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1. REFERENCE FRAMES, FORCE AND
MOMENT EQUATIONS
1.1. Rigid Body Dynamics
The force and moment equilibrium equations of a rigid body can
be written in a inertial reference frame as follows:
=
⋅=
→
IF
IF
dt
HdM
dt
VdmF
FrameReferenceInertial
where H
is the angular momentum vector.
General property of a vector A
(conversion from fixed to moving
frame):
Adt
Ad
dt
Ad
GI
∧ω+
=
where ω
is the angular velocity of the moving frame G referred to
an inertial frame I.
Hence, forces and moments can be written as:
∧ω+
=
∧ω+
⋅=
Hdt
HdM
Vdt
VdmF
GF
GF
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where FG is a generic right-handed reference frame and ω
is the
angular velocity of the generic frame FG referred to the inertial
frame.
1.2. Euler Angles
The Euler angles ( ΦΘψ ,, ) are three independent quantities able to
define the position of a generic frame relative to an inertial one (or
to another reference frame): in the following chapters the frames
are considered Cartesian.
Note that the sequence of rotations is fixed as ( ΦΘψ ,, ) is not abase for a linear space.
Through the three Euler angles, the components of a vector can be
defined for both reference frames.
As a matter of fact, the components of a vector relative to the
frame ( )1111 Z,Y,XF , with unit vectors )k , j,i(
, must be converted
into a second frame ( )2222 Z,Y,XF , with unit vectors )n,m,l(
.
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The three rotations (with anti-clockwise direction considered
positive) are applied to the frame 2F so that it will be aligned to
1F :
- 1st rotation (ψ ): positive rotation about 2Z (anti-clockwise), so
that an intermediate frame is defined )Z'Z,'Y,'X('F 22222 ≡=
with unit vectors n'n,'m,'l
= .
- 2nd
rotation (Θ ): about 'Y2 of 'F2 . Another intermediate frameis defined )''Z,'Y''Y,''X(''F 22222 ≡= with unit vectors
''n,'m''m,i''l
== .
- 3rd rotation (Φ): about 12 X''X ≡ that eventually aligns 2F to
1F .
Right handed reference frames are assumed.
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1.2.1. Transformation algorithms
From a given right-handed coordinates system:
)Z,Y,X(O =
a second reference frame is obtained by means of a generic
positive rotation δ about the axis Z:
)Z'Z,'Y,'X('O ≡=
Hence, according to the previously described procedure,
)Z,Y,X(O = is equivalent to the second reference frame 2F and
)Z'Z,'Y,'X('O ≡= is the first reference system 1F :
≡
≡→
)',','('',','
),,(,,R
Z Y X O R R R
Z Y X O R R R
z y x
z y x
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A relationship between the vector components in the two
reference frames can be established:
++=
+δ+δ=
+δ−δ=
zz
yxy
yxx
'R00R
0cos'Rsin'RR
0sin'Rcos'RR
Converting to matrix form:
⋅
δδ
δ−δ
=
z
y
x
z
y
x
'R
'R
'R
100
0cossin
0sincos
R
R
R
Similar procedures apply for the rotations about X and Y axes.
General properties of transformation matrices
1. the main diagonal of the transformation matrix features cosines
of the generic rotation angle δ and a unit element 1 for the row
referred to the rotational axis only;
2. the unit element 1 identifies a single row of zeros and a singlecolumn of zeros;
3. the other terms in the rotation matrix are sines of the generic
rotation angle (with – in the row following the one referred to
the rotational axis);
4. the transformation matrices are orthogonal: [ ] [ ]
1T
MM
−
= .
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The inverse generic rotational matrix form can be computed as:
⋅
δδ−
δδ
=
z
y
x
z
y
x
R
R
R
100
0cossin
0sincos
'R
'R
'R
Summary of transformation matrices (direct and inverse form):
[ ]
ψψ
ψ−ψ
=Ψ100
0cossin
0sincos
[ ]
ψψ−
ψψ
=Ψ −
100
0cossin
0sincos
1
[ ]
ΘΘ−
ΘΘ
=Θ
cos0sin
010
sin0cos
[ ]
ΘΘ
Θ−Θ
=Θ −
cos0sin
010
sin0cos1
[ ]
ΦΦ
Φ−Φ=Φ
cossin0
sincos0
001
[ ]
ΦΦ−
ΦΦ=Φ −
cossin0
sincos0
0011
The unit vectors can be obtained through transformation matrix:
[ ]
⋅Ψ=
'n
'm
'l
n
m
l
[ ]
⋅Θ=
"n
"m
"l
'n
'm
'l
[ ]
⋅Φ=
k
j
i
"n
"m
"l
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Hence:
[ ] [ ] [ ] [ ]
⋅=
⋅Φ⋅Θ⋅Ψ=
k
j
i
L
k
j
i
n
m
l
21
where [ ] [ ] [ ] [ ]Φ⋅Θ⋅Ψ=21L is the complete transformation matrix
converting the vector components from the first reference frame
1F to the second reference frame 2F .
[ ]
⋅=
1Z
1Y
1X
21
2Z
2Y
2X
R
R
R
L
R
R
R
[ ]
⋅=
−
2Z
2Y
2X1
21
1Z
1Y
1X
R
R
R
L
R
R
R
By multiplying the three rotation matrices, one can explicitly
obtain [ ]21L , which is equal to:
=]L[ 21
ΘΦΘΦΘ−
ψΦ−ψΘΦψΦ+ψΘΦψΘ
ψΦ+ψΘΦψΦ−ψΘΦψΘ
=
coscoscossinsin
cossinsinsincoscoscossinsinsinsincos
sinsincossincossincoscossinsincoscos
=−121]L[
ΘΦψΦ−ψΘΦΦψ+ΦΘψ
ΘΦψΦ+ψΘΦψΦ−ψΘΦ
Θ−ψΘψΘ
=
coscoscossinsinsincossinsincossincos
cossincoscossinsinsinsincoscossinsin
sinsincoscoscos
One can notice that [ ]21L is orthogonal: [ ] [ ]1
21T
21 LL−
=
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1.2.2. Derivatives of Euler angles
Derivatives of Euler angles can be expressed as a function of
Euler angles and components of relative angular velocity betweenthe two frames, expressed with reference to 1F . Hence:
( ) ,,,,,,f ,,1Z1Y1X
ψΘΦωωω=ψΘΦ
( )ΨΘΦψΘΦ=ωωω ,,,,,f ,,1Z1Y1X
[ ]T1Z1Y1X
ωωω are the components referred to 1F reference
frame of the relative angular velocity between 1F and 2F .
i'mnr
⋅Φ+⋅Θ+⋅Ψ=ω
Ψ
= angular velocity of 'F2 referred to 1F
=Θ angular velocity of ''F2 referred to1F
The angular velocity Ψ in the reference frame F1 can be obtained
in the following way:
[ ][ ]
[ ][ ] [ ] [ ]2'F
11
1F1F2'F
1F2''F
2''F2'F
ΨΘΦ=Ψ⇒ΨΦΘ=Ψ⇒
ΨΦ=ΨΨΘ=Ψ −−
The transformation matrix between 'F2 and 1F can be computed
as:
[ ] [ ] [ ] [ ] 1
)0(21
111
1'2LL
−
=ψ
−−− =Θ⋅Φ=
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In the same way, the components of Θ relative to 1F are given by
the transformation matrix [ ] 1−Φ .
