the master equation dr. dimitri mavris boeing associate professor for advanced aerospace systems...

The Master Equation

Dr. Dimitri MavrisBoeing Associate Professor for Advanced Aerospace Systems

Analysis

Compiled by the Loyal FYGS of the Class of 2003.

The Master Equation

• Derived from Energy Considerations.

2

2V

dt

d

g

W

dt

dhWVRDT

o

(2.1)

Rate of Mechanical EnergyInput

Storage RateOf Potential Energy

Storage RateOf Kinetic Energy

The Master Equation

• The master equation is most useful in its dimensionless form.

dt

dz

Vg

Vh

dt

d

VW

RDT e

o

1

2

1 2

(2.2a)

The Master Equation

• The left hand side, when multiplied by Velocity (V), becomes the equation for “weight specific excess power.”

dt

dz

g

Vh

dt

dP e

os

2

2

(2.2b)

The Master Equation

• Correct thrust using lapse rate, – T=TSL

• Correct weight using fuel/payload correction, .– W=WSL

oTOTO

SL

g

Vh

dt

d

VW

RD

W

T

2

1 2

(2.5)

(2.4)

(2.3)

Use Lift Relationships to find CL

SqCnWL L

Where n is the “load factor” (number of g’s). n=1 for straight and level flight.

(2.6)

S

W

q

n

qS

nWC TO

L

(2.8)

Solving for CL and substituting equation 2.4 to get the takeoff wing loading.

Use the Equations for Drag

SqCD D

Equation 2.7 yields the relationship between drag coefficient and drag.

(2.7)

oDLLD CCKCKC 22

1 (2.8)

The parabolic lift-drag polar equation is given above in equation 2.9.

Combine the CL, CD, and D Equations

DqSCD (2.7)

oDLL CCKCKqSD 2

21

Substitute Equation 2.9 into Equation 2.7 for CD.

oDTOTO CS

W

q

nK

S

W

q

nKqSD

2

2

1

Then substitute Equation 2.8 for CL in the above Equation to yield Equation 2.10

(2.9b)

(2.10)

Back to the Master Equation

oD

TOTO

TOTO

SL

g

Vh

dt

d

VqS

RC

S

W

q

nK

S

W

q

nK

W

qS

W

To 2

1 2

2

2

1

(2.11)

oTOTO

SL

g

Vh

dt

d

VW

RD

W

T

2

1 2

(2.5)

Now, substitute Equation 2.10 back into Equation 2.5 (the Master Equation) for D.

The above form of the equation can be manipulated to set the constraints on thedesign plot.

Using the Master Equation• Case 1: Constant Altitude/Speed Cruise• Case 2: Constant Speed Climb• Case 3: Constant Altitude/Speed Turn• Case 4: Horizontal Acceleration• Case 5: Takeoff Ground Roll (lots of Thrust)

• Case 6: Takeoff Ground Roll (not so much thrust)– A: Do NOT Clear the Obstacle During Transition– B: Clear the Obstacle During Transition

• Case 7: Braking Roll• Case 8: Service Ceiling

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

oD

TOTO

TOTO

SL

g

Vh

dt

d

VqS

RC

S

W

qK

S

W

qK

W

qS

W

To 2

111 2

2

2

1

Assumptions:dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightR=0 Not on the groundh & v Values are Given

Assumptions:dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightR=0 Not on the groundh & v Values are Given

oDTOTO

TOTO

SL CS

W

qK

S

W

qK

W

qS

W

T

2

2

1(2.11b)


oDTOTO

TOTO

SL CS

W

qK

S

W

qK

W

qS

W

T

2

2

1 (2.11b)

Equation 2.11b can be simplified by canceling WTO and qS

SW

q

CK

S

W

qK

W

T

TO

DTO

TO

SL o

21 (2.12)

Equation 2.12, a function of wing loading and 1/wing loading, is used for case 1.


For very large and very small values of wing loading, the T/W ratio goes to infinity.To find the point at which the T/W ratio is minimum, take the partial derivative of Equation 2.12 with respect to WTO/S and set it equal to zero.

SW

q

CK

S

W

qK

SW

d

d

TO

DTO

TO

o

210

1K

Cq

S

WoDTO

Wing loading for

minimum T/W


Also, for very large q (when CD=CDo), a hyperbolic relationship between T/W and W/S is observed. If q is extremely large, the first term goes to zero.

SW

q

CK

S

W

qK

W

T

TO

DTO

TO

SL o

21(2.12)

Term 1

SW

q

C

W

T

TO

D

TO

SL o

K2 very small

oDTO

TO

SLqC

S

W

W

T

(2.13)

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

oD

TOTO

TOTO

SL

g

Vh

dt

d

VqS

RC

S

W

qK

S

W

qK

W

qS

W

To 2

111 2

2

2

1

Assumptions:dV/dt =0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundh, dh/dt & v Values are Given


(2.14)

dt

dh

VS

Wq

CK

S

W

qK

W

T

TO

DTO

TO

SL o1

21


(2.14)

dt

dh

VS

Wq

CK

S

W

qK

W

T

TO

DTO

TO

SL o1

21

SW

q

CK

S

W

qK

W

T

TO

DTO

TO

SL o

21 (2.12)

Note that the only difference between Case 1 and Case 2 is the final term, for climb rate.

Case 1:

Case 2:


(2.14)

dt

dh

VS

Wq

CK

S

W

qK

W

T

TO

DTO

TO

SL o1

21

Since the final term is not a function of W/S, taking the derivative as was done in case 1 will yield the same W/S value at which the minimum T/W occurs; however, the value of T/W at this point will be different based on the desired rate of climb.

1K

Cq

S

WoDTO

Wing loading for

minimum T/W

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

oD

TOTO

TOTO

SL

g

Vh

dt

d

VqS

RC

S

W

q

nK

S

W

q

nK

W

qS

W

To 2

1 2

2

2

1

Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundh, n, & V Values are Given

Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundh, n, & V Values are Given

(2.15)

SW

q

CnK

S

W

qnK

W

T

TO

DTO

TO

SL o

22

1


1K

C

n

q

S

WoDTO

Wing loading for

minimum T/W

The behavior of Equation 2.15 is the same as that of Case 1 and Case 2. The point of minimum thrust/weight is now a function of the load factor, n.


The relationships for n in the turn can be calculated from the Pythagorean theorem and centripetal force equations.

2

12

1

og

Vn

2

12

1

coRg

Vn(2.16) (2.17)

Where is the rotation rate and Rc is the radius of curvature for a load factor, n.

RcW

L=nW

Bank angle(W/go)V=(W/go)V2/Rc

Case 4: Horizontal Acceleration

dt

dV

gS

Wq

CK

S

W

qK

W

T

TO

DTO

TO

SL

021

10

dtgVdVPs 0/Assumptions:dV/dt =0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundh, dh/dt & v Values are Given


Under the current above conditions, equation 2.11 becomes:

(2.18)

Rearranged it will yield:

SW

q

CK

S

W

qK

W

T

dt

dV

g TO

DTO

TO

SL

0

210

1

(2.19)

Case 4: Continued

Equation 2.9 must be then integrated to =

Allowable

InitialFinal

t

VV

gdt

dV

g 00

11

Case 5: Takeoff Ground Roll

oD

TOTO

TOTO

SL

g

Vh

dt

d

VqS

RC

S

W

q

nK

S

W

q

nK

W

qS

W

To 2

1 2

2

2

1

Assumptions:dh/dt=0 Constant AltitudeR <> 0 On the ground!sG, , CLmax, kTO GivenTSL >> (D+R) Loads of Thrust

Assumptions:dh/dt=0 Constant AltitudeR <> 0 On the ground!sG, , CLmax, kTO GivenTSL >> (D+R) Loads of Thrust

Note: sG is the distance of the takeoff ground roll in ft.


