This document provides details of the third weight estimation for a small surveillance aircraft model. The total weight from the second estimation is 1045.3g. Design parameters like a NACA 2414 airfoil with 16cm chord, 1m wingspan, and 45.38N/m^2 wing loading are assumed. Balsa wood is selected as the construction material. Component weights like power plant (256g), payload (120g) are known. The third estimation will account for additional structural weights of the wings, fuselage, tail surfaces, and fittings to obtain the final total weight.

1. INDEX
Sr.
No.
Name of Experiments
Page
No.
Signature
1. Mission goal decision & First weight estimation
2. Second weight estimation
3. Wing loading and thrust to weight ratio
4. Third weight estimation
5. Sizing and 3-D layout
6. Propeller design
7. Improved drag polar
8. Calculation of center of gravity
9. Stability analysis
10. Revised Performance Calculations

2. 1. Mission Goal Decision & First Weight Estimation
Introduction
Our mission is to build a small flying aircraft model. To venture in to the art of flying, model
aircraft are comparatively cost effective and easier, hence in general more fascinating. This way
one can develop skills both in building and flying aircrafts (E.g.: Radio Controlled (RC)
aircrafts). Such aircrafts are also being used for military and scientific purposes like weather
monitoring, reconnaissance, surveillance etc.
Powered models contain an onboard power-plant to propel the aircraft such as electric motor,
internal combustion etc. The power from these units is used to rotate the propeller which
generates the required thrust for the aircraft. As the present mission criterion involves a pilotless
small aircraft, the radio control is preferred.
Model aircrafts use lightweight materials such as polystyrene, balsawood, foam and fiber-glass.
The construction of flying models consist of forming the frame of the model using thin strips of
light wood such as balsa, then covering it with fabric and subsequently doping the fabric to
form a light and sturdy frame.
Based on the Systems Engineering approach, an aircraft will be designed during three phases:
1. Conceptual design phase,
2. Preliminary design phase, and
3. Detail design phase.
In the conceptual design phase, the aircraft will be designed in concept without the precise
calculations. In another word, almost all parameters are determined based on a decision making
process and a selection technique. The preliminary design phase tends to employ the outcomes
of a calculation procedure. In the preliminary design phase, the parameters that are determined
are not final and will be altered later.
Missions of Surveillance Aircraft
• Reconnaissance
• Traffic monitoring
Preliminary Design Consideration
It is assumed that the preliminary weight estimation has already been carried out, and the
airplane performance and structural capabilities have been worked out. The preliminary design
parameters taken into consideration for design are given in the table below. These may be
altered at a later stage depending on the better estimate of parameters.

3. Table 1.1
Parameter Preliminary Estimate
Weight ~1.2 kg
Wing Span ~1.25m
Length ~0.9m
Wing area ~0.14 m2
Wing type High wing or Low wing (with dihedral)
Aspect Ratio ~8 to 11
Power plant Battery driven propeller
Control Ailerons, Elevators, Rudders, Speed
Thrust/Weight To be decided
Range ~1.5km
TABLE 1.2 Specifications

4. First Weight Estimation
The payload weights for the chosen existing aircrafts are not specified in the open literature.
Hence, we start with the payload weight is 120 gm. and gross weight as 1200gm.
The preliminary weight estimation is done by assuming the payload to be 10 % of the gross
weight. Table shows the power-plant (WPP) and the empty weights (WS) of the some chosen
models aircrafts.
1) Power-plant weight v/s
Gross weight. [Plot 1.1]
y = 0.2421x + 53.261
R² = 0.9347
0
100
200
300
400
500
600
700
0 500 1000 1500 2000 2500
power-plantweightWPP
(ingram)
gross weight Wo (in gram)
plot 1.1
wpp
Linear (wpp)
Linear (wpp)

5. 2) Empty weight v/s Gross weight. [Plot 1.2]
The power-plant weight to gross weight ratio (WPP/ W0) and the empty weight to gross weight
ratio (WS/W0) can be obtained from the slopes of the Plot 1 and 2 respectively. It should be
noted that the power-plant weight includes the battery, motor as well as the propeller weight.
From these values the gross weight can now be estimated using the following relation.
Conclusion
The initial estimated gross weight is ……..gm for the required payload weight of 120 gm. This
weight estimation is preliminary weight estimation which is likely to change during further
process of design.
y = 0.6578x - 52.838
R² = 0.9906
0
200
400
600
800
1000
1200
1400
1600
0 500 1000 1500 2000 2500
structuralweight(Ws)ingram
gross weight (Wo) in gram
plot 1.2
ws
Linear (ws)

6. 2. Second Weight Estimation
Mission Requirements
For our mission we decided to choose an airfoil which has very low value of stalling speed. To
fulfill this requirement we should have higher value of CLmax. And (L/D) max. These two
parameters are of prime importance for choosing airfoil. Further the high (L/D) max is more
important because higher the value of (L/D) max, lower will be the minimum drag. This drag
multiplied by the cruise velocity will give the power required corresponding to minimum drag
condition. During ascent, the wing is flying at a lower speed and therefore at a higher lift
coefficient. The rate of climb is a function of the excess horsepower available so, the lower the
drag, the more power is available to gain altitude.
For our design and weight estimation consideration we chose different NACA 4digit airfoils
and having stalling angle of attack in the range of 13-15 degrees and under cambered for good
low speed lift.