[ ] [ ]2''F
1
1F1F2''FΘΦ=Θ⇒ΘΦ=Θ −
Hence:
[ ] [ ]
Ψ
⋅+
Θ⋅Φ+
Φ
=
ω
ω
ω−
=ψ
−
0
0
L
0
0
0
01
)0(21
1
1Z
1Y
1X
Φ⋅Θ⋅Ψ
Φ⋅Θ⋅Ψ
Θ⋅Ψ−
+
Φ⋅Θ−
Φ⋅Θ+
Φ
=
coscos
sincos
sin
sin
cos
0
0
0
The explicit form is:
Φ⋅Θ⋅Ψ+Φ⋅Θ−=ω
Φ⋅Θ⋅Ψ+Φ⋅Θ=ω
Θ⋅Ψ−Φ=ω
coscossin
sincoscos
sin
1Z
1Y
1X
Or by using matrix form:
[ ]
Ψ
Θ
Φ
⋅=
ω
ω
ω−
1
1Z
1Y
1X
L
where
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[ ]
Θ⋅ΦΦ−
Θ⋅ΦΦ
Θ−
=−
coscossin0
cossincos0
sin01
L1
Moreover, one can express the inverse transformation matrix:
[ ]
ω
ω
ω
⋅=
Ψ
Θ
Φ
1Z
1Y
1X
L
[ ]
ΘΦΘΦ
Φ−Φ
Θ⋅ΦΘ⋅Φ
=
cos / coscos / sin0
sincos0
tgcostgsin1
L
One can notice that [ ]L matrix may be applied for reference frame
whose relative pitch angles °≠Θ 90 ; more in general, this meansrelations are usually correct only for relatively small values of
pitch angle.
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Summary of basic vector calculus
[ ] v~
vvv
www
k ji
vwu w
zyx
zyx Ω=
=∧=
where
[ ]
−
−
−
=Ω
0ww
w0w
ww0~
xy
xz
yz
w
Hence:
=
z
y
x
v
v
v
v
w~wv vΩ=∧
[ ]
−
−
−
=Ω
0vv
v0v
vv0~
xy
xz
yz
v
=
z
y
x
w
w
w
w
[ ]v~Ω is the cross operator.
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1.2.3. Alternative orientation models
The Euler method is only one possible way to define relative
orientation between two reference frames (with conventions of sign shown in figure below).
This orientation model is broadly applied in a wide range of
scientific fields, such as celestial mechanics, applied mechanics,
molecular and solid state physics, quantum mechanics, nuclear
physics, particle physics, computer graphics and, obviously,
aerospace applications.
However, other methods are viable if one aims to express relative
attitude of two frames using angular velocities; in some cases
these alternative formulations can be useful to implement a flight
dynamics model able to describe a vertical or high-pitch
manoeuvre, fixing a weakness of Euler method in which a division
by zero would occur.One of these alternative methods is the quaternion method.
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Quaternion method
The quaternion method is based on the Euler theorem, which
states that the attitude of a rigid body can be changed from any
given orientation to any other orientation by rotating the body
about an axis that is fixed to the body and stationary in an inertial
reference frame.
In accordance with the Euler theorem, a given fixed reference
frame FI can be aligned with any reference frame FB (a bodyreference frame, for instance) by simply rotating the frame FI of an
angle α about an axis a
fixed on the two reference frames.
If a direction a
which remains unchanged after the application of
the transformation matrix exists, then the components of the
vector a
are the same in both the reference frames:
k a jaianamalaa ˆˆˆˆˆˆ321321 ++=++=
.
The components of a
must satisfy the following relationship:
12
3
2
2
2
1 =++ aaa .
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The four quaternions (also said Euler parameters) are defined as
follows:
;2
sin
;2
sin
;2
sin
;2
cos
33
22
11
0
α
α
α
α
aq
aq
aq
q
=
=
=
=
The quaternions are not independent of each other, but constrained
by the relationship:
12
3
2
2
2
1
2
0 =+++ qqqq
because of the relationship existing between the components of a
.
The matrix 21 L can be written in terms of the components of thevector a
, assuming the following expression:
By knowing that
2sin2
cos2sin α α α =
and
12
cos22
sin212
sin2
coscos 2222 −=−=−=α α α α
α
one can obtain:
−+−−−−
+−−+−−
+−+−−+
=
)cos1(cos)(sin)cos1()(sin)cos1(
)(sin)cos1()cos1(cos)(sin)cos1(
)(sin)cos1()(sin)cos1()cos1(cos
),(2
3132213
132
2
2312
231321
2
1
α α α α α α
α α α α α α
α α α α α α
α
aaaaaaa
aaaaaaa
aaaaaaa
a L
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( ) ( )
( ) ( )
( ) ( )
+−−+
++−−
−++−
=
)(2122
2)(212
22)(21
][
2
2
2
110322031
1032
2
3
2
13021
31203021
2
3
2
2
qqqqqqqqqq
qqqqqqqqqq
qqqqqqqqqq
Q
By comparing the direction cosines matrix related to the Euler
angles and the transformation matrix based on quaternion it is
possible to obtain the relationship between Euler angles and
quaternion:
ΦΘΦΘΘ−
ΦΨ−ΦΘΨΦΨ+ΦΘΨΘΨΦΨ+ΦΘΨΦΨ−ΦΘΨΘΨ
=
coscossincossin
sincoscossinsincoscossinsinsincossinsinsincossincoscossinsinsincoscoscos
21 L
Since ][][ 21 LQ = , one eventually obtains:
( )[ ]
( )
( )
+−−
+=Ψ
−−+
+=Φ
−−=Θ
23
22
21
20
3021
23
22
21
20
3210
2031
qqqq
qqqq2tana
qqqq
qqqq2tana
qqqq2sina
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1.3. Earth Centered Earth Fixed Frame (ECEF
)Z,Y,X(F EEEE =
This terrestrial reference system is mainly used to post-process
results from satellite geodesy.
It defines the position of a generic point through three coordinates
referred to the Centre of Mass of the reference ellipsoid.
