Design Report of STOL transport aircraft

A Report on

INTER THEATER TACTICAL TRANSPORT WITH AUSTERE STOL CAPABILITY

By

N/C SYED HASSAN MAHMOOD WASTI (060901)

Submitted to Sqn Ldr MESSAM ABBAS

In fulfillment of the requirements for the course

AE-441, Aircraft Conceptual Design

Major: Aerospace Engineering

Department of Aerospace Engineering

COLLEGE OF AERONAUTICAL ENGINEERING

PAF Academy, Risalpur

DATED 8TH APRIL 2010

1

CH-69(INTER-THEATER TACTICAL TRANSPORT WITH AUSTERE STOL CAPABILITY)

2

TABLE OF CONTENT

i. Introduction 4

ii. Design requiremets 5

iii. Conceptual sketch 6

iv. Initial weight sizing 8

v. Airfoil geometry selction 10

vi. Thrust to weight and wing loading 12

vii. Revised weight sizing 14

viii. Geometric sizing 16

ix. Special considerations 19

x. Propulsion and fuel integration 23

xi. Aerodynamics 27

xii. Structures and loads 34

xiii. Weight approximation 40

xiv. Stability 46

xv. Performance 50

xvi. Solid edge model 60

xvii. Conclusion 67

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INTRODUCTION

As fighting and insurgencies intensify around the world, different countries continue to send aid, in the form of troops as well as material aid, to stricken areas in the hopes of defending and improving their lives. It is therefore necessary for the military to provide these troops with the best technology to aid in the fighting. This RFP magnifies the needfor tactical warfare mobility to be utilized by Air Forces throughout the world. This short take off and landing concept for a transport aircraft is yet to be used extensively and is still a relatively new concept when it comes to aircraft industry.

The basic requirements for this aircraft state that it must be able to carry payload the dimensions and weight of a future deployable armored vehicle. The aircraft must be capable of short take-off and landings on an unimproved runway. Considerations must be made for the required aircraft range, ensuring that the aircraft can safely be operated in combat situations. The transport must also be able to reach speeds of at least Mach 0.8. While all criteria mentioned in the RFP must be met, several stand out as driving factors of the design. The speed, payload size, range, and conditions at landing are of the most concern. The engine selection is based on factors such as the required Mach of 0.8, as well as the take off and landing distance requirements. The take off and landing requirements will also drive the aerodynamic design of the aircraft. Stability will be concerned. The cargo hold and loading ramp must be designed in order to accommodate the payload size and ease of use in combat conditions. Fuel considerations must be made in order to ensure the aircraft can meet the range requirements for different missions, especially factoring in the differing payload requirements of the missions.

Now I would describe and explain different procedures I have used in aircraft design accompanied by relevant graphs and drawings. Every possible effort has been made to make this report concise and pertinent.

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The procedure followed is shown in the table below

Table 1. design sequence

I. DESIGN REQUIREMENTS

FDAV Mission

Payload

25 tons with additional 5 tons of support equipment which raises the nominal payload to 60000lb

Volume of payload dimension: 56 in. long, 128 in. wide, and 114 in. high Additional 12 in. wide escape path around the vehicle Center of gravity of the vehicle located 28 in longitudinally, 64 in laterally, and 48

in

Crew

light crew: 2 (pilot and copilot) Cabin crew: 1 (loadmaster)

5

DESIGN REQUIREMENTS

MISSION PROFILE

INITIAL WEIGHT SIZING

AIRFOIL AND GEOMETRY SELECTION

INITIAL DESIGN POINT

SELECTION

REVISED WEIGHT SIZING

GEOMETRIC SIZING

3 D MODELLING

SPECIAL CONSIDERATION

S

PROPULSION AND FUEL SYSTEM

AERODYNAMIC ANALYSIS

PROPULSION ANALYSIS

STRUCTURE AND LOAD ANALYSIS

WEIGHT ANALYSIS

STABILITY ANALYSIS

PERFORMANCE

Mission Requirements

Balanced field takeoff length must not exceed 2,500 ft. with the aircraft able to clear 50 ft obstacle in 3000ft.

Cruise/climb to best cruise altitude Cruise at best cruise altitude and M cruise at 0.8 for 500 nm less distance

traveled during climb out Descend to 1000 ft for 100 nm at speed of Mach 0.6 If powered lift not required, use idle power Useful area of 3000 feet by 150 feet 50 foot obstacles at 250 feet from either end of the runway Landing length of 3000 feet Takeoff under combat rules, with mirrored mission segments for return and climb

over a 50 foot obstacle from a standing start

Ferry Mission

Payload

10 tons of bulk cargo density of 20 lb/ft3 when properly packed

Crew

Flight crew: 2 (pilot and copilot) Cabin crew: 1 (loadmaster)

Mission requirements

Balanced field takeoff length must not exceed 2,500 with the aircraft able to clear 50 ft obstacle in 3000ft.

Cruise/climb to best cruise altitude Cruise at best cruise altitude and M cruise > 0.8 for 3200 nm less distance

traveled during climb out Aerial refueling permitted to achieve range Descend to sea level Normal approach to runway of 3000 ft or less If powered lift not required, use idle power Enough reserve fuel for a missed approach plus 150 nm diversion and 45 minute

hold at 5,000 ft

6

II. MISSION PROFILE

FDAV MISSION

0

5000

10000

15000

20000

25000

30000

1 2TAKE OFF

3

CLIMB

4CRUISE @ M 0.8

500NM 5

DESCEND

6CRUISE @ M 0.6

100NM

78LANDING AND TAKE

OFF

910 11

CRUISE @ M0.6 100NM

12

CLIMB

13CRUISE @ M 0.8

500NM 14

DESCEND

15 LANDING

16

Fig.1 FDAV mission

FERRY MISSION

0

5000

10000

15000

20000

25000

30000

35000

CRUISE150 NM

10 11

LOI-TER 45 MIN

Fig. 2 ferry mission

7

III. Conceptual sketch

Concept chosen

8

IV. INITIAL WEIGHT SIZING

In this section a rough weight of the aircraft was to be determined. Here I would like to mention that in rough weight sizing we do not cater payload drop however as payload drop was a major segment of my mission therefore I have included it in my calculation. The calculations are shown below:

FDAV MISSION

flight segment weight fraction weight after flight fuel consumedTO W3/W1 0.97 127757.85 3951.27climb W4/w3 0.98 125841.48 1916.36Cruise W5/W4 0.96 121851.12 3990.35descend W6/W5cruise W7/W6 0.992 120918.53 932.59landing W8/W7 0.995 120313.94 604.59PL drop 60313.940TO W9/W8 0.97 58504.52 1809.41climb W10/w9 0.985 57626.95 877.56cruise W10/W9 0.992 57185.90 441.05Climb W11/W10 0.985 56328.11 857.78cruise W12/W11 0.968 54541.98 1786.13landing W13/W12 0.995 54269.27 272.70

FINAL RESULTS TRANSPORT

W0 131709.12

We/Wo 0.407W empty 53669.

