12 performance of an aircraft with parabolic polar

08/12/15 1

Performance of an Aircraft with Parabolic Polar

SOLO HERMELIN

Updated: 14.03.2004

08/12/15 2

Performance of an Aircraft with Parabolic PolarSOLO

Table of Content

Flat Earth Three Degrees of Freedom Aircraft Equations

Performance of an Aircraft with Parabolic Polar

Aircraft Drag

Energy per unit mass E

Load Factor n

Aircraft Trajectories

Summary

Constraints

Horizontal Plan Trajectory ( )0,0 == γγ

Horizontal Turn Rate as Function of ps, n

Horizontal Turn Rate as Function of nV ,

References

08/12/15 3

SOLO

Assumptions:

•Point mass model.•Flat earth with g = constant.•Three-dimensional aircraft trajectory.•Air density that varies with altitude ρ=ρ(h)•Drag that varies with altitude, Mach number and control effort D = D(h,M,n)•Thrust magnitude is controllable by the throttle. •No sideslip angle.•No wind.

α

T

V

L

D

Bx

Wx

Bz

Wz

Wy

By

Aircraft Coordinate System


4

SOLO

• Rotation Matrix from Earth to Wind Coordinates

[ ] [ ] [ ]321 χγσ=WEC

whereσ – Roll Angleγ – Elevation Angle of the Trajectoryχ – Azimuth Angle of the Trajectory

Force Equation:

amgmTFA

=++

where:

• Aerodynamic Forces (Lift L and Drag D)( )

−

−=

L

D

F WA 0

• Thrust T ( )

=

α

α

sin

0

cos

T

T

T W

• Gravitation acceleration

( ) ( )

−

−

−==

g

cs

sc

cs

sc

cs

scgCg EWE

W 0

0

100

0

0

0

010

0

0

0

001

χχχχ

γγ

γγ

σσσσ

( ) g

cc

cs

s

g W

−=

γσγσ

γ

α

T

V

L

D

Bx

Wx

Bz

Wz

Wy

By


5

SOLO

α

T

V

L

D

Bx

Wx

Bz

Wz

Wy

By

• Aircraft Acceleration

( )( )

( ) ( )WWW

W VVa ×+=

→

ω

where:

( )

=

0

0

V

V W

and( )

=

→

0

0

V

VW

( )

−+

−+

−=

=

χχχχχ

γγγ

γγσ

σσσσω

0

0

100

0

0

0

0

0

010

0

0

0

0

0

001

cs

sc

cs

sc

cs

sc

r

q

p

W

W

W

W

or ( )

+−+

−=

=

γσχσγγσχσγγχσ

ωccs

csc

s

r

q

p

W

W

W

W

therefore( )

( )( ) ( ) ( )

( )

+−+−=

−=×+=

→

γσχσγγσχσγω

cscV

ccsV

V

qV

rV

V

VVa

W

WWW

WW


6

SOLO

α

T

V

L

D

Bx

Wx

Bz

Wz

Wy

By

• Aircraft Acceleration


From the Force equation we obtain:

( )( )

( ) ( ) ( ) ( )( ) ( )WWWA

WWW

W gTFm

VVa ++=×+=

→ 1ω

or

( )( ) ( ) σ

σσσ

γσαγσχσγγσγσχσγ

γα

s

c

c

s

ccgmLTcscVqV

csgccsVrV

sgmDTV

W

W

−−−

++−=+−=−=+−=

−−=

/sin

/)cos(

from which we obtain:

( )( )

+=−+=

−−=

msLTcV

cgmcLTV

sgmDTV

/sin

/sin

/)cos(

σαγχγσαγ

γα

Define the Load Factor

gm

LTn

+= αsin:

7

SOLO

α

T

V

L

D

Bx

Wx

Bz

Wz

Wy

By

• Velocity Equation


( ) ( )

==

=

0

0

V

CVC

h

y

x

V EW

WEW

E

−

−

−=

0

0

0

0

001

0

010

0

100

0

0 V

cs

sc

cs

sc

cs

sc

h

y

x

σσσσ

γγ

γγχχχχ

=

==

γχγχγ

sVh

scVy

ccVx

or

• Energy per unit mass E

g

VhE

2:

2

+=

Let differentiate this equation:( )

