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Justin Oyas – MAE 159 Final Report 1

MAE 159: Aircraft Design – Final Report

Author: Justin Oyas

June 3, 2016

University of California, Irvine

Justin Oyas – MAE 159 Final Report 2 Table of Contents

1) Introduction ................................................................................................................................................................ 5

2) Design Specifications ................................................................................................................................................. 5

3) Design Analysis ......................................................................................................................................................... 6

A. Aspect Ratio .................................................................................................................................................. 6

B. Wing Sweep Angle ....................................................................................................................................... 7

C. Engine Criteria .............................................................................................................................................. 7

D. Seat Configuration ........................................................................................................................................ 8

E. Conventional vs Supercritical Airfoils .......................................................................................................... 9

F. Advanced Technology ................................................................................................................................ 10

i. D.O.C vs Risk ................................................................................................................................ 10

G. Optimum Designs ....................................................................................................................................... 11

H. Final Design Choice ...................................................................................................................................... 3

4) Summary & Conclusion ........................................................................................................................................... 13

A. Comparison of 1970 vs Current Technology .............................................................................................. 13

B. Recommendations for future technologies that will have the greatest impact on commercial airplane

performance ............................................................................................................................................... 13

5) Description of Configurations .................................................................................................................................. 15

6) Aircraft Configuration Drawings ............................................................................................................................. 17

7) References ................................................................................................................................................................ 30


List of Figures

1) D.O.C Ton/Mile vs. Sweep at different aspect ratios ............................................................................................... 6

2) Optimal Wing Sweep Angle – Aluminum & Composite JT9D ............................................................................... 7

3) D.O.C vs Sweep with different number of engines .................................................................................................. 8

4) D.O.C Ton/Mile vs Seats Abreast ............................................................................................................................. 9

5) Conventional vs Super Critical Airfoils .................................................................................................................. 10

6) Aluminum vs Composite ........................................................................................................................................ 11

7) Current Technology Aircraft Drawings .................................................................................................................. 18

8) 1970s Technology Aircraft Drawings ..................................................................................................................... 23


List of Tables

1) Design Specification - Current Technology Airplane ............................................................................................... 5

2) Design Specification - 1970's Technology Airplane ................................................................................................. 5

3) JT9D-7 vs JT8D-9 ..................................................................................................................................................... 7

4) Optimal Design Choices ......................................................................................................................................... 11

5) Final Optimum Airplane Design Specifications ..................................................................................................... 12

6) 1970's vs Optimal Design ....................................................................................................................................... 13

7) Aircraft Configurations ............................................................................................................................................ 16


1. Introduction

The objective of the MAE 159 final report is to study the cost and performance of a subsonic

commercial transport and to choose the optimum design based on the mission requirements.

The optimal design will be supplemented with 3-view drawings of the aircraft configuration

once the final sizing values are obtained. Design specifications are given for an airplane with

current technology and an airplane with 1970s technology. The report will cover the design

analysis, which describes the selection process of preliminary designs and how parameters were

varied for the final design. The report will include the selection of trial designs, parameter

iterations, justification for the use of advanced technology (D.O.C vs Risk), and the final

specifications. A recommendation section will be provided for future technologies as to how

they will make have the greatest impact on commercial airplane performance.

The an optimal 1970s airplane design will be compared with an optimum airplane that uses

current technology and supplemented with their corresponding 3 view drawings. To account for

the various parameters in aircraft design, a MATLAB program will be used to study the cost

and performance. Parameters such as aspect ratio, wing sweep angle, and number of engines are

iterated through the Matlab program for design optimization. Furthermore, the program will

locate the center of gravity for various components as well as the overall airplane center of

gravity. Solidworks will then be utilized to model the final designs.

2. Design Specifications

Current Technology Airplane Number of Passengers (2-class domestic rules) 110

Weight of cargo (10 pounds/ft3) 2000

Range (Still Air) 2500

Takeoff Field Length 6000 feet

Landing Approach Speed (Design Payload plus

100% max fuel)

130 knots

Cruise Mach number 0.80

Initial Cruise Altitude 35,000 ft

Takeoff conditions Sea-Level - Hot Day (84°F) Table 1- Design Specification - Current Technology Airplane

1970’s Technology Airplane – Same Specifications with following

restrictions: Cruise Mach number 0.75

Engine Type Pratt & Whitney JT8D-9

Turbofan

Structure Aluminum

Airfoil Conventional Table 2- Design Specification - 1970's Technology Airplane


3. Design Analysis

The objective of the report is to design a commercial airplane with the lowest achievable

direct operating cost that meets the given specifications. Parameters such as Aspect Ratio,

Sweep Angle, Airfoil Type, Engine Criteria, Seats Abreast, Number of Aisle, and Advanced

Technology are iterated in the Matlab code to attain several optimum designs. After iterating

through the parameters, a comparison of several trial designs will be compared in ordered to

choose the optimum aircraft design.

A. Aspect Ratio – Aspect ratio is important as it directly affects the induced drag of an

aircraft which contributes to the overall drag of the design. Induced Drag is expressed in

Equation 1:

𝐶𝐷𝑖 =𝐶𝐿

2

𝜋𝐴𝑅𝑒 Eqn. 1

It is favorable to have a larger aspect ratio since induced drag will become smaller when

the aspect ratio increases; therefore, total drag will decrease for the overall airplane.

Increasing the aspect ratio to an optimum value is critical; however, it reaches a certain

point where the large aspect ratio contributes heavily on the weight. This increase raises

the direct operating costs. Larger wing span also jeopardizes the structural integrity as the

wings experience large bending moments and forces. Figure 1 below shows that as aspect

ratio increases, the direct operating cost increases as well; furthermore, from this

configuration, the optimum aspect ratio is 7, as it achieves the lowest possible operating

costs at several sweep angles.

Figure 1 - D.O.C Ton/Mile vs. Sweep at different aspect ratios.

0.168

0.17

0.172

0.174

0.176

0.178

0.18

0.182

0.184

0.186

0 10 20 30 40 50

D.O

.C T

on

/Mile

Sweep L

Aluminum JT9D - Current Technology(Abreast=4, Aisle=1, Engines=2)

AR 6 AR 7 AR 8 AR 9 AR 10 AR 11


B. Wing Sweep Angle – Sweeping an aircraft’s wing is extremely important as it can delay

the rise in drag due to compressibility effects which enhances the aircraft’s capability and

performance. In Figure 2, it can be seen that after a sweepback of 30°, the direct operating

cost sharply increases and the sweepback is not desirable; the reason for the sharp increase

is due to aeroelastic problems as stated in the manual.

