skma4253 aircraft design ii group two · 2020. 7. 16. · ahmad amirul amin bin abdul wahab...

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SKMA4253 AIRCRAFT DESIGN II

GROUP TWO

RC-PLANE GROUP PROJECT (FINAL DRAFT)

LEADER NAME:

MOHAMAD ASHRAF BIN AB HAN

LECTURERS NAME:

DR. ING. MOHD. NAZRI BIN MOHD. NASIR

DR. WAN ZAIDI BIN WAN OMAR

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GROUP MEMBERS

1. AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017)

2. AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022)

3. ALEXANDER JONG JIUN WEI (A16KM0033)

4. CHENG WEI QIN (A16KM0062)

5. LEONG JOE YEE (B17KM0011)

6. MOHAMAD ASHRAF BIN AB HAN (A16KM0176)

7. MOHAMAD AZMIL BIN BAHJAM (A16KM0178)

8. MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222)

9. MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224)

10. MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289)

11. MUHAMMAD NASRUL BIN YAZI (A16KM0496)

12. MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330)

13. NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390)

14. SAMALADEWI A/P MURUKAPPAN (A16KM0417)

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TABLE OF CONTENTS

TITLE PAGE

TABLE OF CONTENTS i

CHAPTER 1 INTRODUCTION 1

1.1 Objectives 1

1.2 Gantt chart 2

1.3 Flowchart 3

CHAPTER 2 CONCEPTUAL DESIGN 5

2.1 What is Conceptual Design? 5

2.2 Design Selection 5

2.2.1 Wing Configuration 5

2.2.2 Fuselage Configuration 7

2.2.3 Tail Configuration 8

2.2.4 Landing Gear 9

2.2.5 Propulsion System 10

2.3 Material Selection For Rc Plane 11

2.4 Weight Estimation 12

2.5 Cost Estimation 13

2.5.1 Design Sketches 14

CHAPTER 3 RC PLANE CONFIGURATION 16

CHAPTER 4 PROPULSION AND POWER PLANT SYSTEM 18

4.1 Introduction 18

4.2 Objective 20

4.3 Methodology 20

4.4 Preliminary Design 20

4.5 Preliminary Selection of Motor 20

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4.6 Power Plant Configuration 24

4.7 Aerodynamic Drag 26

4.8 Thrust Required 28

4.9 Power Available 29

4.10 Power Required 31

4.11 Rate of Climb 32

4.12 Endurance 33

4.13 Range 34

4.14 Takeoff Performance 34

4.15 Landing Performance 36

4.16 Flight Path 37

4.17 Flight Power Consumption 38

CHAPTER 5 AVIONICS 40

5.1 Electronic Components Selection 40

5.1.1 Transmitter and Receiver 40

5.1.2 TX-RX Communication Signal 43

5.1.3 Electronic Speed Controller (ESC) 49

5.1.4 Battery 52

5.1.5 Servo 54

5.1.6 Wiring System Schematic Diagram 56

CHAPTER 6 WING 57

6.1 Introduction 57

6.1.1 Function of Wing 57

6.1.2 Aircraft Wing Design 58

6.1.3 Wing Load Analysis 60

6.1.4 Aircraft’s Airworthiness 62

6.1.5 Assumptions on Wing Design 63

6.2 Flight Envelope and Wing Loadings 64

6.2.1 Flight Envelope (V-n Diagram) 64

6.2.2 Wing Loading without Aileron 66

6.2.3 Wing Loading with Aileron 69

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6.2.4 Reaction force of Wing Strut 72

6.2.5 Shear Force Diagram and Bending Moment

Diagram 77

6.2.6 Discussion 89

6.3 Structural Analysis 91

6.3.1 Material Selection 91

6.3.2 Rib Analysis 92

6.3.3 Spar Analysis 100

6.3.4 Skin Analysis 104

6.3.5 Stress Analysis of Multi Cell Sections 108

6.3.5.1 Shear stress due to shear force

(monocoque) 108

6.3.5.2 Shear stress due to shear force (semi-

monocoque) 112

6.3.5.3 Shear stress due to torsion

(monocoque) 116

6.3.5.4 Shear stress due to torsion (semi-

monocoque) 118

6.3.5.5 Shear Stress Summary 121

6.3.6 Part Attachment 123

6.3.6.1 Joint of the part attachment 123

6.3.6.2 Structural Analysis of Wing Strut 128

6.3.6.3 Part attachment between aerofoil and

aileron 130

6.4 Summary of Safety Factor 134

CHAPTER 7 FUSELAGE AND TAIL 142

7.1 Introduction 142

7.2 Monocoque Structure 143

7.3 Material Properties 144

7.4 Shear and Bending Moment Diagram for Fuselage 145

7.4.1 𝑮 = 𝟏 145

7.4.2 𝑮 = 𝟑 146

7.4.3 𝑮 = −𝟏. 𝟓 147

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7.5 Flow Simulation 150

7.6 Stress Analysis 151

7.6.1 Former structure 151

7.6.2 Bulkhead structure 153

7.7 Compressive-Buckling Analysis 155



7.8 Shear-Buckling Analysis 158



7.9 Shear Flow of Fuselage due to Torsion 161

7.10 Structure Analysis on Horizontal Tail 164

7.11 Shear and Bending Moment Diagram for Tail 167

7.11.1.1 G = 1 167

7.11.2 G = 3 168

7.11.3 G = -1.5 169

CHAPTER 8 LANDING GEAR 173

8.1 Introduction 173

8.2 Static Load Analysis 173

8.3 Introduction of FEM 177

8.3.1 Software Used 177

8.3.2 Material of Landing Gear 178

8.3.2.1 Material Property for Landing Gear 178

8.3.3 Applied Load on Front Landing Gear 178

8.3.4 Boundary Conditions for Front Landing Gear 179

8.3.5 Meshing for Front Landing Gear 180

8.4 Results of Finite Element Analysis for Front Landing

Gear 182

8.5 Discussion 184

CHAPTER 9 FLIGHT STABILITY AND CONTROL 185

9.1 Center of Gravity 185

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9.1.1 Empty Weight cg 186

9.1.2 Maximum Weight cg 187

9.1.3 Forward cg 188

9.1.4 Aft cg 189

9.2 Stability Control System 191

9.2.1 Contribution of the Wing to Mcg 191

9.2.2 Contribution of the Wing-Body to Mcg 192

9.2.3 Contribution of the the Tail to Mcg 193

9.2.4 Total Pitching Moment about the centre of

gravity 194

9.2.5 Equations for Longitudinal Static Stability 194

9.3 Static Lateral Roll Stability 205

9.3.1 Influence of roll rate on the rolling moment 206

9.3.2 Influence of deflection angle of aileron on the

roll rate 209

9.3.3 Influence of deflection angle of aileron on the

rolling moment 210

9.3.4 Influence of airspeed on rolling moment 212

9.4 Static Lateral Directional Stability 214

9.4.1 Pure Yawing Motion 217

9.5 Dynamic Longitudinal Stability System 220

9.6 Dynamic Lateral Stability System 226

CHAPTER 10 FLIGHT TESTING PLANNING AND

PREPARATION 232

10.1 Flight test and preparation 232

10.2 Checklist 234

REFERENCES 236

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LIST OF TABLES

TABLE NO. TITLE PAGE

Table 1-1 : List of activities from week 1 to week 14 2

Table 2-1 : Properties of EPP Foam 11

Table 2-2 : Weight & Area estimation (based on SolidWorks): 15

Table 3-1 : The configuration of RC plane 16

Table 4-1 General characteristics or our RC plane 20

Table 4-2 Thrust-to-Weight Ratio for different type of aircraft 20

Table 4-3 The values of airspeed according to the motor Kv 21

Table 4-4 : Different brand of brushless motor 23

Table 4-5 : The time and altitude at which the RC plane will takeoff,

cruising and landing 38

Table 4-6 : Percentage of thrust needed for Takeoff, Cruising and Landing

38

Table 5-1: Transmitter specification 41

Table 5-2: Receiver specification 42

Table 5-3: Sunnysky X2216 Motor Performance Chart 50

Table 5-4: ESC Specifications 51

Table 5-5: Comparison of RC Battery 52

Table 5-6: Battery Specification 54

Table 5-7: Servo Specification 55

Table 6-1 Specification of the wing 60

Table 6-2 Federal Aviation Regulations, Part 23 design requirements and

specifications 63

Table 6-3 Project Requirement on Wing Design 63

Table 6-4 Wing loading in each station for point A (Aileron deflects

downwards) 77

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Table 6-5 Wing loading in each station for point D (Aileron deflects

downwards) 78

Table 6-6 Wing loading in each station for point E (Aileron deflects

downwards) 79

Table 6-7 Wing loading in each station for point G (Aileron deflects

downwards) 80

Table 6-8 Wing loading in each station for point A (Aileron deflects

upwards) 81

Table 6-9 Wing loading in each station for point D (Aileron deflects

upwards) 82

Table 6-10 Wing loading in each station for point E (Aileron deflects

upwards) 83

Table 6-11 Wing loading in each station for point G (Aileron deflects

upwards) 84

Table 6-12 Maximum Shear force and Bending moment in each position 88

Table 6-13 Material Selection of each Wing Component 91

Table 6-14 The chordwise load distribution for each bay 92

Table 6-15 Material properties of plywood 93

Table 6-16 Material Properties of EPP Foam 100

Table 6-17 Factor of safety of upper and lower wing 107

Table 6-18 Dimensions of cell 1 and cell 2 wing section 109

Table 6-19 Integral of cell 1 and cell 2 110

Table 6-20 Total shear flow for monocoque and semi-monocoque wing

structure 122

Table 6-21 Total Shear Stress of Monocoque and Semi-monocoque 122

Table 6-22 Safety Factor of Monocoque and Semi-monocoque 122

Table 6-23 Material Properties of AN Steel (hinge pin) 124

Table 6-24 Properties of AN Standard Steel bolt/pin 124

Table 6-25 Material Properties of Mica Plastic (fixed hinge) 126

Table 6-26 Average limit control surface loading 131

Table 6-27 Factor of Safety of Wing Components 134

Table 7-1: Material properties 144

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Table 7-2: Maximum values of shear force and bending moment for

fuselage 148

Table 7-3: Stress result for former 152

Table 7-4: Stress result for bulkhead 154

Table 7-5: Fuselage parameter 161

Table 7-6: Stress on surface 164

Table 7-7: Maximum values of shear force and bending moment 170

Table 8-1 Material used 178

Table 8-2 Properties of the material used 178

Table 9-1 Empty weight cg calculation 186

Table 9-2 Maximum weight cg calculation 187

Table 9-3 Forward cg calculation 188

Table 9-4 Aft weight cg calculation 189

Table 9-5 Pitching moment at various angle of attack for empty weight 197

Table 9-6: Pitching moment at various angle of attack for maximum

weight 199

Table 9-7: Pitching moment at various angle of attack for forward cg 200

Table 9-8: Pitching moment at various angle of attack for aft cg 201

Table 9-9 Torque varies with the aileron deflection angle 211

Table 9-10 Torque varies with the airspeed 212

Table 9-11: Longitudinal dimensionless derivatives (Nelson, 1998) 220

Table 9-12: Longitudinal dimensional derivatives (Nelson, 1998) 221

Table 9-13: Longitudinal stability coefficients 221

Table 9-14: Longitudinal dimensional derivatives 222

Table 9-15 Parameters of Cessna 172 226

Table 9-16 Flying Qualities 231

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LIST OF FIGURES

FIGURE NO. TITLE PAGE

Figure 1-1 : Flow chart for the project 3

Figure 2-1 : Cessna 172 with rectangular wing design 6

Figure 2-2 : Semi-monocoque configuration 7

Figure 2-3 : Conventional tail on RC plane 8

Figure 2-4 : Tail wheel design on light aircraft 9

Figure 2-5 : Expanded Polypropylene Foam (EPP) 11

Figure 2-6 : Isometric View of RC Plane 14

Figure 2-7 : Front, Side and Upward View of the RC plane 15

Figure 4-1 Aerodynamics forces acting on an airfoil (Anderson Jr, 2001) 18

Figure 4-2 Graph of W/g versus kv 22

Figure 4-3: Detail specification of Sunnysky X2216 23

Figure 4-4 : SunnySky X2216 23

Figure 4-5 Three common methods to mount propellers (and engines). A

is a tractor configuration, B is a pusher configuration, and

C is a tractor mounted on a nacelle configuration. 24

Figure 4-6 : Graph of CL/CD against Velocity 27

Figure 4-7 : Graph of Drag against Velocity 28

Figure 4-8 Dimension of motor 30

Figure 4-9 Propeller Specification 30

Figure 4-10 : Graph or PR, PA against velocity 31

Figure 4-11 : Illustration of climbing flight 32

Figure 4-12 : Rate of Climb against Velocity 33

Figure 4-13 : Flight path for the RC plane 37

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Figure 4-14: Flight Current/Thrust Consumptions 38

Figure 5-1: Flysky FS-iA6 RC Controller 41

Figure 5-2: Flying mode 41

Figure 5-3: Flysky FS-i6 Receiver 42

Figure 5-4: PWM Pulse 44

Figure 5-5: PPM Pulse 44

Figure 5-6: High gain omnidirectional antenna radiation pattern 47

Figure 5-7: Horizontally Polarized RC Radio Antenna (Left) 48

Figure 5-8: Forward TX Antenna Position (Right) 48

Figure 5-9: 90 Degree Polarization Diversity 48

Figure 5-10: Skywalker 30A ESC 51

Figure 5-11: Maxpower Graphene LiPo Battery 54

Figure 5-12: PWM Signal in servo control 54

Figure 5-13: Tower Pro 9g SG90 Servo Motor 55

Figure 6-1 Aerodynamic forces, airspeeds and pressure acting on wing

aerofoil 58

Figure 6-2 Dimension of the rectangular wing 59

Figure 6-3 Wing components 62

Figure 6-4 Flight Envelope 66

Figure 6-5 Resultant aerodynamic force and the components into which it

splits (Anderson, 2010) 67

Figure 6-6 the parameters c, cf and δ in aerofoil with plain aileron

(McCormick, 1995) 69

Figure 6-7 Position and dimension of ailerons 70

Figure 6-8 Aerodynamic characteristics of NACA 2412 airfoil section

(Abbott, 1945) 71

Figure 6-9 Front View of the Aircraft 72

Figure 6-10 Free Body Diagram (before simplification) 72

Figure 6-11 Free Body Diagram (after simplification) 73

Figure 6-12 Release System 74

Figure 6-13 Unit load system 74

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Figure 6-14 Reaction force of wing strut 75

Figure 6-15 Graph of Normal Lift across Wingspan (Zoom in) 85

Figure 6-16 Graph of Normal Lift across Wingspan (Zoom out) 85

Figure 6-17 Graph of Shear Force across Wingspan when the aileron

deflects downwards 86

Figure 6-18 Graph of Shear Force across Wingspan when the aileron

deflects upwards 86

Figure 6-19 Graph of Bending Moment across Wingspan when aileron

deflects downwards 87

Figure 6-20 Graph of Bending Moment across Wingspan when aileron

deflects upwards 87

Figure 6-21 Graph of Torsion across Wingspan 88

Figure 6-22 Location of ribs along the wingspan 92

Figure 6-23 The load distribution on rib 93

Figure 6-24 Cell 1 of Rib 2 94

Figure 6-25 Cell 2 of Rib 2 96

Figure 6-26 Shear force across the chord 98

Figure 6-27 Bending Moment across the chord 99

Figure 6-28 Dimension and location of spar in wing 100

Figure 6-29 Cross section of spar 101

Figure 6-30 Wing’s skin 104

Figure 6-31 Shear flow diagram of 2 cells wing 108

Figure 6-32 Vertical distance from centroid, z of cell 1 and cell 2 109

Figure 6-33 Line integral of wing section 112

Figure 6-34 Part attachment (wing strut) 123

Figure 6-35 Dimension of the connection 124

Figure 6-36 The moment arm for bending moment on hinge pin 125

Figure 6-37 Graph of kt against d/w 127

Figure 6-38 Shear stress on the contact area of the glue 127

Figure 6-39 Combined compression, bending and shear (Bruhn, 1973) 130

Figure 6-40 Average limit control surface loading 131

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Figure 6-41 Chordwise lift distribution of the aerofoil with deployed

aileron 132

Figure 6-42 Wing loading on the deployed aileron 132

Figure 6-43 Aileron attachment area 133

Figure 7-1: Fuselage monocoque structure 143

Figure 7-2: Cross section of maximum bending moment location for

fuselage (cm) 148

Figure 7-3: Pressure contour 150

Figure 7-4: Flow visualization 151

Figure 7-5: Load applied on former 152

Figure 7-6: Pressure contour on former structure for (a) equivalent (Von-

Mises) stress (b) shear stress 153

Figure 7-7: Load applied on bulkhead 154

Figure 7-8: Pressure contour on bulkhead structure for (a) equivalent

(Von-Mises) stress (b) shear stress 155

Figure 7-9: Shear flow of the fuselage 161

Figure 7-10: Flow simulation with 33.6m/s and 860RPM 164

Figure 7-11: Shear-Buckling-Stress Coefficient of Plates as a Function of

a/b 165

Figure 7-12: Chart of Nondimensional Compressive Buckling Stress 166

Figure 8-1 Diagram for nose landing gear load calculation 174

Figure 8-2 Centre of gravity for nose landing gear (front landing gear) 174

Figure 8-3 Centre of gravity for main landing gear (rear landing gear) 175

Figure 8-4 Overview of ABAQUS on Front Landing Gear being modelled

177

Figure 8-5 Loading on Front Landing Gear 179

Figure 8-6 Boundary Condition of Front Landing Gear 179

Figure 8-7 Meshing of Front Landing Gear 180

Figure 8-8 Incrementation of the analysis 181

Figure 8-9 Plot Undeformed Shape of Rear Landing Gear 182

Figure 8-10 Contour Plot of Front Landing Gear 183

Figure 8-11 Magnitude displacement of Front Landing Gear 183

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Figure 9-1 CG location of each weight on the aircraft 185

Figure 9-2 Location of cg empty and payloads 188

Figure 9-3 Location of all cg on the aircraft 190

Figure 9-4 Wing geometry for pitching moment. 191

Figure 9-5 Geometry of wing tail combination. 193

Figure 9-6: Graph of moment coefficient vs angle of attack for empty

weight 198

Figure 9-7: Graph of moment coefficient vs angle of attack for maximum

weight 199

Figure 9-8: Graph of moment coefficient vs angle of attack for forward cg

200

Figure 9-9: Graph of moment coefficient vs angle of attack for aft cg 201

Figure 9-10 Static Lateral Roll Stability 205

Figure 9-11 Wing planform undergoing a rolling motion. 207

Figure 9-12 Flap effectiveness parameter 208

Figure 9-13 Transient reponse of roll rate 209

Figure 9-14 Graph of roll rate vs deflection angle 209

Figure 9-15 Ailerons for roll control 210

Figure 9-16 Graph of torque vs deflection angle 211

Figure 9-17 Graph of torque vs airspeed 213

Figure 9-18 Static directional stability 214

Figure 9-19: Wing Body Interference Factor 215

Figure 9-20: Reynolds number correction factor 216

Figure 9-21: Time response open-loop for longitudinal stability 223

Figure 9-22: Time response for short period 224

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LIST OF APPENDICES

APPENDIX TITLE PAGE

Appendix I Graph to determine aerodynamic coefficients 238

Appendix II McCormick’s method on wing aileron analysis with aileron

239

Appendix III Bruhn’s method on structural analysis 241

Appendix IV EPP foam material properties 244

Appendix V MATLAB coding on dynamic stability and control 245

Appendix VI Minutes of meetings 247

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CHAPTER 1

INTRODUCTION

1.1 Objectives

Upon receiving this project, there are several objectives that we have to achieve. The

objectives are as follows:

i. Design, build and fly unmanned remote control aircraft

ii. Fixed wing

iii. Maximum takeoff weight is less than 5kg

iv. There will be payload with the mass of 500g

v. The flying time is around 5 minutes

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1.2 Gantt chart

The Gantt chart shows the planning for our group activities from week 1 to week 14.

Table 1-1 : List of activities from week 1 to week 14

Project SEMESTER 2 SESSION 2019/2020

W1 W2 W3 W4 W5 W6 W7 W8 W9 W10 W11 W12 W13 W14

Conceptual

Design

Preliminary

Design

Detailed Design

Construction

and

Manufacturing

Flight Testing

and

Competition

Report

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1.3 Flowchart

No

No

No

Yes

Yes

Yes

Performance data

O.K?

Select team members Design

• Fuselage

• Power Plant

• Wing

• Tail

• Landing gear

Calculation and

analysis Is design

O.K?

Construction

• Fuselage

• Power Plant

• Wing

• Tail

• Landing gear

Report Funding

Test Flight

O.K?

Practice Flight

Competition

Yes

Figure 1-1 : Flow chart for the project

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The figure above shows the flow chart for our RC plane project. It started with

the selection of members into subgroups, which will design the fuselage, wing, tail and

the landing gear. The design of the parts must be compatible with other parts according

to some regulations. Some subgroups will select the appropriate type of power plant

and electronics systems so that it will suit the RC plane.

Detailed calculations are made by each groups to avoid any unnecessary errors

when assembling or fabrication of the RC plane. The funding for this RC plane project

is by collecting RM50 from each member to buy the necessary parts. If the theoretical

performance data is acceptable, then we proceed to the flight test for our RC plane and

move on to practice flying and finally compete with the other groups.

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CHAPTER 2

CONCEPTUAL DESIGN

2.1 What is Conceptual Design?

Aircraft conceptual design involves sketching a variety of possible

configurations that meet the required design specifications. By drawing a set of

configurations, designers seek to reach the design configuration that satisfactorily

meets all requirements as well as go hand in hand with factors such as aerodynamics,

propulsion, flight performance, structural and control systems. This is called design

optimization. Fundamental aspects such as fuselage shape, wing configuration and

location, engine size and type are all determined at this stage. Constraints to design

like those mentioned above are all taken into account at this stage as well. The final

product is a conceptual layout of the aircraft configuration on paper or computer

screen, to be reviewed by engineers and other designers.

2.2 Design Selection

In this section, we will discuss the configuration of our RC plane.

2.2.1 Wing Configuration

Aircraft wings are airfoils that create lift when moved rapidly through the air.

Aircraft designers have created a variety of wings with different aerodynamic

properties. Attached to the body of an aircraft at different angles, these wings come in

different shapes. These different types of angle and shape of wing need to be evaluated

to determine the most suitable wing configuration of our aircraft design.

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The wing configuration are choose based on the analysis between different wing types

of aircraft that have been produced. From this, the best design will be used to apply on

our RC plane design. The design that we chose is the rectangular wing design. Below

shows the justification for our selection.

Rectangular Wing (Selected Design)

Figure 2-1 : Cessna 172 with rectangular wing design

The rectangular wing design is one of the most wing configuration among

aircrafts. Besides being the simplest wing to manufacture, it is also a non-tapered,

straight wing that being used in small aircrafts. The wing shape extends out from the

aircraft’s fuselage at right angles approximately.

Rectangular wing is also easy to build, and easy to design. They are also easy

to maintain and can carry more fuel. It can be designed to have one single airfoil profile

from root to tip (no aerodynamic twist) and evenly thick.

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2.2.2 Fuselage Configuration

There are several types of fuselage configuration namely Truss/framework,

monocoque and semi-monocoque. In this project, we chose semi-monocoque

configuration. This is because:

• It relies largely on the strength of the skin to carry the primary loads.

• This construction uses formers, frame assemblies, and bulkheads to

give shape to the fuselage.

• The heaviest of these structural members are located at intervals to

carry concentrated loads and at points where fittings are used to attach

others units such as wings, power plants, and stabilizers.

• The design is simple and ease to fabricate.

Figure 2-2 : Semi-monocoque configuration

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2.2.3 Tail Configuration

For the tail configuration, we had to choose between the conventional tail, T-tail and

the V-tail. After comparison, the tail that we are using is the conventional type. This

is because conventional type has its own advantages, which are:

• Most common form.

• Provides adequate stability and control with lowest structural weight.

• Low risk and ease of control and manufacturability.

• Most efficient for the speed RC planes are expected to fly it.

Figure 2-3 : Conventional tail on RC plane

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2.2.4 Landing Gear

Landing gear is one of the most important part for RC plane because we have to fly it

from the ground and land it to the ground too. Upon consideration, the suitable landing

gear for our RC plane is the tail wheel because we can connect it to our rudder for

ground movement control. The other advantages are:

• Helps slow the aircraft upon landing and provides directional stability.

• Advantageous when operating in and out of non-paved runways.

• A steerable tail wheel, connected by cables to the rudder or rudder

pedals, is also a common design.

Figure 2-4 : Tail wheel design on light aircraft

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2.2.5 Propulsion System

Based on the calculation in motor selection, the power required by the RC

aircraft is 273W. The motor that will be selected should have higher power compared

to required power. For the propeller, it depends on the motor used because different

motor have different recommended propeller. Since there is only one type of motor

that our lecturer recommended hence we chose Sunnysky X2216.

Sunnysky X2216

RPM 1100kv

Battery 2~4S

Max Current 18A

Max Power 385W

Motor Resistance 72mΩ

ESC 30A

Weight 72g

11

2.3 Material Selection For Rc Plane

For the fuselage, tail and wing. We use only one material to fabricate it which

is the EPP Foam. Expanded Polypropylene (EPP) is a highly versatile closed-cell bead

foam that provides a unique range of properties, including outstanding energy

absorption, multiple impact resistance, thermal insulation, buoyancy, water and

chemical resistance, exceptionally high strength to weight ratio and 100%

recyclability. EPP can be made in a wide range of densities, from 15 to 200 grams per

litre, which are transformed by moulding into densities ranging from 18 to 260 grams

per litre. Individual beads are fused into final product form by the steamchest moulding

process resulting in a strong and lightweight shape.

Figure 2-5 : Expanded Polypropylene Foam (EPP)

Table 2-1 : Properties of EPP Foam

EPP density range 15 g/l to 200 g/l

Tensile strength (kPa) 270 to 1930

Tensile elongation (%) 21 to 7.5

Compressive strength (kPa)

25% strain 80 to 2000

50% strain 150 to 3000

75% strain 350 to 9300

12

2.4 Weight Estimation

Below shows the weight estimation for our RC plane. The weight estimation is based

on the information that we got from the supplier and from a reliable source in the

internet. The total estimated weight for our RC plane is 1.467 kg.

i. Propulsion & Engine

Motor Sunnysky X2216 100g

Propeller Blade 35g

ii. Electronics

Servo 9g (x4) 36g

Flysky FS-i6 Receiver 6.4g

Lipo Battery (4S) 200g

ESC 30A 25g

Wires 5g

Pushrod (x4) 60g

iii. Fuselage

Polyfoam (x2 boards) 200g

Plywood (x3) 300g

iv. Payload

Plasticine 500g

13

2.5 Cost Estimation

After contacting all the supplier and surveying all the components cost. This is our

cost estimation for our RC plane.

i. Propulsion & Engine

Motor Sunnysky X2216 + Propeller

blade RM120.00

ii. Electronics

Servo 9g (x4) RM27.60

Flysky FS-i6 Receiver RM66.76

Lipo Battery 3S 2200mah RM59.60

ESC 30A RM27.00

wires RM6.00

Pushrod (x4) RM16.00

iii. Fuselage

Polyfoam (x2 boards) RM40.00

Plywood (x3) RM15.00

iv. Payload

Plasticine RM4.90

Total estimated cost of aircraft = RM382.86

14

2.5.1 Design Sketches

By using the software SolidWorks, we design the 3D sketches for our RC plane. The

purpose of these sketches is to obtain the weight estimation based on the software since

we can change the material for the RC plane based on our preferences and to have the

clear view of the RC plane. We can also calculate the surface are of the RC plane by

using SolidWorks.

Figure 2-6 : Isometric View of RC Plane

15

Figure 2-7 : Front, Side and Upward View of the RC plane

Table 2-2 : Weight & Area estimation (based on SolidWorks):

Fuselage & Tail (Foam only) = 20.97𝑔

Wing (include rib & spar) = 34.46𝑔

Surface area for polyfoam:

Wing Skin + Aileron = 4167.620 𝑐𝑚2

Fuselage = 2083.44 𝑐𝑚2

Horizontal Tail + Elevon = 307.44 𝑐𝑚2

Vertical Tail + Rudder = 124.41 𝑐𝑚2

Total Foam Area = 6682.91 𝑐𝑚2

16

CHAPTER 3

RC PLANE CONFIGURATION

The configuration of our RC plane is discussed in this chapter. Since the estimated

weight is approximately 14.715N (1.5kg) including the payload, then the empty weight

is without the 500g payload which makes it 9.81N.

Table 3-1 : The configuration of RC plane

𝐺𝑟𝑜𝑠𝑠 𝑤𝑒𝑖𝑔ℎ𝑡 14.715 𝑁

𝐸𝑚𝑝𝑡𝑦 𝑤𝑒𝑖𝑔ℎ𝑡 9.81 𝑁

𝑊𝑖𝑛𝑔 𝑎𝑖𝑟𝑓𝑜𝑖𝑙 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑁𝐴𝐶𝐴 2412

𝑊𝑖𝑛𝑔 𝑠𝑝𝑎𝑛, 𝑏 1.2 𝑚

𝑊𝑖𝑛𝑔 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚 𝑎𝑟𝑒𝑎, 𝑆 0.204 𝑚2

𝐴𝑠𝑝𝑒𝑐𝑡 𝑟𝑎𝑡𝑖𝑜, 𝐴𝑅 7.0588

𝐶ℎ𝑜𝑟𝑑 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑐 0.17 𝑚

𝑇𝑎𝑝𝑒𝑟 𝑟𝑎𝑡𝑖𝑜, 𝜆 5.8824

𝐹𝑢𝑠𝑒𝑙𝑎𝑔𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 0.73 𝑚

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑙𝑖𝑓𝑡 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 1.29

Limit factors

Our lecturer had given limiting factors to us and the values are as shown.

𝑛𝑙𝑖𝑚𝑝𝑜𝑠= 3; 𝑛𝑙𝑖𝑚𝑛𝑒𝑔

= −1.5

Stall speed

Stall speed is the slowest speed a plane can fly to maintain level flight. In our case,

by using the formula with the known value, we got:

𝑉𝑠 = √2𝑊𝑀𝑇𝑂𝑊

ρ𝐶𝐿𝑚𝑎𝑥𝑆

= √2(1.5 𝑥 9.81)

(1.225)(1.29)(0.204) = 9.5179 𝑚/𝑠

17

Takeoff speed

Takeoff speed is the speed needed for the plane to take off. Based on our

calculations, the takeoff speed for our RC plane is:

𝑉𝑇𝑂 = 1.2𝑉𝑠𝑡𝑎𝑙𝑙 = 1.2√2𝑊

ρ∝𝑆𝐶𝐿𝑚𝑎𝑥

= 11.4215 𝑚/𝑠

Landing speed

Landing speed is the speed when the aircraft land. The landing speed of our plane is

shown below:

𝑉𝐿0 = 1.3𝑉𝑠𝑡𝑎𝑙𝑙 = 1.3√2𝑊

ρ∝𝑆𝐶𝐿𝑚𝑎𝑥

= 1.3(8.6293) = 12.3733 𝑚/𝑠

18

CHAPTER 4

PROPULSION AND POWER PLANT SYSTEM

4.1 Introduction

According to NASA, the propulsion means to push forward or drive an object

forward while a propulsion system is a machine that produces thrust to push an object

forward. The foundations of powered flight can be set in 1799 when Sir George

Cayley, the father of Aerodynamics, first identified the four basic forces of flying:

weight, lift, drag, and thrust and set forth the concept of the modern aeroplane as a

fixed-wing flying machine (no flapping wings like in bird flight), with separate

systems for lift (tilted planes), propulsion (engine), and controls, designing the first

successful glider to carry a human being aloft. The basic forces of flying or known as

aerodynamic forces can be shown as in Figure 1.

Figure 4-1 Aerodynamics forces acting on an airfoil (Anderson Jr, 2001)

On airplanes, thrust is usually generated through some application of Newton's

third law of action and reaction. A gas, or working fluid, is accelerated by the engine,

and the reaction to this acceleration produces a force on the engine. A general

19

derivation of the thrust equation shows that the amount of thrust generated depends on

the mass flow through the engine and the exit velocity of the gas. Aircraft propulsion

has a wide variety of systems available. Different propulsion systems generate thrust

in slightly different ways. The choice depends on the speed or Mach number required

and the role for which the aircraft is to be used.

Aircraft electric propulsion is not a new, the idea of an all-electric air vehicle

has been researched and implemented since the mid-70’s (Newcome, 2004). Even

though piston engine is one of the most efficient internal combustion engines when

compared to others but electric motor has the potential to reach 80-90% of efficiencies.

This is due to the output of electric motor is relatively not affected by altitude,

temperatures and humidity factors which bring a lot of impact to the power output of

a combustion engine. Other than that, electrical propulsion could increase the

sustainability in aviation since it can be powered by renewable sources such as wind

and solar. The instantaneous torque and silent operation also the advantages of using

electric motor.

There are few things need to be considered when selecting electric motor

depending the application for example Alternating Current (AC) against Direct

Current (DC) and others. Even though the AC motor has a cost and a maximum torque

advantage over the DC motor, but the efficiency loss of converting the DC battery

power to AC current for an aircraft application is a primary reason that a DC motor is

selected for this project. The DC motor has a brushed and brushless construction but

brushless motor offered less maintenance, more controllable speed, no voltage drop

across brushes, high output power, small frame size and high speed range. Thus, for

electric motor, these are the following motor characteristics:

1. Direct Current (DC)

2. Brushless

20

4.2 Objective

The objective is to determine the aircraft’s motor and propeller with the power

available and required to meet the requirement of aircraft in order to produce thrust.