W

VDT

W

DTg

g

VV

g

VVhEps

−=

−−+=+== αγαγ cos

sincos

sin:

Return to Table of Content

8

SOLO Flat Earth Three Degrees of Freedom Aircraft Equations

Summary

γχγχγ

sin

sincos

coscos

Vh

Vy

Vx

=

==

( )

( )

σγ

σγ

αχ

γσγσαγ

αγα

sincos

sincos

sin

coscoscoscossin

cossin

cos

n

V

g

W

LT

V

g

nV

g

W

LT

V

g

W

VDTEor

W

DTgV

=+=

−=

−+=

−=−

−=

wheremgW =

Aircraft Thrust( ) 10, ≤≤= ηη VhTT MAX

( ) ( )MSCVhL L ,2

1 2 αρ=

( ) ( )LD CMSCVhD ,2

1 2ρ=

( ) ( ) ( ) 20, LDLD CMKMCCMC +=

( ) 0/0

hheh −= ρρ

( ) ( ) soundofspeedhaNumberMachMhaVM === &/

Aircraft Weight

Aircraft Lift

Aircraft Drag

Parabolic Drag Polar


9


Constraints:

State Constraints

• Minimum Altitude Limit minhh ≥

• Maximum dynamic pressure limit ( ) ( )hVVorqVhq MAXMAX ≤≤= 2

2

1 ρ

• Maximum Mach Number limit ( ) MAXMha

V ≤

Aerodynamic or heat limitation

Control Constraints

• Maximum Lift Coefficient or Maximum Angle of Attack

( ) ( ) ( )VhorMCMC STALLMAXLL ,, _ ααα ≤≤

• Minimum Load Factor ( )MAXn

W

VhLn ≤= ,

• Maximum Thrust ( )VhTT MAX ,≤

10


MAXα MAXα

αα

LCDC

MAXLC

0DC

( ) ( )αα 20 LDD kCCC +=

Drag and Lift Coefficients as functions of Angle of Attack

11


( ) ( )( )

Limit

Vhor

MCMC

STALL

MAXLL

,

, _

ααα

=

=

( )( )

Limit

hVVor

qVhq

MAX

MAX

=

== 2

2

1 ρ

minhh =

MAXMM =

Mach

Altitude

Flight Envelope of the Aircraft


12


W

LTn

+= αsin:'

W

Ln =:

20

:LD

L

D

L

CkC

C

CSq

CSq

D

Le

+===

We assumed a Parabolic Drag Polar:2

0 LDD CkCC +=

Let find the maximum of e as a function of CL

( ) ( ) 02

220

20

220

220 =

+

−=+

−+=∂∂

LD

LD

LD

LLD

L CkC

CkC

CkC

CkCkC

C

ee

LC*LC

*2

1

LCk

CL/CD as a function of CL

The maximum of e is obtained for

k

CC DL

0* =

( ) 00

02

0 2** DD

DLDD Ck

CkCCkCC =+=+=

Start with

Load Factor

Total Load Number

Lift to Drag Ratio

Climbing Aircraft Performance

13


e

LC*LC

*2

1

LCk

CL/CD as a function of CL

The maximum of e is obtained for

k

CC DL

0* =

( ) 00

02

0 2** DD

DLDD Ck

CkCCkCC =+=+=

*2

1

*2

1

2

1

2*

**

2200

0

LLDD

D

D

L

CkCkCkCkC

C

Ce =====

We have WnCSVCSqL LL === 2

2

1 ρ

Let define for n = 1

=

=

==

2

0

*2

1:*

*:

2

*21

:*

Vq

V

Vu

CS

kW

CS

WV

DL

ρ

ρρ

20

:LD

L

D

L

CkC

C

CSq

CSq

D

Le

+===


14


Using those definitions we obtain

L

L

L

L

C

Cnqq

WCSq

WnCSqL **

**=→

===

22

2

1

21

*21

*

uV

Vn

q

q ==ρ

ρ

2

**

*

u

CnC

q

qnC L

LL ==

( )