C. Engine Criteria – The type of engine, the placement of the engines, as well as the number

of engines used are major factors for the design of a commercial subsonic transport. There

are two different types of engines that are factored into this design analysis which are the

Pratt & Whitney JT9D-7 & the older Pratt & Whitney JT8D-9. A comparison of the

engine specifications is show in Table 3.

Engine Specs JT9D-7 Turbofan Engine JT8D-9

Thrust (lb) SFC

(lb/hr/lb) Thrust (lb)

SFC

(lb/hr/lb)

Takeoff Thrust 45500 0.355 14500 .595

Max Continuous Thrust 38500 0.337 12600 .570

Maximum Climb Thrust 38500 0.337 12600 .570

Maximum Cruise Thrust 35500 0.332 14000 .555

Dry Weight (With Standard

Equipment) 8770 lbs 3218 lbs

Table 3 - JT9D-7 vs JT8D-9

0.18

0.181

0.182

0.183

0.184

0.185

0.186

0.187

0 10 20 30 40 50

D.O

.C. T

on

/Mile

Sweep L

Aluminum JT9D - Current Technology(Abreast=6, Aisle=1, Engines=2)

AR 6 AR 7

0.148

0.1485

0.149

0.1495

0.15

0.1505

0.151

0.1515

0.152

0 10 20 30 40 50D

.O.C

To

n/M

ile

Sweep L

Composite JT9D - Current Technology(Abreast=4, Aisle=1, Engines=3)

AR 6 AR 7

Figure 2- Optimal Wing Sweep Angle – Aluminum & Composite JT9D


For the purposes of this design, the engines will be placed on the wing for which the

configuration’s location factors are accounted for in the weight section. As seen in Figure 3, as

the number of engines increase from 2 to 4 engines on the plane, the weight and direct operating

cost increase. It is optimal to choose the 2 engine configuration for the iterated designs.

Figure 3- D.O.C vs Sweep with different number of engines

D. Seat Configuration – The number of seats abreast and the number of aisle were iterated

though the program which had a direct impact on the length and diameter of the fuselage,

which impacted the overall weight of the aircraft. A general pattern was found which was

that 2 aisles increased the direct operating cost. For the required specifications, it is clear

that having 1 aisle in the optimum design choice as it reduces the cost. As for the number

of seats abreast increasing, the direct operating cost increases due to the expansion of

fuselage length and diameter and its effect on weight. From Figure 4, the 4 seats abreast

configuration had the lowest direct operating cost; however, for the purposes of this

design, 5 seats abreast will be chosen for the consideration of fuselage and airplane layout.

0.146

0.148

0.15

0.152

0.154

0.156

0.158

0.16

0.162

0.164

0 5 10 15 20 25 30 35 40 45

D.O

.C T

on

/Mile

Sweep L

Conventional, JT9D, AR = 8

2 Engines 3 Engines 4 Engines


Figure 4- D.O.C Ton/Mile vs Seats Abreast

E. Conventional vs Supercritical Airfoils – According to Shevell, Supercritical Airfoils are

airfoils that provide a higher MDIV than conventional airfoils and its geometry allows the

reduction of the high adverse pressure gradient at the rear and permits the aft cambering

without excessive pressure drag. Supercritical wings can allow greater speed or less sweep

at a given thickness configuration which reduces wing weight. Compared to the

conventional airfoil which are comprised of the NACA 4,5, and 6 series, supercritical

airfoils prove to be superior as they reduce weight which in turn reduces direct operating

cost. In Figure 5, these parameters were iterated and the data found was parallel to that of

the descriptions given by Shevell. It is therefore a better design choice to use supercritical

airfoils for the airplane design.

0.14

0.145

0.15

0.155

0.16

0.165

0.17

0 1 2 3 4 5 6 7 8 9

D.O

.C T

on

/Mile

Seats Abreast

Direct Operating Cost vs Seats AbreastAR=8 , Sweep 25, JT9D 2 Engines

Aisle=1 Aisle=2


F. Advanced Technology – When designing airplane, advanced technology such as

composites can significantly improve the direct operating cost when compared to more

common aircraft material such as aluminum or titanium and their respective alloys. In

Figure 6, an aluminum and composite based designed are compared and it can be seen that

the composite material reduces the overall direct operating cost.

i. DOC VS Risk – Composites have outstanding material properties as they are

lightweight have a very high stiffness-to-weight ratio; however, due to composites

being made of a matrix which is reinforced by fiber, failure modes are needed to

be extensively studied [2]. Composites are often made of “different ply layers into

a laminate structure...[and] can "delaminate" between layers where they are

weaker” [2]. One cause of delamination is the “Out-of-plane loads perpendicular to

the layers;” therefore, designers have to be aware of all of the potential loads and

paths in the structure to avoid failure. Given that extensive research and testing on

the composite material and out of plane loads are accounted for, it is greatly

beneficial to apply composite technology to aircraft.

0.155

0.16

0.165

0.17

0.175

0.18

0 5 10 15 20 25 30 35 40 45

D.O

.C T

on

/Mile

Sweep L

Conventional vs Supercritical Aluminum, JT9D, 2 Engine, 4 Abreast, 1 Aisle

Conv AR 7 SuperC AR 6 SuperC AR 7

SuperC AR 8 SuperC AR 9 SuperC AR 10

Figure 5 - Conventional vs Super Critical Airfoils.


Figure 6- Aluminum vs Composite

G. Optimum Designs – The optimal design choices reduced down to 5 optimal planes with

various fixed parameters such as number of engines (2), number of seats abreast (5), and

number of aisle (1) and aspect ratio (7). The rest of the parameters vary with each type of

design. The optimal design parameters are listed below in Table 4.

Design Material Airfoil Type Sweep Engine

Weight

Takeoff

(lbs)

D.O.C

Ton/Mile

D.O.C

Pass/Mile

Range

(N.m)

1 Aluminum Conventional 25° JT9D 177376.88 0.1780 0.0208 3062.4

2 Aluminum Supercritical 35° JT9D 162033.24 0.1673 0.0195 3066.0

3 Composite Conventional 25° JT9D 149694.83 0.1588 0.0185 3068.2

4 Composite Supercritical 35° JT9D 127369.71 0.1425 0.0166 3063.1

1970’s Aluminum Conventional 25° JT8D 170215.55 0.1827 0.0213 3041.8

Table 4 - Optimal Design Choices

0.14

0.145

0.15

0.155

0.16

0.165

0.17

0 5 10 15 20 25 30 35 40

D.O

.C T

on

/Mile

Sweep L

Aluminum vs CompositeAR=7, Supercritical, JT9D, 2 Engine, 4 Abreast, 1 Aisle

Composites Aluminim


H. Final Design Choice – In Subsection G, five optimal airplane designs were iterated and

tabulated. All of these airplanes met the required specifications that were initially given.

