skma 4523 aircraft design 2 final project report...4.5.1 schrenk’s approximation method 101 4.5.2...
TRANSCRIPT
SKMA 4523 AIRCRAFT DESIGN 2
FINAL PROJECT REPORT
GROUP THREE
UNIVERSITI TEKNOLOGI MALAYSIA
ii
Group Members:
1. Ten Jia Yee A17KM0432
2. Thevan Tangaraju A16KM0459
3. Shathasivam Parumasivam A16KM0509
4. Muhammad Imran Aiman Idris B17KM0032
5. Athiseshan Balan A16KM0047
6. Nur Amyra Mohd Aseme B17KM0051
7. Nur Aizat Nazihah Azmi B17KM0050
8. Melvin John A16KM0164
9. Siti Mastura Maskor A16KM0492
10. Wee Jun Wei A16KM0474
11. Shakgantan Balakrishnan A16KM0427
Lecturer’s Name:
Ir. Dr.-Ing. M. Nazri M. Nasir
Dr. Wan Zaidi Bin Wan Omar
iii
TABLE OF CONTENTS
TITLE PAGE
TABLE OF CONTENTS iii
LIST OF TABLES viii
LIST OF FIGURES xi
CHAPTER 1 INTRODUCTION 1
1.1 Introduction 1
1.2 Cost Estimation 3
1.3 3D-Modelling Design 3
CHAPTER 2 PROPULSION SYSTEM AND AIRCRAFT
PERFORMANCE 8
2.1 Background Study on Electronic Parts Selection 8
2.1.1 Motor Selection 8
2.1.2 Propeller Selection 10
2.1.3 Battery Selection 14
2.1.4 Electronic Speed Controller (ESC) Selection 15
2.2 Performance 16
2.2.1 Thrust Required, TR 16
2.2.2 Power Available, PA 18
2.2.3 Power Required, PR 18
2.2.4 Rate of Climb 18
2.2.5 Endurance 20
2.2.6 Range 20
2.2.7 Take-off Performance 20
2.2.8 Landing Performance 22
2.2.9 V-n Diagram 22
2.3 Performance Calculation 23
2.3.1 Sample calculation of Thrust Required, TR 23
iv
2.3.2 Sample calculation of Power Available 24
2.3.3 Sample calculation of Power Required 25
2.3.4 Calculation for Rate of Climb, R/C 25
2.3.5 Calculation for Endurance 25
2.3.6 Calculation for Range 26
2.3.7 Calculation for Take-off Performance 26
2.3.8 Calculation for Landing Performance 27
2.3.9 V-n Diagram 30
CHAPTER 3 AVIONICS AND CONTROL 33
3.1 CG Locations 33
3.1.1 Maximum weight CG 36
3.1.2 Empty weight CG 38
3.1.3 Forward CG 39
3.1.4 Aft CG 40
3.2 Stability 42
3.2.1 Longitudinal Stability 42
3.2.1.1 Wing Section Lift-Curve Slope 42
3.2.1.2 Wing Lift-Curve Slope 43
3.2.1.3 Wing Pitching Moment 44
3.2.1.4 Wing-Body Lift-Curve Slope 45
3.2.1.5 Downwash Gradient 47
3.2.1.6 Tail Section Lift-Curve 48
3.2.1.7 Tail Lift Curve Slope 49
3.2.1.8 Maximum Angle of Attack 50
3.2.1.9 Calculation for Equation of
Longitudinal Static Stability 52
Calculation for Static Margin 59
3.2.2 Lateral Stability 60
3.3 Communication Signals (Transmitter-Receiver) 63
3.3.1 Transmitter 63
3.3.1.1 Specifications of Transmitter 64
v
3.3.1.2 Chosen Transmitter 67
3.3.2 Receiver (RX) 71
3.3.2.1 Chosen Receiver 72
3.3.3 Converting Signals from Transmitter to Receiver 76
3.3.4 Signal Interference 77
3.3.4.1 Possible Signal Interferences 78
3.3.4.2 Technology to Reduce Interferences 78
3.3.5 Communication resolution 80
3.3.6 Antenna 81
3.3.6.1 Antenna Type 82
3.3.6.2 Gains 83
CHAPTER 4 WING ANALYSIS 85
4.1 Introduction 85
4.2 Wing Configuration 87
4.3 Shear and Bending Stresses 88
4.3.1 Wing without ailerons 90
4.3.2 Wing with ailerons 93
4.4 Lift Distribution of Aileron 97
4.5 Wing Loading 100
4.5.1 Schrenk’s Approximation Method 101
4.5.2 Sample Calculation 108
4.6 Wing Mounting 109
CHAPTER 5 FUSELAGE, AND LANDING GEAR
ANALYSIS 114
5.1 Fuselage 114
5.1.2 Introduction 114
5.1.3 Shear Force and Bending Moment Diagram 115
5.1.3.1 Shear Flow Diagram (3g) 115
5.1.3.2 Bending Moment Diagram (3g) 116
5.1.3.3 Shear Flow Diagram (-1.5g) 117
vi
5.1.3.4 Bending Moment Diagram (-1.5g) 118
5.1.4 Shear and Flexural Analysis 119
5.1.4.1 Conceptual Structural Analysis 119
5.1.4.2 Flexural Shear Flow 121
5.1.4.3 Fuselage Shear Flow 122
5.1.5 Structural Analysis 123
5.1.6 Compressive- Buckling Analysis 126
5.1.6.1 Former Structure 126
5.1.6.2 Bulkhead Structure 127
5.1.7 Shear- Buckling Analysis 128
5.1.7.1 Former Structure 128
5.1.7.2 Bulkhead Structure 129
5.2 Landing Gear 130
5.2.1 Main landing gear simulation analysis 130
5.2.2 Rear landing gear simulation analysis 131
5.2.3 Front landing gear simulation analysis 135
5.2.4 Discussion 138
CHAPTER 6 EMPENNAGE ANALYSIS 139
6.1 Introduction 139
6.2 Preliminary Horizontal and Vertical Tail Sizing 140
6.2.1 Choice of Empennage Shape 140
6.2.2 Horizontal and Vertical Stabilizer Sizing 141
6.2.3 Theoretical Analysis of Horizontal and Vertical
Stabilizers 142
6.2.3.1 Horizontal and Vertical Stabilizer Effectiveness 142
6.2.3.2 Horizontal and Vertical Stabilizer Strength 145
6.3 Preliminary Control Surfaces Sizing 160
6.3.1 Control Surface Sizing 160
6.3.2 Theoretical Analysis of Control Surfaces 164
vii
6.3.2.1 Theoretical Lift calculation at different angle of deflection 164
6.4 Strength of adhesive on the joint between tail and fuselage 173
CHAPTER 7 FLIGHT PLANNING AND TESTING 175
7.1 Flight Test 175
7.1.1 Pre-Flight Test 175
7.1.2 Preparation 177
7.1.3 Execution 177
7.1.4 Analysis and reporting 178
7.2 Flight Test Location 179
7.3 RC Airplane pre-flight checklist 180
7.4 Performing Range Check 182
7.5 Meteorological Conditions on Site 183
7.6 Flying Site 184
REFERENCES 187
APPENDICES 188
viii
LIST OF TABLES
TABLE NO. TITLE PAGE
Table 1.1 Cost estimation 3
Table 2.1 Part of manufacturer datasheet 11
Table 2.2 Value of thrust and current in three flight conditions 12
Table 2.3 Data for theoretical thrust and efficiency 13
Table 2.4 Performance data 28
Table 3.1 Calculations for maximum weight CG of aircraft 37
Table 3.2 Calculations for empty weight CG of aircraft 38
Table 3.3 Calculations for forward CG of aircraft 39
Table 3.4 Calculations for aft CG of aircraft 41
Table 3.5 Total pitching moment coefficient of gross weight CG with various angle of attack 54
Table 3.6 Total pitching moment coefficient of empty weight CG with various angle of attack 55
Table 3.7 Total pitching moment coefficient of forward CG with various angle of attack 56
Table 3.8 Total pitching moment coefficient of aft CG with various angle of attack 57
Table 3.9 Major sequence of channels 66
Table 3.10 Assigned channels of transmitter 69
Table 3.11 Operation specifications of transmitter 69
Table 3.12 RF modes of transmitter 71
Table 3.13 Operation specifications of receiver 72
Table 3.14 Channels of receiver 76
Table 3.15 Functions of rubber ducky antenna components 82
Table 3.16 Units of gain 84
Table 4.1 Wing Configuration 87
Table 4.2 Aileron Configuration 88
ix
Table 4.3 Shear and bending stresses for load factor -1.5g, 1g and 3g without aileron 93
Table 4.4 Comparison of shear and bending stresses for load factor -1.5g, 1g and 3g 96
Table 4.5 Calculated data at different angle of attack for aileron 99
Table 4.6 Wing specification 101
Table 4.7 Wing Loading Data 102
Table 4.8 Wing Loading Result 103
Table 4.9 Calculation Result 104
Table 5.1 Flange analysis 120
Table 5.2 Flexural shear flow at each stiffener 121
Table 5.3 Constant shear flow at each stiffener 122
Table 5.4 Flexural shear system 122
Table 5.5 Physical properties 123
Table 5.6 Component Weight 123
Table 5.7 Volumetric properties 131
Table 5.8 Material properties 131
Table 5.9 Load and fixtures detail of rear landing gear 132
Table 5.10 Reaction forces and moments 132
Table 5.11 Volumetric properties 135
Table 5.12 Material properties 135
Table 5.13 Load and fixtures detail of front landing gear 136
Table 5.14 Simulation results of front landing gear 136
Table 6.1 NACA 0012 horizontal stabilizer dimensions and sizing 141
Table 6.2 NACA 0012 vertical stabilizer dimensions and sizing 142
Table 6.3 Suggestions for tail volume ratio of horizontal tail by various authors (Raymer 1992, Jenkinson 1999, Roskam 1985, Torenbeek 1982, Nicolai 1975, Schaufele 2007) 144
Table 6.4 Suggestions for tail volume ratio of vertical tail by various authors (Raymer 1992, Jenkinson 1999, Roskam 1985, Torenbeek 1982, Nicolai 1975, Schaufele 2007) 145
x
Table 6.9 Table of lift, lift coefficient and velocity at different angle of attack for elevator 170
Table 6.10 Table of lift, lift coefficient and velocity at different angle of attack for rudder 171
Table 7.1 Flight test checklist 180
xi
LIST OF FIGURES
FIGURE NO. TITLE PAGE
Figure 1.1 Project Flowchart 2
Figure 1.2 3D Design 4
Figure 1.3 Side view 5
Figure 1.4 Front and back view 6
Figure 1.5 Top view 7
Figure 2.1 Brushless Motor SunnySky X2216 KV1100 8
Figure 2.2 Graph power required versus velocity 9
Figure 2.3 Graph TR and TA versus velocity 10
Figure 2.4 Propeller Gemfan APC9045 11
Figure 2.5 TCBWORTH 5200mAh 4S 14.8V 60C Lipo Battery XT60 14
Figure 2.6 The flight power consumption of the aircraft 15
Figure 2.7 Hobbywing Skywalker 40A 16
Figure 2.8 Comparison of lift-induced and zero-lift thrust required 17
Figure 2.9 Free body diagram of a climbing aircraft 19
Figure 2.10 Airplane take-off procedures 21
Figure 2.11 Flight Envelope 31
Figure 3.1 CG for each components 36
Figure 3.2 Maximum CG 37
Figure 3.3 Empty Weight CG 39
Figure 3.4 Forward CG 40
Figure 3.5 Aft CG 41
Figure 3.6 Moment coefficient versus angle of attack for gross weight CG 55
Figure 3.7 Moment coefficient versus angle of attack for empty weight CG 56
Figure 3.8 Moment coefficient versus angle of attack for forward CG 57
xii
Figure 3.9 Moment coefficient versus angle of attack for aft CG 58
Figure 3.10 Positive sideslip angle 61
Figure 3.11 Shape of wing tips 61
Figure 3.12 Graph of rolling moment versus sideslip angle 62
Figure 3.13 Block diagram of radio transmitter [4] 63
Figure 3.14 Spectrum graph analysis of some common ISM systems 65
Figure 3.15 Components of transmitter 66
Figure 3.16 Transmitter modes 67
Figure 3.17 FrSky Taranis Q X7 transmitter 67
Figure 3.18 Channel switches of transmitter 68
Figure 3.19 Spectrum analysis of ACCST system 70
Figure 3.20 Block diagram of radio receiver 71
Figure 3.21 FrSky V8R4-II 2.4Ghz 4CH receiver 72
Figure 3.22 Example of analogue signal and its corresponding PWM and PPM signals 73
Figure 3.23 Wiring Diagram of Flight Testing 77
Figure 3.24 Types of spread-spectrum technology 79
Figure 3.25 Communication resolution 80
Figure 3.26 Relationship between resolution and dynamic range 81
Figure 3.27 Block diagram of typical radio system [4] 81
Figure 3.28 Components of rubber ducky antenna 82
Figure 3.29 Radiation pattern of dipole antenna 83
Figure 3.30 3D radiation pattern of dipole antenna 83
Figure 4.1 Streamline on airfoil surface 85
Figure 4.2 Different types of monoplane 86
Figure 4.3 Full dimension of half wing with aileron 88
Figure 4.4 Front view of the aircraft 89
Figure 4.5 Free body diagram for g = -1.5 without aileron 90
Figure 4.6 Shear force and bending moment diagram for g = -1.5 without aileron 90
xiii
Figure 4.7 Free body diagram for g = 1 without aileron 91
Figure 4.8 Shear force and bending moment diagram for g = 1 without aileron 91
Figure 4.9 Free body diagram for g = 3 without aileron 91
Figure 4.10 Shear force and bending moment diagram for g = 3 without aileron 92
Figure 4.11 Free body diagram for g = -1.5 with aileron 94
Figure 4.12 Shear force and bending moment diagram for g = -1.5 with aileron 94
Figure 4.13 Free body diagram for g = 1 with aileron 94
Figure 4.14 Shear force and bending moment diagram for g = 1 with aileron 95
Figure 4.15 Free body diagram for g = 3 with aileron 95
Figure 4.16 Shear force and bending moment diagram for g = 3 with aileron 95
Figure 4.17 2D body surface of aileron 97
Figure 4.18 Graph of Lift against Angle of Attack 100
Figure 4.19 Graph Cl vs y 105
Figure 4.20 Graph cCl vs y 105
Figure 4.21 Graph Cy vs y 105
Figure 4.22 Graph Ln vs y 106
Figure 4.23 Graph V vs y 106
Figure 4.24 Graph M vs y 106
Figure 4.25 Step of wing mounting using rubber bands 110
Figure 4.26 Rubber band size 110
Figure 4.27 Young's Modulus of several items 111
Figure 5.1 Shear force diagram for fuselage at 3g 115
Figure 5.2 Bending moment diagram for fuselage at 3g 116
Figure 5.3 Shear force diagram for fuselage at -1.5g 117
Figure 5.4 Bending moment diagram for fuselage at -1.5g 118
Figure 5.5 Stringer location at side 120
xiv
Figure 5.6 Component weight 124
Figure 5.7 Side view 124
Figure 5.8 Component weight 125
Figure 5.9 Isometric view 125
Figure 5.10 Bottom part of fuselage 125
Figure 5.11 Rear landing gear 131
Figure 5.12 Static stress 133
Figure 5.13 Static displacement 134
Figure 5.14 Static strain 134
Figure 5.15 Front landing gear 135
Figure 5.16 Static Stress 137
Figure 5.17 Static strain 137
Figure 6.1 Structural configuration of the empennage 139
Figure 6.2 Tail configuration of our design 140
Figure 6.3 Calculation on effectiveness of horizontal stabilizer 143
Figure 6.4 Calculation on effectiveness of vertical stabilizer 144
Figure 6.5 Lift distribution of the front view of the tail plane at G =-1.5 146
Figure 6.6 Shear and bending moment diagram at G = -1.5 146
Figure 6.7 Lift distribution of the front view of the tail pla ne at G = 1.0 147
Figure 6.8 Shear and bending moment diagram at G = 1.0 147
Figure 6.9 Lift distribution of the front view of the tail plane at G = 3.0 148
Figure 6.10 Shear and bending moment diagram at G = 3.0 148
Figure 6.11 Free body diagram of stabilizer under shear and bending conditions 150
Figure 7.1 Location of flight test from Google Map 179
Figure 7.2 UTM Marching Paddock (Padang Kawad UTM) 180
Figure 7.3 Johor average and max wind speed and gust (kmph) 183
Figure 7.4 Average temperature of location selected 184
Figure 7.5 Average rainfall days of location selected 184
Figure 7.6 The designed flying site 186
xv
1
CHAPTER 1
INTRODUCTION
1.1 Introduction
Unmanned aerial vehicle (UAV) can be defined as a flying machine without
pilot onboard. Therefore, it can be operated by human operator with remote control
or fully autonomously by computers onboard [1]. Usually, it is used to complete dull,
dangerous and dirty (3D) missions [2]. This is due to its ease of deployment, ability
to hover and high-mobility [3]. Although UAV is mainly used in military sector at the
beginning, its applications are quickly expanding to other sectors such as agricultural,
logistics, aerial photography and recreational use.
In this project, an UAV airplane is required to be designed. The design
criterions are stated as below:
1. Maximum gross weight is 5 kg.
2. Maximum load factor is 3 whereas the minimum load factor is -1.5.
3. The maximum speed of aircraft is 20 m/s.
4. The aircraft must be able to carry 500g payload.
The flowchart of the design project is displayed in figure 1.1
2
Figure 1.1 Project Flowchart
Start
Selection of group members
Distribution into five sub-groups
Carry out feasibility study
Determine the main aircraft specifications
Electronic
components selection
Aircraft CG
components and stability
Structural analysis
Control system
Flight planning and preparation
End
Yes
Yes
Yes
No
No
No
3
1.2 Cost Estimation
Financial plan is required when fabricating an UAV airplane as the components
have to be bought. The cost estimation of this project is shown in table 1.1.
Item Price
(Rm)
Shipping
Fee (Rm)
Delivery
(Days) Description Link
Battery 165.00 3.50 2-12 14.8/4s
5200mah lazada
Propeller 11.50 4.13 3-21 APC 9045 shopee
Motor 70.00 1.00 3-21 X2216 1100kv shopee
ESC 40.00 3.50 2-15 40A Brushless lazada
Foam 100.00 11.00 2-7 5 pieces
Battery
Checker 5.00 - 2-7 buzzer
Receiver 79.00 - 2-7 2.4Ghz
Sub-Total 470.50 23.13
TOTAL
(RM) 493.63
Table 1.1 Cost estimation
1.3 3D-Modelling Design
First and foremost, the aircraft model has to be designed. The drawings are
drawn in Solidworks to visualize the designed aircraft model. Figure 1.2 shows the 3D
design of the aircraft. At the same time, figures 1.3, 1.4 and 1.5 depicts the side view,
front view and top view of the aircraft respectively.
4
Figure 1.2 3D Design
5
Figure 1.3 Side view
6
Figure 1.4 Front and back view
7
Figure 1.5 Top view
8
CHAPTER 2
PROPULSION SYSTEM AND AIRCRAFT PERFORMANCE
2.1 Background Study on Electronic Parts Selection
2.1.1 Motor Selection
For motor selection, there are several factors that need to be taken into
consideration which is brushless motor KV, size and thrust to weight ratio that depends
on the type of aircraft (Pond, 2009). The first one is brushless motor KV. The motor
KV indicates the number of revolutions per minute spin per volt. The lower the KV,
the stronger the motor. As a result, a larger propeller needed for maximum thrust. The
next factor to considering a motor is size (Remzak, 2017).
Since the aircraft needs to have a higher thrust and not higher speed, the
suitable motor with the KV value between 850 and 1500 is chosen. For our aircraft,
brushless motor SunnySky X2216 KV1100 is chosen as shown in figure 2.1.
Figure 2.1 Brushless Motor SunnySky X2216 KV1100
9
According to FliteTest website, it is better to choose the motor that can produce
more thrust than the weight of airplane. The maximum velocity given for this project
is 20m/s. So, we need to choose motor that can obtain desired velocity less than 20m/s
by plotting graph power required versus velocity.
From graph power required versus velocity as shown in figure 2.2, the
intersection point between power required and power available is the design speed for
our aircraft. The power required can be calculated by referring the motor specification
obtained from the manufacturer. We need to choose a random motor KV in the range
between 850 and 1500 to obtain the design speed of our aircraft not more than 20 m/s.
Since the design speed obtained is 19.3 m/s which is less than 20 m/s, then the motor
KV that we choose which is KV1100 is good enough.
Figure 2.2 Graph power required versus velocity
From the figure, PR indicates the power required whereas PA indicates the
power available. At the same time, 2PR is referred to the value of doubling the power
required. In this case, we only refer the 2PR to obtain the design speed for our aircraft.
The reason we used 2PR is because of it is more realistic. In other words, PR is a very
optimistic assumption where usually aircraft requires more than that.
0
100
200
300
400
500
600
700
0 6
7.5 9
10.
5 12
13
14.
5 16
17.
5 19
20.
5 22
23.
5 25
26.
5 28
29.
5 31
32.
5 34
PR
Velocity (m/s)
Power Required
2PR
PR
PA(1500)
Poly. (2PR)
10
Figure 2.3 Graph TR and TA versus velocity
In steady, level flight, the maximum velocity of the aircraft is determined by
the high speed intersection of the thrust required and thrust available curves. Therefore,
figure 2.3 is plotted due to the thrust available (TA) must be always higher than thrust
required (TR) to ensure that the aircraft can function properly. From figure 2.3, the
maximum velocity is 21.5 m/s, which is more than design speed, 19.3 m/s obtained
from figure 2.2. Thus, the conclusion that the motor used in the aircraft can always
provide enough thrust to the aircraft.
2.1.2 Propeller Selection
After selecting the motor, the propellers that match with the motor is listed.
Table 2.1 shows part of the manufacturer datasheet. From the table, the recommended
propeller is APC 9045 as shown in figure 2.4.
0
2
4
6
8
10
12
14
16
18
20
0 5 10 15 20 25 30 35 40
Thru
st
Velocity
TR and TA vs Velocity
TR
TA
Poly. (TR)
Poly. (TA)
11
Table 2.1 Part of manufacturer datasheet
Prop(inch) Voltage(V) Amps(A) Thrust(gf) Watts(W) Efficiency
(g/W)
APC9045 14.8
0.7 100 10.36 9.652509653
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
Figure 2.4 Propeller Gemfan APC9045
By referring information from table 2.1, the estimated current used for take-off,
cruising and landing can be obtained by using interpolation. The values of estimated
current are displayed in table 2.2.
12
Table 2.2 Value of thrust and current in three flight conditions
Take-off Cruising Landing
Thrust (gf) 770 550 220
Current (A) 10.12 6.5 1.9
Since the propeller diameter expressed in inches, the theoretical thrust can be
calculated by using equation below:
𝑇𝑔𝑟𝑎𝑚 = (𝑃𝐷𝑖𝑛𝑐ℎ
𝐶)
23
where C is the air density dependent coefficient 30° C,1 atm:
𝑇𝑔𝑟𝑎𝑚 =1
0.0127√
𝑔3
2𝜋𝑄𝑎𝑖𝑟= 0.02780
At the same time, the efficiency of the motor and the propeller is given by:
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =𝑡ℎ𝑟𝑢𝑠𝑡
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑡ℎ𝑟𝑢𝑠𝑡× 100%
For sample calculation, the maximum thrust is taken therefore:
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =1100
1824 .226722× 100% = 60.3%
As a result, the efficiency that with unit of percentage can be obtained and
shown in table 2.3.
13
Table 2.3 Data for theoretical thrust and efficiency
Prop
(inch)
Voltage
(V)
Amps
(A)
Thrust
(gf)
Watts
(W)
Efficiency
(g/W) C
Theoretical
Thrust Efficiency (%) Efficiency
APC9045 14.8
0.7 100 10.36 9.652509653 0.028037 222.7998741 44.88332877 0.448833288
1.7 200 25.16 7.949125596 0.028037 402.5460122 49.68376135 0.496837613
2.7 300 39.96 7.507507508 0.028037 547.9724914 54.74727376 0.547472738
4.2 400 62.16 6.435006435 0.028037 735.6689753 54.37228066 0.543722807
5.7 500 84.36 5.926979611 0.028037 901.7776863 55.44603815 0.554460382
7.3 600 108.04 5.553498704 0.028037 1063.485292 56.4182697 0.564182697
9 700 133.2 5.255255255 0.028037 1222.769472 57.24709491 0.572470949
10.6 800 156.88 5.099439062 0.028037 1363.704137 58.66375106 0.586637511
12.5 900 185 4.864864865 0.028037 1522.14581 59.12705566 0.591270557
14.6 1000 216.08 4.627915587 0.028037 1688.177671 59.23547132 0.592354713
16.4 1100 242.72 4.531970995 0.028037 1824.226722 60.29952236 0.602995224
14
2.1.3 Battery Selection
The selection of battery is mainly related to the endurance of the aircraft. The
most important parameter is the battery capacity. Battery capacity shows the amount
of current that stores in the battery. Besides, the C value indicates the discharge rate of
the battery. High C value enables the aircraft to have a short response time. Since the
aircraft is designed to have a relatively low speed, the battery does not require a very
high C value. By researching to the battery, the battery selected is TCBWORTH
5200mAh 4S 14.8V 60C Lipo Battery XT60 as shown in figure 2.5.
Figure 2.5 TCBWORTH 5200mAh 4S 14.8V 60C Lipo Battery XT60
The battery with capacity 5200mAh was chosen by calculating the endurance
for the battery lifetime to fly as one of the design requirements of the aircraft is with 5
minutes of cruising time. Therefore, if we choose low capacity of battery, it is not sure
that our aircraft can fly back and forth to the design location including take-off and
landing time.
𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 =5200
1000×
0.8
20× 60 = 12.48 𝑚𝑖𝑛
The endurance for the battery 5200mAh is 12.48 minutes. By referring to the
flight consumption graph as shown in figure 2.6, we can obtain the estimated time
during flight including take-off and landing.
15
Figure 2.6 The flight power consumption of the aircraft
From the graph above, the time taken for our aircraft to take-off, cruising and
landing can be obtained.
𝑡𝑡𝑎𝑘𝑒 𝑜𝑓𝑓 = 0.35 min
𝑡𝑐𝑟𝑢𝑖𝑠𝑖𝑛𝑔 = 5 min
𝑡𝑙𝑎𝑛𝑑𝑖𝑛𝑔 = 2.67 min
Therefore, total estimated time for our aircraft is 8.02 min. Since the endurance
of the battery is more than the estimated time for our aircraft, it is proved that the
battery capacity that was chosen is suitable for our flight test.
2.1.4 Electronic Speed Controller (ESC) Selection
The maximum ampere value of ESC indicates the maximum ampere that can
withstand by the ESC. In other words, it may be burnt out if the current received by
the ESC is higher than the maximum ampere rating. Therefore, the current rating of
ESC must be higher than the maximum current used by the motor.
In usual case, the maximum ampere of the ESC will be stated in manufacturer
datasheet. Based on manufacturer data for the selected motor, it is recommended to
use ESC 30A. But, to make it safe, we choose Hobbywing Skywalker 40A ESC as
shown on Figure 2.7. This is because it is good to use an ESC rated at a higher
0
2
4
6
8
10
12
0 31 60 91 121 152 182 213 244 274 305 335 366 397 425 456 486 517 547
Cu
rren
t (A
)
Time (s)
Flight Power Consumption
ampere
16
amperage than intend running motor at as an insurance against over stressing the ESC
causing failure and potential damage to aircraft.
Figure 2.7 Hobbywing Skywalker 40A
2.2 Performance
The preliminary performance analysis is a key in determining the feasibility
and effectiveness of the designed aircraft. Basically, parameters for drag polar, power,
thrust, range, endurance and etc. are estimated using equation given from reference
book Anderson (1999).
2.2.1 Thrust Required, TR
In steady flight (unaccelerated flight), the general equation of motions is
derived is such special case giving:
𝑇 = 𝐷
𝐿 = 𝑊
Therefore, to maintain the same amount of speed at certain altitude, the thrust
must be generated to overcome the drag and keep the airplane going. In such case, this
is the thrust required. Thrust required denoted as TR depends on the velocity, altitude,
and the aerodynamic shape, size and weight of the airplane. By using analytical
approach, at steady and level flight, the thrust required can be derived as:
17
𝑇𝑅 = 𝐷 =𝐷
𝑊𝑊 =
𝐷
𝐿𝑊 =
𝑊
𝐿𝐷
Since TR is a relation to L/D, thus minimum TR occurs when L/D is maximum.
