Tensors of Rank One
Tensor Flight Dynamics Tutor
2 Geometry
2.2 Multiplying Tensors of Rank One
2.3 Frames and Coordinate Systems
Displacement tensor (vector)
Coordinate systems
Addition of tensors of rank one
Coordinate systems have no origin
2 Geometry

6
Displacement tensor (vector)
2MaSTech 2020 2 Geometry
Figure 2.1 Rocket ascent
R
L
RLs
Displacement tensor (vector) of rocket R with respect to (wrt) launch pad L: RLs
up
north
east
s
s
s
s
G
RL
G
RL
G
RL
G
RL











3
2
1
)(
)(
)(
][Vector expressed in east, north, up
coordinates, called G
RLs
up
rangedown
rangecross
s
s
s
s
C
RL
C
RL
C
RL
C
RL 












3
2
1
)(
)(
)(
][
Vector expressed in cross, down, up
coordinates, called C
RLs
G
RL
CGC
RL sTs ][][][
133313 

Relationship established by
transformation matrix (TM)
CG
T][































G
RL
G
RL
G
RL
C
RL
C
RL
C
RL
s
s
s
ttt
ttt
ttt
s
s
s
3
2
1
333231
232221
131211
3
2
1
)(
)(
)(
)(
)(
)(
Matrices expressed in their components
are coded for numerical evaluation:
ABAB
xTx ][][][  (2.1)
For any two coordinate systems with their
transformation matrix , the following transformation
holds for any tensor of rank one x
BA
],]
BA
T][
The inverse of this statement is also true: any entity x that transforms
like Equation 2.1 for all coordinate systems is a tensor of rank one 
pp. 2, 3*
* Ref:
“Introduction to
Tensor Flight
Dynamics”,
Amazon 2019,
2020

2 Geometry
7
Coordinate systems
X
2AX2
X3
X1
1A
3A
All coordinate systems are right-handed: turn 1A into 2A to get 3A
Coordinate systems are drawn as coordinate axes











A
A
A
A
x
x
x
x
3
2
1
)(
)(
)(
][
The vector x has three components in the A
coordinate system
The vector x has three components in the B
coordinate system











B
B
B
B
x
x
x
x
3
2
1
)(
)(
)(
][
The components are related by the transformation matrix ; i.e., the transformation
of coordinate system B wrt coordinate system A
BA
T][
ABAB
xTx ][][][ 
Coordinate systems are identified by capital superscripts
Concatenation rule 1: The letters of the respective coordinate system are adjacent
Figure 2.2 Coordinate axes
Brackets indicate matrices (numbers),while boldface lower case letters, like x, are
tensors of rank one (vectors) 
pp. 4, 5

8
Addition of tensors of rank one
R
L
C
Figure 2.3 Cinetheodolite added
A cinetheodolite C is tracking the rocket R; what is givenRLs LCs
But is given, thus:LCs LCRCRL sss 
From the vector triangle CLRCRL sss 
Concatenation rule 2: two displacement vectors are additive if the
inner subscripts are the same and thus cancel out to obtain RLs
Sign reversal rule: reversing the subscript of a displacement
vector changes its sign LCCL ss 
Subscripts are read from left to right; e.g., displacement of
rocket R wrt launch pad L
RLs
pp. 5, 6
Example:
up
north
east
s G
RC











5000
50
1200
][Measurement











20
0
1100
][ G
LCsLocation (2.2)
up
north
east
sss G
LC
G
RC
G
RL

































4800
50
100
20
0
1100
5000
50
1200
][][][


2 Geometry
9
Coordinate systems have no origin
N
E
U
L
C1
C2
Figure 2.4 Launch pad with two
cinetheodolites and upward coordinate axes
We don’t need the origin of a coordinate system
to solve a geometric problem:
1. Solve the physics part of the problem using the
concatenation rule 1212 LCLCCC sss 
The second cinetheodolite has the coordinates  405003000][ 2 
G
LCs
3. Compute the problem by using the coordinate system G in order to convert the
tensors into matrices G
LC
G
LC
G
CC sss ][][][ 1212 
From Eq. 2.2 we already have ; note, I use the transposed
(overbar) to save space
 2001100][ 1 
G
LCs
Solution



































20
500
4100
20
0
1100
40
500
3000
][ 12
G
CCs
No need for an origin; therefore no need for so-called radius vectors (Book, pp 9, 10)
2LCs
22 LCLC ss 
1212 LCLCCC sss 
2. Modify: but because is given, reverse the
subscripts
Tensor flight dynamics’ three step approach: Solve 1, Modify 2, Compute 3 
Given determine and21 and LCLC ss 12CCs
G
CCs ][ 12pp. 7, 8

10
Notable Quotables
 Tensors are mathematical symbols that model physical phenomena
 Nature operates perfectly without coordinate systems
 Coordinate systems are only introduced for numerical computations
 Tensor Flight Dynamics’ Three-Step:
1. Model with tensors
2. Introduce coordinate systems
3. Compute with matrices
 Tensors are recognized by their transformations
 Displacement vectors are tensors of rank one, connecting two physical points
o Subscripts are read from left to right inserting ‘with respect to’ (wrt)
o Subscript reversal changes the sign
o Concatenation rule of subscripts
 Coordinate systems link the Cartesian space to algebraic numbers
o Limited to right-handed coordinate systems
o Coordinate axes are geometrical images with direction and units
o They have no origin
 Coordinate transformations relabel the components in Cartesian space
o They are 3x3 matrices, not tensors
 A vector expressed in a coordinate system is a 3x1 matrix
 Radius vectors are debarred

2 Geometry
11
2 Geometry
Scalar product
Vector product
Dyadic product
Geometric tensors
Notable Quotables

12
Scalar product
pp. 12, 13
Figure 2.7 Falcon 9 landing on barge
 The transposed of a vector multiplied by another vector results in a scalar. Scalars are
tensors of rank zero and are the same (same value) in any coordinate system
Generalized
cos|||| PBRBPBRB ssss
cos|||| yxyx (2.8)Scalar product
Example For computation we introduce coordinate system into Eq. 2.8
A
]
cos||||][][ yxyx AA

  cos2
3
2
2
2
1
2
3
2
2
2
1
3
2
1
321 










yyyxxx
y
y
y
xxx
cos2
3
2
2
2
1
2
3
2
2
2
1332211  yyyxxxyxyxyx
Square of the absolute value of a vector 2
||0cos|||| xxx o
xx

What product results in a scalar?
Horizontal distance of rocket from barge
cos|||| RBPB ss 
cos|| RBPBRB sus (2.6)
(2.5)

2 Geometry
13
Vector product
If is given, what form must have?][
13
x

][
33
X

We start with the matrix product ][]][[
131333 
 zyX (2.11)
Introduced into Eq. 2.11
















































3
2
1
2112
3113
3223
3
2
1
12
13
23
0
0
0
z
z
z
yxyx
yxyx
yxyx
y
y
y
xx
xx
xx
Tensor algebra is similar to the matrix algebra of Eq. 2.11 : zXy  (2.12)
Where is a skew-symmetric tensor of rank twoX
 The skew-symmetric tensor of vector multiplied by vector results in vectorX xx y
p. 14
is the skew-symmetric 3x3 matrix of][X
























0
0
0
12
13
23
3
2
1
xx
xx
xx
x
x
x
][x
What product results in a vector?
Notation: Tensors of rank one (abbreviated: vectors) are lower case bolded; tensors of
rank two (abbreviated: tensors) are upper case bolded


14
Dyadic product
What product results in a tensor?
Figure 2.7 Falcon 9 landing on barge
From Eqs. 2.5 and 2.6 we get the length PBRBPBs us||
Because it is a scalar we can take the transposed RBPBPBs su||
Now is length x unit vectorPBs |||| PBPBPBPBPB ss uus 
What is ? In matrix algebra it would be a (3x1)x(1x3)
product, called a dyadic product which results in a (3x3) matrix
Combined RBPBPBPB suus 
PBPBuu
 Vector multiplied by the transposed of vector results in the tensorx y Z
In our special case, where both vectors are the same unit vector , the tensor is called
the projection tensor
PBu
P
In tensor algebra we have the general dyadic product Zyx  (2.13)
PBPBuuP  (2.14 )
i., e. the projection of onto the barge is the vector RBPB sPs RBs
pp. 15, 16
Example For any Cartesian coordinate system A
 

































333231
232221
131211
332313
322212
312111
321
3
2
1
][][][
zzz
zzz
zzz
yxyxyx
yxyxyx
yxyxyx
yyy
x
x
x
Zyx AAA


2 Geometry
15
Geometric tensors
Figure 2.9 Barge surface orientation
defined by unit vectoru
From the vector triangle B, R, P we get
Previously, we used the projection tensor , formed
from the horizontal unit vector , ,
to get the vector
P
PBu PBPBuuP 
RBPB sPs 
Now we want to use the unit vector normal to
the barge surface to determine
u
PBs
First project onto to get vector usingRBs u r uuP 
RBPsr 
RBRBRBRBRBPB NssPEPssrss  )(
 is the normal projection tensor. It projects a vector into a plane,
whose normal unit vector is given
N
uuEPEN 
u
r
u
t
t'
Figure 2.10 Reflection tensor
The reflection tensor reflects a vector into vector by
the mirror, whose normal unit vector is given
M t t
u
Mtt'
To determine we get from the vector triangle rtt' 2
But , thereforePtr  MttPEPttt'  )2(2
M
pp. 17, 18
u2uE2PEM 


16
Notable Quotables
 Vector multiplications are named after the form of their product
o Scalar product results in a scalar
o Vector product results in a vector
o Dyadic product results in a tensor
 Scalars are tensors of rank zero
o They are the same in all coordinate systems
 Symbiotic relationship between tensors and matrices
o Both are subject to the same multiplication rules
 Tensor notation
o Vectors in lower case, bolded fonts
o Tensors in upper case, bolded fonts
 Tensors of rank two
o Projection tensor
o Normal projection tensor
o Reflection tensor

2 Geometry
17
2 Geometry
Frames
TM from column unit matrices
TM from row unit matrices
TM from direction cosines
Two angle transformation f and 
Notable Quotables

