15 seater commuter aircraft
TRANSCRIPT
DESIGN OF 15 SEATER COMMUTER AIRCRAFT
(PART I AERODYANMIC DESIGN)
By
BALA ABINESH.C (42007101009)
KARTHIK.K (42007101023)
PRADEEP KUMAR.S (42007101033)
SAI SSHRIMAN.M (42007101041)
YADHAVAN.U (42007101306)
Under the guidance of
Dr. K. Padmanaban – Professor,
Department of Aeronautical Engineering,
Tagore Engineering College
A report submitted to the Department of Aeronautical Engineering,
Tagore Engineering College
in partial fulfilment of the requirements for the degree of
Bachelor of Engineering
ANNA UNIVERSITY,
Chennai.
April 2010
BONAFIDE CERTIFICATE
This is to certify that the design project report titled
DESIGN OF 15 SEATER COMMUTER AIRCRAFT being submitted by
BALA ABINESH.C (42007101009)
KARTHIK.K (42007101023)
PRADEEP KUMAR.S (42007101033)
SAI SSHRIMAN.M (42007101041)
YADHAVAN.U (42007101306)
to the Department of Aeronautical Engineering, Tagore Engineering College Chennai, in partial
fulfilment of the requirements for the award of Degree of Bachelor of Engineering
(Aeronautical Engineering) is a bonafide record of the work carried out by this group under my
guidance and supervision in the even semester of the academic year 2009-2010.
Dr.K.PADMANABAN, Dr.P.BASKARAN,
Professor, Professor and Head,
Dept. of Aeronautical Engineering, Dept. Of Aeronautical Engineering,
Tagore Engineering College, Tagore Engineering College,
Chennai - 600 048. Chennai - 600 048.
INTRODUCTION
TYPE:
Twin-turboprop, 15 seater commuter aircraft.
DESIGN FEATURES:
Low mounted unswept wing, circular section pressurised fuselage, conventional tail
with fixed incidence tail plane. Wing section, NACA 631412, incidence at 1.88˚ with a dihedral
of 3˚.
FLYING CONTROLS:
Conventional. Split flaps at the trailing edge of the wing.
LANDING GEAR:
Retractable, tricycle arrangement with nose wheel. All units retract into the fuselage.
Tyre pressure & size: Main wheel = 90 p.s.i
Main wheel = 27 7.25 in2
Nose wheel = 40 p.s.i.
Nose wheel = 19 6.25 in
2
POWERPLANT:
Two PRATT & WHITNEYCANADA PT6A-13A turboprop engines each rated 750 shp,
driving a 3 blade variable pitch propeller of diameter 3m.
ACCOMODATION:
Two pilots in flight deck, Main cabin accommodates one attendant and 15 passengers in
pressurised and air-conditioned environment.
BASIC SPECIFICATIONS
OF
15 SEATER COMMUTER AIRCRAFT
CRUISE VELOCITY : 461 km/hr
PAYLOAD : 15 passengers
RANGE : 2000 km
CRUISE ALTITUDE : 4 km
DETAILED SPECIFICTIONS
DIMENSIONS:
WING SPAN : 15.58 m
LEGNTH OF FUSELAGE : 13.44 m
FUSELAGE DIAMETER : 2.7 m
WING AREA : 27 m2
HORIZONTAL TAIL AREA : 4.79 m2
VERTICAL TAIL AREA : 4.05 m2
WING ASPECT RATIO : 9
ROOT CHORD OF WING : 2.475 m
TIP CHORD OF WING : 0.99 m
MEAN AERODYNAMIC CHORD : 1.732 m
WING LOADING : 2000 N/ m2
TAIL ASPECT RATIO
HORIZONTAL TAIL : 4
VERTICAL TAIL : 1.5
WING AEROFOIL : NACA 631412
TAIL AEROFOIL : NACA 0009 (both vertical and horizontal tails)
PROPELLER DIAMETER : 3m
WEIGHTS:
TAKEOFF WEIGHT : 54,249 N
FUEL WEIGHT : 14,479 N
ENGINE WEIGHT : 3060.72 N
PAYLOAD WEIGHT : 15,000 N
PERFORMANCE:
MAXIMUM SPEED : 128 m/s
CRUISE SPEED : 115 m/s
RANGE : 2000 Km
ENDURANCE : 4.83 hrs
CRUISE ALTITUDE : 4 Km
RUNWAY LENGTH : 900 m
RUNWAY LOADING : 4.85 ton/ft2.
RATE OF CLIMB MAXIMUM : 821 m/min at sea level
TIME TO CLIMB TO CRUISE ALTITUDE : 5.83 minutes
LIFT TO DRAG RATIO AT CRUISE : 13.3
SERVICE CEILING : 6.9 km
ENGINE SPECIFICATION:
S.H.P : 750 hp
WEIGHT : 1530.36 N
S.F.C : 2.7468 (N/hr)/SHP
LENGTH : 1575 mm
DIAMETER : 483 mm
CONTENTS
List of symbols used.
List of tables
List of graphs
Comparative data
1. Initial weight estimation
2. Engine selection
3. Fuel weight estimation
4. Second weight estimation
5. Selection of tip chord and root chord
6. Estimation of thickness to chord ratio of wing aerofoil
7. Aerofoil selection
8. Estimation of landing speed and stalling speed
9. Flap selection
10. Tyre selection
11. Fuselage details
12. Propeller design
13. Configuration layout
14. C.G calculations
15. Drag estimation
16. Drag polar estimation
17. Performance calculations
18. Stability analysis
19. V-n diagram
20. Conclusion
Bibliography
LIST OF SYMBOLS USED
AR Aspect Ratio
a Temperature Lapse Rate ˚C/m
aw Slope of the CL vs. α curve for wing /deg
at Slope of the CL vs. α curve for a horizontal tail. /deg
av Slope of the CL vs. α curve for a vertical tail /deg
a.c aerodynamic centre
b Wing span m
Mean aerodynamic chord m
CD Drag coefficient
CDi Induced drag coefficient
CD0 wing Drag coefficient of wing
CDt Total Drag Coefficient
CL Lift Coefficient
Cmac Pitching moment coefficient at aerodynamic centre
Cmc.g Pitching moment coefficient at centre of gravity
Cmfus,nac Pitching moment coefficient due to fuselage and nacelle
Cn Yawing Moment coefficient.
cr Root Chord m
ct Tip Chord m
Cs Speed Power Coefficient of Propeller
Clβ. Dihedral effect /deg
C.G Centre of gravity m
D Diameter of propeller m
D Drag N
E Endurance Hrs
e Ostwald‟s efficiency factor
it Orientation of tail plane on the fuselage deg
iw Orientation of wing on fuselage deg
J Advance ratio of propeller
K Gust alleviation factor.
le Distance between inoperative engine and centre line of fuselage m
lt Distance between C.G position of aircraft and horizontal stabilizer m
lv Distance between C.G position of aircraft and vertical stabilizer. m
N Rotation per minute. /min
N0 Neutral point m
n Rotation per second /s
n Load factor
R Range km
R/C Rate of climb m/min
Re Reynolds number
S Wing area m2
St Horizontal tail area m2
Sv Vertical tail area m2
Sл Area of individual components m2
S.F.C Specific fuel consumption (N/hr)/SHP
S/L Sea Level
SHP Shaft horse power
THP Thrust horse power
t/c Thickness to chord ratio
Tail volume ratio
U Gust velocity. m/s
Vcruise Cruise velocity m/s
Vs Stalling velocity m/s
Wf Weight of fixed equipment like seats, galleys etc N
Wfuel Weight of the fuel N
Wpayload Weight of payload (passengers) N
Wpilot Weight of the pilot N
Wpowerplant Weight of the power plant N
Wmax Maximum weight of the aircraft N
Wstructure Weight of the structure of the aircraft N
WT.O Takeoff weight N
W/S Wing loading N/ m2
Xa,c Distance between nose of the aircraft to the a.c of Aircraft m
XC.G Distance between nose of the aircraft to the C.G position of the aircraft. m
α Angle of attack deg
β Blade angle. deg
Dihedral angle deg
Floating tendency /deg
Restoring tendency /deg
δe Deflection of the elevator deg
δr Deflection of the rudder deg
Damping ratio
ηt Tail efficiency
ε Angle of downwash deg
Density kg/ m3
Density of air at sea level kg/ m3
ζ Density ratio
λ Taper ratio
μ Viscosity N-s/ m2
Airplane mass parameter. (W/S)/ g
Elevator effectiveness factor
Airplane time parameter (W/S)/ gV s
undamped natural frequency
LIST OF TABLES
1. Comparative data
2. Engines and its specification
3. Aerofoils and their CLMAX & CDMIN
4. Different runways and their runway loadings.
5. Propeller efficiency and blade angle at various condition
6. C.G calculation of fuselage
7. C.G calculation of wing
8. C.G calculation for 10% fuel and full payload condition
9. C.G calculation for 10% fuel and no payload condition
10. Airplane‟s C.G position at various configuration
11. Parasite drag calculation for takeoff condition
12. Parasite drag calculation for cruising condition
13. Parasite drag calculation for landing condition
14. Drag polar estimation for takeoff condition
15. Drag polar estimation for cruising condition
16. Drag polar estimation for landing condition
17. Rate of climb estimation for various altitude
18. Elevator deflection for various CL
19. Yawing moment coefficient for various velocity
20. Load factor limitations for various category of aircraft
21. Velocity at various load factors
LIST OF GRAPHS
1. Aspect ratio Vs velocity
2. Wing loading Vs velocity
3. Span to length ratio Vs velocity
4. Drag polar
5. THP Vs velocity
6. Rate of climb Vs velocity
7. Maximum rate of climb Vs altitude
8. 1/(R/C)max Vs altitude
9. CL Vs α
10. Cm Vs CL for various C.G positions
11. Cm Vs CL for various elevator deflections
12. Elevator deflection Vs equilibrium CL
13. Cn Vs velocity
14. Amplitude Vs time for phugoid oscillation
15. V-n diagram
COMPARATIVE DATA
Aircraft design is both an art and engineering. From the time that an
airplane first materializes as a new thought in the mind of one or more persons to the time that
the finished product rolls out of the manufacturer‟s door, the complete design process has gone
through three distinct phases that are carried out in the sequence. These phases in chronological
order are conceptual design, preliminary design & detail design.