As mention earlier, lift-to-drag ratio is one of the most important parameters affecting
the airplane performance. And since TR=D, therefore:
𝑇𝑅 = 𝐷 = 𝑞∞𝑆𝐶𝐷 = 𝑞∞𝑆(𝐶𝐷,0 + 𝐶𝐷,𝑖)
Note that, 𝐶𝐷,𝑖 = 𝐾𝐶𝐿2 and 𝐿 = 𝑊 so:
𝐾 =1
𝜋𝑒𝐴𝑅 , 𝐶𝐿 = 2𝑊𝜌∞𝑉∞
2𝑆
In other word,
𝑇𝑅 = zero-lift 𝑇𝑅 + lift-induced 𝑇𝑅
where
zero-lift 𝑇𝑅 = thrust required to balance zero-lift drag
lift-induced 𝑇𝑅 = thrust required to balance drag due to lift
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. Figure 2.8 shows the relationship between 𝑇𝑅 and
velocity.
Figure 2.8 Comparison of lift-induced and zero-lift thrust required
18
2.2.2 Power Available, PA
The actual power available supplied by the motor can be calculate using
equation:
𝑃𝐴 = 𝜂 × 𝑃
where,
𝜂 = efficiency of the motor and the propeller
𝑃 = power (W)
2.2.3 Power Required, PR
The general equation for power is the product of force and velocity provided
they are in the same direction as in vector dot product. As the airplane cruising at
certain velocity (free stream velocity), 𝑉∞, times with the thrust required, 𝑇𝑅, yields
the power required, denoted as 𝑃𝑅 (Anderson, 1999).
𝑃𝑅 = 𝑇𝑅𝑉∞
Since 𝑇𝑅 = 𝐷 (for unaccelerated flight) so
𝑃𝑅 = 𝑞∞(𝐶𝐷,0 + 𝐶𝐷,𝑖)𝑉∞
where 𝐶𝐷,𝑖, can be written as 𝐾𝐶𝐿2
Therefore,
𝑃𝑅 = 𝑞∞(𝐶𝐷,0 + 𝐾𝐶𝐿2)𝑉∞
2.2.4 Rate of Climb
Consider an airplane in steady, unaccelerated and climbing flight. The free
body diagram of an aircraft in climbing flight is as shown in figure 2.9.
19
Figure 2.9 Free body diagram of a climbing aircraft
From figure 2.9, thrust (T) is not only working to overcome drag (D), but for
climbing flight, it is also supporting component of weight of aircraft (W). The equation
of motion can be derived as:
𝑇 − 𝐷 − 𝑊 sin 𝜃 = 0
𝐿 − 𝑊 cos 𝜃 = 0
The vertical velocity of the flight is the rate of climb, 𝑅/𝐶
𝑅/𝐶 = 𝑉∞ sin 𝜃
As mentioned earlier, 𝑇𝐴 𝑉∞ is the power available, 𝑃𝐴 and 𝑇𝑅 𝑉∞ or 𝐷𝑉∞ is the
power required, 𝑃𝑅 to overcome the drag. We define
𝑇𝐴𝑉∞ − 𝑇𝑅 𝑉∞ = 𝐸𝑥𝑐𝑒𝑠𝑠 𝑃𝑜𝑤𝑒𝑟
Hence, substituting sin 𝜃 into the rate of climb, 𝑅/𝐶 can be written as;
𝑅/𝐶 =𝐸𝑥𝑐𝑒𝑠𝑠 𝑃𝑜𝑤𝑒𝑟
𝑊
Note: This formula is an approximation as 𝑃𝑅 = 𝑇𝑅𝑉∞ is for a level flight. Thus
this equation only good for small 𝜃.
20
2.2.5 Endurance
By definition, endurance is the total time that an airplane stays in the air on a
tank of fuel. In order to obtain a longer time for the aircraft to stay in the air, minimum
number of pounds of fuel per hour need to be used. In order to get maximum endurance
for an aircraft, it need to fly at the lowest minimum power required such that it is flying
at a velocity such (𝐶𝐿3/2/𝐶𝐷) is at maximum. The following formula is used to find
the endurance.
𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 = 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ×80% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑢𝑠𝑒𝑑
𝑐𝑟𝑢𝑖𝑠𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡× 60
2.2.6 Range
By definition range is the total distance traversed by the airplane on a tank of
fuel. To obtain the maximum range for an aircraft, the lowest possible specific fuel
consumption need to be used and the aircraft is flying at maximum C𝐿/𝐶𝐷 . The
following formula is used to find the range.
𝑅𝑎𝑛𝑔𝑒 = 𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
2.2.7 Take-off Performance
The total take-off distance take-off distance, as defined in the Federal Aviation
Requirements (FAR), is the sum as defined in the Federal Aviation
Requirements(FAR), is the sums LO and the distance the distance (measured along the
ground) to clear a 35-ft height (for jet-powered civilian transports) or a 50-ft height
(for all other airplanes). The aircraft takeoff operation is shown in figure 2.10.
21
Figure 2.10 Airplane take-off procedures
Basically, total take-off distance is covered by ground roll, 𝑆𝑔 and airborne
distance, 𝑆𝑎. Ground roll distance is the distance where the aircraft start moving until
the aircraft reach its lift off speed, 𝑉𝐿𝑂 whereas the airborne distance covers the
distance where the aircraft required to clear an obstacle after becoming airborne. In
this chapter we are only interested on the ground roll for take-off. The equation for
ground roll is as follow
𝑆𝐿𝑂 ≈1.44W2
gρ∞SCLmax[T − [D + μr(W − L)]0.7𝑉𝐿𝑂]
where D:
0.5ρ∞𝑉20.7𝑉𝐿𝑂
𝑆 (𝐶𝐷,0 + 𝜑𝐶𝐿
2
𝜋𝑒𝐴𝑅)
𝜑 =(16ℎ/𝑏)2
1 + (16ℎ/𝑏)2
During take-off, the angle of attack of the airplane is restricted by the
requirement that the tail does not drag the ground, therefore assume that CLmax during
ground roll is limited to 1.3 where:
𝑔 = gravitational force
ρ∞ = sea level air density
S = wing area
CLmax = maximum lift during take-off
T = thrust at take-off configuration
𝜇𝑟 = land friction coefficient
𝑊 = weight of an aircraft
22
2.2.8 Landing Performance
The opposite of the take-off procedure is the landing procedure. Just as in the
take-off, the landing maneuver consists of two parts:
1. The terminal glide over a 50 ft obstacle to touchdown
2. The landing ground run
Ground roll distance is the distance where the aircraft start decelerates until the
velocity goes to zero. Again in this chapter we are only consider about the ground roll
while landing. The equation is as follow.
𝑆𝐿 =1.69W2
gρ∞SCLmax[T𝑟𝑒𝑣 + [D + μr(W − L)]0.7𝑉𝑇𝐷]
where
𝑔 = gravitational force
ρ∞ = sea level air density
S = wing area
CLmax = maximum lift during landing
𝑇𝑟𝑒𝑣 = reversal thrust during landing
𝜇𝑟 = land friction coefficient
𝑊 = weight of an aircraft
2.2.9 V-n Diagram
The V-n diagram plays an important role in aircraft design. The V-n diagram
is a plot between the load factor and the velocity. Load factor is defined as the ratio of
the aerodynamic load to the weight of the aircraft. Aircraft has to perform different
loading conditions at different speeds, controls and high loads due to stormy weather.
But at the same time, it is impossible to investigate all possible loading conditions.
From V-n diagram, we can easily determine the stall region for a particular aircraft at
23
different flight speed. Also we are able to identify the structural limit on the aircraft.
Beyond the limit, structural damage might occur.
The V-n diagram is drawn referring to FAR 23 standards: Airworthiness
Standards for Normal, Utility, Acrobatic and Commuter Category Airplanes. Since our
UAV is deemed as a normal category aircraft, therefore this standard is the appropriate
one to refer to. Flight load factor is an important parameter in this V-n diagram
generation. Flight load factor is the ratio of aerodynamic force acting on the airplane
to the weight of the airplane. For V-n diagram, we only focus on the maneuver
envelope.
Maneuver Diagram illustrates the variation in load factor with airspeed for
maneuver. At low speeds the maximum load factor is constrained by aircraft maximum
CL. At higher speeds the maneuver load factor may be restricted. According to FAR
23, the normal category is limited to airplanes that have a seating configuration,
excluding pilot seats, of nine or less, a maximum certificated takeoff weight of 12,500
pounds or less, and intended for non-acrobatic operation. Non-acrobatic operation
includes:
I. Any maneuver incident to normal flying
II. Stalls (except whip stalls
III. Lazy eights, chandelles, and steep turns, in which the angle of bank is not
more than 60 degrees
2.3 Performance Calculation
2.3.1 Sample calculation of Thrust Required, TR
Calculate the free stream velocity, 𝑉∞
𝑉∞ = √2𝑊
𝜌∞CLmax𝑆
24
𝑉∞ = √2 × 7.27525
0.002377 × 1.3 × 4.2195= 33.4058 𝑓𝑡/𝑠
Thrust required, 𝑇𝑅 can be obtained using equation,
𝑇𝑅 = 𝐷 = 𝑞∞𝑆𝐶𝐷 = 𝑞∞𝑆(𝐶𝐷,0 + 𝐾𝐶𝐿2)
𝑇𝑅 =1
2𝜌∞𝑉∞
2𝑆(𝐶𝐷,0 + 𝐾𝐶𝐿2)
where in this case, the velocity used is not free stream velocity, 𝑉∞ and there are
different velocity value.
𝑇𝑅 =1
2𝜌∞𝑉2𝑆(𝐶𝐷,0 + 𝐾𝐶𝐿
2)
where,
𝑊 = 𝐿 = 0.5ρ∞𝑉2𝑆𝐶𝐿
𝐶𝐿 = (2×3.3×9.81)
1.157(5)2(0.392)= 5.39325 (𝑓𝑜𝑟 𝑉 = 5 𝑚/𝑠)
𝐾 = 1
𝜋𝑒𝐴𝑅= 0.0995
𝐶𝐷,0 = 0.02808
𝑆 = 0.392 𝑚2
𝜌∞ = 1.157 𝑘𝑔/𝑚3
𝑊 = 3.3 𝑘𝑔
Therefore,
𝑇𝑅 =1
2(1.157)(5)2(0.392)(0.02808 + (0.0995)(5.3933) 2) = 17.5408
2.3.2 Sample calculation of Power Available
The power available is related to the power generated by the motor. The
calculation of the value is shown as below:
𝑃𝐴 = 𝜂 × 𝑃 = 0.602995224 × 242.72 = 146.359 𝑊 = 107.95 𝑙𝑏. 𝑓𝑡/𝑠
25
2.3.3 Sample calculation of Power Required
The power required, 𝑃𝑅 at sea level is
𝑃𝑅 = 𝑇𝑅𝑉∞
where in this case, the velocity used is not free stream velocity, 𝑉∞. Since 𝑇𝑅 =
𝐷 (unaccelerated flight)
𝑃𝑅 = 𝑞∞(𝐶𝐷,0 + 𝐾𝐶𝐿2)𝑉
𝑃𝑅 = 1.0680 𝑙𝑏 × 33.4058 𝑓𝑡/𝑠
𝑃𝑅 = 35.6774 𝑙𝑏. 𝑓𝑡/𝑠
2.3.4 Calculation for Rate of Climb, R/C
At sea level,
𝑇𝐴𝑉∞ − 𝑇𝑅𝑉∞ = 𝐸𝑥𝑐𝑒𝑠𝑠 𝑃𝑜𝑤𝑒𝑟
𝑃𝐴 − 𝑃𝑅 = 𝐸𝑥𝑐𝑒𝑠𝑠 𝑃𝑜𝑤𝑒𝑟
107.95 − 35.6774 = 72.2726 𝑙𝑏. 𝑓𝑡/𝑠
𝑅/𝐶 =𝐸𝑥𝑐𝑒𝑠𝑠 𝑃𝑜𝑤𝑒𝑟
𝑊
𝑅/𝐶 =72.2726
7.2725= 9.9378 𝑓𝑡/𝑠
2.3.5 Calculation for Endurance
To calculate the endurance, the following equation is used,
𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 = 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ×80% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑢𝑠𝑒𝑑
𝑐𝑟𝑢𝑖𝑠𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡× 60
26
where,
Battery capacity used = 5200 mAh
Cruising current used = 20 A
𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 =5200 𝑚𝐴ℎ
1000×
0.8
20× 60 = 12.48 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
2.3.6 Calculation for Range
The following equation is used,
𝑅𝑎𝑛𝑔𝑒 = 𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
where velocity = 19.3 m/s
𝑅𝑎𝑛𝑔𝑒 = 12.48 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 × 19.3 𝑚/𝑠 × 60 = 14,451.8 𝑚
2.3.7 Calculation for Take-off Performance
The following equation is used,
𝑆𝐿𝑂 ≈1.44W2
gρ∞SCLmax[T − [D + μr(W − L)]0.7𝑉𝐿𝑂]
where,
𝑉𝑠𝑡𝑎𝑙𝑙 =√2𝑊
𝜌∞CLmax𝑆= √
2×7.27525
0.002377×1.3×4.2195= 33.4058 𝑓𝑡/𝑠
𝑉𝐿𝑂 = 1.2 𝑉𝑠𝑡𝑎𝑙𝑙 = 1.2 (33.4058) = 40.0870 𝑓𝑡/𝑠
0.7𝑉𝐿𝑂 = 0.7(40.0870) = 28.0609 𝑓𝑡/𝑠
𝜑 = (
16ℎ
𝑏)
2
1+(16ℎ
𝑏)
2 =(
16(0.6851)
4.5932)
2
1+((16(0.6851)
4.5932)
2 = 0.8509
27
𝑊 = 𝐿 = 0.5ρ∞𝑉20.7𝑉𝐿𝑂
𝑆𝐶𝐿
7.2753 = 0.5 × 0.002377 × 28.0609 2 × 4.2195 × 𝐶𝐿
𝐶𝐿 = 1.8424
𝐶𝐷,𝑖 = 𝜑𝐶𝐿
2
𝜋𝑒𝐴𝑅= 0.8509 (
1.84242
𝜋×0.8621×5) = 0.2133
𝐷𝐿𝑂 =0.5(0.002377)(28.0609) 2(4.2195)(0.02808 + 0.2133) = 0.9532 𝑙𝑏
𝑔 = 32.2 𝑓𝑡/𝑠2
ρ∞ = 0.002377 𝑠𝑙𝑢𝑔/𝑓𝑡3
𝑆 = 4.2195 𝑓𝑡2
μr = 0.02
CLmax = 1.3
𝐿 = 0.5 × 0.002377 × 28.0609 2 × 4.2195 × 1.3 = 5.1334 𝑙𝑏
𝑇 = 𝑃𝐴
𝑉𝐿𝑂=
146.359
40.0870= 3.6510
Therefore,
𝑆𝐿𝑂
= 1.44(7.2753)2
(32.2)(0.002377)(4.2195)(1.3)[3.6510 − [0.9532 + 0.02(7.2753 − 5.1334)]]
𝑆𝐿𝑂 = 68.378 𝑓𝑡 = 20.84 𝑚
2.3.8 Calculation for Landing Performance
The following equation is used,
𝑆𝐿 =1.69W2
gρ∞SCLmax[T𝑟𝑒𝑣 + [D + μr(W − L)]0.7𝑉𝑇𝐷]
where,
μr = 0.02
𝑔 = 32.2 𝑓𝑡/𝑠2
ρ∞ = 0.002377 𝑠𝑙𝑢𝑔/𝑓𝑡3
𝑆 = 4.2195 𝑓𝑡2
28
𝑊 = 7.2753 𝑙𝑏
V𝑇 = 1.3 𝑉𝑠𝑡𝑎𝑙𝑙 = 1.3 (33.4058) = 43.4275 𝑓𝑡/𝑠
0.7𝑉𝑇 = 0.7(43.4275) = 30.3993 𝑓𝑡/𝑠
𝐿 = 0.5(0.002377) (30.3993) 2(4.2195)(1.3) = 6.0246 𝑙𝑏
𝑊 = 𝑊1
7.2753 = 0.5 × 0.002377 × 30.3993 2 × 4.2195 × 𝐶𝐿
𝐶𝐿 = 1.5699
𝐶𝐷,𝑖 = 𝜑𝐶𝐿
2
𝜋𝑒𝐴𝑅= 0.8509 (
1.56992
𝜋×0.8621×5) = 0.1549
𝐷𝐿𝑂= 0.5(0.002377)(30.3993) 2(4.2195)(0.02808 + 0.1549) = 0.8480 𝑙𝑏
𝑇𝑟𝑒𝑣 = 0 𝑙𝑏
Therefore,
𝑆𝐿 =1.69(7.2753 )2
(32.2)(0.002377)(4.2195)(1.3)[0 + [0.8480 + 0.02(7.2753 − 0 )]0.7𝑉𝑇𝐷]
𝑆𝐿 = 214.45 𝑓𝑡 = 65 𝑚
In short, table 2.4 shows the calculated values by using the equations from the
performance part.
Table 2.4 Performance data
V
(m/s) 𝐶𝐿 𝐶𝐷 2𝐷 𝐷 = 𝑇𝑅 𝑃𝑅 2𝑃𝑅 𝑃𝐴(1500)
5 5.3933 2.9223 35.0817 17.5408 87.7041 175.4083 146.3590
5.5 4.4572 2.0048 29.1224 14.5612 80.0866 160.1732 146.3590
6 3.7453 1.4238 24.6136 12.3068 73.8408 147.6815 146.3590
6.5 3.1913 1.0414 21.1286 10.5643 68.6680 137.3360 146.3590
7 2.7517 0.7815 18.3875 9.1938 64.3564 128.7127 146.3590
7.5 2.3970 0.5998 16.2005 8.1003 60.7519 121.5038 146.3590
8 2.1067 0.4697 14.4351 7.2175 57.7403 115.4806 146.3590
8.5 1.8662 0.3746 12.9966 6.4983 55.2354 110.4708 146.3590
9 1.6646 0.3038 11.8158 5.9079 53.1713 106.3425 146.3590
9.5 1.4940 0.2502 10.8415 5.4207 51.4970 102.9939 146.3590
29
10 1.3483 0.2090 10.0345 5.0173 50.1727 100.3454 146.3590
10.5 1.2230 0.1769 9.3652 4.6826 49.1673 98.3346 146.3590
11 1.1143 0.1516 8.8102 4.4051 48.4561 96.9121 146.3590
11.5 1.0195 0.1315 8.3512 4.1756 48.0196 96.0392 146.3590
12 0.9363 0.1153 7.9737 3.9869 47.8424 95.6849 146.3590
12.22 0.9029 0.1092 7.8304 3.9152 47.8434 95.6869 146.3590
12.5 0.8629 0.1022 7.6660 3.8330 47.9125 95.8251 146.3590
13 0.7978 0.0914 7.4185 3.7093 48.2204 96.4408 146.3590
13.5 0.7398 0.0825 7.2235 3.6118 48.7588 97.5175 146.3590
14 0.6879 0.0752 7.0746 3.5373 49.5220 99.0440 146.3590
14.5 0.6413 0.0690 6.9664 3.4832 50.5061 101.0121 146.3590
15 0.5993 0.0638 6.8944 3.4472 51.7081 103.4162 146.3590
15.5 0.5612 0.0594 6.8550 3.4275 53.1262 106.2524 146.3590
16 0.5267 0.0557 6.8449 3.4225 54.7595 109.5189 146.3590
16.5 0.4952 0.0525 6.8615 3.4308 56.6076 113.2152 146.3590
17 0.4665 0.0497 6.9025 3.4512 58.6710 117.3419 146.3590
17.5 0.4403 0.0474 6.9658 3.4829 60.9505 121.9010 146.3590
18 0.4161 0.0453 7.0497 3.5249 63.4476 126.8951 146.3590
18.5 0.3940 0.0435 7.1529 3.5764 66.1639 132.3278 146.3590
19 0.3735 0.0420 7.2739 3.6369 69.1017 138.2034 146.3590
19.5 0.3546 0.0406 7.4116 3.7058 72.2633 144.5266 146.3590
20 0.3371 0.0394 7.5651 3.7826 75.6514 151.3028 146.3590
20.5 0.3208 0.0383 7.7336 3.8668 79.2689 158.5379 146.3590
21 0.3057 0.0374 7.9161 3.9580 83.1190 166.2380 146.3590
21.5 0.2917 0.0365 8.1121 4.0560 87.2049 174.4099 146.3590
22 0.2786 0.0358 8.3209 4.1605 91.5301 183.0603 146.3590
22.5 0.2663 0.0351 8.5421 4.2710 96.0982 192.1964 146.3590
23 0.2549 0.0345 8.7750 4.3875 100.9129 201.8258 146.3590
23.5 0.2441 0.0340 9.0194 4.5097 105.9781 211.9562 146.3590
24 0.2341 0.0335 9.2748 4.6374 111.2976 222.5953 146.3590
24.5 0.2246 0.0331 9.5409 4.7704 116.8756 233.7513 146.3590
25 0.2157 0.0327 9.8173 4.9086 122.7162 245.4323 146.3590
25.5 0.2074 0.0324 10.1038 5.0519 128.8234 257.6468 146.3590
30
26 0.1995 0.0320 10.4001 5.2001 135.2016 270.4033 146.3590
26.5 0.1920 0.0317 10.7061 5.3530 141.8552 283.7104 146.3590
27 0.1850 0.0315 11.0214 5.5107 148.7884 297.5768 146.3590
27.5 0.1783 0.0312 11.3459 5.6729 156.0057 312.0114 146.3590
28 0.1720 0.0310 11.6794 5.8397 163.5115 327.0230 146.3590
28.5 0.1660 0.0308 12.0218 6.0109 171.3104 342.6209 146.3590
29 0.1603 0.0306 12.3729 6.1864 179.4069 358.8139 146.3590
29.5 0.1549 0.0305 12.7326 6.3663 187.8056 375.6113 146.3590
30 0.1498 0.0303 13.1007 6.5504 196.5111 393.0222 146.3590
30.5 0.1449 0.0302 13.4772 6.7386 205.5280 411.0561 146.3590
31 0.1403 0.0300 13.8620 6.9310 214.8611 429.7221 146.3590
31.5 0.1359 0.0299 14.2549 7.1275 224.5149 449.0298 146.3590
32 0.1317 0.0298 14.6559 7.3279 234.4942 468.9884 146.3590
32.5 0.1277 0.0297 15.0648 7.5324 244.8038 489.6076 146.3590
33 0.1238 0.0296 15.4817 7.7409 255.4484 510.8967 146.3590
33.5 0.1201 0.0295 15.9064 7.9532 266.4327 532.8654 146.3590
34 0.1166 0.0294 16.3389 8.1695 277.7616 555.5232 146.3590
34.5 0.1133 0.0294 16.7791 8.3896 289.4399 578.8798 146.3590
35 0.1101 0.0293 17.2270 8.6135 301.4724 602.9447 146.3590
2.3.9 V-n Diagram
Figure 2.11 shows the flight envelope of the designed aircraft. Based on figure 2.11,
the flight envelope determines aerial platform operating limits for maximum speed and
load factor given a limited atmospheric density. The flight envelope is an area where
the aircraft can operate safely.
31
Figure 2.11 Flight Envelope
Positive Load factory
𝑛𝑚𝑎𝑥𝑝𝑜𝑠 = 2.1 +24000
𝑊 + 10000= 2.1 +
24000
7.2752 + 10000= 4.498
Manoeuvre speed (Positive Limit)
𝑉𝐴 = √ 𝑛𝑚𝑎𝑥 𝑝𝑜𝑠𝑊
0.5ρSCLmax
𝑉𝐴 = √(4.498)(7.2753)
0.5(0.002377)(4.2195)(1.3)= 70.8489 𝑓𝑡/𝑠
Design Dive Speed
𝑉𝐷 = 1.4𝑉𝐶 𝑚𝑖𝑛
where 𝑉𝐶 𝑚𝑖𝑛 is cruising speed
𝑉𝐷 = 1.4(63.3202) = 88.6483 𝑓𝑡/𝑠
Maximum Negative Load factor
𝑛𝑚𝑎𝑥𝑛𝑒𝑔 = 0.4 𝑛𝑚𝑎𝑥𝑝𝑜𝑠 = 0.4(4.498) = 1.7992
-2
-1
0
1
2
3
4
0 1 2 3 4 5 6 7 8 9
10
11
12
13
14
15
16
17
18
18.
5
19.
5
n
V
V-n diagram
n(+) n(-)
32
Manoeuvre speed (Negative Limit)
𝑉𝐺 = √ 𝑛𝑚𝑎𝑥 𝑛𝑒𝑔𝑊
0.5ρS(0.8)CLmax
𝑉𝐺 = √(1.7992 )(7.2753)
0.5(0.002377 )(4.2195)(0.8)(1.3)= 50.0978 𝑓𝑡/𝑠
For the curve part of the V-n diagram, the curve is drawn by varying airspeed
before manoeuvring speed with equation:
𝑛 =0.5ρ𝑉2𝑆CLmax
𝑊 (𝑓𝑜𝑟 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑉 − 𝑛)
For positive V-n region, airspeed varies from 0 to 88.6483 ft/s.
𝑛 =0.5ρ𝑉2𝑆(0.8CLmax)
𝑊 (𝑓𝑜𝑟 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑉 − 𝑛)
For negative V-n region, airspeed varies from 0 to 50.0978 ft/s.
33
CHAPTER 3
AVIONICS AND CONTROL
3.1 CG Locations
An airplane in flight can be maneuvered by the pilot using the aerodynamic control
surfaces; the elevator, rudder, or ailerons. As the control surfaces change the amount
of force that each surface generates, the aircraft rotates about a point called the center of
gravity. The center of gravity is the average location of the weight of the aircraft. The weight
is actually distributed throughout the airplane, and for some problems it is important to know
the distribution. But for total aircraft maneuvering, we need to be concerned with only the total
weight and the location of the center of gravity.
The center of gravity (CG) of an aircraft is the point over which the aircraft would
balance. The center of gravity affects the stability of the aircraft. To ensure the aircraft is safe
to fly, the center of gravity must fall within specified limits.
Center of gravity (CG) is calculated as follows:
Determine the weights and arms of all mass within the aircraft.
Multiply weights by arms for all mass to calculate moments.
Add the moments of all mass together.
Divide the total moment by the total mass of the aircraft to give an overall arm.
The arm that results from this calculation must be within the center of gravity limits
dictated by the aircraft manufacturer. If it is not, weight in the aircraft must be removed, added
(rarely), or redistributed until the center of gravity falls within the prescribed limits.
34
Aircraft center of gravity calculations are only performed along a single axis from the
zero point of the reference datum that represents the longitudinal axis of the aircraft (to
calculate fore-to-aft balance). The weight, moment and arm values of fixed items on the aircraft
do not change.
The plane is a combination of many parts; the wings, engines, fuselage, and tail, plus
the payload and the fuel. Each part has a weight associated with it which the engineer can
estimate, or calculate, using Newton's weight equation:
W = m * g
where W is the weight, m is the mass, and g is the gravitational constant which is 32.2
ft/square sec in English units and 9.8 meters/square sec in metric units. To determine the center
of gravity cg, we choose a reference location, or reference line. The cg is determined relative
to this reference location. The total weight of the aircraft is simply the sum of all the individua l
weights of the components.
Since the center of gravity is an average location of the weight, we can say that the
weight of the entire aircraft W times the location cg of the center of gravity is equal to the sum
of the weight w of each component times the distance d of that component from the reference
location:
𝑊 ∗ 𝑐𝑔 = [𝑊 ∗ 𝑑]𝑓𝑢𝑠𝑒𝑙𝑎𝑔𝑒 + [𝑊 ∗ 𝑑]𝑤𝑖𝑛𝑔 + [𝑊 ∗ 𝑑]𝑒𝑛𝑔𝑖𝑛𝑒𝑠 + ⋯
Therefore, the conclusion that the center of gravity is the mass-weighted average of the
component locations can be made.
We can generalize the technique discussed above. If we had a total of "n" discrete
components, the center of gravity cg of the aircraft times the weight W of the aircraft would be
the sum of the individual i component weight times the distance d from the reference line (w *
d) with the index i going from 1 to n. Mathematicians use the Greek letter sigma to denote this
addition. (Sigma is a zig-zag symbol with the index designation being placed below the bottom
35
bar, the total number of additions placed over the top bar, and the variable to be summed placed
to the right of the sigma with each component designated by the index.)
𝑊 ∗ 𝑐𝑔 = ∑[𝑊 ∗ 𝑑]𝑖
𝑛
𝑖=1
This equation says that the center of gravity times the sum of "n" parts' weight is equal
to the sum of "n" parts' weight times their distance. The discrete equation works for "n" discrete
parts. As the location of the centre of gravity affects the stability of the aircraft, it must fall
within specified limits that are established. Both lateral and longitudinal balance are important,
but the primary concern is longitudinal balance which is the location of the CG along the
longitudinal or lengthwise axis.
Four important centre of gravity locations are determined for the analysis of balancing
and stability as shown below,
1. Maximum weight CG
2. Empty weight CG
3. Aft CG
4. Forward CG
The location of centre of gravity of each components and the calcuations of the four
centre of gravity locations are shown.