18
Frames
pp. 20, 21
Notice, how the concatenation rule also applies here
 A frame is a mathematical model of an object whose
elements are mutually fixed
B
R
P
RBs
1b
2b
3b
3r
1r
2r
Figure 2.12 Frames R and B
represented by the base triads R and B
The elements of a frame extend beyond the material confines, like in the case of the earth-frame E,
which reaches out into all altitudes
It is represented by its base triad, consisting of three
orthogonal base vectors, like for barge, and
for rocket
321 ,, bbb
321 ,, rrr
The rocket frame R is related to the barge frame B by
the rotation tensor , which maps the base vectors
into base vectors
RB
R
321 ,, bbb 321 ,, rrr
is a tensor of rank twoRB
R
If is given, and in addition the rotation tensor of the barge B wrt earth E, , we can
determine the orientation of the rocket wrt the earth
RB
R BE
R
BERBRE
RRR 

If is given instead of you take the transposeEB
R BE
R EBBE
RR 
3,2,1,  ii
RB
i bRr (2.19)

2 Geometry
19
TM from column unit matrices
A
x
a2
1A
3A
2A
1B 3B
2B
a1
a3
Figure 2.15 Vector x decomposed
into the A base triad
ABAB
xTx ][][][ 
 Preferred coordinate systems line up with base vectors,
e.g., coordinate system A with frame A. Base vectors have a
simple expression in the preferred coordinate system
Transformation matrices relate coordinate systems
(2.20)
While frames model physics, coordinate systems convert tensors
into matrices for computation

































1
0
0
][,
0
1
0
][,
0
0
1
][ 321
AAA
aaa
Decompose vector into the three base vectors 321 ,, aaax 332211 aaax
AAA
xxx 
Express in coordinatesx B
] BABABAB
axaxaxx ][][][][ 332211 
A scalar product: where 











A
A
A
BBBB
x
x
x
aaax
3
2
1
321 ][][][][ A
A
A
A
x
x
x
x
][
3
2
1











Substituted   ABBBB
xaaax ][][][][][ 321
The columns of consist of frame A base vectors expressed in coordinatesBA
T][ B
]

pp. 23, 24
Compared with Eq. 2.20  BBBBA
aaaT ][][][][ 321 (2.21)

20
TM from row unit matrices
1B
3B
2B
1A
3A
2A
B
x
b3
b2b1
pp. 24, 25
Now we reverse the process and express the vector
in the base vectors
x
321 ,, bbb
332211 bbbx
BBB
xxx 
Coordinated in the system
A
]
 
  BAAA
B
B
B
AAA
ABABABA
xbbb
x
x
x
bbb
bxbxbxx
][][][][
][][][
][][][][
321
3
2
1
321
332211













Solving for
B
x][ A
A
A
A
B
x
b
b
b
x ][
][
][
][
][
3
2
1











The elements of transformation
matrices can be interpreted as column
3x1 matrices or row 1x3 matrices A
A
A
BA
BBB
b
b
b
ttt
ttt
ttt
T
aaa
][
][
][
][
][][][
3
2
1
333231
232221
131211
321
















Figure 2.16 Vector x decomposed
into the B base triad
Comparing with Eq. 2.20 (previous slide)











A
A
A
BA
b
b
b
T
][
][
][
][
3
2
1
(2.22)

2 Geometry
21
TM from direction cosines
pp. 25, 26, 30
3B
2B
1B
b2
11,ab
b1
b3
a1
Figure 2.17 Direction cosines
Properties of transformation matrices
The elements of the transformation matrix can be
interpreted as the cosine of the angle between two base vectors
ija
BA
T][
    B
B
B
B
BB
a
a
a
a
ab 11
31
21
11
1111 001][][,cos 










 ab
  3,2,1;3,2,1;,cos  kiaB
ikki ab
For any and we can calculate the cosine of the angle accordinglyib ka
Because TMs are orthogonal matrices, their inverse equals their transpose   1
][][

 BABA
TT
Taking the transpose of a TM reverses the sequence of transformation
BAAB
TT ][][ 
Combining TMs is subject to the concatenation rule BACBCA
TTT ][][][ 
Multiplying the TM with its transposed yields the unit matrix ETTTTT BBABBABABA
 ][][][][][
Relationship between rotation tensor and coordinate transformation

Given the rotation tensor of frame B wrt frame A, , and the transformation matrix
of the two preferred coordinate systems wrt , then both are linked together by
BABBAABA
TRR ][][][  (2.28)
BA
R
BA
T][
B
] A
]

22
Two angle transformation f and 
1B
3B2B
1R
3R
2R
Figure 2.18 Coordinate systems of
rocket and barge
f
f

 1B
3B
2B
1R
3R
2R
f
f

 1B
3B
2B
1R
3R
2R
2X
2X
3X
3X
1X 1X
Figure 2.21 Both transformations by angles f and  combined
Again, Eq. (2.22) we
get the second TM











X
X
X
RX
r
r
r
T
][
][
][
][
3
2
1
 
 
 

cos0sin][
010][
sin0cos][
3
2
1



X
X
X
r
r
r









 



cos0sin
010
sin0cos
][ RX
T











B
B
B
XB
x
x
x
T
][
][
][
][
3
2
1
Applying Eq. (2.22)
we get the first TM













ff
ff
cossin0
sincos0
001
][ XB
T
 
 
 ff
ff
cossin0][
sincos0][
001][
3
2
1



B
B
B
x
x
x
































 

ff
ff
ff
ff
ff


coscossincossin
sincos0
cossinsinsincos
cossin0
sincos0
001
cos0sin
010
sin0cos
][ RB
T
XBRXRB
TTT ][][][ Multiplying the two matrices

pp. 27 - 30

2 Geometry
23
Notable Quotables
 Frames and coordinate systems are distinctly separate entities
o Frames model physics
o Coordinate systems link Cartesian space to algebraic numbers
 Frames model rigid bodies and reference systems
 Frames are represented by their base triads consisting of
o Base point
o Base vectors
 The rotation tensor relates the orientation of two frames
o It is a tensor of rank two
 Coordinate systems convert tensors to matrices for computation
 The base triad is real, while the coordinate triad is mathematical
 The preferred coordinate system expresses the base vector in simple matrices
 Transformation matrices are 3x3 orthogonal matrices
 The transformation matrix is not a tensor
 There are three ways to interpret the transformation matrix
 The orange peel method displays the coordinate axes on the unit sphere
 Get to know the many coordinate transformation matrices of flight dynamics
 Remember the notation convention
o Subscripts are points
o Superscripts are frames or coordinate systems
o Two sub- or superscripts are read from left to right inserting ‘wrt’
 Concatenation rule
o Adjacent letters cancel

25
3.1 Rotational Time Derivative
3 Kinematics
3.2 Euler Transformation of Frames
3.3 Attitude Determination
Linear velocity in vector mechanics
Rotational time derivative
Linear velocity in tensor mechanics
Angular velocity
Notable Quotables
3 Kinematics

26
Linear velocity in vector mechanics
2MaSTech 2020 3 Kinematics
pp. 43, 44
B
RBs
R
G
R
B
The linear velocity of the rocket c.m. R with respect to the
barge frame B is
Figure 3.2 Rocket velocity observed from
barge and pier
As observed from the pier frame G, we would write in
vector mechanics
We have two velocities , and an extra term
Because of the extra term, the velocity does not transform like a tensor of rank one
In tensor mechanics they should transform—for any two coordinate systems A and B—like
(3.1)
But Eq. 3.1, written in matrix form, has also this extra term
What do we need to do in order to create a linear velocity vector that is a tensor of rank one?
B
Rv dt
d RBB
R
s
v 
RB
BG
B
RB
G
RB
dt
d
dt
d
sω
ss

where is the angular velocity of frame B wrt frame G
BG
ω
G
RB
dt
ds
B
RB
dt
ds
RB
BG
sω 
 
A
RBBA
B
RB
dt
ds
T
dt
ds








  














 B
RB
BBG
B
RBGB
G
RB
s
dt
ds
T
dt
ds
][][

B
Rv
dt
d RBs

3 Kinematics
27
Rotational time derivative
Start with Eq. 2.1 (Slide 2.1_2) BABA
xTx ][][][ 
Take the ordinary time derivative on both sides B
ABB
AB
A
x
dt
dT
dt
dx
T
dt
dx
][][ 











Modify the last term ; (3.2)


















 B
BA
BA
B
AB
A
x
dt
dT
T
dt
dx
T
dt
dx
][][][ BAAB
TT ][][  ;
Bring to the outside 


















 B
AB
AB
B
AB
A
x
dt
dT
T
dt
dx
T
dt
dx
][][][ ][][][ ETT ABAB
AB
T][ ;
This is our new rotational time derivative operator wrt frame A expressed in B coordinates
(3.3)B
BA
BA
B
BA
x
dt
dT
T
dt
dx
xD ][][][ 








A
D
Substitute into Eq. 3.2 and with Eq. 3.3 BAABAA
xDTxD ][][][ 
pp. 44, 45
 If is a tensor of rank one, then is also a tensor of rank onexA
Dx

Now coordinate Eq. 3.3 in A
A
A
AA
AA
A
AA
dt
dx
x
dt
dT
T
dt
dx
xD 











 ][][][ ]0[







dt
dE
dt
dT
AA
;

28
Linear velocity in tensor mechanics
A
R
·
·
R
ARs
R
A
v
Figure 3.4 Radar tracking an aircraft
The relative velocity of point A wrt frame R is
Evaluated in radar coordinates
The differential acceleration of point A with respect to point M as observed from frame R
Linear acceleration
Determine the differential velocity of the aircraft point
A wrt missile point M as observed from radar frame R
Take the rotational time derivative wrt the radar frame R
Radar R tracks aircraft A : ARs
R
ARR
AR
RR
RR
R
ARR
AR
RRR
A
dt
ds
s
dt
dT
T
dt
ds
sDv 













 ][][][][
A
M
R
·
·
·
R
Figure 3.5 Radar tracking aircraft
and missile
Vector triangle MRARAM sss 
MR
R
AR
R
MRAR
R
AM
R
DDDD sssss  )(
R
M
R
AMR
R
AR
R
AM
RR
AM DDD vvsssv 
The differential velocity is the difference
of the two relative velocities
AM
RR
AM D sv 
MR
RR
MAR
RR
A DD svsv  ,
(3.9)The relative acceleration of aircraft point A wrt the radar frame R
R
A
RR
A D va 
AR
RR
A D sv  (3.7)
R
M
RR
A
RR
AM
R
DDD vvv 
R
M
R
A
R
AM aaa 

pp. 47 - 49

3 Kinematics
29
Angular velocity
Angular velocities are additive and abide by the concatenation rule
Let’s identify two frames A (cylinder) and B (cone), which are rotating
relative to each other. Each frame is represented by its base vector a
and b, respectively.
b (t)

v
.
Figure 3.6 Angular velocity
In vector mechanics we get the linear velocity bωv 
In tensor mechanics is replaced by its skew symmetric formω Ω
(3.12)bΩv 
a b
v
A
B
Figure 3.7 Frame A and B
    
