My Aircraft Design Project - I

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JAWAHARALAL

INSTITUTE OF TECHNOLOGY

(Approved by AICTE & Affiliated to Anna University)

COIMBATORE – 641 105

NAME : J.Dinesh Raja Ruban

REG.NO : 080101134017

SUBJECT : Aircraft Design Lab – I

COURSE : Aeronautical Engineering



JAWAHARLAL INSTITUTE OF TECHNOLOGY

COIMBATORE – 641 105

DEPARTMENT OF AERONAUTICAL ENGINEERING

Certified that this is the bonafide record work done by

J.Dinesh Raja Ruban in the AIRCRAFT DESIGN LAB – I of this institution as

prescribed by the Anna University, Coimbatore for the Sixth semester during the

year 2010 – 2011.

Staff In charge: Head of the Department

University Register No.:080101134017

Submitted for the Practical Examination of the Anna University conducted on

……………

INTERNAL EXAMINER EXTERNAL EXAMINER



ACKNOWLEDGEMENT

Firstly I would like to thank the Almighty God for always being by my side and providing me with

strength and capability to face all types of situations during this project tenure.

I extend my fullest and ever owing thanks to Dr.K.K.Babu, Principal, Jawaharlal Institute of

Technology, Coimbatore, for the academic freedom and inspiration.

With deep sense of gratitude, I extend my earnest & sincere thanks to our guideDr.Rajasekar

M.S., Ph.D, Head Of The Department, Aeronautical Engineering , Jawaharlal institute of technology,for

his systematic guidance, encouragement and for providing valuable insights offered over the course of

this project report.

I also thank everyone who lent us support in the completion of this project.





Introduction:

There 2 classes of fighter aircraft. They are class-1 and class-2 fighter aircrafts. The class

one fighter is officially an air superiority fighter. Most of it can function either as multi role

fighters and ground attack. Air superiority fighter mainly does the function to gain air space

control over the enemy territory so that the bombers can bomb their targets safely, and give

support for the ground units. They literally make the enemy air space home ground for the

invaders aircrafts. Class2 fighters mainly concentrate on electronic warfare and ground attack

along with surveillance.

Today, complex sets of requirements and objectives include specification of airplane

performance, safety, reliability and maintainability, subsystems properties and performance, and

others. Some of these are illustrated in the table below







--Good neighbor in peace --

Dectability in war





Today, complex sets of requirements and objectives include specification of airplane

performance, safety, reliability and maintainability, subsystems properties and performance, and

others. Some of these are illustrated in the table below

Aircraft Design Objectives and Constraints

Issue Military

Dominant design criteria

Mission accomplishment and

survivability

Performance

Adequate range and response

Overall mission accomplishment

Airfield environment

Short-to-moderate runways

All types of runway surfaces

Often Spartan ATC, etc.

Limited space available

System complexity and mechanical

design

Low maintenance- availability

issue

Acceptable system cost

Reliability and survivability

Damage tolerance

Government regulations and community

Military standards



acceptance --Performance and safety --

Reliability oriented

Low noise desirable

--Good neighbor in peace --

Dectability in war







Mission Profile:

3 4 5

6 7

12 8 9

A

L

T

I

T

U

D

E

Range





M ission

SegmentDescription Distance Time Al titude

1-2 Ground run 150 Meters 5 Seconds 0 Meter

2-3 Ascent 185 Meters 6 Minutes 14175 Meters

3-4 Cruising 2000 Meters 4 Seconds 14175 Meters

4-5 Aerobatic 1000 Meters 2 Seconds 14175 Meters

5-6 Nose down 4175 Meters 4 Seconds

14175-10000

Meters

6-9 Descent 200 Km 3 Minutes 10000-0 Meters

9-10 Halt 100 Meters 4 Seconds 0 Meter











[W5/W4]@6 = e-[(700*0.6) / (3186.72*6)]

[W5/W4]@6 = 0.9782

[W5/W4]@8 = e-[(700*0.6) / (3186.72*8)]

[W5/W4]@8 = 0.9836

[W5/W4]@10 = e-[(700*0.6) / (3186.72*10)]

[W5/W4]@10 = 0.9869

[W5/W4]@12 = e-[(700*0.6) / (3186.72*12)]

[W5/W4]@12 = 0.9890

For the nose down maneuver, from the historical data,

W6 / W5 = 0.9860

Next segment in the mission profile is cruise before halting.

Range = 200

G.S.R = 200/1.5

G.S.R = 133.33 Km

Altitude = 9175 m ≈ 10 Km

Pressure = 2.6500×104

W5/W4= e-RC

t/ V×(L/D)

max



Density = 4.1351×10-1

a = √ (1.4×26500×104) / (4.1351×10-1)

a = 299.53 m/s

a = 1078.30 Km/hr

As mach number is 3,

V∞ = a×3

V∞ = 1078.30×3

V∞ = 3234.924 Km/hr

Time = 133.33 / 3234.924

Time = 0.04121 hours

Head wind = 15 m/s

Head wind = 54 Km/hr

Actual additional distance = 54×0.04121

Actual additional distance = 2.2256 Km

Total rate range = 200 + 2.2256

Total rate range = 202.2256 Km

a = γP/ρ



For cruising,

[W7/W6]@6 = e-[(202.22*0.6) / (3234.924*6)]

[W7 / W6]@6 = 0.9937

[W7/W6]@8 = e-[(202.22*0.6) / (3234.924*8)]

[W7 / W6]@8 = 0.9953

[W7/W6]@10 = e-[(202.22*0.6) / (3234.924*10)]

[W7 / W6]@10 = 0.9962

[W7/W6]@12 = e-[(202.22*0.6) / (3234.924*12)]

[W7 / W6]@12 = 0.9968

For descending, the aircraft is assumed to consume less amount of fuel.