7. Selection of Airfoil
After investigating all the above airfoil data and it’s characteristics we decided to choose
NACA-2414 as its features fulfill our requirement. This airfoil has highest value of both Clmax
and (L/D) max compared to the above five airfoils.
Calculation Based On NACA-2414
Assuming:
Chord Length (c): 16 cm.
Wingspan (b): 1.0 m
Density of air estimated in RTU: 1.15 kg/m3
Viscosity of air estimated in RTU: 1.983*10-5 N-sec/m2
Aspect ratio (AR) = b/c
AR= 1/0.16=6.25
Designed cruising speed = 18.65m/s
Reynolds’s number = 173050.93
The following values were obtained at Reynolds no. of 173050.93
Table 2.2 Values obtained at Reynolds no. of 173050.93 for NACA 2414
Sr.No AOA(Degree) Cl Cm Cd
1 -3 -0.093 -0.052 0.013
2 -2 0.028 -0.052 0.0125
3 -1 0.149 -0.053 0.0124
4 0 0.27 -0.053 0.0122
5 1 0.392 -0.054 0.0118
6 2 0.513 -0.054 0.0122
7 3 0.634 -0.054 0.0126
8 4 0.754 -0.055 0.0136
9 5 0.871 -0.055 0.0144
10 6 0.972 -0.056 0.0158
11 7 1.061 -0.056 0.0175
12 8 1.138 -0.057 0.0188
13 9 1.204 -0.057 0.0208
14 10 1.259 -0.058 0.0225
15 11 1.303 -0.058 0.0253

8. Power plant Weight Estimation
From drag polar plot,
Zero lift drag coefficient, Cd0 = 0.012615
Ostwald’s efficiency factor for rectangular wing, e = 0.85
So, K = 0.0599 ≈ 0.06
Therefore, drag polar equation for our airfoil is
Cd = Cd0+ KCl2
Cd= 0.012615+ 0.06Cl2
For (L/D) max condition, we have
Cd0= KCl2
Cl= Cl=0.4585
Also Cd= 2*Cdo
Cd = 0.02523
Cl / Cd = 18.1728
For steady level flight,
Thrust (T) = Drag (D)
& Lift (L) = Weight (W)
W = 1.016*9.81 = 9.96696 N
Thrust required, T= W / (Cl / Cd)
T = 0.54845 N
Power required for cruising speed,
P = T × V
P = 0.54845 × 18.65
P = 10.228 Watts
Motor
Motor selected - TURNIGY AerodriveXp, Model: TR35-30C
Specification: -
• Input Voltage: 7.4V~11.1V (2~3S Li-po)
• KV: 1100 rpm/V within 10%
• Max. Efficiency current: 10~15A
• Dimensions: 35mm X 30mm / 1.38in X 1.18in
• Shaft diameter: 4mm / 0.16in
• Weight: 77g / 2.72oz
• Number of poles/magnets: 14
• Recommended model weight: 600~1300g / 21~45oz
• Power equivalent: .12, 2stroke
• Stator Dimensions: 28-08

9. Battery
• Electricity Power Series LiPo Battery:
• Model: GPMP0504
• Description: 2S 7.4V 1300mAh 25C
• Weight: 2.7 oz (77 gm.)
• Dimension: 3.4 x 1.4 x 0.5
Propeller
• 9x6 (length-228.6 cm, pitch-152.4 mm)
• Weight: 26 g
• Some addition equipments (fittings, etc.) weight = 50gm
Payload-RC Plane Camera
Fly DV Micro Video Camera 2GB for RC Airplane (TCSC30041)
This is smallest fly DV on the world! Weight only 80g, bring no overload to your helicopter or
airplane. The Fly Camera is a light video camera that can be used for Model air planes,
Helicopters, RC cars etc.
Specification:-
• Easy to Install
• High Speed Recording & Quick Light
Response
• Automatic photographing every 15 second
• 25 Frames Per Second for 640x480 Video Output
• Motion JPEG Video Encoding and MP4 Format
• High performance Microphone
• Sound Activated Recording
• Built-In Rechargeable Lipo Battery with 2.5 Hour Battery Life
• Memory Capacity: 2G
• USB 2.0
• DV 640*480@25FPS
• Weight: 0.080kg
• Size: 15x10x7cm