27 Wf

17439.85

FERRY MISSION

flight segment weight fraction

weight after flight fuel consumed

TAKE OFF

w2/w0 0.97 71746.88 2218.97

CLIMB w3/w2 0.985 70670.67 1076.20

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CRUISE w4/w3 0.815 57501.06 13169.61LOITER w5/w4 0.997 57375.10 125.96LANDING w6/w5 0.995 57088.22 286.87

MISSED APPROACH

flight segment weight fraction

weight after flight fuel consumed

CRUISE W7/w6 0.990 56539.01 549.20LOITER W8/W7 0.980 55434.05 1104.96LANDING W10/W9 0.995 55156.88 277.17

FINAL RESULT FERRY MISSION

We/W0

0.424278239

Wo

73965.85823

We

31382.10407

Wf

18808.97185

Note:

As can be seen that the empty load obtained from transport mission is greater than ferry therefore empty weight of transport mission was selected. Secondly, as the ferry mission was consuming more fuel therefore, the fuel storage capacity would be chosen accordingly.

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23%

11%

23%5%

3%

10%

5%

3% 5%

10%

2%

fuel fragment in mission segmentTO W3/W1climb W4/w3cruise W5/W4descend W6/W5cruise W7/W6landing W8/W7PL dropTO W9/W8climb W10/w9cruise W10/W9climb W11/W10cruise W12/W11landing W13/W12

V. AIRFOIL AND GEOMETRY SELECTION

AIRFOIL SELECTION

While considering airfoil it was kept in mind that the airfoil should have

1. A high value of Cl max2. Should have a high critical mach number to avoid shock waves and drag losses

related to it.3. High L/D ratio

Keeping these considerations in mind the following airfoils were analyzed

clmax (mach 0.1) design cl(mach 0.8)

cd at design cl

SC(2)-0614 1.15 0.55 0.0075SC(2)-0710 1.35 0.6 0.007airfoil J 1.36 0.62 0.008SC(2)-0610 1.15 0.7 0.0082

The airfoil selected was bacj ( airfoil J ) primarily because its critical mach was 0.82

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0 0.2 0.4 0.6 0.8 1 1.2

-0.3

-0.2

-0.1

5.55111512312578E-17

0.1

0.2

0.3

This airfoil produces the lift coefficient needed for cruise at roughly -0.3 degrees angle of attack. A drag divergence Mach number occurs around Mach 0.82. This airfoil allows for cruise speeds of Mach 0.8 without a dramatic increase in drag due to shockwave formation. This configuration results in efficient cruise conditions, which cut down on fuel consumption, weight, and cost.

the other selections are summarized below

Λ LE 30Λ/4 18λ 0.25

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Twist 3°WING INCIDENCE 1°

DIHEDRAL -3WING TIPS WINGLET

TAIL GEOMETRYΛ le 35

vertical tail sweep 50HORIZONTAL TAIL

AR 2.94λ 0.5

VERTICAL TAILAR 0.7λ 1.0

LE sweep

The LE sweep selected was 30 from fig 4.20. This would further decrease the velocity component of air at mach 0.8 delaying Mdd to 0.946.

Taper ratio

Taper ratio eliminates undesired effect of constant chord on rectangular wing and thus brings the lift distribution close to the elliptical lift distribution (ideal).Wing taper ratio is the ratio between centreline root chord and tip chord. A taper ratio of 0.25 was selected as it is easier to manufacture and gives good results

Location of Wing:

A high-wing vertical location was selected based on:

Historical trend Clearance for cargo trucks to pass under Improves stability due to effective dihedral effect Fuselage closer to ground so its easier to load it Fuselage only 4-5 feet above the ground for easy loading without ground

equipment Less chances of wing tip striking the ground

Wing Twist:

3° twist is chosen for our swept back aircraft following the historical trends. It delays tip stall and provides elliptical distribution.

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Wing Incidence

a wing incidence of 3° is chosen in accordance with historical data. More over the value of cl also increases contributing in lift.

Wing Tip Arrangement

Winglet was selected. Although it is a trade off between skin friction drag and parasite drag but the overall effect on drag is positive. a winglet is cambered and twisted so that a lift force is created that is in the positive direction hence decreasing drag.

TAIL ARRANGEMENT

Horizontal tail

T tail was selected as the end plate effect of T TAIL allows

1. Smaller vertical tail2. Lifts the horizontal tail clear of wake3. Makes it more efficient and reduces size

HORIZONTAL TAILAR 2.94λ 0.5

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Vertical tail

VERTICAL TAILAR 0.7λ 1.0

VI. INITIAL DESIGN POINT SELECTION

a constraint diagram was made in order to ascertain the acceptable T/W and W/S values the major constraints in my case were

Cruise at 30000ft @ M 0.8 Cruise at 1000ft @ M 0.6 Landing Take off

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0 20 40 60 80 100 120 1400

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

FDAV MISSION

cruise at 30000ftcruise 2nd time 30000cruise 10000cruise 2nd time 10000take offlanding

W/S

T/W

16

0 20 40 60 80 100 120 1400

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

FERRY MISSION

cruise at 30000ftcruise at 5000fttake offlanding

W/S

T/W

Selection point

Initially the design point was selected as

T/W=0.36

W/S=83

However, later on while computing performance I noticed that the aircraft was not fulfilling certain requirements due to less thrust and therefore the revised design point was selected as

T/W=0.42

W/S=83

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VII. REVISED WEIGHT SIZING

Aircraft sizing is the process of determining the take off gross weight and fuel weight

required for an aircraft to do its mission profile, the Initial sizing which is done earlier in

this report was a rough sizing depend on limited to fairly simple design mission, in this

chapter the weight of the aircraft is revised as some of the geometric and aerodynamic

parameters of the aircraft are now known.