The origin of the system is the Earth Centre of Mass. The Z-axis
EZ points towards the North Pole. The direction of X-axis EX is
determined by the intersection of the plane defined by the
Greenwich meridian and the equatorial plane.
The Y-axis EY completes the right handed reference frame; it lies
in the equatorial plane and points 90° East of X-axis direction.
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s / rad10272.7 5E
−⋅=Ω is the angular velocity of the Earth about
EZ (the angular velocity of EF relative to an inertial system).
E
V
is the velocity of the aircraft referred to the planet. Note that
the Earth is moving with velocity R R ×Ω=τ V along the ecliptic
(where RΩ
= 2 10-7
rad/s is the revolution angular velocity of the
Earth).
EFEZEY
EX
EV
V
V
V
=
EFEE
0
0
Ω
=Ω
Two characteristic angles can be defined in this frame:
:λ latitude, defined as the angular distance of a generic point on
Earth measured from the Equator along the meridian which
intercepts the point itself.
:τ angular distance measured along the equator between
Greenwich meridian and meridian which intercepts the given
point.
N.B. angular velocity of the Earth due to its motion around the
Sun, i.e. RΩ , is neglected in further analysis.
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1.3.1. Equations of Forces in ECEF reference
frame
Aircraft C.G. velocity referred to an inertial reference system can
be written as:
IFIF
0
IFdt
Rd
dt
Rd
dt
'RdV
+
=
=
Considering that:
τ=
VdtRd
IF
0
RVRdt
Rd
dt
RdEEE
EFIF
∧Ω+=∧Ω+
=
Where:
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• τV
is the angular velocity of the Earth in the ecliptic plane
• EV
is velocity of the aircraft referred to the Earth.
As a result, V expression can be rearranged as follows:
RVVV EE
∧Ω++= τ
Hence, equation of forces can be computed as:
( ) ( )[ ]
∧Ω++∧Ω+
∧Ω++= ττ RVVRVVdtdmF EEE
EFEE
Assuming angular velocity of the Earth EΩ
as a constant:
( )
∧Ω∧Ω+∧Ω⋅+
∧Ω++
=
τ
τ RV2Vdt
Vd
dt
VdmF
EEEEEEF
E
Neglecting 2nd and 4th terms, one can obtain:
∧Ω⋅+
= EE
EF
E V2dt
VdmF
EEEFEV
~2VmF
⋅Ω⋅+⋅=
Ω
FE acceleration referred to
the inertial frame FI
Centripetal acceleration of the
aircraft C.G. referred to the FE
frame, due to rotation of the
Earth
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As a result, equation of forces can be expressed as:
⋅
Ω
Ω−
⋅+
⋅=
EZ
EY
EX
E
E
EZ
EY
EX
EZ
EY
EX
V
V
V
000
00
00
2
V
V
V
m
F
F
F
If we consider forces acting on the aircraft equal to:
[ ] VEEEFEVFFFE gLmXgmXF +=+=
⋅+
=
+
=
mg
0
0
]L[
Z
Y
X
mg
mg
mg
Z
Y
X
F
F
F
EV
E
E
E
EZ
EY
EX
E
E
E
EZ
EY
EX
Where ]L[ EV is the transformation matrix between ECEF and
NED (Vehicle-Carried reference frame), analysed in subsection
1.5.1, and equal to:
[ ]
λ−λ
τλ−ττλ−τλ−τ−τλ−
=
sin0cos
sincoscossinsin
coscossincossin
LEV
Hence:
Aerodynamic and
propulsive forcesGravity forces
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=
⋅−
=
mg
0
0
]L[
F
F
F
Z
Y
X
EV
EZ
EY
EX
E
E
E
⋅−
⋅
Ω
Ω−
⋅+
⋅=
g
0
0
]L[
V
V
V
000
00
00
2
V
V
V
m EV
EZ
EY
EX
E
E
EZ
EY
EX
⇓
VFEVEEEFE g]L[V
~2VmF ⋅−⋅Ω⋅+⋅= Ω
Rearranging vector equation into the three scalar components, one
finally obtains force equations along ECEF axis:
( )( )( )
λ+=
λτ+Ω+=
λτ+Ω−=
singVmZ
cossingV2VmY
coscosgV2VmX
E
EE
EE
ZE
XEYE
YEXE
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1.4 . Earth Centered Inertial Frame (ECI
)Z,Y,X(F ECIECIECIECI =
The ECI is the so called Inertial Geocentric Reference Frame; it’s
commonly used to study the motion of a body orbiting around the
Earth (for example a satellite) referred to a pseudo geocentric
inertial frame, with the axes oriented to the fixed stars.
The origin of this reference frame is the Centre of the Earth. The
X-axis ECIX is directed to Aries (spring equinox), the Z-axis ECIZ
points toward the North Pole with Y-axis ECIY completing the
right-handed frame.
The actual rotation of the Earth θ referred to the inertial reference
frame ECI can be computed as:
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( )oEo tt −Ω+θ=θ
where oθ is the Earth rotation at the time 0t , t is the actual time
and s / rad10272.7 5E −⋅=Ω is the angular velocity of the Earth
about ECIZ .
1.4.1 Transformation matrix between ECI and
ECEF
The rotation θ for the frame ECIF about ECIZ is sufficient to align
the two reference frames; as a result, θ is equivalent to a ψ rotation (using Euler angles conventions).
Hence:
[ ]
θθ
θ−θ
=
100
0cossin
0sincos
LIE
[ ]
θθ−
θθ
=−
100
0cossin
0sincos
L1
IE
⇓
[ ]
⋅=
E
E
E
IE
ECI
ECI
ECI
Z
Y
X
L
Z
Y
X
[ ]
⋅=
−
ECI
ECI
ECI1
IE
E
E
E
Z
Y
X
L
Z
Y
X
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1.4.2 Force equations in ECI reference frame
Since ECI is assumed as an inertial reference frame, the velocity
of aircraft C.G. is equal to:
ECIFdt
Rd
V
=
⇓
RVRdt
Rd
dt
RdEEE
EFECIF
∧Ω+=∧Ω+
=
Since relative angular velocity between ECIF and EF is EΩ
.