W8/W7 = 0.97

For landing and halting, fuel consumption is very less

W9/W8 = 0.99

[W9/W0]@6 = 0.85*0.960*0.9936*0.9752*0.9782*0.986*0.9937*0.97*0.99

W7 / W6 = e- RC

t/V×(L/D)max

W9/W0=(W1/W0)×(W2/W1)×(W3/W2)×(W4/W3)×

(W5/W4)×(W6/W5)×(W7/W6)×(W8/W7)×(W9/W8)









[We/Wg]@6 = 1.202[2.202*52323.58]-0.06

We/Wg@6 = 0.5068

[We/Wg]@8 = 1.202[2.202*43132.17]-0.06

0

10000

20000

30000

40000

50000

60000

0 2 4 6 8 10 12 14

W g

(L/D)Max

Wg Vs (L/D)Max

Series1

We/Wg = 1.202 [2.202 Wg]-0.06



We/Wg@8= 0.5127

[We/Wg]@10 = 1.202[2.202*41401.87]-0.06

We/Wg@10= 0.5140

[We/Wg]@12 = 1.202[2.202*40331.26]-0.06

We/Wg@12= 0.5148

For L/D = 6,

WPay= 0.1146

Wf /Wg = 0.2889

We/Wg= 0.5068

Wg = 52323.58 Kg

For L/D = 8,

WPay= 0.1391

Wf /Wg = 0.2566

We/Wg= 0.5127

Wg = 43132.17 Kg



For L/D = 10,

WPay= 0.1449

Wf /Wg = 0.2493

We/Wg= 0.5140

Wg = 41401.87 Kg

For L/D = 12,

WPay= 0.1487

Wf /Wg = 0.2445

We/Wg= 0.5148

Wg = 40331.26 Kg





Where,

ρ = 1.225 Kg/m3

(W/S)Land = 1/2×1.225× (31.40)2×

3

(W/S)Land = 1811.70

(W/S)Land = 1/2×ρV2SCLMax





LANDING WEIGHT OF AN AIRCRAFT:

Let the landing weight ratio = 0.62

Landing weight = 32500 Kg

SLand = W/1811.70

SLand = 32500/1811.70

SLand = 17.93 m

Stall velocity, Vs@+10% = 31.40+31.40×(10/100)

Stall velocity, Vs@+10% = 34.54 m/s

(W/S)Land = 1/2×1.225× (34.54)2×

3

(W/S)Land = 2192.15

SLand = 32500 / 2192.15

SLand = 14.82 m

Stall velocity, Vs@-10% = 31.40 - 31.40× (10/100)

Stall velocity, Vs@+10% = 28.26 m/s





(W/S)Land = 1/2×1.225×(28.26)2×

3

(W/S)Land = 1467.47 Kg

SLand = 32500 / 1467.47

SLand = 22.146 m





SELECTION OF WING LOADING:

VMax = 1.1× 3186.72

VMax = 3505.392 Km/hr

Log SWet = {0.5+0.5log10[52323.58×2.202]} / (3.29)2

Swet = 39.506 m2

K = 1/πeA

K = 1/ (π×0.7×3)

K = 0.0757

Where,

CD0 = 0.005× (39.506/17.93)

CD0 = 0.0110

CD = 0.0110 + (0.0757×32)

CD = 0.692

CD = CD0 + KCL2

VMax = 1.1× VCruise

CD0 = Cfe × (Swet/S)

T/W = CD×1/2×ρV2Max / (W/S)



T/W = {0.692×1/2×1.225×(973.72)2}/1811.70

T/W = 221.81

CD0 = 0.005× (39.506/14.82)

CD0 = 0.0133

CD = 0.0133+ (0.0757×32)

CD = 0.694

T/W = {0.694×1/2×1.225×(973.72)2}/2190.15

T/W = 184.01

CD0 = 0.005× (39.506/22.146)

CD0 = 8.91×10-3






CD = 8.91×10-3

+ (0.0757×32)

CD = 0.690

T/W = {0.690×1/2×1.225 × (973.72)2}/1467.47

T/W = 273.14

T/W = 3090, CD = 0.692

3090 = 0.692 × (1/2) ×1.225 × (3505.39)2 × (S/W)

(W/S) = 1.6848×103

T/W = 2595.5, CD = 0.694

2595.5 = 0.694 × (1/2) ×1.225 × (3505.39)2 × (S/W)

(W/S) = 2.012×10

3

T/W = 3753.71, CD = 0.690


(T/W) = CD× 1/2×ρV2Max(S/W)





3753.71 = 0.690 × (1/2) × 1.225 × (3505.39)2 × (S/W)

(W/S) = 6.420×104

WT0 = 52323.58 Kg

C = 0.1, D = 0.5

Log (SWet) = 0.1 + 0.5log(52323.58)

SWet = 39.506 m2

(W/S)take off = 1811.7 / 0.62

(W/S)take off = 2922.07

CD0 =0.005 × (39.506/17.90)

CD0 = 0.0110

Log (SWet) = C + Dlog(WT0)

(W/S)take off = (W/S)landing / 0.62

CD0 = Cfe × (Swet/Stake off )

CD = {Cfe × Swet(W/S)take off }/ {Wtake off × 9.81}



CD = {0.005×21.458×2438.35} / {52323.58 × 9.81}

CD = 5.095×10-4

CD Clean = 0.0110 + (1/πeA)×32 + 0

CD Clean = 0.0110 + (1/π×0.8×6)×32

CD Clean = 0.6078

CD Take off = 0.0110 + (1/πeA)×32

+ 0.01

CD Take off = 0.0110 + (1/π×0.75×6)×32+ 0.01

CD Take off = 0.6576

CD Landing = 0.0110 + (1/πeA)×32 + 0.05

CD Landing = 0.0110 + (1/π×0.7×6)×32+ 0.05

CD Clean = CD0 + KCL2 + ΔCD0

CD Take off = CD0 + KCL2 + ΔCD0

CD Landing = CD0 + KCL2 + ΔCD0



CD Landing = 0.7430

CD Landing Gear = 0.0110 + 0.020

CD Landing Gear = 0.031

F1 = 3×0.005

F1 = 0.015

F2 = (0.0110 – 0.015) / 1811.7

F2 = -2.2078×10-6

F3 = 1.8643×10

-7

CD Landing Gear = CD0 + ΔCD0

F1 = 3×Cfe

F2 = (CD0 – F1) / (W/S)

F3 = K/q2

F3 = 1/[πeA(1/2×ρV2Max)]

2

(W/S)Max = F1/F3



(W/S)Max = √0.015/1.8643×10-7

(W/S)Max = 283187.3 m2

(W/S)Max = 283.18×103 m2





Airfoils:

One of the difficulties in designing a good airfoil is the requirement for acceptable off-

design performance. While a very low drag section is not too hard to design, it may separate at

angles of attack slightly away from its design point. Airfoils with high lift capability may

perform very poorly at lower angles of attack.One can approach the design of airfoil sections

with multiple design points in a well-defined way. Often it is clear that the upper surface will be

critical at one of the points and we can design the upper surface at this condition. The lower

surface can then be designed to make the section behave properly at the second point. Similarly,

constraints such as Cmo are most affected by airfoil trailing edge geometry.When such a

compromise is not possible, variable geometry can be employed (at some expense) as in the case

of high lift systems.