The inputs were

Cruise @ 30000ftT/w 0.42W/s 83AR 8.49600394

1M max 0.8q @ 30000ft 282.0675Cdo 0.015e 0.8Range @ 30000ft 3038057.7TSFC ( c ) 0.00013888

9V @ 30000ft 795.84

FDAV mission

flight segment

Wi-1/Wi weight fraction

weight after flight

fuel consumed

TO W3/W1 0.97 156687.93 4846.01climb W4/w3 0.978 158016.58 1328.65cruise W5/W4 0.966 152725.04 5291.54 W/s 81.1 L/D 15.5descend W6/W5 0.99 151197.79 1527.25cruise W7/W6 0.986 149191.29 2006.49 W/s 77.6 L/D 9.46landing W8/W7 0.995 148445.343 745.956PL drop 88445.3433TO W9/W8 0.97 85791.9830 2653.3climb W10/w9 0.984 84479.00 1312.9

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cruise @ 10000ftV@10000 667.44R 607611.55C 0.000138889q 514.2643521

cruise W10/W9 0.972 82564.81 1914.18 W/s 43.4 L/D 5.51climb W11/W10 0.99 82021.07 543.73cruise W12/W1 0.945 77510.19 4510.8 W/s 42.1 L/D 9.3descend W13/W12 0.99 76735.09 775.10landing and taxi

W 14/W13 0.995 76351.41 383.67

Final results

We/Wo 0.4482812

Wo 161533.95

Wf 27839.838

We 72412.629

FERRY MISSION

W crew 600 lbW payload 22046.24

4lb

V@30000ft 795.84 ftR 1944357

0ft

C 0.0001389

l/s

AR 8.3425365

SWET/SREF

4.8

L/D MAX 19CRUISE L/D

16.454

flight segment

Wi-1/Wi weight fraction

weight after flight

fuel consumed

TO w2/w0 0.97 134355.7 4155.3climb w3/w2 0.97 131426.8 2928.95cruise w4/w3 0.79 104518.9 26907.8 w/s 67.5 l/d 14.8

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MISSED APPROACHv 877.68R@5000FT 911417

.3FT

LOITER@5000FT

2700 SEC

q 785.7286

loiterV 296.18

69DENSITY

2.04E-03

q 117

descend w5/w4 0.99 103473.7 1045.18cruise w6/w5 0.978 101289.4 2184.36 w/s 53.16 l/d 6.75loiter W7/w6 0.97 98998.4 2290.9 w/s 52.04 l/d 16.3descend W8/W7 0.99 98008.4 989.9landing W10/W9 0.995 97518.43 490.04

FINAL RESULTS

W fuel 40992.687

We 62903.821

Wo 138511.12

NOTE:

As the empty load obtained from transport mission is greater than ferry therefore empty weight of transport mission was selected. Secondly, as the ferry mission was consuming more fuel therefore, the fuel storage capacity would be chosen accordingly

VIII. GEOMETRIC SIZING

After the value of gross weight has been estimated and revised, the fuselage, wing, and

vertical tail in addition to the control surfaces can be sized. This chapter deals with the

calculation of the different parameters of fuselage, wing, tail and control surface which

will help us in the later stages of the design process.

The major parts that were to be sized in my case were:

1. Fuselage 2. Wing3. Vertical tail4. Horizontal tail5. Control surfaces

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FUSELAGE

During fuselage sizing I took into consideration the geometry of the payload that it has to carry. That was 56 in. long, 128 in. wide, and 114 in. high. Maximum upsweep of 25 was chosen The dimensions are shown in the figure below

I would like to mention here that the rfp already provided a clearance of 1 ft at all sides hence the total clearance at the the top would be 3.86 ft and a clearance of 2 ft at each side if the payload is loaded centrally

Length of the fuselage was also decided while keeping the rfp in mind.

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WING

SPAN AREA 1936.94b 128.28C root 24.15C tip 6.03Mean Chord ĉ 16.91AR 8.49Λ LE 30Λ/4 18λ 0.25Twist 3°WING INCIDENCE 1°DIHEDRAL -3

WINGLET USED

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HORIZONTAL TAIL

The airfoil used was

LHT 55.33CHt 0.95Cw 16.91Sw 1936.94SHT 562.38AR 3.08λ 0.5b HT 41.66C root 18C tip 9C MEAN 14

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Naca 63-510 t/c 0.1

VERTICAL TAIL

The airfoil used was

naca 63012 t/c 0.12LVT 50.72Cvt 0.076bw 128.28Sw 1936.94SVT 372.31AR 0.67λ 1bvt 15.87C root 23C MID 23C tip 23

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Control surfaces:

Control surfaces are responsible for controlling and stabilizing the aircraft in the same

time according to the mission requirement, as my aircraft is conventional therefore it

requires all ailerons for roll control, elevator for pitch control and rudder for yaw control.

Ailerons and flaps

As my primary requirement is STOL therefore flap requirement was of paramount imortance

Chord Ca/Cw 0.25AILERON span/ bw 0.32b aileron 41.0C aileron 4.22span 456.5

25

ELEVATOR

EXTEND to

90% of span

chord Ce/Cw

0.25

Ce 4.227745237

RUDDERchord Cr/Cw

0.32

Cr 5.411513903

IX. SPECIAL CONSIDERATIONS

Carry through structure

For my aircraft the most important consideration is the wing carry through structure in my situation I have used a wing box configuration. This configuration minimizes the fuselage weight as the fuselage is not taking any bending weight.

Flutter

The solution to the control surface flutter is that do not allow the centre of mass to be behind hinge line .the control surface should never be convex, and have bevelled trailing edge

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Fuel Tanks

The aircraft is designed is such a way to give the least vulnerability. The fly by wire

system gives it high stability even when a portion of the controls is damaged. Stand by

hydraulic system and electric system is also in there .The self-sealing bladder, and their

location away from the engine makes them safe.

Maintainability considerations:

The design aircraft uses systems as modules where ever possible so that in case of

failure the complete module can be changed immediately. Panels are present on the

fuselage to give an easier access to the in inner of the aircraft like engine, instruments

and components.

Aural signatures

As the aircraft might have to operate from civil airports therefore its aural signature

should be low for this well designed engine mounts and mufflers can be used.

