FINAL PRELIMINARY DESIGN
Middle of Market Transport Team 2
Submitted By:
Ryan Burke, Michael Cleek, Paige Dumke, Brandon Goss,
Ryan McGill, Anthony Ragusa, Andrew Willard
The Pennsylvania State University
Aerospace 402A
Dr. James Coder
Contents
1.0 Final Configuration .................................................................................................................................. 2
1.1 Three-View Drawing ........................................................................................................................... 2
1.2 Wing Configuration ............................................................................................................................. 2
1.3 Fuselage Configuration ....................................................................................................................... 4
1.4 Empennage Configuration .................................................................................................................. 6
2.0 Aircraft Specifications ............................................................................................................................. 6
3.0 Weight Analysis ....................................................................................................................................... 9
3.1 Mission Sizing ...................................................................................................................................... 9
3.2 Comprehensive Weight Analysis....................................................................................................... 10
3.3 Composite Structures ....................................................................................................................... 12
4.0 Airfoil & Wing Aerodynamics ................................................................................................................ 14
4.1 Maximum Lift Coefficient and Lift Curve .......................................................................................... 14
4.2 High Lift Devices ................................................................................................................................ 17
5.0 Performance ......................................................................................................................................... 17
5.1 Drag Build-Up .................................................................................................................................... 17
5.2 Thrust Requirements ........................................................................................................................ 21
5.3 Propulsion System............................................................................................................................. 23
5.4 Climb Parameters.............................................................................................................................. 27
5.5 Takeoff and Landing Analysis ............................................................................................................ 28
6.0 Stability Analysis ................................................................................................................................... 31
6.1 Longitudinal Static Stability............................................................................................................... 31
6.2 Stability Requirements ...................................................................................................................... 33
7.0 Cost Estimates ....................................................................................................................................... 34
Concluding Statements ............................................................................................................................... 36
References .................................................................................................................................................. 37
Appendix ..................................................................................................................................................... 39
A.
Additional Figures and Tables ......................................................................................................... 39
B.
Advanced Composite Materials Trade Study .................................................................................. 40
C.
MATLAB Scripts ............................................................................................................................... 42
1
1.0 Final Configuration
1.1 Three-View Drawing
162 ft
142 ft
1.2 Wing Configuration
Wing Area
To size the wing, one looks to the takeoff constraints. In this case, takeoff dictates the minimum wing size.
To obtain a takeoff wing size a few assumptions were made. Assuming the maximum lift coefficient and
approach speed are the same as the 757-200 (giving the same airport reference code) and based on the
weight, wing area and approach speed of a 757-200, one can find its maximum lift coefficient to be 2.8.
Thus, with the two assumptions specified, one can obtain the takeoff wing sizing from Equation 1.1.
2WTO
2
stall cLmax
Slanding = ρV
2
(1.1)
It is important to note that the takeoff weight is used to define the landing wing area in case of emergency
landing immediately after takeoff. The constraint on the wing area for landing is a minimum of S landing =
2240 ft2.
With the landing constraint in mind looking to cruise coefficient can help to size the wing. For maximum
range,
𝐶𝐿 1/2
should be maximized. Differentiating, setting equal to zero, and assuming a drag polar as shown
𝐶𝐷
in Equation 1.2, obtains the optimum lift coefficient expression for cruise, shown in Equation 1.3.
CD = CD0 +
CLopt =
(CL −CL0 )2
(1.2)
πe0 AR
2kCL0 +√4k2 CL0 2 −(16k)(−kCL0 2 −CD0 )
6k
where k =
1
πe0 AR
(1.3)
From Equation 1.3, one can see that the optimum lift coefficient is a strong function of the lift coefficient
at zero angle of attack, 𝐶𝐿0 . Also, from the drag polar, one can see that the drag coefficient decreases with
increasing CL0 . By setting the landing wing area equal to the wing area at cruise, the optimum lift coefficient
in cruise can be obtained. With a cruise wing area of 2240 ft2 and 𝐶𝐿0 = 0.35 (See cruise lift curve) a CLopt
of 0.511 was obtained.
Wing Span
Once the wing area is specified, one can obtain other wing characteristics. By holding the aspect ratio
constant at AR = 9, a span of 142.0 ft can be obtained by applying Equation 1.4.
𝑏 = √𝐴𝑅 ∗ 𝑆
(1.4)
With span specified a mean aerodynamic chord (MAC) can be obtained by using the equation for chord as
a function of span for a wing with a single linear taper (Equation 1.5) and the definition of mean
aerodynamic chord (Equation 1.6). The MAC obtained is 18.11 ft.
2𝑆
𝑐(𝑦) = (1+𝜆)𝑏 [1 −
2
2(1−𝜆)
|𝑦|]
𝑏
𝑏/2 2
𝑀𝐴𝐶 = ∫0
𝑆
𝑐 𝑑𝑦
(1.5)
(1.6)
A typical mid-sized transport taper ratio of λ=0.2 and the MAC can then be used to determine the tip chord
(ct) and root chord (cr) via Equations 1.7 and 1.8, respectively.
𝑐𝑟 =
3𝑀𝐴𝐶
𝜆+1
(𝜆2 +𝜆+1)
2
(1.7)
𝑐𝑡 = 𝜆𝑐𝑟
(1.8)
The values of the obtained root and tip chord respectively are found to be 26.0 ft and 5.0 ft.
3
Aspect Ratio
Based on existing aircrafts, an aspect ratio can be chosen to best match the mission requirements. In this
case, an aspect ratio of 9 was chosen and held constant throughout the wing sizing process. This was based
on the aspect ratios of both Boeing 757-200 and 787. It was also chosen as 9 to reflect use of composites,
much like the Boeing 787.
Fuel Tanks
Since the fuselage has little room to spare after fitting the payload requirements, the fuel for the flight must
be held elsewhere, mainly inside the wing structure. From Nicolai and Carichner Chapter 8.6, they provide
fuel densities and packaging factors for several different fuels and tanks. This transport will presumably
have integrated tanks in the wings and wing box to contain the 103,000 lbs. of fuel. This translates to a fuel
tank volume of 2760 ft3. The fuel used for this calculation was JP-8, a common jet fuel. However, by 2025
and beyond, the use of natural gas and fuels may be limited, so it’s always a possibility a new fuel source
may be utilized. It can be concluded that tanks capable of holding 1380 gallons should be present in both
wings.
1.3 Fuselage Configuration
The final design fuselage configuration remained the same as in previous iterations. It will feature a singleaisle with three seats on each side. Primary reasons for choosing a narrow-bodied aircraft was a belief that
there would be less drag than a wide-body aircraft, which increases performance. Since this design is
carrying 220 passengers it falls within the range of multiple narrow-body passenger capacities.
The length of the fuselage was determined to be 162 Ft. This was the minimum length that will be able to
accommodate all 220 passengers in two classes (First and Economy) and other various necessities such as
lavatories (8), exit aisles (4), the cockpit, nose cone and finally the empennage (see Table 1.1 for fuselage
length build-up). Windows on the fuselage will be located at every 40 inches (Nicolai) down the fuselage
primarily in the seating sections of the aircraft. Figure 1.1 provides an inboard profile view of the fuselage
and demonstrates how the lengths in Table 1.1 were utilized.
Table 1.1 Fuselage Length Build-Up
Component
Length
Quantity
Radome
4 ft
1
Cockpit
8 ft
1
First Class Row (seat Pitch)
38 in.
6
Economy Class Row (seat pitch)
32 in.
33
Exit Areas
6 ft
4
Empennage
20 ft
1
Total Length
4
Total Length [ft]
4
8
19
88
24
20
162
Figure 1.1 Interior View of Fuselage
The cockpit would presumably feature a state-of-the-art avionics and flight control system. Aircraft systems
today are almost fully digital or “fly-by-wire” and the advances in this technology could possibly make the
cockpit smaller or make other aspects of the aircraft more efficient like weight, volume, etc. For now, this
is not a primary concern for the scope of this preliminary design.
Figure 1.2 depicts the designed cross section for the economy class. This is the critical design area because
the economy class is made to be the most volume efficient, and therefore drives cabin diameter for at least
the top of the fuselage. The first class seating arrangement will consist of a 2-2 per row configuration to
have a more comfortable seating option in the aircraft.
Figure 1.2 Aircraft Economy Class Cross Section (From Jetsizer-II Excel Workbook)
5
The cargo hold is located beneath the cabin floor, also seen in Figure 1.2 which must be able to contain
approximately 3520 ft3 for the specified passengers, (based on 225 passenger and cargo assumption, and
volume assumption from Nicolai text). The most volume efficient and lightest option for containers was
found to be (20) LD-3 size containers with a combined volume of 3520 ft3. These containers have overall
dimensions of 5.3 x 6.6 x 5 ft (searates.com) where the width falls within width of the lower fuselage.
1.4 Empennage Configuration
Once the wing design was also completed, the tail design began since the wing drives the tail design. The
tail functions as a counter mechanism to the wing, to create a downforce behind the aircraft’s neutral point
to eliminate the nose up pitching moment the lift of the wing creates. It also allows for control surfaces to
be implemented, elevators on the horizontal tail and a rudder the vertical tail provides aircraft stability and
control in both longitudinal and lateral/directional axes. This preliminary report does not include the
elevator and rudder sizing, but instead the surface areas for the vertical and horizontal tail.
As most passenger transports today, including the chosen parent aircraft use a “conventional”
configuration for the tail surfaces. This includes two horizontal fins on each side of the fuselage somewhere
mid-tail cone, and the vertical tail is located on the fuselage’s centerline on the top surface. This
configuration allows for larger horizontal tails sizes and best control authority because the engines are
mounted below the wings. There is no extra weight added to the current design because the fuselage
already exists for the tail to be mounted on. In addition, a conventional style tail reduces the stress on the
vertical tail as opposed to mounting the horizontal stabilizer onto the vertical stabilizer. In turn, the vertical
tail is smaller which keeps the weight of the aircraft lower.
