14 fixed wing fighter aircraft- flight performance - ii

Fixed Wing Fighter AircraftFlight Performance

Part II

SOLO HERMELIN

Updated: 04.12.12 28.02.15

1

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Table of Content

SOLO

2

Aerodynamics

Introduction to Fixed Wing Aircraft Performance

Earth Atmosphere

Mach Number

Shock & Expansion Waves

Reynolds Number and Boundary Layer

Knudsen Number

Flight Instruments

Aerodynamic Forces

Aerodynamic Drag

Lift and Drag Forces

Wing Parameters

Specific Stabilizer/Tail Configurations

Fixed

Wi

ng

Part

I

Fixed Wing Fighter Aircraft Flight Performance

Table of Content (continue – 1)

SOLO

3

Specific Energy

Aircraft Propulsion Systems

Aircraft Propellers

Aircraft Turbo Engines

Afterburner

Thrust Reversal Operation

Aircraft Propulsion Summary

Vertical Take off and Landing - VTOL

Engine Control System

Aircraft Flight Control

Aircraft Equations of Motion

Aerodynamic Forces (Vectorial)

Three Degrees of Freedom Model in Earth Atmosphere

Fixed

Wi

ng

Part

I



SOLO Fixed Wing Fighter Aircraft Flight Performance

4

Parameters defining Aircraft Performance

Takeoff (no VSTOL capabilities)

Landing (no VSTOL capabilities)

Climbing Aircraft Performance

Gliding Flight

Level Flight

Steady Climb (V, γ = constant)

Optimum Climbing Trajectories using Energy State Approximation (ESA)Minimum Fuel-to- Climb Trajectories using Energy State Approximation (ESA)Maximum Range during Glide using Energy State Approximation (ESA)

Aircraft Turn Performance

Maneuvering Envelope, V – n Diagram


SOLO Fixed Wing Fighter Aircraft Flight Performance

5

Air-to-Air Combat

Energy–Maneuverability Theory

Supermaneuverability

Constraint Analysis

References

Aircraft Combat Performance Comparison

SOLO

This Presentation is about Fixed Wing Aircraft Flight Performance.

The Fixed Wing Aircraft are•Commercial/Transport Aircraft (Passenger and/or Cargo)•Fighter Aircraft


Continue from Part I

7

Fixed Wing Fighter Aircraft Flight PerformanceSOLO

The Aircraft Flight Performance is defined by the following parameters

• Take-off distance• Landing distance• Maximum Endurance and Speed for Maximum Endurance• Maximum Range and Speed for Maximum Range• Ceiling(s)• Climb Performance• Turn Performance• Combat Radius• Maximum Payload

Parameters defining Aircraft Performance

8

Performance of an Aircraft with Parabolic PolarSOLO

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) and is given by a Parabolic Polar.•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

To understand how different parameters affect Aircraft Performance we start with a Simplified Model, where Analytical Solutions can be obtained. Results for real aircraft will then be presented.

Return to Table of Content

9

SOLO Aircraft Flight Performance

Takeoff

The Takeoff distance sTO is divided as the sum of the following distances:sg – Ground Runsr – Rotation Distancest – Transition Distancesc – Climb Distance to reach Screen Height

ctrgTO sssss +++=

Ground RunV = 0

sg

sTO

sr str

V TORotation

Transition

sCL

θ CL

htr

hobs

R

Takeoff htransition < hobstacle

θ CL

Ground RunV = 0

sg

sTO

sr sobs

V TORotation

Transition

hobs

R

Takeoff htransition > hobstacle

We distinguish between two cases of Takeoff •The Aircraft must passes over an obstacle at altitude hobs..•The obstacle is cleared during the transition phase.

Assume no Vertical Takeoff Capability.

10

Takeoff (continue – 1)

During the Ground Run there are additionaleffects than in free flight, that must be considered:-Friction between the tires and the ground during rolling.-Additional drag due to the landing gear fully extended.-Additional Lift Coefficient due to extended flaps.-Ground Effect due to proximity of the wings to the ground, that reduces the Induced Drag and the Lift.

Ground Run


Ground run sg

Transition distance st

Climb distance sc

Stall safety

Take-off possible with one engine

Continue take-off if engine fails

after this point

Stop take-off if engine fails before

this point

Acceleration at full power

γ c

Total take-off if distance

VCRVMCGVTVS

hc

L

W

TD

R

μR

The Aircraft can leave the ground when the velocity reaches the Stall Velocity where Lift equals Weight

max,2

02

1Lstallstall CSVLW ρ==

max,0

12

Lstall CS

WV

ρ=

The Liftoff Velocity is 1.1 to 1.2 Vstall.

11

ReactionGroundLWR

gW

RDT

td

VdV

Vtd

xd

−=

−−==

=

/

µ

( )

( )LWDT

gW

Vd

td

LWDT

gWV

Vd

sd xs

−−−=

−−−==

µ

µ/

/


Average Coefficient of Friction Values μ

Ground Run


Ground run sg


Climb distance sc

Stall safety



after this point


this point


γ c


VCRVMCGVTVS

hc

L

W

R

μRD

T

V

12



Ground run sg


Climb distance sc

Stall safety



after this point


this point


γ c


VCRVMCGVTVS

hc

L

W

R

μRD

T

V

T (Jet)

Lift, Drag

, Thrust, Resistance

–lb

L, D

, T, R

T (Prop)

D +μ R

Ground Speed – ft/s

Texcess(Prop)=T(Prop) -(D+μ R)Texcess(Jet)=T(Jet) -(D+μ R)

Vground

( )

ReactionGroundLWR

RDTg

WV

−=

+−= µ

Ground Run (continue -1)

13

20 VCVBTT ++=

cVbVaVd

td

cVbVa

V

Vd

sd xs

++=

++==

2

2

1


Ground Run (continue – 2)

To obtain an Analytic Solution assume that during the Ground Run the Thrust can be approximated by

Using

=

=

L

D

CSVL

CSVD

2

2

2

12

1

ρ

ρ

( )

−=

=

+−−=

µ

µρ

W

Tgc

W

gBb

W

gCCC

W

Sga LD

0:

2:

2:

where


Ground run sg


Climb distance sc

Stall safety



after this point


this point


γ c


VCRVMCGVTVS

hc

L

W

R

μRD

T

V

14

cVbVaVd

td

cVbVa

V

Vd

sd xs

++=

++==

2

2

1



Integrating those equations between two velocities V1 and V2 gives

−−⋅

++

−+

++++=

2

1

1

2

2

121

222

1

1

1

1ln

42

ln2

1

a

a

a

a

caba

b

cVbVa

cVbVa

asg

−−⋅

++

−=

1

2

2

1

2 1

1

1

1ln

4

1

a

a

a

a

cabtg

where

cab

bVaa

cab

bVaa

4

2:

4

2:

2

22

2

11

−+=

−+=


Ground run sg


Climb distance sc

Stall safety



after this point


this point


γ c


VCRVMCGVTVS

hc

L

W

R

μRD

T

V

20 VCVBTT ++=

15



then

( )( )

−−−−

=

+=

TL

LDLD

g

CWTCCCCg

SW

c

cVa

as

µµµρ

/1

1ln

/

ln2

1

0

22

0,00 01 ==⇐== CBTTV

Assume

where

22:&

/2: VV

V

SWC T

TLT

==ρ

A further simplification, using , givesZZ

Z 1

1

1ln

<<≈

−

−

=µρ

W

TCg

SWs

TL

g0

/


gL sCg

SW

W

T

Tρ

/0 >

Ground run sg


Climb distance sc

Stall safety



after this point


this point


γ c


VCRVMCGVTVS

hc

L

W

R

μRD

T

V

16


Rotation Distance

At the ground roll and just prior to going into transition phase, most aircraft areRotated to achieve an Angle of Attack to obtain the desired Takeoff Lift CoefficientCL. Since the rotation consumes a finite amount of time (1 – 4 seconds), the distancetraveled during rotation sr, must be accounted for by using

where Δt is usually taken as 3 seconds.


tVs tr ∆=

Ground RunV = 0

sg

sTO

sr str

V TORotation

Transition

sCL

θ CL

htr

hobs

R

L

W

R

μRD

T

V

17


Transition Distance

In the Transition Phase the Aircraft is in the Air (μ = 0) and turn to the Climb Angle.The Equation of Motion are:


Ta

Tat

Tat

VV

DT

VV

g

Wt

DT

VV

g

Ws

>

−−=

−−=

2

2

22

DT

gW

Vd

td

DT

gWV

Vd

sd xs

−=

−==

/

/

Assuming T – D = const., we canIntegrate the Equations of Motion(assuming Va > VT)

Ground RunV = 0

sg

sTO

sr str

V TORotation

Transition

sCL

θ CL

htr

hobs

R

18


Climb Distance

The Climb Distance is evaluated from the following (see Figure):


c

c

c

cc

hhs

c

γγγ 1

tan

<<

≈=

For small angles of Climb L = W.We can write

Ground RunV = 0

sg

sTO

sr str

V TORotation

Transition

sCL

θ CL

htr

hobs

R

cL

cLD

c

cc C

CkC

W

T

L

D

W

T

,

2,0 +

−=−=γ

We have

cLcLD

cc CkCCWT

hs

,,0 // −−≈

19



19

ctrgTO sssss +++=

sec41−=∆∆= ttVs tr

Ta

Tat

Tat

VV

DT

VV

g

Wt

DT

VV

g

Ws

>

−−=

−−=

2

2

22

−−⋅

++

−+

++++=

2

1

1

2

2

121

222

1

1

1

1ln

42

ln2

1

a

a

a

a

caba

b

cVbVa

cVbVa

asg

cab

bVaa

cab

bVaa

4

2:

4

2:

2

22

2

11

−+=

−+=

−−⋅

++

−=

1

2

2

1

2 1

1

1

1ln

4

1

a

a

a

a

cabtg

Ground RunV = 0

sg

sTO

sr str

V TORotation

Transition

sCL

θ CL

htr

hobs

R

Takeoff Summary

Rotation Phase

Climb Phase

Transition Phase

Ground Run

cLcLD

cc CkCCWT

hs

,,0 // −−≈

20Minimum required takeoff runway lengths.

Summary of takeoff requirementsIn order to establish the allowable takeoff weight for a transport category airplane, at any airfield, the following must be considered:•Airfield pressure altitude•Temperature•Headwind component•Runway length•Runway gradient or slope•Obstacles in the flight path


21

LandingLanding is similar to Takeoff, but in reverse.We assume again that the Aircraft doesn’t haveVTOL capabilities.The Landing Phase can be divide in the followingPhases:

1. The Final approach when the Aircraft Glides toward the runway at a steady speed and rate of descent.

2. The Flare, or Transition phase. The Pilot attempts to rotate the Aircraft nose up and reduce the Rate of Sink to zero and the forward speed to a minimum, that is larger than Vstall. When entering this phase the velocity is less than 1.3Vstall and 1.15 Vstall at touchdown.

3. The Floating Phase, which is necessary if at the end of Flare phase, when the rate of descent is zero, an additional speed reduction is necessary. The Float occurs when the

Aircraft is subjected to ground effect which requires speed reduction for touchdown.

4. The Ground Run after the Touchdown the Aircraft must reduce the speed to reach a sufficient low one to be able to turn off the runway. For this it can use Thrust Reverse (if available), spoilers or drag parachutes (like F-15 or MIG-21) and brakes are applied.


