Flight Mechanics - Part 1

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Flight Mechanics – Part I (Aircraft Performance)Dr Shuhaimi Mansor, Aeronautical Engineering, Universiti Teknologi Malaysia.

SMF3212

Flight Mechanics

Course Content

1. Introduction Flight Mechanics

2. Basic Aerodynamics and International Standard Atmosphere

3. Aircraft PerformanceStraight and level flightClimbingRange and EnduranceTake-off and LandingTurning Flight

4. Aircraft StabilityLongitudinal Static Stability and ControlLateral Static Stability ControlIntroduction to dynamic stability




Part I:Aircraft Performance

Chapter 1

1. INTRODUCTION TO FLIGHT MECHANICS

Flight Mechanics involves:Performance

StabilityAeroelasticity

Performance:take-off, climb, cruising, range & endurance, decent and landing

Stability:static and dynamic stability, flight control

Aeroelasticity:The effect of structural flesxibility on performance, stability and control

In this course we focus onAircraft PerfomanceStatic Stability

An aircraft is analised as point mass flying under the effect of weight, aerodynamic forces, thrust,atmospheric.

Basic knowledge on:

1. Aircraft components2. Basic aerodynamics3. International Standard Atmosphere (ISA) and altitute4. Engineering mechanics

are required.




1.1 AIRCRAFT MAJOR PARTS

• Fuselage

• Wing

• Emphenage

• Power Plant• Landing Gear

Figure shows an aircraft major part.

The function of the major parts:

FuselageIt is a space to accommodate internal systems and components, payload, pilot and others in a

systematic manner. Should be low in drag, and also function as an attachment for wing, tail and powerplant. Common component seen in the fuselage:

Cockpit – Pilot sit. Also known as `flight-deck’, or `crew-cabin’ for large transport aircraftCanopy – Cockpit cover Tailboom – structure to carry emphennage Nacelle – space for engine Nose –front part of fuselage




WingTo generate lift which able the aircrafte to float. Common component seen in the wing:

Wing aerofoil to give lift force, drag force and pitching moment..Flap – trailing edge control surface use to increase lift and stall angle.Slat – extended control surface located at the leading or trailing ende of wing section use to

modify the wing aerodynamic characteristics.Aileron – control surface to make aircraft roll.

Terms and notation for wing:

S wing area b wing spancr root chord ct tip chord c mean chord

Λ taper ratio (ct/cr)

AR aspect ratio (b

2

/S or b/c)λ sweep angle

EmphenageEmphenage is used to stabilize the aircraft and to control the aircraft motion. Allow aircraft tomove in a control manner and safe. Component of emphenage:

1. Vertical Tail-provide directional stability. Also known as ‘fin’.2. Horizontal Tail –provide pitching stability. Also known as ‘tailplane’ or ‘stabilizer’.3. Elevator-a control surface located at the trailing edge of the horizontal tail use to control the

aircraft angle of attack.4. Trim Tab-a control surface located at the back of an elevator use to reduce stick forceexperience by pilot.

5. Rudder- a control surface located at the back of vertical tail use to control yaw angle.

PowerplantPowerplant generates power to drive aircraft toward its direction. There are four types of

powerplant engine: piston-prop, turbo-prop, turbo-fan dan turbo-jet.

a) Piston-propProston-prop engine is using a reciprocal combustion engine to produce power. Ouput power ismeasured in Horse Power (HP). The power is not depend on aircraft speed but varies with altitudeand throttle. The fuel consumption is proportional to horse power. Propeller converts the shaft power to thrust power. Thrust power is equals to the product of thrust force and aircraft speed.




b) Turbo-JetThe thrust is produced from the expansion of the hot gas combustion through nozzle. The thrust produce is a function of altitude and speed. Normally this type of engine is installed to high speed aircraft.

c) Turbo-FanTurbo-fan is a turbo-jet furnished with fan to increase the driving efficientcy of a low and mediumspeed aircraft. The operasion of the turbo-fan is quite similar to turbo-jet which is a thrust producing engine.

d) Turbo-ProbTurbo-prob is a piston and turbo-fan engine using propeller to convert engine power to thrust. Theoperasion of the turbo-prob is quite similar to piston-prob which is power producing engine.

Fundamentally both the turbo-fan and turbo-prob engine is a turbo-jet engine where the combustion

gas is expanded fully in the turbine in order to produde extra power from what is required to run thecompressor. The excess power is used to run the fan.

To simplify the study of powerplant, the type of powerplant is classified into two types.

1) Power producing engine which comprises of piston-prob, turbo-prob.2) Thrust producing engine which comprises of turb-jet and turbo-fan.




Chapter 2

AIRCRAFT PERFORMANCE

2.1 IntroductionAircraft performance is a measure of the ability of an aircraft to do its specific mission.

Civil Aircraft Operation - focus on operational cost and contribution to economical in operation

Military Aircraft Operation - under combat conditions, manoeuvring time, optimization time totarget, range, manoeuvrability and payload; is a measure of effectiveness and superiority.

Performance is also a measure of flight safety. While maintaining an access thurst in level flight, an

aircraft must be able to increase potential energy to climb. If the performance of an aircraft is notable to maintain both altitude and airspeed in climbing and decent, clearly this will limit the safetymargin and limit the safe operation. The performace aspect is the consideration of airworthiness,airworthiness practices, and performance which they are related.