4.3 Methodology

4.4 Preliminary Design

Table 4-1 General characteristics or our RC plane

Type RC aircraft

Maximum Take-off Weight 1.5 kg

Engine type DC motor engine

4.5 Preliminary Selection of Motor

An important criterion in selecting a motor is a thrust to weight ratio, which

depends on the type of the aircraft.

Table 4-2 Thrust-to-Weight Ratio for different type of aircraft

Type of Aircraft Thrust- to- Weight Ratio

Glider/Trainer 0.35 to 0.55

Scale Flight 0.6 to 0.7

Sport and Slow Acrobatic 0.7 to 0.8

Acrobatic Fast 0.8 to 1.00

Jets and 3D 1.00 to 2.5

Estimate weight of

aircraft

Select motor for

desired

thrust/weight ratio

Select propeller

blade that suitable

for the motor

21

Based on the table, since the aircraft is a scale flight, the suitable thrust-to-weight ratio

that suits our aircraft is from 0.6 to 0.7.

To calculate the necessary thrust of the aircraft, the following formula is used:

𝑇ℎ𝑟𝑢𝑠𝑡 = 𝑊𝑒𝑖𝑔ℎ𝑡 ×𝑇ℎ𝑟𝑢𝑠𝑡

𝑊𝑒𝑖𝑔ℎ𝑡

Since the maximum weight of the aircraft would be not more than 5000g, then the

necessary thrust is:

𝑇ℎ𝑟𝑢𝑠𝑡 = 1500 × 0.6 = 900𝑔

Kv is simply the revolutions per minute (rpm) an electric motor will spin at per volt

when under no load.

To choose the Kv of the motor, the required speed of the aircraft must be determined:

Table 4-3 The values of airspeed according to the motor Kv

Kv Air Speed (Km/h) Air Speed (mph)

1000 70 43

2000 140 87

3000 210 131

4000 280 175

Based on the table above, it can be conclude that the higher the value of Kv, the higher

the air speed. Since the aircraft will be flying in a low air speed, the suitable value of

Kv will be around 1000.

22

Victor (2017) to determine the suitable power/thrust ratio with has done an experiment

regarding the power/thrust ratio given value of kv.

Figure 4-2 Graph of W/g versus kv

Based on the graph plotted, the equation to obtain the power/thrust ratio is :

𝑊

𝑔= 0.17 ×

𝐾𝑣

1000+ 0.09

For this case, the required thrust is 900g and by interpolation, the power/thrust ratio is

approximately 0.26. From there, the needed power can be calculated by:

𝑃𝑜𝑤𝑒𝑟(𝑊) = 900𝑔 ×0.26𝑊

𝑔= 234𝑊

With these calculated values, the motor is chosen from manufacturer’s catalogue. A

higher value of the motor power with 40% to 50% more to fly easily at 50 to 60%

throttle. The power value in parenthesis reflex this consideration.

Currently, there are 2 motors that can be used for this aircraft:

0

0.2

0.4

0.6

0.8

1

0 1000 2000 3000 4000 5000 6000

W/g

kv (rpm/volt)

Watts/gram vs Kv

23

Table 4-4 : Different brand of brushless motor

Motor Turnigy D2836/8 Sunnysky X2216

RPM 1100kv 1100kv

Max Power 336W 385W

Weight 70g 72g

The most suitable motor that the aircraft can use is the Sunnysky X2216 Brushless

Outrunner Motor. The specifications for the motor is as shown in the figure below:

Figure 4-3: Detail specification of Sunnysky X2216

Figure 4-4 : SunnySky X2216

24

4.6 Power Plant Configuration

Propellers can be mounted in a number of ways to an airplane. Three common

methods are shown in Figure x, a tractor (A), pusher (B), and a configuration featuring

the engine and propeller mounted on the wing in nacelles (C). Configuration (C) is a

variation of (A) or (B). The advantages and disadvantages of these installation methods

are listed in Table 14-1. Note that so-called “inline” configurations, which consist of

a tractor and pusher, can simply be treated as a combination of configurations (A) and

(B). Tractor configuration will be use in this propject.

Figure 4-5 Three common methods to mount propellers (and engines). A is a tractor

configuration, B is a pusher configuration, and C is a tractor mounted on a nacelle

configuration.

The incoming air of Tractor Configuration is undisturbed. Ground clearance is

not an issue during rotation at T-O, or flare before touch-down. A Placing the fuselage

behind the propeller allows for a reduced streamtube inflow distortion and less

asymmetric disc loading that would increase blade stresses. There is less chance that

the propeller suffers damage due to FOD, in particular when the aircraft is moving.

Propeller is not subject to excessive heat from the exhaust. Propwash can help with T-

O rotation during soft- or short-field take-offs, by increasing the dynamic pressure at

the horizontal tail.

25

While in pusher configuration, Ground clearance may be an issue during

rotation at T-O, or flare before landing. Fuselage ahead of the propeller may distort

flow inside the streamtube, causing asymmetric disc loading and increased blade

stresses. This distortion may affect the propeller’s performance. Propeller may suffer

FOD, in terms of both pebbles shot by tires and ingestion by ice shedding off a

fuselage. Propeller may be subject to excessive heat from the exhaust. Special

regulatory requirements for pushers are stipulated in 14 CFR 23.905. Results in a

higher fly-over noise and possible propeller corrosion issues.

The nacelle Configuration has many of the same disadvantages as the tractor

configuration; however, since the nacelle usually extends far forward from the wing’s

structural support, the airframe may be subject to structural oscillations that, if ignored,

may cause premature structural failure, and even whirl flutter phenomena.

26

4.7 Aerodynamic Drag

In this section, the theoretical calculation of our aerodynamic drag is shown below;

𝐶𝐿 =𝑊

0.5𝜌𝑉2𝑆

𝐶𝐷 = 𝐶𝐷,0 + 𝛽𝐶𝐿2 = 𝐶𝐷,0 +

𝐶𝐿2

𝜋𝑒𝐴𝑅

𝑤ℎ𝑒𝑟𝑒 𝐶𝐷0= 𝑝𝑎𝑟𝑎𝑠𝑖𝑡𝑒 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑎𝑡 𝑧𝑒𝑟𝑜 𝑙𝑖𝑓𝑡 = 0.01

𝑒 = 𝑜𝑠𝑤𝑎𝑙𝑑 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 =1

1.05 + 0.007𝜋𝐴𝑅= 0.8297

𝐷𝑟𝑎𝑔 =𝐶𝐷

0.5𝜌𝑉2𝑆

Velocity

(m/s)

Lift coefficient,

CL

Induced drag,

CDi

Drag

coefficient, CD CL/CD

Drag /

Thrust

required

4 8.742536 4.661796 14.00549 0.624222 27.9998

6 3.885572 0.920849 2.782646 1.396359 12.5169

8 2.185634 0.291362 0.894187 2.44427 7.1506

10 1.398806 0.119342 0.378126 3.699312 4.7247

12 0.971393 0.057553 0.192759 5.039414 3.4683

14 0.713676 0.031066 0.113297 6.299159 2.7747

16 0.546409 0.01821 0.07473 7.311728 2.3904

18 0.43173 0.011369 0.054206 7.964693 2.1944

20 0.349701 0.007459 0.042477 8.232798 2.1230

22 0.289009 0.005095 0.035384 8.167908 2.1398

24 0.242848 0.003597 0.030891 7.861406 2.2233

26 0.206924 0.002612 0.027935 7.407422 2.3595

28 0.178419 0.001942 0.025925 6.882174 2.5396

30 0.155423 0.001473 0.02452 6.338597 2.7574

32 0.136602 0.001138 0.023514 5.809296 3.0086

27

34 0.121004 0.000893 0.022779 5.312046 3.2903

Figure 4-6 : Graph of CL/CD against Velocity

Based on the graph, we can obtain the L/D max for our RC plane;

(𝐿

𝐷)𝑚𝑎𝑥

= 6.7347

𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 (𝐿

𝐷)𝑚𝑎𝑥

= 18 𝑚/𝑠

0

1

2

3

4

5

6

7

8

0 5 10 15 20 25 30 35 40

CL/

CD

Velocity (m/s)

28

4.8 Thrust Required

Consider an airplane in steady and level flight at a given altitude and a given

velocity. Forces acting:

𝑇 = 𝐷 = 𝑞∞𝑆𝐶𝐷

𝑊 = 𝐿 = 𝑞∞𝑆𝐶𝐿

The thrust required for an airplane to fly at a given velocity in level, unaccelerated

flight is:

𝑇𝑅 = 𝐷 = 𝑞∞𝑆(𝐶𝐷,0 + 𝐶𝐷,𝑖) = 𝑞∞𝑆𝐶𝐷

Figure 4-7 : Graph of Drag against Velocity

At minimum 𝑇𝑅,

𝐶𝐷,0 = 𝐶𝐷,𝑖

𝑧𝑒𝑟𝑜 − 𝑙𝑖𝑓𝑡 𝑑𝑟𝑎𝑔 = 𝑑𝑟𝑎𝑔 𝑑𝑢𝑒 𝑡𝑜 𝑙𝑖𝑓𝑡

This yields an interesting aerodynamic result that at minimum thrust required, zero-

lift drag equals drag due to lift.

𝑡ℎ𝑟𝑢𝑠𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑚𝑖𝑛 = 𝑑𝑟𝑎𝑔 𝑚𝑖𝑛 = 2.5952

𝑉𝑚𝑖𝑛 𝑎𝑡 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑡ℎ𝑟𝑢𝑠𝑡 = 18 𝑚/𝑠

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45

Dra

g

Velocity (m/s)

29

4.9 Power Available

Motor SunnySky X2216 1100kv

Propeller APC9045 (9x4.5)

The maximum power available based on the datasheet of the motor is 143.63W with

the efficiency of 4.628 g/W and 1000gf thrust.

To find the actual power available of the motor, the calculations are based on the

website (STARLINO) where they provide the relevant equations in obtaining the

actual power.

Based on the propeller diameter expressed in inches, the theoretical thrust can be

calculated:

𝑇𝑔𝑟𝑎𝑚 = (𝑃𝐷𝑖𝑛𝑐ℎ

𝐶) = (

143.63(9)

0.0278)

23

= 1316.987 𝑔𝑓

Where C is the air density dependent coefficient 30𝑜𝐶, 1 𝑎𝑡𝑚

= 1

0.0127√

𝑔3

2𝜋𝑄𝑎𝑖𝑟 = 0.0278

The efficiency of the motor and the propeller is given by :

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =𝑡ℎ𝑟𝑢𝑠𝑡

𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑡ℎ𝑟𝑢𝑠𝑡=

924

1316.897= 70%

Hence, the actual power available supplied by the motor is:

𝑃𝐴 = 143.63 × 0.7 = 100.8𝑊

30

Figure 4-8 Dimension of motor

Figure 4-9 Propeller Specification

Efficiency (g/W) Amps (A) Thrust (gf) Watts (W)

Voltage (V)

31

4.10 Power Required

Consider an airplane in level, unaccelerated flight at a given altitude and with

velocity 𝑉∞

The power required:

𝑃𝑅 = 𝑇𝑅𝑉∞

But as 𝑇𝑅 = 𝐷 (for an unaccelerated flight)

𝑃𝑅 = 𝑞∞𝑆(𝐶𝐷,0 + 𝐶𝐷,𝑖)𝑉∞

Or in other word,

𝑃𝑅 = 𝑧𝑒𝑟𝑜 𝑙𝑖𝑓𝑡 𝑝𝑜𝑤𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 + 𝑙𝑖𝑓𝑡 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑝𝑜𝑤𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑

Figure 4-10 : Graph or PR, PA against velocity

At minimum power required,

𝑃𝑅𝑚𝑖𝑛 = 42.24

𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑃𝑅𝑚𝑖𝑛 = 14 𝑚/𝑠

The maximum flight velocity is determined by the intersection of the maximum 𝑃𝐴

and the 𝑃𝑅 curves.

𝑉𝑚𝑎𝑥 = 28 𝑚/𝑠

0

50

100

150

200

250

0 5 10 15 20 25 30 35

Po

wer

(W

Att

)

Velocity (m/s)

Power Required

Power Available

32

4.11 Rate of Climb

Consider an airplane in steady, unaccelerated and climbing flight,

Figure 4-11 : Illustration of climbing flight

Summing forces parallel to the flight path:

𝑇 = 𝐷 + 𝑊𝑠𝑖𝑛휃

And perpendicular to the flight path:

𝐿 = 𝑊𝑐𝑜𝑠휃

The vertical velocity is called the rate of climb, R/C:

𝑅/𝐶 = 𝑉∞𝑠𝑖𝑛휃

As we know,

𝑃𝐴 = 𝑇𝑉∞ (𝑤ℎ𝑒𝑟𝑒 𝑇 = 𝑇𝐴)

𝑃𝑅 = 𝐷𝑉∞ (𝑓𝑜𝑟 𝑎 𝑙𝑒𝑣𝑒𝑙 𝑓𝑙𝑖𝑔ℎ𝑡)

𝑠𝑖𝑛𝑐𝑒 𝐷 = 𝑇𝑅

Hence,

𝑇𝑉∞ − 𝐷𝑉∞ = 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟

And for the rate of climb,

33

𝑅

𝐶= 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟/ 𝑊

Figure 4-12 : Rate of Climb against Velocity

For the maximum R/C:

(𝑅

𝐶)𝑚𝑎𝑥 = 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟𝑚𝑎𝑥/ 𝑊 =

58.56

19.62= 2.9847

From graph power vs V

max 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟 =58.56 𝑎𝑡 14 𝑚/𝑠

4.12 Endurance

RC airplane can’t fly with a LiPo battery until it is completely flat. This is because

there is a limit on how many amp-hours can discharge before the battery isn’t able to

be recharged. This means that the effective capacity of a LiPo is only 80% of your

total amp-hours.

2200𝑚𝐴ℎ 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 ∶ 4 𝑥 (2200𝑚𝐴ℎ

1000)𝑥 0.8 𝑥 60 / 35𝐴 = 12 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

-8

-6

-4

-2

0

2

4

0 5 10 15 20 25 30 35 40

Rat

e o

f cl

imb

(R

/C)

Velocity (m/s)

34

4.13 Range

To calculate the range, we multiply the endurance by the velocity of our RC plane.

𝑅𝑎𝑛𝑔𝑒 = 𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 𝑥 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦

= 12.0686 𝑥 60 𝑥 14 = 10137.624 𝑚𝑒𝑡𝑒𝑟𝑠

4.14 Takeoff Performance

The takeoff performance will estimate the distance that our aircraft needs in before

takeoff. The general formula of takeoff performance is:

𝑆𝐿𝑂 ≈1.44W2

gρ∞SCLmax[T − [D + μr(W − L)]0.7𝑉𝐿𝑂]

Where,

𝑉𝑠𝑡𝑎𝑙𝑙 = √2𝑊

𝜌∞𝐶𝐿𝑀𝑎𝑥𝑆

𝑉𝑠𝑡𝑎𝑙𝑙 = 9.5179 𝑚/𝑠

VLO = 1.2(Vstall) = 1.2(9.5179) = 11.4215 𝑚/𝑠

0.7VLO = 0.7(11.4215) = 7.9951𝑚/𝑠

𝜑 = (16ℎ𝑏

)2

1 + (16ℎ𝑏

)2

)

= (16(0.045)

1.2 )2

1 + (16(0.045)

1.2)2

= 0.2647

L = 1

2𝜌𝑉2𝑆CL

14.715 = 0.5(1.225)(7.99512)(0.204)CL

CL = 1.8424

CDi = 𝜑 (𝐶𝐿

2

𝜋𝑒𝐴𝑅)

35

= 0.2647 (1.84242

𝜋(0.8297)(7.06))

= 0.0488

DLO = 0.5𝜌𝑉0.7𝑉𝐿𝑜2𝑆(CDo + CDi)

= 0.5(1.225)(7.9951)2(0.204)(0.01 + 0.0488)

= 0.4696 𝑁

𝐿 = 0.5𝜌𝑉0.7𝑉𝐿𝑜2𝑆CLmax

= 0.5(1.225)(7.9951)2(0.204)(1.3)

= 10.3831 𝑁

g = 9.81

𝜌 = 1.225

S = 0.204

During take-off, the angle of attack of the airplane is restricted by the

requirement that the tail not drag the ground, and therefore CLmax is assumed

to be limited to 1.3 during ground roll.

CLmax = 1.3

T = 9.8 𝑁

μr = 0.02

W = 14.715 𝑁

𝑆𝐿𝑂 ≈1.44W2

gρ∞SCLmax[T − [D + μr(W − L)]0.7𝑉𝐿𝑂]

𝑆𝐿𝑂

=1.44(14.715)2

(9.81)(1.225)(0.204)(1.3)[9.8 − [0.4696 + 0.02(14.715 − 10.3831)]0.7𝑉𝐿𝑂]

𝑆𝐿𝑂 ≈ 10.58 m

36

4.15 Landing Performance

The landing performance will estimate the landing distance that our RC plane needed.

The formula for calculating the landing performance is given by:

𝑆𝐿 =1.69W2

gp∞SCLmax[T𝑟𝑒𝑣 + [D + μr(W − L)]0.7𝑉𝑇𝐷]

μr = 0.02

W = 14.715 𝑁

g = 9.81

p∞ = 1.225

S = 0.204 m2

VT = 1.3(Vstall) = 1.3(9.5179) = 12.3733 𝑚/𝑠

𝑉0.7𝑉𝑇= 0.7(12.3733) = 8.6613 𝑚/𝑠

𝐿 = 0.5𝜌𝑉0.7𝑉𝐿𝑜2𝑆CLmax

= 0.5(1.225)(8.6613)2(0.204)(1.3)

= 12.1856 𝑁

𝑊 = 𝑊1

L = 1

2𝜌𝑉2𝑆CL

14.715 = 0.5(1.225)(8.6613 2)(0.204)CL

CL = 1.5698

CDi = 𝜑 (𝐶𝐿

2

𝜋𝑒𝐴𝑅)

= 0.2647 (1.56982

𝜋(0.8297)(7.06)) = 0.0354

37

DT = 6.2462N

T𝑟𝑒𝑣 = 0 𝑙𝑏

𝑆𝐿 =1.69W2

gp∞SCLmax[T𝑟𝑒𝑣 + [D + μr(W − L)]0.7𝑉𝑇𝐷]

𝑆𝐿 =1.69(14.715)2

(9.81)(1.225)(0.204)(1.3)[0 + [6.2462 + 0.02(14.715)]0.7𝑉𝑇𝐷]

𝑆𝐿 ≈ 17 m

4.16 Flight Path

The figure below depicts the flight path for our RC plane; the maximum altitude we

estimated that our RC plane could fly is up to 20m. Based on the previous calculations,

we estimated the flight path of our RC plane for the whole time.

Figure 4-13 : Flight path for the RC plane

0

5

10

15

20

25

0 50 100 150 200 250 300 350 400 450

Alt

itu

de

(met

re)

Time(sec)

Flight Path

38

Table 4-5 : The time and altitude at which the RC plane will takeoff, cruising and

landing

Condition Time (sec) Altitude (metre)

Takeoff 0 − 11 0 to 20

Cruising 11 − 186 20

Landing 186 − 300 20 to 0

4.17 Flight Power Consumption

The figure below shows the flight current consumption for the RC plane. The grey line

indicates the thrust while the blue line indicates the current.

Figure 4-14: Flight Current/Thrust Consumptions

Table 4-6 : Percentage of thrust needed for Takeoff, Cruising and Landing

0

100

200

300

400

500

600

700

800

900

1000

0

2

4

6

8

10

12

14

0 50 100 150 200 250 300 350 400 450

Thru

st (

gf)

Cu

rren

t (a

mp

ere)

Time(sec)

Current Thrust

39

Condition Thrust (gf)

*Max thrust=1320gf

Current (A)

Takeoff speed 70% 𝑥 1320 = 924 13.04

Cruising speed 50% 𝑥 1320 = 660 9.314

Landing Speed 20% 𝑥 1320 = 264 2.34

We estimated the flight current and thrust consumption based on the percentage that

we need from the maximum thrust available from the motor. Since we required more

thrust and current during takeoff, we estimated 70% of the thrust, which is equivalent

to 924gf and 13.04A, are needed for our RC to takeoff from the ground. In cruising,

only 50% of the thrust is needed and for the landing it could be less than 20% of the

thrust because we can save more battery consumption if we lessen the thrust needed

for landing.

40

CHAPTER 5

AVIONICS

5.1 Electronic Components Selection

The selection of electronics components is mainly based on the suitability of

project, compatibility with the other components, price and avaibility of components

(preferable to buy them locally than overseas). The main electronic components to

powered an UAV are transmitter, receiver, electronic speed controller (ESC), battery

and servo. Detail description on the selection, including the specifications of the

products is explained below.

5.1.1 Transmitter and Receiver

Transmitter is an electronic device that uses radio signals to transmit

commands wirelessly via a set radio frequency over to the receiver. (All About

Multirotor Drone Radio Transmitters and Receivers, 2018) Pilot control the RC

airplane remotely through transmitter. The chosen transmitter model is the Flysky FS-

i6 RC Controller. The specification of the Flysky is as shown in Table 5-1. 6 channels

are chosen because it is needed to control the aileron, rudder, transmitter and landing

gears. It is recommended for beginners, as it is one of the reputable brand, where all

the tutorials on how to use the transmitter can be easily obtained online. In addition, it

has 20 models memory, which means it can works with 20 different receivers. Then,

the price is consider affordable for a 6 channels transmitter and receiver set (RM165).

41

Table 5-1: Transmitter specification

Figure 5-1: Flysky FS-iA6 RC

Controller

Besides, mode 2 is set as default mode on the transmitter, the stick movement

is as shown in Figure 5-2. Also, this transmitter adapt a 2.4GHz frequency, which is a

newer technology that offers “frequency hopping”. It manages a multiple user

frequency transmitting at the same time by scanning the frequency band and finding

the best available channel during the transmission. (All About Multirotor Drone Radio

Transmitters and Receivers, 2018)

Figure 5-2: Flying mode

Receivers are electric devices that works closely with transmitter. It has built

in antennas that intercept the radio signals from the transmitters, and convert them into

Model Flysky FS-i6 RC Controller

Channel 6

Suitable RC Type Fixed wing / helicopter / glider

Weight 392g

Size 174 × 89 × 190mm

Channel resolution 1024 steps

Battery Voltage 6V 1.5AA × 4

Bandwidth 500 kHz

DSC port PS2

Output PPM, PWM, iBUS

Control range 500m

42

alternating current pulses. (Perlman, 2016) The receiver then relays these signals to

the servos and sends out to the respective channels through Pulse Position Modulation

(PPM) or S-BUS commands. It is important to take note on the frequencies and

protocol during the selection of receiver. Frequencies must be the same on both

receiver and transmitter. For instance; a 2.4GHz transmitter can only work with

2.4GHz receiver. Also, a receiver must be compatible with the transmitter in order to

establish a good communication, thus a same brand of receiver and transmitter is

always recommended.

Therefore, we decided to buy the same brand receiver, which is the Flysky FS-

iA6 6 Channel receiver, where the specifications are shown in Table 5-2.

Table 5-2: Receiver specification

Model Flysky FS-i6 Receiver

Channel 6

RF receiver sensitivity -105dBm

Transmitting Power ≤20dBm

Frequency 2.4GHz

Weight 6.4g

Size 40.4 × 21.2 × 7.35mm

Channel resolution 1024 steps

Antenna length 26 × 2mm (Dual Antenna)

Modulation GFSK

System type AFHDS2A / AFHDS

Blind port Yes

Power port Yes (VCC)

Input Power 4V – 6.5V DC

Figure 5-3: Flysky FS-i6

Receiver

43

5.1.2 TX-RX Communication Signal

Frequency Band

Some commonly use frequencies are 27MHz, 72MHz, 433MHz, 900MHz,

1.3GHz and 2.4Ghz. 433Mhz, 900Mhz and 1.3GHz are typically used in long range

FPV and RC systems, while 27Mhz and 72Mhz are older frequencies which were

being used for many years in RC. Equipment operating on 27Mhz and 72Mhz used

crystals to bind the Transmitter with a Receiver, which can found on a few older

version of RC toys in the market.

2.4GHz is most popular frequency band, which adapted on Flysky FS-iA6 set

as well. It is a newer technology and it offers “frequency hopping” which does the job

of managing multiple users frequency transmitting at the same time. This is done by

scanning the frequency band and finding the best available channel during the

transmission. 2.4GHz antennas are very compact as well. Generally speaking the lower

the frequency, the larger the antenna. For that reason, 2.4GHz quickly became the “go

to” frequency.

Communication Protocol

Different Transmitter and Receiver might be capable of different protocols. There

are 2 major type of communication protocol, which is:

• TX Protocols: Communication between Radio Transmitter and Radio

Receiver.

• RX Protocols: Communications between Radio Receiver and Flight

Transmitter.

For TX Protocols, FS-iA6 transmitter capable to switch between AFHDS and

AFHDS 2A. AFHDS stands for Automatic Frequency Hopping Digital System, and is

a digital protocol that ensures 2 or more radios can operate at the same time without

interfering each other’s respective aircraft (or receiver). AFHDS 2A is the 2nd

generation of the system that added 2-way communication capability and allows for

telemetry.

44

For RX Protocols, FS-iA6 receiver support the universal PWM and PPM, as

well as the brand specific iBUS. PWM stands for Pulse Width Modulation. In PWM

each RC channel has own cable. The value of each channel is represented as a 1

millisecond (ms) to 2ms “ON” signal and this signal repeats (or updates) every 20

milliseconds. When PWM command is selected on TX, the receiver will output each

channel 1-6 respectively. . (FS-i6 Instruction Manual, 2015)

Figure 5-4: PWM Pulse

On the other hand, PPM stands for Pulse Position Modulation.Think of PPM

as several PWM signals lined up back to back. In PPM, the same signalling is used but

each channel is sent successively, then a delay, then it loops back to channel 1. In

normal PWM there are 50 updates sent per second (50Hz) which means each update

takes 20 milliseconds. So if each channel takes up to 2ms, then 10 channels can be

performed within that 20ms before which loop back to channel 1. PPM has less wiring

compared to PWM. When PPM is selected on transmitter, the receiver will output a

standard PPM signal. (Liang, 2013)

Figure 5-5: PPM Pulse

Similar to SBUS, iBUS is a digital serial protocol that allows for up to 18

channels to be supported on a single wire (compared to the 16 channels offered by

SBUS). These digital protocols have less latency than PPM, providing an overall better

solution to traditional analogue RX signals. Another point in favour of iBUS over

45

SBUS is that the signal is un-inverted, so you are not required to invert the signal when

using F1 and F4 boards as you do with SBUS. (Liang, FLYSKY TRANSMITTER &

RECEIVER BUYER’S GUIDE, 2019)

TX and RX Special Features

As the AFHDS 2A protocol is developed on the transmitter, thus the systems features

mentioned below are thus available:

• Bidirectional Communication

Capable of sending and receiving data, each transmitter is capable of receiving data

from temperature, altitude and many other types of sensors, servo calibration and

iBUS support.

• Multi-channel Hopping Frequency

This systems bandwidth range from 2.4055GHz to 2.475GHz. This band is divided in

140 channels. Each transmitter hips between 16 channels in order to reduce

interference from other transmitters.

• Omni-directional Gain Antenna

The high efficiency Omni-directional high gain antenna cuts down on interference,

while using less power and maintaining a strong reliable connection.

• ID Recognition System

Each transmitter and receiver has it’s own unique ID. Once the transmitter and

receiver have been paired, they will only communicate with each other, preventing

other systems accidentally connecting to or interfering with the system operation.

• Low Power Consumption

The system is built using highly sensitive low power components, maintaining high

receiver sensitivity, while consuming as little as little as one tenth the power of a

standard FM system, extending battery life.

Besides, safety features such as Failsafe is available to protect the models and

users if the receiver loses signal and therefore is no longer controllable. In the case of

a loss of signal, the corresponding servo will keep its last received position. If it

displays a percentage, the servo will instead move to the selected position.

46

TX-RX Binding Procedures

The transmitter and receiver have been pre-bound before delivery. The

following steps show the the binding procedures of transmitter and receiver (FS-i6

Instruction Manual, 2015):

1. Connect the supplied bind cable to the B/VCC port on the receiver.

2. Insert power into any other port.

3. Hold the ‘BIND KEY’ while powering on the transmitter to enter the bind mode.

4. Remove the power and bind cable from the receiber. Then connect the power cable

to the B/VCC port.

5. Check the servos’ operation. If not working as expected, restart this procedure from

the beginning.

Range Test Procedures

The limit of range is normally where the receiver can no longer clearly hear

what the transmitter telling it and typically falls in the 500m-600m range in normal

conditions.

In order to know the effective range, range test can be performed. Basically

range test can be done once the binding procedures are completed. The following

procedures shows the range test procedures (FS-i6 Instruction Manual, 2015):

1. Operator/pilot holds the transmitter.

2. Another helper moves the model away from the transmitter.

3. Check the model and mark the distance from where the model starts to lose

control.

47

Antenna Gain

In electromagnetics, an antenna’s power gain is a key performance number

which combines the antenna’s directivity and electrical efficiency. As transmitting

antenna, gain is describes as how well the antenna converts input power into radio

waves headed in a specified direction. As receiving antenna, gain is described as how

well the antenna converts radio waves

arriving from a specified direction into

electrical power.

Figure 5-6: High gain omnidirectional

antenna radiation pattern

High-gain omnidirectional antennas are

built for this Flysky FS-iA6 set. "Higher gain" means that the antenna radiates less

energy at higher and lower elevation angles and more in the horizontal directions.

High-gain omnidirectional antennas are generally realized using collinear dipole

arrays. The gain is increased which in turn, reduces the beamwidth, this can reeduce

interference as well. (mpnxt1, 2019) (Salt, 2019)Figure 5-6 represents the high gain

omnidirectional antenna radiation pattern.

Next, we investigate on the best antenna orientation for both transmitter and

receiver to get the best radio frequency energy. If we observe the 3D pattern from

Figure 5-6, we can see a big round doughnut shaped area of radio waves radiating

outward all around the vertical antenna, which also referred as “RF Doughnut”. The

strongest radio waves will be out to the sides all the way around the antenna (red), and

the weakest (the null zone) in line with both top and bottom tips of the antenna (green).

Therefore, it is advice to avoid having the point/tip of the TX antenna, pointing

directly at your aircraft. In that orientation it's radiating the least amount of RF energy

toward the model. The solution is to tilt the TX antenna from the vertical to the

horizontal (Figure 5-7) RF doughnut is now on its side and no worries about the

doughnut hole null zone out to the front and above. Another even better solution is

48

tilted the antenna up and forward (Figure 5-8) RF doughnut hole null zones are directed

up and out behind the pilot and down forward in front.

Figure 5-7: Horizontally Polarized RC Radio Antenna (Left)

Figure 5-8: Forward TX Antenna Position (Right)

Next, move on to the receiver antenna. Most RC manufacturers suggested to

orientate a 90 Polarization Diversity with vertical and horizontal polarized RC

antenna as shown in Figure 5-8. Polarization Diversity improves the RX's "listening"

ability in both horizontal and vertical planes as well. Bounced and reflected signals

that become polarized can also be detected in some instances by having more than one

plane of antenna polarization. (Salt, 2019)

Figure 5-9: 90 Degree Polarization Diversity

49

5.1.3 Electronic Speed Controller (ESC)

An electronic speed control or ESC is an electronic devices that controls and

regulates the speed of the brushless motor. (Electronic speed control, 2020) It may also

provide reversing of the motor and dynamic braking. The 3 main considerations when

selecting the right ESC are:

• Electric Motor: The size of the motor will greatly dictate the amount of amperage the

ESC must be able to handle.

• Propeller: Choice of propeller will also dictate what amp rating the ESC should be.

Will your drone be spinning 3, 4, 5, or 6 inch propellers? What kind of performance

will you want, and therefore what kind of propeller pitch will you use?

• Battery: ESCs are rated for amperage, along with battery cells.

The battery, motor and propeller selected are the 4S LiPo battery, SunnySky

X2216 1100kV Brushless Motors and APC 9045 Propeller. Therefore, to decide the

most suitable ESC, the motor performance chart (Table 5-3) from official SunnySky

website is referred.

From the chart, we found that with 9 inch APC propeller and 4S 14.8V LiPo

battery, the expected maximum current drawn at maximum throttle is approximately

21A. A suitable ESC need to have a 20% higher amp rating than the maximum current

drawn, which is 25.2A. Thus, a 30A rating ESC would be sufficient to support the

system.

Secondly, generally there are two type of ESC in the market, the programmable

and non programmable ESC. Programmable speed controls generally have user-

specified options which allow setting low voltage cut-off limits, timing, acceleration,

braking and direction of rotation. Reversing the motor's direction may also be

accomplished by switching any two of the three leads from the ESC to the motor.

Skywalker ESC is one of the trusted brand for programmable ESC, thus 30A

Skywalker ESC is selected.

50

Next, we found that a battery eliminator circuit (BEC) built in with the ESC is

a better option than an external BEC. A ESC with BEC onboard is cheaper, lighter

and less complicated wiring. The downside of built-in BEC is most of them has linear

circuit. The linear power supplys step down voltage by turning the extra wattage into

heat, which reduce the efficiency. Also, it is incapable of driving larger loads without

burning up, however 4 servos is still within the BEC capability, thus it is safe. Table

5-4 shows the specification of the selected 30A Skywalker ESC chosen.

Table 5-3: Sunnysky X2216 Motor Performance Chart (SunnySky X2216 Brushless Motors, n.d.)