Airfoil Parameters:

For my Aircraft, the selected airfoil is NACA 6 digit series. That is NACA 64a204.

The taper ratio of the wing is 0.295

Leading edge sweep angle is 46º

Root chord of the wing is 8.61m

Tip chord of the wing is 2.067m

Finess ratio is 0.24

Planform area is 171.098 m

2

Aspect Ratio is 6

Span of the wing is 32.04 m

Half span is 16.02 m



Aerofoil:





Wing Design:

There are essentially two approaches to wing design. In the direct approach, one finds the

planform and twist that minimize some combination of structural weight, drag, and CLmax

constraints. The other approach involves selecting a desirable lift distribution and then

computing the twist, taper, and thickness distributions that are required to achieve this

distribution. The latter approach is generally used to obtain analytic solutions and insight into the

important aspects of the design problem, but is is difficult to incorporate certain constraints and

off-design considerations in this approach. The direct method, often combined with numerical

optimization is often used in the latter stages of wing design, with the starting point established

from simple (even analytic) results.

Wing lift distributions play a key role in wing design. The lift distribution is directly related to

the wing geometry and determines such wing performance characteristics as induced drag,

structural weight, and stalling characteristics. The determination of a reasonable lift and Cl

distribution, combined with a way of relating the wing twist to this distribution provides a good

starting point for a wing design. Subsequent analysis of this baseline design will quickly show

what might be changed in the original design to avoid problems such as high induced drag or

large variations in Cl at off-design conditions.

Parameters:

Span:

Selecting the wing span is one of the most basic decisions to make in the design of a wing. The

span is sometimes constrained by contest rules, hangar size, or ground facilities but when it is not

we might decide to use the largest span consistent with structural dynamic constraints (flutter).

This would reduce the induced drag directly.However, as the span is increased, the wing



structural weight also increases and at some point the weight increase offsets the induced drag

savings. This point is rarely reached, though, for several reasons.

1. The optimum is quite flat and one must stretch the span a great deal to reach the actual

optimum.

2. Concerns about wing bending as it affects stability and flutter mount as span is increased.

3. The cost of the wing itself increases as the structural weight increases. This must be

included so that we do not spend 10% more on the wing in order to save .001% in fuel

consumption.

4. The volume of the wing in which fuel can be stored is reduced. It is more difficult to

locate the main landing gear at the root of the wing.



The selection of my aircraft wing is based on the following. They are

Landing speed/landing distance

Maximum speed VMax

Absolute ceiling

Rate of climb

Based on range





SELECTION OF WING BASED ON

(A)

LANDING SPEED/LANDING DISTANCE:

Landing ground run:

It is the actual distance the airplane travels from the time the wheel first touch to the time

the airplane comes to a halt.

Landing distance:

It is the horizontal distance the airplane covers from being at the screen height (15m) till

comes to a halt.

Sland = Landing distance

Sland = 500m

VA - Approach Velocity

VA= 1.71× √500 m/s

VA= 40.8 m/s

Vs - Stalling Speed

Vs = VA/1.3

Vs = 31.40 m/s

CLmax – Maximum lift coefficient

CLmax – To be taken from various reference airplanes.



The value of CLmax depends on the following.

(a) Wing geometry i.e. aspect ratio (A), taper ratio (λ) andsweep (Λ).

(b)Airfoil shape.

(c) Flap type, ratio of flap area to wing area (Sflap/S) and flap deflection (δflap).

(d) Type of leading edge slat and its deflection.

(e) Reynolds number.

(f) Surface texture.

(g) Interference effect from fuselage, nacelle and pylons.

(h) Influence of propeller slip stream, if present.

Density ρ at the landing airport = 1.225 kg/m3

Wing loading (W/S) based on landing distance = CLmax× (ρ×Vs

2

)/2 (N/m2

)

= (3×1.225×402)/2

= 1811.7 N/m2

To be estimated from mission,

Fuel consumption and disposed weight = 0.62

Wland/Wtake-off = [1- (Wfuel/ Wtake-off ) - (Wdisposible/Wtake-off )]

(W/S)take-off = (W/S)landing / (Wland/Wtake-off )



(W/S)take-off = 1811.7/ (32500/52323.58)

(W/S)take-off = 1811.7/0.62

(W/S)take-off = 2922.09 N/m2

0

5

10

15

20

25

30

35

40

0 500 1000 1500 2000 2500

V S t a l l

W/S

VStall Vs W/S

Series1





B) MAXIMUM SPEED VMax:

The optimization from Vmax consideration aims at finding out the wing loading which will

result in the lowest thrust requirement for a chosen Vmax at Hcr (cruise altitude).

1)

Estimation of Drag polar

(A)

Estimation of CD0

C and d are constant based on type of airplane

Wtake-off = 52323.58 kg

C = 0.1

d = 0.5

log10 Swet = (0.1) + (0.5) log10 (52323.58×2.205)

Swet = 39.506

Cfe = Equivalent skin friction drag coefficient (varies from 0.0025 to 0.0065 based on type of

aero plane) = 0.005

Swet = Wetted area

Swet= 39.506 m2

Log10Swet = C + d log10Wtake-off

CD0 = CfeSwet / S = CfeSwet (W/S)take-off /Wtake-off



CD0= 0.005 × 39.506 × 2922.09 / 52323.58×9.81

CD0 = 0.00124

(B)

Estimation of K:

For low speed airplane K = 1/(π×A×e)

Where,

A = Aspect ratio

Aspect ratio = 6

e = Oswald efficiency factor (lies between 0.8 to 0.88 with unswept wing)

e swept wing = e unswept wing × cos (Λ0-5)

e = 0.7

K = 1/(π×A×e)

K = 1/(π×6×0.7)

K =0.0757

1)

Estimation of drag polar:

CD= 1.2 × 10-3

× 0.0757 × CL2

CD0 = Cfe× Swet / Stake-off = Cfe × Swet × (W/S)take-off / Wtake-off

CD = CD0 + k CL2



CD Clean cleans drag polar can be used while calculating maximum rate of climb and

subsonic cruise cases.