• Rearranging yields:

dVVT

W

gds

SL

TO

o

• The master equation (2.5) can be reduced to:

Vds

dV

gdt

dV

gW

T

ooTO

SL

/


• This is then integrated over s=0, V=0 to takeoff (s=sG, V=VTO):

o

TO

SL

TOG g

V

T

Ws

2

2

• Given the takeoff values of and are used.


• Defining:

STALLTOTO VkV (2.20)

• Where kTO >1 and VSTALL is the stall speed of the aircraft, then:

TOLSTALLL WSCVSqC max2

21

max


• Or….

S

W

C

kVk

V TO

L

TOSTALLTO

TO

max

222

2

22

(2.21)

• And finally:

S

W

Cgs

k

W

T TO

LoG

TO

TO

SL

max

22

(2.22)

• But….


…in this limiting case:

TO

SL

W

TIs proportional to:

S

WTO

And inversely proportional to: Gs

Case 6: Takeoff Ground Roll (sG)

TO

TOTOLTODRD

TO W

WqSCCC

W

RD

Then:

Note: sG is the distance of the takeoff ground roll in ft.

Assumptions:dh/dt=0 Constant AltitudedV/dt=0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundsG, , CLmax, kTO GivenD=qCDS Equation 2.7

Assumptions:dh/dt=0 Constant AltitudedV/dt=0 Constant Speedn1 Lift approximately equals WeightR=0 Not on the groundsG, , CLmax, kTO GivenD=qCDS Equation 2.7

SCqWSCqR LTOTODR


dt

dV

gW

Sq

W

T

oTO

TOTO

TO

SL 1

(2.24)

Eq. 2.5 becomes:

Where:

)( LTODRDTO CCC

(2.23)

kTO

LTO

TO

SL

TO

TOo

TOG

CWTg

SWs

2max

1ln)/(

(2.25)

After rearranging and integrating the equation, the outcome is:

provided that the representative takeoff values for α and β are used.

It is assumed that TSL/WTO increases continuously with WTO/S. The limit, as all terms in ζTO go to zero and ln(1-ε) goes to – ε, is then taken in order to obtain the final result

kTO

L

TO

SL

TO

TOo

TOG

CWTg

SWs

2max

)/(


S

W

CgsW

T TO

LoG

TO

TO

SL kmax

22

(2.22)

From the previous page then we get

On the next slide are the assumptions for Total Takeoff Distance


SG

STO

SR Sobs

hTR

V = 0Ground Roll

Rotation

Transition

Climb

RC

VTO

VTO

VCL

Fig. 2.6b Takeoff Terminology ( )obsTR hh

Takeoff Field Length (A)

hOBS

CL

In Case A, the obstacle is NOT clearedDuring the transition.

Distance to pitch plane to liftoff CL

SWCktVts TOLTORTORR /)/(2 max

Multiply time to pitch and velocity to obtain distance

tR is roughly 3 seconds for modern aircraft – varies with rotational inertia and magnitude of control moments. This is the total aircraft rotation time.

max

2

2

2

1

L

TOTO

TOLTOTO

CS

WV

WSCVL

Take-off velocity determined by speed needed at TO CL to make lift equal to the weight

Multiply take-off velocity by a constant to account for the fact that CL lift-off is not CLmax

(2.26)

Distance to bring plane to reach climb angle

1sin

sin0

2

ng

VRs CLTO

CLCTR

S

W

Ckg

k

kg

Vs TO

LTO

CLTO

TO

CLTOTR

max2

0

2

20

2 2

18.0

sin

18.0

sin

The distance is the horizontal of the climb arc

Climb arc radius is defined by centripedal force needed to move the vehicle through the arc – excess lift tightens the arc

Breaking out velocity again links back to design variables

(2.27)

Distance to clear obstacleOutside the turn to climb, the flight path is a straight line at an angle to the ground defined by the vehicle climb angle

Climb angle defined by climb rate and velocity -> excess power

Height to climb through is the starting point at the arc from previous slide to the obstacle – geometry gives the horizontal distance

S

W

Ckg

k

kg

Vh TO

LTO

CLTO

TO

CLTOTR

max2

0

2

20

2 2

18.0

cos1

18.0

cos1

W

DT

dt

dh

V CL

sin

1CL

TRobsCL

hhs

tan

(2.28)

(2.29)

If hobs>htr:

If hobs>htr:

Case 6: Takeoff Ground Roll (B)

SG

STO

SR Sobs

hobs

V = 0Ground Roll

Rotation

Transition

Climb

RC

VTO

VTO

VCL

Fig. 2.6b Takeoff Terminology ( )obsTR hh

CL

In Case B, the obstacle is clearedDuring the transition.

Takeoff Ground Roll (B)B. (Fig. 2.6b) The distance sG and sR are the same as in case A, but the obstacle is cleared during transition so:

18.0

sinsin 2

0

2

TO

obsTOobsCobs

kg

VRs

obsRGTO ssss

(2.30)

where sobs is the distance from the end of rotation to the point where the height hobs is attained. There follows:

where

C

obsobs R

h1cos 1

2.2.7 Case 7: Braking Roll (sB)

TO

TOLDRD

TO W

WqSCCC

W

RD

(2.31)

Given: (reverse thrust), dh/dt = 0and values of , VTD = kTDVSTALL, D = qCDS,and

Given: (reverse thrust), dh/dt = 0and values of , VTD = kTDVSTALL, D = qCDS,and

0a

CLTOBDR qSWSqCR

Under these conditions:

So that Eq. (2.5) becomes

dt

dV

gW

Sq

W

T

oTOL

TO

SL 1

is defined in Equation 2.32L

Case 7: Braking Roll

• where…

• and after rearranging and integrating…

• as long as representative landing values of α and β are used.

)( LBDRDL CCC

2max0 )(

1ln)/(

TD

L

TO

SLB

L

L

TOB

kC

WTag

SWs

(2.32)

(2.33)

Case 7: Braking Roll• Reverse thrust can be used to an advantage in this case.

If “a” is made very large…

• or…

2max0

)()/(

TD

L

TO

SL

L

L

TOB

kC

WTag

SWs

S

W

Cgs

k

W

T TO

LB

TD

TO

SL

max0

22

(2.34)

Note: Total Landing Distance

hobs

Vobs

Approach

VTDVTD

Free Roll

Braking V=0

sA sFR sB

sL

Figure 2.7 Landing Terminology

• The total landing distance, sL, is the sum of the distance to clear an obstacle, sA, the free roll before the brakes are applied, sFR, and the braking roll, sB.

Case 7: Braking Roll

STALLobsobs VkV

22max

22

22

0

22

TDobs

obs

DRD

L

TDobs

TDobsTO

DRDA kk

h

CC

C

kk

kk

S

W

CCgs

(2.35)

Where hobs is the height of the obstacle and the velocity at the obstacle is

(2.36)

SWCktVts TOLTDFRTDFRFR /)/(2 max (2.37)

where tFR is a total system reaction time based on experience (normally 3 s) that allows for the deployment of a parachute or thrust reverser.

and

Case 8: Service Ceiling (Ps=dh/dt)

dt

dh

VS

Wq

CK

S

W

qK

W

T

TO

DTO

TO

SL 1021

TOL WSqC

(2.38)

Under these conditions, Eq. (2.11) becomes

where or

S

W

qC TO

L

(2.39)

Given:dV/dt = 0 No AccelerationN=1 Straight and Level (L=W)R=0 Not on the grounddh/dt > 0 Value is givenh (i.e., σ) and CL Given

Given:dV/dt = 0 No AccelerationN=1 Straight and Level (L=W)R=0 Not on the grounddh/dt > 0 Value is givenh (i.e., σ) and CL Given

Case 8: Service Ceiling

S

W

C

Vq TO

L

2

2

and

S

W

CV TO

LSL2or

(2.40)

Case 8: Service Ceiling

dt

dh

VC

CKCK

W

T

L

DL

TO

SL 10

21

Equation 2.41 can be used to find

(2.41)

TO

SL

W

Tas a function of

S

WTO

• V may be computed from given conditions and Eq. 2.40

• CL may be computed with V and Eq. 2.39

• α is a given function (Eq. 2.3)

NOTHING BEYOND THIS POINT

HAS BEEN CHECKED OR FORMATTED!!!