This conceptual aerodynamic design project involves the estimation of
weight and choice of the aerodynamic characteristics that will be best suited to the mission
requirements. It also estimates drag, size of the powerplant, the best airframe size to
accommodate the payload, wing and engine placement. This conceptual design locates principal
weight groups in order to satisfy static stability requirements. It also sizes control surfaces to
achieve the degree of manoeuvrability.
The designing process started with the collection of comparative data from
various aircraft of the present requirement existing in market. Data on nearly 12 aircraft were
collected out of which 6 aircraft were selected. From the comparative data parameters like
aspect ratio, span to length ratio, wing loading and maximum velocity were finalised.
The comparative data was obtained from “JANE’S ALL WORLD
AIRCRAFT-2006-07”
NAME OF
AIRCRAFT
Beechcraft
King Air
C90GTi
Beechcraft
King Air
B100
Cessna
441
Dornier
Do 228
Beechcraft
B200
Piaggio
P.180
Avanti
COUNTRY OF ORIGIN
U.K U.K INDIA ITALY
POWERPLANT PWC PT6A-135A
Garrett TPE-331-6-251B
Garrett TPE331
Garrett
TPE331 PWC PT6A-42
PWC PT6A-66
COUNTRY OF ORIGIN
U.S.A U.S.A U.S.A U.S.A U.S.A U.S.A
NO. OF ENGINES
2 2 2 2 2 2
DIMENSIONS: LENGTH 10.82 m 12.17 m 11.89 m 16.56 m 13.34 m 14.41 m
SPAN 15.32 m 14.0 m 15.04 m 16.97 m 16.61 m 14.03 m
HEIGHT 4.34 m 4.7 m 4.01 m 4.86 m 4.57 m 3.97 m
WEIGHTS: W(MAX) 4,580 kg 5,352 kg 4,611 kg 6,200 kg 5,670 kg 5,239 kg
W(EMPTY) 3,150 kg 3,212 kg 2,489 kg 3,687kg 3,520 kg 3,400 kg
W(PAY LOAD) 907kg
W(FUEL) 1653 kg 1271kg
W(LANDING) 5900 kg 5,670 kg 4965 kg
AREAS: WING 27 m² 26.0 m² 23.5 m2 32.0 m² 28.2 m² 16 m²
HORIZ.TAIL 8.33 m² 4.52 m² 3.83m²
VERT.TAIL 1.5 m² 3.46 m² 4.73m²
FLAPS 5.87 m² 4.17 m² 1.60m²
AILERONS 1.67 m² 0.58m²
RUDDER 1.4 m² 0.66m²
ELEVATOR 1.79 m² 1.24m²
SPEEDS:
V(CRUISE) 416 km/h 454 km/h 315 km/h 536 km/h 593 km/h
V(MAX) 500 km/h 491 km/h 480 km/h 433km/h 545 km/h 0.70 M
V(STALL) 145 km/h 148 km/h 139 km/h 172km/h
RANGE/ENDURANCE: WITH FULL PAYLOAD
2,446 km 2,455 km 2,078 km 2445km 3,338 km 2213 km
TAKE OFF DISTANCE
442 m 567 m 869 m
LANDING DISTANCE
457 m 536 m 872 m
ASPECT RATIO 8.7 7.5 9 9.8 12.3
TAPER RATIO 0.34
SWEEP BACK 1°11‘24“
WING LOADING
170 kg/m² 205.84
kg/m²
193.8 kg/m²
201.6
kg/m² 327.4
kg/m²
From the above data, graphs of aspect ratio, wing loading, and span to length ratio
were drawn against the velocity.
0
2
4
6
8
10
12
14
0 200 400 600 800 1000
Velocity (km/hr)
AR
Aspect Ratio = 9
Velocity =461 km/hr
Aspect Ratio Vs Velocity
0
500
1000
1500
2000
2500
3000
3500
0 200 400 600 800 1000
W/S
Velocity (km/hr)
Wing loading Vs Velocity
Velocity = 461 km/hr
Wing Loading =2000 N/m2
THE SPECIFICATIONS FOUND FROM THE COMPARATIVE DATA ARE:
ASPECT RATIO AR = 9
SPAN TO LENGTH RATIO b/l = 1.15
WING LOADING W/S = 2000 N/m2
MAXIMUM VELOCITY VMAX = 128 m/s
CRUISE VELOCITY VCRUISE = 115 m/s
RANGE R = 2000 km
CRUISE ALTITUDE H = 4 km
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 200 400 600 800 1000
b/l = 1.15V = 461 km/h
Velocity (km/hr)
b/l
Span to Length Ratio Vs Velocity
INITIAL WEIGHT ESTIMATION
The weight estimation is an iterative process. There are various ways to subdivide and
categorize the weight components of the aircraft. The standard way is to divide the weight as
1. Crew weight Wcrew : The crew comprises the staff necessary to operate the airplane in
fight. In this design it is assumed that the airplane has a pilot, co-pilot and an attendant.
Hence 3 crew members. The unit weight for crew is assumed as 1000 N.
2. Payload weight Wpayload : The payload is what the airplane is intended to transport.
Thus 15 passengers are the payload of this aircraft. It assumed that the passenger unit weight
is 1000 N.
3. Fuel weight Wfuel : This is the weight of the fuel in the fuel tanks. It is assumed that it is 15%
of the total weight of the aircraft.
4. Powerplant weight Wpowerplant: This is the weight of the engine. The weight of the engines is
assumed as 10% of the total weight of the aircraft.
5. Structural weight Wstructure : It is assumed that structural weight of the aircraft is 30% of the
total weight.
6. Fixed equipment weight: Fixed equipment is the weight of seat, galleys, electronic
equipments ....etc. It is assumed that fixed equipment weight is 4.5% of the total weight.
The total weight of the aircraft is calculated by adding the weight of its components as
shown below.
INITIAL WEIGHT ESTIMATION:
W1 = Wstructure + Wpayload + Wpowerplant + Wfuel + Wcrew + Wfixed equipment
= 0.3W1 + (1000 x NO OF PASSENGERS) + 0.1W1 + 0.15W1 +
(1000 x NO OF CREW) + 0.045W1
= 0.3W1 + 1000*(15+3) + 0.1W1 + 0.15W1 + 0.045W1
0.405W1 = 18000
W1 = 44,444 N
From the comparative data (SHP/W) AVG is calculated to be 0.01506 hp/N.
Therefore SHP = (SHP/W) AVG * W1 hp
= 669.32 hp
An engine with 10% extra power is considered.
ENGINE SELECTION
From the Jane‟s engine data book, engines having shaft horse power in the range
650 – 750 hp were considered.
ENGINE S.H.P WEIGHT
N
SFC
(N/hr)/SHP
PT6A-114A 675 1557.828 0.640
PT6A-15AG 680 1459.728 0.602
PT6A-25C 750 1539.287 0.595
PT6A-27 680 1459.728 0.602
PT6A-34AG 750 1472.579 0.595
PT6A-135 750 1530.360 0.585
PT6A-36 750 1472.579 0.590
From that PT6A-135A PRATT WHITNEY CANADA ENGINE was selected as it had
low weight, low SFC, high SHP. The specification of the airplane engine is
S.H.P = 750 hp
WEIGHT = 156.0 kg = 1530.36 N
S.F.C = 0.585 (lb/h)/ehp = 2.7468 (N/hr)/SHP
LENGTH = 1575 mm
DIAMETER = 483 mm
FUEL WEIGHT ESTIMATION
The weight of the fuel was calculated from the data obtained from the engine configuration
and comparative data collected. The formula used to calculate the fuel weight is
Weight of the Fuel = (SFC x (Range/VCruise) x SHPALTITUDE x No of Engines)
For emergency needs 20% of the fuel is kept as reserve. Thus the weight of the fuel is
multiplied by a factor of 1.2.
Weight of the Fuel = 1.2(SFC x (Range/VCruise) x SHPALTITUDE x No of Engines)
From comparative data
RANGE = 2000 km
CRUISE VELOCITY = 410 km/hr
NO. OF ENGINES = 2
CRUISE ALTITUDE (H) = 4 km
SHPALTITUDE = SHPS/L * 1.2
Where is the density ratio which is approximately determined from the equation
= 20 – H/20 + H = 0.6667 for 4 km altitude.
Therefore
SHPALTITUDE = 750 * 0.66671.2
= 461 hp
SFC = 2.7468 (N/hr)/SHP
Thus the weight of the fuel is
Weight of the Fuel = 1.2(2.7468*(2000/416)*416*2)
= 14,472 N
SECOND WEIGHT ESTIMATION
Since the exact weight of the powerplant and the fuel are known the corresponding
values are substituted in weight equation to obtain the second weight estimation.
W2 = Wstructure + Wpayload + Wpowerplant + Wfuel + Wcrew + Wfixed equipment
WFUEL = 14,472 N
Wpowerplant(2 engines) = 156 * 2 * 9.81
= 3061 N
Therefore W2 = 0.3 W2 + 1000*(15+3) + 3061 + 14,472 + 0.045 W2
0.655 W2 = 35,533 N
W2 = 54,249 N
This is taken as the maximum takeoff weight of the airplane in the subsequent calculations.
SELECTION OF TIP CHORD AND ROOT CHORD
As the wing loading and the weight of the aircraft from the second weight estimation are known
the wing area is found as follows:
W/S 2000 N/m2
S 54,249 / (2000)
S 27 m2
Similarly AR b2 / S = 9
b2 9*27
b 15.58 m
b/l 1.15 (from comparative data)
l 15.58/1.15 = 13.55 m
ct/cr ------------------- (1)
Where, aper Ratio
0.4 (From the comparative Data)
S b*(ct+cr)/2 ------------------- (2)
Solving (1) & (2), we get
Root chord cr 2.475 m
Tip chord ct 0.99 m
The mean aerodynamic chord 1.732 m
ESTIMATION OF THICKNESS TO CHORD RATIO OF WING
AEROFOIL.
