DESIGN OF THE 1903 WRIGHT FLYER REPLICA

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Transcript DESIGN OF THE 1903 WRIGHT FLYER REPLICA

DESIGN OF
THE 1903 WRIGHT FLYER
REPLICA
MADRAS INSTITUE OF TECHNOLOGY
CHENNAI - 44
WEIGHT ESTIMATION
Elements
Structure
Engine
Propeller
Landing gear
Servo motors
Radio Controls
Fuel
Mounting + belt
Fuel tank
Misc.
Quantity
1
2
3
3
All
0.3 litres
2
TOTAL WEIGHT
Weight (N)
7.3
5
1
2.25
0.58
0.212
2.36
3
0.6
2.5
24.802 N
AERODYNAMIC
DESIGN
Lift Calculation
As the t/c ratio of the airfoil is less than 0.05 the classical theory of thin airfoils
can be employed, by using this theory all the parameters other than drag is
forecasted .
CL Vs Alpha curve for inviscid flow
3
2.5
2
1.5
CL
1
0.5
0
-15
-10
-5
0
5
-0.5
-1
-1.5
Alpha
10
15
20
25
Drag Polar
Induced Drag Estimation
AR for a biplane
Span
Chord length
AR
Gap
= 4 b/c
= 5 feet
= 12 inches
= 20
= 9 inches
CDi = 1/(AR)*(1+)CL2
CDi = 0.11136 CL 2profile
Profile Drag Calculation
CD wet /Cf = 1+ 1.5 (t/c)3/2 +7 (t/c)3
CDp/Cf = 60 (t/c CL/5)4
The drag polar of our model is
CD = 0.1303 + 0.1277CL2
Wing warp
Rolling moment for Both wings = 0.56 (k/c) sin  (l+ k cos )2
Where c is the chord of the wing
 is the angle of warp from the undisturbed configuration
k is the length of wing warp
POWER PLANT
SELECTION
Power available
100
90
80
1500
70
2500
3000
power
60
3500
4000
50
4500
5000
40
1000
5500
30
6000
20
10
0
1
2
3
4
5
velocity
6
7
8
9
specifications
From drag calculations the power required
0.25 bHp
Diameter of the propeller ( 2-blade propeller)
10 inches
The diameter is determined from the thrust to be produced.
The ground clearance was also taken into account while determining the
diameter of the propeller.
STRUCTURAL
DESIGN
WING FRONT SPAR
The bending moment about X axis (Mx) = 14.96 Nm
The formula used,
Mxc =(Mx-(My*Ixy/Iyy)) /( 1-Ixy²/ (Ixx*Iyy)) =36.65 Nm
Myc =(My-(Mx*Ixy/Ixx)) / (1-Ixy²/ (Ixx*Iyy)) = -108.04 Nm
The maximum stress on the front spar σz = 32 MPa
The maximum allowable bending stress for spruce wood = 41 MPa
WING REAR SPAR
The maximum stress on the rear spar σz = 40 MPa
The maximum allowable bending stress for spruce wood = 41 MPa
ELEVATOR AND RUDDER SPARS
ELEVATOR FRONT SPAR
REAR SPAR
RUDDER SPAR
Design of truss members
Though the diameter of the truss members are different, for fabrication
simplicity all the members are designed with diameter 5 mm.