Is not the purpose of airwothiness to limit or creates conflict in determining aircraft performanceand flight safety. It gives the code of practice and not to stop an aircraft to have a performace beyond the limit of code of practice. For that the airwothiness code of practice is to determine in a practical way, the safety limit of the aircraft operation for the risk of unsafe operation can bereduced to minimum level. Code of practice varies with aircraft size, number of engine, country of registration, operation requirement and time.

In aircraft performance the study is divided into two: Estimation and Measurement.

Performance Estimation: Estimation of aircraft performance from the design consideration inaerodynamics, powerplant and state of operation. Applicable for a new designed or modification of existing aircraft.

Performance Measurement: Flight performance measurement in true atmospheric, in which the pressure and temperature are different compare to design process, and data variation refer to ISA.

2.2 Atmospheric ModelThe performance of air breathing engine is depended on the combination of temperature, pressureand density of surrounding air. The motion of air mass and climate/season change create a dramaticchange in the distribution of earth atmosphere. For that, a single atmospheric reference is required to simplify the analysis. The common reference is based to the mid-latitute of the NorthernHemisphere, which is known as an ‘International Standard Atmosphere’ or I.S.A




The International Standard Atmosphere (I.S.A) represents the average climate/season atmosphereand geograhical atmosphere. An aircraft is assumed to operate far below the range from a cold articclimate to the hot tropical and the aircraft performance has to be estimated through this range. Theatmospheric model design is describe of giving an 'off-standard' data of the atmosphere. Basically,the atmospheric model can be obtained through an addition of a temperature rise to the element of

I.S.A atmospheric model with an adjustment to give a hot temperature, a cold temperature and thestandard model. Figure 3 shows the model used in the 'airwothiness codes of practice' (JAR).

2.3 Relative Atmosphere

The condition of atmosphere is defined as

RT P ρ =

If this equation is related to sea-level, the state of the ISA of the atmosphere is given by:

ooo T

T

P

P

ρ

ρ =

Or can be written as

σθ δ =

where

oP

P

=δ relative pressure

o ρ

ρ σ = relative density

oT

T =θ relative temperature

Figure 4 shows the properties of relative atmosphere.




Isothermal RegiondhgdP o ρ −=

dh RT

g

RT

dhg

p

dp oo ⎟ ⎠

⎞⎜⎝

⎛ −=−=

ρ

ρ




∫∫ −=h

h

o p

pdh

RT

g

P

dP

11

)(1o

o

hh RT

g

p

p

n −−=l

Or ⎥⎦

⎤⎢⎣

⎡−−

=)(

1

oo hh

RT

g

e p

p

But1111 ρ

ρ

ρ

ρ ==

T

T

p

p

⎥⎦

⎤⎢⎣

⎡−−=

)(

1

oo hh RT

g

e ρ

ρ

Gradient Region

)( 11 hhaT T −+=

dT a

dh1

=

T

dT

aR

g

p

dp o ⎟ ⎠

⎞⎜⎝

⎛ −=

∫∫ −=T

T

o p

p T

dT

aR

g

P

dP

11

11 p

pn

aR

g

p

pn o

ll −=

aR

g o

T

T

p

p −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =11

But111 T

T

p

p

ρ

ρ =




ThenaR

g o

T

T

p

p

T

T −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ==

1111 ρ

ρ

1)(

11

−−

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ =

aR

go

T T

ρ ρ

2.4 Airspeed Measurement

The relative speed between aircraft and the air mass is known as airspeed. This is a very important parameter because it affects many others flight parameters related to performance such as stallingspeed, best climb speed or cruising speed and maximum speed.

The airspeed is measured from the different between the pitot pressure (total) and static pressure of

atmosphere. The measured airspeed by the instrument is known as an ‘indicated airspeed ’, Vi. Asthe pitot-static is located within the air flow around the aircraft, the recorded pressure may bedifferent from the undisturbed free stream pressure. Correction to the air pressure is required (pressure-error correction) and the indicated airspeed is connected to pressure-error, which gives anairspeed is known as ‘calibrated airspeed, Vc. The calibrated airspeed is the measurement of aircpeed referred to an assumption that the atmosphere is having a constant pressure at allaltitude.This assumption is used in the altitude correction scale and gives the value of equivalentequivalent airspeed, Ve. This is based on the dynamic pressure which is given as,

22

2

1

2

1V Veq o ρ ρ ==

Where V is the true airspeed.

So that σ /VeV = gives the true relative velocity between aircraft and air mass.

The characteristic of the airflow is given in Mach number, M , which gives the ratio between trueairspeed and the speed of sound in free stream.

a

V M =

Where is the speed of sound which has the relationship with air temperature as;a

RT a γ =

For that the Mach number is not the value if air velocity measured by the instrumentation installed in the aircraft.




2.5 Airworthiness, Safety and Certification

To ensure aircraft operates with the degree of safety approved by the safety authority. Low risk statistic of 1 in 10 events.

Airworthiness requirement is somehow specific to the country of registration; generally has similar system, format and contents, probably different in code of practice, determination criteria and typeof test to meet the requirement of specific country.

American Federal Aviation Regulation, FAR's – USA and North America certificationEuropean Joint Airwothiness Requirement, JAR’s – European certification

FAR's part of Federal Law – fail to comply - criminal offence.JAR's is advisory body which leads to negotiation – fail to comply – not criminal offence but could be charged under civil action.

Military code of practice - American Mil.Spec. and Def.Stan. It has similar concept with civil code but considering military operation.

Airworthiness performance criteria covers – take-off, climbing and landing – elemen of high risk operations.

2.6 Aircraft Mission Profiles

Aircraft is designed to meet specific missions and requirements. Elements of mission demand the

combination of performance of aircraft and engine. Elements of mission comprise of take-off,climb, cruise, descent and landing.