+=

+=

+=+=

=

2

22

0402

02

*

4

22

022

0

**

**

02

u

nuCSq

u

CnCuSq

u

CnkCuSqCkCSqD

DD

D

CCk

LDLD

DL

*2

1

**** 0

0 eW

C

CCSqCSq

L

DLD ==

+=

2

22

*2 u

nu

e

WD

Therefore



=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

15


We obtained

Let find the minimum of D as function of u.

nu

u

nu

e

W

u

nu

e

W

u

D

=→

=−=

−=

∂∂

2

3

24

3

2

0*

22*2

*2min e

WnDD

nu==

=

Aircraft Drag


u 0

- - - - 0 + + + + +

D ↓ min ↑

n

u

D

∂∂

+=

2

22

*2 u

nu

e

WD

16


Aircraft Drag

( )MAXn

W

VhLn ≤= ,

+=

= 2

22

*2 u

nu

e

WD MAX

nn MAX

Maximum Lift Coefficient or Maximum Angle of Attack( ) ( ) ( )VhorMCMC STALLMAXLL ,, _ ααα ≤≤

We have

uC

C

u

n

u

CnC

q

qnC

L

MAXL

CC

LLL

MAXLL*

**

* _

2

_

=→===

2

2

_

2

2

_2

*1

*2

**2_

uC

C

e

W

uC

Cu

e

WD

L

MAXL

L

MAXL

CC MAXLL

+=

+=

=

Maximum dynamic pressure limit

( ) ( ) MAXMAX

MAXMAX uV

VuhVVorqVhq =<→≤≤= :

*2

1 2ρ

*eW

D

MAXLC _

2

2

_12

1u

C

C

L

MAXL

+

+=

2

22

2

1*

u

nue

W

D MAX

LIMIT

nn MAX=

2min * ueW

D =

+=

2

22

2

1*

u

nue

W

D

MAXuu =MAX

MAXL

LCORNER n

C

Cu

_

*=

n

LIMIT

uMAXnu =

as a function of u*eW

D



Maximum Load Factor

=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

17


Energy per unit mass ELet define Energy per unit mass E: g

VhE

2:

2

+=Let differentiate this equation:

( ) ( )W

VDT

W

VDT

W

DTg

g

VV

g

VVhEps

−≈−=

−−+=+== αγαγ cos

sincos

sin:

*&*2 2

22 VuVu

nu

e

WD =

+=

Define *: eW

Tz

=

We obtain( )

+−=

+−

=−=2

22

2

22

2

1

*

**

2

1*

*

u

nuzu

e

V

W

Vuu

nue

W

T

e

W

W

VDTps

or ( )u

nuzu

e

Vps

224 2

*2

* −+−=

nznzzu

nzzunuzup

constns>

−+=

−−=→=+−→=

=22

2

221224 020

( ) ( )2

224

2

2243 23

*

*244

*

*

u

nuzu

e

V

u

nuzuuuzu

e

V

u

p

constn

s ++−=−+−−+−=∂

∂

=

3

30

22 nzzu

u

pMAX

constn

s ++=→=

∂∂

=

221

2uu

uuMAX <<+

nz >



Energy per unit mass E

g

VhE

2:

2

+=


Energy Height versus Mach NumberEnergy Height versus True Airspeed

( )hV

VM

sound

=:( )00

:T

TV

T

TMhVTAS sound ==

19


Energy per unit mass Esp

2u1u MAXu2

21 uu + u

MAXn

n

1=n

( )u

nuzu

e

Vps

224 2

*

* −+−=

ps as a function of u

( )u

nuzu

e

Vps

224 2

*

* −+−=

uV

peuzunnuzuu

V

pe ss

*

*222

*

*2 242224 −+−=→−+−=

From which uV

peuzun s

*

*22 24 −+−=

( )*

*244 3

2

V

peuzu

u

n s

constps

−+−=∂

∂

=

( )3

0412 22

22 zuzu

u

n

constps

=→=+−=∂

∂

=



=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

20


Load Factor n

u

3

z z

z2z3

z

u

2n

0=sp

0>sp

0<sp

0<sp

0=sp

0>sp

( )u

n

∂∂ 2

( )2

22

u

n

∂∂

3

zu

( ) ( ) 22

2

22

,, nu

n

u

n

∂∂

∂∂ as a function of u

uV

peuzun s

*

*22 24 −+−=

( )3

0412 22

22 zuzu

u

n

constps

=→=+−=∂

∂

=

( )*

*244 3

2

V

peuzu

u

n s

constps

−+−=∂

∂

=


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

21


Load Factor nFor ps = 0 we have

zuuzun 202 24 ≤≤+−=

Let find the maximum of n as function of u.

