Istanbul Technical University Faculty of Aeronautics and Astronautics

Aeronautical Engineering Department

Principles of Aircraft Design UCK451E

Project Report

Conceptual Design of a Two Seat General Aviation

Aircraft

Instructor             :  Prof. Mehmet Şerif Kavsaoğlu

Course Assistant :  Uğur Özdemir

Date                     :  19/01/2009

Prepared by       : GROUP 18

Merve MELEK 110040027 mrv_melek@yahoo.com Seyit Türkmen KOÇ 110040131 trkmnko@yahoo.com

ABSTRACT

General aviation covers a huge range of activities, both commercial and non-

commercial, including private, flight training, police aircraft, air-ambulance, air charter and

etc. General aviation aircrafts generally use piston propeller engine especially single piston

propeller engine. This more traditional class of aircrafts includes nearly all aircraft from

Wright Brothers up through World War II. Today, piston engines are used almost exclusively

on light, general aviation aircraft.

In this project conceptual design of a two seat, piston propeller general aviation

aircraft will be performed. It is expected that aircraft will fly around 130 knots at 8000ft and

to a range of 500 nautical miles. Design study will be performed both by taking design

account in and comparing to competitor data, and with considering historical values. Design

method is mainly based on textbook on Raymer [1].

I

TABLE OF CONTENTS

Title Page Number

CHAPTER 1 INTRODUCTION 1

1.1. Purpose 1

1.2. Requirements And Mission Profile 1

1.3. Description of The Report 2

CHAPTER 2 COMPETITOR STUDY 3

2.1. Introduction 3

2.2. Competitor Study 3

2.3. Summary Of Results 10

2.4. Conclusion 11

CHAPTER 3 FIRST GUESS SIZING 12

3.1. Introduction 12

3.2. Design Requirements&Mission Profile 12

3.3. Conceptual Sketching&Initial Sizing 13

3.4. Calculations 14

3.5. Trade Studies 18

3.6. Summary of Results 26

3.7. Conclusion 27

CHAPTER 4 AIRFOIL and GEOMETRY SELECTION 28

4.1. Introduction 28

4.2. Wing Airfoil Selection 28

4.3. Wing Geometry 31

4.4.Selection of Tail Airfoil 35

4.5. Tail Geometry 38

4.6. Non Dimensional Drawings 40

4.7. Summary Of Results 41

4.8. Conclusion 42

CHAPTER 5 THRUST TO WEIGHT RATIO and WING LOADING 43

5.1. Introduction 43

5.2. Selection of Horsepower to Weight Ratio 43

5.3. Selection of Wing Loading 48

5.4. Analysis and Calculations 55

5.5. Summary Results 67

II

5.6. Conclusion 68

CHAPTER 6 INITIAL SIZING 69

6.1. Introduction 69

6.2. Rubber Engine Sizing 70

6.3. Engine Selection 75

6.4. Fixed Engine Sizing 76

6.5. Geometry Sizing 80

6.6. Summary Results 85

6.7. Conclusion 86

CHAPTER 7 CONFIGURATION LAYOUT and INTERIOR DESIGN 87

7.1. Introduction 87

7.2. Wing and Tail Surfaces 87

7.3. Fuselage and Interior Design 95

7.4. Fuel Tanks Location 98

7.5. Drawings of Top, Front and Side views of the Aircraft 99

7.6. Summary of Results 100

7.7. Conclusion 102

CHAPTER 8 PROPULSION and FUEL SYSTEM INTEGRATION 103

8.1. Introduction 103

8.2. Propulsion Selection 103

8.3. Selected Engine Properties 104

8.4. Propeller-Engine Integration 105

8.5. Fuel System 109

8.6. Summary of Results 112

8.7. Conclusion 113

CHAPTER 9 LANDING GEAR and SUBSYSTEMS 114

9.1. Introduction 114

9.2. Landing Gear Arrangement 114

9.3. Tire Sizing 115

9.4. Shock Absorber 121

9.5. Castoring Wheel Geometry 125

9.6. Gear Retraction Geometry 126

9.7. Subsystems 127

9.8 Summary of Results 129

9.9 Conclusion 131

CHAPTER 10 AERODYNAMICS 132

III

10.1. Introduction 132

10.2. Estimation of Lift 132

10.3. Estimation of Parasite Drag 137

10.4. Calculations 138

10.5. Estimation of Drag Due Lift 142

10.5. Plots 143

10.6. Summary of Results 147

10.7. Conclusion 147

CHAPTER 11 DECRIPTION OF JANE FORMAT 148

CHAPTER 12 CONCLUSION 149

REFERENCES 150

APPENDIX A HISTOGRAMS FROM COMEPTITOR DATA 151

APPENDIX B AIRFOIL DATA 155

1

CHAPTER 1

INTRODUCTION Single piston propeller nose mounted aircrafts are more traditional which are generally

preferred for general aviation or training. These types of aircrafts configuration are important

because they are cheap and have lower fuel consumption according to turboprop for the same

hp. For many years, peoples have flied with general aviation aircrafts, so there are many

different choices for this type if someone wants to buy. By considering this situation, it is

important to design aircraft which has lower cost, lower fuel consumption and better

performance. Thus, it can be found a place in the market.

In this project, conceptual design of a two seat, single piston propeller engine aircraft

will be performed in order to provide the requirements and mission profile below.

1.1. PURPOSE

The project aircraft will be designed for training about 500nm.

1.2. REQUIREMENTS AND MISSION PROFILE

For a two seat, training aircraft with following objectives are given below. The

following requirements are fixed after the Initial Sizing (Chapter 6);

Range : 500 nm = 3040000ft

Vcruise : 130 knots = 219, 44 ft/s

hcruise : 2440m = 8000ft

Wcrew : 100kg = 220, 26lb

Wpayload : 100kg = 220, 26lb

Mission Profile:

Figure 1.1 Mission profile

2

1.3. DESCRIPTION OF THE REPORT

The design is performed mostly by using the textbook, AIRCRAFT DESIGN: A

CONCEPTUAL APPROACH, Raymer D.P, AIAA Education Series, Washington, 1992.

Books that are used stated in REFERENCES part of the project at the end. The detailed

drawings of the design aircraft with all necessary dimensions are given in Appendices and

should be referred in any calculation.

3

CHAPTER 2

COMPETITOR STUDY

2.1 INTRODUCTION

The purpose of this chapter is to show how the competitor study must be executed

according to the required aircraft design. To design an aircraft we need some specifications of

the similar aircrafts. Therefore; first, these specifications are searched and gathered. Then, we

analysis these aircraft’s characteristics in order to design own aircraft.

2.2. COMPETITOR STUDY

After we find the characteristics of the aircrafts, we make a table to compete them. This

table includes the competitor study for single piston propeller engine, one or two seat and

general aviation or trainer aircrafts. The information about existing aircrafts that match our

category is tabulated below:

4

2.2.1. COMPETITOR AIRCRAFT SPECIFICATIONS

1 2 3

AIRCRAFT Cessna-140 Cessna-152 Piper PA-28-180

Cherokee COUNTRY USA USA USA - Florida ACCOMMODATION(CREW) 2 2 4 PAYLOAD WEIGHT,Wp(kg) 255 242,71 430 EMPTY WEIGHT,We(kg) 403 504.394 558 FUEL WEIGHT,Wf(kg) 83,34 88,75 189 MAX. TAKE OFF WEIGHT,Wo(kg) 658 757 1090,9

POWERPLANT 1× Continental C85-12

1*Lycoming O-235 K2C 1× Lycoming O-320-E2A

POWER(h.p.) 85 110 180 WING SPAN(m) 10,25 10,068 9 WING AREA(m2) 15,1 14,864 14,86 WING ASPECT RATIO 6,96 6,82 5,59 WETTED ASPECT RATIO 1,78 1,75 1,43 WING TAPER RATIO 1 1 0,9248 SWEEP ANGLE(I.e.) 0 0 0 SWEEP ANGLE(c/4) 0 0 1,48 WING THICKNESS RATIO(root) 12% 12% 15% WING THICKNESS RATIO(tip) 12% 12% 15% AIRFOIL(root) NACA 2412 NACA 2412 NACA 652-415 AIRFOIL(tip) NACA 2412 NACA 0012 NACA 652-415 FLAPS Fowler Fowler slotted (10,25,40 degree) CRUISING SPEED(knots) 94 100-110 124 STALLING SPEED,FLAPS UP(knots) 39 48 50 STALLING SPEED,FLAPS DOWN(knots) 46 43 47 MAX. WING LOADING,W/S(lb/ft2) 8,68 10 15 MAX.POWER LOADING(h.p. /lb) 0,058 0,0625 0,0752 SERVICE CEILING(m) 4700 4267.13 4998,72 TAKE OFF DISTANCE(m) 154 224.024 219,456 LANDING DISTANCE(m) 70 135,634 182,88 RANGE(nautical mile) 395 690 510 MAX. RATE OF CLIMB(ft/min) 680 715 750

Table 2.1 Competitor Aircrafts’ Specifications

5

4 5 6

AIRCRAFT Aeronca 11AC

Chief Boeing/Stearman

(PT_13) Pitts S-2 C COUNTRY USA USA USA ACCOMMODATION(CREW) 2 2 2 PAYLOAD WEIGHT,Wp(kg) 238,1 200 213,3 EMPTY WEIGHT,We(kg) 354,12 878 520 FUEL WEIGHT,Wf(kg) 56,9 174,25 106,25 MAX. TAKE OFF WEIGHT,Wo(kg) 567 1232 771

POWERPLANT Continental A65-8 Lycoming R-680-5 Lycoming AEIO-

540 POWER(h.p.) 65 220 260 WING SPAN(m) 11 9,8 6,1 WING AREA(m2) 16,3 27,6 11,8 WING ASPECT RATIO 7,42 3,5 5,98 WETTED ASPECT RATIO 1,9 0,89 1,53 WING TAPER RATIO 1 1 1 SWEEP ANGLE(I.e.) 0 0 7 SWEEP ANGLE(c/4) 0 0 7 WING THICKNESS RATIO(root) 12% 13% 15% WING THICKNESS RATIO(tip) 12% 13% 12% AIRFOIL(root) NACA 4412 NACA 2213 NACA 63A015 AIRFOIL(tip) NACA 4412 NACA 2213 NACA0012 FLAPS Cowl No flap double slotted CRUISING SPEED(knots) 72 106 150,33 STALLING SPEED,FLAPS UP(knots) 30,41 48 56 STALLING SPEED,FLAPS DOWN(knots) 33 44 54 MAX. WING LOADING,W/S(lb/ft2) 7,1 9,15 13,3 MAX.POWER LOADING(h.p. /lb) 0,052 0,081 0,152 SERVICE CEILING(m) 3291,9 3415 6400 TAKE OFF DISTANCE(m) 177,7 182,88 169 LANDING DISTANCE(m) 268,224 91,44 - 152,4 229 RANGE(nautical mile) 180 260 300 MAX. RATE OF CLIMB(ft/min) 360 505 2900

Table 2.1 Competitor Aircrafts’ Specifications (continued)

6

7 8 9

AIRCRAFT Yakovlev Yak-52 Zenith CH601 XL Diamond DA 20

Katana COUNTRY Russia Canada Canada ACCOMMODATION(CREW) 2 2 2 PAYLOAD WEIGHT,Wp(kg) 290 282 271 EMPTY WEIGHT,We(kg) 1015 318 529 FUEL WEIGHT,Wf(kg) 125 81,25 83,7 MAX. TAKE OFF WEIGHT,Wo(kg) 1305 595 800

POWERPLANT 1× Vedeneyev M-

14P 1× Jabiru 3300 flat-6

engine 1× Continental IO-

240-B POWER(h.p.) 360 110 125 WING SPAN(m) 9,3 8,23 10,87 WING AREA(m2) 15 12,3 11,61 WING ASPECT RATIO 5,8 5,6 10,18 WETTED ASPECT RATIO 1,49 1,44 2,61 WING TAPER RATIO 0,542 0,875 0,695 SWEEP ANGLE(I.e.) 4,8 0 1 SWEEP ANGLE(c/4) 0 0 0,5 WING THICKNESS RATIO(root) 14,50% 15% WING THICKNESS RATIO(tip) 9,30% 15% AIRFOIL(root) Clark YH Riblett GA-35-A-415 Wortmann FX 63-137 AIRFOIL(tip) Clark YH Riblett GA-35-A-415 Wortmann FX 63-137 FLAPS Split Plain Slotted CRUISING SPEED(knots) 128 120 138 STALLING SPEED,FLAPS UP(knots) 60 44,32 42 STALLING SPEED,FLAPS DOWN(knots) 54-57 or 46-49 38,23 34 MAX. WING LOADING,W/S(lb/ft2) 17,61 9,85 13,2 MAX.POWER LOADING(h.p. /lb) 0,127 0,085 0,0756 SERVICE CEILING(m) 4000 4875 min 4000 TAKE OFF DISTANCE(m) 170 152 390 LANDING DISTANCE(m) 300 152 201,5 RANGE(nautical mile) 290 575 547 MAX. RATE OF CLIMB(ft/min) 1400 1200 1000

Table 2.1 Competitor Aircrafts’ Specifications (continued)

7

10 11 12

AIRCRAFT Aviat Husky A-1B-

200 AMD Alarus

CH2000 De Havilland Chipmunk COUNTRY United states USA-Georgia Canada ACCOMMODATION(CREW) 2 2 2 PAYLOAD WEIGHT,Wp(kg) 172 275,58 310 EMPTY WEIGHT,We(kg) 598 493 533 FUEL WEIGHT,Wf (lt) 137 108 68,2 MAX. TAKE OFF WEIGHT,Wo(kg) 907 769 998

POWERPLANT Lycoming IO-360-

A1D6 Lycoming 0-235-

N2C 1× de Havilland Gipsy

Major POWER(h.p.) 350 116 145 WING SPAN(m) 10,82 8,73 10,47 WING AREA(m2) 17 12,73 16 WING ASPECT RATIO 6,89 5,987 6,85 WETTED ASPECT RSTIO 1,77 1,54 1,76 WING TAPER RATIO 1 0,89 0.53 SWEEP ANGLE(I.e.) 0 1 1 SWEEP ANGLE(c/4) 0 0 0 WING THICKNESS RATIO(root) 11, 72% 18% 15% WING THICKNESS RATIO(tip) 11, 72% 18% 11, 61% AIRFOIL(root) Modified Clark Y NACA 640-18 NACA 1415 AIRFOIL(tip) Modified Clark Y NACA 640-18 USA 35B

FLAPS conventional &

manual Split Slotted (beaver ın) CRUISING SPEED(knots) 138 99 113 STALLING SPEED,FLAPS UP(knots) 53 48 39 STALLING SPEED,FLAPS DOWN(knots) 47 30 35 MAX. WING LOADING,W/S(lb/ft2) 10,9 12,3 10,3 MAX.POWER LOADING(h.p. /lb) 0,1 0,0685 0,072 SERVICE CEILING(m) 6096 5200 TAKE OFF DISTANCE(m) 80 499,872 160 LANDING DISTANCE(m) 121 554,736 RANGE(nautical mile) 828 470 280 MAX. RATE OF CLIMB(ft/min) 1700 750 900

Table 2.1 Competitor Aircrafts’ Specifications (continued)

8

13 14 15

AIRCRAFT Ikarus C 42 Piper PA-38 Tomahawk Sukhoi Su 31

COUNTRY Germany USA Russia ACCOMMODATION(CREW) 2 2 1 PAYLOAD WEIGHT,Wp(kg) 183 200 100 EMPTY WEIGHT,We(kg) 265 512 750 FUEL WEIGHT,Wf(kg) 50 102,5 288,8 MAX. TAKE OFF WEIGHT,Wo(kg) 473 757 1100

POWERPLANT 1× Rotax 912s 1× Avco Lycoming O-

235 1* VOKBM M-14PF

POWER(h.p.) 100 112 360 WING SPAN(m) 9,45 10,36 7,8 WING AREA(m2) 12,5 11,6 11,8 WING ASPECT RATIO 7,14 9,25 5,16 WETTED ASPECT RATIO 1,83 2,37 1,32 WING TAPER RATIO 1 1 0,46 SWEEP ANGLE(I.e.) 0 0 5,2 SWEEP ANGLE(c/4) 0 0 0 WING THICKNESS RATIO(root) 12% 18% WING THICKNESS RATIO(tip) 12% 18% AIRFOIL(root) NACA 2412 NASA GA(W)-1 Symmetrical AIRFOIL(tip) NACA 2412 NASA GA(W)-1 Symmetrical FLAPS Plain Plain No flap CRUISING SPEED(knots) 105 108 113 STALLING SPEED,FLAPS UP(knots) 42 48 66 STALLING SPEED,FLAPS DOWN(knots) 32 46 MAX. WING LOADING,W/S(lb/ft2) 7,6 13,39 1,55 MAX.POWER LOADING(h.p. /lb) 0,1 0,067 0,134 SERVICE CEILING(m) 3658 4000 4000 TAKE OFF DISTANCE(m) 80 250 110 LANDING DISTANCE(m) 150 215,5 300 RANGE(nautical mile) 450 384 156,6 MAX. RATE OF CLIMB(ft/min) 1050 718 3543

Table 2.1 Competitor Aircrafts’ Specifications (continued)

9

When we make a competitor table, we need lots of information about the aircrafts

characteristics. Some of these characteristics were found Jane’s All of the World and the other

sites. Besides, the wing flap types are searched. It can be seen in the table that the competitor

aircrafts has different types of flap such as plain or simple, fowler, slotted and split flap.

According to this information, we formed the table which is shown above. Some

characteristics of aircrafts such as taper ratio, sweep angle were measured from the aircrafts

pictures. Also, while we were calculating the wetted aspect ratio, we used the Sref/ Swet = 3,

9. This ratio was taken from a graph for Sref / Swet in the Raymer. These all of information

will be used in the next part of the study to compare the value of the desired aircraft.

2.2.2 GRAPHS FOR COMPETITOR OF AIRCRAFTS

It is made a table for 15 trainer and general aviation aircrafts specifications which can

be found. According to this table, some specifications for competitor study are plotted below:

Graph 2.1 We/Wo – Wo (kg)

According to this graph, empty weight-takeoff weight ratio changes between the 500-

1300kg. The intensity is seen approximately 700-800kg.

10

Graph 2.2 Wf/Wo – Wo (kg)

In this graph, fuel weight-takeoff weight ratio changes nearly between 400-1400kg.

However, the most intensity region can be said the approximately 700-800kg.

Histograms of competitor aircrafts are replaced in Appendix A.

2.3. SUMMARY OF RESULTS

The results of this competitor study are listed as a table below:

Range 200-600nm Vcruise 100-140knot Wo 750-1000kg Wp 200-300kg Take off distance 100-200m Landing distance 100-200 W/S 10-15 lb/sq ft h.p./W 0,05-0,075

According to this table we can choose the characteristics of aircraft which we will

design. Our aircraft will be single engine piston propeller, two seat, general aviation or trainer

11

aircraft. We use this competitor study to create our won aircraft. Therefore, the competitor

study and the plotted data of 15 aircrafts are very important for us during the designation.

2.4. CONCLUSION During competitor study, we have been interested in 15 aircrafts that matches our

design category. Then, we analyzed these aircrafts and we did tables and graphs to see and

compete better these characteristics of the aircrafts. The results of plotted diagrams give us

much useful introductory information about our design process.

12

CHAPTER 3

FIRST GUESS SIZING

3.1. INTRODUCTION

First guess sizing is an important stage to create an aircraft. In this chapter, the aim is

doing an introduction to the design process. That is to say, this study offers quick method of

estimating takeoff weight, empty weight and fuel weight from a conceptual sketch. In this

direction; first, a conceptual sketch is constituted. Then, with the design requirements which

are given below, it is started to make calculations to estimate takeoff weight.

3.2. DESIGN REQUIREMENTS & MISSION PROFILE For a two seat, training aircraft with following objectives are given below:

Range : 500 nm = 3040000ft

Vcruise : 130 knots = 219, 44 ft/s

hcruise : 2440m = 8000ft

Wcrew : 100kg = 220, 26lb

Wpayload : 100kg = 220, 26lb

Figure 3.1 Mission Segments 0–1: Takeoff 1–2: Climb 2–3: 500 nm Cruise at 2440m at Vcruise=130 knots

2 3

5

6 1

13

3–4: Descent 4–5: 30 min. Loiter 5–6: Landing

3.3. CONCEPTUAL SKETCHING & INITIAL SIZING

14

According to aircrafts which are competed, the aspect ratio is selected as 7.

AR=7 The wetted area ratios, Swet / Sref is chosen 3, 8 from Figure 3.5 [1]. Then, the wetted

aspect ratio is calculated according to this formula: ARwetted = /AR

Swet Sref

After that, from Figure 3.6[1], max. L/D ratio can be found: (L/D) max = 12, 5

For cruise (L/D) = (L/D) max and for loiter (L/D) = 0,866 (L/D) max.

3.4. CALCULATIONS

3.4.1. Takeoff Weight Calculation

Take of weight can be estimated by using formula that given below:

Wo Wcrew Wpayload Wfuel Wempty= + + + (3.1) [1]

Wcrew and Wpayload are given at the mission requirements. According to it;

• Wcrew = 100kg = 220,26lb

• Wpayload = 100kg = 220,26lb

Then, Wfuel and Wempty are written as a function of Wo. Thus; the formula becomes:

(3.2) [1]

WeWo

: Empty Weight Fraction

WfWo

: Fuel Weight Fraction

3.4.2. Empty Weight Estimation

The empty weight fraction (We/W0) can be estimated from the historical data as shown

in Table 3.1[1]. The empty weight fraction (We/W0) is:

00

cevs

W AW KW

= (3.3) [1]

)()(100

0

WW

WW

WWW

ef

payloadcrew

−−

+=

15

According to the Table 3.1., for fixed wing and single engine, the coefficients A, C

and Kvs are determined as the below:

A = 2, 36

C = -0, 18 0.180 0

0

2,05cevs

W AW K WW

−= =

Kvs = 1 3.2.4.3. Fuel Fraction Estimation

3.4.3.1. Mission Segment Weight Fractions

The ratios of W1/W0, W2/W1, W4/W3, W6 /W5 can be estimated from the table of

historical mission segment weight fractions.

Take off and warm up W1/W0 0,97 Climb W2/W1 0,985 Descent W4/W3 1 Landing W6/W5 0,995

Table 3.1 Historical mission segment weight fractions [1]

To determine the W3/W2 (for cruise) and W5/W4 (for loiter), it should be some

calculations. These are shown below.

1-

RVC

LD

nWW

i

i=

−

1

CRUISE

For estimate the cruise segment weight fractions W3/W2, the Brequet’s range equation

[1] is used:

WW

RCV L D

3

2

=−

exp

( / ) (3.4) [1]

R Range (feet) C Specific Fuel Consumption(1/s) V Velocity(ft/s) L/D Lift to drag ratio

Table 3.2

To find the value of C, it is used the formula [1] is shown below:

C = C V550

bhp

pη (3.5) [1]

16

Propeller specific fuel consumption Cbhp = 0, 4 lb/hr/bhp and propeller efficiency p =

0.8 are taken from Table 3.4[1] for cruise conditions. Also, the cruise speed Vcruise = 130 knot =

219.44ft /s is given in mission profile. According to these values;

C = 0,000110828 1/s

For the aspect ratio which is selected for two seat, trainer aircraft, L/D is determined

as 12, 5 from Figure 3.6 [1].

During cruise conditions,

(L/D)max = L/D = 12,5

By using the equation (3.4) [1], the cruise segment fraction is calculated as:

W3/W2 = 0,884415519

2-

EC

LD

nWW

i

i=

−1 1

LOITER

For estimate loiter segment weight fractions W5/W4, the endurance equation [1] must

be considered:

5

4

exp( / )

W ECW L D

−=

(3.6) [1]

E is the endurance time during loiter. This value is given in the mission profile as the

30 minute = 1800 s. Also, for loiter conditions, Cbhp = 0, 5 lb/hr/bhp and propeller efficiency p = 0.7 are

taken from Table 3.4 [1].