Prop(inch) Voltage (V)

Amps (A)

Thrust (gf)

Watts (W)

Efficiency (g/W)

Load temperature in 100% throttle

APC9045 14.8

0.7 100 10.36 9.652509653

54°

1.7 200 25.16 7.949125596

2.7 300 39.96 7.507507508

4.2 400 62.16 6.435006435

5.7 500 84.36 5.926979611

7.3 600 108.04 5.553498704

9 700 133.2 5.255255255

10.6 800 156.88 5.099439062

12.5 900 185 4.864864865

14.6 1000 216.08 4.627915587

16.4 1100 242.72 4.531970995

21 1320 310.8 4.247104247

51

Table 5-4: ESC Specifications

Figure 5-10: Skywalker 30A

ESC

Model Skywalker 30A ESC

Continuous Current 30A

Burst Current 40A (10s)

BEC included

BEC output 5V/ 2A

BEC output capability 4 servos

Size 70 × 25 × 8 mm

Weight 38g

Programmable Yes

52

5.1.4 Battery

Battery plays an essential role in powering the motor, receiver and servo. The

common battery for RC plane are the NiMH (Nickel Metal Hydride), Li-ion (Lithium-

Ion), LiPo (Lithium Polymer) battery. Table 5-5 below are the comparison in term of

properties and capability. (Monti, 2010) (Differences Between NiMH and LiPo

Batteries, n.d.)

Table 5-5: Comparison of RC Battery NiMH battery Li-ion battery LiPo battery

Rechargable Yes

Memory effect exist

Yes

No Memory effect

Yes

No memory effect

Weight Heavy Medium Light

Energy density Least High Medium

Self discharge High High Low

Discharge rate Low High High

Capacitance Low High High

Suitability Simple project More complicated RC

project

More complicated

RC project

Price Lowest Medium Highest

Safety Reliable, common

battery type, safe

choice for beginners.

Not fully mature -

metals and chemicals

are changing on a

continuing basis.

higher resistant to

overcharge and

electrolyte leakage.

*Memory effect: Batteries must be fully discharged in order to keep full capacity

available.

* No memory effect: Batteries don’t have to be fully discharged before recharging.

Next, the major considerations when choosing the suitable battery include the

following:

• Voltage: The higher the voltage, the more power and higher rpm of motor will

achieve.

• Capacity: the current that could be discharged for an hour.

53

• Weight: High power/weight ratio is always preferable.

• Dimension: Able to fit in the compartment (fuselage).

• Connector: Compatible with ESC.

• Discharge rate: The rating of how many amps of power the battery can deliver.

• Safety: Trusted and reliable brand.

By comparing the type of batteries with above factors, we notice that LiPo

battery is the best option, although it comes with a higher price, but it is rechargable,

higher power/weight ratio, higher discharge rate and capacitance, and more safer

option for a RC airplane project. Therefore, a Maxpower LiPo battery is chosen. The

specifications is as below (Table 5-6).

Next, a 4S 14.8V of LiPo battery is chosen as it can produce a relatively more

power required than a 3S LiPo battery for a stable take off. Next, for a 4S 2200maH

35C LiPo pack we obtained:

(2200mAh × 35C) / 1000 = 77A

In other words, a 2200mAh / 35C LiPo can only handle an continuous current

draw of 77A at max. By referring the motor performance chart again (Table 3),

maximum current draw is only 21A, which means a 35C of discharge rate is sufficient.

(Salt, 11 Things to Know About LiPo Batteries to Get the Best Performance, Life,

Value & Fun Out of Them, Whatever You Fly., 2020)

As the requirement for the flight test is a 5 minutes of endurance, thus the

capacity of battery (2200mAh) has ensured to meet the requirement using formula as

below:

((2200mAh/1000) × 0.8 × 60) / 20A = 5.28 minutes

54

Table 5-6: Battery Specification

Figure 5-11: Maxpower Graphene

LiPo Battery

5.1.5 Servo

Servos provide the ability to move control surfaces. A typical servo consists of

an electric motor that moves a rotary actuator via a reduction gearset, normally

providing increased torque as well as clutch mechanism. Theoretically, each servo

output generates a standard PWM signal: the peak of the signal lasts for a period of

between one and two milliseconds, with a resolution control of 1 μs. So when the on-

time is 1ms the motor will be in 0° and when 1.5ms the motor will be 90°, similarly

when it is 2ms it will be 180°. So, by varying the on-time from 1ms to 2ms the motor

can be controlled from 0° to 180°. (Introduction to Servos, 2020) The figure represents

these parameters graphically.

Figure 5-12: PWM Signal in servo control

Model Maxpower Graphene Lipo

2200mAh 4S 14.8V 35C

Capacity 2200 mAh

Voltage 4S 14.8V

Discharge rate 35C

Size 106 × 34 × 33 mm

Weight 233g

Connector Dean Plug

55

The specifications of the servo selected is as shown in table 5-7. The servo

chosen is Arduino Tower Pro 9g SG90 Servo Motor. A total of 4 servos are purchased

for aileron, rudder and elevator respectively. The most important parameter, is the

torque at which the motor operates. The Towerpro SG90 Motor provides a 1.8kg/cm

torque at 4.8V voltage. In other words, this 1.8kg/cm torque means that the motor can

pull a force of 1.8kg when it is suspended at a distance of 1cm through pushrod. So if

the pushrod is 0.5cm long then the motor can pull a load of 3.6kg similarly if the

pushrod is 2cm long then can pull only 0.9kg.

Also, it is important to familiar with the wire configuration. There are 3 wires

with different colour and purposes respectively. To make this motor rotate, we have to

power the motor with +5V using the Red wire, while the Brown wire as ground wire

and PWM signal is given in through this Orange wire to drive the servo motor. (Servo

Motor SG-90, 2017)

Table 5-7: Servo Specification

Figure 5-13: Tower Pro 9g

SG90 Servo Motor

Model Arduino Tower Pro 9g SG90

Servo Motor

Modulation Analog

Operating voltage 4.8V

Stall torque (4.8V) 1.8 kg/cm

Operating speed (4.8V) 0.12 sec/ 60

Size 23 × 12.2 × 29 mm

Weight 9g

Gear type POM gear set

Operating Temperature 0C - 55C

Servo wire length 25cm

56

5.1.6 Wiring System Schematic Diagram

57

CHAPTER 6

WING

6.1 Introduction

6.1.1 Function of Wing

A wing is a type of fin that produces lift, while moving through air or some other fluid. As

such, wings have streamlined cross-sections that are subject to aerodynamic forces and act as

aerofoils. A wing's aerodynamic efficiency is expressed as its lift-to-drag ratio. The lift a wing

generates at a given speed and angle of attack can be one to two orders of magnitude greater than

the total drag on the wing. A high lift-to-drag ratio requires a significantly smaller thrust to propel

the wings through the air at sufficient lift.

The design and analysis of the wings of aircraft is one of the principal applications of the

science of aerodynamics, which is a branch of fluid mechanics. In principle, the properties of the

airflow around any moving object can be found by solving the Navier-Stokes’ equations of fluid

dynamics. However, except for simple geometries these equations are notoriously difficult to solve

and simpler equations are used.

58

Figure 6-1 Aerodynamic forces, airspeeds and pressure acting on wing aerofoil

For a wing to produce lift, it must be oriented at a suitable angle of attack. When this occurs,

the wing deflects the airflow downwards as it passes the wing. Since the wing exerts a force on

the air to change its direction, the air must also exert an equal and opposite force on the wing,

resulting in different air pressures over the surface of the wing. A region of lower-than-normal air

pressure is generated over the top surface of the wing, with a higher pressure on the bottom of the

wing. These air pressure differences can be measured directly using instrumentation or can be

calculated from the airspeed distribution using basic physical principles such as Bernoulli's

principle, which relates changes in air speed to changes in air pressure.

It is possible to calculate lift from: the pressure differences, the different velocities of the

air above and below the wing, or from the total momentum change of the deflected air. Debates

over which mathematical approach is the most convenient to use can be mistaken as differences

of opinion about the basic principles of flight.

6.1.2 Aircraft Wing Design

In our RC place design, we implemented the high wing design by considering some of the

factors. Each wing configuration is beneficial in its unique way for training, performance,

maintenance, and everyday use. High wing aircraft are generally less aerodynamic, slightly easier

to train in for the new pilot, and easier to access for routine maintenance than low wing aircraft.

59

For high and low wing aircraft, major differences will occur in the following regimes:

lateral (roll) stability, low speed handling characteristics, and general aircraft performance (cruise

speed, take-off and landing distances etc.).

For the high wing aircraft, the centre of gravity sits below the wing, meaning the fuselage

of the aircraft acts as a pendulum to increase roll stability relative to the low wing aircraft, whose

centre of gravity is balanced above the wing.

Figure 6-2 Dimension of the rectangular wing

Rectangular wing design is chosen as it can bring a lot of advantages such as act as wing-

root stallers. For the high wing aircraft, the centre of gravity sits below the wing, meaning the

fuselage of the aircraft acts as a pendulum to increase roll stability relative to the low wing aircraft,

whose centre of gravity is balanced above the wing.

According to Part 23 of Federal Aviation Regulations, the criteria of wing design that are

required to abide by are as follows,

60

1. A main wing located closer to the airplane's center of gravity than to the aft, fuselage-

mounted, empennage;

2. A main wing that contains a quarter-chord sweep angle of not more than 15 degrees

fore or aft;

3. A main wing that is equipped with trailing-edge controls (ailerons or flaps, or both);

4. A main wing aspect ratio not greater than 7.

Table 6-1 Specification of the wing

Type High Wing

Airfoil Selection NACA2412

Aspect Ratio 7.06

Chord Length 0.17m

Wing Area 0.20387m

2

Wingspan 1.2m

6.1.3 Wing Load Analysis

In this study, our RC plane is designed and analyzed. The design process starts with a

sketch of how the airplane is envisioned. Weight is estimated based on the sketch and a chosen

design mission profile. A more refined method is conducted based on calculated performance

parameters to achieve a more accurate weight estimate which is used to acquire the external

geometry of the airplane.

In order to obtain a more refined estimation of the airplane weight several performance

parameters must be estimated such as the maximum lift-to-drag ratio, the maximum lift coefficient,

the wing loading and the thrust-to-weight ratio.

According to Federal Aviation Regulations (FAR), aircraft loads are those forces and

loadings applied to the aircraft structural components to establish the strength level of the

61

completed aircraft. Those are the loads that an aircraft structure must be designed to withstand. A

large part of the forces that make up design loads are the forces resulting from the flow of air about

the aircraft surfaces – the same forces that enables flight and control of the aircraft. The loads that

must be determined early in the structural design.

In normal straight and level flight, the wing lift supports the weight of the aircraft.

However, the greatest air loads on an aircraft usually come from the generation of lift during high-

g manoeuvres. The amount of additional loads depends on the severity or turbulence (during gust),

and its magnitude is measures in terms of load factor, n. load factor is a multiplying factor which

defines a load in terms of weight.

At lower speeds, the highest load factor an aircraft may experience is limited by the lift

available, and it is limited to some arbitrary value based upon the operation of the aircraft. At high

angle of attack, the load direction may actually be forward of the aircraft body-axis vertical,

causing a forward load component on the wing structure.

The aircraft maximum speed, or dive speed 𝑉𝐷𝑖𝑣𝑒 represents the maximum dynamic

pressure, q. The point representing maximum q and maximum load factor is vital for structural

sizing. At this condition, the aircraft is at a fairly low angle of attack because of the high dynamic

pressure, so the load is approximately vertical in the body axis.

Compliance with the strength requirements must be shown at any combination of airspeed

and load factor on and within the boundaries of a flight envelope, that represents the flight loading

conditions specified by the manoeuvring (excluding gust) criteria in our case. In manoeuvring

envelope, the aircraft is assumed to be subjected to symmetrical manoeuvres resulting in the

following limit load factors.

1. The positive manoeuvring load factor (+3.0) at speeds up to 𝑉𝐷𝑖𝑣𝑒.

2. The negative manoeuvring load factor (-1.5) at 𝑉𝐶 (Design cruising speed).

62

The process is then followed by the analysis of the wing parts such as spar, skin, rib and

also shear flow of the wing.

Figure 6-3 Wing components

In a fixed-wing aircraft, the spar is often the main structural member of the wing, running

span wise at right angles (or thereabouts depending on wing sweep) to the fuselage. The spar

carries flight loads and the weight of the wings while on the ground. The wing rib of an aircraft is

a very critical part. It provides airfoil contour to the wing. Its principal role in wing structure is

that to transfer load from skin to stringers or other parts of wing.

6.1.4 Aircraft’s Airworthiness

To maintain a high level of safety, the aviation industry is heavily regulated. The design

process of an aircraft is regulated by airworthiness authority to ensure the aircraft’s suitability for

safe flight. In this project, the Federal Aviation Regulation (FAR), Part 23 – Airworthiness

Standards: Normal Category Airplanes, is applied in the design process and summarized in Table

6-2, for utility aircraft. Also, our project requirement is align with the FAR as shown in Table 6-

3.

63

Table 6-2 Federal Aviation Regulations, Part 23 design requirements and specifications

Table 6-3 Project Requirement on Wing Design

Project Requirement

Maximum take-off weight <5kg

Positive load factor, n +3.0

Negative load factor -1.5

6.1.5 Assumptions on Wing Design

There are a few assumptions outlined on the wing design, which can be listed as follows,

1. The wing is assumed to be a 2D wing, at which the factors of 3D wing design will not be

taken into account.

2. The value of maneuvering speed 𝑉𝐴 need not exceed the value of cruising speed 𝑉𝐶 used

in design, as stated in FAR Section 23.335.

Utility Aircraft: FAR 14

Maximum take-off weight 12,500 lbs (5700kg) 14 CFR23.3

Positive load factor, n +4.4 14 CFR23.337

Negative load factor -1.76 (> 0.4 times positive n)

Design airspeeds 𝑉𝐷𝑖𝑣𝑒 > 1.5 𝑉𝑐𝑚𝑖𝑛 14 CFR23.335

64

6.2 Flight Envelope and Wing Loadings

6.2.1 Flight Envelope (V-n Diagram)

In aerodynamics, the flight envelope defines operational limits for an aerial platform with

respect to maximum speed and load factor given a particular atmospheric density. The flight

envelope is the region within which an aircraft can operate safely. If an aircraft flies 'outside the

envelope' it may suffer damage; the limits should therefore never be exceeded.

To construct the flight envelope, the load factor limits are +3.0 and -1.5 according to the

project requirement, as per FAR Part 23. Also, there are various important features of the V-n

diagram at point A, D, E, and G.

The intersection of the positive limit of the load factor and the line of maximum lift (point

A) defines the maximum airspeed that allows full manoeuverability. This point is called the

manoeuver speed or corner speed. At lower speeds, the structure cannot be overstressed as it will

stall before reaching the limit load factor. At the manoeuver airspeed the aircraft's limit load factor

will be reached at the lowest possible airspeed. At higher speeds, possible structural damage may

be caused.

Besides, the intersection of the negative limit load factor and line of maximum negative

lift capability (point G) defines the maximum airspeed that allows full manoeuverability in a

negative lift situation. Airspeeds greater than point G provide sufficient negative lift to damage

the structure.

Additionally, the diagram defines the never exceed speed or diving speed (point D and E).

This is the maximum speed before the aircraft enters the region where structural failure is possible.

65

Positive load factor

𝑛𝑚𝑎𝑥 = 3

Maneuver speed (Positive Limit)

𝑉𝐴 = √𝑛𝑚𝑎𝑥𝑝𝑜𝑠

𝑊

0.5𝜌𝑆𝐶𝐿𝑚𝑎𝑥

𝑉𝐴 = √3(14.715)

0.5(1.225)(0.203871)(1.29) = 16.5545 𝑚/𝑠

Design Dive Speed

𝑉𝐷 = 1.2𝑉𝑚𝑎𝑥

𝑉𝐷 = 1.2(28)

𝑉𝐷 = 33.6 𝑚/𝑠

Maximum Negative Load Factor

𝑛𝑚𝑎𝑥𝑛𝑒𝑔= −1.5

Maneuver Speed (Negative Limit)

𝑉𝐺 = √𝑛𝑚𝑎𝑥𝑛𝑒𝑔

𝑊

0.5𝜌𝑆𝐶𝐿𝑚𝑎𝑥

𝑉𝐺 = √−1.5(14.715)

0.5(1.225)(0.203871)(−0.6647)

𝑉𝐺 = 16.3073 𝑚/𝑠

For the curve part of the V-n diagram, the curve is drawn by varying airspeed before maneuvering

speed with equation:

𝑛 =0.5𝜌𝑉2𝑆𝐶𝐿𝑚𝑎𝑥

𝑊 (for positive V − n)

66

For positive V-n region, airspeed varies from 0 to 16.5545 ft/s.

𝑛 =0.5𝜌𝑉2𝑆(−𝐶𝐿𝑚𝑎𝑥

)

𝑊 (for negative V − n)

For negative V-n region, airspeed varies from 0 to 16.3073 m/s.

Figure 6-4 Flight Envelope

6.2.2 Wing Loading without Aileron

Normal Lift Force

Firstly, the lift coefficient, 𝐶𝐿 is calculated from the equation:

𝐶𝐿 =𝑛𝑊

12⁄ 𝜌𝑉2𝑆

where,

𝑊 : maximum take-off weight

𝑛 : load factor at points A, D, E and G of V-n diagram

𝐶L : lift coefficient

67

𝜌 : air density

𝑉 : stall velocity at points A, D, E and G of V-n diagram

𝑆 : wing area

Since assumption was made that the wing was a 2D wing, all the aerodynamic coefficient

are obtained, based on 2D wing analysis. the section aerodynamic angle of attack, 𝛼 is then

determined from the graph of 𝐶𝐿 𝑣𝑠 𝛼 for NACA 2412 wing section in Appendix I. Then, the

aerofoil section lift coefficient, 𝐶𝑙 and drag coefficient, 𝐶𝑑 are determined from graphs in Appendix

I and substituted into the following equations, which are derived from resolving aerodynamic

forces in Figure 6-5, to find the lift force acting perpendicular to aerofoil, at each station for points

A, D, E and G of V-n diagram.

Figure 6-5 Resultant aerodynamic force and the components into which it splits (Anderson, 2010)

𝐶𝑛=𝐶𝑙𝑐𝑜𝑠𝛼+𝐶𝑑𝑠𝑖𝑛𝛼

𝑁 =1

2𝐶𝑛𝜌𝑉2𝑆

68

Shear Force

The changes in shear force, Δ𝑉𝑛 at each station is given by,

Δ𝑉𝑛 =𝑁𝑛+1 + 𝑁𝑛

2(𝑦𝑛+1 − 𝑦𝑛)

and the corresponding shear force, 𝑉𝑛 at each station is,

𝑉𝑛 = ∆𝑉20 + ∆𝑉19 + ⋯+ ∆𝑉𝑛

Bending Moment

The changes in bending and moment distribution, Δ𝑀𝑛 at each station is given by,

Δ𝑀𝑛 =𝑉𝑛+1 + 𝑉𝑛

2(𝑦𝑛+1 − 𝑦𝑛)

and the corresponding bending moment, 𝑀𝑛 at each station is,

𝑀𝑛 = ∆𝑀20 + ∆𝑀19 + ⋯+ ∆𝑀𝑛

Torsional Moment

The value of moment coefficient (𝐶𝑚) can be obtained in similar way as how angle of

attack is determined. At each station, the 𝐶𝑚 value can be determined from the graph in Appendix

I for points A, D, E and G of V-n diagram.

where,

T : Torsion

𝜌 : air density

𝑉 : stall velocity at points A, D, E and G of V-n diagram

𝑆=cdy : wing area at each station, y

c : chord length at each station, y

𝐶𝑚 : moment coefficient

𝑇 =1

2𝐶𝑚𝜌𝑉2𝑆𝑐

69

The calculated results are tabulated in the following tables for all points A, D, E and G while the

graphs are plotted in figures in section 6.2.5.

6.2.3 Wing Loading with Aileron

Change in Lift Coefficient

Roll control can be achieved by the differential deflection of ailerons. The deflection of

aileron controls the increment of lift and the incremental change in roll moment. Downward aileron

deflection will increase the lift coefficient value while upward aileron deflection will decrease the

lift coefficient value. The deflection of aileron, either upward or downward, will affect the resultant

normal lift acting on the aircraft’s wing. McCormick (1995) gives the following formula for the

section lift coefficient increment due to deflection of ailerons.

Where

𝐶𝑙𝛼 :Airfoil section lift-curve slope

τ :Flap effectiveness factor

η :Correction factor to flap effectiveness factor

δ :Deflection angle

Figure 6-6 the parameters c, cf and δ in aerofoil with plain aileron (McCormick, 1995)

∆𝐶𝐿 = 𝐶𝑙𝛼𝜏η𝛿

70

Figure 6-7 Position and dimension of ailerons

Sample calculation is done for aircraft’s wing at position A. From Figure 1 in Appendix II,

τ = 0.64 for cf/c = 0.24, and from Figure 2 in Appendix II, η = 0.53 for a plain flap deflected 30o.

Hence, from Equation, ∆𝐶𝐿 is equal to


∆𝐶𝐿 = (0.2623)(0.64)(0.53)(30)(𝜋

180)

∆𝐶𝐿 = 0.04659

From Figure 3 in Appendix II, the ratio of ∆𝐶𝐿𝑚𝑎𝑥 to ∆𝐶𝐿 is obtained as 0.75. Hence,

∆𝐶𝐿𝑚𝑎𝑥 = 0.03494

Sample calculation is done for aircraft’s wing at position D. From Figure 1 in Appendix II,

τ = 0.64 for cf/c = 0.24, and from Figure 2 in Appendix II, η = 0.8 for a plain flap deflected 10o.

Hence, from Equation, ∆𝐶𝐿 is equal to


∆𝐶𝐿 = (0.2623)(0.64)(0.8)(10)(𝜋

180)

∆𝐶𝐿 = 0.02344

From Figure 3 in Appendix II, the ratio of ∆𝐶𝐿𝑚𝑎𝑥 to ∆𝐶𝐿 is obtained as 0.75. Hence,

∆𝐶𝐿𝑚𝑎𝑥 = 0.01758

71

Position A and position G will have the same maximum aileron deflection while the

maximum aileron deflection at position D and E is the same. Based on the roll control analysis, to

control the roll, when the critical speed is too high, the approximated maximum aileron deflection

is 10o, while when the critical speed is lower, the maximum aileron deflection is approximated to

be 30o. When the aileron deflects downwards, the total 𝐶𝐿 value for the wing section with aileron

is the addition of value of ∆𝐶𝐿𝑚𝑎𝑥 and the value of 𝐶𝐿 at the angle of attack in position A, D, E and

G respectively. On the other hand, when the aileron deflects upwards, the total 𝐶𝐿 value for the

wing section with aileron is the subtraction of value of ∆𝐶𝐿𝑚𝑎𝑥 and the value of 𝐶𝐿.

Change in Moment Coefficient

The roll control of the aileron also affects the value of moment coefficient 𝐶𝑚. The

estimated change in 𝐶𝑚 value can be determined from Figure 6-8. The figure shows that the graph

shifts down when the aileron deflects with an angle of 60o. As our maximum aileron deflection

angle is 30o and 10o for position A and G as well as position D and E respectively, interpolation

method is used to obtain the approximated change in moment coefficient value.

Figure 6-8 Aerodynamic characteristics of NACA 2412 airfoil section (Abbott, 1945)

72

6.2.4 Reaction force of Wing Strut

The reaction force of the wing strut is necessary when considering the wing loading, as this

affects the shear force diagram and bending moment diagram. To determine the reaction force of

the wing strut, the analysis of the wing loading must first be conducted for half of the wingspan.

The simplification of the problem is shown as follows,

Figure 6-9 Front View of the Aircraft

Figure 6-10 Free Body Diagram (before simplification)

The positive convention of the reaction force and moment is indicated in Figure 6-10. The

aircraft’s wing is assumed to be a cantilevered beam. The reaction force of the wing strut is

indicated at point B, which is located at the middle of the wingspan. The horizontal reaction forces,

RxA and RxB will cancel out each other at the end of the calculation. During initial stage of design,

since the wing weight is smaller, as compared to other component weight of the aircraft, the weight

73

of the aircraft can be assumed to be concentrated at a point A. The applied load, which is indicated

with the orange line, shows the total concentrated weight of the whole aircraft.

Figure 6-11 Free Body Diagram (after simplification)

The simplified problem now is a statically indeterminate structure. The reaction force of

the wing strut at point B can be determined by using the flexibility method, according to the

Castigliano’s Theorem. Since the system has three reaction forces, RyA, RyB and MA, the degree

of statically indeterminate structures can be determined through the equation below,

Q = [Total number of reactions] – [Total number of Static Equations]

Where, Q = Redundancy

The general equations of the flexibility method is shown as follows,

𝛿𝑖𝑛 = 𝛿𝑖𝑛0 + ∑ 𝑋𝑖𝑚

𝑘𝑚=1 ∙ 𝑓𝑚𝑛 n=1,2,3, … k

𝛿𝑖𝑛0 = ∫

𝑀𝐿(𝑥)𝑀𝑢𝑛(𝑥)

𝐸𝐼𝑑𝑥

𝑓𝑚𝑛 = ∫𝑀𝑢𝑚(𝑥)𝑀𝑢𝑛(𝑥)

𝐸𝐼𝑑𝑥

Where,

𝑋𝑖𝑚 = Parameter referring to redundancy

𝛿𝑖𝑛0 = Displacement of statically determinate system

𝑓𝑚𝑛 = flexibility

74

𝑀𝐿(𝑥) = Bending moment from actual load

𝑀𝑢𝑚(𝑥) = Bending moment for unit load in reference to redundancy

𝑀𝑢𝑛(𝑥) = Bending moment for unit load due to deflection

Sample Calculation at Position A

𝑄 = 3 − 2 = 1 𝑟𝑒𝑑𝑢𝑛𝑑𝑎𝑛𝑐𝑦

Take 𝑅𝐵 as the redundancy,

For release system,

Figure 6-12 Release System

𝑀𝐿 = −𝑞𝑥2

2

For unit load,

Figure 6-13 Unit load system

𝑀𝑢1 = 0 𝑓𝑜𝑟 0 < 𝑥 <𝐿

2

𝑀𝑢1 = (𝑥 −𝐿

2) 𝑓𝑜𝑟

𝐿

2< 𝑥 < 𝐿

The general equation,

𝛿𝐵10 = ∫

𝑀𝐿(𝑥)𝑀𝑢𝑛(𝑥)

𝐸𝐼𝑑𝑥

𝐿

𝐿2

75

𝛿𝐵10 = ∫

−𝑞𝑥2

2(𝑥 −

𝐿2)

𝐸𝐼𝑑𝑥

𝐿

𝐿/2

𝛿𝐵10 = −

𝑞𝐿2

8𝐸𝐼+

𝑞 (𝐿2)2

8𝐸1+ [

𝑞𝐿4

12𝐸𝐼−

𝑞𝐿 (𝐿2)3

12𝐸𝐼]

𝛿𝐵10 = −

17𝑞𝐿4

384𝐸𝐼

𝑓𝑚𝑛 = ∫(𝑥 −

𝐿2) 2

𝐸𝐼𝑑𝑥

𝐿

𝐿/2

𝑓𝑚𝑛 =1

𝐸𝐼∫ (𝑥2 − 𝑥𝐿 +

𝐿2

4)𝑑𝑥

𝐿

𝐿/2

𝑓𝑚𝑛 =𝐿3

24𝐸𝐼

Therefore,

𝛿𝐵1 = 𝛿𝐵10 + 𝑥1 ∙ 𝑓11 = 0

−17𝑞𝐿4

384𝐸𝐼+

𝐿3

24𝐸𝐼𝑥1 = 0

𝑥1 = 𝑅𝐵 = 1.0625𝑞𝐿

At position D, q = 44.18931 N/m, L = 0.6m

𝑅𝐵 = 28.1707 𝑁 (𝑎𝑐𝑡𝑖𝑛𝑔 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑)

To determine the reaction force of the wing strut,

Figure 6-14 Reaction force of wing strut

𝑅𝑠𝑡𝑟𝑢𝑡 =𝑅𝐵

sin(19.46)

76

𝑅𝑠𝑡𝑟𝑢𝑡 =27.7721

sin(19.46)= 84.5589

77

6.2.5 Shear Force Diagram and Bending Moment Diagram

Table 6-4 Wing loading in each station for point A (Aileron deflects downwards)

y (m) y/b α (o) Cl Cd Cn Va (N) N (N) Delta V Vn (N) Delta M M (Nm) Cm T (Pa)

0 0.0000 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -1.3938 -0.0669 0.0179 -0.1 -0.58184

0.032 0.0533 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 -2.7879 -0.1074 0.0848 -0.1 -0.58184

0.063 0.1050 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -4.1384 -0.1547 0.1921 -0.1 -0.58184

0.095 0.1583 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 -5.5324 -0.1924 0.3469 -0.1 -0.58184

0.126 0.2100 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -6.8829 -0.2426 0.5393 -0.1 -0.58184

0.158 0.2633 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 -8.2770 -0.2775 0.7819 -0.1 -0.58184

0.189 0.3150 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -9.6275 -0.3304 1.0594 -0.1 -0.58184

0.221 0.3683 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 -11.0215 -0.3750 1.3898 -0.1 -0.58184

0.253 0.4217 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 -12.4156 -0.4058 1.7648 -0.1 -0.58184

0.284 0.4733 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 0.0436 -13.7661 -0.0138 2.1706 -0.1 -0.58184

0.285 0.4750 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 -13.8425 -13.8096 -0.0069 2.1844 -0.1 -0.58184

0.286 0.4767 12.0179 1.2898 0.0543 1.2728 16.5545 -27728.53 -13.8425 0.0329 0.0070 2.1912 -0.1 -0.58184

0.287 0.4783 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.2634 13.8754 0.3841 2.1843 -0.1 -0.58184

0.316 0.5267 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3505 12.6120 0.3700 1.8002 -0.1 -0.58184

0.347 0.5783 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.3941 11.2615 0.3381 1.4302 -0.1 -0.58184

0.379 0.6317 12.0179 1.2898 0.0543 1.2728 16.5545 43.5641 1.4128 9.8675 0.2932 1.0921 -0.1 -0.58184

0.411 0.6850 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.3867 8.4547 0.2406 0.7990 -0.54 -3.14194

0.442 0.7367 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.4315 7.0679 0.2033 0.5584 -0.54 -3.14194

0.474 0.7900 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.3867 5.6365 0.1532 0.3551 -0.54 -3.14194

0.505 0.8417 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.4315 4.2497 0.1131 0.2019 -0.54 -3.14194

0.537 0.8950 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.3867 2.8182 0.0659 0.0888 -0.54 -3.14194

0.568 0.9467 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 1.4315 1.4315 0.0229 0.0229 -0.54 -3.14194

0.6 1.0000 12.0179 1.3248 0.0543 1.3070 16.5545 44.7338 0.0000 0.0000 0.0000 0.0000 -0.54 -3.14194 0.0000 -1.3938 0.0179

78

Table 6-5 Wing loading in each station for point D (Aileron deflects downwards)


0 0.0000 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -1.1490 -0.0594 0.1497 -0.10 -2.3969

0.032 0.0533 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 -2.5631 -0.1007 0.2091 -0.10 -2.3969

0.063 0.1050 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -3.9329 -0.1485 0.3098 -0.10 -2.3969

0.095 0.1583 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 -5.3470 -0.1870 0.4583 -0.10 -2.3969

0.126 0.2100 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -6.7168 -0.2376 0.6453 -0.10 -2.3969

0.158 0.2633 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 -8.1309 -0.2733 0.8828 -0.10 -2.3969

0.189 0.3150 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -9.5008 -0.3266 1.1561 -0.10 -2.3969

0.221 0.3683 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 -10.9148 -0.3719 1.4828 -0.10 -2.3969

0.253 0.4217 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 -12.3289 -0.4034 1.8547 -0.10 -2.3969

0.284 0.4733 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 0.0442 -13.6988 -0.0137 2.2581 -0.10 -2.3969

0.285 0.4750 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 -14.0412 -13.7429 -0.0067 2.2718 -0.10 -2.3969

0.286 0.4767 0.3376 0.3131 0.0543 0.3134 33.6000 -28126.51 -14.0412 0.2982 0.0073 2.2786 -0.10 -2.3969

0.287 0.4783 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.2815 14.3394 0.3973 2.2712 -0.10 -2.3969

0.316 0.5267 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.3699 13.0579 0.3836 1.8740 -0.10 -2.3969

0.347 0.5783 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4141 11.6880 0.3514 1.4904 -0.10 -2.3969

0.379 0.6317 0.3376 0.3131 0.0543 0.3134 33.6000 44.1893 1.4537 10.2740 0.3055 1.1390 -0.10 -2.3969

0.411 0.6850 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4467 8.8202 0.2510 0.8335 -0.27 -6.4716

0.442 0.7367 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4934 7.3735 0.2121 0.5825 -0.27 -6.4716

0.474 0.7900 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4467 5.8802 0.1599 0.3705 -0.27 -6.4716

0.505 0.8417 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4934 4.4335 0.1180 0.2106 -0.27 -6.4716

0.537 0.8950 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4467 2.9401 0.0687 0.0926 -0.27 -6.4716

0.568 0.9467 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 1.4934 1.4934 0.0239 0.0239 -0.27 -6.4716

0.6 1.0000 0.3376 0.3307 0.0543 0.3310 33.6000 46.6679 0.0000 0.0000 0.0000 0.0000 -0.27 -6.4716

0.0000 -1.1490 0.1497

79

Table 6-6 Wing loading in each station for point E (Aileron deflects downwards)


0 0.0000 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 1.3452 0.0545 0.3042 -0.10 -2.3969

0.032 0.0533 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 2.0598 0.0746 0.2497 -0.10 -2.3969

0.063 0.1050 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 2.7521 0.0995 0.1752 -0.10 -2.3969

0.095 0.1583 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 3.4667 0.1182 0.0757 -0.10 -2.3969

0.126 0.2100 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 4.1590 0.1445 -0.0425 -0.10 -2.3969

0.158 0.2633 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 4.8736 0.1618 -0.1871 -0.10 -2.3969

0.189 0.3150 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 5.5659 0.1895 -0.3489 -0.10 -2.3969

0.221 0.3683 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 6.2806 0.2124 -0.5384 -0.10 -2.3969