Ground Run sgr

Transition

Airborne Phase

Total Landing Distance

Float

sfFlare stGlide sg

γ

hg

hf

Touchdown

22

Landing (continue – 1)

Descending Phase


The Aircraft is aligned with the landing runaway at an altitude hg and a gliding angle γ.The Aircraft Glides toward the runway at a steady speed and rate of descent, until it reachesThe altitude ht at which it goes to Transition Phase, turning with a Radius of Turn R. The Descending Range on the ground is :

γγγ

γγ RhRhhh

s ggtgg

−≈

−=

−=

<<1

tan

cos

tan

Ground Run sgr

Transition

Airborne Phase


Float

sfFlare stGlide sg

γ

hg

hf

Touchdown

23


Transition Phase


If γ is the descent angle and R is the turn radiusthen the Aircraft must start the Transition Phase at an altitude ht, above the ground, given by:

( )γcos1−=RhtThe Transition Range on the ground is

γγ RRst ≈= sin

To calculate the turn radius we must use the flight velocity which varies between 1.3 Vstall

at the beginning to 1.1 Vstall at Touchdown. Let use an average velocity

3.11.1 −∈= tstalltt mVmV

If the Transition Turn Acceleration is nt = 1.15 – 1.25 g than the Turn Radius is

( ) gn

VR

t

t

1

2

−=

The Transition Turn time is ( ) gn

V

RVt

t

t

tt 1/ −

== γγ

Ground Run sgr

Transition

Airborne Phase


Float

sfFlare stGlide sg

γ

hg

hf

Touchdown

24


Float Phase


In this phase the Pilot brings the nose wheel to the ground at the touchdown velocity Vt:

tVs tf ∆=

where Δt is between 2 to 3 seconds.

Ground Run sgr

Transition

Airborne Phase


Float

sfFlare stGlide sg

γ

hg

hf

Touchdown

25


Ground Run Phase


The equations of motion are the same as those developed for Takeoff, but with different parameters, adapted for Landing. Those equations are:

cVbVaVd

td

cVbVa

V

Vd

sd xs

++=

++==

2

2

1

( )

−=

=

+−−=

µ

µρ

W

Tgc

W

gBb

W

gCCC

W

Sga grLgrD

0

,,

:

2:

2:

where

−−⋅

++

−+

++++=

2

1

1

2

21

21

222

1

1

1

1ln

42ln

2

1

a

a

a

a

caba

b

cVbVa

cVbVa

asg

−−⋅

++

−=

1

2

2

1

2 1

1

1

1ln

4

1

a

a

a

a

cabtg

wherecab

bVaa

cab

bVaa

4

2:

4

2:

2

22

2

11

−+=

−+=

20 VCVBTT ++=

Assume a constant Thrust T = T0: B = 0, C = 0. V1 = Vtouchdown, V2 = final velocity

cVa

cVa

asg +

+−=22

21ln

2

1

−−⋅

++

−=

1

2

2

1

1

1

1

1ln

4

1

a

a

a

a

catg ( )

−==−−= µµρW

TgcbCC

W

Sga LD

0:,0,2

:

Ground Run sgr

Transition

Airborne Phase


Float

sfFlare stGlide sg

γ

hg

hf

Touchdown

26


Ground Run Phase (continue – 1)


where( )

−=

−−=

µ

µρ

W

Tgc

CCW

Sga grLgrD

0

,,

:

2:

cab

Vaa

ca

Vaa

touchdown

4

2:

4

2:

2

11

−=

−=

Assume a constant Thrust T = T0: B = 0, C = 0. V1 = Vtouchdown, V2 = final velocity

cVa

cVa

asg +

+−=22

21ln

2

1

−−⋅

++

−=

1

2

2

1

1

1

1

1ln

4

1

a

a

a

a

catg

For the Landing Ground Run Phase the following must included:• if Thrust Reversal exists we must change T0 to – T0_reversal .•The Drag Coefficient CD0,gr must consider: - the landing gear fully extended. - spoilers or drag parachutes (if exist)•μ – the friction coefficient must be increased to describe the brakes effect.

Ground Run sgr

Transition

Airborne Phase


Float

sfFlare stGlide sg

γ

hg

hf

Touchdown

27


Summary


where( )

−=

−−=

µ

µρ

W

Tgc

CCW

Sga grLgrD

0

,,

:

2:

cab

Vaa

ca

Vaa

touchdown

4

2:

4

2:

2

11

−=

−=cVa

cVa

asg +

+−=22

21ln

2

1

−−⋅

++

−=

1

2

2

1

1

1

1

1ln

4

1

a

a

a

a

catg

Ground Run Phase

tVs tf ∆=

Float Phase

( ) gn

VRs

t

tt 1

2

−== γγ ( ) gn

V

RVt

t

t

tt 1/ −

== γγTransition Phase

Descent PhaseGround Run sgr

Transition

Airborne Phase


Float

sfFlare stGlide sg

γ

hg

hf

Touchdown

( )γγ

γγ

1/

tan

cos

tan

2 −−=

−=

−= ttggfg

g

nVhRhhhs

28

H.H. Hurt, Jr., “Aerodynamics for Naval Aviators “,NAVAIR 00=80T-80 1-1-1965, pg. 35


29


Level Flight

The forces acting on an airplane in Level Flight are shown in Figure

0=

=

h

Vx

Lift and Drag Forces:

( ) TCkCSVCSVD

WCSVL

LDD

L

=+==

==

20

22

2

2

1

2

12

1

ρρ

ρ2

2

VS

WCL ρ

=

+=

+=

SV

WkCSV

SV

WkCSVD DD 2

2

02

242

2

02 2

2

14

2

1

ρρ

ρρ

Lift

DragThrust

Weight

Equations of motion:

0

0

=−=−

DT

WLQuasi-Static

30


Level Flight

DragInducedDragParasite

D SV

WkCSVD

2

2

02 2

2

1

ρρ +=

Because of opposite trends in Parasite Drag and Induced Drag, with changes in velocity, the Total Drag assumes a minimum at a certain velocity. If we ignore the change in velocity of CD0 and k with velocity we obtain

04

3

2

0 =−=SV

WkCSV

Vd

DdD ρ

ρ

The velocity of minimum TotalDrag is

*4

0

2V

C

k

S

WV

D

==ρ

We see that the velocity of minimum Total Drag is equal to the Reference Velocity.

02

2

1DCSVρ

SV

Wk2

22

ρ*V

Lift

DragThrust

Weight

31


Level Flight

For the velocity, V*, of minimum Total Drag we have

02*

22Di CkW

SV

WkD ==

ρ

DragInducedDragParasite

D SV

WkCSVD

2

2

02 2

2

1

ρρ +=

000min 2 DDD CkWCkWCkWD =+=

and

02

2

1DCSVρ

SV

Wk2

22

ρ*V

Lift

DragThrust

Weight

32M. Corcoran, T. Matthewson, N. W. Lee, S. H. Wong, “Thrust Vectoring”

Comparison of Takeoff Weight and Empty Weight of different Aircraft

33


Level Flight

The Power Required, PR, for Level Flight is

SV

WkCSVVDP DR ρ

ρ2

03 2

2

1 +=⋅=

The Power Required for Level Flight assumes a minimum at a certain velocity Vmp. If we ignore the change in velocity of CD0 and k with velocity we obtain

02

2

32

2

02 =−=

SV

WkCSV

Vd

PdD

R

ρρ

or

*4

0 3

1

3

2V

C

k

S

WV

Dmp ==

ρ

*02, 3

32L

D

mpmpL C

k

C

VS

WC ===

ρ

( )*

000

02

,0

, 866.01

4

3

/3

/3e

CkkCkC

kC

CkC

Ce

DDD

D

mpLD

mpLmp ==

+=

+= *

2

min, 866.03

8

e

VW

SV

WkP mp

mpR ==

ρ

03

2

1DCSVρ

SV

Wk

ρ

22

*

3

1V

min,RP

Lift

DragThrust

Weight

34


Available Aircraft Power and Thrust

• Throttle Effect

10 ≤≤= ηη ATT

• Propeller

airspeedwithvariationsmallVTP propellerA ≈⋅=,V

Pa, propeller

Power

Propeller Aircraft Available Powerat Altitude (h)

At a given Altitude h

• Turbojet

airspeedwithvariationsmallT jetA ≈,

V

Ta, jet

Thrust

Jet Aircraft Available Powerat Altitude h

At a given Altitude h

Lift

DragThrust

Weight

Lift

DragThrust

Weight

Level Flight

35


Vmin Vmax

Pa, propeller

PRPmin

BA

ηaPa, propeller

Propeller Aircraft

Vmin Vmax

Ta, jet

TR

Dmin

η Ta, jet

AB

Jet Aircraft

Level Flight

To have a Level Flight the requirement must be satisfied by the available propulsion performance.•For a Propeller Aircraft, the available power Pa,propeller , at a given altitude h, is almost insensitive with changes in velocity. The Velocity in Level Flight is steady when the graph of Required Power PR intersects the graph of Pa,propeller at points A and B. We get two velocities Vmin (h) at A and Vmax (h) at B. By controlling the Propeller Power ηa Pa,propeller (0< ηa <1) we can reach any velocity between Vmin (h) and Vmax (h).

•For a Jet Aircraft, the available Thrust Ta,jet , at a given altitude h, is almost insensitive with changes in velocity. The Velocity in Level Flight is steady when the graph of Required Thrust TR intersects the graph of Ta,jet at points A and B. We get two velocities Vmin (h) at A and Vmax (h) at B. By controlling the Jet Thrust η Ta,jet (0< η <1) we can reach any velocity between Vmin (h) and Vmax (h).

36


Vmin Vmax

Ta, jet

TR

Dmin

η Ta, jet

AB

Jet Aircraft

Level Flight

We have

Analytical Solution for Jet Aircraft

( )SV

WkCSVCkCSVDT DLD 2

2

022

02 2

2

1

2

1

ρρρ +=+==

Define

0

*

00

*

**

4

0

2:

2*,*,:

2:*,

*:

D

DDD

L

D

L

D

CkW

T

W

eTz

CCk

CC

C

Ce

C

k

S

WV

V

Vu

==

===

==ρ

2

2

/1

20

0

2

2

0

2

2

2

u

D

u

Dz

D V

Ck

SW

Ck

SW

VT

CkW

ρ

ρ

+=012 24 =+− uzu

Lift

DragThrust

Weight

37


Vmin Vmax

Ta, jet

TR

Dmin

η Ta, jet

AB

Jet Aircraft

Level Flight


012 24 =+− uzu Solving we obtain

1

1

2max

2min

−+=

−−=

zzu

zzu

4

0maxmaxmax

4

0minminmin

2*

2*

D

D

C

k

S

WuVuV

C

k

S

WuVuV

ρ

ρ

==

==

Lift

DragThrust

Weight

0

*

00

*

**

4

0

2:

2*,*,:

2:*,

*:

D

DDD

L

D

L

D

CkW

T

W

eTz

CCk

CC

C

Ce

C

k

S

WV

V

Vu

==

===

==ρ

38


Level Flight


1

1

2max

2min

−+=

−−=

zzu

zzu

12min −−= zzu

12max −+= zzu

At the absolute Ceiling (when is only one possible velocity) we have umax = umin, thereforez = 1.

max,

2

Lstall CS

WV

ρ=

Lift

DragThrust

Weight

0

*

00

*

**

4

0

2:

2*,*,:

2:*,

*:

D

DDD

L

D

L

D

CkW

T

W

eTz

CCk

CC

C

Ce

C

k

S

WV

V

Vu

==

===

==ρ

39

Drag Characteristics


40


Level Flight

Aircraft Range in Level Flight

Lift

DragThrust

Weight

Range in Level Flight of Jet Aircraft


0

0

=−=−

DT

WL

0=

=

h

Vx

We add the equation of fuel consumption

TcW −= c – specific fuel consumption

We assume that fuel consumption is constant for a given altitude.