0
aRaRaRbv ABABAABAAA
DDDD 
To get the tip speed of b we apply the rotational derivative wrt
frame A, and use Eq. 2.19 (Slide 2.3_2) aRb BA

Using Eq. 2.19 again aEaaRRbR  BABABA
  bRRv BABAA
D:
 In tensor mechanics the angular velocity is a skew-symmetric tensor of rank two and
obtained from the rotation tensor according to Eq. 3.14
BA
Ω
BA
R
The skew-symmetric tensor of rank two can be contracted to a tensor
of rank one , called the angular velocity vector
BA
Ω
BA
ω
BACBCA
ωωω 
pp. 50, 51
Comparing with Eq. 3.12 BABAABA
D RRΩ )( (3.14)
Reversing the direction of the angular velocity changes the sign or the order of
the superscripts (doing both returns to the original)ABBA
ωω 


30
Notable Quotables
 The ordinary time operator destroys the tensor property
o If the two frames are revolving
 The rotational time operator maintains the tensor property
o Of tensors of rank one
o And also of tensors of rank two (not covered here)
 The rotational time operator is a linear operator
 The linear velocity of a point is referred to a frame
o Not to another point
 Relative velocity is one point relative to one frame
 Differential velocity is two independent points relative to one frame
 Linear acceleration of a point is referred to one frame
o Not to another point
 Relative acceleration is one point relative to one frame
 Differential acceleration is two independent points relative to one frame
 Angular velocity relates the motion of one frame relative to another frame
o No points or axes are involved
o Only in calculations will axes of rotation be needed
 Angular velocity is a skew-symmetric tensor of rank two
 Angular velocity tensor can be contracted to a vector
o Assume right-handedness
 The concatenation rule applies to angular velocity tensors and vectors

3 Kinematics
31
3 Kinematics
Transformation of reference frames
Aircraft example
Ground radar example: Physics
Ground radar example: Computation
Coriolis transformation
Grubin transformation
Notable Quotables

32
Transformation of reference frames
pp. 55, 56
From Eq. 3.3 (Slide 3.1_3)
To derive Euler’s transformation of frames we start again with vector mechanics
xω
xx
 BA
BA dt
d
dt
d
And express it in matrices BBBA
BA
x
dt
dx
dt
dx
][][







To establish the equal sign we need to introduce the TM
AB
T][















 BBBA
B
AB
A
x
dt
dx
T
dt
dx
][][][
BB
B
AA
A
xD
dt
dx
xD
dt
dx
][and][ 







Substituted  BBBABBABAA
xxDTxD ][][][][][ 
This holds for all Cartesian coordinate systems, and is therefore a tensor concept
(3.16)xΩxx BABA
DD 

Given the rotational time derivative wrt frame B of vector , , it can be shifted over to
frame A, , if we add the compensation term
xB
D
xA
D xΩBA
x

3 Kinematics
33
Aircraft example
A
G
G
A
AGs
Figure 3.11 Aircraft approach
An aircraft is approaching an airport. The tower
tracks the velocity of the aircraft A relative to ground
G, (A is a point, G is a frame) . Because landing
on a runway requires accurate velocity, the tower
requests that the aircraft transmits its measured
velocity for verification.
G
Av
G
Av
Tower measurement AG
GG
A D sv 
Aircraft measurement GA
AA
G D sv 
To determine the velocity as the tower sees it, the aircraft uses the Euler transformation
to transfer its toA
GGA
A
D vs  G
AAG
G
D vs 
GA
AGA
G
G
A
GA
AG
GA
A
AG
G
AG
AG
AG
A
AG
G
DD
DD
sΩvv
sΩss
sΩss



The aircraft sends the velocity to the tower as the tower perceives it by tracking a
point on the ground and making corrections for aircraft rotations
G
Av
GAs AG
Ω
pp. 56, 57
So far we only solved the physics part of the problem using tensors exclusively. To be
programmed in the flight computer, coordinate systems must be introduced so that
the tensors are converted to matrices


34
Ground radar example: Physics
pp. 57 - 59 R
G
R
BBGs
RBsRGs
B
G
2B
3B
1B
Figure 3.12 Ground radar directing the booster
For the ground radar G to direct the rocket R to
land on the barge B it must know in barge
coordinates
B
Rv
BB
Rv ][
(3.18)Time derivative wrt G BG
G
RG
G
RB
G
DDD sss 
But is the differential velocityG
RBRB
G
D vs 
What we need is the relative velocity RB
BB
R D sv 
Use Euler transformation RB
GB
RB
G
RB
B
DD sΩss 
With Eq. 3.18 RB
GB
BG
G
RG
G
RB
BB
R DDD sΩsssv 
Vector triangle BGRGRB sss  (3.17)
Where and are directly tracked by the radar, and from Eq. 3.17BG
GG
B D sv RG
GG
R D sv  RBs
Changing the sequence of the superscripts BGBGGBGGB
D RRΩΩ )(
We have solved the physical part of the problem
(3.20))()( BGRG
BGBGG
BG
G
RG
GB
R DDD ssRRssv 

The remaining is acquired as using Eq. 3.14GB
Ω BG
Ω
BGBGGBG
D RRΩ )((Slide 3.1_5)

3 Kinematics
35
Ground radar example: Computation
pp. 59, 60
 Determine BB
Rv ][
Eq. 3.20 in radar ground coordinates
)][]([][][][][][ G
BG
G
RG
GBGGBGGG
BG
GG
RG
GGB
R ssRRDsDsDv 
Where all rotational time derivatives have become ordinary time derivatives
)][]([][][ G
BG
G
RG
GBG
GBGG
BG
G
RGGB
R ssR
dt
dR
dt
ds
dt
ds
v 














And finally we transform to the barge coordinates






















 )][]([][][][ G
BG
G
RG
GBG
GBGG
BG
G
RGBGBB
R ssR
dt
dR
dt
ds
dt
ds
Tv
The radar tracks/measures (Eq. 2.28, Slide 2.3_5)
BGGBGG
BG
G
RG TRss ][][,][,][ 
To land the rocket perpendicular to the surface of
the barge the two lateral velocity components have
to be reduced to zero, and for touch-down, the
third component has to vanish
 

































0
0
0
)(
0
0
)(
)(
)(
][
down-touch
3
ebartonormal
3
2
1
BB
R
g
BB
R
BB
R
BB
R
BB
R
vv
v
v
v

)()( BGRG
BGBGG
BG
G
RG
GB
R DDD ssRRssv  (3.20)From previous slide

36
Coriolis transformation
pp. 61, 62
I
R
R
B
BRs
Figure 3.13 Coriolis Transformation
between two reference frames I and R
What are the additional terms?
Determine the vehicle’s B acceleration wrt inertial frame I,
when its acceleration wrt reference frame R is given, while,
in turn, frame R is moving wrt frame I.
termsadditional BR
RR
BR
II
DDDD ss
Transfer the second derivative from I to R using Euler’s
transformation
 BR
RI
BR
RI
BR
II
DDDD sΩss 
Again transfer the derivative I to R    BR
RI
BR
RRI
BR
RI
BR
RR
BR
II
DDDDD sΩsΩsΩss 
Multiply out terms   BR
RIRI
BR
RRI
BR
RIR
BR
RR
BR
II
DDDDDD sΩΩsΩsΩss 
Use chain rule on second term on the right hand side     BR
RRI
BR
RIR
BR
RIR
DDD sΩsΩsΩ 
And insert it above    
termsadditional
2 BR
RIR
BR
RIRI
BR
RRI
BR
RR
BR
II
DDDDDD sΩsΩΩsΩss 
R
BBR
R
D vs With
(3.23)
  













onacceleratiangular
onacceleratilcentripeta
onacceleratiCoriolis2
BR
RIR
BR
RIRI
R
B
RI
BR
RR
BR
II
D
DDDD
sΩ
sΩΩ
vΩ
ss

, we are dealing only with Coriolis and centripetal accelerationsIf R is earth E,   0sΩ BR
EIE
D

3 Kinematics
37
Grubin transformation
pp. 64, 65
B
Br B
II
Figure 3.16 Asymmetric satellite
Determine the inertial acceleration of the main geometrical
point Br , given the inertial acceleration of the center of mass B
Vector triangle BIBBIB rr
sss 
?What is BB
II
r
DD s
whereUse the Euler transformation  BB
BI
BB
BI
BB
II
rrr
DDDD sss  0s BB
B
r
D
Another Euler transformation     BB
BIBI
BB
BIB
BB
BII
BB
II
rrrr
DDDD ssss 
Twice the derivative wrt I: BI
II
BB
II
IB
II
DDDDDD rr
sss  (3.25)
Evaluate first term on right-hand side       BB
BIB
BB
BBI
BB
BIB
BB
BIB
rrrr
DDDD ssss 
And substituting   BB
BIBI
BB
BIB
BB
II
rrr
DDD sss 
we want; is given
I
BIB
II
rr
DD as  I
BBI
II
DD as 
Into Eq. 3.25   BI
II
BB
BIBI
BB
BIB
IB
II
DDDDD rrr
ssss 
Re-written in acceleration terminology   I
BBB
BIBI
BB
BIBI
B rrr
D assa 
given;
I
Ba angular acceleration  BB
BIB
r
D scentripetal acceleration;BB
BIBI
r
s


38
Notable Quotables
 The Euler transformation is the second pillar of tensor flight dynamics
o It is a tensor relationship
o It transforms the reference frame of the rotational time derivative
o It maintains the tensorial properties of time operations
 The Euler transformation is used to prove two properties of angular
velocities
o Angular velocities are additive
o The rotational time derivative of angular velocities is equal in either
frame
 The Coriolis transformation relates the accelerations of rotating frames
o The additional terms are
 Coriolis acceleration
 Centripetal acceleration
 Angular acceleration
 The Grubin transformation relates the accelerations of accelerated points
o The additional terms are
 Centripetal acceleration
 Angular acceleration

3 Kinematics
39
3 Kinematics
Euler angle differential equations
Direction cosines differential equations
Quaternion differential equations
Summary
Notable Quotables