0.253 0.4217 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 6.9952 0.2276 -0.7508 -0.10 -2.3969

0.284 0.4733 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.0223 7.6875 0.0077 -0.9784 -0.10 -2.3969

0.285 0.4750 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 7.0959 7.7098 0.0042 -0.9861 -0.10 -2.3969

0.286 0.4767 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 14214.17 7.0959 0.6139 -0.0029 -0.9903 -0.10 -2.3969

0.287 0.4783 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6476 -6.4820 -0.1786 -0.9873 -0.10 -2.3969

0.316 0.5267 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6923 -5.8344 -0.1701 -0.8087 -0.10 -2.3969

0.347 0.5783 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.7146 -5.1421 -0.1531 -0.6386 -0.10 -2.3969

0.379 0.6317 -2.0480 -0.1565 0.0543 -0.1584 -33.6000 -22.3318 -0.6750 -4.4275 -0.1309 -0.4855 -0.10 -2.3969

0.411 0.6850 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6155 -3.7525 -0.1068 -0.3546 -0.25 -5.9923

0.442 0.7367 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6354 -3.1370 -0.0902 -0.2478 -0.25 -5.9923

0.474 0.7900 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6155 -2.5017 -0.0680 -0.1576 -0.25 -5.9923

0.505 0.8417 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6354 -1.8862 -0.0502 -0.0896 -0.25 -5.9923

0.537 0.8950 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6155 -1.2508 -0.0292 -0.0394 -0.25 -5.9923

0.568 0.9467 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 -0.6354 -0.6354 -0.0102 -0.0102 -0.25 -5.9923

0.6 1.0000 -2.0480 -0.1390 0.0543 -0.1408 -33.6000 -19.8547 0.0000 0.0000 0.0000 0.0000 -0.25 -5.9923

0.0000 1.3452 0.3042

80

Table 6-7 Wing loading in each station for point G (Aileron deflects downwards)


0 0.0000 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 1.0646 0.0454 0.1684 -0.10 -0.5646

0.032 0.0533 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 1.7720 0.0656 0.1230 -0.10 -0.5646

0.063 0.1050 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 2.4574 0.0900 0.0575 -0.10 -0.5646

0.095 0.1583 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 3.1649 0.1087 -0.0325 -0.10 -0.5646

0.126 0.2100 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 3.8502 0.1345 -0.1412 -0.10 -0.5646

0.158 0.2633 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 4.5577 0.1519 -0.2757 -0.10 -0.5646

0.189 0.3150 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 5.2431 0.1791 -0.4276 -0.10 -0.5646

0.221 0.3683 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 5.9506 0.2017 -0.6067 -0.10 -0.5646

0.253 0.4217 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 6.6580 0.2170 -0.8085 -0.10 -0.5646

0.284 0.4733 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.0221 7.3434 0.0074 -1.0255 -0.10 -0.5646

0.285 0.4750 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 7.0249 7.3655 0.0039 -1.0329 -0.10 -0.5646

0.286 0.4767 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 14071.99 7.0249 0.3406 -0.0032 -1.0367 -0.10 -0.5646

0.287 0.4783 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6411 -6.6844 -0.1846 -1.0335 -0.10 -0.5646

0.316 0.5267 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6854 -6.0432 -0.1767 -0.8490 -0.10 -0.5646

0.347 0.5783 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.7075 -5.3579 -0.1601 -0.6723 -0.10 -0.5646

0.379 0.6317 -8.0000 -0.6646 0.0543 -0.6657 -16.3073 -22.1085 -0.6891 -4.6504 -0.1378 -0.5121 -0.10 -0.5646

0.411 0.6850 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6497 -3.9613 -0.1127 -0.3743 -0.505 -2.8512

0.442 0.7367 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6707 -3.3116 -0.0952 -0.2616 -0.505 -2.8512

0.474 0.7900 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6497 -2.6409 -0.0718 -0.1664 -0.505 -2.8512

0.505 0.8417 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6707 -1.9911 -0.0530 -0.0946 -0.505 -2.8512

0.537 0.8950 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6497 -1.3204 -0.0309 -0.0416 -0.505 -2.8512

0.568 0.9467 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 -0.6707 -0.6707 -0.0107 -0.0107 -0.505 -2.8512

0.6 1.0000 -8.0000 -0.6297 0.0543 -0.6311 -16.3073 -20.9594 0.0000 0.0000 0.0000 0.0000 -0.505 -2.8512

0.0000 1.0646 0.1684

81

Table 6-8 Wing loading in each station for point A (Aileron deflects upwards)

y (m) y/b Cl Cn N (N) Delta V Vn (N) Delta M M (N) Cm T (Pa)

0 0.0000 1.2898 1.2728 43.5641 1.3941 -1.8734 -0.0823 -0.2204 -0.100 -0.5818

0.032 0.0533 1.2898 1.2728 43.5641 1.3505 -3.2674 -0.1222 -0.1382 -0.100 -0.5818

0.063 0.1050 1.2898 1.2728 43.5641 1.3941 -4.6179 -0.1701 -0.0159 -0.100 -0.5818

0.095 0.1583 1.2898 1.2728 43.5641 1.3505 -6.0120 -0.2073 0.1541 -0.100 -0.5818

0.126 0.2100 1.2898 1.2728 43.5641 1.3941 -7.3625 -0.2579 0.3614 -0.100 -0.5818

0.158 0.2633 1.2898 1.2728 43.5641 1.3505 -8.7565 -0.2924 0.6193 -0.100 -0.5818

0.189 0.3150 1.2898 1.2728 43.5641 1.3941 -10.1070 -0.3457 0.9117 -0.100 -0.5818

0.221 0.3683 1.2898 1.2728 43.5641 1.3941 -11.5011 -0.3903 1.2575 -0.100 -0.5818

0.253 0.4217 1.2898 1.2728 43.5641 1.3505 -12.8951 -0.4207 1.6478 -0.100 -0.5818

0.284 0.4733 1.2898 1.2728 43.5641 0.0436 -14.2456 -0.0143 2.0685 -0.100 -0.5818

0.285 0.4750 1.2898 1.2728 43.5641 -13.8425 -14.2892 -0.0074 2.0828 -0.100 -0.5818

0.286 0.4767 1.2898 1.2728 -27728.5359 -13.8425 -0.4467 0.0065 2.0901 -0.100 -0.5818

0.287 0.4783 1.2898 1.2728 43.5641 1.2634 13.3958 0.3702 2.0836 -0.100 -0.5818

0.316 0.5267 1.2898 1.2728 43.5641 1.3505 12.1324 0.3552 1.7135 -0.100 -0.5818

0.347 0.5783 1.2898 1.2728 43.5641 1.3941 10.7819 0.3227 1.3583 -0.100 -0.5818

0.379 0.6317 1.2898 1.2728 43.5641 1.3753 9.3879 0.2784 1.0356 -0.100 -0.5818

0.411 0.6850 1.2549 1.2387 42.3945 1.3142 8.0126 0.2280 0.7572 0.340 1.9783

0.442 0.7367 1.2549 1.2387 42.3945 1.3566 6.6983 0.1926 0.5292 0.340 1.9783

0.474 0.7900 1.2549 1.2387 42.3945 1.3142 5.3417 0.1452 0.3365 0.340 1.9783

0.505 0.8417 1.2549 1.2387 42.3945 1.3566 4.0275 0.1072 0.1913 0.340 1.9783

0.537 0.8950 1.2549 1.2387 42.3945 1.3142 2.6709 0.0624 0.0841 0.340 1.9783

0.568 0.9467 1.2549 1.2387 42.3945 1.3566 1.3566 0.0217 0.0217 0.340 1.9783

0.6 1.0000 1.2549 1.2387 42.3945 0.0000 0.0000 0.0000 0.0000 0.340 1.9783

0.0000 -1.8734 -0.2204

82

Table 6-9 Wing loading in each station for point D (Aileron deflects upwards)


0 0.0000 0.3131 0.3134 44.1893 1.4141 -2.1652 -0.0919 -0.3552 -0.1000 -2.3969

0.032 0.0533 0.3131 0.3134 44.1893 1.3699 -3.5793 -0.1322 -0.2633 -0.1000 -2.3969

0.063 0.1050 0.3131 0.3134 44.1893 1.4141 -4.9492 -0.1810 -0.1311 -0.1000 -2.3969

0.095 0.1583 0.3131 0.3134 44.1893 1.3699 -6.3632 -0.2185 0.0499 -0.1000 -2.3969

0.126 0.2100 0.3131 0.3134 44.1893 1.4141 -7.7331 -0.2701 0.2684 -0.1000 -2.3969

0.158 0.2633 0.3131 0.3134 44.1893 1.3699 -9.1471 -0.3048 0.5385 -0.1000 -2.3969

0.189 0.3150 0.3131 0.3134 44.1893 1.4141 -10.5170 -0.3592 0.8433 -0.1000 -2.3969

0.221 0.3683 0.3131 0.3134 44.1893 1.4141 -11.9311 -0.4044 1.2024 -0.1000 -2.3969

0.253 0.4217 0.3131 0.3134 44.1893 1.3699 -13.3451 -0.4349 1.6068 -0.1000 -2.3969

0.284 0.4733 0.3131 0.3134 44.1893 0.0442 -14.7150 -0.0147 2.0418 -0.1000 -2.3969

0.285 0.4750 0.3131 0.3134 44.1893 -14.0412 -14.7592 -0.0077 2.0565 -0.1000 -2.3969

0.286 0.4767 0.3131 0.3134 -28126.5107 -14.0412 -0.7180 0.0063 2.0643 -0.1000 -2.3969

0.287 0.4783 0.3131 0.3134 44.1893 1.2815 13.3231 0.3678 2.0579 -0.1000 -2.3969

0.316 0.5267 0.3131 0.3134 44.1893 1.3699 12.0416 0.3521 1.6902 -0.1000 -2.3969

0.347 0.5783 0.3131 0.3134 44.1893 1.4141 10.6718 0.3189 1.3381 -0.1000 -2.3969

0.379 0.6317 0.3131 0.3134 44.1893 1.3744 9.2577 0.2743 1.0192 -0.1000 -2.3969

0.411 0.6850 0.2955 0.2958 41.7107 1.2930 7.8833 0.2243 0.7450 0.0700 1.6778

0.442 0.7367 0.2955 0.2958 41.7107 1.3347 6.5903 0.1895 0.5206 0.0700 1.6778

0.474 0.7900 0.2955 0.2958 41.7107 1.2930 5.2555 0.1429 0.3311 0.0700 1.6778

0.505 0.8417 0.2955 0.2958 41.7107 1.3347 3.9625 0.1054 0.1882 0.0700 1.6778

0.537 0.8950 0.2955 0.2958 41.7107 1.2930 2.6278 0.0614 0.0828 0.0700 1.6778

0.568 0.9467 0.2955 0.2958 41.7107 1.3347 1.3347 0.0214 0.0214 0.0700 1.6778

0.6 1.0000 0.2955 0.2958 41.7107 0.0000 0.0000 0.0000 0.0000 0.0700 1.6778

0.0000 -2.1652 -0.3552

83

Table 6-10 Wing loading in each station for point E (Aileron deflects upwards)


0 0.0000 -0.1565 -0.1584 -22.3318 -0.7146 0.3296 0.0220 -0.2004 -0.1000 -2.3969

0.032 0.0533 -0.1565 -0.1584 -22.3318 -0.6923 1.0442 0.0431 -0.2224 -0.1000 -2.3969

0.063 0.1050 -0.1565 -0.1584 -22.3318 -0.7146 1.7365 0.0670 -0.2655 -0.1000 -2.3969

0.095 0.1583 -0.1565 -0.1584 -22.3318 -0.6923 2.4511 0.0867 -0.3325 -0.1000 -2.3969

0.126 0.2100 -0.1565 -0.1584 -22.3318 -0.7146 3.1434 0.1120 -0.4192 -0.1000 -2.3969

0.158 0.2633 -0.1565 -0.1584 -22.3318 -0.6923 3.8580 0.1303 -0.5312 -0.1000 -2.3969

0.189 0.3150 -0.1565 -0.1584 -22.3318 -0.7146 4.5503 0.1570 -0.6616 -0.1000 -2.3969

0.221 0.3683 -0.1565 -0.1584 -22.3318 -0.7146 5.2649 0.1799 -0.8186 -0.1000 -2.3969

0.253 0.4217 -0.1565 -0.1584 -22.3318 -0.6923 5.9796 0.1961 -0.9985 -0.1000 -2.3969

0.284 0.4733 -0.1565 -0.1584 -22.3318 -0.0223 6.6718 0.0067 -1.1946 -0.1000 -2.3969

0.285 0.4750 -0.1565 -0.1584 -22.3318 7.0959 6.6942 0.0031 -1.2013 -0.1000 -2.3969

0.286 0.4767 -0.1565 -0.1584 14214.1682 7.0959 -0.4017 -0.0039 -1.2044 -0.1000 -2.3969

0.287 0.4783 -0.1565 -0.1584 -22.3318 -0.6476 -7.4977 -0.2080 -1.2005 -0.1000 -2.3969

0.316 0.5267 -0.1565 -0.1584 -22.3318 -0.6923 -6.8500 -0.2016 -0.9924 -0.1000 -2.3969

0.347 0.5783 -0.1565 -0.1584 -22.3318 -0.7146 -6.1578 -0.1856 -0.7908 -0.1000 -2.3969

0.379 0.6317 -0.1565 -0.1584 -22.3318 -0.7543 -5.4431 -0.1621 -0.6052 -0.1000 -2.3969

0.411 0.6850 -0.1741 -0.1760 -24.8089 -0.7691 -4.6889 -0.1334 -0.4431 0.0500 1.1985

0.442 0.7367 -0.1741 -0.1760 -24.8089 -0.7939 -3.9198 -0.1127 -0.3097 0.0500 1.1985

0.474 0.7900 -0.1741 -0.1760 -24.8089 -0.7691 -3.1259 -0.0850 -0.1969 0.0500 1.1985

0.505 0.8417 -0.1741 -0.1760 -24.8089 -0.7939 -2.3568 -0.0627 -0.1120 0.0500 1.1985

0.537 0.8950 -0.1741 -0.1760 -24.8089 -0.7691 -1.5630 -0.0365 -0.0492 0.0500 1.1985

0.568 0.9467 -0.1741 -0.1760 -24.8089 -0.7939 -0.7939 -0.0127 -0.0127 0.0500 1.1985

0.6 1.0000 -0.1741 -0.1760 -24.8089 0.0000 0.0000 0.0000 0.0000 0.0500 1.1985

0.0000 0.3296 -0.2004

84

Table 6-11 Wing loading in each station for point G (Aileron deflects upwards)


0 0.0000 -0.6646 -0.6657 -22.1085 -0.7075 0.5934 0.0303 -0.0657 -0.1000 -0.5646

0.032 0.0533 -0.6646 -0.6657 -22.1085 -0.6854 1.3009 0.0510 -0.0960 -0.1000 -0.5646

0.063 0.1050 -0.6646 -0.6657 -22.1085 -0.7075 1.9863 0.0749 -0.1469 -0.1000 -0.5646

0.095 0.1583 -0.6646 -0.6657 -22.1085 -0.6854 2.6937 0.0941 -0.2218 -0.1000 -0.5646

0.126 0.2100 -0.6646 -0.6657 -22.1085 -0.7075 3.3791 0.1195 -0.3159 -0.1000 -0.5646

0.158 0.2633 -0.6646 -0.6657 -22.1085 -0.6854 4.0866 0.1373 -0.4354 -0.1000 -0.5646

0.189 0.3150 -0.6646 -0.6657 -22.1085 -0.7075 4.7719 0.1640 -0.5727 -0.1000 -0.5646

0.221 0.3683 -0.6646 -0.6657 -22.1085 -0.7075 5.4794 0.1867 -0.7367 -0.1000 -0.5646

0.253 0.4217 -0.6646 -0.6657 -22.1085 -0.6854 6.1869 0.2024 -0.9234 -0.1000 -0.5646

0.284 0.4733 -0.6646 -0.6657 -22.1085 -0.0221 6.8722 0.0069 -1.1258 -0.1000 -0.5646

0.285 0.4750 -0.6646 -0.6657 -22.1085 7.0249 6.8944 0.0034 -1.1327 -0.1000 -0.5646

0.286 0.4767 -0.6646 -0.6657 14071.9915 7.0249 -0.1306 -0.0036 -1.1361 -0.1000 -0.5646

0.287 0.4783 -0.6646 -0.6657 -22.1085 -0.6411 -7.1555 -0.1982 -1.1324 -0.1000 -0.5646

0.316 0.5267 -0.6646 -0.6657 -22.1085 -0.6854 -6.5144 -0.1913 -0.9342 -0.1000 -0.5646

0.347 0.5783 -0.6646 -0.6657 -22.1085 -0.7075 -5.8290 -0.1752 -0.7429 -0.1000 -0.5646

0.379 0.6317 -0.6646 -0.6657 -22.1085 -0.7259 -5.1215 -0.1523 -0.5677 -0.1000 -0.5646

0.411 0.6850 -0.6995 -0.7003 -23.2576 -0.7210 -4.3957 -0.1251 -0.4154 0.3050 1.7220

0.442 0.7367 -0.6995 -0.7003 -23.2576 -0.7442 -3.6747 -0.1057 -0.2903 0.3050 1.7220

0.474 0.7900 -0.6995 -0.7003 -23.2576 -0.7210 -2.9305 -0.0797 -0.1846 0.3050 1.7220

0.505 0.8417 -0.6995 -0.7003 -23.2576 -0.7442 -2.2095 -0.0588 -0.1049 0.3050 1.7220

0.537 0.8950 -0.6995 -0.7003 -23.2576 -0.7210 -1.4652 -0.0342 -0.0462 0.3050 1.7220

0.568 0.9467 -0.6995 -0.7003 -23.2576 -0.7442 -0.7442 -0.0119 -0.0119 0.3050 1.7220

0.6 1.0000 -0.6995 -0.7003 -23.2576 0.0000 0.0000 0.0000 0.0000 0.3050 1.7220

0.0000 0.5934 -0.0657

85

Figure 6-15 Graph of Normal Lift across Wingspan (Zoom in)

Figure 6-16 Graph of Normal Lift across Wingspan (Zoom out)

86

Figure 6-17 Graph of Shear Force across Wingspan when the aileron deflects downwards

Figure 6-18 Graph of Shear Force across Wingspan when the aileron deflects upwards

87

Figure 6-19 Graph of Bending Moment across Wingspan when aileron deflects downwards

Figure 6-20 Graph of Bending Moment across Wingspan when aileron deflects upwards

88

Figure 6-21 Graph of Torsion across Wingspan

Based on the all shear force diagrams and bending moment diagrams, the maximum shear

force and maximum bending moment when the aileron deflects upwards and deflects downwards

can be summarised in Table 6-12..

Table 6-12 Maximum Shear force and Bending moment in each position

Position Deflection of Aileron downwards Deflection of Aileron upwards

V (N) M (Nm) T (Nm) V (N) M (Nm) T (Nm)

A 13.8754 2.1912 -3.1419 -14.2892 2.0901 1.9783

D 14.3394 2.2786 -6.4716 13.3231 2.0643 -2.3969

E 7.7098 -0.9903 -5.9923 -7.4977 -1.2044 -2.3969

G 7.3655 -1.0367 -2.8512 -7.1555 -1.1361 1.7220

-8

-6

-4

-2

0

2

4

0 0.2 0.4 0.6 0.8 1 1.2

T (N

m)

2y/b

T vs y

Position A (Down)

Position D (Down)

Position E (Down)

Position G (Down)

Position A (Up)

Position D (Up)

Position E (Up)

Position G (Up)

89

6.2.6 Discussion

After taking into account the lift coefficient for the part without aileron and for the part

with the deflected aileron, the normal lift force is calculated, tabulated and presented in graph

shown in Figure 6-15 when it was zoomed in to a smaller range of normal lift force axis, and in

Figure 6-16 when it was zoomed out into a larger range of normal lift force axis. The sudden spike

of the graph is due to the reaction force due to the wing strut, which we assume to be a concentrated

load force at a small area. In Figure 6-15, the normal lift force is a constant straight line, since the

wing is assumed to have no twist angle due to its small size, and the angle of attack is constant

across the wingspan without aileron. However, there is a slight change in the normal lift force at a

distance of 0.4m away from the wing root. That is where the aileron is located on the wing. The

change of the normal lift force due to the aileron, which is resulted from the change in lift

coefficient, has been discussed in previous section. As the aileron deflects downwards, the lift

coefficient increases, hence the increase in normal lift force, and vice-versa.

Figure 6-17 and Figure 6-18 show the shear force across the wingspan when the aileron

deflects downwards and deflects upwards respectively. The graphs are almost similar due to the

reaction force of the wing strut, that supports the wing. The sudden change in the shear force in

the graph indicates where the wing strut is located. As a result, the shear force diagram differs

from the normal shear force diagram that shows a graph with a decreasing slope along the

wingspan. The difference between Figure 6-17 and Figure 6-18 is that the maximum shear force

is relatively smaller, when the aileron deflects upwards in Figure 6-18. This is because the change

in lift coefficient when the aileron deflects downwards is larger.

Furthermore, Figure 6-19 and Figure 6-20 show that the bending moment diagrams are

affected by the reaction force of the wing strut too. The bending moment diagram and shear force

diagram are derived from the normal lift force diagram in Figure 6-15 and Figure 6-16. As shown

in Figure 6-19 and Figure 6-20, the bending moment is negative when the load factor is negative

and vice-versa. Also, difference between Figure 6-19 and Figure 6-20 is that when the aileron

deflects downward, the moment of reaction at the wing root is positive and when the aileron

deflects upward, the moment of reaction at the wing root is negative. This is to counter the opposite

direction of the applied moment on the wing.

90

Figure 6-21 shows the graph of torsion across the wingspan. As explained in the previous

section, the value of torsion depends on the value of moment coefficient. The change in the

moment coefficient is due to the deflection of the aileron. The change in moment coefficient

depends on the position of the wing during flight. Position A and G experience smaller angle

deflection of aileron than position D and E due to maneuverability and control of the aircraft at

lower speed of the aircraft, therefore the change of torsion is relatively smaller. When the aileron

deflects downward, the torsion decreases as the moment coefficient decreases. The amount of

change in moment coefficient depends on how much the graph in Figure 6-8 is shifting downwards,

as explained in section 6.2.3.

Based on Table 6-12, results show that the maximum shear force is 14.3394 N at position

D, when the aileron deflects downwards. As the shear force is the highest, the value of both the

bending moment and torsion at position D is also the highest. The highest maximum bending

moment and maximum torsion is 2.2786 Nm and -6.4716 Nm respectively. This is because the

speed and the load factor are the highest at position D, as compared to other positions. The shear

force and bending moment at position A and D are higher, as compared to position E and G because

the magnitude of positive load factor of aircraft at position A and D is larger than the negative load

factor of aircraft at position E and G. The maximum shear force, maximum bending moment and

maximum torsion is used in structural analysis conducted in the next section in order to determine

if the structure is safe.

91

6.3 Structural Analysis

6.3.1 Material Selection

The material used in our aircraft’s wing construction is mainly Expanded Polypropylene

(EPP) Foam, and a minority of the components are made up of plywood.

The reason why Expanded Polypropylene (EPP) is selected because it is a highly versatile

closed-cell bead foam that provides a unique range of properties, including outstanding energy

absorption, multiple impact resistance, thermal insulation, buoyancy, exceptionally high strength

to weight ratio and 100% recyclability. EPP can be made in a wide range of densities. In

manufacturing, individual beads are fused into final product form by the steamchest moulding

process resulting in a strong and lightweight shape.

On the other hand, plywood is chosen for some part of the wing, such as rib and wing strut

because it offers increased stability, high impact resistance, high surface dimensional stability,

high strength to weight ratio, and effective resistance in shear. These properties are beneficial in

the construction of our lightweight wing.

Table 6-13 Material Selection of each Wing Component

Wing Components Material

Rib, Wing strut Plywood

Skin, Spar, Aileron Expanded Polyproplyene (EPP) Foam

92

6.3.2 Rib Analysis

All of the four ribs along the wingspan are of the same dimension and thickness. Only the analysis

of the rib which carries the critical load is carried out, since if the rib which carries the critical load is safe,

then the other ribs will be safe as well. The critical position of the aircraft during flight, which is position

D, is selected because the maximum bending moment and the shear force is the highest, as compared to

other positions. The analysis of the bending moment as well as the shear is performed with the design factor

of 1.5. Also, each of the rib cells is compared with the plate buckling strength to determine the margin of

safety and the safety factor.

Based on the graph of normal lifting force across the wingspan, the area under the graph, which is

the normal lifting force is obtained to determine the load distribution along the rib, which is indicated as

the value of w. Since bay 1 and bay 3 have the same lift distribution force, therefore only one of them is

taken for analysis. The material selected for the rib is plywood and its material properties are listed in Table

6-15.

Figure 6-22 Location of ribs along the wingspan

Table 6-14 The chordwise load distribution for each bay

Bay Normal lifting force (N) w (N/m)

Bay 1 13.8393 46.5187

Bay 2 26.5136 89.1213

93

Table 6-15 Material properties of plywood

Material properties Plywood

Ultimate tensile strength 5000psi

Ultimate shear strength 1000psi

Shear Modulus 110ksi

Modulus of Rupture 10ksi

Modulus of Elasticity 1000ksi

Figure 6-23 The load distribution on rib

Total area of the load distribution based on our rib design as shown in Figure 6-23 can be

obtained as follows,

𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎, 𝐴 = (0.15𝑐 × 3𝑤) + (0.45𝑐 × 2𝑤) + (0.4𝑐 × 𝑤) = 1.75𝑐𝑤

1.75𝑐𝑤 = 26.5136

𝑤 = 89.1213 𝑁/𝑚

Cell 1 Analysis

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑖𝑏′𝑠 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛, 𝐴1 = 0.000648𝑚2

𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑟𝑖𝑏, 𝑡1 = 0.003𝑚

ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑎 = 0.0415𝑚

𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑏 = 0.0202𝑚

94

𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑, 𝑧 = 0.012𝑚

𝐹0.7 = 700𝑘𝑠𝑖

𝑛 = 10

Figure 6-24 Cell 1 of Rib 2

The compressive stress of cell 1 can be determined by obtaining the bending moment due

to the leading edge’s air load acting on the rib, as shown below,

𝑀1 = [3𝑤 × 0.15𝑐 × (0.15𝑐

2+ 0.15𝑐)] + [2𝑤 × 0.15𝑐 ×

0.15𝑐

2]

𝑀1 = 0.3187 𝑁𝑚

𝐼 = 𝐴1(𝑧1)2 = 9.3312 × 10−8𝑚4

𝜎 =𝑀𝑧

𝐼= 0.040985 𝑀𝑃𝑎

By assuming the perimeter of the rib web is simply supported, and obtaining the 𝑘𝑏value

from Figure 3 of Appendix III, the web buckling strength is

𝜎𝑐𝑟 =𝜋2𝑘𝑏𝐸

12(1 − 𝑣2)(𝑡

𝑏)2

𝜎𝑐𝑟 =𝜋2(36)(1000 × 103)

12(1 − 0.32)(0.003

0.0202)2

𝜎𝑐𝑟 = 7.177 × 105 𝑝𝑠𝑖 = 4948.0988𝑀𝑃𝑎

𝜎𝑐𝑟

𝐹0.7= 1.0253

Since the 𝜎𝑐𝑟value is larger than the 𝐹0.7 value, correction needs to be done. Based on

Figure 5 of Appendix III, the new ratio will be as follows,

95

𝜎𝑐𝑟

𝐹0.7= 0.845

𝜎𝑐𝑟 = 0.845 × 700000

𝜎𝑐𝑟 = 591500𝑝𝑠𝑖 = 4078.24894 𝑀𝑃𝑎

Thus,

𝑅𝑏 =1.5 × 0.040985

4078.24894= 1.5074 × 10−5

𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 =1

𝑅𝑏− 1 = 66336.26

The shear stress due to the total air load of cell 1 can be determined as follows. Then, the

shear is divided by the area of the cross section to obtain the shear stress.

𝑉1 = (3𝑤)(0.15𝑐) + (2𝑤)(0.15𝑐) = 11.363𝑁

𝜏 =𝑉1

𝐴𝑟𝑒𝑎

𝜏 = 17535.494𝑃𝑎

Assuming that the perimeter of the rib web is simply supported, and obtaining the 𝑘𝑠value

from Figure 2 of Appendix III, the buckling strength is

𝜎𝑐𝑟 =𝜋2𝑘𝑠𝐸

12(1 − 𝑣2)(𝑡

𝑏)2

𝜎𝑐𝑟 =𝜋2(6.4)(1000 × 103)

12(1 − 0.32)(0.003

0.0202)2

𝜎𝑐𝑟 = 1.27584 × 105 𝑝𝑠𝑖 = 879.662 𝑀𝑃𝑎

Since the 𝜎𝑐𝑟value is much smaller than the 𝐹0.7 value, correction is not required.

Thus,

𝑅𝑠 =1.5 × 0.017535494

879.662= 2.9902 × 10−5


𝑅𝑠− 1 = 33442.103

Therefore, the combined bending and shear is

96

𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 = 1

√𝑅𝑏2 + 𝑅𝑠

2

− 1


√(1.5074 × 10−5)2 + (2.9902 × 10−5)2− 1 = 29861.655

𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 29861.655 + 1 = 29862.655

Since the factor of safety for cell 1 is more than one, cell 1 of rib 2, which carries the

critical load is safe.

Cell 2 Analysis

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑖𝑏′𝑠 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛, 𝐴2 = 0.001729𝑚2

𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑟𝑖𝑏, 𝑡1 = 0.003𝑚

ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑎 = 0.1285𝑚

𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑏 = 0.0202𝑚

𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑, 𝑧 = 0.0099𝑚

𝐹0.7 = 700𝑘𝑠𝑖

𝑛 = 10

Figure 6-25 Cell 2 of Rib 2

The compressive stress of cell 2 can be determined by obtaining the bending moment due

to the leading edge’s air load acting on the rib, as shown below,

𝑀2 = [2𝑤 × 0.3𝑐 × (0.3𝑐

2)] + [𝑤 × 0.4𝑐 × (

0.4𝑐

2+ 0.3𝑐)] = 0.7469 𝑁𝑚

𝐼 = 𝐴1(𝑧1)2 = 1.6946 × 10−7𝑚4

97

𝜎 =𝑀𝑧

𝐼= 0.043635 𝑀𝑃𝑎

Assuming the perimeter of the rib web is simply supported, and obtaining the 𝑘𝑏value from

Figure 3 of Appendix III, the web buckling strength is


12(1 − 𝑣2)(𝑡

𝑏)2

𝜎𝑐𝑟 =𝜋2(24)(1000 × 103)

12(1 − 0.32)(0.003

0.0202)2

𝜎𝑐𝑟 = 0.478441 × 106 𝑝𝑠𝑖 = 3298.7325 𝑀𝑃𝑎


Thus,

𝑅𝑏 =1.5 × 0.043635

3298.7325= 1.9842 × 10−5


𝑅𝑏− 1 = 50397.9

The shear stress due to the total air load of cell 2 can be determined as follows. Then, the

shear is divided by the area of the cross section to obtain the shear stress.

𝑉2 = (𝑤)(0.4𝑐) + (2𝑤)(0.3𝑐) = 15.1506𝑁

𝜏 =𝑉1

𝐴𝑟𝑒𝑎

𝜏 = 8762.65𝑃𝑎

Assuming that the perimeter of the rib web is simply supported, and obtaining the 𝑘𝑠value

from Figure 2 of Appendix III, the buckling strength is


12(1 − 𝑣2)(𝑡

𝑏)2

𝜎𝑐𝑟 =𝜋2(5.5)(1000 × 103)

12(1 − 0.32)(0.003

0.0202)2

𝜎𝑐𝑟 = 0.109642 × 106 𝑝𝑠𝑖 = 755.9595 𝑀𝑃𝑎


Thus,

98

𝑅𝑠 =1.5 × 8762.65

755959536.5537= 1.73872 × 10−5


𝑅𝑠− 1 = 57512.77

Therefore, the combined bending and shear is


√𝑅𝑏2 + 𝑅𝑠

2

− 1


√(1.9842 × 10−5)2 + (1.73872 × 10−5)2− 1 = 37903.366

𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 37903.366 + 1 = 37904.366

Since the factor of safety for cell 2 is more than one, cell 2 of rib 2, which carries the

critical load is safe.

Figure 6-26 Shear force across the chord

0

5

10

15

20

25

30

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18

Shea

r fo

rce,

V (

N)

Length, x (m)

Shear force across the chord

99

Figure 6-27 Bending Moment across the chord

Based on the chordwise shear force diagram and chordwise bending moment diagram, we

can conclude that the shear force and bending moment decreases across the chord. The trend of

both diagrams depends on the lift distribution across the chord, which is in reference to FAR Part

23. Stepwise lift distribution is selected due to the type and size of the aircraft.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18

Ben

din

g M

om

ent

, M (

Nm

)

Length, x (m)

Bending Moment across the chord

100

6.3.3 Spar Analysis

Spar is the wing structure that supports an aircraft on ground and allows it to taxi, take-off,

and land. In fact, Spar design tends to have several interferences with the aircraft structural design.

The strength and the weight of spar have become important factors. Efforts are being made to

reduce the weight of the aircraft and consequently increase the payload. The aerodynamic lift load

acting on the top and bottom surfaces of the wing gets transferred to the rib structures, which in-

turn gets transferred to the spar web as concentrated shear loads. These shear loads cause bending

of the spars which are the major loading on the spars. The spar is made up of Expanded

Polypropylene (EPP) foam, which its material properties are listed in Table 6-16.