Fire, Air, Icing Systems

The aircraft’s fire protection is composed to two components. The Ansul Cleanguard handheld extinguisher will provide electrically non-conductive and bio-friendly protection for the cockpit. The rest of the aircraft will be automatically monitored by the Ansul INERGEN system which utilizes inert agents, and a quick detection and reaction system that are both electrically safe and people friendly as well.

The air conditioning and pressurization system will consist of the typical systems in place on cargo and commercial aircraft. Bleed air will be directed from different stages of the four engines into two heat exchangers. From there it will be led through an air cycle Machine underneath the flight deck which will perform a refrigeration cycle on the air and send it through a mix manifold. From there the air will be circulated by fans through the flight deck and the cargo area. The flight deck and cargo area will both be pressurized to 8,000 ft atmospheric conditions. In addition, the flight deck will contain a 25 L oxygen converter for pilot and co-pilot, and a separate 25 L converter will be placed in the cargo area for the load master.

The aircraft will feature an electric de-icing system from Cox & Company Inc. This system, called an Electro-Mechanical Expulsion Deicing System (EMEDS), is a low powered ice protection system that utilizes electronically triggered actuators that work

27

with the composite structure of the aircraft to remove ice from leading edge slats, engine inlet cowls and cockpit windows.

The light system on the aircraft will feature MIL Spec and FAR light systems. Green, red, and white position lights will be placed on the right, left and rear surfaces of the aircraft. In addition, the aircraft will include anti-collision strobe lights on the top and bottom of the center of the fuselage.

Counter Measure Systems

Several pre-existing counter measure systems were selected to provide both electronic and physical protection from various missile based threats. These systems’ primary purposes are to warn the crew and help defeat both infrared and radar based missile systems.

The countermeasure dispensing unit assembly consists of 3 AN/ALE-47 Countermeasure Dispensor Systems, CMDS, set up with one facing downwards from the underside of the aircraft’s fuselage and one on either side of the aircraft’s fuselage. This setup provides a highly effective cover pattern by dispensing countermeasures both behind the fuselage and behind the wing mounted engines to best mask the heat signature of the aircraft . These dispensing units were chosen based on their current use in other comparable USAF aircraft and for their ability to dispense both infrared defeating flares and radar defeating chafes in both manual and automated modes.

X. CREW STATION, PASSENGER AND PAYLOAD

The cockpit layout is same as that used in conventional cargo aircrafts

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CARGO DOOR

Three types of configurations were available

1. Rear door

2. Side opening door

3. Nose opening door

Keeping in mind that the aircraft would have to operate from unprepared fields the first

option was chosen as

It is easier and simpler to manufacture

Requires lesser time to open and close

Requires no extra equipment

XI. PROPULSION AND FUEL SYSTEM

The sizing of the engines, their characteristics, intakes nozzles, geometry are analyzed

in this section.

While deciding the propulsion system different options were available like

Turboprop Axial flow turbojet Centrifugal turbojet High by pass Turbo fan engine

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Low by pass turbo fan engine

The engine selected was high by pass turbofan as it was providing the best fuel economy and was operatable at the given altitude and MACH number.

Engine selection

After deciding to use high by pass engine different engines were analyzed and the TF-39 was selected. It was decided to use 2 engines to cater for engine failure The total thrust requirement was

THRUST 2 engines 67521.742761 engine 33760.87138

Engines Variants Thrust

Aviadvigatel PS-90 I. PS-90AII. PS-90A-76

III. PS-90A2

35300 lbf32000 lbf39600 lbf

Pratt & Whitney PW2000 I. PW2000 38400-43800 lbf

Rolls-Royce RB211 I. RB211-535CII. RB211-535E4

37400 lbf40100 lbf

Tf – 39 41 100 lbf

V 2500 I. V2500-A1II. V2533-A5

III. V2524-A5IV. V2527-A5V. V2528-D5

25000 lbf33,000 lbf23500 lbf27000 lbf28000 lbf

Another major reason for TF-39 selection was the availability of thrust variation with

altitude data.

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The final specifications and dimensions as inferred from the available data were:

altitude (max ) 41000

31

weight 8000length 22.58fan tip diameter 8.33

The scale factor was coming out to be 0.821 and after catering for this the specifications and dimension were:

L 16.69D 7.93W 4832.59A max 49.40

CAPTURE AREA

Next the capture area was calculated

Rough estimation (fig 12.26)

45.2 ft^2

Accurate calculation 38.3 ft^2

FUEL SYSTEM

According to calculations the fuel required was

Transport mission 752 ft^3Ferry mission 890 ft^3

Accordingly the volume of wing was taken out from solid edge and was found out to be 1265. Using bladder tanks and leaving 20% space for control surfaces still allowed all the fuel to be filled in the wing

VI. AERODYNAMICS

The initial sizing of the aircraft was based upon rough estimates of the aircraft aerodynamics, weights and propulsion characteristic. At that time we couldn’t calculate the actual characteristics of the design because the aircraft had not been designed yet, now the aircraft design can be analyzed as it is drawn to see it actually meets the required mission range.

In this section I have calculated lift curve slope, zero lift drag, lift induced drag coefficient,

32

LIFT CURVE SLOPE

Although my aircraft requirement is only mach 0.8 however, as I do not know the actual achievable mach right now so I have calculated lift curve slope at supersonic speed as well.

0 0.5 1 1.5 2 2.50

1

2

3

4

5

6

7

CL α vs Mach

SUBSONICSUPERSONIC

Mach

CL α

Discussion

In the transonic regime there is no formula available and therefore these values were

interpolated. The shape at transonic regime is flat in accordance to theory for thick

airfoils. Subsonic Lift Curve Slope is function of aspect ratio, sweep at maximum cross

sectional area and increases with the mach no. gradually until the transonic region

starts. There is an abrupt increase in drag at transonic speeds so for a steady level

flight the same amount of thrust should also be produced so we expect a lift increase in

this region which is shown in the graph. The supersonic Lift curve slope decreases with

the mach no. as the aircraft flies at smaller angle of attack so the amount of lift

produced by the lifting device is reduced. CLmax is a function of Reynolds no. With the

increase in mach no, the Reynolds no. is increasing and has Inverse effect on Clmax,

therefore Clmax is decreasing with mach no.