36
Figure 3.1 CG for each components
3.1.1 Maximum weight CG
First and foremost, the weight and location of each component are estimated
and then the multiplication of the parameters are obtained as shown in table 3.1. After
that, the maximum weight CG is calculated by dividing the sum of calculated values
by the sum of weights of components. As a result, figure 3.2 shows the location of the
maximum weight CG of the aircraft.
37
Table 3.1 Calculations for maximum weight CG of aircraft
No Items Weight(N) Distance(m) Weight*Distance
1 Wing 3.9981 0.3900 1.5593
2 Vertical tail 0.1962 0.9800 0.1923
3 Horizontal tail 0.0481 0.9800 0.0471
4 Fuselage 4.5097 0.5700 2.5705
5 Front landing gear 1.2469 0.2000 0.2494
6 Back landing gear 0.5346 0.8000 0.4277
7 Engine 0.7063 0.1000 0.0706
8 Battery 5.3465 0.2500 1.3366
9 Servo(aileron) 0.1766 0.4700 0.0830
10 Servo(rudder) 0.0883 0.9600 0.0848
11 Servo(elevator) 0.0883 1.0700 0.0945
12 Receiver 0.0343 0.1500 0.0051
13 Payload 4.9050 0.4500 2.2073
14 Propeller 0.1736 0.0000 0.0000
15 ESC 0.4218 0.1500 0.0633
Total(max) 22.4743 8.9914
CG location in horizontal, XCG = Ʃ𝑊𝑥
Ʃ𝑊=
8.9914
22.4743
= 0.4 m
Figure 3.2 Maximum CG
38
3.1.2 Empty weight CG
First and foremost, the weight and location of each component that contributes
to empty weight of the aircraft are listed and then the multiplication of the parameters
are obtained as shown in table 3.2. After that, the empty weight CG is calculated by
dividing the sum of calculated values by the sum of weights of the components. As a
result, figure 3.3 shows the location of the empty weight CG of the aircraft.
Table 3.2 Calculations for empty weight CG of aircraft
No Items Weight(N) Distance(m) Weight*Distance
1 Wing 3.9981 0.3900 1.5593
2 Vertical tail 0.1962 0.9800 0.1923
3 Horizontal tail 0.0481 0.9800 0.0471
4 Fuselage 4.5097 0.5700 2.5705
5 Front landing gear 1.2469 0.2000 0.2494
6 Back landing gear 0.5346 0.8000 0.4277
7 Engine 0.7063 0.1000 0.0706
8 Battery 5.3465 0.2500 1.3366
9 Servo(aileron) 0.1766 0.4700 0.0830
10 Servo(rudder) 0.0883 0.9600 0.0848
11 Servo(elevator) 0.0883 1.0700 0.0945
12 Receiver 0.0343 0.1500 0.0051
13 Propeller 0.1736 0.0000 0.0000
14 ESC 0.4218 0.1500 0.0633
Total(max) 17.5693 6.7842
CG location in horizontal, XCG = Ʃ𝑊𝑥
Ʃ𝑊=
6.7842
17.5693
= 0.39 m
39
Figure 3.3 Empty Weight CG
3.1.3 Forward CG
First and foremost, the weight and location of each component that located in
front of empty weight CG of the aircraft are listed and then the multiplication of the
parameters are obtained as shown in table 3.3. After that, the forward CG is calculated
by dividing the sum of calculated values by the sum of weights of the components. As
a result, figure 3.4 shows the location of the forward CG of the aircraft.
Table 3.3 Calculations for forward CG of aircraft
No Items Weight(N) Distance(m) Weight*Distance
1 Wing 3.9981 0.3900 1.5593
2 Vertical tail 0.1962 0.9800 0.1923
3 Horizontal tail 0.0481 0.9800 0.0471
4 Fuselage 4.5097 0.5700 2.5705
5 Front landing gear 1.2469 0.2000 0.2494
6 Back landing gear 0.5346 0.8000 0.4277
7 Engine 0.7063 0.1000 0.0706
8 Battery 5.3465 0.2500 1.3366
9 Servo(aileron) 0.1766 0.4700 0.0830
10 Servo(rudder) 0.0883 0.9600 0.0848
11 Servo(elevator) 0.0883 1.0700 0.0945
12 Receiver 0.0343 0.1500 0.0051
13 Payload 4.9050 0.4500 2.2073
14 Propeller 0.1736 0.0000 0.0000
15 ESC 0.4218 0.1500 0.0633
Total(max) 22.4743 8.9914
40
CG location in horizontal, XCG = Ʃ𝑊𝑥
Ʃ𝑊=
8.9914
22.4743
= 0.4 m
Figure 3.4 Forward CG
3.1.4 Aft CG
First and foremost, the weight and location of each component that behind
empty weight CG of the aircraft are listed and then the multiplication of the parameters
are obtained as shown in table 3.4. After that, the aft CG is calculated by dividing the
sum of calculated values by the sum of weights of the components. As a result, figure
3.5 shows the location of the aft CG of the aircraft.
41
Table 3.4 Calculations for aft CG of aircraft
No Items Weight(N) Distance(m) Weight*Distance
1 Wing 3.9981 0.3900 1.5593
2 Vertical tail 0.1962 0.9800 0.1923
3 Horizontal tail 0.0481 0.9800 0.0471
4 Fuselage 4.5097 0.5700 2.5705
5 Front landing gear 1.2469 0.2000 0.2494
6 Back landing gear 0.5346 0.8000 0.4277
7 Engine 0.7063 0.1000 0.0706
8 Battery 5.3465 0.2500 1.3366
9 Servo(aileron) 0.1766 0.4700 0.0830
10 Servo(rudder) 0.0883 0.9600 0.0848
11 Servo(elevator) 0.0883 1.0700 0.0945
12 Receiver 0.0343 0.1500 0.0051
13 Propeller 0.1736 0.0000 0.0000
14 ESC 0.4218 0.1500 0.0633
Total(max) 17.5693 6.7842
CG location in horizontal, XCG = Ʃ𝑊𝑥
Ʃ𝑊=
6.7842
17.5693
= 0.39 m
Figure 3.5 Aft CG
42
3.2 Stability
3.2.1 Longitudinal Stability
3.2.1.1 Wing Section Lift-Curve Slope
The method used is basically a reference of Anon: Royal Aeronautical Society
Data sheets- Aerodynamics, Vol. II (Wings 01.01.06), 1955 according to DATCOM.
This method accounts for the development of the boundary layer for airfoils with
transition fixed at the leading edge and with maximum thickness less than
approximately 20%. The airfoil section lift curve slope at Mach numbers up to critical
Mach number is given by
𝑐𝑙α =1.05
β[
clα
(clα)theory](clα)theory
From appendix A, airfoil M15 with M < 0.15, then the value of 𝛽=0.99
[clα
(clα)theory] = empirical correction factor
From appendix B,
tan1
2∅′TE =
Y902
−Y99
29
From appendix A,
NACA M15 with M < 0.15, then the value of Y90
2 = 1.825 and
Y99
2 = 0.335
tan1
2∅′TE =
Y902
−Y99
29
= 1.825 − 0.335
9
= 0.16556
43
From appendix B,
tan1
2∅′𝑇𝐸 = 0.16556, and Reynolds number of 6x106, Thus,
[clα
(clα)theory] = 0.78
The chord length is 0.28m, and for NACA M15, the thickness is 12% of the
chord length. Thus, the thickness is
𝑡 =12
100𝑥0.28 = 0.0336𝑚
From appendix C, Figure 4.1.1.2 – 8b,
For the thickness ratio, 𝑡
𝑐=
0.0336
0.28= 0.12, the value of (𝑐𝑙𝛼)ℎ𝑒𝑜𝑟𝑦 = 6.88 𝑝𝑒𝑟 𝑟𝑎𝑑
After we put all the value into the formula, we get,
𝑐𝑙α =1.05
β[
clα
(clα)theory] (clα)theory
𝑐𝑙α = 1.05
0.99(0.78)(6.88)
= 5.6916 𝑝𝑒𝑟 𝑟𝑎𝑑
= 0.0993 per deg
3.2.1.2 Wing Lift-Curve Slope
The wing lift curve slope can be calculated by using two methods which are
using the formula as written in the DATCOM book or using the graph in figure 4.1.3.2-
49 in DATCOM book. The formula is as follows:
44
𝑐𝐿α
𝐴=
2𝜋
2 + √𝐴2β2
𝑘2 (1 +𝑡𝑎𝑛2Λ𝑐
2β2 ) + 4
The factor k is the ratio of the two-dimensional lift-curve slope (per radian) at
the appropriate Mach number to 2𝜋
𝛽 where,
𝑘 =𝑐𝑙𝛼(𝑝𝑒𝑟 𝑟𝑎𝑑)
2𝜋𝛽
= 5.6916
2𝜋0.99
= 0.8968
A= Aspect Ratio = 5
Based on the reference of DATCOM book, the value of sweep angle, Λ is zero
if the wing is a straight rectangular wing. Thus,
𝑐𝐿α
𝐴=
2𝜋
2 + √(52)(0.992 )
0.89682 (1 +𝑡𝑎𝑛200.992 ) + 4
= 0.7983 per rad
Therefore,
𝐶𝐿𝛼 = (𝐶𝐿𝛼
𝐴)(𝐴) = 0.7983𝑥5 = 3.9915 𝑝𝑒𝑟 𝑟𝑎𝑑 = 0.0697 𝑝𝑒𝑟 𝑑𝑒𝑔
3.2.1.3 Wing Pitching Moment
The low-speed zero- lift pitching moment for untwisted, constant section wings
with elliptical loading may be approximated by
45
(C𝑚𝑜)(θ=0) =𝐴𝑐𝑜𝑠2Λ𝑐/4
𝐴 + 2𝑐𝑜𝑠Λ𝑐/4C𝑚𝑜
where, C𝑚𝑜 = section pitching-moment coefficient at zero lift. From the airfoil
information,
C𝑚𝑜 = −0.072
Thus, the equation 3.6 can be simplified as
(C𝑚𝑜)(θ=0) =𝐴
𝐴 + 2C𝑚𝑜
= 5
5 + 2(−0.072)
= −0.0514
3.2.1.4 Wing-Body Lift-Curve Slope
According US DATCOM Section 4.3.1.2, the equation for wing-body lift-
curve slope based on the total projected wing area, including that intercepted by the
fuselage is
(𝐶𝐿𝛼)𝑊𝐵 = [𝐾𝑁 + 𝐾𝑊(𝐵) + 𝐾𝐵(𝑊) ](𝐶𝐿𝛼)𝑒
𝑆𝑒
𝑆𝑤
where,
𝐾𝑁 is the ratios of the nose lift
𝐾𝑊(𝐵) is the wing lift in the presence of the body
𝐾𝐵(𝑊) is the body lift in the presence of the wing respectively, to the wing –
alone lift
𝑆𝑒 is exposed wing area
𝑆𝑤 is total projected wing area
(𝐶𝐿𝛼)𝑒 is lift-curve slope of exposed wing based on exposed wing area and
exposed aspect ratio
46
Calculations:
𝑑
𝑏=
0.1
1.4= 0.0714
where,
𝑑 = fuselage diameter
𝑏 = wingspan
From appendix D, using the 𝑑
𝑏 =0.0714 value, we obtained the value of
𝐾𝑊(𝐵) = 1.07, 𝐾𝐵(𝑊) = 0.1
From previous calculation, (𝐶𝐿𝛼)𝑒 = 0.0697 per deg, 3.9915 per rad
𝑆𝑒
𝑆𝑤=
0.3354
0.392= 0.8556
where,
From Solidworks, 𝑆𝑤𝑒𝑡 = 0.684𝑚2
𝑆𝑤𝑒𝑡 = 𝑆𝑒[1.977 + 0.52 (𝑡
𝑐)] , 𝑓𝑜𝑟
𝑡
𝑐> 0.05
0.684 = 𝑆𝑒[1.977 + 0.52(0.12)]
𝑆𝑒 = 0.3354𝑚2
𝑆𝑤 = reference area = 0.392 𝑚2
𝐾𝑁 =(𝐶𝐿𝛼
)𝑁𝑆𝑁𝑟𝑒𝑓
(𝐶𝐿𝛼)𝑒𝑆𝑒
=(0.0349)(7.854 × 10−3)
(0.0697)(0.3354)
= 0.0117
where,
47
(𝐶𝐿𝛼)𝑁 is nose lift-curve slope, usually used 2 per rad = 0.0349 per deg
𝑆𝑁𝑟𝑒𝑓 is reference area for nose lift-curve slope, usually 𝜋𝑟2 = 𝜋(0.1
2)2 =
7.854 × 10−3𝑚2
(𝐶𝐿𝛼)𝑊𝐵 = [𝐾𝑁 + 𝐾𝑊(𝐵) + 𝐾𝐵(𝑊) ](𝐶𝐿𝛼)𝑒𝑆𝑒
𝑆𝑤
= [0.0117 + 1.07 + 0.1](3.9915)(0.8556)
= 4.0357 per rad = 0.0704 per deg
3.2.1.5 Downwash Gradient
(𝜕𝜖̅
𝜕𝛼) = 4.44[𝐾𝐴 𝐾𝜆 𝐾𝐻 (cos 𝛬𝑐/4)1/2]1.19
2ℎ𝐻
𝑏=
2(0)
1.4= 0
2𝑙𝐻
𝑏=
2(0.6525)
1.4= 0.9321
From appendix E,
𝐾𝐻 = 1.06
KA =1
𝐴−
1
1 + 𝐴1.7
=1
5−
1
1 + 51.7
= 0.1391
𝐾𝜆 =10 − 3𝜆
7
=10 − 3(0)
7
= 1.4286
Therefore,
(𝜕𝜖̅
𝜕𝛼) = 4.44 [0.1391(1.4286)(1.06)(cos0)
12]
1.19
= 0.6957 𝑝𝑒𝑟 𝑑𝑒𝑔
48
3.2.1.6 Tail Section Lift-Curve
𝑐𝑙α =1.05
β[
clα
(clα)theory](clα)theory
From appendix A, NACA 0012 with M < 0.15, then the value of 𝛽=0.99
[clα
(clα)theory] = empirical correction factor
From appendix B,
tan1
2∅′TE =
Y902
−Y99
29
From appendic A, NACA 0012 with M < 0.15, then the value of Y90
2 = 1.448
and Y99
2 = 0.260
tan1
2∅′TE =
Y902
−Y99
29
= 1.448 − 0.260
9
= 0.132
From appendix B,
tan1
2∅′𝑇𝐸 = 0.132, and Reynolds number of 3x106,
Thus,
[clα
(clα)theory] = 0.76
The chord length is 0.28m, and for NACA 0012, the thickness is 12% of the
chord length. Thus, the thickness is
𝑡 =12
100𝑥0.21 = 0.0252𝑚
49
From appendix C, for the thickness ratio, 𝑡
𝑐=
0.0252
0.21= 0.12, the value of
(𝑐𝑙𝛼)ℎ𝑒𝑜𝑟𝑦 = 6.88 𝑝𝑒𝑟 𝑟𝑎𝑑. After we put all the value into the formula, we get:
𝑐𝑙α =1.05
β[
clα
(clα)theory] (clα)theory
𝑐𝑙α = 1.05
0.99(0.76)(6.88)
= 5.5457 𝑝𝑒𝑟 𝑟𝑎𝑑
= 0.0968 per deg
3.2.1.7 Tail Lift Curve Slope
𝑐𝐿α
𝐴=
2𝜋
2 + √𝐴2β2
𝑘2 (1 +𝑡𝑎𝑛2Λ𝑐
2β2 ) + 4
The factor k is the ratio of the two-dimensional lift-curve slope (per radian) at
the appropriate Mach number to 2𝜋
𝛽 where,
𝑘 =𝑐𝑙𝛼(𝑝𝑒𝑟 𝑟𝑎𝑑)
2𝜋𝛽
= 5.5457
2𝜋0.99
= 0.8738
A= Aspect Ratio = 2.86
Based on the reference of DATCOM book, the value of sweep angle, Λ is zero
if the wing is a straight rectangular wing. Thus,
𝑐𝐿α
𝐴=
2𝜋
2 + √(2.862)(0.992)
0.87382 (1 +𝑡𝑎𝑛200.992 ) + 4
= 1.0818 per rad
50
Therefore,
𝐶𝐿𝛼 = (𝐶𝐿𝛼
𝐴)(𝐴) = 1.0818𝑥2.86 = 3.9039 𝑝𝑒𝑟 𝑟𝑎𝑑 = 0.054 𝑝𝑒𝑟 𝑑𝑒𝑔
3.2.1.8 Maximum Angle of Attack
For untwisted, constant section wings zero lift angle of attack based on
DATCOM book is:
(α0)(θ=0) = α𝑖 −𝑐𝑙𝑖
𝑐𝑙α
where,
𝑐𝑙𝑖 = section design lift coefficient
α𝑖 = angle of attack for design lift coefficient
𝑐𝑙α = section lift-curve slope
From appendix F,
𝑐𝑙𝑖 = 0.76𝑥4
6= 0.5067 𝑝𝑒𝑟 𝑑𝑒𝑔
α𝑖 = 0.74𝑥4
6= 0.4933 𝑑𝑒𝑔
And from section 2.2.1.1,
𝑐𝑙α = 0.0993 𝑝𝑒𝑟 𝑑𝑒𝑔
Therefore,
(α0)(θ=0) = 0.4933 −0.5067
0.105
= −4.3324 𝑑𝑒𝑔
51
From appendix G in DATCOM book,
For Λ𝐿𝐸=0 ,
𝐶𝐿𝑚𝑎𝑥
𝐶𝑙𝑚𝑎𝑥= 0.9
From airfoil information,
CLmax = 1.3308
From appendix H,
ΔCLmax =−0.1
𝑇𝑜𝑡𝑎𝑙 CLmax =1.3308−0.1
=1.2308
From appendix I,
For Λ𝐿𝐸 = 0 and M < 0.15,
ΔCLmax = 0
𝐶𝐿𝑚𝑎𝑥 = (𝐶𝐿𝑚𝑎𝑥
𝐶𝑙𝑚𝑎𝑥
)𝐶𝑙𝑚𝑎𝑥 + Δ𝐶𝐿𝑚𝑎𝑥
= (0.9)(1.2308) + 0
= 1.1077 𝑝𝑒𝑟 𝑑𝑒𝑔
From appendix J,
ΔαCLmax = 2.10
Therefore, maximum angle of attack is
52
𝛼𝐶𝐿𝑚𝑎𝑥 =𝐶𝐿𝑚𝑎𝑥
𝐶𝐿𝛼+ 𝛼0 + Δ𝛼𝐶𝐿𝑚𝑎𝑥
=1.1077
0.0697 + (−4.3324) + 2.1
= 13.66 deg
3.2.1.9 Calculation for Equation of Longitudinal Static Stability
Slope of moment coefficient curve
In Anderson’s book,
𝜕𝐶𝑀,𝑐𝑔
𝜕𝛼= 𝑎 [ℎ − ℎ𝑎𝑐𝑤𝑏
− 𝑉𝐻
𝑎𝑡
𝑎(1 −
𝜕𝜀
𝜕𝛼)]
where in our case,
𝑎 = 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.0704 𝑝𝑒𝑟 𝑑𝑒𝑔
ℎ = 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
ℎ𝑎𝑐𝑤𝑏= 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑎𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑒 = 0.39 𝑚
𝑉𝐻 = 𝑡𝑎𝑖𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑟𝑎𝑡𝑖𝑜 = 0.5
𝑎𝑡 = 𝑡𝑎𝑖𝑙 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.054 𝑝𝑒𝑟 𝑑𝑒𝑔
𝜕𝜀
𝜕𝛼= 0.6957
Therefore,
i. Empty Weight Centre of Gravity
ℎ = 0.39 𝑚
𝜕𝐶𝑀,𝑐𝑔
𝜕𝛼= 0.0704 [0.39 − 0.39 − 0.5
0.054
0.0704(1 − 0.6957)]
𝜕𝐶𝑀,𝑐𝑔
𝜕𝛼= −0.00822
53
ii. Gross Weight Centre of Gravity
ℎ = 0.4 𝑚
𝜕𝐶𝑀,𝑐𝑔
𝜕𝛼= 0.0704 [0.4 − 0.39 − 0.5
0.054
0.0704(1 − 0.6957)]
𝜕𝐶𝑀,𝑐𝑔
𝜕𝛼= −0.00751
iii. Forward Centre of Gravity
ℎ = 0.4 𝑚
𝜕𝐶𝑀,𝑐𝑔
𝜕𝛼= 0.0704 [0.4 − 0.39 − 0.5
0.054
0.0704(1 − 0.6957)]
𝜕𝐶𝑀,𝑐𝑔
𝜕𝛼= −0.00751
iv. Aft Centre of Gravity
ℎ = 0.39 𝑚
𝜕𝐶𝑀,𝑐𝑔
𝜕𝛼= 0.0704 [0.39 − 0.39 − 0.5
0.054
0.0704(1 − 0.6957)]
𝜕𝐶𝑀,𝑐𝑔
𝜕𝛼= −0.00822
Moment coefficient at zero angle of attack
𝐶𝑀,0 = 𝐶𝑀,𝑎𝑐𝑤𝑏+ 𝑉𝐻𝑎𝑡(𝑖𝑡 + 𝜀0)
where in our case,
𝐶𝑀,𝑎𝑐𝑤𝑏= 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑜𝑒𝑓𝑓𝑐𝑖𝑒𝑛𝑡 𝑎𝑡 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑎𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑒 = −0.0514
𝑉𝐻 = 𝑡𝑎𝑖𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑟𝑎𝑡𝑖𝑜 = 0.5
𝑎𝑡 = 𝑡𝑎𝑖𝑙 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.054 𝑝𝑒𝑟 𝑑𝑒𝑔
𝑖𝑡 = 𝑡𝑎𝑖𝑙 𝑠𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒 = 2.7
𝜀0 = 𝑑𝑜𝑤𝑛𝑤𝑎𝑠ℎ 𝑎𝑛𝑔𝑙𝑒 𝑎𝑡 𝑧𝑒𝑟𝑜 𝑙𝑖𝑓𝑡 = 0
Therefore,
54
𝐶𝑀,0 = −0.0514 + 0.5(0.054)(2.7 + 0)
𝐶𝑀,0 = 0.0215
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.
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,
i. Gross Weight Centre of Gravity
𝐶𝑀,𝑐𝑔 = −0.00751𝛼 + 0.0215
Table 3.5 Total pitching moment coefficient of gross weight CG with various angle
of attack
Angle () 𝑪𝑴,𝒄𝒈
0 0.0215
2 0.0065
4 -0.0085
6 -0.0236
8 -0.0386
10 -0.0536
12 -0.0686
14 -0.0837
16 -0.0987
55
The moment coefficient curve is displayed in Figure 3.6.
Figure 3.6 Moment coefficient versus angle of attack for gross weight CG
Equilibrium angle of attack is obtained from,
0 = −0.00751𝛼𝑒 + 0.0215
𝛼𝑒 = 2.8628° (trim angle)
ii. Empty Weight Centre of Gravity
𝐶𝑀,𝑐𝑔 = −0.00822𝛼 + 0.0215
Table 3.6 Total pitching moment coefficient of empty weight CG with various angle
of attack
Angle () 𝑪𝑴,𝒄𝒈
0 0.0215
2 0.0051
4 -0.0114
6 -0.0278
8 -0.0442
10 -0.0607
12 -0.0771
14 -0.0935
16 -0.1100
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0 2 4 6 8 10 12 14 16
Mo
men
t Co
effc
ien
t
Angle of Attack
Moment coefficient versus Angle of Attack for Gross Weight Centre of Gravity
56
The moment coefficient curve is displayed in Figure 3.7.
Figure 3.7 Moment coefficient versus angle of attack for empty weight CG
Equilibrium angle of attack is obtained from,
0 = −0.00822 𝛼𝑒 + 0.0215
𝛼𝑒 = 2.6156° (trim angle)
iii. Forward Centre of Gravity
𝐶𝑀,𝑐𝑔 = −0.00751𝛼 + 0.0215
Table 3.7 Total pitching moment coefficient of forward CG with various angle of
attack
Angle () 𝑪𝑴,𝒄𝒈
0 0.0215
2 0.0065
4 -0.0085
6 -0.0236
8 -0.0386
10 -0.0536
12 -0.0686
14 -0.0837
16 -0.0987
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0 2 4 6 8 10 12 14 16
Mo
men
t Co
effi
cien
t
Angle of Attack
Moment coefficient versus Angle of Attack for Empty Weight Centre of Gravity
57
The moment coefficient curve is displayed in Figure 3.8.
Figure 3.8 Moment coefficient versus angle of attack for forward CG
Equilibrium angle of attack is obtained from,
0 = −0.00751𝛼𝑒 + 0.0215
𝛼𝑒 = 2.8628° (trim angle)
iv. Aft Centre of Gravity
𝐶𝑀,𝑐𝑔 = −0.00822𝛼 + 0.0215
Table 3.8 Total pitching moment coefficient of aft CG with various angle of attack
Angle () 𝑪𝑴,𝒄𝒈
0 0.0215
2 0.0051
4 -0.0114
6 -0.0278
8 -0.0442
10 -0.0607
12 -0.0771
14 -0.0935
16 -0.1100
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0 2 4 6 8 10 12 14 16
Mo
men
t Co
effc
ien
t
Angle of Attack
Moment coefficient versus Angle of Attack for Forward Centre of Gravity
58
The moment coefficient curve is displayed in Figure 3.9.
Figure 3.9 Moment coefficient versus angle of attack for aft CG
Equilibrium angle of attack is obtained from,
0 = −0.00822 𝛼𝑒 + 0.0215
𝛼𝑒 = 2.6156° (trim angle)
Apparently, the angle of attack falls within the reasonable flight range, 𝛼𝑚𝑖𝑛 ≤
𝛼𝑒 ≤ 𝛼𝑚𝑎𝑥 (13.66˚). Therefore, the aircraft is longitudinally balanced as well as
statically stable.
Calculation for Neutral Point
ℎ𝑛 = ℎ𝑎𝑐,𝑤𝑏 + 𝑉𝐻
𝑎𝑡
𝑎(1 −
𝜕𝜀
𝜕𝛼)
where in our case,
ℎ𝑎𝑐𝑤𝑏= 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑎𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑒 = 0.39 𝑚
𝑉𝐻 = 𝑡𝑎𝑖𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑟𝑎𝑡𝑖𝑜 = 0.5
𝑎𝑡 = 𝑡𝑎𝑖𝑙 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.054 𝑝𝑒𝑟 𝑑𝑒𝑔
𝑎 = 𝑤𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 = 0.0704 𝑝𝑒𝑟 𝑑𝑒𝑔
𝜕𝜀
𝜕𝛼= 0.6957
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0 2 4 6 8 10 12 14 16
Mo
men
t Co
effi
cien
t
Angle of Attack
Moment coefficient versus Angle of Attack for Aft Centre of Gravity
59
Therefore,
ℎ𝑛 = 0.39 + 0.50.054
0.0704(1 − 0.6957)
ℎ𝑛 = 0.5067 𝑚
Calculation for Static Margin
𝑆𝑡𝑎𝑡𝑖𝑐 𝑀𝑎𝑟𝑔𝑖𝑛 = ℎ𝑛 − ℎ
Therefore,
i. Empty Weight Centre of Gravity
ℎ = 0.39 𝑚
𝑆𝑡𝑎𝑡𝑖𝑐 𝑀𝑎𝑟𝑔𝑖𝑛 = 0.5067 − 0.39
𝑆𝑡𝑎𝑡𝑖𝑐 𝑀𝑎𝑟𝑔𝑖𝑛 = 0.1167 𝑚
ii. Gross Weight Centre of Gravity
ℎ = 0.4 𝑚
𝑆𝑡𝑎𝑡𝑖𝑐 𝑀𝑎𝑟𝑔𝑖𝑛 = 0.5067 − 0.4
𝑆𝑡𝑎𝑡𝑖𝑐 𝑀𝑎𝑟𝑔𝑖𝑛 = 0.1067 𝑚
iii. Forward Centre of Gravity
ℎ = 0.4 𝑚
𝑆𝑡𝑎𝑡𝑖𝑐 𝑀𝑎𝑟𝑔𝑖𝑛 = 0.5067 − 0.4
𝑆𝑡𝑎𝑡𝑖𝑐 𝑀𝑎𝑟𝑔𝑖𝑛 = 0.1067 𝑚
iv. Aft Centre of Gravity
ℎ = 0.39 𝑚
𝑆𝑡𝑎𝑡𝑖𝑐 𝑀𝑎𝑟𝑔𝑖𝑛 = 0.5067 − 0.39
𝑆𝑡𝑎𝑡𝑖𝑐 𝑀𝑎𝑟𝑔𝑖𝑛 = 0.1167 𝑚
60
From the calculation of static margin, the value is positive in all centre of
gravity. Hence, the aircraft is statically stable .