The thickness to chord ratio of the aerofoil for wing is selected such that the wing
has sufficient space for storing the fuel. Part of the fuel can also be stored in fuselage if
necessary. But, for the enhanced safety of the occupants, it is extremely desirable to store the
entire fuel in the wing rather than in the fuselage. Also, with the fuel storage in wings, the shift in
the airplane‟s centre of gravity as fuel is consumed is usually much less than with fuel in the
fuselage.
The front spar is located at 15% of the chord from the leading edge and the rear spar
is at 65 % of the chord. The fuel must be stored within this region. The volume of the fuel can be
calculated from the weight of the fuel used and the specific gravity of the fuel. Once the volume
of the fuel tank is calculated (t/c)max can be estimated by equating the volume available in the
wing.
Specific gravity of the fuel (aviation kerosene) = 0.8
Weight of the fuel = 14,472 N
Volume of the fuel = 14,472/ (0.8*1000*9.81)
= 1.844 m3
Area available for holding
fuel
Front spar at 0.15
7.8 m
¾ of b/2 Rear spar at 0.65
The volume available in both the wings = 2*((t/c)max *area* )
Area available = 5.61 m2
Mean chord for area shown = 1.577m
Thus on equating the volume of the fuel and volume available in the wing the (t/c)max
is found to be 10.42 %
A 12% aerofoil is selected.
ESTIMATION OF CL cruise:
The cruising CL was calculated by knowing the air density at cruising altitude and cruising
velocity
*0.667
kg/m3
CL cruise =2(W/S)/ V2
cruise
= 2*2000/(0.8173*1152)
CL cruise = 0.37
AIRFOIL SELECTION
The airfoil selection is one of the crucial steps in airplane design. Criteria for
the selection of airfoil are that, the airfoil should have high lift coefficient and the drag
coefficient should be as low as possible. Various conventional airfoils like NACA 4 digit, 5 digit
airfoils & laminar flow airfoils were considered.
The airfoil data were collected from the book “THEORY OF WING
SECTIONS” by Abbott and von Doenhoff. The graphs in the book were given for various
Reynolds numbers. As a first step, the Reynolds number at which the aircraft was flying was
found.
T = To – aH
a = Temperature Lapse Rate
= 0.0065 K/m
T = To – aH
= 288.15 – 0.0065*4000
= 262.15 K
*(262.15/288.15)
= 1.628 * 10-5
N-s/m2
Re = V /
cruising altitude = 4 km
Data on laminar flow NACA airfoils with 12 % thickness were collected at Re = 9*106 and
compared.
NACA 631–412 was selected, as it had high CL as well as low CD0 min compared to other airfoils.
The digits in the airfoil nomenclature indicate the nature of the airfoil. The last
two digits indicate that the thickness of the airfoil is 12% of the chord. The third digit stands for
the design lift coefficient of airfoil viz. 0.4. The fourth digit indicates the location of the
maximum pressure which is at 30% of the chord. The subscript 1 indicates that minimum drag
extends on either side of design CL by 0.1 in CL vs.CD graph.
AEROFOIL CL MAX CD MIN
63-012 1.42 0.004
631212 1.6 0.004
631412 1.8 0.004
63015 1.5 0.0045
63215 1.6 0.005
ESTIMATION OF LANDING SPEED & STALLING SPEED
Since the aircraft is to be operated between small cities, it should be designed such
that it lands in small runways. Hence the runway length was chosen as 900 m.
The CAR (Civil Air Regulation) states that the landing distance should not exceed
60% of the runway length.
Therefore landing distance = runway length *0.6
= 900*0.6
= 540 m
From the Newton‟s laws of motion, it is known that
v2 – u
2 = 2aSL where SL is the landing distance.
The final velocity of the aircraft is zero. Also, deceleration while landing is maintained
at 20% of the acceleration due to gravity. Substituting these values and solving for u gives the
value of the landing speed. It is found that the landing speed is equal to 46 m/s.
Also the landing speed is usually 15% more than the stalling speed.
Thus the stalling speed = 46/1.15
= 40 m/s at sea level.
FLAP SELECTION
ESTIMATION OF CL max:
The required CL max) was calculated by computing the difference between required CL max and
available CL max.
CLmax(reqd) = 2(W/S)/ VS2
= 2*2000/ (1.226*402)
= 2.04
CL max reqd = CL max reqd – CLmax available
=2.04–1.8
=0.24
This CL max reqd value must be made available using part span flaps.
From the flap data book CL is found to be 0.98 for the t/c ratio of 12%.For Cf/C=0.3
the flap data given is for the full span. Hence it is converted for the part span using correction
factor of 0.6
( CL)avg from part span flap = 0.98*0.6
= 0.59
Thus the flap can provide the necessary CL max) for the aircraft.
S.NO t FULL SPAN
CL
PART SPAN
CL x 0.6
1. 30o 0.69 0.414
2. 45o o.88 0.530
3. 60o 0.98 0.590
TYRE SELECTION
Various landing gear arrangements were observed, like tail dragger, bicycle, tricycle
arrangements. Tricycle arrangement with nose wheel was chosen because of the following
advantages of the arrangement.
1. Cabin floor for the passengers is horizontal when the aircraft is on the ground
2. Forward visibility is improved for the pilot on the ground
3. The tricycle landing gear requires the center of gravity of the airplane be ahead of
the main landing gear and this enhances the stability of the plane during ground
roll, allowing the airplane to “crab” into a cross wing, i.e. the fuselage does not
have to be aligned parallel to the runway
Usually in tricycle arrangement 90% of the total load of aircraft is taken by the main
landing gear and remaining 10% is taken by the nose wheel on the ground.
The landing gear should be selected such that the runway loading should be acceptable for
the type of runway chosen. For different runways the allowable loadings are
GRASS 2 ton/ft2
GRASS STRIP 3.5 ton/ft2
ASPHALT(TAR) 7 ton/ft2
CONCRETE 11 ton/ft2
Since the 15 seater aircraft is a commuter aircraft, it should be able to land on the small airports
which usually have runway loading 7 ton/ft2 & below. Thus the selected tyre should satisfy three
main criteria:
1. It should have low runway loading.
2. It should be able to carry the aircraft‟s total load.
3. It should be compact for retraction into fuselage/wing.
WT.O = 54,249 N = 12,193.5 lb=5.529 tons.
Load taken by main landing gear = 0.9 WT.O = 10,974.224 lb
Load taken by nose landing gear = 0.1 WT.O = 1219.3 lb
Main Landing Gear: The main wheel is designed to have 1 tyre per leg. Therefore the load on
the main landing gear is withstood by these two tyres equally.
Load taken by each leg = 5487.112 lb
From the “DUNLOP tyre manual” the tyre with 90 psi tyre pressure was chosen. The tyre
dimensions are:
1. Inflation pressure = 90 p.s.i
2. Diameter of the tyre (2R) = 27 in
3. Width of tyre (b) = 7.25 in
4. Radius at maximum deflection (r) = 9.35 in
The runway loading of the aircraft can be calculated by knowing the area of contact of the tyre
at the maximum deflection under the load.
The area of contact is approximated as ellipse whose area = a*b
Where a = (R2 – r
2)
0.5= (11.8
2–9.35
2)
0.5= 7.2 in
b = ½(width of tyre) =3.625 in
Hence the area of the contact = a*b
=
in2 = 0.5694 ft
2
The runway loading is found by dividing the aircraft‟s weight by the total area of contact of the
tyres.
Hence the runway loading is = 5.529 ton/ (0.5694*2) ft2
= 4.85 ton/ft2
This loading is acceptable as it is below 7 ton/ft2
Nose Landing Gear:
Load taken by nose landing gear = 1219.3 lb
Inflation pressure = 40 p.s.i
Diameter of tyre (D) = 19 in
Width of the tyre = 6.25 in
Radius at maximum deflection = 6.15 in
No of wheels = 1
LANDING GEAR POSITIONING:
The nose wheel is located at a distance of 0.75 m from the nose of the aircraft.
The track length i.e. the distance between the legs of the main landing gear is
taken as one third of the wing span i.e. 15.58*(1/3) = 5.2 m
PROPELLER DESIGN
The selected engine for the 15 seater aircraft is PT6A-135A turboprop engine. In the
turboprop engine more than 80% of the thrust is produced by the propeller. Thus an efficient
propeller is needed for the aircraft.
The choice of propeller depends on various parameters like the propeller with high
aerodynamic efficiency, sufficient thrust for cruise and high static thrust for the take off. Other
conditions are low noise level, ground clearance, etc.,
The propeller parameters here are calculated by estimating a non-dimensional quantity called
speed power coefficient. The propeller rotates at 2000 rpm.
The speed power coefficient Cs is calculated by the formula
Cs = {( V5)/Pn2}1/5
Where density kg/m3
P = shaft horse power = SHP * 746
n = rotation per second = RPM/60
V= velocity m/s
The Cs is calculated for both take off and cruising speed. Substituting the appropriate values the
Cs for both the cases is found to be
Cs takeoff = 1.0839
Cs cruise = 2.1227
From the propeller charts of Cs Vs J (advance ratio), the value of the J and blade angle β are
read on maximum efficiency line.
J take off = 0.6 βT.O = 15˚
J cruise = 1.25 βcruise = 30˚
J = V/n*D
Where n = rotation per second
D = diameter of the propeller m
V = velocity m/s
Since the values of the J, V & n are known from the above formula, the diameter of the propeller
is calculated to be 3m (for takeoff) & 2.76m (for cruising).
The propeller with diameter 3m is selected for the aircraft. The blade angle of the propeller
which is calculated at 75% of the radius is found to be 15˚ for takeoff and 30˚ for cruise.
Efficiency of the propeller at takeoff and cruising are found to be 80% and 85% respectively.
Propeller diameter = 3.0 m
CONDITION BLADE ANGLE β EFFICIENCY η
TAKE OFF 15˚ 80%
LANDING 30˚ 85%
CONFIGURATION LAYOUT
Based on the values calculated so far, an initial 3-view diagram of the aircraft is drawn.
Even though the data acquired so far can clearly define a certain type of aircraft, a large number
of size and shape could satisfy these data.