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 25
Hence:
( ) ( )[ ]
∧Ω+∧Ω+
∧Ω+= RVRVdt
dmF EEE
EFEE
( )
∧Ω∧Ω+∧Ω⋅+
= RV2
dt
VdmF EEEE
EF
E
In analogy to what already done in section 1.3.1, forces can berearranged separating aerodynamic and propulsive contribution
from gravity forces; hence:
+
=
EZ
EY
EX
E
E
E
EZ
EY
EX
mg
mg
mg
Z
Y
X
F
F
F
As a result, aerodynamic and propulsive forces can be computed
as follows:
Centripetal acceleration of the
aircraft C.G. referred to the FE
frame, due to rotation of the
Earth
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 26
[ ] [ ] =
⋅−
−⋅⋅
Ω
Ω−
⋅
Ω
Ω−
+
⋅
Ω
Ω−
⋅+
⋅=
=
=
g R
V
V
V
V
V
V
m
Z
Y
X
EV EV E
E
E
E
Z
Y
X
E
E
Z
Y
X
E
E
E
F
E
E
E
E
E
E
E
0
0
L0
0
L
000
00
00
000
00
00
000
00
00
2
X
[ ] [ ] [ ] [ ] )V V E
E E
E
E E
E E
F EV F EV F F
F E F F E
R
V V m
gLL~~
~2
⋅−⋅⋅Ω⋅Ω+
+⋅Ω⋅+⋅=
ΩΩ
Ω
Rearranging the equation into the three scalar components:
( )( )
λ+=
λτ+λτΩ−Ω+=
λτ+λτΩ−Ω−=
singVmZ
cossingcossinRV2VmY
coscosgcoscosRV2VmX
EZE
2E
EYE
EYE
2EEYEEXE
In the traditional approach, the latter system of equations may
undergo further simplifications due to the hypothesis that
centripetal component ( )REE
∧Ω∧Ω is usually negligible.
Hence:
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 27
[ ] =
⋅−
⋅
Ω
Ω−
⋅+
⋅=
=
=
gV
V
V
V
V
V
m
Z
Y
X
EV
Z
Y
X
E
E
Z
Y
X
E
E
E
F
E
E
E
E
E
E
E
0
0
L
000
00
00
2
X
[ ] [ ] ⋅−⋅Ω⋅+⋅= Ω VFEVEFEEFEEFE gLV~2Vm
⇓
( )( )
( )
λ+=
λτ+Ω+=
λτ+Ω−=
singVmZ
cossingV2VmY
coscosgV2VmX
EZE
EYEEYE
EYEEXE
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1.5 Vehicle-Carried Vertical Frame (NED)
)Z,Y,X(F VVVV =
The origin of NED reference frame is at the aircraft C.G.
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 29
The vertical axis VZ is directed along the local gravity
acceleration vector; VX and VY are in a plane parallel to the one
tangent to the surface of the Earth for h=0.
VX aims to the North, while VY is oriented Eastwards (right-handed frame).
The velocity components for FV are:
VFD
E
N
E
V
V
V
V
=
+Ω=ω+Ω=ω
V
V
V
VFEVE
R
Q
P
• VE is the velocity of the aircraft referred to the Earth,
expressed inVF
• ω
is the angular velocity of V
F referred to an inertial
reference system
• Vω
is the angular velocity of the frame VF relative to EF .
1.5.1 Transformation matrix between NED and
ECEF
To align ECEF reference frame with NED two rotations are
necessary (3rd rotation: 0=Φ ):
1st rotation: τ=ψ (angle of azimuth), about EZ axis aligning EY
with VY .
2nd rotation: )90( λ+°−=Θ (angle of elevation), about EY
aligning the reference frameE
F withVF .
As a result, Euler angles set for transformation between EF and
VF is:
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 30
( )
=Φ
λ+°−=Θ
τ=ψ
0
90
Thus leading to the following transformation matrix:
[ ]
λ−λ
τλ−ττλ− τλ−τ−τλ−=
sin0cos
sincoscossinsincoscossincossin
LEV
[ ]
λ−τλ−τλ−
ττ−
λτλ−τλ−
=−
sinsincoscoscos
0cossin
cossinsincossin
L1
EV
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The components of EΩ
with respect toVF are:
[ ] [ ]
λΩ−
λΩ
=
Ω⋅=Ω −
sin
0
cos
0
0
L
E
E
E
1EVVFE
vω
components can be expressed as:
⋅−
−
⋅
=
−⋅=
= −
=
λ τ
λ
λ τ
τ
λ ω τ
sin
cos0
][ 1
0
EV
V
V
V
V L
R
Q
P
Hence:
[ ]
sin
cos
sin
0
cos
E
E
VF
λτ−
λ−
λτ
+
λΩ−
λΩ
=ω
⇓
=−=λ
λ=λ=τ
RVQ
)cosR(Vcos / P
NV
EV
where hRR E += ( ER is the mean radius of the Earth and h isthe altitude).
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 32
1.5.2 Force equations in NED reference frame
Aircraft C.G. velocity referred to an inertial reference system canbe written as:
IFIF
0
IFdt
Rd
dt
Rd
dt
'RdV
+
=
=
Considering that:
τ=
VdtRd
IF
0
=∧Ω+∧ω+
=∧ω+
=
RR
dt
RdR
dt
Rd
dt
RdEV
VFVFIF
( ) RV E
V
FE
∧Ω+=
Where:
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 33
• τV
is the linear velocity of the Earth in the ecliptic plane
• ( )VFEV
is velocity of the aircraft referred to the Earth,
expressed in NED reference frame.
As a result:
RVVV EE
∧Ω++= τ
Neglecting τV
contribution, one can obtain:
( ) ( ) ( )
∧Ω+∧ω+Ω+∧Ω+⋅= RVRVdt
dmF EEVEVFEE
Considering that:
( ) ( ) ( )
( ) ( ) ( ) ( )
∧Ω∧ω+Ω+∧Ω=∧Ω
∧Ω∧Ω+∧Ω=∧Ω
RRdt
dR
dt
d
RR
dt
dR
dt
d
EVEVFEIFE
EEEFEIFE
Moreover, since ( ) 0dt
d
EFE =Ω
, one can compute:
( ) EEEFE VRdtd
∧Ω=∧Ω
( ) ( )RVRdt
dEVEEVFE
∧Ω∧ω−∧Ω=∧Ω
Substituting in equation of forces, one obtains:
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 34
( )
∧ω+∧Ω∧Ω+∧Ω+
⋅= EVEEEE
VF
E VRV2dt
VdmF
As a result, aerodynamic and propulsive forces can be written as:
[ ] [ ] [ ]
[ ] −⋅Ω+
+
⋅Ω⋅Ω+⋅Ω+⋅=
ω
ΩΩΩ
VFVFEVFV
VFVFEVFEVFEVFEVFEVF
gV~
R~~
V~
2VmX
Explicitating the matrix form of previous equation:
−
⋅
λτλ
λτ−λτ−
λ−λτ
+
+
−
⋅
λΩ
λΩ−λΩ−
λΩ
⋅
λΩ
λΩ−λΩ−λΩ
+
+
⋅
λΩ
λΩ−λΩ−
λΩ
+
⋅=
g
0
0
V
V
V
0cos
cos0sin
sin0
R
0
0
0cos0
cos0sin
0sin0
0cos0
cos0sin
0sin0
V
V
V
0cos0
cos0sin
0sin0
2
V
V
V
m
Z
Y
X
D
E
N
E
EE
E
E
EE
E
D
E
N
E
EE
E
D
E
N
V
V
V
Rearranging into the three scalar components, one eventually
obtains:
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 35
( )( )
( )
−λτ+λ+λΩ+λΩ+⋅=
λτ−λτ−λ+λΩ−⋅=
λ−λτ+λλΩ+λΩ+⋅=
gcosVVcosRcosV2VmZ
cosVsinVcosVsinV2VmY
VsinVcossinRsinV2VmX
EN22
EEEDV
DNDNEEV
DE2
EEENV
1.5.3 Simplified equations of Forces in NEDreference frame
The acceleration terms in the scalar equations can be rewritten by
including the expression of τ and λ as a function of EV and NV :
( )
−++λΩ+λΩ+=
−λ−λ+λΩ−=
−λ+λλΩ+λΩ+=
gR
V
R
VcosRcosV2Va
R
VVtan
R
VVcosVsinV2Va
R
VVtan
R
VcossinRsinV2Va
2E
2N22
EEEDVZ
DNNEDNEEVY
DN2E2
EEENVX
VXa ,
VYa ,VZa are measured by the accelerometers on the inertial
platform oriented according to VX , VY , VZ . After the measurement of
VXa ,
VYa ,VZa the velocity
components of the aircraft referred to NED axes and its position
relative to the Earth can be estimated as shown in scheme 1.