Vtd

Wd

Wd

xd

td

xd ==

Dc

V

Tc

V

W

V

Wd

xd DT

−=−===

41


Level Flight


Lift

DragThrust

Weight


Dc

V

Wd

xd −=

The quantity dx/dW is called the “Instantaneous Range” and is equal to the Horizontal Range traveled per unit load of fuel or the “Specific Range”. Multiply and divide by L = W

Wc

V

C

C

Wc

V

D

L

Wd

xd

D

L

−=

−=

Integrating we obtain

∫

−=−= f

i

W

WD

Lif W

WdV

cC

CxxR

1:

42


Level Flight


Lift

DragThrust

Weight


To perform the integration we must specify the variation of CL, CD and V. Let consider two cases:

∫

−=−= f

i

W

WD

Lif W

WdV

cC

CxxR

1:

a. Range at Constant Altitude of Jet Aircraft

We have LCVSLW 2

2

1 ρ==LCS

WV

ρ2=

The velocity changes (decreases) since the weight W decreases due to fuel consumption.

[ ]ifD

LW

WD

L WWC

C

cW

Wd

ScC

CR

f

i

−

=

−= ∫

221

ρ

43


Level Flight


Lift

DragThrust

Weight

a. Range at Constant Altitude of Jet Aircraft

[ ]ifD

LW

WD

L WWC

C

cW

Wd

ScC

CR

f

i

−

=

−= ∫

221

ρ

The maximum range is obtained when

[ ]ifD

L WWC

C

cR −

=

max

max

2

max

20max

+

=

LD

L

D

L

CkC

C

C

C

( ) 030

221

2022

0

20

20

=−⇒=+

−+

=

+ LD

LD

LL

L

LD

LD

L

L

CkCCkC

CkCC

CkC

CkC

C

Cd

d

The maximum range is obtained when*0

3

1

3

1L

DL C

k

CC ==

The Velocity at maximum range is ( ) ( ) ( ) ( )tVCS

tW

CS

tWtV

LL

*4*

4*

32

33/

2 ===ρρ

44


Level Flight


Lift

DragThrust

Weight

b. Range at Constant Velocity of Jet Aircraft

=

−= ∫

f

i

D

LW

WD

L

W

W

C

C

c

V

W

WdV

cC

CR

f

i

ln1

The Velocity V is constant and equal to V* corresponding to initial weight Wi.

4

0*

* 22

D

i

L

i

C

k

S

W

CS

WV

ρρ==

The maximum range is obtained when

=

=

f

i

f

i

D

L

W

We

c

V

W

W

C

CV

cR lnln

1 **

max

max

To keep Velocity V constant when weight W decreases, the air density ρ must also decrease, hence the Aircraft will gain (qvasistatic) altitude

( )Pc

td

Wd

td

hde

td

hdp

hh −==−= − 0/0ρρ

45


Level Flight


Range in Level Flight of Propeller Aircraft

Lift

DragThrust

Weight

The equation of fuel consumption

PcW P−=

cp – specific fuel consumption (consumed per unit power developed by the engine per unit time

We assume that fuel consumption is constant for a given altitude.

Vtd

Wd

Wd

xd

td

xd ==

Pc

V

W

V

Wd

xd

p

−==

- Required PowerVDPR ⋅=

PP pA ⋅=η - Available Power

ηp – propulsive efficiency

AR PP =p

VDP

η⋅=

46


Level Flight



Lift

DragThrust

Weight

WcC

C

WcD

L

DcPc

V

Wd

xd

p

p

D

L

p

pWL

p

p

p

ηηη

−=−=−=−=

=

Integration gives ∫

−=−= f

i

W

Wp

p

D

Lff W

Wd

cC

CxxR

η:

We assume• Angle of Attack is kept constant throughout cruise, therefore e = CL/CD is constant•ηp is independent on flight velocity

f

i

p

p

W

We

cR ln

η= Bréguet Range Equation

The maximum range of Propeller Aircraft in Level Flight is

f

i

Dp

p

f

i

p

p

W

W

CkcW

We

cR ln

2

1ln

0

*max

ηη==

47

Louis Charles Bréguet(1880 – 1955)

The Bréguet Range Equation

The Bréguet range equation determines the maximum flight distance. The key assumptions are that SFC, L/D, and flight speed, V are constant, and therefore take-off, climb, and descend portions of flights are not well modeled (McCormick, 1979; Houghton, 1982). ( )

⋅

=final

initial

W

W

SFCg

DLVRange ln

/

Winitial = Wfuel + Wpayload + Wstructure + Wreserve

Wfinal = Wpayload + Wstructure + Wreserve

where

( )

++

+⋅

=reservestructurepayload

fuel

WWW

W

SFCg

DLVRange 1ln

/

where SFC, L/D, and Wstructure are technology parameters while Wfuel, Wpayload, and Wreserve are operability parameters.


48


Level Flight



Lift

DragThrust

Weight

Let assume that the flight to maximum range is performed in one of two ways

1. Propeller Aircraft Flight at Constant Altitude

In Constant Altitude Flight the velocity changes with the decrease of weight such that

( ) ( )4

0

* 2

DC

k

S

tWVtV

ρ==

2. Propeller Aircraft Flight with Constant Velocity

In Constant Velocity Flight the velocity is the V* velocity based on the initial weight of the Aircraft

.2

4

0

* constC

k

S

WVV

D

i ===ρ

49


Level Flight

Aircraft Endurance in Level Flight

Lift

DragThrust

Weight

The Endurance of an Airplane remains in the air and is usually expressed in hours.

Endurance of Jet Aircraft in Level Flight

We have TcW −=

c – specific fuel consumption

W

Wd

c

e

W

Wd

D

L

cDc

Wd

Tc

Wdtd

WLDT

−=−=−=−=== 1

Integrating we obtain ∫−= f

i

W

W W

Wd

c

et

Assuming that the Angle of Attack is held constant throughout the flight, e =CL/CD is constant

f

i

W

W

c

et ln=

f

i

Df

i

W

W

CkcW

W

c

et ln

2

1ln

0

*

max ==

The Maximum Endurance for Jet Aircraft occurs for e = e*, CL = CL*, V = V*, D = Dmin.

50


Level Flight

Aircraft Endurance in Level Flight

The Endurance of an Airplane remains in the air and is usually expressed in hours.

Endurance of Propeller Aircraft in Level Flight

We have ppp VDcPcW η/⋅−=−=

W

Wd

Ve

cW

Wd

VD

L

cVD

Wd

ctd

p

p

p

pWL

p

p 11 ηηη−=−=

⋅−=

=

Assuming that the Angle of Attack is held constant throughout the flight, e =CL/CD is constant

Lift

DragThrust

Weight cp – specific fuel consumption (consumed per unit power developed by the engine per unit time.ηp – propulsive efficiency

Integrating we obtain

∫−= f

i

W

Wp

p

W

Wd

Ve

ct

1η

The Endurance of Propeller Aircraft depends on Velocity, therefore we will assume two cases1.Flight at Constant Altitude2.Flight with Constant Velocity

51


Level Flight


Lift

DragThrust

WeightThe velocity will change to compensate for the decrease in weight

∫ =−= f

i

W

WD

L

p

p

C

Ce

W

Wd

Ve

ct

1η

1. Propeller Aircraft Flight at Constant Altitude

We haveLCVSLW 2

2

1 ρ==LCS

WV

ρ2=

−

=

ifD

L

p

p

WW

S

C

C

ct

11

2

2 2/3 ρη

For Maximum Endurance Propeller Aircraft has to fly at that Angle of Attack such that (CL

3/2/CD) is maximum, which occurs when CL=√3 CL* and V = 0.76 V*.

−

=

ifDp

p

WW

S

Ckct

11

2

27

4

12

03max

ρη

52


Level Flight


Lift

DragThrust

Weight

∫ =−= f

i

W

WD

L

p

p

C

Ce

W

Wd

Ve

ct

1η

2. Propeller Aircraft Flight with Constant Velocity

f

i

p

p

W

W

Ve

ct ln

1η=

For Maximum Endurance Propeller Aircraft has to fly at a velocity such that e=(CL/CD) is maximum, which occurs when CL=CL

* and V = V*, which is based on initial weight Wi

4

0*

* 22

D

i

L

i

C

k

S

W

CS

WV

ρρ==

0

*

2

1

DCke =

f

i

D

i

p

p

f

i

D

i

Dp

p

f

i

p

p

W

W

CkS

W

cW

W

C

k

S

W

CkcW

W

Ve

ct ln

1

2ln

2

2

1ln

14 3

0

4

00*

*max ρ

ηρ

ηη===

53

D=TR

V

V*tmax

Slope min(PR/V)

Bréguet

Velocities for Maximum Range and Maximum Endurance of Propeller Aircraft


Graphical Finding of Maximum Range and Endurance of Jet Aircraft in Level Flight

=

⇔

V

D

D

VR

VVminmaxmax

Maximum Range

From Figure we can see that min (D/V) is obtained by taking the tangent to D graph that passes through origin.The point of tangency will give D and V for (D)min.

Maximum Endurance

∫∫<

=

<

−=−=00

111Wd

DcWd

Tct

DT

( )DD

tVVmin

1maxmax =

⇔

From Figure we can see that min (PR) is obtained by taking the PR and V for (PR)min.

Lift

DragThrust

Weight

∫∫<

−==0

WdDc

VxdR

54

PR

V

V*Rmax

0.866 V*tmax

Slope min(PR/V)

Velocities for Maximum Range and Maximum Endurance of Propeller Aircraft

Lift

DragThrust

Weight


Graphical Finding of Maximum Range and Endurance of Propeller Aircraft in Level Flight

∫∫∫∫>>>

⋅−=−=−==

000

WdVD

V

cWd

P

V

cWd

Pc

VxdR

p

p

Rp

p

p

ηη

DV

P

P

VR

V

R

VR

Vminminmaxmax =

=

⇔

Maximum Range

From Figure we can see that min (PR/V) is obtained by taking the tangent to PR graph that passes through origin.The point of tangency will give PR and V for (PR/V)min.

Maximum Endurance

∫∫<<

−=−=00

11Wd

PcWd

Pct

Rp

p

p

η

( ) ( )VDPP

tV

RV

RV

⋅==

⇔ minmin

1maxmax

From Figure we can see that min (PR) is obtained by taking the PR and V for (PR)min.

55

H.H. Hurt, Jr., “Aerodynamics for Naval Aviators “,NAVAIR 00-80T-80 1-1-1965, pg. 35


56



57



58

Flight Ceiling by the available Climb Rate- Absolute 0 ft/min- Service 100 ft/min- Performance 200 ft/min

True Airspeed

Altitude

Absolute CeilingService CeilingPerformance Ceiling

Excess Thrust provides the ability to accelerate or climb

True Airspeed

Thrust AvailableThrust

RequiredThrust

True Airspeed

Thrust

AvailableThrust

RequiredThrust

A AB B

C D

E

E

Thrust

True Airspeed

AvailableThrust

RequiredThrust

C D

Jet Aircraft Flight Envelope Determined by Available ThrustFlight Envelope: Encompasses all Altitudes

and Airspeeds at which Aircraft can Fly

Stengel, MAE331, Lecture 7, Gliding, Climbing and Turning Performance


Lift

DragThrust

Weight

Changes in Jet AircraftThrust with Altitude

59

Propeller Aircraft Ceiling Determined by Available Power

To find graphically the maximum Flight Altitude (Ceiling) for a Propeller Aircraft we use the PR (Power Required) versus V (Velocity) graph. The maximum Flight Altitude corresponds to maximum Range Rmax.

We have shown that to find Rmax we draw the Tangent Line to PR Graph, passing trough the origin.


Lift

DragThrust

Weight

Changes in Propeller Aircraft Powerand Thrust with Altitude

VC

Pa, propeller

PR

hcruiseA

h2

h1

h0

h0 < h1 <h2 < hcruise

The intersection point A with PR Graph defines the Ceiling Velocity VC, and the Pa (Available Power – function of Altitude) with this point defines the Ceiling Altitude.


60


Gliding Flight

A Glider is an unpowered airplane.