2.3 Preliminary Estimates For Constraint Analysis

• Preliminary Estimates for aerodynamic characteristics of the airframe and the installed engine thrust lapse are needed for constraint analysis

• 2.3.1 Aerodynamics– Maximum coefficient of lift (CLmax) is important in constraint analysis

during takeoff and landing

– For Fighter aircraft, a clean wing has CLmax between 1.0 and 1.2, and a wing with a leading edge slat has a CLmax between 1.2 and 1.6

– Table 2.1 has typical CLmax values for cargo and passenger aircraft

2.3 Preliminary Estimates For Constraint Analysis

High Lift Device Typical Flap Angle (deg) CLmax/cos(Λc/4)

Trailing Leading Edge Takeoff Landing Takeoff Landing

Plain -- 20 60 14-16 17-20

Single Slot -- 20 40 15-17 18-22

Fowler -- 15 40 20-22 25-29

Double sltd -- 20 50 17-20 23-27

Double sltd slat 20 50 23-26 28-32

Triple sltd slat 20 40 24-27 32-35

Λc/4 is the sweep angle at the quarter chord

Table 2.1 CLmax for High Lift Device

Lift-drag polar estimation• The lift-drag polar for most large cargo and passenger aircraft we can be

approximate use Fig.2.8 and Eq. (2.9) with

03.0001.0 K

eARnK

1

3.01.0 min LC

where wing planform efficiency factor (e) is between 0.75 and 0.85 and AR (wing aspect ratio) is between 7 and 10.

Lift-drag polar estimation

KKK 1

2minmin LDDo CKCC

2minmin LDDo CKCC


0

0.01

0.02

0.03

0.04

minDC

oM

0.2 0.4 0.6 0.8 1.0

Turboprop

Turbofan/jet

Fig 2.8 CDmin for Cargo and Passenger Aircraft


Fig 2.9 K1 for Fighter Aircraft

The lift-drag polar for high-performance fighter type aircraft can be

approximate use Eq.(2.9), K2=0, and Figs 2.9 and 2.10

0

0.2

0.4

0.6

1K

oM

0.5 1.0 1.5 2.0

Future

Current

Future

Current


Fig. 2.10 CDo for Fighter Aircraft

0

0.01

0.02

0.03

0.04

DoC

oM

0.5 1.0 1.5 2.0

Current

Future

PropulsionTrust of installed engine varies with Mach number, altitude, and afterburner operation can be described with simple algebraic equation that has been fit to either existing data of company-published performance curves or predicted data based on the output of off-design cycle analysis with estimates made for installation losses. The following algebraic equations for installed engine trust lapse are based on expected performance of advanced engines in the 1990 era and beyond

(2.43a)

6.032.125.0568.0 Ma (2.42)

For a High bypass ratio turbofan engine (M<0.9):

For a Low bypass ratio mixed turbofan engine with afterburner:

7.04.16.0245.088.072.0 Maa mil

7.02max 4.038.094.0 Maawet (2.43b)

(2.45a)

7.05.15.0262.0907.076.0 Maa mildry

(2.44a)For an advanced turbojet with afterburning:

For an advanced turboprop:

7.02max 4.03.0952.0 Maawet

(2.44b)

a For M ≤ 0.1

02.0

12.0

Ma For 0.1 < M < 0.8 (2.45b)

WEIGHT FRACTION

• Instantaneous weight fraction , , is needed in computing the constraints. is a fuel/payload correction.– Initial value for can be based on experience– Typical values for different mission phases are shown on the

mission profile figures on the following slides.• Fig 2.11: for Typical Fighter Aircraft• Fig 2.12: for Typical Cargo and Passenger Aircraft

Cruise

Descend

Descend & Land

Warm-up & Takeoff

Descend

Combat Air Patrol

Penetration

Subsonic Cruise Climb

= 1

= 0.85

= 0.9

= 0.78 = 0.68

= 0.60

Loiter

Accelerate & Climb

Instantaneous weight fraction, , for a Typical Fighter Aircraft

Combat

Deliver Expendables

Climb

Escape Dash

(Figure 2.11)

Cruise

Descend

Descend & Land

Warm-up & Takeoff

Descend & Land

Warm-up & Takeoff

Climb

Cruise

= 1

= 0.95 = 0.86

= 0.81 = 0.73

= 0.70

Loiter

Climb

Instantaneous weight fraction, , for a Typical Cargo and Passenger Aircraft

(Figure 2.12)

2.4 Example Constraint Analysis

• Creating constraint boundary on a diagram of takeoff thrust loading (TSL/WTO) vs wing loading (WTO/S)

• Based on RFP for Air-to-Air Fighter given in Ch. 1

• Done by creating equation:

f T W W SS L T O T O{ ( / ), ( / )} 0

Table 2.E1 Selected AAF SpecificationsMission Phases & Segments

Performance Requirements

1-2 Takeoff

Acceleration

Rotation

2000 ft PA, 100oF, STO = SG + SR ≤ 1500 ft

kTO = 1.2, TO=0.05, max power

VTO, tR = 3 s, max power

6-7 Supersonic

Penetration

And Escape Dash

1.5 M/ 30 k ft, no afterburning (if possible)

7-8 Combat

Turn 1

Turn 2

Acceleration

30,000 ft

1.6M, one 360 deg 5g sustained turn, with afterburning

0.9M, two 360 deg 5g sustained turns, with afterburning

0.8 1.6M, μt ≤ 50 s, max power

13-14 Landing

Free roll

Braking

2000 ft PA, 100˚F, sL = sFR + sB ≤ 1500 ft

kTD = 1.15, tFR = 3 s, B = 0.18

Drag chute diameter 15.6 ft, deployment ≤ 2.5 s

Max Mach Number

RFP para. C.1

2.0M/40k, max power

In order to proceed, you need:• CLmax • Lift drag polar• Engine data (fig. 2.E1)

These are obtained from the figures and the equation to come…

00.4 0.8 1.2 1.6 2.0 0.4 0.8 1.2 1.6 2.0

0

0.1

0.2

0.3

0.4K1

Mach Number

0.01

0.02

0.03

0.04CDo

Mach Number

CD = K1CL2 + CDo

CLmax = 2.0

(a) Aerodynamic Data

00.4 0.8 1.2 1.6 2.0 0.4 0.8 1.2 1.6 2.0

0

0.2

0.4

0.6

0.8adry

Mach Number

0.01

0.02

0.03

0.04awet

Mach Number(b) Installed Thrust Lapse

40k

30k

20k

10k

SL

40k

30k

20k

10kSL

T

Ta Md ry

S Ld ry 0 7 2 0 8 8 0 2 4 5 0 6 1 4 0 7. { . . ( | . | ) } *. .

