Notes on Aircraft Performance 2016v2

8/17/2019 Notes on Aircraft Performance 2016v2

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Introduction Aerofoil Geometry

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4. Glide Performance...................................................................................................................... 29

4.1 Glide Angle.............................................................................................................................. 29

4.2 Rate of Descent ...................................................................................................................... 30

5. Cruise Performance.................................................................................................................... 30

5.1 Range...................................................................................................................................... 30

5.2 Cruise Calculations – Jet Aircraft............................................................................................ 31

5.3 Range and Endurance of Jet Aircraft...................................................................................... 33

5.4 Jet Case: SAR in terms of C L and C D ..................................................................................... 33

5.5 Jet Case: SE in terms of C L and C D ........................................................................................ 33

5.6 Cruise Climb............................................................................................................................ 34

5.7 Program 2 Constant Lift Coefficient and Density.................................................................... 37

5.8 Program 3 Constant Speed and Density ................................................................................ 38

6. Correctly Banked Level Turn ...................................................................................................... 41

6.1 Load Factor n .......................................................................................................................... 41

6.2 CL in Correctly Banked Level Turn.......................................................................................... 41

6.3 Maximum Turn Rate................................................................................................................ 42

6.4 Tightest Turn ........................................................................................................................... 43

6.5 Centripetal Acceleration .......................................................................................................... 43

6.6 Drag During Turn .................................................................................................................... 44 6.7 Minimum Turning Radius ........................................................................................................ 46

7. Steady Banked Climb - Jet Aircraft............................................................................................. 47

7.1 Steepest Climb........................................................................................................................ 48

7.2 Best Rate of Climb .................................................................................................................. 48

8. Turn at Constant Thrust and C L.................................................................................................. 51

9. Take-off Performance ................................................................................................................. 52

9.1 Basic Estimate of Takeoff Run................................................................................................ 52

9.2 Refined Estimate..................................................................................................................... 55

9.3 Multi-Engine Aircraft Take-Off Performance Considerations – Balance Field Length............ 56

Notation

a Local sonic speed. For an idea gas, a = √(γ airRT).

a Lift curve slope (a = dC L /d α)b Wing span

c Chordc f Thrust specific fuel consumption


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CL Lift coefficientCLmax Maximum lift coefficient (dependent on aerofoil section).CD Drag coefficient

CM Pitching moment coefficientD Drag

h AltitudeHe Energy Height

L LiftM, m Aircraft massm f Fuel mass

M = V a Mach Number

M Pitching moment

R Gas constant for air ( = 287.1 J.kg -1.K-1)

Re = Vc Vc ρ µ ν = Reynolds NumberS Reference area: wing plan-form (projected)

P 0 ISA sea level static pressure

T0 ISA sea-level air temperature

V True airspeed

Vstall Stall speed

W Aircraft weight (W = mg)

T Air temperature (absolute)

T Thrust

WAT Weight, altitude, temperature

w w = Mg/S = wing loading

α Angle of attack (AoA) β Angle of sideslip

air γ Ratio of specific heats for air ( = 1.4)

γ Flight path angle

ρ Air density

0 ρ ISA sea-level air density

τ Thrust-to-weight ratio T Mgτ =

0 µ ISA sea-level dynamic viscosity

µ Dynamic viscosity of air

ν Kinematic viscosity of air ( ν = µ/ρ)

Ω Turn rate


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1. Introduction

Aircraft Performance involves the prediction of the capabilities of aircraft based on:

• Aerodynamics and key airframe design parameters• Power-plant characteristics• Atmospheric conditions• Operating environment

Performance estimation is important in the specification & design of new aircraft, in the modification ofexisting types, and in determining how a given type can be operated. Key performance parametersinclude:

• Maximum speed• Maximum rate of climb, maximum climb angle, ceiling altitude• Range

• Endurance• Maximum rate of turn, minimum turn radius• Runway requirements

The atmosphere is modelled using the “International Standard Atmosphere” (ISA). For example, byinternational convention, at ISA sea level, pressure P 0=101325 N m

-2, temperature T 0 = 288.16 K,density ρ0 = 1.2256 kg m -3, and viscosity µ0 = 1.783x10 -5 Ns m -2.

This course of lectures will deal with the basic theory as it applies to sub-sonic fixed-wing aircraft.Extensions of the theory are required to allow supersonic aircraft to be dealt with. Rotorcraft(helicopters, tilt-rotors etc) require an altogether different approach.

1.1 Aerofoil Geometry

Figure 1-1 Symmetric Aerofoil

Figure 1-2 Aerofoil Section Parameters

• Draw Chord Line• Camber line drawn with respect to the chord line.


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Introduction Aspect Ratio, Taper, and Sweepback

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• Thickness Distribution which is added to the camber line, normal to the camber line.

1.2 Aspect Ratio, Taper, and Sweepback

Figure 1-3 Wing Geometry

Gross wing area S = plan area of wing including part within fuselage

Mean chord c S b=

Aspect Ratio2

: b b

ARS c

= =

Taper = ct/c0

Trapezoidal wing S = 0.5 ×(c0 + ct) ×b

1.3 Fundamentals of Lift and DragFigure 1-4 shows the four main forces acting on an aircraft flying straight-and-level with a velocity V

through still air. The forces are: Lift (L), Drag (D), Thrust (T) and Weight (mg)

Figure 1-4 Force in Level Flight


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Introduction Lift Coefficient

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The following points should be noted.1. So far as rigid-body motion is concerned, which is the case in Aircraft Performance

analysis, the normal pressure and shear-stress distributions caused by the flow can bereplaced by a single equivalent force acting at the centre of pressure (CP).

2. The lift is the component of the resultant aerodynamic force perpendicular to V.3. The drag is the component of the resultant aerodynamic force parallel to V.4. The CP is not a fixed point; it shifts with AoA, Reynolds number, Mach number etc.5. Thrust is assumed to act parallel to V. 6. For performance analysis, the aircraft is assumed to be in trim; moments balance, and

one can treat all the forces as though they act at the centre of gravity (CG).Lift and drag forces will be functions of many variables, principally:

• angle of attack α • angle of sideslip β • airspeed V• Mach number M

• vehicle shape and size (characterized by appropriate length d)• local air properties (density ρ, temperature T, viscosity µ)

1.4 Lift CoefficientIt is useful to deal with a non-dimensional Lift Coefficient, defined as

212

: L L

C V S ρ

=

Using Dimensional Analysis it can be deduced that the Lift Coefficient

( )Re, , LC f M shapeα β = , , It is the dependence of C L on incidence ( α ) that is the most important, at least in performanceestimation. A typical “C L vs. Alpha” relation for an aerofoil is shown in Figure 1-5. The C L-α curve for acambered asymmetric section is shown in Figure 1-6.

Figure 1-5 C L vs. Alpha (Symmetric Section)

Figure 1-6 C L vs. Alpha (cambered section)

The reference area S, whether dealing with wings or an entire aircraft, is the wing plan-form area. Thelift-curve slope dC L /dα is typically about 2 π rad -1. Stall typically occurs at an incidence of


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Introduction Lift Coefficient

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approximately 15 deg. The maximum lift coefficient C Lmax is approximately 0.9 for thin symmetricaerofoils; modern wing sections with high-lift devices (flaps, slats) may have a value C Lmax of 2.5 ormore.

1.4.1 Reynolds and Mach Number Ratio Effects

These important effects, although outside the scope of this course, do nonetheless merit someremarks. The Reynolds Number determines whether flow is laminar or turbulent, and hence whereflow separates. Thus even at fixed incidence, C L is to some extent dependent on the Reynoldsnumber. This must be taken into account when inferences based on scale-model wind-tunnel tests aremade during the design process. The same is true of Mach Number; Figure 1-7 shows have the lift-curve slope may depend on Mach Number M.

Figure 1-7 Lift curve slope vs. Mach Number

1.4.2 Influence of Aspect Ratio and Aerofoil Thickness

Aspect ratio (AR) has a very strong influence on wing performance. For unswept wings at low Machnumber the following approximate relationship holds (see Figure 1-8):

2( ) 4

a ARa

a AR

π

π ∞

∞

=+ +

.

Aerofoil thickness is also a key influencing factor. Thin sections (t/c < 0.08) => separation from leadingedge, rapid increase in drag. Thick sections (t/c) > 0.15 => separation from trailing edge

Figure 1-8 Lift curve slope vs. Aspect Ratio


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Straight and Level Flight Drag Coefficient and Drag Polar

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Figure 1-9 Influence of Aerofoil Thickness on C L

1.5 Drag Coefficient and Drag PolarIt is also more convenient to work with drag in a non-dimensional form. The drag coefficient is definedas the drag force D divided by the product of the free-stream dynamic pressure and the wingreference area S, viz.

212

D

DC

V S ρ

∆= .

As in the case of lift, DC is Reynolds and Mach number dependent, i.e.

( ),Re, D DC C M α = but in performance estimation, it is usually sufficient to assume that

( ) D DC C α = .According to the theory of aerodynamics, the drag coefficient is well approximated by the so-calledDrag Polar

20 D D L

k C C C

ARπ = +

in which k > 1 and typically around 1.1. We shall use the alternative notation

2 D LC a bC = +

The coefficients a and b are Reynolds and Mach number dependent, but in performance estimationthey are usually treated as constants.

2. Straight and Level Flight

In straight and level flight, Lift = Weight, i.e. L mg=

2.1 Lift to Drag Ratio (aerodynamic efficiency)Wings are designed to generate lift. Drag is an inescapable side-effect of lift generation. The lift-to-drag ratio (L/D) can be regarded as a measure of Aerodynamic Efficiency (E).

