Critical Design Review
AAE451 – Team 3
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December 9, 2003
Brian Chesko
Brian Hronchek
Ted Light
Doug Mousseau
Brent Robbins
Emil Tchilian
AAE 451
Team 3
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Aircraft Name
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av·a·tar - n. - 1. <chat, virtual
reality> An image representing a
user in a multi-user virtual reality.
Source: The Free On-line Dictionary of Computing
http://wombat.doc.ic.ac.uk/foldoc/
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Introduction
Walk Around
Design Requirements and Objectives
Sizing
Propulsion
Aerodynamics
Dynamics and Controls
Structures
Performance
Cost
Summary
Questions
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Aircraft Walk Around
•Wing Span = 14.4 ft
•Wing Chord = 2.9 ft
•T-Tail – NACA 0012
•A/C Length = 10 ft
•Pusher
•Internal Pod
•Tricycle Gear
•Low wing – Clark Y
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Design Requirements & Objectives
• Maximum weight < 55 lbs
• Cruise speed > 50 ft/sec
• Stall speed < 30 ft/sec
• Climb angle > 5.5°
• Operating ceiling > 1000 ft
• Flight time > 30 minutes
• Payload of 20 lbs in 14”x6”x20” pod
• Carry pitot-static boom
• Spending limit < $300
• T.O. distance < 106 ft (~60% of McAllister Park runway length)
• Rough field capabilities
• Detachable wing
• Easy construction
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Constraint Diagram
Aircraft Constraint Diagram
Cruise Speed
40
Stall Speed
Climb
36
T/O dist
32
Landing dist
Ceiling
28
W/P (lbf/shp)
Endurance
Minimum Structure
24
Minimum Power
20
16
12
8
Power Loading = 15.5 lbf/shp
Wing Loading = 1.28 lbf/ft^2
4
0
0
0.25
0.5
0.75
1
W/S (lbf/ft^2)
1.25
1.5
1.75
2
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Propulsion
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Chosen Engine
• O.S. Max 1.60 FX-FI
–
–
–
–
3.7 BHP @ 8500 RPM
1,800-9,000 RPM
2.08 lbs
Fuel Injected
Ref. www.towerhobbies.com
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Chosen Propeller 4-blades
• Zinger 16X7 Wood Pusher Propeller
– 16 inches in diameter with 7 inch pitch
– 4 blades
Ref. www.zingerpropeller.com
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Chosen Fuel Tank
1
3 (in)
2
• Fuel tank chosen
is:
– Du-Bro 50 oz. fuel
tank
– Available from
Tower Hobbies
– Located at the C.G.
of aircraft
– Good for up to 32
min. of flight time
(when completely
full).
3
4 (in)deep
8
3
8 (in)
8
Ref. www.towerhobbies.com
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Takeoff EOM Integration
Drag + Rolling Friction
Thrust
W
a  T  D  W
g
vs. Position
at Takeoff
Velocity vs.Velocity
Position
at Takeoff
35
30
Velocity [ft/s]
[ft/s]
Velocity
25
Takeoff Distance
Within Constraint
20
15
10
5
0
0
20
40
60
Position [ft]
Position
[ft]
80
100
120
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Max Velocity
Maximum Velocity
20
18
16
Thrust/Drag [lbf]
14
Thrust
12
10
8
6
Drag
4
2
0
30
40
50
60
70
Flying Velocity [ft/s]
80
90
100
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Aerodynamics
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Wing Dimensions
• Prandtl’s Lifting line theory used for
aerodynamic modeling of the lifting
components
• Input parameters: AR, a0, aL=0, a.
• Lifting Line Model Gives CL, CDi at prescribed a
• CDvisc found using Xfoil which was used to obtain
CD = CDi+CDvisc
5° Dihedral
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Airfoil Selection
Drag Polar for Candidate Airfoils
0.015
NACA 4412
4415
4425
2418
23018
ClarkY
0.014
0.013
Section Drag Coefficient, c
d
0.012
0.011
0.01
0.009
0.008
0.007
Region of
Interest
Clark Y
0.006
0.005
-1
-0.5
0
0.5
Section Lift Coefficient, c l
1
1.5
Clark Y Airfoil has low drag over range of interest
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Airfoil Selection
Clark Y Airfoil
0.2
Y/C
0.1
0
-0.1
-0.2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
X/C
2
Coefficient Cd
Drag
SectionSection
Drag Coefficient, c
1.5
1
d
l
Cl
LiftLiftCoefficient
SectionSection
Coefficient, c
0.05
0.5
0
-0.5
0.045
0.04
0.035
0.03
0.025
0.02
0.015
0.01
-1
-10
-8
-6
-4
-2
0
Angle of Attack
2
4
Angle of Attack (AOA)
6
8
10
0.005
-1
-0.5
0
0.5
Section Lift Coefficient, c l
1
Section Lift Coefficient Cl
1.5
2
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1.4
1.2
1
CL no flap
CL 10 deg flap
CL 15 deg flap
Required CL
0.8
0.6
CL
•
•
CL vs aoa for different flap deflections
CL needed = 1.19
Wing without flaps
reaches CL at a=13°
Wing stall possible
Wing with 15° flap
deflection reaches
CL at 11°
CL
•
•
Wing Stall Performance
0.4
0.2
0
-0.2
-0.4
-10
-5
0
5
10
15
aoa (deg)
Angle of Attack (degrees)
Flaperons necessary to meet stall requirements
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Wing Performance
CD vs CL for different flap deflections
0.14
CDvsCL no flap
CDvsCL 10 deg flap
CDvsCL 15 deg flap
0.12
0.1
CD
CD
0.08
0.06
0.04
0.02
Required CL at stall
0
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
CL
CL
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Drag Build Up At Cruise
Component
CD
Drag
Wing
0.018
2.6 lbf
Fuselage
0.0045
0.6 lbf
Horizontal
Tail
0.0043
0.6 lbf
Vertical Tail
0.0017
0.04 lbf
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Wing Operating Parameters
CL
a (of wing)
Flaperon
Deflection
CD
L/D
Stall
1.19
11°
15°
0.119
10
T/O
0.989
8°
15°
0.084
12
Cruise
0.44
2.8°
0°
0.018
24
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Dynamics and Controls
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Center of Gravity & Aerodynamic Center
• Aircraft Center of Gravity is 3.2 ft from nose.
– Calculated from CAD program Pro-E
• Aircraft Aerodynamic Center is 3.7 ft from nose.
– Position where pitching moment of aircraft doesn’t change with
angle of attack
– Calculated using Lift from Wing and Horizontal Tail
Aerodynamic Center
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Center of Gravity
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Static Margin
• Desired Static Margin is 15% - 20%
– Dependent on C.G. and A.C. location
• Static Margin is 15%
• Contributes to
Horizontal Tail Sizing
X