Configuration ΔCD0 e

Landing gear 0.015 to 0.025 No effect

Landing flaps 0.05 to 0.075 0.7 to 0.75

Take-off flaps 0.05 to 0.075 0.75 to 0.8

Clean - 0.8 to 0.85

Take-off flaps,

ΔCD0 (1) =0.01

e(1) = 0.75

Landing flaps,

ΔCD0 (2) = 0.05

e(2) = 0.7

Landing gears,

ΔCD0(3) = 0.015

K 1 = 1/(π×6×0.75)

K 1 = 1/(π×A×e)



K 1 = 0.0707

K 2 = 1/(π×6×0.7)

K 2 = 0.075

Take-off flaps, landing gear up

ΔCD0 (4) = 0.01


ΔCD0(5) = 0.025

Landing flaps, gears up

ΔCD0(6) = 0.05

Landing flaps, gears down

ΔCD0(7) = 0.065

DRAG POLAR FOR DIFFERENT CONFIGURATION:

K 2 = 1/(π×A×e)

ΔCD0 (4) = ΔCD0 (1)

ΔCD0 (5) = ΔCD0 (1) + ΔCD0(3)

ΔCD0 (6) = ΔCD0 (2)

ΔCD0 (7) = ΔCD0 (2) + ΔCD0 (3)



Clean Configuration,

CD = 1.2 ×10-3+0.0757CL2


CD = 0.011 + 0.0707 CL2

Take-off flaps, landing gear down

CD = 0.02612 + 0.0707 CL2

Landing flaps, gear up

CD= 0.05112 + 0.075 CL2

Landing flaps, gear down

CD = 0.01112 + 0.07 CL2

BREAK-UP OF DRAG POLAR:

CD = CD0+KCL2

CD = CD0 + ΔCD0 (4) + K 1 × CL2

CD = CD0 + ΔCD0 (5) + K 1×CL2

CD = CD0 + ΔCD0 (6) + K 2×CL2

CD = CD0 + ΔCD0 (7) + K2×CL2

CD = F1 + F2 × (W/S) + F3 × (W/S)2





Mmax = 3.04

Vmax = 1033.6 m/s

F3=1/[(π×6×0.7)×(0.5×1.225×1033.62)

2]

F3=1.77×10-13

Wing loading for the lowest thrust requirement (TVmax) for chosen at a given ceiling Hcr

(W/S) = (0.5×ρ×Vmax2)×√(F1× π A e)

(W/S) = (0.5×1.2165×10-1

×88.192

) × (F1× πAe)

(W/S) = 22662.38 N/m2

Vmax= Mmax × speed of sound

F3 =(π×A×e(0.5×ρ×Vmax2)

-1

(W/S) = CD × qmax/(TVmax/W)

(W/S) = (F1/F3)



0

50

100

150

200

250

300

350

400

450

0 1000 2000 3000 4000 5000 6000

V m a x

W/S

Vmax Vs (W/S)

Series1





C) ABSOLUTE CEILING (Hmax):

At absolute ceiling the flight is possible only at one speed at which

Thrust required = Thrust minimum = Drag minimum

If the flight velocity (VHmax) at the absolute ceiling

Thrust Loading,

Both VHmax and Hmax are prescribed, then

CD0 =1.12×10-3

K =0.0757

ρHmax = 1.2165×10-1

kg/m3

= (√1.12×10-2

/0.0757)×(0.5×1.2165×10-1

×885.192)

= 7.3984×10-3

×VHmax2

Thrust Loading (T/W) = √(4CD0 K) = √(4k(F1+F2(w/s))

(T/W) = (0.5×ρHmax×VHmax2)×2CD0/(W/S)

(W/S) = (CD0/K)×(0.5×ρHmax×VHmax2)

(W/S) = (CD0/k)×(0.5×ρHmax×VHmax2)



0

100

200

300

400

500

600

700

800

900

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

V H M a x

W/S

VH Max Vs (W/S)

Series1





(D) RATE OF CLIMB:

Wing loading is such that thrust required is minimum for the specified rate of climb (Vc).

For chosen flight velocity V, the optimum wing loading = (W/S)opt = (F1/F3) 0.5

(tR/C)v = (Vc/V) + q(F1/PR/C + F3 + F3PR/C)

For each V, we have an optimum ρ

Each curve corresponding to each V has a minimum. To get the minimum of the minima,

we draw an envelope which is tangential to all the curves. Minimum of this envelop gives the

optimum wing loading from the rate of climb consideration and the corresponding minimum

thrust loading (tR/Cmax)min.

For Jet Airplanes,

(L/D)max = (4 CD0 K)-0.5

(L/D)max = (4×1.12×10-3

× 0.0757)-0.5

(L/D)max = 54.3016

For propeller Airplanes,

V(R/C)max ={(T/W)(W/S)/3ρ∞CD.0[1+ (1+(3/(L/D)max2(T/W)

2)]}

1/2

V(R/C)max = {2/ ρ∞ ((K/3CD0)(W/S))}1/2



V(R/C)max = [(1*(W/S))/ (3*0.22785*1.12*10-3) {1+√ (1+ (3/54.30162

))}]1/2

V(R/C)max = [2613.32 (W/S)]1/2

(W/S) = V(R/C)2/2613.32

0

100

200

300

400

500

600

700

800

900

0 50 100 150 200 250 300

V R / C M

a x

W/S

VR/C Max Vs (W/S)

Series1







(E) BASED ON RANGE:

For jet airplanes,

The density on altitude and the TSFC depends on flight velocity and altitude. Optimum

wing loading for a chosen Vcr is obtained at different altitude. Minimum of this curve gives the

optimum wing loading and corresponding cruising altitude.