MAXIMUM LIFT

33

Next maximum lift was calculated for both clean and flapped conditions. This was an important parameter as far as our design was concerned. Flaps and slats were incorporated in the design to get a high CL max.

TRIPPLE SLOTTED FLAP (ΔCL MAX)LANDING TAKE OFF

FLAPS

1.59 1.275

SLATS

0.299 0.239

TOTAL

1.892 1.514

-0.8 0.2 1.20

0.5

1

1.5

2

2.5

3

3.5

CL max vs MACH #

cleanflapped

Mach #

CL m

ax

α CL MAX

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90

5

10

15

20

25

α CL max vs Mach

mach #

α C

L m

ax

34

DRAG POLAR

Drag polar is the standard presentation format for aerodynamic data used in performance calculation. It is simply the plot of the coefficient of lift vs. coefficient of drag. Virtually all aerodynamic information of the aircraft is wrapped up in the drag polar.

PARASITE (ZERO LIFT) DRAG

The zero lift drag consists of:

Skin friction drag Form drag Interference drag Miscellaneous drag Wave drag

For the calculation of skin friction drag wetted areas were needed. They were calculated on solid edge. The values were

Fuselage 3562 ft^2Nacelle 658 ft^2Wing 3463 ft^2Vertical tail 721 ft^2Horizontal tail 1012 ft^2

The flow was assumed to be laminar till 0.15c for subsonic flow. As is in the case of subsonic aircrafts.

In the miscellaneous drag calculations the following drags were included:

Upsweep drag Landing gear drag Flap drag

Next wave drag was calculated using the transonic drag rise estimation as given in the book.

These calculations were done on different altitude

35

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Cdo @ sea level

Σ CdoCd)miscCd)l&pCd)waveCd)total

mach #

Cdo

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Cdo @ 1000ft


mach #

Cdo

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Cdo @ 10000ft


mach #

Cdo

36

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Cdo @ 20000ft


mach #

Cdo

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Cdo @ 30000ft


mach #

Cdo

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Cdo @ 35000ft


mach #

Cdo

37

The combined Cdo vs mach curves were

0 0.5 1 1.5 2 2.50

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Cdo vs Mach

S.L1000ft10000ft20000ft30000ft35000ft40000ft

Mach

Cdo

DRAG DUE TO LIFT (INDUCED DRAG):

The induced Drag coefficient at moderate angles of attack is proportional to the square of the lift co-efficient with a proportionality factor called the drag due to lift factor.

There are two methods for the calculation of K.• Oswald Span Efficiency Method• Leading Edge Suction Method

I have used the second method for calculations.

38

0 0.5 1 1.5 2 2.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0% to 100% K vs Mach

KoK 100

mach #

K

0 0.5 1 1.5 2 2.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

K values at different mach #

CL stall 2.89K100 / koCL 0.33CL 1.6

Mach #

K

Discussion on graphs

As my aircraft was subsonic therefore, according to raymer the value of leading edge suction factor would vary from 0.8 at design lift coefficient to 0.8 at stall lift coefficient.

The value of K could lie 0.23 and 0.04 in this case The variation in K value is extremely less at different MACH #. The variation in K at different Cl values is also very low The value of induced drag increases with Cl

Drag polar

39

0 0.1 0.2 0.3 0.4 0.5 0.60

0.5

1

1.5

2

2.5

3

Drag Polar at 30,000 ft

MACH 0.2MACH 0.4MACH 0.6MACH 0.8

Cd

Cl

Discussion on graphs

Due to the increase of velocity the Re# increases , subsequently the skin friction coefficient decreases , now we know as the coefficient decreases the drag due to this coefficient decreases which in turn reduces the Cdo in subsonic flow only with increase in mach number.

In transonic region the drag rise is because of the presence of the shock waves it is exclusively the pressure drag effect due to adverse pressure gradient.

In supersonic flow Cdo also tends to reduce with the increase of Mach number but it is still more than the drag in subsonic flow because of the wave drag.

CDo increases with increasing altitude: because density decreases, it is clear from the graph that parasite drag coefficient is different at different altitudes and it varies with mach numbers

Parasite drag coefficient is different at different altitudes and it varies with mach

numbers. The drag in the subsonic region is mainly due to the friction and due to

the compressibility effect. As the mach increases above Mdd, Cdo starts

increasing and the increment in supersonic regime is purely due to wave drag.

CD=CDo+kCL2

Tangent line from the origin to the drag polar locates the point of maximum L/D

40

ratio of aircraft.

With increase in CL there is a corresponding increase in CD for the same Mach No.

For the same values of Cl there is a corresponding decrease in Cd with increase in Mach number

Subsonic drag-coefficient rise with an increase in lift coefficient is more prominent as the leading edge sweep is quite high. So for the same amount of drag produced the Lift generated by the wing is less.

VII. PROPULSION ANALYSIS

Till now we had only calculated the uninstalled thrust now I am going to calculate the installed thrust which is less than uninstalled thrust

Reference Engine Thrust:The following graph shows the data of original thrust as provided by roskam.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5

10

15

20

25

30

35

40

45

UNINSTALLED ENGINE THRUST

SEA LEVEL15000 FT35000FT41000FT

MACH #

THRU

ST (I

N 1

000

lbs)

Figure 1 Original Uninstalled thrust

Scaled down Thrust:

41

Using a scale factor of 0.832 the following data was achieved:

0 0.5 1 1.5 2 2.50

10000

20000

30000

40000

50000

60000

70000

80000

Scaled up Uninstalled Thrust

Sea Level10000 ft20000 ft30000 ft36000 ft40000 ft50000 ft

Mach No.

Thrust (lbf)

Figure 2 Scaled-up Uninstalled thrust

Installed Engine Thrust Corrections:

The manufacturer’s un-installed engine thrust is based upon an assumed inlet pressure recovery. For a supersonic engine, it is approximated that the inlet has the pressure recovery of approximately 85% to 90%.

The short duct of a subsonic-podded nacelle will have a pressure recovery of 0.98 or better. Reducing inlet pressure recovery has a greater than proportional effect on the engine thrust.

The following corrections are applied to the scaled-up thrust values:

Loss due to incomplete pressure recovery across intake Loss due to air bleed Loss due to Inlet drag Loss due to nozzle drag

Loss due to Air Bleed:

42

Cbleed is the “bleed correction factor”. This is provided by the manufacturer for various flight conditions. For initial analysis, Cbleed can be approximated as 2.0. And the bleed mass flow has been taken as 5% of the engine mass flow, using equation 13.8 of the text book which is shown below.