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 36
1
1 Carlo Casarosa, "Meccanica del Volo", Ed. Plus 2004.
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 37
Moreover, an analysis of the orders of magnitude of acceleration
terms can be performed assuming constant altitude ( DV = 0) and
equatorial trajectory ( NV = 0; λ = 0):
−=−+Ω+Ω=
=
=
gag
R
VRV2a
Va
0a
1ZV
2E2
EEEZ
EY
X
V
V
V
The 1st and 2nd term of aZV1
are contribution due to the Earth
rotation (Coriolis acceleration + frame translation acceleration)
while the 3rd term derives from motion of NED frame around the
planet (centripetal acceleration). Orders of magnitude of each
acceleration term are showed in figure 1.
The terms due to the Coriolis acceleration and to the effects of
frame translation are generally dropped: their overall order of
magnitude is g1002 ⋅ for OE V
101V ⋅= where the orbital speed is
s / m7800gRVO ≅= at sea level.
Hence, the Earth is assumed to be flat ( )
∞→E
R and its rotation
in space is neglected ( )0E →Ω for aircraft performance and
stability analysis. This assumption cannot be extended to long
range navigation problems.
Anyway, when flying at very high speeds (1000 m/s) a moderate
decrease of lift (L = 0.965 W) required for equilibrium is observed
which implies a minimal correction of elevator deflection.
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 38
Figure 1
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 39
1.6 Air-Trajectory Reference Frame (Wind axes
)Z,Y,X(F WWWW =
The reference frame has its origin fixed at the aircraft C.G.
The longitudinal axis WX is aligned with the airspeed direction V
(aircraft velocity relative to the atmosphere).
wVVE
+=
where EV
is the velocity of the aircraft referred to the Earth and
w
is the velocity of the wind.
The lateral axis WY is orthogonal to WX oriented from left to
right, with respect to C.G. trajectory.
WZ lies in the plane of symmetry of the aircraft, directed from
upper to lower surface of wing airfoil.
C.G. trajectory
C.G. trajectory
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 40
The velocity components for WF are:
WF
W
E
0
0
V
V
=
+
+
Ω
Ω
Ω
=ω+ω+Ω=ω
W
W
W
VW
VW
VW
WZ
WY
WX
WVE
R
Q
P
R
Q
P
where:
• ω
is the angular velocity of WF referred to an inertial
reference system
• EΩ
is the angular velocity of the Earth
• Vω
is the angular velocity of the frame WF due to thecurvature of the Earth
• Wω
is the angular velocity of the aircraft along the trajectory
resolved intoW
F .
1.6.1 Force equations in wind axes reference
frame
The force equations in wind axes can be written starting from 1st
cardinal equation:
∧ω+
⋅= E
WF
E Vdt
VdmF
Where:
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 41
WVE ω+ω+Ω=ω
Assuming Earth is flat and fixed in space, one can notice that:
Wω≅ω
As a result, equations of forces can be written as follows:
∧ω+
⋅= EW
WF
E V
dt
VdmF
[ ] [ ]
⋅−⋅Ω+⋅= −
ω VF1
VWWFEWFWWFEWF
gLV~
VmX
Where [ ] 1VWL−
is the transformation matrix between WF and VF :
[ ] [ ] 121
1VW LL
−− = with
Ψ=Ψ
Θ=Θ
Φ=Φ
W
W
W
Using matrix form to rearrange previous equation, one obtains:
[ ]
⋅−
⋅
−
−−
+
⋅=
−
VF
1VW
W
WW
WW
WWW
W
W
W
g
0
0
L
0
0
V
0PQ
P0R
QR0
0
0
V
m
Z
Y
X
Leading to the following three scalar equations:
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 42
ΦΘ−−⋅=
ΦΘ−⋅=
Θ+⋅=
)coscosgVQ(mZ
)sincosgVR(mY
)singV(mX
WWWWW
WWWWW
WWW
1.7 Body Reference Frame
The main feature of this frame is to be a body-fixed reference,
with axes bound to the aircraft during its motion.
The origin of Body Reference frame is set at the aircraft C.G.
There are three typologies of Body axes:
• Generic Body Axes
• Principal Axes of Inertia
• Stability Axes
Principal Axes of Inertia
)Z,Y,X(F .)I.P(B.)I.P(B.)I.P(B.)I.P(B =
The P.I. reference frame has the axes directed along the aircraftprincipal axes of the inertia. The .)I.P(BZ direction is considered
positive from upper to lower surface of wing airfoil; .)I.P(BX is
orthogonal to .)I.P(BZ , pointing in forward direction, whereas
.)I.P(BY completes the right handed reference frame.
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 43
Generic Body Axes
)Z,Y,X(F BBBB =
BX and BZ lie in the aircraft plane of symmetry, with BX
generally parallel to the fuselage reference line and BZ directed
from upper to lower surface of wing airfoil. BY axis is selected so
that the frame is right-handed.
The velocity components for BF are:
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BFB
B
B
E
W
V
U
V
=
+
+
Ω
Ω
Ω
=ω+ω+Ω=ω
B
B
B
VB
VB
VB
BZ
BY
BX
BVE
R
Q
P
R
Q
P
Where:
• ω
is the angular velocity of BF referred to an inertial
reference system
• EΩ
is the angular velocity of the Earth
• Vω
is the angular velocity of the frame BF due to the
curvature of the Earth
• Bω is the angular velocity of the aircraft due to the fact that
BF is bound to the airframe.