40
Euler angle differential equations
2B
3B
1B
r
q
p
Figure 3.17 Aircraft attitude motions
2X
2Y
NoseofVehicle
1G
2G
1X
3G
3X
=
1Y
3Y
2B
1B
3B
North
Right
Wing


f
Figure 3.21 Euler Angle Transformation
NoseofVehicle
1G
2G
1X
2X
3G
3X
=
1Y
3Y
2Y
2B
1B
3B
North
Right
Wing


f
1yf
3x
2y
Figure 3.22 Euler angular velocities
(2) Euler angle differential equations integrate the angular
velocity vector of body B wrt ground G
 Given the body rates
calculate the Euler angles f, , 
 rqpBBG
][ω
(1) Euler angle transformation XGYXBYBG
TTTT )]([)]([)]([][ f
consists of three individual transformations BG
T )]([ f 
(Note, the ground coordinates used in the introduction are
replaced by the geographic coordinates North, East, down)
G
]
G
]











100
0cossin
0sincos
][ 

XG
T









 



cos0sin
010
sin0cos
][ YX
T












ff
ff
cossin0
sincos0
001
][ BY
T














fffff
fffff

coscossincoscossinsinsinsincossincos
sincoscoscossinsinsincossinsinsincos
sincossincoscos
][ BG
T
321 xyyω f  BG
In body coordinates BBBBBG
xyy ][][][][ 321 f  
pp. 67 - 71
(3.26)































r
q
p
ff
ff
ff


f
cos/coscos/sin0
sincos0
tancostansin1





(3.27)

3 Kinematics
41
Direction cosines differential equations
pp. 71 - 73
The transformation matrix can also be obtained by integrating the direction
cosines differential equations
BG
T][
Use the definition of the rotational time derivative Eq. 3.3 (Slide 3.1_3)
B
BG
BG
B
BG
x
dt
dT
T
dt
dx
xD ][][][ 








Express the Euler transformation Eq. 3.16 (Slide 3.2_2) in body coordinates B
BBBGBBBG
xxDxD ][][][][  BBBG
B
BG
x
dt
dx
xD ][][][ 




Compare
BBG
BG
BG
dt
dT
T ][][ 



(3.28)Solve for
BG
dt
dT



 BG
B
BG
BG
T
dt
dT
][][



Nine linear differential equations


































333231
232221
131211
333231
232221
131211
0
0
0
ttt
ttt
ttt
pq
pr
qr
ttt
ttt
ttt




Euler angles )sgn(
cos
arccos,)arcsin(,)sgn(
cos
arccos 12
11
1323
33
t
t
tt
t
















f


42
Quaternion differential equations
pp. 73, 74
Quaternions are vectors in four dimensions 3210 qqqq kjiq 
Written in 4x1 matrix



















3
2
1
0
0
][
q
q
q
q
q
q
Given ][][ rqpBBG






















][][][
][0
2
1
][
00
q
q
q
q
BBGBBG
BBG




Linear differential equations (3.29)















































3
2
1
0
3
2
1
0
0
0
0
0
2
1
q
q
q
q
pqr
prq
qrp
rqp
q
q
q
q




Initializing with Euler angles
)2/sin()2/sin()2/cos()2/cos()2/cos()2/sin(
)2/sin()2/cos()2/sin()2/cos()2/sin()2/cos(
)2/cos()2/sin()2/sin()2/sin()2/cos()2/cos(
)2/sin()2/sin()2/sin()2/cos()2/cos()2/cos(
3
2
1
0
ff
ff
ff
ff




q
q
q
q

Euler angles 2
3
2
2
2
1
2
0
1032
20312
3
2
2
2
1
2
0
3021 )(2
tan,)(2sin,
)(2
tan
qqqq
qqqq
qqqq
qqqq
qqqq





 f
Euler transformation TM
   
   
    













2
3
2
2
2
1
2
010322031
1032
2
3
2
2
2
1
2
03021
20313021
2
3
2
2
2
1
2
0
22
22
22
][
qqqqqqqqqqqq
qqqqqqqqqqqq
qqqqqqqqqqqq
T BG

3 Kinematics
43
Summary
EULER ANGLES DIRECTION COSINES QUATERNIONS
Given ][][ rqpBBG
 from Euler's Attitude Equations
Differential
Equations 



























r
q
p
f )],,([ f


f
BGBBGBG
TdtdT ][][]/[ 















































3
2
1
0
3
2
1
0
0
0
0
0
2
1
q
q
q
q
pqr
prq
qrp
rqp
q
q
q
q




Initialization Directly by 000 ,, f BG
tT )]0([  ),,()}0({ 000 fftq 
TM ),,(][ ffT BG
 Directly calculated  3210 ,,,][ qqqqfT BG

Euler Angles Directly calculated From BG
T][ ),,,( 3210 qqqqf












f
Advantage
Three differential equations. An-
gular attitude calculated directly.
Direct initialization
Transformation matrix calculated
directly. Six linear differential equa-
tions.
Four linear differential equations.
Simple orthogonality condition.
Disadvantage
Singularity at 2/  Nonlinear
differential equations. Transfor-
mation matrix not directly avail-
able
Computationally ineffective. Euler
angles not directly available. Initial
calculations necessary.
Transformation matrix and Euler
angles not directly available. Ini-
tial calculations necessary.
Application Legacy Spacecraft, boosters Airplanes, missiles

44
Notable Quotables
 The fundamental kinematic problem in flight dynamics is:
o Given the body rates yaw, pitch, and roll
o What are the Euler angles yaw, pitch, and roll
 The Euler transformation matrix consists of three individual transformations
o Yaw transformation
o Pitch transformation
o Roll transformation
 Euler angle differential equations
o Consist of three nonlinear first order differential equations
 Singularity at pitch angle vertical up or down
o Only found in legacy simulations
 Direction cosine differential equations
o Consist of nine linear first order differential equations
 No singularities
o Used in orbital and planetary simulations
 Quaternion differential equations
o Consists of four linear first order differential equations
 No singularities
o Used in terrestrial aircraft and missile simulations

45
4.1 Newton’s Law
4.1 Newton’s Law
4 Point-Mass Dynamics
4.1 Newton’s Law
4.2 Vehicles requiring Inertial Frame
4.3 Vehicles using Earth Frame
Trajectory Equations
Gravitational force
Aerodynamic force
Propulsion
Notable Quotables

46
4.1 Newton’s Law
Trajectory Equations
2MaSTech 2020 4 Point-Mass Dynamics
pp. 88 - 93
The time derivative is to be taken wrt the inertial frame I
If earth E is our non-inertial reference frame, Newton’s law is with Eq. 3.23 (Slide 3.2_6)
Newton’ Second Law: The time-rate-of change of linear
momentum equals the impressed force and is in the
direction in which this force acts
fp I
B
I
D
I
I
·
B
E
E
Figure 4.3 Earth and inertial frames
Linear momentum: I
B
I
B mvp 
(4.7)Newton’s law for us fav  I
B
I
B
I
mmD
    I
B
II
B
II
B
II
B
II
B
I
mDmDmDmDD vvvvp 
 2E
2I
1E
1I
3E3I
i3

i1
i2
I
 
e1
e2
e3
Greenwich Meridian
E
Figure 4.2
centripetal accelerationsfsΩΩsΩa 









  
onacceleratilcentripeta
BE
EIEI
BE
EEIE
B Dm 2Inertial frame perspective

(4.3)













100
0cossin
0sincos
][ EI
T
Transformation matrix of earth wrt inertial
coordinates through hour angle 
  
forcesapparentlcentrifuga
BE
EIEI
BE
EEIE
B mDmm sΩΩsΩfa  2Earth frame perspective centrifugal apparent forces

47
4.1 Newton’s Law
Gravitational force
pp. 94 - 96
But gravitational acceleration is opposed by centrifugal force
Newton’s inverse square law of gravitation 2
r
Mm
Gf 
M mass of earth
m mass of vehicle
r distance from earth center
G universal gravitational constant
Gravitational acceleration (4.9)2
r
M
G
m
f
g 
The TM of geographic wrt earth coordinates is obtained by
multiplying three transformations: longitude, complement of
latitude, and a 180 deg flip
On earth the centrifugal acceleration opposes the gravitational acceleration
BE
EIEI
BE
BE
BE
EIEI
g GM sΩΩ
s
s
sΩΩgg  3
||
Gravity acceleration is a function of latitude, besides altitude; mean value is
2
/8066.9 smgg 
Express Eq. 4.9 in tensors (vectors) 3
|| BE
BE
GM
s
s
g  displacement of vehicle c.m.
wrt to earth center
BEs
1E
2E
3E
1X
2X
3X
1Y
2Y
3Y
3G
2G
1G
l


 
Equator
Greenwich Meridian
Figure 4.7 Geographic wrt Earth
transformation by longitude l and latitude 
XEYXoGYoGE
lTTTT )]([)]90([)]180([][ 

















sinsincoscoscos
0cossin
cossinsincossin
][
ll
ll
ll
T GE
(4.10)

48
4.1 Newton’s Law
Aerodynamic force
pp. 99 - 102
Most important aerodynamic forces are lift L and drag D
 shapeon/off,powerattack,ofangleMach,, fCC DL 
Sq
D
C
Sq
L
C DL  :tCoefficienDrag,:tCoefficienLift
dynamic pressure; S reference area
2
2
Vq


1B
3B

E
Bv
L
D
3V
1V
Figure 4.14 Angle-of-attack, Lift,
and Drag Aerodynamic force


































 





cossin
0
sincos
0
cos0sin
010
sin0cos
][][][
LD
LD
L
D
V
a
BVB
a
CC
CC
Sq
C
C
SqfTf
CD
CL
CLo
CDo
Offset drag polar
Fig 4.10 Offset drag polar
K induced drag coefficient 2
00 LLDD CCkCC  (4.11)
Linear
range

CL Buffeting
onset
Fig 4.12 Linear lift slope
Linear lift slope

 aLLL CCC  0
lift slopeLC(4.13)
Lift and drag are in velocity
coordinates V and force is
expressed in body coordinates B 








 



cos0sin
010
sin0cos
][ BV
T (4.14)

49
4.1 Newton’s Law
Propulsion
pp. 103, 104
Rocket propulsion
cmt fp 
Thrust = fuel-mass-rate ejected through the nozzle with exit velocity cfm
Specific impulse is the measure of merit for rocket thrust. It is the ratio of impulse
delivered over propellant weight consumed
of
p
of
p
sp
gm
t
tgm
tt
I