Table 6-16 Material Properties of EPP Foam

Material Properties EPP Foam

Modulus of Elasticity, E 1000ksi

Yield Stress 4000psi

Density 1.3 lb/ft2

Figure 6-28 Dimension and location of spar in wing

101

Figure 6-29 Cross section of spar

𝑎 = 0.6𝑚, 𝑏 = 0.01𝑚, 𝑡 = 0.005𝑚

E = 6895MPa

𝑀𝑚𝑎𝑥 = 20.97266𝑁𝑚

𝑉 = 52.71447 𝑁

𝑆𝑎𝑓𝑒𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟 = 1.5

𝐹0.7 = 4826.295 𝑀𝑃𝑎

n = 10

𝐼 =𝑏ℎ3

12=

(0.005)(0.01)3

12= 4.1667 × 10−10 𝑚4

Ultimate Bending Moment

𝜎 =𝑀𝑦

𝐼=

20.97266 × 0.005 × 1.5

4.1667 × 10−10= 377.50558 𝑀𝑃𝑎

By assuming the spar is simply supported, and obtaining kb value from Figure 3 in

Appendix III, the critical buckling strength due to compressive force is


12(1 − 𝑣2)(𝑡

𝑏)2

=𝜋2 × 24 × 6895𝑀

12(1 − 0.32)(0.005

0.01)2

= 37390.6𝑀𝑃𝑎

𝜎𝑐𝑟

𝐹0.7=

37390.6

4826.295= 7.7473

102

Based on Figure 4 in Appendix III, as 𝜎𝑐𝑟 value is greater than 𝐹0.7 value, correction must

be done, where the new ratio is given as follows,

𝜎𝑐𝑟

𝐹0.7= 1.05

𝜎𝑐𝑟 = 1.05 × 4826.295𝑀𝑃𝑎 = 5067.6𝑀𝑃𝑎

Therefore,

𝑅𝑐 =377.50558

5067.6= 0.07449

𝑀𝑆 =1

0.07449− 1 = 12.424

Ultimate Shear Force

τ =𝑉

𝐴=

52.71447

0.005 × 0.01 × 2= 527144.7 𝑃𝑎

By assuming the spar is simply supported, and obtaining ks value from the Figure 2 in

Appendix III, the critical buckling strength due to compressive force is


12(1 − 𝑣2)(𝑡

𝑏)2

=𝜋2 × 5.5 × 6895𝑀

12(1 − 0.32)(0.005

0.01)2

= 8568.683𝑀𝑃𝑎

𝜎𝑐𝑟

𝐹0.7=

8568.683

4826.295= 1.7754

Based on Figure 5 in Appendix III, as 𝜎𝑐𝑟 value is greater than 𝐹0.7 value, correction must

be done, where the new ratio is given as follows,

𝜎𝑐𝑟

𝐹0.7= 0.55

𝜎𝑐𝑟 = 0.55 × 4826.295𝑀𝑃𝑎 = 2654.462𝑀𝑃𝑎

Therefore,

103

𝑅𝑠 =527144.7

2654462250= 1.9859 × 10−4

𝑀𝑆 =1

1.655 × 10−6− 1 = 604264.7

𝑀𝑆 =1

√(𝑅𝑏)2 + (𝑅𝑠)

2=

1

√(0.07449)2 + (1.9859 × 10−4)2= 13.4246

𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 13.4246 + 1 = 14.4246

Since the factor of safety is more than 1, the spar will not buckle and it is under a safe condition.

104

6.3.4 Skin Analysis

According to wing loading calculation, the highest value of shear and bending moment are

when the flight is in position D. Therefore, the maximum bending moment and maximum shear

force can be used to determine the suitable thickness of wing skin to withstand the applied load.

The case will be analysed with the condition of compression force acting on the upper wing and

with the tension force acting on the lower skin. Shear force is exerted on both side of the wing at

the same time.

Figure 6-30 Wing’s skin Figure 3.9:

The calculation is being divided into two parts, which is upper wing and lower wing

to find the factor of safety for the skin. For the upper wing, the buckling strength can be calculated

first as below:

𝐹𝑐𝑟 =𝑘𝜋2𝐸

12(1 − 𝑣2)(𝑡

𝑏)2

Upper skin

Lower skin

105

Shear load

Based on Figure 2 in Appendix III, the value of ks is obtained. Buckling strength due to

shear force can be determined as follows,

𝐹𝑠,𝑐𝑟 =𝑘𝑠𝜋

2𝐸

12(1 − 𝑣2)(𝑡

𝑏)2

=(9.6)𝜋2(6.9 𝑥 106)

12(1 − 0.32)(0.005

0.25)2

= 23.95𝑘𝑃𝑎

𝐹𝑠,𝑐𝑟

𝐹0.7=

23.95𝑘𝑃𝑎

4826.3𝑀𝑃𝑎= 4.96𝑥10−6

The ratio value of Fs/F0.7 is so small that the correction can be neglected.

The applied shear loads have been obtained by taking the value of maximum positive and

maximum negative shear stress:

𝜏1 =𝑉

𝐴 𝜏2 =

𝑉

𝐴

=14.3394

(0.25)(1.2) =

−14.2892

(0.25)(1.2)

= 47.798𝑃𝑎 = −47.631𝑃𝑎

𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦 = |𝐹𝑠,𝑐𝑟

𝜏1| 𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦 = |

𝐹𝑠,𝑐𝑟

𝜏2|

= |23.95𝑥103

47.798| = |

23.95𝑥103

−47.631|

= 501.07 = 502.82

106

We can say that the minimum factor of safety for upper wing due to shear load is 501.07. Hence,

it will not buckle due to the shear load.

Compressive Force

For compression force acting on upper wing, the same step used to determine the factor of

safety. Based on Figure 1 in Appendix III, the value of kc is obtained. Buckling strength due to

compressive force can be determined as follows,

𝐹𝑐,𝑐𝑟 =𝑘𝑐𝜋

2𝐸

12(1 − 𝑣2)(𝑡

𝑏)2

=(4)𝜋2(6.9 𝑥 106)

12(1 − 0.32)(0.005

0.25)2

= 9.98𝑘𝑃𝑎

𝐹𝑐,𝑐𝑟

𝐹0.7=

9.98𝑘𝑃𝑎

4826.3𝑀𝑃𝑎= 2.07𝑥10−6

Again, the ratio of Fc/F0.7 is so small that the correction can be neglected.

The applied compression loads have been obtained by taking the value of maximum positive and

maximum negative axial stress:

𝜎1 =𝑀𝑦

𝐼 𝜎2 =

𝑀𝑦

𝐼

=(2.2786)(0.6)

((0.25)(1.23)

12 ) =

(2.0901)(0.6)

((0.25)(1.23)

12 )

107

= 37.98𝑃𝑎 = 34.84𝑃𝑎

𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦 = |𝐹𝑐,𝑐𝑟

𝜎1| 𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦 = |

𝐹𝑐,𝑐𝑟

𝜎2|

= |9.98𝑥103

37.98| = |

9.98𝑥103

34.84|

= 262.77 = 286.45

The smallest factor of safety due to the compression load is 262.77. Based on the observation, the

lowest factor of safety among both shear and compression loads for upper side of the wing is

262.77, showing that the skin will not buckle due to the forces applied. The same steps have been

used to find the factor of safety for lower wing, and the results are being tabulated as below:

Table 6-17 Factor of safety of upper and lower wing

Wing Part Shear Compression Bending Lowest Factor of Safety

Upper Wing 501.07 262.77 - 262.77

Lower Wing 544.62 - 241.72 241.72

Based on the Table 6-17, the lowest factor of safety at the upper and lower side of the wing is

above 1, showing that it will not fail due to the load applied of the aircraft wing’s skin.

108

6.3.5 Stress Analysis of Multi Cell Sections

The shear force at any chordwise section along the span produce shear stresses that are

resisted by the skin and the spars. The shear force at any section obtained from the shear force

diagram evaluated. The resulting shear flow distribution is due to the combined effects of shear

and torsion. In this analysis, two type of wing structure which are monocoque structure and semi-

monocoque structure is taking into account. Both positive (upward) deflection and negative

(downward) deflection of aileron are considered by taking the maximum shear force and torsion

from each case.

6.3.5.1 Shear stress due to shear force (monocoque)

The RC plane wing is supported by one spar which then make it to have 2 cells of wing.

Figure 6-31 shows the RC plane 2 cells wing section with shear flow diagram.

Figure 6-31 Shear flow diagram of 2 cells wing

To simulates monocoque structure, panel AB were cut in order to analyses them as open cell.

Figure 6-32 indicate cell 1 and cell 2 separately.

109

Figure 6-32 Vertical distance from centroid, z of cell 1 and cell 2

Table 6-18 shows the properties of the wing including the area cell 1 and cell 2, vertical

distance from centroid, z, the thickness of the skin and the thickness of the spar.

Table 6-18 Dimensions of cell 1 and cell 2 wing section

𝐴1 = 0.648 × 10−3 𝑚2 𝐴2 = 1.729 × 10−3 𝑚2

𝑧1 = 0.012 𝑚 𝑧1 = 0.0099 𝑚

𝑡𝑠𝑘 = 0.003 𝑚 𝑡𝑠𝑝 = 0.011 𝑚

Calculation of Moment of Inertia

𝐼𝑥𝑥 = Σ𝐴𝑧2

= 2.628 × 10−7 𝑚4

Data from Material Properties

𝐹𝑠 = 13.785 × 106 𝑃𝑎

𝐸 = 6894.76 × 106 𝑃𝑎

110

Calculating Integral part:

Table 6-19 Integral of cell 1 and cell 2

z A ∫𝑧 𝑑𝐴

BA 0.012 0.648 × 10−3 𝑚2 7.776 × 10−6 𝑚3

AB 0.0099 1.729 × 10−3 𝑚2 1.712 × 10−5 𝑚3

Positive (+ve) deflection analysis

The maximum shear force applied for upward deflection for the aircraft is 14.2892 N. The shear

flow is calculated followed by the shear stress and safety factor.

Maximum Shear, 𝑉𝑆 = 14.2892 𝑁

Calculating Shear Flow

𝑞𝑏 ⇒ open section

𝑞𝑏 =𝑉𝑆

𝐼𝑥𝑥∫𝑧 𝑑𝐴

𝑞𝑏𝐵𝐴= 422.8499 𝑁/𝑚

𝑞𝑏𝐴𝐵= 930.8082 𝑁/𝑚

Calculating Shear Stress

𝜏𝑏 =𝑞𝑏

𝑡

Where 𝑡 = 0.003 𝑚

𝜏𝑏𝐵𝐴= 0.1409 × 106

𝜏𝑏𝐴𝐵= 0.3103 × 106

111

Calculating Safety Factor

𝑆𝐹 =𝐹𝑠

𝜏𝑏𝑚𝑎𝑥

=13.785 × 106

0.3103 × 106

= 44.43

According to the safety factor obtained, the structure of the wing is safe and undergo any failure.

Negative (-ve) deflection analysis

The maximum shear force applied for downward deflection for the aircraft is 14.3394 N. The shear


Maximum Shear, 𝑉𝑆 = 14.3394 𝑁

Calculating Shear Flow

𝑞𝑏 =𝑉𝑆

𝐼𝑥𝑥∫𝑧 𝑑𝐴

𝑞𝑏𝐵𝐴= 424.3355 𝑁/𝑚

𝑞𝑏𝐴𝐵= 934.0782 𝑁/𝑚

Calculating Shear Stress

𝜏𝑏 =𝑞𝑏

𝑡

Where 𝑡 = 0.003 𝑚

𝜏𝑏𝐵𝐴= 0.1415 × 106

𝜏𝑏𝐴𝐵= 0.3114 × 106

112

Calculating Safety Factor

𝑆𝐹 =𝐹𝑠

𝜏𝑏𝑚𝑎𝑥

=13.785 × 106

0.3114 × 106

= 44.27

The value shows a positive value with greater than 1. This means a safe condition for the RC plane

to fly with this applied force.

6.3.5.2 Shear stress due to shear force (semi-monocoque)

Shear flow for each cell

𝑞𝑠 = 𝑞𝑏 + 𝑞𝑠,0

𝑞𝐵𝐴 = 𝑞𝑏𝐵𝐴+ 𝑞1 = 1559.9419 + 𝑞1

𝑞𝐴𝐵 = 𝑞𝑏𝐴𝐵+ 𝑞2 = 3433.8583 + 𝑞2

𝑞𝐴𝐵𝑖𝑛= 𝑞1 + 𝑞2

Figure 6-33 shows the line integral is obtained from the wing section.

Figure 6-33 Line integral of wing section

113

𝑙𝐵𝐴 = 0.0888 𝑚

𝑙𝐴𝐵 = 0.2582 𝑚

𝑙𝐴𝐵𝑖𝑛= 0.0202 𝑚

𝑙𝐵𝐴

𝑡𝑠𝑘= 29.6

𝑙𝐴𝐵

𝑡𝑠𝑘= 86.0667

𝑙𝐴𝐵𝑖𝑛

𝑡𝑠𝑝= 1.8364

Modulus of Rigidity

𝐺 =𝐸

2(1 + 𝑣)

=6894.76 × 106

2(1 + 0.3)

= 2651.8308 × 106

Angular displacement

휃𝑖 =1

2𝐴𝑖𝐺𝑖∫

𝑞

𝑡𝑑𝑠

𝑖

Angular displacement for each cell is the same

Cell 1:

휃1𝐺 =1

2𝐴1[𝑞1 (

𝑙𝐵𝐴

𝑡𝑠𝑘) + 𝑞12 (


𝑡𝑠𝑝)]

29.6𝑞1 + 1.8364𝑞1 − 1.8364𝑞2 − 2𝐴1𝐺휃 = 0

31.4364𝑞1 − 1.8364𝑞2 − 3.4368 × 106휃 = 0

Cell 2:

휃2𝐺 =1

2𝐴2[𝑞2 (

𝑙𝐴𝐵

𝑡𝑠𝑘) − 𝑞12 (


𝑡𝑠𝑝)]

114

86.0667𝑞2 − 1.8364𝑞1 + 1.8364𝑞2 − 2𝐴1𝐺휃 = 0

−1.8364𝑞1 + 87.9031𝑞2 − 9.17 × 106휃 = 0


The moment, M produced by the total shear flow about any convenient moment centre is given

by below equation. Taking the moment at point A.

Equilibrium of Moment

𝑀 = Σ2𝐴𝑖𝑞𝑖

2𝐴1𝑞𝐵𝐴 + 2𝐴2𝑞𝐴𝐵 = 0

2𝐴1(1559.9419 + 𝑞1) + 2𝐴2(3433.8583 + 𝑞2) = 0

1.296 × 10−3𝑞1 + 3.458 × 10−3𝑞2 + 0휃 = −3.7667

Solve simultaneously for Cell 1, Cell 2 and Equilibrium of Torque equation

𝑞1 = −838.9335

𝑞2 = −774.8531

휃 = −0.0073

Calculate Shear Stress

𝜏1 =𝑞1

𝑡𝑠𝑘+

(𝑞1 + 𝑞2)

𝑡𝑠𝑝= −0.4264 × 106

𝜏2 =𝑞2

𝑡𝑠𝑘+

(𝑞2 − 𝑞1)

𝑡𝑠𝑝= −0.2525 × 106

115

Calculate Safety Factor

𝑆𝐹1 =𝐹𝑠

𝜏1

=13.785 × 106

0.4264 × 106

= 32.33

𝑆𝐹2 =𝐹𝑠

𝜏2

=13.785 × 106

0.2525 × 106

= 51.60

Therefore, cell 1 and cell 2 both have a safe structural system.


The calculation of moment for downward deflection is same as upward deflection scenario.

Equilibrium of Moment

𝑀 = Σ2𝐴𝑖𝑞𝑖

2𝐴1𝑞𝐵𝐴 + 2𝐴2𝑞𝐴𝐵 = 0

2𝐴1(−1171.3717 + 𝑞1) + 2𝐴2(−2578.5091 + 𝑞2) = 0

1.296 × 10−3𝑞1 + 3.458 × 10−3𝑞2 + 0휃 = −3.78


𝑞1 = −841.8957

𝑞2 = −777.5891

휃 = −0.0073


𝜏1 =𝑞1

𝑡𝑠𝑘+

(𝑞1 + 𝑞2)

𝑡𝑠𝑝= −0.4279 × 106

𝜏2 =𝑞2

𝑡𝑠𝑘+

(𝑞2 − 𝑞1)

𝑡𝑠𝑝= −0.2534 × 106

116


𝑆𝐹1 =𝐹𝑠

𝜏1

=13.785 × 106

0.4279 × 106

= 32.22

𝑆𝐹2 =𝐹𝑠

𝜏2

=13.785 × 106

0.2534 × 106

= 54.41

This highlighted that the applied shear force does not fail the wing.

6.3.5.3 Shear stress due to torsion (monocoque)

The problems involving torsion are common with aircraft structures. The material covered

wing of the airplane are basically thin walled tubular structures subjected to large torsion moments

under many flight and landing conditions. In this analysis, both upward and downward deflection

of aileron is analyses. The monocoque structure of the wing is being analyse. In monocoque

structure, the spar is not taking into account as the support system. Only one cell is investigated

for the shear stress analysis.


The maximum torsion applied for upward deflection for the aircraft is 2.3969 Nm. The shear


Maximum Torsion, 𝑇 = 2.3969 𝑁𝑚

Area, 𝐴 = 𝐴1 + 𝐴2 = 2.377 × 10−3 𝑚2

117

Calculate Shear Flow

𝑞 =𝑇

2𝐴

= 504.1859 𝑁/𝑚


𝜏 =𝑞

𝑡

= 168061.9829


𝑆𝐹 =𝐹𝑠

𝜏

=13.785 × 106

168.0620 × 103

= 82.02

The safety factor obtain is greater than 1. Thus, this means that the structure of the wing is safe to

be operate.


The maximum torsion applied for downward deflection for the aircraft is 6.4716 Nm. The shear

flow is calculated, followed by the shear stress and safety factor.


Area, 𝐴 = 𝐴1 + 𝐴2 = 2.377 × 10−3 𝑚2

118

Calculate Shear Flow

𝑞 =𝑇

2𝐴

= 1361.2958 𝑁/𝑚


𝜏 =𝑞

𝑡

= 453765.2503


𝑆𝐹 =𝐹𝑠

𝜏

=13.785 × 106

453.7653 × 103

= 30.38

This proved that the structure will not fail when there is downward aileron deflection applied to

the aircraft.

6.3.5.4 Shear stress due to torsion (semi-monocoque)

In this part, the investigation is done for the wing semi-monocoque structure. This type of

structure taking the spar into account in the analysis as the support system. The aircraft wing

consists of only one spar thus the analysis will be done with two cell system.

The line integral for shear stress due to torsion is same as in the shear stress due to shear

flow part. The angular displacement in each Cell 1 and Cell 2 equation also same as in the shear

119

stress due to shear flow part. The equilibrium of torque is different in this part for both positive

and negative deflection.

Cell 1:

31.4364𝑞1 − 1.8364𝑞2 − 3.4368 × 106휃 = 0

Cell 2:

−1.8364𝑞1 + 87.9031𝑞2 − 9.17 × 106휃 = 0


The maximum torsion applied for upward deflection for the aircraft is 2.3969 Nm. The shear



The wing section comprises 2 cells and carries a torque of 2.3969 Nm which generate individual

but unknown torques in each of the cells. Each cell therefore develops a constant shear flow.

Then the equation of torque is

Equilibrium of Torque

𝑇 = Σ2𝐴𝑖𝑞𝑖

2𝐴1𝑞1 + 2𝐴2𝑞2 = 2.3969

1.296 × 10−3𝑞1 + 3.458 × 10−3𝑞2 + 0휃 = 2.3969


𝑞1 = 533.8465

𝑞2 = 493.0697

휃 = 0.0046

120


𝜏1 =𝑞1

𝑡𝑠𝑘+

(𝑞1 + 𝑞2)

𝑡𝑠𝑝= 0.2713 × 106

𝜏2 =𝑞2

𝑡𝑠𝑘+

(𝑞2 − 𝑞1)

𝑡𝑠𝑝= 0.1607 × 106


𝑆𝐹1 =𝐹𝑠

𝜏1

=13.785 × 106

0.2713 × 106

= 50.81

𝑆𝐹2 =𝐹𝑠

𝜏2

=13.785 × 106

0.1607 × 106

= 85.81

Both value of calculated safety factor shows that the structure is able to support the applied torsion.

Thus, it is considered as a safe structure.


The maximum torsion applied for downward deflection for the aircraft is 6.4716 Nm. The shear

flow is calculated followed by the shear stress and safety factor. The calculation and steps are same

as the upward deflection.


Equilibrium of Torque

𝑇 = Σ2𝐴𝑖𝑞𝑖

2𝐴1𝑞1 + 2𝐴2𝑞2 = 6.4716

1.296 × 10−3𝑞1 + 3.458 × 10−3𝑞2 + 0휃 = 6.4716

121


𝑞1 = 1.4414 × 103

𝑞2 = 1.3313 × 103

휃 = 0


𝜏1 =𝑞1

𝑡𝑠𝑘+

(𝑞1 + 𝑞2)

𝑡𝑠𝑝= 0.7325 × 106

𝜏2 =𝑞2

𝑡𝑠𝑘+

(𝑞2 − 𝑞1)

𝑡𝑠𝑝= 0.4338 × 106


𝑆𝐹1 =𝐹𝑠

𝜏1

=13.785 × 106

0.7325 × 106

= 18.82

𝑆𝐹2 =𝐹𝑠

𝜏2

=13.785 × 106

0.4338 × 106

= 31.78

The safety factor for both cells shows a value of more than 1. Thus, this prove that there will be

no structural damage or fail during the flight operation.

6.3.5.5 Shear Stress Summary

From the analysis and calculation of the shear stresses due to shear force and torsion, the total

shear flow for monocoque and semi-monocoque structure is obtained. The maximum shear flow

value of each cases is taken into consideration. Table 6-20 shows the total shear flow obtained.

122

Table 6-20 Total shear flow for monocoque and semi-monocoque wing structure

Shear Flow, 𝑞

Monocoque Semi-monocoque

Cell Cell Cell 1 Cell 2

Due to Shear Force Torsion Shear Force Torsion Shear Force Torsion

Maximum q 934.078 1361.296 -838.93 1441.400 -774.853 1331.300

Total 𝑞 2295.374 602.467 556.447

The combine total shear stress due to shear force and due to torsion is calculated and tabulated in

Table 6-21 below.

Table 6-21 Total Shear Stress of Monocoque and Semi-monocoque

Total Shear Stress


Cell 1 Cell 1 Cell 2

765124.663 306177.930 181298.700

The safety factor for the combine stresses is shown in Table 6-22. The value of the safety factor

indicate that the structure is safe for operation with the applied maximum shear force and torsion.

Table 6-22 Safety Factor of Monocoque and Semi-monocoque

Safety Factor


Cell 1 Cell 1 Cell 2

18.02 45.02 76.03

123

6.3.6 Part Attachment

The part attachment between the wing and the fuselage is the wing strut. The wing strut

supports the wing structure. Structural analysis of the wing strut is carried out to determine if the

it is able to support the load on the wing. Furthermore, the analysis of the joint between the wing’s

wall and the wing strut as well as that between the fuselage’s wall are carried out. Also, the analysis

of the part attachment between the aileron and the aerofoil is executed.

6.3.6.1 Joint of the part attachment

The wing strut is attached to the wing and the fuselage by applying glue and using the hinge

with a hinge pin. The material selected for hinge and the hinge pin are mica plastic and AN Steel

respectively. The moment at the joint is assumed to be zero at the initial design stage, and thus,

the wing strut is a two force-member with pin joints at both ends. Since the wing strut is a two-

force member, the result of the analysis at the joint between the wing and the wing strut is the same

as at the joint between the fuselage and the wing strut. Thus, only the joint between the wing and

the wing strut is performed. The hinge pin is used to resist the vertical reaction force of the wing

strut while the glue joint aims to resist the horizontal shear force between the surfaces of the wing

and the wing strut.

Figure 6-34 Part attachment (wing strut)

124

Hinge pin

Table 6-23 Material Properties of AN Steel (hinge pin)

Material Properties AN Steel

Ultimate Tensile Strength, Ftu 125000 psi

Ultimate Shear Strength. Fsu 75000 psi

Bearing Strength, Fb 180000 psi

Safety factor of fitting 1.15

According to Bruhn (1973), the section properties and the ultimate shear, tension and

bending strengths for AN Standard Steel bolt/pin at room temperature are as follows,

Table 6-24 Properties of AN Standard Steel bolt/pin

Size of pin 0.19

Area of solid section (in2) 0.02835

Moment of inertia of solid (in4) 0.0000640

Ultimate single shear strength at full D, lb 2126

Ultimate tensile strength (in thread), lb 2210

Ultimate bending moment in lbs 121

Figure 6-35 Dimension of the connection

125

Dimensions

Diameter of hole, D = 0.19 in

Thickness, t = 0.08 in

Width of hinge, 2R = 0.4 in

From wing loading calculation,

Vertical reaction force on Wing strut, Rstrut = 28.1707 𝑁 (6.333 lb)

𝐷𝑒𝑠𝑖𝑔𝑛 𝑓𝑖𝑡𝑡𝑖𝑛𝑔 𝑙𝑜𝑎𝑑, 𝑃 = 1.15 × 6.333 = 7.283 𝑙𝑏

Check Shear Strength of Hinge Pin

The bolt is double shear. From Table 6-24, the ultimate single shear strength is 2126 lb.

Hence, 𝑃𝑢 = 2 × 2126 = 4252 𝑙𝑏

𝑆𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 =𝑃𝑢

𝑃=

4252

7.283= 583.83

Check Bending Strength of Hinge Pin

Figure 6-36 The moment arm for bending moment on hinge pin

126

Dimensions

𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑤𝑎𝑙𝑙, 𝑡1 = 0.1969 𝑖𝑛

𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 ℎ𝑖𝑛𝑔𝑒, 𝑡2 = 0.08 𝑖𝑛

𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑔 = 0.004 𝑖𝑛

𝑏 =𝑡12

+𝑡24

+ 𝑔

𝑏 = 0.12245

Therefore, bending moment on the pin,

𝑀 = 0.5𝑃𝑏 = 0.5(7.283)(0.12245) = 0.4459 𝑙𝑏 ∙ 𝑖𝑛

𝑓𝑏 =𝑀𝑟

𝐼=

(0.4459) (0.192 )

0.000064= 661.883 𝑝𝑠𝑖

As 𝐹𝑏 = 180000𝑝𝑠𝑖, then

𝑆𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝐹𝑏

𝑓𝑏= 271.9515

Hinge

Table 6-25 Material Properties of Mica Plastic (fixed hinge)

Material Properties Mica Plastic

Ultimate Tensile Strength, Ftu 25000 psi

Ultimate Shear Strength. Fsu 38400 psi

Safety factor of fitting 1.15

Check Tensile Strength of Pin Hole

𝑃𝑢 = 𝑘𝑡𝐹𝑡𝑢𝐴𝑡

𝑘𝑡 is the tension efficiency factor to take into account of the stress concentration due to

the pin hole, and its value can be obtained from the graph shown in Figure 6-37.

127

Figure 6-37 Graph of kt against d/w

𝑃𝑢 = 𝑘𝑡𝐹𝑡𝑢(2𝑅 − 𝐷)𝑡

𝑃𝑢 = 2.19(25000)(0.4 − 0.19)(0.08) = 919.8 𝑙𝑏

Therefore,

𝑆𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 =𝑃𝑢

𝑃=

919.8

7.283= 126.2941

Check Shear Out Strength of Hinge

𝑃𝑢 = 𝐹𝑠𝑢𝐴𝑠

𝑃𝑢 = 38400(0.4 − 0.19)(0.08) = 645.12 lb

The value is over the design fitting load of 7.283 lb

Glue

Check Shear Strength of Glue

Figure 6-38 Shear stress on the contact area of the glue

128

𝜏 =𝑉

𝐴

𝜏 =𝑉

𝐴𝑠𝑡𝑟𝑢𝑡 + 𝐴ℎ𝑖𝑛𝑔𝑒

𝜏 =84.5589cos (19.46)

(0.01)(0.002) + (0.01)(0.02)

𝜏 = 362401.88 𝑃𝑎

The shear strength of the glue is assumed as 6000 psi (41.36854 MPa), therefore

𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 41.36854

0.3624= 114.151

Since all the values of safety factor are greater than 1, we can conclude that the all the

joints of the wing strut to the wall of the wing and the wall of the fuselage are safe.

6.3.6.2 Structural Analysis of Wing Strut

The wing strut can buckle due to tension and compression, depending on the direction of

the wing’s deflection. The cross section of the wing strut also experiences shear.

Axial force

The applied stress due to the axial force acted on the wing strut can be calculated as follows,

𝜎 =𝐹

𝐴

𝜎 =84.5589

(0.01)(0.002)= 4.2279 𝑀𝑃𝑎

The critical buckling strength of the wing strut can also be obtained, based on the material

properties of the wing strut (plywood). Assuming that the wing strut is simply supported, the

buckling strength is

𝜎𝑐𝑟 =𝜋2𝑘𝑐𝐸

12(1 − 𝑣2)(𝑡

𝑏)2

129

𝜎𝑐𝑟 =𝜋2(4)(1000 × 103)

12(1 − 0.32)(0.002

0.01)2

𝜎𝑐𝑟 = 1.4461 × 105 𝑝𝑠𝑖 = 997.048 𝑀𝑃𝑎

The 𝐹0.7value is 700ksi. Since the 𝜎𝑐𝑟value is much smaller than the 𝐹0.7 value, correction

is not required. Therefore,

𝑅𝑐 = 𝜎

𝜎𝑐𝑟= 4.2404 × 10−3

Shear force

The applied stress due to the shear force acted on the wing strut can be calculated as follows,

𝜏 =𝑉

𝐴

𝜎 =84.5589 sin(19.46)

(0.01)(0.002)= 1.40853 𝑀𝑃𝑎

Assuming that the wing strut is simply supported, the buckling strength is


12(1 − 𝑣2)(𝑡

𝑏)2

𝜎𝑐𝑟 =𝜋2(5.5)(1000 × 103)

12(1 − 0.32)(0.002

0.01)2

𝜎𝑐𝑟 = 1.9884 × 105 𝑝𝑠𝑖 = 1370.9410 𝑀𝑃𝑎

The 𝐹0.7value is 700ksi. Since the 𝜎𝑐𝑟value is much smaller than the 𝐹0.7 value, correction

is not required. Therefore,

𝑅𝑠 = 𝜎

𝜎𝑐𝑟= 1.0274 × 10−3

130

Figure 6-39 Combined compression, bending and shear (Bruhn, 1973)

𝑅𝑠

𝑅𝑐= 0.2423, 𝑅𝑏 = 0

From Figure 6-39, 𝑅𝐶𝐴 = 0.91

𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑅𝐶𝐴

𝑅𝐶=

0.91

4.2404 × 10−3= 214.602

Since the safety factor for the loading of wing strut due to axial load and shear load is larger

than 1, we can conclude that the wing strut is safe.

6.3.6.3 Part attachment between aerofoil and aileron

According to Federal Aviation Regulations (FAR) Appendix A to Part 23 (Simplified Design Load

Criteria), the simplified limit surface loadings for the aileron is specified as follows.

131

Table 6-26 Average limit control surface loading

Figure 6-40 Average limit control surface loading

132

Based on Figure 6-40,

𝑤𝑎𝑟𝑒𝑎 = 0.466 (𝑛𝑊

𝑆)

Where,

𝑊𝑒𝑖𝑔ℎ𝑡,𝑊 = 14.715 𝑁 (3.30806 𝑙𝑏)

𝑊𝑖𝑛𝑔𝑠𝑝𝑎𝑛 𝑎𝑟𝑒𝑎, 𝑆 = 0.20387 𝑚2 (2.19444 𝑓𝑡2)

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑖𝑙𝑒𝑟𝑜𝑛 = 0.235𝑚

𝑤𝑎𝑟𝑒𝑎 = 0.466 (3(3.30806)

2.19444) = 2.10745 𝑙𝑏/𝑓𝑡2 (100.905𝑁/𝑚2)

𝑤 = 100.905 × 0.235 = 23.7127𝑁/𝑚

Figure 6-41 Chordwise lift distribution of the aerofoil with deployed aileron

Figure 6-42 Wing loading on the deployed aileron

𝑉 =1

2(0.04)(23.7127) = 0.4743 𝑁

𝑀 =1

2(0.04)(23.7127) (

1

3) (0.04) = 0.0063234 𝑁𝑚

133

The result shows that the amount of the applied load is very small. A fibre tape is used to

withstand the shear and bending moment of the part attachment between the aileron and the

aerofoil. Fibre tape is chosen due to its strong bonding properties. The fibre tape is applied to one

side of the aileron only with the upper part of the wing so that it can be deflect upward and

downward freely. The tape function like a hinge to the aileron. The area of tape use for the

attachment in each aileron is shown in Figure 6-43. It is assumed that the fibre tape is so strong

that it can withstand 4kg load and is able to withstand shear load of 83.65N.

Figure 6-43 Aileron attachment area

134

6.4 Summary of Safety Factor

Therefore, the safety factor obtained from the structural analysis can be summed up in

Table 6-27. Based on Table 6-27, as all the factor of safety is more than 1, the structure of the

wing components is safe.

Table 6-27 Factor of Safety of Wing Components

Structure Factor of safety

Spar Rectangular spar 14.4

Skin Upper Skin 262.8

Lower Skin 241.7

Rib Cell 1 29862.7

Cell 2 37904.4

Part Attachment

Wing Strut 214.6

Hinge Pin 272

Hinge 126.3

Glue 114.2

142

CHAPTER 7

FUSELAGE AND TAIL

7.1 Introduction

The basic functions of an aircraft’s fuselage structure are to transmit and resist applied

load, to provide an aerodynamics shape and to protect passengers, payload, systems and others

from the environmental conditions encountered in flight. These requirements in most aircraft,

result in thin shell structures where the outer surface or skin of the shell is usually supported by

longitudinal stiffening members and transverse frames to enable it to resist bending, compressive

and tensional loads without buckling. Such structures are known as semi-monocoque, while thin

shells which rely entirely on their skins for their capacity to resist loads are referred to as

monocoque.