Stability Body Axes
)Z,Y,X(F SSSS =
XS
lies on the projection of EV (at initial reference time) in the
plane of symmetry, ZS
is orthogonal to XS
and positive from the
upper to the lower side of the wing airfoil. YS
axis completes the
right handed reference frame.
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The velocity components for SF are:
SSF
0S
0S
)0t(E
FS
S
S
E
0
V
U
V
W
V
U
V
=
= =
+
+
ΩΩ
Ω
=ω+ω+Ω=ωS
S
S
VS
VS
VS
SZ
SY
SX
SVE
R
Q
P
R
Q
P
Where:
• ω is the angular velocity of SF referred to an inertial
reference system• EΩ
is the angular velocity of the Earth
• Vω
is the angular velocity of the frame SF due to the
curvature of the Earth
• Sω
is the angular velocity of the aircraft due to the fact that
SF is bound to the airframe.
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1.7.1 Transformation matrix between Wind axes
and Generic Body Axes
The frame WF is rotated to obtain BF (wind to body axes
transformation) according to the following sequence:
1st rotation: β−=ψ (angle of sideslip), about WZ axis aligning
WY with BY .
2nd rotation: α=Θ (angle of attack ), about BW YY ≡ aligning thereference frame
WF with BF .
3rd rotation: 0=Φ
As a result, Euler angles set for this transformation are:
=Φ
α=Θ
β−=ψ
0
Hence transformation matrix can be computed as:
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 47
[ ]
αα−
βα−ββα−
βαββα
=
cos0sin
sinsincossincos
cossinsincoscos
LWB
[ ]
αβα−βα
ββ
α−βα−βα
=−
cossinsincossin
0cossin
sinsincoscoscos
L1
WB
One can exploit transformation between the two frames to get
values of sideslip angle and angle of attack as a function of linear
velocity components; hence:
⇒
⋅=
−
0
0
V
]L[
W
V
U W1
WB
B
B
B
βα⋅=
β⋅=
βα⋅=
cossinVW
sinVV
coscosVU
WB
WB
WB
Assuming small values for aerodynamic angles α and β the
following simplified equations apply:
≅=β
≅=α
−
−
V
V
V
Vsin
V
W
U
Wtan
BB1
B
B
B1
BWWBrel j n
⋅α+⋅β−=ω−ω=ω
Where:
• relω is the relative angular velocity between BF and WF
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F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 48
• WWW n,m,l
are the unit vectors of WF and BBB k , j,i
are the
unit vectors of BF .
The components of relω referred to WF frame are:
[ ] [ ] [ ]=ω−ω⋅ WBWBL
[ ] [ ]
β
−
α⋅=
−
⋅ =α
0
0
0
0
L
R
Q
P
R
Q
P
L0WB
W
W
W
B
B
B
WB
[ ]
β
+
α−⋅=
0
0
R
Q
P
L
R
Q
P
B
B
B
WB
W
W
W
Converting to scalar equations:
β+α+α−=
βα−βα−+βα−=
βα+βα−+βα=
cosRsinPR
sinsinRcos)Q(sincosPQ
cossinRsin)Q(coscosPP
BBW
BBBW
BBBW
Solving eventually for α and β :
α−α+=β
βα−βα−β
−=α
cosRsinPR
tansinRtancosPcos
BBW
BBW
B
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1.7.2 Transformation matrix between NED
reference frame and Generic Body Axes BF
In some cases (e.g. navigation purposes or flight data display) itmay be useful to express some variables referred to vehicle carried
vertical frame, hence computing transformation matrix between
BF and VF reference frames.
The Euler angles for the transformation [ ]VBL are:
Φ=Φ
Θ=Θψ=ψ
B
B
B
• 1st rotation: BΨ (yaw angle).
The axis 'XV of the intermediate frame 'FV lies in the local
horizontal plane. Hence, BΨ measures the heading between thelongitudinal axis BX and a reference heading represented by
VX direction.
Neglecting the effects of magnetic deviation, BΨ is the so
called magnetic course indicated by the compass.
• 2nd rotation:B
Θ (pitch angle)
The rotation about 'YV aligns 'XV with BX ; as a result, pitch
angle is a measure of aircraft inclination referred to the horizon.
• 3rd rotation: BΦ (roll angle)
The rotation about BX eventually aligns the two reference
frames. This angle is the measure of lateral bank.
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BΨ , BΘ , BΦ are the angles of gymbal of the onboard inertial
platform.
As a result, transformation matrix can be expressed as:
[ ] ][][][L BBBVB Φ⋅Θ⋅Ψ=
1.7.3 Transformation matrix between Stability
Body Axes SF and Generic Body Axes BF
Assuming for initial conditions symmetric flight, i.e. 0V 0S = , SF is aligned with WF .
However, during aircraft motion, SF remains fixed to this position
while WF follows the evolution of the trajectory of the C.G. in
terms of axes orientation (notably, WX is tangent to the flight
path).
As a result, SF is rotated of the angle of attack α with respect toBF :
[ ]
−
=Θ=
00
00
cos0sin
010
sin0cos
][L
α α
α α
SB
[ ]
−
=Θ= −−
00
00
11
cos0sin
010
sin0cos
][L
α α
α α
SB
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1.7.4 Equations of Forces in Generic Body Axes
BF reference frame
Equations of forces in BF reference frame can be written usingcardinal equation of force; notably:
∧ω+
⋅= E
BF
E Vdt
VdmF
Where:
BVE ω+ω+Ω=ω
Assuming Earth is flat and fixed in space:
Bω≅ω
As a result, force equations can be rewritten as:
∧ω+
⋅= EB
BF
E Vdt
VdmF
[ ] [ ] V F VB BF E
BF B BF E BF m gLV~VX 1 ⋅−⋅Ω+⋅= −
ω
Rearranging in matrix form:
[ ]
⋅−
⋅
−
−
−
+
⋅=
−
VF
1VB
B
B
B
BB
BB
BB
B
B
B
B
B
B
g
0
0
L
W
V
U
0PQ
P0R
QR0
W
V
U
m
Z
Y
X
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Separating into scalar equations:
ΦΘ−+−⋅=ΦΘ−−+⋅=
Θ++−⋅=
)coscosgVPUQW(mZ
)sincosgWPURV(mY
)singWQVRU(mX
BBBBBBBB
BBBBBBBB
BBBBBBB
1.7.5 Moment equations
Considering the aircraft as a rigid body, the second cardinal law of
mechanics is applied to determine moment equations:
H dt
H d
dt
H d M
GF I F
∧ω+
=
=
[ ] ω⋅=GF GF
I H
With:
• [ ]GF
I : inertia matrix referred to GF reference frame
• ω : angular velocity of the frame referred to an Inertial
Reference Frame.