SM 
ac
 X cg 
cwing
 15%  20%
Position on Aircraft vs. Horizontal Tail Area
4
Aerodynamic Center of Aircraft
Distance from Nose of Aircraft => x(ft)
3.8
3.6
Static Margin = 15%
Center of
Gravity
3.4
3.2
3
2.8
Static Margin = 20%
8
9
10
11
12
Horizontal Tail Area (ft 2)
13
14
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Horizontal Tail Sizing
• Tail sized based on desired static margin for static
stability and take-off rotation ability
–  double-dot should be at least 10 deg/sec2
16
Ref. Roskam, Airplane Flight Dynamics
Theta Double Dot at Instant of Rotation (deg/sec2)
15
14
13
Long Grass
Area
12 ft2
 g  0.10
Span
6 ft
Chord
2 ft
12
11
2 ft
10
Concrete
 g  0.02
9
8
7
6
10.1
10.15
10.2
10.25
10.3
10.35
10.4
10.45
6 ft
Horizontal Tail Area (ft 2)
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Vertical Tail Sizing
• Value of yawing coefficient due to sideslip angle should
Ref. Roskam, Airplane Design
be approximately 0.001 = 10e-4
• Tail area should be ~2 ft2
-4
20
Coefficient of Yaw Moment due to Sideslip vs Vertical Tail Area
x 10
15
Area
2 ft2
Span
1 ft
Chord
2 ft
C
Nb
[deg-1]
10
5
2 ft
1 ft
0
-5
0
0.5
1
1.5
Vertical Tail Area (ft 2)
2
2.5
3
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Dihedral Angle
Recommendations
• Survey of Roskam data on homebuilt &
agricultural low-wing aircraft: ~5°
• “Wing and Tail Dihedral for Models” - McCombs
– RC w/ailerons (for max maneuverability, low wing): 02° EVD (Equivalent V-Dihedral ≈ dihedral)
– Free Flight Scale model low wing: 3-8° EVD
5° dihedral is a good compromise
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Control Surface Sizing
• Sizes calculate from traditional lifting device
Ref. Roskam, Airplane Design
percentages.
Flaperon
Elevator
Rudder
Chord
0.58 ft
0.6 ft
0.6 ft
Inboard Position
0.95 ft
0.2 ft
0.1 ft
Outboard Position
7.2 ft
3 ft
1 ft
0.6 ft
0.58 ft
0.9 ft
6.25 ft
0.6 ft
2.8 ft
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Trimming
• Incidence of Horizontal Tail calculated from trimmed
flight during cruise (0 Angle of Attack)
• Analysis set
incidence at
-2
Cm of Aircraft vs. Alpha of Aircraft at Cruise with 0 Degree Flaperon Deflection
1.2
Elevator Deflection = -25 [degrees]
-20
-15
-10
-5
0
5
1
0.8
0.4
0.2
m
C of Aircraft C.G.
0.6
0
-0.2
-0.4
-0.6
-8
-6
-4
-2
0
2
4
Alpha of Aircraft
6
8
10
12
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Structures
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Wing Spar Design
2 Spar Design (at .15 & .60 chord):
• Resist Bending
• Assuming 5-g loading
• 53 lbf weight
• Safety factor of 1.5
• Resist Torsion
• Less than 1o twist at tip under normal
flight conditions
Spar Results:
• Material of Choice: Bass or Spruce Wood
• Front Spar:
• 3.6” high (based on airfoil)
• 0.37” thick (0.73” at root)
• Rear Spar:
• 3” high (based on airfoil)
• 0.16” thick (0.25” at root)
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Longitudinal Beam Design
2 Beam Design:
• Resist Bending from:
• 20 lbf payload
• Horizontal tail loads
• Resist Torsion from:
• Rudder deflections
• Prop wash over tail
Beam Results:
• Material of Choice: Bass or Spruce Wood
• Beam Dimensions:
• 3” high
• 0.25” thick
• 8” between the beams
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Tail Structures
Foam core with
carbon fiber shell
• Horizontal and vertical tails comprised of carbon fiber w/ foam core
• Possible to make two foam cores, and cure entire tail at one time
• Control surfaces just need to be cut out of tail structure
• Tail spars allow attach points and transfer load to beams
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Rear Gear Design