: L D

L C E D C

= =


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Straight and Level Flight Stall Speed

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The L/D ratio has a maximum when( )

2 0 D L

L L

d C C ab

dC C = − + = .

i.e. when( / ) max L D L L Dmd

C C C a b= = = .

The corresponding drag coefficient is ( / ) max 2 L D DC a= . Note that we have also called this C Dmd whichdenotes C D for minimum drag in straight-and-level flight. The maximum (L/D) ratio is

( )max 1 2 L DC C ab=

Typical (L/D) max valuesGliders 35 (approaching 60 for competition-class)Transport 18Light aircraft 14Fighter 10 (M1)

2.2 Stall SpeedTo maintain lift as speed is reduced, the aircraft must be flown at an increased incidence. Stall occurswhen C L reaches C Lmax .

max 212

Lstall

mgC

V S ρ = =>

1max2

stall L

mgV

C S ρ =

2.3 Minimum Drag and Minimum Drag Speed (V md)

Drag21

2 D D V SC ρ = Using the drag polar,

( )2 2 2

2 2 2 21 1 12 2 22 21 1

2 2 L

mg m g D V S a bC V S a b V Sa b

V S V S ρ ρ ρ

ρ ρ

= + = + = +

The drag force comprises two components:

Profile/Zero-Lift or Form Drag: 210 2 D V Sa ρ = ; this increases as the square of speed

Induced/Lift-Dependent) Drag:

2

212

i

bL D V S ρ = ; this decreases as 1/V

2

for given L

Example: An aircraft has wing loading 2300 N/m 2 and C Lmax = 1.4. Find the stalling speed at (i) sealevel ( ρ =1.225 kg/m 3) and (ii) 5000m ( ρ = 0.737 kg/m 3).

Solution: (i) At sea level,2300

0.5 1.225 1.4stallV =

× × = 51.8 m/s, (

51.80.514

= 100.8 kts).

(ii) At 5000 m,2300

0.5 0.737 1.4stallV =

× × = 66.8 m/s (

66.80.514

= 130.0kts)

(NB 1 knot = 0.514 m/s)


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Straight and Level Flight Minimum Drag and Minimum Drag Speed (Vmd)

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Figure 2-1 Total Drag and its Components Figure 2-2 Flight at varying C L and V

Dmin can be found as follows.

( )2

b mg D ax x

= + , where 212: x V S ρ =

Minimum occurs when( )2

2 0b mgdD

adx x

= − = ; i.e. when :md x x mg b a= = .

min 2 D mg ab= .

Note that D min is independent of density (i.e. altitude) but is directly proportional to aircraft weight.Hence the heavier the aircraft, the greater will be the drag. The speed at which the drag force is

minimum is denoted V md and is given by

2md

mgV b a

S ρ = .

Vmd depends on wing loading w = mg/S and on altitude; as the latter increases, density decreases,

and V md increases. This must be so, in order to maintain21

2 md V S mg b a ρ = . From the above it canbe deduced that

( )2md

md

b mgax

x= ;

i.e. at the minimum drag speed, profile drag = induced drag. The lift coefficient at the minimum dragcondition is given by 212 Lmd md md C L V S mg x a b ρ = = = , i.e.

Lmd C a b=

Figure 2-3 (vmdplot.m) shows a typical drag versus speed variation. Shown alongside in Figure 2-4 isthe required power: i.e. the product of drag and speed. Power is discussed in more detail in Section2.6.


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Straight and Level Flight Minimum Drag in terms of CL

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Figure 2-3 Drag vs. Speed vs. Altitude Figure 2-4 Required Power vs. Speed vs.Altitude

2.4 Minimum Drag in terms of C L In straight-and-level flight, drag can be written: ( ) ( ) D L D D L mg C C mg= = . For a given vehicleweight, straight-and-level flight can be maintained over range of different speeds, each with acorresponding value of C L (or equivalently, of incidence α ). What is required is that 212 Lmg V SC ρ = .For a given vehicle weight, drag is minimized by minimizing D LC C . If one assumes that

2 D LC a bC = + , it follows that D L L LC C a C bC = + which is minimized by solving

( )0 D L

L

d C C

dC = .

The details follow:

( )2 0

D L

L L

d C C ab

dC C = − + = => Lmd

aC

b= as before, and 2 Dmd C a= .

At the minimum, ( )min 2 D LC C ab= and the minimum drag can be written( )min min 2 D L D mg C C mg ab= = as calculated before.

2.5 Equivalent Air-SpeedThe Equivalent Air Speed (EAS) denoted V E is defined such that

2 21 102 2 E V V ρ ρ = where 0 ρ is the

sea-level density. Wherever dynamic pressure occurs in formulae, use of V E enables two variables tobe replaced by one; for example, the drag force, if written in terms of EAS, viz.

( ) ( )( ) ( ) ( )( )

2 2

2 21 102 22 21 1

02 2

E

E

b mg b mg D a V S a V S

V S V S ρ ρ

ρ ρ = + = +

is independent of altitude (i.e. density).

2.6 Power Required and Minimum Power Speed (V mp)The power P required to overcome the drag D at speed V is given by P =DV. Hence

( )2 2

3 2 3 31 1 12 2 221 1

2 2

( ) L

mg b mgP V S a bC V S a b V Sa

V S VS ρ ρ ρ

ρ ρ

= + = + = +

Power can also be expressed in terms of C L rather than V as follows:


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Straight and Level Flight Maximum Speed – Jet Aircraft

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2 D L L

C mgP mg

C SC ρ = × =>

( )

32

32

2( ) D

L

C P mg

S C ρ = × ×

In a similar to manner to drag, a minimum power condition can be found in terms either of V or of C L.

For given weight and altitude, it is32

D LC C that must be minimized. Assuming the standard drag

polar,3 12

D 2 23 32 2

C( ) ( )

( ) ( )

L L L

L L

a bC a C b C

C C

−+= = +

whose minimum occurs when5 1

3 12 22 2( ) ( ) 0 L La C b C

− −− + = , i.e.

5 12 23 ( ) ( ) L La C b C

− −= => 3: L Lmp

aC C

b= =

The drag coefficient at minimum power is 4 DmpC a= .

Hence

( ) ( )

1 3312 4 4

L 3min

C 4 DC a b= .

The minimum power is given by

33 1 42 4

min

24( )

3b

P mg aS ρ

= × × ×

and the speed for minimum power is142

3mpmg b

V S a ρ

= .

The following can also be useful: 3 31min 2 2mp mp mp Dmp mpP D V V SC a V S ρ ρ = = =

Typical variation of P with V over a range of densities is shown in Figure 2-4 above. Note that P min andVmp both increase with altitude. (See dashed line on graph.) Note, too, the reduction with increasingaltitude in the required power for high-speed straight-and-level flight. It has been shown that

142

md

mg bV

S a ρ =

and

142

3mpmg b

V S a ρ

=

and from this, it follows that

( )141 3 0.76mp md md V V V = = .

Vmp is typically relative low, and sustained flight at V mp may be undesirable, due to the onset of stalland buffet, but also partly due to the inherent speed instability below V md ; i.e. “on the back side of thedrag curve.” In terms of lift-to drag ratio (aerodynamic efficiency) one finds that

mp md md max

30.87 0.87

2 E E E E = ≈ =

2.7 Maximum Speed – Jet AircraftIn many performance calculations, the maximum thrust available from a jet engine is assumed to beindependent of speed (see Figure 2-5) and to vary as a function of relative density in the followingmanner

( )max, 0 x

SLT T ρ ρ =


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Straight and Level Flight Maximum Speed – Jet Aircraft

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where max, SLT is the maximum thrust at sea level and x is a positive constant, often close to unity; see

Figure 2-6. (Note that in practice, thrust may be a function of forward speed for a turbo jet.)As long as minT D≥ , steady flight is possible. In general, two speeds V can be found satisfying

2 221

2 21

2

bm gT V Sa

V S ρ

ρ = + .

and these are indicated in Figure 2-5. Rearranging, one obtains following quadratic equation:

( ) ( )22 2 2 2 21 12 2 0V S a V TS bm g ρ ρ − + = the roots to which are

( )2

2

41 1

T abV

Sa T mg ρ

= ± −

.

Three possibilities arise:

1. 2T mg ab> Flight possible over range of speeds;

The maximum speed is given by( )

2max 2

41 1

T abV

Sa T mg ρ = + −

.

2. 2T mg ab= Flight possible at one speed, namely, md V V = ;This corresponds to the absolute ceiling of the aircraft.

3. 2T mg ab< Steady flight not possible; aircraft is above absolute ceiling.

Figure 2-5 Drag and Maximum Thrust

Example: Estimate the maximum speed at sea-level of a jet-powered aircraft with the following data:

Thrust-to-weight ratio at SL = 0.20; E max = 14. Thrust variation: ( )0.7

max, 0SLT T ρ ρ = . Drag polar: a =0.015, b = 0.085. S = 85 m2; Mass m = 33000 kg.

( ) ( )

2

max 2 2

4 0.2 33000 9.81 (1 14)1 1 1 1

1.225 85 0.015 0.2

T abV

Sa T mg ρ × ×= + − = + − × ×

m/s = 283m/s (550kts)

The maximum speed for the above jet aircraft is plotted in Figure 2-7 as a function of relative density.


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Straight and Level Flight Absolute Ceiling for Jet Aircraft

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Example: Estimate the absolute ceiling of a jet-powered aircraft with the following data: Thrust-to-

weight ratio at SL = 0.20; E max = 14. Thrust variation: ( )0.7

max, 0SLT T ρ ρ = . Drag polar: a = 0.015, b= 0.085. S = 85 m2; Mass m = 33000 kg.