Thus by considering the merits and demerits of various layouts, a basic configuration is
decided.
Basic configuration:
1. Unswept low wing.
2. Conventional horizontal and vertical stabilizers.
3. Wing mounted turboprop engine
4. Landing gear with tricycle arrangement, main wheel retracting into the
fuselage.
5. The propeller is of tractor type.
C.G CALCULATIONS
The weight of an airplane changes during the flight due to consumption of fuel, also
the payload and the amount of fuel carried may vary from flight to flight. All these factors lead
to change in the location of the centre of gravity (C.G) of the airplane.
The shift in the C.G location affects the stability and controllability of the airplane.
Thus the C.G calculation is one of the crucial steps in the design process. For conceptual design
it is assumed that C.G position lies at 30% of mean aerodynamic chord of the wing.
The variation of C.G for different cases such as
1. C.G position with full payload and full fuel.
2. C.G position with full payload and 90% of fuel emptied.
3. C.G position with no payload and 90% of fuel emptied.
4. C.G position with no payload and full fuel.
are calculated.
It is acceptable if the variation is within 5% from the 30% of mean aerodynamic chord.
WEIGHT BREAKUP:
To calculate C.G it is mandatory to know the weight of the various components of the
aircraft. From the second weight estimation, the takeoff weight had been calculated. Hence the
component weights are assumed as some fraction of the takeoff weight. The weight breakage is
listed below
WT.O = 54,249 N Wstructure= 0.3WT.O= 16274.7 N
Wwing = 0.100 WT.O = 5,424.9 N
WFus structure = 0.100 WT.O = 5,424.9 N
WV.T = 0.020 WT.O = 1085 N
WH.T = 0.030 WT.O = 1628 N
WNose (U.C) = 0.010 WT.O = 542.49 N
WMain(U.C) = 0.040 WT.O = 2170 N
TOTAL = 0.300 WT.O = 16274.7 N
WFixed Equipment = 0.045 WT.O = 2441.205 N
C.G OF THE WEIGHTS IN FUSELAGE
The C.G of the weights in the fuselage is calculated as follows. The fuselage
houses the payload, crew, electronic equipments, horizontal tail, vertical tail and nose wheel. For
calculating C.G of the fuselage, weight of these components and the distances at which they are
placed from the nose must be taken into account.
C.G of the fuselage = ΣWX/ ΣW, where ΣW is the summation of the weights of
all the fuselage components, X is the distance of the components from the nose of the fuselage;
ΣWX is the summation of the moment created by the components about the nose of the aircraft.
The distance X is assigned for all the components.
SL.NO COMPONENTS WEIGHT
N
X
m
W*X
N-m
1. PILOT-2 nos 2000 1 2,000
2. PASSENGER-2 nos 2000 3.38 6,760
3. PASSENGER-2 nos 2000 4.26 8,520
4. PASSENGER-2 nos 2000 5.14 10,280
5. PASSENGER-2 nos 2000 6.02 12,040
6. PASSENGER-2 nos 2000 6.90 13,800
7. PASSENGER-2 nos 2000 7.78 15,560
8. PASSENGER-2 nos 2000 8.66 17,320
9. PASSENGER-2 nos 2000 9.54 19,080
10. HORIZONTAL TAIL 1628 11.00 17,908
11. VERTICAL TAIL 1058 10.25 10,844.5
12. NOSE WHEEL 542.5 0.75 406.875
13. FUSELAGE + FIXED
EQUIPMENT
7866.205 5.42 42,634.83
TOTAL 29094.705 177154.2061
XC.G OF FUSELAGE = ΣWX/ ΣW
= 177154.2061/29094.705 = 6.08 m. from the nose of the aircraft.
C.G OF WING
The wing houses the fuel, powerplant and main landing gear. The C.G of the
wing is also calculated in the same manner as that of the fuselage by considering the weight of
the wing components and their respective distances from the leading edge of the wing root chord.
The tabular column below is for half span of wing.
XC.G OF WING = ΣW*X/ΣW
= 12387.19/12561.5
= 0.986 m. from the nose of the leading edge of wing root chord.
The C.G of wing structure is taken at 35 % of the mean aerodynamic chord of the
wing.
The distance of wing from the nose of the aircraft (x) can be calculated by equating
the sum of moments of fuselage and wing about the nose to the total moment of the aircraft about
nose. The Xfinal is assumed to be at 35% of the mean aerodynamic chord of the wing.
(WFus. Xfuselage) + Wwing (X+ Xwing) = (X + XFinal)Wtotal
177154.2061 + 25123*(x + 0.986) = (x+ 0.9777)*54217.5
From above equation the value of x is found to be 5.118 m from the nose of the aircraft.
SL.NO COMPONENTS WEIGHT
N
X
m
W*X
N-m
1. Fuel 7236 0.9906 7167.98
2. Powerplant 1530 0.5910 904.23
3. Main landing gear 1085 1.5345 1664.9325
4. Wing structure 2710.5 0.9777 2650.0385
TOTAL 12561.5 12387.19
CASE 1: C.G WHEN 90% OF THE FUEL IS EMPTIED:
XC.G OF WING = ΣW*X/ΣW
= 5936.016/6049.1 = 0.9813 m from leading edge of wing root chord.
Finding C.G:
(WFus. Xfuselage) + Wwing (X + Xwing) = (X+ XFinal)Wtotal
177154.2061 + 12098.2*(5.118+0.9813) = 41192.905*(5.118 + XFinal)
Solving the above equation, the value of XFinal is found to be 0.9739 m from the leading
edge of the wing.
This shifts the airplane C.G to 34.7% of the aerodynamic chord.
SL.NO COMPONENTS WEIGHT
N
X
m
W*X
N-m
1. Fuel 723.6 0.9906 716.798
2. Powerplant 1530 0.5910 904.23
3. Main landing gear 1085 1.5345 1664.9325
4. Wing structure 2710.5 0.9777 2650.0385
TOTAL 6049.1 5936.016
CASE 2: FUSELAGE WITH NO PAYLOAD & 90% 0F THE FUEL
SL.NO COMPONENTS WEIGHT
N
X
m
W*X
N-m
1. Pilot-2 2000 1.00 2000
2. Horizontal tail 1628 11.00 17908
3. Vertical tail 1058 10.25 10844.5
4. Nose wheel 542.5 0.75 406.875
5. Fuselage + fixed
equipment.
7866.205 5.42 42634.83
Σ 13094.705 73794.205
Xfuselage = ΣW*X/ΣW
= 73794.205/13094.705
= 5.635 m from the leading edge of the wing root chord.
Finding the C.G :
(WFus. Xfuselage) + Wwing (X + Xwing) = (X + XFinal)Wtotal
73794.205 + 12098.2*(5.118+0.9813) = 25192.905*(5.118+Xfinal)
Solving the equation the value of Xfinal is found to be 0.74018 m, i.e. the C.G is found to be
at 21.2% of the mean aerodynamic chord.
CASE 3: FUSELAGE WITH NO PAYLOAD & FULL FUEL
WFus = 13094.705 N Xfuselage = 5.635 m
Wwing = 25123 N Xwing = 0.
(WFus. Xfuselage) + Wwing (X+ Xwing) = (X + XFinal)Wtotal
73794.205+25123*(5.118+0.986) = 38217.705*(5.118+ XFinal)
Solving the above equation, the XFinal is found to be 0.82549 m, thus the C.G is at 26.2% of the
mean aerodynamic chord of the wing.
The C.G variation for the above cases is listed below.
SL.NO CONFIGURATIONS C.G. % of mean chord
1. Airplane when fully loaded 35
2. Airplane with only 10% fuel 34.7
3. Airplane with 10% fuel & no payload 21.2
4. Airplane with full fuel &no payload 36.2
This C.G range is acceptable.
DRAG ESTIMATION
The drag is estimated by proper area method. This method evaluates the drag of
the aircraft by considering the appropriate areas & adding the drag of each component. Drag
coefficients for various components like fuselage, powerplant, etc. were obtained from the book
“ FLUID-DYNAMIC DRAG” by S.F.HOERNER.
The total drag of an airplane is sum of wing drag, parasite drag and induced drag.
Wing drag is the drag produced by the wing. Dwing= ½( V2SCD0W)
Induced drag is the drag due to lift. CDi= CL2/ eAR
Parasite drag is the drag of all the non-lifting surfaces (including horizontal &
vertical tail)
Sum of Parasite drag and wing drag is called parasite drag of airplane.
Parasite drag = drag of wing + drag of fuselage + drag of horizontal tail + drag of
vertical tail +drag of powerplant +drag of propeller + drag of
nose, main wheel + drag of flaps.
CD0 a/p = 1/2 V2S (CD0W + CD0F SF/S + CD0HTSHT/S + CD0VT SVT/S + …….)
Where S is the wing planform area, SF is the frontal area of fuselage, SHT
horizontal tail area, SVT is the vertical tail area etc.
The areas of various components of the aircraft are calculate as follows
Fuselage Area = d2
max / 4
= 2/ 4 = 5.72 m
2
Powerplant Frontal Area = d2max / 4 x No of Engines
= x 2 = 0.366 m2
Main Landing Gear Area = (Diameter x Width) x No of Tyres
= (0.6858*0.18415)*2 = 0.25258 m2
Nose Wheel Area = (Diameter x Width) x No of Tyres
= (0.4826*0.15875)*1 = 0.0766 m2
Flap Area:
To calculate the area of the flap, first the length of the flap is calculated by assuming
aileron area to be 5% of the total wing area. Thus the length of the flap is total wing span
subtracted by fuselage diameter and the length of the aileron.
i.e. LFlaps = Wing Span(b) – Laileron –Fuselage Diameter(dmax)
= 15.58 – 0.779 – 2.6 = 12.201 m
The rear spar is located at the 65% of the chord. Thus the remaining 35% of the mean
aerodynamic chord would be the flap width
bFlaps = 0.35*1.732 = 0.6062 m
Therefore the area of the flap is = LFlaps * bFlaps = 7.72 m2
PROPELLER SWEPT AREA:
The area swept by the propeller is D2/ 4. Where D is the diameter of the propeller.