• Blue lines represent pin joints
• Black tie-downs absorb energy from landing
• Up to a 33 ft/sec “crash” from 5 feet high
• Need 18” relaxed length tie-down
• Square aluminum tube transfers landing load
to tie-downs and surrounding structure
• 1” x 1” x 0.065” thick – 6063-T6
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Front Gear Design
Aluminum Bolt
•Provides pivot for gear (does not break)
Elastic Band & Nylon Bolt
• Elastic Band Absorbs some
energy from landing
• Nylon bolt breaks during
hard landing
Front Gear Aluminum Tube
• Designed not to break
• Designed not to bend
• Al tube:
1” x 1” x 0.065” thick
6063-T6
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Other Odds and Ends
• Covering for Wing:
– Coverite 21st Century Iron
on Fabric
– 0.34 oz/ft2
– Stronger, and resists tears
better than MonoKote
• Covering for Fuselage:
Ref. www.towerhobbies.com
– Fiberglass
• Either mold or foam core
• Not conductive – won’t
interfere with internal
electronics
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Final Weight Estimate
Part Description
Weight (lbf)
Propulsion
50 oz fuel tank
O.S. 1.60 FX
Sliencer E-5010
Fuel (30 min. based on O.S. info)
16X7 4-Blade Prop
0.38
2.04
0.66
2.20
0.47
Structures
Bass/Spruce Wing Spars
21st Century Fabric
Ribs
Bass/Spruce Tail Beam
Fuselage Skin (1 ply of E-glass)
V-Stab Foam Core
H-Stab Foam Core
V-Stab Carbon Fiber Covering (1 ply Carbon-Fiber)
H-Stab Carbon Fiber Covering (1 ply Carbon-Fiber)
Engine Supports
Tail Spars
4.40
1.74
0.70
2.15
2.78
0.55
2.77
0.37
1.69
0.15
0.35
Landing Gear
Rear Gear
Front Gear
Smooth Wheels 4"
Bungees
2.90
0.42
0.61
0.20
Electronics
Pod
9 channel R149DP PCM
MP-2000 (includes all components)
Servos - S3104
Wires
Battery for Receiver FUTM 1280
20.00
0.08
0.40
1.26
0.25
0.50
Miscellaneous
Total
3.00
53.01
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Performance
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Aircraft Performance
PARAMETER
Endurance (at cruise) (with 2.2lbf fuel)
Range (at cruise)
Minimum Flight Velocity
Rate of Climb (at takeoff conditions)
Maximum Velocity
Climb Angle
VALUE (Approx)
30 min
17 miles
30 ft/sec
7.5 ft/sec
90 ft/sec
mph
90
13.1 deg
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Cost
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Airframe Cost
Part Description
Brand
Bass/Spruce Wing Spars
21st Century Fabric
Ribs
Bass/Spruce Tail Beam
Fuselage Skin (1 ply of E-glass)
V-Stab Foam Core
H-Stab Foam Core
V-Stab Carbon Fiber Covering (1 ply)
H-Stab Carbon Fiber Covering (1 ply)
Engine Supports
Tail Spars
n/a
Coverite
n/a
n/a
n/a
n/a
n/a
n/a
n/a
n/a
2
4
21
2
$15.00
$39.99
$0.98
$10.00
1
2
$15.00
$15.00
1
2
$5.00
$5.00
Rear Gear
Front Gear
Smooth Wheels 4"
Bungees
Misc (Bolts, Nuts, Washers, etc)
n/a
n/a
Sullivan Skylite
Tool Shop
4
1
3
2
$4.74
$4.74
$12.39
$1.50
TOTAL
Quantity Cost/unit
Total Cost
$30.00
$159.96
$20.58
$20.00
HAVE
$15.00
$30.00
HAVE
HAVE
$5.00
$10
$18.96
$4.74
$37.17
$3.00
$25.00
$379.41
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Electronics Cost
Part Description
POD
Onboard Laptop Computer
MIDG
MIDG Power Supply uINS Power Supply
Camera
National Instrument PCMCIA DAQ Card
Wireless Network Card
Vehicle Mount Antenna -- Enterasys
Vehicle Mount Antenna Cable
Range Extending Antenna
AVIONICS
9 Channel R149DP PCM (Included w/Trans)
MP-2000 (includes all components)
Servos - S3104
Wires
Battery for Receiver FUTM
TOTAL
Brand
Quantity
Cost/unit
Total Cost
$2,566.80
$6,750.00
not determined
$1,500.00
$2,566.80
$6,750.00
Canon PowerShot G5
1
1
1
1
1
Enterasys (CSICD-AA-128)
CSICD-AA-128
CSIES-AA-M05
CSIES-AA-PT250
CSIBB-IA
1
1
1
1
1
$1,195.00
$80.00