Alternate thrust equation (4.19)ofspp gmIt  m/s28066.9og
Ae exhaust nozzle areaBackpressure correction with altitude eAltSLSLAlt Apptt )( 
Turbojet propulsion
Thrust = exit air mass flow minus entry air mass floweaVm Vma
)( VVmt eap  
Thrust depends on several parameters
 AttackofAngleSettingPowerAltitudeMachftp  ,,,
Specific fuel consumption is the measure of merit for turbojets
pfF tmb / Usually given in kg/(dN hr),


50
4.1 Newton’s Law
Notable Quotables

 Newton’s Second Law governs all trajectories
o We call it ‘F=ma’
o Its reference frame is the inertial frame
o Non-inertial frames incur correction terms that must be assessed
 Newton’s Inverse Square Law governs gravitational acceleration
o On earth we speak of gravity acceleration
 Centrifugal force opposes gravitational force
 Atmosphere is divided into troposphere and stratosphere
o Standard atmospheres
 US 1976
 ISO 2533 1962 & 1975
o Mach regimes
 M<1: subsonic
 1<M>5: supersonic
 M>5: hypersonic
 Drag polar relates drag and lift parabolically
 Newton’s Second Law also governs propulsion
o Rocket thrust equals rate-of-change of fuel mass x exhaust velocity
o Turbojet thrust equals air mass x rate-of-change of velocity

51
4.2 Vehicles Requiring Inertial Frame
4.1 Newton’s Law
Satellite orbiting the earth
Astronaut in space station
Rocket’s inertial equations of motion
Polar equations in tensors
Polar equations in matrices
Hypersonic vehicles
Notable Quotables

52
Satellite orbiting the earth
pp. 106 - 108
The satellite’s trajectory is independent of its mass!
In this special case the gravitational acceleration points towards the center of the earth along the 1I axis

If the satellite were stationary, it would fall towards the earth center, but its orbital speed prevents
this from happening
I ·
B
E
E
I
Figure 4.15 Satellite B orbiting the earth
In inertial coordinates   III
B
II
B
I
gmdtdvmvDm ][][ 
(4.24)
Gravitational acceleration best given in geographic axes
GIGII
B gTdtdv ][][][ 

















sin0cos
cossincossincoscoscoscossinsincossinsinsinsinsin
sinsincoscoscoscossincoscossinsinsinsincoscossin
][ llllll
llllll
T IG
Simple case: 0,0,0  l






























 






0
00
0
001
010
100
][][
g
g
gT
dt
dv GIG
II
B
With
Eq. 4.3 (Slide 4.1_2)EI
T][
Eq. 4.10 (Slide 4.1_3)GE
T][
GEEIEGIEIG
TTTTT ][][][][][ The TM is obtained from
Newton’s Law, Eq. 4.7 (Slide 4.1_2), with gravitational force only
gv mmD I
B
I
 (4.21)
G
g][

53
Astronaut in space station
An astronaut’s “weightlessness” demands an explanation!
Shift the reference frame of the derivative from I to frame B (satellite) twice (Coriolis transformation)
The astronauts do not experience their weight because it is opposed by the centrifugal force
(the astronaut B tracks the point I)Introduce in Eq. 4.21 (previous slide) andBI
II
B D sv  BIIB ss 
gs mDmD IB
II

 
    
  
   gsΩΩsΩsΩs
gsΩΩsΩsΩs
gsΩsΩsΩs
gsΩs
gs
mDDDDm
mDDDDm
mDDDm
mDmD
mDmD
IB
BIBI
IB
BBI
IB
BIB
BI
BB
IB
BIBI
IB
BBI
IB
BIB
IB
BB
IB
BI
IB
BBI
IB
BI
IB
BB
IB
BI
IB
BI
IB
II





2
Circular orbit: ,0vs  I
BIB
B
D,constIB s   0Ω BIB
D,0as  I
BIB
BB
DD
0gsΩΩ  mm IB
BIBI
Difference between inertial velocity and earth (geographic) velocityI
Bv E
Bv
Start with and transform the derivative to earth frame EBI
II
B sDv BI
EI
BI
E
BI
I
DD sΩss 
Pick any point E on the earth and substituteEIBEBI sss  BI
EI
EI
E
BE
E
BI
I
DDD sΩsss 
But because is fixed in the earth, therefore0s EI
E
D EIs BI
EIE
B
I
B sΩvv 
BI
EII
B
E
B sΩvv 
satellite to earth center
earth angular velocityEI
Ω
BIs

(4.28)
pp. 108 - 110

54
Rocket’s inertial equations of motion
pp. 111, 112
The right hand side of Eq. 4.7 (Slide 4.1_2) consists of aerodynamic and propulsive forces, and
weight
Computation is in inertial coordinates
But aerodynamic forces are given in velocity V coordinates
Propulsive force is given in body B coordinates
Gravitational weight is given in geographic coordinates G
(4.29)gpa
I
B
I
mD fffv 
I
g
I
p
I
a
II
B
I
fffvDm ][][][][ 
G
g
IGB
p
IBV
a
IVII
B fTfTfTdtdvm ][][][][][][][ 
We incur 5 coordinate systems and 4 coordinate transformations
GB
GV
EGIE
T
T
TT
][
][
][][ 
Computed in inertial coordinates
II
B
I
BI vdtds ][][ 

Once is obtained from Eq. 4.29 another integration yields the displacement vectorI
Bv
I
BBI
I
D vs 
BIs

55
Polar equations in tensors
pp. 112 - 114
The inertial equations of motion are transferred to the velocity frame V
First transfer the derivative of Eq. 4.7 (Slide 4.1_2) to the earth frame E I
B
EII
B
EI
B
I
vDD Ωvv 
Then solve Eq. 4.28 (Slide 3) for and substitute twiceBI
EIE
B
I
B sΩvv 
  BI
EIEIE
B
EI
BI
EIEE
B
EI
B
I
DDD sΩΩvΩsΩvv 
Apply chain rule to the second term on the right side, and note 
(earth’s angular velocity is constant)
BEBI ss  E
BBE
E
BI
E
DD vss 
    E
B
EI
BE
EEI
BE
EIE
BI
EIE
DDD vΩsΩsΩsΩ 
Finally substitute into Newton’s law Eq. 4.7 (Slide 4.1_2) to get the polar equations of motion
(4.36)BE
EIEIE
B
EIE
B
VEE
B
V
mmDm sΩΩvΩfvΩv  2)(
(4.32)
Transformation of velocity axes wrt geographic axes


1G
2G
1V
2V
3V
1X
2X
3X
3G=

































 








cossinsincossin
0cossin
sinsincoscoscos
100
0cossin
0sincos
cos0sin
010
sin0cos
][ VG
T
Figure 4.16
Substituted and change derivativeBE
EIEIE
B
EIE
B
EI
B
I
DD sΩΩvΩvv  2 E
B
VEE
B
VE
B
E
DD vΩvv 
BE
EIEIE
B
EIE
B
VEE
B
VI
B
I
DD sΩΩvΩvΩvv  2

From which we get, given ][][ GGG
GE
B wvuv 
Heading angle 






G
G
u
v
arctan Flight path angle










22
arctan
GG
G
vu
w


56
Polar equations in matrices
pp. 115 - 116
Express Eq. 4.36 (previous slide) in V coordinates
(4.40)
E
BE
EEIEEIVEVE
B
VEEEIVEVVE
B
VVEVE
B sTvTTf
m
vdtdv ][][][][][][][][2][
1
][][]/[ 
The geographic velocity vector is in V coordinates with]00[][ Vv VE
B  || E
BV v
Another integration yields the displacement vector of vehicle c.m. wrt earth center E
(4.42)
E
BEs ][
VE
B
VEE
BE vTdtds ][][]/[ 
·


1G
2G
1V
2V
3V
1X
2X
3X
3G
·
Figure 4.17 Heading and
flight-path angle rates
Special treatment of in Eq. 4.40VVE
][
Fig. 4.17  23 vxω   VE VXVXVVE
vxT ][][][][ 23   









































 








cos
sin
0
1
0
1
0
0
cos0sin
010
sin0cos
][



VVE
TMs: Eq. 4.14, Slide 4.1_4, Eq. 4.32, Slide 4.2_5, Eq. 4.10, Slide 4.1_3BV
T )]([  VG
T )],([  GE
lT )],([ 
The forces are
GVGB
p
VBV
a
V
gTfTff ][][][][][][  (4.41)
The left side of Eq. 4.40 contains the integration variables V, ,  :

























































V
V
VVV
vdtdv VE
B
VVEVE
B cos
0
0
0sin
sin0cos
cos0
0
0][][]/[ (4.44)


57
Hypersonic vehicles
pp. 116 - 121
Aerodynamics are applied in the maneuver plane (which coincides with the
symmetry plane of the airframe) and transformed to the velocity coordinates
We use Eqs. 4.40 and 4.42 from the previous slide to fly the NASA X30 hypersonic aircraft with
the externally applied forces of Eq. 4.41
2V
f
3V3B
2B
L
horizontal
Figure 4.20 Aircraft
maneuver plane
 )(0)(][ MCMCSqf LD
M
a 
(4.46);












ff
fff
cossin0
sincos0
001
)]([ MV
T













f
f
cos)(
sin)(
)(
][][][
MC
MC
MC
SqfTf
L
L
D
M
a
MVV
a
Propulsive force is given in body coordinates B  00][ ofsp
B
p gmIf 
To transform to we need two TMs
B
pf ][
V
pf ][ B
p
MBVMB
p
VBV
p fTTfTf ][)]([)]([][][][ f
































 

ff
ff
ff
ff
ff


coscossincossin
sincos0
cossinsinsincos
cossin0
sincos0
001
cos0sin
010
sin0cos
][][][ MVBMBV
TTT


































f
f

fff
fff

cossin
sinsin
cos
0
0
coscossincossin
sincoscossinsin
sin0cos
][][][ ofsp
ofsp
B
p
BVV
p gmI
gmI
fTf 

Gravitational force is given in geographic coordinates  mgf G
g 00][  G
g
VGV
g fTf ][][][ 
( see Eq. 4.32 (Slide 4.2_5)







































cos
0
sin
0
0
cossinsincossin
0cossin
sinsincoscoscos
][ mg
mg
f V
g

Summary






































f
f

f
f
cos
0
sin
||
cossin
sinsin
cos
cos)(
sin)(
)(
][][][][ 2
BE
ofsp
L
L
D
V
g
V
p
V
a
V
s
GM
mgmI
MC
MC
MC
Sqffff 
VG
T][
(4.51)