1 10.7 0.7

0 max

10.23

14 0.20SL

mg E T

ρ ρ

= = = ×

At the ceiling,T

V Sa ρ

= where T = 0.2 x (33000g) x (0.23) 0.7 = 23143N and ρ = 1.225 x 0.23 =

0.282kg.m -3. Therefore23143

0.282 85 0.015V =

× ×m/s = 254 m/s (493kts)

Figure 2-6 Thrust Dependence on Air Density

2.8 Absolute Ceiling for Jet AircraftUnder the above assumptions, the density corresponding to the absolute ceiling altitude is defined by

the condition ( )max, 0 2 x

SLT T mg ab ρ ρ = = .

i.e. ( )1

0 max,2 x

SLmg ab T ρ ρ = .The greater the thrust-to-weight ratio and the peak aerodynamic efficiency, the greater the ceiling. The

so-called service ceiling , defined as the altitude at which the best rate of climb has fallen to someprescribed value such as 100 ft min -1, is of greater practical importance.


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Straight and Level Flight Maximum Speed – Prop Aircraft

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Figure 2-7 Maximum Speed vs. Relative Density: Jet Aircraft

2.9 Maximum Speed – Prop AircraftThe standard approach for prop-driven aircraft is to assume that engine and propeller togetherproduce a power (as opposed to thrust) output that is independent of speed. Calculations analogousto those carried out above for the jet aircraft, but based on power rather than drag, lead to formulae formaximum speed, absolute ceiling, and so on. The resulting equations do not however lend themselvesto simple closed-form solutions, and generally, iterative solutions need to be found.Suppose the maximum engine shaft power output is P max . The propeller will convert this to usefulpropulsive power with an efficiency η ; the useful available propulsive power is thus given by

maxavP Pη = The propeller efficiency, though in practice a function of many variables including speed, will betreated as constant.


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Climb Performance Steady Climb

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Example: calculate the maximum speed at sea level of a turbo-prop aircraft with the followingparameters: m = 42000kg, S = 100m 2, shaft power output 8000kW, propeller efficiency 0.85, dragpolar a = 0.015, b = 0.055.The equation that needs to be solved is

2

31max 2 12

( )b mgP V Sa VS η ρ ρ = +

i.e. maximum available power = required power. In general there will be two solutions; it is the greaterof the two that is required, the other corresponding to low-speed flight on the back side of the dragcurve, where buffet and stall rather than power considerations will be the limiting factor. The terms onthe RHS of the equation are, respectively, the power losses due to form drag and due to induced drag.An effective iterative approach is first to estimate V max based on form drag only: i.e.

6max 33max,1

2 2 0.85 8 101.225 100 0.015

PV

Saη ρ

× × ×= =× ×

=194.9m/s.

This must be greater than the true maximum speed since the induced power loss has been neglected.

The procedure now it to calculate the induced power component ,1iP at max,1V and subtract it from themaximum available power, then form a new estimate max,2V based on this lower available net power.

One finds that2 2

,1 1 1,12 2

( ) 0.055 (42000 9.81)1.225 194.9 100i i

b mgP

V S ρ × ×= =

× × ×W = 782kW.

=>6

max 33max,2

2 2 0.85 (8 0.782) 101.225 100 0.015

PV

Saη ρ

× × − ×= =× ×

=188.3m/s

2 2

,2 1 1,22 2

( ) 0.055 (42000 9.81)

1.225 188.3 100i

i

b mgP

V S ρ

× ×= =× × ×

=810kW

=>6

max 33max,3

2 2 0.85 (8 0.810) 101.225 100 0.015

PV

Saη ρ

× × − ×= =× ×

=188.1m/s

After three iterations, the scheme has converged to within 1m/s of the required value.

3. Climb Performance

Two types of climb will be considered. The first is the so-called steady climb. This is a climb in whichthe all acceleration components (i.e. parallel and perpendicular to the flight-path) are assumednegligible. Then, in Section 3.2 we consider the case of climb with acceleration which is particularlyrelevant to the analysis of high-performance aircraft.

3.1 Steady ClimbIn a steady climb, the forces of lift, drag, thrust, and weight are in balance; see Figure 3-1. The flight-path angle, denoted γ, is the angle between the inertial velocity vector and the horizontal plane . It isassumed that the thrust acts along the flight path, i.e. parallel to V.


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Example: Let the drag polar be a = 0.014, b = 0.06, weight mg = 1.4MN, wing area S=285m 2.Conditions: ISA sea-level. Maximum thrust = 0.2 x mg.

( ) ( )max max min D LT mg C C γ = − .Lift coefficient for steepest climb: L Lmd C C a b= = =0.483 and ( )min 2 D LC C ab= =0.058.Steepest Possible Climb: maxγ = 0.2 – 0.058 = 0.142rad = 8.1deg.

Corresponding True Air Speed:6

max2 2 1.4 10

1.225 285 0.483md Lmd

mgV V

SC γ ρ × ×= = =× ×

m/s = 129m/s

Corresponding rate of climb: max max max max maxsinh V V γ γ γ γ γ = ≈ɺ

=129m/s x 0.142 =18.3m/s

Figure 3-1 Aircraft in Steady ClimbBy resolving the forces parallel and perpendicular to the flight path, the following equilibrium equationsare obtained (see Figure 3-1):

cos L mg γ =

sinT D mg γ = + From these, we deduce that:

D

sin

C - cos

L

T Dmg

T mg C

γ

γ

−=

=

We initially assume that the flight-path angle is small. This will be a good approximation for all but themost high-performance aircraft and it leads to considerable simplification. We can then write:-

DC - L

T mg C

γ ≈

.

3.1.1 Steepest Climb (shallow)

To maximize the flight path angle, the thrust (T) needs to be maximum and the aircraft must be flownat the incidence which maximises L DC C , i.e. at the minimum drag speed.

3.1.2 Excess Power and Maximum Climb Rate

Also of interest is the rate of climb hɺ , given by

sin T Dh V V mg

γ −= =

ɺ

.


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The rate of climb will be maximized when the excess power ( )V T D− is maximum. If the throttle isat maximum setting, the thrust T (in the case of an ideal jet) or the propulsive power P = VT (in thecase of an ideal prop) will be maximum and the corresponding quantity

max : sT D

V Pmg

− =

or, in the propeller case,

( )max : sVT VD

Pmg

− =

is called the specific excess power . At any given WAT there will be a speed at which P s is maximumand this is the speed at which the aircraft can gain height fastest in a steady climb. (From this point on

we shall present only the equations for an ideal jet aircraft.) The formula for hɺ can be re-expressed interms of speed V by substituting for drag D,

( )2

2 2 2 21 1 12 2 2 21

2

D L

mg D V SC V S a bC V S a b

V S

ρ ρ ρ

ρ

= = + ≈ +

where ‘ ≈’ is used because we are assuming small γ . Hence the rate of climb is approximately:23

12 21

2

VT V S mgh a b

mg mg V S ρ

ρ

≈ − + ɺ

Figure 3-2 shows typical rate-of-climb and flight-path angle versus speed variations for a jet aircraft atone WAT. Distinct speeds corresponding to best rate-of-climb and steepest climb are seen to exist. Interms of the lift coefficient, the rate of climb can be written:

2 2 2sin sin L

L L L

a bC mg mg T h V

SC SC mg C γ γ

ρ ρ

+= = = −

ɺ

The lift coefficient corresponding to best rate-of-climb is found by solving2

1/ 2 3/ 2

b - 0

L

L L L L

a C dh d T dC dC mg C C

+= =

ɺ

For a jet aircraft where T is assumed independent of V, the solution is

( ) ( )2max

+ 12

2s L LPT mg T mg ab

C C b

− += = .

The speed for maximum rate of climb is

max

max

2

s

h

LP

mgV

SC ρ

=ɺ

Example: Let the drag polar be a = 0.014, b = 0.06, weight mg = 1.4MN, wing area S=285m 2.Conditions: ISA sea-level. Maximum thrust = 0.2 x mg. (see also Figure 3-2):-

Lift coefficient for maximum rate of climb:( ) 2

max

0.2 + 0.2 12 0.014 0.06

2 0.06s LPC

− + × ×=×

= 0.198

True Air speed for maximum rate of climb:3

max

2 4.91 101.225 0.198h

V × ×=

×ɺ = 201m/s.

Corresponding flight-path angle: max D

h

L

C T mg C

γ

= −

ɺ =0.2–(0.016/0.198)rad=0.119rad = 6.8deg.

Maximum rate of climb: max max maxh hh V γ ≈ ɺ ɺɺ

= 201 x 0.119m/s = 23.9m/s.


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Figure 3-2 Variation of rate-of-climb and climb angle with speed at given T/W

3.1.3 Maximum Climb Rate versus AltitudeThe above analysis shows that the climb performance is dependent on altitude via ρ and T, both of

which enter the various equation. Using the data from the previous example we can investigate thedrop in climb performance with altitude. We assume that the maximum available thrust varies with

relative density according to ( )nSL ρ ρ with n=0.7. The results are tabulated in Table 3-1. Note thatthe maximum rate of climb falls with altitude.

h [m] ρ [kg/m 3] T/W C L Psmax maxhV ɺ (m/s) CD maxhɺ

(m/s) γ (deg)0 1.225 0.200 0.198 201.2 0.016 23.63 6.7

2000 0.967 0.169 0.229 210.5 0.017 19.93 5.74000 0.780 0.146 0.260 220.0 0.018 16.81 4.86000 0.629 0.125 0.294 230.6 0.019 13.87 4.08000 0.507 0.108 0.329 242.6 0.021 11.06 3.2

10000 0.409 0.093 0.366 256.2 0.022 8.35 2.412000 0.330 0.080 0.404 271.6 0.024 5.69 1.6

Table 3-1: Maximum rate of climb as a Function of Altitude (see "rate of climb.xls")

It is now possible to estimate the minimum time t min to climb from height h 1 to h 2 by plotting max1 hɺ

against h and integrating:2

1

minmax

h

h h

dht

h== ∫ ɺ

The word estimate is stressed, because the analysis so far is quasi-steady; i.e. assumes there to beno acceleration along the flight path at any altitude. In fact during the climb schedule in Table 3-1 theTAS increases by 70m/s by 12000m.