Military operation: combat and attack manoeuvres.

Civil aircraft mission: carry payload from one place to other place, mission has to be planned toallow change of path and destination.

Military aircraft mission: more than civil, from transportation to interception. Normally return to based and carrying enough fuel for return flight although airborne refueling is allowed to increaserange or increase payload as long not exceed take-off weight.




Basic element of mission can be considered based on the engine thrust and fuel mission.

Table 2.1: Aircraft weight distribution (Ref: Ashelby M, Cranfield IT Lecture Notes)

Subsonic Transport Fighter Long Range Short Range

% Weight F/mg % Weight F/mg % Weight F/mgAccessory Airframe 40.0 45.0 45.0Powerplant 10.0 10.0 20.0Fuel

Takeoff 0.5 0.2-0.3 0.5 0.2-0.3 1.0 1.0Climb 2.5 0.15-0.25 2.5 0.2-0.25 - 0.7-1.0Cruise 26.0 0.06 11.0 0.08 21.0 0.15-0.25Decent 0.5 0.02-0.05 0.5 0.02-0.05 - 0.1Landing 0.5 0.15 0.5 0.15 0.5 0.4Reserve 5.0 5.0 2.5

Total Fuel 35.0 20.0 25.0

Payload 15.0 25.0 10.0Total 100.0 100.0 100.0Typical Weight (kg) 15,000- 30,000- 15,000-

400,000 200,000 30,000

Reserve fuel depends on company policy and operation regulation. Typical for civil aircraft, reserve fuel is enough for 45 minutes loiter plus 10% of total fuel weight.

Takeoff

Accelerate to reach takeoff speed within available takeoff distance. Normally required maximumthrust available.

For military aircraft, full power is usually apply during takeoff and climb to reach operation altitudeand March number in minimum time. Thrust to weight ratio is high around 1:1 and required 1% of aircraft weight for fuel burned (or 4% of total fuel).

For transport aircraft require around 0.2 to 0.3 thrust-weight ratio and depend on number of

engines. Four engine aircraft require two enjin memerlukan kurang dari 2 enjin dan pesawat 1kerana kegagalan salah satu enjin menyebabkan kehilangan tujah di dalam kadar yang kecil.Typically around 5% of aircraft weight is used during takeoff as fuel burned.

Climbing Civil transport aircraft climbs at slow rate and acceleration can be neglected when compare tomilitary fighter aircraft. Typically between 5 to 10% of fuel is burned during takeoff.




Cruise

Cruise is normally the longest segment of flight mission where large amunt of fuel is used. Lowthrust-wieght ratio, low specific fuel consumption is required in order to gain optimum range.Optimization process is complex which has to consider aerodynamic quality and powerplant.

Decent Descent at low thrust rate which normally close to idle thrust. However, powerplant should be ableto produce electrical power, hydraulic pressure and air bleed to limit the minimum engine speed.Fuel requires is typically low.

Landing

Thrust rate is considered high. This happened due to high drag configurationin order to avoid dragend part of drag curve fenomena. High thrust is necessary to ensure overall power required is ableto produce.

2.7 Aerodynamic ForcesResultant or vectored aerodynamic force is produced from the aircraft motion in atmosphere isresolved in wind-axis component. Component of forces along x-axis is called drag, D. The dragopposes the aircraft motion and function of velocity square. Component of forces along z-axis iscalled lift, L (normal to velocity). Lift force acting upward against aircraft aircraft weight and tomake aircraft floating in the air. Component along y-axis is called side force generates dueassymtrical motion or or velocity vector of symmetry aircraft. Sideslip angle is generated.

All aircraft external parts generate aerodynamic forces. Wing influences the aerodynamic forcessignificantly. The coss-section of the wing is called aerofoil. Symetrical aerofoil has symmetryshape about aerofoil cross-section axis. Assymetrial aerofoil id called cambered aerofoil. Positif

cambered aerofoil generates negative pitching moment which makes the occurred at negativeangle of attack. A vice versa criteria occur for a case of negative camber. Figure shows a positiveand negative camber.

oCL

Lift force L L SC V qSC L 2

2

1 ρ ==

Drag force L L SC V qSC D 2

2

1 ρ ==

L

T D

W




q - dynamic pressure (N/m2

),

ρ - air density (kg/m )V - true aircpeed (m/s)

LC - lift coefficient

DC - drag coefficient

Lift and drag coefficient is a function of aircraft angle of attack, Mach number, Reynolds number

and aerofoil shape. The typical curve is shown in figureα CL

The equation for drag polar is given by:

Re

2

A

C C C L

Do Dπ

+=

DC is drag coeffficient, is total lift coefficient contribution of wing, fuselage and horizontal

tail. is known as parasite drag coefficient at zero lift (contribution of wing profile drag, friction

and pressure drag of tail, fuselage, engine, undercarriage, and other parts expose to air flow.

LC

DoC

Term is called induced drag which depends on lift force. While e is known as Oswald

efficientcy factor, typical value of e is around 0.7 to 0.9.

Re/2

AC L π

At low Mach number (i.e. M<0.4), three types of drag are generated, profil drag, boundary layer drag and trailing edge vortex.

At high Mach number (i.e. M>1), drag due shock wave exist.