T

Ta Mw et

S Lw e t { . . ( . ) } * .0 9 4 0 3 8 0 4 2 0 7

NOTE: TSL = Std.sea level max power static thrust

Fig. 2.E1 Preliminary AAF Data

Procedure with Takeoff• The airplane accelerated by thrust with no resisting forces in the ground roll

• Thrust is balanced by drag forces during the constant velocity rotation

Given these conditions and eqn 2.2 becomes:

sk

g C aT

w

w

St k

C

w

ST OT O

o L w e tS L

T O

T OR T O

L

T O { }* * *

{ *

m a xm a x( )

( ) ( )2 2

1

22

(2.E1)

Where:

a W S b W S cT O T O( / ) / 0

From table 2.E1, Fig 2.E1 and standard atmosphere of Appendix B

kTO = 1.2β = 1.0ρ = 0.002047 sl/ft3

go = 32.17 ft/s2

CLmax = 2.0σ = 0.8613

(awet)M=1.0 = 0.8775tR = 3.0 ssTO = 1500 ft

sk

g C aT

w

w

St k

C

w

ST OT O

o L w e tS L

T O

T OR T O

L

T O { }* * *

{ *

m a xm a x( )

( ) ( )2 2

1

22

(2.E1)

Evaluate coefficients a, b, and c from eqn 2.E1

Rewritten eqn 2.E1:

( ) { }w

S

b b a c

aT O

224

2(2.E2)

at wS L T O

1 2 4 7.

/

b = 79.57

c = 1500

To obtain the data:

TSL/WTO 0.4 0.8 1.2 1.6 2.0 2.4

WTO/S (lb/ft2) 33.4

57.5

77.1

93.7

108

121

Maximum Mach Number Cruise Constraint for Fighter Aircraft

Assumptions:dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightT = D Thrust equals DragR=0 Not on the groundh & v Values are GivenK2 Small

Assumptions:dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightT = D Thrust equals DragR=0 Not on the groundh & v Values are GivenK2 Small

Pertinent Data:= 0.78(wet)M=2 = 0.7189q = 1101 lb/ft2

K1 = 0.36CDo = 0.028

Pertinent Data:= 0.78(wet)M=2 = 0.7189q = 1101 lb/ft2

K1 = 0.36CDo = 0.028

SW

q

C

S

W

qK

W

T

TO

DTO1

TO

SL o

(2.12)

Begin with equation 2.12

Maximum Mach Number Cruise Constraint for Fighter Aircraft

SWS

W

W

T

TO

TO

TO

SL 42.88102.767 4

Substitution of pertinent data into equation 2.12 yields the following

(2.E3)

The maximum Mach number cruise constraint combined with the takeoff constraint enclose a solution space as illustrated in figure 2.E2.

Figure 2.E2 Constraints for Takeoff and Max Mach Number

Selection of the Air-to-Air Fighter Design Point

2TO ft/lb64S

W

Figure 2.E3 AAF Design Constraints

The requirements defined for a new air-to-air fighter result in the constraint diagram shown in figure 2.E3.

Based on the problem assumptions, the constraints that are dominate the design are the takeoff, landing, and supercruise constraints.

To keep cost and practicality of the AAF design under control, thrust loading is chosen near the region of previous experience.

Reasonable first choice:

2.1W

T

TO

SL

• Capability requirements for short takeoff and landing and nonafterburning supercruise drive design

• Relaxing these constraints would not significantly change the design choice because other constraints are important just outside of the takeoff, landing, and supercruise constraints.

• For many flight segments, the aircraft will be operating at full thrust; thus, no one requirement will cause the size to become excessively large. This is a result of careful consideration and balancing of requirements.



If thrust loading and wing loading are selected, it is possible to determine weight specific excess power (Ps) for a given flight condition by combining equations 2.2b and 2.11. Taking R = 0, this combination yields the following relationship:

S

W

q

CK

S

W

qK

W

TVP

TO

D2

TO1

TO

SLs

0

(2.E4)


This relationship is used to generate the contours of constant Ps in figure 2.E5 for specified loadings, , and flight conditions. This figure can be used to determine the minimum time-to-climb flight path; this flight path corresponds to the maximum Ps attainable for a given line of constant energy height, ze. This time-to-climb relationship is a modified version of equation 2.2b:

Figure 2.E5 AAF Ps Contours 2e

1e

z

zs

e2

1 P

dzdtt

Mission Phase 1-2: Takeoff

22

2

4

a

acbb

S

WTO

2

1ln

TO

LTO

TO

SL

TO

TOo

k

C

WTg

aMAX

MAXLTOR C

ktb

2

Assumptions:2000 ft. PA100ºFkTO=1.2µTO=0.05tR=3 ssTO=sG+sR1500 ftMax Power

From (2.25) and (2.26), with STO=sG+sR:

where:

TOsc

Mission Phase 1-2: Takeoff

• =1.0 • K1=0.18

• =0.002047 slug/ft3 • =0.8613

• CLmax=2.0 • (wet)M=0.1=0.8775

• kTO=1.2 • µTO=0.05

• CDo=0.014 • tR=3.0s

• TO=0.36 • sTO=1500 ft

05.08775.0

2601.01ln03.42

TO

SL

WT

a

57.79b

We estimate and to be constant at appropriate mean values. A conservative estimate for TO for the AAF is obtained by assuming (CDR - µTOCL) =0 and evaluating CD at CL=CLmax/kTO

2.

With: Then:

1500c

TSL/WTO 0.4 0.8 1.2 1.6 2.0

WTO/S (lb/ft2) 14.3 45.1 67.2 85.3 101

whence:

Mission Phases 6-7 and 8-9: Supersonic Penetration and Escape Death

2

1

SW

q

C

S

W

qK

W

T

TO

DoTO

TO

SL

Assumptions:1.5M/30k ftNo afterburning

From (2.12):

Mission Phase 6-7: Supersonic Penetration and Escape Dash

oD

TOTO

TOTO

SL

g

Vh

dt

d

VqS

RC

S

W

q

nK

S

W

q

nK

W

qS

W

To 2

1 2

2

2

1

SW

q

C

S

W

qK

W

T

TO

DTO

TO

SL o

1

Assumptions (most from Case1):dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightR=0 Not on the groundh & v Values are GivenK2=0 Pure Parabolic Drag Polar

Assumptions (most from Case1):dh/dt = 0 Constant AltitudedV/dt =0 Constant Speedn=1 Lift equals WeightR=0 Not on the groundh & v Values are GivenK2=0 Pure Parabolic Drag Polar

1

Supersonic Penetration and Escape Dash Example

WTO/S 20 40 60 80 100 120

TSL/WTO 2.35 1.77 1.2 0.913 0.746 0.638

SWS

W

W

T

TO

TO

TO

SL 25.7010*345.4 4

With:

=0.78 dry=0.3953 q=991.8 psf

=0.3747 K1=0.28 CDo=0.028

Mission Phase 7-8: Combat Turn 1

oD

TOTO

TOTO

SL

g

Vh

dt

d

VqS

RC

S

W

q

nK

S

W

q

nK

W

qS

W

To 2

1 2

2

2

1

Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundh, n, & V Values are GivenK2=0 Pure Parabolic Drag Polar

Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundh, n, & V Values are GivenK2=0 Pure Parabolic Drag Polar

SW

q

C

S

W

qnK

W

T

TO

DTO

TO

SL o

2

1

Combat Turn 1 Example

SWS

W

W

T

TO

TO

TO

SL 22.4210*407.5 3

With:

=0.78 wet=0.7481 q=1128 psf n=5

=0.3747 K1=0.30 CDo=0.028

WTO/S 20 40 60 80 100 120

TSL/WTO 2.22 1.27 1.03 0.96 0.963 1

2.4.3.4 Mission Phase 7-8: Combat Turn 2

--0.9 M/30K ft, two 360 degree 5g sustained turns, with afterburning.

From Eq. (2.15)

SWS

W

W

T

TO

TO

TO

SL 34.1201473.0

SW

q

C

S

W

qnK

W

T

TO

DoTO

TO

SL

2

1

with β = 0.78 αwet = 0.5206 n = 5 CDo = 0.018

σ = 0.3747 K1 = 0.18 q = 357.0 lb/ft2

then

whence

WTO/S (lb/ft2) 20 40 60 80 100 120

TSL/WTO 0.910 0.900 1.09 1.33 1.60 1.87

2.4.3.5 Mission Phase 7-8: Horizontal Acceleration

--0.8 → 1.6 M/30K ft, Δt ≤ 50s, max power.