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Climb Performance Accelerative (none-steady) Climb: Energy Height

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Figure 3-3: Minimum time to climb from h1 to h2 as area under (1/ maxhɺ

) vs. h graph

In Figure 3-3 each grid square represents 40seconds. So to climb from 0 to 2000m will takeapproximately 90s. From Table 3-1, note that in climbing the first 2000m, the speed would haveincreased from 201 to 211 m/s during the climb: an average acceleration of (211 - 201)/90 = 0.11m/s 2.Note that as the ceiling altitude is approached, the function (and the area under it) rapidly increases.To climb from sea level to 8000m will take approximately 480 s (8 minutes)

3.2 Accelerative (none-steady) Climb: Energy HeightEnergy height ( e H ) is defined as the aircraft’s total energy per unit weight:-

2 212 : =

2emgh mV V

H hmg g+= +

Energy height is useful for analysing the climb of high-performance aircraft undergoing rapidacceleration and climb, e.g. jet fighters with high thrust/weight ratio, for which the previous methods ofanalysis based on quasi-steady motion are inadequate.

Contours of constant H e are shown in the (V, h) plane in Figure 3-4; the curves are parabolae whoseparametric equation is 2 2eh H V g= − . An aircraft diving from condition ‘A’ along a constant energyline (i.e. flown with thrust = drag) would have a crash speed of V A. The equation of motion of the (jet)aircraft along the flight path is

max - - sin dV

T D mg mdt

γ =

This can be rearranged to give

max( - ) 1sinV T D dV

V V mg g dt

γ = +

The term on the LHS was encountered earlier, it is the Specific Excess Power (SEP), denoted sP .


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SEP = max( - )

sV T D

Pmg

= .

The SEP is the excess power (per kg of aircraft) after drag has been balanced at the given throttlecondition, speed and height, available to accelerate the aircraft and/or to make it climb. The term onthe RHS is the rate of change (derivative) of the energy height, i.e.

sin edH V dV V g dt dt

γ + =

Therefore

. esdH dh V dV

Pdt g dt dt

= + =

The dimensions of sP are the same as those of velocity; the SI units of SEP are m/s.

Figure 3-4: (V,h) diagram with contours of constant energy height (h e)In the previous quasi-steady state analysis this excess power was considered only available forclimbing. However, in general the excess power can be used to either accelerate in level flight, or to

climb or do both. Suppose the aim is to go from energy state A to energy state B. One could firstperform a constant velocity climb, followed by constant altitude acceleration, as shown by dotted linesin Figure 3-4. Alternatively, one could accelerate at constant altitude, then climb at constant speed.One could also do anything in between. In fact, there are infinite possible flight paths from A to B.However there will be one strategy which is quickest . There will be another which results in minimumfuel burn. Suppose an aircraft flying at (V1, h1) is required to climb to (V2, h2). What sort of flight pathshould it follow and what is the minimum time to accomplish the manoeuvre? (The situation isdepicted in Figure 3-5.) Exact solution of such problems generally requires application of the Calculusof Variations. We will consider some approximate solutions.

Figure 3-5: Possible climb strategies to change height and speed

We will consider the former.


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Figure 3-6: V-h envelope showing contours of constant SEP

Figure 3-7: 1/(dH E/dt) max vs. H E

3.2.1 Maximum Energy-Rate Climb Schedule

For each energy level there will be one point in the envelope (i.e. one V, h combination) correspondingto maximum P S. The set of these points is precisely the set of points at which the families“Ps=constant” and “H E=constant” curves are tangent to one another. The locus of these points (V

*, h *)is shown by the solid line with intermittent ‘*’ in Figure 3-6. The value of P S at these points is maximumand is equal to the maximum rate at which H E can be increased. It follows that this locus defines the

quickest path between different energy levels, i.e.


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2 2

max max1 1

min e e

e e

H H

e e

e S H H

dH dH t

H P= =∫ ∫ɺ

Thus for example, according to Figure 3-7 the minimum time for transfer from 3km to 8km energyheight is approximately 3.5 x 0.05 x 2000 seconds = 350seconds (~6 minutes).

Knowledge of the time-optimal trajectory allows the climb schedule to be computed as follows. Alongthe optimal trajectory, h and P S are know functions of V, i.e. ( )h h V = and ( )S S P P V = .

Figure 3-8: Optimal Schedule of (V, h) for fastest gain on total energy.

The gradientdh

dV

can be computed by differentiating the function in Figure 3-8.

Also, by the chain rule,dh dV

hdV dt

=ɺ .

Since

S

VV P h

g

dh VV dh V V V

dV g dV g

= +

= + = +

ɺ

ɺ

ɺ

ɺ ɺ

This gives the required acceleration schedule:-

S PV dh V dV g

=+

ɺ .

The required flight path schedule is computed as follows:21 1 1 1 1

sin e S S S S dH PdV dV dh dh V

P P P V dh dV dV gV dt g dt V g dt V g V dV

γ = − = − = − = + +

3.2.2 Maximum Acceleration in Level Flight

Maximum acceleration is determined in level flight at three separate altitudes and over a range of

speeds.


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Energy height2

2eV

h hg

= +

For constant hV dV

.g dt

edhdt

=

V (m/s) 210 230 240 260 270 290 300dV dt

(m/s 2)- 4.26 4.18 3.90 3.65 2.65 1.49

edhdt

(m/s)- 99.8 102.3 103.4 100.4 78.3 45.6

eh (m) 5248 5696 5936 6445 6716 7286 7587

h = 3000 m

dV dt

(m/s 2)- 3.69 3.61 3.31 3.08 1.85 -

edhdt

(m/s) - 86.5 88.3 87.7 84.8 54.7 -

eh (m) 8248 8696 8936 9445 9716 10286 10587

h = 6000 m

dV dt

(m/s 2)3.05 3.02 2.90 2.77 2.56 1.49

edhdt

(m/s)65.3 70.8 70.9 73.4 70.4 44.0

eh (m) 11248 11696 11936 12445 12716 13286 13587

h = 9000 m

The above data could be used to estimate the minimum time to climb from 3000m and a speed of290m/s to a height of 9000m at 230m/s, assuming the aircraft weight remained constant.

0 50 100 150 200 250 3000

2

4

6

8

10

12

14

V [m/s]

H e i g h t [ k m

]

A

B

Ps=103m/s

Ps=88m/s

Ps=73m/s

Figure 3-9: Optimal Schedule of (V, h) for fastest gain on total energy.

Points A and B in Figure 3-9 represent the start and end points respectively. The three diamondsrepresent the locations of maximum P s at 3km, 6km and 9km altitudes.


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Climb Performance Removing the small-angle assumption

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3.3 Removing the small-angle assumptionThe analysis of the climb in Section 3 was mostly based on small-angle approximations. The analysiscan be extended to the large angle case, as is now shown. The equations are more cumbersome andanalytical solutions have to give way to numerical ones. Returning to the “large angle” formula:

sin cos D L

C T T

C mg W γ γ

+ = =

,

(T/W) is plotted in Figure 3-10 against γ for three different values of C L /C D. The dashed linesrepresent the small-angle linear approximation. The function (T/W) has a maximum

( ) ( )2max 1 D LT W C C = + at tan L DC C γ = ; for practical values of C L and C D (T/W) max is approximately 1 and it is achieved

close to γ =90°.Thus an aircraft with a thrust-to-weight ratio of around unity will be able to climb almost vertically. Forvalues of (T/W) < 1, the value of the flight-path angle can be calculated in terms of C L /C D as follows.

( )

( ) ( )1 1

2sin tan1

D L

D L

T W C C

C C γ

− −

= − + .

This is tabulated below for four different T/W ratios for the case (L/D) = 16 and plotted in Figure 3-11.

( )T W 0.1 0.25 0.6 1.0γ 2.2° 11° 33° 83°

Table 3-2 Maximum climb angle for varying (T/W) at (L/D) = 16

N.B. We assume that W ≅ constant during the climb. Also, T decreases as h increases. Therefore theclimb angle will also decrease, becoming zero at the ceiling condition.

Figure 3-10 Flight Path Angle versus Thrust-to-Weight Ratio


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Climb Performance Removing the small-angle assumption

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Figure 3-11 Maximum Climb Angle versus TWR and L/D[The above calculations are performed using m-file plot_gam_steep_climb.m]

3.3.1 Steady Climb Rate

The rate of climb can be written as follows

( )

221

2 2

2 cossin

T S bh V V V a

mg mg V S mgγ

γ ρ ρ

= = − −

ɺ

This equation is awkward to deal with, since γ and V cannot be separated algebraically. But by fixingV, the corresponding value(s) of gamma can be computed numerically, and by repeating the process,a graph of v versus γ constructed.