In the interim design, it was assumed that the aerodynamic centers of both the horizontal and vertical tail
airfoils were located at the same distance in respect to the wing aerodynamic center. This was later dealt
with when performing C.G and neutral point calculations, and for stability it was determined the horizontal
moment arm would be equal to 80 ft and the vertical tail moment arm equal to 75 ft. This changed the
required tail areas and therefore multiple other parameters. See Section 2.0 for more information on the
tail sizing.
2.0 Aircraft Specifications
Below is a list of all relevant parameters for this design. Note that much of these values are subject to
change once the next phase of design is started, so values are given with little precision and are to be
treated as estimates.
Table 2.1 Aircraft Specifications
Variable
Value
Units
WTO
Wfuel
Wfixed
Wempty
lbs.
lbs.
lbs.
lbs.
Parameter
Weight
Gross Weight (Maximum Take-Off)
Fuel Weight
Fixed Weight
Empty Weight
6
281005
103480
51075
126451
Wing Sizing
Wing Area - Landing
Wing Area - Cruise
Wing Span
Root Chord
Tip Chord
Mean Aerodynamic Chord
Local Airfoil Aerodynamic
Center
Taper Ratio
Sweep Angle
Thickness Ratio
Effective Cruise Mach
Landing Max Clean Lift Coefficient
Cruise Max Clean Lift Coefficient
Landing Lift Curve Slope
Cruise Lift Curve Slope
Wing Sizing
Aspect Ratio
Efficiency
Cruise Wing Loading
Take-Off Wing Loading
Optimum Lift Coefficient
Average Cruise Weight
High Lift Devices
Double Slotted Fowler
c'/c
Sflapped/S
Sweep Angle, High Lift
Delta Max Lift Coefficient
Krueger
Sflapped/S
Delta Max Lift Coefficient
Slanding
Scruise
b
𝑐𝑟
𝑐𝑡
MAC
XAC
2240
2076
142
26
5
18.11
4.53
ft2
ft2
ft
ft
ft
ft
ft

Λ
t/c
Meff_cr
0.20
30
0.14
0.69
1.74
1.4
0.10
0.30
degrees
degrees-1
degrees-1
9
0.6
109.25
psf
CLmax, landing
CLmax, cruise
CLα, landing
CLα, cruise
AR
eo
𝑊
𝑆 𝑐𝑟𝑢𝑖𝑠𝑒
𝑊
𝑆 𝑇𝑂
𝐶𝐿𝑜𝑝𝑡
𝑊𝑐𝑟𝑢𝑖𝑠𝑒𝑎𝑣𝑔
135.38
psf
0.319
226758
lbs.
Λ HL
CLmax
1.3
0.5
20.3
0.878
degrees
-
CLmax
0.8
0.203
-
2
52,200
0.312
93.6
132.2
97
lbs.
(lbs._m/hr)/lbs._f
in
in
in
Engines
Number of Engines
Power (Thrust) Rating
Specific Fuel Consumption
Fan Diameter
Length
Width/Diameter
SFC
7
Horizontal Tail
Area
Span
Root Chord
Tip Chord
Mean Aerodynamic Chord
Taper Ratio
Aspect Ratio
SHT
bHT
𝑐𝑟𝐻𝑇
𝑐𝑡𝐻𝑇
MAC

AR
561
50.26
11.17
3.69
8.43
0.33
4.5
ft2
ft
ft
ft
ft
-
Vertical Tail
Area
Span
Root Chord
Tip Chord
Mean Aerodynamic Chord
Taper Ratio
Aspect Ratio
SVT
bVT
𝑐𝑟𝑉𝑇
𝑐𝑡𝑉𝑇
MAC

AR
385
24.02
16.01
5.28
12.55
0.33
1.5
ft2
ft
ft
ft
ft
-
29194
17.33
14.42
162.83
1
38
33
6
8.33
20
3.5
ft3
ft
ft
ft
ft
ft
ft
ft/s
Vstall, landing
Vstall, cruise
774.17
0.8
177.78
401.6
Vmin power, SL
Vmin power, cruise
366
717
ft/s
ft/s
593
5897
43627
45217
16
ft/s
nmi
ft
ft
min
Fuselage Sizing
Pressurized Volume
Fuselage Width
Fuselage Height
Fuselage Length
Number of Aisles
Total Number of Rows
Economy Rows
First Class Rows
Cockpit Length
Empennage
Radome
Performance
Cruise Speed
Cruise Mach Number
Stall Speed (Landing)
Stall Speed (Cruise)
Maximum Endurance
Velocity at Min. Power, Sea Level
Velocity at Min. Power, Cruise
Maximum Range
Max Velocity, Sea Level
Maximum Cruise Range
Service Ceiling
Absolute Ceiling
Time to Climb
Vcruise
Mcruise
8
ft/s
ft/s
3.0 Weight Analysis
3.1 Mission Sizing
As previously stated in the initial design report, maximum take-off weight, WTO, was determined by
summing the Wfixed, Wempty, and Wfuel. Fixed weight consists of the weight of the 7 crew members and 220
passengers, assuming a weight of 225 lbs. for each individual, which yields 51,075 lbs.. This weight is viewed
as constant through our calculations. For fuel weight, calculations can be seen below. Ultimately, fuel
weight was found to be a function of take-off weight, which was found through iterations to be
0.3498*WTO. Regarding empty weight, Equation 3.1 is also found to be a function of take-off weight. From
Table 5.1 in the Nicolai and Carichner text, constants a and b for a transport aircraft are 0.911 and 0.947,
respectively.
Wempty = aWTO b
(3.1)
Fuel weight is expressed as several ratios at each segment during the mission profile. This is because fuel
is burned consistently during flight and the ratios at take-off will not equal that of the ratios at descent or
landing. (Nicolai) For example, the ratio of the weight after take-off to the initial weight is equal to 0.975,
while the ratio of weight after ascending to 32,000 feet to the weight after the take-off is equivalent to
0.97. For a turbojet, the weight ratio for cruise can be found below in Equation 3.2.
R=C
V
TSFC
L
W
ln(Wi )
D
f
(3.2)
Figure 3.1 Designed Mission Profile
From the Mission Profile in Figure 3.1 above, the range is close to 5000 feet and each phase of the mission
requires different amounts of fuel. In Table 3.1 below, the fuel burn during segments can be seen using the
range equation to compare the initial weight with the final weight of each segment.
9
Table 3.1 Mission Profile
Mission Segment
Fuel Burn (W_n+1/W_n)
Segment
Weight (lbs..)
n=1
n=2
0.975
n=1
281005
n=2
n=3
0.97
n=2
273980
n=3
n=6
0.70649
n=3
265760
n=6
n=7
1
n=6
187756
n=7
n=8
0.98251
n=7
187756
n=8
n=9
1
n=8
184472
n=9
n=10
1
n=9
184472
n=10
184472
Final Weight
3.2 Comprehensive Weight Analysis
More Detailed weights include Landing Gear, Starting Systems, Surface Controls, Flight Instruments,
Furnishings, and AC and Icing instruments. These weight estimations came from the several following
empirical Equations (3.3-3.7) which were found in the Nicolai & Carichner text, Chapter 20.2.1.
WLanding Gear = 62.21 ∗ (. 01WTO )0.84
(3.3)
WStarting system = 38.93(2 ∗ WENG ∗ .01)0.918
(3.4)
WSurface Controls = 56.01(WTO ∗ qMAX ∗ .0001)0.576
(3.5)
WFlight Instruments = 2 ∗ (15 + .032 ∗ (. 01 ∗ WTO ) + 4.8 + .006 ∗ (. 01 ∗ WTO ) + .15 ∗ (. 01 ∗ WTO ))
WAC and Anti−Icing = 469.3(Volume ∗ 227 ∗ .001)0.419
(3.6)
(3.7)
The other detailed weights include the weight of the furnishing on the aircraft included the flight deck,
passenger seats, lavatories, buffet stations, oxygen system, windows, cargo holdings, and miscellaneous.
Structure weights included the planform weight, fuselage weight, and the tail weight, all adjusted for using
advanced composite materials (see section 3.3 for more details of how this was done). After calculating
these, Table 3.2 below shows the detailed weights of the furnishings and the sum of them all. Engine weight
is found from manufacturing data.
Table 3.2 Comprehensive Weight Analysis
Component
Weight (lbs.)
Systems
10
% TOGW
Starting System
719
0.26%
Integrated Fuel Tanks
3347
1.19%
C.G Control System
304
0.11%
Engine Controls
67
0.02%
6662
2.37%
Flight Instruments
51
0.02%
Engine Instruments
14
0.005%
Miscellaneous Instruments
49
0.02%
Electrical
2503
0.89%
Air Conditioning & Anti-Icing
7138
2.54%
Flight Deck
275
0.26%
Passenger Seats
7047
1.19%
Lavatories
1448
0.11%
Food Provisions
2387
0.02%
O2 System
316
2.37%
Windows
6954
0.02%
Cargo Provisions
813
0.005%
Cargo Containers
3748
0.02%
Miscellaneous Furnishings
254
0.89%
Fuselage
18552
6.6%
Landing Gear
8095
2.9%
Planform
20505
7.3%
Empennage
11250
4.0%
Control Surface Hydraulics &
Pneumatics
Furnishings
Structures
11
Engines & Nacelles
23954
8.5%
Crew & Attendants
1575
0.6%
Passengers
49500
17.6%
Mission
97271
34.6%
Reserve
5174
1.8%
Trapped
1035
0.4%
281005
100%
Payload
Fuel
TOTAL
3.3 Composite Structures
When designing an advanced mid-sized transport aircraft, it is important to examine the use of advanced
materials to reduce weight of the airplane structures. A trade study on the benefits and penalties of the
use of composites in the construction of a modern aircraft was conducted and found that there are three
main advantages to the use of composite structures. They are weight savings, maintenance savings, and
increased lifespan.