10. Formula
• Wo = Wpl+Wpp+Wstr.
• W02 = Wpl + Wpp + (Wstr/Wo) Wo.
Total Power plant Weight (Wpp)
• Wpp = Wmotor + Wbattery + Wprop.+Wextra = 100+80+26+50
The total power plant weight (Wpp = 256 gm.).
Second Weight Estimation
The takeoff gross weight, Wo of an aircraft can be summarized as
The above equation can be simplified for the calculation of structural weight as the fraction of
total take-off weight as,
= (Wpl+Wpp)/ (1-Wstr/Wo)
= 1045.521gm
Conclusion
1. NACA 2414 Airfoil is chosen for our aircraft mission.
2. Second weight estimation is done by approximating the ratio of structural weight to total
weight based on previous report.
3. Second Weight estimation, = 1045.521gm.

11. 3. Wing Loading and Thrust to Weight Ratio
Introduction
Wing loading is defined as the loaded weight of the aircraft divided by the area of the wing. In
most aircraft designs, wing loading is determined by consideration of Vstall and landing distance.
However W/S also plays the role in the maximum velocity of the airplane .Vmax increases as
W/S increases.
The primary constraints on W/S will be Vstall and landing and we will take that approach The
faster an aircraft flies, the more lift is produced by each unit area of wing, so a smaller wing can
carry the same weight in level flight, operating at a higher wing loading. Correspondingly, the
landing and take-off speeds will be higher. The high wing loading also decreases
maneuverability. The wing loading determines the design lift coefficient and influences drag
through its effect upon wetted area and wingspan.
Wing Loading for Stall Conditions
The lift force L on a wing of area S is defined as,
For steady level flight,
L = W
To account for the 3-D effect for the finite aspect ratio, Raymer suggest that, for finite wings
with aspect ratio > 5,
Wing Loading for Take-Off Performance

12. (L/D)VLO =CL/CD = 15.1852
Thrust, T = W*(CD/CL)
= 0.675229 N
Radius of take-off,
R = (6.96 × Vstall2)/g
= 62.423 m
Flight path angle
θOB = cos-1(1 – hOB/R)
= 10.26 degrees
Approach distance, Sa = R sin (θOB)
=11.11 m
Sg = S – Sa
= 50 – 11.11 => 38.881 m
At velocity, V∞ = 0.7VLO
=0.7*10.787 => 7.5509
CL (0.7LO) =1.735
Using drag polar
CD = 0.012615 + 0.06CL2
CD (0.7LO) = 0.012615 + 0.06 (CL (0.7LO)) 2
= 0.19322
(T/W) 0.7VLO = (CD/CL) 0.7VLO =>0.11154
Sg 1.21(W/s)/ (g × ρ × CLmax× (T/W) 0.7VLO)
W/S = 45.385 N/m2
Assuming height = 1 m
We are going to use the equation below for the wing loading determination
Sg 1.21*W/S)/(g × ρ × CLmax× (T/W)0.7VLO]
Total take-off distance is given by
S=Sg+ Sa
Where, Sg= Ground roll
Sa = Approach distance
Designed take-off distance, S = 50 m
Liftoff velocity, VLO = 1.15Vstall
= 1.15 × 9.38
= 10.787 m/sec
L = 0.5 × ρ × VLO2 × CL
For L W during take-off ground run
(VLO /Vstall) = Sqrt[CL(max)/ ]/1.15 )2 = 0.85029
Using drag polar
CD = 0.012615 + 0.06CL2
= 0.012615 + 0.06 (0.85029)2
= 0.05599

13. 4. Third Weight Estimation
Introduction
We have to estimate the weight of various parts of the airplane and add them to obtain total
weight. In the second weight estimation the total weight is obtained as 1045.3 g. The airfoil
chosen for the mission requirements is NACA 2414. The chord length of this airfoil is
assumed as 0.16 m and the aspect ratio was fixed as 6.25. With these values the span length
obtained was 1.0 m. The minimum wing loading carried out in the last report as 45.38 N/m 2.
Since wing sweep is used primarily to reduce the adverse effect of transonic and supersonic
flow. An elliptical wing plan form is difficult and expensive to build. We assume in this report
that no dihedral, no sweep and no taper and a box cross-section for the fuselage. To mount the
motor and battery in the aircraft, provision is planned to be made near the nose of the aircraft is
done. The uncambered four digit airfoil is commonly used for tail surface of subsonic aircraft.
Tails commonly used for trim, stability and control. An aft horizontal tail inclined to balance
the wing pitching moment.
Power plant weight = 256 gm
Payload weight = 120 gm
Material selection
Material selected is medium density BALSA WOOD. Balsa wood is a very interesting material
and is typically used for construction of model airplanes. Balsa is technically a hard wood,
because of the shape of the leaves. It is not the lightest wood, but it is the lightest wood which
has some strength. Actually Balsa wood’s strength to weight ratio and stiffness to weight ratio
is very good, better than any man made material that I am aware of. It is not considered to be a
great material to use for the design of most things because its properties vary greatly.