During loiter conditions,

(L/D)loiter =0,866(L/D)max = 10,825

It should be known Vloiter, to calculate specific fuel consumption.

loiter

cruisecruiseloiter VV

ρρ76.0= (3.7) [1]

- loiter = 1,048 kg/m3 (@ 5246ft = 1600m)

- cruise = 0,963 kg/m3 (@ 8000ft = 2440m) [2]

From these data;

Vloiter = 159,8681 ft/s

17

Then, specific fuel consumption, C is determined as the below:

C = 0,000115345 1/s

Therefore, by using the equation (2.6), the loiter weight fraction is calculated as:

W5/W4 = 0,981002997

After the calculating all of weight fractions, the total weigh fraction W6/W0 is required to find

(3.8) [1]

From this formula the total weight fraction;

W6 / W0 = (0.97) (0.985) (0, 888603528) (1) (0.981002997) (0.995) = 0.778693

W6/W0 = 0,824817

Then, the fuel weight fraction is calculated from the formula is given below.

)1(06.10

6

0 WW

WW f −= (3.9) [1]

According to this, fuel weight fraction: Wf / W0 = 0,185694

From take-off weight calculation equation which is given (2.2);

• Wcrew and Wpayload values are given at the design requirements as 100kg = 220, 26lb.

• Wf / W0 = 0,185694

• We / W0 = 0.1802,36W −

To find the value of W0, it must to do iteration. For doing this, the iterative equation is got as given: After ordering this equation, the formula becomes:

5

6

4

5

3

4

2

3

1

2

0

1

0

6

WW

WW

WW

WW

WW

WW

WW

=

)()(100

0

WW

WW

WWW

ef

payloadcrew

−−

+=

0 0.180

220,26 220,261 0.185694 2,36

WW −

+=

− −

0 0.180

440,520,814306 2,36

WW −=

−

18

0W = 540,976 + 2, 8982 0W 0, 82

Then; the iterative process was done to calculate W0. Thus; some solution was found. These

are given below:

W0 (Guess) (lb) W0 (Calculated) (lb) 2000 2016,59 2017 2026,6 2030 2036,3 2038 2040,67

2040,67 2041,881858 Table 32.3 Calculation of Wo

Wo = 2041,881858lb = 927kg

In conclusion, the results are estimated:

W0 (lb) 2041,88 We (lb) 1222,19 Wf (lb) 379,165

Table 3.4 All weights

3.5. TRADE STUDIES

3.5.1. Cruising Speed Trade Off 3.5.1.1. Trade Off for Vcruise = 120 knots = 202, 56 ft/s

For trade study, first cruise speed was changed as 120 knots = 202, 56 ft/s. Therefore,

some data must be calculated again for new cruise speed. Here are the variable specifications:

(For cruise)

3

2

WW

is the same.

0,963(0.76)(202,56) 147,57061,048loiter

ftVs

= =

(For loiter)

C = 0,000102303 1/s

C = 0,000106472 1/s

19

5

4

WW

= 0,982451404

New total weight fraction and fuel fraction:

and Wf / W0 = 0,184403

Then; using the new data, the iterative process was done again to calculate W0. The

solutions are found:

W0 (Guess) (lb) W0 (Calculated) (lb) 2000 2013,4 2015 2021,5 2022 2029,3 2030 2033,16

2033,16 2033,744675 Table 3.5

W0 (lb) 2033,74 We (lb) 1218,19 Wf (lb) 375,028

Table 3.6

Wo = 2033,744675lb = 923,32kg 3.5.1.2. Trade Off for Vcruise = 140 knots = 236, 32 ft/s

In this part, there are some change for Vcruise = 140 knots = 236.32 ft/s. Therefore, some

datas must be calculated again for new cruise speed. Here are the variable specifications:

(For cruise)

3

2

WW

is the same.

0,963(0.76)(236,32) 172,16571,048loiter

ftVs

= =

(For loiter)

5

4

WW

= 0,979556726

W6/W0 = 0,8326035

C = 0,000119354 1/s

C = 0,000124218 1/s

20

New total weight fraction and fuel fraction:

and Wf / W0 = 0,186983

Then; using the new data, the iterative process was done again to calculate W0. The

solutions are found:

W0 (Guess) (lb) W0 (Calculated) (lb) 2000 2019,7 2020 2031,7 2032 2038,9 2040 2045,9 2046 2049,73

2049,73 2049,921731 Table 3.7

Wo = 2049,921731lb = 930,66kg

W0 (kg) 2049,92 We (kg) 1226,13 Wf (kg) 383,3

Table 3.8

Here, there are the graphs of trade off for cruise speed:

Vcruise-We Graph

y = 0,2352x + 1170,6R2 = 1

1217

1218

1219

1220

1221

1222

1223

1224

1225

1226

1227

200 205 210 215 220 225 230 235 240

Vcruise(knot)

We(

kg)

Vcruise-We

Doğrusal (Vcruise-We)

W6/W0 = 0,823601

21

Figure 3.2 Cruise Velocity Trade wrt We

In this graph, it can be seen that the cruise speed changes with the empty weight

linearly. While cruise speed is increasing, the empty weight is increasing, too. However, just

only for this graph, the change is linearly. For another graph, it couldn’t be linear.

Vcruise-Wf Graph

y = 0,245x + 325,4

R 2 = 1

374

375

376

377

378

379

380

381

382

383

384

200 210 220 230 240

Vcruise(ft/s)

Wf(

kg)

Vcruis e-WfG raph

ğ l

Figure 3.3 Cruise Velocity Trade wrt Wf

In this graph, it can be seen that the cruise speed changes with the fuel weight linearly.

While cruise speed is increasing, the fuel weight is increasing, too like the empty weight.

Again, this linear change is only for this graph.

22

Vcruise-Wo Graph

y = 0,4792x + 1936,7R2 = 1

2032

2034

2036

2038

2040

2042

2044

2046

2048

2050

2052

200 210 220 230 240

Vcruise(ft/s)

Wo

(kg)

Vcruise_Wo GraphDoğrusal (Vcruise_Wo Graph)

Figure 3.4 Cruise Velocity Trade wrt Wo

In this graph, it can be seen that the cruise speed changes with the takeoff weight

linearly only for this graph. While cruise speed is increasing, the takeoff weight is increasing,

too like empty and fuel weight. The results are obtained for takeoff weight between the 900 –

1000kg as seen in the graph.

* Cruise speed trade of graphs shows that the change of weight (Wo, We, Wf) with cruise

speed is not seriously. Therefore, the selected value for cruise speed, 130 knots seems to be

suitable for the design.

3.5.1. Range Trade Off 3.5.1.1. Trade Off for Range = 400 nm = 2432000 ft

In this part, only cruise weight fraction changes. This changing affects the total weight

fraction and fuel weight fraction.

3

2

WW

= 0,906410825

Thus, new total weight fraction and fuel weight fraction are obtained as the below:

and Wf / W0 = 0,16395 W6/W0 = 0,84533

23

Then; using the new data, the iterative process was done again to calculate W0. The

solutions are found:

W0 (Guess) (lb) W0 (Calculated) (lb) 2000 1964,125 1960 1942,95 1940 1930,42 1918 1915,97 1914 1912,6

1912,6 1912,152341 Table 3.9

Wo = 1912,152341lb = 868,117kg

W0 (lb) 1912,15 We (lb) 1158,14 Wf (lb) 313,497

Table 3.10 3.5.1.1. Trade Off for Range = 600 nm = 3648000ft

Same calculations are done for the trade off range = 3648000ft. According to these

calculations, some data are changed:

3

2

WW

= 0,862953959

Thus, new total weight fraction and fuel weight fraction are obtained as the below:

and Wf / W0 = 0,20691

Then; using the new data, the iterative process was done again to calculate W0. The

solutions are found:

W0 (Guess) (lb) W0 (Calculated) (lb) 2000 2070,52 2071 2114,2 2115 2141,1 2142 2157,63 2167 2177,77 2178 2181,52 2182 2183,24

2183,24 2183,754761 Table 3.11

W6/W0 = 0,804802

24

Wo = 2183,754761lb = 991,425kg

W0 (lb) 2183,75 We (lb) 1291,4 Wf (lb) 451,841

Table 2.12 Here, there are the graphs of trade off for range:

Range-Wo Graph

y = 0,0002x + 1366,9R2 = 0,9993

1850

1900

1950

2000

2050

2100

2150

2200

0 500000 1E+06 2E+06 2E+06 3E+06 3E+06 4E+06 4E+06

Range (feet)

Wo

(kg)

Range-Wo GraphDoğrusal (Range-Wo Graph)

Figure 3.5 Range Trade wrt We

In this graph, it can be seen that the curve of empty weight variation with the range is

a linear variation. While range is increasing from 400 nm (2432000ft) to 600 nm (3648000ft),

the empty weight is increasing linearly just only for this study.

25

Range-Wf Graph

y = 0,0001x + 35,642R2 = 0,9991

0

50

100

150

200

250

300

350

400

450

500

0 500000 1000000

1500000

2000000

2500000

3000000

3500000

4000000

Range(feet)

Wf (

kg)

Range-Wf Graph

Doğrusal (Range-Wf Graph)

Figure 3.6 Range Trade wrt Wf

In this graph, it can be seen that the curve of fuel weight - range shows a linear

variation. While range is increasing, the fuel weight is increasing linearly for this study.

26

Range-We Graph

y = 0,0001x + 890,76R2 = 0,9995

1140

1160

1180

1200

1220

1240

1260

1280

1300

0 500000 1E+06 2E+06 2E+06 3E+06 3E+06 4E+06 4E+06

Range (feet)

We

(kg)

Range-We GraphDoğrusal (Range-We Graph)

Figure 3.7 Range Trade wrt Wo

In this graph, it can be seen that the curve of takeoff weight variation with the range is

a linear variation. While range is increasing, the takeoff weight is increasing linearly for this

study.

* On the contrary cruise speed; the results as to range vary much.

3.6. SUMMARY OF THE RESULTS

Briefly, it is aimed to summarize the studies in this chapter; first, it is chosen an aspect

ratio and L/D ratio for designation of a conceptual sketch. According to these data, some

weight fraction are calculated and done estimation of takeoff weight, empty weight and fuel

weight. The results of these weights are found nearly the specifications of the aircrafts which

were searched in competitor study. According to it, the takeoff weight is determined as 927kg

(approximately 2041lb).

The takeoff weight (W0), empty weight (We) and fuel weight (Wf) of the designed aircraft are

estimated from the calculations as follows:

W0 (lb) 2041,88 We (lb) 1222,19 Wf (lb) 379,165

27

Table 3.13

After the estimation of these weights, trade studies are done for cruise speed and

range. According to the trade study, the variations of We, Wo and Wf is calculated and plotted

in trade off parts.

3.7. CONCLUSION

First guess study has an important role to enter the design process. In order to start this

study, first, a conceptual sketch is formed according to selected some characteristics such as

aspect ratio. Then, n the direct of the given mission profile data, weight fractions for the

mission profile was calculated. Thus, the wanted specifications (takeoff weight, empty weight

and fuel weight) are estimated. The results are found agreeable, when we compare the trainer

or general aviation aircrafts.

In trade study, it can be observed how the variation of takeoff weight and the other

weights as to cruise speed and range is. While the weight changing is not very much for cruise

speed, it has a serious change for range.

28

CHAPTER 4

AIRFOIL AND GEOMETRY SELECTION 4.1. INTRODUCTION

The purpose of this chapter is to select the appropriate airfoil and the geometries of

wing and tail. Airfoil determines cruise and stall speeds, so choosing airfoil type has very

important role on aircraft performance. In this study wing geometry; aspect ratio, wing

sweep, taper ratio, twist, wing incidence, dihedral, wing vertical location and wing tips, wing

and tail airfoils were selected. Finally, airfoil geometries were plotted and non-dimensional

sketches of the reference wing and tail platforms were drawn showing the mean aerodynamic

chord, the quarter chord point and the quarter chord line.

4.2. WING AIRFOIL SELECTION

Airfoil selection is the most important part of the aircraft design process because the

Airfoil has a major effect on aerodynamic efficiency during all phases of flight. For example,

cruise speed, takeoff and landing distances and stall speed are affected by the airfoil.

Therefore, an airfoil with suitable flight characteristics must be chosen out of hundreds of

available sections when designing a plane.

The purpose in the airfoil selection is to obtain the lowest drag at low angles of attack

while obtaining a high CLmax. According to this, airfoil varies as to lots of characteristics such

as Reynolds number, design Mach number, design lift coefficient, stall characteristics,

thickness ratio etc. Most important data are Reynolds number and design Mach number. Each

airfoil is designed for a certain Reynolds number. Before deciding to an airfoil, it is calculated

the Reynolds number according to the given cruise conditions. Then, the design Mach number

is found and searched for the thickness ratio from Fig. 4.14 [1].

To calculate Reynolds number, it is determined a mean aerodynamic chord from the

competitor aircrafts:

29

Aircraft C(root) C(tip) Taper Ratio c Cessna 140 1,47317 1,47317 1 1,47317 Cessna 152 1,47636 1,47636 1 1,47636 Cherokee 1,71562 1,5866 0,9248 1,65195 Aeronca 1,48182 1,48182 1 1,48182 Stearman 2,81633 2,81633 1 2,81633

Pitts 1,93443 1,93443 1 1,93443 Yak-52 2,09196 1,13384 0,542 1,66033 Zenith 1,59417 1,3949 0,875 1,49675

Diamond 1,25918 0,87513 0,695 1,07868 Aviat 1,57117 1,57116 1 1,57116 Alarus 1,54306 1,37332 0,89 1,45984

Chipmunk 1,99762 1,05874 0,53 1,57624 Ikarus 1,32275 1,32275 1 1,32275

Tomahawk 1,11969 1,11969 1 1,11969 Su-31 2,07236 0,95328 0,46 1,5818

Table 4.1 Competitor study data

It can be seen the root and tip chord, taper ratio and mean aerodynamic chord ( c ) in

the table above. These values are calculated some formulas using Excel. These calculations

are made in order to obtain average of mean aerodynamic chord. Thus, it can be found mean

aerodynamic chord using for our airplane.

c = 1,580087m

After mean aerodynamic chord is calculated, the Reynolds number is found at a given

cruise conditions.

6Re 6 10V cρ× ×= ≅ ∗

µ

Cruise conditions (@2440 m): ρcruise = 0,963 kg/m3

μcruise = 0,000017124 kg/ms

Vcruise = 130 knots = 66, 82 m/s

acruise = 330,74 m/s

According to the given cruise conditions, our Reynolds number are calculated

approximately 6000000.

Mdesign = V / a = 0,2

30

From this formula, our design Mach number is calculated as 0, 2.

The thickness ratio for this design Mach number is found between 12% - 15% [1]. In

the table which is given below, there are some airfoils varies thickness from 12 to 15.

Greater t/c tends to increase Clmax up to a point. In order to it, greater t/c increases fuel

volume. Because of these reasons, we would like to make the t/c as large as possible to reduce

wing weight. Therefore, the airfoils which have 15 % thickness are thought to prefer. The

airfoils are chosen NACA airfoils because they are more available than others.

Profile NACA 4412 NACA 23012 NACA 652415 NACA 2412 NACA 2415 NACA 23015

Re 600X104 600X104 600X104 570X104 600X104 600X104 Roughness - - - - - -

Clmax behavior D B D D D D

Clmax 1,63 1,75 1,6 1,68 1,59 1,7 ao -3,9 -1,2 -2,8 -2 -1,9 -1,2

dCl/dα 0,11 0,105 0,113 0,104 0,107 0,103 Design Cl 0,5 0,19 0,3 0,41 0,21 0,09

CD min 0,0062 0,0061 0,0041 0,0061 0,0063 0,0062 Cma -0,094 -0,013 -0,062 -0,051 -0,049 -0,005 xa / c 0,246 0,241 0,266 0,243 0,246 0,239 ya / c -0,051 0,035 -0,065 -0,004 -0,013 -0,043

ρ0 1,1 1,1 0,669 1,1 1,1 1,1 f / c 0,018 0,018 0,022 0,02 0,02 0,018 xf / c 0,15 0,15 0,5 0,4 0,4 0,15 t / c 0,12 0,12 0,15 0,12 0,15 0,15 xt / c 0,3 0,3 0,41 0,3 0,3 0,3

Table 4.2 Airfoils for wing [3]

According to the Table3.2, NACA 23015 is selected because of fact that moment

coefficient of NACA 23015 is the smallest. It can be seen in the Table 3.2 cm = - 0.005. Also,

it is 5-digit airfoil, so it is chosen between the other profiles. The advantage of the 5-digit

airfoil using is that it has higher maximum lift coefficient. Another thing about 5-digit NACA

profile, due to their high maximum lift coefficient, they have good stall characteristics.

NACA 23015 airfoils’ stall behavior is classified as “D” [Figure 3.1.]. It can be seen the curve

of maximum lift coefficient is not sharp.

Our competitor aircrafts generally have 4-digit airfoil. However, we don’t want to

select a 4-digit airfoil because of its low maximum lift coefficient and high moment

coefficient. Beside, we don’t select a 6-digit airfoil owing to the fact that it is laminar airfoils.

31

Figure 4.1 Behaviour of Clmax

Figure 4.2 NACA 23015 airfoil

For NACA 23015 airfoil, according to the some given Reynolds numbers C l-α , C d-

α, Cm-α graphs and C d-C l curves are indicated in Appendix B.

4.3. WING GEOMETRY

4.3.1. Aspect Ratio

The aspect ratio of a wing is defined as the square of the span divided by the wing

area. It is a measure of how long and slender a wing is from tip to tip. Aspect ratio and

planform are powerful indicators of the general performance of a wing, although the aspect

32

ratio as such is only a secondary indicator. The wingspan is the crucial component of the

performance for a rectangular wing; this reduces to the ratio of the wings span to the chord

length. High aspect ratio wings have long spans while low aspect ratio wings have either short

spans or thick chords.

While high aspect wings create less induced drag, they have greater parasite drag.

Also, higher aspect ratio wing have heavier wing than low aspect ratio wing.

An important effect of changing aspect ratio is a change in stalling angle. For

aerodynamic efficiency, higher aspect ratio is better, but it is worse for stall. Low aspect ratio

wing reaches stall later than higher aspect ratio wing.

From the competitor study, the average aspect ratio is calculated as 6, 6. Beside, it can

be seen the from Table 4.1[1] equivalent aspect ratio is 7, 6 for the general aviation - single

engine. According to these values, aspect ratio of our design is selected as 7. These are shown

in the table is given below.

Competitor Study : 6,6 Table 4.1 : 7,6 Our aspect ratio : 7

Table 4.3

4.3.2. Wing Sweep

Wing sweep is important for the pitch-up characteristics of the plane. There are two

sweep angles; one is leading edge sweep angle (ΛL.E.) which is important for transonic and

supersonic speeds; the other is quarter-chord line sweep angle (Λc/4) which is important for

subsonic speeds. It is better for stability. A swept wing has a natural dihedral effect. To

prevent excessive stability, generally sweep and dihedral are not used together. Wing sweep

increases the structural weight both because of the increased tip loading and because of the

increased structural span.

The selected aircraft is subsonic, there is nearly 4 degree quarter-chord line sweep

angle (Λc/4) and it has a sweep from leading edge. To determine leading edge sweep angle, it

is calculated the wing span average of competitor studies aircrafts. The wing span is found as

9. And the mean aerodynamic chord c has found as nearly 1, 58m. Than, the leading edge

wing sweep is calculated from the wing geometry (the angle of between the span and leading

edge). According to these calculations, the leading edge wing sweep angle is found as 8, 6.

(ΛL.E.) = 8, 6

(Λc/ 4) = 4

33

4.3.3. Taper Ratio

Taper ratio is the ratio of the length of the chord at the wing tip the length of the chord

at the wing root. Taper affects the distribution of lift along the span of the wing. To obtain

nearly elliptical wing, wings are tapered because the lift distribution on the tapered wing

approaches to elliptical lift distribution. Thus, the induced drag on the wing is low effective.

Also, lower taper ratios lead to lower wing weight.

To give a taper to the wing is better than the rectangular wing which is no taper (Taper

ratio = 1). Rectangular wing is the basic wing shape. Because of low cost and easy building,

many single-engine planes use rectangular wings. In this type of wings, chord doesn’t vary

from root to tip, so there isn’t any approach to elliptical wing. Thus, lift at the wing tips is

higher. However, lift at the tapered wing tips has essentially zero.

For most low swept wings, taper ratio generally is between 0, 4 – 0, 5 [1]. Therefore, in our

design taper ratio is selected as 0, 4.

= 0, 4

4.3.4. Twist

Wing twist is used to prevent tip stall and to revise the lift distribution to approximate

an ellipse. Typically, wings are twisted between zero and five degrees. However, we can not

use any twist in our design to do a simple design. Twist makes the design complex.

4.3.5. Wing Incidence

Angle of incidence at which the wing or horizontal tail of an airplane is installed on

the fuselage, measured relative to the axis of the fuselage. Wing incidence minimizes drag

especially during cruise.

If the wing is untwisted, the incidence is simply the angle between the fuselage axis

and the wing’s airfoil chord lines. Usually, to obtain maximum L/D at cruise condition for

general aviation aircraft have an incidence of about 2 degree.

Angle of Wing Incidence = 2

4.3.6. Dihedral

Wing dihedral is the angle of the wing with respect to the horizontal when seen from

the front. Dihedral tends to roll the aircraft level whenever it is banked. That is to say, a

dihedral on the wing helps to stabilize the airplane for minor rolling motions from side to side.

Also, dihedral works to stabilize the plane, there is a minor loss in the lift of the wing. The

34

dihedral angle is selected from Table 4.2, as 0 degrees for the subsonic swept angle, mid wing

aircraft.

For easily production, it is chosen no dihedral angle because the wing is not

monolithic. The wing is formed from two parts.

Dihedral Angle = 0 deg

4.3.7. Wing Vertical Location

Aircraft operation conditions determine the wing vertical location with respect to the

fuselage. For small aircraft, the high wing arrangement can block the pilot's visibility in a turn

and in a climb. If the fuselage is roughly circular and fairings are not used, the mid-wing

arrangement provides the lowest drag. High and low wing arrangements must use fairings to

attain acceptable interferences drag with a circular fuselage. The mid wing offers some of the

ground clearance benefits of the high wing. Also, mid wing has better maneuver capability

than high and low wing. It can be seen the mid wing airplane in the figure which is given

below:

Figure 4.3 Mid wing [1]

4.3.8. Wing Tips

Wing tip shape has two effects upon subsonic aerodynamics performance. The tip

shape affects the aircraft wetted area. A far more important effect is the influence the tip

shape has upon the lateral spacing of the tip vortices. This is largely determined by the ease

with which the higher pressure air on the bottom of the wing can escape around the tip to the

top of the wing. It is selected in the design sharp edge wing tips because a tip with a sharp

edge reduces the induced drag.

35

Figure 4.4 Wing tips [1]

4.4. SELECTION OF TAIL AIRFOIL

4.4.1. Tail Arrangement

For most aircraft designs, the conventional tail will usually provide adequate stability

and control at the lightest weight and also 70% or more of the aircraft in service use

conventional type. Although T-tail is effective because of not being in the wake of wing, it is

heavier than the conventional type. Therefore, conventional tail is selected for tail

arrangement from Figure 4.30 Aft Tail Variations table [1].

Figure 4.5 Conventional tail

First mission of the tails is not provides to lift. Most of lift is generated by wings. It

can be said that tails are little wings. Tails’ priority function is provides for trim, stability and

control. Therefore, it is important to become the stall speed of tail later than the wing but it

isn’t the most crucial parameter for design because tail doesn’t produce more lift, it is only a

fraction of its lift potential. Thus, it is considered another specification of tails. These are

36

pitch moment and wing incidence. To provide balance of the moment, horizontal tail has

negatively incidence angle of about 2-3 degree and thus, the balance the wing pitching

moment is stabilized by negative incidence of horizontal tail. According to it, wing incidence

angle of tail is chosen as - 2 degree.

To select the tail airfoil, it is needed some specifications. For instance, the thickness

ratio of tails is usually similar to the wing thickness ratio, according to the historical data.

Therefore, it is selected in this study both of horizontal and vertical tail thickness ratio is the

same of the wing thickness ratio as 15 %.

Vertical and horizontal tail airfoils are not the same. It is selected in this design that

vertical and horizontal tail airfoils are different. To prevent unbalanced yawing moment, most

aircraft use symmetric airfoil for vertical airfoil because any lift obtaining from vertical tail is

not needed. In this case, it isn’t required any camber of tails airfoil. For four digit and five

digit NACA airfoils, it couldn’t any necessary information about same thickness ratio.

Although 6-digit airfoils are laminar, it can be selected a six digit airfoil NACA 652015. This

airfoil is selected because it is symmetric and its drag coefficient is lower than other NACA

symmetric airfoils.