3.2.2 Lateral Stability
Lateral stability is the stability about the airplane's longitudinal axis, which
extends form nose to tail. This helps to stabilize the lateral or rolling effect when one
wing gets lower than the wing on the opposite side of the airplane. There are four main
design factors which make an airplane stable laterally - dihedral, keel effect,
sweepback, and weight distribution.
The static lateral stability of an aircraft involves consideration of rolling
moments due to sideslip. If an aircraft has favorable rolling moment due to a sideslip,
a lateral displacement from wing level flight produces a sideslip, and the sideslip
creates a rolling moment tending to return the aircraft to wing level flight. By this
action, static lateral stability will be evident. Of course, a sideslip will produce yawing
moments depending on the nature of the static directional stability, but the
consideration of static lateral stability will involve only the relationship of rolling
moments and sideslip.
The axis system of an aircraft defines a positive rolling as a moment about the
longitudinal axis which tends to rotate the right wing down. As in other aerodynamic
considerations, it is convenient to consider rolling moments in the coefficient form so
that lateral stability can be evaluated independent of weight, altitude, speeds, etc.
The sideslip angle relates the displacement of the aircraft centre line from the
relative airflow. Sideslip angle is provided the symbol β (beta) and is positive when
the relative wind is displaced to the right of the aircraft centre line as shown in the
figure. The sideslip angle, β, is essentially the “directional angle of attack” of the
aircraft and is the primary reference in directional stability as well as lateral stability
considerations.
61
Figure 3.10 Positive sideslip angle
The principal surface contributing to the lateral stability of an aircraft is the
wing. The effect of geometric dihedral is a powerful contribution to lateral stability.
Dihedral angle is the angle between the plane of each wing and the horizontal when
the aircraft is level. Wing position also greatly effects the lateral static stability. A high
wing location gives a stable contribution. The direction of relative airflow increases
the effective angle of attack of the wing into wind and decreases the effective angle of
attack of the wing out of wind, tending to decrease the rolling moment. Therefore, a
high wing position usually requires no geometric dihedral. In our case, the RC aircraft
uses a high wing and there is no dihedral angle.
The rolling moment derivatives due to the side slip is given as:
𝑐𝑙𝛽 = (𝑐𝑙𝛽
𝜃)𝜃 + Δ𝑐𝑙𝛽
The wing tip shapes are shown in figure 3.11.
Figure 3.11 Shape of wing tips
62
Since the chosen Δ𝑐𝑙𝛽 =−0.0002
𝑟𝑎𝑑 and dihedral, 𝜃 = 0,
𝑐𝑙𝛽 = (𝑐𝑙𝛽
𝜃)𝜃 + Δ𝑐𝑙𝛽
𝑐𝑙𝛽 = 0 +−0.0002
𝑟𝑎𝑑
𝑐𝑙𝛽 = −0.0002/𝑟𝑎𝑑
Static lateral stability can be illustrated by a graph of rolling moment
coefficient, Cl, versus sideslip angle, β. Clβ is the slope of the graph.
The equation of the graph is as followed,
𝑦 = −0.0002𝑥
The graph is plotted based on the equation as shown in figure.
Figure 3.12 Graph of rolling moment versus sideslip angle
From figure 3.12, it can be said that the aircraft laterally stable. When the
aircraft is subjected to positive sideslip angle, lateral stability is evident when negative
rolling moment coefficient results. Thus, when the relative airflow comes from the
right (+β), a rolling moment to the left (-Cl) will be created which tends to roll the
airplane to the left. Lateral stability exist when the curve of Cl versus β has a negative
slope and the degree of stability will be a function of the slope of this curve.
63
3.3 Communication Signals (Transmitter-Receiver)
3.3.1 Transmitter
There are a few elements in a radio transmitter to generate radio waves that
consists of information for examples audio, video or digital data. First and foremost,
required electrical power for the operation of transmitter is supplied by battery. After
that, alternating current at transmitting frequency will be created by oscillator. The
carrier wave (generated wave) usually is a sine wave. Then, some useful information
is added to the carrier wave by modulator. This information adding process can be
done by two methods, which are amplitude modulation (AM) and frequency
modulation (FM). The amplitude of the carrier wave is slightly modified if AM is
applied whereas application of FM modifies the frequency of the carrier wave. The
power of the carrier wave is increased by amplifying the wave with amplifier. In short,
powerfulness of the broadcast depends on the power of amplitude. Last but not least,
the amplitude signal is converted to radio waves. Figure 3.13 shows the block diagram
of radio transmitter.
Figure 3.13 Block diagram of radio transmitter [4]
64
3.3.1.1 Specifications of Transmitter
Band Type and Frequency Range
Basically, there are two major band types, that is, Citizen Band (CB) and
Independent, Scientific, Medical (ISM) band. Some fixed frequencies that range from
27MHz to 75MHz are included in the CB band. However, the frequency band is
localized. In other words, the frequency band is different from countries. For example,
model aircraft usually has 72MHz frequency band with 50 channels (11-60) in USA
whereas model aircraft in most European countries has 35MHz frequency band with
36 channels (55-90) [5], [6].
Nowadays, most of the RC hobbyist use ISM 2.4GHz band. The frequency
range in this band is between 2.4000GHz to 2.4835GHz. The main advantage of using
this frequency band is frequency hopping spread spectrum (FHSS) is used. The system
allows the transmitted frequency to jump between whole range of different channels.
In this case, data transmission occurs when both of the transmitter and receiver hop to
a common random sequence of frequencies. As a result, one or more packets of data
are transmitted between each hop. Figure 3.14 shows spectrum graph analysis of some
common ISM systems.
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Figure 3.14 Spectrum graph analysis of some common ISM systems
Channels
Generally, channel is referring to each separate controllable function of an
airplane. A typical aircraft consists of 4 channels, which is, aileron, elevator, rudder
and throttle. However, a complex RC airplane may require more channels that control
flaps, retractable landing gear and so on. Usually, the channels are controlled by
switches as shown in figure 3.15. At the same time, table 3.9 shows two major channel
sequences of primary control of RC airplane for transmitter and receiver.
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Figure 3.15 Components of transmitter
Table 3.9 Major sequence of channels
No Sequence Protocol Channel Control
1 TAER Spektrum
1 Throttle (THR)
2 Aileron (AIL)
3 Elevator (ELE)
4 Rudder (RUD)
2 AETR Futaba
1 Aileron (AIL)
2 Elevator (ELE)
3 Throttle (THR)
4 Rudder (RUD)
Modes
Generally, there are four modes in the transmitter as shown in figure 3.16.
However, most of the transmitter are using modes 1 and 2. In our case, mode 2 is used
due to it is most commonly used so that the pilot can get used to the mode easily.
Antenna
Channel
switches
Display
Main
on/off
switch
Control
sticks
Trims
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Figure 3.16 Transmitter modes
3.3.1.2 Chosen Transmitter
In our case, FrSky Taranis Q X7 with 2.4GHz frequency transmitter is used.
Figure 3.17 displays the transmitter chosen.
Figure 3.17 FrSky Taranis Q X7 transmitter
Advantages and Disadvantages of Chosen Band Type
This transmitter has a frequency band of 2.4GHz. This frequency band is
chosen mainly due to the application of “frequency hopping” that replacing the “crystal”
technology which consisted of the MHz systems. As a result, the RC airplane can be
operated in more frequencies. At the same time, the frequency band produces a lower
interference. This is because the frequencies of any sorts of noise are automatically
separated by the radio. Besides, it provides a higher performance level as more
responsive control is offered as compared to 35MHz system. This is mainly due to its
high data rate. The 2.4GHz system consumes much lesser power consumption than
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35MHz system. For example, for the control range up to 250m, about 4MW of power
is consumed in 2.4GHz system whereas 35MHz system consumes 759MW of power.
Last but not least, it requires a smaller antenna. In other words, the length of antenna
required on the receiver is only 3cm.
However, there are some disadvantages of using the frequency band. For
example, it is more susceptible to bad installations such as weak batteries. Besides, it
is considered as a free-for-all band as various of other things such as Wi-Fi, wireless
video senders and other data-links are shared in the band. Using 2.4GHz system causes
a higher cost as compared to 35MHz system.
Channels
A total of 16 channels can be assigned with this transmitter. The locations of
switches A, B, C, D, F and H are shown in figure 3.18. There are 3 positions in switches
A, B, C and D whereas switches F and H possesses 2 positions per switch. Therefore,
the conclusion that this transmitter has enough channels for our flight testing.
Figure 3.18 Channel switches of transmitter
In our flight testing, only four channels are used, which are, throttle, aileron,
elevator and rudder, to complete the primary control of the airplane. The sequence of
the channels included is AETR, which is the Futaba protocol. In this case, the assigned
channels are depicted in table 3.10.
SA SB
SF
SD SC
SH
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Table 3.10 Assigned channels of transmitter
No Control Switch Position
1 Aileron SB
1
2 Elevator 2
3 Throttle SC
1
4 Rudder 2
Operation
The power required for operation of transmitter is supplied by battery. The
battery comes in a few options, which is NiMH and LiPo battery. The operation
specifications of the transmitter are stated in table 3.11.
Table 3.11 Operation specifications of transmitter
No Specifications Values
1 Operating voltage 6 to 15 V
2 Operating current 210 mA maximum
3 Output power 60 mW
From the table, the operating voltage indicates that 2 or 3 cells LiPo batteries
are accepted. This is because operating voltage of 2S and 3S LiPo batteries indicates
7.4V and 11.1V. At the same time, the operating current is essential when connecting
to external module such as XJT, DJT and R9M. Moreover, the output power indicates
the actual power of a radio frequency (RF) that produces by the transmitter.
Features
First and foremost, the transmitter has quad ball bearing gimbals instead of hall
sensors as control sticks. They provide a smooth feeling to the pilot when moving the
sticks. At the same time, it possesses the audio speech outputs. This means that the
values, alarms and setting such as changing mode can be heard by the pilot to ease the
flight testing process. Moreover, its full telemetry capability allows the pilot to access
to the real-time flight data easily such as battery voltage reading whereas Receiver
Signal Strength Indicator (RSSI) alert helps pilot to eliminate the problems of lost
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connection between transmitter and receiver by monitoring the reception quality of
signals continuously and producing alert before the quality becomes critical. The super
low latency in the transmitter ensures the data messages are transmitter with minimal
delay to minimize the probability of crashing the airplane. Furthermore, the transmitter
alerts the pilot if there is too much vibration as it may affect the stability of the airplane.
Last but not least, the OpenTx firmware that allows editing settings and running radio
simulations is installed in the transmitter.
TX Protocol
The system used in FrSky is Advanced Continuous Channel Shifting
Technology (ACCST). Basically, this system allows the frequency to be shifted
hundreds of times per second. As a result, signal conflicts and interruptions are not
involved in this system. In other words, the system has a robust frequency agility.
Figure 3.19 shows the spectrum analysis of the system. From the figure, around 50
different frequencies are switched between a pseudo-random basis.
Figure 3.19 Spectrum analysis of ACCST system
Generally, the transmitter is able to compatible with FrSky X series, D series
and V8-II series receiver. Besides, the compatible receivers may be increase if an
external module is used. At the same time, there are three RF modes that supported by
the transmitter. Table 3.12 shows the mode of RF of the transmitter. At the same time,
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the table states the receivers that are compatible and number of output channels. In this
case, the RF mode is set as D8 due to the receiver used is in V8-II series.
Table 3.12 RF modes of transmitter
No Mode Compatible Receivers Number of Output Channels
1 D8 V8-II series in D mode, D series 8
2 D16 X series Up to 16
3 LR12 L series 12
3.3.2 Receiver (RX)
On the other hand, radio receiver can be considered as opposite of a radio
transmitter. This is because antenna is used to capture radio waves in receiver.
Generally, it is only a length of wire. However, a very small AC current will be induced
in the antenna when the wire is exposed to radio waves. Then, the radio frequency (RF)
signal with very weak is amplified by RF amplifier to enable the processing of the
signal by a tuner. Basically, the tuner can be defined as a circuit that responsible to
process the signal by extracting signals of resonant frequency. In other words, any AC
signals at frequency other than resonant frequency will be blocked. After that, the
useful information is extracted from the carrier wave by detector. Last but not least, an
amplifier is used to amplify the weak signal so that it can be executed. Figure 12
depicts the block diagram of radio receiver.
Figure 3.20 Block diagram of radio receiver
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3.3.2.1 Chosen Receiver
In our case, FrSky V8R4-II 2.4Ghz 4CH receiver is used in the flight testing.
Figure 3.21 displays the receiver chosen.
Figure 3.21 FrSky V8R4-II 2.4Ghz 4CH receiver
This receiver is selected due to a few reasons. The main reason is the frequency
band of the transmitter and receiver must be the same, that is, 2.4 GHz in this case to
receive signals wirelessly from transmitter successfully.
Operation
The power required for operation of receiver is supplied by battery. The battery
used in the flight testing is Li-Po battery. The operation specifications of the receiver
are stated in table 9.
Table 3.13 Operation specifications of receiver
No Specifications Values
1 Operating voltage 3 to 16 V (HV version)
2 Operating current 30 mA
3 Specified Range >1km
Although the receiver can operate at a large range of voltage, the voltage
supplied to the receiver is only 5V. This is because the Battery Elimination Circuit
(BEC) in Electronic Speed Controller (ESC) reduces the battery voltage to 5V needed
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to prevent short circuit from receiver. At the same time, the operating current is
essential when calculating the endurance of the airplane. Moreover, the operating
range indicates the distance of the transmitted signal that can be received. In short, the
larger the range, the more effective the receiver.
RX Protocol
Generally, there are some universal RX protocols including Pulse Width
Modulation (PWM), Pulse Position Modulation (PPM) and Pulse Code Modulation
(PCM). On the other hand, some of the protocols are exclusive to certain brands. For
example, FrSky uses SBUS as RX protocols. The example of analog signal and its
corresponding PWM and PPM signals are depicted in figure 3.22.
Figure 3.22 Example of analogue signal and its corresponding PWM and PPM
signals
PWM can be defined as an analogue signal where the pulse length indicates
the servo output or throttle position. Usually, the length of the signal normal varies
between 1000s (minimum) to 2000s (maximum). However, this protocol is
becoming less popular due to its low cost. This is due to messy wiring especially when
more channels are required.
Therefore, PPM is introduced with only one signal wire is used for several
channels. In this case, the series of PWM signals are sent one after another on the same
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wire. As a result, the received signals may be not as accurate as the serial
communications. However, it is more popular and supported by many flight controllers.
Furthermore, PCM is a data type similar to PPM. However, it is a digital signal.
In other words, the value is between 0 and 1. The advantages of this protocol is it
enables the signal error detection and even error correction. Therefore, the conclusion
that PCM is more reliable can be mode. Nevertheless, addition conversion of the signal
is required and leads to a more expensive model.
During our flight testing, since we are using the FrSky communication system,
its exclusive communication protocol is required to be investigated as well. The serial
communication protocol used is Serial BUS (SBUS). In this case, up to 18 channels
are supported by using only one signal cable. Basically, it is the inversion of UART
communication signal. However, an inverter may be required as some of the flight
controller cannot accept the signal such as Naze32 Rev5.
Features
The receiver is able to compatible with al FrSky modules that is in V8_mode
and D_mode. Referring to the transmitter manual, the D8 mode allows it to compatible
with the receiver. Therefore, the conclusion that this pair of transmitter and receiver
can be bind. However, it allows only one-way communication. In other words, it can
only receive the transmitted signal but cannot send the information back to transmitter.
At the same time, it has a latency of 22ms. This value is considered small enough
because the flight speed is limited to 20 m/s in our flight testing. Therefore, there is
acceptable time gap for the pilot to recover the airplane if there is any incident happens.
Compatibility and Binding Mode
The transmitter manual states the D8 mode is used in the RF module. The
D_mode used indicates the telemetry information is involved. After that, the
procedures of putting the receiver into the binding mode is described. Firstly, make
sure that the Switch 1 and Switch 2 are OFF. After that, connect the Channels 1 and 2
signal pins of the receiver by the provided jumper. Then, the receiver is connected to
the battery directly. Note that the F/S button on receiver is no need to be hold.
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Binding Procedures
Before conducting the flight testing, the receiver must be bind with transmitter
to allow the information transmission. The binding procedures are stated as below:
1. The transmitter is turned on and switched to PPM mode.
2. The transmitter is turned off.
3. The transmitter is turned on while holding the F/S button on transmitter module.
4. The button is released. In this case, the flashing of RED LED on the transmitter
module accompanied with the beeper sound indicate the transmitter is ready to
bind to the receiver.
5. The receiver is put in the binding mode. Then, the flashing RED LED on
receiver indicates the binding process is completed.
6. Both of the transmitter and receiver are turned off. The jumper is disconnected.
7. The transmitted is turned on and the receiver is connected to battery. As a result,
the blinking GREEN LED on receiver indicates the received signal strength.
Operating Range
Before the flight testing, the pre-flight range check is done. This is because
nearby metal fences, concrete building or trees may reflect the signals. As a result, loss
of signal may occur both during range check and during the flight test. The checking
procedures are stated as below:
1. The model is placed at least 60cm (two feet) above non-metal contaminated
ground.
2. The receiver antennas are separated in the model. Note that they should not
touch the ground.
3. The antenna of the transmitter is placed in a vertical position.
4. The transmitter and the receiver are turned on.
5. The F/S button of the transmitter module is pressed for 4 seconds to enter range
check mode. As a result, the RED LED of the transmitter module is off and the
GREEN LED is flashed rapidly. The effective distance is decreased to 1/30 of
full range.
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6. The checker is walked away from the model. At the same time, the controls on
the transmitter are operated. This is to confirm that all of the controls operate
normally to at least 30 meters.
7. The F/S button of the transmitter module is pressed again to exit the range
check mode. As a result, the RED LED is on and indicates it is back to normal
operation.
3.3.3 Converting Signals from Transmitter to Receiver
The electromagnetic signals are generated by the pilot by adjusting the gimbals.
From that, the digital data is transmitted to the radio receiver. After that, the data is
interpreted by the radio receiver. The stack of data is then sent to the particular
destinations depending on the channels. The channels connecting with receiver are
shown in table 3.14 whereas the wiring diagram is portrayed in figure 3.23.
Table 3.14 Channels of receiver
Channel Flying Mode
Channel 1 Roll
Channel 2 Pitch
Channel 3 Throttle
Channel 4 Yaw
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Figure 3.23 Wiring Diagram of Flight Testing
For example, if the gimbal at the right hand side of transmitter is pushed
downwards, the digital data is received by the receiver and transferred to servo motor
connecting to the elevator through channel 2. As a result, the aircraft pitches
downwards. Besides, if the gimbal at the left hand side of transmitter is pushed
upwards, electromagnetic waves are transmitted to receiver and then transferred to
ESC that connects with brushless motor. Therefore, the speed of the motor can be
controller precisely.
3.3.4 Signal Interference
Since the signals are travelled wirelessly, there are some types of interference
that susceptible to the airplane. The levels of interference should be minimized as it
weakens the wireless signals. If the signals are weakened or completely prevented, the
airplane will loss of control and accidents or incidents might occur.
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3.3.4.1 Possible Signal Interferences
First and foremost, the most common interference source is physical objects
such as trees, buildings and other physical structures. The amount of RF signal that
can pass through the walls is determined by the density of the wall. For example, a
signal is difficult to pass through a concrete wall. As a result, the signal is weakened
or interfered.
At the same time, radio frequency interference my cause interferences to signal.
In this case, RF range of 2.4 GHz system is used. However, there are many other
devices used this channel such as microwaves, cordless phones and Wi-Fi devices. As
consequences, the devices may cause noise and weaken the signals transmitted.
Furthermore, electrical interference serves as one of the sources of signal
interference. This interference comes from devices such as computers, fans or any
other motorized devices. The interference level from this source is depending on the
proximity of the electrical device to the wireless access point. However, impact of this
interference type on wireless transmission is greatly reduced
Moreover, the integrity of signals is greatly depended on environmental factors.
For example, electrical interference could be caused by lightning. Besides, the signals
can be weakened as they passed through a fog.
3.3.4.2 Technology to Reduce Interferences
The signal interference is a big issue and an attention should be given in this
part. Therefore, there are a few technology is researched to reduce the interference.
Spread-Spectrum Technology
First and foremost, the most popular technology is spread-spectrum technology.
Originally, the data is travelled straight through a single RF band due to it is the
shortest distance between two points. This type of transmission is called narrowband
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transmission. On the other hand, spread spectrum technology requires the data signals
either changing the data pattern constantly or alternate between two carrier frequencies.
Therefore, the data is clearer and easier to be detected. Figure 3.24 shows two types of
spread-spectrum radio.
Figure 3.24 Types of spread-spectrum technology
In FHSS technology, the frequencies are changed in a predictable pattern by
using narrowband signals. The term frequency hopping is referred when the data
signals hopping between narrow channels following a predetermined cyclical pattern.
Therefore, this technology is strongly resistant to interference and environmental
factors.
On the other hand, the spreading of signal is over a full transmission frequency
spectrum with DSSS transmissions. The main characteristics of this technology is
redundant bit pattern is sent with sending of every bit of data in order to increase both
safety and reliability. In this case, the impact of interference and background noise can
be reduced. Although this technology provides greater better security and signal
delivery, it is a sensitive technology that may be affected by many environmental
factors.
Spread-Spectrum Technology
Frequency-Hopping Spread-Spectrum (FHSS) Technology
Direct-Sequence Spread-Spectrum (DSSS) Technology
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Orthogonal Frequency Division Multiplexing (OFDM)
This technique allows transferring of large amounts of data over 52 separate
and evenly spaced frequencies. In other words, the radio signal is split into these
frequencies and at the same time transmitted to the receiver. As a result, crosstalk
interference is greatly reduced by splitting the signal and transmitted over different
frequencies.
3.3.5 Communication resolution
Basically, communication resolution can be defined as the magnitude
difference between adjacent steps as shown in figure 3.25. Therefore, it is also called
quantization interval or quantum. At the same time, the resolution is equal to the
voltage of the minimum step size, which is equal to the voltage of the least significant
bit. By referring to the manufacturer’s datasheet, both of the transmitters and receivers
have the channel resolution of 3072 steps.
Figure 3.25 Communication resolution
From the figure, it is obvious that the received and transmitted signals are
distorted with decreasing in the communication resolution. On the other hand, a large
value of communication resolution reduces the dynamic range of the signal as shown
in figure 3.26. Nevertheless, a signal interference can be detected easily with the
resolution. As a result, the accidents or incidents can be eliminated.
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Figure 3.26 Relationship between resolution and dynamic range
3.3.6 Antenna
Generally, the process of converting oscillating electrical energy into
electromagnetic radiation is done by using an antenna. At the same time, antenna also
do the opposite work (conversion of electromagnetic radiation into oscillations of
electrical energy). The block diagram of a typical radio system is depicted in figure
3.27.
Figure 3.27 Block diagram of typical radio system [4]
From the figure, allocation of radio signal with a RF carrier wave along a
defined channel width. After that, the carrier wave is modulated and transmitted into
space by an antenna. These waves are propagating in space until meeting a receiving
antenna that receives them. Then, the waves are demodulated and converted into
oscillating electrical energy.
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3.3.6.1 Antenna Type
Basically, most of the radio control system uses rubber ducky antenna. This is
mainly due to it is compact and very robust in the same time. At the same time, figure
3.28 depicts the components in the antenna whereas table 3.15 shows the functions of
the components.
Figure 3.28 Components of rubber ducky antenna
Table 3.15 Functions of rubber ducky antenna components
No Components Function
1 Element Transmit the oscillating electrical signal to the air as
electromagnetic waves
2 Ground Plane Amplify the radio signal transmitted or received by the
element
3 Coaxial Transfer electrical signals without transmitting radio
signals
4 Structure Provide physical support for the antenna
5 Connector Connects the components to electrical signal source
(TX) or sink (RX)
This antenna consists of dipole antennas and are omni-directional. In other
words, the radio frequency (RF) energy can be propagated in 360 in the horizontal
plane. The radiation pattern when the antenna is oriented vertically is shown in the
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figure 3.29. The figure shows a radiation pattern resembling a donut shape that are
symmetrical 360 in the horizontal plane. At the same time, the vertical beamwidth of
approximately 75 is portrayed in the figure. Therefore, a 3D radiation pattern is
depicted in figure 3.30.
Figure 3.29 Radiation pattern of dipole antenna
Figure 3.30 3D radiation pattern of dipole antenna
3.3.6.2 Gains
Antenna gain serves as one of the performance measurement by considering
the combination of directivity and electrical efficiency of antenna. In general,
transmitting gain displays the efficiency of the antenna converting power input into
radio waves in a specific direction. On the other hand, the efficiency of converting
arriving radio waves into electrical power is described by receiving gain. Basically,
there are three units of the antenna gain as shown in table 3.16.
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Table 3.16 Units of gain
Unit Description
dB decibels
Eg: 2 dB indicates 2 times the energy relative to isotropic antenna in the
peak direction of radiation
dBi decibels relative to an isotropic antenna
Eg: 3 dBi shows twice the power relative to isotropic antenna in the peak
direction
𝐺𝑑𝐵𝑖= 10 log10 𝐺
dBd decibels relative to a dipole antenna
Eg: 7.85 dBd depicts the peak gain
𝐺𝑑𝐵𝑑= 𝐺𝑑𝐵𝑖
− 2.15𝑑𝐵
For both of the transmitter and receiver selected, there are a few instructions
given by the manufacturer to obtain the best signal. The procedures for receiver
antenna to obtain the best signal are as followed:
1. Keep the two antennas as straight as possible to prevent reduction of the
effective range
2. The two antennas should be placed perpendicular to each other.
3. Keep the antennas away from conductive materials by at least half inch.
4. Keep the antennas away from ESC, motor and other noise sources as far as
possible.
At the same time, the instructions for module antenna to obtain the best signal
are stated as below:
1. Avoid transmitter antenna from pointing directly to the model during the flight
as it will create a weak signal for receiver.
2. Keep the antenna perpendicular to the transmitter’s face. This is because a
better RF condition can be created for the receiver.
3. Never grip the antenna during flight due to degradation of the RF quality.
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CHAPTER 4
WING ANALYSIS
4.1 Introduction
Wings are the most important component in an aircraft because wings generate
most of the lift while moving rapidly through the air or some other fluid. The shape of
aircraft wings are known as airfoil shape. Wings come in different configurations ,
depending on the aircraft purposes. They are also attached to fuselage at different
angles and different position. In order for the wing to generate lift, it needs to be
positioned at a correct value of angle of attack.
Figure 4.1 Streamline on airfoil surface
As according to figure 4.1, the air streamline coming from the leading edge of
the airfoil will split into two parts, the upper streamline and the lower streamline. The
upper streamline will have higher velocity than the lower part therefore created a
region with a lower air pressure on top of the airfoil and vice versa. This phenomenon
caused the difference in air pressure between those two streamlines. The differences
can be measured directly by using suitable equipment or instrumentation or by using
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the basic physics principal, Bernoulli’s principle. In short, lift produced can be
determined from the different in pressure and velocity for the top surface of the airfoil
and its bottom surface.
Figure 4.2 Different types of monoplane
A fixed-wing aircraft can have different number of wings; monoplane, biplane,
quadruplane and multiplane. Monoplane is a one wing plane. It is also the simplest one
to be built. There are several wing positions based on Figure 4.2 that distinguish every
type of monoplane which are low wing, mid wing, shoulder wing, high wing and
parasol wing.
A low wing is the type of aircraft that has wings positioned on or almost at the
bottom of the fuselage. By locating the wing down low, it allows a better visibility and
releases the central fuselage from carrying wing spar. A mid wing is position midway
up the fuselage. The fuselage is now bound to carry the wing spar and hence reduce
the useful fuselage volume near its centre of gravity, where space is often in most
demand. A shoulder wing is a category that is located in between mid wing and high
wing. The configuration for this type is that the wing is situated close to the top surface
of the fuselage but not exactly on the top of the fuselage. A tall wing has its wing
mounted on top of the upper surface of the fuselage floor. This shares many benefits
and drawbacks with the shoulder wing, but the high wing has less visibility upwards
on a light aircraft. A parasol wing is the type of wing that is not being attached directly
to the fuselage but instead is being supported by cabane struts or pylon. This
configuration often needs additional bracing.
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4.2 Wing Configuration
As for the wing configuration, the type of fixed wing monoplane has been
chosen since it is the simplest configuration to be fabricated. In addition to that, a
monoplane is the most common configuration since it has the highest efficiency and
lowest drag. Apart from that, the high wing is being picked instead of low wing
because low wing aircraft is prone to incur more ground effect during landing than
high wing. Shown below in table 4.1 is the tabulated data for the detail specification
for our aircraft.