$85.00
$65.00
$80.00
$1,195.00
$80.00
$85.00
$65.00
$80.00
Futaba
Micropilot
Futaba
1
$139.95
6
6
1
$32.99
$4.00
$44.99
$139.95
HAVE
$197.94
$24.00
$44.99
Dell Latitude C610
Futaba
$1,500.00
$12,728.68
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Propulsion Cost
Part Description
Brand
50 oz fuel tank
O.S. 1.60 FX-FI
Sliencer E-5010
Engine Mount
16X7 4-Blade Prop
Dubro
O.S.
Bisson-Pitts
O.S.
Zinger
TOTAL
Quantity
Cost/unit
Total Cost
1
1
1
1
1
$11.49
$714.99
$49.99
$26.99
$54.60
$11.49
$714.99
$49.99
$26.99
$54.60
$858.06
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Total Aircraft Cost
Airframe Cost
Electronics Cost
Propulsion Cost
$379.41
$12,728.68
$858.06
TOTAL
$13,966.15
What Purdue Will Pay For This Project
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Total Aircraft Value
• Total Aircraft Value = (Engineering Pay) + (Cost) + (Value
of Already Possessed Parts)
• Engineering Pay = 823.75 hr x $100/hour = $82,375
• Aircraft Cost = $13,966.15
• Value of Already Possessed Parts = $10,000
– Micropilot = $5,000
– Carbon Fiber & E-Glass = $5,000 (estimate)
TOTAL AIRCRAFT VALUE = $106,341.15
What Purdue Would Pay to Outsource This Project
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Summary
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Summary – Internal View
Internal Pod
Camera View
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Summary – 3-View
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Summary -Major Design Points
• Aircraft Description
–
–
–
–
Aspect Ratio = 5
Wing Span = 14.4 ft
Wing Area ~ 42 ft2
Aircraft Length = 10 ft
(not including air data
boom)
– Engine = 3.7 hp O.S.
1.60 FX-FI – Fuel
Injected
– Weight = 53 lbf
• Aircraft Configuration
–
–
–
–
–
–
T-Tail
Low Wing
Pusher
High Engine
Tricycle Gear
Internal Pod
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Questions?
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References (I)
•[1] MATLAB. PC Vers 6.0. Computer Software. Mathworks, INC. 2001
•[2] Raymer, Daniel P., Aircraft Design: A Conceptual Approach, AIAA Education Series, 1989.
•[3] Roskam, Jan., Airplane Flight Dynamics and Automatic Flight Controls. Part I. DAR Corporation, Kansas. 2001
•[4] Gere, James M., Mechanics of Materials. Brooks/Cole, Pacific Grove, CA. 2001
•[5] Tower Hobbies. 9 December 2003. http://www.towerhobbies.com
•[6] XFoil. PC Vers. 6.94. Computer Software. Mark Drela. 2001.
•[7] Niu, Michael C., Airframe Structural Design, Conmilit Press Ltd. Hong Kong. 1995.
•[8] Halliday, et al., Fundamentals of Physics, John Wiley & Sons. New York. 1997.
•[9] Roskam, Jan, Airplane Design (Parts I-VIII), Roskam Aviation and Engineering Corp. Ottawa KS. 1988.
•[10] Kuhn, P., “Analysis of 2-Spar Cantilever Wings with Special Reference to Torsion and Load Transference”. NACA Report
No. 508.
•[11] McMaster-Carr. 9 December 2003. http://www.mcmaster.com
•[12] Pro/ENGINEER. PC Release 2001. PTC Corporation.
•[13] Roskam, Jan., Methods for Estimating Stability and Control Derivatives of Conventional Subsonic Airplanes. Publisher Jan
Roskam. Lawrence, KS. 1977.
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References (II)
•[14] Zinger Propeller. 9 December 2003. http://www.zingerpropeller.com
•[15] McCombs, William F., “Wing and Tail Dihedral for Models”, Model Aviation. Dec. 1994. 104-112.
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Appendix
52
SIZING
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Cruise Speed
Equating Pout to thrust ti mes velocity, and equating thrust to drag gives
Pout  T  V  550  (0.75)  SHP  p 
1
3
Vcruise
CD
2
rearrangin g to give
W
(0.75)  550  2  p SHP
S