022

4424

3

=+−+−=

∂∂

uzu

uzu

u

n

Therefore the maximum value for n is achieved for zu =

( ) znMAXps

==0

u 0 √z √2z

∂ n/∂u | + + + 0 - - - - | - -

n ↑ Max ↓

z2z

u

n

0=sp

0>sp

0<sp

MAXn

z

MAXMAXL

L nC

C

_

*

n as a function of u



=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

22


−+=

=

γσα

γσ

coscossin

cossin

V

g

Vm

LTq

V

gr

W

W

nW

L

W

LTn =≈+= αsin

:'

Therefore

( )

−=

=

γσ

γσ

coscos'

cossin

nV

gq

V

gr

W

W

γσγσγσω 2222222 coscoscoscos'2'cossin +−+=+= nnV

gqr WW

or

γγσω 22 coscoscos'2' +−= nnV

g

γγσω 22

2

coscoscos'2'

1

+−==

nng

VVR

Aircraft Trajectories

We found

Aircraft Turn Performance

23


( ) ( )

( )γσφ

γαχ

γσγσαγ

cos

sinsin

cos

sin

coscos'coscossin

V

gLT

nV

g

V

g

Vm

LT

=+=

−=−+=

2. Horizontal Plan Trajectory ( )0,0 == γγ

( )

1'

1

1''

11'sin'

cos

1'01cos'

2

2

22

−=

−=

−==

=→=−=

ng

VR

nV

g

nn

V

gn

V

g

nnV

g

σχ

σσγ


1. Vertical Plan Trajectory (σ = 0)

( )

γ

γγ

χ

cos'

1

cos'

0

2

−=

−=

=

ng

VR

nV

g

24

Vertical Plan Trajectory (σ = 0) SOLO


25



We can see that for n > 1

1

1

1'

1

11'

2

2

2

2

22

−≈

−=

−≈−=

ng

V

ng

VR

nV

gn

V

gχ

We found that2

2 *

*u

C

Cn

u

CnC

L

LLL =→=

n

1n

2n

MAXn

u u

LC

MAXLC _

1_

nC

C

MAXL

LMAX

MAXL

Lcorner n

C

Cu

_

*=

*2 L

MAXL C

u

nC =

MAXMAXL

Lcorner n

C

Cu

_

*= MAXL

L nC

C

1

*

MAXLC _

2LC

1LC

2

*1 u

C

Cn

L

L=

MAXn

n, CL as a function of u


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

26

R

V=:χ1'2 −= nV

gχ

Contours of Constant n and Contours of Constant Turn Radiusin Turn-Rate in Horizontal Plan versus Mach coordinates

Horizontal Plan Trajectory SOLO

27


MAX

MAX

L

MAXL

n

n

C

C

V

g 1

**

2_ −

MAXMAXL

Lcorner n

C

C

V

gu

_

*

*=

MAXL

L

C

C

V

gu

_1

*

*=

MAXn

2n1n

MAXLC _

2LC

1LC

u

χ

MAXu

a function of u, with n and CL as parameters χ

We defined 2

*&

*: u

C

Cn

V

Vu

L

L==

We found 22

2

22 1

**1

*1

uu

C

C

V

gn

Vu

gn

V

g

L

L −

=−=−=χ

This is defined for 1:**

1__

<=≥≥= uC

Cun

C

Cu

MAXL

LMAX

MAXL

Lcorner



=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

28


From2

2

2

22 1

**1

*1

uu

C

C

V

gn

Vu

gn

V

g

L

L −

=−=−=χ

4

2

2

2

22

1

*

1*

1

*:

uC

Cg

V

n

u

g

VVR

L

L −

=

−==

χ

Therefore

cornerMAXMAXL

L

MAXL

L

L

MAXL

Cun

C

Cu

C

Cu

uC

Cg

VR

MAXL=≤≤=

−

=

__1

4

2

_

2 **

1

*

1*_

cornerMAXMAXL

L

MAX

nun

C

Cu

n

u

g

VR

MAX=≥

−=

_2

22 *

1

*

MAXL

L

L

L

L

L

Cn

C

Cu

C

Cu

uC

Cg

VR

L

**

1

*

1*1

4

2

2

≤≤=

−

=

nC

Cu

n

u

g

VR

MAXL

Ln

_2

22 *

1

* ≥−

=



29


R (Radius of Turn) a function of u, with n and CL as parameters

1

**2

_

2

−MAX

MAX

MAXL

L

n

n

C

C

g

V

MAXMAXL

Lcorner n

C

C

V

gu

_

*

*=

MAXL

L

C

C

V

gu

_1

*

*=

MAXn

2n

1nMAXLC _

2LC 1L

C

u

R

4

2

2

2

22

1

*

1*

1

*:

uC

Cg

V

n

u

g

VVR

L

L −

=

−==

χ




=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

30


Horizontal Turn Rate as Function of ps, n ( )