PROPELLER SHAFT DESIGN
The formula used to calculate the diameter of the shaft
Me = (M +√(M²+T²)) / 2 = 0.15306 Nm
Te = √(M²+T²)
= 0.7938 Nm
Maximum bending strength of the balsa wood σb = 1.18934*10^7 N/m
τ = 2482113 N/m²
Dmoment =7.15 mm
Dtorque =7.95 mm
Therefore the required diameter for the propeller shafts = 8 mm
MATERIALS TO BE USED
S.NO
COMPONENT
MATERIAL
1
WING SPARS
SPRUCE
2
OTHER STRUCTURAL
COMPONENTS
BALSA
3
SKIN
REYNOLDS PLASTIC
4
FUEL TANK
PLASTIC
PERFORMANCE
CALCULATION
INTRODUCTION
The performance design covers the five major calculations which are listed
below
Steady level flight performance
Climb performance
Range & Endurance
Take – Off & Landing
Turn Performance
LEVEL FLIGHT PERFORMANCE
Cruising Velocity
Stalling Velocity
VminD
Dmin
Pmin
VminP
= 4.7 m/s
= 2.35 m/s (CLmax = 2.04)
= 2.64 m/s
= 2.423 m/s
= 6.09 W
= 2.06 m/s
Range
= 1.616 km (for cruise condition)
Endurance
= 5 minutes 54 seconds
CLIMB PERFORMANCE
R/C = Excess Power / Weight
Excess Power = Power Available – Power Required
Maximum rate of climb occurs at 6 m/s
Velocity
Power
Available
Power
Required
Excess
Power
R/C max
Angle of
Climb
m/s
W
W
W
m/s
degree
2
8
6.108897
1.891103
0.075644
2.167557
3
12
7.83886
4.16114
0.166446
3.180502
4
30
13.4841
16.5159
0.660636
9.50645
5
60
22.52976
37.47024
1.49881
17.44327
6
90
36.55183
53.44817
2.137927
20.87438
7
90
60.97091
29.02909
1.161164
9.548366
8
91
90.17925
0.820751
0.03283
0.235128
EXCESS POWER
100
90
80
POWER W
70
60
POWER
AVAILABLE
50
POWER
REQUIRED
40
30
20
10
0
2
4
6
VELOCITY m/s
8
Take – Off
The take-off is curved up into 3 phases
They are ground run, transition and initial climb upto 2 m and the same is
repeated for landing
Ground run
Vavg = 0.7 VLO (lift off velocity)
= 0.84 Vstall
r = 0.1 for grass land
VLO = 2.82 m/s
CLLO= 0.8 CLmax
Ground Run
Ground Run in transition
Ground Run in climb
Total take off distance
= 6.3 m
= 2.1 m
= 4.48 m
= 12.88 m
Ground Transition Climb
Run
Landing & Turning performance
Landing distance total = 17.11 m
Minimum turn radius = 0.4 m
Corresponding time taken = 1.15 seconds
V-n diagram is a plot between the velocity and load factor ( n = L/W)
It gives the structural limit (max) of the aircraft and the highest and lowest
possible velocity that can be reached by the aircraft
The maximum load factor = 275/25 = 11
V-n DIAGRAM
12
11
10
9
8
7
n
stall limit
6
Structural limit
Max velocity
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
Velocity
From the v-n diagram it is clear that n is maximum for the velocity of 8 m/s and
the maximum velocity can be 35.75 m/s for the n value less than 11
STABILITY ANALYSIS
LONGITUDINAL STATIC STABILITY
DIRECTIONAL STATIC STABILITY
CROSS COUPLING EFFECT
Change in yaw co-efficient for different pitch rates (in rad/s)
At cruising velocity of 4 m/s
0.003
Incremental Yaw co-efficient CN
0.0025
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.5708
0.002
0.0015
0.001
0.0005
0
0
10
20
30
40
50
60
Wing warp deflection angle (deg)
70
80
90
100
COST ESTIMATION
Item
Cost
4 channel radio control (with transmitter, receiver, 4- servos, Connectors etc.) 15000
Engine (0.25 bhp)
4000
Balsa Wood
2500
propellers
700
Fabrication cost
1000
Skin, belt, pulley, wires, LG etc.
1500
Total
24 700
RADIO CONTROL COMPONENTS
Engine throttle is controlled by servo motor.
Four channel receiver set with 4 servo motors and connectors are used.
The R/C unit weighs about 0.75 N.
The R/C unit is placed just below the wing so that it reduces the bending moment caused by the
lift.
POSITION OF SERVOS
POSITION OF RECEIVER
PROBLEMS
We are amateur designers
But we are confident that we can overcome this problem after taking part
in this workshop
Since the stability of the aircraft is at a high risk we feel that flying the
aircraft safely would require a lot of training