Wing Loading

Wing loading, is the ratio of aircraft weight to wingw

S

mgw = (N/m2)

Stall Speed

Stall speed is given as

max

2

LSC

mgVs

ρ =




General equation for aircraft performance

V

dt dH /

mg

L

D

T

γ

V mmg DT

kosmg L

&++=

=

γ

γ

sin

=− DT Excess Thrust =⎥⎦

⎤⎢⎣

⎡ +=g

V

V

dt dH mg

&/Potential Energy +Kinetic Energy

Consider aircraft is flying in steady straight and level flight, so that γ = 0, V = 0&

and mg L = DT =

T

mg

D

L= and Lift to drag ratio = E

E is known as aero efficiency.

pressuredynamic

loadingwing

V

S mg

S V

mgC L ===

2

212

21

/

ρ ρ

Equation for drag polar

2

L Do D KC C C +=

then2

L Do

L

D

L

KC C C

C C

D L

+==

For D

Lmaximum

)(2

0

2

L D

Do L

L KC C

C KC

dC

D

Ld

−

−=

⎟ ⎠

⎞⎜⎝

⎛




or which Do L C KC =22

12

min

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

K

C C Do

dragimum L

So the maximum aero coefficient, Do

mak KC E 2

1

=




Chapter 3

STEADY FLIGHT PERFORMANCE

3.1 Straight and Level Flight

Straight and level flight atconstant velocity is the simplest case in performance analysis. Forces thatacting on the aircraft is lift force, L and aircraft weight, W in which they are acting vertically, thrustforce,T and drag, D acting horizontally.

In equilibrium, thrust equals drag DT = 3.1.1and lift equals weight W L = 3.1.2

Lift is given by L

SC V L 2

2

1 ρ = 3.1.3

Then for lift equals weight LSC V LW 2

21 ρ == 3.1.4

3.2 Stalling Speed

From 3.1.4, aircraft speed V is given as

LSC

W V

ρ

2= 3.2.1

This equation is true if assumed the aircraft is flying weight W and wing area S at specific altitude.The stall conditions occur at maximum lift coefficientCLmax. Stall speed is given by

max

2

LSC

W V

ρ = 3.2.2

V , is the minimum speed for the aircraft to maintain steady fligh. Value of C Lmax is also depend onflap and landing gear. Stalling speed is normally determined from flight test.

3.3 Equivalent speed in level flight

In level flight W is equals to lift, L and can be written as

LSC V W 2

21 ρ =




From equivalent air speed,2

212

21

E oV V ρ ρ =

Then L E o SC V W 2

21 ρ = 3.3.1

Equivalent airspeed is given as, Lo SC

W V ρ

2= 3.3.2

For a given weight, W and wing area, S for an aircraft, is proportional toV LC

1

With the weight and wing area the equivalent airspeed for straight and level flight is only depend on lift coefficient and not depend on air density and altitude.

If the aircraft is flying at constant altitude or constant angle of attack, for that C L is constant, the

estimation of indicated airspeed is equals to equivalent airspeed will indicates constant value.

3.4 Minimum Drag, V md

Minimum drag is important for jet engine aircraft. At manimum drag, the aircraft speed givesmaximum endurance for jet engine aircraft. While maximum range occur when the aircraft is flyingat a speed a little higher than V md .

Figure shows the variation of drag with aircraft speed.

Determine the relationship between drag force and minimum drag condition.

⎥⎦

⎤⎢⎣

⎡= L

D L D

But in steady straight and level flight L = W , then

⎥⎦

⎤⎢⎣

⎡= L

D

W D

For given weight, the minimum drag occurs at ⎥⎦

⎤⎢⎣

⎡ L

Dminimum or at ⎥⎦

⎤⎢⎣

⎡ D

Lmaxsimum.

L

L Do

L

D

L

D

C

KC C

C

C

SC V

SC V

L

D2

2

21

2

21 +

=== ρ

ρ 3.4.2




Differentiate (3.4.2) with respect to C L

2

22)2(

L

L L L Do

L

L Do

L C

KC C KC C

C

KC C

dC

d −+=⎥

⎦

⎤⎢⎣

⎡ +

0222 =−+ L L Do KC KC C

Di L Do C KC C == 23.4.3

At minimum drag condition value of C Do is equals to C Di. For value of C L at mimimum drag isgiven by:

K

C C Do

Lmd =

Lmd

md SC

W V

ρ 21

= 3.4.4

Can be written as,4

12

1

21 ⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

Do

md C

K

S

W V

ρ 3.4.5

Also from (3.4.3)

3.4.6 Do L Do D C KC C C 22 =+=

Lift to drag ratio

Do Do

Do

D

L

KC C K

C

C

C

D

L

2

1

2

1=⎥

⎦

⎤⎢⎣

⎡== 3.4.7

3.5 Thrust Required

Assume aircraft is flying in steady straight and level flight at a given constant altitude. Thrustrequired for enjin turbojet is equal to drag force.