D2/ 4 = 3
2/ 4 = 7.07 m
2
Horizontal Tail Area & vertical tail area:
The primary purpose of the tail is to counter the moments produced by the wing. Thus the
relationship between the size of the tail and wing can be used to calculate the tail area.
The relationship is obtained from the equilibrium equation for the longitudinal stability.
From the equation, the horizontal tail volume ratio CHT=SHTLHT/S*
The vertical tail volume ratio CVT=SVTLVT/S*b
From the book “AIRCRAFT DESIGN A CONCEPTUAL APPROACH” by
DANIEL.P.RAYMER, it is learnt that for the twin turboprop engine the tail volume ratio
CHT =0.48 & CVT = 0.05.
The vertical tail area and horizontal tail area are calculated to be 15% and 17% of the
wing area respectively. I.e. vertical tail area is 4.05 m2 and horizontal tail area is 4.79 m
2.
“NACA 0009” is chosen as tail aerofoil.
The aspect ratio of horizontal tail & vertical tail are chosen as 4 and 1.5 respectively.
The parasite drag for various cases like cruise, takeoff & landing are found as follow
FOR TAKEOFF:
S.NO COMPONENT CD
S m2
CD S m2
1 FUSELAGE 0.03 5.72 0.1716
2 POWER PLANT 0.03 2*0.18 0.0108
3 PROPELLER 0.015 2*7.07 0.2121
4 HORIZONTALTAIL 0.007 4.59 0.03213
5 VERTICAL TAIL 0.0065 4.05 0.02633
6 NOSE WHEEL 0.12 0.0766 0.009192
7 MAIN WHEEL 0.12 2*.12629 0.03096
8 FLAP(3/4) 0.03 2*3.86 0.231747
TOTAL 0.711908
(CD0)OTHERS = Σ CD S S = 0.026367
5% of the (CD0)OTHERS is taken as interference drag.
Therefore the (CD0)OTHERS for takeoff condition is 0.02769
FOR CRUISE:
S.NO COMPONENT CD
S m2 CD S m
2
1 FUSELAGE 0.03 5.72 0.1716
2 POWER PLANT 0.03 2*0.18 0.0108
3 PROPELLER 0.015 2*7.07 0.2121
4 HORIZONTAL TAIL 0.007 4.59 0.03213
5 VERTICAL TAIL 0.0065 4.05 0.02633
TOTAL 0.44066
(CD0)OTHERS = Σ CD S S = 0.44066/27 = 0.01632
5% of the (CD0) OTHERS is taken as the interference drag.
Therefore the (CD0) OTHERS for cruise condition is 0.01714.
FOR LANDING:
S.NO
COMPONENT
CD
S m2 CD S m
2
1 FUSELAGE 0.03 5.72 0.1716
2 POWER PLANT 0.03 2*0.18 0.0108
3 PROPELLER 0.015 2*7.07 0.2121
4 HORIZONTAL TAIL 0.007 4.59 0.03213
5 VERTICAL TAIL 0.0065 4.05 0.02633
6 NOSE WHEEL 0.12 0.0766 0.009192
7 MAIN WHEEL 0.12 2*12629 0.0303096
8 FLAP(600) 0.04 2*3.86 0.3088
TOTAL 0.78896
(CD0)OTHERS = Σ CD S S = 0.78896/27 = 0.029221
5% of the (CD0) OTHERS is taken as the interference drag.
Therefore the (CD0) OTHERS for landing condition is 0.03068.
DRAG POLAR ESTIMATION
Drag polar is the graph between the CL Vs CDt . where CDt is the total drag
coefficients of the aircraft which is the sum of the parasite drag of airplane and induced drag
coefficients. As stated earlier Induced drag is the drag due to lift. CDi= CL2/ eAR or CDi = KCL
2.
The „e‟ is known Ostwald‟s efficiency factor and it is taken as 0.85.
FOR TAKEOFF:
CL CD0W CD0 others K CL2
CDt = KCL2 + CD0
0 0.0060 0.02769 0.000 0.03369
0.3 0.0045 0.02769 0.00381 0.036
0.6 0.0055 0.02769 0.0149792 0.04817
0.9 0.0082 0.02769 0.0337032 0.0696
1.2 0.0100 0.02769 0.05991696 0.0976
1.5 0.014 0.02769 0.09362025 0.1353
1.8 0.020 0.02769 0.13481316 0.1825
FOR CRUISE:
CL CD0W CD0 others KCL2 CDt = KCL
2 + CD0
0 0.0060 0.01714 0.000 0.02314
0.3 0.0045 0.01714 0.00381 0.0254
0.6 0.0055 0.01714 0.0149792 0.0376
0.9 0.0082 0.01714 0.0337032 0.0590
1.2 0.0100 0.01714 0.05991696 0.0871
1.5 0.014 0.01714 0.09362025 0.1248
1.8 0.020 0.01714 0.13481316 0.17195
FOR LANDING:
CL CD0W CD0 others KCL2 CDt = KCL
2 + CD0
0.0 0.0060 0.03068 0.00000 0.03668
0.3 0.0045 0.03068 0.00374481 0.0389
0.6 0.0055 0.03068 0.01497924 0.05116
0.9 0.0082 0.03068 0.03370329 0.0726
1.2 0.0100 0.03068 0.05991696 0.1006
1.5 0.014 0.03068 0.09362025 0.1383
1.8 0.020 0.03068 0..13481316 0.1855
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0 0.05 0.1 0.15 0.2
Take off
Cruise
Landing
DRAG POLAR
CDt
CL
PERFORMANCE CALCULATIONS
The performance analysis is important to assess the capabilities of an airplane.
This analysis would give the thrust or power required, maximum speed of the airplane, etc. to
achieve a desired performance. Performance analysis also points out as to what new
developments are possible in airplane aerodynamics or engine selected.
The performance parameters calculated are,
THPreqd = D*V/746 hp
Where D = (CDt/CL)*WT.O N , CDt = CD0 + KCL2.
THPavailable = SHP*efficiency of propeller at cruise
Rate of climb can be calculated as follows
R/C = (THPavailable- THPreqd)*746*60/WT.O m/min
THPavailable & THPreqd Vs velocity graphs are plotted to calculate the maximum speed of the
aircraft at various altitudes.
(R/C) max Vs altitude plot gives the absolute and service ceilings of the aircraft.
The area under the curve of 1/ ((R/C)max) Vs altitude gives the time to climb to the cruising
altitude.
Performance parameters for various altitudes are calculated below,
At sea level : kg/m3
S.
NO
V
m/s
CL CD0W CD0 other CDt D
N
THPREQD THPAVA R/C
m/min
1 42.57 1.8 0.02 0.01714 0.1720 5.184 295.82 1275 807.82
2 50 1.3047 0.0123 0.01714 0.1003 4.17 279.49 1275 821.05
3 60 0.9064 0.0083 0.01714 0.0596 3.57 287.13 1275 814.99
4 70 0.6661 0.0058 0.01714 0.0414 3.37 316.22 1275 790.99
5 80 0.5101 0.005 0.01714 0.0330 3.51 376.41 1275 741.34
6 90 0.4028 0.0045 0.01714 0.0284 3.82 460.86 1275 671.66
7 100 0.3267 0.0044 0.01714 0.0260 4.317 578.69 1275 574.45
8 110 0.2698 0.0053 0.01714 0.0255 5.127 755.99 1275 519.01
9 115 0.2467 0.0053 0.01714 0.0250 5.497 847.39 1275 352.78
10 120 0.2265 0.0053 0.01714 0.0245 5.867 943.75 1275 273.28
At 1 km altitude : ρ = 1.111kg/m3
S.
NO
V
m/s
CL CD0W CD0 other CDt D
N
THPREQD THPAVA R/C
m/min
1 44.72 1.8 0.02 0.01714 0.1720 5.18 310.52 1132.86 678.43
2 50 1.44 0.0128 0.01714 0.1162 4.38 293.56 1132.86 692.42
3 60 1.00 0.0088 0.01714 0.0676 3.67 295.17 1132.86 691.09
4 70 0.73 0.0064 0.01714 0.0457 3.40 319.03 1132.86 671.41
5 80 0.56 0.005 0.01714 0.0352 3.41 365.68 1132.86 632.92
6 90 0.44 0.005 0.01714 0.0301 3.71 447.59 1132.86 565.34
7 100 0.36 0.005 0.01714 0.0261 3.93 526.81 1132.86 499.99
8 110 0.30 0.005 0.01714 0.0254 4.59 676.81 1132.86 456.05
9 115 0.27 0.0051 0.01714 0.025 5.02 773.86 1132.86 296.175
10 120 0.25 0.0051 0.01714 0.0245 5.316 855.12 1132.86 229.135
At 2 Km altitude: kg/m3
S.
NO
V
m/s
CL CD0W CD0 other CDt D
N
THPREQD THPAVA R/C
m/min
1 46.9 1.8 0.02 0.01714 0.17195 5.18 326.2 1006.35 561.12
2 50 1.59 0.0145 0.01714 0.13683 4.66 312.3 1006.35 572.59
3 60 1.104 0.0086 0.01714 0.07645 3.75 301.6 1006.35 581.41
4 70 0.811 0.0072 0.01714 0.05170 3.45 323.7 1006.35 563.18
5 80 0.621 0.0055 0.01714 0.03868 3.38 362.3 1006.35 531.34
6 90 0.490 0.005 0.01714 0.03203 3.55 427.8 1006.35 477.30
7 100 0.397 0.005 0.01714 0.02859 3.91 523.7 1006.35 398.18
8 110 0.328 0.005 0.01714 0.02651 4.38 645.8 1006.35 294.97
9 115 0.30 0.005 0.01714 0.02579 4.65 717.5 1006.35 238.30
10 120 0.276 0.0052 0.01714 0.02511 4.74 762.46 1006.35 201.20
At 3Km altitude: kg/m3
S.