0

sin

cos

==

=

W

Vh

Vx

γγ

1<<γ0=+

=γWD

WL

.constW

Vh

Vx

==

=

γ

Lift and Drag Forces:

( ) γρρ

ρ

WCkCSVCSVD

WCSVL

LDD

L

−=+==

==

20

22

2

2

1

2

12

1

LCS

WV

ρ2=

eC

C

L

D

W

D

L

DLW 1−=−=−=−=

=γ


0sin

0cos

=+=−

γγ

WD

WLQuasi-SteadyFlight

61


Gliding Flight

We found

LCS

WV

ρ2= eC

C

L

D

W

D

L

DLW 1−=−=−=−=

=γ

Flattest Glide” (γ = γmin)

The Flattest Glide (γ = γmin) is given by:

0*

max

minmin 22

1DL CkCk

eW

D −=−=−=−=γ

e

LC*LC

*2

1

LCk

CL/CD as a function of CL

k

CC DL

0* =

The flight velocity for the Flattest Glide is given by:

4

0*..

2*

2

DL

GF C

k

S

WV

CS

WV

ρρ===

The flight velocity for the Flattest Glide is equal to the reference velocity V* or u = 1.The Flattest Glide is conducted at constant dynamic pressure.

.2

1

0*

2.. const

C

kW

C

WVSq

DL

GFG ==== ρ

62


Gliding Flight

CLEAR CONFIGURATION

LANDING CONFIGURATION

LIFT to DRAGRATIO

L/D(L/D)max

LIFT COEFFICIENT, CL

CLEAR CONFIGURATION

LANDING CONFIGURATION

RATE OFSINK

VELOCITY

(L/D)max

TANGENT TO RATE OF SINK GRAPH AT THE ORIGIN

Gliding Performance


63


Gliding Flight

We have:

Distance Covered with respect to Ground

The maximum Ground Range is covered for the Flattest Glide at the reference velocity V* or u = 1.

γVtd

hd

Vtd

xd

=

=

D

Le

V

V

hd

xd −=−===γγ1

Assuming a constant Angle of Attack during Glide, e is constant and the Ground Range R, to descend from altitude hi to altitude hf is given by:

( ) hehhehdexxR fi

h

hif

f

i

∆=−=−=−= ∫:

and

0

maxmax2 DCk

hheR

∆=∆=

e

LC*LC

*2

1

LCk


64


Gliding Flight

Rate of sink is defined as:

Rate of Sink

==⋅=−=−=

=

=2/32

22

L

D

L

D

L

L

D

W

D

CS

WV

sC

C

S

W

C

C

CS

W

W

VDV

td

hdh

L

ρργ

ρ

The term DV = PR represents the Power Required to sustain the Gliding Flight.Therefore the Rate of Sink is minimum when the Power Required is minimum, or(CD/CL

3/2) is minimum

( ) ( )0

2

3

2

3423

2

2/50

2

2/5

20

2

3

20

2/12/3

2/3

20

2/3 =−=+−=+−

=

+=

L

DL

L

LDL

L

LDLLL

L

LD

LL

D

L C

CCk

C

CkCCk

C

CkCCCCk

C

CkC

Cd

d

C

C

Cd

d

Denote by CL,m the value of Lift Coefficient CL for which (CD/CL3/2) is minimum

*0, 3

30*

L

k

CC

DmL C

k

CC

DL =

== 274

3

3

03

2/3

0

00

min

2/3D

D

DD

L

D Ck

kC

kC

kC

C

C =

+=

65


Gliding Flight

Rate of Sink

*0, 3

30*

L

k

CC

DmL C

k

CC

DL =

==0

3

max

2/3 27

4

1

DD

L

CkC

C =

We found:

The velocity Vm for glide with minimum sink rate is given by:

*4

0

76.0~

4

4

0,

76.02

3

1

3

22

*

VC

k

S

W

C

k

S

W

CS

WV

V

D

DmLm

≈

=

==

ρ

ρρ

S

CkW

C

C

S

Wh D

L

Ds ρρ 27

22 03

min

2/3min, =

=

The minimum sink rate is given by:

66


Gliding Flight

Endurance

The Endurance is the total time the glider remains in the air.

MinimumSink Rate

tmax

FlatestGlideRmax

−== 2/3

2

L

D

C

C

S

WV

td

hd

ργ

−==

D

L

C

C

W

S

V

hdtd

2/3

2

ργ

( )fiD

Lh

hD

L hhC

C

W

Shd

C

C

W

St

f

i

−

=

−= ∫

2/32/3

22

ρρ

Assuming that the Angle of Attack is held constant during the glide and ignoring the variation in density as function of altitude, we have

For Maximum Endurance the Glider has to fly at that Angle of Attack such that (CL

3/2/CD) is maximum, which occurs when CL=√3 CL

* and V = 0.76 V*.

−=

4

27

24

03max

fi

D

hh

CkW

St

ρReturn to Table of Content

67


W

LTn

+= αsin:'

W

Ln =:

20

:LD

L

D

L

CkC

C

CSq

CSq

D

Le

+===

We assume 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

e e

LC*LC

*2

1

LCk


The maximum of e is obtained for

k

CC D

L0* =

( ) 00

02

0 2** DD

DLDD Ck

CkCCkCC =+=+=

Start with

Load Factor

Total Load Number

Lift to Drag Ratio


68


e

LC*LC

*2

1

LCk


The maximum of e is obtained for

k

CC D

L0* =

( ) 00

02

0 2** DD

DLDD Ck

CkCCkCC =+=+=

*2

1

*2

1

2

1

2*

**

2200

0

LLDD

D

D

L

CkCkCkCk

C

C

Ce =====

We have WnCSVCSqL LL === 2

2

1 ρ

Let define for n = 1

=

=

==

2

4

0

*2

1:*

*:

2

*21

:*

Vq

V

Vu

C

k

S

W

CS

WV

DL

ρ

ρρ

20

:LD

L

D

L

CkC

C

CSq

CSq

D

Le

+===


69


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

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

*2

1

**** 0

0 eW

C

CCSqCSq

L

DLD ==

+=

2

22

*2 u

nu

e

WD

Therefore



70


We obtained

+=

2

22

*2 u

nu

e

WD

u 0

- - - - 0 + + + + +

D ↓ min ↑

n

u

D

∂∂

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


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

71


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 haveu

C

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

72


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 VuV

u

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

* −+−=

020 224 =+−→==

nuzupconstns

( ) ( )2

224

2

2243 23

*

*244

*

*

u

nuzu

e

V

u

nuzuuuzu

e

V

u

p

constn

s ++−=−+−−+−=∂

∂

=

0=∂∂

=constn

s

u

p 221

2uu

uuMAX <<+

nz >


nznzzu

nzzu>

−+=

−−=22

2

221

3

3 22 nzzuMAX

++=

=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

73


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

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

=→=+−=∂

∂

=

( )u

nuzu

e

Vps

224 2

*2

* −+−=


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

74


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

( )3

0412 22

22 zuzu

u

n

constps

=→=+−=∂

∂

=

( )*

*244 3

2

V

peuzu

u

n s

constps

−+−=∂

∂

=

Integrating once

uV

peuzun s

*

*22 24 −+−=

Integrating twice


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

75


Load Factor n

For 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


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 ==


77




0sin

0cos

==−−

==−

td

Vd

g

WWDT

td

dV

g

WWL

γ

γγ

Equation of Motion for Steady Climb:

γγ

sin

cos

Vh

Vx

=

=

Define the Rate of Climb:( )

sRa

C pW

PP

W

DTVVh =−=−⋅== γsin

wherePa = V T - available powerPR = V D - required powerps - excess power per unit weight

Weight

ThrustExcess

W

DT =−=γsin

C

C

WL

const

γγγ

cos

.

===

Lift

Drag

Thrust

Weight

78

Performance of an Aircraft with Parabolic PolarSOLOClimbing Aircraft Performance

LC CSVW 2

2

1cos ργ =

( )s

CD

LD

C pSV

WkCSVVT

WW

CkCSVVTh =

−−=+−

=ρ

γρρ

21

cos

2

1121

22

03

20

3

Let find the velocity V for which the Rate of Climb is maximum, for the Propeller Aircraft:

0cos2

2

312

22

02 =

+−==

SV

WkCSV

Wtd

pd

td

hdC

DsC,Prop

ργρ


For a Propeller Aircraft we assume that Pa=T V= constant.

or **

44

04

76.03

12

3

1VV

C

k

S

WV

DClimb.Prop ===

ρ

sC

DaPropC pSV

WkCSVP

Wh =

−−=

ργρ

22

03

,

cos2

2

11

We can see that the velocity at which the Rate of Climb of Propeller Aircraft is maximum is the same as the velocity at which the Required Power in Level Flight is maximum.

Lift

Drag

Thrust

Weight

79


LC CSVW 2

2

1cos ργ =

( )

−−=+−

=SV

WkCSVVT

WW

CkCSVVTh C

D

LD

C

ρ

γρρ

21

cos

2

1121

22

03

20

3

Let find the velocity V for which the Rate of Climb is maximum, for the Jet Aircraft:

0cos2

2

312

22

02 =

+−=

SV

WkCSVT

Wtd

hd CD

C

ργρ


For a Jet Aircraft we assume that T = constant.

Define

0

*

00

*

**

4

0

2:2*,*,:

2:*,

*:

D

DDD

L

D

L

D CkW

T

W

eTzCC

k

CC

C

Ce

C

k

S

WV

V

Vu =======

ρ

0cos

2

23

2 2

/1

20

0

2

2

0

2

2

=+− C

u

D

u

Dz

D V

Ck

SW

Ck

SW

VT

CkWγρ

ρ

0cos23 224 =−− Cuzu γ

Czzu γ22 cos3++=

80



ps versus the nondimensional velocity u

ps versus the velocity V

0sin ==−−td

Vd

g

WWDT γ

1

2

22

*2 =

+=

nu

nu

e

WD

Define

0

*

00

*

**

4

0

2:2*,*

,:2

:*,*

:

D

DDD

L

D

L

D

CkW

T

W

eTzCC

k

CC

C

Ce

C

k

S

WV

V

Vu

====

===ρ

+−==−=

22

*

12

2

1sin

uuz

eV

p

W

DT sγ

To find the maximum γ we must have

02

22

1sin3*

=

−−=

uu

eud

d γ

4

0

2*

max

DC

k

S

WVV

ργ ==

( ) ( )1*

*2

*2

*

11

224

, max−=−+−=

==

ze

V

u

nuzu

e

Vp

un

s γ *

,max

1sin

max

max

e

z

V

ps −==γ

γγ

1max

=γu

SOLO

81

Aircraft Flight Performance

Construction of the Specific Excess Power contours ps in the Altitude-Mach Number map for a Subsonic Aircraft below the Drag-divergence Mach Number. These contour are constructed for a fixed load factor W/S and Thrust factor T/S, if the load or thrust factor change, the ps contours will shift.

( ) ( )W

VDT

W

VDT

W

DTg

g

VV

g

VVhEps

−≈−=


sincos

sin:

In Figure (a) is a graph of Specific Excess Power contours ps versus Mach Number. Each curve is for a specific altitude h. In Figure (b) each curve is for a given Specific Excess Power ps in Altitude versus Mach Number coordinates. The points a, b, c, d, e, f for ps = 0 in Figure (a) are plotted on the curve for ps = 0 in Figure (b). Similarly all points ps = 200 ft/sec in Figure (a) on the line AB are projected on the curve ps = 200 ft/sec in Figure (b).

Specific Excess Power contours ps for a Subsonic Aircraft

Specific Excess Power contours ps

SOLO

82


Specific Excess Power contours ps for a Supersonic Aircraft

In the graphs of Specific Excess Power ps versus Mach Number Figure (a) for a Supersonic Aircraft we see a “dent” in h contour in the Transonic Region. This is due to the increase in Drag in this region.

2

In Figure (b) the graphs of Altitude versus Mach Number we see a “closed” ps = 400 ft/sec contour due to the increase in Drag in this Transonic Region.