Table 4.1 Wing Configuration
Aircraft Type Fixed-wing Monoplane
Wing Type High Wing
Airfoil Type M15
Wingspan 1.4 m
Wing Area 0.392 m2
Wing Chord 0.28 m
Aspect Ratio 5
Coefficient of Lift Cruising – 0.7173
Clmax – 1.3308
According to Mohammad (2012), aileron takes about 5 to 10% of the wing area,
the ratio of aileron chord to wing chord is about 15 to 25%, the ratio of aileron span to
wing span is about 20 to 30% and the inboard aileron span is about 60 to 80%. Shown
in Table 4.2 is the dimension of aileron designed by this group.
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Table 4.2 Aileron Configuration
Half-Wing Aileron
0.196 m2 Area (5%) 0.0098 m2
0.28 m Chord (20%) 0.056 m
0.7 m Span (25%) 0.175 m
Inboard Span (70%) 0.49 m
Figure 4.3 Full dimension of half wing with aileron
Figure 4.3 illustrated the dimension of half wing for this aircraft. The tabulated
dimension can be referred in Table 4.1 and 4.2
4.3 Shear and Bending Stresses
A structure has the purpose of carrying loads that it was design for. In order to
do so, it needs to be able to transfer the load from one point to another. For an instance,
the load Is being transmitted from the location of the load to the supports. This is done
by developing a system called internal force system and the distribution of these
internal forces must be identified before the stress distributions and displacements can
be calculated respectively. In structural design, knowledge about stress is very
important since any members in the structure must be able to withstand the maximum
amount of load exerted on its cross sectional area so that will not cause breakdown in
the crystalline structure of the material or simply said, a structural failure. On top of
that, strains and displacements are also needed to be calculated in order to make sure
that the member in the structural system retains adequate rigidity or stiffness as well
as strength to avoid distortions that can affect the surrounding of the structure.
89
Now that the dimension of the aileron has been determined, the next step is to
identify the strength of the wing with and without ailerons so that the material can be
chose properly. The stresses can be obtained by drawing the shear force and bending
moment diagram for the mentioned structures at cruising speed with different load
factors which are g = -1.5, 1, and 3. Since the analysis is taken for cruising state
therefore the lift is assumed to be equal to the weight of the aircraft.
Figure 4.4 Front view of the aircraft
The analysis is done on half wing since they are symmetrical as can be seen
from Figure 4.4. Now the free body diagram can be illustrated as a cantilever beam
and therefore the weight of the wing and the fuselage, the wingspan and the width of
the fuselage need to be divided into half. Only then the weight distribution can be
calculated.
𝑊𝑖𝑛𝑔: 1
2⁄ 𝑊𝑤
12⁄ 𝑏𝑤
=1.99905 𝑁
0.7 𝑚= 2.8558 𝑁/𝑚
𝐹𝑢𝑠𝑒𝑙𝑎𝑔𝑒: 1
2⁄ 𝑊𝑓
12⁄ 𝑏𝑓
=2.2549 𝑁
0.05 𝑚= 45.098 𝑁/𝑚
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4.3.1 Wing without ailerons
A. CASE 1: -1.5g
𝐿𝑖𝑓𝑡 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 =−1.5(3.3 𝑘𝑔 × 9.81
𝑚𝑠2)
1.4 𝑚= −34.6854 𝑁/𝑚
Figure 4.5 Free body diagram for g = -1.5 without aileron
Figure 4.6 Shear force and bending moment diagram for g = -1.5 without aileron
B. CASE 2: 1g
𝐿𝑖𝑓𝑡 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 =1(3.3 𝑘𝑔 × 9.81
𝑚𝑠2)
1.4 𝑚= 23.1236 𝑁/𝑚
91
Figure 4.7 Free body diagram for g = 1 without aileron
Figure 4.8 Shear force and bending moment diagram for g = 1 without aileron
C. CASE 3: 3g
𝐿𝑖𝑓𝑡 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 =3(3.3 𝑘𝑔 × 9.81
𝑚𝑠2)
1.4 𝑚= 69.3707 𝑁/𝑚
Figure 4.9 Free body diagram for g = 3 without aileron
92
Figure 4.10 Shear force and bending moment diagram for g = 3 without aileron
In designing aircraft, there must be a limit set to the flying qualities that
depends on the capability of the structure to withstand certain loads without
experiencing damage or failure. This limit is normally calculated by first identif ying
the load factors. As for this case, the minimum load factor is -1.5 while the maximum
is 3. Shear force and bending moment diagram has been plotted for each case to
determine the maximum stress the structure can withstand before failing.
For the first case, g = -1.5, as refer to Figure 4.5 and 4.6, the maximum shear
force exerted on the structure is at the central fuselage. The force is then decrease
substantially until reaching 24.4018 N at the wing-fuselage attachment part before
becoming zero at the wing tip. The same trend can be seen for the bending moment
where maximum bending moment occurs at the central fuselage before increasing
slowly until it finally reaches zero at wing tip. As for the third case, g = 3, by referring
to Figure 4.9 and 4.10, the central fuselage also becomes the most crucial part as it
needs to withstand the highest amount of shear force and bending moment before
finally reaching zero. The reason why central fuselage needs to withstand the largest
amount of force and moment is because it acts as support system to the cantilever beam.
However, since the analysis done in this part is for the wing, therefore our main
concern is at the wing-fuselage attachment part. This is because, this part needs to be
able to support whole weight of the wing together with the lift created without
93
damaging the wing structure. The shear and bending stresses that have been calculated
earlier for their respective load factor at the wing-fuselage attachment part are
tabulated as shown in Table 4.3 below:
Table 4.3 Shear and bending stresses for load factor -1.5g, 1g and 3g without aileron
Load Factor Shear Stress, τ (kPa) Bending Stress, σ(kPa)
-1.5 4.304 -310.817
1 -2.3234 167.801
3 -7.038 469.237
From the values obtained from the theoretical analysis, it can be seen that the
structure is supposed to be able to withstand shear stress in the range of −7.038 kPa ≤
τ ≤ 4.304 kPa while the bending stress is in between the range of −310.817 kPa ≤
σ ≤ 469.237 kPa. Therefore, in order to ensure a safe flight, the aircraft must be
operated within this range only. Note that this analysis is only applicable for the wing
without ailerons.
4.3.2 Wing with ailerons
The first thing to do for analysis of wing with ailerons is to find the lift
generated by the aileron. The lift can be calculated by using the usual lift formula, 𝐿 =
1
2𝜌𝑣2𝑆𝐶𝐿. Since this analysis is done during the cruising state, therefore CL during
cruising needs to be determined by finding out its respective angle of attack In this
case, the angle of attack during cruising is 𝛼 = 2.8 so 𝐶𝐿 ≈ 0.7173
𝐿𝑎𝑖𝑙𝑒𝑟𝑜𝑛 =1
2× 1.225 × 19.32 × 0.0098 × 0.7173 = 1.6038 𝑁
The lift distribution on aileron is then can be also calculated since the lift
generated by aileron is known.
𝐿𝑎𝑖𝑙𝑒𝑟𝑜𝑛
𝑏𝑎𝑖𝑙𝑒𝑟𝑜𝑛=
1.6038 𝑁
0.175 𝑚= 9.1645 𝑁/𝑚
94
A. CASE 1: -1.5g
Figure 4.11 Free body diagram for g = -1.5 with aileron
Figure 4.12 Shear force and bending moment diagram for g = -1.5 with aileron
B. CASE 2: 1g
Figure 4.13 Free body diagram for g = 1 with aileron
95
Figure 4.14 Shear force and bending moment diagram for g = 1 with aileron
C. CASE 3: 3g
Figure 4.15 Free body diagram for g = 3 with aileron
Figure 4.16 Shear force and bending moment diagram for g = 3 with aileron
96
The trend as for the wings with aileron seems to be similar to without aileron
but with different values of stresses. For an instance, by looking at Figure 4.11 and
4.12, the maximum shear stress is 4.304kPa while maximum bending stress is 277.666
kPa. This data is then compared with the second case which is referring to Figure 4.13
and 4.14. from here we can see that the maximum shear stress should be exerted on
the wing is 2.6064 kPa while bending stress is 200.9617 kPa. Finally for the third case,
as illustrated in Figure 4.15 and 4.16, the wing should withstand the highest stresses
which are 7.908 kPa for shear stress and 582.526 kPa for bending stress. Shown in
table 4.4 below are the data collected between without aileron and with aileron.
Table 4.4 Comparison of shear and bending stresses for load factor -1.5g, 1g and 3g
Load Factor
Shear Stress, τ (kPa) Bending Stress, σ(kPa)
Without
aileron
With aileron Without
aileron
With aileron
-1.5 4.0208 4.304 -310.817 -277.666
1 -2.3234 -2.6064 167.801 200.9617
3 -7.038 -7.908 469.237 582.526
We can see here from Table 4.4 that wing with aileron should be able to
withstand a slightly more stresses than the one without aileron or in simpler word,
wing with aileron is stronger. This is because aileron just like flaps, when we lower
the aileron, it means that the chord line of the wing is also changed. This results in
producing a higher angle of attack. Therefore, as the angle of attack increases, the lift
is also increases. The increase in lift will simultaneously cause the increase in induced
drag. In short, the wing with aileron should be stronger than without aileron is because
it needs to withstand the addition lift and drag generated by the ailerons.
97
4.4 Lift Distribution of Aileron
The lift distribution of aileron is calculated at different angle of attack and its
respective velocity is identified afterwards The lift is calculated by using formula:
L = N cos 𝛼 − A sinα
where
𝑁 = − ∫ (𝑃𝑢𝑝𝑝𝑒𝑟
𝑇𝐸
𝐿𝐸𝑐𝑜𝑠𝜃 + 𝜏𝑢𝑝𝑝𝑒𝑟 𝑠𝑖𝑛𝜃)𝑑𝑆𝑢𝑝𝑝𝑒𝑟 + ∫ (𝑃𝑙𝑜𝑤𝑒𝑟𝑐𝑜𝑠𝜃 − 𝜏𝑙𝑜𝑤𝑒𝑟 𝑠𝑖𝑛𝜃)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
𝑇𝐸
𝐿𝐸
𝐴 = ∫ (−𝑃𝑢𝑝𝑝𝑒𝑟
𝑇𝐸
𝐿𝐸𝑠𝑖𝑛𝜃 + 𝜏𝑢𝑝𝑝𝑒𝑟 𝑐𝑜𝑠𝜃)𝑑𝑆𝑢𝑝𝑝𝑒𝑟 + ∫ (𝑃𝑙𝑜𝑤𝑒𝑟𝑠𝑖𝑛𝜃 + 𝜏𝑙𝑜𝑤𝑒𝑟𝑐𝑜𝑠𝜃)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
𝑇𝐸
𝐿𝐸
The geometrical relations for the integration of pressure and shear stress
distributions over a two-dimensional body surface of aileron is shown in Figure 4.17.
Figure 4.17 2D body surface of aileron
To obtain the normal force, the values of pressure, shear stress and angle are
obtained as follows:
Angle, θ = tan−13 × 10−3
0.056= 3.066°
98
Based on the elevation of our aircraft (10m) the following parameters are
identified
Pressure , plower and pupper = 1.01 × 105 Pa
Temperature , T = 288.09K
Density , ρ = 1.225kg /𝑚3
Dynamic viscocity, μ = 1.78869 × 10−5𝑁𝑠/𝑚2
Freestream velocity,𝑈∞ =0.2
√1.4×287×288.15= 68.059m/s
τupper surface = τlower surface = μdu
dy
= (1.78869 × 10−5) (𝑈∞
0.025𝑐)
= (1.78869 × 10−5) (68.059
0.025(0.056))
= 0.8695Pa
N = − ∫ ((1.01 × 105)cos3.066° + (0.8695)sin3.066°)𝑑𝑆𝑢𝑝𝑝𝑒𝑟
0.056
0
+ ∫ ((1.01 × 105)cos3.066° − (0.8695)sin3.066°)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
0.056
0
= -0.0065216 N
A = ∫ ((−1.01× 105)sin3.066° + (0.8695)cos3.066°)𝑑𝑆𝑢𝑝𝑝𝑒𝑟
0.056
0
+ ∫ ((1.01 × 105)sin3.066° + (0.86951)cos3.066°)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
0.056
0
=0.097216 𝑁
Taking α = -12° as sample,
L = (−0.0065216) cos(−12°) − (0.097216) sin(−12°)
= 0.013833 N
99
The lift coefficient can be obtained based on the classical thin airfoil theory:
the symmetrical airfoil. To do so, the following formulae is used to determine
theoretically which is as follows:
𝐶𝐿 = 2𝜋𝛼
Upon obtaining value of lift coefficient, the respective velocity can be
calculated afterwards by using equation:
𝑉 = √2(𝐿)
(𝜌)(𝑆)(𝐶𝐿)
Table 4.5 Calculated data at different angle of attack for aileron
Angle of Attack, α
(°) Lift Force, L (N) Coefficient of Lift, CL Velocity, V (m/s)
-12 0.0138 -1.3160 1.3234
-10 0.0105 -1.0966 1.2605
-8 0.0071 -0.8773 1.1588
-6 0.0037 -0.6580 0.9648
-4 0.0003 -0.4387 0.3236
-2 -0.0031 -0.2193 1.5407
0 -0.0065 0.0000 0.0000
2 -0.0099 0.2193 2.7437
4 -0.0133 0.4386 2.2464
6 -0.0167 0.6580 2.0531
8 -0.0200 0.8773 1.9483
10 -0.0233 1.0966 1.8816
12 -0.0266 1.3159 1.8348
Based on Table 4.5, it can be seen that the coefficient of lift is negative as the
angle of attack is negative. However, as the angle of attack increases, the coefficient
of lift also increases. This trend differs from the velocity trend. The highest velocity is
when the aileron is deflected at 2° with 2.7437 m/s while the lowest velocity is when
the aileron is not being deflected.
100
Figure 4.18 Graph of Lift against Angle of Attack
Now as refer to Figure 4.18, it can be seen that the coefficient of the lift and
angle of attack has a linear relationship as mentioned previously, the lift increases as
the angle of attack increases.
4.5 Wing Loading
Wing loading is the total mass of an aircraft divided by the area of the wing.
The level flight of an aircraft straight stalling speed is determined by wing loading. An
aircraft with larger wing area relative to mass have low wing loading compared to an
aircraft with smaller wing area. The faster an aircraft fly, the more lift produced by the
wing area. Therefore, the faster aircraft usually obtained higher wing loading
compared to the slower aircraft. The higher wing loading also increases takeoff landing
distances and decreases the maneuverability. According to Peery, Schrenk’s
approximation assumes that the load distribution on an untwisted wing or tail has a
shape that is the average of the actual platform shape and an elliptic shape of the same
span and area. It is simple approximation for the spanwise lift distribution that has
been accepted by the Civil Aeronautics Administration (CAA) as satisfactory for civil
airplanes.
-0.02
-0.015
-0.01
-0.005
0
0.005
0.01
0.015
0.02
0.025
0.03
-15 -10 -5 0 5 10 15
Lift against Angle of Attack
101
4.5.1 Schrenk’s Approximation Method
The wing specification is shown in table 4.6.
Table 4.6 Wing specification
Parameters Value
Wing section 18
Wingspan, b 1.4 m
Wing Area (S) 0.392 𝑚2
Aspect Ratio, AR 5
Chord Length 0.28m
Taper Ratio 1
Cl max 2D 1.3
Wing area;
𝑆 = 𝑡𝑖𝑝 𝑐ℎ𝑜𝑟𝑑 × 𝑏 = 0.392 𝑚2
Aspect ratio;
𝐴𝑅 =𝑏2
𝑆= 5
Chord length;
𝑐 =𝑏
𝐴𝑅= 0.28𝑚
Taper ratio;
𝜆 =𝑏
𝑆= 0.48
Limit load factor;
𝑙𝑖𝑚𝑖𝑡 𝑙𝑜𝑎𝑑 𝑓𝑎𝑐𝑡𝑜𝑟 =𝑙𝑖𝑚𝑖𝑡 𝑙𝑜𝑎𝑑
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑎𝑖𝑟𝑐𝑟𝑎𝑓𝑡=
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑡𝑎𝑘𝑒𝑜𝑓𝑓 𝑤𝑒𝑖𝑔ℎ𝑡
𝑒𝑚𝑝𝑡𝑦 𝑤𝑒𝑖𝑔ℎ𝑡
= 1.179
102
Table 4.7 Wing Loading Data
Station y (m) 2y/b √(1-(2y/b)^2) 4S/πb*√(1-(2y/b)^2)
1 0.0 0 1.0000 0.356
2 0.04 0.1 0.9983 0.356
3 0.08 0.1 0.9930 0.354
4 0.12 0.2 0.9843 0.351
5 0.16 0.2 0.9719 0.346
6 0.21 0.3 0.9557 0.341
7 0.25 0.4 0.9356 0.333
8 0.29 0.4 0.9112 0.325
9 0.33 0.5 0.8822 0.314
10 0.37 0.5 0.8482 0.302
11 0.41 0.6 0.8084 0.288
12 0.45 0.6 0.7621 0.272
13 0.49 0.7 0.7079 0.252
14 0.54 0.8 0.6439 0.230
15 0.58 0.8 0.5666 0.202
16 0.62 0.9 0.4696 0.167
17 0.66 0.9 0.3364 0.120
18 0.70 1.0 0.0000 0.000
103
Table 4.8 Wing Loading Result
n Wo Station y (m) c (m) cCl Cl cClΔy ΔV
1.179 3.3 1
0.0000 0.28 0.318 1.137 0.0131 2.60
1.179 3.3 2
0.0412 0.28 0.318 1.135 0.0131 2.59
1.179 3.3 3
0.0824 0.28 0.317 1.132 0.0130 2.58
1.179 3.3 4
0.1236 0.28 0.315 1.127 0.0130 2.57
1.179 3.3 5
0.1648 0.28 0.313 1.119 0.0128 2.54
1.179 3.3 6
0.2060 0.28 0.310 1.108 0.0127 2.52
1.179 3.3 7
0.2472 0.28 0.307 1.096 0.0125 2.49
1.179 3.3 8
0.2884 0.28 0.302 1.080 0.0124 2.45
1.179 3.3 9
0.3296 0.28 0.297 1.062 0.0121 2.40
1.179 3.3 10
0.3708 0.28 0.291 1.040 0.0119 2.35
1.179 3.3 11
0.4120 0.28 0.284 1.015 0.0115 2.28
1.179 3.3 12
0.4532 0.28 0.276 0.985 0.0112 2.21
1.179 3.3 13
0.4944 0.28 0.266 0.951 0.0107 2.13
1.179 3.3 14 0.5356 0.28 0.255 0.910 0.0102 2.02
1.179 3.3 15
0.5768 0.28 0.241 0.861 0.0096 1.90
1.179 3.3 16
0.6180 0.28 0.224 0.799 0.0087 1.73
1.179 3.3 17
0.6592 0.28 0.200 0.714 0.0041 0.82
1.179 3.3 18
0.7004 0.28 0.000 0.000 0.0000 0.00
104
Table 4.9 Calculation Result
Station y (m) V ΔM M (N.m) Ln(N)
1 0.0000 38.16 1.52 12.12 63.03
2 0.0412 35.56 1.41 10.60 62.97
3 0.0824 32.97 1.31 9.19 62.79
4 0.1236 30.39 1.20 7.88 62.48
5 0.1648 27.83 1.09 6.68 62.04
6 0.2060 25.28 0.99 5.59 61.47
7 0.2472 22.76 0.89 4.60 60.76
8 0.2884 20.28 0.79 3.71 59.90
9 0.3296 17.83 0.69 2.93 58.87
10 0.3708 15.43 0.59 2.24 57.67
11 0.4120 13.08 0.49 1.66 56.27
12 0.4532 10.80 0.40 1.16 54.63
13 0.4944 8.59 0.31 0.76 52.72
14 0.5356 6.46 0.22 0.45 50.46
15 0.5768 4.44 0.14 0.23 47.73
16 0.6180 2.54 0.07 0.09 44.31
17 0.6592 0.82 0.02 0.02 39.61
18 0.7004 0.00 0.00 0.00 0.00
105
Figure 4.19 Graph Cl vs y
Figure 4.20 Graph cCl vs y
Figure 4.21 Graph Cy vs y
0.000
0.200
0.400
0.600
0.800
1.000
1.200
0.0000 0.2000 0.4000 0.6000 0.8000
Cl
y (m)
0.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.0000 0.2000 0.4000 0.6000 0.8000
cC
L
y(m)
0.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.400
0.0 0.2 0.4 0.6 0.8
Cy
y(m)
106
Figure 4.22 Graph Ln vs y
Figure 4.23 Graph V vs y
Figure 4.24 Graph M vs y
For the wing loading, the Schrenk’s approximation method has been used to
calculate the wing loading for this aircraft. In this cases, the calculation is done on the
0.00
10.00
20.00
30.00
40.00
50.00
60.00
70.00
0.0000 0.2000 0.4000 0.6000 0.8000
Ln
y (m)
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
0.0000 0.2000 0.4000 0.6000 0.8000
Sh
ear F
orc
e, (
N)
y(m)
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
0.0000 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000
Ben
din
g M
om
ent,
N (m
)
y(m)
107
half of the wing. The wing is divided into 18 station. There are several important data
to be used in this analysis which are wing span (b), wing area (S), aspect ratio (AR),
chord length (c), taper ratio, and maximum lift coefficient (Cl max).
Based on data obtained, there are several graph has been plotted to observe the
relationship between the lift coefficient, lift forces, shear forces and bending moment
versus y station of the wing. Figure 4.17 shows the graph of lift coefficient versus y
station of the wing. Based on the graph, the lift coefficient slightly decreases in each
station and fall to zero lift coefficient on tip chord. Similar with the graph cCl vs y, Cy
vs y, Ln vs y graph. The maximum lift coefficient produces at the root chord which is
1.137. For the wing load distribution coefficient, cCl, the maximum value also
obtained at the root chord of the wing which is 0.318. logically, the lift force also
produced the similar type of graph with the lift coefficient. The higher lift force is also
obtained at the root chord of the wing which is 63.03N.
Figure 4.21 shows the graph of shear force versus y station. Based on the result,
the graph shows the decreasing linear type of graph. From that, we can observe that
the shear force decreasing toward the tip chord of the wing. The higher shear force
produces also produce at the root chord which is 38.16 N. For the bending moment
versus station y graph, the bending moment also decreasing when it reach the tip chord
of the wing. The higher bending moment value is 12.12 Nm and zero at the tip chord.
108
4.5.2 Sample Calculation
a) Calculate wing load distribution coefficient, 𝑐𝐶𝐿 in order to determine the
spanwise shear forces values.
𝑐𝐶𝐿 =4𝑆√1 − (
2𝑦𝑏
)2
2𝜋𝑏+
𝑐
2
𝑐𝐶𝐿 =
4(0.392)√1 − (2(0)1.4
)2
2𝜋 (1.4)+
0.28
2
= 0.318
b) Calculate the local lift coefficient, 𝐶𝐿
𝐶𝐿 =𝑐𝐶𝐿
𝑐ℎ𝑜𝑟𝑑
𝐶𝐿 =0.318
0.28
= 1.135
c) Calculate the value of (𝑐𝐶𝐿∆𝑦)𝑛
(𝑐𝐶𝐿∆𝑦)𝑛 =𝑐𝐶𝐿𝑛 + 𝑐𝐶𝐿(𝑛+1)
2[𝑦𝑛+1 − 𝑦𝑛]
(𝑐𝐶𝐿∆𝑦)𝑛 =0.318 + 0.318
2[0.0412 − 0]
= 0.0131
d) The change in shear force distribution, ∆𝑉𝑛 on each section
∆𝑉𝑛 =(𝑐𝐶𝐿∆𝑦)𝑛 × 𝑛𝑔 × 𝑊𝑜 × 𝑔
Σ(𝑐𝐶𝐿∆𝑦)
∆𝑉𝑛 =0.0131 × 1.179 × 3.3 × 9.81
0.1927
= 2.60 𝑁
109
e) Compute the corresponding shear force, 𝑉𝑛 value at each station
𝑉𝑛 = Δ𝑉18 + Δ𝑉17 + ⋯ Δ𝑉𝑛
𝑉𝑛 = 0 + 0.82 + ⋯ 2.60
= 38.16 N
f) The changes in bending moment distribution, Δ𝑀𝑛 at each station
Δ𝑀𝑛 =𝑉𝑛+1 + 𝑉𝑛
2(𝑦𝑛+1 − 𝑦𝑛)
Δ𝑀𝑛 =35.56 + 38.16
2(0.0412 − 0)
= 1.52 Nm
g) Corresponding bending moment, 𝑀𝑛 at each station
𝑀𝑛 = Δ𝑀18 + Δ𝑀17 + ⋯ Δ𝑀𝑛
𝑀𝑛 = 0 + 0.02 + ⋯ 1.52
= 12.12 𝑁𝑚2
h) The local lifting force at specified station, 𝐿𝑛
𝐿𝑛 =(𝑐𝐶𝐿)𝑛
Σ𝑐𝐶𝐿Δ𝑦× 𝑛𝑔 × 𝑊0 × 𝑔
𝐿𝑛 =0.318
0.1927× 1.179 × 3.3 × 9.81
= 63.03N
4.6 Wing Mounting
Mounting the wing is one of the most critical steps in building the model.
Depending on the methods used, it can be tedious, but a methodical approach with an
experienced helper will make it much easier. Failure to get this right will result in a
model that will not fly straight or trim properly. The goal is to mount the wing
absolutely square to the fuselage centerline in all respects. Additionally, the incidence
must be correct. The wing should be centered (both tips equal distance from the
fuselage) and perpendicular to the fuselage centerline. The tips should be equal height
110
above the building board as well. Not counting incidence settings, any adjustment
made affects the others. Therefore, squaring the wing is a matter of "dialing in" more
than anything else. In our project, we plan to mount the wing by using rubber bands as
shown in figure 4.25.
Figure 4.25 Step of wing mounting using rubber bands
By referring to the rubber band sized as shown in figure 4.26, the rubber band
with size 31 was chosen.
Figure 4.26 Rubber band size
Therefore, the extension of the rubber band is calculated as below:
Extension of rubber band, s = 0.28 − 0.0635
= 0.2165 m
111
The equation to find the ideal of elasticity of rubber band is introduced as below:
∆E =1
2𝑘𝑠2
where
𝑘 = 𝑠𝑝𝑟𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑓 𝑟𝑢𝑏𝑏𝑒𝑟 𝑏𝑎𝑛𝑑 = 17.38 𝑁/𝑚
𝑠 = 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑟𝑢𝑏𝑏𝑒𝑟 𝑏𝑎𝑛𝑑 = 0.2165 𝑚
Therefore,
∆E =1
2(17.38)(0.2165) 2
= 0.4073 𝐽
As the simple calculation above, we can know one single rubber band tie
around the wing will produce 0.4073 Joules.
By considering the Young’s Modulus of the rubber band, the allowable stress
can be obtained with the equation shown as follow:
E =Allowable Stress
Strain
where
E = Young′s Modulus
Strain =Elongation
Length=
0.2165
0.0635= 3.4094
From figure 4.27, the Young’s Modulus value of the rubber band is assumed
as 1 MPa.
Figure 4.27 Young's Modulus of several items
112
Therefore, the maximum allowable stress of a single rubber band is calculated
as below:
Allowable Stress = E × Strain
= 1 × 3.4094
= 3.4094 MPa
Since there are 15 rubber bands tied at each side of the wing, the total allowable
stress that can be withstand by the rubber band is assumed as below:
Total Allowable Stress = 3.4094 × 15
= 51.141 MPa
By considering the applied shear stress and bending stress with different flight
case, the safety factor for each flight case is calculated. Since larger shear and bending
stresses are applied without ailerons, the calculations involved only wing without
ailerons. At the same time, the flight cases with -1.5g and 3g produce a higher applied
stresses, the calculations involved only for -1.5g and 3g.
Flight case 1: G = -1.5
From the section above, the applied shear and bending stresses are calculated as 4.304
kPa and -310.817 kPa respectively. Therefore, the safety factor is calculated as follow:
SF for shear stress =0.5 × 51141
4.304
= 5941.1013
SF for bending stress =51141
310.817
= 164.5373
Flight case 2: G = 3
From the section above, the applied shear and bending stresses are calculated as -
7.0384 kPa and 469.2567 kPa respectively. Therefore, the safety factor is calculated
as follow:
113
SF for shear stress =0.5 × 51141
7.0384
= 3632.999
SF for bending stress =51141
469.2567
= 108.983
From the calculations above, it is shown that the rubber band can withstand
with the applied shear and bending stresses from the flight cases. Therefore, the rubber
band is shown to be can be used in our project to allow us to fly the UAV within the
flight envelope.
114
CHAPTER 5
FUSELAGE, AND LANDING GEAR ANALYSIS
5.1 Fuselage
5.1.2 Introduction
Fuselage contributes very little to lift and produces more drag but it is an
important structural component. It is the connecting member to all load producing
components such as wing, horizontal tail, vertical tail, landing gear and thus
redistributes the load. It also serves the purpose of housing or accommodating
practically all equipment, accessories and systems in addition to carrying the payload.