The conclusion of the trade study shows that using composite materials has more benefits than using
typical aluminum materials for the airplane structure. Continuing concerns over the current cost of
production for advanced materials have been raised, but the team feels as though cost will decrease as
production increases, so that composite materials will significantly benefit our airplane design. For
reference, the trade study can be found in Appendix B.
Previous calculations for the weight of the structure were adjusted for the implementation of composites
materials. The empty weight was reduced by 17% and the gross takeoff weight was reduced by 15%. Table
3.3 below shows in detail how the empty weight was adjusted for composites. The reduction percentages
in Table 3.3 were chosen directly from chapter 20 of Nicolai & Carichner’s text.
Components
Planform
Empennage
Fuselage
Landing Gear
Table 3.3 Weight Corrections for Advanced Composite Materials
%Reduction
Weight (lbs.)
Without Composites
0%
25631
With Composites
20%
20505
Without Composites
0%
15000
With Composites
25%
11250
Without Composites
0%
24736
With Composites
25%
18552
Without Composites
0%
8742
12
All Else Empty
Engines
With Composites
Without Composites
With Composites
Without Composites
8%
0%
2%
0%
Revised Empty Weight
Empty Weight Reduction
Takeoff Gross Weight Reduction
8095
44757
44095
23954
126450
17%
15%
Composites are a combination of materials mixed together to achieve specific structural properties, the
independent materials in composites do not dissolve or merge completely but rather act together as one.
These materials generally consist of a fibrous material embedded in a resin matrix that is laminated with
fibers oriented in alternating directions. This structure helps establish strength. Different strength for a
desired direction can be produced by changing the fiber orientations for layers in a laminate.
Carbon Fiber Reinforced Polymer is an extremely strong and light fiber-reinforced plastic. It is often
comprised of carbon fibers and a binding polymer. It can be made by heating the fibers, typically aramid
(Kevlar), to ~2000ºC in an oxygen-deprived oven. These fiber end up being approximately 6μm (six
micrometers) in diameter and spun into a thread. These threads are then woven into sheets and mixed
with hardening resins, such as epoxy or silica, to form the various material needs. The manufactured
composite laminate is known to have the high strength to weight ratio as well as rigidity, which lends itself
to be very applicable to the aerospace industry.
Looking at our parent aircraft, the Boeing 787, carbon fiber reinforced polymer is used in the fuselage,
wings, empennage, and other various parts of the aircraft as seen in Figure 3.2. The Boeing 787 uses
approximately 50% composite materials by weight. This percentage is the largest by far of any Boeing
commercial airplane use in composite materials. Compared to an all-aluminum approach, the Boeing 787
saves about 20% of weight while full.
The composite material usage does not only cut down on the weight saved, but also on the maintenance
required, both scheduled and nonscheduled. This can be seen in the more experiences Boeing 777, which
has approximately 12% composites. Compared to the 767 aluminum tail, the 777 composite tail is
approximately 25% larger, while requiring about 35% less maintenance. Apart from the tail, the 777
incorporates a composite floor beam in all 565 airplanes. In the 10 years of flying, none of the 777’s needed
a composite floor beam to be replaced. Even if there is repair needed on a composite component, the same
process is used – with bolted repairs. There is also the option to perform bonded composite repairs, which
keeps the airplane’s aerodynamic and aesthetic finish. This rapid composite repair technique will allow
these airplanes to get back into the air faster than an aluminum airplane. Another technique that Airbus is
implementing is the use of composite panels, instead of full composite bodies. This allows for a whole new
panel to be installed for the broken panel. This method is faster in many cases.
Issues with composite materials are that they do not have visible cracks unlike metals and often can be
missed by simple inspection. In addition, composite materials are very porous in nature and collect
moisture caused stresses and strains inside the material. The stresses can cause delamination, or composite
layer separation. Many non-destructive inspection are used to see if the material is close to failure. Such
procedures include ultrasonic, X-ray, moisture detector, and audible sonic testing.
13
Figure 3.2 Boeing 787 Material Composition [modernairliners.com]
4.0 Airfoil & Wing Aerodynamics
4.1 Maximum Lift Coefficient and Lift Curve
To obtain the maximum lift coefficients at cruise and landing conditions, two-dimensional lift curves were
obtained for each case. The cruise lift curve was obtained using Xfoil at landing conditions, which
correspond to an effective Mach number of 0.1747 (a landing speed of 137 knots) and a Reynold’s number
of 26,007,742. The resulting two-dimensional lift curve can be seen in Figure 4.1.
Figure 4.1 Lift curve of SC2-0714 at sea level conditions.
It can be noted that Xfoil likely overestimated the higher end of the lift curve. Estimating that the maximum
lift coefficient probably occurs somewhere around a Cd=0.0250, a corrected Clmax of 2.21 can be obtained. This
maximum lift coefficient occurs at α=15 degrees. From this value, a three-dimensional CLmax of 1.72 was
obtained using Equation (4.1).
14
𝐶𝐿𝑚𝑎𝑥 = 0.9𝐶𝑙𝑚𝑎𝑥 cos (𝛬𝑐 )
(4.1)
4
Likewise, the two-dimensional lift curve was able to be obtained for cruise conditions from figures of NASA
Technical Memorandum 4044. Noting that for small angles the airfoil lift coefficient, Cl, is approximately equal
to the normal force coefficient, Cn, the lift curve at cruise conditions (Meff=0.70 and Re=40 x 106) is given and
can be seen in Figure 4.2
Figure 4.2 Lift curve of SC2-0714 at cruise conditions.
To find CLmax at cruise altitude, a Mach number for CLmax was obtained from Equation (4.2), where the
subscript “cruise” indicates conditions at cruise speed.
MCLmax =
Mcruise √CLcruise
1.2
(4.2)
The Mach number for maximum lift coefficient is 0.417, which corresponds to a Reynold’s number of
18,048,022. This condition was then used in Xfoil to obtain a Clmax of 1.79. Again, using Equation (4.2), a
CLmax of 1.40 was obtained. To approximate the full wing lift curve, the wing lift curve slope was found by
Equation (4.3).
AR
CLα = Clα (AR+2)
(4.3)
The two dimensional lift curve slope was obtained from the linear region of the airfoil lift curve and was
0.118 deg-1 for sea level and 0.155 deg-1 in cruise. Also, by fitting a line to the linear region of the lift curve,
an angle of attack for zero lift can be obtained, which is the same for infinite aspect ratio as it is for a finite
aspect ratio. The angle of attack for zero lift, α0L, is approximated to be -4.95 degrees. Using this
15
information, three-dimensional lift curve slopes were obtained, their values being 0.096 deg-1 and 0.127
deg-1 respectively. The effect of aspect ratio on the wing lift curve slope can be seen in Figures 4.3 and 4.4.
Figure 4.3 Lift curve of SC2-0714 with Infinite AR vs. AR=9 at Sea Level.
Figure 4.4 Lift curve of SC2-0714 with Infinite AR vs. AR=9 in Cruise.
16
4.2 High Lift Devices
By adding double slotted Fowler flaps, ΔCLmax at sea level increases by 0.878, which was calculated using
Equation (4.4) and noting that Equation (4.5) gives the ΔCLmax for double slotted Fowler flaps. Here, c’ is
the new chord with flaps extended, which was chosen to be a 30 percent increase from the base chord
(c’/c=1.3).
c’
ΔCLmax = 1.6 ∗ c
ΔCLmax = 0.9ΔClmax
Sflapped
Splanform
(4.4)
cos(ΛHL )
(4.5)
The sweep angle of the fowler flaps was calculated to be 20.3 degrees by Equation (4.6) to satisfy the 30
percent increase in chord. Also, it was assumed that the flapped area was approximately 50 percent of the
planform area.
4
1−λ
tan(Λ HL ) = tan(Λ LE ) − [0.7
(4.6)
]
AR
1+λ
Likewise, by adding Kreuger flaps along 80 percent of the span and applying Equation (4.5) with Kreuger
flaps contributing a ΔClmax of 0.3, CLmax was increased by 0.203. With the high lift devices deployed the new
ΔCLmax at sea level is 2.82.
5.0 Performance
5.1 Drag Build-Up
Drag contributions were contributed to profile drag for the wing, fuselage, and tail, the induced drag cause
by the wing, and finally trim drag from the tail. Each component of drag was calculated individually across
the speed range at both sea level and cruise altitudes. Then, they were summed to find the total drag across
the aircraft, Figs. 5.1 and 5.2 are the resulting drag force curves with respect to velocity. For Figures of drag
contributions to the total drag, see Appendix A.
17
Figure 5.1 All Drag versus Velocity Curves across Sea Level Speed Range
Figure 5.2 All Drag versus Velocity Curves across Cruise Altitude Speed Range
18
Profile Drag
The profile drag at sea level conditions was obtain using Xfoil and varying Mach number and Reynold’s
number. These values were then run at the corresponding lift coefficient needed for that speed to obtain
the speed’s section drag coefficient. Using the drag Equation (5.1), the profile drag was obtained. The
results of this can be seen in Figures 5.1.
1
Dprofile = 2 ρV 2 SCdprofile
(5.1)
The profile drag at a cruise altitude of 35,000 feet was obtained from NASA Technical Paper 2890. Again,
speed was varied and matched with a given Cn (Cn ≈ Cl for small α) to obtain the section drag coefficients.
Equation (5.1) was applied to obtain the total profile drag. The results can be seen in Figure 5.2.
Fuselage Drag
The fuselage drag was one of the more complicated drags to calculate, considering changing Reynolds
number along the length of the fuselage and different equations for skin friction coefficients due to
changing Reynolds number. A MATLAB script was written to deal with the complexities of the fuselage drag,
but also for the other drags. See Appendix B for the complete drag build-up script. The total length and
correct cross section of the aircraft was inputted into the Jetsizer-II workbook and produced nine different
points along the fuselage, dividing it into ten sections. This was done to use surface areas between the ten
sections of the aircraft to calculate the drag.