Profile NACA 652415 NACA 642015 NACA 652015 Re 600X104 600X104 600X104

Roughness - - - Clmax

behavior D D D Clmax 1,6 1,45 1,41

ao -2, 8 0 0 dCl/dα 0,113 1,112 1,111

Design Cl 0,3 0 0 CD min 0,0041 0,0048 0,0042 Cma -0, 062 0 0 t / c 0,15 0,15 0,15

Table 4.4 Airfoils for vertical tail [3]

37

Figure 4.6 NACA 652015 Airfoil

For NACA 652015 airfoil, according to the some given Reynolds numbers C l-α, C d-

α, Cm-α graphs and C d-C l curves are indicated in Appendix B

An airplane must be in balance longitudinally in order to fly. This means that the net

affect of all. Because the horizontal the forces acting on the airplane produces no overall

pitching moment about the centre of gravity. The horizontal tail provides a balancing force to

maintain equilibrium for different speeds and center of gravity positions tail is located some

distance from the center of gravity, even the small amount of lift it produces can generate a

large pitching moment at the centre of gravity. Generally, four-digit NACA profile is used for

horizontal tail. In the table which is given below, there are three NACA profile have same

thickness ratio with the wing. Ix digit airfoil is not chosen because of laminar effects.

Horizontal tail’s mission is not lift, it provides control. Thus symmetric profile is more

suitable for horizontal tail. Because of these reason, it can be selected NACA 0015 for the

horizontal tail.

Profile NACA 4415 NACA 0015 NACA 652415

Re 600X104 861X104 600X104 Roughness - - -

Clmax behavior D A D

Clmax 1,56 1, 66 1,6 ao -4, 2 -0, 9 -2,8

dCl/dα 0,106 0,97 0,113 Design Cl 0,4 0,3 0,3

CD min 0,065 0,0064 0,0041 Cma -0, 095 0 -0, 062 xa / c 0,241 0,238 0,266 ya / c -0, 04 -0,4 -0, 065

ρ0 1,1 1,1 0,669 f / c 0,4 0 0,22 xf / c 0,4 - 0,5 t / c 0,15 0,15 0,15 xt / c 0,3 0,3 0,41

Table 4.5 Airfoils for horizontal tail [3]

38

Figure 4.7 NACA 0015 Airfoil

For NACA 0015 airfoil, according to the some given Reynolds numbers C l-α, C d-α,

Cm-α graphs and C d-C l curves are indicated in Appendix B.

4.5. TAIL GEOMETRY

4.5.1 Aspect Ratio

For the selected horizontal and vertical tail, it is selected from Table 3 Tail aspect

ratio and taper ratio [1]. In this table, the aspect ratio for horizontal tail changes between 3-5

and for vertical tail changes between 1, 3-2. According to this table, the aspect ratio of

horizontal tail is chosen 5 and for vertical tail is chosen 1, 5 because preferring large control

surfaces are better.

4.5.2. Taper Ratio

In the same table, it can be seen the taper ratio for horizontal and vertical tail. As to

this table, taper ratio for horizontal tail is selected as 0, 6 and vertical tails taper ratio is

selected as 0, 5.

4.5.3. Sweep Angle Leading edge sweep angle of horizontal tail is generally wanted to be 5 degree more

than the wing sweep to make the tails stall after than the wing. Since the sweep angle of wing

is 8, 6, the tail sweep angle is considered nearly 5 degree more than it. Thus, tail sweep angle

is found 13, 6.

4.5.4. Tail Vertical Location

Spin recovery is the important situation for the aircraft because of resulting stall.

Vertical tail has a big role in spin recovery. An aircraft in a spin, one wing tends to stall more

39

deeply than the other. The wing that stalls first will fall, increasing its angle of attack and

become deep the stall. During these circumstances, the other wing will rise, decreasing its

angle of attack, and the aircraft will yaw towards to more deeply-stalled wing. The difference

in lift between the two wings the aircraft to roll and the difference in drag causes the aircraft

the yaw. The yaw axis is perpendicular to the wings and lies in the plane of the aircraft

centerline. A yaw motion is a side to side movement of the nose of the aircraft. This yawing

moment is prevented by using rudder which is hinged at the aft of the vertical stabilizer.

Because of this, vertical tail is important for spin recovery. Also, tail geometry for spin

recovery, location of the horizontal tail and rudder, is essential parameter.

In this study, the tail is chosen whose horizontal tail is at the forward with respect to

the vertical tail and also, horizontal tail vertical location is chosen high. In the figure, shown

below, the selected location of horizontal tail is given above for spin recovery. The “uncover”

parts of the rudder provide to increase rudder control.

Figure 4.8Tail geometry for spin recovery [1]

4.5.5. Twist and Dihedral

There is no twist and dihedral angle for horizontal and vertical tail. These parameters

are not very important for this aircraft. It isn’t used any twist because of cost. And tails are

control surfaces. Therefore, it isn’t requiring any thing to provide stability.

40

4.6. NONDIMENSIONAL DRAWINGS Nondimensional drawing of Wing

c = 0, 74 Nondimensional drawing of Horizontal Tail

c = 0, 79

Croot=1

Y=0,91

Ctip=0,4

Y=0,89

Croot=1

Ctip=0,6

41

Nondimensional drawing of Vertical Tail

c = 0, 77 4.7. SUMMARY OF RESULTS

Characteristics WING HORIZONTALTAIL VERTICAL TAIL

Airfoil NACA 23015 NACA 2415 NACA 652-015 Aspect Ratio 7 5 1,5

Wing Sweep(l.e.) 8,6° 13,6° 41,6°

Wing Sweep(c/4) 4° 6,76° 40° Taper ratio ( λ ) 0,4 0,6 0,5

Twist untwist untwist untwist Dihedral 0° - - Incidence 2° -2° - Wing Tip Sharp Sharp -

Wing Vertical Location Mid High -

Table 4.6 Characteristics of wing and tails

Croot=1

Ctip=0,5

42

4.8. CONCLUSION

In this study, firstly airfoils are selected which satisfies the required characteristics of

the wing in terms of lift coefficient, thickness ratio, etc. After that the initial decisions on

aspect ratio, sweep, taper ratio, twist, incidence, dihedral, wing vertical location and tips are

determined. During all selections, aircraft’s improved performances are considered. Actually,

it is important to point out that, the competitor study was played an important role in the

determination of these parameters. However, these are only initial estimations and likely to be

revised in the following chapters in order to achieve an optimal design.

43

CHAPTER 5

THRUST TO WEIGHT RATIO AND WING LOADING 5.1 INTRODUCTION

Horsepower-to-weight ratio (hp/W) and wing loading (W/S) properties of an aircraft

plays an important role for the performance of the propeller aircraft. By using the hp/W and

W/S values, it can be estimated the other performance parameters. Therefore, correctly

estimation of these values is very essential for performance calculations. It is defined the

hp/W as the T/W (thrust to weight ratio) for propeller aircraft. By using a formula, T/W can

be converted to hp/w.

In this study, hp/w and W/S is selected by considering the competitors average,

historical value and the calculations for the aircraft performance. Then, according to obtained

value, the final selection for hp/W and wing loading is done. Also, other performance

properties of the aircraft will be evaluated from the selected hp/W and W/S. To make this

estimations, it is used the given performance requirements.

The performance requirements are given below:

- Stall Speed( stallV ) : 50 knots @ sea level, standard day, flaps down full down

- Stall Speed( stallV ) : 60 knots @ 5000ft hot day flaps full down (landing)

- Take off Field Length (FAR 23): 600m (seal level, standard day)

- Landing Field Length (FAR 23): 600m (sea level, standard day)

- Cruising Speed: 130 knots (with 75% power engine and reversible @ 8000ft

Altitude)

5.2. SELECTION OF HORSEPOWER-TO-WEIGHT RATIO

5.2.1. Competitor Study

Hp/W is inverse of power loading of the aircrafts which exist in competitor study is given below as a table.

44

AIRCRAFT Hp/W(hp/lb) Cessna–140 0,058 Cessna–152 0,0625

Piper PA–28–180 Cherokee 0,0752 Aeronca 11AC Chief 0,052

Boeing/ Stearman Model 75(PT_13) 0,081 Pitts S–2 C 0,152

Yakovlev Yak–52 0,127 Zenith CH601 XL 0,085

Diamond DA 20 Katana 0,0756 Aviat HUSKY 0,1

AMD Alarus CH2000 0,0685 De Havilland Chipmunk 0,072

Ikarus C 42 0,1 Piper PA–38 Tomahawk 0,067

SU–31 0,134 Average 0,08732

Table 5.1 HP/W

It can be seen an average HP/W value of the competitor aircrafts is approximately 0,087.

0.087hpW

=

5.2.2. Statistical Estimation

For general aviation-single engine, piston propeller aircrafts, power-to-weight ratio is

0.07 as to Table 5.2 [1] historical average values for power-to-weight ratio.

0.07hpW

=

There is another approach for first estimate horsepower-to-weight ratio. According to

this approach, it is considered that the HP/W is related to maximum speed. To apply this

process, it can be benefited from Table 5.4[1] for general aviation-single engine aircrafts. To

determine the power-to-ratio is used the formula [1] given below:

45

max0

cP aVW

=

(5.1)

Vmax is given in the performance requirements as 130 knots = 219, 44 ft/s. Also,

according to the Table 5.4:

a = 0, 025

C = 0, 22

Thus;

0,22

0

0,025(219,44) 0,0818PW

= =

5.2.3. Thrust Matching

Another method to estimate horsepower-to-weight ratio is “thrust matching”. Thrust

matching is used to make a better initial of the required T/W for aircraft designed primarily,

for efficiency during cruise. This refers to the comparison of the selected engine’s thrust

available during cruise to the estimated aircraft drag.

During level flight, when the thrust must equal the drag and weight must equal the lift.

Therefore, it can be written;

( )

1/cruise cruise

TW L D =

(5.2)[1]

To get the equivalent take-off T/W, the required cruise T/W must be regulated because

maximum take-off thrust at sea level much than the cruise thrust at altitude. To do this

estimation is used the equation (5.3):

takeoffcruise

takeoff cruise takeoff cruise

TWT TW W W T

= (5.3)[1]

For a propeller aircraft, the required takeoff P/W can be found by solving in Eq. (5.4)

[1] at cruise conditions and adjusting weight and power back to takeoff conditions.

1550. ( / )

takeoffcruise cruise

takeoff p cruise takeoff cruise

hpV WhpW L D W hpη

= ∗ ∗ ∗ (5.4)[1]

46

1550. ( / )

cruise

cruise p cruise

VhpW L Dη

= ∗

In order to find (hp/W) for take-off, it should be known some parameters’ value and

approaches. It is known for propeller, 0,8pη ≅ and maxcruise

L LD D

=

To estimatemax

LD

, it is used the formula which indicates the relation between lift and drag

coefficients.

L DC C eARπ=0

(5.5)[1]

The value of parasite drag coefficients generally varies between 0, 02 and 0, 03 for

single engine-general aviation aircrafts. In this study, parasite drag coefficient CDo is selected

as 0, 024. Because it is important the performance of aircraft for this project, CDo value is

chosen for smaller drag. The better choice may be the most little value 0, 02, but this is very

high cost for this aircraft. Therefore, it is not selected.

* CDo = 0.024 (for single engine light aircrafts no strut) [4]

* e (Oswald efficiency) = 0.8

* AR = 7 (It is determined at previous chapter)

According to these values, CL is calculated as the below:

CL =0, 649

D = qSCD and CD = 2CDo, so D = 2qSCD0

CD =0, 048

max

0,649 13,520,048

L

cruise D

CL LD D C = = = =

It is known from Table (3.2) [1];

lim

0.97cruise

c b

WW

=

and lim 0.985c b

takeoff

WW

=

47

Thus;

lim

lim

0.97 0.985 0.956cruise cruise c b

takeoff c b takeoff

W W WW W W

= × = × =

For a piston-powered, propeller-driven aircraft, the power available varies with the

density of the air provided to the intake manifold. Therefore, piston-powered aircraft typically

cruise at about % 75 of take-off power [1]. And also, in the performance requirement, it is

given the cruise speed is with 75% power at 8000ft altitude.

. . 1 1.33. . 0.75

takeoff

cruise

h ph p

= =

Vcruise is given 130 knots = 219.44 ft/s in the performance requirements.

After all needed parameters were calculated, it can be found (hp/W)take-off :

( ). . 219,44 1 0.956 (1.33) 0,0469550 0,8 13,52takeoff

h pW

= = ×

It is found by thrust matching;

0,0469hpW

=

5.2.4. Final Selection

In order to find hp/W, some different methods were considered. Here, these

results are given as a table:

Method Result for hp/W Competitor Study 0,087

Statistical Estimation 1 0,07 Statistical Estimation 2 0,0818

Thrust Matching 0,0469 Table 5.2

The final selection of the horsepower-to-weight ratio was made according to these

different values. It can be seen that the thrust matching results are very low according to the

other results such as competitors’ aircrafts and historical value. Therefore, thrust matching

value of hp/W is not considered. If the thrust matching value is selected, it would be made

very large wing to provide the lift. During selecting the hp/W, to consider the reducing fuel

48

consumption is important because W0 becomes lower while reducing fuel consumption.

Finally; the hp/W is chosen 0, 085.

0,085hpW

=

5.3. SELECTION OF WING LOADING 5.3.1. Competitor Study

Wing loading (W/S) of the aircrafts which exist in competitor study is given below as

a table.

AIRCRAFT W/S (lb/ft^2) Cessna-140 8,68 Cessna-152 10

Piper PA-28-180 Cherokee 15 Aeronca 11AC Chief 7,1

Boeing/ Stearman Model 75(PT_13) 9,15 Pitts S-2C 13,3

Yakovlev Yak-52 17,61 Zenith CH601 XL 9,85

Diamond DA 20 Katana 13,2 Aviat HUSKY 10,9

AMD Alarus CH2000 12,3 De Havilland Chipmunk 11,709

Ikarus C 42 7,6 Piper PA-38 Tomahawk 13,39

SU-31 1,55 Average 10,75593333

Table 5.2 Wing Loading 5.3.2. Statistical Estimation

From Table 5.5[1], for a single engine aircraft, wing loading is 17 lb/ft2.

aircraft takeoff

W WS S

=

217 /W lb ftS=

49

5.3.3. Stall Speed

From the formula which is given Equation (5.6) [1], to calculate wing loading, it is

found some unknowns. One of the most parameter all of them is stall speed. Airplanes can be

equipped with devices to prevent or postpone a stall or to make it less (or in some cases more)

severe, or to make recovery easier. Therefore, stall speed is a major provider to safety flying.

Stall speed is directly related with the lift coefficient. After the aircrafts enter stall state, lift of

aircrafts is decrease suddenly. There is a lift coefficient at stall speed which is the maximum

lift coefficient of the aircraft.

5.3.3. a. Stall Speed for 50 knots at sea level, standard day, flaps full down

In this part, Vstall = 50 knots @ sea level, standard day for flaps full down and landing

conditions. Also, at sea level, standard day conditions the air density is 0, 00238 slug/ ft3.

W= L = qstall S CL max = 12

ρ V2 stall S CL max (5.6a) [1]

WS

= 12

ρ V2 stall CL max (5.6b) [1]

To determine wing loading, there is still an unknown, maximum lift coefficient, is the

more difficult to estimate than other parameters. This value can vary from about 1.2 to 1.5 for

a plain wing with no flaps to as much as 5.0 for a wing with large flaps affected by prop wash.

It is directly related to use of flaps. When some flaps conditions or no-flap, maximum lift

coefficient is changing. To determine this change, it is benefited from Figure 5.3[1]. This

figure shows maximum lift coefficient values for different flap types. The aircrafts in the

competitor study, there is generally aircrafts which have no-flap, fowler flap and single slotted

flap.

Firstly, it is found the Clmax of airfoil towards to the quarter chord line sweep angle.

The (Λc/4) is designed as 4 degree. In respect of this sweep angle, the maximum lift coefficient

is chosen 1, 7. However, this value of maximum lift coefficient is for airfoil. To find the

maximum lift coefficient for wing there is an assumption: (CLmax) wing = 0.9(Clmax)airfoil.

According to this information, the (CLmax) wing = 1, 53 is calculated. Than, this value of the

maximum lift coefficient of the wing is shifted to the graph which is given at Figure (5.3) [1].

It is searched for the maximum lift coefficient of wing with flaps because flap reduces the

50

stall speed by increasing the camber of the wing thereby increasing the maximum lift

coefficient. For different flap configurations the Clmax values are given below:

* Clmax for no flap : 1, 53

* Clmax for fowler flap : 2, 52

* Clmax for single-slotted flap: 2, 15

* Clmax for plain flap : 1, 8

From the formula (5.6b) [1], for Vstall = 50 knots = 88, 4 ft/s and = 0, 00238 slug/ft3.

According to this information, the wing loading W/S is calculated for the flap types:

* For no flap :

* For fowler flap :

* For single-slotted flap:

* For plain flap :

5.3.3. b. Stall Speed for 60 knots at 5000ft, hot, flaps full down

In this part, the stall speed Vstall = 60 knots at 5000ft altitude, hot day for flaps full

down and landing conditions. Also, @5000ft hot day conditions the air density is 0, 00189

slug/ ft3.

If the same considerations and assumptions are made for this stall speed, the maximum

lift coefficient of the wing for no-flap, plain, fowler and slotted flap remain the same. The

different parameter is air density and stall speed for this part. Because of this, by using the

same formula which is given Equation (5.6b) the wing loading can be calculated.

Vstall = 60 knots = 101, 28 ft/s

= 0, 002043 slug/ft3

* For no flap :

W/S = 12, 97lb /ft2

W/S = 18, 23 lb/ft2

W/S = 21, 36 lb/ft2

W/S = 16,74 lb/ft2

W/S = 16,03 lb/ft2

51

* For fowler flap :

* For single-slotted flap:

* For plain flap :

5.3.4. Takeoff Distance

Takeoff is the phase of flight in which an aircraft goes through a transition from

moving along the ground (taxiing) to flying in the air. A number of different values are

referred to as “takeoff distance”. The most noticeable definition of takeoff distance or

“balance field length (BFL)”, is the distance that pilot can stop the aircraft safely if a

problem occurs at one of the engines or to continue takeoff on the remaining engines at

decision speed. Balance field length is sometimes called the “FAR 23 takeoff field length”.

However, Far 23 doesn’t require meeting balance field length. There are some parameters

which affect the takeoff distance such as wing loading and thrust-to-weight ratio. This relation

can be seen in Equation (5.7). The following equation gives the maximum allowable wing

loading for the given takeoff distance of a propeller aircraft.

( ) ( )W hpTOP CS W

σ= LTAKEOFF (5.7)[1]

In the performance requirements, take of field length is given 600m = 1968, 5ft

for sea level, standard day. Then, take off parameter (TOP) is found for the take off distance

1968, 5ft from Figure (5.4) [1].

• TOP (ground roll) = 210

• TOP (over 50ft) = 280

For ground roll line and over 50ft line, the take of parameter TOP is different from each other.

Also, the Figure (5.4) [1] doesn’t give an accurate value for TOP, because of this TOP is

chosen between these two values as 250.

TOP = 250

• = 1 ( for sea level)

In the formula, the lift coefficient is not maximum lift coefficient. Here, it is the

takeoff lift coefficient, is definition of stall speed. Lift off speed is generally 1, 1 times of stall

W/S = 26,41 lb/ft2

W/S = 22,53 lb/ft2

W/S = 18,86 lb/ft2

52

speed. With another saying, maximum lift coefficient for take of is not same value for the

maximum lift coefficient. CLmax is valid for landing configuration (flaps full down) and CLmax

for takeoff is nearly 80% of CLmax value.

0,80C C=L LTAKEOFF MAX

From these values, the wing loading for different flap types is calculated. * For no flap : 0,80 0,80 1,53 1,224C C= = × =L LTAKEOFF MAX

(250)(1)(1,224)(0,085)WS= = 26,01 lb/ ft2

* For fowler flap: 0,80 0,80 2,52 2,016C C= = × =L LTAKEOFF MAX

(250)(1)(2,016)(0,085)WS= = 42,84 lb/ ft2

* For single-slotted flap: 0,80 0,80 2,15 1,72C C= = × =L LTAKEOFF MAX

(250)(1)(1,72)(0,085)WS= = 36,55 lb/ ft2

* For plain flap : 0,80 0,80 1,8 1,44C C= = × =L LTAKEOFF MAX

(250)(1)(1,44)(0,085)WS= = 30,6 lb/ ft2

5.3.5. Landing Distance

Landing distances consist basically of two segments: the air run from a height of 50ft

to the surface accompanied by a slight deceleration and flare which is defined as “FAR 23

landing field length”, and the ground deceleration from the touchdown speed to a stop as

shown in the figure below. Landing distance is largely determined by wing loading. Wing

loading affects the approach speed, which must be certain multiple of stall speed. This is

1,15*Vstall for the general aviation aircraft. Therefore, for better approximation of the landing

distance this can be used to estimate the maximum landing wing loading.

53

SWS C

SlandingL

a=

⋅

+80

1σ

max

(5.8)[1]

In this formula, the first term represents to ground roll to absorb the kinetic energy at

touchdown speed. The constant term, Sa, represents to obstacle-clearance distance.

First of all, it is computed the competitors’ average landing distance 600m = 1968, 5ft.

Also;

• = 1 ( for sea level)

• Sa = 600ft /for general aviation-type power-off approach)

• CLmax = 1,53 (for no flap)

• CLmax = 2,52 (for fowler flap)

• CLmax = 2,15 (for single-slotted flap)

• CLmax = 1,8 (for plain flap)

* For no flap : 11968,5 80 6001 1,53

WS

= + ×

* For fowler flap: 11968,5 80 6001 2,52

WS

= + ×

* For single-slotted flap: 11968,5 80 6001 2,15

WS

= + ×

* For plain flap: 11968,5 80 6001 1,8

WS

= + ×

5.3.6. Wing Loading for Cruise

To optimize cruise range, wing loading should be selected to provide a high L/D at the

cruise conditions. This range gives the best range for cruise. A propeller aircraft, which loses

thrust efficiency as speed goes up, gets the maximum range when flying at the speed for best

L/D. To maximize range, parasite drag equals to induce drag. For maximum propeller range;

0

21 . . .2L cruise cruise D

W qC V A e CS

ρ π= = ⋅

CL was found as 0, 649.

26,17WS= lb/ft2

43,11WS= lb/ft2

36,78WS= lb/ft2

30,79WS= lb/ft2

54

For 8000ft altitude cruise conditions, the performance requirements are given:

= 0, 001869 slug/ft3

Vcruise = 219, 44 ft/s

21 (0.001869)(219,44) 7 0.8 0.02

2WS

π= × × × 29,2cruise

WS

=

Lb/ft2

This wing loading value is made for cruise conditions. It must be adjusted to sea level

conditions (takeoff conditions); it must be multiplied by weight fraction. 1

11 2

0 1

29,2.(0.97 0.985)Takeoff cruise

W WW WS S W W

−

− = ⋅ = ×

230,57 /

takeoff

W Ib ftS

=

5.3.7. Final Selection of Wing Loading After all methods for wing loading estimation are made, it is required to make a final

selection of wing loading. The results of the all methods are shown as a table which is given

below. For takeoff and landing estimation, it is selected single-slotted flap values for wing

loading.

Method W/S for plain flap Competitor Study 10,76

Statically Estimation 17 Stall Speed 1(50kt) 16,74 Stall Speed 2(60kt) 18,86 Takeoff Distance 30, 6 Landing Distance 30,79

Cruise Speed 30,57 Table 5.3 Final selection of wing loading

It can be seen the results of the method different fro each other. From the wing

loadings estimated above, the lowest value is selected to ensure that the wing is large enough

for all flight conditions. As a lowest value, stall speed for standard day gives better solution

for wing loading compare to the other method. It can be seen from the Table 5.3. wing

loading becomes higher by using the flap.

55

The estimations are made for three types of flap, the plain flap is chosen. The plain

flap is simply a section of the trailing edge that is hinged to bend down. Because of simple

mechanism and lower cost of plain flap, it is selected between the other flap types.

Also, the calculated values of wing loading for stall speed especially for 50 knots, at

sea level, standard day conditions are better for both competitors and historical values. It is

higher than competitors and approximately equals to historical value. According to these

results, 17 lb/ft2 is chosen for wing loading estimation.