Fuselage while of different shape to the aerodynamics surfaces, comprise members which

perform similar functions to their counterparts in the wings and tail plane. However, there are

differences in the generation of the various types of load. Aerodynamic forces on the fuselage skin

are relatively low, on the other hand, the fuselage supports large concentrated loads such as wing

reactions, tail plane reactions, undercarriage reactions and it carries payloads of varying size and

weight, which may cause large inertia forces.

Furthermore, aircraft designed for high altitude flight must withstand internal pressure. The

shape of the fuselage cross-section is determined by operational requirements. For example, the

most efficient sectional shape for a pressurized is circular or a combination of circular elements.

Irrespective of shape, the basic fuselage structure is essentially a single cell thin-walled tube

comprising skin, transverse frames and stringers; transverse frames which extend completely

across the fuselage are known as bulkheads.

143

7.2 Monocoque Structure

Monocoque fuselage design as shown in Figure 7-1 relies on the strength of the skin (also

known as the shell or covering) to carry the various loads. True monocoque construction does not

use formers, frame assemblies, or bulkheads to give shape to the fuselage. Instead, the skin carries

all fuselage stresses. Since no bracing members are present, the skin must be strong enough to

keep the fuselage rigid. Thus, the biggest challenge in monocoque design is maintaining enough

strength while keeping the weight within allowable limit.

Figure 7-1: Fuselage monocoque structure

For our fuselage monocoque design, the construction is done by using the two former and

one bulkhead to give shape to the fuselage. The material used for both bulkhead and formers is

plywood with thickness of 3mm. Then, between the two formers, there is a servo platform (blue

colour) where both servo for rudder and elevator was placed. This servo platform was made of

EPP foam with the thickness of 5mm. There is also the battery tray (red colour), made from EPP

foam where the battery is going to be and it will be hold by the battery holder (black colour) which

made from the Velcro tape.

144

7.3 Material Properties

The Medium Density Fiberboard (MDF) with 3mm thickness use to fabricate both formers and

bulkhead. The material properties for the MDF as shown in Table 7-1 below.

Table 7-1: Material properties

Mechanical Properties

Compressive strength 10 Mpa

Elastic (Young’s Tensile) Modulus 4.0 Gpa

Modulus of rupture 35.85 Mpa

Elongation at break 0.5%

Poisson’s ratio 0.25

Shear modulus 1.6 Gpa

Ultimate tensile strength 18 Mpa

Thermal Properties

Specific heat capacity 1700 J/kg-K

Thermal conductivity 0.3 W/m-K

Thermal expansion 12 µm/m-K

Other Properties

Density 0.75 g/cm3

Dielectric strength 0.5 kV/mm

145

7.4 Shear and Bending Moment Diagram for Fuselage

Analysis of the fuselage is taken when the plane flying in cruising condition with

differences G with -1.5, 1 and 3.

7.4.1 𝑮 = 𝟏

The maximum shear force and bending moment are 6.4 N and -0.48 Nm respectively.

146

7.4.2 𝑮 = 𝟑

The maximum shear force and bending moment are 19.24 N and -1.44 Nm respectively.

147

7.4.3 𝑮 = −𝟏. 𝟓

The maximum shear force and bending moment are -9.61 N and 0.7 Nm respectively.

148

Table 7-2: Maximum values of shear force and bending moment for fuselage

G = -1.5 G = 1 G = 3

Shear Force (N) -9.61 6.4 19.24

Bending moment

(Nm) 0.7 -0.48 -1.44

Material used for fuselage skin is EPP foam with material properties of ultimate stress

27.37 MPa with the thickness of 0.005 m.

𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎𝑏

27.57 𝑀𝑃𝑎 = 𝑀𝑦

𝐼

Where

M bending moment (Nm)

y Vertical distance away from the neutral axis (m)

I Second moment of inertia (m4)

The second moment of inertia, 𝐼 = 𝑏𝑑3

12 is assume as figure below:

Figure 7-2: Cross section of maximum bending moment location for fuselage (cm)

149

𝐼 = (0.12)(0.105)3

12−

(0.11)(0.095)3

12

𝐼 = 3.717 × 10−6 𝑚4

From the bending stress, the maximum bending moment can be stand by the foam is given as:

𝑀𝑚𝑎𝑥 =27.57 × 106 × 3.717 × 10−6

0.0525

𝑀𝑚𝑎𝑥 = 1.951 𝑘𝑁𝑚

For the maximum shear stress is assumed to be the same as ultimate stress:

𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠, 𝑣𝑚𝑎𝑥

27.57 𝑀𝑃𝑎 =𝑉𝑄

𝐼𝑏

Where

Q Calculated statically moment (m3)

V Shear force (N)

I Moment of inertia (m4)

b Width of the fuselage (m)

Calculated statically moment, 𝑄 = 𝐴𝑦

Where

A Cross section area of fuselage


𝑄 = (0.12 × 0.105 × 0.06 ) − (0.11 × 0.095 × 0.055)

𝑄 = 1.813 × 10−4 𝑚3

From the Shear stress, the maximum load force able to resist from damage:

𝑉 = 27.57 × 106 × 3.717 × 10−6 × 0.12

1.813 × 10−4

𝑉 = 67.83 𝑘𝑁

150

Therefore, the ultimate stress from the foam show that maximum shear force and bending

moment can be accept are 67.83 kN and 1.951 kNm respectively. The result from the analysis of

shear force and bending moment diagram having the maximum shear force with 19.24 and

maximum bending moment -1.44 Nm when G is 3. Both of it is within the range of maximum

shear force and bending moment of the foam. Overall, during the cruising condition the EPP foam

will not damage easily.

7.5 Flow Simulation

In order to see the pressure acting on our rc plane, Solidwork flow simulation was used to

see the pressure especially at the area of formers and bulkhead as shown in Figure 7-3; thus it will

be used as an estimation of pressure applied on the internal structure in order to do the structural

analysis. The simulation was run with the maximum speed of 33.6m/s based on the v-n diagram

that we already plotted.

Figure 7-3: Pressure contour

Figure 7-4 shown the air flow on the rc plane and there is some vorticity flow (blue) occur

on the top surface at the middle of the wing where the attachment between the wing and fuselage

is located.

Area of interest

151

Figure 7-4: Flow visualization

7.6 Stress Analysis

7.6.1 Former structure

For the structural analysis, we used Ansys as a tool to do the analyze the structure by

assuming the load applied on the former as shown in Figure 7-5. There is an additional load from

on top of the former due to the wing load and we assume the bottom surface to be fixed support.

The value of pressure applied is based on the result of flow simulation.

152

Figure 7-5: Load applied on former

The result of the stress analysis for former is shown in Table 7-3. We analyze based on two

difference type of stresses which is equivalent (Von-Mises) stress and shear stress at the condition

when the rc plane at the maximum speed.

Table 7-3: Stress result for former

Velocity

(m/s)

Load (MPa) Equivalent (Von-Mises)

Stress (MPa)

Shear Stress (MPa)

Max Min Max Min

33.6 0.10187 5.0187 0.0141 2.2434 -2.2418

153

To identify the location of maximum and minimum of both type of stresses acting on the

former, refer to Figure 7-6 where it show the pressure contour when the load is applied. For

equivalent (Von-Mises) stress, we can see that most of the area has low stress and very small are

has the high stress. Besides, most area of the former has the approximately zero shear stress area.

Figure 7-6: Pressure contour on former structure for (a) equivalent (Von-Mises) stress (b) shear

stress

7.6.2 Bulkhead structure

Figure 7-7 show the assumption on how the pressure acting on the bulkhead. The pressure

was obtained from the flow simulation and no additional load is acting on the structure. For the

analysis, the Ansys tool was used to identify the stress on the structure.

(a) (b)

154

Figure 7-7: Load applied on bulkhead

The result of the stress analysis for bulkhead is shown in Table 7-4. We analyze based on

two difference type of stresses which is equivalent (Von-Mises) stress and shear stress at the

condition when the rc plane at the maximum speed.

Table 7-4: Stress result for bulkhead

Velocity

(m/s)

Load (MPa) Equivalent (Von-Mises)

Stress (Pa)

Shear Stress (Pa)

Max Min Max Min

33.6 0.10105 0.3923 0.0699 0.1680 -0.1680

To identify the location of maximum and minimum on both type of stresses acting on the

bulkhead, refer to Figure 7-8 where it show the pressure contour when the load is applied. For

equivalent (Von-Mises) stress, we can see that most of the area has low stress and very small area

has the high stress. Besides, most area of the bulkhead has the approximately zero shear stress

area.

155

Figure 7-8: Pressure contour on bulkhead structure for (a) equivalent (Von-Mises) stress (b) shear

stress

7.7 Compressive-Buckling Analysis

Based on Bruhn, the equation for the elastic instability of flat sheet in compression is,

𝜎𝑐𝑟 =𝜋2𝑘𝑐𝐸

12(1 − 𝜐2)(𝑡

𝑏)2

Where;

𝑘𝑐 is buckling coefficient which depends on edge boundary conditions and sheet aspect ratio

(a/b). Refer graph of compressive-buckling coefficient for flat rectangular plate in Bruhn.

𝐸 is modulus of elasticity

𝜐 is elastic Poisson’s ratio

𝑏 is short dimension of plate or loaded edge

𝑡 is sheet thickness

(a) (b)

156


The former aspect ratio is,

𝑎

𝑏=

110

107= 1.03

From graph compressive-buckling coefficient for flat rectangular plate

𝑘𝑐 = 6.0

The critical elastic compression buckling stress,

𝜎𝑐𝑟 =𝜋2(6.0)(4 × 109)

12(1 − 0.252)(

3 × 10−3

107 × 10−3)

2

𝜎𝑐𝑟 = 16.5513 𝑀𝑃𝑎

To find safety factor,

𝑆𝐹 = (𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑏𝑢𝑐𝑘𝑙𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠 ) = (

16.5513

5.0187) = 3.30

To find the margin of safety,

𝑀𝑆 = 𝑆𝐹 − 1 = 3.30 − 1 = 2.30

Since 𝜎𝑐𝑟 = 16.5513 𝑀𝑃𝑎 the buckling stress is greater than the maximum stress

experience by the structure, which is 5.0187 MPa, the formers structure will not buckle. Besides,

the positive value of margin of safety show that the structure is safe.

157



𝑎

𝑏=

750

550= 1.36


𝑘𝑐 = 5.8


𝜎𝑐𝑟 =𝜋2(5.8)(4 × 109)

12(1 − 0.252)(

3 × 10−3

550 × 10−3)

2

𝜎𝑐𝑟 = 0.6056 𝑀𝑃𝑎



𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠) = (

0.6056

0.3923) = 1.54


𝑀𝑆 = 𝑆𝐹 − 1 = 1.54 − 1 = 0.54

Since 𝜎𝑐𝑟 = 0.6056 MPa the buckling stress is greater than the maximum stress experience

by the structure, which is 0.3923 MPa the bulkhead structure will not buckle. Besides, the positive

value of margin of safety show that the structure is safe.

158

7.8 Shear-Buckling Analysis

The critical elastic shear buckling stress for flat plates with various boundary conditions is

given by the following equation,

𝜎𝑠𝑐𝑟=

𝜋2𝑘𝑠𝐸

12(1 − 𝜐2)(𝑡

𝑏)2

Where;

𝑘𝑠 is shear buckling coefficient and is plotted as a function of the plate aspect ratio a/b.

Refer graph of shear-buckling-stress coefficient of plates in Bruhn.

𝐸 is modulus of elasticity

𝜐 is elastic Poisson’s ratio

𝑏 is short dimension of plate or loaded edge

𝑡 is sheet thickness

159



𝑎

𝑏=

110

107= 1.03

From graph shear-buckling-stress coefficient of plates plate

𝑘𝑠 = 9.8


𝜎𝑠𝑐𝑟=

𝜋2(9.8)(4 × 109)

12(1 − 0.252)(

3 × 10−3

107 × 10−3)

2

𝜎𝑠𝑐𝑟= 27.0339 𝑀𝑃𝑎




27.0339

2.2434) = 12.05


𝑀𝑆 = 𝑆𝐹 − 1 = 12.05 − 1 = 11.05

Since 𝜎𝑠𝑐𝑟= 27.0339 MPa the shear buckling stress is greater than the maximum shear

stress experience by the structure which is 2.2434 MPa , the structure will not buckle. The positive

value of margin of safety show that the structure is safe.

160



𝑎

𝑏=

750

550= 1.36


𝑘𝑠 = 7.8


𝜎𝑠𝑐𝑟=

𝜋2(7.8)(4 × 109)

12(1 − 0.252)(

3 × 10−3

550 × 10−3)

2

𝜎𝑠𝑐𝑟= 0.8144 𝑀𝑃𝑎




0.8144

0.1680) = 4.85


𝑀𝑆 = 𝑆𝐹 − 1 = 4.85 − 1 = 3.85

Since 𝜎𝑠𝑐𝑟= 0.8144 MPa the shear buckling stress is greater than the maximum stress

experience by the structure, which is 0.1680 MPa the bulkhead structure will not buckle. Besides,

the positive value of margin of safety show that the structure is safe.

161

7.9 Shear Flow of Fuselage due to Torsion

In this analysis, the semi-monocoque structure of the fuselage is being analyse. This type

of structure have former that will be taking account in the analysis as the support system. This

aircraft fuselage consist of two formers. So, the analysis will be consisting of three cell system.

The maximum torsion applied to the aircraft when in cruising speed is 1.119 Nm.

Figure 7-9: Shear flow of the fuselage

Table 7-5: Fuselage parameter

Wall Length (m) Thickness

(m) G (MPa) Cell Area

120 0.45 0.005 10.6 𝐴1 = 0.01569

121 0.12 0.003 291.7

13 0.19 0.005 10.6 𝐴2 = 0.020412

24 0.19 0.005 10.6

341 0.08 0.003 291.7 𝐴3 = 0.019263

340 0.75 0.005 10.6

162

Using equation:

𝑇 = ∑ 2

𝑛

𝑅=1

𝐴𝑅𝑞𝑅

Then,

𝑇 = 2𝐴1𝑞1 + 2𝐴2𝑞2 + 2𝐴3𝑞3

So,

1.119 = 2(0.01569)𝑞1 + 2(0.020412)𝑞2 + 2(0.019263)𝑞3

Equation 1:

1.119 = 0.03138𝑞1 + 0.040824𝑞2 + 0.038526𝑞3

Known that,

휃 =𝑑휃

𝑑𝑧=

1

2𝐴𝑅∫𝑞

𝑑𝑠

𝐺𝑡

Then,

휃1 =1

2𝐴1[

𝑞1

10.6 × 106(

0.45

0.005) +

𝑞12

291.7 × 106(

0.12

0.003)]

휃2 =1

2𝐴2[

𝑞2

10.6 × 106(0.19 + 0.19

0.005) −

𝑞12

291.7 × 106(

0.12

0.003) +

𝑞23

291.7 × 106(

0.08

0.003)]

휃3 =1

2𝐴3[

𝑞3

10.6 × 106(

0.75

0.005) −

𝑞23

291.7 × 106(

0.08

0.003)]

휃1 = 휃2

휃1 = 휃3

𝑞12 = 𝑞1 − 𝑞2

𝑞23 = 𝑞2 − 𝑞3

So, Equation 2:

2.783 × 10−4𝑞1 − 1.856 × 10−4𝑞2 + 2.239 × 10−6𝑞3 = 0

Equation 3:

163

2.7494 × 10−4𝑞1 − 1.997 × 10−6𝑞2 − 3.6967 × 10−4𝑞3 = 0

Solving Equation (1) to (3) simultaneously gives

𝑞1 = 9.225𝑁

𝑚 𝑞2 = 13.915

𝑁

𝑚 𝑞3 = 6.786

𝑁

𝑚

To calculate shear stress using formula:

𝜏1 =𝑞1

𝑡𝑠𝑘+

(𝑞1 + 𝑞2)

𝑡𝑓𝑜𝑟

𝜏2 =𝑞2

𝑡𝑠𝑘+

(𝑞2 − 𝑞1)

𝑡𝑓𝑜𝑟+

(𝑞2 + 𝑞3)

𝑡𝑓𝑜𝑟

𝜏3 =𝑞3

𝑡𝑠𝑘+

(𝑞3 − 𝑞2)

𝑡𝑓𝑜𝑟

So, shear stress for each cell as follow:

𝜏1 = 9.558 𝑘𝑁 𝜏2 = 11.247 𝑘𝑁 𝜏3 = −1.019 𝑘𝑁

To calculate safety factor using formula:

𝑆𝐹 =𝐹𝑆

𝜏

𝐹𝑆 = 13.785 𝑀𝑃𝑎

𝑆𝐹1 = 1442 𝑆𝐹2 = 1226 𝑆𝐹3 = 13528

164

Based on the calculation of safety factor, all cell obtain safety factor that exceed 1. Thus,

the skin of the aircraft fuselage can withstand the applied torsion.

7.10 Structure Analysis on Horizontal Tail

Figure 7-10: Flow simulation with 33.6m/s and 860RPM

Table 7-6: Stress on surface

maximum minimum

Horizontal tail 101476.71 Pa 101215.83 Pa

Vertical tail 101364.90 Pa 101300.00 Pa

The analysis taken from a small element for the maximum and minimum stresses on the

horizontal tail since the values of vertical tail within the range of maximum and minimum for

horizontal tail.

a

b

fixed

free

a

b

165

Element taken with the dimension of a, b, and t are 0.025m, 0.01m, 0.005 respectively.

Where t is the thickness of the horizontal tail. by taking of 1.5 as the safety factor:

maximum minimum

𝐹𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒𝑚𝑎𝑥= 𝐹𝑚𝑎𝑥 × 1.5

= 101476.71 × 1.5

= 152215.07 𝑃𝑎

𝐹𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒𝑚𝑖𝑛= 𝐹𝑚𝑖𝑛 × 1.5

= 101215.83 × 1.5

= 151823.75 𝑃𝑎

Young modulus of the foam is 27.57 MPa. According to Bruhn, Figure 7-11 𝑎

𝑏 is 2.5. so 𝑘𝑠 = 6

and 𝑣𝑒 = 0.3.

Figure 7-11: Shear-Buckling-Stress Coefficient of Plates as a Function of a/b

Critical stress, 𝐹𝑐𝑟 = 𝜋2𝑘𝑠𝐸

12 (1− 𝑣𝑒2)

(𝑡

𝑏)2

𝐹𝑐𝑟 = 𝜋2(6)(27.57 × 106)

12 (1 − (0.3)2)(0.005

0.01)2

= 37.38 𝑀𝑃𝑎

𝐹0.7 = 0.7 𝐸

166

= 0.7 × 27.57 × 106

= 19.30 𝑀𝑃𝑎

Figure 7-12: Chart of Nondimensional Compressive Buckling Stress

Then, correction method must be done. The element taken is assume as one is simply

supported and another one is free while the minimum number of n is 2.

𝐹𝑐𝑟

𝐹0.7=

37.38 𝑀𝑃𝑎

19.30 𝑀𝑃𝑎= 1.94

As referred in figure 3, 𝐹𝑐𝑟

𝐹0.7= 1.3 after applying the correction method.

𝐹𝑐𝑟𝑝= 1.3 × 27.57 𝑀𝑃𝑎

= 35.84 𝑀𝑃𝑎

The value of safety factor and margin of safety must be positive. If values are less than zero it

means the elements will buckle due to the stresses.

Safety factor of horizontal tail given by:

𝑆𝐹 = 𝐹𝑐𝑟𝑝

𝐹𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒𝑚𝑎𝑥/𝑚𝑖𝑛

Maximum Minimum

35.84 𝑀𝑃𝑎

152215.07 𝑃𝑎= 0.234 × 103

35.84 𝑀𝑃𝑎

151823.75 𝑃𝑎= 0.236 × 103

Margin of safety, 𝑀𝑆 = 𝑆𝐹 − 1

167

𝑀𝑆 = 0.234 × 103 − 1

= 233

𝑀𝑆 = 0.236 × 103 − 1

= 235

Therefore, the stresses on the surface will not damage the tail of the plane

7.11 Shear and Bending Moment Diagram for Tail

Analysis taken when the plane flying in cruising condition with differences G with -1.5, 1,

3.

7.11.1.1 G = 1

The lift generated by the horizontal tail is 1.14 N/m along the tail while the weight of

horizontal tail is 0.04 kg and length of tail, L is 0.35m.

168

Maximum shear force and bending moment are 0.1995 N and -0.0512 Nm respectively.

7.11.2 G = 3

The lift generated by the horizontal tail is 3.42 N/m along the tail while the weight of


169

Maximum shear force and bending moment are 0.5985 N and 0.1408 Nm respectively.

7.11.3 G = -1.5

The lift generated by the horizontal tail is -1.71 N/m along the tail while the weight of


170

Maximum shear force and bending moment are -0.9909 N and -0.1734 Nm respectively.

Overall the maximum values of shear force and bending moment according to the analysis

are illustrated as below:

Table 7-7: Maximum values of shear force and bending moment

G = -1.5 G = 1 G = 3

Shear Force (N) -0.9909 0.1995 N 0.5985

Bending moment

(Nm) -0.1734 -0.0512 0.1408

The material used is EPP foam with the material properties of ultimate stress of 27.57 MPa

with the thickness of 0.005m.

171

𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎𝑏

27.57 𝑀𝑃𝑎 = 𝑀𝑦

𝐼

Where

M bending moment (Nm)


I Second moment of inertia (m4)

Second moment of inertia, 𝐼 = 𝑏𝑑3

12

𝐼 = (0.1)(0.005)3

12+

(0.135)(0.005)3

12

𝐼 = 5.052 × 10−9 𝑚4

From the bending stress, the maximum bending moment can be stand by the foam is given as:

𝑀𝑚𝑎𝑥 =27.57 × 106 × 5.052 × 10−9

0.0675

𝑀𝑚𝑎𝑥 = 2.06 𝑁𝑚

For the maximum shear stress is assumed to be the same as ultimate stress:

𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠, 𝑣𝑚𝑎𝑥

27.57 𝑀𝑃𝑎 =𝑉𝑄

𝐼𝑏

Where

Q Calculated statically moment (m3)

V Shear force (N)

I Moment of inertia (m4)

b Width of the tail (m)

Calculated statically moment, 𝑄 = 𝐴𝑦

Where

A Cross section area of tail

y length of the tail

172

𝑄 = (0.35 × 0.005 × 0.1) + (0.135 × 0.005 × 0.1)

𝑄 = 0.2425 × 10−3 𝑚3

From the Shear stress, the maximum load force able to resist from damage:

𝑉 = 27.57 × 106 × 5.052 × 10−9 × 0.35

0.2425 × 10−3

𝑉 = 201.03 𝑁

Therefore, the ultimate stress from the foam show that maximum shear force and bending

moment can be accept are 201.03 N and 2.06 Nm respectively. The result from the analysis of

shear force and bending moment diagram having the maximum shear force of 0.5985N when G is

3 and maximum bending moment 0.1734Nm when G is -1.5. Both of it is within the range of

maximum shear force and bending moment of the foam. Overall, during the cruising condition the

EPP foam will not damage easily.

173

CHAPTER 8

LANDING GEAR

8.1 Introduction

The landing gear part for our RC-plane is tailwheel-type landing gear. It is also known as

conventional landing gear which is an aircraft undercarriage consisting of two main wheels

forward of the center of gravity and a small wheel or skid to support the tail. The term

"conventional" persists for historical reasons, but all modern jet aircraft and most modern propeller

aircraft use tricycle gear with a single nose wheel in the front and two or more main wheels aft.

The tailwheel configuration offers several advantages over the tricycle landing gear

arrangement, which make tailwheel aircraft less expensive to manufacture and maintain. Due to

its position much further from the center of gravity, a tailwheel supports a smaller part of the

aircraft's weight allowing it to be made much smaller and lighter than a nosewheel. As a result, the

smaller wheel weighs less and causes less parasitic drag. Because of the way airframe loads are

distributed while operating on rough ground, tailwheel aircraft are better able to sustain this type

of use over a long period of time, without cumulative airframe damage occurring. If a tailwheel

fails on landing, the damage to the aircraft will be minimal. Tailwheel aircraft are easier to fit into

and maneuver inside some hangars.

8.2 Static Load Analysis

The calculation of nose gear load uses the diagram shown in Figure 8-1 and the following

appropriate formulas based on American Institute of Aeronautics & Astronautics (1988):

𝑀𝑎𝑥 𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑖𝑛 𝑔𝑒𝑎𝑟 𝑙𝑜𝑎𝑑 (𝑝𝑒𝑟 𝑠𝑡𝑟𝑢𝑡) = W(F − M)

2𝐹

174

𝑀𝑎𝑥 𝑠𝑡𝑎𝑡𝑖𝑐 𝑛𝑜𝑠𝑒 𝑔𝑒𝑎𝑟 𝑙𝑜𝑎𝑑 = W(F − L)

𝐹

Where W is the maximum gross weight and the other quantities are defined in Figure 8-1.

Figure 8-1 Diagram for nose landing gear load calculation

Figure 8-2 Centre of gravity for nose landing gear (front landing gear)

175

Figure 8-3 Centre of gravity for main landing gear (rear landing gear)

176

𝑊 = 1.5𝑘𝑔

𝐹 = 452.15𝑚𝑚

𝑁 = 19.16mm

𝑀 = 432.89𝑚𝑚

𝐿 = 442.76𝑚𝑚

𝐽 = 103.75𝑚𝑚

𝑀𝑎𝑥 𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑖𝑛 𝑔𝑒𝑎𝑟 𝑙𝑜𝑎𝑑 (𝑝𝑒𝑟 𝑠𝑡𝑟𝑢𝑡)

= W(F − M)

2𝐹

=1.5(0.45215 − 0.43289)

2(0.45215)

= 0.03195𝑁

Since the front landing gear is consist of two tyres and struts;

𝐿𝑟𝑒𝑎𝑟,𝑒𝑎𝑐ℎ =0.03195

2= 0.01598𝑁

𝑀𝑎𝑥 𝑠𝑡𝑎𝑡𝑖𝑐 𝑛𝑜𝑠𝑒 𝑔𝑒𝑎𝑟 𝑙𝑜𝑎𝑑

= W(F − L)

𝐹

=1.5(0.45215 − 0.44276)

(0.45215)

= 0.03115N

177

8.3 Introduction of FEM

Finite element method (FEM) is a fundamental numerical procedure for the analysis of

structure. It is a numerical method for solving problems of engineering and mathematical physics.

The use of numerical methods to solve mathematical models of complex structures has become

essential criteria in design process. Finite element method is also a popular method used and a

powerful simulation tool to analyse structural problems. It has been widely used to solve

engineering problems in visualizing the stresses and displacement of a model. It provides an

outcome of finite element analysis which is able in turn help in producing a better products to an

industry company. Then, an improvement to the product will reduce the spending on prototypes

by optimizing the design through the finite element analysis.

8.3.1 Software Used

ABAQUS is commercial software to be used in order to conduct the simulation in finite

element analysis. ABAQUS can solve problems ranging from relatively simple linear analyses to

the most challenging nonlinear simulations based on finite element method. It offers a wide range

of capabilities for simulation of linear and nonlinear applications. It also contains a vast library of

elements that can model virtually any geometry. It has an equally extensive list of material models

that can simulate the behaviour of most typical engineering materials.

Figure 8-4 Overview of ABAQUS on Front Landing Gear being modelled

178

8.3.2 Material of Landing Gear

Table 8-1 Material used

Components Material Used

Attachment of Fuselage Plywood

Supports 18/10 Stainless Steel

Wheel EPP Foam

8.3.2.1 Material Property for Landing Gear

Table 8-2 Properties of the material used

Specification Parameter

Material Used 18/10 Stainless Steel Plywood

Mass density 0.0077kg/cm3 0.00063kg/m3

Young Modulus 2e7 N/cm2 860000 N/cm2

Poisson ratio 0.265 0.28

8.3.3 Applied Load on Front Landing Gear

The weight of the designed aircraft is 1.5kg. The weight is considered as the only loading

that will exert on the front landing gear per struts is 0.01598N. The load exerted on the surface is

uniform distributed pressure of 5.811E-4N/cm2 since the surface area for front landing gear

is(5.5cm × 5cm). Figure 8-5 illustrate the load exerted on the surface which attached of the

fuselage.

179

Figure 8-5 Loading on Front Landing Gear

8.3.4 Boundary Conditions for Front Landing Gear

Boundary condition for rear landing gear is the main concern of the finite element analysis.

From Figure 8-6, left bottom and right bottom are the parts that attached with the wheel. In our

case, we just considered the analysis is static deformation, and then both of the parts that attached

with wheels are considered as fixed support. The boundary condition of the bottom parts is fixed

in displacement and rotation in the analysis. Beside, only the y direction of displacement will vary

with the loading for the upper part which attached with fuselage. The other boundary conditions

for the upper part are also fixed in displacement and rotation. Figure 8-6 show all the boundary

conditions of the front landing gear.

Figure 8-6 Boundary Condition of Front Landing Gear

180

8.3.5 Meshing for Front Landing Gear

The meshing of the element is the most significant step in the analysis. Meshing of the

element will show every single element of rear landing gear that being to deform and support all

of the conditions. The approximate global size of the meshing is 2 and the meshing control of the

element shape is tetrahedron as shown in Figure 8-7.

Figure 8-7 Meshing of Front Landing Gear

The incrementation of the step is 100 number of increment which is the process of

increasing in size of the element from 0.0001 to 1. Figure 8-8 show the number of increment of

the analysis.

181

Figure 8-8 Incrementation of the analysis

182

8.4 Results of Finite Element Analysis for Front Landing Gear

Front landing gear in the analysis has 1292 number of element after obtaining the result.

Figure 8-9 show the shape of the rear landing gear before deformation that load exerted and

boundary conditions applied.

Figure 8-9 Plot Undeformed Shape of Rear Landing Gear

Contour plot in ABAQUS is based on mises stress of the structure. Contour plot display

the variation of a variable across the surface of the model. From Figure 8-10, the red colour contour

plot is the most critical stress of the element whereas the blue colour is the lowest stress of the

element that followed by the mises stress. The elements that near the load exerted and boundary

conditions applied will experience the highest stress compared to others. The curved area will also

exert more stress compared to others. Figure 8-10 illustrate the contour plot of the element.

183

Figure 8-10 Contour Plot of Front Landing Gear

Figure 8-11 shows the contour plot for the magnitude displacement for front landing gear,

the red colour contour plot is the most critical displacement of the element whereas the blue colour

is the lowest magnitude displacement of the element that followed by the contour plot. Figure 8-

11 illustrate the magnitude displacement of element.

Figure 8-11 Magnitude displacement of Front Landing Gear

184

8.5 Discussion

The main function of the nose landing gear is to balance the aircraft and provide it with the

ability to steer when plane is grounded. The landing gear must be fuselage mounted, but must also

have necessary track width for the aircraft to operate as required. The major material of use in the

landing gear is 18/10 stainless steel because the characteristics of itself have higher yield stress to

withstand the 90% of load.

From the result obtained by using ABAQUS in Figure 8-9, undeformed shape of front

landing gear was plotted. This undeformed shape is the situation when there is no force acting on

it. When 0.01598N of load was acting on the front landing gear, it began to deflect downward and

deformed shape of front landing gear.

Figure 8-11 shows the highest magnitude displacement acting on the front landing gear is

9.539cm whereas Figure 8-10 shows contour plot of rear landing gear. Based on this figure 8-10,

the biggest stress acting on the front landing gear are about 1.127 N/cm2 which acting at near to

the load applied and at contact surface between front landing gear with tyres since the assumption

of contact surface between front landing gear with tyres are fixed. The result can be more accurate

when the numbers of elements is larger by decreasing the size of elements.

185

CHAPTER 9

FLIGHT STABILITY AND CONTROL

9.1 Center of Gravity

There are four types of cg that we need to find:

1. Empty weight cg

2. Maximum weight cg

3. Forward cg

4. Aft cg

Figure 9-1 CG location of each weight on the aircraft

Based on Figure 9.1, we can proceed to find all the four types of cg.

186

9.1.1 Empty Weight cg

Table 9-1 Empty weight cg calculation

Items Mass (g) Weight,W

(N)

Horizontal Vertical

Arm,x

(mm)

Moment Wx

(Nmm)

Arm,y

(mm)

Moment Wy

(Nmm)

Propeller 20 0.1962 24.05 4.71861 0 0

Motor 72 0.70632 74 52.26768 -9.52 -6.7241664

Wing 210 2.0601 273.74 563.931774 48 98.8848

Horizontal Tail 20 0.1962 771.34 151.336908 12.17 2.387754

Vertical Tail 20 0.1962 753.41 147.819042 58.55 11.48751

Fuselage 140 1.3734 347.51 477.270234 -6.91 -9.490194

front landing gear 20 0.1962 263.84 51.765408 -92.02 -18.054324

back landing fear 10 0.0981 718.58 70.492698 -34.46 -3.380526

ESC 38 0.37278 161.21 60.0958638 -28.04 -10.4527512

LiPo battery 229 2.24649 242.96 545.8072104 -27.14 -60.9697386

Servo (aileron) 18 0.17658 323.57 57.1359906 44.67 7.8878286

Servo (elevator) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792

Servo (rudder) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792

Receiver 6.4 0.062784 192.94 12.11354496 -28.04 -1.76046336

∑W=8.058 ∑Wx=2246.7 ∑Wy=2.6677

Xcg= 278.818

Ycg=0.331

187

9.1.2 Maximum Weight cg

Table 9-2 Maximum weight cg calculation


(N)

Horizontal Vertical

Arm,x

(mm)

Moment Wx

(Nmm)

Arm,y

(mm)

Moment Wy

(Nmm)

Propeller 20 0.1962 24.05 4.71861 0 0

Motor 72 0.70632 74 52.26768 -9.52 -6.7241664

Wing 210 2.0601 273.74 563.931774 48 98.8848

Horizontal Tail 20 0.1962 771.34 151.336908 12.17 2.387754

Vertical Tail 20 0.1962 753.41 147.819042 58.55 11.48751

Fuselage 140 1.3734 347.51 477.270234 -6.91 -9.490194



ESC 38 0.37278 161.21 60.0958638 -28.04 -10.4527512

LiPo battery 229 2.24649 242.96 545.8072104 -27.14 -60.9697386

Servo (aileron) 18 0.17658 323.57 57.1359906 44.67 7.8878286

Servo (elevator) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792

Servo (rudder) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792

Receiver 6.4 0.062784 192.94 12.11354496 -28.04 -1.76046336

Payload 500 4.905 290 1422.45 -3.57 -17.51085

∑W=8.058 ∑Wx=3669.15 ∑Wy=-14.843

Xcg= 283.05

Ycg=-1.145

188

9.1.3 Forward cg

Figure 9-2 Location of cg empty and payloads

From Figure 9.2, all payloads are located behind empty weight cg on the x axis, thus position of

forward cg is equal to empty weight cg and aft cg located at same position as maximum weight

cg.