58
Notable Quotables
 Satellites and space stations orbit the earth
o By their weight they would fall towards the earth
o Except the centrifugal force keeps them in orbit
o Inside you are in free-fall, i.e., weightless
 Newton’s law requires for high fliers an inertial reference frame
o Inertial equations of motion
 Used for high fidelity simulations
o Polar equations of motion
 Used for simpler simulations
 Hypersonic airplanes use bank to turn to maneuver
o Coordinated turn is without side-slipping
 Phugoid is the long period flight mode
o Exchanges of kinetic and potential energy
o At constant angle-of-attack

59
4.3 Vehicles Using Earth Frame
4.1 Newton’s Law
Equations of motion
Rockets
Rocket Example
UAV & Aircraft
UAV Example
Notable Quotables

60
Equations of motion
pp. 125 - 127
These are the so-called flat earth, point-mass equations of motion in tensor form
The left side we already have from Eq. 4.44 (Slide 4.2_6)
With earth serving as the ‘inertial’ reference frame, the Coriolis and centrifugal terms in Eq. 4.36 vanish
(4.38)fvΩv  )( E
B
VEE
B
V
Dm
To get the displacement vector of the vehicle c.m. B wrt to an earth reference point E we integrate again
E
BBE
E
D vs  (4.39)
The flat earth assumption uses a local level plane, tangent to the
earth at the point of interest E, as the inertial reference frame
1E
2E
3E
3L
2L
1L
l

.
Local Level Plane
E
Figure 4.25 Local level plane
tangential to point E
Instead of the earlier inertial coordinates we now have the
local level coordinates
I
]
L
]
Express Eq. 4.38 in velocity coordinates V
VVE
B
VVEVE
B
V
fvvDm ][)][][]([ 
But now  and  are with respect to the local level coordinates L

















V
V
V
mvmdtdvm VE
B
VVEVE
B cos][][]/[

And with integrate again 00][ Vv VE
B 
VE
B
VL
L
BE
vT
dt
ds
][][



(4.56)
is the same as Eq. 4.32 (Slide 4.2_5)
(4.55)
With the added, Eq. 4.38 is in matrix form
  LVLV
p
V
a gTff
m
V
V
V
][][][][
1
cos 










 




V
f ][
VL
T][

61
Rockets
pp. 128 - 130
Aerodynamic drag only











0
0
)(
][
MC
Sqf
D
V
a Propulsive thrust











0
0][
ofsp
V
p
gmI
f

Forces applied to the equation of motion, Eq. 4.55 (previous slide); Rocket flies ballistically at 0
Substituted into Eq. 4.55
(previous slide)


























 



cos
0
sin)(
cos
g
g
m
gmI
MC
m
Sq
V
V
V
ofsp
D




Ballistic rocket equations (4.60)
  
















 cos
sin)(
Vg
g
m
gmI
MC
m
SqV ofsp
D




Integrated












































sin
sincos
coscos
0
0
cos0sin
sinsincossincos
cossinsincoscos
][][ V
V
vT
dt
ds VE
B
VL
L
BE
Gravity







































cos
0
sin
0
0
cossinsincossin
0cossin
sinsincoscoscos
][][][ g
g
gTg LVLV
(4.59)

62
Rocket Example
pp. 128 - 131
Initial conditions:
Altitude 100 m
Velocity 90 m/s
Elevation 85.9 deg
Note:
Ballistic equations
only allow for very
steep launches,
because no
aerodynamic lift is
available at low
launch speeds to
counteract gravity

0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 5 10 15
CD vs Mach for MRBM Mach CD
0 0.292
0.5 0.311
0.8 0.645
1.05 1.241
1.5 1.096
2 0.957
3 0.833
4 0.776
8 0.723
12 0.651
Ref area = 0.636 m2
G&CWarhead
Oxygen and Kerosene
12 m
0.9m
0.6 m
Launch mass 6000 kg
Empty mass 2000 kg
Specific impulse 230 sec
Burn time 70 sec
CD vs. Mach for SRBM
SCILAB

63
UAV & Aircraft
pp. 132 - 135
We use again Eqs. 4.55 & 4.56 (Slide 2)
Propulsion is given in body coordinates B
UAVs and aircraft, modeled in point-mass, are steered directly by
angle-of-attack  and bank angle f
1B
3B
1V
3V
1M
3M

f
L
D
E
Bv
Figure 4.29 Maneuver plane of aircraft
Aerodynamics is given in the maneuver plane 1M, 3M
   LD
M
a CCSqLDf  00][
Gravity contribution is taken from Eq. 4.59 (Slide 3)
Converted to velocity coordinates with Eq.5.51 (Slide 4.2_7)
Converted to velocity V coordinates with Eq. 4.46 (Slide 4.2_7)




































f
f
ff
ff
cos
sin0
cossin0
sincos0
001
][][][
L
L
D
L
D
M
a
MVV
a
C
C
C
Sq
C
C
SqfTf
 00][ p
B
p tf 


































f
f

fff
fff

cossin
sinsin
cos
0
0
coscossincossin
sincoscossinsin
sin0cos
][][][ p
p
B
p
BVV
p t
t
fTf
(4.66)
Collecting the external
forces and substituting
into Eq. 4.55 (Slide 2)






























ff
f

f




coscossincos
sinsin
cos
sin
cos
sincos
V
g
mV
t
C
mV
Sq
mV
t
C
mV
Sq
g
m
t
C
m
Sq
V
p
L
p
L
p
D




And integrating


















sin
sincos
coscos
V
dt
ds
L
BE
(4.67)

64
UAV Example
pp. 136 - 137
Let’s cruise a UAV at
Mach =0.65 with these
aerodynamics and
mass= 850 kg
We use Eqs. 4.66 and 4.67 (previous slide) and program them with our favorite matrix tool box
For cruise we set thrust = drag and pick an angle-of-attack that maintains constant altitude
The oscillations
are called
phugoids
Write

Ref Area = 0.929 m2
Lift Coefficient
0 0.5 1-0.5
0
0.05
0.1
0.15
0.2
0.25
DragCoefficient Mach = 0.65
Mach=0.65
LiftCoefficient
Angle-of attack - deg
0 2 4 6 8
-0.5
0
0.5
1
Figure 4.30 UAV Trajectories
The initial conditions are altitude = 500 m, heading = 20 deg, velocity = 221 m/s (Mach 0.65)

65
Notable Quotables
 Flat earth equations of motion use the earth as reference frame
o Neglecting the Coriolis and centrifugal terms
o Tangential plane to earth becomes the flat plane
o Geographic coordinates become the local level coordinates
 Rockets fly ballistically
o They have rotational symmetry
o They fly at zero angle-of-attack
o Equations of motion have 5 integration variables
o Don’t forget the backpressure correction
 UAVs and aircraft maneuver
o They have planar symmetry
o They maneuver with angle of attack and bank angle
o Exhibit the phugoid mode

67
5.1 Newton's and Euler's Laws
5 Rigid Body Dynamics
5.2 Missiles and Rockets
5.3 Aircraft and UAV
Equations of motion
Moment of inertia
Elliptical earth
Earth as reference frame
6 DoF Aerodynamics
Linearization
Linear equations of motion
Notable Quotables

68
Equations of motion
2MaSTech 2020 5 Rigid Body Dynamics
pp. 144 - 146
i3

i1
i2
I
1I
2I
3I
Figure 5.1 J2000 Inertial reference frame
Newton’s law (Slide 4.1_2) governs the translational motions:
The inertial time rate of the angular momentum equals the
external moment applied at the center-of-mass B
Attitude dynamics are governed by Euler’s law
B
BI
B
I
D ml 
Shift from I to B via Euler transformation B
BIB
B
BIBIB
B
BBIB
B
I
DD mωIΩωIωI  )()(
Chain rule applied B
BIB
B
BIBIBB
B
neglected
BIB
B
BBIB
B
BIBIB
B
B
DDD mωIΩωIωIωIΩωI 

)()(
B
BIB
B
I
D mωI )(

6 DoF equations of
motion in matrix form
B
B
BBIBB
B
BBI
BBI
BB
B mI
dt
d
I ][][][][][ 







(5.7)
(5.6)
BBI
II
B
fT
dt
dv
m ][][





Angular momentum of body B wrt inertial frame I, referred to
the center of mass B: BIB
B
BI
B ωIl  moment of inertiaB
BI
(5.4)Six-degrees-of-freedom
equations of motion
B
BIB
B
BIBIBB
B D mωIΩωI 
fv I
B
I
mD
(5.5)
(also Eq. 4.7 (Slide 4.1_2))
fp I
B
I
D

69
Moment of inertia
pp. 148 - 150
205
Dimension - m
B
B1
B2
R
i
2
Figure 5.2 Booster
Determining the center-of-mass B wrt arbitrary
reference point R, , of i bodies with mass




 n
i
i
n
i
RBi
BR
m
m i
1
1
s
s
BRs im
(see figure)Moment of inertia of point-mass i wrt c.m. B iBiBi
i
B mI ss
Elevated to tensor of rank two
iBiBi
i
B m SSI  is the skew-symmetric form ofiBS iBs
Summing over entire body B (5.13) 
i
iBiBi
i
i
B
B
B m SSII
In body coordinates  
i
B
iB
B
iBi
i
Bi
B
BB
B SSmII ][][][][
Symmetric real matrix; diagonal elements are the principal moments of
inertia, off-diagonal elements are the products of inertia





















i
iBiBi
i
iBiBi
i
iBiBi
i
iBiBi
i
iBiBi
i
iBiBi
i
iBiBi
i
iBiBi
i
iBiBi
BB
B
ssmssmssm
ssmssmssm
ssmssmssm
I
)(
)(
)(
][
22
22
22
213231
323121
312132
(5.14)
 The moment of inertia (MOI) of body B referred to an arbitrary point R is equal to
the MOI referred to its center-of-mass B plus a term calculated as if all mass of body B
mB were concentrated at its center-of-mass B

Huygen’s Rule BRBR
BB
B
B
R m SSII  (5.15)

70
Elliptical earth
pp. 151 -152
The external shape of the earth is a geoid. The first approximation is a sphere, the second and
better approximation is an ellipsoid
What we called the geographic system is now renamed the
geocentric system G
The WGS84 model requires a new coordinate system, called the
geodetic system D, where the 3D axis is still locally perpendicular
to the earth surface but does not, in general, point to the center
of the earth
Figure 4.7 (Slide 4.1_3) gave us the geographic coordinate system
of a spherical earth, where the 3G axis was perpendicular to the
earth surface and pointed to the center of the earth
 GPS and most test ranges use the WGS84 ellipsoidal earth model
c d
BIs .
.
B
I
h