217 /W lb ft

S=

5.4. ANALYSIS AND CALCULATIONS

The purpose of this part is to make an estimation of some performance parameters

such as stall speed, takeoff and landing distance for different conditions by using the

determined horsepower-to-weight ratio and wing loading.

5.4.1. Stall Speed for Landing Configuration

The landing conditions are considered for the stall speed to calculate in this part. These

landing conditions are that the landing becomes at sea level, standard day and flaps full down.

Because of that it is decided to use the plain flap in the design, the maximum lift

coefficient of the plain flap for landing configuration is taken. It is (CLmax)landing =1,8

According to this maximum lift coefficient value, the stall speed for landing configuration is

determined by using the equation (5.6b):

W/S = ½ ρ V2stall CLmax and

landing takeoff

W WS S

=

=17

Also, for standard day, at sea level, the air density is = 0, 00238 slug/ft3.

max

2( / ) 2 (17) 89,08 / 52,770.00238 1,8stall

L

W SV ft s knotsCρ

×= = = =

⋅ ×

56

5.4.2. Stall Speed for Takeoff Configuration

To estimate the stall speed for takeoff configuration, it is assumed for sea level,

standard day and takeoff configurations. Flaps contribute to takeoff stall speed because of

they are used about half the maximum angle configuration. Therefore, an aircraft uses the

80% CLmax at the take off conditions ( 0,80C C=L LTAKEOFF MAX). Again, the calculation of Ctakeoff

is made for the plain flap. By using the same formula;

max

2( / ) 2 (17) 99,6 / 590.00238 1.44stall

L

W SV ft s knotsCρ

×= = = =

⋅ ×

5.4.3. Stall Speed for Cruise Configuration

To calculate the cruise configuration of stall speed, the assumptions are for sea level,

standard day and no flap deflection. Because of flap off configuration, maximum lift

coefficient is taken from Figure 5.3 for no flap curve. It is shown above as (CLmax) cruise=1, 53.

21 2

0 1

17 (0.97 0.985) 16,24 /cruise takeoff

W WW W lb ftS S W W

= × = × × =

max

2( / ) 2 (16,24) 94,44 / 55,950.00238 1.53stall

L

W SV ft s knotsCρ

×= = = =

⋅ ×

5.4.4. Stall Speed for 5000ft, Hot Day, Landing Configuration

In this part, it is wanted to estimate the stall speed for 5000ft altitude, hot day and

landing configuration. For 5000ft altitude, the density is 5000ρ =0, 00189 slug /ft3. Also, it is

known the maximum lift coefficient for landing from the above calculations. For plain flap,

(CLmax) landing = 1, 8. While the same calculations are made for this stall speed;

landing takeoff

W WS S

=

= 17

57

max

2( / ) 2 (17) 99,97 / 59,220.00189 1,8stall

L

W SV ft s knotsCρ

×= = = =

⋅ ×

5.4.5. Takeoff Ground Roll

For takeoff ground roll, the assumptions are for standard day, sea level. To determine

ground roll distance, it is benefited from Figure 5.4[1] for the takeoff parameter (TOP) which

is computed according to the takeoff lift coefficient CL TO and sea level density ratio σ=1. For

plain flap, 0,80C C=L LTAKEOFF MAX= 1, 44

2( / ) 17 138,89 /( / ) 1 (1,44) 0,085LTO

W STOP lb ftC hp Wσ

⋅

= = =⋅ × ×

From Figure 5.4[1] for propeller aircraft, grounds roll distance which is the intersection

with TOP and grounds roll line, is found. Thus; ground roll is estimated as nearly 1000ft .for

138, 89 lb/ft2 TOP.

Takeoff Ground Roll = 1000ft = 304, 8m

5.4.6. Takeoff Field Length

It is made some assumptions for the takeoff ground roll. Only one thing is different

than ground roll which takeoff field length is for 50ft obstacle. TOP for takeoff field length is

the same with ground roll because the terms in the TOP equation are not change. For the same

takeoff parameter, TOP =138, 89 lb/ft2, takeoff field length is found from the Figure 5.4[1],

over 50ft line. According to it, takeoff field length is found as nearly 1300ft.

Takeoff Field Length = 1300ft = 396, 24m

5.4.7. Landing Ground Roll

The calculation of landing ground roll is made for sea level, standard day

configuration.

σ=1

(CLmax)landing =1,8 landing takeoff

W WS S

=

=17 lb/ft2

58

SWS C

SlandingL

a=

⋅

+80

1σ

max

Landing Ground Roll = max

1 180 80 17 755,56 230,31 1,8L

W ft mS Cσ

= × × = = ⋅ ×

Landing Ground Roll = 755, 56ft = 230, 3m

5.4.8. Landing Field Length

The landing field length is estimated for sea level, standard day conditions.

SWS C

SlandingL

a=

⋅

+80

1σ

max

= . .landing aS L G R S= +

Landing ground length is found above as 755, 56 ft. Thus; landing field length is

calculated by added LGR to Sa.

Slanding (Landing Field Length) = 755, 56 + 600 = 1355, 56ft = 413, 18 m

5.4.9. The Altitude for Best Range

The assumption for the altitude estimation is cruise speed which is the speed for the

best range is 130 knots = 219, 44ft/s.

1 2

0 1

17 (0.97 0.985) 16,24bestrange cruise takeoff

W WW W WS S S W W

= = × × = × × =

lb/ft2

It is calculated the wing loading for the best range conditions 16, 24 lb/ft2. Now, it is

known both the wing loading and speed for best range conditions. According to the best range

speed equation (5.9) [1], it can be calculated the altitude through the air density.

.2 1

b rDo

WVS C eARρ π

=×

(5.9) [1]

2 17 1219,44

0,024 0,8 7ρ π×

=× × ×

= 0, 001087 slug/ft3

By using the linear interpolation method, the altitude for = 0, 001087 slug/ft3 is

determined from (Table B.1-1) [1]. Thus;

59

h = 24447, 43ft = 7451, 6m

5.4.10. The Mach number

In this part, the Mach number is found for the calculated altitude and best range speed

130 knots.

From (Table B.1-2) [1], by using the linear interpolation for h = 7451, 6 ≅ 7, 5km the

temperature T = 242, 71 K. To determine the Mach number, first the speed of sound must be

calculated.

1, 4 287 242,71 312,28 /a RT m sγ= = × × = = 607, 548 knots

Then, the Mach number for the best range;

Mb.r. = 130 0,214607,548

brVa

= =

5.4. 11. Maximum value of L/D

For maximum L/D condition the lift coefficient is calculated by using the equation

(5.10) [1]. This lift coefficient is for minimum drag or thrust in level flight. In level flight

thrust equals to the drag.

CL, min thrust or drag = DC eARπ0

(5.10) [1]

Thus; CL, min thrust or drag = 0,02 0,8 7π× × × = 0, 649

For the level flight equilibrium;

L = W = qSCL (5.11) [1]

When Eq. (5.10) [1] is substituted into Eq. (5.11) [1], it is obtained lift for minimum

thrust or drag conditions.

L = W = qS DC eARπ0

= qS (0, 649)

At any given weight, the aircraft can be flown at the optimal lift coefficient for

minimum drag by varying velocity or altitude.

To calculate maximum L/D value, also, it can be needed the estimation of the drag for

minimum thrust or drag. For level flight, drag equals to thrust.

T = D = qS (CD0 +2

LCeARπ

) (5.12) [1]

60

By using the assumption for lift coefficient which is given in Eq. (5.10) [1], the

(5.12) [1] formula is formed:

D min thrust or drag = ( )2

( ) 2D

D D D D

C eARqS C qS C C C qS

eAR

π

π

+ = + =

0

0 0 0 0 (5.13) [1]

According to this formula; D min thrust or drag = 2(0, 02) qS = 0, 048.qS

After the calculations of the lift and drag for minimum thrust or drag conditions which

gives the maximum value of the L/D, it is estimated (L/D) max.

(L/D) max = min

min

(0,649) 13,52(0,048)

thrustordrag

thrustordrag

L qSD qS

= =

(L/D) max = 13, 52 5.4. 12-13. Hpcruise / hptakeoff value

max

L LD D

= = cruise

13, 52 1 2

0 1

0,97 0,985 0,956cruise

takeoff

W W WW W W

= = × =

( )1 1 0,074

/ 13,52cruise cruise

TW L D = = =

When the Equation (5.4) [1] is rearranged, it is obtained the formula for

cruise

takeoff

hphp

which is given below:

0.415takeoff

cruisehphp

=

5.4. 14. Loiter Speed for Best Endurance at Sea Level

Best endurance speed is the same loiter speed.

( )1 219,44 1. . 0,956550 ( / ) 550 0,8 13,52 0,415

0,085

cruise cruise

p cruise takeoffcruise

takeoff

takeoff

V WL D WhphphpW

η ⋅ ⋅ × × = = =

61

For a propeller aircraft; ( )

.

2 / 13b e loiter

SL Do

loiterW SV V

C eARρ π= = (5.14) [1]

To calculate the loiter speed; it should be known the wing loading for loiter.

Therefore, loiter wing loading is found by multiplying the weight fractions to takeoff wing

loading. For loiter, the wing loading weight fraction is assumed initial of loiter.

234 2 1

3 2 1 0

17 (1 0.8886 0.985 0.97) 14,43 /Loiter takeoff

WW W WW W lb ftS S W W W W

= = × × × × =

If the found wing loading puts in to the Equation (5. 14)[1], loiter speed for best

endurance at sea level determines like the below:

.2 114,43 103,75 / 61,5

0.00238 3 0.024 0,8 7b e loiterV V ft s knotsπ

= = × × = =× × × ×

Vb.e = 103, 75 ft/s = 61, 5 knots

5.4. 15. Instantaneous Turn Rate at Cruising Speed and Altitude The calculations for instantaneous turn rate are made at cruising speed and altitude.

According to this condition, cruse speed Vcruise = 130 knots = 219, 44 ft/s and the altitude for

cruise is hcruise = 8000ft.

The turn rate is a function of load factor “n”. Because of this; firstly, it is determined

the load factor at cruise conditions. The load factor formula is given below:

(5. 15) [1]

Instantaneous turn rate estimation is made for the maximum lift coefficient for no flap

condition. According to this information, the maximum lift coefficient for no flap condition is

found as 1, 53 above estimations.

CL max =1.53

ρ8000ft=0, 001869slug/ft3

( )/L L

cruise

qSC qCLnW W W S

= = =

62

1 2

0 1

17 (0.97 0.985) 16,24cruise takeoff

W WW WS S W W

= × ⋅ = × × =

After all values of terms in the equation are determined, the load factor is calculated

from Equation (5. 11) [1].

( ) ( )

2 2max

max

1 1. 0.001869 (219,44) (1.53)2 2

/ / 16,24

cruise LL

cruise cruise

V CqCnW S W S

ρ × ×= = = = 4,24

Then, the instantaneous turn rate is estimated by using the turn rate formula which

is given below as Equation (5. 16) [1].

RadialAccelerationTurnRateVelocity

= ; 2 1

cruise

g nV

ψ• −= (5. 16) [1]

* g = gravitational acceleration is 32, 2 ft/s2.

232, 2 4, 24 1 0.605 /219,44

rad sψ −= = = 34, 658 /s

ψ = 0, 605 rad/s = 34, 658 /s 5.4.16. Climb Gradient and Rate of Climb at the Beginning of the Climb

Climb gradient, “G”, is the ratio between vertical and horizontal distance traveled. The

calculations of the climb gradient and rate of climb at the beginning of climb are made for sea

level and V = 1,2Vstall conditions. Climb gradient description;

G = (T-D) / W or D T GW W

= −

To find the climb gradient, first of all, it can be determined the D/W and T/W terms.

D/W can be expressed as in Eq. (5. 17) [1]:

(5. 17) [1]

In order to solve this equation, it is needed to find the wing loading for climb

conditions.

2( / ) 1/

D L DqSC qS C Ae qCD WW W W S S q Ae

ππ

+= = +0 o

63

17 (0.97) 16,49WW WS S W

= × = × =

climb

climb takeoff takeoff

Also, Vclimb is equal to 1, 2 times Vstall. Thus; Vclimb = 1, 2. (99, 6 ft/s) = 119, 52 ft/s.

And the dynamic pressure is;

q = 2 21 1 0,00238 (119,52) 16,999 172 2

Vρ = × × = ≅ psf

21

1 17 (0,024) 12 16,49 0,0798/ 16,49 17 7 0,8

oDV CD WW W S S q Ae

ρ

π π×

= + = + × =× × ×

After D/W ratio is found, to estimate the climb gradient “G”, it is needed to calculate

the thrust to weight ratio for climb conditions:

lim

550 . 550 0.8 0.085 0.313119,52

p

takeoff c b

T T h pW W V W

η × = = = × =

Climb gradient;

G = T DW W

− = 0, 313 – 0, 0798 G = 0, 2332

Climb gradient is a description for the ratio between the vertical and horizontal speeds.

Horizontal speed Vx defines the climb velocity and the vertical velocity refers to rate of climb.

y

x

VG

V= Vy = G. Vx = G. Vclimb = (0, 2332).119, 52ft/s = 27, 875 ft/s

Rate of climb = 27, 87 ft/s

5.4.17. Climb Gradient and Rate of Climb at the End of the Climb

For this part, it is made the same calculations; however at different flight conditions.

The assumptions are that it becomes at cruising altitude and cruising speed.

hcruise = 8000ft

Vcruise = 219, 44 ft/s

cruise = 0, 001869 slug/ft3

64

17 (0.97) (0,985) 16,24W WW WS S W W

= × × = × × =

climb cruise

climbclimb takeoff takeoff

q = 2 21 1 0,001869 (219,44) 44,999 452 2

Vρ = × × = ≅ psi

21

1 45 (0,024) 12 16,24 0,0865/ 16,24 45 7 0,8

oDV CD WW W S S q Ae

ρ

π π×

= + = + × =× × ×

After D/W ratio is found, to estimate the climb gradient “G”, it is needed to calculate

the thrust to weight ratio for climb conditions:

550 . 550 0.8 0.085 0.17

219,44p

takeoff cruise

T T h pW W V W

η × = = = × =

Climb gradient;

G = T DW W

− = 0, 17 – 0, 0865 G = 0, 0835

Rate of climb estimation:

y

x

VG

V= Vy = G. Vx = G. Vclimb = (0, 0835). (219, 44) ft/s = 18, 65 ft/s

Rate of climb = 18, 65 ft/s

5.4.17. Maximum Ceiling

The maximum ceiling altitude is the altitude at which the aircraft will fly with design

lift coefficient. From this assumption, some calculations are done for maximum ceiling value.

First, it is calculated design CL and q form by using the following equations.

W = L = q design ×S × CL design

(W/S)cruise = q design × CL design

WW WS S W

= ×

climb

climb takeoff takeoff

65

It is known the values of terms in order to compute the dynamic pressure for design

from performance requirements:

Vcruise = 130 knots = 219, 44ft/s (at 75% power at 8000ft)

cruise = 0, 001869 slug/ft3

According to this value, it can be dynamic pressure;

q = 2 21 1 0,001869 (219,44) 44,999 452 2design designVρ = × × = ≅

2

4Re

2Re

Do

cruise

CT TG GW W AW

Sq A

π

π

− ± − − =

(5. 18) [1]

By using equation (5. 18) [1], with G = 0, the equation becomes;

2 2 0.0244 4Re 45 7 0.8

2 2Re 45 7 0.8

Do

cruise

CT T T TW W q A W WW

Sq A

π π

π π

± − ± − × × × = =

× × ×

16,24cruise

WS

=

;

2

0.0001213 0.041T TW W ± − =

After solving this equation, it is calculated (T/W) value for ceiling.

0.022ceiling

TW =

.

.,

550 . .p T O h

cruise T Oh cruise cruise o

WT h pW V W W

η ρρ

= × × ×

66

max,

0 max,

cruise

takeoff

hphp

ρρ

= and ,

2 designh cruise

qV

ρ×

=

550 0,8 10,022 0,0750,97 0,985 0,002382

h

design

h

qρ

ρ

×= × × ×

××

In the above formula, there are two air density for ceiling altitude “h”. The value of ρh is:

ρh = 0, 000591 slug/ft3

According to this air density value, from Table (B.1-1)[1] the maximum ceiling altitude

is estimated:

hmax, ceiling =39128, 5ft=11926, 4m

67

5.5. SUMMARY OF RESULTS

Horsepower To Weight Ratio (hp/W)takeoff 0, 085

Wing Loading (W/S)takeoff 17 lb/ft2

Clmax (Plain Flaps) 1,8

Vstall (sea level, standard day, landing conf.) 52,77 knots

Vstall (sea level, standard day, takeoff conf.) 59 knots

Vstall (sea level, standard day, cruise) 55,95 knots

Vstall (5000ft, hot day, landing conf.) 59,22 knots

Take off Ground Roll (sea level, standard day)

1000ft(304, 8)

Take off field length (seal level, standard day)

1300ft(396, 24m)

Landing Ground Roll (seal level ,standard day)

755,56ft(230,3m)

Landing Field Length (seal level ,standard day)

1355,56ft(413,18m)

Altitude (for best range) 24447, 43ft(7451, 6m)

Mbest range (sea level) 0, 214

(L/D)max 13, 52

hpcruise/hptakeoff 0,415

Vbest endurance (loiter) (sea level) 61, 5 knots

Instantaneous Turn Rate (cruising altitude) 34,658 deg/s

Climb Gradient (G) (Beginning of Climb) 0, 2332

Rate of climb (Vy) (Beginning of Climb) 27, 87 ft/s

Climb Gradient (G) (End of Climb) 0,0835

Rate of climb (Vy) (End of Climb) 18, 65 ft/s

Maximum Ceiling 39128, 5ft(11926, 4m)

68

5.6. CONCLUSION

In this study, the final selection of wing loading and horsepower-to-weight ratio were

performed by using various methods by considering some approaches. Wing loading is

chosen according to the plain flap, stall speed conditions because of the using flap increases

the wing loading. If the wing loading is higher, the wing area would be smaller. Also, plain

flap is selected because the simple design and low cost. Also, while making a decision for

hp/W value, it is considered the competitors average, historical dates and the calculations.

According to these obtained value; it is selected the suitable hp/W for the design.

After performing the selection of horsepower-to-weight ratio and wing loading, many

performance criterions are determined and reported under performance analysis section for

various conditions.

69

CHAPTER 6

INITIAL SIZING

6.1. INTRODUCTION

Purpose of this chapter is to make a sizing study on aircraft. Aircraft’s takeoff gross

weight and fuel weight will be calculated in sizing study. The first guess sizing which has

been done previously was limited to fairly simple, at this stage we have more information

about the aircraft so a better analysis can be done.

An aircraft can be sized using some existing engine or a new design engine. The

existing engine is fixed in size and thrust, and referred as a “fixed engine”. The new design

engine can be built in any size and for any thrust required, and it is called a “rubber engine”.

We will start by doing a rubber engine sizing; after requirements are met we will continue

with a fixed engine sizing. Finally geometric sizing will be done for fuselage, wings and

control surfaces

6.1.1. Design Requirements

The requirements for the design project aircraft are given below:

Range : 500 nm =3040000 ft Vcruise : 130 knots = 219, 4 ft/s hcruise : 2440m = 8000ft Wcrew : 100kg (1 pilot) = 220, 26lb Wpayload : 100kg (1 passengers) = 220, 26lb The design mission is:

0–1: Takeoff 1–2: Climb 2–3: 500 nm Cruise at 2440m at Vcruise=130 knots 3–4: Descent 4–5: 30 min. Loiter 5–6: Landing

2 3 5

6 1

70

6.2. RUBBER-ENGINE SIZING

In first guess sizing study, it is determined the takeoff weight by making iteration from

the equation which is shown.

0

0 0

1

crew payload

fe

W WW

WWW W

+=

− −

(6.1) [1]

Both payload and crew weight are given in the design requirements. The unknowns in

the formula are fuel weight and empty weight. To calculate the takeoff weight, it must be

firstly estimation of fuel and empty weight fractions. It can be seen the fuel weight and empty

weight are dependent on the total takeoff weight, so they are defined the fractions of total

weight. However, this approach is made for first guess size study. For initial sizing study, this

equation is not valid. The equation of takeoff weight is refined for this chapter as seen below:

WoWoWeWfuelWpayloadWcrewWo ⋅

+++= (6.2) [1]

Here, empty weight is again expressed as an empty weight fraction (We/W0).W0, but

the fuel weight is determined directly.

6.2.1. Empty Weight Estimation

From the Table 6.2[1], empty weight fraction can be estimated statically for the general

aviation-single engine aircrafts. It can be easily seen the empty weight fraction is related to

the takeoff gross weight, aspect ratio, horsepower to weight ratio, wing loading and maximum

cruise speed. The aspect ratio was determined as 7 at the airfoil selection part. Also,

horsepower-to-weight ratio and wing loading was calculated as 0, 075 and 17 lb/ft2 in the

previous chapter “Horsepower to Weight Ratio and Wing Loading”. And the maximum cruise

speed is given n the design requirements as 130 knots. According to these data, it can be

written empty weight fraction from the formula (6.3) [1] by using the terms which are given

Table 6.2 [1]:

3 4

1 2 5max

0

. .C C

C C Ce oo

o

W Wh pa bW AR VW W S

= +

(6.3) [1]

0.20 0.08 0.05 0.05 0.270

0

0, 25 1.18 (7) (0.075) (17) (130)eW WW

− −= − +

71

0.200

0

0, 25 3,913eW WW

−= − +

6.2.2. Fuel Fraction Estimation

The fuel weight fraction estimation depends on the mission profile of the selected

aircraft. To find the total fuel weight fraction, it should be calculated at all mission segments’

fuel weight fractions. The calculations in this part are similar to the first guess sizing

calculations. After that the all fuel weight fractions are found, the total fuel weight fraction is

calculated by multiplying all of them.

6.2.2.1. Engine Start, Taxi and Take off

The engine start, taxi and takeoff weight fraction is estimated historically [1]. This

value can be selected between 0, 97 – 0, 99.A reasonable estimate is chosen as;

1

0

0,98WW

=

6.2.2.2. Climb and Accelerate

The climb weight fraction of piston-propeller aircraft can be estimated from the Table

2.1[1] of historical mission segment weight fractions like the fist guess sizing study.

2

1

0,985WW

=

6.2.2.3. Cruise

For estimate the cruise segment weight fractions W3/W2, the Brequet’s Range

Equation [1] is used:

RVC

LD

nWW

i

i=

−

1

WW

RCV L D

3

2

=−

exp

( / ) (6.4) [1]

0

11D

cruise

cruise

LqCD W

W S q eAS

π

= +

(6.5) [1]

72

To calculate (L/D);

Range = 500nm = 3040000ft

hcruise = 2440m = 8000ft

= 0, 001869 slug/ft3 (for 8000ft)

Cbhp = 0, 4lb/hr/hp = 0, 0001111 lb/s/hp (for cruise)

p = 0.8

Vcruise = 130knots = 219, 44ft/s

CD0 = 0, 024 (in the previous chapter is selected)

From these data, firstly it should be calculated wing loading for cruise:

21 2

0 1

17 0.98 0.985 16,41 /cruise TO

W WW W lb ftS S W W

= × × = × × =

Then, dynamic pressure for cruise altitude:

2 2 21 1 0.001869 (219,44) 45 /2 2cruise cruiseq V lb ftρ= ⋅ = × × ≅

According to above values, it can be computed the (L/D) from equation (6.5) [1]:

1 11,5645 0,024 116,4116,41 45 0,8 7

LD

π

= =×

+ ×× × ×

After the determination of L/D, it is calculated weight fraction for cruise by using the (6.4) [1].

3

2

(3040000)(0.0001111)exp 0.936550 (0.8) (11,56)

WW

−= = × ×

3

2

0,936WW

=

6.2.2.4. Descent

The descent weight fraction is estimated historically [1]. This value can be selected

between 0, 99 – 0, 995.A reasonable estimate is chosen as;

4

3

0,995WW

=

73

6.2.2.4. Loiter

To calculate loiter weight fraction, it is used Brequet’s Endurance Equation [1] for

propeller aircraft:

−=

)/(550exp

4

5

DLEVC

WW

p

bhp

η (6.6) [1]

Before the estimation of weight fraction for loiter, it should be found the (L/D) for

loiter. To calculate (L/D) loiter:

E = 30 min = 1800s

Cbhp = 0, 5lb/hr/hp = 0, 0001389 lb/s/hp

loiter = 1, 048 kg/m3 (for 1600m, from first guess sizing study)

cruise = 0, 963 kg/m3

Vloiter = 0,76 cruisecruise

loiter

V ρρ

= 159, 868 ft/s

qloiter = 26 lb/ft2

Wing loading for loiter:

231 2 4

0 1 2 3

17 0.98 0.985 0,936 0,995 15,28 /loiter TO

WW W WW W lb ftS S W W W W

= × × × × = × × × × =

1 13,4726 0,024 115,2815,28 26 0,8 7

LD

π

= =×

+ ×× × ×

After the determination of L/D, it is calculated weight fraction for cruise by using the (6.6) [1].

