Dispense Cap1

5/11/2018 Dispense Cap1 - slidepdf.com

http://slidepdf.com/reader/full/dispense-cap1 1/71

F.Quagliotti – Meccanica del Volo – A.A. 2011-2012 1

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




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(

.




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.




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




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

−

= .




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




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−

=




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=




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




[ ]

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.




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.




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.




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 .




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




( ) ( )

( ) ( )

( ) ( )

+−−+

++−−

−++−

=

)(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




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.




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.




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:




• 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




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




=

⋅−

=

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




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:




( )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




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

.




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




[ ] [ ] =

⋅−

−⋅⋅

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:




[ ] =

⋅−

⋅

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




1.5 Vehicle-Carried Vertical Frame (NED)

)Z,Y,X(F VVVV =

The origin of NED reference frame is at the aircraft C.G.




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:




( )

=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




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).




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:




• 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:




( )

∧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:




( )( )

( )

−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.




1

1 Carlo Casarosa, "Meccanica del Volo", Ed. Plus 2004.




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.




Figure 1




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




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:




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:




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.




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:




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


• 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.




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


• 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.




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:




[ ]

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




• 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

QQ

BBW

BBW

B




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.




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




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




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




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:




⋅

−

−

⋅

−

−

−

+

+

⋅

−

−

=

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




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




• 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




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.




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.




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




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.




2





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.




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




=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

QQ

BBW

BBW

B

Pattern for complete mathematical model in wind axes is shown in

scheme 4.




4





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




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.




5





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




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




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

Dispense Cap1

Documents

Transcript of Dispense Cap1