Assuming that:
GFω≡ω
As a result, equations of moments can be rewritten as:
[ ] [ ] [ ] [ ] ( )GFGFGFGFGFGFGFGF
I~
dt
dII
dt
dM ω⋅⋅Ω+ω⋅+ω⋅= ω
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Usually, moment equations are solved in Body Axes, since in this
reference frame the inertia matrixBF
I is time invariant; as a
result, substituting BG FF = :
[ ] [ ] [ ] BFBBFBFBFBBFBF
I~
IM ω⋅⋅Ω+ω⋅= ω
Concerning the definition of inertia tensor, a commonly adopted
assumption is that BBZX is the plane of symmetry of the aircraft;
hence, componentsBYBX
I andBZBY
I are dropped:
[ ]
−
−
=
BZBXBZ
BY
BZBXBX
BF
I0I
0I0
I0I
I
Aircraft angular velocity, given the hypothesis of fixed Earth
)0( E =Ω
and flat Earth )0( V =ω
, can be simply expressed as:
=ω
B
B
B
B
R
Q
P
Hence:
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⋅
−
−
⋅
−
−
−
+
+
⋅
−
−
=
B
B
B
BZBXBZ
BY
BZBXBX
BB
BB
BB
B
B
B
BZBXBZ
BY
BZBXBX
B
B
B
R
Q
P
I0I
0I0
I0I
0PQ
P0R
QR0
R
Q
P
I0I
0I0
I0I
N
M
L
⇓
−+−+=
−+−+=
+−−+=
)PQR(IQP)II(RIN
)RP(IRP)II(QIM
)QPR(IRQ)II(PIL
BBBBZBXBBBXBYBBZB
2B
2BBZBXBBBZBXBBYB
BBBBZBXBBBYBZBBXB
Coupling effects giving rise to gyroscopic phenomena are taken
into account in the following components:
BBBYBZRQ)II( ⋅− Roll inertial coupling
BBBZBXRP)II( ⋅− Pitch inertial coupling
BBBXBYQP)II( ⋅− Yaw inertial coupling
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1.8 Sun Centred Inertial Reference Frame
)Z,Y,X(F CICICICI =
The origin of this reference frame is the centre of the solar system.
The X-axis CIX lies in the ecliptic plane and is oriented to Aries;
the Z-axis CIZ is perpendicular to the orbit plane of the Earth(ecliptic), the Y-axis CIY is selected so that the frame is right-
handed.
Note that CIX and ECIX are parallel, whereas ECICI YY − and
ECICI ZZ − are respectively rotated of 23.27°, i.e. the inclination
of the Earth equatorial plane referred to the ecliptic plane.
1.8.1 Transformation matrix between CI and
ECI reference frames
Since X-axis of the two reference frames are already parallel, a
simple rotation about CIX is required to align the two frames:
• 1st rotation: 0=Ψ
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• 2nd rotation: 0=Θ
• 3rd rotation: i≡Φ
i is the inclination of the Earth equatorial plane referred to theecliptic, as shown in figure below.
Moreover, if one needs not only an alignment but also a
coincidence of these frames, ECIF must be translated to take into
account for the displacement of the Earth relative to the Sun.
If solar longitude ( SΛ ) and the distance from the Sun to the ECIF
origin (S
d ) are known, one can finally compute the transformation
algorithm:
Λ
Λ
+
⋅
−=
0
sind
cosd
Z
Y
X
icosisin0
isinicos0
001
Z
Y
X
SS
SS
ECI
ECI
ECI
CI
CI
CI
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Λ
Λ
−
⋅
−
=
0
sind
cosd
Z
Y
X
icosisin0
isinicos0
001
Z
Y
X
SS
SS
CI
CI
CI
ECI
ECI
ECI
Neglecting the inclination of the Earth orbit, SΛ and Sd are
calculated from the ephemeredes.
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1.9 Complete mathematical models
Before proceeding in further analysis, it may be useful to
summarize force and moment equations for the whole referenceframes set:
• ECEF
[ ] VFEVEFEEF
gLVmX ⋅−⋅=
• NED
VFVFEVF
gVmX −⋅=
• Wind Axes
[ ] [ ]
⋅−⋅Ω+⋅=
−
ω VF
1
VWWFEWFWWFEWF gLV
~
VmX
• Body axes
[ ] [ ] V F VB BF E
BF B BF E BF g LV
~V m X ⋅−⋅Ω+⋅= −
ω
1
[ ] [ ] [ ] ( )BFBBFBF
BF
BBFBFI~
.IM ω⋅⋅Ω+
ω⋅= ω
N.B. moment equations are obtained only for body axes due to
time-invariant inertia matrix enabling a simplification in the
equations.
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1.9.1 Mathematical model in Earth Centered
Earth Fixed (ECEF) reference frame
Force equations are:
λ+⋅=
λτ+⋅=
λτ+⋅=
)singV(mZ
)cossingV(mY
)sincosgV(mX
EZE
EYE
EXE
By integrating the equations of forces, one can obtain the aircraftvelocity components expressed in the Earth frame; to transform
these into vehicle carried frame components, one can write
[ ]
⋅=
−
EZ
EY
EX1
EV
D
E
N
V
V
V
L
V
V
V
λ−τλ+λτ−=
τ−τ−=
λ+λτ−λτ−=
sinVsincosVcoscosVV
cosVsinVV
cosVsinsinVsincosVV
EZEYEXD
EYEXE
EZEYEXN
=λ
λ=τ
R
V.cosRV
.
N
E
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Equations of moments are referred in any case to body axes,
independently on the reference frame used to solve forces
equations; hence:
−+−+=
−+−+=
+−−+=
)PQR(IQP)II(RIN
)RP(IRP)II(QIM
)QPR(IRQ)II(PIL
BBBBZBXBBBXBYBBZB
2B
2BBZBXBBBZBXBBYB
BBBBZBXBBBYBZBBXB
The integration of moment equations ( )BBB R,Q,P gives as a
result the angular rates )R,Q,P( BBB . Then we obtain
BBB ,, ΨΘΦ , and finally BBB ,, ΨΘΦ .
A procedure to implement this process is described in scheme 2.
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2
2 Carlo Casarosa, "Meccanica del Volo", Ed. Plus 2004.
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1.9.2 Mathematical model in Vehicle Carried
Vertical Frame (NED
−⋅=
⋅=
⋅=
)gV(mZ
VmY
VmX
DV
EV
NV
No transformation matrix, obviously, is required to obtain velocity
components in vehicle carried frame; moment equations areexpressed in body axes. Calculation pattern for this frame is
shown in scheme 3.