3
W
SHP
Vcruise
CD
S
 (0.75)  550  2  p  W
W

 
3
SHP 
Vcruise
CD
 S
CD  0.0275 ( Aero )
  0.002309
slug
ft 3
(1000 ft )
ft
( Andrisani )
s
 p  0.67 (Pr opulsion )
Vcruise  50
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Stall Speed
Starting with the equation for lift
1
2
L  Vstall
CL max S
2
and rearrangin g gives
W 1
2
 Vstall
CL max
S
2
CL max  1.2 ( flaps )
  0.002378
Vstall  30
ft
s
slug
ft 3
( sea  level )
( Andrisani )
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Climb Angle
sin   
Thrust
Weight

1
Lift
Drag
Thrust
SHP  p  550
V
 p  0.67
SHP
CD  0.0275
550  p
Weight
 sin   



1
C Lmax
CD
  V  1.1
stall



ft
( Andrisani )
s
 1.2( flaps )
Vstall  30
CL max
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Ceiling
Starting with the equation for specific excess power :
 T qC D0
KW
Ps  V  
 n2

q S
W W / S
and rearrangin g gives
W
(1100) p  Vcruise
W
S

2
SHP
W 
4
2
C D0  Vcruise  4 Kn 
S
 p  0.67
  0.002309
Vcruise  50
ft
s
slug
ft 3
(1000 ft )
( Andrisani )
n  1 .0
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Endurance
Starting with endurance equation given in class :
W

Pcruise
1
W
2
S

1
4C D0
 3C D0

 k



3
4
 550 


 g


C D0  0.025
AR  5
Where :
k
1
 ( AR)(e)
e  1.78(1  0.045 AR 0.68 )  0.64
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Takeoff
Using Equation (3.2) from Roskam gives
W   CL max T / O  TOP23

W
P
S
where TOP23 is defined by Roskam (Equation 3.4) as
s TOG  4.9 TOP23  0.009 TOP232
and sTOG is 105 ft. Solving quadratica lly for TOP23 gives TOP23  21.
CLTO  CL max / 1.21  ( Roskam)
 
TO
 0.98
 SL
sTOG  105 ft
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Landing Distance
Starting with equation for landing distance :
Dlanding  0.265Vstall
2
Inserting previous equation for stall speed :
0.5
 1  W
 
Dlanding  0.265
 2 /  * CLmax  
1
.
687
 S
 



2
CL max  1.2 ( flaps )
  0.002378
slug
ft 3
( sea  level )
60
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Appendix
• OS 1.60 FX-FI
• Consistency: The Fuel Injection system
constantly supplies the correct air/fuel
mixture to the engine, regardless of speed,
altitude, or attitude.
• Recommended is a 450-550cc fuel tank
that allows approximately 10 to 12 minute
flights. = 30 min. with 50 oz. tank.
62
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Aerodynamic Modeling
Prandtl’s Lifting line theory used for aerodynamic modeling of the lifting components
Solving Prandt’s equation
cl
2

 a  a L 0  a i
a0 a0V c
sin n
a i ( )   nAn
sin 
n 1
N
Substituting:
Equation to solve:
( )  2bV
N
4b N
sin n
A
sin
n