Then the Reynolds number was calculated for every location of the fuselage at every velocity in the speed
range. The script was written to determine if either Equations (5.2) or (5.3) should be used for skin friction
coefficient. Finally, the drag at every velocity was found using Equation (5.4).
1
CfLaminar = 1.328Re−2
(5.2)
Cf Turbulent = 0.455Re−2.58
(5.3)
Where 𝑅e =
ρVx
μ
and transiton Re ≅ 500,000
DFuselage = ∑𝑖 0.5ρV 2 Swi Cfi
(5.4)
Tail Profile Drag
Difficult to determine without a tail airfoil selected, the profile drag of the horizontal and vertical stabilizers
was estimated by taking the ratio of surface area between the tail and wing. This served as a correction
factor to the wing profile drag coefficient and it was decided to use these to find the tail drag. Of course,
this is an extremely crude approximation, but comparing to the Jetsizer-II workbook, the tail drags were of
similar magnitude.
Induced Drag
Induced drag is a result of the wing creating the necessary pressure difference between upper and lower
surfaces in order to produce lift, hence the term ‘induced’ drag. Furthermore, it is directly related to the
lift coefficient of the aircraft as seen in Equation 5.5.
19
C2
L
CDi = πeAR
(5.5)
Since CL is already a function of velocity squared, it is squared again in the induced drag coefficient. Then,
CDi is multiplied by velocity squared when finding total induced drag, therefore, the total induced drag
should be inversely proportional to velocity squared and this is evident in Figures 5.1 and 5.2. The induced
drag is the largest contributor to the aircraft drag.
Trim Drag
Trim drag is the consequence of the tail providing a down force with a backwards component to cancel the
aircraft nose-up pitching moment. Although it is a necessary evil, it should only account for 1-10%) of the
total drag of the aircraft (Nicolai & Carichner). Using a free body diagram of the aircraft in trimmed flight,
the lift coefficient for the tail required to obtain trimmed flight can be find as seen in Equation 5.6.
S
CLt = S l (c̅Cmac + xCL )
(5.6)
t t
Where the reference areas of the wing and tail appear, along with tail moment arm, lt , mean aerodynamic
chord, c̅, and moment coefficient about the wing aerodynamic center, Cmac. Trim drag coefficients were
then calculated from the derived formula of Equation 5.7.
S CLt
CDtrim = S
t
[
CLt eAR
CL CL et ARt
− 2] CDi
(5.7)
It is clear the trim drag is dependent on wing and tail parameters, and more importantly the induced drag
coefficient which was found beforehand. The trim drag at both altitudes were 5% and 4% of the total drag
and therefore were considered to be within reasonable magnitude.
Total Drag
Each component of drag were combined and the result is shown in Figures 5.1 and 5.2. Critical information
L
from these figures include the D
max
L
at both altitudes as well as the D at cruise velocity. Lift to drag ratios
are at a maximum at minimum drag, therefore the lowest points on the drag curve are where the values in
Table 5.1 are derived from. The maximum lift to drag ratios were considered reasonable for an advanced
mid-size transport, but it is thought that these values still have potential to increase, especially the cruise
L/D which reflects the efficiency of the aircraft and should be optimized.
Table 5.1 Lift to Drag Ratios Taken From Drag Curve
Sea Level
L/D max=23.7
L/D max=20.4
Cruise (35,000 Ft.)
L/D cruise=15.9
There are still many other forms of drag that were not accounted for in this preliminary analysis. Base,
interference, and transonic wave drag are some of the other forms of drag that could have a significant
impact on the values in Table 5.1, and therefore the design. Although the drag force will most likely
increase, with further analysis and detail design the wings may be able to produce more lift than expected,
or be optimized to reduce induced drag, trim drag, or others.
20
5.2 Thrust Requirements
For a constant altitude, the thrust of the aircraft is limited by the thrust available while being required to
overcome the drag at certain velocities. The thrust required for the aircraft is dependent on the total drag
build up from Figure 5.1 and 5.2. The drag buildup for a third altitude was also calculated to show a better
analysis of altitude variation for thrust. The mid-cruise altitude of 10,000 feet was used to find a third drag
build up, which was expected to fall in between the sea level and cruise curves.
The thrust available for this aircraft is dependent on the constraint diagram and thrust needed to take off.
Therefore, the thrust available to ensure the mission can be executed was 52,200 lbs. per engine at sea
level. The static thrust available for a constant altitude is equal to the original thrust available multiplied
by the ratio of the air density at that altitude over the air density of sea level. So the original 104,000 lbs.
at takeoff and sea level decreases to 74% that value at 10,000 feet and 31% that value at 35,000 feet. The
actual thrust available would experience a slight decrease in this value as airspeeds increase, until it
approaches the maximum speed wherein it increases again.
From the calculated thrust required from the drag build up and the thrust available given by the engine and
air density, the thrust curves for each of the altitudes can be found in Figures 5.3, 5.4 and 5.5.
Figure 5.3 Thrust Requirement Curve for Sea Level
21
Figure 5.4 Thrust Requirement Curve for 10,000 ft
Figure 5.5 Thrust Requirement Curve for 35,000 ft
22
5.3 Propulsion System
To size the propulsion system many parameters must be taken into account. To start the thrust to weight
ratio for the take-off, cruise, landing stall speed and service ceiling need to be found and graph compared
to wing loading to find the ideal engine size. Stall speed is the first parameter found and the calculation is
relatively simply. To find the stall speed we took Equation (5.8) and manipulated it to solve for wing loading
at stall speed, which produced Equation (5.9).
1
2
L = W = 2 ρVmax
SCLmax
W
S VS
( )
(5.8)
1
2
2
= ρVmax
SCLmax
(5.9)
This value will set the right side limit of the constraint diagram. Everything to the left of this wing loading
value will be in the acceptable range and meet the requirements set forth by the limiting stall speed.
Next the thrust to weight ratio for cruise is calculated to and graphed against wing loading as wing loading
increases. Starting with Equation (5.10), where 𝑉𝑚𝑎𝑥 is the velocity during cruise, and breaking down 𝐶𝐷
into component form we can get the following Equation (5.11) which is a manipulated thrust to weight
equation for cruise.
1
TSL σ = 2 ρV2max SCD
1
2
(5.10)
2
𝑇𝑆𝐿 𝜎 = 𝜌𝑉𝑚𝑎𝑥
𝑆 (𝐶𝐷0 + 𝐾 ∗ [
2
2𝑊
)
]
2
𝜌𝑉𝑚𝑎𝑥 𝑆
(5.11)
Finally after rearranging Equation (5.12), the thrust to weight ratio during cruise can be compared to the
wing loading.
TSL
)
W Vmax
(
2
= ρ0 Vmax
CD0
1
W
S
2( )
+
2K
W
( )
ρσV2max S
(5.12)
Next, the thrust to weight ratio during take-off needs to be accounted for. This will most likely set the
lower limit of the thrust to weight ratio needed to properly size the engines. Starting with the modified
take-off distance Equation (5.13), found from Mohammad Sadraey's report "Preliminary Aircraft Design"
where 𝐶𝐷𝐺 = (𝐶𝐷𝑇𝑂 − 𝜇𝐶𝐿𝑇𝑂 ) and 𝐶𝐿𝑅 =
𝐶𝐷𝑇𝑂 = 𝐶𝐷0 + 𝐶𝐷0 𝑇𝑂 + 𝐶𝐷 0
𝑓𝑙𝑎𝑝𝑠
2𝑚𝑔
2
𝜌𝑆𝑉𝑐𝑟𝑢𝑖𝑠𝑒
, the T/W value for take-off can be found. Noting that
and keeping in mind the average values for 𝐶𝐷0 𝑇𝑂 and 𝐶𝐷 0
𝑓𝑙𝑎𝑝𝑠
are 0.008
and 0.005 respectively, also found in Sadraeys's report, the equation can be manipulated to solve for T/W
and we get Equation (5.14).
1.65W
STO = ρgSC
DG
23
T
−μ
W
CD
−μ− G
W
CL
R
ln[ T
]
(5.13)
T
(W)
STO
(μ−(μ+
=
CD
1
0 )[exp(0.6ρgC S
D0 TO W )])
CL
R
S
1
(5.14)
1−exp(0.6ρgCD0 STO W )
S
This gives us a close to linear line when graphed and becomes the lower limit for operational thrust to
weight limits allowed within the design constraints.
Finally the thrust to weight ratio for the service ceiling is calculated. Using the rate of climb, Equation
(5.15) and solving for 𝑇/𝑊 we get Equation (5.16).
ROC =
T
(W) =
Pavl −Preq
ROC
2
W
( )
√ √CD0 S
ρ
(5.15)
W
+
1
(5.16)
L
( )
D max
K
To relate this to the service ceiling we use a 𝑅𝑂𝐶 equal to 100 ft/min and alter the equation to include
the ratio or densities at the giving service ceiling altitude, 𝜎. This now gives us Equation (5.17). This is
another critical equation in deciding the thrust to weight ratio needed for engine size selection as the ratio
needs to be above this curve for the aircraft to meet the requirements.
T
(W)
sc
=
ROC
2
W
σ
( )
√ √CD0 S
ρ
+
1
L
σ( )
D max
(5.17)
K
Pulling all these equations together and using the known values we are able to plot these four equations,
via Matlab, all on the same graph and therefore find the ideal thrust to weight ratio needed, seen in Figure
5.6.
24
Figure 5.6 Propulsion Design Constraint Diagram
The region in Figure 5.6 that is in between the four different curves is the ideal thrust to weight ratio region
and the designated design space for our engine. We chose a value for 𝑇/𝑊 of 0.325, the upper limit above
where the service ceiling and take-off 𝑇/𝑊 intersect.