5

4

(1800)(159,868)(0.0001389)exp 0.993550 (0.8) (13,47)

WW

−= = × ×

5

4

0,993WW

=

6.2.2.4. Landing and Taxi Back

The weight fraction for landing and tax back is taken from again from historical data.

Historical data change between 0, 992 – 0, 997.In this project, it is selected as 0, 997.

6

5

0,997WW

=

74

To obtain total weight fractions, all of these mission segment weight fractions are

multiplied together. Therefore;

6 3 5 61 2 4

0 0 1 2 3 4 5

0.98 0.985 0.936 0.995 0.993 0.997 0,89W W W WW W WW W W W W W W

= = × × × × × =

Total fuel includes, 100% mission fuel + 5% reserve fuel + 1% trapped fuel.

According to these proportions total fuel weight fraction is estimated as shown:

)1(06.10

6

0 WW

WW f −= (6.7) [1]

0

1.06(1 0,89) 0,1166fWW

= − =

6.2.3. Takeoff Weight Calculation

From take-off weight calculation equation which is given (6.1) [1];

• Wcrew and Wpayload values are given at the design requirements as 100kg = 220, 26lb.

• Wf / W0 = 0,1166

• We / W0 = -0, 25 + 3,913W0-0,20

To find the value of W0, it must to do iteration. For doing this, the iterative equation is got as given:

After ordering this equation, the formula becomes:

W = 388, 67 + 3, 4524 W0, 8

Then; the iterative process was done to calculate W0. Thus; some solution was found.

These are given below:

0

0,1166fWW

=

)()(100

0

WW

WW

WWW

ef

payloadcrew

−−

+=

0 0.200

220,26 220,261 0.1166 ( 0,25 3,913 )

WW −

+=

− − − +

0 0.200

440,521,1334 3,913

WW −=

−

75

W0 (Guess) (lb) W0 (Calculated) (lb)

1600 1651, 71 1684, 26 1704, 65 1717, 38 1725, 31 1735, 22 1736, 406

1737 1738, 3506 Table 6.1

Wo = 1738, 3506lb = 789, 21kg

Wf = 202, 69lb = 92, 02kg

We = 1095, 14lb = 497, 2kg

6.3. ENGINE SELECTION

After calculating the take off gross weight, required engine horsepower can be found

simply by using the horsepower to weight ratio (h.p. /W0) which has been calculated

previously and the calculated take off gross weight (W0).

The horsepower-to-weight ratio was 0, 075. Firstly, all calculations are made for this

hp/W0 value. However, as a result of this value, the range was found as 923nm. This range

was very higher than the design requirements 500nm. To correct it, it is changed the hp/W0

value as 0, 085. Thus, the range value is nearly the design requirements. Beside, with

increasing the horsepower-to-weight ratio not only takeoff weight decreases, but also

performance better as to old value of hp/W0.

0

hpW

= 0, 085

Required power is =0

hpW

. W0 = (0, 085). (1738, 3506lb) = 147, 76 hp

According to the design requirements, the aircraft has one piston propeller engine.

This engine must be provide the required power is determined as 147, 76 hp. After searching

Jane’s and web pages of engine companies, it is decided to select an appropriate engine which

provides 150 hp powers. This value is nearly same with required power 147, 76hp.

The selected engine is used generally light general aviation and trainer aircrafts.

76

Figure 6.1Lycoming O-320 E2A [5]

MANUFACTURER LYCOMING ENGINE (Pennsylvania) MODEL O-320 E2A Type Four cylinder air-cooled horizontal opposed engine

Propeller Drive Two blade fixed pitch McCauley propeller Max Diameter 74in = 1879, 6mm Weight, Dry 244lb Power (at T-O) 150 hp Max Continuous 160 Max Cruise Rating 140 Specific Fuel Consumption (at T-O) 0, 42 lb/h/lb Core RPM 2700 rpm

Compression Ratio 07:01

Combustion Chamber Aluminum-alloy casting and a fully machined

combustion chamber

Table 6.2 Engine Specifications [5-6]

6.4. FIXED-ENGINE SIZING Fixed-engine sizing procedure is similar to the rubber engine sizing with several

exceptions. These exceptions result from the fact that either the mission range or the

performance must be considered a fallout parameter, and allowed to vary as the aircraft is

sized.

It is selected an engine’s power output is higher than the required engine power, so

aircraft’s takeoff gross weight will change:

(6. 8) [1]

00/

eachN hpWhp W×

=

77

“N” points out the number of engine. Here, it is had one engine; therefore, N equals to

“one”.

01 150 1764,7050.085

W lb×= =

With the takeoff weight known, the range capability can be determined from equation

(6.1) [1] using a modified iteration technique. The known takeoff weight is repeatedly used as

the “guess” W0, and the range for one or more cruise legs is varied until the calculated W0

equals the unknown W0. In this method, during calculate the takeoff weight; it is considered

the range is variable by keeping the horsepower-to-weight ratio constant.

In order to find total weight fraction, it should be determined the weight fractions of

all missions. Weight fractions remain same except cruise and loiter mission segments. The

weight fractions for takeoff, climb and accelerate, descent and landing are defined above:

4

3

6

5

0.995

0.997

WWWW

=

=

Now, it is calculated the weight fractions for cruise and loiter which they change if

range is variable.

* The estimation of weight fraction for cruise;

The specific fuel consumption (SFC) is 0, 42 lb/hp/lb = 0, 0001167 lb/s/lb for

selected engine. By using the equation (6. 4) [1];

3

2

0,0001167exp550 0.8 11,56

W RW

λ× = − = × ×

* The estimation of weight fraction for loiter;

The specific fuel consumption (SFC) is 0, 50 lb/hr/lb = 0, 0001389 .By using the

equation (6. 6) [1];

5

4

(1800)(159,868)(0,0001389)exp(550)(0.8)( / )loiter

WW L D

ϕ

= − =

During iteration process, cruise and loiter weight fractions are called as and for

easiness.

W0 = 1764, 705lb = 801, 176kg

1

0

2

1

0.98

0.985

WWWW

=

=

78

We have to recalculate (L/D)loiter because wing loading at loiter had changed due to

change at cruise weight fraction

231 2 4

0 1 2 3

17 0.98 0.985 ( ) 0.995 (16,328 ) /loiter TO

WW W WW W lb ftS S W W W W

λ λ = × × × × = × × × × =

2

126 0,024 1 0,0357 0,03816,32816,328 26 0,8 7

loiter

LD

λλλ

λ π

= =× ++ ×

× × ×

Then, this expression substitutes into weight fraction equation for loiter:

2 25

4

0,0908(0,0357 0,038) 0,00345 0,00324exp expWW

λ λϕλ λ

+ − −= = − =

Then, total mission weight fraction is;

6 3 5 61 2 4

0 0 1 2 3 4 5

0.98 0.985 ( ) 0.995 ( ) 0.997 0.9576W W W WW W WW W W W W W W

λ ϕ λϕ= = × × × × × =

After determination of mission weight fraction, it is calculated the fuel weight

fraction:

6

0 0

1.06(1 ) 1.06(1 0,9576 ) 1,06 1.015fW WW W

λϕ λϕ= − = − = −

These are put into the takeoff weight equation given (6.1) [1]; it is obtained the new

takeoff weight equation:

0 0.200

440,520,19 1.015 3,913

WWλϕ −=

+ −

It should be do the iteration process to find the new range. During this iteration

process, we need follow the some considerations. According to this iteration process, first it is

guessed a range value. For this guessed range value, it is found the variable “”. After that, it

0 0.200

220,26 220,261 (1,06 1,015 ) ( 0,25 3,913 )

WWλϕ −

+=

− − − − +

79

is calculated the “” from value of . Then, it is calculated the takeoff weight from weight

equation. At the last, it is done the iteration until the calculated W0 equals the unknown W0.

Range (nm) Wo Guessed (lb)

Wo Calculated (lb)

500 0, 9326 0, 9933 1764 , 705 1759, 98 510 0, 9313 0, 9933 1764 , 705 1761, 63 520 0, 93 0, 99329 1764 , 705 1763, 66 525 0, 9294 0, 99329 1764 , 705 1764, 66

Table 6.3 Range Calculation

As it can be seen in the table, the new range for fixed-engine sizing study is found as:

R = 525nm = 3192000ft

The new weight fractions are as follows:

W1/W0 0, 98

W2/W1 0, 985

W3/W2 0, 9294

W4/W3 0, 995

W5/W4 0, 99329

W6/W5 0, 997

W6/W0 0, 884

Wf/W0 0, 12296

Table 6.4 Weight Fractions For W0 = 1764, 705lb;

Wf = 216, 988lb

We = 1107, 197lb

After the calculations as to range, it is found the range value for fixed-engine sizing

study. If some range requirement must be satisfied, then performance must be the secondary

parameter. The takeoff weight will be set by fuel requirements, and the fixed-sized engine

may not necessarily provide the thrust-to-weight ratio desired for performance considerations.

80

6.5. GEOMETRY SIZING

Since the takeoff gross weight is known, the fuselage, wing and tails can be sized.

Most of these sizing procedures based on historical approaches given in Ref [1].

6.5.1. Fuselage

The fuselage size is determined by real-world constraints. The fuselage length can be

found from Table 6.3[1] by using the statistical equation L aWC= 0 . According to this formula,

fuselage length is related to the takeoff gross weight. The value of a and C are taken from the

table. For general aviation-single engine aircraft, a = 4, 37 and C = 0, 23. Thus, the fuselage

length can be calculated by using the takeoff gross weight W0 = 1764, 705lb

Fuselage Length = (4, 37).(1764, 705)0, 23 = 24, 4ft =7, 437m

Fuselage diameter can be found from the fuselage fineness ratio (FFR) which is

basically;

FFR=FuselageLength

FuselageMaximumDiameter (6. 9) [1]

For subsonic aircraft, fuselage fineness ratio changes from 3 to 8. Subsonic drag is

minimized by a fineness ratio of about 3. However, fuselage fineness ratio is 3, may not

provide enough tail moment arm, so tail boom can be added, with a smooth fairing from the

front part of fuselage. This creates the streamlined “tadpole” shape characteristic of several

newer small airplanes. According to all of them, the FFR is selected;

FFR = 5

Maximum diameter of fuselage is: Length / FFR = 7, 437/5

Dmax = 1, 4874m

6.5.2. Wing

The actual wing area can be calculated simply as the takeoff weight divided by the

takeoff wing loading. But this wing area is the reference area of the theoretical, trapezoidal

wing and includes the area extending into the aircraft centreline. Both values are found above

and in previous study. Therefore;

81

( )20

0

1764,705 103,806/ 17wing

WS ftW S

= = = = 9, 644m2

The aspect ratio has been chosen as 7 before. The span of the wing can be found now:

26,96 8,217wing wingb AR S ft m= × = =

We can find the root and tip chords by using the geometric equations below:

Taper ratio (λ) has been chosen as 0.4 before;

Croot = 2S / b(1+λ) =5, 5ft= 1, 677m

Ctip = λ Croot =0, 6708m

6.5.3. Tail Sizing

Previously, for the estimation of tail size it is used historical approaches. Here, it is

considered the tail size would be in some way the wing size because the primarily purpose of

a tail is to counter the moments produced by the wing. To estimate tail size is more difficult

because of importance of both area and tail moment arms. The effectiveness of a tail in

generating the necessary moment about the center of gravity is proportional to the lift

produced by the tail and to the tail moment arm.

First, it is required a method for initial estimation of tail size which is called “tail

volume coefficient”. It is shown below the tail volume coefficient both vertical and horizontal

tails.

VT VTVT

w w

L Scb S

= (6.10) [1]

HT HTHT

w w

L Scc S

= (6.11) [1]

For a vertical tail, the wing yawing moments which must be countered are most

directly related to the wing span (bw). This leads to the “vertical tail volume coefficient” as

can be seen Eq. (6.10) [1]. For a horizontal tail, the pitching moments which must be countered

are most directly related to the wing mean aerodynamic chord ( wC ). This leads to the

“horizontal tail volume coefficient” as shown by Eq. (6.11) [1].

82

Mean aerodynamic chord of the wing is:

wC =2/3 Croot (1+λ+λ2)/ (1+λ) = 1, 246m = 4, 088ft

From Table 6.4[1], for a general aviation - single engine aircraft, vertical tail volume

coefficient cVT is determined as 0.04. The tail arm is taken about 60% for an aircraft with a

front-mounted propeller engine.

LVT = 0.6L = (0.6) (7, 437m) =4, 4622m = 14, 64ft

After estimation of the vertical tail length, it is calculated the area of vertical tail. The

value of win span and wing area are found above. Also, vertical tail volume coefficient is

determined from Table (6. 4) [1]. According to these values, it can be found the area of vertical

tail by using the Eq. (6.10) [1] as given the below:

SVT = (26, 96) (103, 806) (0, 04)/ (14, 64) = 7, 65ft2

SVT = 7, 65ft2= 0, 71m2

To calculate the span and chord of tails, it is required aspect ratios of tails. The aspect

ratio values for horizontal tail are determined 5, and for vertical tail 1, 5m2.

Wing span for vertical tail;

bVT = 1.5 0,71 1,032VT VTA S m= × =

Mean chord length for vertical tail:

cVT = 1, 032/1.5 = 0, 688m

Also, the same estimations and calculations are made for the horizontal tail. From

Table 6.4[1], for a general aviation - single engine aircraft, horizontal tail volume coefficient

cHT is determined as 0, 7. The tail arm is taken again about 60% of the fuselage length.

LVT = 0.6L = (0.6) (7, 437m) =4, 4622m = 14, 64ft After estimation of the horizontal tail length, it is calculated the area of horizontal tail

by using the Eq. (6.10) [1] as given the below:

SHT = (4, 088) (103, 806) (0, 7)/ (14, 64) = 20, 29ft2

SHT = 20, 29ft2 = 1, 885m2

Wing span for horizontal tail;

bHT = 5 1,885 3,07HT HTA S m= × =

Mean chord length for horizontal tail:

cHT = 3, 07/5 = 0, 614m

83

6.5.4. Control Surface Sizing

The primary control surfaces are the ailerons (roll), elevator (pitch) and rudder (yaw).

Final sizing of these surfaces is based upon dynamic analysis of control effectiveness,

including structural bending and control system effects.

- Aileron;

Ailerons are typically about 15-25% of the wing chord. In this project, this ratio is

chosen as 20%. According to this value, it is estimated the aileron span over wing span ratio

from Fig (6.3) [1] which can be seen below. This historical guideline shows that the aileron

span over wing span ratio is between the ratios of 40-50% for 0, 20 aileron chord over wing

chord. From that, it is selected as 45%.

Figure 6.3 Aileron Guidelines [1]

For ca / cw = 0.20 and ba / bw = 0.45;

cw = 1, 246m caileron = ca = 1, 246×0, 20 = 0, 2492m

bw = 8, 217m baileron = ba = 8, 217 ×0, 45 = 3, 697m

- Flaps;

Flaps are typically about 15-25% of the wing chord, too. If it is performed the same

procedure for the flaps, it is chosen flap chord length to be %24 of wing chord length for plain

flap. Also, flap span over wing span ratio is taken as 40%.

84

For cf / cw = 0.24 and bf / bw = 0.40;

cw = 1, 246m cflap = cf = 1, 246×0, 24 = 0, 299m

bw = 8, 217m bflap = bf = 8, 217 ×0, 40 = 3, 287m

- Rudder and Elevators;;

Rudders and elevators are typically about 25-50% of the tail chord. Elevators and

rudders generally begin at the side of the fuselage and extend to the tip of the tail or to about

90% of the tail span. According to this information, Rudder chord length over vertical tail

chord length was chosen %35. Also, elevator chord length over horizontal tail chord length

was chosen %40.

For cr / cVT = 0, 40 and br / bVT = 0.90;

cVT = 0, 688m crudder = cr = 0, 688×0, 40 = 0, 2752m

bVT = 1, 032m brudder = br = 1, 032 ×0, 90 = 0, 9288m

For ce / cHT = 0.40 and be / bHT = 1;

cHT = 0, 614m celevator = ce = 0, 614×0, 40 = 0, 2456m

bHT = 3, 07m belevator = be = 3, 07 ×1 = 3, 07m

6.6. SUMMARY OF THE RESULTS

RUBBER ENGINE SIZING W1/W0 (Takeoff) 0.98 W2/W1 (Climb) 0.985 W3/W2 (Cruise) 0.936

W4/W3 (Descent) 0.995 W5/W4 (Loiter) 0.993

W6/W5 (Landing) 0.997 W0 1738, 3506lb We 1095, 14lb Wf 202, 6916lb

Required Horsepower 147, 76 hp

Table 6.5 Summary of Rubber Engine Sizing

85

FIXED ENGINE SIZING

Selected Engine Lycoming O-235 E2A Horsepower 150 hp

W1/W0 (Takeoff) 0.98 W2/W1 (Climb) 0.985 W3/W2 (Cruise) 0.9294

W4/W3 (Descent) 0.995 W5/W4 (Loiter) 0.99329

W6/W5 (Landing) 0.997 W0 1764, 705lb We 1107, 197lb Wf 216, 988lb

Range 525nm

Table 6.6 Summary of Fixed Engine Sizing

Table 6.7 Summary of Geometry Sizing

GEOMETRY SIZING Fuselage L = 7, 437m Dmax= 1, 4874m Wing Sw= 9, 644m2 bw = 8, 217m Croot = 1, 677 m Ctip = 0, 6708m Taper ratio = 0.4 Aspect ratio = 7 Horizontal tail SHT = 1, 885m2 bHT = 3, 07m cHT = 0, 614m Aspect ratio = 5 Vertical tail SVT = 0, 71m2 bVT = 1, 032m cVT= 0, 688m Aspect ratio = 1, 5 Ailerons ca / cW = 0, 20 ba / bW = 0, 45 ca = 0, 2492m ba = 3, 697m Elevators ce/cHT=0, 40 be/bHT= 1.0 ce = 0, 2456m be = 3, 07m Rudder cr/cVT=0, 40 br/bVT= 0, 9 cr = 0, 2752m br = 0, 9288m Flaps (plain) cf/cw=0, 24 bf/bw=0, 4 cf = 0, 299m bf = 3, 287m

86

6.7. CONCLUSION

In this part, the aim was to do calculations for initial sizing of general aviation-single

engine aircraft. In this sense, firstly, rubber engine sizing is performed and aircraft takeoff

weight is calculated. Then, the required power is found by using the horsepower-to-weight

ratio. Previously, the hp/W value was determined as 0, 075, but here it is decided to change

because of range calculation. According to 0, 075 hp/ lb, the range is found very higher than

design requirements. This means that the aircraft empty weight increases unnecessarily. The

higher range does not provide better performance. Therefore, the new hp/W value is estimated

as 0, 085. Thus, the range is found nearly design requirements. From hp/W ratio, an engine

which can support this power was selected. From that engine, a fixed engine sizing study was

done and new takeoff gross weight and new range value was calculated for this engine.

Finally, geometry sizing of aircraft is done. All areas and lengths for main parts of aircraft

like fuselage wings and tails were calculated. Same calculations were done for control

surfaces.

87

CHAPTER 7

CONFIGURATION LAYOUT AND INTERIOR DESIGN

7.1. INTRODUCTION The configuration layout and interior design is very important part of the design. In

this part includes all drawings of the aircraft such as wing, tail, control surfaces and whole

aircraft. Drawing means of the product of the aircraft. All of the calculations or analytical

tasks are very important, too. Drawing comes after these calculations of the aircraft. While the

analytical tasks are vitally important, the designer must remember that these tasks serve only

to influence the drawing, for it is the drawing alone that ultimately will be used to fabricate

the aircraft.

The purpose of this study is to make necessary calculations for determining the aircraft

configuration layout and draw the external geometry and interior design includes cockpit of

the aircraft.

7.2. WING AND TAIL SURFACES

From the previous studies, we can tabulate a table including the parameters for wing

and tail surfaces in order to make the necessary calculations. This table is shown below.

Table 7.1 General specifications of wing and tails

The following equations will be used to make the calculations for wing, horizontal and

vertical tails.

Span; b AS= (1) [1]

Root chord; CS

broot = +21( )λ

(2) [1]

Tip chord; C Ctip root= λ (3) [1]

Characteristics WING HORIZONTAL TAIL

VERTICAL TAIL

Airfoil NACA 23015 NACA 0015 NACA 652-015 AR (Aspect Ratio) 7 5 1,5

Λc/4 4° 6, 76° 40° Taper ratio ( λ ) 0,4 0, 6 0, 5

Twist - - - Area (S) (m2) 9, 644 1, 885 0, 71

88

Mean aerodynamic chord (M.A.C); C Croot=+ ++

23

11

2λ λλ

(4) [1]

Location of M.A.C; Yb

=++6

1 21

λλ

(5) [1]

After the calculations of these terms, initial sizing study is done. In this part, some

dimensions are found such as span, chords, main aerodynamic chord and location of M.A.C.

•

Wing ⇒ b = 8, 217m Horizontal Tail ⇒ b = 3, 07m Vertical Tail ⇒ b = 1, 032m

Span:

• Wing ⇒ Croot = 1, 677m; Ctip = 0.6708m Horizontal Tail ⇒ Croot 0, 614m;

Chords:

0,6 0,614 0,3684tip rootC C mλ= = × =

Vertical Tail ⇒ 2 2 0,71 0,917(1 ) 1,032 (1 0,5)root

SCb λ

×= = =

+ × +

0,5 0,917 0,458tip rootC Cλ= = × =

•

Mean Aerodynamic Chord (Ĉ):

( ) ( )22 1 / 13 rootc C λ λ λ= + + +

Wing ⇒ Ĉ = 1, 246m Horizontal tail ⇒ Ĉ = 0, 501m Vertical Tail ⇒ Ĉ = 0, 713m •

Wing ⇒

Location of M.A.C (Ŷ):

1 2 8,217 1 0,8 1,761 5,7786 1 6 1 0,4bY m ftλ

λ+ + = × = × = = + +

Horizontal Tail ⇒ 1 2 3,07 1 1,2 0,703 2,3066 1 6 1 0,6bY m ftλ

λ+ + = × = × = = + +

Vertical Tail ⇒ 2 1 2 2 1,032 1 1 0,4586 1,5056 1 6 1 0,5bY m ftλ

λ+ × + = × = × = = + +

89

The results are tabulated below;

Geometry b (m) Croot (m) Ctip (m) Ĉ (m) Ŷ (m) Wing 8, 217 1, 677 0, 6708 1, 246 1, 761

Horizontal Tail 3, 07 0, 614 0, 3684 0, 501 0, 703 Vertical Tail 1, 032 0, 917 0, 458 0, 713 0, 4586

Table 7.2 Main dimensions of the wing and tails

According to these parameters, we can draw the top view of the wing and tail surfaces.

Figure – 7.1 Top view of the horizontal tail

0.614

0.704

0.501

1.535

0.368

90

Figure 7.2 Side view of the vertical tail

Figure 7.3 Top view of the wing

0.458

0.713

0.917

1.032

0.67

1.761

4.109

1.677

1.246

91

When the wing is designed, it is considered that the length of the entering part to

fuselage as nearly 0, 30m. This part exists at the location of the maximum fuselage diameter.

However, the aileron and flap lengths sums are approximately 3, 4m. After flap, there is 0,

42m to the fuselage. In order to not close the flap to fuselage, it is selected the distance

between the flap and fuselage on wing is 0, 30m.

Flaps, ailerons, elevator and rudder dimensions are determined in previous study

“Initial Sizing”.