DSC V DT 2

2

1 ρ == 3.5.1

and LSC V W L 2

2

1 ρ == 3.5.2




devided Eq.(3.5.1) by Eq.(3.5.2) L

D

C

C

W

T = 3.5.3

Thrust required

D

L

W

C

C

W T

D

L

== 3.5.4

T is usually label as T R

For a minimum drag condition, thrust required is mimimum where the aircraft is flying at minimumdrag speed, V md

3.6 Power Available and Power Required

Thrust required gives the power required of an aircraft

L

L

D

L

R R

SC W V

SC V W L

V

C C

W V T P

ρ

ρ

2

2

1 2

=

==

==

(L/D)max

Thrust

TR (N)

V (m/s)

Vmd




Then

( )

Re

2

2

1

2

1

Re

2

2

1

2

1

Re2

1

2

1

)2

1

2

2

33

2

2

33

2

22

22

A

VS W

S V SC V

A

S V W

S V SC V

A

C SV V V SC V

V KC C S V DV V T P

SC

W

C C

W P

Do

Do

L Do

L Do R R

L

D

L

R

π

ρ ρ ρ

π

ρ ρ ρ

π ρ ρ

ρ

ρ

+=

+=

+=

+===

=

3.6.1

3.6.1 Pistonprop and Turboprop

TA PA

V V

Power produce by a shaft engine can be assumed constant with airspeed. For a propeller drivenaircraft the performance is analyzed in the form of power available. At minimum drag the relationof is applied. Maximum endurance of a fan engine can be achived if the aircraft is flying

at minimum power required speed, typically a little less than V md . While the thrust produce by thefan is inversely propotional with aircraft speed.

DL Do KC C =

Relationship of shaft power to thrust

746

..

V T THP BHP ==η

then, 746..

746.V

BHP

V

THPT

η ==




where η is the fan efficiency (typically between 0.75~0.85) and 746 is the conversion unit fromHorsepower to watt (i.e. 1 HP = 746 Watt )

3.6.2 Jet Engine (turbojet dan turbofan)

Thrust produce by jet engine is constant with aircraft speed. Power produce is proportional tospeed.

TA PA

V

For a jet angine aircraft, the performance depend on excess thrust (i.e. the different between thrustforce and drag force). Maximum endurance is achieved when flying at minimum drag speed,V md .At this condition the drag is minimum, and relationshipC Do = KC DL is used. The maximum rangeis achieved when the aircraft is flying at a speed little more than V md because at minimum drag, thevalue of drag is increase a little with speed where that the change of distance increases with fuelconsumption.

V

Power available and power required is a function of altitude.




Figure ## Altitude Effects on Maximum Excess Power

3.7 Minimum Power Required

At minimum power requirement condition,

Re

2

2

3

0

222

A

S V W SC V

dV

dP

dV

dP

Do R

R

π

ρ ρ +=

=

⎥⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤

⎢⎣

⎡−=

Re3

1

2

3

Re

2

2

3

2

2

242

432

2

A

C C S V

A

S V W C S V

L Do

Do

π ρ

π

ρ ρ

For that a minimum power required occur when

2

3

1 L Do KC C =

Power available is determined by the powerplant characteristics

As a conclusion for a maximum range and endurance, a jet engine aircraft needs a minimum dragand correspondence speed in a straight level flight. While for fan drive engine needs minimum power required in a straight and level flight.




Chapter 4

CLIMBING FLIGHT

Figure 4.1 Forces Acting on an Aircraft in a Climb

When aircraft is in steady climbing at a climb angle, γ with horizontal, the equilibrium equation of force parallel to the flight path is given

0sin =−−− dt dV mg DT γ 4.1

The change of altitude with time is given by

γ sinV dt

dH = 4.2

substitute sinγ term from Eq. 4.2 in Eq.4.1, gives

dt

dV

mdt

dH

V

mg

DT +=− )( 4.3

rearranged

dt

dV

g

V

dt

dH

mg

V DT +=− )(




If the aircraft is climbing at steady speed,dt

dV becomes zero, then the climb rate can be written as

mg

V DT

dt

dH )( −=

and TV – DV is known as excess power dan

mg

V DT )( −is known as specific excess power

In general the climb rate can be defined as

W

Power Excess

dt

dH =

a) Propeller Driven Engine

b) Jet Engine

Then the maximum climb rate

W

Power Excess Maximum

dt

dH =⎟

⎠

⎞⎜⎝

⎛

max




MINIMUM TAKE-OFF THRUST REQUIREMENT

Climb gradient, sin γ =mg

D

mg

F

mg

thrustexcess N −=

Typicallymg

D= 0.1 for the take-off configuration, u/c up

Assumemg

F N = 0.24; sin γ = 0.24 – 0.1 = 0.14; γ = 8.05o

or Grad = 14.1% (grad% = 100 tan γ)

All engines operating

F N/mg 0.24

Grad γ % 14.1

Min grad-oei γ %

One engine inoperative4 eng. 3 eng. 2 eng.

0.18 0.16 0.12

8 6 2

3.0 2.7 2.4

JAR 25.121(b)MIN GRAD %

F N/mgMin. aeo γ %

F N/mgequiv. γ %

F N/mgISA-15, γ %

(ISA + 15oC5,000’)

ISA, SL SL

4 eng. 3.0

3 eng. 2.7

2 eng. 2.4

0.203

0.221

0.284

10.4

12.2

18.7

0.234

0.254

0.327

13.5

15.6

23.3

0.243

0.264

0.339

14.4

16.4

24.6

(NB Drag of inoperative engine not included – required F N/mg is pessimistic)

EXAMPLE

An aircraft with 16,380 kg weight, wing area S = 42 m2

and wing span 16 m has a drag polar CD =0.014 + 0.05 CL

2. This aircraft is installed with turboprop engine. Maximum speed at sea level is

270 m/s. Power available PAV is assumed not at maxium at the speed when it is occurred. Calculatethe maximum rate of climb.

Maximum speed occur when

REQ AV PP =

222 /7.44687)270)(226.1(2

1

2

1mskgV q === ρ




427.44687

81.916380

××

==qS

W C L

0144.0)086.0(05.0014.0 2 =+= DC

N qSC D D 4.26276)42)(7.44687)(0144.0( ===

W xV DP REQ

51095.70)270)(4.26272(. ===

At maximum rate of climb

AR

C C L

Doπ

2

3

1=

2)05.0(

3

1014.0 LC = 0.014 ⇒ 9165.0= LC

smS C

W V

L

/5.82)42)(9165.0)(226.1(

)81.9)(16380)(2(

..