From an economical perspective, the design with lowest direct operating cost is airplane

design 4, which is the final optimum airplane design. The final optimized 1970’s airplane

specifications are also calculated and compared to with design 4. Detailed specifications

are listed below in Table 5.

Specifications 1970’s Technology

Design

Final Optimum Airplane

Design (Current)

Wing Specifications

Sweep 25° 35°

Airfoil Type Conventional Supercritical

Aspect Ratio 7 7

Wing Area 1737.8 ft 1195.4 ft

Wing Span 110.29 ft 91.47 ft

Taper Ratio 0.35 0.35

Thickness-Chord Ratio

(t/c)

.1293 0.2119

Fuselage Specifications

Number of Aisles 1 1

Number of Seats

Abreast

5 5

Fuselage Diameter 12.46 ft 12.46 ft

Fuselage Length 127.51 ft 127.51 ft

Engine Specifications

Engine Type JT8D-9 JT9D-7 Turbofan Engine

Number of Engines 2 2

Engine Thrust 41,305 lbs 31,666 lbs

Engine Diameter 6.04 ft2 7.07 ft2

Engine Length 10 ft 10.6 ft

Engine Technology Low Bypass,

Standard Equipment

Higher Bypass, Standard

Equipment

Advance Technology

Material Aluminum Composites

Weight Specifications

Weight Takeoff 170215.55 lbs 134857.86 lbs

Weight Fuel 46,173 lbs 30,977 lbs

Weight Payload 25,650 lbs 25,650 lbs

Direct Operating Cost

D.O.C 0.1827 $/ton mile 0.1492 $/ton mile

D.O.C pass/mile 0.0213 $/pass mile 0.0174 $/pass mile Table 5- Final Optimum Airplane Design Specifications


Figure 7- Payload vs Range

4. Summary & Conclusion

A. Comparison of 1970 vs Current Technology – As shown in Table 4, the 1970’s

technology plane has the highest direct operating cost compared to the Current

Technology airplanes (Design 1-4). When compared a current aluminum aircraft with

a conventional airfoil (Design 1), the values are fairly close with the exception of

D.O.C as it is much higher. Supercritical airfoils, higher bypass engines, and

composite materials have truly revolutionized the aircraft industry as weight and

direct operating cost values have significantly improved. The final chosen design vs

the 1970’s design is compared below in Table 6 to exemplify some of the

improvements from advanced technology. Similar to Boeing’s 787, which makes

greater use of composite material, the overall aircraft has reduced about 20 % of its

weight [1].

1970’s Technology

Design

Current Technology

Optimized Design

Percent Change

(Improvement)

Weight Takeoff 170215.55 134857.86 lbs 20.770 %

Weight Fuel 46,173 lbs 30,977 lbs 32.900 %

D.O.C $/ pass

mile 0.0213 $/pass mile 0.0174 $/pass mile 18.309%

Table 6 - 1970's vs Optimal Design

B. Recommendations for future technologies that will have the greatest impact on

commercial airplane performance – The next generation of commercial airplanes will

need to use lighter and stronger materials in order to expand aspect ratios for drag

reduction and reduce weight for aircraft performance. The higher the aspect ratio, the

0

5000

10000

15000

20000

25000

30000

35000

0 5000 10000 15000 20000 25000 30000 35000 40000

Pay

load

(lb

s)

Range (Nautical Miles)

Payload vs. Range

Current

1970s


greater the bending moment and forces on the wing structures; therefore, composites

are an ideal for such scenarios. Moreover, similar to the Boeing 787, aircraft designers

will need to incorporate composites all throughout the aircraft in order to significantly

reduce weight. In addition, jet engine technology will have to evolve and make use of

composites and nanocomposite technology. Rolls Royce recently installed a jet engine

that uses Titanium Carbide composite technology that reduces up to 1,500 lb of the

total aircraft. Since the engines are lighter, there will be less weight on wing structure

and the overall aircraft [3]. One negative aspect of composites is that the data available

is not comprehensive compared to aluminum and titanium which are typically used in

aircraft. For the future of aerospace and aircraft technologies to improve, thorough

research will need to be conducted on composites on areas of failure, fatigue, life

cycle and delamination. It is up to the next generation of material scientists and

engineers to improve composites and nanocomposites for aerospace applications. The

research of other materials and combinations of materials that provide high strength

properties and high temperature properties will also be a critical aspect as many

materials used in aerospace are subjected to extreme conditions.

5. Descriptions of Configuration

In the previous sections, the final sizing values were determined through a MATLAB program

for the current airplane technology design as well as the 1970’s airplane design. With the

preliminary sizing values, placement of the horizontal and vertical tail and their corresponding

geometries were configured such that the center of gravity would fall within ± 10% from the

mean aerodynamic chord. Conditions are examined in order to configure the wing and tail

placement for both 1970s and Current Tech Airplane: Fuel, No Fuel, Cargo, No Cargo. For a

commercial aircraft, the fuselage is made up of 3 distinct sections: Constant Section Passenger

Compartment, Nose Section, and Afterbody Section or tail cone (Schaufele).

5.1. Nose Section

5.1.1. The nose section must be designed to accommodate for the crew seats, the

instrument panel, glare shield, windows and a variety of levers, knobs, and pedals.

The crew must be able to comfortably reach and operate all the instruments instilled

in the crew station. The pilot must be able to see other aircraft as well as slightly

below his path flight; therefore, the Pilot Eye Position must be within 18°-21° for a

jet transport. Integrating the requirements along with the aerodynamic requirements

for low drag, the nose shape requires an average value of (Lnose/Dfuselage) to be 1.48

from the top and side.

5.2. Interior Configuration


5.2.1. The length of the passenger compartment must be sufficient to accommodate the

designed passenger payload. Seat pitch, entrance and exit doors, galley space,

lavatory space, and coat room space are taken into consideration. Details for such

specifications are defined by the Federal Air Regulations.