NO
V
m/s
CL CD0W CD0 other CDt D
N
THPREQD THPAVA R/C
m/min
1 49.4 1.8 0.02 0.01714 0.17195 5.18 343.22 890.78 451.73
2 60 1.22 0.010 0.01714 0.08907 3.96 318.49 890.78 472.14
3 70 0.898 0.008 0.01714 0.05869 3.55 332.64 890.78 460.46
4 80 0.687 0.0057 0.01714 0.04247 3.35 359.57 890.78 438.24
5 90 0.543 0.0053 0.01714 0.03470 3.47 418.27 890.78 389.82
6 100 0.440 0.005 0.01714 0.03018 3.72 498.92 890.78 323.28
7 110 0.364 0.005 0.01714 0.02763 4.12 607.9 890.78 233.38
8 115 0.3326 0.005 0.01714 0.02674 4.36 672.27 890.78 180.27
9 120 0.3054 0.005 0.01714 0.02602 4.622 743.48 890.78 121.52
At 4Km altitude: kg/m3
S.
NO
V m/s
CL CD0W CD0 other CDt D N
THPREQD THPAVA R/C m/min
1 52.08 1.8 0.02 0.01714 0.17195 5.18 361.62 786.05 350.15
2 60 1.356 0.012 0.01714 0.10564 4.23 339.89 786.05 368.08
3 70 0.996 0.0088 0.01714 0.06721 3.67 344.08 786.05 364.62
4 80 0.763 0.0065 0.01714 0.04785 3.40 364.93 786.05 347.42
5 90 0.63 0.0056 0.01714 0.03775 3.39 408.98 786.05 311.08
6 100 0.488 0.005 0.01714 0.03206 3.56 477.34 786.05 254.68
7 110 0.403 0.005 0.01714 0.02891 3.89 573.15 786.05 175.64
8 115 0.369 0.005 0.01714 0.02782 4.09 630.18 786.05 128.59
9 120 0.339 0.005 0.01714 0.02692 4.307 692.81 786.05 76.923
0
500
1000
1500
2000
2500
3000
3500
0 50 100 150 200
THP REQIURED AT S/L
THP AVAILABLE AT S/L
THP REQUIRED AT 1km
THP AVAILABLE AT 1 km
THP Vs VELOCITY ( S/L & 1 km)
Velocity (m/s)
THP
0
500
1000
1500
2000
2500
3000
0 50 100 150 200
THP REQUIRED AT 2km
THP AVAILABLE AT 2km
THP REQUIRED AT 3 km
THP AVAILABLE AT 3 km
THP Vs VELOCITY ( 2km & 3km)
Velocity (m/s)
THP
0
500
1000
1500
2000
2500
0 50 100 150 200
THP REQUIRED AT 4 km
THP AVAILABLE AT 4km
THP Vs VELOCITY ( 4km)
Velocity (m/s)
THP
0100200300400500600700800900
0 50 100 150
AT SEA LEVEL
AT 1 Km
RATE OF CLIMB Vs VELOCITY (S/L & 1km)
Velocity (m/s)
R/C
(m/min)
0
100
200
300
400
500
600
700
0 50 100 150
AT 2 Km
AT 3 km
RATE OF CLIMB Vs VELOCITY ( 2km & 3km)
Velocity (m/s)
R/C
(m/min)
0
50
100
150
200
250
300
350
400
0 50 100 150
RATE OF CLIMB VS VELOCITY (4km)
Velocity (m/s)
R/C
(m/min)
SERVICE CEILING:
Service ceiling is that altitude at which the maximum rate of climb is 30m/min. The
service ceiling can be calculated from the graph of maximum rate of climb Vs altitude.
The service ceiling of the aircraft is found to be 6.9 km.
The absolute ceiling of the aircraft is found to be 7.2 km.
0
100
200
300
400
500
600
700
800
900
0 2 4 6 8Altitude (km)
max.R/C(m/min)
MAXIMUM RATE OF CLIMB Vs ALTITUDE
SERVICE CEILING = 6.9 Km
Time to climb:
Time to climb the cruising altitude is found out by calculating the area under the curve of
the graph 1/(R/C)MAX Vs altitude.
The area under the curve of the above graph upto 4km gives the time to climb to the
cruising altitude as 5.83 minutes.
0
0.5
1
1.5
2
2.5
3
3.5
4
0 1 2 3 4 5 6Altitude (km)
1/(R/C)max
x 10-3
(min/m)
1/(R/C)max Vs Altitude
Time to climb = 5.83 min
RANGE AND ENDURANCE
The total distance (measured with respect to ground) traversed by an airplane on one
load of the fuel is defined as the range R of the aircraft.
The amount of time that an airplane can stay in the air on one load of the fuel is
defined as the endurance.
The range & endurance of the aircraft can be obtained from the Breguet equation,
which assumes that at flight the velocity, lift coefficient CL, & lift to drag ratio are constant.
The specific fuel consumption (C) of a propeller driven aircraft is defined as the
weight of the fuel consumed per unit time per unit power.
C = (dW/dt)/(η*SHP)
Therefore integrating the equation dt = dW/ (C* η*SHP) the endurance can be obtained as
E = (CL/CD)*( η*746/(V*C))*ln (Wi/Wf) Hrs
Where CL is cruise lift coefficient & CD is the total drag at cruise speed V
dW/dt is the rate of change of fuel weight. The efficiency of the propeller is given by η p.
Wi is the initial weight, Wf is the final weight after all the fuel has been consumed.
E = (0.369/0.02782)*(0.85*3600*746/0.28*9.81*115)*ln (54249/39777)
= 8.27 hrs
The range R = V * E km
= 8.27*115*3600/1000
= 3422.87 km
STABILITY ANALYSIS
If an airplane is to remain in steady uniform flight, the resultant forces as
well as the resultant moments about the center of gravity must both be equal to zero. An airplane
satisfying this requirement is said to be in a state of equilibrium or flying at a trim condition.
The aircraft stability is generally divided into static and dynamic stability.
The aircraft is said to be statically stable, if it shows a tendency to return to its equilibrium
position on its own when disturbed. It doesn‟t matter how much time it takes. The aircraft is said
to possess dynamic stability, if it returns to equilibrium position within a finite period of time
after a disturbance.
The aircraft possessing the static stability need not possess dynamic
stability. But the aircraft possessing dynamic stability must possess static stability. For the
stability analysis, equilibrium equations must be written and solved for the unknowns. The
airplane has 6 degrees of freedom with control surfaces locked and 9 degrees of freedom with
control surfaces free to move. These 9 degrees of freedom are divided into,
Longitudinal Degrees Of Freedom:
Translations along X and Z axis, rotation about Y axis and deflection of the
elevator about its hinge line.
Lateral Degrees Of Freedom:
Translations along Y and Z axis, rotation about X axis and deflection of the aileron
about its hinge line.
Directional Degrees Of Freedom:
Translations along X and Y axis, rotation about Z axis and deflection of the rudder
about its hinge line.
For each degree of freedom one equilibrium equation has to be written, thus forming
four equations and 4 unknowns in longitudinal, lateral and directional mode respectively.
For the ease of initial analysis, it is assumed that the controls are locked (control stick is
fixed) in each mode, reducing the equilibrium equation to 3 equations and 3 unknowns.
If the stability analysis of the airplane is carried out with the control sticks locked, it is
called stick fixed stability.
If the analysis includes the effect of freeing the controls it is known as stick free stability
analysis.
STATIC LONGITUDINAL STABILITY
The study of longitudinal equilibrium and static stability of the airplane requires an
investigation into moments about the airplane‟s Y axis through the C.G and their variation with
the airplane‟s lift co-efficient.
Equilibrium demands that the summation of these moments equal zero and the static
stability demands that a diving moment accompany an increase in lift coefficient and a stalling
moment accompany a decrease in lift coefficient from the equilibrium.
It is assumed that wing and tail surfaces can be represented by a mean aerodynamic
chord, the forces and moments on which represent all the forces and moments operating on the
surface. It is also assumed that there exists an aerodynamic center on this mean aerodynamic
chord about which the wing pitching moment co-efficient is invariant with lift coefficient.
Considering the contribution of wing, tail and fuselage, a moment equation in a non-
dimensionalised form can be written as
Cmc.g = CL*–
Cmac+ Cm fus,nac at* t* * (αw iw + it) ……… (1)
Where,
Cmc.g = moment co-efficient about C.G of the airplane.
CL = lift co-efficient at cruise. = 0.37
Xc.g = C.G location with respect to mean aerodynamic chord. = 0.35
Xa.c = aerodynamic center location with respect to the aerodynamic chord.
= 0.270C for NACA 631412 [from “THEORY OF WING SECTIONS”]
Cmac = moment coefficient about aerodynamic centre.
= 0.075 for NACA 631412 [from “THEORY OF WING SECTIONS”]
Cm fus,nac = = 0.035*0.37 = 0.01295
at = horizontal tail lift slope
From “THEORY OF WING SECTIONS” at = 0.1/deg for NACA 0009, also AR = 4. Thus
at for aspect ratio corrected can be calculated from the following equation,
at = 0.0686/deg
t = tail efficiency = 0.9
= = tail volume ratio = 0.481
Where St is tail area, lt is the distance between the C.G of tail to C.G of airplane. S is wing
area and is the mean aerodynamic chord
αw = absolute angle of attack.
= downwash angle = (d /dα)* αw
iw = wing incidence angle = 1.88˚
it = tail incidence angle.
-0.5
0
0.5
1
1.5
2
-10 -5 0 5 10 15 20 25
CL VS ALPHA
CL VS ALPHA FOR AR CORRECTED
CL VS α FOR WING AEROFOIL NACA 631412
α (deg)
CL
FOR AR = INFINITE
FOR AR = 9
TAIL INCIDENCE ANGLE :
At level flight, the aircraft is trimmed and hence Cmcg = 0.
Therefore the equation (1) becomes
CL*–
Cmac+ Cm fus,nac at* t*( )*( )* (αw iw + it) = 0
αw = 4.88˚ , (d /dα) = 0.4, = * αw =0.4*4.88 =1.952˚ , iw = 1.88˚
Therefore
0.47*(0.008) – 0.075 + 0.01295 – 0.0686*0.9*0.481*(4.88-1.88-1.952- it) = 0
it = 2.215˚
STICK FIXED STATIC LONGITUDINAL STABILITY:
By differentiating the equation (1), the equation for stick fixed static stability can be obtained
as
c.g = –
fus – *( )*( )* t* ...........(2)
= 0.08 + 0.035 –
= 0.09967 s
The negative sign of c.g indicates that the aircraft is stable in stick fixed condition.