From Eq. (2.18)

6484.079.28

10704.3 4

S

WS

W

W

T

TO

TO

TO

SL

tg

Ma

SW

q

C

S

W

qK

W

T

oTO

DoTO

TO

SL

1

We obtain approximate constant values of ρ, K1, CDo, and α at a mean Mach number of 1.2.

with β = 0.78 K1 = 0.23 a = 994.8 ft/s

σ = 0.3747 q = 634.7 lb/ft2 ΔM = 0.8 αwet M=1.2= 0.5952 CDo = 0.025 Δt = 50 s

then

Constituted from p. 60

whence,

20 40 60 80 100 120

2.10 1.38 1.15 1.04 0.973 0.933

)/(/ 2ftlbSWTO

TOSL WT /

Mission Phase 13-14: Landing

From Eqs. (2.33) and (2.37) with2

2

2

4

a

acbb

S

WTO

s

sss

st

k

BFRL

BFR

TD

2.5 deployment

ft,15.6diameter chute, drag GFE

ft1500

,18.0,3

,15.1 100F, PA,2000ft

BFRL sss

Mission Phase 13-14: Landing(cont.)

Where,

2

0 max)(

1ln

TD

L

TO

STB

L

L

k

C

WTag

a

max/2 LTDFR Cktb Lsc

Assumptions


5348.0

500ft area wingairplane Estimate

ft 191 area chute RFP

4.1 chute Drag

2

2

DRc

D

C

C

A conservative estimate of from at 0.8 of touchdown lift coefficient and by assuming

L DC0)( LBDRcDR CCC )/( 2

max TDL kC

0.18 0.18,K

,1500s ,8123.0 ,014.0C ,0.2

,3 ,5348.0C ,210.1C ,slug/ft002047.0

0a ,2775.0C ,15.1 0.56,

B1

LLD

DRcL3

D

0max

ftC

st

k

L

FR

TD


2/5.70/ ftlbSWTO

whence

1500c 57.06b 47.14 athen

Section 2.4.3.7

• This gives

S

WS

W

W

T

TO

TO

TO

SL 88.4210*767.2 4

W TO /S (lb/ft 2 ) 20 40 60 80 100 120

T SL /W TO 2.14 1.07 0.713 0.535 0.428 0.357

The table below shows a few calculations

Chapter 3 – Mission Analysis

• Section 3.1 - Concept– With the values of thrust loading (TSL/WTO) and wing

loading (WTO/S) determined, now the focus shifts to the estimation of the gross takeoff weight, which is a sum of component weight:

– Wp is specified in the RFP and is separated into payload weight (WPE) that consists of the cargo being delivered during a mission (ie ammunition) and the permanent payload weight (WPP), which consists of the crew, passengers, their equipment, and anything else that is carried the entire mission

FEPTO WWWW (Eqn 3.1)


– WE consists of the basic aircraft structure and any permanently attached equipment

– WF represents the required fuel weight to fly the mission


• WF depends on the rate of fuel consumption, which depends on the thrust specific fuel consumption (TSFC)

• Thrust can be found from (Eqn 2.1) while the TSFC depends on the engine cycle, flight conditions (altitude, mach #, etc…)


• Fuel consumption analysis results in multiple benefits:– Little information needed for calculation– Reveals optimum solution (minimum fuel) for

flying certain segments of the mission– Shows the fuel consumed during each mission

segment as a fraction of the aircraft weight at the beginning of the mission segment, which makes WF a calculable fraction of WTO

Chapter 3 – Mission AnalysisFigure 3.1 Weight Fractions of Cargo and Passenger Aircrafts

Figure 3.2 Weight Fractions of Fighter Aircrafts


• Section 3.2 – Design Tools

• This section deals with the thermodynamics of the flight and the way the thrust work of the engine is used by the aircraft

• Begin by considered the change in weight of aircraft from fuel burn

TTSFCdt

dW

dt

dW F *

(Eqn 3.2)


• (Eqn 3.2) can be rewritten as:

• Note that T is the installed thrust and TSFC is the installed thrust specific fuel consumption

• Also, since:

dtW

TTSFC

W

dW** (Eqn 3.3)

VW

Tdsds

ds

dt

W

Tdt

W

T

*


• The above equation shows the incremental weight-velocity specific engine thrust work done as an amount of fuel is consumed (dWF)

• This work is partly used in the mechanical energy (PE + KE) of the airplane mass and partly dissipated into the atmosphere (wing tip vortices, heating, etc…)

Weight Fractions

• The goal of subsequent section is to integrate Eq.(3.3) to get “weight fractions”

• Requires knowledge of TSFC behavior and “instaneous thrust loading” as a function of time along flight path.

• Two distinct classes of integration– Ps > 0 (Type A)

– Ps > 0 (Type B)

i

f

initial

final

W

W

W

W

TO

SL

W

T

W

T

Instantaneous Thrust Loading Behavior: Type A

• Ps > 0; examples of Type A found in cases1. Constant speed climb2. Horizontal acceleration3. Climb and acceleration4. Takeoff acceleration

• Using Eq.(2.2a) to move towards solution:

• Combining Eqs.(3.3) and (3.4)

• u = how total engine thrust work is distributed between mechanical energy and dissipation

• Eq. 3.5 shows u is fraction of thrust work dissipated

edzuV

TSFC

W

dWor

g

Vhd

uV

TSFC

W

dW

121 0

2

T

RDuwhere

u

dz

u

gVhdds

W

TVdt

W

T e

11

2 02

(3.4), (3.5)

(3.6a), (3.6b)

Instantaneous Thrust Loading Behavior: Type A (continued)

• Integration of Eq.(3.6a) which largely depends on variation of {TSFC/V(1-u)} with altitude and velocity. When that is relatively constant:

• Δze = total change in energy height

• Eqs. (3.6) and (3.7) show how potential and kinetic energy can be “traded” within certain limits

• Example of when bad: No drag (u = 0) and constant ze

– Unpowered dive or zoom climb

– No fuel consumed and dW = 0 or Wf = Wi

ei

f

i

f

zuV

TSFC

W

Wor

g

Vh

uV

TSFC

W

W

1exp

21exp

0

2

(3.7a)

(3.7b)

(from Type A, cont.)Note: Minimum Fuel Path

• Fuel consumption analysis also shows the supposed “best” way to fly certain legs for minimum fuel used.

• For Type A legs, Eqs. (3.2) (2.2b) and (2.3) produce

es

SLF dzP

aTSFCTdtTSFCTdW

Pg 58

VuW

TV

W

RDTPs )1(

)(

Defining the fuel consumed specific work (fs) as:

then:

Eq. 3.9 shows that the minimum fuel-to-climb flight from ze1(h1,V1) to ze2(h2,V2)Corresponds to a flight path that produces the max. thrust work per unit weight of fuelAt each energy height level in the climb (i.e. max value of fs at each ze)

8.3TSFCa

Pf s

s

2

1

2

1

219.3

e

e

z

z s

eSLFF f

dzTdWW

Pg 58Graphical method for finding max. fs at any ze (thus min. fuel-to-climb path from ze1 to ze2)

• constant fs and ze contours in the altitude-velocity plane (fig. 3.E5)

• Analogous to using the Ps and ze contours (fig. 2.E5) to find min time-to-climb path (Fig. 3.E2) in conjunction with the time-to-climb equation from eq. 2.2b.