Figure 3-12 Climb performance based on numerical solution to full equations


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Climb Performance Steady Shallow Climb with Head-Wind

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3.4 Steady Shallow Climb with Head-Wind

Figure 3-13: Climb into headwind: forces and velocity components

Key variablesV = Velocity of plane relative to moving air massVw = Velocity of air mass relative to groundVi = Velocity of plane relative to ground (inertial velocity).γ = Flight path angle

Assumptions1) Horizontal steady head-wind2) Steady climb (i.e. forces balanced)3) Thrust parallel to drag4) Vw Rate of climb sin sin 'idh

V V dt

γ γ = =

Since angles are small, wV

bV

γ ≈ and ' 1 wV bV

γ γ γ = − ≈ −

Equilibrium Equations L mg≈

sin 'T D mg γ = + Equation for flight-path angle ( γ )

Since sin ' ' 1 wV T D T Dmg mg mg L V

γ γ γ = − ≈ − ≈ ≈ −

it follows that1 w

T Dmg L

V V

γ

−

≈−

.

In terms of lift and drag coefficients,

2 21 12 2

L

L mgC

V S V S ρ ρ = ≈

212

D

DC V S ρ =


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Climb Performance Steady Shallow Climb with Head-Wind

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the flight path angle is given by

( )

2121

D

L

w L

C T mg C

V C

mg S

γ ρ

−

≈−

In terms of the variable

( )

212: w

V mg S ρ

β = ,

which we see is simply the dynamic pressure due to wind speed normalized with respect to wingloading, we can write:

1

D

L

L

C T mg C

C γ

β

−

≈−

.

Note that in the case of a tail wind, this becomes

1

D

L

L

C T mg C

C γ

β

−

≈+

.

Figure 3-14: Flight-path angle vs. lift coefficient: headwind

Zero headwind

25m/s headwind


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Glide Performance Glide Angle

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Figure 3-15: Flight-path angle vs. lift coefficient: tailwind

4. Glide Performance

Assumptions: glide angle γ small (i.e. shallow glide)Relevant to sports gliding and to analysis of performance following engine failure

Note: no thrust force appears in analysis!Equilibrium equations:-

(i) cos L mg γ = (ii) sin D mg γ =

4.1 Glide Angle

From (ii), => sin D L

C D Dmg L C

γ = ≈ = .

Hence to minimize glide angle (and thereby maximize range) we require the minimum drag condition;i.e. fly at V md .

Figure 4-1: Forces acting on aircraft during glide

Zero tailwind

25m/s tailwind


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Cruise Performance Rate of Descent

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4.2 Rate of DescentRate of descent = sinV γ

= D

V mg

D

L

C DV V

L C ≈ =

Now the lift coefficient 2 21 12 2

2 L

L

L mg mgC V V S V S SC ρ ρ ρ = ≈ ⇒ ≈

Therefore

Rate of descent = 1.52 D

L

C mgS C ρ

The quantity on the right-hand side has a minimum at3

: L Lmpa

C C b

= =

The corresponding airspeed is mpV (which is approximately 0.76 md V )

Minimizing the rate of descent maximizes the time in the air from a given starting height.

5. Cruise Performance

Aircraft spend much of their flight in the cruise . The majority of fuel carried is likely to be consumed

during cruise. In Civilian operations, cruise mode has strong influence on economics of operation: costof fuel vs. cost of time. In Military operation, cruise determines

–radius of action –need for in-flight refuelling –endurance for patrol, surveillance,

It is important to be able to calculate: • how far a given aircraft can fly at given flight condition• for how long aircraft can fly at given flight condition• optimal conditions for maximum range or endurance given certain amount of fuel

Assumptions• Steady level straight flight with no acceleration • Forces and moments in balance (i.e. in trim)• In practice, cruising may involve very low levels of climb or acceleration• Adjustments to course neglected (corrections can be applied if necessary)

5.1 Range

5.1.1 Safe Operating Range

The SOR is the max distance between airfields which the aircraft can fly with full allowance forheadwind, diversion, stacking.

• Distance flown in climb included• Fuel for take-off, climb, landing included• Allowances made in standardized manner according to type of operation

5.1.2 Gross Still-Air Range This is what we shall focus on. It assumes that the aircraft starts at cruising altitude

• No allowance made for climb, descent etc• Important to know how to calculate it


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Cruise Performance Cruise Calculations – Jet Aircraft

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Changes in all-up mass (m) during cruise are not negligible because large quantities – 100s of tons –of fuel can be consumed.

M+m f = total mass (aircraft + fuel) at start of cruisem f = mass of fuel burned during cruiseω = (M+m f)/Mζ = m f /(M+m f) = 1 – (1/ ω)

Typical values: very long range: ω = 1.5; mid/long range: ω=1.3; short range: ω=1.1

5.2 Cruise Calculations – Jet AircraftDuring cruise, weight mg falls; in the case of long range aircraft, this can be by a very considerableproportion. During cruise, we assume forces are balanced, i.e. negligible accelerations. Therefore

212 L L mg V C ρ = =

T = D

Note that as m decreases, ρ and/or V and/or C L must be adjusted to maintain L = mg. Determining

optimal ways of doing this, say to minimize fuel burn on a given route, is complicated because range isa function of many variables, and its true “optimization” requires application of the calculus ofvariations. The following three flight programmes are however of considerable practical importance:

1. Cruise Climb. Fix V and C L and climb gradually so density decreases in proportion with m.2. Cruise at constant altitude and constant speed, gradually reducing C L.3. Cruise at constant altitude and constant C L, gradually reducing speed.

The cruise climb is of particular practical importance in commercial aviation. The advantage of fixingCL is that the lift-to-drag ratio can be kept near to some sort of “optimum”.

Let m = total “all-up” mass at time t. The mass dynamic due to fuel burn can be expressed:

f

dmm

dt = − ɺ

where f mɺ denotes the fuel burn rate. Jet and prop aircraft have to be treated separately due todifferent fuel consumption characteristics, which we model as

Jet aircraft f F m c T =ɺ Prop aircraft f m cP=ɺ

Jet: c F = TSFC (assumed constant) Prop: c = SFC (assumed constant)

We focus on the jet case. Noting that T = D, we can write range and endurance as integrals, first withrespect to time (t), then w.r.t. mass (m).

Range =( ) ( )0

0 (0) ( )

f f M m M mm t mt

f f f f t m m t M M

dm dm dm VdmVdt V V V

m m m c D

+ +

=

= − = = =

∫ ∫ ∫ ∫ ∫ɺ ɺ ɺ

Endurance =0

f M mT

f t M

dmdt c D

+

==∫ ∫


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Cruise Performance Cruise Calculations – Jet Aircraft

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m denotes the mass of fuel burned after t seconds of cruise.

5.2.1 Cruise ClimbSince V and CL are constant, range is given by

ln f f f M m M m M m

f L L

f f f D f D M M M

M mVC VC V dm V L dm dm Rc D c mg D c gC m c gC M

+ + +

+ = = = = ∫ ∫ ∫

ln f L f D

M mVC R

c gC M

+ =

("Breguet Range Equation")

Eliminating V, this can also be written as

( ) 0.51

21ln

f f L

F D

M m g M mC R gc S C M ρ

+ + = × ×

where ρ1 denotes the density at which the cruise climb commences. Note how cruise climb range is

• proportional to 0.5 L DC C , hence can be maximized by flying at some optimal incidence;• inversely proportional to the square root of the “starting” density, hence higher altitude should

lead to greater range, subject of course to ceiling limit.• inversely proportional to c f.

The value of C L for which cruise climb range is maximum can easily be shown to be C L = 3a b ; thecorresponding drag coefficient is C D = 4a/3.

It is useful to derive an expression for drag. When the all-up mass has reduced to m and the density toρ, the following equations must hold:

1

f M mm ρ ρ

+= .

By the time all the cruise fuel has been used, m = M and 2 ρ ρ = where

2 1

M m M ρ ρ

+= => 21 f

M M m

ρ ρ

=+

.

In other words, density will vary in the range

1 1 f

M M m

ρ ρ ρ

≥ ≥ +

On the other hand, drag

( )1

D D f

L L

C C D mg M m gC C

ρ ρ

= = +


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Cruise Performance Range and Endurance of Jet Aircraft

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Engine performance will diminish as altitude increases, so at all times we must have

max

x

SLSL

T D ρ

ρ

>

In other words we must have

( )max1

x

DSL f

SL L

C T M m g

C ρ ρ

ρ ρ

> +

5.3 Range and Endurance of Jet AircraftFuel flow rate governs thrust T: f F m c T =ɺ

f F

V V SAR

m c T ⇒ = =

ɺ

and1 1

f F

SE m c T

= =ɺ

Since, in level flight, T = D and L = mg

F

V LSAR

c D mg= × 1

F

LSE

c D mg= ×

5.4 Jet Case: SAR in terms of C L and C D

In straight and level flight 212 Lmg V SC ρ = so2

L

mgV

SC ρ ⇒ = .

122 1 2 L L

F L F D F D

C C V L mgSAR

c D mg SC c C mg c Smg C ρ ρ

∴ = × = × =

Max SAR obtained by flying at AoA corresponding to ( )12max

L DC C .

Was an exercise to show ( ) ( )11 42 3

max0.75 1 3 L DC C a b= at 3 LC a b= .

This gives optimum airspeed for max range V=

14

4 23 1.316md mg b

V S a ρ

= .

Note:Vmd is a function of m and ρ For practical reasons this optimum is not usually employed

5.5 Jet Case: SE in terms of C L and C D 1 1 LF D

C SE

c mg C = × ×

We know that ( )max1

2 L DC C ab= and that the corresponding lift coefficient


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Cruise Performance Cruise Climb

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Lmd

aC

b=

=> For maximum SE, jet aircraft should be flown at V mdHowever, this speed is on the edge of the region of speed instability so it may not be desirable actuallyto maximize endurance.