nA
 n

n
a0c n1
sin 
n 1
N
 A sin n
n 1
n
 a  a L 0
2
Main Results CL = πAR*A1*(α- αLo)
CDi
C
 L (1   )
A
where
2
A 
   n  n   0
n2
 A1 
N
•System of N equations with N unknowns (Solve N  N matix)
•Take N different spanwise locations on the wing where
the equation is to be satisfied: 1, 2, .. N; (but not at the tips, so: 0 <  < )
•The wing is symmetrical  A2, A4,… are zero
•Take only A1, A3,… as unknowns
•Take only control points on half of the wing: 0 < i  /2
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Choice of main wing airfoil
From lifting line with Initial
parameters:
0.7
Cl
Cd*10
alphai*10
0.6
•Rectangular planform, 1000 ft
•a0 = 2pi,
•αL0 = 0,
•AR = 5;
•W/S = 1.28 (from sizing)
•CL = 0.4437
Cl distribution found at cruise
Cl varies :0 to 0.58
Taking into account the Cl
variation above, the need of
an airfoil with a drag bucket
at the specified Cl’s
Xfoil utilized for different foils
at the above conditions
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
65
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Airfoil Selection
Drag Polar for Candidate Airfoils
0.015
NACA 4412
4415
4425
2418
23018
ClarkY
0.014
0.013
Section Drag Coefficient, c
d
0.012
0.011
0.01
0.009
0.008
0.007
Region of
Interest
Clark Y
0.006
0.005
-1
-0.5
0
0.5
Section Lift Coefficient, c l
1
1.5
Clark Y Airfoil Drag Bucket location fits best
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ClarkY foil
0.2
0.1
y/c
Xfoil runs of ClarkY foil
at cruise and take-off
0
-0.1
-0.2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x/c
Cruise:
Takeoff no flap:
Takeoff 10deg flap:
Takeoff 15deg flap:
αL= -3.5deg
αL= -3.8deg
αL= -7deg
αL= -7.8deg
2
1.5
1
0.5
no flap
10deg
15 deg
Cruise
0
-0.5
-10
-5
0
5
10
15
5
10
15
Cl vs aoa
2
1.5
Cl
1
In lifting Line Equation:
a0 – updated depending on condition
αL - updated according to above
0.5
0
-0.5
-10
-5
0
aoa (deg)
67
Stall Performance
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CL vs aoa for different flap deflections
•
•
CL needed = 1.19
Wing without flaps
reaches CL at 13 deg
aoa
Wing stall possible
Wing with 15 deg flap
deflection reaches CL
at 11 degrees
1.2
1
CL no flap
CL 10 deg flap
CL 15 deg flap
Required CL
0.8
0.6
CL
•
•
1.4
0.4
0.2
0
-0.2
-0.4
-10
-5
0
5
10
15
aoa (deg)
Flaperons necessary to meet stall requirements
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Stall Performance Drag Calculation
CDtotal = CDinduced+CDvisc
CD vs CL for different flap deflections
0.14
CDvsCL no flap
CDvsCL 10 deg flap
CDvsCL 15 deg flap
CDinduced – from Lifting line
0.12
CD visc – integrated at the found Cls
0.1
Viscous Drag Coefficient calculaton
0.035
0.08
0.03
CD
original data
polyfit function
values for our cl from polyfit funtion
0.06
cd
0.025
0.04
0.02
0.02
Required CL
0.015
0
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
CL
0.01
0.005
-0.4
-0.2
0
0.2
0.4
0.6
cl
0.8
1
1.2
1.4
1.6
CD = 0.119 at
required CL
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Cruise Performance
CL needed = 0.44
Total Lift produced = 57lbf
CL achieved at 2.8 deg
Total Drag = 2.6 lbf, L/D =21
CL,cl vs aoa
Cd,CDtotal vs aoa
1.6
0.12
3D CD total
Cd visc 2d
3D CL
2D cl
1.4
0.1
1.2
1
0.08
Cd,CD
0.8
0.06
0.6
0.4
0.04
0.2
0.02
0
-0.2
-6
-4
-2
0
2
4
aoa deg
6
8
10
12
14
0
-6
-4
-2
0
2
4
aoa (deg)
6
8
10
12
14
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Operating Parameters
CL
Aoa
Flap
Deflection
CD
L/D
Stall
1.19
11 deg 15 deg
0.119
10
T/O
0.989
8. deg 15 deg
0.084
12
Cruise
0.44
2.8 deg 0 deg
0.018
24
71
D&C
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Center of Gravity
• Center of Gravity of Aircraft
– Weight of Horizontal Tail changes with area
9
WHT
lbs
 0.44 2  AreaHT
ft
CGAircraft 
W x
i 1
i i
WTotal
Note: 0.44 lbs/ft2 based on
aircraft sizing code
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Aerodynamic Center
• Aerodynamic Center as a function of Horizontal Tail
Area


 d  h  S h 
X acAircraft
 X acwing  CLah 1  da  S  X ach 

 



 d  h  Sh  
1  CLah 1  da  S  

  

X ac  X ac  cwing
d h
 0.49
da
Roskam Eq 11.1
Raymer Fig 16.12
CLa  3.6rad 1
h
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Takeoff Rotation Equation
• This sizing based on angular acceleration
during take-off rotation
W (x
Sh ( ft ) 
2

cg g

 xmgg   g zcgg   g zmgg )  Dg ( zDg  zcgg )  T ( zcgg  zTg )
 Lwf g ( xmgg  xacwf   g zcgg   g zmgg )  M acwf  I yymg 
g
(CLmax
hground
g

hg qrotate )( xach  xmgg   g zmgg   g zcgg )
g
Ref. Roskam 421 book, pg 288-290
Variable definitions found in above reference
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Yaw Moment due to Sideslip
• Vertical Tail sized from Coefficient of Yaw
Moment due to Sideslip
Cn  Cn wb  CLa  SV S  xV b 
v
Due to Wing and Fuselage:

Roskam Eq 11.8
Vol 2
Cn wb  Cn  Cn f  0  57.3  K N  K R1 S f s l f Sb
W
CLa  3.6rad 1
S  41.76 ft 2
xV  5.7 ft
b  14.4 ft
v

Roskam Eq 10.42
Vol 6
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Dihedral Angle
EVD = A + kB
CL
X
A = 0°
B
k = f(x/(b/2)) = 0.98
B = EVD / k ≈ EVD
A=0°
Ref. McCombs, William F. “Wing and Tail Dihedral for Models.”
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Dynamics
Short Period Mode
Phugoid Mode
Pole
-14.391 ± 1.0079i
Pole
-0.078823 ± 0.71828i
Natural
Frequency
14.431 (rad/s)
Natural
Frequency
0.72259 (rad/s)
Damping
Ratio
0.99721
Damping
Ratio
0.10908
Dutch Roll Mode
Pole
-1.1607 ± 2.4427i
Natural
Frequency
2.7045 (rad/s)
Damping
Ratio
0.42918
Spiral Mode
Pole
0.29086
Roll Mode
Pole
Ref. Purdue University AAE565, Matlab Predator Code
-25.748
78
STRUCTURES
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What Materials to Use
Titanium
Bass / Spruce
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Material Properties
Titanium = difficult to obtain
Wood = not difficult to obtain
Bass
Spruce
Young's Modulus
[E] (psi)
1.46E+06
1.43E+06
Torsional Stiffness Max Compression Stress
[G] (psi)
[s yc] (psi)
2.92E+05
4730
2.86E+05
5180
Max Tension Stress
[s yt] (psi)
8700
9400
Ref. 1999 Forest Products Laboratory Wood
Handbook
Ref. www.towerhobbies.com
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Twist Constraint (<1o)
1.5
(rad ) 
Where
TL  tanh( L) 
1 