2

1

2

1

e

e

z

z s

e

P

dzdtt

Pg 58/9Instantaneous Thrust Loading Behavior: Type B

When Ps=0, it is generally true that the speed and altitude are essentially constant and specific information is given regarding the total amount of time (Δt or Δs/V) which elapses. Examples of type B cases are:

constant speed cruiseconstant speed turn

Best cruise Mach number and altitudeLoiterwarm-up

Take off rotationConstant energy height maneuver

For this type of flight, the thrust work is completely dissipated and the required thrust may not be known in advance. The thrust is modulated or throttled so that T=(D+R) or u=1 by Eq. 3.5 and dze=0 by eq. 3.4

Using Eq. 2.2 it follows that

)10.3(dtW

RDTSFC

W

dW

Pg. 59Often TSFC (D+R)/W remains relatively constant over the flight leg

So eq. 3.10 can be integrated

If desired, the required thrust may be computed after the fact.

timeflighttotalt

tW

RDTSFC

W

W

i

f

11.3exp

TSFC Behavior (Thrust Specific Fuel Consumption)

•Complex function:

•Altitude

•Speed

•Throttle setting (especially if has afterburner)

Pg. 59Good starting point at this stage is the assumption:

C = constant

θ = represents the usual thermodynamic cycle improvement due to a lower ambient temperature at higher altitude.

Basis of this assumption is the observation that:

TSFC is not a strong function of speed & throttle setting

C could be selected differently for different legs of the mission.

NOTE:

Assumptions for TSFC other than eq. 3.12 can be made and the weigh fraction analysis continued to completion but this model is a good for turbojet and turbofan. Different assumptions would be required for turboprop engines which TSFC is primarily proportional to flight Mach # (eq. 1.25). The same process could be applied.

12.3CTSFC

Summary of Weight Fraction Equations

• The equations previously mentioned for dW/W and Wf/Wi (Eqs. 3.6, 3.7, 3.10, 3.11) are combined with the equation for TSFC (Eq. 3.12) in order to get the equations for the Type A and Type B situations.

Summary of Weight Fraction Equations (cont.)

• TYPE A (Ps >0, T=αTSL given)

oi

f

o

g

Vh

uV

C

W

W

g

Vhd

uV

C

W

dW

2)1(exp

2)1(

2

2

Instantaneously exact

(3.13)

Exact when

sqrt(θ)/ V(1-u) constant

(3.14)

Summary of Weight Fraction Equations (cont.)

• TYPE B (Ps =0, T=D+R given)

tW

RDC

W

W

dtW

RDC

W

dW

i

f

exp

Instantaneously exact

(3.15)

Exact when

sqrt(θ)(D+R)/ W constant

(3.16)

Special Cases

• The following slides will discuss a various special cases to give both specific design tools and more insight into weight fraction analysis.

• Assumptions– α, β, K1, K2, CDo are known

– n=1 and L=W • except for Cases 4,8 (n>1), and 10

– R=0• except where the aircraft is on the ground (Cases 4 and 10)

CASE 1: Constant Speed Climb (Ps=dh/dt)

• Given:– dV/dt = 0– n=1(L=W)– R=0

– T = αTSL (Full Thrust)

– hinitial

– hfinal

– V

• Conditions

SL

TO

L

D

T

W

C

Cu

Case 1.

• Eq. (3.14) becomes Eq. (3.17)

where the level flight relationships (3.18)

have been used, and the change in altitude is

SLTOLDi

f

TWCC

h

V

C

W

W

/1exp

L

D

L

D

C

C

SqC

SqC

L

D

W

D

initialfinal hhh

Case 1.

• Notes on using Eq. 3.17• Typically, variables values are averaged for the altitude interval used in the

analysis although the variance in the quantity

should be checked to verify the accuracy of this averaging approach. If the accuracy is poor, the climb should be broken into intervals and Eq. (3.17) should be applied to each interval. The overall Wf / Wi the then the product of the interval results.

• Note that the fraction of engine thrust dissipated is

while is the fraction of the engine thrust invested

in potential energy.

u1

SLTOLD TWCC /

SLTOLD TWCC /1

Case 2.

• Horizontal Acceleration

• Under these conditions

so that Eq. (3.14) becomes Eq. (3.19)

where Eq. (3.18) has been used and

Similar to the previous analysis, average values for the speed intervals should be used unless the quantity V(1-u) varies too much.

dtg

VdVP

os

Given: dh/dt = 0, n =1 (L = W), R = 0, full thrust (T = aTSL) and the values of h, Vinitial, and Vfinal.

SL

TO

L

D

T

W

C

Cu

SLTOLDi

f

TWCC

gV

V

C

W

W

/1

2exp 0

2

222initialfinal VVV

Case 4:

Where u can be obtained from Eq. (2.21) as

and VTO is as given by Eq. (2.20)

SL

TOTO

TOTO T

W

W

Squ

(3.22)


Given:

Under these conditions, Eqs. (3.16) and (3.18) become

sVh

R

WLndt

dV

dt

dh

range cruise and, , of valuesand

,0

),(1

0 ,0

s

C

C

V

C

W

W

L

D

i

f exp (3.23)


Where CL is obtained by setting lift equal to weight and CD is computed from Eq. (2.9). Since CD/CL will vary as fuel is consumed and the weight of the aircraft decreases, this result is not, strictly speaking, exact. Hence, it might become advisable to break the cruise into several intervals and apply Eq. (3.23) to each.The Type B case 6-10 that follow are, however, exact integrals.


Given:

This situation can be treated much the same as Case 5, except that L=nW. The duration of the turning can be shown, with the help of Eq. (2.17), to equal

Under these conditions, Eqs. (3,16) and (3.24) become

NnVh

Rdt

dV

dt

dh

turnsofnumber and ,1, , of valuesand

,0

0 ,0

1

222

0

ng

nNV

V

NnRt c

1

2exp

20 ng

nNV

C

nCC

W

W

L

D

i

f (3.25)

(3.24)


Given: Under these conditions, Eqs. (3,16) and (3.24) become

1

2exp

20 ng

nNV

C

nCC

W

W

L

D

i

f

Equations 3.26-3.28

S

W

qC TO

L

SLL

d

a

Cds

MC

C

W

dW

1

3.2.7 Case 7: Best Cruise Mach Number and Altitude (BCM/BCA) (Ps = 0)

Given: dh/dt = 0, dV/dt = 0, n=1, R=0, and value of cruise range is Δs

Replacing dt with ds/V, Eq. (3.15) becomes:

(3.26)

3.27

L

DoLL

L

D

C

CCKCK

C

C 2

21

From the above equation, a minimum CD/CL can be found by differentiating the equation with respect to CL and setting the result equal to zero. This leads to:

21

*

4 KKCC

CDo

L

D

(3.27)

From Eq. (3.26) weight reduction can be minimized by operating at lowest value of (1/M)(CD/CL). This is known as the best cruise condition and the parameters associated with it contain a * superscript. The below equation shows that drag and lift do not depend on M below critical drag rise Mach number:

3.28

• Insert equation 3.28 here because I can’t see it because it is cut off. It is very short though!

Mission Analysis

1

0202*

K

CKCC D

DD

LD CC /*)//()/1( LD CCM

(3.29)

Since the lowest achievable value of is constant below the critical M, it follows that decreases as M increases. Once MCRIT is reached, the situation changes. Further increases in M cause to increase and finally cause to increase. Rather than search for the exact at which the minimum occurs, it is more practical to take , then

*)/( LD CC

BCMMM CRIT *

CRIT

D

L

D

M

KKC

MC

C 210* 4)

1(

CRITMM

*)//()/1( LD CCM

*)//()/1( LD CCM

(3.30)

This may be substituted into Eq. (3.27) to yield

Mission Analysiswhich may be substituted into Eq. (3.26) to yield

dsa

C

M

KKC

W

dW

SLCRIT

D

2104

(3.31)

This may be directly integrated to yield the equivalent of Eq. (3.16),

sa

C

M

KKC

W

W

SLCRIT

D

i

f 24exp 10

(3.32)

WSCPM

SqC LL 2

2

S

W

KCMPP

P TO

DCRITSLSL 102 /

12

Finally, since the aircraft must sustain its weight under this condition,

or,

(3.33)

Mission Analysis

Which may be used to find the altitude of “best” cruise or BCA. Since must gradually diminish as the mission progresses, this last result shows that the altitude must gradually increase. Since the flight Mach number is fixed at , it also follows that the speed, which is proportional to , must gradually decrease. In most cases, these changes occur so slowly that the assumption of PS = 0 remains true.