This 𝑇/𝑊 ratio of 0.325 gives us a needed thrust of 91,326 lbs. for our take-off weight of 281,005 lbs. Using
a two engine set up this means we need about 45,663 lbs. of thrust for each engine. This value now means
we are going to look for a lightweight fuel-efficient engine that can provide at least this amount of thrust
per engine.
After research, the Pratt & Whitney PW4152 is the engine that most fits our requirements. With 52,200
lbs. of thrust per engine this is plenty of power to fly our plane. Intentionally oversizing the engines slightly
leads to greater safety in case one were to fail and the SFC is actually lower for this engine than many other
engines with a thrust closer to our ideal value. Overall this engine will provide the thrust required for all
elements of flight and provides our team with a resourceful and cost effective efficient turbofan engine.
25
Engine Specs
The engine chosen for our aircraft design is the Pratt & Whitney PW4152. Two engines will be
implemented into our design, each with a thrust of 52,200 lbs. and specific fuel consumption (SFC) of
0.312. More specification of the PW4152 can be found in Table 5.2, as well as an image of the engine in
Figure 5.7.
Table 5.2 Engine Specifications
Manufacturer
Pratt & Whitney
Units
Model
PW4152
-
Applications
A310-324/-324ET/-324F
-
Weight
9,213
[lb]
Thrust
52,200
[lbf]
SFC
0.312
[lb/lbf hr]
Fan Diameter
93.6
[in]
Length
132.2
[in]
Width/Diameter
97
[in]
Figure 5.7 PW4152 Engine Image
26
5.4 Climb Parameters
From the thrust requirement curves for the three different altitudes, the rate of climb of the aircraft can
be found. The rate of climb is a function of excess power, which is found by the velocity multiplied by the
difference between the thrust available and the thrust required. The high available thrust available of sea
level allows for the most excess power at a speed above take off velocity. Therefore, at sea level the rate
of climb is around 2939 ft/min. As altitude increases, the rate of climb decreases to 708 ft/min at the
cruise altitude. This data is plotted in Figure 5.8, in order to find the service ceiling and absolute ceiling
limits.
Rate of Climb vs. Altitude
3500
Rate of Climb (ft/min)
3000
2500
2000
1500
1000
500
0
0
5
10
15
20
25
30
35
40
45
50
Altitude (1e10 ft)
Figure 5.8 Rate of Climb
From this data in Figure 5.8, the service ceiling can be found as the altitude at which the rate of climb is
100 ft/min. Thus, the service ceiling is defined at 43,627 feet. The absolute ceiling is similarly found as
the altitude for which the rate of climb is 0 ft/min. The absolute ceiling was found at 45,217 feet. Both
these ceilings are limits and consequently larger than the mission requirements of 43,000 feet ceiling. Yet,
these values being relatively close to the mission requirements ensures the aircraft is operating efficiently.
The time of climb to the cruise altitude from sea level uses the rate of climb and distance needed to be
climb. From Figure 5.8, the time to climb to cruise without any other maneuvers or restrictions would be
approximately 16 minutes. Although this is relatively short, the excess power granted from the engines at
the altitudes leading up to the desired 35,000 feet allows for the higher rate of climb. A shorter time of
climb value is optimal for decreasing profile drag, but is limited by the rate of climb to prevent excess fuel
consumption.
27
5.5 Takeoff and Landing Analysis
Take off distance is the sum of ground distance, rotation distance, transition distance, and climb distance.
These values all depend on various aircraft properties such as stall velocity, coefficient of lift, coefficient
of drag, etc. The following Equations (5.18-5.21) pertain to values used in the takeoff estimation for this
aircraft.
g
(T − D − μ(WTO − L))
WTO
acceleration =
VTO = 1.2Vstall = 1.2 ∗ √
2 ∗ WTO
ρSREF CLmax
∆CDG = ∆CDflaps + ∆CDgear + ∆CDi
VTO
SG = ∫
0
2
d(V 2 )
VTO
≈ 0.5
2a
a
(5.18)
(5.19)
(5.20)
(5.21)
There are several ways to estimate the takeoff ground distance for an airplane. This estimation procedure
assumes the acceleration is constant, and equal to the value at V = 0.707 V_TO. Using Equation (5.18), it
is possible to find the ground acceleration of the airplane, which is found to be 9.5 feet per second, when
the airplane is at 0.707 velocity take-off. The takeoff velocity is found using Equation (5.19), incorporating
the stall speed at sea level. For this aircraft, the maximum lift coefficient during takeoff is 2.6, then the
takeoff speed is calculated to be 240 feet per seconds. This speed is consistent with our parent aircrafts.
The lift and drag are changing as the velocity changes, so to calculate these values, these coefficients are
needed. The initial lift coefficient is 0.49 and incorporating double fowler flaps, we can increase this value
to 1.37 as a maximum lift coefficient. The drag coefficient is calculated using the initial drag coefficient
and changes due to flaps, gear, and induced drag. This is represented in Equation 5(5.20). These values
can be found in figure 9.25 and 10.5 in Nicolai for the flaps and gear respectively. The induced drag can
be calculated using the new value of coefficient of lift with the given flap setting. The coefficient of drag
is found to be 0.2 for this aircraft during takeoff at sea level. Equation (5.21) is then used to calculate the
ground roll distance which was found to be 5896 feet.
The next distance can be calculated using Equation (5.22) below. This is the rotation distance in which the
aircraft is (still on the ground) rotated to an angle of attack in such the coefficient of lift is now 0.8 ΔCLmax .
After this rotation, the coefficient of lift will have a value of 2.1 for this aircraft. This distance is estimated
to take 2 seconds and is calculated to be 482 feet.
SR = (2 seconds) ∗ VTO
(5.22)
STR = Rsin(θCL )
(5.23)
R=
2
VTO
0.152g
Rate of Climb
θCL = sin−1 (
)
VTO
28
(5.24)
(5.25)
SCL =
(hobsticle − hTR )
tan(θCL )
(5.26)
Equation (5.23) estimates the transition distance. This is the distance that the aircraft flies a constantvelocity arc with a radius R. The radius can be calculated in Equation (5.24) and was found to be almost
12000 feet. The angle of climb can be estimated from the rate of climb, which is 3400 feet per minute.
The angle of climb is calculated to be about 7 degrees, using Equation (5.25). Using Equation (5.23), the
distance for transition is calculated to be 1378 feet. At the end of this takeoff phase, the airplane is at a
height of 160 feet. This value is greater than the given FAA requirement for an obstacle, 35 feet. Figure
5.9 describes a takeoff trajectory.
Figure 5.9 Schematic for Takeoff Analysis (Nicolai)
Therefore, Equation (5.26) is not needed, as we are above the climb needed and the climbing distance is
zero feet. The takeoff distance can be seen in Table 5.3 below. The sum of the takeoff distances gives the
total distance to be 7756 feet. This value is just under the 7800 feet takeoff requirement.
Table 5.3 Take off Distance for different thrust amounts
Thrust Amount
Full
SG
5896 feet
SR
482 feet
STR
1378 feet
SCL
0 feet
STOTAL
7756 feet
The landing distance is defined as the horizontal distance required to clear a 50 foot obstacle, free roll,
and then break to complete stop. Figure 5.10 below show the schematic for landing. The velocity over the
50 foot obstacle is assumed to be 1.3 Vstall, or 228 feet per second. The touchdown velocity is assumed to
be 1.15 Vstall, or 202 feet per second. The weight of the aircraft is assumed to be when half the fuel is
burned, or 231,172 pounds. The coefficient of lift for landing is assumed to be the maximum value with
engaged flaps and slats in the landing configuration. The lift coefficient is 2.82 for the landing coefficient.
29
Figure 5.10 Schematic for Landing Analysis (Nicolai)
The sum of the air distance, free roll distance, and braking distance is the estimation of the total landing
distance. The air distance uses the relationship between kinetic and potential energy. It can be calculated
using Equation (5.27) below. This Equation utilizes the lift and drag of the airplane. The lift can be assumed
as the weight of the landing airplane. The drag can be calculated using Equation (5.28). The coefficient of
drag must be altered due to drag effects of the flaps, gears, and changing induced drag. The coefficient of
drag is calculated to be 0.346, and this is represented in Equation (5.29) below. The air distance can now
be calculated to be 774 feet.
2
2
L V50
− VTD
(5.27)
SA = (
− hobsticle )
D
g
D = q50 SREF CD
(5.28)
CD = CD0 + CDi + ∆CDflaps + ∆CDgear
(5.29)
The free roll distance is the distance covered while the pilot reduces the power to idle, retracts the flaps,
deploys the spoilers, and applies the brakes. These adjustments change the values of the thrust and the
drag coefficient. The thrust can then be assumes to be zero and the coefficient of drag is updated to be
about 0.353. Free roll is assumed to last 3 seconds and it is represented in Equation (5.30) below.
SFR = (3 seconds)VTD
(5.30)
The braking distance can be calculated using Equation (5.31) below. The acceleration can be calculated
using Equation (5.32), where R is the reverse thrust. For this estimation, thrust is assumed to be zero while
breaking, and the reverse thrust is neglected for now. The coefficient of drag is also calculated to be 0.353
for a landing configuration that does not retract the flaps during free roll. This is done using Equation
(5.33). The deceleration can then be calculated as 6 feet per second. If the pilot does retract the flaps
during free roll, the coefficient of drag is 0.17. The deceleration can be calculated to be 3 feet per second.