(a) Assume 6.5 square feet of cabin floor area per passenger for all-economy class

seating and short range (excludes the cockpit)

(b) Passenger Weight is 180lb plus 40 lb baggage which totals 220 lb

(c) Allow 5 cubic feet per passenger for under-floor storage

(d) Lavatory: 1 lav (1st class) & 50 passengers/lavs in (Econ Class)

(e) Galley: Height and width are required & 15” per cart (wall thickness &

clearance are included)

(f) Galley-Carts/Passenger: .30 carts/pass (1st Class) & .075 carts/pass (Econ Class)

(g) Coat Room: 1.5 in/pass (1st Class) & 0 in/pass

5.3. Aft Body Section Configuration

5.3.1. The aft fuselage section is to be designed with the conflicting requirements of

aerodynamics, structural weight, pitch attitude during take of or landing, and the

upsweep angles that minimized drag penalties. In order to meet such requirements,

the average value of the top (Laft/Dfuselage) is 2.35, the side (Laft/Dfuselage) is 3.06, and

the average upsweep is 4.75°

5.4. Fuel and Payload Configuration

5.4.1. Single Row LD-W containers are used for the fuel box

5.4.2. Significant amount of fuel (60% of chord) is placed in the main wings, and the left

over amount of fuel is placed in the LD-W containers which are located right ahead

of the main wings (1970s Design)

5.4.3. Due to the supercritical airfoil shape, amount of fuel needed fits spaciously in the

main wing; therefore, fuel box containers are not needed (Current Design)

5.5. Engine Placement

5.5.1. 33% Percent of the Half Span

5.5.2. In Front of the Leading Edge

5.5.3. Tilted inward (2°)

5.6. Landing Gear Configuration (Manual/Schaufele).

5.6.1. Jet Transport Configuration

5.6.2. Type VII Tire – Extra High Pressure Tire

5.6.3. Main Gear: Assume 100% T.O.G.W

5.6.4. Main Gear: Rm/W= 0.94

5.6.5. Nose Gear: Assume 15% T.O.G.W

5.6.6. Nose Gear: Rn/W=0.06


5.7. Aircraft Configurations

Specification 1970’s Design Current Technology Design

Body Configuration

Fuselage Diameter 12.46 ft 12.46 ft

Fuselage Length 127.51 ft 127.51 ft

Overall Fineness Ratio (L/D) 10.23 10.23

Nose Fineness Ratio 1.14 1.14

Tail Fineness Ratio (Top) 1.87 1.87

Tail Fineness Ratio (Side) 2.54 2.54

Upsweep 4.5° 4.5°

Pilot Eye View 20° 20°

Interior Configuration

Passenger Mix (2 Class)

(1st Class/Econ)

9.5%

(10 pass / 100 pass)

9.5%

(10 pass / 100 pass)

Passenger Seats 110 110

Seat Pitch (1st Class/Econ) (38 in/ 32 in) (38 in/ 32 in)

Seat Depth (1st Class/Econ) (25 in / 25 in) (25 in / 25 in)

Seat Recline (1st Class/Econ) (36 in / 32 in) (36 in / 32 in)

Bulkhead to Seat Nose

(1st Class/Econ) (22 in / 20 in) (22 in / 20 in)

Aisle Width (1st Class/Econ) (22.00 in / 21.31in) (22.00 in / 21.31in)

Fuselage Ceiling Height 80.65 in 80.65

Overhead Compartment Height 63.32 in 63.32 in

Galley Carts/Passenger

(1st Class / Econ) (.30/pass / .075 pass) (.30/pass / .075 pass)

Cart Size 12 in x 34 in 12 in x 34 in

Galley Area (1st Class / Econ) 11.875 ft2 /37.92 ft2 11.875 ft2 /37.92 ft2

Lavatory/Passenger

(1st Class / Econ)

1 required / 50pass/lav

2 lavs (Econ)

1 required / 50pass/lav

2 lavs (Econ)

Lavatory Area 38 in x 40 in 38 in x 40 in

Coat Room 15 in x 15 in (1st Class)

None (Econ Class)

15 in x 15 in (1st Class)

None (Econ Class)

Entry Door Type A (w x h) 42 in x 72 in 42 in x 72 in

Door Type 2 (w x h) 20 in x 44 in 20 in x 44 in

Fuel & Payload

Wing – Fuel Capacity 3740.26 gallons 4562.1 gallons

Fuel Box Capacity 166 ft3 None

Number of Fuel Boxes 2 (Depth: 42.3 in) None

Fuel Box Size (LD-W) A: 3400 in2 L: 42.3 in A: 3400 in2 L: 42.3 in

Cargo Area/Volume 200 ft3 200 ft3


Passenger Baggage Volume 550 ft3 550 ft3

Cargo Container Size (LD-W) A: 3400 in2 L: 42.3 in A: 3400 in2 L: 42.3 in

Number of Cargo Containers 9 9

Tail Configurations 1970s Current

Horizontal Tail

Span 47.28 ft 41.42 ft

Root 14.57 ft 11.24 ft

Tip 3.06 3.90 ft

Tail Location (Fuselage to LE) 110.52 ft 110.42 ft

Vertical Tail

Half Span 14.30 ft 16.42 ft

Root 13.39 ft 11.8 ft

Tip 7.06 ft 5.48 ft

Tail Location (Fuselage to LE) 107.26 ft 114.55 ft Table 7 - Aircraft Configurations

6. Aircraft Configuration Drawings – (Solid Works Drawings)

6.1. Current Technology Airplane (As Labeled & Attached)

6.1.1. Side View

6.1.2. Side View with Wing

6.1.3. Fuselage Interior Top View

6.1.4. Fuselage Cross Section View

6.1.5. Front View

6.1.6. Top View

6.2. 1970’s Airplane (As Labeled & Attached)

6.2.1. Side View

6.2.2. Side View with Wing

6.2.3. Fuselage Interior Top View

6.2.4. Fuselage Cross Section View

6.2.5. Front View

6.2.6. Top View


7. References

1. AERO - Boeing 787 from the Ground Up. (n.d.). Retrieved May 06, 2016, from

http://www.boeing.com/commercial/aeromagazine/articles/qtr_4_06/article_04_2.h

tml

2. Schaufele, R. D. (2007). The Elements of Aircraft Preliminary Design. Santa Ana, CA:

ARIES Publication.

3. Shevell, R. S. (1983). Fundamentals of flight. Englewood Cliffs, NJ: Prentice-Hall.

4. Tovey, A. (2014, September 1). Rolls-Royce powers ahead with advanced engine

designs. Retrieved

May 06, 2016, from

http://www.telegraph.co.uk/finance/newsbysector/industry/engineering/11066706/

Rolls-Royce-powers-ahead-with-advanced-engine-designs.html

5. Yancy, R. (2012, June 11). How Composites Are Strengthening the Aviation Industry.

Retrieved May

06, 2016, from http://www.industryweek.com/none/how-composites-are-

strengthening-aviation-industry

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Justin Oyas

ID #43026527

MAE 159 – Aircraft Design – Final Report

Calculations

Sample Mission Specifications

Number of Passengers (2-class domestic rules) 110

Weight of cargo (10 pounds/ft3) 2000

Range (Still Air) 2500

Takeoff Field Length 6000 feet

Landing Approach Speed (Design Payload plus 100% max fuel) 130 knots

Cruise Mach number 0.80

Initial Cruise Altitude 35,000 ft

Takeoff conditions Sea-Level - Hot Day (84°F)