2===

ρ

N C

C W D

L

D 3.9818)81.9(9165.0

056.0)16380( ===

W xV DP REQ

5101.8)5.82)(3.9818(. ===

Maximum Rate of ClimbW

PP RC

REQ AV −=max = sm x x

/2.39)81.9)(16380(

101.81095.70 55

=−

EXAMPLE

Given the characteristics of a jet engine aircraft

M = 16380 kg, S = 42 m2, CL = 0.2352

CD = 0.014 + 0.05 CL2

If the thrust given by the engine is 26699 N at sea level. Calculate the maximum rate of climb

RCmax and the related speed. Determine the climb angle.




Solution

smS C

wV

L

/163)42)(2352.0)(226.1(

)81.9)(16380)(2(2===

ρ

01676.0)2352.0(05.0014.0 2 =+= DC

N C

C W D

L

D 4.114542352.0

01676.0)81.9)(16380( ===

,0954.0sin =−

=W

DT AV γ = 0.0954, o5.5=γ

smV RC /55.15)0954.0)(163(sinmax === γ

W

DT AV −=γ sin , L

D

C

C W D =min and

2

L D KC C =

205.0014.0 LC = 529.0= LC

028.0)529.0(05.0014.0 2 =+= DC

N C

C W D

L

D 2.8505529.0

028.0)81.9)(16380( ===

1132.0)81.9)(16380(

2.850526699sin max =

−=

−=

W

DT AV γ

Sin γmax =81.916380

2.850511454

×−

= 0.1132

o5.6max =γ

smC S

W V

L

/7.108..

2==

ρ




Chapter 5

TAKEOFF AND LANDING

5.1 Introduction

Take-off is the most critical flight phase and should pay more attention. The control system should

be angle to rotate the aircraft at good climb position and can be trimmed. An aircraft should be able

to demonstrate it ability to avoid yawing and maintain it direction. For example during crosswind

disturbance and lost of engine power. Excess power is required for handling the operation and

optimum climb speed.

Similar case is considered during landing. Aicraft should be able to rotate to at ‘touchdown’

position where the aircraft nose can be lift off to required speed. Aircraft has to maintain wing level

and crosswind landing. Aircraft has to demonstrate it ability to balance if engine losing power. In

term of performance, optimum speed during approach and power for required slope. Touch down

speed has to be known and other operation consideration at speed and configuration

5.2 Take-off

Take-off distance is the distance required for an aircraft to gain a lift-off speed and reach 35 ft (10.5

m) or 50 ft (15 m). Take-off distance is divided into two parts. First is the ground run distance, is a

required distance to gain lift-off speed, and second is the airborne distance, is the distance from lift-

off speed to reach height of 50 or 35 ft.




5.2.1 Take-off Ground Run Distance

Assume an aircraft is under a summation of force F from zero speed to lift-off speed, V LO.

Where

max

22.12.1

L

LOSC

W VsV

ρ == 5.1

The change in distance the product of aircraft forward speed and time

5.2Vdt dS =

Acceleration is the change in velocity

dt dV a = 5.3

Substituting the expression of dt from Eq(5.3) in Eq(5.2), gives

dV a

V dS = 5.4

Consider forces acting on aircraft during take-off are thrust (T ), drag ( D), lift ( L), weight (mg) and

friction (μ R). Figure shows forces acting on aircraft where γ is a runway slope which typically

small around 5o, R is the reaction force on wheel μ is the friction coefficient.

Figure




Equilibrium force parallel to flight path

ma Rmg DT =−−− γ sin (5.5)

Equilibrium force normal to flight path

γ kosmg L R =+ (5.6)

substitute R term from Eq.(5.6) in Eq.(5.5)

ma Lkosmgmg DT =−−−− )(sin γ γ

Assumed γ is very small

ma Lmg DT =−−− )( (5.7)

Total distance is the integration of Eq.(5.4)

⎥⎥⎦

⎤

⎢⎢⎣

⎡=∫=

a

V dV

a

V LOS 2

2

(5.8)

Substitute term a from Eq.(5.7) in Eq.(5.8) gives

)]([2

2

Lmg DT mV S LO

−−−= μ

Generally T is constant (especially jet aircraft), W is also constant. Except L and D is a function of

speed.

LSC V L2

21 ρ =

⎥

⎦

⎤⎢

⎣

⎡+=

Re

2

2

21

A

C C S V D L

Doπ

φ ρ

φ - is known as ground effect, exist when aircraft flow very low to ground and the effect of

trailling vortex is reduced.




2

2

161

16

⎟ ⎠

⎞⎜⎝

⎛ +

⎟ ⎠

⎞⎜⎝

⎛

=

b

h

b

h

φ

– height from wing level to ground h

– wing spanb

For simplicity assume t is constant and average value of drag and resistant is reduced to

[ ]avelW D )( −+

The effective force [ ] t conslW DT F aveeff tan)( =−+−=

substitute F = F eff

{ }ave

LO

G LW DT

mV S

)]([2

2

−+−=

μ

and [ ] [ ] Lovave

lW DlW D 7.0)()( −+=−+

Conclusion

1. Lift-off distance is sensitive to aircraft weight (reduce with w2). If double the weight, the

takeoff distance increase 4 times.