B0
.
Figure 5.3 Geocentric and
geodetic latitudes
The angular difference is called the deflection angle cd  
(5.20)

And the transformation of geodetic wrt
the earth coordinates














ddd
ddd
DE
ll
ll
ll
T


sinsincoscoscos
0cossin
cossinsincossin
][
With the transformation matrix














cos0sin
010
sin0cos
][ DG
T (5.19)

71
Earth as reference frame
pp. 155 - 157
Many applications allow the use of the earth as the reference frame in Newton’s and Euler’s
equations
Relate Eq. 5.4 (Slide 2) to the earth frame E fv E
B
E
mD
(5.21)And shift the derivative to the body frame B gffvΩv mmDm pa
E
B
BEE
B
B

Replace frame I by frame E in Eq. 5.5 (Slide 2) pa BB
BEB
B
BEBEBB
B D mmωIΩωI  (5.22)
Eq. 5.21 expressed in body coordinates B with gravity in local level coordinates L
LBLB
p
B
a
BE
B
BBEBE
B gTmffvmdtdvm ][][][][][][]/[  (5.23)
Where with its vector form comes from the attitude equations
BBE
][  rqpBBE
][
Eq. 5.22 expressed in body coordinates B
(5.25)
B
B
B
B
BBEBB
B
BBEBBEBB
B pa
mmIdtdI ][][][][][]/[][  
Once is obtained by integration, another integration yields wvuv BE
B ][
L
BEs ][
BE
B
BL
L
BE
vT
dt
ds
][][



(5.24)

Using the quaternion integration, Slide 3.3_4, we get the TM and the Euler angles
BL
T][ f ,,

72
6 DoF Aerodynamics
pp. 158 - 162
Aerodynamic forces and moments are represented by
tables and derivatives and expressed in body axes
Aerodynamics are modeled in body axes B
Aerodynamics in 6 DoF equations of motion depends on angle-
of-attack  and sideslip angle b relating to the wind axes W
A
Bv

b
1B
2B
3B
1W
1S
Figure 5.7 Attitude of aircraft B wrt
atmosphere A
1B
2B
3B
1W
2W
3W
1S
2S
3S

b
Figure 5.8 TM of wind wrt body
coordinates











100
0cossin
0sincos
][ bb
bb
WS
T









 



cos0sin
010
sin0cos
][ BS
T
Assuming that the atmospheric motion wrt the earth can be neglected
 wvuvv BE
B
BA
B  ][][ 






u
w
arctan 






V
v
arcsinb 

where a aileron, e elevator, r rudder deflections











),,,(
),,,(
),(
][
eMC
rMC
MC
Sqf
Z
Y
X
B
a
b
b












.).,,,,,(
.).,,,,,(
),,,,(
][
mcrrMC
mceqMC
apMC
lSqm
n
m
l
B
B a
b
b
b













bbb
bbb
cos0sin
sinsincossincos
cossinsincoscos
][][][ BSWSWB
TTT (5.28)

73
Linearization
pp. 162 - 165
Linearization of the equations of motion is relative to a steady flight, which is unaccelerated and
without maneuvers
 constwconstvconstuv rrr
BE
Br
][  000][  rrr
BEB
rqpr

The TM of Eq. 3.26 (Slide 3.3_2) for small Euler angles is with its perturbation matrix
BL
T][














1
1
1
][
f
f

BL
T














0
0
0
][
f
f

BL
T
The TM of Eq. 5.26 (Slide 6) for small  and b is with its perturbation matrixWB
T][












10
01
1
][

b
b
WB
T












00
00
0
][

b
b
WB
T
If reference flight is at r and perturbed flight at p,

Can be approximated by the slope at)(mC mC )( rmC 
 
 mrmpm CCC )()(
Aerodynamic derivatives: pitching moment example

Cm

r






)()( rmpm
m
CC
C
p
mC
The perturbations are the difference between the perturbed flight and the reference flight
E
B
E
B
E
B rp
vvv 



































r
r
r
p
p
p
w
v
u
w
v
u
w
v
u

EBEBEBBE prp
ωωωω 
0
























p
p
p
r
q
p
r
q
p

(5.32)

74
Linear equations of motion
pp. 166 - 167
We linearize the left side of Eq. 5.23 (Slide 5)
(5.23)LBLB
p
B
a
BE
B
BBEBE
B gTmffvmdtdvm ][][][][][][]/[ 
Second term:
)[0]][][(because][][
)[0]][flightsteadyin(because][]([][
)][]([][)][]([][][][
BBE
B
BEBBE
B
BEB
BBEBBE
B
BE
B
BBE
BE
B
BE
B
BBEBE
B
BE
B
BEBBE
B
BEB
vvm
vvm
vvmvvmvm
r
r
r
rr
r
p
p



(steady reference flight)Time derivative
BE
B
BE
B
BE
B
BE
B dtvddtvddtdvdtdv rp
]/)([]/)([]/[]/[ 
With force perturbations
LBLLBL
r
B
p
B
p
B
a
B
a
BE
B
BEBBE
B gTmgTmffffvmdtvdm rrr
][][][][][][][][][][]/)([ 
Because in reference flight forces are balanced BLBL
r
B
p
B
a gTmff rr
]0[][][][][ 
(5.37)LBLB
p
B
a
BE
B
BEBBE
B gTmffvmdtvdm r
][][][][][][]/)([ 
(no reference moments)B
B
B
B
BBEBB
B
BBEBBEBB
B pa
mmIdtdI ][][][][][]/)([][
ordersecondtosmall

  


B
B
B
B
BBEBB
B pa
mmdtdI ][][]/)([][  (5.39)
BBEBBEBEBBEB
BBEBBEBEBBEB
rp
rp
dtddtddtddtd
][][][][
]/)([]/)([]/[]/[

 
We linearize the left side of Eq. 5.25 (Slide 5) with the angular velocity perturbations
(steady flight )
BBEBr
]0[][ 

75
Notable Quotables
 Newton’s law governs the translational equations of motion
 Euler’s law governs the attitude equations of motion
 A massive body has three mass moments
o 0-th: scalar mass
o 1-st: center-of-mass
o 2-nd: moment of inertia
 Moment of inertia matrix is a real symmetric matrix
o Diagonal elements are the axial moments of inertia
o Off-diagonal elements are the products of inertia
 WGS84 is the ellipsoidal earth model used in GPS
o It uses the geodetic coordinate system (instead of geocentric)
o Latitude is called geodetic latitude
 Low fliers can use Newton’s and Euler’s laws referenced to the earth frame
o Tactical missiles and rockets
o Aircraft and UAVs
 Aerodynamic forces and moments are mainly a function of
o Angle-of-attack
o Sideslip angle
o But let’s not forget the dynamic pressure and Mach number
 Linearization is accomplished with small perturbations
 Steady flight means
o Constant velocity without maneuvers
 Aerodynamic derivative
o Slope to the curve of an aerodynamic coefficient
o At the reference flight condition
 Linearized equations of motion
o Six linear differential equations
o Coupled

77
Translational equations of motion
Attitude equations of motion
Moving center-of-mass
State-space equations of motion
Surface-to-air missile example
Notable Quotables

78
pp. 169 -171
We start with Eq. 5.37 (Slide 5.1_8) and neglect the gravity term (small compared to aerodynamics)
B
p
B
a
BE
B
BEBBE
B ffvmdtvdm r
][][][][]/)([ 
Displaying the state variables u, v, w, and neglecting the thrust perturbations
B
a
r
rr
r
f
m
qu
rupw
qw
w
v
u
][
1



























Aerodynamic forces are in body coordinates











),,(
),,(
),(
][
qMC
rMC
qMC
Sqf
Z
Y
X
B
a

b
rVw
rVvb
 22
rrr wuV 
(rates r, q are
nondimensionalized)
Aerodynamics expressed as derivatives
 
 
 











rZZ
rYY
rx
B
a
VqlCC
VrlCC
VqlC
Sqf
q
r
q
2
2
2
][

b

b

(5.47)
Substituting and
replacing  and b
 
   
   


































rZrZ
rYrY
rx
r
rr
r
VqlCVwC
VrlCVvC
VqlC
m
Sq
qu
rupw
qw
w
v
u
q
r
q
2
2
2

b



(dropping Where  wvuv BE
B  ][














0
0
0
][
pq
pr
qr
BBE
 rr
BE
B wuv r
0][ ;;

79
pp. 172 -173
Now we have to contend with the 3x3 MOI matrix, which fortunately is a diagonal matrix for
missiles and rockets and therefore can easily be inverted
We recall Eq. 5.39 (Slide 5.1_8)
B
B
B
B
BBEBB
B pa
mmdtdI ][][]/)([][ 
 
























3
2
1
1
3
2
1
1
100
010
001
00
00
00
][
I
I
I
I
I
I
I BB
B
With the perturbation variables and again dropping the  rqpBBE
 ][ 
because of symmetry I3 = I2
































),,(
),,(
),(
100
010
001
2
2
1
rMC
qMC
pMC
lSq
I
I
I
r
q
p
n
m
l
b




The aerodynamics are modeled by derivatives
 
 
 





















rnn
rmm
rl
n
m
l
VrlCC
VqlCC
VplC
C
C
C
r
q
p
2
2
2
b

b

(5.49)

The attitude equations are
 
    
    





























rnrn
rmrm
rl
VrlCVvC
I
lSq
VqlCVuC
I
lSq
VplC
I
lSq
r
q
p
r
q
p
2
2
2
2
2
1
b





80
Moving center-of-mass
pp. 173 -174
Missiles and rockets expel much of their mass as thrust, resulting in significant center-of-mass shift
Aerodynamic moments are measured in the wind tunnel relative to a fixed reference point refmcx ..
ref
mc ....mc
mC
1B
3B
refmcmC ..;
ZC
x
refmcx ..
..mcx
Figure 5.11 Effect of c.m. shift on pitch
coefficient
The actual c.m. location is at ..mcx
Difference .... mcmc xx ref

The force (coefficient) measured at generate a
moment at the actual c.m.
refmc ..
lCxx Zmcmc ref
/)( .... 
Correction for pitching moment and yawing moment
coefficients
lCxxCC
lCxxCC
Ymcmcmcnn
Zmcmcmcmm
refref
refref
/)(
/)(
......;
......;