Table 9-3 Forward cg calculation


(N)

Horizontal Vertical

Arm,x

(mm)

Moment Wx

(Nmm)

Arm,y

(mm)

Moment Wy

(Nmm)

Propeller 20 0.1962 24.05 4.71861 0 0

Motor 72 0.70632 74 52.26768 -9.52 -6.7241664

Wing 210 2.0601 273.74 563.931774 48 98.8848

Horizontal Tail 20 0.1962 771.34 151.336908 12.17 2.387754

Vertical Tail 20 0.1962 753.41 147.819042 58.55 11.48751

Fuselage 140 1.3734 347.51 477.270234 -6.91 -9.490194



ESC 38 0.37278 161.21 60.0958638 -28.04 -10.4527512

LiPo battery 229 2.24649 242.96 545.8072104 -27.14 -60.9697386

Servo (aileron) 18 0.17658 323.57 57.1359906 44.67 7.8878286

Servo (elevator) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792

Servo (rudder) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792

Receiver 6.4 0.062784 192.94 12.11354496 -28.04 -1.76046336

∑W=8.058 ∑Wx=2246.7 ∑Wy=2.6677

Xcg= 278.818

Ycg=0.331

189

9.1.4 Aft cg

Table 9-4 Aft weight cg calculation


(N)

Horizontal Vertical

Arm,x

(mm)

Moment Wx

(Nmm)

Arm,y

(mm)

Moment Wy

(Nmm)

Propeller 20 0.1962 24.05 4.71861 0 0

Motor 72 0.70632 74 52.26768 -9.52 -6.7241664

Wing 210 2.0601 273.74 563.931774 48 98.8848

Horizontal Tail 20 0.1962 771.34 151.336908 12.17 2.387754

Vertical Tail 20 0.1962 753.41 147.819042 58.55 11.48751

Fuselage 140 1.3734 347.51 477.270234 -6.91 -9.490194



ESC 38 0.37278 161.21 60.0958638 -28.04 -10.4527512

LiPo battery 229 2.24649 242.96 545.8072104 -27.14 -60.9697386

Servo (aileron) 18 0.17658 323.57 57.1359906 44.67 7.8878286

Servo (elevator) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792

Servo (rudder) 9 0.08829 294.17 25.9722693 -40.48 -3.5739792

Receiver 6.4 0.062784 192.94 12.11354496 -28.04 -1.76046336

Payload 500 4.905 290 1422.45 -3.57 -17.51085

∑W=8.058 ∑Wx=3669.15 ∑Wy=-14.843

Xcg= 283.05

Ycg=-1.145

190

After locating empty weight cg, maximum weight cg, forward cg and aft cg from all

the tables, then plot the cg location on the aircraft as illustrated in Figure 9.3.

Figure 9-3 Location of all cg on the aircraft

191

9.2 Stability Control System

Stability is the ability of an aircraft to correct for conditions that act on it, like

turbulence or flight control inputs. For aircraft, there are two general types of stability:

static and dynamic. Static stability is the initial tendency of an aircraft to return to its

original position when it's disturbed. There are three kinds of static stability which is

positive, natural and negative. Stability of aircraft is the short- and intermediate-time

response of the attitude and velocity of the vehicle. Stability considers the response of

the vehicle to perturbations in flight conditions from some dynamic equilibrium, while

control considers the response of the vehicle to control inputs.

9.2.1 Contribution of the Wing to Mcg

Figure 9-4 Wing geometry for pitching moment.

The moment about the aerodynamic center of the wing denoted by Mac,w and

the wing lift and drag are, Lw and Dw. Those Mac,w, Lw and Dw all contribute to the

moments about the center of gravity. Consider the forces and moments on the wing

only, as shown in Figure 9.4. The relative wind is inclined at the angle αw with respect

to the zero life line, where αw is the absolute angle of attack of the wing. The c is the

mean zero lift chord of the wing. The center of gravity is located a distance hc behind

the leading edge, and the zc above the zero lift line. Hence, h and z are coordinates of

the center of gravity in fractions of chord length. The aerodynamic center is distance

192

hac,wc from the leading edge. Referring to Figure 9.4, the moments about center of

gravity on the aircraft due to wing contributions only is shown as equation below.

𝑀𝑐𝑔𝑤 = 𝑀𝑎𝑐𝑤 + 𝐿𝑤 𝑐𝑜𝑠 𝛼𝑤(ℎ𝑐 − ℎ𝑎𝑐𝑤𝑐) + 𝐷𝑤 𝑠𝑖𝑛 𝛼𝑤(ℎ𝑐 − ℎ𝑎𝑐𝑤𝑐) + 𝐿𝑤

𝑠𝑖𝑛 𝛼𝑤 𝑧𝑐 − 𝐷𝑤 𝑐𝑜𝑠 𝛼𝑤 𝑧𝑐

For normal flight range of an aircraft, αw is small, thus 𝑠𝑖𝑛 𝛼𝑤 ≈ 𝛼𝑤 and 𝑐𝑜𝑠

𝛼𝑤 ≈ 1.

𝑀𝑐𝑔𝑤 = 𝑀𝑎𝑐𝑤 + (𝐿𝑤 + 𝐷𝑤 𝛼𝑤)(ℎ𝑐 − ℎ𝑎𝑐𝑤𝑐) + (𝐿𝑤𝛼𝑤 − 𝐷𝑤) 𝑧𝑐

Dividing the equation above by q∞Sc, Thus, the moment coefficient about

the center of gravity are as equation below.

𝐶𝑀,𝑐𝑔𝑤 = 𝐶𝑀,𝑎𝑐𝑤 + (𝐶𝐿,𝑤 + 𝐷𝐷,𝑤 𝛼𝑤)(ℎ − ℎ𝑎𝑐𝑤) + (𝐶𝐿,𝑤𝛼𝑤 − 𝐶𝐷,𝑤) 𝑧

𝐶𝑀,𝑐𝑔𝑤 = 𝐶𝑀,𝑎𝑐𝑤 + 𝐶𝐿,𝑤(ℎ − ℎ𝑎𝑐𝑤)

9.2.2 Contribution of the Wing-Body to Mcg

When consideration of fuselage is added into contribution of wing to moment

about center gravity, body of aircraft experiences a moment about its aerodynamic

center, because of some lift force and drag force due to the airflow around it. The

airflow about the wing-body combination is different from that over the wing and body

separately. Aerodynamic interference occurs where the flow over the wing affects the

body flow. Due to the interference, the moment due to the wing-body combination is

not the summation of wing and body moments. Similarly, the lift and drag of the wing-

body combination are affected by aerodynamic interference. By modify the equation

for the moment coefficient for wing only, the contribution of the wing-body

combination to Mcg is:-

𝐶𝑀,𝑐𝑔𝑤𝑏 = 𝐶𝑀,𝑎𝑐𝑤𝑏 + 𝐶𝐿,𝑤𝑏(ℎ − ℎ𝑎𝑐𝑤𝑏)

𝐶𝑀,𝑐𝑔𝑤𝑏 = 𝐶𝑀,𝑎𝑐𝑤𝑏 + 𝑎𝑤𝑏𝛼𝑤𝑏(ℎ − ℎ𝑎𝑐𝑤𝑏)

193

9.2.3 Contribution of the the Tail to Mcg

Figure 9-5 Geometry of wing tail combination.

From the Figure 9.5, the summation of moments about the centre of gravity

due to lift, drag and moment of the tail is

𝑀𝑐𝑔𝑡 = −𝑙[𝐿𝑡 cos(𝛼𝑤𝑏 − 휀) + 𝐷𝑡 sin(𝛼𝑤𝑏 − 휀)]

+ 𝑧𝑡𝐿𝑡 sin(𝛼𝑤𝑏 − 휀) + 𝑧𝑡𝐷𝑡 cos(𝛼𝑤𝑏 − 휀) + 𝑀𝑎𝑐,𝑡

𝑧𝑡 ≪ 𝑙𝑡

𝐷𝑡 ≪ 𝐿𝑡

𝛼𝑤𝑏 − 휀 is small, hence sin(𝛼𝑤𝑏 − 휀) ≈ 0 and cos(𝛼𝑤𝑏 − 휀) = 1

𝑀𝑎𝑐, is small in magnitude

𝑀𝑐𝑔𝑡 = −𝑙𝑡𝐿𝑡

𝑀𝑐𝑔𝑡 = −𝑙𝑡𝑞∞𝑆𝑡𝐶𝐿,𝑡

Tail volume ratio,

𝐶𝑀,𝑐𝑔,𝑡 = −𝑉𝐻𝐶𝐿,𝑡

194

9.2.4 Total Pitching Moment about the centre of gravity

Total pitching moment is use the consideration of the aircraft as whole which

is due to contribution of wing body combination and the tail.

9.2.5 Equations for Longitudinal Static Stability

1. Slope of moment coefficient curve

The criteria necessary for longitudinal balance and static stability which discussed

before are 𝐶𝑀,0 must be positive and must be negative, both conditions with

implicit assumption that αe falls within the practical flight range of angle of attack. By

substitute the αa = 0, the equation is obtained shown as below.

𝐶𝑀,0 = (𝐶𝑀,𝑐𝑔)𝐿=0 = 𝐶𝑀,𝑎𝑐,𝑤𝑏 + 𝑉𝐻𝑎𝑡(𝑖𝑡 + 휀0)

While the equation for the slope of moment coefficient curve are shown as

below.

𝜕𝐶𝑚.𝑐𝑔

𝜕𝛼= 𝛼 [ℎ − ℎ𝑎𝑐𝑤𝑏

− 𝑉𝐻

𝑎𝑡

𝑎(1 −

𝜕휀

𝜕𝛼)]

where

𝑎 = 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.1066 𝑝𝑒𝑟 𝑑𝑒𝑔𝑟𝑒𝑒

ℎ = 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑦

ℎ𝑎𝑐𝑤𝑏= 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑎𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑒 = 0.274 𝑚

195

𝑉𝐻 = 𝑡𝑎𝑖𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑟𝑎𝑡𝑖𝑜 =𝑙𝑡𝑆𝑡

𝑐𝑆=

(0.59)(0.075)

(0.17)(0.204)= 1.276

𝑎𝑡 = 𝑡𝑎𝑖𝑙 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.106 𝑝𝑒𝑟 𝑑𝑒𝑔𝑟𝑒𝑒

For subsonic flow,

𝜕휀

𝜕𝛼=

2𝐶𝐿𝛼𝑤

𝜋𝐴𝑅𝑤=

2(0.113)

𝜋(7.059)= 0.0102

Therefore,

Empty Weight Centre of Gravity

ℎ = 0.278 𝑚



− 𝑉𝐻

𝑎𝑡

𝑎(1 −

𝜕휀

𝜕𝛼)]


𝜕𝛼= −0.133

Maximum Weight Centre of Gravity

ℎ = 0.283 𝑚



− 𝑉𝐻

𝑎𝑡

𝑎(1 −

𝜕휀

𝜕𝛼)]


𝜕𝛼= −0.132

Forward Centre of Gravity

ℎ = 0.279 𝑚



− 𝑉𝐻

𝑎𝑡

𝑎(1 −

𝜕휀

𝜕𝛼)]

196


𝜕𝛼= −0.133

Aft Centre of Gravity

ℎ = 0.283 𝑚



− 𝑉𝐻

𝑎𝑡

𝑎(1 −

𝜕휀

𝜕𝛼)]


𝜕𝛼= −0.132

2. Moment coefficient at zero angle of attack

𝐶𝑀,0 = 𝐶𝑀,𝑎𝑐𝑤𝑏+ 𝑉𝐻𝑎𝑡(𝑖𝑡 + 휀0)

where in our case,

𝐶𝑀,𝑎𝑐𝑤𝑏= 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑎𝑡 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑎𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑒 = 0.228


𝑐𝑆= 1.276


𝑖𝑡 = 𝑡𝑎𝑖𝑙 𝑠𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒 = 7°

휀0 = 𝑑𝑜𝑤𝑛𝑤𝑎𝑠ℎ 𝑎𝑛𝑔𝑙𝑒 𝑎𝑡 𝑧𝑒𝑟𝑜 𝑙𝑖𝑓𝑡 = 0°

Therefore,

𝐶𝑀,0 = 𝐶𝑀,𝑎𝑐𝑤𝑏+ 𝑉𝐻𝑎𝑡(𝑖𝑡 + 휀0)

𝐶𝑀,0 = 1.175

197

From the calculation of last two sections, all of the slope of the pitching moment

coefficient curve is negative and the 𝐶𝑚.𝑐𝑔 is positive. Hence, the aircraft model is

statically stable.

3. Total pitching moment about centre of gravity

Assuming linear curve for the moment coefficient versus alpha graph,

𝐶𝑀,𝑐𝑔 =𝐶𝑀,𝑐𝑔

𝜕𝛼𝛼 + 𝐶𝑀,0

Therefore,


𝐶𝑀,𝑐𝑔 = −0.133𝛼 + 1.175

Table 9-5 Pitching moment at various angle of attack for empty weight

Angle (rad) 𝐶𝑀,𝑐𝑔

0 1.175

2 0.909

4 0.643

6 0.377

8 0.111

10 -0.155

12 -0.421

14 -0.687

16 -0.953

198

Figure 9-6: Graph of moment coefficient vs angle of attack for empty weight

Equilibrium angle of attack was obtained from,

0 = −0.133𝛼 + 1.175

𝛼 = 8.83° (𝑡𝑟𝑖𝑚 𝑎𝑛𝑔𝑙𝑒)

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8 10 12 14 16 18

Mom

ent

Coef

fici

ent

Angle of Attack

Moment Coefficient vs Angle of Attack

for Empty Weight

199


𝐶𝑀,𝑐𝑔 = −0.132𝛼 + 1.175

Table 9-6: Pitching moment at various angle of attack for maximum weight


0 1.175

2 0.911

4 0.647

6 0.383

8 0.119

10 -0.145

12 -0.409

14 -0.673

16 -0.937

Figure 9-7: Graph of moment coefficient vs angle of attack for maximum weight


0 = −0.132𝛼 + 1.175


-1.5

-1

-0.5

0

0.5

1

1.5

0 5 10 15 20

Mom

ent

Coef

fici

ent

Angle of Attack


for Maximum Weight

200


𝐶𝑀,𝑐𝑔 = −0.133𝛼 + 1.175

Table 9-7: Pitching moment at various angle of attack for forward cg


0 1.175

2 0.909

4 0.643

6 0.377

8 0.111

10 -0.155

12 -0.421

14 -0.687

16 -0.953

Figure 9-8: Graph of moment coefficient vs angle of attack for forward cg


0 = −0.133𝛼 + 1.175


-1.5

-1

-0.5

0

0.5

1

1.5

0 5 10 15 20

Mom

ent

Coef

fici

ent

Angle of Attack


for Forward CG

201


𝐶𝑀,𝑐𝑔 = −0.132𝛼 + 1.175

Table 9-8: Pitching moment at various angle of attack for aft cg


0 1.175

2 0.911

4 0.647

6 0.383

8 0.119

10 -0.145

12 -0.409

14 -0.673

16 -0.937

Figure 9-9: Graph of moment coefficient vs angle of attack for aft cg


0 = −0.132𝛼 + 1.175


-1.5

-1

-0.5

0

0.5

1

1.5

0 5 10 15 20

Mom

ent

Coef

fici

ent

Angle of Attack


for Aft CG

202

Apparently, the angle of attack falls within the reasonable flight range, 𝛼𝑚𝑖𝑛 ≤ 𝛼 ≤

𝛼𝑚𝑖𝑛 (12.02°). Therefore, the aircraft is longitudinally balanced as well as statically

stable.

4. Calculation of neutral point

ℎ𝑛 = ℎ𝑎𝑐,𝑤𝑏 + 𝑉𝐻

𝑎𝑡

𝑎(1 −

𝜕휀

𝜕𝛼)

where in our case,

ℎ𝑎𝑐𝑤𝑏= 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑎𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑒 = 0.274 𝑚


𝑐𝑆= 1.276


𝑎 = 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.1066 𝑝𝑒𝑟 𝑑𝑒𝑔𝑟𝑒𝑒

For subsonic flow,

𝜕휀

𝜕𝛼=

2𝐶𝐿𝛼𝑤

𝜋𝐴𝑅𝑤= 0.0102

Therefore,

ℎ𝑛 = ℎ𝑎𝑐,𝑤𝑏 + 𝑉𝐻

𝑎𝑡

𝑎(1 −

𝜕휀

𝜕𝛼)

ℎ𝑛 = 1.5298 𝑚

203

5. Calculation of static margin

𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = ℎ𝑛 − ℎ

Therefore,


ℎ = 0.278 𝑚

𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.5298 − 0.278

𝑠𝑡𝑎𝑡𝑖𝑐 𝑚𝑎𝑟𝑔𝑖𝑛 = 1.252 𝑚


ℎ = 0.283 𝑚




ℎ = 0.279 𝑚



204


ℎ = 0.283 𝑚



From the calculation of static margin, the value is positive in all centre of gravity.

Hence, the aircraft is statically stable. We can also conclude that the degree of static

stability is high due to the large value of static margin.

205

9.3 Static Lateral Roll Stability

An airplane possesses static roll stability if a restoring moment is developed

when it is disturbed from a wings-level attitude. The restoring rolling moment can be

shown to be a function of the sideslip angle 𝛽 as illustrated in Figure 9.10. The

requirement for stability is that 𝐶𝑙𝛽 < 0. The roll moment created on an airplane

Figure 9-10 Static Lateral Roll Stability

The major contributor to 𝐶𝑙𝛽 is the wing dihedral angle Γ. The dihedral angle

is defined as the spanwise inclination of the wing with respect to the horizontal. If the

wing tip is higher than the root section, then the dihedral angle is positive; if the wing

tip is lower than the root section, then the dihedral angle is negative. The equation of

𝐶𝑙𝛽 can be written as follows,

𝐶𝑙𝛽= (

𝐶𝑙𝛽

𝛤)𝛤 + ∆𝐶𝑙𝛽

Where,

𝛤: 𝑑𝑖ℎ𝑒𝑑𝑟𝑎𝑙 𝑎𝑛𝑔𝑙𝑒

𝐶𝑙𝛽: 𝑠𝑖𝑑𝑒𝑠𝑙𝑖𝑝 𝑙𝑖𝑓𝑡 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

As our aircraft’s dimension follows strictly on the model of Cessna 172, we

assume the 𝐶𝑙𝛽value to be the same as the model, which is -0.089. Based on Figure

206

9.10, the static roll stability is achieved since the slope of the graph, which is 𝐶𝑙𝛽value

is negative.

9.3.1 Influence of roll rate on the rolling moment

In this section we will examine how the roll rate creates a rolling moment.

When an airplane rolls about its longitudinal axis, the roll rate creates a linear velocity

distribution over the vertical, horizontal, and wing surfaces. The velocity distribution

causes a local change in angle of attack over each of these surfaces that results in a

change in the lift distribution and, consequently, the moment about the center of

gravity. Figure 9.11 shows a wing planform rolling with a positive rolling velocity. On

the portion of the wing rolling down, an increase in angle of attack is created by the

rolling motion. This results in an increase in the lift distribution over the downward-

moving wing. If we examine the upward-moving part of the wing we observe that the

rolling velocity causes a decrease in the local angle of attack and the lift distribution

decreases. The change in the lift distribution across the wing produces a rolling

moment that opposes the rolling motion and is proportional to the roll rate, p.

The control derivative 𝐶𝑙𝛿𝑎, is a measure of the power of the aileron control; it

represents the change in moment per unit of aileron deflection. The larger 𝐶𝑙𝛿𝑎the more

effective the control is at producing a roll moment. The equation of derivative 𝐶𝑙𝛿𝑎 is

given by,

𝐶𝑙𝛿𝑎= 2

𝐶𝑙∝𝜏

𝑆𝑏𝑤∫ 𝑐𝑦 𝑑𝑦

𝑦2

𝑦1

Wings of large span or high aspect ratio will have larger roll damping than low

aspect ratio wings of small wing span. An estimate of the rolling damping derivative,

𝐶𝑙𝑝, due to the wing surface can be expressed as follows,

𝐶𝑙𝑝 = −𝐶𝑙∝

(1 + 3𝜆)

12(1 + 𝜆)

207

Figure 9-11 Wing planform undergoing a rolling motion.

𝑤𝑖𝑛𝑔𝑠𝑝𝑎𝑛, 𝑏𝑤 = 1.2𝑚

𝑎𝑖𝑙𝑒𝑟𝑜𝑛 𝑐ℎ𝑜𝑟𝑑, 𝑐𝑎 = 0.17𝑚

𝑤𝑖𝑛𝑔 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒, 𝐶𝑙∝ = 4.9 𝑟𝑎𝑑−1

𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝑞 = 120.05 𝑁/𝑚2

𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎, 𝐼𝑥𝑥 = 0.14𝑘𝑔/𝑚2

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑎𝑖𝑙𝑒𝑟𝑜𝑛, 𝑆𝑎 = 0.0188𝑚2

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑤𝑖𝑛𝑔, 𝑆𝑤 = 0204𝑚2

𝑇𝑎𝑝𝑒𝑟 𝑟𝑎𝑡𝑖𝑜, 𝜆 = 1

𝐶𝑙𝑝 = −𝐶𝑙∝

(1 + 3𝜆)

12(1 + 𝜆)= −0.8167

𝐿𝑝 =𝑄𝑆𝑏2𝐶𝑙𝑝

2𝐼𝑥𝑢𝑜

𝐿𝑝 =(120.05)(0.204)(1.2)2(−0.8167)

2(0.14)𝑢𝑜=

−102.863

𝑢𝑜

𝑆𝑎

𝑆𝑤=

0.0188

0.204= 0.0922

𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ 𝑖𝑛 𝑓𝑖𝑔𝑢𝑟𝑒 , 𝜏 = 0.2

208

Figure 9-12 Flap effectiveness parameter

𝐶𝑙𝛿𝑎= 2

𝐶𝑙∝𝜏

𝑆𝑏𝑤∫ 𝑐𝑦 𝑑𝑦

𝑦2

𝑦1

𝐶𝑙𝛿𝑎= 2

𝐶𝑙∝𝜏𝑐

𝑆𝑏𝑤[𝑦2

2]𝑦2

𝑦1

𝐶𝑙𝛿𝑎= 2

4.9(0.2)(0.17)

(0.204)(1.2)[𝑦2

2]

0.590.355

𝐶𝑙𝛿𝑎= 0.1511

𝐿𝛿𝑎=

𝑄𝑆𝑏𝐶𝑙𝛿𝑎

𝐼𝑥=

120.05(0.0204)(1.2)(0.1511)

0.14= 3.1718

𝑅𝑜𝑙𝑙 𝑟𝑎𝑡𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛,

𝑝

𝛿𝑎=

𝐿𝛿𝑎

𝑠 − 𝐿𝑝=

3.1718

𝑠 +102.863

𝑢𝑜

𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 𝑤ℎ𝑒𝑛 𝑠 → 0, 𝑡ℎ𝑒𝑛

𝑝

𝛿𝑎=

3.1718𝑢𝑜

102.863

Therefore, the value of roll rate is influenced by the airspeed as well as the deflection

angle.

If the deflection angle, 𝛿𝑎 and the airspeed, 𝑢𝑜 are 30o and 14 m/s, respectively, the

transient response of the roll rate can be obtained as follows,

𝑝 =3.1718(14)

102.863(30𝜋

180) = 0.226

209

The Figure 9.13 shows the transient response of roll rate, at which when the

aircraft reaches steady state, the roll rate is levelled off to a value.

Figure 9-13 Transient reponse of roll rate

9.3.2 Influence of deflection angle of aileron on the roll rate

At cruising speed of 14 m/s, the graph of roll rate against the deflection angle

can be plotted based on the equation of roll rate p as follows,

Figure 9-14 Graph of roll rate vs deflection angle

Figure 9-14 shows that the roll rate varies linearly with the deflection angle.

This is because the change in the lift distribution across the wing, controlled by the

0

0.05

0.1

0.15

0.2

0.25

0 5 10 15 20 25 30 35

Ro

ll ra

te (

rad

/s)

Deflection angle (deg)

Graph of roll rate vs deflection angle

210

deflection angle of aileron produces a rolling moment that opposes the rolling motion

and is proportional to the roll rate, p.

9.3.3 Influence of deflection angle of aileron on the rolling moment

Roll control is achieved by the differential deflection of small flaps called

ailerons which are located outboard on the wings, or by the use of spoilers. Figure 9.15

is a sketch showing both types of roll control devices. The basic principle behind these

devices is to modify the spanwise lift distribution so that a moment is created about

the x axis.

Figure 9-15 Ailerons for roll control

The change of lift coefficient has been discussed in wing loading with aileron

section. The variation of the aileron deflection angle will result in change of lift

coefficient, and hence change in the lift force acting on the wing. The torque due to

the change in lift force about the x-axis is therefore calculated as follows,

𝑇 = ∆𝐿 × 𝑥

where,

∆𝐿 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑖𝑓𝑡 𝑓𝑜𝑟𝑐𝑒

𝑥 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑙𝑖𝑓𝑡 𝑓𝑜𝑟𝑐𝑒𝑠 𝑜𝑓 𝑎𝑖𝑙𝑒𝑟𝑜𝑛

211

Table 9-9 Torque varies with the aileron deflection angle

Deflection angle

(deg)

n Delta Cl Delta L Torque

(Nm)

-30 0.53 -0.10678 -2.61345 -2.46971

-25 0.58 -0.09738 -2.38334 -2.25226

-20 0.69 -0.09268 -2.26828 -2.14353

-15 0.76 -0.07656 -1.8738 -1.77074

-10 0.8 -0.05373 -1.31495 -1.24262

-5 0.8 -0.02686 -0.65747 -0.62131

0 0.8 0 0 0

5 0.8 0.026863 0.657473 0.621312

10 0.8 0.053727 1.314946 1.242624

15 0.76 0.076561 1.873798 1.770739

20 0.69 0.092679 2.268281 2.143526

25 0.58 0.09738 2.383339 2.252255

30 0.53 0.106782 2.613454 2.469714

Figure 9-16 Graph of torque vs deflection angle

-3

-2

-1

0

1

2

3

-40 -30 -20 -10 0 10 20 30 40

Torq

ue

(Nm

)

Deflection angle (Deg)

Graph of Torque vs Deflection Angle

212

Figure 9-16 shows that the graph varies linearly at small angle but eventually

levels off at high angle of deflection. This is because the flap effectiveness factor 𝜏 of

the aileron limits the continuous rising of the torque.

9.3.4 Influence of airspeed on rolling moment

Table 9-10 Torque varies with the airspeed

V (m/s) q (N/m2) Cl Delta Cl Torque (Nm)

4 9.8 7.365138 0.981062 0.927104

6 22.05 3.273395 0.436028 0.412046

8 39.2 1.841284 0.245266 0.231776

10 61.25 1.178422 0.15697 0.148337

11.42 79.88005 0.903584 0.120361 0.113741

14 120.05 0.601236 0.080087 0.075682

16 156.8 0.460321 0.061316 0.057944

18 198.45 0.363711 0.048448 0.045783

20 245 0.294606 0.039242 0.037084

22 296.45 0.243476 0.032432 0.030648

24 352.8 0.204587 0.027252 0.025753

26 414.05 0.174323 0.02322 0.021943

28 480.2 0.150309 0.020022 0.01892

30 551.25 0.130936 0.017441 0.016482

32 627.2 0.11508 0.015329 0.014486

34 708.05 0.10194 0.013579 0.012832

36 793.8 0.090928 0.012112 0.011446

38 884.45 0.081608 0.01087 0.010273

213

Figure 9-17 Graph of torque vs airspeed

At 30o aileron deflection angle, the torque varies with the velocity as shown in

Figure 9.17. Since the value of lift coefficient varies with the dynamic pressure, which

depends on the velocity, the torque that varies with the change in lift coefficient varies

too. The graph in Figure 9.17 shows that the torque decreases drastically when the

speed approaches the stalling speed. When the airspeed increases further, the torque

will level off and approach to zero.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 5 10 15 20 25 30 35 40

Torq

ue

(Nm

)

Airspeed (m/s)

Graph of Torque vs Airspeed

214

9.4 Static Lateral Directional Stability

Directional stability is related with the static stability of the airplane about the z axis,

which is the tendency of the airplane return to an equilibrium condition when subjected

to yawing disturbance. In order to achieve positive static directional stability, the slope

of the yawing moment curve must be positive (𝐶𝑛𝛽> 0), as shown in Figure 9.18.

Figure 9-18 Static directional stability

The fuselage and engine nacelles contribute more to destabilize to directional

stability, compared to wing. The wing fuselage contribution can be calculated from the

following empirical expression as below:

𝐶𝑛𝛽𝑤𝑓= −𝑘𝑛𝑘𝑅𝑙

𝑆𝑓𝑠𝑙𝑓

𝑆𝑤𝑏 𝑝𝑒𝑟 𝑑𝑒𝑔

Where

𝑘𝑛= empirical wing-body interference factor. (Determine from Figure 9.19)

𝑘𝑅𝑙 = empirical correction factor (Determine from Figure 9.20)

𝑆𝑓𝑠= the projected side area of the fuselage

𝑙𝑓 = the length of the fuselage

In order to complete the get the data, the following data is obtained beforehand.

215

√ℎ1

ℎ2= √

0.12

0.05= 1.549

𝑙𝑓2

𝑆𝑓𝑠=

0.732

0.055= 9.689

𝑥𝑚

𝑙𝑓=

0.153

0.73= 0.2096

𝑅𝑙𝑓 =𝑉𝑙𝑓

𝛾=

9.314 × 0.73

1.46 × 10−5= 4.657 × 105

ℎ

𝑤𝑓= 1

Figure 9-19: Wing Body Interference Factor

216

Figure 9-20: Reynolds number correction factor

Thus from Figure 9.19 and figure 9.20, we obtained that the value of 𝑘𝑛 and

𝑘𝑅𝑙 are 0.007 and 1 respectively. Put the value obtained back to the equation, we get:

𝐶𝑛𝛽𝑤𝑓= −0.007(1)

0.055(0.73)

(0.204)(1.2)= −1.148 × 10−4

Since the wing-fuselage contribution to directional stability is destabilizing, the

vertical tail must be properly sized to ensure that the airplane has directional stability.

The contribution of the vertical tail to directional stability now can be obtained

𝐶𝑛𝛽𝑡= 𝑉𝜈휂𝜈𝐶𝐿𝛼𝑡

(1 +𝑑𝜎

𝑑𝛽)

Where 𝑉𝜈 is the vertical tail volume ratio and 휂𝜈 = 𝑄𝜈/𝑄𝑤 is the ratio of the

dynamic pressure at the vertical tail to the dynamic pressure at the wing. Also, a simple

algebraic equation for estimating the combined sidewash and tail efficiency factor 휂𝜈

is presented in and reproduced here:

휂𝜈 (1 +𝑑𝜎

𝑑𝛽) = 0.724 + 3.06

𝑆𝜈/𝑆

1 + cos Λ𝑐/4𝑤+ 0.4

𝑧𝑤

𝑑+ 0.009𝐴𝑅𝑤

= 0.724 + 3.06130.66 × 10−4/1.2

1 + cos 0+ 0.4

0.065

0.12+ 0.009(7.0588)

= 1.02085

217

𝑉𝜈 =𝑙𝑣𝑆𝑣

𝑆𝑏=

0.08(130.66 × 10−4)

(0.204)(1.2)= 4.27 × 10−3

𝐶𝑛𝛽𝑡= 4.27 × 10−3(0.9198)(1.02085) = 4.0094 × 10−3

Therefore, summing up both contribution of wing fuselage and vertical tail

𝐶𝑛𝛽= 𝐶𝑛𝛽𝑤𝑓

+ 𝐶𝑛𝛽𝑡

𝐶𝑛𝛽= −1.148 × 10−4 + 4.0094 × 10−3 = 3.89 × 10−3

As the 𝐶𝑛𝛽> 0, thus we can conclude that the airplane is directionally stable

with a positive static directional stability.