Wing Sizing

AR=7

Sweep=35°

CL is initially guess as 0.5 and iterated for

CLIC = (

𝑊

𝑆)𝐼𝐶

1481∗.2360∗𝑀2 until |CL−CLIC

CLIC| ≤ 0.0000000001

CL=0.4872 & CLIC = 0.4872

∆𝑀𝑑𝑖𝑣 = −0.2088 ∗ (𝐶𝐿)2 − 0.1048 ∗ 𝐶𝐿 + 0.1206

𝑀𝑑𝑖𝑣 = 𝑀 + .004 − ∆𝑀𝑑𝑖𝑣 = 0.7840

𝑡

𝑐= −36.767 ∗ (𝑀𝑑𝑖𝑣)3 + 98.73 ∗ (𝑀𝑑𝑖𝑣)2 − 89.063 ∗ (𝑀𝑑𝑖𝑣) + 27.07 = 0.2119

𝑄 = 𝑐𝑜𝑠2(𝑆𝑤𝑒𝑒𝑝) ∗ (𝑡

𝑐)2 ∗ 𝐴𝑅 = 0.2109

𝐶𝐿𝑚𝑎𝑥𝑇𝑂| = −30.631 ∗ (𝑄)2 + 12.95 ∗ (𝑄) + 1.1749 = 2.5

𝐶𝐿𝑚𝑎𝑥𝐿𝐷𝐺| = −24.489 ∗ (𝑄)^2 + 11.544 ∗ (𝑄) + 2.1581 =3.5

𝑊

𝑆|

𝐿𝐷𝐺= (

130

1.3)2 ∗

0.9532 ∗ 𝐶𝐿𝑚𝑎𝑥𝐿𝐷𝐺|

296= 112.8163

𝑊𝑙𝑜𝑎𝑑𝑡𝑜 =112.8163

1 − 0 ∗ .390= 112.81

𝑉𝑐𝑟𝑢𝑖𝑠𝑒 = 𝑀 ∗ 576.6 = 461.12 𝑘𝑛𝑜𝑡𝑠

𝑅𝑎𝑙𝑙𝑜𝑢𝑡 = 𝑅𝑎𝑛𝑔𝑒 + 200 + (0.75 ∗ 𝑉𝑐𝑟𝑢𝑖𝑠𝑒) = 3045.8 𝑛𝑎𝑢𝑡𝑖𝑐𝑎𝑙 𝑚𝑖𝑙𝑒𝑠

TOFL at 6000ft

From digitized TOFL: 𝐾 = −0.3417 ∗ (𝑇𝑂𝐹𝐿)^2 + 39.332 ∗ 𝑇𝑂𝐹𝐿 − 103.81=119

𝑊

𝑇|

0.7𝑉𝑙𝑜=

119

112.12∗ 2.5 ∗ .953 = 2.5759

𝑉𝑙𝑜 = 1.2 ∗ (296 ∗ 𝑊𝑙𝑜𝑎𝑑𝑡𝑜

. 953 ∗ 𝐶𝐿𝑚𝑎𝑥𝑇𝑂|)

0.5

= 140.844

MLO= 0.2183 & 0.7MLO=0.1528

JT9D Chart TSLST=45000 Tm=37200

𝑊

𝑇= 2.57 ∗

37200

45000= 2.12

Weight

𝑊𝑤 =. 00945 ∗ 𝐴𝑅0.8 ∗ (1 + 𝜆)0.25 ∗ 𝐾𝑤 ∗ 𝑛0.5

(𝑡̅𝑐

)0.4

∗ cos(𝑠𝑤𝑒𝑒𝑝) ∗ (𝑊𝑙𝑜𝑎𝑑𝑡𝑜).695

= .0076

𝑙𝑓𝑢𝑠 = (3.76 ∗ (𝑃𝐴𝑆𝑆

𝐴𝑏𝑟𝑒𝑎𝑠𝑡) + 33.2) ∗ 1.10 = 150.26

𝑑𝑓𝑢𝑠 = (1.75 ∗ 𝐴𝑏𝑟𝑒𝑎𝑠𝑡 + 1.58 ∗ 𝐴𝑖𝑠𝑙𝑒 + 1.0) ∗ 1.10 = 10.538

𝑊𝑓𝑢𝑠 = .6727 ∗ 11.5 ∗ (𝑙_𝑓𝑢𝑠)^.6 ∗ ((𝑑_𝑓𝑢𝑠)^.72) ∗ (3.75)^.3= 1268.3

𝑊𝑙𝑔 = .04

𝑊𝑛𝑝 = (.0555/2.12)*0.8= .0209 (Composite Factor)

𝑊𝑡𝑠 = (. 17 + (. 08

𝑛𝑢𝑚𝑒𝑛𝑔

)) ∗ 𝑊𝑤 = 0.0016

𝑊𝑤𝑡𝑠= (𝑊𝑤 + 𝑊𝑡𝑠) ∗ .70 = 0.0087 (𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝐹𝑎𝑐𝑡𝑜𝑟)

𝑊𝑝𝑝 = (1

3.58 ∗ 2.12) = 0.1312

𝑊𝑓 = 1.0275 ∗ .240 ∗ 0.85 = 0.2096 (𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝐹𝑎𝑐𝑡𝑜𝑟)

𝑁𝐹𝐶 = 2 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐹𝑙𝑖𝑔ℎ𝑡 𝐶𝑟𝑒𝑤

𝑁𝐶𝐴 = 6 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶𝑎𝑏𝑖𝑛 𝐴𝑡𝑡𝑒𝑛𝑑𝑎𝑛𝑡𝑠

𝑊𝑃𝐿 = 215 ∗ 𝑃𝐴𝑆𝑆 + 𝑊𝑐𝑎𝑟𝑔𝑜 = 25650

𝑊𝐹𝐸 = (132 ∗ 𝑃𝐴𝑆𝑆) + ((300 ∗ 𝑛𝑢𝑚_𝑒𝑛𝑔)) + (260 ∗ 𝑁_𝐹𝐶) + (170 ∗ 6) = (𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝐹𝑎𝑐𝑡𝑜𝑟)

(𝑊𝑤𝑡𝑠∗ 𝑊𝑇𝑂

1.195) + (𝑊𝑓𝑢𝑠 ∗ 𝑊𝑇𝑂.235) + ((𝑊𝑙𝑔 + 𝑊𝑛𝑝

+ 𝑊𝑝𝑝 + 𝑊𝑓 + .035 − 1) ∗ 𝑊𝑇𝑂) + (𝑊𝐹𝐸 + 𝑊𝑃𝐿) = 0

𝑊𝑇𝑂 = 127369.712716844 (𝐼𝑡𝑒𝑟𝑎𝑡𝑒𝑑 𝑤𝑖𝑡ℎ 𝑣𝑝𝑎𝑠𝑜𝑙𝑣𝑒)