(5.51)

Compensations added to Eq. 5.49 (Slide 3)
 
    
    



































lCxxVrlCVvC
I
lSq
lCxxVqlCVwC
I
lSq
VplC
I
lSq
r
q
p
Ymcmcrnrn
Zmcmcrmrm
rl
refr
refq
p
/)(2
/)(2
2
....
2
....
2
1
b





81
pp. 174 -175
But first we simplify the equations further:
Eqs. 5.47 (Slide 2) and 5.51 (Slide 4) can be rearranged into two uncoupled sets of equations
Neglecting small
No roll excursions: p = f = 0
Horizontal flight only: (sufficient to study the dynamics, then )0rw
qxC
(5.56)
State-space formulation of the pitch equations for dynamic analysis






















































q
w
u
C
xx
u
lC
I
lSq
lu
C
xx
u
C
I
lSq
C
mu
Slq
uC
mu
Sq
q
w
r
Z
mcmc
r
m
r
Z
mcmc
r
m
Z
r
rZ
r
q
ref
q
ref
q
2
)(
2
)(
2
....
2
....
2




Both equation give the same dynamic response.
Pitch equations
    
        







































luqlCuwCxxuqlCuwC
I
lSq
uqlCuwC
m
Sq
qu
q
w
Z
qrefq
q
C
rZrZmcmcrmrm
rZrZr
/2)(2
2
....
2   




    
        







































lurlCuvCxxurlCuvC
I
lSq
urlCuvC
m
Sq
ru
r
v
Y
rrefr
r
C
rYrYmcmcrnrn
rYrYr
/2)(2
2
....
2   


bb
b
Yaw equations
rr uV 

82
Surface-to-air missile example
pp. 176 -181

Initial conditions: w0 = 11 m/s (0 = 1 deg)
Flight conditions: Altitude = 6000 m, Mach = 2
5 m
Rocket MotorWarheadG&CSeeker
Area S=0.0491 m2; Length l=0.25 m; c.m. = 3 m from tip
Mass Properties
Mass: m-kg MOI: I2-kgm2 c.m.: x-m
Launch 300 440 2.9
Burn-out 118 180 2.16
Flight Conditions; r=0 deg Aerodynamic Derivatives (per radian)
Mach
Altitude
m
Dyn. Pres.
Pa
0.5 1000 15,729 -16.2 -206 -18.3 -837
2 6000 132,253 -19.0 -255 -20.0 -1932
3 9000 194,107 -18.5 -255 -9.07 -2064
qZC mCZC qmC Static margin:


Z
m
C
C
stm 
Characteristic Polynomial
74.573442 +2.0578264s +s2
Roots
-1.0289132 + 8.574076i
-1.0289132 - 8.574076i
Frequency (rad/sec)
8.6355916
8.6355916
Damping
0.119148
0.119148

83
Notable Quotables

 Missiles and rockets exhibit tetragonal symmetry
o Their outer mold line is duplicated every 90 deg rotation
o The moment of inertia matrix is diagonal
o The cross principal moments of inertia are equal
 The gravity term in the translational equations is neglected
 Thrust perturbations are neglected
 The linearized equations of motion can be put into the state-variable format
o To take advantage of modern control techniques
 Static stability implies a negative moment derivative of the incidence angle
o The c.m. lies forward of the neutral point
o Also called weather cock stability
 It is desirable for air vehicles to exhibit static stability throughout the flight
 Dynamic stability implies the dampening of disturbances

85
5.3 Aircraft and UAVs
Aircraft example
Boeing 747 Data
Dutch-roll dynamics
Notable Quotables

86
pp. 183 -186
+a: positive rolling moment
+e: negative pitching moment
+r: negative yawing moment
Positive control deflection:
Aerodynamics
We start again with Eq. 5.37 (Slide 5.1_8) and keep the gravity term:





































0
0
0
0
0
0
][][ g
g
m
g
mgTm LBL
f

f
f

from Eq. 5.32 (Slide 5.1_7)BL
T][
Limiting to horizontal flight:
No propulsion perturbations
 00][ r
BE
B uv r

































0
][
1
0
f

gf
m
q
ru
w
v
u
B
ar



Small angle assumption leads to ; ;  rr uwuw  arcsin rr uwuw  )(arcsin ruw 

























eCC
rCaCC
eCC
Sq
eMC
raMC
eMC
Sqf
e
ra
e
ZZ
YYY
XX
Z
Y
X
B
a

b


b


b

),,(
),,,(
),,(
][
2B
1B
3B
CY
Cm
Cl
Cn
CZ
CX
e
a
r
Figure 5.17 Positive control surface
deflection
(5.61)

 
 
  



































eCuwC
m
Sq
qu
grCaCuvC
m
Sq
ru
uwgeCuwC
m
Sq
w
v
u
e
ra
e
ZrZr
YYrYr
rXrX

f


b

)(
)(
)()(



With , ruwrr uvuw  b ,

87
pp. 186 -188
Fortunately I13 is usually small and can be neglected
very messy!To solve for requires the inversion of
BBE
dtd ]/)([ 











331
2
131
0
00
0
][
II
I
II
I BB
B
Aerodynamics  
 
 



























aCrCrubCCb
eCqucCCc
rCaCpubCCb
Sq
arrMCb
eqMCc
rapMCb
Sqm
arr
eq
rap
a
nnrnn
mrmm
llrll
n
m
l
B
B
b

b
b

b
b

b
)2(
)2(
)2(
),,,,(
),,,(
),,,,(
][
For aircraft and UAV dynamic analysis it is assumed that the c.m. does not shift, and that the
aerodynamic moments are referred to the c.m.
Equations of motion
 
 
 


































aCrCrubCCb
eCqucCCc
rCaCpubCCb
Sq
I
I
I
r
q
p
arr
eq
rap
nnrnn
mrmm
llrll
b

b
b

b
)2(
)2(
)2(
100
010
001
3
2
1



(5.64)

 
 
 





































aCrCrubCuvC
I
Sbq
eCqucCuwC
I
Scq
rCaCpubCuvC
I
Sbq
r
q
p
arr
eq
rap
nnrnrn
mrmrm
llrlrl



b

b
)2(
)2(
)2(
3
2
1



Starting with Eq. 5.39 (Slide 5.1_8):
B
B
B
B
BBEBB
B pa
mmdtdI ][][]/)([][  (5.39)

88
pp. 188 -189
State-space equations use dimensional derivatives
Dimensional force derivatives
e
ra
e
ZeZ
YrYaY
XeX
C
m
Sq
ZC
m
Sq
Z
C
m
Sq
YC
m
Sq
YC
m
Sq
Y
C
m
Sq
XC
m
Sq
X

b


b




,
,,
,
(5.67)Pitch
State-space equations of motion (uncoupled)
e
M
Z
X
q
w
u
MuM
uuZ
uguX
q
w
u
e
e
e
qr
rr
rr





































 











0
0
00




Yaw 









































 













r
a
NN
LL
YY
r
p
v
NuN
LuL
guuY
r
p
v
ra
ra
ra
rr
pr
rr


ff



b
b
b
000010
00
00
0




(5.68)
Dimensional moment derivatives
arr
eq
rap
nanrn
r
rn
mem
r
qm
lrlal
r
pl
C
I
Sbq
NC
I
Sbq
NC
Iu
Sbq
NC
I
Sbq
N
C
I
Scq
MC
Iu
Scq
MC
I
Scq
M
C
I
Sbq
LC
I
Sbq
LC
Iu
Sbq
LC
I
Sbq
L
b

b
b

b
333
2
3
22
2
2
111
2
1
,,
2
,
,
2
,
,,
2
,




89
Aircraft example
pp. 190-193
70 m
60 m
8.3 m
Figure 5.18 Boeing 747
The static margin expressed in dimensional
derivatives as fraction of chord c (data next Slide)
cZ
M
m
I
C
C
stm
Z
m 12





225.0
3.8
1
108
30.1
288714
10488.4 7
20 




kftstm
292.0
3.8
1
103
61.1
288714
10488.4 7
40 




kftstm
1.5912239 +1.0793134s +s2
Roots
-0.5396567 + 1.140173i
-0.5396567 - 1.140173i
Natural frequency (rad/sec)
1.2614372
1.2614372
Damping
0.427811
0.427811

Initially, the upward velocity is caused by
the lift increase due to the positive
elevator deflection ( )79.7eZ

Figure 5.19 Pitch response to 10 deg elevator input (20 kft)

90
Boeing 747 Data
Page 191


91
Dutch-roll dynamics
pp. 194 -195

20 kft case with aileron input
0.0337074+0.9956639s+1.3296459s2
+1.2059552s3+s4
Roots
-0.1322316 + 1.0153929i
-0.1322316 - 1.0153929i
-0.9060091
-0.035483
Natural frequency (rad/sec)
1.0239668
1.0239668
Damping
0.1291366
0.1291366
Figure 5.20 Dutch-roll mode response to 10 deg aileron input
20 kft case with rudder input



92
Notable Quotables

 Aircraft and UAVs exhibit planar symmetry
o The symmetry plane lies in their 1B and 3B axes
o The moment of inertia matrix has three diagonal elements
o The moment of inertia matrix has also one off- diagonal element
 The gravity term in the translational equations is not neglected
 Thrust perturbations are neglected
 Remember the ‘weird’ definition of positive elevator and rudder
 The linearized equations of motion can be put into the state-variable format
o To take advantage of modern control techniques
 The equations of motion are separated into two uncoupled groups
o Longitudinal equations of motion
o Lateral equations of motions
 Aircraft aerodynamics uses two reference length
o The mean chord for the pitching moment
o The wing span for the rolling and yawing moments
 In aircraft dynamics it is common practice to use dimensional derivatives
 Aircraft modes
o Short period mode - longitudinal
o Dutch-roll mode - lateral
o Spiral mode - lateral

Technical Publications by AIAA

Text Books by MaSTech
Undergraduate
textbook
Practical guide with
lots of code

Work Books by MaSTech
Basic Flight Dynamics
C++
Programming
C++
Simulations

Inspirational Publications by GloryGram
Called to live and seek
God’s glory
A mountaineering
adventure
How to live
God’s glory

The tensor flight dynamics tutor

The tensor flight dynamics tutor

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to The tensor flight dynamics tutor

Similar to The tensor flight dynamics tutor (20)

Recently uploaded

Recently uploaded (20)

The tensor flight dynamics tutor