u

nuzu

e

Vps

224 2

*2

* −+−=

upV

euzun s*

*22 242 −+−=

2

24

2

2 1*

*22

*

1

* u

upV

euzu

V

g

u

n

V

g s −−+−=−=χ

2

24

4

2423

1*

*22

2

1*

*222

*

*244

*

u

upV

euzu

u

upV

euzuuup

V

euzu

V

g

us

ss

−−+−

−−+−−

−+−

=∂∂ χ

Therefore

−−+−

++−=

∂∂

1*

*22

1*

*

*244

4

upV

euzuu

upV

eu

V

g

us

sχ

For ps = 0

222

12

24

011

12

*uzzuzzu

u

uzu

V

gsp

=−+<<−−=−+−==

χ

( ) 222

1244

4

0

1112

1

*uzzuzzu

uzuu

u

V

g

usp

=−+<<−−=−+−

+−=∂∂

=

χ


31



For ps = 0

222

12

24

011

12

*uzzuzzu

u

uzu

V

gsp

=−+<<−−=−+−==

χ

( ) 222

1244

4

0

1112

1

*uzzuzzu

uzuu

u

V

g

usp

=−+<<−−=−+−

+−=∂∂

=

χ

Let find the maximum of as a function of u χ

( )12

1

* 244

4

0 −+−

+−=∂∂

= uzuu

u

V

g

usp

χ

( ) ( )12*

100

−=====

zV

gu

ss ppMAX χχ

u 0 u1 1 (u1+u2)/2 u2

∞ + + 0 - - - - - - -∞

↑ Max ↓

u∂∂ χ

χ

From

2

24 1*

*22

* u

upV

euzu

V

g s −−+−=χ

−−+−

++−=

∂∂

1*

*22

1*

*

*244

4

upV

euzuu

upV

eu

V

g

us

sχ


32



u

u

0<sp

0<sp

0=sp

0=sp 0>sp

0>sp

χ

u∂∂ χ

( )12*

−zV

g

1=u1u2u

as a function of u with ps asparameter

u∂∂ χχ

,

−−+−

++−=

∂∂

1**2

2

1**

* 244

4

upVe

uzuu

upVe

u

V

g

us

sχ

2

24 1**2

2

* u

upVe

uzu

V

g s −−+−=χ

Because ,we have0*

* >uV

e

000 >=<>>

sss pppχχχ

01

01

01

0>

==

=<

= ∂∂<=

∂∂<

∂∂

sss pu

pu

pu uuu

χχχ


33



u

0<sp

0=sp

0>sp

χ

( )12*

−zV

g

1=u1u2u

2

24 1**2

2

* u

upVe

uzu

V

g s −−+−=χ

1*

2 −MAXnuV

g

22

2

_ 1

** uu

C

C

V

g

L

MAXL −

MAXL

L

C

C

_

*

MAXMAXL

L nC

C

_

*

LIMIT

nMAXLIMIT

C MAXL_

MAX

MAX

L

MAXL

n

n

C

C

V

g 1

**

2_ −

a function of u, with ps as parameter

χ

2

24 1**2

2

* u

upVe

uzu

V

g s −−+−=χ


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

34



2

24 1*

*22

* u

upV

euzu

V

g s −−+−=χ

( ) ( )ss

s

puupuup

V

euzu

u

g

VVR 21

24

42

1*

*22

* <<−−+−

==χ

3242

232

24

4

224

34243

2

1*

*222

2*

*322

*

1*

*22

2

1*

*22

*

*2441

*

*224

*

−−+−

−−

=

−−+−

−−+−

−+−−

−−+−

=∂∂

upV

euzuu

upV

euzu

g

V

upV

euzu

u

upV

euzu

pV

euzuuup

V

euzuu

g

V

u

R

s

s

s

s

ss


35



324

22

1*

*22

2*

*32

*

−−+−

−−

=∂∂

upV

euzu

upV

euzu

g

V

u

R

s

s

or

We have

>+

+

=

<+

−

=→=

∂∂

04

16*

*9

*

*3

04

16*

*9

*

*3

02

2

2

1

z

zpV

eup

V

e

u

z

zpV

eup

V

e

u

u

R

ss

R

ss

R

u 0 u1 uR2 u2

∞ - - - 0 + + ∞

↓ min ↑

u

R

∂∂

R

222

124

42

011

12

*uzzuzzu

uzu

u

g

VR

sp=−+<<−−=

−+−=

=

( )( ) 2

221324

22

0

1112

1*2uzzuzzu

uzu

uzu

g

V

u

R

sp

=−+<<−−=−+−

−=∂∂

=


36



u

R

0>sp0=sp

0<sp

MAXL

L

C

C

_

*

1

**2

_ −MAX

MAX

MAXL

L

n

n

C

C

g

V

1

1*2 −zg

V

4

2

_

1*

1*

uC

Cg

V

MAXL

L −

1

*2

22

−MAXn

u

g

V

MAXMAXL

L nC

C

_

*

LIMIT

C MAXL_

LIMIT

nMAX

z

1

12 −− zz 12 −+ zz

1**2

2

*

24

42

−−+−=

upVe

uzu

u

g

VR

s

The minimum of R is obtained for zu /1=

1

1*2

2

0 −=

=zg

VR

sp

R (Radius of Turn) a function of u, with ps as parameter

( ) ( )ss

s

puupu

upVe

uzu

u

g

VVR

21

24

42

1**2

2

*

<<

−−+−==

χ


Because ,we have0*

* >uV

e000 >=<

<<sss ppp

RRR000 minminmin >=<

<<sss pRpRpR uuu


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

37



( )W

VDT

g

VVhEps

−≈+==

:

For an horizontal turn 0=h

Vg

Vu

g

VVps

*==

We found2

24 1*

*22

* u

upV

euzu

V

g s −−+−=χ

from which2

24 1*2

* u

uegV

zu

V

g−

−+−

=

χ

defined for

2

22

1 :1**1**: ueg

Vze

g

Vzue

g

Vze

g

Vzu =−

−+

−≤≤−

−−

−=


38



Let compute

2

24

4

2423

1*2

2

1*22*44

*

u

ueg

Vzu

u

ueg

Vzuuuue

g

Vzu

V

g

u−

−+−

−

−+−−

−+−

=∂∂

χ

−

−+−

+−=∂∂

1*2

1

*244

4

ueg

Vzuu

u

V

g

u

χ

or u 0 u1 1 (u1+u2)/2 u2

∞ + + 0 - - - - - - -∞

↑ Max ↓

u∂∂ χ

χ

−−= 1*2

*e

g

Vz

V

gMAX

χ


39



u

0<V

0=V

0>V

χ

( )12*

−zV

g

1=u1u2u

2

24 1*2

* u

uegV

zu

V

g−

−+−

=

χ

1*

2 −MAXnuV

g

22

2

_ 1

** uu

C

C

V

g

L

MAXL −

MAXL

L

C

C

_

*

MAXMAXL

L nC

C

_

*

LIMIT

nMAXLIMIT

C MAXL _

MAX

MAX

L

MAXL

n

n

C

C

V

g 1

**

2_ −

as function of uand as parameterχ

V



=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

08/12/15 40


References

Angelo Miele, “Flight Mechanics, Volume I, Theory of Flight Paths”, Addison-Wesley, 1962


S. Hermelin, “3 DOF Model of Aircraft in Earth Atmosphere”

41

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

12 performance of an aircraft with parabolic polar

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