𝑆 =𝑊𝑇𝑂

𝑊𝑙𝑜𝑎𝑑𝑡𝑜

= 1129

𝑏 = (𝐴𝑅 ∗ 𝑆)2 = 88.89

𝑚𝑎𝑐 =𝑆

𝑏= 12.699

T =WTO

WT

= 59815

𝑇𝑒 =𝑇


= 29907

Drag

RNK = 2.852E6 ∗ .5 = 1426000

𝑅𝑁𝑤𝑖𝑛𝑔 = 𝑅𝑁𝐾 ∗ 𝑚𝑎𝑐 = 1.81𝐸7

𝐶𝑓𝑤𝑖𝑛𝑔 = .0027

𝑅𝑁𝑓𝑢𝑠 = 𝑅𝑁𝐾 ∗ 𝑙𝑓𝑢𝑠= 2.14E8

𝐶𝑓𝑓𝑢𝑠 = .0018

Z =(2−M2)∗cos(Sweep)

sqrt((1−M2∗(cos(Sweep)2)))= 1.4748

𝐾𝑤 = 1 + (𝑍 ∗ (𝑡

𝑐)) + (100 ∗ (

𝑡

𝑐)

4

)= 1.5530

𝑆𝑤𝑒𝑡𝑤𝑖𝑛𝑔 = 2 ∗ (𝑆 − (20 ∗ 30)) ∗ 1.02 = 1079.1

𝑓𝑤𝑖𝑛𝑔 = 𝑆𝑤𝑒𝑡𝑤𝑖𝑛𝑔 ∗ 𝐶𝑓𝑤𝑖𝑛𝑔 ∗ 𝐾𝑤 = 4.5253

𝑆𝑤𝑒𝑡𝑓𝑢𝑠 = 0.9 ∗ 𝑝𝑖 ∗ 𝑑𝑓𝑢𝑠 ∗ 𝑙𝑓𝑢𝑠 = 4477

𝐿

𝐷=

𝑙𝑓𝑢𝑠

𝑑𝑓𝑢𝑠

= 17.34

𝑓𝑓𝑢𝑠 = 𝑆𝑤𝑒𝑡𝑓𝑢𝑠 ∗ 𝐶𝑓𝑓𝑢𝑠 ∗ 𝐾𝑓𝑢𝑠 = 0.79769

𝑓𝑡𝑠 = .38 ∗ 𝑓𝑤𝑖𝑛𝑔 = 1.7196

𝑆𝑤𝑒𝑡𝑁𝑎𝑐 = 2.1 ∗ 𝑠𝑞𝑟𝑡(𝑇𝑒) ∗ 𝑛𝑢𝑚𝑒𝑛𝑔 = 726.3

𝐾𝑁𝑎𝑐 = 1.25

𝑓𝑁𝑎𝑐 = 𝐶𝑓𝑤𝑖𝑛𝑔 ∗ 𝑆𝑤𝑒𝑡𝑁𝑎𝑐 ∗ 𝐾𝑁𝑎𝑐 = 2.451399

𝑓𝑝𝑦𝑙 = .20 ∗ 𝑓𝑁𝑎𝑐 = 0.49027

𝑓𝑡𝑜𝑡 = (𝑓𝑤𝑖𝑛𝑔 + 𝑓𝑓𝑢𝑠 + 𝑓𝑡𝑠 + 𝑓𝑁𝑎𝑐 + 𝑓𝑝𝑦𝑙) ∗ 1.06 = 16.5519

𝐶𝐷𝑜 =𝑓𝑡𝑜𝑡

𝑆= 0.014660

𝑒𝑓𝑓 =1

1.035 + (0.38 ∗ 𝐶𝐷𝑜 ∗ 𝑝𝑖 ∗ 𝐴𝑅)= 0.863920

Climb

𝑆𝑖𝑔𝑚𝑎 = .5702

𝑊𝑐𝑙𝑖𝑚𝑏 = (1 + .965

2) ∗ 𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓 = 125140

𝑉𝑐𝑙 = 1.3 ∗ (12.9

(𝑓𝑡𝑜𝑡 ∗ 𝑒𝑓𝑓)14

) ∗ (𝑊𝑐𝑙𝑖𝑚𝑏

𝑆𝑖𝑔𝑚𝑎 ∗ 𝑏)

.5

= 428.48

aclimb = 576.54

𝑀𝑐𝑙𝑖𝑚𝑏 =𝑉𝑐𝑙

𝑎𝑐𝑙𝑖𝑚𝑏

= 0.74

𝑇𝑟𝑐𝑙= (

𝑆𝑖𝑔𝑚𝑎 ∗ 𝑓𝑡𝑜𝑡 ∗ (𝑉𝑐𝑙2)

296) + (

94.1

𝑆𝑖𝑔𝑚𝑎 ∗ 𝑒𝑓𝑓) ∗ ((

𝑊𝑐𝑙𝑖𝑚𝑏

𝑏)

2

∗ (1

𝑉𝑐𝑙

)2

) = 7915.8

𝑇15 = 17.413 ∗ 𝑀𝑐𝑙𝑖𝑚𝑏4 − 44.292 ∗ 𝑀𝑐𝑙𝑖𝑚𝑏

3 + 45.427 ∗ 𝑀𝑐𝑙𝑖𝑚𝑏2 − 30.119 ∗ 𝑀𝑐𝑙𝑖𝑚𝑏 + 27.441 = 17.278

𝑇25 = −14.58919 ∗ 𝑀𝑐𝑙𝑖𝑚𝑏4 + 32.77303 ∗ 𝑀𝑐𝑙𝑖𝑚𝑏

3 − 23.49468 ∗ 𝑀𝑐𝑙𝑖𝑚𝑏2 + 3.82968 ∗ 𝑀𝑐𝑙𝑖𝑚𝑏 + 15.36777

𝑇25 = 14.2392

𝑇20 = ((20000 − 15000) ∗ ( 𝑇25 − 𝑇15)

25000 − 15000+ 𝑇15) ∗ 1000 = 15758.8 (𝐼𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑒𝑑)

𝑐15 = (−2.645𝐸 − 04 ∗ 𝑇153 + 1.991𝐸 − 02 ∗ 𝑇152 − 5.188𝐸 − 01 ∗ 𝑇15 + 5.022) = 0.63757

𝑐25 = (−0.01471 ∗ 𝑇253 + 0.77220 ∗ 𝑇252 − 13.33893 ∗ 𝑇25 + 76.43786) = 0.600669

𝑐20 =(20000 − 15000) ∗ (𝑐25 − 𝑐15)

25000 − 15000+ 𝑐15 = 0.6191240 (𝐼𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑒𝑑)