Specific Excess Power contours ps

( ) ( )W

VDT

W

VDT

W

DTg

g

VV

g

VVhEps

−≈−=


sincos

sin:


83


Optimum Climbing Trajectories using Energy State Approximation (ESA)

We defined the Energy per unit mass E (Specific Energy):g

VhE

2:

2

+=Differentiate this equation:

( ) ( )W

VDT

W

VDT

W

DTg

g

VV

g

VVh

td

Edps

−≈−=


sincos

sin:

Minimum Time-to-Climb

The time to reach a given Energy Height Ef is computed as follows

E

Edtd

= ∫= fE

Ef E

Edt

0

The minimum time to reach the given Energy Height Ef is obtained by using at each level.

( )∫= fE

Ef E

Edt

0max

max,

( )maxE

84



Minimum Time Climb Profiles for Subsonic Speed

( ) ( )W

VDT

W

VDT

W

DTg

g

VV

g

VVhEps

−≈−=


sincos

sin:



( )maxE

Energy can be converted from potential to kinetic or vice versa along lines of constant energy in zero time with zero fuel expended. This is physically not possible so the method gives only an approximation of real paths.

SOLO

85



Minimum Time Climb Profiles for Supersonic Speed

( ) ( )W

VDT

W

VDT

W

DTg

g

VV

g

VVhEps

−≈−=


sincos

sin:


( )maxE

The optimum flight profile for the fastest time to altitude or time to speed involves climbing to maximal altitude at subsonic speed, then diving in order to get through the transonic speed range as quickly as possible, and than climbing at supersonic speeds again using .( )maxE

86

SOLO



Shaw, “Fighter Combats – Tactics and Maneuvering”

Minimum Time Climb Profiles


The minimum time to reach the given Energy Height Ef is obtained by using at each level .

( )maxE

87

SOLOClimbing Aircraft Performance


A.E. Bryson, Course “Performance Analysis of Flight Vehicles”, AA200, Stanford University, Winter 1977-1978


A.E. Bryson, Jr., “Energy-State Approximation in Performance Optimization of Supersonic Aircraft”, Journal of Aircraft, Vol.6, No. 5, Nov-Dec 1969, pp. 481-488

Approximate (ESA) Solutions.

Implicit to ESA Approximation is the possibility of instantaneous jump between kinetic to potential energy (from A to B ).This non physical situation is called a “zoom climb” or “zoom dive”.

A

B


( )maxE



“Exact” calculated using Optimization MethodsComputations


Comparison between “Exact” and Approximate(ESA) Solutions.

Implicit to ESA Approximation is the possibility of instantaneous jump between kinetic to potential energy (fromA to B , and from C to D).This non physical situation is called a “zoom climb” or “zoom dive”. We can see the “exact” solution in those cases.

A

B

C

D


( )maxE

88

A.E. Bryson, Course “Performance Analysis of Flight Vehicles”, AA200, Stanford University, Winter 1977-1978


89http://msflights.net/forum/showthread.php?1184-Supersonic-Level-Flight-Envelopes-in-FSX

F-15 Streak Eagle Time to Climb Records, which follow the ideal path to reach set altitudes in a minimal amount of time. The Streak Eagle could break the sound barrier in a vertical climb, so the ideal flightpath to 30000m involved a large Immelmann.

https://www.youtube.com/watch?v=HLka4GoUbLo https://www.youtube.com/watch?v=S7YAN9--3MAF-15 Streak Eagle Record Flights part 2F-15 Streak Eagle Record Flights part 1




https://www.youtube.com/watch?v=HLka4GoUbLo

https://www.youtube.com/watch?v=S7YAN9--3MA

https://www.youtube.com/watch?v=S7YAN9--3MA

https://www.youtube.com/watch?v=HLka4GoUbLo

90

How to climb as fast as possible

Takeoff and pull up: You want to build energy (kinetic or potential) as quickly as you can. Peak acceleration is at mach 0.9, which is the speed that energy is gained the fastest. You should first accelerate to near that speed. Avoid bleeding off energy in a high-g pull up. Start a smooth pull up before at mach 0.7-0.8 and accelerate to mach 0.9 during the pull.

http://msflights.net/forum/showthread.php?1184-Supersonic-Level-Flight-Envelopes-in-FSX



F-15 Streak Eagle Time to Climb Records, which follow the ideal path to reach set altitudes in a minimal amount of time. The Streak Eagle could break the sound barrier in a vertical climb, so the ideal flightpath to 30000m involved a large Immelmann.


Climb again: to 36000ft for maximum speed, or higher as to not exceed design limits or to save fuel for a longer run

Climb: Adjust your climb angle to maintain mach 0.9. In a modern fighter, the climb angle may be 45-60 degrees. If you need a heading change, during the pull and climb is a good time to make it.

Level off: between 25000 and 36000ft by rolling inverted. Maximum speed is reached at 36000, but remember the engines produce more thrust at higher KIAS, so slightly denser air may not hurt acceleration through the sound barrier.

Break the mach barrier: Accelerate to mach 1.25 with minimal wing loading (don't turn, try to set 0AoA)


91




Minimum Fuel-to- Climb Trajectories using Energy State Approximation (ESA)

The Rate of Fuel consumed by the Aircraft is given by:

=−=AircraftJetTc

AircraftPropellerPc

td

Wd

td

fd

T

p

We can write ( )DTV

EdW

E

Edtd

−==

The fuel consumed in a flight time , tf for a Jet Aircraft is:

( )∫∫∫ −=== fff t

Tt

T

t

f EdTDV

Wc

E

EdTctd

td

fdf

000 /1

The minimum fuel consumed in a flight time tf is obtained when using Maximum Thrust and the Mach Number that minimize the integrand:

( )∫ −= ft T

Mf Ed

TDV

Wcf

0max

min, /1minarg

for each level of E.

92




Minimum Fuel-to- Climb Trajectories using Energy State Approximation (ESA)

Assuming W nearly constant, during the climb period, contours of constant( )

max

max

Tc

DTV

T

− can be computed, as we see in the Figure

A.E. Bryson, Course “Performance Analysis of Flight Vehicles”, AA200, Stanford University, Winter 1977-1978A.E. Bryson, Jr., “Energy-State Approximation in Performance Optimization of Supersonic Aircraft”, Journal of Aircraft, Vol.6, No. 5, Nov-Dec 1969, pp. 481-488

The Minimum Fuel-to- Climb Trajectory is obtained by choosing at each state.

( )max

max

Tc

DTV

T

−

The Minimum Time-to- ClimbPath is also displayed.

Implicit to ESA Approximation is the possibility of instantaneous jump between kinetic to potential energy(from A to B) where the Total Energy is constant.

A

B


93




Maximum Range during Glide using Energy State Approximation (ESA)

Equations of motion


γ

γγ

sin

0cos

WDTVg

W

WLtd

dV

g

W

−−=

≈−=

( )W

VDTEps

−== :

g

V

W

DT −−=γsin

γγ

sin

cos

Vh

Vx

=

=

−−=g

V

W

DTVh

γγ cos

1

cos

−−===g

V

W

DT

V

h

x

h

xd

hd

During Glide we have: T = 0, W = constant, dE≤0, |γ| <<1, therefore

+−=

g

V

W

D

xd

hd

( )γcos

1

VW

DT

xd

Ed −=

2

2

1: VhE +=

( ) ( )( )EL

VED

W

VED

td

Ed −≈−=

Vtd

xd = ( )( )( )ED

EL

ED

W

Ed

xd −≈−= ( )( )( )∫∫∫ −≈−== EdED

ELEd

ED

WxdR

94




Maximum Range during Glide using Energy State Approximation (ESA)

We found


( )( )( )∫∫ −≈−= EdED

ELEd

ED

WR

Using the first integral we see that to maximize R we must choose the path that minimizes the drag D (E). The approximate optimal trajectory can be divided in:1.If the initial conditions are not on the maximum range glide path the Aircraft shall either “zoom dive” or “zoom climb” at constant E0, A to B path in Figure .2.The Aircraft will dive on the min D (E) until it reaches the altitude h = 0 at a velocity V and Specific Energy E1=V2/2, B to C in the Figure.3.Since h=0 no optimization is possible and to stay airborne one must keep the drag such that L = W, by increasing the Angle of Attack and decreasing velocity until it reaches Vstall and Es=Vstall

2/2, C to D in Figure Since h=0, d E=V dV.

( ) ( ) ( )[ ]∫ ∫∫=

=

−−−= 1

0 1

0

00min

10

max

E

E

E

Eh

pathon

E

E

s

VdVD

VWEd

ED

WEd

ED

WR


95


−+=

=

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


96


( ) ( )

( )γσσ

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

98

Vertical Plan Trajectory (σ = 0) SOLO

Prof. Earll Murman, “Introduction to Aircraft Performance and Static Stability”, September 18, 2003

99

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

100Maneuverability Diagram

R

V=:χ1'2 −= n

V

gχ

Horizontal Plan Trajectory

101F-5E Turn Performance

Horizontal Plan Trajectory

102



We can see that for n > 1

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

ρ1

1

1'

1

11'

2

2

2

2

22

−≈

−=

−≈−=

ng

V

ng

VR

nV

gn

V

gχ Horizontal Turn Rate

Horizontal Turn Radius

103


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

Horizontal Turn Rate as 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

104


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

* ≥−

=



=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

105


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

106


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


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

107


Horizontal Turn Rate as Function of ps, n 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**2

2

* u

upVe

uzu

V

g s −−+−=χ

−−+−

++−=

∂∂

1**2

2

1**

* 244

4

upVe

uzuu

upVe

u

V

g

us

sχ


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

108


Horizontal Turn Rate as Function of ps, n

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


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

109



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

SustainedTurn

InstantaneousTurn

110



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


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

111



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

=−+<<−−=−+−

−=∂∂

=


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

112



u

R

0>sp

0=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

Minimum Radius of Turn 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

113


Horizontal Turn Rate as Function of nV ,

( )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 =−

−+

−≤≤−

−−

−=


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

114



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


=

=

=

2

0

*2

1:*

*:

2:*

Vq

V

Vu

CS

kWV

D

ρ

ρ

115



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

116http://forum.keypublishing.com/showthread.php?69698-Canards-and-the-4-Gen-aircraft/page11

Example of Horizontal Turn, versus Mach, Performance of an Aircraft


117

Mirage 2000 at 15000ft.http://forums.eagle.ru/showthread.php?t=98497

Max sustained rate (at around 6.5G on the 0 Ps line) occurring at around 0.9M/450KCASlooking at around 12.5 deg sec

9G Vc (Max instant. Rate) is around 0.65M/320KCAS looking at 23.5 deg sec


http://en.wikipedia.org/wiki/File:Mirage_2000C_3-view.gif

118http://n631s.blogspot.co.il/2011/03/book-review-boyd-fighter-pilot-who.html

Example of Horizontal Turn, versus Mach, Performance of MiG-21


http://2.bp.blogspot.com/-UpcwT52fKb8/TWGFbfBdDTI/AAAAAAAABCY/l3bGEY2T_T0/s1600/e-m%2Bdiagram%2Bmig-21.jpg

SOLO

119


Comparison of Sustained ( ) Turn Performance of three Fightry AircraftsF-16, F-4 and MiG-21 at Altitude h = 11 km = 36000 ft

0=V

120


121

The black lines are the F-4D, the dark orange lines are the heavy F-4E, and the blue lines are the lightweight F-4E (same weight as F-4D). Up to low transonic mach numbers and up to medium altitudes, the F-4E is about 7% better than the F-4D (15% better with the same weight). At higher mach numbers, the F-4 doesn't have to pull as much AoA to get the same lift, so the slats actually cause a drag penalty that allows the F-4D to perform better. For reference, the F-14 is known to turn about 20% better than the unslatted F-4J. So, if the slats made the F-4S turn about 15% better, sustained turn rates would almost be pretty close between the F-14 and F-4S. The F-4E, being heavier, would still be significantly under the F-14. However, with numbers this close, pilot quality is everything rather than precise performance figures.

http://combatace.com/topic/71161-beating-a-dead-horse-us-fighter-turn-performance/

F-4


http://en.wikipedia.org/wiki/File:McDONNELL_DOUGLAS_F-4_PHANTOM_II.png

122http://www.worldaffairsboard.com/military-aviation/62863-comparing-fighter-performance-same-generations-important-factor-war-2.html

F-15F-4


smb://upload.wikimedia.org/wikipedia/commons/d/d6/F-15_Eagle_drawing.svg


123

http://www.airliners.net/aviation-forums/military/print.main?id=153429



124

Corner Speed

MaximumPositive

Capability(CL) max

MaximumNegative

Capability(CL) min

Load Factor -

n

Structural Limit

Structural Limit

Limit Airspeed

Area ofStructural Damage of

Failure

Vmin V

n

OperationalLoad Limit


StructuralLoad Limit


Typical Maneuvering EnvelopeV – n Diagram

Maneuvering Envelope:Limits on Normal Load Factor andAllowable Equivalent Airspeed-Structural Factor-Maximum and Minimum allowable Lift Coefficient-Maximum and Minimum Airspeeds-Corner Velocity: Intersection of Maximum Lift Coefficient and Maximum Load Factor


125



126R.W. Pratt, Ed., “Flight Control Systems, Practical issues in design and implementation”,AIAA Publication, 2000



127

Air-to-Air Combat

Destroy Enemy Aircraft to achieve Air Supremacy in order to prevent the enemyto perform their missions and enable to achieve tactical goals.