B0 
L 
Chord-wise Lift Resultant
Ref. Kuhn pg. 49
1
Rib Cap (not shown)
0.5
T = Torque (in-lbf)
l = f(B0, A0)
Y (ft)
L = Length (in)
(ref. Appendix)
Rib
Rib
0
0
0.5
1
Rib
1.5
2
2.5
3
-0.5
A0 = f(E, I) (ref. Appendix)
Shear Center
-1
Rear Spar
Forward Spar
B0 = f(G,J) (ref. Appendix)
E = Young’s Modulus (psi)
I = Moment of Inertia (in4)
G = Torsional Stiffness (psi)
J = Polar Moment of Inertia (in4)
T (in  lbf )  ( Force)( Distance )
(base)( height )( height 2  base 2 )
J (in ) 
12
4
Ref. Gere
-1.5
X (ft)
Assumptions:
Small Deflections
Spars & Ribs Carry all Torsion
Span ~ 14.4 ft
Chord ~ 2.9 ft
Safety Factor = 1.5
G-Loading = 5.0
Weight = 53 lbs
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Twist at Tip
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Twist at Tip (Zoom)
Chosen Front Spar = 0.73” thick
Chosen Rear Spar = 0.25” thick
(note, this doesn’t include the
step)
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Deflection at Tip
( Load )( a 2 )
 tip (in ) 
3( L  a)
6( E )( I front  I rear )
Load (lbf)
Ref. Gere pg. 892
Where
Load = Weight*SF*G-loading (lbf)
L = Length (in)
a (in)
L (in)
E = Young’s Modulus (psi)
I = Moment of Inertia (in4)
For this design:
a ~ 3 ft or 36 in
(based on span-wise lift distribution)
Assumptions:
Small Deflections
NO TORSION
Span ~ 14.4 ft
Chord ~ 2.9 ft
Safety Factor = 1.5
G-loading = 5.0
Weight = 53 lbs
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Deflection at Tip
Chosen Spar
Configuration
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Is Stress too High?
( M )( y)
 ( psi ) 
( I front  I rear )
Load (lbf)
Ref. Gere pg. 323
Where
M = Weight*SF*G-loading*a (in-lbf)
y = Maximum Dist from Neutral Axis (in)
a (in)
L (in)
I = Moment of Inertia (in4)
For this design:
a = 3 ft or 36 in
(based on span-wise lift distribution)
Assumptions:
Span ~ 14.4 ft
Chord ~ 2.9 ft
Safety Factor = 1.5
G-loading = 5.0
Weight = 53 lbs
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Max Tension Stress
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Max Compression Stress
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Covering
• Traditional Monocote
may not be strong
enough for these
large aircraft
• Coverite 21st Century
Iron on Fabric is
stronger, and resists
tears much better
– 0.34 oz/ft2
– Approx. 2 lbs for entire
wing
Ref. www.towerhobbies.com
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Summary
• Main Wing
– Spruce or Bass wood
– Front Spar
h
• 0.73” thick by 3.6” high
– Rear Spar
• 3/8” thick by 3” high
t
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Rear View of Tail
•NOTES
•Torsion can effectively be reduced
with appropriate beam spacing
•Bending can be reduced by
increasing moment of inertia of
beams (not spacing)
Side force from Vstab creates torsion
effect on beams
•Some torsion is inherent, torsion
can not be negated as it could in
wing
Downward force from H-stab
creates bending moment on
beams
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Deflection at Tip (Rear of Tail)
Load (lbf)
( Load )( S .F .)(Gloading )( L3 )
 tip (in ) 
3( E )( I right  I left )
Ref. Gere pg. 892
L (in)
Where
Load = (lbf)
L = Length (in)
E = Young’s Modulus (psi)
I = Moment of Inertia (in4)
Moment of inertia of rectangular beam:
I
(in4)
=
(t)(h3)/12
t and h shown on next slide
Assumptions:
Small Deflections
Safety Factor = 1.5
G-loading = 3.0
Rectangular Beams
Current Known Values:
L = 6.2 ft
Load ~ 8 lbf
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Deflection at Tip (Rear of Tail)
Green = spruce
Black = bass
h
h=2 in
t
h=3 in
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Deflection at Tip (Rear of Tail)
Green = spruce
Black = bass
h
h=2 in
t
h=3 in
Required t
~0.55 in
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Landing Gear Placement (I)
θ = tipback angle =
Landing gear placement based on
guidelines found in Raymer
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Landing Gear Placement (II)
γ = overturn angle =
Landing gear placement based on
guidelines found in Raymer
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Easily Obtainable Square Tubing