CRITM

Case 8: Loiter (Ps=0)Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundn=1 (L=W)

Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=0 Not on the groundn=1 (L=W)

t Flight Duration

dtC

CC

W

dW

L

D

(3.34)

Putting these assumptions into Equation 3.15, we get:

dtKKCCW

dWDo 214 (3.35)

From inspection in case 7, we know that the optimum CD/CL for minimum fuel consumption is equal to:

214 KKCCC DoLD

Thus,

Case 8: Loiter (Ps=0)

tKKCCW

WDo

i

f 214exp (3.36)

S

W

KCMPTO

DoSL 12 /

12

(3.37)

Integrating equation 3.35 for weight from Wi to Wf and for the time interval Δt, we get:

Variation in pressure altitude can be found as a function of Mach number for loiter by equating Lift and Weight.

)(

Case 9: Warm-Up (Ps=0)Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=T=αTSL

V=0D=0Given:h AltitudeΔt Warm-up time

Assumptions:dh/dt=0 Constant AltitudedV/dt =0 Constant SpeedR=T=αTSL

V=0D=0Given:h AltitudeΔt Warm-up time

Equation (3.15) under these assumptions becomes:

dtTCdW SL

Mission Analysis

• Derived from ¿¿PAGE 66??

tW

TC

W

W

TO

SL

i

f

1 (3.39)

Weight Fraction Where β is evaluated atthe beginning of warm-up

CASE 10: TAKEOFF ROTATION (PS = 0)

• Given: dh/dt = 0, dV/dt = 0, (D+R) = T = αTSL and values of h, V, and rotation time tR

(3.40) dtTCdW SL

(3.41)

RTO

SL

i

f tW

TC

W

W

1

Where β is evaluated at the beginning of rotation

Under these conditions, Eq. 3.15 becomes

CASE 11: CONSTANT ENERGY HEIGHT MANEUVER (PS = 0)

• Given: Δze = 0, n = 1 (L=W), R = 0 and thevalues of hinitial, hfinal, Vinitial, and Vfinal

(3.42)

t

C

CC

W

W

L

D

i

f exp

Where CL is obtained by setting lift equal to weight and CD is computed from Eq. 2.9. The maneuver time interval t is best [continued in next page]

In this case potential energy is exchanged for kinetic energy and any loss to aerodynamic drag is balanced by the engine thrust.Under these conditions, Eqs. 3.16 and 3.18 become

Calculation of Takeoff Weight

• Weight fraction for each mission segment is the ratio of final weight to the initial weight for that segment

1 fi

i

f

W

W(3.43)

• For the weight ratio of mission legs where only fuel is consumed

Calculation of Takeoff Weight (cont).

• For example, for several consecutive mission segments:

525443324

5

3

4

2

3

2

5

W

W

W

W

W

W

W

W

Expendable payload case

• For the special case where the expendable payload is delivered at some point j in the mission, we write instead:

j

PE

j

PEj

W

W

W

WW

1 (3.44)

Takeoff Weight

• Recall that takeoff weight is defined as

PPPEEFTO WWWWW (3.45a)

Fuel Weight• For a mission with n junction and payload

delivery at junction j (j<n), is expressed in terms of by the sum of the aircraft weight decrements between takeoff and junction j and between junction j (after delivery of expendable payload) and landing in the following manner:

FW

TOW

njPE

jTOPE

jTO

jTOTOF WWWWWWW

111

which reduces to

njPE

nTOF WWW 11

1(3.45b)

Historical Correlations• Figures 3.1 and 3.2 relate empty weight to

takeoff weight via

)( TOWf

)( TOTO

E WfW

W

(3.45c)

where is given by Eq-ns (3.47)-(3.50).

• Combining Eq-ns (3.45a) – (3.45c) and solving for yieldsTOW

n

njPEPP

TO

WWW

1

(3.46)

Summary• Takeoff weight is proportional to and

(where matters more because it is carried the whole way)

• Improved structural materials, which reduce (and therefore ), can reduce dramatically

• An initial estimate of must be made in order to place values upon the of the payload deliveries and of the structural weight.

• If the initial estimate of takeoff weight is far from the value given by the Eq-n (3.46), an iterative solution of Eq-n (3.46) is will be necessary.

PPW

PEW PEW

EW TOW

TOW

TO

PEW

W

Pages 70-73 go here

• Arvind

Pages 74-77 go here

• Joan

Table 3.E3 Summary of Results – Mission Analysis

Mission Phases & Segments β=W/WTO End of Leg

Wf/Wi=∏ if

1-2 Warm-up and Takeoff 0.9759 0.9759 2-3 Accelerate and Climb 0.9445 0.9678 3-4 Subsonic Cruise Climb 0.9141 0.9678 4-5 Descend 0.9141 1.0000 5-6 Combat Air Patrol 0.8780 0.9605 6-7 Supersonic Penetration 0.8035 0.9152 7-8 Combat 0.7441 0.9261

Deliver Expendables 0.6917 ------- 8-9 Escape Dash 0.6757 0.9769 9-10 Minimum Time Climb 0.6743 0.9979

10-11 Subsonic Cruise Climb 0.6487 0.9620 11-12 Descend 0.6487 1.0000 12-13 Loiter 0.6210 0.9573 13-14 Descend and Land 0.6210 1.0000

Figure 3.E3 Fraction of Takeoff Weight vs Mission Phase

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1 2 3 4 5 6 7 D 8 9 10 11 12 13 14

Mission Phase Number

W/W

TO

3.4.2 Determination of WTO, TSL, and S

• Now compute the takeoff weight for the AAF (Air-to-Air Fighter) using Equation 3.46 shown below.

148

141

PP PETO

W WW

Where,

WPP= 200 lb Pilot plus equipment270 Cannon405 Ammunition feed system275 Returning ammunition198 Casings weight

1,348 lb

(3.46)

3.4.2 ContinuedWPE = 382 lb Sidewinder missiles

652 AMRAAMs275 Spent Ammunition

1,309 lb

0.6273E

TO

W

W

14 9 148 8 13

14 2 141 1 13

Source: AAF RFP Sec. D

Source: Eq. (3.49)Assumed WTO = 25,000 lb

Thus,1348 (1309)(0.8978)

62,0000.6680 0.6273TOW lb

Source: Table 3.E3

Example Mission Analysis (cont.)

Fig. 3.E4 Weight Fractions of Fighter Aircraft

0.3

0.4

0.5

0.6

0.7

0.8

10 20 30 40 50 60 70 80 90 100

WTO (1000 lbs)

WE/W

TO

X-29

F-102A

F-16C/ D

F-16A/ B

F-106AF-15

F-15C

F-105GF-14A

F-4E

F-111F

F-101B

F/ A-18

F-100D

F-104C

YF16YF17

Notice (Figure 3.E4) that this conventional metal version of the AAF would be considerably heavier than existing lightweight fighters (e.g., F-16).

Example Mission Analysis (cont.)

lbsWTO 400,24)90.0(6273.06680.0

)8978.0)(1309(1348

Reducing the payload provides little relief because the large WTO is the result of the very small denominator.