STICK FIXED NEUTRAL POINT:
The neutral point N0 is the point, where the aircraft is neutrally stable. i.e. c.g 0
= = fus *( )*( )* t* ………..(3)
0.27 – 0.035 + 0.2146
0.4496.
The neutral point gives the most aft location at which the C.G can be placed before making
the airplane unstable. Therefore the airplane‟s permissible C.G travel is limited to the point
0.4496 . If a stable airplane is desired, the airplane should never be balanced aft of this point.
Once the neutral point is known, the stability at any other C.G position can be obtained.
c.g - N0
STICK FREE STATIC STABILITY:
The effect of freeing the elevator is also considered in the analysis. Thus the equation
c.g changes to,
c.g –
+ fus - *( )*( )* t* * ……(4)
Where,
floating tendency , restoring tendency
= elevator effectiveness factor.
From the book “AIRPLANE PERFORMANCE STABILITY AND CONTROL” by
Perkins and Hage, the values of , are taken as 0.008 and – 0.012 respectively. is
taken as 0.5. Therefore
c.g 0.08 + 0.035 –
0.08 + 0.035 0.1431 0.0281
STICK FREE NEUTRAL POINT:
= = fus *( )*( )* t* * ……(5)
= 0.270 – 0.035 + 0.1431 = 0. 3781
ELEVATOR POWER :
Deflecting the elevator effectively changes the angle of attack of the whole horizontal tail,
thereby changing its lift and producing a control moment about the airplane‟s centre of gravity.
The magnitude of the moment coefficient obtained per degree deflection of the elevator is
termed the elevator power and is analytically written as
Cmδ = at*( )*( )* ήt* ……………..(6)
= 0.0686*0.481*0.9*0.5
= 0.01485.
-0.1
-0.08
-0.06
-0.04
-0.02
5E-17
0.02
0.04
0.06
0.08
0.1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
AT C.G 21.2%
AT C.G 26.2%
AT C.G 34.7%
AT C.G 35%
C.G AT 45%
Cm Vs CL FOR VARIOUS C.G
CL
Cm
ELEVATOR ANGLE VERSUS EQUILIBRIUM LIFT COEFFICIENT :
The equilibrium equation for the stick free condition is
Cmc.g = CL*–
Cmac+ Cm fus,nac at* t*( )*( )* (αw iw + it +τδe)………(7)
The control of the equilibrium lift co-efficient is affected through the influence of
the term τδe.
A change in the elevator deflection and lift co-efficient will not change the slope of
the pitching moment curve , for η is independent of the lift co-efficient and the term
vanishes when the derivative is taken with respect to CL.
For every CL, the elevator deflection required to trim the airplane is found by
equating the equation (7) to zero and substituting various CL. αw , , Cm fus,nac are the functions of
CL. Hence both αw and changes with change in CL.
S.NO CL aw δe
1. 0.10 1.203 0.4812 +2.1750
2. 0.30 3.607 1.4408 +0.8468
3. 0.37 4.880 1.8800 0.0000
4. 0.50 6.013 2.4052 -0.4760
5. 0.70 8.415 3.3660 -1.7970
6. 1.00 11.020 4.4080 -2.5804
7. 1.30 15.626 6.2504 -5.7689
8. 1.50 18.030 7.2120 -7.0930
9. 1.80 22.636 9.0544 -10.280
The elevator deflection required at most forward C.G position at maximum CL :
Most forward C.G position = 0.212
CLmax = 1.8
δe =
––
....(8)
Where = (αw iw + it).
δe = (0.212-0.270)*1.8 – 0.075 + 0.035*1.8 – 0.02943*(22.63-9.05-1.88-2.215)/{0.02943*0.5)
= - 27˚
`
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0 0.5 1 1.5 2
Cm
CL
Cm vsCL at C.G 35%
ELEVATOR DEFLECTION = +2.5 deg
ELEVATOR DEFLECTION = +0.85 deg
ELEVATOR DEFLECTION = 0 deg
ELEVATOR DEFLECTION = - 1.8 deg
ELEVATOR DEFLECTION = -5.8 deg
ELEVATOR DEFLECTION = - 10.3 deg
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
0 0.5 1 1.5 2
AT C.G 35 %
AT NEUTRAL POINT
de Vs CL
de
CL
+
-
ESTIMATION OF THE AIRPLANE DIHEDRAL EFFECT
Control over the angle of bank is necessary to provide a force to accelerate
the flight path in the horizontal plane. With simple rudder control the airplane can be made to
sideslip, thereby creating a cross wind or side force that can accelerate the flight path in the
horizontal plane. However this force is small for modern aircraft and totally inadequate for the
rate of turn required.
The problem of holding the wings level or of maintaining some angle of
bank is one of the controls over the rolling moments about the airplane‟s longitudinal axis. The
phenomenon of rolling moment due to sideslip is termed as dihedral effect.
An airplane is said to have stable dihedral effect if a negative rolling
moment is created as the result of positive sideslip β. The rolling moment due to sideslip is
mainly created by wing dihedral angle , which is positive for tip chord above the root chord. In
a sideslip, the angle of attack of the forward wing will be higher than the angle of attack of the
trailing wing, thereby creating a rolling moment about the axis.
The desired dihedral effect (Clβ) of the airplane is 0.0014.The dihedral
effect of the complete airplane may be given as,
(Clβ)airplane = (Clβ)wing + (Clβ)vertical tail + ( Clβ)1 + ( Clβ)2..........(9)
( Clβ)1 = dihedral effect due to wing- fuselage interference.
( Clβ)2 = dihedral effect due to wing-vertical tail interference.
From the book “AIRPLANE PERFORMANCE STABILITY AND CONTROL” by Perkins and
Hage, the value of ( Clβ)1 and ( Clβ)2 for a low wing is found to be -0.0008 and 0.00016.
( Clβ)1 = -0.0008
( Clβ)2 = 0.00016
(Clβ)vertical tail = av*(sv/s)*(Zv/b)*ηt = 0.045*0.15*(3.73/15.58)*0.9
= 0.001454
Where, Zv is the vertical distance from C.G of wing to the C.G of vertical tail.
Therefore, (Clβ)wing = 0.0014 +0.0008-0.00016-0.001454 = 0.000586.
But (Clβ)wing = 0.0002 wing = 0.000586 wing = 2.93 3
Thus the dihedral angle of the wing is 3 .
ONE ENGINE INOPERATIVE CONDITION (OEI)
In an airplane with more than one engine, if one engine is not functioning, the
other working engine thrust will produce yaw which is the product of the engine thrust and
distance of the engine from airplane centreline.
The problem is severe for wing mounted engine arrangements than in airplane
with rear mounted engines. Also the outermost engine failure will be critical as the moment arm
will be the largest & more rudder power is required to produce yaw. Again the problem is more
in low forward speed as the large rudder area is required to overcome this yaw.
T.H.P = F*V/746 = S.H.P*ηP
F = S.H.P*ηP *746/V
The moment arm produced is . Where, le is the distance of engine from the
airplane centreline.
Thus the yawing moment coefficient Cn due to OEI = =
S.H.P = S.H.Palt = 461.0535
ηP = efficiency of the propeller at cruise = 0.85
le = 3.3m
Cn due to OEI = 5598.38/V3
S.NO V m/s Cn due to OEI
1. 40 0.0874
2. 50 0.0447
3. 60 0.0259
4. 70 0.0163
5. 80 0.0109
6. 90 0.0076
7. 115 0.0036
Cn DUE TO RUDDER DEFLECTION
Cn = *δr
= av*η*(Sv/S)*(lv/b)* = 0.045*0.15*(4.1543/15.58)*0.9*0.5
= 0.00081
The maximum rudder deflection is 30˚.
Therefore, the Cn at full rudder deflection = *30˚ = 0.00081*30
= 0.0243
The critical velocity at maximum rudder deflection (Cn = 0.0243) is 61.30 m/s.
The graph indicates that below the critical speed , the engine induced yaw is more than the
rudder induced yaw (due to full rudder). Thus during OEI condition, the aircraft must never be
flown below this critical speed.
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0 25 50 75 100 125
Cn
V(m/s)
Cn Vs V (OEI)
Engine Induced Yaw (OEI)
Yaw due to full Rudder
Critical speed = 61.3 m/s
DYNAMIC STABILITY ANALYSIS
Longitudinal stick fixed dynamics:
The equations of equilibrium governing the longitudinal stick fixed dynamics is,
(CD + d)*u + ½*(CDa – CL)*α + CLθ/2 = 0--------------(10)
CL*u + (1/2* CLα + d)* α – dθ = 0................. (11)
(Cmα + Cmdα*d) *α + (Cmdψ*d – hd2) θ = 0-------------- (12)
These equations can be written in matrix form as shown below,
–
– ....................(13)
These equations are first order linear differential equations. The solutions for these equations are
of the form
X(t) = X0 , where X0 is the Eigen vector , is the eigen value , is the
airplane time parameter, seconds. is the dimensionless time.
These equations are homogeneous algebraic equations. Hence there is a non zero
solution for the equation only when the determinant of the equation vanishes.
–
– = 0 ................ (14)
By expanding this determinant a quartic equation in is obtained. The general form of the
equation is
A4 + B
3 + C
2 + D + E = 0 ..................(15)
The solution for this quartic, for a statically stable airplane, in almost all cases combines into
two complex pairs,
1,2 = 1 1
3,4 = 2 2
The corresponding eigen vector could be
a1 + a2
a3 + a4
Substituting the value of & the solution can be expanded as
a1 + a2
The above equation can be rearranged as (A1cos + A2sin ).
This indicates that airplane‟s longitudinal motion, when a disturbance with elevator locked,
have two oscillatory modes.
The period and damping of these modes can be obtained as follows,
Period = seconds
Time to damp to ½ amplitude = seconds = seconds
Logarithmic decrement =
Where is the damping ratio equal to . is the „ undamped‟ circular frequency
.