Flaps: Flaps are placed on the wing which is 40% of the span and 24% of the chord.

cf / cw = 0, 24; bf / bw = 0, 40

cw = 1, 246m and cf = 0, 24× 1, 246 = 0, 299m

bw = 8, 217m and bf = 0, 40× 8, 217 = 3, 2868m ( For each wing = 1, 6434m )

Ailerons: Ailerons are placed on the wing which is 45% of the span and 20% of the chord.

ca / cw = 0, 25; ba / bw = 0, 45

cw = 1, 246m and ca = 0, 20× 1, 246 = 0, 2492m

bw = 8, 217m and ba = 0, 45× 8, 217 = 3, 697m ( For each wing = 1, 84 m )

Elevator: Elevators are placed on the horizontal tail which is 100% of the span and 40% of

the chord.

ce / cw = 0, 40; be / bw = 1

ce = 0, 40× 0, 614 = 0, 2456m

be = 0, 100× 3, 07 = 3, 07m ( For each horizontal tail = 1, 535m )

Rudder: Rudder is placed on the vertical tail which is 90% of the span and 35% of the chord.

cr/cVT=0, 35; br / bw = 0, 90

cr = 0, 40× 0, 688 = 0, 2752m

br = 0, 90× 1, 032= 0, 9288m

92

Figure 7.5 The sketch of the wing with flap and aileron

Figure 7.6 The sketch of the horizontal tail with elevator

1.8485

1.643

0.2456

153.5

93

Figure 7.7 The sketch of the vertical tail with rudder

As explained in previous studies, NACA 23015 airfoil is used for the wing, both at the

root and tip. NACA 652015 airfoil is used for the vertical tail and NACA 0015 airfoil is used

for the horizontal tail, both at the root and tip. The wing has constant camber and thickness;

hence there is no need to show any auxiliary control lines. For horizontal and vertical tail

there is no camber. Wing has 2 degree and horizontal tail has -2 degree incidence angle.

Also, there is no twist distribution for wing and tails.

The plotting of the airfoils for wing and tails are shown below:

Figure 7.8 NACA 23015 profile

0.9288

0.2752

94

Figure 7.9 NACA 0015 profile

Figure 7.10 NACA 652015 Airfoil

Location of Center of Gravity Point:

For a stable aircraft, the wing should be initially located such that the aircraft center of

gravity is at about 30% of the mean aerodynamic chord. When the effects of the fuselage and

tail are considered, the center of gravity would be about 25% of the total subsonic

aerodynamic center of the aircraft [1].

The mean aerodynamic chord of wing is 1, 246m. According to above information, the

location cg point is 0, 3115m. The desired location of the cg point is shown in Figure 7.9

below.

95

Figure 7.11 The location of center of gravity point for wing 7.3. FUSELAGE AND INTERIOR DESIGN 7.3.1 Fuselage From previous study, fuselage length and maximum diameter have been found as;

L = 24, 4ft = 7, 437m and Dmax = 1, 4874m

7.3.2. Cockpit Design For cockpit design there are four important points to be considered:

The pilots and other crew members can reach all the controls comfortably.

The pilots and other crew members must be able to see all flight essential instruments

without undue effort.

Communication by voice and by touch must be possible without undue effort.

Visibility from cockpit must adhere to certain minimum requirements.

Cockpit will be designed for two crew members, pilot and co-pilot. For two crew member

including configurations, 100inch (2.54m) cockpit length can be considered. [1] Pilots are

96

located by observing requirements for pilot visibility and for pilot ability to reach important

control and instruments. It is also considered a smooth exterior, 20º unobstructed vision

angle above the horizontal axis considering the pilot’s eye and 15º vision over the nose from

the front windshield. Besides, Lc (length between the pilot’s eye and the windshield) value is

generally between the 0, 5 and 0, 6m. According to this interval, Lc is considered about 0.5m.

Cockpit width is considered as 1,08m. For cockpit seats, economy class ones are chosen and

seat width is considered as 0, 45m table 9.1[1]. Also, there is 0, 2m space between the seats.

Figure 7.12 Front view of cockpit 7.3.3. Wetted Area Determination

Aircraft wetted area, is the total exposed surface area, and it must be calculated for drag

estimation. In order to find the wetted area first the exposed area must be found, referring to

section 7.10[1], it is mentioned that the exposed area is found by dividing the reference wing

area by the cosine of the dihedral angle. After the exposed area is found the wetted area is

found by using equation 7.11[1] for t/c > 0, 05. All of thickness ratios are t/c = 0, 15 for this

project. Therefore;

* Wing Wetted Area

Sexposed = Sreference / cos (dihedral) = 9, 644 / cos (0°) (dihedral angle = 0°) = 9, 644m2

0.45

0.45

1.08

0.40

FLOOR

SEAT

SEAT

0.2

97

Because of no dihedral, exposed area is determined only by considering that the wings

reference area minus the part of the wing covered by the fuselage. Thus;

Sexposed = 8, 87m2

Swet = Sexposed [1.977+0, 52(t/c)] = 8, 87 [1.977+0, 52(0, 15)]

Swet, wing = 18, 23m2

* Horizontal Tail Wetted Area

Sexposed = Sreference / cos (dihedral) = 1, 885 / cos (0°) (dihedral angle = 0°) = 1, 885

Sexposed = 1, 73m2

Swet = Sexposed [1.977+0, 52(t/c)] = 1, 73 [1.977+0, 52(0, 15)]

Swet, horizontal tail=3, 56m2

*

Vertical Tail Wetted Area

Sexposed = Sreference / cos (dihedral) = 0, 71 / cos (0°) (dihedral angle = 0°) = 0, 71

Sexposed = 0, 65m2

Swet = Sexposed [1.977+0, 52(t/c)] = 0, 65 [1.977+0, 52(0, 15)]

Swet, vertical tail = 1, 33m2

*

3, 42

top sidewet

A AS

+ ≅

Fuselage

For the fuselage the wetted area was found by using equation 7.12 [1]. By using Catia,

for Fig. 12 the top area (Atop) of the fuselage is found to be 8, 05m2 and the side area

(Aside) is 7, 226m2. Therefore,

Swet, fuselage = 25, 969m2

Total Wetted Area

Swet, total = 18, 22 + 3, 56 + 1, 33 + 25, 969

Swet, total = 49, 08m2

98

Aircraft Internal Volume

Aircraft internal volume is determined by using equation 7.13 [1]. The top area and side

area are found above. These data are put into the equation, it is calculated the internal volume.

The calculation is given below:

3,44

top sideA AVol

L×

≅

Vol = 6, 648m3

7.4. FUEL TANKS LOCATION For the aircrafts of competitor study, the fuel tanks are usually being in the wings as a

wing tanks. Therefore, it is considered the fuel tanks in each wing.

The power-plant for this project; one 150 hp Lycoming O-320 E2A four cylinder air-

cooled horizontal opposed engine driving a two blade fixed pitch McCauley propeller. Fuel

contained in two integral wing tanks.

Figure 7.13 Fuel tanks location in wings

99

7.5. DRAWINGS OF THE TOP, FRONT AND SIDE VİEWS OF THE AIRCRAFT

Figure 7.14 Top view of the aircraft

Figure 7.15 Front view of the aircraft

100

Figure 7.16 Side view of the aircraft 7.6. SUMMARY OF RESULTS

Wetted Area Determination Swet,wings 19, 82m2

Swet,horizontal tail 3, 87m2 Swet, vertical tail 1, 46m2 Swet, fuselage 25, 969m2 Swet,total 51, 12m2 Atop 8, 05m2 Aside 7, 226m2 Volume 6, 648m3

Table 7.3 Wetted Area Determinations

101

Table 7.4 General specifications of wing and tails

Table 7.5 Geometry sizing

Characteristics WING HORIZONTAL TAIL

VERTICAL TAIL

Airfoil NACA 23015 NACA 0015 NACA 652-015 AR (Aspect Ratio) 7 5 1,5

Λc/4 4° 6, 76° 40° Taper ratio ( λ ) 0,4 0, 6 0, 5

Twist - - - Dihedral 0° - - Incidence 2° - 2° - Wing Tip Sharp Sharp -

Wing Vertical Location Mid High -

GEOMETRY SIZING Fuselage L = 7, 437m Dmax= 1, 4874m Wing Sw= 9, 644m2 bw = 8, 217m Croot = 1, 677m Ctip = 0, 6708m Taper ratio = 0.4 Aspect ratio = 7 Horizontal tail SHT = 1, 885m2 bHT = 3, 07m CHT(mean) = 0, 501m Aspect ratio = 5 Vertical tail SVT = 0, 71m2 bVT = 1, 032m CVT(mean) = 0, 713m Aspect ratio = 1, 5 Ailerons ca / cW = 0, 25 ba / bW = 0, 45 ca = 0, 2492m ba = 3, 697m Elevators (for horizontal tai) ce/cC=0, 40 be/bHT= 1, 0 ce = 0, 2456m be = 3, 07m Rudder (for vertical tail) cr/cVT=0, 35 br/bVT= 0, 90 cr = 0, 2752m br = 0, 9288m Flaps (plain) cf/cw=0, 24 bf/bw=0, 40 cf = 0, 299m bf = 3, 2868m Location of cg 0, 3115m

102

7.7. CONCLUSION

In this chapter, it is done the configuration layout and interior design for the aircraft.

To be formed whole aircraft, geometry sizing is made. For geometry sizing, it is determined

some dimensions of the aircraft such as wing and tails chords, locations of center of gravity of

the aircraft etc. Almost all of the geometric parameters were shown on technical drawings of

wing and tail surfaces. Then, fuselage and interior design or cockpit design are done basically.

The drawings are not complex and difficult; on the contrary, all of them are made simply.

During the drawings, it is considered to reach the best possible configuration for the aircraft.

After that, wetted area and aircraft internal volume are determined which is important in drag

calculations. Finally, it is contemplated the fuel tanks exist in the wings.

103

CHAPTER 8

PROPULSION AND FUEL SYSTEM INTEGRATION

8.1. INTRODUCTION

In this chapter, it is intended to analyze the propulsion system for desired aircraft. It is

wanted to select the most suitable propulsion system. Also, it will be determined the propeller

sizing and tried to select the propeller for the aircraft according to the found specifications.

Then, the propulsion system and propeller will be installed to the aircraft. Eventually, the fuel

system will be analyzed and the fuel tank will be defined.

8.2. PROPULSION SYSTEM SELECTION

There are many options for aircraft propulsion. These propulsion system options are

shown in the Figure 8.1 [1]. According to this figure, the most suitable one should be selected

for desired aircraft at given conditions. All aircraft engines operate by compressing outside

air, mixing it with fuel, burning the mixture, and extracting energy from the resulting high-

pressure hot gases. However, the engine types have some advantages and disadvantages with

respect to each other and their specific properties determine the application areas of each

engine. Therefore, here, it is discussed which engine is selected and why selected because the

propulsion system selection is important for the aircraft. Firstly, when it is considered the

Figure 8.2[1], it can be seen easily that aircraft design Mach number determines the engine

type. For lowest Mach number (for the project, M is found nearly 0, 2), the best engines are

both piston-prop and turboprop. The engine choice is reduced from ten to two. It is preferred

the piston-propeller engine instead of turboprop. Piston-props generally are cheaper than the

turboprops. Also, piston-props use lowest fuel consumption. However, turboprops use more

fuel for the same horsepower. Although producing a lot of noise and vibration, it is selected

the piston-prop engine for the design project because still the piston-props are the best choice

for the lighter aircrafts.

104

Figure 8.1 Propulsion system options [1]

Figure 8.2 Propulsion system speed limits [1]

8.3. SELECTED ENGINE PROPERTIES

The engine has been chosen as Textron Lycoming O -320 E2A for the aircrafts in

previous study. The selected engine is still the same. The detailed properties of this engine

which was taken from Jane’s All the World’s Aircraft and Wikipedia are tabulated below:

105

MANUFACTURER LYCOMING ENGINE (Pennsylvania) MODEL O-320 E2A

Type Four cylinder air-cooled horizontal

opposed engine

Propeller Drive Two blade fixed pitch McCauley

propeller Max Diameter 74in = 1879, 6mm Bore 130mm (5, 118in) Weight, Dry 244lb (110, 7kg) Height 22,99in = 0, 58m = 1, 92ft Width 32,24in = 0,81m = 2,69ft Length 29, 56in = 0,75m = 2,46ft Cylinders (stroke) 98, 4mm Cylinders (capacity) 5200cc Power (at T-O) 150 hp Max Continuous 160 Max Cruise Rating 140 Specific Fuel Consumption (at T-O) 0, 42 lb/h/lb Core RPM 2700 rpm Compression Ratio 07:01

Combustion Chamber Aluminum-alloy casting and a fully

machined combustion chamber

Table 8.1 Engine Specifications [6, 7]

8.4. PROPELLER – ENGINE INTEGRATION

In this part, the propeller and engine dimensions are determined. Then, propeller

location and engine installations are defined. Also, required inlets and exhausts are

determined.

8.4.1. Propeller Sizing The actual details of the propeller design such as the blade shape and twist are not

required to lay out a propeller-engine aircraft. Instead of, it should be determined the propeller

diameter, engine dimensions and inlet, exhausts. In propeller sizing part, it is found the

propeller diameter for two conditions which are tip speed and hp. Propeller diameter is

function of hp. Also, tip speed is related to the length of the propeller blade. It should be kept

106

below sonic speed [1]. According to these; the smallest one is chosen for the propeller

diameter. Generally, larger propeller diameter, the more efficient the propeller will be [1].

However, if the propeller is larger, it is increased the drag and weight. Therefore, it is selected

the two blade.

Propeller diameter estimation for tip speed (Vtip) static = πnd / 60 (ft/s) (8.1) [1]

n = rotational rate (rpm)

d = diameter

:

( )2

2

60tip forwardhelical

ndV Vπ = +

(8.2)[1]

To calculate the helical speed, first it should be found the forward speed which is

related to the cruise speed. The cruise speed for design requirements is 130 knots = 219,

44ft/s. Also, other cruise condition is the altitude of the cruise. The cruise altitude has been

given as 8000ft in the previous study.

1/2 ρcruise Vcruise = 1/2 ρSL VSL2

ρcruise = 0.001869 slug/ft3 (@ 8000ft)

ρSL = 0.00238 slug/ft3

It is determined the forward speed which is the same sea level speed by using the

above equation:

VSL = VForward = 194, 5 ft/s

The metal is thought as a propeller material. At sea level the helical tip speed of a

metal propeller for low noise should not exceed 700ft / s.

From (8.2) [1], it is determined the required propeller diameter. For n = 2700rpm;

( )2

2.2700.700 194,560

dπ = +

By solving this equation, the propeller diameter is found as 1, 45m.

d = 4, 756ft = 1, 45m

107

441,83 1,83 150 6,4 1,95d Hp ft m= = × = =

Propeller diameter estimation for HP(for two blade):

d = 6, 4ft = 1, 95m

From these two metal blades propeller diameters, the smaller one is selected because

of the tip speed limitation. Therefore; the propeller has two blades and diameter is;

d = 4, 756ft = 1, 45m

Now, it is required to decide that the propeller has fixed or variable pitch. A variable-

pitch propeller can be used to improve thrust. Therefore, variable pitch makes positive effects

for different cruise speeds, but in this design the variable pitch is unnecessary because of the

constant cruise speed. Thus, it is decided to use for the aircraft a fixed pitch propeller is

called a cruise prop or climb prop. Also, the selected engine Lycoming O-320 E2A uses the

two blades fixed pitch McCauley propeller. This information is considered in the decisions,

too.

Spinner exists on the most of the aircrafts. The shape of spinner is generally cone or

bullet. And spinner mission is pushing the air out to where the propeller is more efficient.

Ideally, the spinner should cover the propeller out to about 25% of the radius [1]

Rspinner = 0.25Rpropeller = (0, 25) (1, 45/2) Rspinner = 0, 18125m = 0, 055ft

To sum up, the specifications of the propeller are: two metal blades, fixed pitch

propeller with spinner.

8.4.2. Propeller Location

The propeller locations matrix is shown in the figure below. According to it, it can be

seen there is different location for the propeller. For example, the propeller can be settled on

the fuselage, wing, tail or front, behind, top etc. For this project, it is decided the tractor

location. Tractor engine location offers a fuselage mounted engine configuration. Tractor

location is the standard for most of the history of aviation. The heavy engine is in the front for

the tractor location. This causes the shorter fore body, and allowing smaller tail area and

improved stability. Also, whether water or air cooled, front placement offers the most

efficient position for cooling for tractor location. Because of these causes, for this general

aviation aircraft, the conventional tractor engine location is selected.

108

Figure 8.3 Propeller locations [1]

8.4.3. Engine – Size Estimation

In this part, it is made engine-size study and the results are compared with the real

engine size (Lycoming O-320 E2A) which has been selected before. To product of an engine,

it is considered four different type propeller powerplant. One of them is the horizontally-

opposed piston engine. The selected engine (Lycoming O-320) is an opposed engine, too.

Opposed piston engine sees most use today.

The estimation of the engine size is found by using Table 10.4[1], for a piston-prop

engine. The calculations are given below:

X= a (bhp) b (8.3) [1]

• Weight: X=W= (5, 47)(150)0, 780 = 272, 48lb

• Length: X=L= (3, 86)(150)0, 424 = 32, 3in = 0, 82m = 2, 7ft

• Diameter: X=D= 23in = 1, 92ft = 0, 58m

By comparing the estimated engine values and real engine values;

Estimated Value Real Value (from table 8.1)

Weight, kg 272, 48lb = 123, 71kg 244lb = 110, 7kg Length, m 32, 3in = 2, 7ft = 0, 82m 29, 56in = 2,46ft = 0,75m

Diameter, m 23in = 0.58m 22, 99in = 0, 58m

Table 8.2 Comparing of estimated values and real values

109

It can be seen that the obtained values for engine are very close to the selected real

engine. Even, the diameters of both are the same. The estimated values of the engine length

and weight are a little much than the selected engine.

8.4.4. Piston Engine Installation

There are special installation requirements for the piston engines. These installation

requirements have many effects on the configuration layout. For this design, tractor engine

location is selected. Therefore, the engine exists on the front part of the fuselage. Thus, the

cooling-air intake is located directly in front of the engine cylinders. By baffles which are flat

sheets of metal, the air diverted over top of the engine to engine compartment. There are two

types of the air-cooling: down-draft and up-draft. Down-draft cooling is selected for this

aircraft because up-draft cooling can heat up the cabin by dumping the hot air in front of the

windscreen. Also, windscreen can be covered by the engine oil leak. And, exists air is into a

high pressure region for down-draft cooling. Because of these factors, down-draft cooling is

chosen although it has a poor place for the exit air.

Figure 8.4 Tractor locations [1]

Figure 8.5 Down-draft cooling [1]

8.5. FUEL SYSTEM

An aircraft fuel system includes the fuel tanks, fuel lines, fuel pumps, vents and fuel-

management controls.

There are three types of fuel tanks: discrete, bladder, integral. Discrete tanks are

normally used only for small general aviation and homebuilt aircrafts. Bladder tanks are made

110

by stuffing a shaped rubber bag into a cavity in the structure. Rubber bag of the bladder tank

is thick, so 10% of the available fuel volume is lost. And, bladder tank is used for self-sealing.

Integral tanks are areas inside the aircraft structure that have been sealed to allow fuel storage.

Integral tanks are part of the aircrafts structure; therefore, they cannot be removed for service

or control. The most commonly used and efficient fuel tank for the competitor aircrafts are the

integral tanks. Therefore, integral fuel tank configuration will be used for the design of the

aircraft. The wings are used for the fuel tanks. Required fuel tank volume was determined by

the mission fuel. Mission fuel weight is;

Wf = 216, 988lb = 98, 5kg

Total fuel weight will be stored in both each sides of the wings. Each wing includes

two different fuel tanks because of the landing gear location. If only one tank would be in the

wing, there was no location for the landing gear. For each side of the wing, the mission fuel

weight is the half of the total weight:

Mf1=Mf2 =Wf / 2 = (216, 988)/2 = 108, 5lb = 49, 3kg

To find the capacity of the fuel tanks, the areas are separated on airfoils A1 and A2, by

distance H, the following formula is used;

Ai = 0.9(0.5c) (t/c) (c) (8.4) [1]

According to the formula, in total it is used the half of the airfoil chord length for fuel.

(From 0.15c to 0.65c, other parts are for flaps, ailerons, actuators etc.). Also, for each tank,

the cross sectional area loss due to airfoil curvature is taken as 10% of the rectangular area

shown.

By defining two fuel tanks for each side of the wing; • Volume of the tank from airfoil (A1) to airfoil (A2), which is H1 distance apart, is;

11 1 1 2 2( )

3HV A A A A= + + (8.5)[1]

When fuel tank volume is calculated for tank, the thickness ratio is 15% due to the

wing airfoil. Each wing has one fuel tank. It should not divide each tank into two separate

parts because the landing gear is considered on the fuselage. Therefore, it is not require

making a room for the landing gear.

111

13,7 (0,0076 (0,0076) (0,0384) 0,0384) 0,0783

V = × + × + =

Fuel tank:

The chord length at the tip of the wing (fuel tank end section) is 0, 335m. And, the

chord length of the fuel tank beginning section is 0, 754m. From equation (8.4) [1];

A1 = 0, 9 (0, 1675) (0, 15) (0, 335) = 0, 0076m2

A2 = 0, 9 (0, 377) (0, 15) (0, 754) = 0, 0384m2

H1 (perpendicular distance between A1 and A2) = 3, 7m

m3

The total volume of the fuel tanks; Vtotal =2Vwing= 0, 156m3

The density of the fuel is taken as 785 kg/m3, also assuming that 15% volume for wing

tanks is lost due to structural elements inside the tank, and the weight is;

MTotal = (0, 85) (785) (0, 156) = 104, 1kg

This obtained value is really good value for this design. The required mission fuel

weight is 98, 5kg. However, for integral tanks, there is foam in the wing to prevent fire and

leak. This foam covers 2.5% of the volume. Also, 2.5% of the fuel is absorbed by the foam.

When these losses are considered, remaining fuel is nearly the same the mission fuel.

112

Figure 8.4 Fuel tank in the wing

8.6. SUMMARY OF RESULTS

PROPELLER Diameter 1, 45m

Spinner radius 0, 18125m Blade number 2

Type fixed - pitch Location tractor

ENGINE Weight 123, 71kg Length 0, 82m

Diameter 0, 58m Installation down-draft

FUEL SYSTEM Type integral

Volume 0, 156m3

Table 8.2

0,335m

0,754m

3,7m

113

8.7. CONCLUSION

In this study, propulsion system and fuel integration trainings are done. The propeller

designations and location are determined. Also, engine and fuel system are defined. Engine

dimensions and installation are made. And, the selections and decisions for the fuel system

are discussed. In order to do this, wing fuel tank volume is calculated by considering the

technical drawings. Thus, it is checked the wing fuel tank is sufficient. Finally, the estimated

engine dimensions are compared with the actual values such as the aircrafts in the competitor

study.

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CHAPTER 9

LANDING GEAR AND SUBSYSTEMS

9.1. INTRODUCTION

In aviation, the landing gear is the structure that supports an aircraft on the ground and

allows it to taxi. Landing gear usually includes wheels. The design of the landing gear is one

of the more fundamental aspects of aircraft design. The design and integration process

encompasses numerous engineering disciplines, e.g., structures, weights, runway design and

economics. Landing gear must not cause any trouble for the structure of aircraft and it must

provide a safe landing. In this chapter, it will be discussed that the design and installation of

landing gear and other subsystems

9.2. LANDING GEAR ARRANGEMENT

Since many years, various arrangements have been used for wheels and structures to

connect them to the airplane. For example, tail dragger, single main, bicycle, tricycle or multi-

bogey. Single main landing gear is especially for sail planes due to its simplicity. Another

type is multi-bogey which is for heavy aircraft. This type is not considered because desired

aircraft is lighter. Now, the most popular landing gear arrangements are the tricycle landing

gear arrangement. The tricycle arrangement has one gear strut in front, called the nose wheel,

and two or more main gear struts slightly aft of the center of gravity. This type of landing gear

makes the aircraft easier to handle on the ground and it also makes landings much safer. The

main advantage of this layout is that it eliminates the ground loop problem of the tail dragger.

Also, although tail dragger is conventional type and has less drag and weight, it is unstable on

the ground because the center of gravity locates behind of the wheel. However, tricycle

landing gear arrangement is instead a stable design because of the location of the main gear

with respect to the center of gravity. It also improves forward visibility and permits a flat

cabin design. Because of these causes, in this design project, the tricycle landing gear

arrangements with one main unit (strut), two wheels on each side and one wheel on the

forward (steer able) is selected. Tricycle arrangement is shown in figure [1] below:

115

Figure 9.1 Tricycle landing gear arrangements [1]

9. 3. TIRE SIZING The tires are sized by considering the aircraft weight. It should be carried the weight of

the aircraft. Generally main wheels carry the most of the aircraft total weight about 90%.