After substitution and manipulations, Equation (5.31) can be simplified to Equation (5.34) to find the
braking distance. The breaking distance without retracting the flaps is 8070 feet. The breaking distance
with retracting the flaps is 9274 feet.
d(V 2 )
2a
V2
TD
0
SB = ∫
30
(5.31)
1
a = ( ) (T − D − μ(L − WL ) − R)
m
(5.32)
CD = CD0 + CDi + ∆CDflaps + ∆CDgear + ∆CDmisc + CDspoilers
(5.33)
WL
SB =
gμρSREF ((
CD
) − CLG )
μ
ln (1 +
ρ SREF CD
2
( − CLG ) VTD
)
2 WL μ
(5.34)
The total landing distance is the sum of the air distance, free roll distance, and braking distance. During
the free roll the pilot has the option to retract the flaps and slats to decrease lift, so that the total landing
distance is calculated to be 10980. If the pilot does not, the coefficient of drag remains high and the total
braking distance is 9777 feet. While these both exceed the takeoff requirement, it is still a reasonable
value for various international airport runways worldwide. While these estimates can hold close to actual
values, there may be more drag available to slow the airplane down as the drag build up at sea level close
to stall speed has a coefficient of drag of 0.443. Other forms of deceleration could be from reverse thrust
or high energy absorption brakes. Table 5.4 below shows the landing distances below.
Flaps
SA
SFR
SB
STOTAL
Table 5.4 Landing Distance
Retracted
1102 feet
605 feet
9274 feet
10981 feet
Engaged
1102 Feet
605 Feet
8070 Feet
9777 Feet
6.0 Stability Analysis
6.1 Longitudinal Static Stability
For the design to be considered airworthy, it must be able to fly safely and adhere to stability requirements
set by the Federal Aviation Administration. In this preliminary design, only critical longitudinal static stability
requirements were considered. This analysis required calculating the center of gravity (C.G.) location, C.G.
limits, neutral point location, Static Margin, and pitch stiffness.
Center of Gravity
The C.G. location for this aircraft was determined from the different components' weights and moment
arms in relation to the location of the mean aerodynamic chord of the wing. The weights of the wing
planform, fuselage, engines, landing gears, empennage, and all else empty were found from the
comprehensive weight analysis summarized in Table 3.2.
31
The moment arms for each component were dependent on the location of the component to the leading
edge of the mean aerodynamic chord. The weight of the wing and engines, positioned at 55% the length
of the plane, were acting at a moment arm of 40% the mean aerodynamic chord aft the leading edge.
For the landing gear, the nose gear was positioned 15% the length aft the nose of the plane, with the main
gear acting at 85% the total length. The weights of the fuselage and all else were positioned at 50% the
length of the aircraft. For the empennage, the horizontal tail mean aerodynamic center was determined
to be 80.2 feet from the mean aerodynamic center of the wing, and the distance for the vertical tail was
75.3 feet.
Using these moment arms, the C.G. location was found by the ratio of each component's weight and
moment arm product over the total weight of the aircraft. Therefore, the C.G. location was 4.2 feet aft the
leading edge of the mean aerodynamic chord, or 23% the chord. To see the MATLAB script used to find the
C.G location, refer to Appendix C.
Neutral Point
𝑑𝐶
The neutral point is effectively the aerodynamic center of the entire aircraft or where 𝑚 = 0. Therefore,
𝑑𝛼
the neutral point is where the pitch moment of the entire aircraft does not change when the angle of attack
changes. For most aircraft, the neutral point location in respect to the center of gravity is critical for a design
𝑑𝐶
because it relates to a very important stability derivative which also happens to be 𝑚 or the “pitch
𝑑𝛼
stiffness”.
How the neutral point was found is by virtue of summing the longitudinal moments about the aircraft,
differentiating them with respect to 𝛼, setting the result equal to 0, and finally solve for the X-location of
the neutral point. This process gives the result found in Equation 6.1. For this design, Equation 6.1 yields a
neutral point location that is 6 Ft. aft of the leading of the mean aerodynamic chord, or 33% of the MAC.
Xnp =
S
1 d
Xacw CLα HT CLαHT +
[Mfuselage +Mnacelles ]
S
qSdα
S
CLα + HT ClαHT
S
(6.1)
The moments of the fuselage and nacelles derivatives with respect to the angle of attack were found using
two simple derived formulas that were provided. They are Equations 6.2 and 6.3, for the fuselage and
nacelle respectively.
Cmαfuselage =Kf
Cmα
W2fusemax Lfuse
Sc̅
2∀
nacelles = sc̅
(6.2)
(6.3)
The length and width and wing characteristics are variables of the fuselage pitch moment derivative. The
K f coefficient is a factor from NASA Technical Report TR711 Figure 8, and estimated to be 1.6 for this design.
For the nacelles the ∀ parameter is the enclosed volume of the nacelle assembly.
32
6.2 Stability Requirements
After finding the neutral point location, the next step is to calculate the static margin, which is defined as
the distance per percent chord of the center of gravity and the neutral point, see Equation 6.4. Usual values
of static margin for a transport aircraft are between 4% and 7% (Nicolai 616). This design’s calculated static
margin is equal to 9.8%, which is slightly higher than expected, and translates to a more stable, but less
maneuverable aircraft. In future iterations it will be desired to reduce the static margin to increase aircraft
maneuverability.
SM = (hn − h)
(6.4)
Cmα =CLα (h−hn)
(6.5)
The real problem with the C.G and neutral point locations are that they are believed to be too far forward
on the MAC. Typical values usually range between 25-35% of the MAC for the C.G. This is because the
aircraft wants to be almost balanced about the wing aerodynamic center so that when the C.G moves it
does so only slightly.
The pitch stiffness is proportional to negative static margin, and is shown in Equation 6.5. For stability, the
pitch stiffness must be negative along with a positive Cm0 to allow the aircraft to be stable and trim-able.
Thus a positive static margin is required for stability, and ultimately the C.G may not exceed the neutral
point along the MAC.
Center of Gravity Limits
The Limits for the center of gravity forward and aft of the aircraft can be determined by two criteria. The
simpler of the two is the aft C.G. limit, which is equal to the neutral point location of the aircraft, and does
not change. This is to retain a positive static margin for longitudinal stability requirements. Therefore h may
not exceed 33% of the mean aerodynamic chord. .
The forward limit was approximated by using a free-body diagram approach, where the C.G was moved
forward and the lift forces to trim were calculated. Once the lift force on the horizontal tail became
extremely large, so much that it obviously must be enlarged in order to produce such a force that was the
determined forward limit of the C.G . This was about -10% of the mean aerodynamic chord, where the
negative percentage means it in in fact forward of the wing. Figure 6.1 depicts these limits along with the
center of gravity and neutral point locations.
33
Figure 6.1 Center of Gravity, Neutral Point and C.G. Limits at Maximum Takeoff Weight
In the graphic above, 6.1, the red dashed lines indicated the forward and aft limits of the center of gravity.
These seem very close, in fact the C.G is within 2 Ft. of the aft limit, and 6 Ft. of the forward limit. However,
by keeping the C.G on top of the wing and almost centered on the entire aircraft, it should not move as
much as the fuel is depleted. Also, this aircraft would include a water pump system to control the C.G
movement as the fuel (almost a third of the weight) is removed.
7.0 Cost Estimates
Table 7.1 Cost Estimation Breakdown for 100 Aircraft Produced (2 Test Aircraft)
2015 Calculations
Values
Engineering ($/hr)
132
Engineering
2954762015
Development Support (With Inflation)
446147015
Flight Testing (With Inflation)
43396196
Tooling ($/hr)
143
Tooling
1856700398
Manufacturing ($/hr, 2015)
114
Manufacturing
4747049450
34
Quality Control
360775758
Material and Equipment
2108005139
Engine and Avionics
2574194
Total Cost
12519410166
Cost per Aircraft
122739315
Table 7.2 Production Cost with varying Aircraft Quantity
Number of Aircraft Produced
Production Cost ($)
Cost per Aircraft ($)
100
12,519,410,166
122,739,315
500
28,864,315,492
57,498,636
1000
43,343,200,034
43,256,686
The production cost estimation of the aircraft was calculated using the DAPCA IV Equations, taken from
Nicolai and Carichner chapter 24. Items such as engineering, tooling, manufacturing, and materials have all
experienced increases in “wrap rate” since 1998.
To estimate this increase, Figure 24.4 from Nicolai and Carichner was utilized. Taking the linear curve fits,
and entering the year 2015, the adjusted rates were able to be estimated reasonably. In doing so, it was
apparent that the most expensive item, in terms of rates, was tooling. Fortunately this is a nonrecurring
cost, and therefore does not drive the cost significantly higher in the long term. Factors such as
development support and flight testing have been adjusted using an inflation rate of 1.46%, taken from a
basic inflation calculator.
In evaluating the cost in the long term, for larger orders such as 1000 production aircraft, it became
apparent that the cost per aircraft largely decreases as compared to producing only 100 aircraft. This can
be attributed to the nonrecurring costs such as engineering and tooling, as previously stated. In the end,
the most significant factor which drives the cost of production for the aircraft is ultimately manufacturing
cost. Manufacturing cost consumes an increasingly large percent of production cost as the number of
orders increases. It is to be noted that “fudge factors” have not been included in this cost estimation, as it
is apparent that various different composites are used in highly specific locations on the aircraft, something
outside the scope of this preliminary design. With this said, the cost per aircraft will be inherently higher
than the given aluminum predictions seen above.
35
Concluding Statements
Team 2 started this design almost three months ago. It is fulfilling to have closed out the preliminary
design phase with this report. This preliminary design served as a learning experience for seven young
engineers pursuing undergraduate degrees in Aerospace Engineering. Much of the figures and values of
the design presented in this report went through several iterations until finalizing the design. The project
began with numerous assumptions, many which would need to be corrected and have been since, but for
the means of this project, they were the only possible direction to take to begin the calculations.
There is still, however, much more work to be done and a more detailed design is needed in order to
have a flyable aircraft. In this detailed design, some of the prominent aircraft parameters will change such
as the gross takeoff weight, length, width, and wing area of the aircraft. It is understood by Team 2 that
the aircraft in its current state is most likely not at its optimum design, and has potential to be more
efficient in several areas. One such area is the engine sizing; it is a general consensus among the group
that the engines are too large for the mission requirements, but were chosen because no others in the
market fit the design better.