For Jet airplanes,

V(Cl1/2

/CD)max = [2/0.227885√[(3×0.0757)/(1.12×10-3

)] (W/S)]1/2

V(Cl1/2

/CD)max= 117.17 × (W/S)1/2

W/S = 3000 N/m2

WTake off =52323.58 Kg

L/D = CL/CD = 6

CLMax = 3

CD0 = 112×10-3

K = 0.0757

WFuel/WTake off = 0.288

W/S = (F1+CD0)×(R/3.6× (ρ×q/2)×(TSFC)/Wtake-off ×(Wtake-off /Wf)

V(Cl1/2

/CD)max = {2/ ρ∞× (3K/CD0) × (W/S)}1/2



Range = 2000 Km

Mach number = 3

Cruise altitude = 14000 m

Absolute ceiling = 18000 m

Rate of climb, V∞ = 720 m/s

Rate of climb angle = sin-1

(Ve/V∞) = 14º

Climb velocity = 174.18 m/s

Landing distance = 600 m

ηMax = (3×1/2×0.2278×(720)2)/3000

ηMax = 59.058

Where,

C - Constant (C= 0.6)

Maximum efficiency,

ηMax = CL Max×1/2×ρV2/(W/S)

WWing = C×Sw×AR×(t/c)-0.4 (1+taper ratio)0.1



Sw - Wing area

AR - Aspect ratio

t/c - Finess ratio

W/S = 3000

W = 52323.58 Kg

W = 513294.31 N

S = 513294.31/3000

S = 171.09 m

2

Aspect ratio = 6

Span2 = 171.09×6

Span2 = 1026.54

Span = 32.04

Sweep angle = 60º

Aspect ratio = span2/plan form area

Taper ratio = Tip chord/Root chord

Span2 = Plan form area × Aspect ratio



Cos(Sweep)-1

= Cos(60º)-1

Cos(Sweep)-1

= 2

Mean chord =171.09/32.04

Mean chord = 5.33

(Ct+Cr )/2 = 10.67

Ct+Cr = 10.67

Ct/Cr = 0.24

Ct = 0.24Cr

0.24Cr +Cr = 10.67

1.24Cr = 10.67

Cr = 8.61m

Ct = 0.24 × 8.61

Ct = 2.067m

Plan form area = Span × Mean chord

Mean chord = (Ct+Cr)/2

Taper ratio = Ct/Cr



0.30 × 52323.58= 0.6×28.14×6 (t/c)-0.4

×(1.24)0.1

×2

(t/c)-0.4

= 151.66

t/c = 7.06×10-6

WWing = C×Sw×AR×(t/c)-0.4

(1+taper



0

100

200

300

400

500

600

700

800

900

1000

0 1000 2000 3000 4000 5000 6000 7000

( V C L / C D ) 1 / 2

WS

(VCL/CD)1/2 Vs WS

Series1











Wing with centre of gravity







Cabin Layout and Fuselage Geometry:

The design of the fuselage is based on payload requirements, aerodynamics, and

structures. The overall dimensions of the fuselage affect the drag through several factors.

Fuselages with smaller fineness ratios have less wetted area to enclose a given volume, but more

wetted area when the diameter and length of the cabin are fixed. The higher Reynolds number

and increased tail length generally lead to improved aerodynamics for long, thin fuselages, at the

expense of structural weight. Selection of the best layout requires a detailed study of these trade-

offs, but to start the design process, something must be chosen. This is generally done by

selecting a value not too different from existing aircraft with similar requirements, for which

such a detailed study has presumably been done. In the absence of such guidance, one selects an

initial layout that satisfies the payload requirements.The following sections are divided into

several parts: the selection of cabin cross-section dimensions, determination of fuselage length

and shape, FAR's related to fuselage design and seating, and finally considerations related to

supersonic aircraft.

Cross-Section Shape:

It is often reasonable to start the fuselage layout with a specification of the cross-section:

its shape and dimensions.

Most fuselage cross-sections are relatively circular in shape. This is done for two reasons

By eliminating corners, the flow will not separate at moderate angles of attack or

sideslip

When the fuselage is pressurized, a circular fuselage can resist the loads with

tension stresses, rather than the more severe bending loads that arise on non-

circular shapes.



Fuselage sizing:

We can fine the size of the fuselage by using the formula,

where,

a – 1.0 – 1.8

b – 0.5 – 0.25

lf = 0.4 × (52323.58)0.4

lf = 0.4 × 77.17

lf = 30.87 feet

lf = 9.4092

lf = 9.41

Cockpit:

The cockpit must be designed in such a way that, the pilot can visible the runway clearly.

The angle given for our aircraft is 11º. This is because the pilot will not get the sight problem to

look the ground while taxing.

lf = aW0

c



Seats in the cockpit:

The seats used for our aircraft is the ejection type seat.

As this is a fighter aircraft to save the life of the pilot we are using the ejector seat model.

As it is important to give him enough comfort to the pilot we have to select the space in

cockpit and other things as per the comfortness of pilot.

The seat pitch is 90cm with the seat width of 55cm.

The head room must be in 165cm above.

The aisle height is greater than 193cm.

Volume per passenger is 0.14m

3.

Tail sizing:

We are using the following formula to find the horizontal tail size,

Cht = lht×sht

Formula to find the horizontal tail size is,

Cvt = lvt×svt

Cht/cw = (lht/lw) × (sht/sw)

Cvt/cw = (lvt/lw) × (svt/sw)

Cht/cw+ Cvt/cw = (lht/lw) × (sht/sw) + (lvt/lw) × (svt/sw)

Sht/Sw = (Cht/Cw) × (lw/lht)



Svt/Sw = (Cvt/Cw) × (lv/lvt)

For horizontal tail,

Aspect ratio, AR = 3

Taper ratio, TR = 0.3

For vertical tail,

Aspect ratio, AR = 1.5

Taper ratio, TR = 0.5

(t/c)Tail = 6.354×10-6

Weight:

The structural weight of our aircraft is 1150Kg

The propulsion weight of our aircraft is 1500Kg

The fixed equipment weight of our aircraft is 350Kg

The empty weight of our aircraft is 3000Kg

The weapons weight of our aircraft is 1500 Kg

(t/c)Tail = 0.9×(t/c)Wing













Engine Selection:

As our aircraft is the fighter aircraft flying at a speed of Mach 3, we selected the turbo fan

engines

Two engines are located in the wing

The dry thrust is assumed to be of 76.4 KN (17,185 lbf)

The thrust with after burner is assumed to be of 109.8 KN (24,675 lbf)

Engine Placement:

The arrangement of engines influences the aircraft in many important ways. Safety,

structural weight, flutter, drag, control, maximum lift, propulsive efficiency, maintainability, and

aircraft growth potential are all affected. Engines may be placed in the wings, on the wings,

above the wings, or suspended on pylons below the wings. They may be mounted on the aft

fuselage, on top of the fuselage, or on the sides of the fuselage. Wherever the nacelles are placed,

the detailed spacing with respect to wing, tail, fuselage, or other nacelles is crucial. Engines

buried in the wing root have minimum parasite drag and probably minimum weight. Their

inboard location minimizes the yawing moment due to asymmetric thrust after engine failure.

However, they pose a threat to the basic wing structure in the event of a blade or turbine disk

failure, make it very difficult to maximize inlet efficiency, and make accessibility for

maintenance more difficult.