Because of large amount of equipment inside the fuselage, it is necessary to provide
sufficient number of cutouts in the fuselage for access and inspection purposes. These
cutouts and discontinuities result in fuselage design being more complicated, less
precise and often more less efficient in design.
As a common member to which other components are attached, thereby
transmitting the loads, fuselage can be considered as a hollow beam. The reactions
produced by the wing, tail or landing gear may be considered as concentrated loads at
the respective attachment points. The balancing reactions are provided by the inertia
forces contributed by the weight of the fuselage structure and the various components
inside the fuselage. These reaction forces are distributed all along the length of the
fuselage, though need not be uniformly. Unlike the wing, which is subjected to mainly
unsymmetrical load, the fuselage is much simpler for structural analysis due to its
symmetrical cross-section and symmetrical loading.
The main load in the case of fuselage is shear load because the load acting on
the wing is transferred to the fuselage skin in the form of shear only. The structural
design of fuselage begins with shear force and bending moment diagrams for the
115
respective members. The maximum bending stress produced in each of them is
checked to be less than the yield stress of the material chosen for the respective
member.
5.1.3 Shear Force and Bending Moment Diagram
5.1.3.1 Shear Flow Diagram (3g)
Figure 5.1 Shear force diagram for fuselage at 3g
Figure 5.1 shows the shear flow acting across the fuselage for 3g case. The
maximum shear force is 80.81. The crucial part at this diagram is at the wing section
because need to withstand the highest value of force while the lowest is at the nose
with just 0.5208 shear force need to withstand. Thus;
𝜏 =𝐹
𝐴=
80.8122
0.28×0.18
= 1603.4187 Pa
116
Safety factor = σultimate / σmax
= 2400
1603.4187
=1.497
5.1.3.2 Bending Moment Diagram (3g)
Figure 5.2 Bending moment diagram for fuselage at 3g
Figure 5.2 shows bending moment for 3g case. The maximum bending moment
is 36.4026 Pa. The crucial part at this diagram is at the wing section because need to
withstand the highest value of force while the lowest is at the nose with just 0.5208
bending moment force need to withstand.
I =𝑏𝑑3
12 =
0.28×0.183
12
= 1.3608× 10−4𝑚4
117
𝜎 =36.4026 × 0.09
1.3608 × 10−4
= 24.076 kPa
5.1.3.3 Shear Flow Diagram (-1.5g)
Figure 5.3 Shear force diagram for fuselage at -1.5g
Figure 5.3 shows the shear flow acting across the fuselage for -1.5g case. The
maximum shear force is 40.406 Pa. The crucial part at this diagram is at the wing
section because need to withstand the highest value of force while the lowest is at the
nose with just 0.5208 shear force need to withstand.
Thus, 𝜏 =𝐹
𝐴=
40.406
0.28×0.18
= 801.7063 Pa
118
Safety factor = σultimate / σmax
= 2400
801.7063
= 2.99
5.1.3.4 Bending Moment Diagram (-1.5g)
Figure 5.4 Bending moment diagram for fuselage at -1.5g
Figure 5.4 shows bending moment for -1.5g case. The maximum bending
moment is 18.2019 Pa. The crucial part at this diagram is at the wing section because
need to withstand the highest value of force while the lowest is at the nose with just
0.5208 bending moment force need to withstand.
Thus, I =𝑏𝑑3
12 =
0.28×0.183
12
= 1.3608× 10−4𝑚4
119
𝜎 =18.2019 × 0.09
1.3608 × 10−4
= 12.038 kPa
In aerodynamics, the maximum load factor (at given bank angle) is a
proportion between lift and weight and has a trigonometric relationship. The load
factor is measured in Gs (acceleration of gravity), a unit of force equal to the force
exerted by gravity on a body at rest and indicates the force to which a body is subjected
when it is accelerated. Any force applied to an aircraft to deflect its flight from a
straight line produces a stress on its structure. The amount of this force is the load
factor. For example, a load factor of 3 means the total load on an aircraft’s structure is
three times its weight. Since load factors are expressed in terms of Gs, a load factor of
3 may be spoken of as 3 Gs, or a load factor of 4 as 4 Gs. Load factors are important
for two reasons:
1. It is possible for a pilot to impose a dangerous overload on the aircraft
structures.
2. An increased load factor increases the stalling speed and makes stalls possible
at seemingly safe flight speed.
Based on the title 14 Code of Federal Regulations (14 CFR), part 23, for
certification of the airframe (Subpart C and portions of Subpart D) of normal, utility ,
acrobatic, and commuter category airplanes and airships, the minimum requirement
for safety factor of fuselage is 1.5. The safety factor for 3g and -1.5g is 1.497 and 2.99.
Thus, the safety factor fulfils the requirement for a safe flight.
5.1.4 Shear and Flexural Analysis
5.1.4.1 Conceptual Structural Analysis
By considering the fuselage to be symmetrical, the conceptual structural
analysis is considered to be as shown in Figure 5.5.
120
Figure 5.5 Stringer location at side
Moment of inertia at station (O) about centroid Y axis:
𝐼𝑦 =𝐼𝑜+𝐴ℎ2
where
𝐼𝑜 is neglected because the value is too small
A: Area of stringer (0.1𝑖𝑛2)
h: height of stringer from centroid
Thus,
𝐼𝑦= 3.54352 ×4×0.1 = 5.0226 𝑖𝑛4
Table 5.1 give the necessary calculation in order to determine the flange
bending stress and net total shear load to be taken by the cell skin.
Table 5.1 Flange analysis
Stiffener
no. Arm z Area 𝝈𝒃
𝑷𝑿=𝝈𝒃 ×
𝒂 dz/dx dy/dx Pz Py
1 3.54 0.01 -969.78 -9.70 -0.30 0.00 2.91 0.00
2 -3.54 0.01 969.78 9.70 0.30 0.00 2.91 0.00
Shear taken by stringers= 5.82
121
The stringer has a z component, thus the stringer axial load help resist the
external shear load gives 5.82 lb for half the fuselage section. Hence, net web shear at
station (O) equals:
𝑉𝑤𝑒𝑏= 𝑉𝑒𝑥𝑡+ 𝑉𝑓𝑙𝑎𝑛𝑔𝑒
= 1178.2896 + (2×5.82)
= 1189.9296 lb
5.1.4.2 Flexural Shear Flow
Calculation of Flexural Shear Flow,
q= 𝑞𝑜 - 𝑉𝑧 (𝑤𝑒𝑏) ∑ 𝑎𝑧
𝐼𝑦
q= 𝑞𝑜 - 1189.9296∑ 𝑎𝑧
5.0226
= 𝑞𝑜 – 236.92 ∑ 𝑎𝑧 ……… (A)
Due to symmetry of the section about z axis, the flexural shear flow in the web
at the centre line is zero. So, 𝑞𝑜 will be taken as zero and the summation in equation
(A) will be start with stinger (1)
𝑞12 = 0- 236.92 (3.54 × 0.01) = - 8.39 lb/in
Table 5.2 Flexural shear flow at each stiffener
Stiffener no 𝑽𝒁/ 𝑰𝒀 az q (Ib/in)
1 236.92 0.04 -8.39
2 236.92 -0.04 0.00
3 236.92 -0.04 8.34
4 236.92 0.04 -0.04
The torsional moment T about the centroid of the section at station (O) equals
5 × 1178.29= 5891.45 Ib (clockwise when looking toward station 150). Due to
symmetry of section at station (O), in-plane component of the stringer load produces
122
zero moment about the section centroid. For equilibrium a constant shear flow 𝑞1 is
necessary to make ∑ 𝑀𝑥 = 0
𝑞1= 𝑇
2𝐴 =
5891.45
2×27.900 = -104.58 lb/in
Table 5.3 Constant shear flow at each stiffener
Stiffener no q (lb/in)
3 -112.97
7 -104.58
11 -96.24
15 -104.62
5.1.4.3 Fuselage Shear Flow
Shear Flow by Change in Stringer Load between Adjacent Stations, ∆P Method.
The shear flow will be calculated by considering the change in the axial load in the
longitudinal stringers between fuselage sections at station (O) and (30). As the cross
section is the same between station (O) and (30) so the calculation is simplified. Table
5.4 show the calculation for the flexural shear system.
Table 5.4 Flexural shear system
Stif
no
Area
0
Area
30
Arm
0
Arm
30 𝝈𝒃 0 𝝈𝒃 30
𝑷𝑿=
𝝈𝒃 ×
𝒂 (O)
𝑷𝑿=
𝝈𝒃 ×
𝒂 (30)
∆𝒑
/𝟑𝟎
Panel
Taper
corr, K
∆𝑷𝑲
/𝟑𝟎
Shear
flow
1 0.01 0.01 3.54 1.62 -969.78 -351.21 -9.70 -3.51 0.21 0.80 0.16 0.16
2 0.01 0.01 -3.54 -1.62 969.78 351.21 9.70 3.51 -0.21 0.80 -0.16 -0.16
Since the section is symmetrical, there are no moments induced by the in-plane
component of the stringer force at station (O). By considering material as Magnesium
Alloy (HK31A-0 Sheet t=0.016 to 0.250 in) where the yield stress= 12000
Safety factor = 𝜎(𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒) / 𝜎 𝑚𝑎𝑥
= 12000/ 969.78
= 12.37
123
Table 5.5 Physical properties
MS = SF-1
= 12.37 -1
= 11.37 (Approved)
5.1.5 Structural Analysis
The fuselage is been fabricated by the foam and the value of characteristic is
based on the test that been run to get the exact result. This physical properties, is
needed because the FEM Analysis of the fuselage is based on the material properties.
The contour lines represent a typical von Mises stress distribution along the solid.
Table 5.6 Component Weight
Component Part Weight(kg) Weight(N)
Tail 0.41 4.02
Wing 0.9 8.83
Fuselage 0.46 4.51
Overall Fuselage 3.3 32.36
Every component part of weight is been setup as a load at their location on the
fuselage and the value is from the weight of the part because this is static analysis.
Firstly, we need to setup the meshing size at 0.779 in size to ensure the FEM Analysis
can been proceed. The mesh is more precise when in small size.
Physical
Properties Test Method Units Test Results
Density ASTM-D3575 Grams/Litre 20
Compressive Strength
ASTM-D3575 Mpa 0.31
Tensile Strength ASTM-D3575 Mpa 0.26
Flexural Strength ASTM-D790 Mpa 0.21
Flexural Modulus ASTM-D790 Mpa 9.6
124
Figure 5.6 Component weight
Figure 5.7 Side view
125
Figure 5.8 Component weight
Figure 5.9 Isometric view
Figure 5.10 Bottom part of fuselage
The maximum value of stress is at the bottom part of the fuselage with 16.66
pa where the location of the both landing gear attachment with the fuselage. The wing
126
and tail value with 8.83 and 4.02 respectively is not show any harmful effect toward
the surface of the fuselage where the pressure is just around 8 to 12 Pa.
5.1.6 Compressive- Buckling Analysis
The equation for elastic instability of flat sheet in compression is,
where,
Kc = buckling coefficient which is referred to graph of compressive-buckling
coefficient for flat rectangular plate in Bruhn.
E = modulus of elasticity
v = elastic poisson’s ratio
b= short dimension of plate
t= sheet thickness
5.1.6.1 Former Structure
The aspect ratio is,
𝑎
𝑏 =
168
94 = 1.78
From graph compressive-buckling coefficient for flat rectangular plate,
Kc = 5.8
The critical elastic compression buckling stress,
𝜎= 𝜋2(5.8)(4×109 )
12(1−0.252)(
3×10−3
94×10−3 )2
= 21.446 MPa
127
Then safety factor,
SF= (critical buckling stress
maximum stress) = (
21.446
16.6) = 1.29 Mpa
Margin of Safety,
MS= SF -1 = 0.29
Since the buckling stress is 21.446 MPa which is greater than the maximum stress
experience by the structure, which is 16.6 MPa. Thus, the formers structure will not
buckle. The positive value of margin of safety shows that the structure is safe.
5.1.6.2 Bulkhead Structure
The aspect ratio is,
𝑎
𝑏 =
100
100 = 1
From graph compressive-buckling coefficient for flat rectangular plate,
Kc = 6
The critical elastic compression buckling stress,
𝜎= 𝜋2(6)(4×109 )
12(1−0.252)(
3×10−3
100×10−3 )2
= 18.9496 MPa
Then safety factor,
SF= (critical buckling stress
maximum stress) = (
18.9496
16.6) = 1.14 Mpa
Margin of Safety,
MS= SF -1 = 0.14
128
Since the buckling stress is 18.9496 MPa which is greater than the maximum stress
experience by the structure, which is 16.6 MPa. Thus, the formers structure will not
buckle. The positive value of margin of safety shows that the structure is safe.
5.1.7 Shear- Buckling Analysis
The above equation shows the critical elastic shear buckling stress for flat plates with
various boundary conditions.
Ks = shear buckling coefficient from graph of shear-buckling stress coefficient of
plates in Bruhn
E = modulus of elasticity
v= elastic poisson’s ratio
b= short dimension of plate
t= sheet thickness
5.1.7.1 Former Structure
The aspect ratio is,
𝑎
𝑏 =
168
94 = 1.78
From graph shear-buckling stress coefficient for flat rectangular plate,
Ks = 6.7
The critical elastic shear buckling stress,
𝜎= 𝜋2(6.7)(4×109 )
12(1−0.252)(
3×10−3
94×10−3 )2
= 23.9480 MPa
129
Then safety factor,
SF= (critical buckling stress
maximum stress) = (
23.9480
16.6) = 1.44 Mpa
Margin of Safety,
MS= SF -1 = 0.44
Since the shear buckling stress is 23.9480 MPa which is greater than the maximum
shear stress experience by the structure, which is 16.6 MPa. Thus, the formers structure
will not buckle. The positive value of margin of safety shows that the structure is safe.
5.1.7.2 Bulkhead Structure
The aspect ratio is,
𝑎
𝑏 =
100
100 = 1
From graph shear-buckling coefficient for flat rectangular plate,
Kc = 9.8
The critical elastic shear buckling stress,
𝜎= 𝜋2(9.8)(4×109 )
12(1−0.252)(
3×10−3
100×10−3 )2
= 30.9511 MPa
Then safety factor,
SF= (critical buckling stress
maximum stress) = (
18.9496
16.6) = 1.14 Mpa
Margin of Safety,
MS= SF -1 = 0.14
130
Since the shear buckling stress is 18.9496 MPa which is greater than the maximum
stress experience by the structure, which is 16.6 MPa. Thus, the bulkhead will not
buckle. The positive value of margin of safety shows that the structure is safe.
5.2 Landing Gear
5.2.1 Main landing gear simulation analysis
The Solidworks 3D model is used to create this simulation analysis. However,
a precaution is to ensure the consistency in the unit’s system. Throughout the
simulation, we will adopt the International System of Units (SI) unit system. Selecting
a material requests a lot of investigation in their physical properties, i.e. strength,
ductility, corrosion resistance. Based on the findings, the main landing gear is made of
stainless steel.
In structural analysis, boundary conditions are applied to those regions of the
model where the displacements and/or rotations are known. Such regions may be
constrained to remain fixed (have zero displacements and/or rotation) during the
simulation or may have specified, non-zero displacement and/or rotation. A mesh is a
physical discretization of a domain existing in one, two or three dimensions. Higher
quality mesh is synonymous with smaller mesh. It can often be achieved with careful
partitioning and edge seeds.
131
5.2.2 Rear landing gear simulation analysis
Figure 5.11 Rear landing gear
Table 5.7 Volumetric properties
Volumetric Properties
Mass:0.00584563 kg
Volume:7.4944e-007 m^3
Density:7800 kg/m^3
Weight:0.0572872 N
Table 5.8 Material properties
132
Table 5.9 Load and fixtures detail of rear landing gear
Table 5.9 shows the fixed part will be at the place where tyre will be attached and the
the force will be added at part where the landing gear is attached to the fuselage.
Table 5.10 Reaction forces and moments
Study Results
The coloured and contoured von Mises stress result helps to identify where is
the most critical part and which part might require strength-enhancement. The stress
result indicates that maximum magnitude is located at the holes which will be attached
to the tyres. In order to further improve the structural, its diameter need to be made
thicker. Other than that, the green area also shown the higher stress.
Boundary condition for rear landing gear is crucial. In our case, we just
considered the analysis is static deformation. The boundary condition of the bottom
133
part is fixed in displacement and rotation in the analysis. Besides, only the y-direction
of displacement will vary with the loading for the upper part which attached to the
fuselage. Below figures shows all the analysis including static stress, static
displacement and static strain.
Figure 5.12 Static stress
134
Figure 5.13 Static displacement
Figure 5.14 Static strain
135
5.2.3 Front landing gear simulation analysis
Figure 5.15 Front landing gear
Table 5.11 Volumetric properties
Volumetric Properties
Mass:0.283454 kg
Volume:3.54318e-005 m^3
Density:8000 kg/m^3
Weight:2.77785 N
Table 5.12 Material properties
136
Table 5.13 Load and fixtures detail of front landing gear
Fixture name Fixture Image Fixture Details
Fixed-1
Entities: 2 edge(s), 4 face(s)
Type: Fixed Geometry
Load name Load Image Load Details
Force-1
Entities: 1 face(s)
Type: Apply normal
force
Value: 33 N
Table 5.14 Simulation results of front landing gear
Components X Y Z Resultant
Reaction force(N) -0.248924 -0.00208807 33.0006 33.0016
Reaction Moment(N.m) 0 0 0 0
In table 5.13 shows the fixed part will be at the place where tyre will be attached and
the the force will be added at part where the landing gear is attached to the fuselage is
displayed in table 5.14.
Study Results
The coloured and contoured von Mises stress result helps to identify where is
the most critical part and which part might require strength-enhancement. The stress
result indicates that maximum magnitude is located at the holes which will be attached
to the tyres. In order to further improve the structural, its diameter need to be made
thicker. Other than that, the green area also shown the higher stress.
137
In our case, we just considered the analysis is static deformation. The
boundary condition of the bottom part is fixed in displacement and rotation in the
analysis. Besides, only the y-direction of displacement will vary with the loading for
the upper part which attached to the fuselage. Below figures shows all the analysis
including static stress, static displacement and static strain.
Figure 5.16 Static Stress
Figure 5.17 Static strain
138
5.2.4 Discussion
Tailwheel, or conventional landing gear, is a term referencing the
undercarriage of an aircraft consisting of two wheels positioned forward of the center
of gravity. A small skid or wheel is located at the tail of the aircraft to support the tail.
The aviation term tail dragger is another common term for tailwheel aircraft, also
known as the conventional gear fd configuration. This is considered “conventional”
because traditionally aircraft were configured only with tailwheel; tricycle gear had
not yet been invented.
There are many advantages in the tailwheel aircraft that are not seen in tricycle
gear configuration. Thanks to smaller tires, the induced drag on the aircraft is lower
for the same power settings. Most tailwheel aircraft are cheaper to buy and maintain
in comparison to nosewheel airplanes. Tailwheel aircraft are also easy to handle and
manoeuvre on the ground thanks to its lightweight tail which can release and free-
caster, allowing 'flat spins', something tricycle gear planes can't do! With the nose
high attitude of a tailwheel airplane, less chips and stone damage occur due to the
increased propeller clearance with the ground. There are certainly benefits to a
taildragger, as well. The nose-high attitude on the ground means that the propellers on
tailwheel aircraft often have more clearance from the ground. The extra clearance
makes these aircraft better suited for grass or dirt runways.The tailwheel craft is often
designed and configured for slow flight. This slower speed makes them easier to land
on short runways. Many are high-design and better suited for backcountry flying than
nosewheel aircraft are. These are the reasons we chosen taildragger landing gear in our
aircraft.
139
CHAPTER 6
EMPENNAGE ANALYSIS
6.1 Introduction
The empennage is the whole tail unit at the extreme rear of the fuselage and it
provides the stability and directional control of the aircraft (Fig 6.1). Structurally, the
empennage consists of the entire tail assembly, including the vertical stabiliser,
horizontal stabilisers, rudder, elevators, and the rear section of the fuselage to which
they are attached. The stabilisers are fixed wing sections which provide stability for
the aircraft to keep it flying straight. The horizontal stabiliser prevents the up-and-
down, or pitching, motion of the aircraft nose. The rudder is used to control yaw, which
is the side-to-side movement of the aircraft nose. The elevator is the small moving
section at the rear of the horizontal stabiliser used to generate and control the pitching
motion. The loads on the rudder and elevator are smaller than those acting on the
vertical and horizontal stabilisers, although properties such as stiffness, strength and
toughness are still critically important.
Figure 6.1 Structural configuration of the empennage
140
6.2 Preliminary Horizontal and Vertical Tail Sizing
Information such as location of the centre of gravity(c.g.) of airplane, shift in
(c.g). location during flight and the desirable level of stability are needed to fulfil the
requirements of proper horizontal and vertical tail surfaces design. However, to obtain
the (c.g). location, the weights of horizontal and vertical stabilizer are needed which
depend on their size. Hence, preliminary sizing of the two stabilizers are carried out
with the help of the following steps. To do so, our design mission which is carrying a
500g payload by means of an unmanned aerial aircraft (UAV) that utilises single motor
operated engine driven propellers together with the gross weight of the entire aircraft
has been taken into consider while designing both the sizing of the tails.
6.2.1 Choice of Empennage Shape
The conventional configuration with a low horizontal tail is a natural choice
since roots of both horizontal and vertical surfaces are conveniently attached directly
to the fuselage. In this design, the effectiveness of the vertical tail is large because
interference with the fuselage and horizontal tail increase its effective aspect ratio.
Large areas of the stabilizer are affected by the converging fuselage flow, however,
which can reduce the local dynamic pressure. The design of the structure is shown in
Figure 6.2.
Figure 6.2 Tail configuration of our design
141
6.2.2 Horizontal and Vertical Stabilizer Sizing
Basically, prior to the horizontal and vertical tail sizing, the first step needs to
be determined was the careful selection of the airfoil as elevator and rudder are
attached to the surface to provide stability controls. The elevator and rudder have
deflections on both sides of the undeflected positions. Hence, horizontal and vertical
stabilizer invariably have symmetric airfoil sections. National Advisory Commilis for
Aeronautica (NACA) generated a large amount of data on the aerodynamic
characteristics (Cl vs α, Cα vs Cl and Cm vs Cl) at different Reynolds numbers. Hence,
the airfoils which are commonly used for stabilizer of airplanes flying at low and
medium subsonic Mach numbers is NACA 0012. Upon selection of the NACA the
stabilizer sizing is further processed by means of determining the dimensions of the
both tails. These parameters are determined based on the current design available of
the commercial aircraft that utilises the application of single piston operated engine
driven propellers. Even though, our design is mainly concentrated on UAV aircraft but
the application of the operation is almost similar to the single piston operated engine
driven propellers. The dimensions and sizing of the horizontal and vertical stabilizer
are as shown in Table 6.1 and Table 6.2 and are as follows:
Table 6.1 NACA 0012 horizontal stabilizer dimensions and sizing
Chord 0.21 m
Span 0.6 m
Tail Area 0.6 × 0.21 = 0.126𝑚2
Aspect Ratio 0.62
0.126 = 2.86
Aerodynamic Centre (A.C) 1
4×chord =
1
4×0.21m = 0.05 m
Centre of Gravity (C.G) 0.48 m from the centre of the propeller
Horizontal Tail Weight
Area of Tail × Maximum Thickness of Chord
× Density of Foam(EPP)
= 0.13 × 0.0252 × 20.8 = 0.07kg
Maximum Coefficient of Lift
(Based on Airfoil Generator) 1.1 at 14°
142
Table 6.2 NACA 0012 vertical stabilizer dimensions and sizing
Chord 0.16 m
Span 0.27 m
Tail Area 0.16 × 0.27 = 0.04 𝑚2
Aspect Ratio 0.272
0.04 = 1.69
Aerodynamic Centre (A.C) 1
4×chord =
1
4×0.16m = 0.04 m
Centre of Gravity (C.G) 1.02 m from the centre of the propeller
Vertical Tail Weight
Area of Tail × Maximum Thickness of Chord
× Density of Foam(EPP)
= 0.04 × 0.0192 × 20.8 = 0.02𝑘𝑔
Maximum Coefficient of Lift
(Based on Airfoil Generator) 1.1 at 14°
6.2.3 Theoretical Analysis of Horizontal and Vertical Stabilizers
The theoretical analysis of horizontal and vertical stabilizer is mainly
emphasized on determining the effectiveness of both the surfaces as well as strength
of the materials that are used to design the components. The following parts explains
the ways on how such theoretical analysis are being approached which are as follows:
6.2.3.1 Horizontal and Vertical Stabilizer Effectiveness
In order to analyse theoretically the effectiveness of the horizontal and vertical
stabilizers, they are different types of methods and approaches are being carried out.
One of the best methods that could almost estimate the effectiveness of the stabilizers
is by calculating the tail volume ratios of both horizontal and vertical stabilizers. The
method of solution are as follows:
143
Figure 6.3 Calculation on effectiveness of horizontal stabilizer
L = 3
4× wing chord + 0.39 +
1
4tail chord
= (3
4× 0.28) + 0.39 + (
1
4× 0.21) = 0.6525 m
D = 0.6525 ×0.126
0.392+0.126 =0.158716 m
MAC = 0.28m (since the wing is rectangular)
Safety Margin: 0.28
10 = 0.028 m (distance between the Cg and NP)
Horizontal Tail Volume Ratio = 0.126×(0.6525−0.158716+0.028)
0.392×0.28
= 0.50
To justify the results obtained above, the following extraction of the Table 6.3
from the sources (Priyanka Barua, Tahir Sousa & Dieter Scholz,2013) is referred. The
details of the table are as follows:
144
Table 6.3 Suggestions for tail volume ratio of horizontal tail by various authors
(Raymer 1992, Jenkinson 1999, Roskam 1985, Torenbeek 1982, Nicolai 1975,
Schaufele 2007)
Based on the table above, by referring to the Raymer 1992 sources, the tail
volume ratio for the horizontal stabilizer which is 0.5 is much reasonable and effective
as it can be compared from the table above under the civil propeller aircraft type and
the category of homebuilt where homebuilt can be defined an amateur-built aircraft or
kit planes that are constructed by persons for whom this is not a professional activity.
Figure 6.4 Calculation on effectiveness of vertical stabilizer
L: 1.07 − 0.43 − 3
4×0.16m = 0.52m
Vertical Tail Volume Ratio= (0.16×0.27)×0.52
0.392 ×1.4
= 0.04
145
To justify the results obtained above, the following extraction of the Table 6.4
from the sources (Priyanka Barua, Tahir Sousa & Dieter Scholz,2013) is referred. The
details of the table are as follows:
Table 6.4 Suggestions for tail volume ratio of vertical tail by various authors (Raymer
1992, Jenkinson 1999, Roskam 1985, Torenbeek 1982, Nicolai 1975, Schaufele 2007)
Based on the table above, by referring to the Raymer 1992 sources, the tail
volume ratio for the vertical stabilizer which is 0.04 is much reasonable and effective
as it can be compared from the table above under the civil propeller aircraft type and
the category of homebuilt.
6.2.3.2 Horizontal and Vertical Stabilizer Strength
Upon completion of a proper sizing of the stabilizers based on the available
criteria, the strength of the stabilizers is also need to be theoretically determined for
the proper material selection of the design which depends on the physical and
mechanical properties of the materials. Hence, this can be achieved by comparing the
local stresses of the stabilizers at cruising speed by the different load factor such as G
= -1.5, 1 and 3.0. The lift is assumed to be equal to the gross weight of the aircraft due
to the calculations are made at cruising speed. The graphs of three load factors are
plotted and shear and bending stresses are calculated as follows:
146
Horizontal Stabilizer Strength
𝐆 = −𝟏. 𝟓 𝐚𝐭 𝐜𝐫𝐮𝐬𝐢𝐧𝐠 𝐬𝐩𝐞𝐞𝐝
Figure 6.5 Lift distribution of the front view of the tail plane at G =-1.5
Figure 6.6 Shear and bending moment diagram at G = -1.5
L = W
L = (0.07) kg × 9.81 = 0.6867 𝑁
Since Lift Distribution of the Horizontal Tail = 0.6867
0.6 =1.1445 N/m
147
𝐆 = 𝟏. 𝟎 𝐚𝐭 𝐜𝐫𝐮𝐬𝐢𝐧𝐠 𝐬𝐩𝐞𝐞𝐝
Figure 6.7 Lift distribution of the front view of the tail plane at G = 1.0
Figure 6.8 Shear and bending moment diagram at G = 1.0
148
𝐆 = 𝟑. 𝟎 𝐚𝐭 𝐜𝐫𝐮𝐬𝐢𝐧𝐠 𝐬𝐩𝐞𝐞𝐝
Figure 6.9 Lift distribution of the front view of the tail plane at G = 3.0
Figure 6.10 Shear and bending moment diagram at G = 3.0
149
To determine the structural integrity of the horizontal stabilizer as well as
selection of a proper materials that suits the plan, it is vital to calculate the load factors
at different flight conditions in order to determine the limits of the flying qualities.