% age thrust loss = C bleed(bleed mass flow / engine mass flow) x 100%

% age thrust loss = 8 %

Drag due to Inlet:

Thrust loss due to inlet comprises of the following factors:

loss due to bleed loss due to bypass loss due to additive drag (spillage)

INLET RECOVERY LOSS (inlet duct)C ram 1.35(P1/P0) ref 1(p1/P0) actual 0.98% THRUST LOSS 2.7

Drag due to Nozzle:

Nozzle drag varies with nozzle position as well as with the flight conditions .to properly determine nozzle drag the actual geometry as a function of throttle setting and flight condition must be known and the drag calculated but taking into account the overall aircraft flow field. As an initial approximation, the effect of nozzle position may be ignored and the nozzle drag estimated by typical subsonic values shown in table 13.1

D*Afuselage/q 0.038Afuselage 190.15q 282.06drag 0.056

43

After calculating the values of drag for nozzle we subtract it from the thrust obtained after applying inlet corrections for each altitude and mach no’s. We get the new graph’s these values are the net propulsive thrust available.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5

10

15

20

25

30

35

40

corrected installed thrust

sl15000ft35000 ft41000ft30000ft

mach

thru

st

Figure 3 Corrected Installed Thrust

Discussion:

Thrust is a direct function of density and with increase in altitude, the thrust should decrease as it can be observed from the graph. Moreover, with the increase in velocity, the ram effect of the engine decreases and thus thrust increases with the mach number. T/TO =ρ/ρo, Where T is the thrust at the required altitude and TO is the thrust at the sea level, and similarly the ρ is the density at the required altitude, and the ρo is the density at the sea level.

44

TSFC Curves

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

TSFC

SEA LEVEL TSFC15000 FT35000FT41000FT

Mach #

TSFC

Discussions

TSFC is a ratio of the amount of fuel required to produce unit thrust and therefore it will not be scaled up or corrected.

VIII. STRUCTURES

Before the actual members can be sized and analyzed, the loads they will sustain must

be determined. Aircraft load estimation, a separate discipline of aerospace engineering,

combines aerodynamics, structures and weights.

Loads estimation remains a critical area because an error or faulty assumption will

make the aircraft too heavy or will result in structure failure when the real loads are

encountered in flight. This chapter introduces the concepts of load estimation and

summarizes the subjects of aircraft materials and structural analysis.

Air loads:

Basically there are two types of air loads on aircraft:

Maneuver loads Gust Loads

45

NOTE: I have made V-n diagrams for both transport and ferry missions and have selected the load factor accordingly.

The requirements for the calculation of V-n diagram were

V g 82.02 ft^2V cruise 893.12 ft^2V LIMIT/DIVE 1328.5ft^2

The V-n diagrams were then made

Transport mission

0 200 400 600 800 1000 1200 1400

-3

-2

-1

0

1

2

3

4

5

V-n @ SL MACH 0.8

stalln positiven negativenegative up slopedynamic pressure limitgust Vggust V cruisegust V diveVg gustcruisedive

V

n

At V dive At V cruise At V gU 17.16 U 34.33 U 44.63Δn

2.23 Δn

2.998 Δn

0.357

46

0 200 400 600 800 1000 1200 1400

-3

-2

-1

0

1

2

3

4

5

V-n with gust (transport mission)

stallpositive nn negativeto v divedynamic pressure limit

V

n

Ferry mission

0 200 400 600 800 1000 1200 1400

-4

-3

-2

-1

0

1

2

3

4

5

6

V n with gust ferry mission

stalln positiven negativeto V divedynamic pressure limitgust vggust v cruiseVgV cruiseV diveV dive

v

n

At V dive At V cruise At V gU 14.89606 U 29.792116 U 38.72975Δn

3.278324 Δn

4.4077709 Δn

0.526226

47

0 200 400 600 800 1000 1200 1400

-4

-3

-2

-1

0

1

2

3

4

5

6

v n with gust ferry

stallposi nnega nup1175 1328.533975Series12

V

n

AFTER COMBINING BOTH THE V-N DIAGRAMS

0 200 400 600 800 1000 1200 1400

-4

-3

-2

-1

0

1

2

3

4

5

6

V-n mixed

stalln posin negito v diveSeries10

V

n

Discussion on V-n diagram:

The aircraft maximum speed, or dive speed (Vdive) , at the right of v-n diagram represents the maximum dynamic pressure q. The point representing maximum q and maximum load factor is clearly important for structural sizing. At this condition, the aircraft is fairly at a low AOA because of high q, so the load is approximately vertical in the body axis.

48

The ultimate load factor is the maximum load factor the aircraft can withstand without breaking. It cannot be found experimentally. It is found by multiplying the limit load factor by a factor 1.5. The value of Limit load factor is obtained from historical trend.

The lower line (n=-2) represent the negative load factor limit of the aircraft and corresponds to the amount of load sustained when the aircraft is pulling negative g’s..

Max lift line as it shown in the figure , represent the stall velocity at positive Clmax , for positive load factor , we can see increase in the stall velocity as the load factor increase.

The gust load was coming greater for then the maneuver load in the ferry mission case however theses gust loads are not unusual.

49

IX. WEIGHT ESTIMATION

The estimation of weights of a conceptual aircraft is the critical part of the design process. There are many ways of weight analysis. The previous weight estimation of the aircraft was based on crude analysis techniques. Most sophisticated weights methods estimate the weight of the various components of the aircraft and sum for total empty weight. This chapter has two different methods of weight estimation. One is the “Approximate group weight method” and the other is the “statistical equation using method”. The later technique is sufficient to provide a credible estimate of the major component group.