Sukhoi Su-27

Manufacturer Soviet Union

First flight 20 May 1977

Introduced

ENGINE

Model Saturn/ Lyul'ka AL-31F afterburning turbofansThrust

No. Of engines 2

Dry thrust 33,510 lb (149.06 kN)

Thrust with afterburner 55,116 lb (245.18 kN)

Fuel capacity

ACCOMMODATION

Crew 1

WEIGHT RATIOS

Empty Weight 51,015 lb (23,140 kg)

Gross Weight



Loaded Weight 62,390 lb (28,300 kg)

Maximum Take Off Weight 72,750 lb (33,000 kg)

Max Payload 8,820 lb (4,000 kg)

Armaments

Fuel Capacity 20,725 lb (9,400 kg)

FUSELAGE

Length 71.92 ft (21.94 m)

Height 19.42 ft (5.92 m)

WING

Area (m2) 667.8 ft² (62.04 m

2)

Span 48.17 ft (14.70 m)

PERFORMANCE

Maximum SpeedMach 2.35 1,555 mph (2,500 km/h) at 36,090

ft (11,000 m)

Endurance

Service Ceiling 18,500 m (62,523 ft)



Range 3,530 Km (2,070 mi)

Rate of climb 300 m/s

Wing loading 371 Kg/m2

Thrust/weight 1.09

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

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

http://en.wikipedia.org/wiki/Thrust-to-weight_ratio






Sukhoi Su-35


First flight -

Introduced -

ENGINE

Model Saturn 117S with TVC nozzle turbofan

No. Of engines 2

Dry thrust 8,800 kgf (86.3 kN, 19,400 lbf) each

Thrust with afterburner 14,500 kgf (142 kN, 31,900 lbf) each

Fuel capacity -

ACCOMMODATION

Crew 1

WEIGHT RATIOS

Empty Weight 18,400 kg (40,570 lb)

Gross Weight

Loaded Weight 25,300 kg (56,660 lb)

Maximum Take Off Weight 34,500 kg (76,060 lb)

Armaments -

FUSELAGE

Length 21.9 m (72.9 ft)

Height 5.90 m (19.4 ft)

http://en.wikipedia.org/wiki/Saturn_AL-31#117S

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

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


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

http://en.wikipedia.org/wiki/Saturn_AL-31#117S



WING

Area 62.0 m² (667 ft²)

Span 15.3 m (50.2 ft)

PERFORMANCE

Maximum Speed Mach 2.25 (2,390 km/h, 1,490 mph) at altitude

Endurance -


Range3,600 km (1,940 nmi) ; (1,580 km, 850 nmi

near ground level)

Ferry Range 4,500 km (2,430 nmi) with external fuel tanks

Rate of climb >280 m/s (>55,100 ft/min)

Wing loading 408 kg/m² (84.9 lb/ft²)

Thrust/weight 1.1









Sukhoi Su-25


First flight 22 February 1975

Introduced -

ENGINE

Model Tumansky R-195 turbojets

No. Of engines 2

Thrust 44.18 kN (9,480 lbf) each

ACCOMMODATION

Crew 1

WEIGHT RATIOS


Gross Weight -



Armaments 4,400 kg (9,700 lb)

FUSELAGE

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

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

http://en.wikipedia.org/wiki/Pound-force



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



Length 15.33 m (50 ft 11)

Height 4.80 m (15 ft 9 in)

WING

Area 30.1 m² (324 ft²)

Span 14.36 m (47 ft 1 in)

PERFORMANCE

Maximum Speed 950 km/h (590 mph, Mach 0.77)

Endurance -


Range 2,500 km (1,553 mi)

Rate of climb 58 m/s (11,400 ft/min)

Wing loading 584 kg/m² (119 lb/ft²)

Thrust/weight 0.51

Combat radius 375 km (235 mi)




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

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






Sukhoi Su-24


First flight -

Introduced -

ENGINE

Model

Saturn/Lyulka AL-21F-3A

afterburningturbojet engines

No. Of engines 2

Dry thrust 75 kN (16,860 lbf) each

Thrust with afterburner 109.8 kN (24,675 lbf) each

Fuel capacity 11,100 kg (24,470 lb)

ACCOMMODATION

Crew Two (pilot and weapons system operator)

WEIGHT RATIOS


Gross Weight -



Armaments Up to 8,000 kg (17,640 lb) hard points

FUSELAGE

Length 22.53 m (73 ft 11 in)


http://en.wikipedia.org/wiki/Lyulka_AL-21

http://en.wikipedia.org/wiki/Afterburner_%28engine%29






WING

Area 55.2 m² (594 ft²)

Span

17.64 m extended, 10.37 m maximum sweep

(57 ft 10 in / 34 ft 0 in)

PERFORMANCE

Maximum Speed 2,320 km/h (1,440 mph)

Endurance


Range 2,775 km (1,500 nm, 1,725 mi)



Thrust/weight 0.60

Takeoff roll 1,550 m (5,085 ft

Landing roll 1,100 m (3,610 ft)

G-Force limit 6









Sukhoi Su-17


First flight -

Introduced 1970

ENGINE

Model Lyulka AL-21F-3 afterburningturbojet

No. Of engines 1

Dry thrust 76.4 kN (17,185 lbf)

Thrust with afterburner 109.8 kN (24,675 lbf)


ACCOMMODATION

Crew 1

WEIGHT RATIOS


Gross Weight


Maximum Take Off Weight

Armaments Up to 4000 kg (8,820 lb)

FUSELAGE

Length 19.02 m (62 ft 5 in)


WING


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






Area

Spread: 38.5 m² (415 ft²)

Swept: 34.5 m² (370 ft²)

Span

Spread: 13.68 m (44 ft 11 in)

Swept: 10.02 m (32 ft 10 in)

PERFORMANCE

Maximum Speed

Sea level: 1,400 km/h (755 knots, 870

mph)

Altitude: 1,860 km/h (1,005 knots,

1,155 mph, Mach 1.7)

Endurance -


Ferry Range 2,300 km (1,240 nmi, 1,430 mi)

Combat Range1,150 km (620 nm, 715 mi) in hi-lo-hi attack

with 2,000 kg (4,410 lb) war load



Thrust/weight 0.68

G-force limit 7

Airframe lifespan 2,000 flying hours, 20 years









Sukhoi Su-7


First flight -

Introduced 1955

ENGINE

Model Lyulka AL-7

No. Of engines 1

Dry thrust 66.6 kN (14,980 lbf)