This can be achieved by means of estimating “limit load factors” in which the problem
of load factors in airplane design then reduces to that of determining the highest load
factors that can be expected in normal operation under various operational situations.
For the reasons of safety, it is required that the airplane be designed to withstand these
load factors without any structural damage.
To do so, Figure 6.6, Figure 6.8 and Figure 6.10 are plotted to determine the
maximum stresses of the structures at different flight conditions. From the graphs of
shear and bending moment diagram above we can deduce that the shear force can be
mathematically expressed as 272.5Pa ≤ τ ≤ 1.36KPa whereas the bending moment
is 60.08KPa ≤ σ ≤ 510.74KPa . Hence, for a safe flight such amounts of forces
should hold the shear and bending stresses of the stabilizers without damaging the
structures. This in turn leads to a proper selection of materials as our design of
stabilizers are constructed using the EPP foam. To determine whether the structures
can withstand above stresses, the buckling stresses under shear and bending loads of
the selected material, EPP Foam at a density of 20.8𝑘𝑔
𝑚3 is calculated based on the
available data of the selected material. The calculation of the buckling under shear and
bending loads are as follows:
Collection of variables (refer Appendix, Appendix P-K)
E = 79.67Mpa ; ν = 0.001 (variables are interpolated based on Appendix L)
F0.7 = 400KPa ( based on Appendix K)
Calculation of buckling stress under shear and bending loads
150
Figure 6.11 Free body diagram of stabilizer under shear and bending conditions
Buckling under bending loads
𝐹𝑏𝑒𝑛𝑑𝑖𝑛𝑔,𝑏 =𝐾𝑏 𝜋2𝐸
12(1 − 𝑣2)(
𝑡
𝑏)2
=Kbπ2(79.67 × 106 )
12(1 − (0.001)2)(
0.006
0.3)2
= Kb(26.21 × 103 )
𝑎
𝑏=
0.1(one equal portion)
6 × 10−3
= 16.67
𝐾𝑏 = 24.4
𝐹𝑏 = 24.4(26.21 × 103 )
Plastic Correction;
𝐹𝑏
𝐹0.7=
639.52 × 103
400 × 103
= 1.6
Based on Appendix O;
𝐹𝑏
𝐹0.7= 1.0
𝐹𝑏 = 1.0 × 400 × 103
= 400 kPa
151
The buckling stress due to bending loads calculated is for one portion only.
Since the stabilizer is divided into three equal portions the total buckling stress can be
sum up as follows:
Fb = 3 × 400 kPa
= 1.2 MPa
Buckling under shear loads
𝐹𝑠ℎ𝑒𝑎𝑟,𝑠 =𝐾𝑠𝜋2𝐸
12(1 − 𝑣2)(
𝑡
𝑏)2
=Ksπ2(79.67 × 106)
12(1 − (0.001)2)(0.006
0.21)2
= Ks(53.49 × 103)
𝑎
𝑏=
0.03(one equal portion)
6 × 10−3
= 5
𝐾𝑠 = 5.6
𝐹𝑠 = 5.6(53.49 × 103)
Plastic Correction;
𝐹𝑠
𝐹0.7=
316.34 × 103
400 × 103
= 0.79
Based on Figure Appendix P;
𝐹𝑠
𝐹0.7= 0.79
𝐹𝑠 = 0.79 × 400 × 103
= 316 kPa
152
The buckling stress due to shear loads calculated is for one portion only. Since
the stabilizer is divided into three equal portions the total buckling stress can be sum
up as follows:
Fs = 7 × 316 kPa
= 2.21 MPa
Comparison between theoretical results with buckling stress under shear and
bending loads for Horizontal Stabilizer
As discussed in earlier part on the mathematical expression of the shear and
bending stresses on the stabilizers, we can make a prediction that the maximum shear
force acting on the surface is 1.36 kPa in which is lower that the buckling stress under
shear loading which is 2.21MPa. On the other hand, while looking on the bending
stress, the maximum bending stress acting on the surface is 510.74 kPa which is also
lower to the buckling stress acting on the surface due to bending loading which is
1.2MPa. Hence, from this comparison we could predict in terms of structural integrity
that the structure can manage to sustain both the shear and bending loading acting on
the surface of stabilizers at different load factors as discussed in earlier part.
Besides that, to measure of how well a material can withstand the effects
of tearing, the calculated tearing stresses for different flight conditions are compared
with the tear strength. The following steps will explain how to determine the tearing
stress. These are as follows:
Tear Strength =Distributed Lift Force(N)
Thickness of Foam(m)
Tear strength =1.445
0.006 for G = 1.0
By approaching above formula, the tear stresses at different load factors can be
mathematically expressed as 190.75𝑁
𝑚≤ Tear Strength ≤ 572.25
𝑁
𝑚. Based on the
table, the tear strength at 20.8𝑘𝑔
𝑚3 is 1.77kPa. Thus, the calculated values of tear stresses
are not exceeding the expected tear strength.
153
Vertical Stabilizer Strength
𝐆 = −𝟏. 𝟓 𝐚𝐭 𝐜𝐫𝐮𝐬𝐢𝐧𝐠 𝐬𝐩𝐞𝐞𝐝
Figure 6.12 Lift distribution of the Horizontally Projected Vertical Stabilizer at G = -1.5
Figure 6.13 Shear and bending moment diagram at G = -1.5
L = W
L = (0.02) kg × 9.81 = 0.1962 N
Since Lift Distribution of the vertical tail = 0.1962
0.27 =0.7267 N/m
154
𝐆 = 𝟏. 𝟎 𝐚𝐭 𝐜𝐫𝐮𝐬𝐢𝐧𝐠 𝐬𝐩𝐞𝐞𝐝
Figure 6.14 Lift distribution of the Horizontally Projected Vertical Stabilizer at G = 1.0
Figure 6.15 Shear and bending moment diagram at G = 1.0
155
𝐆 = 𝟑. 𝟎 𝐚𝐭 𝐜𝐫𝐮𝐬𝐢𝐧𝐠 𝐬𝐩𝐞𝐞𝐝
Figure 6.16 Lift distribution of Horizontally Projected Vertical Stabilizer at G = 3.0
Figure 6.17 Shear and bending moment diagram at G = 3.0
156
To determine the structural integrity of the vertical stabilizers as well as
selection of a proper materials that suits the plan, it is vital to calculate the load factors
at different flight conditions in order to determine the limits of the flying qualities.
This can be achieved by means of estimating “limit load factors” in which the problem
of load factors in airplane design then reduces to that of determining the highest load
factors that can be expected in normal operation under various operational situations.
For the reasons of safety, it is required that the airplane be designed to withstand these
load factors without any structural damage.
To do so, Figure 6.13, Figure 6.15 and Figure 6.17 are plotted to determine the
maximum stresses of the structures at different flight conditions. From the graphs of
shear and bending moment diagram above we can deduce that the shear force can be
mathematically expressed as 102.19Pa ≤ τ ≤ 510.96Pa whereas the bending
moment is 17.17KPa ≤ σ ≤ 145.93KPa. Hence, for a safe flight such amounts of
forces should hold the shear and bending stresses of the stabilizers without damaging
the structures. This in turn leads to a proper selection of materials as our design of
stabilizers are constructed using the EPP foam. To determine whether the structures
can withstand above stresses, the buckling stresses under shear and bending loads of
the selected material, EPP Foam at a density of 20.8𝑘𝑔
𝑚3 is calculated based on the
available data of the selected material. The calculation of the buckling under shear and
bending loads are as follows:
Collection of variables (refer Appendix P - K)
E = 79.67Mpa ; ν = 0.001 (variables are interpolated based on Appendix L)
F0.7 = 400KPa ( based on Appendix L)
Calculation of buckling stress under shear and bending loads
157
Figure 6.18 Free body diagram of stabilizer under shear and bending conditions
Buckling under bending loads
𝐹𝑏𝑒𝑛𝑑𝑖𝑛𝑔,𝑏 =𝐾𝑏 𝜋2𝐸
12(1 − 𝑣2)(
𝑡
𝑏)2
=Kbπ2(79.67 × 106 )
12(1 − (0.001)2)(
0.006
0.27)2
= Kb(35.36 × 103 )
𝑎
𝑏=
0.03(one equal portion)
6 × 10−3
= 5
𝐾𝑏 = 26
𝐹𝑏 = 26(35.36 × 103 )
Plastic Correction;
𝐹𝑏
𝐹0.7=
919.36 × 103
400 × 103
= 2.3
Based on Appendix O;
𝐹𝑏
𝐹0.7= 1.0
𝐹𝑏 = 1.0 × 400 × 103
= 400KPa
0.27m
158
The buckling stress due to bending loads calculated is for one portion only.
Since the stabilizer is divided into three equal portions the total buckling stress can be
sum up as follows:
Fb = 9 × 400 kPa
= 3.6 MPa
Buckling under shear loads
𝐹𝑠ℎ𝑒𝑎𝑟,𝑠 =𝐾𝑠𝜋2𝐸
12(1 − 𝑣2)(
𝑡
𝑏)2
=Ksπ2(79.67 × 106)
12(1 − (0.001)2)(0.006
0.16)2
= Ks(92.15 × 103)
𝑎
𝑏=
0.02(one equal portion)
6 × 10−3
= 3.33
𝐾𝑠 = 5.8
𝐹𝑠 = 5.8(92.15 × 103 )
Plastic Correction;
𝐹𝑠
𝐹0.7=
534.47 × 103
400 × 103
= 1.3
Based on Appendix P;
𝐹𝑠
𝐹0.7= 1.3
𝐹𝑠 = 1.3 × 400 × 103
= 520 kPa
159
The buckling stress due to shear loads calculated is for one portion only. Since
the stabilizer is divided into three equal portions the total buckling stress can be sum
up as follows:
Fs = 8 × 520 kPa
= 4.16 MPa
Comparison between theoretical results with buckling stress under shear and
bending loads for Vertical Stabilizer
As discussed in earlier part on the mathematical expression of the shear and
bending stresses on the vertical stabilizer, we can make a prediction that the maximum
shear force acting on the surface is 510.96Pa in which is lower that the buckling stress
under shear loading which is 4.16MPa. On the other hand, while looking on the
bending stress, the maximum bending stress acting on the surface is 145.93 kPa which
is also lower to the buckling stress acting on the surface due to bending loading which
is 3.6MPa. Hence, from this comparison we could predict in terms of structural
integrity that the structure can manage to sustain both the shear and bending loading
acting on the surface of stabilizers at different load factors as discussed in earlier part.
Besides that, to measure of how well a material can withstand the effects
of tearing, the calculated tearing stresses for different flight conditions are compared
with the tear strength. The following steps will explain how to determine the tearing
stress. These are as follows:
Tear Strength =Distributed Lift Force(N)
Thickness of Foam(m)
Tear strength =0.1962
0.006 for G = 1.0
By approaching above formula, the tear stresses at different load factors can be
mathematically expressed as 181.68𝑁
𝑚≤ Tear Strength ≤ 363.35
𝑁
𝑚. Based on the
table, the tear strength at 20.8𝑘𝑔
𝑚3 is 1.77kPa. Thus, the calculated values of tear stresses
are not exceeding the expected tear strength.
160
6.3 Preliminary Control Surfaces Sizing
Generally, control surfaces namely rudder and elevator are designed to
provide stability and control for the aircraft manoeuvres during flight, landing and
take-off. Hence, a proper design analysis should thoroughly emphasize to the structure
as well as selection of materials that can withstand the shear and bending stresses on
the surfaces. Based on Raymer 1992, about 90% percent of the tail span, elevators and
rudders design and construction are beginning from the side of the fuselage and extend
to the tip of the tail. Besides that, typical construction of the rudder and elevator are
about 25-50% of the tail chord due to equal distribution of lift force that could be
balanced by the hinges when the control surfaces are deflected to various angles. It is
also important that the hinge axis should not farther aft than about 20% of the average
chord of the control surfaces. For the purpose of analysis, the integrity of the structure
can be evaluated based on theoretical calculation of shear and bending stresses at
different angle of deflection of control surface as well as comparing the ability of the
structure to withstand such forces and stresses. To do so, the following design
techniques are approached that almost predict the behaviour and performance of the
control surfaces.
6.3.1 Control Surface Sizing
In designing of the control surfaces both rudder and elevator, the processes are
exactly similar as we imposed on the designing of horizontal and vertical stabilizer.
This process includes selection of airfoil in which the NACA selection should be same
as imposed on stabilizers (NACA0012) and determination of control surface sizes
based on the available current aircraft design that matches the mission of the aircraft.
However, the theoretical analysis is different since the control surfaces can deflect up
to certain angle. To ensure that the control surfaces can perform at different angle of
deflection and load factor, proper dimension or sizing of the control surfaces are vital
that have proper justification. The dimension and sizing of the control surfaces are
shown in Table 6.5 and Table 6.6 which are as follows.
161
Table 6.5 NACA 0012 elevator dimensions and sizing
Chord 0.09 m
Span 0.54 m
Elevator Area 0.09 × 0.54 = 0.05𝑚2
Aspect Ratio 0.542
0.05 = 5.83
Aerodynamic Centre (A.C) 1
4×chord =
1
4× 0.09m = 0.03 m
Centre of Gravity (C.G) 1.1 m from the centre of the propeller
Elevator Weight
Area of Elevator × Maximum Thickness of Chord
× Density of Foam(EPP)
= 0.05 × 0.0252 × 20.8 = 0.03kg
Maximum Coefficient of Lift
(Based on Airfoil Generator)
1.1 at 14°
Hinge Axis Length 0.02m from the leading of elevator
Length of Rod 0.18m (between cg of horizontal stabilizer and
elevator)
Elevator chord to horizontal tail
chord Ratio 0.45
Maximum angle of deflection ±25°
162
Table 6.6 NACA 0012 rudder dimensions and sizing
Chord 0.08 m
Span 0.24 m
Rudder Area 0.08 × 0.24 = 0.02 𝑚2
Aspect Ratio 0.242
0.02 = 2.88
Aerodynamic Centre (A.C) 1
4×chord =
1
4×0.08m = 0.02 m
Centre of Gravity (C.G) 1.00 m from the centre of the propeller
Rudder Weight Area of rudder × Maximum Thickness of Chord
× Density of Foam(EPP)
= 0.02 × 0.0192 × 20.8 = 0.007𝑘𝑔
Maximum Coefficient of Lift
(Based on Airfoil Generator)
1.1 at 14°
Hinge Axis Length 0.02m from the leading of rudder
Length of Rod 0.12m (between cg of vertical stabilizer and
rudder)
Rudder chord to vertical tail
chord Ratio
0.47
Maximum angle of deflection ±25°
Based on the Table 6.5 and Table 6.6, the values for the chord can be
determined by referring to Table 6.7 and Table 6.8 below.
163
Table 6.7 Suggestions for C_E/C_H as given by various authors (Roskam 1985,
Schaufele 2007, Torenbeek 1982 and Sadraey 2013)
Table 6.8 Suggestions for the C_R/C_V as given by various authors ((Roskam
1985, Schaufele 2007, Torenbeek 1982 and Sadraey 2013)
From the tables above, the length of chord for elevator (0.09m) and rudder
(0.08m) can be obtained by calculating the ratio between the control surfaces over
stabilizers. To do so, the ratio between the control surfaces over stabilizers are set to
the average values as shown from the reference above which are 0.45(elevator) and
0.47(Rudder) for the homebuilt aircraft category based on Roskam 1985. On the other
hand, from the reference of Raymer 1992 about 90% percent of the tail span, elevators
and rudders design and construction are beginning from the side of the fuselage and
164
extend to the tip of the tail. Hence, by mathematically express the statement (20%×tail
chord), we can obtain the value for the elevator and rudder span which are 0.54m for
elevator and 0.24m for rudder respectively. The hinge distance for the elevator and
rudder to assemble with its horizontal and vertical stabilizers are also calculated by
20%× chord length of the control surfaces where the values from the control surfaces
leading edges are depicted as in Table 6.19 and Table 6.20.
6.3.2 Theoretical Analysis of Control Surfaces
The theoretical analysis of control surfaces is mainly emphasized on
determining the amount of lift created at various angle of attack and the torque/moment
generated can be withstand by the servo motor. The following parts explains the ways
on how such theoretical analysis are being approached which are as follows:
6.3.2.1 Theoretical Lift calculation at different angle of deflection
The purpose of this analysis is to determine the maximum lift generated by
control surfaces at an angle of attack. This will ensure that the maximum force can be
withstand by the servo motor torque especially during take-off and landing. The
analysis of lift is mainly determined by the spanwise distribution of lift of both elevator
and rudder and the angle of deflection is limited to 25°. These steps of the calculations
are as follows:
Figure 6.19 Velocity profile of the rudder and elevator as view from sideview
165
Figure 6.20 Resultant aerodynamic force and the components into which its splits
In order to obtain the lift at different angle of attack, the cross section of the
control surface and free body diagram are drawn which are shown in Figure 6.19 and
Figure 6.20. To calculate the lift force and coefficient of lift, the following formulae
are used.
(I) Calculations on Spanwise distribution of lift at different angle of attack
The geometrical relation between these two sets of components as referred Figure
6.20 is as follows:
L = N cos𝛼 − A sinα
To obtained the normal force(N) and axial force(A), the following formula is
used:
N = − ∫ (𝑝𝑢𝑝𝑝𝑒𝑟 cosθ + 𝜏𝑢𝑝𝑝𝑒𝑟 sinθ)𝑑𝑆𝑢𝑝𝑝𝑒𝑟 + ∫ (𝑝𝑙𝑜𝑤𝑒𝑟cosθ − 𝜏𝑙𝑜𝑤𝑒𝑟sinθ)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
𝑇𝐸
𝐿𝐸
TE
LE
A = ∫ (−𝑝𝑢𝑝𝑝𝑒𝑟 sinθ + 𝜏𝑢𝑝𝑝𝑒𝑟 cosθ)𝑑𝑆𝑢𝑝𝑝𝑒𝑟 + ∫ (𝑝𝑙𝑜𝑤𝑒𝑟sinθ + 𝜏𝑙𝑜𝑤𝑒𝑟cosθ)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
𝑇𝐸
𝐿𝐸
TE
LE
In which;
𝑝 = pressure of upper and lower surface of the airfoil
𝜏 = shear stress of the upper and lower surface of the airfoil
166
Sample calculations for elevator at 𝟓°
The geometrical relations for the integration of pressure and shear stress
distributions over a two-dimensional body surface of elevator is shown in Figure 6.21.
Figure 6.21 Integration of pressure and shear stress distributions over a two-
dimensional body surface of elevator
To obtain the normal force, the values of pressure, shear stress and angle are
obtained as follows:
Angle , θ = tan−13 × 10−3
0.09= 1.9°
Based on the elevation of our aircraft (10m) the following parameters are
calculated by interpolating the atmospheric tables.
Pressure , plower and pupper = 1.01 × 105 Pa
Temperature , T = 288.09K
Density , ρ = 1.224kg /𝑚3
Dynamic viscocity, μ = 1.78869 × 10−5𝑁𝑠/𝑚2
Freestream velocity,𝑈∞ =0.2
√1.4×287×288.15= 68.059m/s
167
Based on Figure 6.19, the shear stress is calculated as follows:
τupper surface = τlower surface = μdu
dy
= (1.78869 × 10−5) (𝑈∞
0.025𝑐)
= (1.78869 × 10−5) (68.059
0.025(0.09))
= 0.54105Pa
N = − ∫ ((1.01 × 105 )cos1.9° + (0.5411)sin1.9°)𝑑𝑆𝑢𝑝𝑝𝑒𝑟 0.09
0
+ ∫ ((1.01 × 105 )cos1.9° − (0.5411)sin1.9°)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
0.09
0
= -3.229× 10−3N
A = ∫ ((−1.01 × 105)sin1 .9° + (0.5411) cos1.9°)𝑑𝑆𝑢𝑝𝑝𝑒𝑟
0.09
0
+ ∫ ((1.01 × 105 )sin1.9° + (0.5411)cos1.9°)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
0.09
0
=97.336 × 10−3N
L = (−3.229 × 10−3) cos(5°) − ( 97.336 × 10−3) sin(5°)
= −0.0117N
Sample calculation for rudder at 𝟓°
To obtain the normal force, the values of pressure, shear stress and angle are
obtained as follows:
Angle , θ = tan−13 × 10−3
0.08= 2.1°
Based on the elevation of our aircraft(10m) the following parameters are
calculated by interpolating the atmospheric tables.
168
Pressure , plower and pupper = 1.01 × 105 Pa
Temperature , T = 288.09K
Density , ρ = 1.224kg /𝑚3
Kinematic viscocity, υ = 1.78869 × 10−5𝑁𝑠/𝑚2
Freestream velocity,𝑈∞ =0.2
√1.4×287×288.15= 68.059m/s
Based on Figure 6.19, the shear stress is calculated as follows:
τupper surface = τlower surface = μdu
dy
= (1.78869 × 10−5) (𝑈∞
0.025𝑐)
= (1.78869 × 10−5) (68.059
0.025(0.08))
= 0.6087Pa
N = − ∫ ((1.01 × 105)cos2.1° + (0.6087)sin2 .1°)𝑑𝑆𝑢𝑝𝑝𝑒𝑟
0.08
0
+ ∫ ((1.01 × 105)cos2.1° − (0.6087)sin2 .1°)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
0.08
0
= −3.5690 × 10−3N
A = ∫ ((−1.01 × 105 )sin2.1° + (0.6087)cos2.1°)𝑑𝑆𝑢𝑝𝑝𝑒𝑟
0.08
0
+ ∫ ((1.01 × 105)sin2.1° − (0.6087)cos2.1°)𝑑𝑆𝑙𝑜𝑤𝑒𝑟
0.08
0
= 97.327 × 10−3N
L = (−3.5690 × 10−3 ) cos(5°) − ( 97.327 × 10−3 ) sin(5°)
= −0.0044N
169
(II) Lift coefficient at different angle of attack
The lift coefficient can be obtained based on the classical thin airfoil theory:
the symmetrical airfoil. To do so, the following formulae is used to determine
theoretically which is as follows:
Sample calculations for elevator at 𝟓°
𝐶𝐿 = 2𝜋 (5 ×𝜋
180) = 0.1097
Sample calculation for rudder at 𝟓°
𝐶𝐿 = 2𝜋 (5 ×𝜋
180) = 0.1097
Velocity at different angle of attack
Sample calculations for elevator at 𝟓°
𝑉 = √2(0.0053)
(1.224)(0.05)(0.5483)
𝑉 = 0.5603
Sample calculation for rudder at 𝟓°
V = √2(0.0049)
(1.224) (0.24)(0.5483)
𝑉 = 0.8568 m/s
𝐶𝐿 = 2𝜋𝛼
𝑉 = √2𝐿
𝜌𝑆𝐶𝐿
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Table 6.5 Table of lift, lift coefficient and velocity at different angle of attack for
elevator
Figure 6.22 Graph of lift against angle of attack for elevator
Figure 6.23 Graph of velocity against lift for elevator
171
Table 6.6 Table of lift, lift coefficient and velocity at different angle of attack for
rudder
Angle of Attack (Degree) Lift Force(N) Lift Coefficient Velocity(m/s)
-25 -0.0444 -2.7416 1.1498
-20 -0.0366 -2.1932 1.1683
-15 -0.0286 -1.6449 1.1926
-10 -0.0204 -1.0966 1.2333
-5 -0.0120 -0.5483 1.3393
0 -0.0036 0.0000 0.0000
5 0.0049 0.5483 0.8568
10 0.0134 1.0966 0.9986
15 0.0217 1.6449 1.0392
20 0.0299 2.1932 1.0560
25 0.0379 2.7416 1.0627
Figure 6.24 Graph of lift against angle of attack for rudder
Figure 6.25 Graph of velocity against lift for elevator
172
Based on the analysis of lift vs angle of attack and velocity vs lift, we can
observe that increasing in angle of attack causes increasing of lift as well as lift
coefficient. On the other hand, deflecting the control surfaces downwards from the
stationary position causes the velocity to increase whereas on the opposite direction of
the deflection of the control surfaces(upwards), the velocity drops gradually. The
maximum lift exerted on the elevator and rudder at an angle of attack of -25° is
0.0441N and 0.0444N which have the velocity of 0.7247m/s (elevator) and 1.1498m/s
respectively. Hence, to determine the torque generated by the control surfaces, we can
compare the maximum lift generation on both control surfaces multiplying with the
length of the push and pull rod which is connected to the servo motor. These values
are eventually with the servo torque in order to determine the effectivity and efficiency
of the servo motor. The mathematical expression to calculate the torque for rudder and
elevator are as follows:
Figure 6.26 Free body diagram of torque calculation
Elevator
Torque, τ = r × Fsinθ
= (0.18) × Fsinθ
= (0.18) × 0.0441sin25°
= 3.3547 × 10−3N.m
Rudder
Torque, τ = r × Fsinθ
= (0.12) × Fsinθ
= (0.12) × 0.0441sin25°
= 2.2365 × 10−3N. m
173
Based on the results obtained, the servo torque is greater than the calculated
torque. Hence, we can conclude that servo motor torque is sufficient to withstand the
lift of both the rudder and elevator surfaces.
6.4 Strength of adhesive on the joint between tail and fuselage
Generally, the construction of the tail part and fuselage are done separately.
These will then be assembled as per the plan or drawing. To do so, a proper selection
of glue is vital as well as analysis on the strength of the glue to withstand the force
generated by the tail on the attachment(fuselage) is closely monitored as to prevent
structural damage due to stress accumulation and overstressing the fuselage. On the
other hand, adhesive bond strength is usually measured by the simple single lap shear
test. Since the stress distribution in the adhesive is not uniform over the bond area, the
reported shear stress is lower than the true ultimate strength of the adhesive. Hence,
for our design, an analysis on the shear strength is done through plotting a shear
diagrams for horizontal and vertical stabilizers at three different load factors. Such
results are compared with the shear strength of the selected glue based on the technical
manual. The steps of the analysis are as follows:
6.4.1 Selection of Glue and Comparison on Theoretical Strength against Glue
Strength based on Technical Data
For our design purpose, the proposed glue for the attachment between fuselage and
horizontal and vertical tails are by means of using Gorilla Glue. The following Figure
6.27 shows the physical properties of the glue as well as its benefits. The details are as
follows:
174
Figure 6.27 Features and benefits of proposed glue
From the Figure 6.27, the theoretical shear stress of horizontal stabilizer is
272.5Pa ≤ τ ≤ 1.36KPa whereas the theoretical shear stress of vertical stabilizer is
102.19Pa ≤ τ ≤ 510.96Pa. By comparing the theoretical shear stress with the shear
strength of the glue (stated above) which is 3500 psi(24.13MPa), the glue has high
tendency to withstand the calculated shear stress and this is turn can made a conclusion
that the structural integrity of the attachment between tails and fuselage are maintained
to an airworthy condition.
175
CHAPTER 7
FLIGHT PLANNING AND TESTING
7.1 Flight Test
Flight testing is developed and gathers data during flight of an aircraft, or
atmospheric testing of launch vehicles and reusable spacecraft, and then analyzes the
data to evaluate the aerodynamic flight characteristics of the vehicle in order to
validate the design, including safety aspects.
The flight test phase accomplishes two major tasks:
1) finding and fixing any design problems
2) verifying and documenting the vehicle capabilities
The flight test phase can range from the test of a single new system for an
existing vehicle to the complete development and certification of a new aircraft.
7.1.1 Pre-Flight Test
The Aircraft
Always do a detailed check on the plane and take time over it. There have been
many examples of jacking pads left on aircraft and tapes covering hinges. If the aircraft
has been cleaned or painted, pay careful attention as these activities can give rise to
numerous “knock on” technical issues such as pitot or sensor damage.
Remember also all those systems that may have been required to be put into
the Ground Test position to allow certain ground checks to be completed prior to flight
clearance. Know what they are and make sure that they are all correctly re-positioned
176
to the flight position prior to flight. Apply the principle that if it can happen, it will
happen. Your job as checker is to ensure that there is no adverse effect on the flight.
You will also need to think carefully about the weight and Centre of Gravity
(CG) for the check flight. An advice would be to try to put the aircraft into a weight
and loading situation with which you feel comfortable and use it as a standard for all
subsequent similar flights. Set up a mid CG if possible, avoid being on the limits and
do consider the effect of the weight and CG on the expected “feel” of the controls.
Expect that the aircraft will inevitably be much lighter than the aircraft on the line. No
big problem there, but think about it and consider the speeds to be used in relation to
stall speed and Minimum Control Speed. It may be that whilst you would normally be
stall speed limited, you may now be on or near the Minimum Control Speed in the Air
(Vmca) limits.