5.6 Cruise Climb

5.6.1 Range

Have shown previously that F

V LSAR

c D mg= ×

Therefore ( )11

M mf M mf

F M M

VL dm R SAR dm

gc D m

+ + = = ×

∫ ∫

Since C L is constant => (L/D) is constant, hence

1

1 1ln

M mf

F F M

VL dm VL M mf R gc D m gc D M

+ + = × × = × × ∫

1

1ln

F

VL M mf R

gc D M

+ = × × ("Breguet Range Equation")

Substituting 2

L

mgV

SC ρ = ,

( ) ( )12 0.253max

0.75 1 3 L DC C a b= (has maximum at V = 1.316 V md ) one obtains

( )0.5

1

1 2ln L

F D

C mg R

gc S C ω

ρ

= × ×

(Note, mg/ ρ is constant).

So ( )0.5

1max

max

1 2ln L

F D

C mg R

gc S C ω

ρ

= × ×

Optimum airspeed for max range V max-range =

14

4 23 1.316md mg b

V S a ρ

=

Lift coefficient 3 0.577 L Lmd C a b C = = . Note, however, that it is not always possible or desirable to fly at V = 1.316 V md Therefore we need to determine effect on range of cruising jet at other speeds It is helpful to userelative airspeed u = V/V md where

0.252

md

mg bV

S a ρ =

Note too that in cruise climb, Vmd will be constant. So

( )0.5

1

1 2ln L

F D

C mg R

gc S C ω

ρ

= × ×


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Now ( )0.252 md mg

a b V S ρ

= ,2

2 2

2 md L

V mgC a b

SV V ρ = = ,

22 4

2 41md md

D

V V C a b a b a

V V

= + = +

=>( )

( )( )

3

1 4

1ln

1md md

F md

V V V R

gc ab V V ω

= × × ×

+

Figure 5-1: Range vs. speed in cruise-climb


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5.6.2 Endurance

1 1 LF D

C SE

c mg C = × ×

( )11 1 M mf M mf M mf L L

F D F D M M M

C C dm dm E SAR dm

gc C m gc C m

+ + + = = × = ×

∫ ∫ ∫

(Remember, C L is constant in cruise climb). Therefore,

1

1ln L

F D

C M mf E

gc C M + = ×

.

Max endurance achieved at C L = C Lmd corresponding tomax

1

2 L

D

C C ab

=

. That is,

1max

1 1ln

2F

M mf E

gc M ab

+ = ×

However, flying at V md is unlikely to be desirable. What will be the endurance at other (suboptimal)

speeds? It can be shown, by writing0.52

2md

L

V aC

V b

= , that

( )

( )

2

1 4

1 1ln

1

md

F md

V V M mf E

gc M ab V V

+ = × × × +

Example Turbojet aircraft, S = 75 m 2 starts cruise at mass 18 000 kg of which 3500 kg is usable fuel.Drag Polar: a = 0.025, b = 0.065, c F = 2.8 × 10 -5 kg N -1 s -1Starts cruise at altitude where ( ρ / ρ0) = 0.53 Determine maximum gross range in an ideal cruise climb.

Solution: Fuel ratio ω = 18000/(18000-3500) = 1.241

( )0.5

1

1 2ln L

F D

C mg R

gc S C ω

ρ

= × ×

( )2 2 18000 9.81

0.53 1.225 75mgS ρ

× ×=× ×

(Evaluated at start of cruise, but constant throughout)

( ) ( )( )

12

0.2530.253max

0.750.75 1 3 17.95

3 0.025 0.065 L DC C a b= = =

× ×

( ) ( )1max 51 2 18000 9.81 17.95 ln 1.2419.81 2.8 10 0.53 1.225 75

R − × ×= × × ×× × × ×

= 1.2017 × 10 6 m = 1201.7 km

As for the speed, we know that 3 0.025 3 0.065 0.358 LC a b= = × = 2 18000 9.81

2 142.3 / 0.53 1.225 75 0.358 L

V mg SC m s ρ × ×

⇒ = = =× × ×


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Cruise Performance Program 2 Constant Lift Coefficient and Density

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Method 1 (cruise climb) yields best range for given quantity of fuel. Air traffic control may not allowaircraft to keep gradually climbing. Aircraft on long flights are usually allowed to perform ‘step climb’during cruise; this involves discrete steps. e.g. discrete steps of 2000 ft after 2 hours to bring mg/ ρ closer to its optimum value. The climb angle is clearly not zero, but so small that it does not seriouslyaffect assumption that flight is level

Figure 5-2: Endurance vs. speed assuming cruise-climb

5.7 Program 2 Constant Lift Coefficient and Density

In this case, true airspeed is reduced as fuel is burned to satisfy 212 Lmg V SC ρ =

It can be shown that C L constant => relative airspeed constant: in fact

21

md L

V aV C b

=

.

Note that reducing TAS during cruise is not normally practical or desirable . Also, there is the needcontinually to compute the optimal airspeed and reduce throttle setting to maintain C L constant. Under this assumption,Error! Objects cannot be created from editing field codes.

0.5

1

1 2

1 2

M mf L

F D M

M mf L

F D L M

M mf L

F D M

C dmV

gc C m

C mg dmgc C SC m

C dmc gS C m

ρ

ρ

+

+

+

= × ×

= × ×

= × × ×

∫

∫

∫


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Cruise Performance Program 3 Constant Speed and Density

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( )

0.5

2

0.5

0.5

0.5

1 2

1 22

1 2 2

221

M mf L

F D M

M mf L

M F D

L

F D

L

F D

C dm R

c gS C m

C m

c gS C

C M mf M c gS C

M mf C M c gS C M mf

ρ

ρ

ρ

ρ

+

+

= × × ×

= × × ×

= × × × + −

+= × × × − +

∫

Now ( ) ( )12 0.253max

0.75 1 3 L DC C a b= .

5.8 Program 3 Constant Speed and Density

Assume level flight with L = mg , ρ = constant . Thrust = drag2 2

212 21

2

bm gT D V Sa

V S ρ

ρ = = +

Fuel burn rate f F m c T =ɺ .2 2

212 21

2 f F bm gm c V Sa V S ρ ρ

∴ = +

ɺ

3 2 221

2 212

M mf

M F

V R m

bm gc V Sa

V S

δ ρ

ρ

+

=

+

∫

For constant V and ρ this integral can be evaluated analytically by reducing it to a standard integral ofthe form:

221

2 21 1

1tan

x x

x x

dx x xα α α

− = + ∫

as follows...

Example Same Turbojet aircraft as previous example, S = 75m starts cruise with mass 18000 kg ofwhich 3500kg is usable fuel. Drag Polar: a = 0.025, b = 0.065, cF = 2.8 x 10 -5 kg N -1 s -1 Starts cruise at altitude where ( 0 / ρ ρ ) = 0.53Determine maximum gross range under cruise method 2

( ) 0.52max

max

221 L

F D

M mf C M R

c gS C M mf ρ

+ = × × × − +

2max 5

2 2 18000 18 3.517.95 1

2.8 10 0.53 1.225 9.81 75 18 R −

× −= × × × − × × × ×

= 1140575 m = 1140 km (cf 1201 km with cruise climb)


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212

3 22 212 2 2 21 21

2 2 2 212

212

2 22122

211 12

2

1

1

1tan tan

M mf M mf

F M M F

M mf

F M

F

V S V V R m m

c bgbg V S c V Sa m V Sa mV S bg

V S V m

c bg V S am

b g

V S V M mf M c bg

ρ δ δ

ρ ρ ρ ρ

ρ δ

ρ

ρ α α α

+ +

+

− −

= = + +

= +

+ = × −

∫ ∫

∫

where221

2 2: V S a

b g ρ

α =

.

The range can be simplified to

where : mf M mf

ς =+

.

There is no simple, closed-form expression for the maximum range R with respect to V at given heightwhen R is of the above form. Using the data from the previous example, we plot R 3(V) vs. V and findthe optimum cruise speed numerically: Maximum range = 1139 km. (See Figure 5-3)

Figure 5-3: Range (R3) vs. Speed at constant altitude and constant speed

( )

1 1 13

12

tan tan tan1

tan

1 1

F F

F

mf V M mf M V

R M mf M gc ab gc ab

M mf V

gc ab M mf

α α α

α α

ς α

ς α

− − −

−

+ = − = + +

+ = + + −


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Figure 5-4: Cruise Range (R3) (V and ρρρρ constant)showing theoretical advantage of flying faster and higher

Figure 5-5: Effect of varying amount of fuel The above analysis has led to expressions for the gross still-air range of a jet aircraft cruising at

• R1: Constant V and C L • R2: Constant r and C L

• R3: Constant V and r


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Correctly Banked Level Turn Load Factor n

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6. Correctly Banked Level Turn

For an aircraft flying at a speed V to execute a level turn of radius-of-curvature R and turn-rate Ω =V/R requires a centripetal force in the horizontal plane. The force must be of magnitude

2 2F mV R mR mV = = Ω = Ω.

The best way to provide such a force is to roll the aircraft about its longitudinal axis, thereby rotatingthe lift vector inwards to produce a horizontal component. Denote this roll (bank) angle φ . Then theequations of motion for a correctly banked level turn are:

cos L mgφ = 2

2sin V

L m mR mV R

φ = = Ω = Ω

From this we can deduce the following important relationships

2tan V Rgφ =

2 21 ( ) sec L mg V Rg mg φ = + =

2 2 4 2 21 ( ) L m g R mg R g= + Ω = + Ω

2 2 2 21 ( ) L m g V mg V g= + Ω = + Ω

Note that the required lift force is greater than that required in straight-and-level flight to support theweight.