14. Mechanical properties of balsa wood
Weight of Wing
Weight of Spar
Figure4.1: Lift force and lift coefficient distribution across an elliptical, rectangular
and triangular wing planform. Since our planform is rectangular the lift distribution is
approximated as uniformly distributed load.
Figure4.2 load distribution on wing
Maximum bending moment = 0.5*(W/S)*planform area*(span/4)
= 0.5*25.692*0.25*2*0.5
= 3.21Nm
From Table 1,
Maximum compressive strength = 12.1 MPa
The imposes load on wing 1.41 greater than level flight load .So Assuming factor of
safety = 3[3*]
Allowable stress, = 12.1/3
= 4.033MPa
From Euler’s equation of Bending,

15. 9
2.23 10
M
I Y
I

−
=
= 
3
12
b d
I

=
Breadth b = 3.37
Breadth b ≈ 4mm
Area of the spar=0.004*0.02
= 0.00008
Weight of the SPAR = density*volume
= 150*(area*span)
= 150*0.00008*1
= 0.012 Kg
Weight of Ribs
Weight of Fuselage
Table 4.2 Rules of thumb for determining the parameters for a model aircraft[5]

16. Weight of Horizontal Tail
Weight of Vertical Tail
Skin
Assuming the skin thickness to be 1mm
Skin weight for wing
Skin Weight for Fuselage
Skin Weight for Horizontal Tail
Skin Weight for Horizontal Tail
COMPONENT WEIGHT(gm)
WING 185.28
FUSELAGE 123.75
HORIZONTAL TAIL 187
VERTICAL TAIL 131.25
SKIN WEIGHT 58.73
SERVO 45
CASING 50
RECIEVER 6.17
TRI CYCLE LANDING GEAR 70
EMPTY VOLUME FILLING 55
OTHER (glue, bolt, etc.) 20
Table 4.3.component and corresponding weight
Total structural weight = 814.88gm
CONCLUSION
➢ The estimated total structural weight is 814.88 gm.
➢ Third weight estimation,
W03 =WPL+WPP+WS
W03 =120+256+814.88 = 1190.88gm

17. 5. Sizing And 3d Layout
Introduction
• The basic geometric details of the aircraft are estimated.
• 3D layout has been drawn.
• High wing is used in this model because the wing stays out of ground effect most of the
time.
• High wing has better stability than low wing aircraft.
Horizontal and vertical tail volume coefficient for some aircrafts:-
Horizontal tail
Horizontal tail length can be calculated by
Vertical tail
Vertical tail length from nose of the main wing is equal to the length of the tail wing distance
from the main wing.
Three Dimensional Layouts
A 3D layout on the basis of previous calculation is drawn in CATIA and here shown the
front, side and top view of the modal aircraft. All dimensions in mm.

18. Front view of the modal aircraft.
Side view of the model aircraft.
384

19. Top view of the modal aircraft.
Conclusion
• A 3D layout is drawn in CATIA and shown the front, side and top view of the modal
aircraft.
384

20. 6. Propeller Design
Introduction
• The function of the propeller is to convert brake horse power from the engine into thrust.
• The primary purpose is to convert engine power to axial thrust through torque transfer to
the propeller.
Two types of propeller are in use:-
1) Fixed pitch
2) Constant speed
Fixed pitch propellers:-
• This propeller is made in one piece.
• Only one pitch setting is possible.
• Usually two blades propeller and is often made of wood or metal.
Pitch:-
• The pitch is defined as the distance traveled forward in one revolution if there
were no slippage.
• Propellers are defined in terms of their diameter and pitch.
• By convention the pitch is defined in either inches or millimeters at 75%of the
blade radius.
High Pitch Propeller properties:
(1) High speed flight
(2) Poor Acceleration
(3) Poor Climb
(4) It can be difficult to slow down for landing

21. Low Pitch Propeller properties:
(1) Low speed flight
(2) Good Acceleration
(3) Good Climb
(4) Finer speed control throughout throttle range — particularly at low throttle
settings.
Conclusion
• Propeller size has been chosen as 10” X6’’
• Propeller Power has been calculated as 22.3971watt