Although the nose wheel carries only about 10% of the static load, it is exposed to higher

dynamic loadings at landing. Static load on landing gear are shown in the Figure 9.2[1]:

Figure 9.2 Wheel load geometry [1]

Calculation of the static loads and dynamic load are indicated by the following

equations:

main(Max Static Load) aNWB

= ⋅ (9.1)[1]

nose(Max Static Load) fMW

B= ⋅ (9.2)[1]

nose(Min Static Load) aMWB

= ⋅ (9.3)[1]

116

nose10(Dynamic Breaking Load) H W

g B⋅ ⋅

=⋅

(9.4)[1]

By using the above equations, it is calculated the static loads on the tires and dynamic

load on the nose tires under a 10ft2/s braking deceleration. Na, Nf, Ma, Mf, H and B are

dimensions for gear landing geometry are shown in Fig. 9.2[1]. These static and dynamic loads

are divided by the total number of main or nose tires to get the load per tire “WW (weight on

wheel)” which is used for tire selections.

From the historical data, it is known:

Mf – Ma = 0.2 c ( c is the mean aerodynamic chord)

Mf = 0.15 B Ma = 0.08 B

B = wheel base

H = Ground and cg distance

Mf = the distance between main wheel and forward cg point

Ma = the distance between main wheel and aft cg point

Also, the mean aerodynamic chord (M.A.C) was found in early studies as 1,246m = 4, 088ft.

Mf – Ma = 0.15B – 0.08B = (0.2)(4, 088)

Thus;

H is chosen as 2, 20m = 7, 22ft, by considering the center of gravity is on the

centerline of the fuselage. This value is a normal value for this type of aircraft because the

average of height overall for the competitor aircrafts is determined as 2, 38m.

8-15% of the aircraft’s weight should be carried by the nose wheel for most-aft and

most-forward cg points. For this aircraft, the weight was found as 1764, 205lb = 800, 1kg

before. Now, the static and dynamic loads are determined as given below. It can be seen that

it is assumed nearly 92% of the weight for main maximum static load. Also, for nose, the

maximum static load is 15% and the minimum static load is 8% of the aircraft’s weight.

main(Max Static Load) 0.92aNW WB

= ⋅ = ⇒ Na= 0.92(11, 68) Na = 10, 75ft = 3, 28m

nose(Max Static Load) 0.15fMW W

B= ⋅ = ⇒ Mf = 0.15(11, 68) Mf = 1, 752ft = 0, 53m

B = 11, 68ft = 3, 56m

117

nose(Min Static Load) 0.08aMW WB

= = ⇒ Ma= 0.08(11, 68) Ma = 0, 93ft = 0, 285m

Nf + Mf = B ⇒ Nf = B – Mf = 11, 68 – 1, 752 Nf = 9, 928ft = 3, 026m

The values of the dimensions for gear landing geometry are shown in Fig. 8.3:

Figure 9.3 Wheel load geometry with dimensions

To add %25 additional load to gears for later growth of the design process is very

common. The aircraft the off gross weight was found as 1764, 705lb. If it is wanted to make a

recalculation the loads for this 25% additional load to gears:

(Max Static Load) main =1,25 1,251764,705 (0,92)

2 2aNW

B= × × = 1014, 7lb

(Max Static Load) nose = 1,25 1764,705 (0,15) 1,25fMW

B= × × =330, 8lb

(Min Static Load) nose = 1,25 1764,705 (0,08) 1,25aMWB

= × × = 176, 47lb

(Dynamic Braking Load) nose=10 10(7,22)(1764,705)1,25 (1,25)

(32,2)(11,68)HWgB

= = 423, 47lb

The statistical tire sizing can be determined from Table 11.1[1] for general aviation

aircrafts:

118

Ww is the weight carried by tire. It is the maximum static load for whole wheels.

• Diameter: A = 1, 51

B = 0, 349 Dmain = 16, 91in. = 0, 4295m

WW = 1014, 7lb

D = AWWB

• Width: A = 0, 7150

B = 0, 312 wmain = 6, 2in. = 0, 1575m

WW = 1014, 7lb

Nose tires can be assumed to be about 80% the size of the main tires.

Dnose = 0.8Dmain = (0.8) (0, 4295) Dnose = 0, 3436m = 12in

wnose = 0.8 wmain = (0.8) (0, 1575) wnose = 0, 126m = 4, 96in

In landing, brakes absorb kinetic energy from aircraft. The brake kinetic energy

absorption requirements must during landing at the design landing weight. For this design,

landing weight is 100% of the takeoff weight (Wlanding = 1764, 705lb). Brake conditions are

important for the estimation of the rim diameter. It is assumed that brakes are applied when

aircraft has stall speed. Thus, by assuming that the brakes are applied when the aircraft goes

to stall speed, it can be ignored that the energy loses because of aerodynamic drag and thrust

reversing. Stall speed was found in previous study for landing configuration as 89, 07ft/s.

w = AWWB

21 12 2

stallbraking

WVKEg

= (9.5)[1]

21 1764,705 (89,07) 108697

4 32,2brakingKE ft lbf×= × = ⋅

The wheel diameter is related to the kinetic energy per braked wheel. By using Figure

9.4 [1], for general aviation, the wheel diameter can be found as nearly 7in. = 0, 178m.

119

Figure 9.4 Wheel diameters for breaking [1]

Table 9.1 Tire data [1]

By using the Table 9.1 [1], it is done the more suitable selection of tire. According to

this table, it can be seen the Type III tire data supplies all conditions for the gear loads and

dimensions. For desired aircraft, it is chosen the 7.00-8 size categories for main tires.

120

The total dynamic braking load for nose should be divided by 1, 3 to find total

dynamic nose wheel load.

Total dynamic nose wheel load = (423, 47) / 1.3 = 325, 75lb

According to this value; the Type III, 5.00-4 tire is selected for nose wheel. The

properties of these tires are tabulated below:

Tire Max Load (lb)

Max Width

(in)

Max Diameter

(in)

Rolling Radius

(in)

Wheel Diameter

(in)

Inflation Pressure

(psi) Main 7.00-8 2400 7.30 20.85 8.3 8 46

Nose 5.00-4 1200 5.05 13.25 5.2 4 55

Table 9.2 Selected tire properties

After the selection of main and nose tires, the pressure checks for them must be

performed. To do this, it is calculated foot print area Ad for tires by using the given formula:

2,32p rdA wd R = −

(9.6)[1]

The terms in this formula: w is tire width, d is tire diameter, and Rr is rolling radius.

Also, weight carried by tire is found by inflation pressure times contact area (foot print area).

These parameters are taken like in the table such as new diameter is 20, 85m for main tires

and 13, 25m for nose tires. Then, pressure is found and compared with the value given in the

table above.

w PW PA= (9.7)[1]

• 20,852,3 2,3 7,30 20,85 8,3

2 2p rdA wd R = − = × − =

Main tires;

60, 3in2

P = Ww/Ap = 1014, 7 / 60, 3 = 16, 82psi < 46psi

P = 16, 82psi is lower than the inflation pressure of main tires which is given in the

Figure 9.4[1] as 46psi. Also, the determined pressure value is compared with the value which

is given in the Table 11.3[1] for major civil airfield. According to this table, the inflation

121

pressure should not be passed the 120psi. For desired aircraft, it is found the inflation pressure

is 16, 82psi. Therefore, the pressure check is ok.

•

13,252,3 2,3 5,05 13,25 5,22 2p rdA wd R = − = × − =

Nose tire;

26, 81in2

P = Ww/Ap = 423, 47 / 26, 81 = 15, 8psi < 55psi

It can be said the same thing for the nose tire. The inflation pressure of nose tire is

lower than the value of given in the table which is 55psi. Also, this value is smaller than the

120psi explained above for the main tires.

9.4. SHOCK ABSORBER

Shock absorbers are an important part of aircraft landing gear. The landing gear must

absorb the occurring shock during the bad landing and smooth out the ride when taxiing. A

shock absorber is named “damper” in technical use because it is a mechanical device designed

to smooth out or damp shock impulse, and dissipate kinetic energy.

There are many different types gear shock absorber. One of them is the solid spring

gear arrangement. Solid spring gear is used in much general aviation. In this design, it is

decided to use the solid-spring gear for shock absorber. This type is very simple and

economic although slightly heavier than other types of gear.

Figure 9.5 Solid-Spring gear/shock arrangements [1]

9.4.1. Stroke Determination

The stroke is directly related to the some parameters such as the vertical velocity of

aircraft at touchdown, shock absorber material and amount of wing stall after touchdown.

The vertical velocity at touchdown is established in various specifications for different types

of aircrafts. Most aircrafts require 10ft/s vertical velocity capability. To determine the stroke it

is used the following equation:

122

2

2vertical T

Tgear

VS Sg N

ηη η

= − (9.8)[1]

where, (from table 11.4[1]) η(Shock-absorber efficiency) = 0, 62 (steel coil spring) ηT (tire efficiency) = 0, 47 N gear (gear load factor) = 3 (from table 11.5[1] for general aviation) Vvertical = 10 ft/s = 120 in/s

For tires, it is assumed that the tires deflect only to its rolling radius; the stroke for

tires (ST) is calculated as given below:

For main tires: ST = (d/2-Rr) = (20, 85/2 – 8, 3) = 2, 125in = 0, 177ft For nose tire: ST = (d/2-Rr) = (13, 25/2 – 5, 2) = 1, 425in = 1, 119ft

Main tires:2 210 0,47 0,177 0,70 8,4 .

2 2 32,2 0,62 3 0,62vertical T

Tgear

VS S ft ing N

ηη η

= − = − × = =× × ×

Smain = 8, 4in. = 0, 7ft = 0, 21m

Nose tire:2 210 0,47 1,119 0,745 8,93 .

2 2 32,2 0,62 3 0,62vertical T

Tgear

VS S ft ing N

ηη η

= − = − × = =× × ×

Snose = 8, 93in = 0, 745ft = 0, 226m

The calculated value for stroke supply that the minimum condition which is said that a

stroke of 8in, is usually considered a minimum, and at least 10-12in is desirable for most

aircraft. The stroke value for both tires are higher than 8in. This is good result. Also, nose

wheel stroke is generally either equal or slightly larger than main tires. This situation provides

a smooth ride while taxiing. The estimated nose stroke value is 8, 93in > 8, 4in (main tires).

This result is good, too.

9.4.2. Solid-Spring Gear Sizing

The deflection geometry for a solid-spring gear leg is shown in the Figure 9.6.

2T rdS R = −

123

Figure 9.6 Solid spring gear deflection [1]

The purpose of this part is to calculate the length of the gear leg. To do this, it should

be make some calculations. First of all, in the previous part, the stroke was found. Stroke is

the vertical component of the deflection of the gear leg. The gear leg is considered as a

constant-cross-section bending beam using the average values of beam width (w) and

thickness (t). Therefore; by assuming the gear leg and height, it can be tried to estimate the

most suitable w and t for desired gear leg.

After some researches about the competitors and other trainer aircrafts, it is decided to

take the height of the gear is 27, 5in = 0, 7m and is 40deg. From these values, it can be

found the gear length:

l= H / cos= 27, 5/cos (40) = 36in = 0, 91m

l (length of gear leg) = 0, 91m

Figure 9.7 Reference dimensions for landing gear under no load condition [8]

124

By using these all of the values, it can be estimated the gear leg width and thickness.

To estimate the w and t, it will be calculated the moment of inertia. Here, there are some

calculations for it.

Because of there are two gear legs, the force of each leg is defined as given below:

2gear

s

WNF = , thus; 1764,705 3 2646,31

2sF ×= = lb (9.9) [1]

The component of the load on the gear that is perpendicular the gear leg is;

F = Fs (sin) (9.10) [1]

F = 2646, 31. (sin40) = 1700lb

The deflection which is perpendicular to the gear leg is related to the stroke. This

relation can be shown as following:

S = y.sin () (9.11) [1]

The gear leg is assumed a constant-cross-section bending beam. Thus, it is calculated

the deflection “y” from the structural bending-beam equation:

3

3FIyEI

= (9.12) [1]

If the Eqs. (9.9 – 9.11) are put into the deflection “y” equation (9.12), it is obtained the

stroke “S” of a solid-spring gear leg:

( )3

2sin3sIS FEI

θ= (9.13) [1]

• E = Elasticity modulus (~ 30 million psi for steel)

• I = beam’s moment of inertia

For a rectangular-cross-section gear leg, the moment of inertia is;

3

12wtI = (9.14) [1]

To determine the moment of inertia it is used the Eqs. 9.13;

( )3

2sin3sIS FEI

θ= = 8, 4in

( ) ( )332 2 4

6

36sin (2646,31)(sin 40) 0,0675

3 3 30 10 8,4slI F inES

θ= = =× × ×

125

If the moment of inertia is I = 0, 0675in4, it can be assumed width and thickness of

gear leg: 3

12wtI = = 0, 0675in4 wt3 = 0, 81 in4

Now, the suitable w and t are estimated according to the above results. Thus, if t is

assumed as 0, 6in, the width becomes 3, 75in. It is thought that these values are suitable for

desired aircraft. Now, all dimensions for the gear and gear leg are shown below:

Height = 27, 5in Length = 36in

w = 3, 75in t = 0, 6in

Also, it is required to calculate the length of the nose tire. It is known the mean tire

diameter and the height.

D = 20, 85in = 0, 53m

H = 0, 7m

If it is added to height of main gear the main gear radius, it is found the distance from

the ground. Then, when it is subtracted the nose gear radius, it can be calculated the height of

the nose gear. The length of the nose gear is equal to the height. Therefore;

l(nose) = 0, 795m = 31, 3in

9.5. CASTORING WHEEL GEOMETRY

It is important especially for ground steering that the tire wheels and nose wheels must

be capable of being castored. Castoring wheel adds static and dynamic loads to the system

and can cause stability problems like wheel shimmy. Wheel shimmy should be prevented.

Therefore, rake angle and trail are used to prevent it.

126

For most tricycle-geared aircraft, a steering linkage is connected to rudder pedals or a

separate steering wheel, providing positive control of the turning angle. A key objective in the

design of a steerable nose wheel is to reduce the required control forces while retaining

dynamic stability. The desired for this project is the landing gear arrangement is tricycle.

Because of this, it is considered providing positive control of a turning angle.

Also, because of that the aircraft is a small aircraft, a rake angle is estimated as 15

and trail is about 20% of tire radius [1].

Castoring wheel geometry and dimensions are shown in figure:

Figure 9.6Castoring wheel geometry

9.6. GEAR – RETRACTION GEOMETRY

Landing gear including struts, wheels, and brakes are the major part of the

undercarriage. It can be fixed or retractable. Retracting the landing gear helps reduce drag. An

aircraft landing gear is movable between a retracted position and an extended position. There

are many different options for main-landing gear retracted positions. For instance, location of

the gear may be in the fuselage, in the wing, in the wing-fuselage junctions or in the nacelle.

However, in this project, there is no nacelle. If gears are in the fuselage or wing-fuselage

junction is better for aerodynamics, but bad for structure because it tends to chop up the

structure. Location of the gears plays important role in aerodynamics of aircraft. In this

project, the aircraft retract the gear into the fuselage. Landing gear consists of the wheels that

stick out below the fuselage so that the airplane can roll down the runway during landing and

takeoff. It is selected the fuselage retraction system because, it is important that the gear leg is

minimum length. If it would be used the wing or wing-fuselage retraction system, it is

required the height of the landing gear is higher than 27, 5in. This means increasing of the

gear leg. Therefore, it is decided to the fuselage retraction system for this aircraft.

127

Four bar linkage configuration is selected for main landings gear. This four bar linkage

mechanism is valid for most landing gear. This mechanism makes the gear simple and

lightweight because the loads pass through the rigid members and simple pivots.

Also, forward retracting four bar linkage type is selected for nose landing gear.

This type of retraction system is more preferable because even under hydraulic failures, air

loads will be able to pull the gear down. As a result, Main gear retraction is into the fuselage

fairings and the nose gear folds forward into the fuselage.

9.7. SUBSYSTEMS

Hydraulic, electrical, pneumatic and auxiliary/emergency power systems are basic

subsystems of the aircrafts. Also, avionics can be considered a subsystem. For this project, it

is tried to do initial design/conceptual design layout of an aircraft. In this point, subsystems do

not have a major impact. It should be only considered that an efficient place for these

subsystems’ installation. Thus, there will be meet no problem in later design steps. In this

part, the subsystems are defined shortly as given below:

9.7.1. Hydraulics

Aircraft Hydraulics is a means of transmitting energy or power from one place to

another efficiently. Some systems operated by hydraulics are flight controls such as flaps

ailerons, etc., landing gear, speed brakes, fixed-wing and rotary-wing folding

mechanisms, auxiliary systems, and wheel brakes. Therefore, hydraulic systems are important

for the aircraft. It can be seen in the below a simplified hydraulic system.

Figure 9.7 Simplified hydraulic systems [1]

128

9.7.2. Electrical System

The primary function of an aircraft electrical system is to generate, regulate and

distribute electrical power throughout the aircraft. An aircraft electrical system provides

electrical power to the avionics, hydraulics, environmental control, lighting and other

subsystems. The electrical system consists of batteries, generators, transformer-rectifiers,

electrical controls, circuit breakers and cables.

9.7.3. Pneumatic System

Pneumatic systems, commonly known as vacuum or pressure systems, power the

heading and attitude indicators in most general aviation (GA) aircraft, and in some aircraft,

also power the autopilot and de-ice systems. For pilots who regularly fly at night or in

instrument meteorological conditions (IMC) these systems are essential. The pneumatic

systems use engine as a compressed air resource.

9.7.4. Auxiliary/Emergency Power

If there would be a problem at the hydraulic systems of aircraft, flight will be

uncontrollable. To prevent the uncontrollable flight, emergency power becomes active. There

are three different emergency systems: ram air turbine (RAT), monopropellant emergency

power unit (EPU), and jet-fuel APU (Auxiliary power unit). An emergency power unit (EPU)

is a device for energy production in case of failure of the primary systems. An APU is much

like an EPU but is designed and installed to allow continuous operation. Also, APU provides

energy for air conditioning, lighting and engine starting.

9.7.5. Avionics

Avionics means “aviation electronics”. The cockpit of an aircraft is a major location

for avionic equipment, including control, monitoring, communication, navigation, weather,

radar, and anti-collision systems. For initial layout, it is necessary to provide sufficient

volume in the avionics section. Also, the nose of the aircraft should be designed to hold the

radar.

On the average, avionics has a density of about 30-45 lb/ft3. It can be estimated the

required avionics weight by using the Table 11.6[1], for general aviation-single engine aircraft.

According to this table, the required avionic weight divided by the empty weight is between 0,

01 and 0, 03. For this project, this proportion is taken as 0, 02. It is known the empty weight

Wempty = 1107, 197lb from previous chapter. Thus;

129

0,02avionics

empty

WW

=

0,02 (1107,197) 22,14 10avionicsW lb kg= × = ≅

9.8. SUMMARY OF RESULTS

Gear Type Tricycle retractable Gear Main Nose

Specifications One strut, one wheel One strut, one wheel Tires Type III Type III

Speed (mph) 120 120 Max load (lb) 2400 1200

Inflation pressure (psi) 46 55 Max width (in.) 7, 30 5, 05

Max diameter (in.) 20, 85 13, 25 Rolling radius (in.) 8, 3 5, 2

Wheel diameter (in.) 8 4 Number of plies 6 6 Shock Absorber Solid-Spring Gear/Shock Arrangement Gear stroke (in.) 8, 4 8, 93

Length(in.) 36in 31, 3in

Gear Retraction Geometry Fuselage

rearward retracting four bar linkage

forward retracting four-bar linkage

Figure 9.8 Location of main landing gears (Bottom view)

130

Figure 9.9 Orientation of main and nose landing gears (front view)

Figure 9.10 Orientation of main and nose landing gears (side view)

1.93

40

131

9.9. CONCLUSION

It is purposed in this study, to make the better design for the landing gear dimensions

and loads and select the some specifications such as tire sizing because it is important that the

aircraft has stable landing. The landing gear should be carrying the aircraft carefully. By these

considerations, it is made the calculations of the landing gear for a good design. To do this,

some assumptions are made by comparing the competitor’s aircrafts. Also, to reduce drag it is

decided to the aircraft retracts the landing gear into the fuselage.

After performing the landing gear, the subsystems of aircraft and their main functions

are can be explained shortly. The subsystems are not examined much, as they don’t affect the

initial layout of the aircraft much except for the space they require.

132

CHAPTER 10

AERODYNAMICS

10.1. INTRODUCTION

The previous studies have presented methods for design layout of a credible aircraft

configuration. Initial sizing, wing geometry, engine installation, tail geometry, fuselage

internal arrangement, and numerous other topics have been discussed. In this chapter, the

aerodynamic properties of the aircraft to be designed are discussed and examined.

Therefore, the lift curve slope ( )LCα

, maximum lift ( )maxLC with and without high lift

devices for cruise, landing and takeoff conditions, angle of attack for maximum lift ( )maxLCα ,

parasite ( )0DC and induced drags ( )iDC are estimated below. LC α− and L DC C− curves are

plotted according to the estimated values for cruise, takeoff and landing configurations.

The calculations done in this chapter are performed for the following configurations, Takeoff / Landing : clean (no flaps) at M = 0.2

Takeoff / Landing : flapped at M = 0.2

10.2. ESTIMATION OF LIFT

In this section, it is estimated the lift curve slope, maximum lift and angles of both for

clean and with high devices at the takeoff and landing configurations. These conditions are

analyzed for M = 0, 2.

10.2.1 Lift Curve Slope

Lift curve slope ( CLα ) can be found from the following relation for a swept wing with M<1 condition.

( )exp

22 2max,

2 2

2

tan2 4 1

osedL

reft

SARC FSAR

α

π

βη β

⋅= Λ + + +

(10.1)[1]

where, β2= 1-M2 and 2 /

LCαη

π β= (airfoil efficiency) (10.2)[1]

• Λmax, t : the sweep of the wing at the chord location where the airfoil is the

133

thickest (xt/c). In this project, the wing profile was selected NACA23015. For NACA 23015,

maximum thickness is 0, 3. This value is nearly the same quarter chord point. Therefore,

maximum thickness sweep angle is assumed the same with ΛC/4 = 4 = 0, 698rad.

• η : Airfoil efficiency is approximated as 0, 95. • AR : 7

• F : is the fuselage lift factor which accounts for the fact that the fuselage creates

some lift due to the “spill over” of lift from the wing and is found by using equation (for D =

1, 4874m and b =8, 217m)

22 1, 48741,07 1 1,07 1 1,49

8,217dFb

= + = + = (10.3)[1]

• Sexposed : is the wings reference area minus the part of the wing covered by the

fuselage, from the previous chapters;

Sreference = 9, 644m2

Sexposed = 8, 87m2 (from previous drawings)

Swet = 18, 22m2

These values are put into the LC α equation; LC α is calculated for M = 0, 2 as given below:

β2 = (1- M2) = (1 - 0.22) = 0, 96 By using equation (10.1);

( )2 2 0

2

2 7 8,87 1,499,6447 (0,96) tan 42 4 1

0,95 0,96

LCα

π × =

+ + +

CLα = 6, 346 rad-1 = 0, 11 deg-1

10.2.2. Maximum Lift (Clean)

The maximum lift coefficient of the wing is generally related to the wing area. This

strongly affects the aircraft takeoff weight to perform the design mission. For high aspect

ratio wings with sweep, maximum lift is related with the following equation;

CL max = 0.9Cl max cos Λ0.25c (10.4) [1]

• Cl max = 1, 7 • CL max = 1, 53 (from previous study, for NACA 23015)

CL max = 0.9Cl max cos Λ0.25c = 0.9(1, 7).cos (4) = 1, 53.cos (4)

134

CL max (clean) = 1, 526 Another way to calculate the CL max is the high Mach number approach. Therefore, for

high Mach number:

ΛL.E. = 8, 6

maxmax max max

max

LL l L

l

CC C CC

= + ∆

(10.5) [1]

maxLC∆ = 0 CL max = 1, 513

According to these CL max values are nearly the same with the CL max value for NACA

23015 (1, 53). However, CL max, clean is closer to actual value than the value for high Mach

number. Because of it, CL max is taken as 1, 526.