𝑇𝑎 = (𝑇𝑒

𝑇𝑆𝐿𝑆𝑇) ∗ 𝑇20 = 10473.5

𝑅𝐶 =101 ∗ ((𝑛𝑢𝑚𝑒𝑛𝑔 ∗ 𝑇𝑎) − 𝑇𝑟𝑐𝑙

) ∗ 𝑉𝑐𝑙

𝑊𝑐𝑙𝑖𝑚𝑏

= 4506.5

𝑇𝑖𝑚𝑒𝑐𝑙 =𝐻

𝑅𝐶= 7.76638 𝑀𝑖𝑛

𝑅𝑎𝑛𝑔𝑒𝑐𝑙 = 𝑉𝑐𝑙 ∗ (𝑇𝑖𝑚𝑒𝑐𝑙

60) = 55.46355

𝑊𝑓𝑐𝑙 = (𝑛𝑢𝑚𝑒𝑛𝑔 ∗ 𝑇𝑎) ∗ 𝑐20 ∗ (𝑇𝑖𝑚𝑒𝑐𝑙

60) = 1678.6

Range

𝑊0 = 𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓 − 𝑊𝑓𝑐𝑙 = 125691

𝑊1 = (1 − 𝑊𝑓𝑊𝑡𝑜𝐽𝑇9𝐷𝑓𝑖𝑥) ∗ 𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓; = 96801

𝐶𝐿𝑎𝑣𝑔 =

𝑊0 + 𝑊12 ∗ 𝑆

1481 ∗ 𝑃𝑅 ∗ 𝑀2= 0.4409

𝐶𝐷𝑖 =(𝐶𝐿𝑎𝑣𝑔)

2

𝑝𝑖 ∗ 𝐴𝑅 ∗ 𝑒𝑓𝑓= 0.01023

𝑑𝑒𝑙𝑡𝑎𝐶𝐷_𝐶 = .0010

𝐶𝐷 = 𝐶𝐷𝑜 + 𝐶𝐷𝑖 + 𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝐶 = 0.025896

𝐿𝐷 =𝐶𝐿𝑎𝑣𝑔

𝐶𝐷= 17.02865

𝑇𝑟𝑟𝑎𝑛𝑔𝑒=

𝑊0 + 𝑊12

𝐿𝐷= 6532.8

𝑇𝑟𝑟𝑎𝑛𝑔𝑒𝐽𝑇9𝐷= 𝑇𝑟𝑟𝑎𝑛𝑔𝑒

∗ (𝑇𝑆𝐿𝑆𝑇

𝑇𝑒) = 9829.57

𝑇𝑟𝑟𝑎𝑛𝑔𝑒𝑝𝑒𝑟=

𝑇𝑟𝑟𝑎𝑛𝑔𝑒𝐽𝑇9𝐷


= 4914.78

Call curve fit read and interpolate function

𝑐𝑟𝑎𝑛𝑔𝑒 =𝑑𝑖𝑓𝑓

100∗

𝑀 − 𝑚𝑎𝑐ℎ(𝑖1)

𝑚𝑎𝑐ℎ(𝑖2) − 𝑚𝑎𝑐ℎ(𝑖1)+

𝑡𝑠𝑓𝑐(𝑖1)

100= 0.633611

𝑅𝑐𝑟𝑢𝑖𝑠𝑒 = (𝑉𝑐𝑙

𝑐𝑟𝑎𝑛𝑔𝑒

) ∗ 𝐿𝐷 ∗ 𝑙𝑜𝑔 (𝑊0

𝑊1) = 3007.59

𝑅 = 𝑅𝑎𝑛𝑔𝑒𝑐𝑙 + 𝑅𝑐𝑟𝑢𝑖𝑠𝑒 = 3063.06

Iteration

𝑅𝑒𝑟𝑟𝑜𝑟 =𝑎𝑏𝑠(𝑅 − 𝑅𝑎𝑜)

𝑅𝑎𝑜

< .01

Check at Top of Climb

𝐶𝐿𝑖𝑐 =

𝑊0𝑆

1481 ∗ 𝑃𝑅 ∗ 𝑀2= 0.49824

𝐶𝑑𝑖 =𝐶𝐿𝑖𝑐2

𝑝𝑖 ∗ 𝐴𝑅 ∗ 𝑒𝑓𝑓= 0.01306

𝐶𝐷 = 𝐶𝐷𝑜 + 𝐶𝑑𝑖 + .0010 = 0.0287272

𝐿𝐷 =𝐶𝐿𝑖𝑐

𝐶𝐷

= 17.3439065

𝑇𝑟𝑒𝑞 =𝑊0

𝐿𝐷

= 7246.98

𝑇𝑟𝑒𝑞𝑒𝑛𝑔=

𝑇𝑟𝑒𝑞


= 3623.49

𝑇𝑟𝑒𝑞𝐽𝑇9𝐷= 𝑇𝑟𝑒𝑞𝑒𝑛𝑔

∗ (𝑇𝑆𝐿𝑆𝑇

𝑇𝑒) = 5452.02

Climb Gradients

1st Segment

𝐶𝐿𝑡𝑜 =𝐶𝑙𝑚𝑎𝑥𝑡𝑜

(1.2)2= 1.766

𝐶𝐿𝑡𝑜𝑚𝑎𝑥𝑡𝑜=

1

(1.2)2= 0.69444444

𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜 = 0.16 ∗ (𝐶𝐿𝑡𝑜𝑚𝑎𝑥𝑡𝑜)

4− 0.1489 ∗ (𝐶𝐿𝑡𝑜𝑚𝑎𝑥𝑡𝑜

)3

+ 0.0894 ∗ (𝐶𝐿𝑡𝑜𝑚𝑎𝑥𝑡𝑜)

2− 0.0707 ∗ (𝐶𝐿𝑡𝑜𝑚𝑎𝑥𝑡𝑜

) + 0.0328

𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜 = 0.014160

𝐶𝐷𝑔𝑒𝑎𝑟 = 𝐶_𝐷𝑜

𝐶𝐷1 = 𝐶𝐷𝑜 + 𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜 + 𝐶𝐷𝑔𝑒𝑎𝑟 + ((𝐶𝐿𝑡𝑜)2