22. 7. Improved Drag Polar
Introduction
For simplicity in calculation, we assume that drag is a function of 2-D area which is
called references area. This area includes wing, tail, fuselage area etc.
In this report we are calculating the improved drag polar by component build up
method.
Parasite Drag
Total drag coefficient is defined as,
There are two methods by which we can calculate parasite drag .
Equivalent Skin Friction Method
This method is used for subsonic aircraft in cruising. While cruising, the parasite drag
is mainly due to skin friction.
The estimation of parasite drag coefficient can be calculated from equivalent skin
friction coefficient of similar type of aircraft, multiplied with wetted area of present
aircraft to the references area.
Table 7.1: Equivalent skin friction coefficient [1]
From the table 7.1,
= 0.0025
From design foil software,
Surface length of NACA 2414 =204*x/c%
= 204.8*0.20/100 = 0.4096 m
Sref= 0.4096*1 m2
= 0.4096 m2
(wing plan form area)

23. Component Characteristic length (m) Swet (m2)
Wing 0.16 0.3832
Horizontal tail 0.100 0.1620
Vertical Tail 0.100 0.1136
Fuselage 0.75 0.18
Total 0.8388
CD0 = Cfe *( SWet/ Sref) …………… (7.1)
CD0 = 0.0025 *(0.8388 / 0.4096)
CD0 = 0.05119
Component Build Up Method
In this method parasite drag is calculated by use flat-plate skin–friction drag
coefficient (Cf) and a component form factor (FF).
CD0 = ……… (7.2)
For calculation of Swet we have followed the method as mentioned below:
If t/c < 0.05
Swet = 2.003* Sexposed……… (7.3)
If t/c >0.05
Swet = Sexposed [1.977 + 0.52(t/c)] ……… (7.4)
Main Wing
The R/C model will typically have Reynolds Numbers less than 500,000 which gives
the wing a predominately laminar boundary layer.
Re
Re = 1.73*105

24. Cf,w ……(for laminar boundary layer )
= 0.00319
FFw … (7.5)
Here = 0, as the present wing and horizontal wing rectangular shape is
considered.
t/c = 0.14
M =Vcruise/speed of sound
M = 19.38 / 348.92
M = 0.0555
FFw =1.0498
Sexposed=b * c (top view area)
= 1.0*0.16
= 0.16 m2
Swet,w= Sexposed [1.977 + 0.52(t/c)]
Swet,w = 0.328 m2
Horizontal tail
Re
= 1.1237*105
Cf,ht
= 0.00396 t/c = 0.12
…….for NACA 0012

25. FFht
FFht =1.0039
Sexposed=0.3846*0.104
= 0.04 m2
Swet,ht= Sexposed [1.977 + 0.52(t/c)]
Swet,ht = 0.08157 m2
Vertical tail
Re =
= 1.0815*105
Cf,vt
= 0.004
FFvt
=1.0039
Sexposed=0.056 m2
Swet,vt= Sexposed [1.977 + 0.52(t/c)]
Swet,vt = 0.1136 m2
Fuselage
Re =
= 8.112*105
> 5*105
→ turbulent flow.

26. Cf,f
= 0.00465 t/c = 6 /
75= 0.08
Sexposed= df* lf = 0.06*2*0.75 = 0.090 m2
Swet,f= Sexposed [1.977 + 0.52(t/c)]
= 0.1816m2
Fig 7.1 description of df and lf
f =
FFf …………(7.6)
FFf = 1.06197
Calculation of of each component
Componen
t
Characteristi
c length (m)
Re Swet
(m2
)
FFc Qc Cfc
Wing 0.16 0.14 1.73*105
0.328 1.0498 1.806
1
0.0031
9
1.9838*10-3
Horizontal
tail
0.104 0.10
4
1.1237*10
5
0.0815
7
1.0039 0.449
1
0.0039
6
0.1456*10-3
Vertical
Tail
0.100 0.10
0
1.0815*10
5
0.1136 1.0039 0.625
5
0.004 0.2853*10-3
fuselage 0.75 0.08 8.112*105
0.1816 1.0619
7
1.0 0.0046
5
0.89676*10-
3
Total 3.31146*10-
3