The weights of different structural parts are as follows:

components Weight (lb)

Wing 17810.89

Vertical Tail 1845.084Fusalge 19308.61Main Landing Gear

3038.22

Nose Landing Gear

492.89

Nacelle group 3084.47Engine Controls

34

Electrical 931.23Avionics 1840.26Fuel systems 3434.05Furnishings 2974.56

Anti-icing 166.56Handling Gear 24.98Pneumatics 167.83Instruments 290.83Flight Controls 948.54Engines 9665.6total 59885

50

The weight distribution is as follows

WingVertical TailFusalgeMain Landing GearNose Landing GearNacelle group Engine ControlsElectrical AvionicsFuel systemsFurnishingsAnti-icingHandling GearPneumaticsInstruments Flight Controlsengines

Afterwards this empty weight was used in calculation and the total weight was iterated to get

FDAV mission

Wo 160766.05Wf 28882.76We 59885.13

Ferry mission

W fuel 38804.79758We 59885.13994wo 133329.6562

51

The empty weight is almost 6 % lesser than that calculated earlier (in refined weight

sizing). The reason may be that these weights are calculated from statistical equations

which may not yield good results for every aircraft. Moreover there may be an error in

the equation used earlier in refined weight analysis.

X. Stability analysis

The customer requirement is of stable and controllable aircraft. The two are opposite of

each other, so the best combination is a good tradeoff between stability and

controllability. For transport aircrfats, the stability requirement is comparatively higher

than the controllability requirements because the aircraft is not supposed to go in

extreme type of maneuvers.

The requirement for good stability, control and handling quantities are addressed

through the use of tail volume coefficient method and through location of aircraft centre

of gravity at some percent of wing mean aerodynamic chord.

Static Stability:

If a system in an equilibrium state returns to equilibrium following a small disturbance,

the state is said to be a stable equilibrium. On the other hand if the system diverges,

the state is said to be an unstable equilibrium.

Categories of Static Stability

There are mainly three types of static stability in an aircraft,

Longitudinal Static stability Lateral static stability Yaw static stability

Calculation of Centre of gravity:

The stability analysis requires that CG location should be known or calculated. After

obtaining the results from previous chapters about the individual weights of the

components, the CG location was estimated.

52

All the distances were calculated from the nose. The CG location was 61.8 ft from the

back. i.e 45.2 ft from the front

Static longitudinal stability

Ideally for a transport aircraft the Cmα value should be negative. The major contributors

toward longitudinal stability are:

wing

horizontal tail and canard ( these are not included in this aircraft)

engine location

fuselage

Another rough estimation of the stability, the centre of gravity should be ahead of the

mean aerodynamic centre of the aircraft. The stability of the aircraft changes with the

location of the centre of gravity. The point where the aircraft if disturbed by any external

force moves to some angle of attack will remain at that angle of attack and there will no

effect of longitudinal stability is called neutral point.

From my calculations the stability at the two mach numbers of 0.6 and 0.8 were

mach 0.6 mach 0.8

from solid edge X/c from solid edge X/c

X acw 67 3.961923 X acw 67 3.961923

X ach 9 0.532199 X ach 9 0.532199

X p 72 4.257589 X p 72 4.257589

Xw 65 3.843656 Xw 65 3.843656

MACH 0.6 MACH 0.8

CLα4.93621387 CLα 5.699314

Cmα FUSELAG

0.00768719

Cmα FUSELAG 0.007687

ηh 0.9 ηh 0.9

Sh 562.38 Sh 562.38

53

Sw 1936.94 Sw 1936.94

Sh/Sw 0.29 Sh/Sw 0.290

Clαh 4.0588 Clαh 4.50

dαh/dα 0.657 dαh/dα 0.60

Fpα 2883.642 Fpα 1581.64

dαp/dα 1.2 dαp/dα 1.2

q 514.264 q 282.06

X np 59.80 X np 60.55

static margin 0.11 static margin 0.073

X cg 61.8 X cg 61.8

Cmα -0.6655 PER RADIANS Cmα -0.474PER RADIANS

-0.011616 PER DEGREE -0.00827PER DEGREE

Hence the static margin was acceptable as was the value of Cmα is negative

Static Lateral Stability:

The lateral – directional analysis consists of two very closely coupled moments, these

are the yaw and roll. It is important here to consider that both these stabilities are driven

by the yaw angle “” because the restoring force in case of the roll disturbance is also

generated by the yaw angle. Also the deflection of rudder or aileron will produce

moments in both yaw and roll.

The major contributors towards the lateral-directional stability are:

wing dihedral wing configuration wing sweep dorsal fin ventral fin

54

The aircraft is said to be laterally stabile if the rolling moment derivative with respect to

sideslip angle is negative, and directionally stable, if the yawing moment derivative with

respect to sideslip angle is positive. The formula used for calculating Cnβ is:

Cnβ=Cnβw+CnβFUSE+CnβVT

CnβVT @ 0.6 -0.171 CnβFUSE @ 0.6 -0.072 Cnβw @ 0.6 -0.029CnβVT @ 0.8 -0.176 CnβFUSE @ 0.8 -0.072 Cnβw @ 0.8 -0.039

Total Cnβ Mach 0.6 0.274Mach 0.8 0.289

Static Directional Stability:

A point of consideration over here is that we don’t want that much high lateral stability

than directional stability because it will end in unwanted phenomena like Dutch roll while

on the other hand if directional stability is marginally higher than lateral stability then we

may encounter undesired phenomena like spiral roll from dynamic stability point of view

Clβ=ClβW+ClβVT

ClβW @ 0.6

-0.1337032 Clβvt @ 0.6

-0.05298

ClβW @ 0.8

-0.1416944 Clβvt @ 0.8

-0.05451

total Clβ @ 0.6 -0.1867

total Clβ @ 0.8 -0.1962

Summary

Mach 0.6 Mach 0.8Cm α -0.665 -0.474

55

Cn β 0.274 0.289Cl β -0.187 -0.196Static margin 9.8% 7.32%

For an aircraft to be statically stable the following requirements should be met: Cmα is positive Clβ is negative Cnβ is positive

The above mentioned requirements are met. so it can be concluded that The aircraft is a Statically Stable.

XI. PERFORMANCE

After having determined all the aerodynamic and structural parameters now we want to ascertain whether the designed aircraft would be able to fulfill all the requirements. This section deals with the performance parameters such as turn rate, rate of climb, takeoff and landing.

I have only calculated those performance parameters that were pertinent to the aircraft requirements. These are as follows

Landing and Take off distance Specific power and fuel specific energy (Ps and Fs) Turn rate Minimum turning radius Range

LANDING AND TAKE OFF DISTANCE

As the design of the aircraft was based on short take off and landing therefore this performance parameter was of paramount importance. The take off segment was broken down into different parts and the distance for each part was calculated before summing them together. The different parts were:

56

Ground roll Transition Climb

FDAV MISSION

Ground roll

Ground roll was calculated for different surfaces as the aircraft should be able to operate from different surfaces. The calculated values are as follows:

dry concrete

1686.74

wet concrete

1737.06

icy concrete

1663.21

hard turf 1737.06firm turf 1711.33soft turf 1792.29wet grass 1822.00

The difference in values was obviously because of different friction coefficient.