SOLO

See S. Hermelin, “Air Combat”, Presentation, http://www.solohermelin.com


128

http://forum.warthunder.com/index.php?/topic/110779-taktik-ve-manevralar-hakk%C4%B1ndaki-e%C4%9Fitim-g%C3%B6rselleri-oz/page-2

Air-to-Air Combat

Before the introduction of all-aspect Air-to-Air Missiles destroying an Enemy Aircraft was effective only from the tail zone of the Enemy Aircraft, so the pilots had to maneuver to reach this position, for the minimum time necessary to activate the guns or launch a Missile.



SOLO

129

Energy–Maneuverability Theory


Energy–maneuverability theory is a model of aircraft performance. It was promulgated by Col. John Boyd, and is useful in describing an aircraft's performance as the total of kinetic and potential energies or aircraft specific energy. It relates the thrust, weight, drag, wing area, and other flight characteristics of an aircraft into a quantitative model. This allows combat capabilities of various aircraft or prospective design trade-offs to be predicted and compared.

Colonel John Richard Boyd (1927 –1997)

Boyd, a skilled U.S. jet fighter pilot in the Korean War, began developing the theory in the early 1960s. He teamed with mathematician Thomas Christie at Eglin Air Force Base to use the base's high-speed computer to compare the performance envelopes of U.S. and Soviet aircraft from the Korean and Vietnam Wars. They completed a two-volume report on their studies in 1964. Energy Maneuverability came to be accepted within the U.S. Air Force and brought about improvements in the requirements for the F-15 Eagle and later the F-16 Fighting Falcon fighters

http://en.wikipedia.org/wiki/File:Boyd56.jpg

131

Turning Capability Comparison of F4E and MiG21 at Sea Level

http://forum.keypublishing.com/showthread.php?96201-fighter-maneuverability-comparison

F-4E

MiG-21



132

http://www.aviationforum.org/military-aviation/16335-fighter-maneuverability-comparison.html

F4 _Phantom versus MIG 21

MiG-21


http://www.aviationforum.org/military-aviation/16335-fighter-maneuverability-comparison.html

http://www.aviationforum.org/attachments/military-aviation/10318d1261471457-fighter-maneuverability-comparison-013b.jpg

http://www.aviationforum.org/attachments/military-aviation/10319d1261471487-fighter-maneuverability-comparison-.jpg

http://www.f-16.net/forum/download/file.php?id=4438&sid=37ceb4ccbc4e45c73b517c083bd356c4&mode=view


133


SOLO

136


In combat, a pilot is faced with a variety of limiting factors. Some limitations are constant, such as gravity, drag, and thrust-to-weight ratio. Other limitations vary with speed and altitude, such as turn radius, turn rate, and the specific energy of the aircraft. The fighter pilot uses Basic Fighter Maneuvers (BFM) to turn these limitations into tactical advantages. A faster, heavier aircraft may not be able to evade a more maneuverable aircraft in a turning battle, but can often choose to break off the fight and escape by diving or using its thrust to provide a speed advantage. A lighter, more maneuverable aircraft can not usually choose to escape, but must use its smaller turning radius at higher speeds to evade the attacker's guns, and to try to circle around behind the attacker.[13]

BFM are a constant series of trade-offs between these limitations to conserve the specific energy state of the aircraft. Even if there is no great difference between the energy states of combating aircraft, there will be as soon as the attacker accelerates to catch up with the defender. Instead of applying thrust, a pilot may use gravity to provide a sudden increase in kinetic energy (speed), by diving, at a cost in the potential energy that was stored in the form of altitude. Similarly, by climbing the pilot can use gravity to provide a decrease in speed, conserving the aircraft's kinetic energy by changing it into altitude. This can help an attacker to prevent an overshoot, while keeping the energy available in case one does occur

Energy Management

SOLO

137


Energy Management

Colonel J. R. Boyd:

In an air-to-air battle offensive maneuvering advantage will belong to the pilot who can enter an engagement at a higher energy level and maintain more energy than his opponent while locked into a maneuver and counter-maneuver duel. Maneuvering advantage will also belong to the pilot who enters an air-to-air battle at a lower energy level, but can gain more energy than his opponent during the course of the battle, From a performance standpoint, such an advantage is clear because the pilot with the most energy has a better opportunity to engage or disengage at his own choosing. On the other hand, energy-loss maneuvers can be employed defensively to nullify an attack or to gain a temporary offensive maneuvering position.

http://www.ausairpower.net/JRB/fast_transients.pdf

“New Conception for Air-to-Air Combat”, J. Boyd, 4 Aug. 1976

http://www.ausairpower.net/JRB/fast_transients.pdf

138http://www.alr-aerospace.ch/Performance_Mission_Analysis.php

F-16


smb://upload.wikimedia.org/wikipedia/commons/9/91/GENERAL_DYNAMICS_F-16_FIGHTING_FALCON.svg

139Comparative Ps Diagram for Aircraft A and Aircraft B. Two Multi-Role Jet Fighters


140

http://www.simhq.com/_air/air_065a.html

http://en.wikipedia.org/wiki/Lavochkin_La-5

Comparison of Turn Performance of two WWII Fighter Aircraft: Russian Lavockin La5 vs German Messershmitt Bf 109

http://en.wikipedia.org/wiki/Messerschmitt_Bf_109



141

Comparison of Turn Performance of two WWII Fighter Aircraft: Russian Lavockin La5 vs German Messershmitt Bf 109

http://en.wikipedia.org/wiki/Lavochkin_La-5http://en.wikipedia.org/wiki/Messerschmitt_Bf_109




142F-86F Sabre and MiG-15 performance comparison

North American F-86 Sabre

MiG-15


http://en.wikipedia.org/wiki/File:North_American_F86-01.JPG

143Falcon F-16C versus Fulcrum MIG 29,

left is w/o afterburner, right is with it, fuel reserves 50%

http://forum.keypublishing.com/showthread.php?47529-MiG-29-kontra-F-16-(aerodynamics-)

FulcrumMiG-29F-16


http://en.wikipedia.org/wiki/File:Serbian_mig-29_missiles.jpg

http://www.google.co.uk/url?sa=i&source=images&cd=&cad=rja&uact=8&ved=0CAgQjRw&url=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FGeneral_Dynamics_F-16_Fighting_Falcon&ei=2X15VP2OIeXMygOMiYCoDQ&psig=AFQjCNECyaZ92yhtL8ohXOYjLwejtTPJBQ&ust=1417334617656380

144

Comparison of Turn Performance of two Modern Fighter Aircraft: Russian MiG-29 vs USA F-16

FulcrumMiG-29

F-16


http://www.evac-fr.net/forums/lofiversion/index.php?t984.html





145


FulcrumMiG-29

F-16







146



Fulcrum MiG-29

F-16




http://www.simhq.com/_air/air_012d.html



147http://www.simhq.com/_air3/air_117e.html

While the turn radius of both aircraft is very similar, the MiG-29 has gained a significant angular advantage.


MiG-29F-16


http://www.simhq.com/_air3/air_117e.html

http://www.simhq.com/_air3/air_117j.html

http://www.simhq.com/_air3/air_117k.html



148



With afterburner, fuel reserves 50%Without afterburner, fuel reserves 50%

MiG-29

F-16





149

http://forums.eagle.ru/showthread.php?t=30263


http://forums.eagle.ru/showthread.php?t=30263

150

An assessment is made of the applicability of Energy Maneuverability techniques (EM)to flight path optimization. A series of minimum time and fuel maneuvers using the F-4Caircraft were established to progressively violate the assumptions inherent in the EM programand comparisons were made with the Air Force Flight Dynamics Laboratory's (AFFDL)Three-Degree-of-Freedom Trajectory Optimization Program and a point mass option of theSix-Degree-of-Freedom flight path program. It was found the EM results were always optimisticin the value of the payoff functions with the optimism increasing as the percentageof the maneuver involving constant energy transitions Increases. For the minimum timepaths the resulting optimism was less than 27%f1o r the maneuvers where the constant energypercentage was less than 35.',", followed by a rather steeply rising curve approaching in thelimit 100% error for paths which are comprised entirely of constant energy transitions. Twonew extensions are developed in the report; the first is a varying throttle technique for useon minimum fuel paths and the second a turning analysis that can be applied in conjunctionwith a Rutowski path. Both extensions were applied to F-4C maneuvers in conjunction with'Rutowski’s paths generated from the Air Force Armament Laboratory's Energy Maneuverabilityprogram. The study findings are that energy methods offer a tool especially useful inthe early stages of preliminary design and functional performance studies where rapidresults with reasonable accuracy are adequate. If the analyst uses good judgment in its applicationsto maneuvers the results provide a good qualitative insight for comparative purposes.The paths should not, however, be used as a source of maneuver design or flightschedule without verification especially on relatively dynamic maneuvers where the accuracyand optimality of the method decreases.