27. Therefore /Sref= 3.31146*10-3
/(0.32768) = 0.010105
For CDmisc the strut and landing gear are the only components considered
CDmisc. = (D/q)/(frontal area) …… (7.7)
Landing gear component Drag [1]
Component (D/q)/ Frontal area CD misc
Regular wheel tire 0.25 3.205 × 10-4
Second wheel and tire in tandem
(2 wheels)
0.15 1.9231 × 10-4
(twice
of this)
Round strut or wire 0.30 18.571 × 10-4
Total 2.5622 × 10-3
For propeller aircraft
CDL&P = 5 % of the parasite drag co-efficient [1]
CDL&P = 0.05*(0.010105)
= 0.50525*10-3
CD0 = 0.010105 + 2.5622*10-3
+0.50525*10-3
CD0 =0.013172
Drag Polar Determination
Drag polar is given by the expression below,
And the drag due to lift is given by,
For a straight wing span the efficiency factor can be estimated by using the
expression below,

28. e = 1.78 * (1 -0.045*AR0.68
) – 0.64 ……..… (7.8)
AR = 6.25, e = 0.86149
K = 1/ (3.14*AR*e)
= 0.05914
CDi = 0.05914 CL
2
Hence the drag polar for the present aircraft is
Conclusions
• Improved Drag profile is calculated as .

29. 8. Calculation of Centre of Gravity
Introduction
The CG is the point at which the total weight of the aircraft is assumed to be
concentrated; its location depends on the distribution of weight in the airplane. We
are considering only half of the section of the model aircraft. If the CG is too far aft, it
will be too near the center of lift and the airplane will be unstable, and difficult to
recover from a stall.
Figure 8.1 Effect of CG locations on stability
C.G. Calculation for Wing
From design foil software,
Xrib = 42.19%*x/c
= 0.4219*16
Xrib = 6.7504 cm ……. From leading edge
From previous report,
Xspar= 0.3*c
Xspar= 4.8 c
Since we are using 20 ribs over the whole wing. This means that there will be 10 ribs
in each side.

30. Weight of the SPAR = 12 gm
Weight of each rib=4.92 gm
……… from the leading edge
……… from the nose tip of the fuselage
From design foil software,
+ 6
= 6.502 cm …………from fuselage base
C.G. Calculation for Fuselage
Our fuselage is tapered at the end and at the beginning .Assuming equal cross section
throughout the length of the fuselage as it will not affect much on the fuselage c.g.
Hence,
……… from the nose tip of the fuselage
…………from fuselage base.
C.G. Calculation of Horizontal and Vertical Tail

31. (a)For horizontal tail
C = (S (A+2B)) / (3(A+B))
= (0(10.4+2*10.4)) / (3(10.4+10.4))
C = 0
MAC = A-(2(A-B)(0.5A+B) / (3(A+B)))
= 10.4-(2(10.4-10.4)(0.5-10.4+10.4) / (3(10.4+10.4)))
MAC = 10.4
= %MAC B.P.*(MAC) + C
The balance point (B.P.) is not the Center of Gravity. Generally it is about 25% of
the chord back from the leading edge.
= 0.25*10.4 + 0
= 2.6 cm …………… From the horizontal tail leading edge
= 15 + 46.2 + 2.6
= 63.8 cm ……………….. From the nose tip of the fuselage
……………….. From fuselage base
(b)Vertical tail
C = (S(A+2B)) / (3(A+B))
= (8.57(14.28+2*5.71)) / (3(14.18+10.4))
C = 3.67
MAC = A-(2(A-B)(0.5A+B) / (3(A+B)))
= 14.28-(2(14.28-5.71)(0.5*14.28+5.71) / (3(14.28+5.71)))
MAC = 10.60
= %MAC B.P.*(MAC) + C
=0.25.*10.6+ 3.67
= 6.32cm ……. from the horizontal tail leading edge
= 15 + 46.2 + 6.32
= 67.52 cm ……. from the nose tip of the fuselage

32. cm
cm …………….from fuselage base
Table 1: mass and CG distance
Component Mass (gm) X(cm) from nose tip of
the fuselage
Z(cm) from the
fuselage base
Fuselage 61.875 37.5 3
Wing 55.2 31.368 6.502
Horizontal Tail 30 63.8 3
Vertical Tail 21 67.52 12
Landing Gear 35 45 -6

33. Propeller 13 -2 3
Motor 50 1.08 1.08
Battery 40 10 1
Payload 50 15 1.4
……. from the nose tip of the fuselage
………………. from the fuselage base
Since all the loadings are symmetric about X-axis. Therefore will lie on X-axis.
Conclusion
• ……. from the nose tip of the fuselage
• ……. from the fuselage base
Correction Note
• To obtain the stable condition of aircraft we have shifted our wing to 250 mm from
the nose tip of the fuselage. Earlier it was 150mm from the nose tip of the fuselage
along X-axis.