6.1 Load Factor n

lift forcein turnn

weight

∆=

Hence2 2 2 2 2

sec 1 ( ) 1 ( ) 1 ( ) L

n V Rg R g V gmg φ = = = + = + Ω = + Ω andError! Objects cannot be created from editing field codes.

φ (deg) 0 5 10 20 30 45 60 70N 1 1.004 1.015 1.064 1.155 1.414 2 2.9

Table 6-1: Load factor n as a function of bank angle φφφφ

It is customary to speak of an n-g turn: one in which the normal load factor is n .

6.2 C L in Correctly Banked Level Turn

2 21 12 2

L L nmgC V S V S ρ ρ

= =


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Correctly Banked Level Turn Maximum Turn Rate

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Suppose that an aircraft is flying straight-and-level at speed V and that it then starts to execute an n-gturn, maintaining speed constant. The lift coefficient must increase by a factor n.

=> Drag coefficient C D must also increase (because C D = a + bC L2)

Note that C L cannot increase beyond C Lmax so there is a danger of stalling if n is too large.

Example Light aircraft: V = 50 m/s, w = 1000 N/m 2, C Lmax = 2.

max212

L L

nmgC C

V S ρ = < =>

21max2 LV C n

w ρ < = 3.06 (at ISA sea level)

This says nothing about whether the aeroplane structure is designed to withstand this load, nor aboutwhether the power plant can provide enough thrust or power at the corresponding C L. (We return tothis point later.)

6.3 Maximum Turn RateThe lift coefficient in a correctly banked level turn can be written,

2 2 2

212

L

m g V C

V S ρ + Ω=

Hence the drag force is given by:

2 2 2 2 221

2 212

m g m V D V Sa b

V S ρ

ρ + Ω= + .

This simplifies to

2 2 221

2 21 12 2

( )mg m D V Sa b b

V S S ρ

ρ ρ Ω= + + .

The above variation of drag with speed differs from the straight-and-level flight case only by theaddition of a constant offset term proportional to 2Ω ; for any Ωsufficiently small that the aircraft hassufficient thrust to maintain a turn, the minimum drag occurs at

2turnmd md

mg bV V

S a ρ = = .

The minimum drag in a turn at Ωrad sec -1 is given by

2 2

min 12

2turn

m D mg ab b

S ρ Ω= + .

From this it follows that the thrust-limited maximum rate of turn is given by

maxmax 2

2

T S g ab

b mg mg

ρ Ω = −


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Correctly Banked Level Turn Tightest Turn

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where maxT denotes the maximum thrust available from the jet engines.

6.4 Tightest TurnFor a given load factor (equivalently, for given bank angle φ), the turn radius is

2 2

2tan 1

V V R g g nφ = = −

For given bank angle, 2 R V ∝ => turn radius large for fast aircraft.

The drag force in a banked level turn of radius R and speed V can be written

( )2 2212 21

2

4b mgm

D V S a b RS V S

ρ ρ ρ

= + +

which is of the form2

212 21

2

( )b mg D V Sa

V S ρ

ρ ′= +

For a given value of R , the minimum drag occurs when2

2 212

( )0

( )b mg

aV S ρ

′ − = , i.e. when

212 2

4

bV S mg b a mg

ma b

RS

ρ

ρ

′= = +

and the minimum drag force itself is

2

2min 2 2 4 m D mg a b mg ab b RS ρ

′= = + .

The tightest turn achievable by a jet aircraft occurs at the radius R min for which

min max D T =

i.e.

2

2max

min

2 4 m

mg ab b T R S ρ

+ =

.

(This is assuming thrust to be independent of V.) Rearranging,

min 2

max

4 /

4

bm S R

T ab

mg

ρ =

−

which is achieved at a speedmin

max

2 Rb mg mg

V S T ρ

=

6.5 Centripetal Acceleration2

2 1 y

V a n g

R= = −

Example: load factor n = 2 (L = 2mg)


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Correctly Banked Level Turn Drag During Turn

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22 1 3 17 ya g g⇒ = − = ≈ m/s 2.

If V = 200 m/s,2 2

2 2

200

1 9.81 2 1

V R

g n

= = ≈

− × −

2354 m

6.6 Drag During Turn

( )

2 2 22 21 1

2 2 2212

D

n m g D V SC V S a b

V S ρ ρ

ρ

= = +

T (thrust) = D (drag)2 2 2 2nmg n m g

T x a b ax b x x

⇒ = + = +

where 212 x V S ρ ∆=

2 2 2 2

0ax Tx bn m g⇒

− + = Can solve for x, and hence V as a function of n,

2 22 0

x x mga bn

T T T

− + = =>

221 1 4

2

mgabn

T xT a

± − =

=>2

2 21 1 4T mg

V abnSa T ρ

= ± − .

Alternatively, for n 2 as a function of V ...

=>( ) ( )22 21 12 22

2 2

T V S a V S n

bm g

ρ ρ −= .

The condition2

21 4 0mg

abnT

− = defines the maximum load factor possible at the given thrust-to-

weight ratio, and the corresponding speed is given by: max nT

V Sa ρ

= . The load factor is then

max

1

2

T n

mgab

=

Exercise: show that the corresponding lift coefficient C Lnmax = a b = C Lmd.

The following graph shows the n-V envelope for a particular aircraft. Check that n max = 4.74.


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Correctly Banked Level Turn Drag During Turn

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0 50 100 150 200 250 300

1

1.5

2

2.5

3

3.5

4

4.5

5

V [m/s]

l o a

d f

a c

t o r n

CD = 0.02 + 0.05*C L2, w = 3161 N/m 2, TWR = 0.3, C Lmax = 2.5, S = 90 m

2

upper power limitlower power limitlower stall limit

Figure 6-1: Sea-level n-V envelope based on thrust and drag, showing lower limit due to stall

Note: The intersection of the stall limit and the lower load-factor limit defines the Corner Speed . Thisis the minimum speed at which the aircraft can reach its maximum load factor without stalling

Figure 6-2: Speed-Load Factor envelope showing contours of constant turn rate increasingfrom 2 to 16deg/s. Maximum rate of turn for given configuration occurs at approximately n=3,

v=100m/s

Corner speed


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Correctly Banked Level Turn Minimum Turning Radius

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Figure 6-3: V-n envelope showing contours of decreasing turn radii from 4km to 0.2km

6.7 Minimum Turning Radius

It has been shown that2

2 1

V R

g n=

−. For a given V, we need the largest possible n. This is

precisely what the previous figure gives us. We then need to minimize the resulting function of V. Wecan do this as follows. Let us assume first that we are to the right of the stall boundary.

( )2

min 2max 1

V R V

g n=

− where n max is a function of V and where ( ) ( )

22 21 1

2 22max 2 2

T V S a V S n

bm g ρ ρ −= .

Differentiate R min with respect to V.

( )

( )32

2 2 maxmax max

min

2max

2 1

1

dnV n V ndR dV

dV g n

− −=

−. This derivative equals zero when ( )2 maxmax max2 1

dnn Vn

dV − = .

But by differentiating the n max - V relationship, we have: ( ) ( )21

2maxmax 2 2

22

T VS a VS V S dnndV bm g

ρ ρ ρ −= .

Therefore, we need to solve:

( ) ( ) ( )2122

max 2 2

22 1

2

T VS a VS V S n V

bm g

ρ ρ ρ −− =

( ) ( ) ( ) ( )22 2 21 1 12 2 22 2 2 2

21

4

T V S a V S T VS a VS V S V

bm g bm g

ρ ρ ρ ρ ρ − −− =


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Steady Banked Climb - Jet Aircraft Minimum Turning Radius

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( )21 12 22 2 1

T V S

bm g

ρ =

2 22 4 4bm g b mg T V

S mgST ρ ρ = =

2tt b mg T

V S mg ρ

= .

NB: need to check that this is greater than stall speed!

Assuming it is, the corresponding (thrust-limited) minimum-radius turn is obtained by back-substitution:

2

4

4

tt

b mgg S

R T ab

mg

ρ

= −

For small turning radius, need small b, large ρ (i.e. low altitude), low wing-loading , and highthrust/weight ratio.

7. Steady Banked Climb - Jet Aircraft

We consider the idealized case in which a jet aircraft banks about the velocity vector by an angle φ and the flight path angle is γ, and the latter is assumed small enough that sin γ = γ . The forces in theXZ-plane are as depicted in Figure 7-1. The horizontal component of lift Lsin φ produces a centripetal

acceleration which turns the vehicle.2sin

V L M

Rφ = (7.1)

R denotes the turn radius of curvature.

Figure 7-1: Forces in XZ-plane During Banked ClimbThe remaining equations of motion, assuming steady flight, are

cos cos L Mg Mgφ γ = ≈ (7.2)sinT D Mg γ − = (7.3)


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Steady Banked Climb - Jet Aircraft Steepest Climb

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From these, we can deduce that the flight path angle and rate of climb (dh/dt) are

sin D L

C T n

Mg C γ = − (7.4)

and

2 D L L

C dh nMg T ndt SC Mg C ρ

= −

(7.5)

where n = 1/cos φ is the normal load factor.

For a conventional aircraft, T/Mg, C D /C L, and n are each controlled by the pilot, via throttle, elevatorand ailerons respectively, roughly speaking.

Any turn (n>1) will degrade the climb performance. Let us quantify this in terms of steepest climb andbest rate of climb. Note that maximum climb rate is not achieved at the steepest possible climb, butrather at a somewhat shallower climb gradient, offset by a higher airspeed.