David T. Johnson, “Evaluation of Energy Maneuverability Procedures in Aircraft Flight Path Optimization and Performance Estimation”, November 1972, AFFDL-TR-72-53


151

Lockheed F-104 Starfighter


152Typical Ps Plot for Lockheed F-104 Starfighter



153


F-104 Flight Envelope


154F-104A flight envelope




155

http://defence.pk/threads/design-characteristics-of-canard-non-canard-fighters.178592/



http://defence.pk/threads/design-characteristics-of-canard-non-canard-fighters.178592

156

http://defence.pk/threads/design-characteristics-of-canard-non-canard-fighters.178592/


http://defence.pk/threads/design-characteristics-of-canard-non-canard-fighters.178592

157http://img138.imageshack.us/img138/4146/image4u.jpg


http://img138.imageshack.us/img138/4146/image4u.jpg

158

https://s3-eu-west-1.amazonaws.com/rbi-blogs/wp-content/uploads/mt/flightglobalweb/blogs/the-dewline/assets_c/2011/05/chart%20combat%20radius-thumb-500x375-125731.jpg




https://s3-eu-west-1.amazonaws.com/rbi-blogs/wp-content/uploads/mt/flightglobalweb/blogs/the-dewline/assets_c/2011/05/chart%20combat%20radius-thumb-500x375-125731.jpg

http://www.flightglobal.com/blogs/the-dewline/2011/05/12/chart%20combat%20radius.jpg



159

Supermaneuverability is defined as the ability of an aircraft to perform high alpha maneuvers that are impossible for most aircraft is evidence of the aircraft's supermaneuverability. Such maneuvers include Pugachev's Cobra and the Herbst maneuver (also known as the "J-turn").Some aircraft are capable of performing Pugachev's Cobra without the aid of features that normally provide post-stall maneuvering such as thrust vectoring. Advanced fourth generation fighters such as the Su-27, MiG-29 along with their variants have been documented as capable of performing this maneuver using normal, non-thrust vectoring engines. The ability of these aircraft to perform this maneuver is based in inherent instability like that of the F-16; the MiG-29 and Su-27 families of jets are designed for desirable post-stall behavior. Thus, when performing a maneuver like Pugachev's Cobra the aircraft will stall as the nose pitches up and the airflow over the wing becomes separated, but naturally nose down even from a partially inverted position, allowing the pilot to recover complete control.

http://en.wikipedia.org/wiki/Supermaneuverability

Supermaneuverability


http://en.wikipedia.org/wiki/Supermaneuverability

160


161

Sukhoi Su-30MKI


http://vayu-sena.tripod.com/interview-simonov1.html

http://vayu-sena.tripod.com/interview-simonov1.html

162


The Herbst maneuver or "J-Turn" named after Wolfgang Herbst is the only thrust vector post stall maneuver that can be used in actual combat but very few air frames can sustain the stress of this violent maneuver.

Herbst Maneuver

http://en.wikipedia.org/wiki/Herbst_maneuver


http://en.wikipedia.org/wiki/Herbst_maneuver

163

Constraint AnalysisSOLO Aircraft Flight Performance

The Performance Requirements can be translated into functional relationship between the Thrust-to-Weight or Thrust Loading at Sea Level Takeoff (TSL/WTO) and the Wing Loading at Takeoff (WTO/S). The keys to the development are•Reasonable assumption hor Aircraft Lift-to-Drag Polar.•The low sensibility of Engine Thrust with Flight Altitude and Mach Number.

The minimum of TSL/WTO as functions of WTO/S are required for:•Takeoff from a Runway of a specified length.•Flight at a given Altitude and Required Speed.•Climb at a Required Speed.•Turn at a given Altitude, Speed and a required Rate.•Acceleration capability at constant Altitude.•Landing without reverse Thrust on a Runway of a given length.

164

Energy per unit mass E

Let define Energy per unit mass E:g

VhE

2:

2

+=Let differentiate this equation:

( ) ( )W

VDT

W

VDT

W

DTg

g

VV

g

VVhEps

−≈−=


sincos

sin:

define

10 ≤<= ββ TOWW WTO – Take-off Weight

( ) ( ) ( ) SLThThhT αα === 0 TSL – Sea Level Thrust

V

p

W

D

W

T s+=

Load FactorW

CSq

W

Ln L==:


TOL WSq

nW

Sq

nC β==

+=

V

p

W

D

W

T s

TO

SL

αβ

Constraint Analysis

165


General Mission Description of a Typical Fighter Aircraft

10: ≤<= ββTOW

W

WTO – Take-off Weight

W – Aircraft Weight during Flight

Constraint Analysis

166

Assume a General Lift-to-Drag Polar Relationship

Total DragRD CSqCSqRD +=+

D, CD - Clean Aircraft Drag and Drag Coefficient

R, CR – Additional Aircraft Drag and Additional Drag Coefficient caused by External Stores, Bracking Parachute, Flaps, External Hardware

02

2

1022

1 DTOTO

DLLD CS

W

q

nK

S

W

q

nKCCKCKC +

+

=++= ββ

TOL WSq

nW

Sq

nC β==

( )

++=V

pCC

W

Sq

W

T sRD

TOTO

SL

βαβ

+

++

+

=

V

pCC

S

W

q

nK

S

W

q

nK

W

Sq

W

T sDRD

TOTO

TOTO

SL02

2

1

βββα

β


Constraint Analysis

167

( )WLntd

Vd

td

hd ==== ,1,0,0

Case 1: Constant Altitude/Speed Cruise (ps = 0)

Given:

+++

=

SW

q

CCK

S

W

qK

W

T

TO

DRDTO

TO

SL

ββ

αβ 0

21

We obtain:

We can see that TSL/WTO → ∞ for WTO/S → 0 and WTO/S→∞, therefore a minimum exist. By differentiating TSL/WTO with respect to WTO/S and setting the result equal to zero, we obtain:

1

0

/min K

CCq

S

W DRD

WT

TO +=

β

( )[ ]210

min

2 KKCCW

TDRD

TO

SL ++=

αβ

Lift

DragThrust

Weight


Constraint Analysis


Case 1: Constant Altitude/Speed Cruise (ps = 0)


Constraint Analysis

169

( )WLntd

hd ≈≈= ,1,0

Case 2: Constant Speed Climb (ps = dh/dt)

Given:


1

0

/min K

CCq

S

W DRD

WT

TO +=

β

( )

+++=

td

hd

VKKCC

W

TDRD

TO

SL 12 210

minαβ

We obtain:

+

+++

=

td

hd

VSW

q

CCK

S

W

qK

W

T

TO

DRDTO

TO

SL 1021 β

βαβ


170

,1,0,0,,

>== ntd

hd

td

Vd

givenhVgivenhV

Case 3: Constant Altitude/Speed Turn (ps = 0)

Given:


1

0

/min K

CC

n

q

S

W DRD

WT

TO +=

β

( )[ ]210

min

2 KKCCn

W

TDRD

TO

SL ++=

αβ

We obtain:

+

+++

=

td

hd

VSW

q

CCnK

S

W

qnK

W

T

TO

DRDTO

TO

SL 102

21 β

βαβ

2

0

22

0

11

+=

Ω+=cRg

V

g

Vn


Constraint Analysis

171

( )WLntd

hd

givenh

=== ,1,0

Case 4: Horizontal Acceleration (ps = (V/g0) (dV/dt) )

Given:

We obtain:

+

+++

=

td

Vd

gSW

q

CCK

S

W

qK

W

T

TO

DRDTO

TO

SL

0

021

1β

βαβ


Lift

DragThrust

Weight

This can be rearranged to give:

+++

=

SW

q

CCK

S

W

qK

W

T

td

Vd

g TO

DRDTO

TO

SL

ββ

βα 0

210

1

Constraint Analysis

172

( )WLntd

hd

givenh

=== ,1,0

Case 4: Horizontal Acceleration (ps = (V/g0) (dV/dt) ) (continue – 1)

Given:


Lift

DragThrust

Weight

We obtain:

+++

=

SW

q

CCK

S

W

qK

W

T

td

Vd

g TO

DRDTO

TO

SL

ββ

βα 0

210

1

This equation can be integrated from initial velocity V0 to final velocity Vf, from initial t0 to final tf times.

( )∫=− fV

Vs

f Vp

VdV

gtt

00

0

1

where

+++

−=

SW

q

CCK

S

W

qK

W

TVp

TO

DRDTO

TO

SLs β

ββα 0

21

The solutions of TSL/WTO for different WTO/S are obtained iteratively.

Constraint Analysis

173http://elpdefensenews.blogspot.co.il/2013_04_01_archive.html

Constraint AnalysisSOLO Aircraft Flight Performance

http://2.bp.blogspot.com/-y15BW5Vx-Fw/UWQsdWchS7I/AAAAAAAAC9w/UBNSOe5zG0c/s1600/JSCFADT-2012-Sub-2-p16.png

174

0=givenh

td

hd

Case 5: Takeoff (sg given and TSL >> (D+R) )

Given:


Ground RunV = 0

sg

sTO

sr str

V TORotation

Transition

sCL

θ CL

htr

hobs

R

Start from:

( )

TO

T

SL

s W

VRDT

td

Vd

g

V

td

hdp

SL

β

αα

+−

≈+=

≈

0

==

TO

SL

V

W

Tg

td

sd

sd

Vd

td

Vd

βα 0

/1

VdVT

W

gsd

SL

TO

=

0αβ

max,2

2

0max,2

0 2

1

2

1L

TO

TOLstallstallTO CS

k

VCSVLW ρρβ ===

The take-off velocity VTO is VTO = kTO Vstall

Where Vstall is the minimum velocity at at which Lift equals weight and kTO ≈ 1.1 to 1.2:

==S

W

C

kVk

V TO

L

TOstallTO

TO

max,0

222

2

22 ρβ

Integration from:

s = 0 to s = sg

V = 0 to V = VTO

2

2

0

TO

SL

TOg

V

T

W

gs

=

αβ

sg – Ground Run

Constraint Analysis

175

Case 5: Takeoff (sg given and TSL >> (D+R) ) (continue – 1)


Ground RunV = 0

sg

sTO

sr str

V TORotation

Transition

sCL

θ CL

htr

hobs

R

2

2

0

TO

SL

TOg

V

T

W

gs

=

αβ

==S

W

C

kVk

V TO

L

TOstallTO

TO

max,0

222

2

22 ρβ

=S

W

Cgs

k

W

T TO

Lg

TO

TO

SL

max,00

22

ρβ

αβ

We obtained:

from which:

=

S

W

C

k

T

W

gs TO

L

TO

SL

TOg

max,0

2

0 ρβ

αβ

We have a Linear Relation between TSL/WTO and WTO/S

Constraint Analysis

176

Case 6: Landing


where ( ) ( )

−=

−−=

µ

µβ

ρ

W

Tgc

CCSW

ga grLgrD

TO

0

,,

:

/2:

cab

Vaa

ca

Vaa

touchdown

4

2:

4

2:

2

11

−=

−=cVa

cVa

asg +

+−=2

2

21ln

2

1

−−⋅

++

−=

1

2

2

1

1

1

1

1ln

4

1

a

a

a

a

catg

Ground Run Phase

We foundGround Run sgr

Transition

Airborne Phase


Float

sfFlare stGlide sg

γ

hg

hf

Touchdown

20 VCVBTT ++=

For a given value of sg , there is only one value of WTO/S that satisfies this equation.

( )gTO sfSW =/

This constraint is represented in the TSL/WTO versus WTO/S plane as a vertical line, at WTO/S corresponding to the required sg.