Very important aspects of the design vehicle not taken into any consideration are the control surfaces:
elevators, rudder, and ailerons. These will be critical systems to incorporate into the aircraft and could
potentially require changes in the tail and wing design.
Despite these concerns, Team 2 believes strongly in this concept aircraft and has confidence that it can be
a very efficient aircraft after more design features are implemented. As for the overall experience itself, it
is safe to say Team 2 has obtained a great amount of knowledge about general aircraft design. If starting
the design over again from right now, the design could have taken a slightly different course, and be
overall better due to the newly gained experience of the team.
As it turns out, the most challenging aspect of this project was dealing with assumptions and choosing
parameters that were unknown for the design. Whether the values came from the parent aircraft, the
textbook, or some other source, it always became a question of if these assumptions were valid to use.
There was very limited test data, and much of aircraft parameters are proprietary data held by real
aircraft manufacturers. This was dealt with in any way possible, and usually included one of the sources
listed below. Another frequent concern was determining if calculated values were of reasonable
magnitude for this aircraft. The general design philosophy assumed that it is a mid-size transport,
therefore it should have roughly mid-sized values to other transport aircraft.
One last note on the project challenges, and the most important one, was how to make this design a
competitive solution for a 2025 entry date. As always, it is difficult to predict where the industry will head
towards in the next decade in terms of advances in technology and market demand. These considerations
are truly what drives an aircraft design and determines what makes it marketable.
Middle of Market Team 2 would like to thank Dr. James Coder and teaching assistant Nick Grasser for
their instruction, guidance, time, and consideration throughout the entirety of the development process.
36
References
"AERO - Boeing 787 from the Ground Up." AERO - Boeing 787 from the Ground Up. N.p., n.d.
Web. 20 Oct. 2015.
"Air Cargo ULD Containers: LD-3 Dimensions." SeaRates. N.p., n.d. Web. 12 Dec. 2015.
"Boeing Considers Single, Twin Aisle, Co-development 757/767 Style for next New Airplane –
Leeham News and Comment." Leeham News and Comment. N.p., 02 Nov. 2014. Web.
21 Oct. 2015.
"Boeing Sees Airlines Craving Longest-Ever Narrow-Body Plane." Bloomberg.com. Bloomberg,
n.d. Web. 21 Oct. 2015.
"Boeing 787 Dreamliner: Analysis." Lissys Ltd., 2006. Web. 20 Sept. 2015.
"Circular Economy Could save Billions of Dollars." New Scientist 213.2854 (2012): 5. Web.
"Civil Jet Aircraft Design - Boeing Aircraft." BookSite. Butterworth-Heinemann, 2001. Web. 22 Oct.
2015.
"Delta Seat Maps Boeing 757-200." SeatGuru. Tripadvisor.com, 2015. Web. 22 Oct. 2015.
Fuchte, Joerg, Bjoern Nagel, and Volker Gollnick. "Twin Aisle for Short Range Operations - An
Economically Attractive Alternative?" 12th AIAA Aviation Technology, Integration, and
Operations (ATIO) Conference and 14th AIAA/ISSMO Multidisciplinary Analysis and
Optimization Conference (2012): n. pag. Web.
Gilruth, R. R., and M. D. White. "Analysis and Prediction of Longitudinal Stability of Airplanes." NACA-TR711 (1941): 137-45. Web.
"How Do Winglets Work?" Flite Test. N.p., n.d. Web. 23 Oct. 2015.
"Innovative Materials | Airbus, a Leading Aircraft Manufacturer." Airbus. N.p., n.d. Web. 20
Oct. 2015.
Jenkinson, L., P. Simpkin, and D. Rhodes. "Butterworth-Heinemann - Civil Jet Aircraft Design - Aircraft Data
File - Boeing Aircraft."Butterworth-Heinemann - Civil Jet Aircraft Design - Aircraft Data File Boeing Aircraft. N.p., n.d. Web. 13 Dec. 2015.
Jenkins, Renaldo V., and Acquilla S. Hill. "Aerodynamic Force." Aerodynamic Performance and Pressure
Distributions for a NASA SC(2)-0714 Airfoil (1988): n. pag. Web.
Jenkins, Renaldo V. NASA SC(2)-07 14 Airfoil Data Corrected for Sidewall Boundary-Layer Effects.
Washington, D.C.?: National Aeronautics and Space Administration, 1989. Web.
Kroo, I. "Vki Lecture Series On Innovative Configurations And Advanced Concepts For Future Civil
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Aircraft." NONPLANAR WING CONCEPTS FOR INCREASED AIRCRAFT EFFICIENCY (2005): n. pag.
Aero.stanford.edu. Stanford University. Web.
Liebeck, R.H. "Design of the Blended Wing Body Subsonic Transport." Journal of Aircraft 41.1 (2004): n.
pag. Vicomplex.hu. Web. 22 Oct. 2015.
"Longitudinal Static Stability." Longitudinal Static Stability. N.p., n.d. Web. 23 Nov. 2015.
Mizrahi, Joe. "Blended Wing Body - New Concept in Passenger Aircraft." Twitt.org. N.p., n.d. Web. 23
Oct. 2015.
Nicolai, Leland M., Grant Carichner, and Leland M. Nicolai. Fundamentals of Aircraft and
Airship Design. Reston, VA: American Institute of Aeronautics and Astronautics, 2010.
Print.
Robert Faye Robert Laprete Michael Winter. BLENDED WINGLETS (n.d.): n. pag. Web.
Sadraey, Mohammad. "Chapter 4. Preliminary Design." Dxc.edu. N.p., n.d. Web. 1 Dec. 2015.
"Supercritical Airfoil Design." Integral Equation Method in Transonic Flow Lecture Notes in Physics (1982):
135-49. Web.
"The Future of the Narrowbody Airplane Market." Seeking Alpha. N.p., n.d. Web. 21 Oct. 2015.
"Their New Materials." WSJ. N.p., n.d. Web. 20 Oct. 2015.
"The Spiroid Winglet." Tails Through Time: The Spiroid Winglet. N.p., 07 Aug. 2010. Web. 23 Oct. 2015.
"UIUC Airfoil Data Site." UIUC Applied Aerodynamics Group, 2015. Web. 22 Oct. 2015.
"US Airways Seat Maps Boeing 757-200 (752) V1." SeatGuru. N.p., 2015. Web. 22 Oct. 2015.
"Winglets - A Triumph of Marketing over Reality »." AirInsight. N.p., 10 Apr. 2012. Web. 22
Oct. 2015.
"Winglets: Making Their Presence Felt." (n.d.): n.
pag. Http://www.aviationpartnersboeing.com/pdf/news/at69winglets.pdf. Apr.-May 2004.
Web. Oct. 2015.
38
Appendix
A. Additional Figures and Tables
Detailed Weight Breakdown By Percentage
0.9%
1.3%
Structure
2.5%
2.4%
8.3%
Fuel
17.9%
0.3%
Payload
2.9%
Propulsion
8.5%
Landing Gear
Start System
18.2%
36.8%
Control Surface
Gear
Figure A.1 Comprehensive Weight Analysis
Empty Weight Reduction By Including Advanced Composite Materials
180000
160000
[Weight [lbs..]
140000
120000
100000
80000
Without Composites
60000
With Composites
40000
20000
0
Planform Empennage Fuselage
Landing
Gear
All Else
Empty
Total
Components
Figure A.2 Empty Weight Reductions
39
Sea Level Drag Contributions by Component
Profile
21.2%
25.1%
1.6%
Induced
Trim
5.5%
Tail Profile
Fuselage
46.5%
Figure A.1 Drag Contributions at Sea Level, values taken at L/Dmax.
Cruise Altitude Drag Contribution by Component
18.4%
28.8%
1.8%
Profile
Induced
4.41%
Trim
Tail Profile
Fuselage
46.6%
Figure A.1 Drag Contributions at 35,000 Ft., values taken at L/Dmax.
B. Advanced Composite Materials Trade Study
In researching the benefits and penalties of the use of composites in construction of modern aircraft, three
glaring advantages were recognized; weight savings, maintenance savings, and increased lifespan.
Both Airbus and Boeing discuss the implementation of advanced composites on their new aircraft. Boeing
states that, “The result is an airframe comprising nearly half carbon fiber reinforced plastic and other
composites. This approach offers weight savings on average of 20 percent compared to more conventional
aluminum designs”, when discussing the new 787 Dreamliner in their website magazine. Similarly, Airbus
40
claims to have seen a 20% weight reduction as a result of the new A350 XWB being constructed from 53%
advanced composite materials. From this, the design team feels comfortable making the assertion that if
composites are used in appropriate aspects of construction, the total weight will be able to be significantly
reduced, and thus the fuel consumption as well (if all things are held equal).
Along with weight reduction, advanced composite materials experience a reduced risk of corrosion and
fatigue. Airbus asserts that, “By applying composites on the A350 XWB, Airbus has increased the service
intervals for the aircraft from six years to 12, which significantly reduces maintenance costs for customers.
The high percentage of composites also reduces the need for fatigue-related inspections required on more
traditional aluminum jetliners, and lessens the requirement for corrosion-related maintenance checks.” On
their company website. Boeing has also seen a significant savings on maintenance, particularly on the 777
where the composite tail is 25% larger, yet requires 35% fewer scheduled maintenance labor hours as a
result of reduced risk of corrosion and fatigue of composites compared to metal. When considering the
design from cradle to grave, a benefit such as this is very appealing to the design process, but overall is not
included in the design requirements for this aircraft.
Benefits in maintenance cost will not only lower the overall market price of the new aircraft being designed,
but will also increase the overall lifespan of the aircraft. The use of advanced composites in structural
components that typically see service for fatigue and corrosion repair ultimately leads to a more enduring
vehicle. While the increased lifespan is not a design requirement, it certainly adds to the argument against
composites: the cost. Certain composite materials can be extremely expensive, and often not worth the
investment for the potential savings. The Wall Street Journal reported in 2013 that, “Robert Schafrik,
general manager of materials and process engineering at GE Aviation, jokes that an advanced material is
"one that costs 10 times as much as the current material." If that can be cut to just twice the price, he says,
other savings in weight and maintenance costs can offset the difference.”