The angle of attack for maximum lift is states as;

max max

maxL L

LC oL C

L

CC α

α α α= + + ∆ (10.6)[1]

Here, oLα represents the airfoil zero lift angle and maxLCα∆ represents the angle of

attack increment. For wing airfoil NACA 23015, it is known from previous chapters that;

Cl max = 1, 7 ; αoL = -1, 2º • ∆αCLmax

should be determined to calculate the αCLmax. To find ∆αCLmax

, firstly, it is

required to estimate the by using ∆y (the leading edge sharpness parameter). ∆y for

common airfoils are given in the Table 12.1[1]. According to this table, For NACA 5

digit airfoils, ∆y = 26(t/c).

∆y = 26(t/c) = 26(0.15) = 3.9 Also, it is needed to know the leading edge sweep angle. The leading edge sweep

angle was found in previous chpater as ΛL.E.= 8, 6º

From Figure 12.10[1], for ∆y = 3.9 and ΛL.E.= 8, 6º; the angle of attack increment is nearly 2.

max2

LCοα∆ =

max2

LCοα∆ =

Maximum lift angle (clean) ;

Maximum lift angle of attack can be found by using equation (10.6). It is found as . αoL = -1, 2º and lift curve slope CLα = 0, 11 deg-1:

max max

max 1,526 ( 1,2) 2 14,670,11L L

LC oL C

L

CC

ο

α

α α α= + + ∆ = + − + = αCL max = 14, 67°

135

10.2.3. Maximum Lift with High Lift Devices

It was decided to use plain flap for desired design. Plain flap rotates on a simple hinge.

Also, it has typically a flap chord “Cf” of 30% of the airfoil chord. The plain flap increases lift

by increasing the camber. Maximum lift occurs around 40-45 degrees of deflection.

It can be seen a figure below. This figure illustrates the effects the high-lift devices

have upon the lift curve of the wing. From this figure, plain flap moves the angle of zero-lift

to the left and increases the maximum lift. The slope of the lift curve remains unchanged, and

the angle of stall is somewhat reduced.

Figure 10.1 Effects of high lift devices [1]

Increase in maximum lift and change in the zero lift angles equations are given:

max max . .0.9 cosflappedL l H L

ref

SC C

S∆ = ∆ Λ (10.7)[1]

( ) . .cosflappedoL oL H Lairfoil

ref

SS

α α∆ = ∆ Λ (10.8)[1]

For plain flap, the CL max value was found before as 1, 8. This value CL max = 1, 8 is

valid for landing configuration. Also, the information to determine the CL max and oLα∆ are

given below:

• ∆Cl max = 0, 9 for plain flaps (from Table 12.2[1])

• ΛH.L. is the sweep angle at the hinge line of the high-lift surface. ΛH.L. is assumed as

nearly -2 because this value’s effect is the same during the calculations.

(CosΛH.L.≅ 1).

• Sflapped can be estimated from previous drawings as Sflapped = 4m2, Sreference = 9, 644m2

136

maxLC∆

At the take off setting:

For take off flap settings, lift increments of about 70% of value should be

used. Besides, the change in zero lift angles for flaps in 2-D case is approximately -10 deg at

the take off setting (half flap). (∆α0L) airfoil= -10º

From equation (10.7);

max40,7 0,9 0,9 cos( 2) 0,235

9,644LC∆ = × × − =

At the take off setting;

CL max = CL max) clean + ∆CL max = 1, 526 + 0, 235 = 1, 761

CL max = 1, 761

Also, by using equation (10.8); 410 cos( 2) 4,15

9,644oLοα∆ = − − = −

At the take off setting the zero lift angle of attack is;

αoL = αoL) clean + ∆αoL = -1, 2 – 4, 15 = -5, 35°

αoL = -5, 35° Maximum lift angle of attack can be found by using equation (10.5) according

tomax

2LC

οα∆ = , and lift curve slope CLα = 0.0929deg-1:

max

1,761 ( 5,35) 2 12,660,11LC

οα = + − + =

αCL max = 12, 66°

max40,9 0,9 1 0,336

9,644LC∆ = × × =

At the landing setting:

The change in zero lift angles for flaps in 2-D case is approximately -15 deg at the

landing setting (full flap).

(∆α0L) airfoil= -15º From equation (10.7);

max 0,336LC∆ =

At the landing settings;

CL max = CL max) clean + ∆CL max = 1, 526+0, 336

137

CL max = 1, 862 Also, by using equation (10.8);

415 cos( 2) 6,22

9,644oLοα∆ = − − = −

At the landing setting the zero lift angle of attack is;

αoL = αoL) clean + ∆αoL = -1, 2 – 6, 22 = -7, 42°

αoL = -7, 42 ° And again by using equation (10.5), it is determined the maximum lift angle of attack;

max

1,862 ( 7,42) 2 11,510,11LC

οα = + − + =

αCL max = 11, 51°

10.3. ESTIMATION OF PARASITE (ZERO-LIFT) DRAG

The following two methods are used for determination of the parasite drag such as

equivalent skin friction method and component buildup method.

10.3.1. Equivalent Skin-Friction Method

For subsonic aircrafts parasite drag consists of skin friction drag and separation drag.

We can estimate the skin friction drag by using the formula:

0

wet

refD fe

SC CS

=

(10.9)[1]

* 0.0045feC = (From Table 12.3[1] for light aircraft-single engine) * Swet,total = 19, 82m2

* Sref = 9, 644m2

0

19,820.0055 0,01139,644DC = =

00,0113DC =

10.3.2. Component Buildup Method

Component buildup method estimates the subsonic parasite drag of each component of

the aircraft using a calculated flat-plate skin-friction drag coefficient (Cf) and a component

“form factor” (FF) that estimates the pressure drag due to viscous separation. Then the

interference effects on the component drag are estimated as a factor “Q” and the total

component drag is determined as the product of the wetted area, Cf, FF, Q [1]. Total drag on

each component is determined by following equation:

138

) ( )0 &

( )c c wetf ccD D Dmisc L Dsubsonic ref

C FF Q SC C C

S= + +∑

(10.10)[1]

Where flat-plate skin-friction coefficient ( )fC for turbulent, subsonic flow and skin

friction coefficient for turbulent flow is given by:

2.58 2 0.6510

0, 455(log ) (1 0.144 )fC

R M=

+ (10.11)[1]

and skin roughness “R” for subsonic flow is,

( )1.05338.21cutoff

lR R k= = (10.12)[1]

“l” is the characteristic length of the component. For fuselage, it is the total length

(l=L) and for wing or tail, it is the mean aerodynamic chord.

“k” is the skin roughness value and for smooth paint, it is 0, 634x10-5m from Table

12.4[1].

After the calculations of the skin roughness and skin friction coefficient, it can be find

the total drag on each component by determination of the FF and estimation of the Q.

10.4. CALCULATIONS 10.4.1. Takeoff and Landing (Clean at M = 0.2)

The following table shows the calculated R and Cf values for each component.

L (m) R Cf Wing 1, 246 14254437 0,00283

Fuselage 7, 437 94000000 0,00214 Vertical

Tail 0, 688 7665803 0,00312

Horizontal Tail 0, 614 6800148 0,0032

Table 10.1 R and Cf values

Component form factors and interference factors

:

Fuselage

3

601400

fFFf

= + +

(10.13)[1]

139

7, 437 51,4874

lfd

= = = (10.14)[1]

3 3

60 60 51 1 1,5400 5 400

fFFf

= + + = + + =

Fuselage interference factor is negligible; Q =1

Wing

40.18 0.280.61 100 1.34 ( )

( / ) mm

t tFF M Cosx c c c

= + + Λ (10.15)[1

(x/c) m=0, 3 (t/c) m=0.15 and Λm=4º for NACA 23015 airfoil;

( ) ( )4 0.18 0.280,61 0,15 100 0,15 1,34(0,2) ( 4 ) 1,350,3

oFF Cos = + + =

For a mid-wing, the interference factor will be negligible so the Q =1

(x/c) m=0, 3 (t/c) m=0, 015 and Λm=40º for NACA 0015

Vertical Tail

( ) ( )4 0.18 0.280,61 0,15 100 0,15 1,34(0,2) ( 40 ) 1,270,3

oFF Cos = + + =

Vertical tail interference factor is assumed as Q =1, 05 for conventional tail.

(x/c)m=0, 3 (t/c)m=0, 15 and Λm=6, 76º for NACA 66-009 laminar airfoil.

Horizontal Tail

( ) ( )4 0.18 0.280,61 0,15 100 0,15 1,34(0,2) ( 6,76 ) 1,350,3

oFF Cos = + + =

Horizontal tail interference factor is taken as Q =1, 04

The following table shows the component drags and total component drag by using

Eqn. (10.10):

140

Mach = 0.2 Cf Swet (m2) FF Q Component CD0 Wing 0,00283 19,82 1,35 1 0,00785

Fuselage 0,00214 25,969 1,5 1 0,00864 Vertical Tail 0,00312 1,46 1,27 1,05 0,000629

Horizontal Tail 0,0032 3,87 1,35 1,04 0,001802 Total Component Drag Coefficient 0,018921

Table 10.2 Component drags for M=0, 2

• Miscellaneous Drags

Due to Upsweep

2.5max3.83

upsweep

D u Aq

= (10.17)[1]

Amax is the maximum cross-sectional area of the fuselage;

Amax= 2

4d = π(3.48)2= 1, 72m2

u is the upsweep angle of the fuselage centerline. From drawings,

u = 12, 68º =0, 22rad

D/q = 3, 83(0, 22)2.5(1, 72) = 0, 15m2

)0

/ 0,15 0,015559,644D upsweep

ref

D qCS

= = = (10.18)[1]

For light aircraft windshields has smoothly far into the fuselage and D/q = 0, 07

(Windshield Frontal Area) is suggested:

D/q = 0, 07 (0, 86) = 0, 0602m2

Due to Canopy

)0

/ 0,0602 0,006249,644ref

D canopy

D qCS

= = =

There is no base drag for desired aircraft. Because of this, there is no base of drag.

) ) )0 0 00,0113 0,00624 0 0,01754D D Dupsweep canopy baseDmisc

C C C C= + + = + + =

141

• Leakage and Protuberance Drag

The effect of leakage & protuberance on CD0 is changing between 5 and 10% of total

parasite drag for propeller aircraft. In this design it is assumed as 6%.

&DL PC =0, 06 (0, 018921+0, 01754) = 0, 00219

Then, total parasite drag at M = 0.2 for clean configuration is:

) ) ) )&0 0 0 0

0,018921 0,01754 0,00219 0,038total component misc L P

D D D DC C C C= + + = + + =

)00,038

totalDC =

10.4.3. Takeoff and Landing (With High Lift Devices at M = 0.2) The flap contribution to the parasite drag is caused by the separated flow above the flap.

For takeoff

δFlap =30º is taken for takeoff. (bflap/bwing is determined as 0, 40 before)

:

00.0023 flap

flapwing

Dflap

bC

bδ∆ = (10.19)[1]

00,0023 0,4 30o

Dflap

C∆ = × × =0, 0276

0DC ) takeoff = 0DC ) clean +∆ 0DC ) flap =0, 038+0, 0276

) ) ),0 0 0

0,0656takeoff flap clean flap

D D DC C C= + ∆ =

For landing

δFlap=60º is taken for landing.

:

00.0023 0.4 60 0,0552o

Dflap

C∆ = × × =

CD0

) landing = CD0) clean +∆CD0

) flap =0, 038 +0, 0552

) ) )0 0 0,0,0932D D Dlanding flap clean flap

C C C= + ∆ =

142

10.5. ESTIMATION OF DRAG DUE TO LIFT (INDUCED DRAG)

The drag polar of the NACA 23015 airfoil given in Chapter 4 has shown that the wing

has approximately a moderate camber and 0DC and

minDC of the wing are very close to each

other, so the following relation between lift and drag coefficient can be used.

20D D LC C KC= + (10.20)[1]

K is the “drag-due-to-lift factor” and defined as,

1KAReπ

= (10.21)[1]

For leading edge sweep angle is smaller than 30º, Oswald efficiency factor “e” can be

calculated from the straight-wing aircraft relation;

0.681,78(1 0,045 ) 0,64e AR= − − (10.22) [1]

0.681,78(1 0,045(7) ) 0,64e = − − = 0, 839

Then, 1 0,0542

7 0,839K

π= =

× ×

So, Di

C =KCL2 = 0, 0542 CL

2

2L0,0542 CDi

C =

When ground effect is considered, Keffective must be determined.

1.5

1.5

33

1 33

effective

hK b

K hb

= +

(10.23)[1]

It is known that; b = 8, 217m and h = 2, 20m. By using the eq. (10.23);

1.5

1.5

2, 20338,2170,0542 0,0444

2,201 338,217

effectiveK

= = +

143

10.6. CL - α AND CL - CD CURVES

CL - α and CL - CD curves will be plotted for the configurations below:

• Clean at M = 0.2 Configuration

20,038 0,0542D LC C= +

• Takeoff at M = 0.2 Configuration

20,0656 0,0542D LC C= +

• Landing at M = 0.2 Configuration

20,0932 0,0542D LC C= +

• Takeoff at M = 0.2 With Ground Effect Configuration

20,0656 0,0444D LC C= + Curves are plotted below: CL – α Curves

Figure 10.2 CL– α graph for clean configuration

144

Figure 10.3 CL –α graph for takeoff-flapped configuration

Figure 10.4 CL–α graph for landing-flapped configuration

145

Cl-Cd Graph (clean configuration at M=0.2)

0,15995

0,0922

0,051550,038

0,05155

0,0922

0,15995

00,020,040,060,08

0,10,120,140,160,18

-1,5 -1 -0,5 0 0,5 1 1,5

Cl

Cd

CL – CD Curves

Figure 10.5 Cl – Cd graph for clean configuration

Cl-Cd Graph (takeoff flapped configuration)

0,18755

0,1198

0,079150,0656

0,07915

0,1198

0,18755

00,020,040,060,08

0,10,120,140,160,18

0,2

-1,5 -1 -0,5 0 0,5 1 1,5

Cl

Cd

Figure 10.6 Cl – Cd graph for takeoff flapped configuration

146

Cl-Cd Graph (Landing Flapped Configuration)

0,21515

0,1474

0,106750,0932

0,10675

0,1474

0,21515

0

0,05

0,1

0,15

0,2

0,25

-1,5 -1 -0,5 0 0,5 1 1,5

Cl

Cd

Figure 10.7 Cl – Cd graph for landing flapped configuration

Cl-Cd Graph (Takeoff Flapped Configuration with Ground Effect)

0,1655

0,11

0,07670,0656

0,0767

0,11

0,1655

00,020,040,060,08

0,10,120,140,160,18

-1,5 -1 -0,5 0 0,5 1 1,5

Cl

Cd

Figure 10.8 Cl – Cd graph for takeoff flapped configuration with geound effect

147

10.7. SUMMARY OF RESULTS Results are tabulated below:

Clean at M = 0.2 Takeoff Flapped at M = 0.2

Landing Flapped at M = 0.2

CL - α

Clα (1/deg) 0,0929 0,11 0,11 CLmax 1,526 1,761 1,862

αL=0 (deg) -1,2 -5,35 -7,42 αClmax (deg) 14,67 12,66 11,51 ΔαClmax (deg) 2 2 2

Table 10.3 Lift calculations

Clean at M = 0.2 Takeoff Flapped at M = 0.2

Landing Flapped at M = 0.2

CL - CD CD 0,038+0,0542CL2 0,0656+0,0542CL

2 0,0932+0,0542CL2

Table 10.4 Drag calculations

10.8. CONCLUSION

Various aerodynamic properties of the aircraft have been calculated in this chapter.

Two configuration has considered for this purpose namely clean at M=0.2 and flapped at

M=0.2 for takeoff/landing. Table 10.3 shows that maximum lift increase and Table 10.4

shows that parasite drag increase considerably with the use of flaps which is expected for

both takeoff and landing. Also, drag analysis is done based upon different types of drag is

estimated at the final.

Although the 2-D airfoil characteristics and 3-D wing characteristics are different,

recalling from Chapter 4, CL can be taken as Cl. Therefore, the wing CL - α and CL - CD

curves are seemed to be good when compared to the NACA 23015 airfoil though this

comparison is a very rough estimate.

148

CHAPTER 11

DESCRIPTION IN JANE’S FORMAT

:

MarTı MT- 1505

Faculty of Aeronautics and Astronautics (ITU) TYPE: Two seat training aircraft. DESIGN FEATURES: Tapered, 8.6 leading edge sweep, mid-mounted, from root to tip the same airfoil NACA 23015; no twist and dihedral, also 2 wing incidence. Section of NACA 0015 is at vertical tail and NACA 652-015 at horizontal tail. Tails have no twist and dihedral. FLYING CONTROLS: Ailerons, elevators and rudder statically and dynamically balanced, with mechanical actuation. Electrohydraulically actuated plain flaps. Plain aileron on the wing. plain type rudder and elevator; all are sealed. STRUCTURE: Fuselage frame with aluminum alloys, and wing and tail carbon composites. LANDING GEAR: Hydraulically retractable tricycle type. Solid-spring gear arrangement for main wheels and steerable nose-wheel. Main gear; fuselage rearward retracting four bar linkage. Nose gear; forward retracting four-bar linkage. Tires used in the main gear 7.00-8 (46 psi), in the nose gear 5.00-4 (55 psi). POWER PLANT: One 150 hp Textron Lycoming O-320 2A four cylinder air-cooled horizontal opposed engine, driving a two blade fixed pitch

McCauley propeller. Fuel tanks are integrated on wings. ACCOMMODATIONS: Two seat for crew. SYSTEMS: Hydraulic Systems, APU, Electrical System, Pneumatic System. DIMENSIONS EXTERNAL: Wing span 8.217m (26.96ft) Wing chord: at root 1.677m (5.5ft) at tip 0.6708m(2.2ft) Wing aspect ratio 7 Horizontal tail span 3.07m (10.07ft) Horizontal tail chord 0.501m (1.64ft) Horizontal tail aspect ratio 5 Length overall 7.437m (24.4ft) Fuselage width 1.487m (4.88ft) Vertical tail span 1.032m (3.385ft) Vertical tail aspect ratio 1.5 Wheel base 3.56m (11.86ft) Height overall 2.2m (7.22ft) DIMENSIONS INTERNAL: Cockpit: Length 1.6m (5.25ft) AREAS: Wings, gross 9.644m2 (104.35 sq ft) Horizontal tail 1.885m2 (20.29sq ft) Vertical tail 0.71m2 (7,64sq ft) WEIGHTS AND LOADING: Weight empty 497.32kg (1095.14lb) Max fuel 92kg (202.69lb) Max T-O 789.2kg (1738.350lb) Wing loading 17 lb/sq ft Hp/W 0.085 PERFORMANCE: Max. Cruising Speed 136.5 knots Stalling Speed 52.77 knots Max. R/C (knots) 40.43 knots Range with max. Payload 525 nm Max. Ceiling 12206.4m (40047.3ft) Takeoff distance (sls) 335.28m (1100ft) Landing distance (sls) 230.3m (755.56ft) L/D) max 14.75 Loiter speed (at 5000m) 64.36 knots

149

CHAPTER 12

CONCLUSION

Conceptual design of the two seat general aviation aircraft provided insight four us

between final layout of the aircraft and mission requirements. Aircraft was designed to meet

mission requirements, and it is seen that after final layout it satisfies almost all of them.

However, range is found as 525nm, this value is a bit higher than the requirements value

500nm. This result is not very important because the difference is not very high, and aircraft

can meet the design requirement.

The visual appearance of the aircraft is similar to this type aircraft, but most of the

lighter aircrafts have low or high wing. In this project, mid-wing design was selected for

desired aircraft because it was wanted to design an aerobatic aircraft by considering the cost.

Also, the landing gear arrangement is not oleo, although oleo is common type today. It was

decided to use solid spring landing gear arrangement. This type is both simple and economic.

Except wing location and landing gear arrangement, this new design is similar to its

competitors with the general appearance.

The project gives sufficient and reasonable results that can be proceed by a

preliminary design based on this study.

About the course, it can be said that this project and lecture gave us skills and chance

to use and gain knowledge on engineering and real problems one can face during design.

150

REFERENCES

[1] Raymer D. P., “Aircraft Design: A Conceptual Approach”, AIAA Education Series, 1992

[2] NACA Report 824

[3] Riegels F. W., “Airfoil Sections”, 1961

[4] Acar H., Flight Mechanics Lecture Notes

[5] www.lycomig.com

[6] en.wikipedia.org/wiki/Lycoming_O-320

[7] Jane’s all of the World’s Aircraft

[8] Grove Aircraft Landing System Inc, Part Catalog 110

151

APPENDIX A

HISTOGRAM OF EXISTING COMPETITOR AIRCRAFTS

# of Aircraft 5 SU-31 AVIAT

4 TOMAHAWK DIAMOND

3 CESSNA140 STEARMAN CHEROKEE

2 CHIPMUNK IKARUS YAK-52

1 AERONCA ALARUS CESSNA152 ZENITH PITTS

60 80 100 120 140 160 Vcruise(knot)

Table A.1 # of Aircraft – Vcruise (knot)

# Of Aircraft

YAK-52

CESSNA 140

ZENITH

AERONCA

CHIPMUNK

TOMAHAWK

ALARUS DIAMOND

SU-31 MAN PITTS IKARUS

CHEROKEE

CESSNA152

RANGE (m)

0 100 200 300 400 500 600 700

Table A.2 # of Aircraft – Range (m)

152

# of Aircraft

CHIPMUNK

6 AVIAT

5 DIAMOND

4 PITTS

3 CESSNA140 ALARUS STEARMAN

2 ZENITH TOMAHAWK SU-31

1 IKARUS AERONCA CESSNA152 CHEROKEE YAK-52

250 500 750 1000 1250 1500 Wo (kg)

Table A.3 # of Aircraft – Wo (kg)

# of Aircraft

6

5 CESSNA152 YAK-52

4 AERONCA ALARUS

3 PITTS ZENITH

2 IKARUS STEARMAN DIAMNOND

1 SU-31 AVIAT TOMAHAWK CESSNA140 CHIPMUNK CHEROKEE

100 150 200 250 300 350 400 450Wpl (kg)

Table A.4 # of Aircraft – Wpl (kg)

153

# of Aircraft

8 STEARMAN

7 AERONCA

6 YAK-52

5 PITTS

4 CHIPMUNK

3 CESSNA140 TOMAHAWK

2 IKARUS ZENITH CESSNA152

1 AVIAT SU-31 CHEROKEE DIAMOND ALARUS

0 100 200 300 400 500 T.O.D.(m)

Table A.5 # of Aircraft – Take off Distance (m)

#of Aircraft

7 Chipmunk

6 Aviat

5 Ikarus

4 Zenith Aeronca

3 Stearman Pitts

2 Cherokee Tomahawk Su 31

1 Cessna-140 Cessna-152 Diamond Yak-52 Alarus

0 100 200 300 400 500 600 L.D. (m)

Table A.6 # of Aircraft – Landing Distance (m)

154

# of Aircraft

7 Tomahawk

6 Pitts

5 Zenith Diamond

4 Stearman Alarus

3 Cessna-140 Chipmunk

2 Ikarus Aviat Yak-52

1 Su 31 Aeronca Cessna-152 Cherokee

0 5 10 15 20 W/S (lb/ft^2)

Table A.7# of Aircraft – Wing Loading (lb/ft2)

# of Aircraft

6 Chipmunk

5 Alarus

4 Tomahawk Zenith

3 Cessna-152 Stearman

2 Cessna-140 Diamond Aviat Su 31

1 Aeronca Cherokee Ikarus Yak-52 Pitts

0,05 0,075 0,1 0,125 0,15 0,175 Hp/W (hp/lb)

Table A.8 # of Aircraft – Power Loading (hp /lb)

155

APPENDIX B

AIRFOIL DATA

Figure B.1 Wing Airfoil (NACA 23015) characteristics [2]

156

Figure B.2 Horizontal Tail Airfoil (NACA 652-015) characteristics [2]

157

Figure B.3 Horizontal Tail Airfoil (NACA 652-015) characteristics [2]

158

Figure B.4 Vertical Tail Airfoil (NACA 0015) characteristics [2]

Figure B.5 Vertical Tail Airfoil (NACA 0015) characteristics [2]

159

Figure B.6 Vertical Tail Airfoil (NACA 0015) characteristics [2]