Constraint Analysis

177Constraint Diagram


+++

=

S

W

q

CCK

S

W

qK

W

T

TO

DRDTO

TO

SL

ββ

αβ 0

21

+

+++

=

td

hd

VS

W

q

CCnK

S

W

qnK

W

T

TO

DRDTO

TO

SL 102

21 β

βαβ

=S

W

Cgs

k

W

T TO

Lg

TO

TO

SL

max,00

22

ρβ

αβ

( )gTO sfSW =/

Constraint Analysis

178

Comparison of Fighter Aircraft Propulsion SystemsSOLO

179

Comparison of Fighter Aircraft Propulsion SystemsSOLO

180


Composite Thrust Loading versus Wing Loading – for different Aircraft

Constraint Analysis

181Constraint Diagram for F-16


Constraint Analysis


182Weapon System Agility

Weapon System Agility


183

References

SOLO

Miele, A., “Flight Mechanics , Theory of Flight Paths, Vol I”, Addison Wesley, 1962


J.D. Anderson, Jr., “Introduction to Flight”, McGraw Hill, 1978, Ch. 6, “Elements of Airplane Performance”

A. Filippone, “Flight Performance of Fixed and Rotary Wing Aircraft”, Elsevier, 2006

M. Saarlas, “Aircraft Performance”, John Wiley & Sons, 2007

Stengel, MAE 331, Aircraft Flight Dynamics, Princeton University

J.D. Anderson, Jr., “Aircraft Performance and Design”, McGraw Hill, 1999

N.X. Vinh, “Flight Mechanics of High-Performance Aircraft”, Cambridge University Press, 1993

F.O. Smetana, “Flight Vehicle Performance and Aerodynamic Control”, AIAA Education Series, 2001

L. George, J.F. Vernet, “La Mécanique du Vol, Performances des Avions et des Engines”, Librairie Polytechnique Ch. Béranger, 1960

L.J. Clancy, “Aerodynamics”, Pitman International Text, 1975

184

Brandt, “Introduction to Aerodynamics – A Design Perspective”, Ch. 5 , Performance and Constraint Analysis


J.D. Mattingly, W.H. Heiser, D.T. Pratt, “Aircraft Engine Design”, 2nd Ed., AIAA Education Series, 2002

Prof. Earll Murman, “Introduction to Aircraft Performance and Static Stability”, September 18, 2003

Naval Air Training Command, “Air Combat Maneuvering”, CNATRA P-1289 (Rev. 08-09)

Patrick Le Blaye, “Agility: Definitions, Basic Concepts, History”, ONERA

Randal K. Liefer, John Valasek, David P. Eggold, “Fighter Aircraft Metrics, Research , and Test”, Phase I Report, KU-FRL-831-2

References (continue – 1)

B. N. Pamadi, “Performance, Stability, Dynamics, and Control of Airplanes”, AIAA Educational Series, 1998, Ch. 2 , Aircraft Performance

L.E. Miller, P.G. Koch, “Aircraft Flight Performance”, July 1978, AD-A018 547,AFFDL-TR-75-89

185

Courtland_D._Perkins,_Robert_E._Hage, “Airplane Performance Stability and Control”, John Wiley & Sons, 1949

SOLO

Asselin, M., “Introduction to Aircraft Aerodynamics”, AIAA Education Series, 1997

Aircraft Flight PerformanceReferences (continue – 2)

Donald R. Crawford, “A Practical Guide to Airplane Performance and Design”,Crawford Aviation, 1981

Francis J. Hale, “ Introduction to aircraft performance, Selection and Design”, John Wiley & Sons, 1984

J. Russell, ‘Performance and Stability of Aircraft“, Butterworth-Heinemann, 1996

Jan Roskam, C. T. Lan, “Airplane Aerodynamics and Performance”, DARcorporation, 1997

Nono Le Rouje, “Performances of light aircraft”, AIAA, 1999

Peter J. Swatton, “Aircraft performance theory for Pilots”, Blackwell Science, 2000

S. K. Ojha, “Flight Performance of Aircraft “, AIAA, 1995

W. Austyn Mair, David L._Birdsall, “Aircraft Performance”, Cambridge University Press, 1992

186

SOLO

E.S. Rutowski, “Energy Approach to the General Aircraft Performance Problem”, Journal of the Aeronautical Sciences, March 1954, pp. 187-195

Aircraft Flight PerformanceReferences (continue – 3)

A.E. Bryson, Jr., “Applications of Optimal Control Theory in Aerospace Engineering”, Journal of Spacecraft and Rockets, Vol. 4, No.5, May 1967, pp. 553

W.C. Hoffman, A.E. Bryson, Jr., “A Study of Techniques for Real-Time, On-Line Optimum Flight Path Control”, Aerospace System Inc., ASI-TR-73-21, January 1973, AD 758799

A.E. Bryson, Jr., “A Study of Techniques for Real-Time, On-Line Optimum Flight Path Control. Algorithms for Three-Dimensional Minimum-Time Flight Paths with Two State Variables”, AD-A008 985, December 1974

M.G. Parsons, A.E. Bryson, Jr., W.C. Hoffman, “Long-Range Energy-State Maneuvers for Minimum Time to Specified Terminal Conditions”, Journal of Optimization Theory and Applications, Vol.17, No. 5-6, Dec 1975, pp. 447-463

A.E. Bryson, Jr., M.N, Desai, W.C. Hoffman, “Energy-State Approximation in Performance Optimization of Supersonic Aircraft”, Journal of Aircraft, Vol.6, No. 6, Nov-Dec 1969, pp. 481-488

187


References (continue – 4)

Solo Hermelin Presentations http://www.solohermelin.com

• Aerodynamics Folder

• Propulsion Folder

• Aircraft Systems Folder




188

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 –

Stanford University1983 – 1986 PhD AA

http://upload.wikimedia.org/wikipedia/en/a/ae/RAFAEL_logo.png

SOLO

189

OODA loop


Colonel John Richard Boyd (1927 –1997)

The OODA loop (for Observe, Orient, Decide, and Act) is a concept originally applied to the combat operations process, often at the strategic level in military operations. The concept was developed by military strategist and USAF Colonel John Boyd.

http://en.wikipedia.org/wiki/File:OODA.Boyd.svg

http://en.wikipedia.org/wiki/File:JohnBoyd_Pilot.jpg

190

http://www.google.com/url?sa=i&source=images&cd=&docid=Ngbzu7gHUmsu0M&tbnid=pALgqMeZqugo2M:&ved=0CAgQjRw4KA&url=https%3A%2F%2Fwww.grc.nasa.gov%2Fwww%2FBGH%2Frealgas.html&ei=4YzWUo_HIomShQe3loHABg&psig=AFQjCNHTT4CGp_B4R-aq7sCET5laPuMsIQ&ust=1389878881773908


Comparison Tables

193


SOLO

194

Aircraft Avionics

195Ray Whitford, “Design for Air Combat”

R.W. Pratt, Ed., “Flight Control Systems, Practical issues in design and implementation”,AIAA Publication, 2000

SOLO

196


199

Ray Whitford, “Design for Air Combat”Northrop F-5 Freedom Fighter

The Northrop F-20 Tigershark (initially F-5G) was a privately financed light fighter, designed and built by Northrop. Its development began in 1975 as a further evolution of Northrop's F-5E Tiger II, featuring a new engine that greatly improved overall performance, and a modern avionics suite including a powerful and flexible radar. Compared with the F-5E, the F-20 was much faster,

F-20 Powerplant: 1 × General Electric F404-GE-100 turbofan, 17,000 lbf (76 kN)

•F-5 Powerplant: 2 × General Electric J85-GE-21B turbojet • Dry thrust: 3,500 lbf (15.5 kN) each• Thrust with afterburner: 5,000 lbf (22.2 kN) each

The rear fuselage presented a problem, however, since the F-5 is, along with other twin-engined aircraft, characterised by a wide, very flat belly. This also contributes favourably to high-AOA performance. The question of how to reconcile this with a single engine basically circular in section was solved by adding shelves, not unlike those on the F-16, aft of the wing trailing edge to flatten the aft underbody. The increased skin friction drag was a small price to pay to lessen the risks of the radical change represented by the switch from a twin to a single-engined layout. The shelves house the horizontal tail control runs.

The diagram shows flight envelopes for two aircraft, the Northrop F-5E and F-20, at two load factors, 1g and 4g. Several points stand out:1 The F-20 is a Mach 2 aircraft and displays a significantly extended high-speed envelope whereas the F-5E is limited to Mach 1.64 at typical combat weight, albeit at the same 11,000 m altitude.2 Sustained maneuvering (the 4g load factor case is shown) greatly curtails the flight envelope for both aircraft as a result of the large increase in lift-dependent drag.3 The primary air battle zone, shown shaded, is limited to subsonic speeds. This is because the pilots tend to fly their aircraft to maximize turn rate — so as to gain an angular position on their opponents — and this can currently be achieved only at subsonic speed. In Vietnam and subsequent conflicts in which adversaries had Mach 2+ aircraft at their command, the combat speed range was predominantly Mach 0.5–0.9, with very little time being spent above Mach 1.1.

http://en.wikipedia.org/wiki/Turbojet

200http://www.simhq.com/_air3/air_117c.html

F-16

http://www.simhq.com/_air3/air_117e.html


201http://www.simhq.com/_air3/air_117c.html

F-16

http://www.simhq.com/_air3/air_117f.html


202

http://forum.warthunder.com/index.php?/topic/174942-wing-loading-and-turning/

The three most important (but far from the only) things to consider about an aircraft's turning performance are shown and explained in relation to the P-51's EM (energy/maneuverability) diagram below;

203

http://forum.keypublishing.com/showthread.php?129077-A-quot-Rough-quot-F-35-Kinematics-Analysis/page2

F-15 FlightF-15


204

http://forum.keypublishing.com/showthread.php?129077-A-quot-Rough-quot-F-35-Kinematics-Analysis/page2

F-15 Drag

F-15


206


207


208


209


210


211


212


213


214


215


216


Level Flight

217Stengel, MAE331, Lecture 7, Gliding, Climbing and Turning Performance

220

Corner Speed

MaximumPositive

Capability(CL) max

MaximumNegative

Capability(CL) min

Load Factor - n

Structural Limit

Structural Limit

Limit Airspeed

Area ofStructural Damage of

Failure

Vmin V

n







Corner Velocity Turn

• Corner Velocity

SC

WnV

Lcorner ρ

max

max2=

• For Steady Climbing or Diving Flight

W

DT −= maxsin γ

• Turning Radius

γγ22

max

22

maxcos

cos

−=

ng

VR

• Turning Rate

( )

γ

γγ

χ

cos'

1

cos'

0

2

−=

−=

=

ng

VR

nV

g

( )γ

γcos

cos22max

V

ngcorner

−=Ω

• Time to Complete a Full Circle

γ

γπ

22max

2cos

cos

−=

ng

Vt

221http://www.worldaffairsboard.com/military-aviation/62863-comparing-fighter-performance-same-generations-important-factor-war-2.html

222

http://www.iitk.ac.in/aero/fltlab/cruise.html

http://www.zweefportaal.nl/main/forum/viewthread.php?thread_id=2537&rowstart=0

223

http://selair.selkirk.bc.ca/training/aerodynamics/range_prop.htm

Effect of Altitude on Specific Range and Endurance

Maximum Range at L/Dmax

How Wind Affects Range and Optimum Cruise Speed

SOLO

224


Drag

SOLO

225


Drag

SOLO

227


Drag

228

http://elementsofpower.blogspot.co.il/2013_04_01_archive.html


229http://elementsofpower.blogspot.co.il/2013_04_01_archive.html


230



http://4.bp.blogspot.com/-KYwVF2CxCdA/UW4f60wMCtI/AAAAAAAACR4/JuYWUU3soB8/s1600/f18_turn_rate-web.jpg

231http://www.f-16.net/forum/viewtopic.php?t=5487

F-16

Comparison of Climb Performance of F-16 and F-$





232http://indiandefence.com/threads/comparing-modern-western-fighters.41124/page-16

F-15


233http://indiandefence.com/threads/comparing-modern-western-fighters.41124/page-16

234

Cobra Turn

http://defence.pk/threads/supermaneuverability.39916/

Immelmann turn

Split S

Roller

Scissors

235

http://www.f-16.net/forum/viewtopic.php?t=13114

F-15

F-22

F-15 versus F-22

236

F-15

F-22

237http://forum.keypublishing.com/showthread.php?72673-Boyd-s-E-M-Theory

238http://defence.pk/threads/cope-india-how-the-iaf-rewrote-the-rules-of-air-combat.300282/page-3

239

http://www.f-16.net/forum/viewtopic.php?t=5487

240

F-16

241

F-16

243

F-16

SOLO

244


Performance in Level Flight

SOLO

245


SOLO

246


Determination of Maximum Flight Altitude in Level Flight

247

http://selair.selkirk.bc.ca/training/aerodynamics/range_jet.htm

248


Air-to-Air Combat

249


Air-to-Air Combat

250


Air-to-Air Combat

251Configurations Evolution

252

North American P-51 Mustang

253Ps Diagram for a Multi-Role Fighter Aircraft at n = 1

254Ps Diagram for a Multi-Role Fighter Aircraft at n = 5

255


256


257

Generic E/M Diagram

14 fixed wing fighter aircraft- flight performance - ii

Science

Transcript of 14 fixed wing fighter aircraft- flight performance - ii