Transition

In the transition phase the aircraft was able to clear the 50ft obstacle. The value of STR was 981 ft. while the value of HTR was100.1ft. now as the RFP had given obstacle

57

clear off distance of 3000 ft therefore, trigonometric ratio was used to calculate the horizontal distance.

ST 981.8Htr 100.19

The total take off distance was:

dry concrete 2909.789wet concrete 2960.109icy concrete 2886.259hard turf 2960.109firm turf 2934.379soft turf 3015.342wet grass 3045.051

Hence the aircraft was able to fulfill the take off requirement.

FERRY MISSION

Ground roll

Ground roll was calculated for different surfaces as the aircraft should be able to operate from different surfaces. The calculated values are as follows:

dry concrete

1064.479

wet concrete

1090.05

icy concrete

1052.361

hard turf 1090.05firm turf 1077.03

3soft turf 1117.58

6wet grass 1132.16

9Transition

Again it was seen that the aircraft had cleared the obstacle distance

ST 1110.367

58

Htr 155.94

The same procedure was used to calculate the distance. The final distances were:

dry concrete

2246.431

wet concrete

2272.002

icy concrete

2234.313

hard turf 2272.002

firm turf 2258.986

soft turf 2299.539

wet grass

2314.122

Again the aircraft was able to satisfy the requirement.

LANDING PERFORMANCE

The portions involved in landing are:

Approach distance Ground roll Flare distance

FDAV MISSION

Braking distance

In the calculation thrust reversal was used and hence the distance was broken into two parts

i. With thrust reversalii. Without thrust reversal

The calculated distances were:

With thrust reversal

Without thrust reversal

59

dry concrete

671.2205

wet concrete

707.9553

hard turf 686.7624firm turf 707.9553soft turf 736.2856wet grass

736.2856

dry concrete

377.3617

wet concrete

847.7544


Approach distance

The approach distance for the aircraft was

S approach359.5722688

Flare distance

The flare distance was

S F 424.670

Free roll

the free roll distance was

S fr 530.76

The total distance was

dry concrete 2350.8wet concrete

2870.716


As can be seen that the aircraft was unable to fulfill requirements for soft turf and wet grass and therefore drag chutes should be deployed.

60

FERRY MISSION

Same procedure was used and the final results were

S approach 395.740385S F 352.1958434S fr 483.3571306

SGTR

dry concrete

524.4476

wet concrete

546.2895

hard turf 533.5345

firm turf 546.2895

soft turf 563.4719

wet grass 563.4719

SGdry concrete

387.1251

wet concrete

847.3106

hard turf 530.6709

firm turf 847.3106

soft turf 2247.095

wet grass 2247.095

TOTAL

61

dry concrete

2142.866

wet concrete

2624.893

hard turf 2295.499

firm turf 2624.893

soft turf 4041.86wet grass 4041.86

In the case of soft turf and wet grass drag chutes have to employed of an area of 350 ft^2 the distance reduced to 3542 ft.

SPECIFIC POWER

It is basically the time rate of change of energy height is equal to the specific excess power. The aircraft can increase its energy height simply by the application of excess power.

It is time rate of change of energy height & is equal to the specific excess power, graph below shows Ps contours on Mach # Vs Altitudes graph.

Ps=(

dhedt )=(

dhdt ) + (

V / g∗dvdt )

Ps=

V∗(T−D)W

Specific excess power PS actually equals the thrust available into velocity minus the

thrust required into velocity and complete term divided by the weight , as the thrust is

the function of velocity and it is at the same time the function of the altitude so the

specific excess power varies in the same manner with the altitude and the Mach# and

this is the only reason for its variation with Mach# and altitude and that is why a same

aircraft with all the same specifications can go for only a specific number of g’s at

different altitudes depending on the amount of excess power the aircraft has at a

specific altitude and at a specific Mach no.

Both the graphs were made in excel first a graph of Ps vs Mach was plotted and then

taking values from this graph. Ps and Fs curves were plotted

FDAV MISSION

Ps

62

Fs

Ferry mission

-0.8 0.2 1.20

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

Ps FERRY

0 1020 3040 5060 7080 90100 110120 1000020000 3000040000 50000stall limit

MACH

ALT

ITU

DE

Fs

63

MIN. TIME =7.41 sec

MIN. FUEL = 1375 LB

MIN TIME= 8.52 sec

-0.8 0.2 1.20

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

Fs FERRY0369121518201000020000300004000050000stall limit

MACH

ALT

ITU

DE

It was observed that both the time and fuel requirements decreased for ferry mission

Turning performance

Although turning performance was not part of our requirement but as it is necessary to calculate it. While designing an aircraft the turning performance was calculated.

64

MIN. FUEL = 1215 lb

0 200 400 600 800 1000 12000

5

10

15

20

25

30

turning performance for FDAV

23456Sustained turn5.38STALL

velocity

ψ

0 200 400 600 800 1000 12000

5

10

15

20

25

30

Turning Performance For ferry

23456Sustained turnCORNER VSTALL

velocity

ψ

Minimum turning radius

65

The minimum turn radius and the load factor at which they are achieved were also calculated for cruise at 1000ft @ M 0.6 and cruise at 30000ft @ M 0.8. the values are as follows

For cruise at 1000ft @ M 0.6

R min 604.9419 ftn Rmin 1.405603 ft

For cruise at 30000ft @ M 0.8

R min 1570.046 ftn Rmin 1.404499 ft

XII. Solid edge views

66

ISOMETRIC

TOP

SIDE

67

FRONT

DIAMETRIC

68

TRIMETRIC

XIII. CONCLUSION

69

The results of the aircraft were satisfactory

The aircraft could be further improved by optimizing it. The aircraft is also low cost as no expensive material was used. This aircraft if manufactured would be the first of its kind as at present there are no STOL tactical transport aircrafts.

Specifications Result

TO Distance Achieved

Max Mach no Achieved

Absolute Ceiling Achieved

Range Achieved

Landing distance Partially achieved

70

Documents

Design Report of STOL transport aircraft