ACCOMMODATION

Crew 1

WEIGHT RATIOS

Empty Weight 8937 kg (lb)

Gross Weight

Loaded Weight 13,570 kg (29,915)


Armaments 2,500 kg (4,410 lb)

FUSELAGE



WING

http://en.wikipedia.org/wiki/Pound_%28mass%29

http://en.wikipedia.org/wiki/Foot_%28length%29

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






Area (m ) 34 m² (366 ft²)

Span 9.31 m (30 ft 7 in)

PERFORMANCE

Maximum Speed

1,150 km/h (620 kn, 715 mph, Mach

0.94) at sea level

2,150 km/h (1,160 kn, 1,335 mph) at

high altitude

Endurance


Range 1,650 km (890 nmi, 1,025 mi)


Wing loading 434.8 kg/m² (89.05 lb/ft²)

Thrust/weight 0.71

Takeoff roll 950 m (3,120 ft)

Landing roll 700 m (2,300 ft)

http://en.wikipedia.org/wiki/Knot_%28speed%29

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

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

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

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











http://en.wikipedia.org/wiki/Knot_%28speed%29



Sukhoi Su-2


First flight -

Introduced -

ENGINE

Model Lyulka AL-21F-3 afterburningturbofan

No. Of engines 2


Thrust with afterburner 89 kN (20,000 lbf) each

Fuel capacity 4,500 kg (9,900 lb) internal

ACCOMMODATION

Crew 1

WEIGHT RATIOS


Gross Weight -



Armaments 5,700 Kg (12450 lb)

FUSELAGE

Length 15.96 m (52.4 ft)

Height 5.28 m (17.3 ft)

WING








Area 51.2 m (551 sq ft)

Span 10.95 m (35.9 ft)

PERFORMANCE

Maximum Speed

At altitude: Mach 2

(2,495 km/h/1,550 mph)

At sea level: Mach 1.2

(1,470 km/h/910 mph)

Endurance -


Range 2,900 km (1,800 mi)


Wing loading 312 kg/m (64.0 lb/ft )

Thrust/weight 1.15









Sukhoi PAK FA


First flight -

Introduced -

ENGINE

Model

New unnamed engine by NPO Saturn and

FNPTS MMPP

No. Of engines 2

Dry thrust AL-41F1 of 147 kN

Thrust with afterburner 157 kN


ACCOMMODATION

Crew 1

WEIGHT RATIOS


Gross Weight -



Armaments

FUSELAGE

Length 19.8 m (65.9 ft)

Height 6.05 m (19.8 ft)

http://en.wikipedia.org/wiki/Saturn_AL-41

http://en.wikipedia.org/wiki/Saturn_AL-41



WING

Area 78.8 m (848.1 ft )

Span 14 m (46.6 ft)

PERFORMANCE

Maximum Speed2,100 – 2,500 km/h (Mach 2+) (1,300 –

1,560 mph) ; at 17,000 m (45,000 ft) altitude

Endurance -


Range -

Rate of climb 350 m/sec (68,900 ft/min)

Wing loading

330(normal) - 470(maximum) kg/m

(67(normal) - 96(maximum) lb/ft2)

Thrust/weight 1.19

http://en.wikipedia.org/wiki/Sukhoi_PAK_FA#cite_note-warfare-63











Rafale B

Manufacturer Dassault Aviation

First flight 4 July 1986

Introduced 4 December 2000

ENGINE

Model Snecma M88-2 turbofans

No. Of engines 2

Dry thrust 50.04 kN (11,250 lbf) each


Fuel capacity -

ACCOMMODATION

Crew 2

WEIGHT RATIOS


Gross Weight



Armaments 13,350 Kg (29,370 lb)

FUSELAGE

Length 15.27 m (50.1 ft)

Height 5.34 m (17.5 ft)

WING



Area 45.7 m² (492 ft²)

Span 10.80 m (35.4 ft)

Airfoil -

PERFORMANCE

Maximum Speed

High altitude: Mach 2 (2,390 km/h,

1,290 knots)

Low altitude: 1,390 km/h, 750 knots

Endurance


Combat Radius 1,852+ km

Range 3,700+ km (2,000+ nmi)

Ferry Range -

Rate of climb 304.8+ m/s (60,000+ ft/min)


Thrust/weight 1.10









Mirage G8-02

Manufacturer Dassault Aviation

First flight 18 November, 1967

Introduced 1960

ENGINE

Model SNECMA Atar 9K50turbojets

No. Of engines 2


Thrust with afterburner -

Fuel capacity -

ACCOMMODATION

Crew 1

WEIGHT RATIOS


Gross Weight -

Loaded Weight -

Maximum Take Off Weight 19,340 Kg (42,548)


FUSELAGE



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



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



WING

Area -

Span

Extended: 15.40 m (50 ft 6 in)

Swept: 8.70 m (28 ft 7 in)

PERFORMANCE

Maximum Speed 2.2 Mach

Endurance -


Range 3,850 km (2,080 nm, 2,390 mi)



MiG-31


First flight -

Introduced -

ENGINE

Model Soloviev D-30F6 afterburning turbofans

No. Of engines 2


Thrust with afterburner 152 kN (34,172 lbf) each

Fuel capacity -

ACCOMMODATION

Crew 2

WEIGHT RATIOS


Gross Weight -



Armaments 15,600 Kg (34,320lb)

FUSELAGE



WING

http://en.wikipedia.org/wiki/Soloviev_D-30



http://en.wikipedia.org/wiki/Soloviev_D-30



Area 61.6 m² (663 ft²)

Span 13.46 m (44 ft 2 in)

PERFORMANCE

Maximum Speed

High altitude: Mach 2.83 (3,000 km/h,

1,860 mph)

Low altitude: Mach 1.2 (1,500 km/h,

930 mph)

Endurance -


Range



Thrust/weight 0.85









MiG-15bis


First flight -

Introduced -

ENGINE

Model Klimov VK-1turbojet

No. Of engines 1

Dry thrust -


Fuel capacity 1,400 L (364 US gal)

ACCOMMODATION

Crew 1

WEIGHT RATIOS


Gross Weight -




FUSELAGE



WING

http://en.wikipedia.org/wiki/Klimov_VK-1


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

http://en.wikipedia.org/wiki/Pound_(mass)

http://en.wikipedia.org/wiki/Pound_(mass)