2. Takeoff distance is depend on air density

S Lo α 2

1

ρ

3. Takeoff distance reduces with increase in wing area.




5.2.2 Take-off Airborne Distance

Aircraft has to accelerate to reach climb speed and rotate to climb position. Takeoff reach the final

stage when eaches the screen height of typically 50 or 35 ft. Airborne distance can obtained from

energy balance.

Change of Energy = Excess Thrust x Distance

(5.9)∫ −=−= Aave S DT ds DT dE )()(

where is the airborne distance. AS

The change of energy between lift-off point to 50 feet is the change of potential and kinetic energy,

Energy Change = (K.E + P.E)50’ – (K.E + P.E) LO

[ ] 02

212

21 +−+= LOmV mghmVh

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

22

22

LOh V mgh

g

V mg

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡+

−= h

g

V V mg LOh

2

22

(5.10)

combine Eq.(5.10) and Eq.(5.9), gives

Aave LOh S DT h

g

V V mg )(

2

22

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡+

−

then the airborne distance

⎭⎬

⎫

⎩⎨

⎧

+

−

−= hg

V V

DT

mg

S

TD

ave A 2)(

22

2

where h is the screen height.

A LT S S S +=




5.3 Landing

5.3.1 Landing Ground Distance

Ground distance is a distance required for an aircraft to slowing from touchdown, V TD speed until

stop.

max

23.13.1

L

sTDSC

W V V

ρ ==

At touchdown, assumed thrust, T=0

ma LW D =−−− )( (5.11)

a

V dV

a

V S

2

2

−=∫−= (5.12)

Substitute a from Eq.(5.11) in Eq.(5.12)

[ ]ave

TD L

LW D

mV S

)(2

2

−+=

μ

where [ ] [ ]TDV ave

LW D LW D7.0)()( −−−=−−

If reverse thrust is applied during landing.

T = -T B

- T B – D - μ (W – L) = ma

[ ]{ }ave B

TD L

LW DT

mV S

)(2

2

−++=

μ

Example

Estimate the takeoff distance of aircraft B on tarmac of μ = 0.02. During takeoff the CL maximum

is not higher than 1.0. The wing level ground clearance is 1.83 m.

Solution:

17.091.10

83.1==

b

h




∴ φ =( )

( )2

2

161

16

bh

bh

+=

( )[ ][ ]2

2

)17.0(161

17.016

+= 0.88

75.83

15.29225.1

81.98984.22.1

22.12.1

max

=

⋅⋅

⋅===

∞ L

s Lo

SC

W V V

ρ

m/s

63.587.0 = LoV m/s

67.82186)1)(54.29)(63.58)(225.1( 2

212

21 === LSC V L ρ N

74.3149)]1(044.0)(08.0(02.0)[54.29()63.58)(225.1( 2

2122

21 =+=+= L Do KC C S V D φ ρ N

32472)2(16236 ==T

94.854.29

25.16 22

===s

b AR 044.0

)81.0)(94.8(

11==

⋅=

π π e ARK

[ ]{ } LoV ave

Lo

Lo LW DT

V g

W

S

7.0

2

)(2 −+−

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

=μ

=

[ ]{ }67.6218681.98984(02.074.3649324722

)75.83)(8984( 2

−×+−

=745.56606

63014338

= 1113.2 meters

Example

Estimate landing distance at sea level for aircraft B. Reverse thrust is not use during (assume T =

0). Spoiler is used during landing which result of L = 0 and zero lift drag increase by 10%. C Lmax

during landing at maximum flap 2.5. Assume C Lmax is used during landing. Assume fuel tank in

landing is at zero state and pilot apply brake which give effect of μ = 0.4.

Solution:

Fuel tank is empty during landing. Weight of fuel is negligible.




Chapter 6

RANGE AND ENDURANCE

6.1 Range

Range covers climb, cruise and decent distance.

The cruise range of an aircraft is equal to the total range covers with respect to fuel quantity.

Specific range is given by range per unit weight in meter per unit kilogram.

kg

mrangespecific =

6.1.1 For Jet Engine and Turbofan Aircraft

Specific range or distance can be written as

( )( ) N hr N kg

hr m

CD

V

CT

V

hr kg flowrate fuel

hr mhour per meter

kg

m

)../(

/==== (6.1)

C is the specific fuel consumption (sfc) in weight per unit thrust per unit hour (kg/N.hr )

and W L

D

D L

W D ==

/

Substitute in Eq.(6.1)

W D

L

C

V

W L DC

V

kg

m 1

)/(== (6.2)

⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡ D L

C V is known as range factor, measure range efficiency from aerodynamic and propulsion

system.

For jet engine, if the average value C and D

Lcan be chosen from aircraft design, the cruise range




W

dW

D

L

C

V dW

kg

m R iw

f w

iw

f w ∫∫ ==

f

ie

w

w

D

L

C

V R log= (6.3)

6.1.2 Propeller Driven Aircraft

The specific distance can be written as

BHPC

V

kg

m= (6.4)

C is the specific fuel consumption in kg fuel per horsepower per hour (kg/BHP. hr )

⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡=

DV

V

C THPC

V

kg

m η

η

W D

L

C DC kg

m 1.

1 η η == (6.5)

BHP – Break horse power

THP – Thrust horse power

[1 horse power (HP) = 746 Watt]

Total range

dW kg

m Range iw

f w∫=

For propeller driven aircraft

W

dW

D

L

C R

iw

f w

η ∫=

f

ie

w

w

D

L

C R log

η = (6.6)

Eq.(6.6) is known as Brequet formula whereη , C and D

Lis assumed constant during flight.