7.1 Steepest ClimbFor given n, maximum sin γ will occur at maximum thrust and maximum (L/D). This means that, for anidealized jet aircraft, the steepest climb will be achieved at the minimum drag condition C L=C Lmd irrespective of n. Speed and rate of climb will or course depend on n.

7.2 Best Rate of ClimbAt a given load factor and thrust-weight ratio T/Mg = τ, the lift coefficient C L giving best rate of climbcan be found by setting to zero the derivative with respect to C L of (7.5), i.e.

12

3

2

0 D L L L

C d C n

dC C τ −

− =

This leads to the quadratic

2 3 0 L LnbC C naτ + − =

which can be solved for, given n and τ, to yield C L in much the same way as the “n=1” case in section3.1.2. Using the following parameters

a=0.04; b=0.024; % Drag Polar M=350000; % All-up mass [kg]

S=500; % Wing reference area [m^2] CLmax=2.5; % Max lift coefficient Tmax_sl=0.9e6; % Sea-level max thrust [N]rho=1.225; % Ambient density [kg/m^3]

we find the C L variation shown in Figure 7-2. An m-file (banked_climb_performance_jet.m ) has beenwritten to solve the equations and plot the results. For example, at n=2, a maximum climb rate of21m/s can be achieved, this at a 7 deg flight-path angle and an airspeed of 165m/s. At the same loadfactor (and hence same bank angle), an 8 deg climb could be achieved at just over 130 m/s airspeedand the climb rate would then be 18 m/s.


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Steady Banked Climb - Jet Aircraft Best Rate of Climb

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1 1.5 2 2.5 3 3.50.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

Load Factor n

C L

Jet: Banked Climb: Full Throttle, CL optimized for Best Rate of Climb

Figure 7-2: Lift Coefficient for Best Climb Rate vs Load Factor

1 1.5 2 2.5 3 3.56

8

10

12

14

16

18

20

22

24

26

Load Factor n

C

l i m b R a

t e [ m / s ]

Jet: Banked Climb TWR=0.3; CD=0.040+0.024CL 2; w=6867N/m 2; ρ=1.225kg/m 3

Best Climb RateSteepest Climb

Figure 7-3: Climb Rate vs Load Factor


50/58

Steady Banked Climb - Jet Aircraft Best Rate of Climb

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1 1.5 2 2.5 3 3.52

3

4

5

6

7

8

9

10

11

12

Load Factor n

F l i g h t P a

t h A n g

l e [ d e g

]



Figure 7-4: Flight-Path Angle vs Load Factor

1 1.5 2 2.5 3 3.590

100

110

120

130

140

150

160

170

180

190

Load Factor n

A i r S p e e

d [ m / s ]



Figure 7-5: Airspeed vs Load Factor


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Turn at Constant Thrust and CL Best Rate of Climb

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8. Turn at Constant Thrust and C L

In this section, we consider what happens to a jet if, when flying straight-and-level, the pilot banks theaircraft to a load factor n while keeping thrust and angle-of-attack constant. The aircraft will start toturn and to descend. The equations of motion in the subsequent turn are

cos L nMg γ = (8.1)sinT Mg Dγ + = (8.2).

where we take γ as positive for a descending flight path; moreover will assume that γ is small so thatsin γ = γ and cos γ = 1. Since AoA is assumed not to change through the manoeuvre, neither too will C L or C D, so based on the pre-turn condition,

D

L

C T Mg C

= (8.3)

Meanwhile during the turn,

cossin

L

D

C L nMg D T Mg C

γ γ

= =+

Hence

cos 1 sin L D

C n

C γ γ = + .

Since cos 1γ ≈ , =>( )

1sin L D

nC C

γ −≈

The airspeed is given by2

L

nMgV

SC ρ =

(Note how the speed increases.)

The rate of descent is( )

( )1.512

sin L D

n n Mgv V

S C C γ

ρ

−= =

The turn radius is2

2

1 L

nw R

gC n ρ =

−

Example: M=350000kg, S=500m 2, a=0.04, b=0.024, C L=0.4, rho=1.0kg/m3. 20 deg bank, constant

throttle and AoA. CD=0.0438.

Rate of descent=( )4

1.5

sec 20 sec 20 12 35 10 9.811.0 500 0.4 0.0438

−× × ××

m/s = 1.34 m/s

Turn Radius = 10.23km


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Take-off Performance Basic Estimate of Takeoff Run

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9. Take-off Performance

Figure 9-1: Important take-off variables

Vstall Stall-speed. Aircraft capable of flight; continues accelerating on ground

Vmcg Minimum control speed on ground. Enough speed for aerodynamic

control on ground in event of engine failure for multi-engined a/c

Vmca Minimum control speed in air. Were the plane in the air (which it isn’t) there would be enough speed for aerodynamic controlin event of engine failure for multi-engined a/c

V1 Critical engine failure speed (multi-engined a/c). Decision speed. Engine failure before V 1 => takeoff must be abortedEngine failure after V 1 => takeoff can proceed

VR Rotation speed. Plane rotated to increase angle of attack (AOA),

hence lift. V R can equal V 1 but must haveVR > 1.05 V mc

Vmu Minimum unstick speed. If AOA is limited by ground clearance, plane must continue to accelerateon ground in rotated stated.(Definition assumes that plane is rotated to maximum extent possibleas limited by ground clearance).

VLO Lift-off speed, generally > V mu because for safety, rotation is limited to a value somewhat less than the ground-clearance limit.

These key speeds all typically lie in the range V stall < V < 1.1V stall

Once airborne, aircraft deemed to have taken off by the time it has reached Screen Height h s , whichmay be specified or laid down by regulation. For civil operations, h s = 35 ft (10.7m)

Takeoff run includes additional distance x 2 over ground while aircraft climbs to screen height.

9.1 Basic Estimate of Takeoff Run Although our initial assumptions are strong, they provide insight into some of the key parameters thatinfluence take-off distance.

Assumptions:1. Aircraft mass (m) constant


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2. Thrust (T) constant3. Drag (D) negligible4. Rolling resistance negligible5. V LO= 1.1V stall 6. C L = 0.9 C Lmax during initial climb (safety margin against stall)

Acceleration along runway: Basic DynamicsdV

m T dt

=

Acceleration can be writtendV dV dx dV

V dt dx dt dx

= =

Therefore,dV

mV T dx

⇒ = and Initial condition: V=0 when x=0 .1

0 0

LOV V x x

V x

m VdV Tdx= =

= =

⇒ =∫ ∫ 21

12 LOmV Tx⇒ =

9.1.1 Effect of Key Parameters on x1

From assumption VLO= 1.1V stall

( )22max

21.1 LO

L

mgV

SC ρ ⇒ = ×

( )

( ) ( )

( ) ( )

( )

21 122 2

1max

22

max

2

max

21.1

1.1

1.1

LO

L

L

L

mV m mg x

T T SC

mg

gSC T

mg S

gC T mg

ρ

ρ

ρ

= = × ×

=

=

ExampleGulfstream-like a/c (values taken from Anderson, J.D. “Aircraft performance and Design”

Wing-loading: 3679 N/m 2 Thrust: 110 920 N (at 0.7V LO)

Wing area: 88m 2Mass: 33 113 kg CLmax : 1.86

Assume V LO=1.1V stall (Vstall = 56.9 m/s)

Wing loading = 3691 N/m 2

Thrust/weight ratio = 0.34 Take density = 1.225 kg/m 3


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( ) ( )( )

2

1max

1.1 586 L

mg S x m

gC T mg ρ = = = (= 1920 ft)

9.1.2 Calculation of x2Once airborne, aircraft CG initially follows approximately circular path.

Change in flight-path angle: γ (rad) Radius of curvature of flight path: R (m)

Figure 9-2: Initial climb to screen height

2V L mg m

R≈ +

Load factor: 2

1 L V

nmg Rg

= ≈ +

Enforcing stall safety margin: Lift coefficient max212

( ) 0.9 L L L

C C V S ρ

= <

If one assumes V = V LO= throughout the initial circular climb, it follows that

( )21max 20.9 1.1 L stall L nmg C V S ρ ⇒ = < ×

( )20.9 1.11.09

mg

mg

= × ×

=

1.09n⇒ <

( )( )( )

2221.1 1.37

1 1stall

stall

V V R V

n g n g⇒ > = =

− − (R in metres, V stall in m/s)

( )2

2

1 cos2

sin

sh R R

x R R

γ γ

γ γ

= − ≈

= ≈


55/58

Take-off Performance Refined Estimate

(55/58)

( )2

2

1 cos2

22 1.17 2

s

ss stall s

h R R

h x R R h R V h

R

γ γ

γ

= − ≈

⇒ ≈ = × = = ×

Figure 9-3: Simplified geometry of initial flight-path acquisition

ExampleTo clear 10.7m screen height

2 1.17 2 1.17 56.9 2 10.7 308stall s x V h≈ × × = × × × = (= 1010 ft)

9.2 Refined EstimateIn practice,

a) resistance to motion is not negligible;b) thrust is not independent of speed

The following mathematical models are commonly used to describe the variation of resistive andpropulsive forces:

Drag: 212 D D V SC ρ = (Drag coefficient constant)Rolling resistance: ( ) N mg L µ µ = = −

Lift: 212 L L V SC ρ = (Lift coefficient also constant)Thrust: 20T T kV = − The equation of motion (level runway, no headwind) becomes:

( )212 D LdV

T V S C C mg mV dx

ρ µ µ − − − =

This can be written

2 dV a bV V dx

− = where 0T mgam µ −= and ( )12 D L

S b C k C

m ρ

µ = + − .

Documents

Notes on Aircraft Performance 2016v2