PHUGOID ANALYSIS
The characteristic modes of the stick fixed longitudinal motion for nearly all the
airplanes are two oscillations one of long period poor damping(PHUGOID) and the other of
short period with heavy damping.
Under the assumption of no change in angle of attack, no damping, no inertia, the
equations (10), (11), (12) reduce to,
(CD + d)*u + CLθ/2 = 0-----------(16)
CL*u – dθ = 0------------(17)
The determinant of the coefficient of the above equation must be zero as the
equations are homogeneous.
0 -------------(18)
Therefore expanding the determinant of the algebraic equation in gives
2 + CD + = 0 ------------(19)
The root of the quadratic equation is given by,
Where, CL& CD are the lift coefficient and total drag coefficient at cruising speed.
CL = 0.37
CD = 0.02782
Therefore
The airplane time parameter seconds
= = 2.1737 seconds.
The period and time to damp for the PHUGOID mode are,
Period = seconds = = 52.32 seconds.
Time to damp to ½ amplitude = seconds = = 108.3 seconds.
„Undamped‟ circular frequency = = 0.12 /s
Damping Ratio = = = 0.0532.
Logarithmic decrement = = 0.03
The longitudinal stick fixed dynamic behaviour of the aircraft is of second order system
and hence can be described in terms of two parameters namely and .
The transfer function of the second order system is given by
=
For phugoid mode, and is found to be 0.0053 & 0.12 respectively. Therefore the
transfer function becomes,
= =
For the sake of analysis, an unit impulse deflection on the elevator is assumed. This causes
pitch angle to increase causing the aircraft to go upward. This leads to decrease in velocity,
because of which the lift is reduced.
Slowly pitch angle will decrease again causing the aircraft to go downward leading to
increase in velocity. This in turn increases the lift.
The pitch angle will increase again. The whole process repeats itself until the motion is
damped out.
The unit impulse response curve for this phugoid mode can be obtained from the transfer
function.
=
The response curve (amplitude Vs time) was obtained using the „MATLAB‟ software.
In the „MATLAB‟, the transfer function of the system is represented as two arrays each
containing the co-efficient of the polynomials in decreasing powers of s as follows
num = [ 0 0 0.0144]
den = [ 0 0.01272 0.0144].
The command “impulse (num, den)” plots the unit –impulse response curve.
The matlab program for amplitude Vs time is given below,
% unit impulse response curve for the phugoid
num = [ 0 0 0.0144];
den = [ 0 0.001272 0.0144];
impulse(num , den);
grid
title (‘ impulse response( phugoid)’)
0 100 200 300 400 500 600 700 800 900-0.1
-0.05
0
0.05
0.1
0.15
Impulse Response (Phugoid)
Time (sec)
Am
plit
ude
-0.1
-0.05
0
0.05
0.1
0.15
0 50 100 150
amp
litu
de
time (sec)
Amplitude Vs Time (PHUGOID)
period=52.5 s
t1/2 = 108 s
1/2 amplitude
period = 52.5 s
t1/2 = 108 s
V-n DIAGRAM
The control of weight in aircraft design is of extreme importance. Increase in weight
requires stronger structures to support them, which in turn lead to further increase in weight & so
on. Excess of structural weight means lesser amounts of payload, affecting the economic
viability of the aircraft.
Therefore there is a need to reduce aircraft‟s weight to the minimum compatible with
safety. Thus to ensure general minimum standards of strength & safety, airworthiness regulations
lay down several factors which the primary structures of the aircraft must satisfy.
These are
1. LIMIT LOAD: the maximum load that the aircraft is expected to experience in normal
operation.
2. PROOF LOAD: product of the limit load and proof factor(1.0-1.25)
3. ULTIMATE LOAD : product of limit load and ultimate factor(1.0-1.5)
The aircraft‟s structure must withstand the proof load without detrimental distortion & should
not fail until the ultimate load has been reached..
The manoeuvrability of the aircraft is also dictated by the loads falling on the structure during
the manoeuvres.
Both the aerodynamic and structural limitations for a given airplane are illustrated in the V-n
diagram, a plot of load factor versus flight velocity.
A V-n diagram is a type of flight envelope for the aircraft establishing the manoeuvre
boundaries.
The BCAR (British Civil Airworthiness Requirements) has given the basic strength and
flight performance limits of various categories of the aircraft. They are listed below
Category Positive load factor (n+) Negative load factor(n-)
normal 2.5 -1
Semi aerobatic 4.0 -2
Fully aerobatic 6.0 -3
The 15 seater commuter aircraft comes under the normal category. Therefore the load factor
limit for the aircraft is 2.5 & -1.
The V-n diagram for the aircraft is drawn for the two cases namely
1. Intentional manoeuvre( pilot induced manoeuvre )
2. Unintentional manoeuvre( gusts)
INTENTIONAL MANOEUVRE:
Intentional manoeuvres are induced by the pilot during climb, pull up or dive, banking
the plane etc...
The load factor is a function of velocity. The expression relating the load factor and
the velocity is given by
nmax = (V/Vs)2
Where nmax is the maximum load factor, V is the speed of the aircraft, Vs is the stalling
speed of the aircraft.
The stalling speed of the aircraft is given by Vs 2 = (2W/S)/ CLmax
Vs= 52.07 m/s at 4km altitude.
For various values of V , nmax is calculated and tabulated below,
V nmax=(V/Vs )2
52.070 1.00
62.490 1.44
78.105 2.25
82.726 2.50
93.726 3.24
104.14 4.00
The cruising speed of the aircraft is 115 m/s.
The dive speed of the aircraft is the maximum speed of the aircraft. The dive speed is taken
as 60 knots above Vcruise as per BCAR.
VD = 115 + 60 knots = 115 +30.55 m/s = 145.55 m/s
V nmax=(V/Vs )2
60.13 -1
72.16 -1.44
UNINTENTIONAL MANOEUVRE:
The movement of air in turbulence is known as gusts. It produces changes in wing
incidence, thereby subjecting the aircraft to sudden or gradual increase or decrease in lift from
which normal accelerations result.
These may be critical for large, high speed aircraft and may possibly cause higher
loads than control initiated manoeuvres.
Thus in the gust analysis, the change in load factor due to the gust is calculated. The
BCAR has given standard gust velocities for stall, cruise & dive speeds as 66, 50, 25 ft/s
respectively. The small change in load factor n due to the gust is calculated by assuming a
sharp gust.
The change in load factor n = aUV/2(W/S)
Where is the density at cruising altitude kg/m3
aw is the lift slope, in radians
U is the gust velocity in m/s
V is the velocity of the aircraft in m/s
W/S is the wing loading in N/m2
In the above formula, gusts are assumed to be sharp but it is usually graded, hence
a relief factor called gust alleviation factor K is introduced in the term.
The value of the K is obtained from the book “AIRPLANE AERODYNAMICS
AND PERFORMANCE” by ROSKAM
Where K = 0.88 , 2(W/S)/ gCLα
Where is the density, is the mean aerodynamic chord, g is the acceleration due to gravity, CLα
is the lift slope in CL Vs graph for the wing aerofoil NACA 631412.
The CLα (corrected for aspect ratio) is 0.083/deg or 4.75/rad.
K = 0.88 = 0.809
Therefore n =K aUV/2(W/S)
For stall speed V= 52.07 m/s, U= 20m/s
n = 0.8197
For cruise speed V= 115 m/s, U= 15m/s
n = 1.357
For dive speed V= 145.55 m/s, U= 7.5 m/s
n = 0.8592
V 1+ n 1- n
52.07 1.8197 0.1803
115 2.3570 -0.3570
145.55 1.8592 0.1408
-3
-2
-1
0
1
2
3
4
5
0 50 100 150 200 250U = + 20 m/s
U = + 15 m/s
U = + 7.5 m/s
U = - 20 m/s
U = - 15 m/s
U = - 7.5 m/s
V-n Diagram
Velocity (m/s)
n
+
-
Vs+ = 52 m/s
Vs- = 60 m/s
Vcruise = 115 m/s
Vdive = 145.55 m/s
load factor limits:
n+ = 2.5n- = -1
CONCLUSION
The aerodynamic design of the 15 seater aircraft was completed with the
calculation of V-n diagram, with which the part II structural design starts.
The parameters obtained in the design are not the final parameters. These
should be refined through numerous computer simulation and wind tunnel testing.
The aircraft design is more often an evolutionary process than revolutionary
one. As Donald W. Douglas said, it is just a matter of development. What we have
got today is the Wright brothers‟ airplane developed and refined. But the basic
principles are just what they always were.
Thus the successful design lies in understanding these basic principles and
applying it in an innovative way satisfying the customer‟s requirement.
BIBLIOGRAPHY
1. Jackson, P. (Editor) “Jane’s All the World’s Aircraft 2006-2007” , Jane‟s
information group ltd., Surrey , UK, 2006.
2. Roskam, J “Airplane design Vol. II &V” Roskam aviation and Engg. Corp.
Ottawa, Kansas 1989.
3. Raymer, D.P. “Aircraft Design - a Conceptual Approach” AIAA` educational
series second edition 1992.
4. Perkins, C.D. & Hage, R.E., “Aircraft Performance, Stability and Control”,
John Wiley 1949.
5. Anderson, Jr. J.D. “Introduction to Flight” McGraw Hill 2005.
6. Anderson, Jr. J.D. “Fundamentals of Aerodynamics” McGraw Hill 2006.
7. Anderson, Jr. J.D “Aircraft performance and design” McGraw Hill
International edition 2006.
8. Hoerner, S.F. “Fluid dynamic drag” published by Hoerner Fluid Dynamics,
Brick Town, NJ, 1965
9. Abbott I. H. & Von Doenhoff A. E. “Theory of wing sections”, Dover, 1959.
10. Etkin , B. and Reid L.D. “Dynamics of Flight –Stability and Control” 3rd
edition John Wiley 1996
11. Ogata K, “Modern Control Engineering” prentice-hall, India.
12. DUNLOP tyre manual.
13. Flap data book(RAeS Data sheets).
14. Propeller charts