Aluminum Alloy 6061
Width (in)
1
1 1/2
2
Height (in)
1
1 1/2
2
Thickness (in)
0.125
0.125
0.125
X-Section Area
0.5
0.75
1
P/N
6546K21
6546K22
6546K23
1 foot
$6.23
$9.00
$10.50
3 feet
$12.43
$20.75
$25.25
6 feet
$22.33
$39.00
$48.00
Thickness (in)
0.125
0.062
0.125
0.125
0.125
0.125
0.125
X-Section Area
0.375
0.248
0.5
0.625
0.75
0.875
1
P/N
88875K52
88875K51
88875K54
88875K58
88875K61
88875K64
88875K67
1 foot
$2.11
$1.57
$3.35
$3.83
$4.10
$4.68
$6.03
3 feet
$6.34
$4.74
$10.00
$11.49
$12.28
$13.97
$18.05
6 feet
$11.09
$11.09
$23.39
$26.77
$28.63
$32.59
$42.14
Aluminum Alloy 6063
Width (in)
3/4
1
1
1 1/4
1 1/2
1 3/4
2
Height (in)
3/4
1
1
1 1/4
1 1/2
1 3/4
2
Ref. www.mcmaster.com
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Buckling of Rear Gear
Pcr (lb f ) 
 cr ( psi ) 
 2 EI
2
Load
Ref. Gere pg. 763
L
Pcr
A
L
Where
L = Length (in)
E = Young’s Modulus (psi)
I = Moment of Inertia (in4)
A = Cross Sectional Area (in2)
For Rear Gear:
L ~ 15.3 in
Load
Assumptions:
Pinned-Pinned Column
1st Mode Buckling
No Eccentricity
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Compressive Failure of Rear Gear
Load
 c ( psi ) 
Load
A
 cy ( psi )  34,000
Where
Ref. MIL-HDBK-5H: 3-255
Load = (Weight)(S.F.)(Gloading)
A = Cross Sectional Area (in2)
L
Load
Assumptions:
Weight = 53 lbf
Gloading = 10
S.F. = 1.5
Aluminum 6061-T6
No Buckling
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Stress on Rear Gear
Smallest easily obtainable
tubing: 1” x 1” x 0.062”
t=0.062”
t=0.125”
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Great, what about the bungee?
• Consider worst reasonable landing
situation
– Moving at (1.1)Vstall
– 5 feet above ground
– Aircraft falls out of the sky
• Can the bungee absorb the energy
associated with this landing?
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Great, what about the bungee?
1
mV 2
2
PE  (m)( g )( altitude)
KE 
Etotal  KE  PE
F (lb f )  k ( x)
1
Wbungee   Energy   kx2
2
•Don’t want x to exceed 3 inches
(beyond initial stretch) on landing
Assumptions:
Weight = 53 lbf
Vstall = 30 ft/sec
Altitude = 5 ft
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What Spring Constant is Needed?
Required k ~ 3.75 lbf/in
1/k ~ 0.266 in/lbf
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What is the Spring Constant?
Inverse of Spring Constant versus Relaxed Length
0.5
0.45
1/(Spring Constant) (in/lbf)
0.4
0.35
y = 0.0152x
0.3
0.25
0.2
Relaxed Length ~18 inches
0.15
0.1
0.05
0
0
2
4
6
8
10
12
14
16
18
Relaxed Length (in)
20
22
24
26
28
30
32
105
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How Big is the Bolt?
 ult
nylon
( psi )  9,000
Ref. Gere pg 900
 moments  0
Load
(Reaction) (3.1" )  (Load)(6.9" )
If load = (Weight)(S.F.)(Gloading) = 795 lbf
Reaction = 1770 lbf (instantaneous)
Reaction
Need cross sectional area of bolt to be 0.197 in2
Diameter of nylon bolt = 0.5 in
Assumptions:
Weight = 53 lbf
Gloading = 10
S.F. = 1.5
3.1”
6.9”
106
PERFORMANCE
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Endurance
• Endurance = Fuel / Consumptionfuel
• Avg. Engine Fuel Consumption =
45.455 mL/min
• Endurance = 30 min
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Range
R
550  p L  W i 
  ln 

C bhp D
Wf


Since this is RC, assume almost instaneous cruise conditions
L/D = 19
Cbhp = 1.5 lb/hr/bhp
Prop eff = .67
Fuel Frac = 1.043
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Minimum Flight Velocity
Velocity min
Weight

Clmax * q
Velocitymin= 29.95 ft/sec
Weight = 53 lbf
CLmax = 1.19
q =1.067 lbf/ft^2
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Rate of Climb
Vv
550 hp engine  p
W

D V
W
Vv= 7.5 ft/sec
D = 6.5lbf
hpengine = 3.7 hp
W = 53 lbf
V = 33 ft/sec
Prop Eff = .3
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Maximum Velocity
Maximum Velocity
20
18
16
Thrust/Drag [lbf]
14
Thrust
12
10
8
6
Drag
4
2
0
30
40
50
60
70
Flying Velocity [ft/s]
80
90
100
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Climb Angle
V
 v
  asin 

 V 
Vv = 7.5 ft/sec
V = 33 ft/sec
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