3
3Carlo Casarosa, "Meccanica del Volo", Ed. Plus 2004.
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1.9.3 Mathematical model in Air-trajectory
Reference Frame (Wind axes)
Force equations are:
ΦΘ−−⋅=
ΦΘ−⋅=
Θ+⋅=
)coscosgVQ(mZ
)sincosgVR(mY
)singV(mX
WWWWW
WWWWW
WWW
By integrating system equations, one obtains WV and twocomponents of angular velocity about wind axes, ( WW Q,R ).
Hence, with the transformation matrix, one can express velocity
components in vehicle carried vertical frame:
[ ]
⋅=
0
0
V
L
V
V
V W
VW
D
E
N
Θ−=
ΨΘ=
ΨΘ=
WWD
WWWE
WWWN
cosVV
sincosVV
coscosVV
To update coefficients of transformation matrix, instantaneousvalues of Euler angles between the two frames are required;
recalling the expression of the derivative of Euler angles:
[ ]
ω
ω
ω
⋅=
Ψ
Θ
Φ
1Z
1Y
1X
L
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=ω
=ω
=ω
W1Z
W1Y
W1X
R
Q
P
Ψ=Ψ
Θ=Θ
Φ=Φ
W
W
W
⇓
Θ
Φ+
Θ
Φ=Ψ
Φ−Φ=Θ
ΘΦ+ΘΦ+=Φ
W
WW
W
WWW
WWWWW
WWWWWWWW
cos
cosR
cos
sinQ
sinRcosQ
tgcosRtgsinQP
Since WQ and WR are known from equations of forces, then the
remaining undetermined component WP is obtained by means of
),,R,P,Q(f P BBBW βα= :
βα+βα−+βα= cossinRsin)Q(coscosPP BBBW
α−α+=β
βα−βα−β
−=α
cosRsinPR
tansinRtancosPcos
BBW
BBW
B
Pattern for complete mathematical model in wind axes is shown in
scheme 4.
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4
4 Carlo Casarosa, "Meccanica del Volo", Ed. Plus 2004.
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1.9.4 Mathematical model in Generic Body Axes
From force and moment equation, one can obtain the three
velocity components BB,B W,VU and angular velocities aboutbody axes BBB R,Q,P .
ΦΘ−+−⋅=
ΦΘ−−+⋅=
Θ++−⋅=
)coscosgVPUQW(mZ
)sincosgWPURV(mY
)singWQVRU(mX
BBBBBBBB
BBBBBBBB
BBBBBBB
−+−+=
−+−+=
+−−+=
)PQR(IQP)II(RIN
)RP(IRP)II(QIM
)QPR(IRQ)II(PIL
BBBBZBXBBBXBYBBZB
2B
2BBZBXBBBZBXBBYB
BBBBZBXBBBYBZBBXB
Vehicle carried frame velocity components can be obtained with
the transformation matrix:
[ ]
⋅=
B
B
B
VB
D
E
N
W
V
U
L
V
V
V
To update Euler angles in [ ]VBL transformation matrix, one canexploit derivatives of the Euler angles; hence:
=ω
=ω
=ω
B1Z
B1Y
B1X
R
Q
P
Ψ=Ψ
Θ=Θ
Φ=Φ
B
B
B
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ΘΦ+
ΘΦ=Ψ
Φ−Φ=Θ
ΘΦ+ΘΦ+=Φ
B
BB
B
BBB
BBBBB
BBBBBBBB
coscosR
cossinQ
sinRcosQ
tgcosRtgsinQP
By integrating the three equations, one can obtain instantaneous
values of Euler angles; as a result, velocities in vehicle carried
vertical frame can be computed as:
ΘΦ+ΦΘ+Θ−=
ΨΦ−ΨΘΦ+
+ΨΦ+ΨΘΦ+ΨΘ=
ΨΦ+ΨΘΦ++ΨΦ−ΨΘΦ+ΨΘ=
BBBBBBBBD
BBBBBB
BBBBBBBBBE
BBBBBB
BBBBBBBBBN
coscosWsincosVsinUV
)cossinsinsin(cosW
)coscossinsin(sinVsincosUV
)sinsincossin(cosW
)sincoscossin(sinVcoscosUV
Mathematical model is shown in scheme 5.
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5
5 Carlo Casarosa, "Meccanica del Volo", Ed. Plus 2004.
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1.9.5 Mathematical model in Stability Body Axes
The mathematical model is similar to the one obtained for generic
body axes, with changes affecting initial conditions which,anyway, are not clearly visible in equations; hence:
ΦΘ−+−⋅=
ΦΘ−−+⋅=
Θ++−⋅=
)coscosgVPUQW(mZ
)sincosgWPURV(mY
)singWQVRU(mX
SSSSSSSS
SSSSSSSS
SSSSSSS
−+−+=
−+−+=
+−−+=
)PQR(IQP)II(RIN
)RP(IRP)II(QIM
)QPR(IRQ)II(PIL
SSSSZSXSSSXSYSSZS
2S
2SSZSXSSSZSXSSYS
SSSSZSXSSSYSZSSXS
Expressions of velocity components in NED can be computed
through transformation matrix between the two frames:
[ ]
⋅=
S
S
S
VS
D
E
N
W
V
U
L
V
V
V
As a result, an updating of matrix coefficients is required; it can be
obtained referring to Euler angle derivatives:
ΘΦ
+ΘΦ
=Ψ
Φ−Φ=Θ
ΘΦ+ΘΦ+=Φ
S
SS
S
SSS
SSSSS
SSSSSSSS
cos
cosR
cos
sinQ
sinRcosQ
tgcosRtgsinQP
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By integrating the three angular velocities, one eventually obtains
instantaneous values of Euler angles; as a result, velocities in
vehicle carried vertical frame can be computed as:
ΘΦ+ΘΦ+Θ−=
ΨΦ−ΨΘΦ+
+ΨΦ+ΨΘΦ+ΨΘ=
ΨΦ+ΨΘΦ+
+ΨΦ−ΨΘΦ+ΨΘ=
SSSSSSSSD
SSSSSS
SSSSSSSSSE
SSSSSS
SSSSSSSSSN
coscosWcossinVsinUV
)cossinsinsin(cosW
)coscossinsin(sinVsincosUV
)sinsincossin(cosW
)sincoscossin(sinVcoscosUV
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1.9.6 Summary of Mathematical Models
Force Moment
(FB)
Position Attitude Kinematics # Equations
ECEF 3 3 3 3 2 14
NED 3 3 - 3 - 9
Wind
Axes
3 3 3 3 3 15
Body
Axes
3 3 3 3 - 12