9.4.1 Pure Yawing Motion

The aircraft data are given as follows,

Moment of inertia, 𝐼𝑧𝑧 : 0.07 kg𝑚2

Vertical tail side force slope, 𝐶𝑦𝛽(𝑉𝑇)∶ 0.21

Fuselage diameter, X : 0.115m

Cruising speed, 𝑈𝑜 : 14 m/s

Body side force slope, 𝐶𝑦𝛽(𝐵𝑜𝑑𝑦): 0.31

Area of vertical tail (side view), 𝑆𝑉𝑇 : 0.0139 𝑚2

Area of fuselage (side view), 𝑆𝐵𝑜𝑑𝑦 : 0.0556 𝑚2

Distance between the cg and the cg of the vertical tail, 𝐿𝑉𝑇 : 0.499m

**Remark: Body = Fuselage

Estimation 𝑁𝛽 and 𝑁𝑟

Body Contribution, 𝑁𝛽𝐵

𝑁𝛽𝐵𝑜𝑑𝑦= −𝑌𝑏𝑜𝑑𝑦𝑋 = −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)𝛽𝑄𝑆𝐵𝑜𝑑𝑦𝑋

(𝜕𝑁

𝜕𝛽)

𝐵𝑜𝑑𝑦

= −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)𝑄𝑆𝐵𝑜𝑑𝑦𝑋

218

𝑁𝛽𝐵𝑜𝑑𝑦 = (𝜕𝑁

𝜕𝛽/𝐼𝑧𝑧 )

𝐵𝑜𝑑𝑦= −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)𝑄𝑆𝐵𝑜𝑑𝑦𝑋/𝐼𝑧𝑧

𝑁𝛽𝐵𝑜𝑑𝑦 = -

(0.31)(1

2(1.225)(14)2)(0.0556)(0.115)

0.07= −3.399

Horizontal Tail Contribution, 𝑁𝛽𝐻𝑇

𝑁𝛽𝑉𝑇= −𝑌𝑉𝑇𝐿𝑉𝑇 = −𝐶𝑦𝛽(𝑉𝑇)

𝛽𝑄𝑆𝑉𝑇𝐿𝑉𝑇

(𝜕𝑁

𝜕𝛽)𝑉𝑇

= −𝐶𝑦𝛽(𝑉𝑇)𝑄𝑆𝑉𝑇𝐿𝑉𝑇

𝑁𝛽𝑉𝑇= (

𝜕𝑁

𝜕𝛽/𝐼𝑧𝑧 )

𝑉𝑇= −𝐶𝑦𝛽(𝑉𝑇)

𝑄𝑆𝑉𝑇𝐿𝑉𝑇/𝐼𝑧𝑧

𝑁𝛽𝐵 = -

(0.21)(1

2(1.225)(14)2)(0.0139)(0.449)

0.07= -2.2477

Body Contribution, 𝑁𝑟(𝐵𝑜𝑑𝑦)

𝛽 = 𝑟|𝑋|/𝑈𝑜

𝑁𝑟(𝐵𝑜𝑑𝑦) = −𝑟|𝑋|

𝑈𝑜𝐶𝑦𝛽(𝐵𝑜𝑑𝑦) 𝑄𝑆𝐵𝑜𝑑𝑦𝑋

(𝜕𝑁

𝜕𝑟)

𝐵𝑜𝑑𝑦= −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)

|𝑋|

𝑈𝑜 𝑄𝑆𝐵𝑜𝑑𝑦𝑋

𝑁𝑟𝐵𝑜𝑑𝑦= (

𝜕𝑁

𝜕𝑟/𝐼𝑧𝑧 )

𝐵𝑜𝑑𝑦= −𝐶𝑦𝛽(𝐵𝑜𝑑𝑦)

|𝑋|

𝑈𝑜 𝑄𝑆𝐵𝑜𝑑𝑦𝑋 /𝐼𝑧𝑧

𝑁𝑟𝐵𝑜𝑑𝑦 = -

(−0.31)(0.115

14)(

1

2(1.225)(14)2)(0.0556)(0.115)

0.07= 0.0279

Horizontal Tail Contribution, 𝑁𝑟𝑉𝑇

𝛽 = 𝑟|𝐿𝑉𝑇|/𝑈𝑜

𝑁𝑟𝐻𝑇= −𝐶𝑦𝛽(𝑉𝑇)

𝑟|𝐿𝑉𝑇|

𝑈𝑜 𝑄𝑆𝑉𝑇𝐿𝑉𝑇

(𝜕𝑁

𝜕𝑟)𝑉𝑇

= −𝐶𝑦𝛽(𝑉𝑇)

|𝐿𝑉𝑇|

𝑈𝑜 𝑄𝑆𝑉𝑇𝐿𝑉𝑇

𝑁𝑟𝑉𝑇= (

𝜕𝑁

𝜕𝑟/𝐼𝑧𝑧 )

𝑉𝑇= −𝐶𝑦𝛽(𝑉𝑇)

|𝐿𝑉𝑇|

𝑈𝑜 𝑄𝑆𝑉𝑇𝐿𝑉𝑇 /𝐼𝑧𝑧

𝑁𝑟𝐻𝑇 = -

(0.21)(0.499

14)(

1

2(1.225)(14)2)(0.0139)(0.499)

0.07= −0.089

Characteristic equation

219

𝑠2 − (𝑁𝑟𝐵𝑜𝑑𝑦+ 𝑁𝑟𝑉𝑇

) 𝑠 − (𝑁𝛽𝐵𝑜𝑑𝑦+ 𝑁𝛽𝑉𝑇

) = 0

𝑠2 + 0.0611𝑠 + 5.6467 = 0

Roots 𝑠 = −0.0306 ± 2.3761𝑖

Natural frequency, 𝜔𝑛 = √5.6467 = 2.3763 rad/s

Damping ratio, 휁 = 0.0611

2𝜔𝑛= 0.013

Based on the result of pure yawing motion, the aircraft has good directional stability

𝑁𝛽, but has a very low yaw damping 𝑁𝑟. This makes its behavior close to neutral stable.

220

9.5 Dynamic Longitudinal Stability System

Table 9-11: Longitudinal dimensionless derivatives (Nelson, 1998)

221

Table 9-12: Longitudinal dimensional derivatives (Nelson, 1998)

The values of dimensionless and dimensional derivatives are presented in

Table 9-13 and Table 9-14 as below:

Table 9-13: Longitudinal stability coefficients

X-force derivatives Z-force derivatives Pitching moment

derivatives

𝐶𝑋𝑢 = −0.062 𝐶𝑍𝑢 = −0.440 𝐶𝑀𝑢 = 0

𝐶𝑋𝛼 = −0.168 𝐶𝑍𝛼 = −6.508 𝐶𝑀𝛼 = −2.212

𝐶𝑋�̇� = 0 𝐶𝑍�̇� = −2.650 𝐶𝑀�̇� = −7.426

𝐶𝑋𝑞 = 0 𝐶𝑍𝑞 = −5.885 𝐶𝑀𝑞 = −16.490

𝐶𝑋𝛼𝑒= 0 𝐶𝑍𝛿𝑒

= −0.315 𝐶𝑀𝛿𝑒= −0.883

222

Table 9-14: Longitudinal dimensional derivatives

𝑋𝑢 = −0.073 𝑍𝑢 = −0.525 𝑀𝑢 = 0

𝑋𝛼 = −2.814 𝑍𝛼 = −108.65 𝑀𝛼 = −76.744

𝑋𝑤 = −0.211 𝑍�̇� = −0.268 𝑀�̇� = −0.111

− 𝑍𝑞 = −0.596 𝑀𝑞 = −3.473

− 𝑍𝛿𝑒= 5.260 𝑀𝛿𝑒

= −30.630

− 𝑍𝑤 = −7.723 𝑀𝑤 = −5.481

− − 𝑀�̇� = −0.111

By taking a flight case for our RC plane flying at sea level with the trimmed

velocity at 14 m/s, the state-space matrix for longitudinal motion due to elevator

deflection are presented as below:

[

𝛥�̇�𝛥�̇�𝛥�̇�

𝛥휃̇

] = [

−0.0739 −0.2111 0 −9.8100−0.5251 −7.7234 14.0000 00.0587 −4.6187 −5.0380 0

0 0 1 0

] [

𝛥𝑢𝛥𝑤𝛥𝑞𝛥휃

] + [

05.2605

−31.21830

] [𝛥𝛿𝑒]

The state-space matrix can be written in the form of transfer function:

𝛥𝑢

𝛥𝛿𝑒=

−1.11𝑠2 + 392.9𝑠 + 2604

𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24

𝛥𝑤

𝛥𝛿𝑒=

5.26𝑠3 − 410.2𝑠2 − 30.35𝑠 − 157.8

𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24

𝛥𝑞

𝛥𝛿𝑒=

−31.22𝑠3 − 267.7𝑠2 − 16.23𝑠

𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24

223

𝛥휃

𝛥𝛿𝑒=

−31.22𝑠2 − 267.7𝑠 − 16.23

𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24

Where the characteristic equation is:

𝑠4 + 12.84𝑠3 + 104.4𝑠2 + 7.848𝑠 + 28.24 = 0

From the characteristic equation, all the sub-value of representing the phugoid and

short period is positive in value, means it still in a stable condition. The time response

of the open loop transfer function of longitudinal motion was plotted by using

MATLAB in time interval between 0 to 50s.

Figure 9-21: Time response open-loop for longitudinal stability

As observed, the forward velocity in phugoid mode, 𝑢 is having a large damping ratio

due to long period oscillation while w and q are having a lower damping ratio due to

short period.

224

State-Space Matrix and transfer function for Short Period Motion

The state-space matrix then reduced to short period motion, the matrix and

transfer functions are shown below:

[𝛥�̇�𝛥�̇�

] = [−7.7604 1−75.8770 −3.5854

] [𝛥𝛼𝛥𝑞

] + [0.3757

−30.6725] [𝛥𝛿𝑒]

𝛥𝛼

𝛥𝛿𝑒=

0.3757𝑠 − 29.33

𝑠2 + 11.35𝑠 + 103.7

𝛥𝑞

𝛥𝛿𝑒=

−30.67𝑠 − 266.5

𝑠2 + 11.35𝑠 + 103.7

The time response of the open loop transfer function for short period are as shown:

Figure 9-22: Time response for short period

225

Pole Damping Frequency

(rad/TimeUnit)

Time Constant

(TimeUnit)

-5.67e+00 +

8.46e+00i

0.557 10.2 0.176

-5.67e+00 -

8.46e+00i

0.557 10.2 0.176

From the open loop transfer function of short period motion, it shown that the system

is having a low stability condition where the damping ratio is 0.557 and natural

frequency of 10.2.

226

9.6 Dynamic Lateral Stability System

Table 9-15 Parameters of Cessna 172

Parameters Values Side Force Coefficient

Derivatives

Density, 𝜌 2.3769 ×

10−3𝑠𝑙𝑢𝑔

𝑓𝑡3

1.225 𝑘𝑔/𝑚3

𝐶𝑦𝛽 -0.31

Mach number, M 0.04 𝐶𝑦𝛿𝑟 0.187

Speed of sound, a 1125.33 ft/s @ 343 m/s 𝐶𝑦𝑝 -0.037

Reference flight

speed, 𝑈𝑜

45.9318 ft/s @ 14 m/s 𝐶𝑦𝑟 0.21

Wing area, S 2.1958 𝑓𝑡2 @ 0.204

𝑚2

𝐶𝑦𝛿𝑎 0

Mean chord, 𝑐̅ 2.2966 ft @ 0.17 m Yawing Moment Coefficient

Derivatives

Wing span, b 3.9370 ft @ 1.2m 𝐶𝑛𝛽 0.065

Aspect ratio, AR 7.06 𝐶𝑛𝛿𝑟 -0.0657

Rolling moment of

inertia, 𝐼𝑥

0.14 kg𝑚2 @ 0.1032

𝑠𝑙𝑢𝑔𝑓𝑡2

𝐶𝑛𝑝 -0.03

Pitching moment

of inertia, 𝐼𝑦

0.12 kg𝑚2 @

0.0885 𝑠𝑙𝑢𝑔𝑓𝑡2

𝐶𝑛𝑟 -0.099

Yawing moment of

inertia, 𝐼𝑧

0.07 kg𝑚2 @


𝐶𝑛𝛿𝑎 -0.053

Product of inertia

about xz, 𝐼𝑥𝑧

0.04 kg𝑚2 @


Rolling Moment Coefficient

Derivatives

227

Product of inertia

about xy, 𝐼𝑥𝑦

0.03 kg𝑚2 @


𝐶𝑙𝛽 -0.089

Gravitational

acceleration, g

32. 171 𝑓𝑡2/𝑠 @ 9.81

𝑚2/𝑠

𝐶𝑙𝛿𝑟 0.0147

Weight, W 104.0474 Ibf @ 14.39

N

𝐶𝑙𝑝 -0.47

Mass, m 3.2342 𝐼𝑏 @ 1.467 kg 𝐶𝑙𝑟 0.096

Dynamic pressure,

Q

2.5073 psi @

17287.225 Pascal

𝐶𝑙𝛿𝑎 -0.178

Oswald efficiency

factor, e

0.7

228

Lateral Directional Derivatives

𝑌𝛽 -0.1582

𝑁𝛽 0.7732

𝐿𝛽 -0.5293

𝑌𝑝 −3.8632 × 10−4

𝑁𝑝 -0.0047

𝐿𝑝 -0.0365

𝑌𝑟 0.0272

𝑁𝑟 -0.0154

𝐿𝑟 0.0075

𝑌𝛿𝑎 0

𝑌𝛿𝑟 0.0296

𝑁𝛿𝑎 -0.6304

𝑁𝛿𝑟 -0.7815

𝐿𝛿𝑎 -1.0587

𝐿𝛿𝑟 0.0874

For lateral state-space equation,

[

∆�̇�∆�̇�∆�̇�∆∅̇

] =

[ 𝑌𝛽

𝑢𝑜

𝑌𝑝

𝑢𝑜−(1 −

𝑌𝑟

𝑢𝑜)

𝐿𝛽 𝐿𝑝 𝐿𝑟

𝑁𝛽

0

𝑁𝑝

1

𝑁𝑟

0

𝑔

𝑢𝑜cos 휃0

000 ]

[

∆𝛽∆𝑝∆𝑟∆∅

] +

[ 0

𝑌𝛿𝑟

𝑢𝑜

𝐿𝛿𝑎𝐿𝛿𝑟

𝑁𝛿𝑎

0

𝑁𝛿𝑎

0 ]

[∆𝛿𝑎

∆𝛿𝑟]

229

From the state-space equation, by using MATLAB program, the transfer function of

the lateral motion is obtained. The stability of the aircraft is able to be determined from

the transfer function.

�̇� = 𝐴𝑥 + 𝐵𝑢

𝑨 = [

−0.0034 0 −0.9994−0.5293 −0.0365 −0.00750.7732

0−0.00471.000

−0.01540

0.7004000

]

𝐵 = [

0 0.0006−1.0587 0.0874−0.6304

0−0.7815

0

]

[

∆�̇�∆�̇�∆�̇�∆∅̇

] =[

−0.0034 0 −0.9994−0.5293 −0.0365 −0.00750.7732

0−0.00471.000

−0.01540

0.7004000

] [

∆𝛽∆𝑝∆𝑟∆∅

]

+[

0 0.0006−1.0587 0.0874−0.6304

0−0.7815

0

] [∆𝛿𝛼

∆𝛿𝑟]

Transfer Function of Lateral Motion

Aileron Input

𝛽

𝛿𝛼

=0.63𝑠2 − 0.7235𝑠 − 0.0081

𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098

𝑝

𝛿𝛼

=−1.0587𝑠3 − 0.0152𝑠2 − 1.1516𝑠

𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098

𝑟

𝛿𝛼

=−0.6304𝑠3 − 0.0202𝑠2 − 0.0001𝑠 − 0.8070

𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098

230

∅

𝛿𝛼

=−1.0587𝑠2 − 0.0152𝑠 − 1.1516

𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098

Rudder Input

𝛽

𝛿𝑟

=0.006𝑠3 + 0.7811𝑠2 + 0.0901𝑠 + 0.0050

𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098

𝑝

𝛿𝑟

=0.0874𝑠3 + 0.0072𝑠2 − 0.3458𝑠

𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098

𝑟

𝛿𝑟

=−0.7815𝑠3 − 0.0311𝑠2 − 0.0001𝑠 − 0.2424

𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098

∅

𝛿𝑟

=0.0874𝑠2 + 0.0072𝑠 − 0.3458

𝑠4 + 0.0553𝑠3 + 0.7734𝑠2 + 0.4014𝑠 + 0.0098

Flying Qualities

Cessna 172 is classified as Class I airplanes as it is small, light

airplanes, such as light utility, primary trainer and light observation craft. This

airplane is categorized as non-terminal flight under category C which is a

gradual maneuver, accurate flight path.

Lateral Motion

𝜆1,2(𝐷𝑢𝑡𝑐ℎ 𝑟𝑜𝑙𝑙) = −0.191 ± 0.945𝑖

𝜆1,2(𝑆𝑝𝑟𝑖𝑟𝑎𝑙) = −0.0256

𝜆1,2(𝑅𝑜𝑙𝑙) = −0.411

231

Table 9-16 Flying Qualities

Modes Damping

( ξ )

Natural

Frequency

(𝜔𝑛)

Time Constant

(s)

Flying

Qualities

Dutch Roll 0.198 0.945 5.25 Level 1

Spiral - 0.0256 39.1 Level 1

Roll - 0.411 2.43 Level 2

Based on Table 9-16, both spiral and dutch mode has the same flying

qualities, which is Level 1 where the flying qualities clearly adequate for the

mission flight phase. For roll mode, it is a Level 2 flying qualities where flying

qualities adequate to accomplish the mission flight phase but with some

increase in pilot workload and degradation in mission effectiveness.

232

CHAPTER 10

FLIGHT TESTING PLANNING AND PREPARATION

10.1 Flight test and preparation

1. Events and activities during the day of flight test.

Flight test of RC plane

• Understand the nature

• Calibrate electronics part and sensors

• Check the control surfaces of RC plane

• Set the location for pilot to stand

• Record the meteorological conditions

• Record data of RC plane and pilot

• Record the flight path and flight time of RC plane

2. Approximation of the nature and environment of the testing field

Area Padang Ragbi UTM

Obstacles Goal post, trees, net

Crowd Around 50 people

Ambient Temperature 32°C

Atmospheric Pressure 1 atm

Wind Speed 3.08 m/s = 11.1 km/hr

3. Battery consumption of each cell.

1 𝑐𝑒𝑙𝑙 = 35𝑐

233

4. Flight path at its altitude and the location of pilot to stand.

Flight Path :

Condition Altitude (metre)

Takeoff 0-15

Cruising 15-20

Landing 0

Location of pilot:

• Pilot should stand 3 metre when the RC plane on ground before take-

off

• RC plane should not fly more than 92 metre from pilot since the

frequency is 2.4GHz

5. Strategies when the engine suddenly cut-off

Maneuverability Control

Glide angle 1.25°

Glide speed Stall speed – 9.52 m/s

Movement of control

surface

Adjust manually by pilot (aileron, rudder,

elevator)

0

5

10

15

20

25

0 100 200 300 400 500

Alt

itu

de

(met

re)

Time(sec)

Flight Path

234

10.2 Checklist

1. Electronics and sensors calibration checklist including its condition during

the day of testing

Battery capacity

Motor

Propeller

• Clockwise / anticlockwise

Servo

• Aileron

• Rudder

• Elevator

Push rod

• Aileron

• Rudder

• Elevator

2. Control surfaces checklist (accurate angles, movements and directions)

Accurate angle 0° angle

Movements and directions

Parts Movements

Elevator (left and

right)

Upward and downward

Aileron (left and

right)

Upward and downward

Rudder Left and right

235

3. Record of meteorological conditions onsite.

Area

Obstacles

Crowd

Ambient Temperature

Atmospheric Pressure

Wind Speed

Wind Direction

4. Testing form

Trial number

Type of aircraft

Pilots’s name

Location

Date

Time

Purpose of test

Requirement to be met 5 minutes flight time

Ambient airspeed

Ambient temperature

Wind direction

MTOW

Flight path

Flight time

Operating altitude

Battery voltage in each

cell

Battery voltage of

transmitter

Comments

Name of reporter

Signature

236

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2 [Video file]. Retrieved from:

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motors?variant=45677872079

Starlino. (2017, September 18). How much power is needed to hover ? Retrieved from

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All About Multirotor Drone Radio Transmitters and Receivers. (2018, February 8).

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multirotor-fpv-drone-radio-transmitter-and-receiver/

Differences Between NiMH and LiPo Batteries. (n.d.). Retrieved from Euro RC:

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batteries

Electronic speed control. (2020, March 4). Retrieved from Wikipedia:

https://en.wikipedia.org/wiki/Electronic_speed_control

FS-i6 Instruction Manual. (2015). Retrieved from Fly SKy:

https://static1.squarespace.com/static/5bc852d6b9144934c40d499c/t/5c0787e

10e2e721a7f17c998/1543997593953/FS-i6+User+manual+20160819.pdf

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https://www.uavnavigation.com/support/kb/general/general-system-

info/introduction-servos

Liang, O. (2013, November 5). PWM AND PPM DIFFERENCE AND

CONVERSION. Retrieved from OscarLiang: https://oscarliang.com/pwm-

ppm-difference-conversion/

https://www.youtube.com/watch?v=ug4zKgNV6zY

https://sunnyskyusa.com/products/sunnysky-x2216-brushless-motors?variant=45677872079

https://sunnyskyusa.com/products/sunnysky-x2216-brushless-motors?variant=45677872079

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237

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GUIDE. Retrieved from Oscarliang: https://oscarliang.com/flysky-tx-rx-

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mpnxt1. (2019, December 20). Omnidirectional Antenna Radiation Patterns

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York: Wiley.

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238

Appendix I Graph to determine aerodynamic coefficients

Figure 1: Graph of Cm and Cl vs angle of attack (left), Graph of Cd and Cm vs Cl (right) (Abbott 1994)

239

Appendix II McCormick’s method on wing aileron analysis with aileron

Figure 1: Flap effectiveness factor (McCormick, 1995)

Figure 2: Correction factor to flap effectiveness factor 𝝉 (McCormick, 1995)

240

Figure 3: 𝑪𝒍𝒎𝒂𝒙increment ratio as a function of flap chord ratio (McCormick, 1995)

241

Appendix III Bruhn’s method on structural analysis

Figure 1: Compressive-buckling coefficients, kc for flat rectangular plates (Bruhn,

1973)

Figure 2: Shear-Buckling-Stress Coefficient ks of Plates as a Function of /b for

Clamped and Hinged Edges (Bruhn, 1973)

242

Figure 3: Bending-Buckling Coefficient kb of Plates as a Function of a/b for Various

Amounts of Edge Rotational Restraint. (Bruhn, 1973)

Figure 4: Chart of Nondimensional Compressive Buckling Stress for Long Clamped

Flanges and for Supported Plates with Edge Rotational Restraint. (Bruhn, 1973)

243

Figure 5: Chart of Nondimensional Shear Buckling Stress for Panels with Edge-

Rotational Restraint (Bruhn, 1973)

244

Appendix IV EPP foam material properties

Figure 1: Static and dynamic loading of 80grams/litre EPP Foam

Figure 2: EPP Foam properties at different densities

245

Appendix V MATLAB coding on dynamic stability and control

%%% Cessna 172 rho = 2.3769e-3; %slug/ft^3 M = 0.6; a = 1125.33; %ft/s Vo = 45.9318 ; %ft/s uo = Vo; %ft/s S = 0.204; %ft^2 c = 0.17; %ft b = 1.2; % ft AR = 7.06; Ix = 0.1032; %Slug.ft^2 Iy = 0.0885; %Slug.ft^2 Iz = 0.0516; %Slug.ft^2 Ixz = 0.0295; %Slug.ft^2 Ixy = 0.0221; Iyz = 0.0516; %Slug.ft^2 g = 32.17095; m = 3.2342; %Ibf Weight = m*g %Ib Q = (1/2)*rho*(Vo^2) e = 0.7;

%Lateral Data %Side Force Coefficient Derivatives CYbeta = -1.0; CYdeltar = 0.187; %due to rudder CYp = -0.187; CYr = 0.21; CYdeltaa = 0; %Yawing Moment Coefficient Derivatives CNbeta = 0.065; CNp = -0.03; CNr = -0.099; CNdeltaa = -0.053; %due to aileron CNdeltar = -0.0657; %due to rudder %Rolling Moment Coefficient Derivatives CLbeta = -0.089; CLp = -0.47; CLr = 0.096; CLdeltaa = -0.178; %due to aileron CLdeltar = 0.0147; %due to rudder %Matrix Lateral Derivatives Ybeta = (Q*S*CYbeta)/m Nbeta = (Q*S*b*CNbeta)/Iz Lbeta = (Q*S*b*CLbeta)/Ix Yp = (Q*S*b*CYp)/(2*m*uo) Np = (Q*S*(b^2)*CNp)/(2*Iz*uo) Lp = (Q*S*(b^2)*CLp)/(2*Ix*uo) Yr = (Q*S*b*CYr)/(2*Iz*uo) Nr = (Q*S*(b^2)*CNr)/(2*Iz*uo) Lr = (Q*S*(b^2)*CLr)/(2*Ix*uo) Ydeltaa = (Q*S*CYdeltaa)/m Ydeltar = (Q*S*CYdeltar)/m Ndeltaa = (Q*S*b*CNdeltaa)/Iz Ndeltar = (Q*S*b*CNdeltar)/Iz Ldeltaa = (Q*S*b*CLdeltaa)/Ix Ldeltar = (Q*S*b*CLdeltar)/Ix thetanot = 0;

246

%Matrix A Lateral Alat = [Ybeta/uo, Yp/uo, -(1-(Yr/uo)),(g*cos(thetanot))/uo; Lbeta, Lp, Lr, 0; Nbeta, Np, Nr, 0; 0, 1, 0, 0]

%Matrix B Lateral Blat = [0, Ydeltar/uo; Ldeltaa, Ldeltar; Ndeltaa, Ndeltar; 0,0]

%% %Aileron Alat_a = [-0.0034 0 -0.9994 0.7004;-0.5293 -0.0365 -0.0075 0;0.7732

-0.0047 -0.0154 0;0 1.00 0 0]; Blat_a = [0;-1.0587;-0.6304;0]; Clat_a = eye(size(Alat_a)); Dlat_a = zeros(size(Blat_a)); [numlat_a, denlat_a]=ss2tf(Alat_a,Blat_a,Clat_a,Dlat_a,1); numlat_a denlat_a

%% %Rudder Alat_r = [-0.0034 0 -0.9994 0.7004;-0.5293 -0.0365 -0.0075 0;0.7732

-0.0047 -0.0154 0;0 1.00 0 0]; Blat_r = [0.0006;0.0874;-0.7815;0]; Clat_r = eye(size(Alat_r)); Dlat_r = zeros(size(Blat_r)); [numlat_r,denlat_r]=ss2tf(Alat_r,Blat_r,Clat_r,Dlat_r,1); numlat_r denlat_r damp(Alat_r)

247

Appendix VI Minutes of meetings

FACULTY OF MECHANICAL ENGINEERING

UNIVERSITI TEKNOLOGI MALAYSIA

12 FEBRUARY 2020 MINUTES OF MEETING NO. 1/2020 AIRCRAFT DESIGN II (SKMA 4523)

Date: 12 February, 2020

Time: 9am-12pm Venue: P20-215

Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening speech from lecturers

1.1 Introduction to the subject of Aircraft Design 2 1.2 Discussion on project done in Aircraft Design 1 1.3 Students divided into groups 1.4 Task given to present the project done in previous class

Action: All students 2.0 Approval of meeting minutes

2.1 First week of meeting Action: All lecturers and students

3.0 Matters arising 3.1 Formation of sub-group on propulsion, wing, fuselage and tail. Landing

gear, avionics and flight testing. 3.2 Conceptual design of aircraft. 3.3 Details of the aircraft are required, such as the v-n diagram, wing NACA,

Cruise Speed, etc. 4.0 Adjournment of meeting

The meeting was adjourned at 12pm Prepared by: Checked and approved by ____________________ ____________________ ____________________

Alexander Jong Jiun Wei Dr. Wan Zaidi bin Wan Omar

Dr. Ing. Mohd Nazri bin Mohd Nasir

248

FACULTY OF MECHANICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA



Time: 4-6pm Venue: P20-215


1.1 Lecture on calculation of flight envelope and load factor 1.2 Lecture on wing loading distribution


2.1 Task distribution on each group. Action: All lecturers and students

3.0 Matters arising 3.1 Conversion of scale. 3.2 Need to finalise the components and weight of aircraft and its

components. 3.3 Final Solidworks drawing of aircraft is needed. 3.4 Material selection and material properties must be determined. 3.5 Determine preliminary cost estimation to build our rc plane.

4.0 Adjournment of meeting The meeting was adjourned at 6pm

Prepared by: Checked and approved by ____________________ ____________________ ____________________



249






1.1 Lecture on calculation of flight envelope and load factor 1.2 Lecture on wing loading distribution


2.1 Mistake made in calculation of the flight envelope were corrected Action: All lecturers and students

3.0 Matters arising 3.1 Discussion on the method used to construct the shear load, bending

moment and torsional diagram. 3.2 Task distribution to complete the construction of the diagrams. 3.3 Suggestion of where to buy materials for rc plane.





250

FACULTY OF MECHANICAL ENGINEERING

UNIVERSITI TEKNOLOGI MALAYSIA



Time: 4pm-6pm Venue: P20-215


1.1 Lecture on avionics component selection Action: All students

2.0 Approval of meeting minutes 2.1 Dr Wan Zaidi introduced the application of aileron in wing loading

analysis. Action: All lecturers and students

3.0 Matters arising 3.1 Calculate the aerodynamic coefficients with and without aileron. 3.2 Calculate the weight distribution, Cg.





251






1.1 Lecture on material selection and stress calculation Action: All students

2.0 Approval of meeting minutes 2.1 Task distribution in each group

Action: All lecturers and students 3.0 Matters arising

3.1 Preliminary selection of motor 3.2 Engine selection and the power curve must be determined 3.3 Internal sketching of the aircraft structure and the part attachment. 3.4 Continue the construction of shear, bending moment and lift distribution

diagrams. 3.5 Electronics components selection for avionics.

4.0 Adjournment of meeting




252


2 MARCH 2020 MINUTES OF MEETING NO. 6/2020 AIRCRAFT DESIGN II (SKMA 4523)

Date: 2 MARCH, 2020

Time: 4-6pm Venue: P20-215


1.1 Dr Wan Zaidi requested students to sketch the wing structure design of their own aircraft.

1.2 Lecture on the wing loading diagram where greater load needed to be at the root of the wing.


2.1 Continue the work assigned during last meeting. Action: All lecturers and students

3.0 Matters arising 3.1 Calculate aircraft performance. 3.2 Determine flight path and flight power consumption. 3.3 Calculate the factor of safety and the stress in the wing components. 3.4 Deciding the best price and place to buy electronic components.





253



Date: 3 MARCH, 2020

Time: 9-11am Venue: E07


1.1 Dr Wan Zaidi explained about the wing load analysis. Action: All students

2.0 Approval of meeting minutes 2.1 Continue the work assigned during last meeting.


3.1 Calculate aircraft performance with the correct data chosen. 3.2 Convinced to use the right maximum and minimum value of forces for the

load analysis. 4.0 Adjournment of meeting

The meeting was adjourned at 11am Prepared by: Checked and approved by ____________________ ____________________ ____________________



254



Date: 11 MARCH, 2020



1.1 No lecture Action: All students

2.0 Approval of meeting minutes 2.1 Information of the Presentation 1 was distributed to the students.


3.1 Discuss the task distribution and compiling slides of presentation 1. 3.2 Discussion on static and dynamic stability system. 3.3 Completion on shear and bending moment diagram for fuselage and tail.





255


8 APRIL 2020 MINUTES OF MEETING NO. 9/2020 AIRCRAFT DESIGN II (SKMA 4523)

Date: 8 APRIL, 2020

Time: 9am-12pm Online Discussion

Attendance AHMAD AMIRUL AMIN BIN ABDUL WAHAB (A16KM0017) AHMAD FAKHRURRAZI BIN ABDUL LATIF (A16KM0022) ALEXANDER JONG JIUN WEI (A16KM0033) CHENG WEI QIN (A16KM0062) LEONG JOE YEE (B17KM0011) MOHAMAD ASHRAF BIN AB HAN (A16KM0176) MOHAMAD AZMIL BIN BAHJAM (A16KM0178) MUHAMAD AZAMUDDIN BIN MOHD TAHIR (A16KM0222) MUHAMAD HANIF SHAH BIN KHAIRUL ANUAR (A16KM0224) MUHAMMAD HAFIZ AKMAL BIN HARIS FADZILAH (A16KM0289) MUHAMMAD NASRUL BIN YAZI (A16KM0496) MUHAMMAD SHAFRIEZ BIN MD SHAIFUL (A16KM0330) NURUL ‘AFIFAH BINTI ZULKEFLI (A16KM0390) SAMALADEWI A/P MURUKAPPAN (A16KM0417) 1.0 Opening

1.1 Thank the members for coming. Action: Chairperson

2.0 Approval of meeting minutes 2.1 Rectifying mistakes made in Presentation 1 2.2 Discussion on the comments made by Dr. Nazri on the aircraft design. 2.3 Task distribution of each each group on each comment.


3.1 Assign the work to each group for each comment by Dr. Nazri. 3.2 Discussion on FEM analysis of landing gear system.





256


19 APRIL 2020 MINUTES OF MEETING NO. 10/2020

AIRCRAFT DESIGN II (SKMA 4523)

Date: 19 APRIL, 2020 Time: 9am-12pm Online Discussion



2.0 Approval of meeting minutes 2.1 Follow-up on the assigned task of each sub-group on Dr Nazri’s comment. 2.2 Task distribution on each group for unassigned task.


3.1 Recheck any missing part for final report. 3.2 Assign the unassigned work to relevant group for the comment made by Dr. Nazri.





257


8 JUNE 2020 MINUTES OF MEETING NO. 11/2020

AIRCRAFT DESIGN II (SKMA 4523)

Date: 8 JUNE, 2020 Time: 9am-12pm Online Discussion



2.0 Approval of meeting minutes 2.1 Follow-up on the assigned task of each sub-group on Dr Nazri’s comment. 2.2 Task distribution on Presentation 2 2.3 Discussion on the format and template of video recording and compiling.


3.1 Preparation of flight test. 3.2 Discussion on the challenges, weakness of our design activities and how

confident we are with the aircraft. 3.3 Choose compiler for our presentation video.

4.0 Adjournment of meeting




skma4253 aircraft design ii group two · 2020. 7. 16. · ahmad amirul amin bin abdul wahab...

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