Another assumption can be the reducing the fuel usage, for example by shortening the range or duration of mission. Yet, another assumption would be reducing the empty weight, for example by using advanced lightweight construction materials as composites.

With the latter assumption, the aircraft’s Γ will be reduced by approximately 10%. Therefore, with this reduction, WTO is recalculated for the assumed WTO of 25,000 lbs,

Example Mission Analysis (cont.)The conclusion is that AAF can have a practical size. This is true only if reliable non metallics with competitive strength, durability, and repairability become available according to schedule. With the choices of TSL/WTO = 1.20 and WTO/S = 64.0 made, the description of the AAF at this stage of design, using WF/WTO = 0.3265, is:

WTO = 24,400 lb

TSL = 29,300 lb

S = 381 ft2

WP = 2,660 lb

WE= 13,800 lb

WF = 7,970 lb

This information will give perspective and insight regarding the nature and shape of the corresponding aircraft.

3.4.3 FIRST REPRISEHere, the key assumptions made so far will be evaluated. Hopefully, the assumptions made are accurate and the results acceptable. If not, the whole process will have to be reiterate with new assumptions.

Earlier, the wing area (S) was assumed to have a value of 500 ft2, whereas the recently derived wing area is 381 ft2. This in turn will increase the solution space in Fig. 2.E3, but it is important to note that it wont provide any design points superior to the one already selected.

Warm-up

tW

TC

TO

SLupwarm

1

Assumptions: 2000 ft Pressure Altitude, 100 °F Temp, 60 second duration, min power

Δt = 60 sec, σ = 0.8613, β = 1.0, TSL/WTO = 1.2,

αdry = 0.6484, θ = 1.08, C = 1.35 h-1

9818.0

9818.0

upwarm

Takeoff AccelerationAssumptions: 2000 ft Pressure Altitude, 100 °F Temp, kTO = 1.2,

μTO = 0.05, max power

SL

TOTO

TOTO

TOAccelTakeoff

T

W

W

Squ

g

V

u

C

01exp

Takeoff Acceleration

β = 0.9818, a = 1160 ft/s, μTO = 0.05, WTO/S = 64 lb/ft2, MTO = 0.1812,

q = 11.31 lb/ft2, σ = 0.8613, C = 2.0 h-1, ξTO = 0.36 (Estimated from 2.4.3.1),

CL max = 2.0, θ = 1.08, u = 0.1067, kTO = 1.2, TSL/WTO = 1.2, VTO = 210.2 ft/s, αwet = 0.8795

9777.0

9958.0

acceltakeoff

Pages 84-85 go here

• Matt daniell

∆ h

(ft)

∆(V2/2go)

(ft)

∆ ze

(ft)

T∆s/W

(ft)

∆t

(min)

∆s

(nmi)

c

(ft/s)-1

∏

a 14,000 2208 16,210 23,250 0.6452 5.683 0.001557 0.9906

b 14,000 3.105 14,000 20,420 0.7589 6.749 0.001485 0.9922

c 13,000 -661.6 12,340 18,520 0.9810 8.437 0.001494 0.9931

∑ 41,000 1550 42,550 62,190 2.385 20.87 n/a n/a

Segment E – Climb/Acceleration Results

9761.0 cbaE

The results do NOT differ insignificantly, therefore the phase should be broken down into finer intervals. In summary, Mission Phase 2-3:

9445.0

9678.0

min892.2

43.23

23

23

23

ED

ED

ED

ttt

nmisss

Example Mission Analysis

Mission Phase 3-4: Subsonic Cruise Climb

BCM/BCA, Δs23+Δs34 = 150 nmi

Assume: (Ps = 0 , T = D+R , K2 = 0)

SLCRIT

DO

a

sC

M

KC341

34

4exp

Where: Δs34 = 126.6 nmi K1 = 0.18

C = 1.35 h-1 Cdo = 0.018

aSL = 1116 ft/s Mcrit = 0.9

Then:

9141.0

9678.034

Mission Phase 5-6: Combat Air Patrol

30k ft , 20 min.

Assume: (K2 = 0)

tKCC DO 156 4exp

Where: Δt = 1200 sec θ = 0.7940

K1 = 0.18 C = 1.35 h-1

Cdo = 0.014

Then:

8780.0

9605.056

Mission Phase 4-5: Descend

BCM/BCA -> Mcap/30k ft

9141.0

0.145

Example Mission Analysis

starts here

• ragini

Mission Phase 8-9 Continued

98

9769.0

6757.0

Mission Phase 9-10: Minimum Time Climb

109

exp tC

Cc

L

D

BCM/BCA = 1.5M/30kftAirplane climbs from 30,00 ft to 50,000 ft and reduces speed from 1.5 to 0.9 MaFrom Eq 3.42:

With (assuming vertical speed = 0.7 Vavg)

avgV

ht

7.0

mideo

mid hzg

Vi

2

2

Mission Phase 9-10: Minimum Time Climb

Given information:hi = 30,00 ftMi = 1.5(a/aSL)i= 0.8911Vi = 1492 ft/sZei = 64,600 ftHf = 50,000 fth = 20,000 ftMf = 0.9

(a/aSL)f= 0.8671Vf = 870.9 ft/sVavg = 1181 ft/st = 24.19 sHmid = 40,000 ftVmid = 1258 ft/s(a/aSL)mid= 0.8671Mmid = 1.3WTO/S = 64 lb/ft2

= 0.6757= 0.1858CL = 0.093K1 = 0.23K2 = 0CD0 = 0.023CD/CL = 0.2687= 0.7519C = 1.35 h-1

109

9979.0

Calculation from given information

6743.0

Mission Phase 10-11: Subsonic Cruise Climb

1110

9620.

BCM/BCA, s10-11 = 150 n.mi.

6487.0

Mission Phase 11-12: DescendBCM/BCA = Mloiter/10kft

6487.0

1211

1

Mission Phase 12-13: Loiter

1312

9573.

BCM/BCA = Mloiter/10kft, 20 min

6210.0

Mission Phase 13-14: Descend and Land

Mloiter/10kft --> 2000 ft PA, 100oF

6210.0

1413

1

95

• Kc and someone else

Case X: Minimum Fuel-to-Climb Path

NOTE: AAF is not required to fly a climb path for minimum fuel burn.

Fuel Consumed Specific Work (fs) is calculated with Eq. (3.8) at military power using:

145.1

135.1

97.0

/64

2.1

01

01

2

MforhC

MforhC

ftlbSW

WT

TO

TOSL


Contours of constant fs and constant energy heights (ze) are shown in the velocity-altitude space (figure 3.E5).

From energy height ze1(h1,V1) to ze2(h2,V2), the min fuel-to-climb path has a maximum value of fs at each ze.

ze = 12 kft @ SL to ze = 100 kft @ 57 kft is also shown in figure 3.E5


By comparing figures 3.E2 and 3.E5:

• Ps = 0 and fs = 0 are the same [from Eq (3.8)]

• Due to the step change in C across M = 1, each fs contour has a discontinuity– Not including fs = 0

• Minimum time-to-climb path occurs at about 50 fps greater than subsonic minimum fuel-to-climb path

•Minimum time-to-climb paths reach transonic and then supersonic speeds at a lower altitude than minimum fuel-to-climb paths

By dividing Eq. (3.9) by W1 = βWTO and given WF1-2 = W1-W2 & 1Π2 = W2/W1

The aircraft weight fraction (1Π2) results in:

In figure 3.E5 (ze1 = 12 kft, ze2 = 64 kft):


2

1

21

211

1e

e

z

z s

e

TO

SLF

f

dz

W

T

W

W

2

1

121

e

e

z

z s

e

TO

SL

f

dz

W

T

9685.021

Note: No equation number were given in the text

the master equation dr. dimitri mavris boeing associate professor for advanced aerospace systems...

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