34. 9. Stability Analysis
Introduction
Aircraft motion consists of translations and rotations about the center of gravity the
motion include six degrees of freedom: forward and aft motion, vertical plunging,
lateral translations, pitch, roll, and yaw.
Figure 9.1: Axis notation
Two types of stability or instability are important.
(A) Static stability.
(B) Dynamic stability.
Static stability is defined as initial tendency towards equilibrium point. The
equilibrium position is usually called the trim position and is adjusted using the trim
tabs.
Dynamic stability is defined as long term tendency towards equilibrium point.
Fuselage
The pitching moment coefficient of fuselage,
fuselsge per degree
Where,
Kf= Empirical pitching moment factor

35. Figure 9.2: Position of ¼ root chord
Position of 1/4th root chord in terms of % fuselage length = ((160/4 +
250)/1000)*100
=29mm
Kf from the above Graph = 0.012
fuselsge
= 0.00061798 per degree
= 0.0354 per radian
Wing
Location of cg of the aircraft from the fuselage tip = 2815 mm cg
= 0.398 (Position of C.G. in terms of fraction of wing chord)
Position of aerodynamic center of wing, acw =
For NACA 2414 airfoil,

36. a
= 0.11 per degree (from previous report)
= 6.277 per radian
a =
a = 4.5781 per radian
Tail Effect
Xach= 388 mm (From wing leading edge)
Aspect ratio =0.3846/0.104
= 3.7
For NACA 0012 airfoil
a0 = 6.283 per radian
a = / (1+
)
a = 3.8606 per radian
Downwash Effect on the Tail

37. =0.38619
0.3638
Power Plant Effect
The dynamic pressure ratio,
)
=0.9 [1]
T = Thrust at takeoff
= 0.11152*0.882*9.81
=0.9649 N
Propeller disk area=Ap
=129717 mm2
0.93724
Location of Neutral Point of Aircraft
=0.4750
=0.4750 ………..from wing leading-

38. edge
Static- margin= cg
= 0.4750 -0.37533
Static- margin= 0.09967
Static-margin = 0.09967
= 4.5781 (
=
Since is negative, which implies that our aircraft is stable.
Trim Analysis
Trim implies that the total moment about c.g. equals zero. For static trim condition
the total pitching moment must be equal to zero. The pitching – moment coefficient
about c.g.of the aircraft is given as [1]
The moment produced by propeller is very small as compared to other terms,
0 = 0.60917 0.45986
= ( )
= ( )
Where, = zero lift angle of attack for wing
= zero lift angle of attack for horizontal tail

39. = downwash effect
The change in zero lift angles due to a plain flap is given as,
Considering chord of flap Cf= 25 mm
Cf/C =25/160 =0.1562 t/C =0.14
Figure9.3: theoretical lift increment for plain Figure9.
4: Empirical correction for
Flaps plain lift increment
From the above graph, we find that =2.7per radian and =1 Therefore
o
=0.1682
=
= 3.8606 )
= 3.8606

40. Using above trim equation, we get
= 0.7352 0.5017(3.8606
= -1.20166 -0.01708
Based on above trim equation we can plot the as following:
α(degrees)
δf = -30
δf= 00 δf= 30
-3 -0.26852 0.063781 -0.26749 0.062887 -0.26647 0.061993
-2 -0.17935 0.042818 -0.17833 0.041925 -0.1773 0.041031
-1 -0.09019 0.021856 -0.08916 0.020962 -0.08814 0.020068
0 -0.00102 0.000894 0 0 0.001024 -0.00089
1 0.08814 -0.02007 0.089164 -0.02096 0.090187 -0.02186
2 0.177304 -0.04103 0.178328 -0.04192 0.179351 -0.04282
3 0.266468 -0.06199 0.267491 -0.06289 0.268515 -0.06378
Conclusion
➢ Static Margin is 0.09967 (9.967%), which is within the limit.
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
-0.3 -0.2 -0.1 0 0.1 0.2 0.3
C
M
cg
CLtotal
CMcg v/s CLtotal

41. 10. Revised Performance Calculations
Introduction
Aircraft design calculation is an iterative process in which we have to update our old
values with the new values so that we will get a good accuracy and nice modal of the
aircraft. In this report we are going to calculate revised performance parameters on
the basis of our revised weight and drag polar calculated data in earlier reports.
From previous reports revised weight and drag polar is as below,
W = 965.75gm
= 9.474N
= 0.01317+ 0.05914
S = 0.16
= 1.245
Stall Speed
Stall velocity
= 9.094 m/s 10.3
Cruise Speed
Cruise Velocity VCruise = 2 *
= 2*9.094
VCruise= 18.189 m/s
Climbing Conditional Minimum
power condition
=0.8173
= 4*
= 4*0.01317
=0.05268