6.2 Endurance

Endurance is a measure of the flight duration

fuel N

hour endurancespecific =

For a maximum endurance, the minimum fuel flow rate per unit time is required. Assfc is assumed

constant, the drag should be at minimum for jet engine. Minimum value of thrust horse power THP

is required for a propeller driven aircraft.

6.2.1 Turbojet atau Turbofan Aircraft

Endurance dW DC

fuel N

hour iw

f w

iw

f w E .

1∫∫= =

( )[ ] W dW

D L

C dW

D LW C iw

f wiw

f w E 1.

1 ∫∫= =

f

ie

w

w

D

L

C E log

1= (6.7)

C and D

Lis assumed constant during flight or base on average value.

An aircraft is required to fly at minimum drag condition to gain a high endurance, i.e. ,wi /w f is

high, ⎟ ⎠

⎞

⎜⎝

⎛ D

L

maximum.

6.2.2 Propeller Driven Aircraft

Endurance:

dW W DV

L

C dW

DV C dW

C THP

E iw

f w

iw

f w

iw

f w

11⋅=

⋅=

⋅= ∫∫∫

η η

η

DV

Lis a ratio of lift to thrust power required and the value is not constant.

but L = W =21 ρ V

2SC L




then V = LSC

W

ρ

2

232 W

dW SC

C

C

C

L

D

Liw

f w E ρ η

∫=

Assumed CD, CL, η , C and ρ are constant at certain altitude, then

[ ] iw

f w

D

L W S

C

C

C E

212123

2

2 −⎥⎦⎤

⎢⎣

⎡−=ρ η

For maximum endurance

1. High propeller efficiency

2. Low sfc

3. High W f , where W o = W i + W f

4. Fly at D

L

C

C 23

maximum

5. Fly at sea level E α ρ 1/2

Range 1) JET C = s.f.c xg

1

2) PROP C = s.f.c xg

1

Endurance 1) JET C = s.f.c xg

1

2) PROP C = s.f.c x2

3g

1




= 1940772.5 m

= 1940.7 km

Endurance, ( ) ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−= ∞2

12

12

12/3

11

2i f

D

L

W W S C

C

C E ρ

η

8.12

max

23

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

CD

C L

Maximum endurance can be achieved if aircraft is flying at sea levelρ∞ = 1.225 kg/m3

E =7

10456.7

8.0−

×

(12.8)(2(1.225)(16.17))1/2 ⎥

⎦

⎤⎢

⎣

⎡−

63.1337

1

41.1171

1

= 164242.56 secs




AIRCRAFT DATA

AIRCRAFT A (BRITISH AEROSPACE JETSTREAM)Power plant = 2 engine propeller drivenPower Rating = 900 horse power per engineWing span, b = 15.85Wing area, S = 25.1 mWing chord, c = 1.71 mCDo = 0.0175Oswold efficiency factor, e = 0.82Fan efficiency = 0.82Aircraft gross weight = 5570 kgFuel Capacity = 635 kg gasoline typeSpecific fuel consumption, s.f.c = 0.204 kg/HPhr

AIRCRAFT B (CESSNA 650 CITATION III)Type = Executive Jet AircraftPowerplant = 2 turbofan enginePower Rating = 16236 N per engine

Wing Span, b = 16.52 mWing Area, S = 29.54 mGross Weight = 8984 kgFuel Capacity = 2862 kg kerosinSpecific fuel consumption, s.f.c = 0.272 kg/N hour CDo = 0.02Oswald Efficiency Factor, e = 0.81

AIRCRAFT C (CESSNA SKYLANE)Type = Private Light AircraftPowerplant = 1 propeller driven engine

Power Rating = Single Piston 230 HP at sea levelWing Span = 10.91 mWing Area = 16.17 mGross Weight = 1337.6 kgFuel Capacity = 166.22 kg (gasoline)Specific Fuel Consumption, s.f.c = 0.204 kg/HP hour CDo = 0.025Oswald Efficiency Factor, e = 0.8Blade Efficiency = 0.8




AICRAFT D (Scruggs-Plummmet SP10-99 Tri-cruiser)

WEIGHTGross Weight 240,000 kgEmpty Weight 122,000 kgMaximum Payload 46,000 kgMaximum Fuel Weight 96,000 kg

DIMENSIONWing Area (gross) 415 m2 Wing Span 54 mHorizontal Tail Area 99 m2 Overal Length 56 m

AERODYNAMIC DATACruise Takeoff/Climb Landing

CDo 0.0145 0.0180 0.0470

CDL 0.0540 0.0585 0.0620

oα fuselage -4 deg -6 deg -8 deg

α d dCL / 5.4/rad 5.4/rad 5.4/rad

Undecarriage drag coef - 0.0225 0.0225Maximum CL 1.6 2.0 2.5

POWERPLANT3 Engine Aircraft, Pratt & Witney NBG-20

Maximum Thrust, Sea Level I.S.A 179 KNSpecific Fuel Consumption, sfc, Sea Level I.S.A 10.6 mg/Ns

Cruise Thrust, 35000 ft, Mach 0.85 48 KNSpecific Fuel Consumption, 35000 ft, Mach 0.85 18.2 mg/Ns

At Sea Level

Air density, o ρ = 1.225 kg/m3

Speed of Sound, ao = 340.29 m/s (661.5 knots)Gas Constant, R = 287.05 Nm/kgK

Conversion Unit1 Horse Power = 746 Watt1 knots = 0.5144 m/s

Flight Mechanics - Part 1

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