Airfield
The airfield to be used is rarely a choice matter but it is wise to consider any
implications stemming from the airfield itself. The runway capability and its effect on
performance, high ground and obstacles, all need to be considered as well as the
general operational situation.
Weather
During certification development flight testing, the weather criteria often drive
the ability to carry out a given test. However, in the check flight world it is rare to have
the privilege of waiting for perfect weather. That said, it is certainly wise to know what
the bottom line is for the checks to be undertaken.
Checklists
The check crew will need to be able to think and work the standard checklist
(whilst still understanding and recognising its importance) and be comfortable doing
so. Checklists should still be used but they should be used for guidance and not treated
as if they are the Law. No checklist can cover all check situations.
177
7.1.2 Preparation
Flight test preparation begins well before the particular aircraft is ready to fly.
Each single test is known as a Test Point. The document used to prepare a single test
flight for an aircraft is known as a Test Card. This will consist of a description of the
Test Points to be flown. Once the flight test data requirements are established, the RC
plane is instrumented to record that data for analysis. The instrumentation parameters
recorded during a flight test for this project are:
Atmospheric (static) pressure and temperature;
Dynamic pressure and temperature, measured at various positions around the
fuselage;
Structural loads in the wings and fuselage
Aircraft attitude, angle of attack, and angle of sideslip;
Accelerations in all six degrees of freedom, measured with accelerometers at
different positions in the aircraft
Battery performance parameters (pressure and temperature at various stages).
Specific calibration instruments, whose behavior has been determined from
previous tests, may be brought on board to supplement the aircraft's in-built probes.
7.1.3 Execution
When the RC plane is completely assembled, ground testing is then conducted.
This allows exploring multiple aspects: basic aircraft vehicle operation, battery
performance and provides a first look at structural loads. The aircraft can then proceed
with its maiden flight, a major milestone in any aircraft.
There are several aspects to a flight test program, among which:
Handling qualities, which evaluates the aircraft's controllability and response
to pilot inputs throughout the range of flight;
178
Performance testing evaluates aircraft in relation to its projected abilities,
such as speed, range, power available, drag and so forth;
Avionics/systems testing verifies all electronic systems perform as designed;
Structural loads measure the stresses on the airframe, dynamic components,
and controls to verify structural integrity in all flight regimes.
Emergency situations are evaluated as a normal part of all flight test program.
For example, battery failure during various phases of flight (takeoff, cruise, landing)
and systems failures. The primary goal of a flight test program is to gather accurate
engineering data, often on a design that is not fully proven.
7.1.4 Analysis and reporting
It analyze the internal and outer part of the flight by checking all parts.
Reporting includes the analyzed data result. Aircraft performance has various missions
such as takeoff, climb, cruise, descent and landing. Performance charts allow a pilot
to predict the takeoff, climb, cruise, and landing performance of an aircraft. We could
record the flight data and create performance charts based on the behavior of the
aircraft during the test flights. By using these performance charts, a pilot can determine
the runway length needed to take off and land and the time required to arrive at the
destination.
It is important to remember that the data from the charts will not be accurate
if the aircraft is not in good working order or when operating under adverse conditions.
Each aircraft performs differently and, therefore, has different performance numbers.
Compute the performance of the aircraft prior to every flight, as every flight is different.
Every chart is based on certain conditions and contains notes on how to adapt the
information for flight conditions.
179
7.2 Flight Test Location
The flight test location would be the defined total airspace that model aircraft
should always stay within while in the air. This area should be clear of unprotected
people, vessels, vehicles or structures. Any obstacles, structures, or areas where people
could be within this defined area should be clearly marked so that pilots know to not
overfly them. The location selected for flight testing would be the Marching
Paddock in UTM which is also known as Padang Kawad UTM to the locals there.
The particular location is selected because it has a very open environment which
is perfect for many students to test their plane. The area has an approximate
measurement of 200m x 100m. Furthermore, it could sustain a large crowd.
Since the class comprises of 39 individuals including two lecturers, there would
not be any problem in terms of over -crowding the flight test location.
Figure 7.1 Location of flight test from Google Map
180
Figure 7.2 UTM Marching Paddock (Padang Kawad UTM)
However, the only obstacle encountered is the flying surface that is made of
tar. It could damage the plane when it belly lands or crashes if anything goes wrong
during the flight test. It could scrape off the surface of the plane making it less efficient.
The potholes present could also damage the plane severely.
7.3 RC Airplane pre-flight checklist
Table 7.1 Flight test checklist
Item Description Completion
Covering and
surface
No punctures in the covering or exterior surface
of the plane (any damage may indicate structural
damage or damaged linkage or wiring)
Landing gear Do the tires roll freely?
Do the tail wheel connected well with the rudder?
Propeller
Propeller secured properly
No scratches, bends, nicks on the propeller
No debris in the propeller unit
Motor
Does it spin freely without making any scraping
or rattling noises?
Does it operate smoothly without excessive
vibration?
181
Battery
Is the battery securely fastened in place?
Is the battery fully charged and registering the
proper maximum voltage?
Is the battery free of damage and the pack does
not look puffy?
Electrical
connections
Are all receiver, servo, ESC and battery are fully
plugged‐in and connected properly?
Are wires affixed firmly in place by tie wraps/zip
ties, or fasteners
Are electrical connections routed away from
servo linkage and servo horn travel paths?
Servos and
linkages
All servos are secure, and linkages to servo and
control surfaces are secure
Servo horns and control horns are secure and not
loose
Servo linkages are able to move freely and are
not binding
Receiver
Is the receiver fully intact and connected?
Is the receiver antenna located away from other
electrical wiring inside the fuselage?
Transmitter
Is the transmitter battery fully charged?
The receiver is connected with transmitter to
allow the information transmission
Range check is completed successfully
Balance and
centre of
gravity
The airplane does not roll aggressively to one
side when held in the centre of the nose (by the
prop) and by the center of the tail?
Does the model balance properly when supported
at the C.G. point?
Control
surfaces
All control surface hinges are secured properly to
its respective flying surface
Make sure pushrods, linkages, clevises and
hinges are not loose
182
Aileron
Are the ailerons moving as they should in
opposite directions?
When you push the right gimbal right, does the
right aileron (looking from the rear) move up and
the left move down as they should?
Elevator
Do both elevators (if applicable) move
uniformly?
When you push the right gimbal up, does the
elevator move downward?
When you pull down on the right gimbal does the
elevator move upward?
Rudder
Does the rudder move left when you move the
left gimbal to the left as it should?
Does it move right when you move the left
gimbal to the right as it should?
7.4 Performing Range Check
The purpose of the range check is to make sure the radio signal from transmitter
to receiver is strong, so that RC airplane can be operated at a normal distance away
without it going out of radio range. This is very important because if the plane does go
out of range, then all control will be lost resulting in crashing of the plane. The
procedures are as followed:
1. Turn on the transmitter and then the receiver.
2. Make sure the radio is in range check mode. This is done by pressing the F/S
button of the transmitter module for 4 seconds. As a result, the RED LED of
the transmitter module will be off and the GREEN LED will be flashing rapidly.
The effective distance is decreased to 1/30 of full range.
3. The range checker slowly walk away at least 30 metres from the airplane while
operating all the controls on the transmitter. All the controls should operate
without any problems. Do not fly the airplane if the control surface response
becomes unreliable before reaching at least 30 metres away from the plane.
183
4. In order to exit the range check mode, the F/S button of the transmitter module
is pressed again. As a result, the RED LED is on to indicate that it is back to
normal operation.
7.5 Meteorological Conditions on Site
It is always necessary to check weather conditions at the site on arrival,
particulary windy and be aware that they can change. The maximum permissible wind
speed including gusts of 13.4 m/s is recommended. The pilot should be mindful of
complex wind profiles as it can occur in all types of terrain. If wind blows a model
with landing gear backwards on a smooth surface it is considered too windy to enjoy
flying. Carrying a handheld anemometer is suggested to check the wind conditions if
they are within operational ranges. However, without anemometer common sense
would tell us if the weather is too windy for flying just by feeling it.
Figure 7.3 Johor average and max wind speed and gust (kmph)
The visibility here is also very clear. It would go around 10 km which is
great for flight testing. The atmospheric pressure would be 1010 mb. The
daytime temperature is going to reach 32 °C and the temperature is going to dip to
25 °C at night. Hence, the testing would be done during day time when the sky is clear.
184
Figure 7.4 Average temperature of location selected
Figure 7.5 Average rainfall days of location selected
In theory, a model aircraft weighing less than 7kg could be flown to any
altitude, with the only limitation being the requirement to keep it in sight. From 7 kg
to 20 kg the altitude is 400 ft, unless it meets the airspace requirements. In practical
terms, the requirement to keep the aircraft in sight "unaided" is the main limitation and
that restricts the altitude of smaller models in particular. Hence, we have decided to
opt to fly our plane at an altitude of 10m so that it is clearly visible in sight and ensure
that it does not collide with anything, especially other aircraft.
7.6 Flying Site
Airspace of Flying Site is the flight area, or “box”, inside of which all flying is
to take place. The flight area would be the defined total airspace that model aircraft
185
should always stay within while in the air. This area should be clear of unprotected
people, vessels, vehicles or structures. Any obstacles, structures, or areas where people
could be within this defined area should be clearly marked so that pilots know to not
overfly them. The size of this “box” should be the first consideration, based on the
type of aircraft anticipated to operate on this site.
For RC scale models up to approximately 72” wingspan and is a typical size
found at many club flying sites. These models will be able to fly in a much smaller
area, about 300’ x 150’, roughly the size of a soccer field.
The flying site should consist of:
Barrier:
Designed to stop models from veering into pilots’ and/or spectators’ positions.
For example, plastic or chain-link fencing, shrubbery, etc. These may run the length
of the flight line or be short to protect a single pilot station. If using metal fencing
(chainlink) consider that the transmitter antenna needs to be in the clear of any metal
that could cause loss of signal to the aircraft.
Runway:
A runway should be designated within the overfly area on the site. This can be
grass, dirt, geotextile or hard-surface. If space allows, it might be desirable to have two
runways to be able to handle most all wind directions.
Safety Line:
Establishes the area in front of which all model flying must occur. Only
personnel associated with flying the model aircraft are allowed at or in front of the
safety line. This line can be straight, segmented, curved, or even box-shaped so that
the pilot and spectators are behind the safety line while aircraft are flying. Under
certain conditions it may be possible to achieve a flying area covering almost 360°.
186
Pilot Line/Station(s):
Where all pilots will stand while flying model aircraft.
Pilot Pit Area:
Where pilots and the team stage and service their models.
Spectator line:
Where spectators can view the action. This can be a simple line of separation
or a nice viewing area complete.
Safety equipments:
First-aid kit and fire extinguisher with appropriate ratings (especially for Li-
PO fires) and sand bucket for Li-PO batteries.
Figure 7.6 The designed flying site
187
REFERENCES
[1] International Civil Aviation Organization (ICAO), “Unmanned Aircraft
Systems (UAS),” Canada, 2011.
[2] D. Irvine, “Doing military’s dangerous, dull and dirty work,” Cable News Network (CNN), 16-Feb-2012.
[3] S. Hayat, E. Yanmaz, and R. Muzaffar, “Survey on Unmanned Aerial Vehicle
Networks for Civil Applications: A Communications Viewpoint,” IEEE Commun. Surv. Tutorials, vol. 18, no. 4, pp. 2624–2661, 2016.
[4] D. Lowe, “Radio Electronics: Transmitters and Receivers,” dummies: A Wiley Brand, 2016. [Online]. Available:
https://www.dummies.com/programming/electronics/components/radio-electronics-transmitters-and-receivers/. [Accessed: 05-Apr-2020].
[5] FAI Aeromodelling Commision (CIAM), “Frequencies | World Air Sports Federation,” The World Air Sports Federation, 2017. [Online]. Available:
https://www.fai.org/page/frequencies. [Accessed: 12-Apr-2020].
[6] B. Nadler, “Electric RC Flying for Cheapskates,” f/22 Press, 2008. [Online]. Available: https://books.google.com.my/books?id=spmYA7TNq10C&pg=PA56&lpg=P
A56&dq=72MHz+is+usually+used+for+model+aircraft+in+USA&source=bl&ots=SeAyfP-HDc&sig=ACfU3U0Smj85CMfFrid4og5NCh89jcPAAA&hl=en&sa=X&ved=2ahUKEwij8sqD5uHoAhUFxTgGHawMAUsQ6AEwD3oECA0QKQ#v=on
epage&. [Accessed: 12-Apr-2020].
[7] Anderson, Anderson, J. D., & Anderson, J. D. (1970, January 1).
Fundamentals of Aerodynamics. Retrieved from
https://www.abebooks.com/book-search/isbn/0072373350/ [8] Frederick, G., Kaepp, G. A., Kudelko, C. M., Schuster, P. J., Domas, F.,
Haardt, U. G., & Lenz, W. (1995). Optimization of expanded polypropylene
foam coring to improve bumper foam core energy absorbing capability. SAE transactions, 394-400.
[9] Barua, P., Sousa, T., & Scholz, D. (2013). Empennage statistics and sizing
methods for dorsal fins. Hamburg University of Applied Sciences, Hamburg, Germany.
188
APPENDICES
Appendix A Subsonic airfoil selection lift-curve slope
Appendix B Two dimensional lift-curve slope
189
Appendix C Two dimensional lift-curve slope
Appendix D Lift ratios – slender body theory
190
Appendix E Horizontal tail location factor
Appendix F Theoretical low speed aerodynamic characteristic of various airfoil mean lines
191
Appendix G Subsonic maximum lift of high aspect ratio wings
Appendix H Effect of Reynolds number on section maximum lift
Appendix I Mach number correction for subsonic maximum lift of high aspect ratio wings
192
Appendix J Angle of attack increment for subsonic maximum lift of high aspect ratio wings
Appendix K Static and dynamic loading of 80grams/litre EPP Foam
Appendix L EPP Foam properties at different densities
193
Appendix M Bending-Buckling Coefficient of Plates as a Function of a/b for
Various Amounts of Edge Rotational Restraint.
Appendix N Shear-Buckling-Stress Coefficient of Plates as a Function of /b
for Clamped and Hinged Edges
194
Appendix O Chart of Nondimensional Compressive Buckling Stress for Long Clamped
Flanges and for Supported Plates with Edge Rotational Restraint.
Appendix P Chart of Nondimensional Shear Buckling Stress for Panels with Edge-
Rotational Restraint.
195
Appendix Q Landing gear drawing
196
Appendix R Aircraft wheel drawing
197
APPENDIX: MINUTE MEETING
MINUTES OF MEETING 1
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 17 & 20 February 2020
TIME : 8.00 p.m. – 10.00 p.m.
PLACE : KO1 KTR
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 8.00 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Group formation
198
ii. Discussion on project title
iii. Task assignation to members
3.0 Discussion
i. Group formation
A group of 11 members are divided into five sub-
group and chairperson and secretary for each sub-
group are chosen based majority agreement
between the group members.
ii. Discussion on project title
First of all, team members went through the project
title as a whole and get some basic knowledge
regarding the RC plane aircrafts. After that, the
process and importance of feasibility study were
researched.
iii. Task assignation to members
- Task is assigned by chairperson to all team
members for information searching
purpose and to distribute the tasks to be
completed.
- Gather information on existing on the RC
plane design as follows:
o Group 1: Propulsion system
o Group 2: Avionics/Control
o Group 3: Wing
o Group 4: Tail, Fuselage and
Landing Gear
o Group 5: Piloting
All members
All members
All members
4.0 Closing
199
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
iii. Announced the meeting will be held on 24
February 2020 (Monday) at 2.00 p.m. at
P20 Aerolab.
Chairperson
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
200
MINUTES OF MEETING 2
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 24 & 27 February 2020
TIME : 2.00 p.m. – 4.00 p.m
PLACE : P20 Aerolab
Group Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 2.00 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Discussion on type of RC plane to be
designed
201
ii. Analyzing the appropriate sizing and
dimension for the selected RC plane
iii. Determination of electronic parts
3.0 Discussion
i. Discussion on type of RC plane to be
designed
Types of RC planes designed were analyzed and
only one type of aircraft designed were chosen
based on the majority agreement from the group
members.
ii. Analyzing the appropriate sizing and
dimension for the selected RC plane
The entire group members were split into sub-
groups to analyze detailly every structure of the
aircraft. Besides that, the selected RC plane models
is scaled down based on our theoretical analyses.
iii. Task assignation to members
(a) Task is assigned by sub-group chairpersons to their
team members for information searching purpose
and to distribute the tasks to be completed.
- Detail explanation for an appropriate
sizing and dimension of the structures
All members
Sub – group
members
Sub-group
members
4.0 Closing
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
Chairperson
202
iii. Announced the meeting will be held on 02
March 2020 (Monday) at 8.30 p.m. at
KTR K01
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
203
MINUTES OF MEETING 3
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 02 & 04 March 2020
TIME : 8.30 p.m. – 10.45 p.m.
PLACE : KTR K01
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 8.30 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Discussion on sizing and dimension of the
structures
204
ii. Study on material properties of the
selected material
iii. Progress from sub-group members
3.0 Discussion
i. Discussion on sizing and dimension of the
structures
All the sub-group members were shared their
current information on their parts for sizing and
dimension. Some of the dimensions need to be
modified based on the lecturer advice.
ii. Study on material properties of the selected
material
Three of the members were asked to do a study on
the material properties to compare the strength of
the materials with the calculated values. This
include yield stress, ultimate yield stress and so
forth of the selected materials.
iii. Update from sub-group members:
o Group 1: Background Study on
Electronic Parts Selection which includes
motor, propeller, battery and electronic
speed controller selection.
o Group 2: Determination of centre of
gravity (C.G)
o Group 3: Determination of wing
configuration and wing loading
o Group 4: Theoretical analysis of fuselage
which includes shear and bending
moment diagram for different load
factors. Similar concepts also performed
for empennage. Studies on the proper
All members
All members
Sub-group
members
Sub-group
members
205
selection of the configuration of the
landing gear.
o Group 5: Pilots are selected for the
training purpose by the team members
agreement.
4.0 Closing
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
iii. Announced the meeting will be held on 09
March 2020 (Monday) at 8.15 p.m. at
KTR K01.
Chairperson
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
206
MINUTES OF MEETING 4
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 09 & 11 March 2020
TIME : 8.15 p.m. – 11.25 p.m.
PLACE : KTR K01
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 8.15 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Progress of the sub-group members
3.0 Discussion
207
i. Progress of the sub-group members:
o Group 1: Final Selection of Electronic
Parts which includes motor, propeller,
battery and electronic speed controller.
o Group 2: Confirmation of centre of
gravity (C.G) upon fixing the sizing and
dimensions of the structures. Analysis on
stability were theoretical calculated.
o Group 3: Confirmation of wing
configuration and wing loading.
Theoretical analysis on the lift
distribution of the control surfaces were
estimated based on the proved methods.
o Group 4: Addition of theoretical analysis
of fuselage which includes shear and
bending moment diagram for different
load factors. Similar concepts also
performed for empennage. Confirmation
on the proper selection of the
configuration of the landing gear upon
agreement from all the members
o Group 5: Progress of the training is
discussed on the selected pilots who have
undergone simulation training with
lecturer.
All members
All members
All members
4.0 Closing
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
Chairperson
208
iii. Announced the meeting will be held on 15
March 2020 (Sunday) at 2.00 p.m. at P20
Aerolab and KTR K01.
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
209
MINUTES OF MEETING 5
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 15 March 2020
TIME : 2.00 p.m. – 4.00 p.m. & 8.15p.m -12.30 a.m.
PLACE : P20 Aerolab & KTR K 01
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 2.00 p.m.&8.15p.m.
and thanked the committee members for being
present
Chairperson
2.0 Outline of the meeting
i. Compiling all the team members work for
first draft submission
210
ii. Preparation for video shooting of the
presentation
3.0 Discussion
i. Compiling all the team members work for
first draft submission.
All the team members work is combined and
rearrange according to the team members
agreement. The flow of the report was also
discussed and assigned before compiling all the
materials into the first draft report for submission.
ii. Preparation for video shooting of the
presentation
Location for the video shooting were determined
and all the props were prepared before
commencing the video shooting.
All members
All members
4.0 Closing
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
Chairperson
211
iii. Announced the meeting will be held on 16
March 2020 (Monday) at 2.00 p.m. at P20
Aerolab for video shooting
iv. Report and video have been sent to
lecturer on 17th March 2020
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
212
MINUTES OF MEETING 6
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 01 April 2020
TIME : 10.15 a.m. – 11.35a.m.
PLACE : WEBEX
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 10.15 a.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Discussion on report corrected by lecturer
3.0 Discussion
213
i. Discussion on report corrected by lecturer
A detail discussion is done on every part that need
to be modified based on the lecturer advice. Every
members of the sub-group are advised to create
their own what’ sup group to discussed the
corrections. The corrections that need to be done
are as follows:
1. For location of CG, give the dimension/
distance
2. Support the wing, fuselage and tail
configuration with value, not only description.
3. Explain reasons or background study of
electronic parts selection
4. In aircraft specification slide, not only state but
also explain the value and its meaning.
5. Include sizing calculation of control surface
and show actual 3D drawing with colour and
render.
6. Argument /references to confirm that this CG
is correct and stable for aircraft
7. Calculate moment/torque produce by the
control surfaces at different airspeed and
deflection angle and relate with the servo
strength.
8. Confirm the obtained value with current
regulations and procedure.
9. Missing information:
i. Communication signals(transmitter-
receiver)
ii. Control and stability analysis (lateral and
longitudinal stability)
iii. Electronics and wirings
All sub-group
members
214
iv. Flight testing and planning and preparation
v. Project milestone
vi. Structural analysis of attachment part
vii. Thrust available and thrust required.
4.0 Closing
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
iii. Announced a duration of two weeks
interval has been given to modify all the
details and next meeting will be held on 15
April 2020 (Wednesday) at 2.00 p.m. at
WEBEX.
Chairperson
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
215
MINUTES OF MEETING 7
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 15 April 2020
TIME : 2.00 p.m. – 3.15p.m.
PLACE : WEBEX
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 2.13 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Discussion of the progress of work from
all the sub-group members and creation of
one new sub-group for flight testing and
216
planning (combination of piloting and
flight testing and planning)
3.0 Discussion
i. Discussion of the progress of work from all the
sub-group members and creation of one new
sub-group for flight testing and planning:
o Group 1: Updates and modifications on the
aircraft performance and related
calculations of it. Proper selection of
motor, propeller, battery and electronic
speed controller.
o Group 2: Updates and modifications on the
CG locations, stability, communication
signals (transmitter- receiver),
communication resolution and antennas
o Group 3: Updates and modifications on the
shear and bending stress of wing with and
without ailerons and wing loading and lift
distribution on the aileron at different angle
of attack.
o Group 4: Updates and modifications on the
analysis of fuselage, empennage that
consist of shear and bending moment
diagram and flexural analysis. Landing
gear structural analysis also done.
o Group 5: Updates on the flight test that
include pre-flight test, planning, execution
and analysis and reporting. Flight Test
Location is also chosen.
All sub-group
members
4.0 Closing
217
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
iii. Announced a duration of two-week
interval has been given to modify all the
details and next meeting will be held on 29
April 2020 (Wednesday) at 2.00 p.m. at
WEBEX.
Chairperson
Prepared by: Approved by:
Shatha Ten (Shathasivam) (Ten Jia Yee)
218
MINUTES OF MEETING 8
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 29 April 2020
TIME : 2.00 p.m. – 3.25p.m.
PLACE : WEBEX
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 2.10 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Final updates from the sub-group members
3.0 Discussion
219
i. Final updates from the sub-group members
(b) All the sub-group members were asked to confirm
the works that has been done by them. This include
of arrange the titles, sub-title and other information
in orderly manner to ease the work of compiling. It
also emphasized to them to justify the paragraph
and choosing the same font size to standardize the
report. Only some additional update is update by
flight testing and planning sub-group such as RC
plane pre-flight checklist, performing range check,
meteorological condition and site as well as flying
site.
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
All sub-group
members
4.0 Closing
220
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
iii. Announced the meeting will be held on 3
May 2020 (Friday) at 2.30 p.m. at
WEBEX.
Chairperson
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
221
MINUTES OF MEETING 9
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 03 May 2020
TIME : 2.30 p.m. – 3.15 p.m.
PLACE : WEBEX
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 2.35 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Compiling all the team members work for
first draft submission.
222
3.0 Discussion
i. Compiling all the team members work for
first draft submission.
All the team members work is combined and
rearrange according to the team members
agreement. The flow of the report was also
discussed and assigned before compiling all the
materials into the first draft report for submission.
All sub-group
members
4.0 Closing
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
iii. Report has been sent to lecturer on 4th
May 2020 through whatsup
Chairperson
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
223
MINUTES OF MEETING 10
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 11 May 2020
TIME : 9.00 p.m.-10.00p.m.
PLACE : WEBEX
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 9.08 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Correction on the report based on lecturer
advice.
224
3.0 Discussion
i. Correction on the report based on lecturer
advice.
Graphs and diagrams left without explanations and
discussions. Interpretation and descriptions are
validated with comparisons on the current
available aviation laws and regulations. All the
sub-groups are advised to interpret the graphs and
diagrams with proper validations based on the
current available laws and regulations.
4.0 Closing
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
iii. Announced a duration of two-weeks
interval has been given to modify all the
details and next meeting will be held on 25
May 2020 (Monday) at 2.00 p.m. at
WEBEX.
Chairperson
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
225
MINUTES OF MEETING 11
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 25 May 2020
TIME : 2.00 p.m. – 3.45 p.m.
PLACE : WEBEX
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 2.10 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Final updates from the sub-group members
3.0 Discussion
226
i. Final updates from the sub-group members
All the team members work is combined and
rearrange according to the team members
agreement. The flow of the report was also
discussed and assigned before compiling all the
materials into the first draft report for submission.
Members are also advised to check none of the
graphs and diagrams are left unexplained and all
the graphs and diagrams are checked by
exchanging the files among other sub-group
members. Once done, the files are copied and
pasted into new folder and it is arranged based on
the thesis format.
All sub-group
members
4.0 Closing
i. Summarized the decisions made at the
meeting and members are give four days
to check the entire report.
ii. Thanked the committee members for
contributing their suggestions during the
meeting
iii. Announced the meeting will be held on 30
May 2020 (Tuesday) at 12.30p.m at
WEBEX.
Chairperson
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
227
MINUTES OF MEETING 12
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 01 June 2020
TIME : 12.30 p.m. – 2.10 p.m.
PLACE : WEBEX & WHATSAPP
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 12.35 p.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Confirmation on the checking of
unexplained graphs and diagrams of other
sub-group members
228
3.0 Discussion
i. Confirmation on the checking of unexplained
graphs and diagrams of other sub-group
members.
(k) All the members were confirmed of their checking
process before compiling is executed. The files are
then compiled based on the suggested thesis
format. Proper page number, font, font size,
justified and space were emphasized before the
submission of the report to the lecturer.
All sub-group
members
4.0 Closing
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting
iii. Completed report is submitted to the
lecturer through whatsup on 01 June 2020.
Chairperson
Prepared by: Approved by:
Shatha Ten
(Shathasivam) (Ten Jia Yee)
229
MINUTES OF MEETING 13
AIRCRAFT DESIGN 2
SCHOOL OF MECHANICAL ENGINEERING
DATE : 17 June 2020
TIME : 10.30 a.m. – 11.10 a.m.
PLACE : WHATSAPP
Members Present:
1. Ten Jia Yee (Chairperson)
2. Shathasivam a/l Parumasivam (Secretary)
3. Muhammad Imran Aiman Idris
4. Siti Mastura binti Maskor
5. Nur Amyra Mohd Aseme
6. Nur Aizat Nazihah Azmi
7. Melvin John
8. Thevan Tangaraju
9. Shakgantan Balakrishnan
10. Wee Jun Wee
11. Athiseshan Balan
Members Apologies: -
NO SUBJECT ACTION BY FEEDBAC
K
1.0 Chairperson Address
Called the meeting to order at 10.30 a.m. and
thanked the committee members for being present Chairperson
2.0 Outline of the meeting
i. Video preparation for each sub-group
ii. Preparation for submission of final report
3.0 Discussion
230
i. Video preparation for each sub-group (l)
(m) All the sub-group members were advised to do
their parts presentation based on their discussion
within the group and the shared video is compiled
by Thevan A/L Thangaraju.
ii. Preparation for submission of final report
All the sub-group members are instructed by the
chairperson to check all the particulars are
available before the report is compiled. Secretariat
(Shatha) is instructed to compiled the final report
with the accumulated minutes of meeting.
All sub-group
members
4.0 Closing
i. Summarized the decisions made at the
meeting
ii. Thanked the committee members for
contributing their suggestions during the
meeting and their cooperation in
performing the duties for this subject to be
accomplished.
iii. Completed Final report is submitted to the
lecturer through whatsup on 30 June 2020.
Chairperson
Prepared by: Approved by:
_Shatha___________ Ten
(Shathasivam) (Ten Jia Yee)