42. Velocity at minimum drag
=14.772 m/s
Velocity at minimum power,
=0.76*
=11.22 m/s
=6.853 watt
= 22.22 watt
Pavis calculated using a calculator available in the website mentioned below
Rate of climbing V
VV(max) =1.622 m/s
W =
Horizontal velocity at VV(max) ,VH(V
= 1.6537 * 16.975
VH(VV(max)) = 28.0715 m/s
Gliding
For minimum sink rate is same as that required for minimum power Condition.
= 0.7469
Corresponding Velocity is given by,

43. Vmin for sink = 11.744m/s (L/D)min
for sink = 16.975
The equilibrium glide angle can be calculated by,
=3.371o
Take Off Performance
We are going to use the equation below for the wing loading determination
Sg 1.21[W/s)/ (g × ρ × CL(max)× (T/W)0.7VLO] (Source: J.D.Anderson)
Total take-off distance is given by
S=Sg+ Sa
Where, Sg= Ground roll
Sa = Approach distance
Designed take-off distance, S = 50 m
Liftoff velocity, VLO = 1.15Vstall
= 1.15 × 9.094
= 10.4581 m/s
L = 0.5 × ρ × VLO
2
× CL
For L W during take-off ground run
(VLO/Vstall) =Sqrt[(CL(max)/(CL(LO)]=1.15
) 2
= 0.9414
Using drag polar

44. CD = 0.01317+ 0.05914CL
2
= 0. 01317 + 0.05914( )
= 0.06558
(L/D)VLO =CL/CD = 0.9414/0.0658=14.35
Thrust, T = W*(CD/CL)
= 0.6381N
Radius of take-off,
R = (6.96 × Vstall
2
)/g
= 58.674 m
Flight path angle θOB =
cos-1
(1 – hOB/R)
= 10.593 degrees
Approach distance, Sa = R sin(θOB)
=10.786 m
Sg = S – Sa
= 50 – 10.786 m
= 39.214 m
At velocity, V∞ = 0.7VLO
=0.7*10.4581
= 7.32067 m/s
(V0.7LO/Vstall) = Sqrt[(CL(max)/(CL(0.7LO)]
CL(0.7VLO) =1.921
Using drag polar
CD = 0.01317 + 0.05914CL
2
CD(0.7VLO) = 0. 0.01317 + 0.05914 (CL(0.7VLO))2

45. = 0.2292
(T/W) 0.7VLO = CD(0.7VLO) / CL(0.7VLO)
=0.1193
Sg 1.21(W/s)/(g × ρ × CL(max)× (T/W)0.7VLO)
= 54.30 N/m2
Wing Loading for Cruising Conditions
To maximize Range during cruise for a propeller aircraft, wing loading is selected to
provide a high (L/D) at cruise conditions.
So wing loading is calculated for maximum range i.e. maximum (L/D) condition.
CDo = k .
CL .
.
= 0.5*1.15* *
=81.509
) = 81.509 N/m2
Wing Loading for Design Landing Distance
Assuming total landing distance as 50 m (Grass Runway)
Total Landing distance is given by,
S = Sa+ Sf +Sg.
Where Sa = Approach distance,
Sf = Flare distance,
Sg= Ground roll.

46. Flight path radius during flare R, n=1.2
for flare (Ref. Raymer)
R =
Where V∞ = Vf = Flare velocity = 1.23* Vstall
=1.23*9.094
=11.1856 m/s
Flight path radius, R = (11.1856)2
/ (9.81*0.2) = 63.770 m.
Design approach angle, a = 30
Flare height, hf = R (1-cos a )
= 63.770 * (1- cos3o
)
= 0.0873 m.
Approach distance, Sa
= 17.439 m.
Assume ~
Flare distance, Sf= R sin
= 3.337 m
Hence,
Sg = S - Sa - Sf
= 50 – 17.439– 3.337 =
29.224 m.
Sg ≈ 30m.
Sg

47. N = 3 for large aircraft and N = 1
for small aircraft.
Taking N=1 and µ=0.35 and solving for as below
Solving the above quadratic equation we get,
) = 68.530 N/m2
Wing Loading for Turning Performance Calculations
Load Factor, n = sec = 1.2
n = ( .
.
) = 78.98 N/m2
From the above calculations the minimum wing loading is 54.30 N/m2
obtained for
takeoff condition.
The minimum wing loading is used to calculate thrust to weight ratio for different
flight conditions as done below.
Calculations
As mentioned above thrust to weight ratio is calculated using ) = 54.30 N/m2
For Cruise Condition

48. For Takeoff Condition
Sg
.
For Climb Condition
) .
For Turning Condition
.
Conclusion
• Using the revised weight and drag polar calculated in earlier reports all the
performance parameters have been calculated at different flight conditions.