Another issue adding to the cost of composites is the difficulty of production. As stated in the Boeing
website magazine, “Boeing and its suppliers have also struggled with the Dreamliner's carbon-fiber and
polymer structures. The plane's fuselage, for example, consists of barrels made from tape wound around
cylindrical molds and then baked. Mastering the process was tough, and several of the first sections
produced were rejected. In 2009, Boeing conceded it had to fix wrinkles in the skin of the first 23 barrels.”
In making light of this, the team recognized that the actual determination of where and when advanced
composite materials should be used is just as important as their actual implementation. Both Boeing and
Airbus stress this when discussing these new materials, and also make light of the advancements in
replacing aluminum with titanium and other new alloys.
After conducting this trade study, it becomes clear that the best choice of material four our
design is advanced composites. For the scope of this design project and provided the design parameters, it
is quite obvious that a primarily advanced composite construction would be greatly beneficial in reducing
the aircraft weight and ultimately reducing fuel burn. While the implementation of these materials may be
costly, the design requirements do not stress cost as a restriction. Inherent concern obviously exists, but if
the material placement is properly and carefully done, the design team feels that overall composites will
significantly benefit in the design of an aircraft to replace the Boeing 757.
41
C. MATLAB Scripts
Drag Build Up Calculations and Plotter
%DRAG BUILD_UP
%M-o-M TEAM 2
clear
clc
%Sea Level Velocity Points
Vsl=[177 202 227 252 277 302 327 352 377 402 427 452 460 490 520 550];
%Cruise Alt. Velocity Points
Vc=[427 452 477 502 527 552 577 602 627 652 677 702 727 752 777 802];
%A/C Parameters
rho=.002378;
rho2=.000738;
qsl=0.5*rho*(Vsl.^2);
qc=0.5*rho2*(Vc.^2);
e=.98;
AR=9;
k=1/(pi()*e*AR);
W_sl=281005;
W_c=226758;
S=2240;
%==========================================================================
%Induced drag==============================================================
Cl_sl=W_sl./(qsl*S);
Cdi_sl=(Cl_sl.^2)*k;
Di_sl=qsl.*S.*Cdi_sl;
Cl_c=W_c./(qc*S);
Cdi_c=(Cl_c.^2)*k;
Di_c=qc.*S.*Cdi_c;
%==========================================================================
%Profile drag==============================================================
%XFOIL drag coeff's
Cdp_sl=[.02438 .01145 .00899 .00768 .00692 .0068 .0068 .00669 .00666 .0066
.00653 .00656 .00647 .00643 .00638 .007];
Cdp_c=[.008 .008 .008 .008 .008 .008 .0082 .0084 .0089 .0093 .0099 .01 .0115
.012 .013 .014];
Dp_sl=0.5*(rho)*(Vsl.^2)*S.*Cdp_sl;
Dp_c=0.5*(rho2)*(Vc.^2)*S.*Cdp_c;
%==========================================================================
%Fuselage drag=============================================================
%X locations along fuselage
X=[0.0000 6.5970 11.9503 18.9469 25.2194 126.3652 140.0341 154.7051
169.1080];
42
Sf=6294.026059; %Total Surface Area of Fuselage
msl=3.737e-7;
mc=2.995e-7;
%SEA LEVEL----------------------------------------------------------------Re_sl=zeros(9,16);
i=1;
j=1;
for i=1:16
%Reynolds number based on X location
for j=1:9
Re_sl(j,i)=(rho*Vsl(i)*X(j))/msl;
end
end
Cf_sl=zeros(9,16);
m=1;
n=1;
for m=1:16
%Coefficient of Friction
for n=2:9
if (Re_sl(n,m) < 500000)
Cf_sl(n,m)=1.328*(Re_sl(n,m)^(-0.5));
else
Cf_sl(n,m)=0.455*((log(Re_sl(n,m))^(-2.58)));
end
end
end
Dfsl=zeros(1,16);
y=1;
for y=1:16
Dfsl(y)=qsl(y)*Sf*sum(Cf_sl(:,y)); %Sea Level Fuselage Drag
end
%CRUISE ALTITUDE----------------------------------------------------------Re_c=zeros(9,16);
i=1;
j=1;
for i=1:16
%Reynolds Number based on X location
for j=1:9
Re_c(j,i)=(rho2*Vc(i)*X(j))/mc;
end
43
end
Cf_c=zeros(9,16);
m=1;
n=1;
for m=1:16
%Coefficient Friction
for n=2:9
if (Re_c(n,m) < 500000)
Cf_c(n,m)=1.328*(Re_c(n,m)^(-0.5));
else
Cf_c(n,m)=0.455*((log(Re_c(n,m))^(-2.58)));
end
end
end
Dfc=zeros(1,16); %Cruise Fuselage Drag
y=1;
for y=1:16
Dfc(y)=qc(y)*Sf*sum(Cf_c(:,y));
end
%==========================================================================
%Trim drag ================================================================
%Tail Parameters
St=561;
ARt=4.5;
et=0.97;
lt=80.23;
c_bar=18.11;
x=4.2086-.25*c_bar;
Clt_sl=(S/(St*lt))*(15*c_bar+x*Cl_sl);
Clt_c=(S/(St*lt))*(15*c_bar+x*Cl_c);
Cd_trim_sl=(St/S).*(Clt_sl./Cl_sl).*(((Clt_sl*e*AR)./(Cl_sl*et*ARt))2).*Cdi_sl;
Cd_trim_c=(St/S).*(Clt_c./Cl_c).*(((Clt_c*e*AR)./(Cl_c*et*ARt))-2).*Cdi_c;
Dtrim_sl=qsl.*Cd_trim_sl;
Dtrim_c=qc.*Cd_trim_c;
%==========================================================================
%Tail drag ================================================================
Cdt_sl=Cdp_sl*St/S;
Cdt_c=Cdp_c*St/S;
Dt_sl=0.5*rho*(Vsl.^2).*Cdt_sl*St;
Dt_c=0.5*rho2*(Vc.^2).*Cdt_c*St;
%==========================================================================
44
%TOTAL DRAG ===============================================================
Dtot_c=Dp_c+Dfc+Di_c+Dt_c+Dtrim_c;
figure
hold on
plot(Vc,Dp_c,Vc,Dfc,Vc,Di_c,Vc,Dt_c,Vc,Dtrim_c,Vc,Dtot_c,'k');
title('Drag Build up: Cruise Altitude=35000 ft.')
xlabel('Airspeed [Ft/s]');
ylabel('Drag Force [lbs.]');
legend('Profile','Fuselage','Induced','Tail Profile','Trim',
'Total','location', 'eastoutside')
hold off
Dtot_sl=Dp_sl+Dfsl+Di_sl+Dt_sl+Dtrim_sl;
figure(2)
hold on
plot(Vsl,Dp_sl,Vsl,Dfsl,Vsl,Di_sl,Vsl,Dt_sl,Vsl,Dtrim_sl,Vsl,Dtot_sl,'k');
title('Drag Build up: Sea Level Altitude')
xlabel('Airspeed [Ft/s]');
ylabel('Drag Force [lbs.]');
legend('Profile','Fuselage','Induced','Tail Profile','Trim','Total')
hold off
Engine Sizing Calculations and Constraint Diagram
% Propulsion system sizing / constraint diagram
% M-o-M Team 2
clear
clc
%% variables for thrust, weight, take off, and cruise parameters
mu = 0.05;
ws = [20:140];
sto = 7800;
cdg = -0.3;
cd0 = 0.018;
clr = 0.325;
rho = 0.002377;
g = 32.174;
w = 281005;
s = 2213;
rho_cr = 0.000738;
e0 = 0.6;
ar = 9;
k = 1 / (pi*e0*ar);
i = 0;
j = 0;
q = 0;
vcr = 774.165333333;
h = 42000;
roc = 100/60;
ldmax = 17;
vst = 177.7838462;
clmax = 2.82;
45
%% while loop to find the relation of the takeoff weight to the wing loading
while i < 121
i = i +1;
tw_num(i) = mu - ((mu+(cdg/clr))*(exp(0.6*rho*g*sto*cdg/ws(i))));
tw_den(i) = 1 - exp(0.6*rho*g*cdg*sto/ws(i));
tw(i) = tw_num(i)/tw_den(i);
end
%% while loop to relate the cruise thrust to weight to take off thrust to
weight and wing loading
sigma = rho / rho_cr;
a = rho*cd0/2;
b = (2*k) / (rho_cr*sigma);
while j < 121
j = j+1;
tw_cr(j) = (a*vcr^2/ws(j))+(b*ws(j)/vcr^2);
end
%% while loop to relate the service ceiling thrust to weight to take off
thrust to weight and wing loading
sigmac = 0.2967*exp(1.7355-(h*4.8075*(10^-5)));
rho_sc = rho*sigmac;
while q < 121
q = q+1;
tw_sc1(q) = roc / (sigmac*sqrt(2*ws(q)*sqrt(k/cd0)/rho_sc));
tw_sc2(q) = 1 / (sigmac*ldmax);
tw_sc(q) = tw_sc1(q) + tw_sc2(q);
end
ws_st = 0.5*rho*vst^2*clmax;
tw_stall = linspace(0,1.4,100);
ws_stall = linspace(ws_st,ws_st,100);
plot(ws,tw,ws,tw_cr,ws,tw_sc,ws_stall,tw_stall)
title('Constraint Diagram')
xlabel('Wing Loading at Take off (W/S)')
ylabel('Thrust to Weight Ratio at Take off (T/W)')
legend('Takeoff','Cruise','Service Ceiling','Stall')
46