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






Area 20.6 m² (221.74 ft²)

Span 10.08 m (33 ft 1 in)

PERFORMANCE

Maximum Speed 1,075 km/h (668 mph)

Endurance -


Range

1,200 km, 1,975 km with external tanks (745

mi / 1,225 mi)


Wing loading 240.8 kg/m² (49.3 lb/ft²)

Thrust/weight 0.54









Lockheed Martin F -22 Raptor

Manufacturer Lockheed Martin

First flight 2009

Introduced 2004

ENGINE

Model Pratt & Whitney F119-PW-100

No. Of engines 2

Dry thrust 23,500 lb (104 kN) each

Thrust with afterburner 35,000+ lb (156+ kN) each

Fuel capacity

18,000 lb (8,200 kg) internally, or 26,000 lb

(11,900 kg) with two external fuel tanks

ACCOMMODATION

Crew 1

WEIGHT RATIOS


Gross Weight



Armaments -

FUSELAGE

Length 62 ft 1 in (18.90 m)

Height 16 ft 8 in (5.08 m)



WING

Area 840 ft² (78.04 m²)

Span 44 ft 6 in (13.56 m)

Airfoil NACA 64A05.92 root, NACA 64A?04.29 tip

PERFORMANCE

Maximum Speed

At altitude:Mach 2.25 (1,500 mph, 2,410

km/h)

Super cruise: Mach 1.82 (1,220 mph, 1,963

km/h)

Endurance -

Service Ceiling 65,000 ft (19,812 m)

Combat Radius 410 nmi (471 mi, 759 km)

Range

1,600 nmi (1,840 mi, 2,960 km) with 2

external fuel tanks

Ferry Range 2,000 mi (1,738 nmi, 3,219 km)

Rate of climb -

Wing loading 77 lb/ft² (375 kg/m²)

Thrust/weight 1.08 (1.26 with loaded weight & 50% fuel)

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



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







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





Jaguar A

Manufacturer SEPECAT (Breguet/BAC)

First flight 8sep1968

Introduced 1973

ENGINE

Model Rolls-Royce/Turbomeca Adour Mk

102turbofans

No. Of engines 2



Fuel capacity -

ACCOMMODATION

Crew 1

WEIGHT RATIOS


Gross Weight -




FUSELAGE

Length 16.83 m (55 ft 2½ in)

Height 4.89 m (16 ft 0½ in)

WING

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

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

http://en.wikipedia.org/wiki/Rolls-Royce_Turbomeca_Adour






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

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



Area 24.2 m² (220 ft²)

Span 8.69 m (28 ft 6 in)

Aspect Ratio 3.12:1

PERFORMANCE

Maximum Speed

Mach 1.6 (1,699 km/h, 917 knots, 1,056 mph)

at 11,000 m (36,000 ft)

Endurance -


Combat Radius

908 km (490 nmi, 564 mi) (lo-lo-lo, externalfuel)

Range -

Ferry Range 3,524 km (1,902 nmi, 2,190 mi)

Rate of climb -

Wing loading -

Thrust/weight -











F-111 Aardvark

Manufacturer General Dynamics

First flight 21 December, 1964

Introduced 18 July, 1967

ENGINE

Model Pratt & Whitney TF30-P-100turbofans

No. Of engines 2

Dry thrust 17,900 lbf (79.6 kN) each

Thrust with afterburner 25,100 lbf (112 kN) each

Fuel capacity -

ACCOMMODATION

Crew 2 (pilot and weapons system operator)

WEIGHT RATIOS


Gross Weight -



Armaments 13,050 lb (5919.38 Kg)

FUSELAGE

Length 73 ft 6 in (22.4 m)

Height 17.13 ft (5.22 m)

WING

http://en.wikipedia.org/wiki/Pratt_%26_Whitney_TF30













Area

Spread: 657.4 ft² (61.07 m²)

Swept: 525 ft² (48.77 m²)

Span

Spread: 63 ft (19.2 m)

Swept: 32 ft (9.75 m)

Aspect Ratio spread: 7.56, swept: 1.95

Drag Area 9.36 ft² (0.87 m²)

Zero-Lift Drag Coefficient 0.0186

Airfoil NACA 64-210.68 root, NACA 64-209.80 tip

PERFORMANCE

Maximum Speed Mach 2.5 (1,650 mph, 2,655 km/h)

Endurance

Combat radius 1,330 mi (1,160 nmi, 2,140 km)

Service Ceiling 66,000 ft (20,100 m)

Ferry Range 4,200 mi (3,700 nmi, 6,760 km)

Rate of climb 25,890 ft/min (131.5 m/s)

Wing loading

Spread: 126.0 lb/ft² (615.2 kg/m²)

Swept: 158 lb/ft² (771 kg/m²)

Thrust/weight 0.61

Lift-to-drag ratio 15.8



















MiG-35


First flight -

Introduced -

ENGINE

Model Klimov RD-33MKafterburningturbofans

No. Of engines 2

Dry thrust 5,400 kgf, 53.0 kN (11,900 lbf) each

Thrust with afterburner 9,000 kgf, 88.3 kN (19,800 lbf) each

Fuel capacity -

ACCOMMODATION

Crew 1 or 2

WEIGHT RATIOS


Gross Weight



Armaments 8,300 Kg (18,260lb)

FUSELAGE



WING

http://en.wikipedia.org/wiki/Klimov_RD-33


http://en.wikipedia.org/wiki/Kilogram-force



http://en.wikipedia.org/wiki/Kilogram-force






Area 38 m (124 ft )

Span 12 m (39 ft 4 in)

PERFORMANCE

Maximum Speed

Mach 2.25 (2,400 km/h, 1,491 mph) at

altitude;[20]

1,450 km/h (901 mph) at low-level

Endurance -


Range 2,000 km (1,240 mi)


Wing loading -

Thrust/weight 1.14


http://en.wikipedia.org/wiki/Mikoyan_MiG-35#cite_note-af_tech-19



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


http://en.wikipedia.org/wiki/M/s





http://en.wikipedia.org/wiki/M/s


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







Conclusion:

Thus the Fighter Aircraft, Class-1 is studied for it performance. The aerofoil has been

selected and the wing has been drawn. The various weight ratios are determined. The centre of

gravity the wing and the fuselage are determined. The center of gravity for the aircraft with the

engine mounted are also determined. And the various aircraft reference data also placed here.

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My Aircraft Design Project - I