𝑝𝑖 ∗ 𝐴𝑅 ∗ 𝑒𝑓𝑓) = 0.2077146

𝐿𝐷𝑡𝑜 =𝐶𝐿𝑡𝑜

𝐶𝐷1= 8.5039964

𝑇𝑟𝑒𝑞1 =𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓

𝐿𝐷𝑡𝑜

= 14977.63

𝑇𝑎𝑒𝑛𝑔1 = (𝑇𝑒

𝑇𝑆𝐿𝑆𝑇) ∗ 34500 = %34500 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ

𝐺𝑟𝑎𝑑1 =(𝑛𝑢𝑚𝑒𝑛𝑔 − 1) ∗ 𝑇𝑎𝑒𝑛𝑔1 − 𝑇𝑟𝑒𝑞1

𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓

∗ 100 = 6.2428

Second Segment

𝐶𝐷2 = 𝐶𝐷𝑜 + 𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜 + ((𝐶𝐿𝑡𝑜)2

𝑝𝑖 ∗ 𝐴𝑅 ∗ 𝑒𝑓𝑓) = 0.193053

𝐿𝐷2 =𝐶𝐿𝑡𝑜

𝐶𝐷2= 9.14979

𝑇𝑟𝑒𝑞2 =𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓

𝐿𝐷2

= 13920.49



∗ 100 = 7.07287

Third Segment

𝐶𝑙35𝑐𝑙𝑒𝑎𝑛 = −303.8 ∗ (𝑡𝑐)3 + 79.898 ∗ (𝑡𝑐)2 − 2.522 ∗ (𝑡𝑐) + 0.859 = 1.021814

𝑉3 = 1.2 ∗ 𝑠𝑞𝑟𝑡 (296 ∗ 𝑊𝑙𝑜𝑎𝑑𝑡𝑜

. 925 ∗ 1.10) = 217.3930𝐾𝑛𝑜𝑡𝑠

𝑀3 =𝑉3

659= 0.32988

𝐶𝑙3 =𝐶𝑙35𝑐𝑙𝑒𝑎𝑛

(1.2)2= 0.709593

𝐶𝐷3 = 𝐶𝐷𝑜 + ((𝐶𝑙3)2

𝑝𝑖 ∗ 𝐴𝑅 ∗ 𝑒𝑓𝑓) = 0.041163

𝐿𝐷3 =𝐶𝑙3

𝐶𝐷3= 17.23825

𝑇𝑟𝑒𝑞3 =𝑊0

𝐿𝐷3= 7291.40

𝑇𝑎𝑒𝑛𝑔3 = (𝑇𝑒

𝑇𝑆𝐿𝑆𝑇) ∗ 26000 = 17279.96



∗ 100 = 7.84218 % 𝑝𝑒𝑟𝑐𝑒𝑛𝑡

Approach

𝐶𝑙𝐴𝑝 =𝐶𝑙𝑚𝑎𝑥𝑡𝑜

(1.3)2= 1.5051

𝐶𝑙𝐴𝑝𝑚𝑎𝑥=

1

(1.3)2= 0.59171

𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜𝐴𝑝 = 0.16 ∗ (𝐶𝑙𝐴𝑝𝑚𝑎𝑥)

4− 0.1489 ∗ (𝐶𝑙𝐴𝑝𝑚𝑎𝑥

)3

+ 0.0894 ∗ (𝐶𝑙𝐴𝑝𝑚𝑎𝑥)

2− 0.0707 ∗ (𝐶𝑙𝐴𝑝𝑚𝑎𝑥

) + 0.0328 =

𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜𝐴𝑝 = 0.01103

𝐶𝐷𝐴𝑝 = 𝐶𝐷𝑜 + 𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜𝐴𝑝 + ((𝐶𝑙𝐴𝑝)

2

𝑝𝑖 ∗ 𝐴𝑅 ∗ 𝑒𝑓𝑓) = 0.1449304

𝐿𝐷𝐴𝑝 =𝐶𝑙𝐴𝑝

𝐶𝐷𝐴𝑝

= 10.3849

𝑊𝑙𝑎𝑛𝑑 = 𝑊𝑙𝑜𝑎𝑑𝑙𝑑𝑔 ∗ 𝑆 = 127369.71

𝑇𝑟𝑒𝑞𝐴𝑝 =𝑊𝑙𝑎𝑛𝑑

𝐿𝐷𝐴𝑝

= 12264.782

𝑉𝐴𝑝 = 𝑠𝑞𝑟𝑡 (296 ∗ 𝑊𝑙𝑜𝑎𝑑𝑙𝑑𝑔

. 953 ∗ 𝐶𝑙𝐴𝑝

) = 152.58

TaAp = (Te

TSLST) ∗ 29500 = 19606.11

𝐺𝑟𝑎𝑑𝐴𝑝 =(𝑛𝑢𝑚𝑒𝑛𝑔 − 1) ∗ 𝑇𝑎𝐴𝑝 − 𝑇𝑟𝑒𝑞𝐴𝑝

𝑊𝑙𝑎𝑛𝑑∗ 100 = 5.76380

Landing

𝐶𝑙𝑙𝑎𝑛𝑑 =𝐶𝑙𝑚𝑎𝑥𝑙𝑑𝑔

(1.3)2= 2.0730400

𝐶𝑙𝐶𝑙𝑚𝑎𝑥 =1

(1.3)2= 0.591715

𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜𝐿𝑎𝑛𝑑 = 0.16 ∗ (𝐶𝑙𝐶𝑙𝑚𝑎𝑥)4 − 0.1489 ∗ (𝐶𝑙𝐶𝑙𝑚𝑎𝑥)3 + 0.0894 ∗ (𝐶𝑙𝐶𝑙𝑚𝑎𝑥)2 − 0.0707 ∗ (𝐶𝑙𝐶𝑙𝑚𝑎𝑥) + 0.0328

𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜𝐿𝑎𝑛𝑑 = 0.0110328

𝐶𝐷𝑙𝑎𝑛𝑑 = 𝐶𝐷𝑜 + 𝑑𝑒𝑙𝑡𝑎𝐶𝐷𝑜𝐿𝑎𝑛𝑑 + 𝐶𝐷𝑔𝑒𝑎𝑟 + ((𝐶𝑙𝑙𝑎𝑛𝑑)2

𝑝𝑖 ∗ 𝐴𝑅 ∗ 𝑒𝑓𝑓) = 0.266554

𝐿𝐷𝑙𝑎𝑛𝑑 =𝐶𝑙𝑙𝑎𝑛𝑑

𝐶𝐷𝑙𝑎𝑛𝑑= 7.77715

𝑇𝑟𝑒𝑞𝑙𝑎𝑛𝑑 =𝑊𝑙𝑎𝑛𝑑

𝐿𝐷𝑙𝑎𝑛𝑑

= 16377.411

𝑇𝑎𝐿𝑎𝑛𝑑 = (𝑇𝑒

𝑇𝑆𝐿𝑆𝑇) ∗ 37200 = 24723.647

𝐺𝑟𝑎𝑑𝐿𝑎𝑛𝑑 =𝑛𝑢𝑚𝑒𝑛𝑔 ∗ 𝑇𝑎𝐿𝑎𝑛𝑑 − 𝑇𝑟𝑒𝑞𝑙𝑎𝑛𝑑

𝑊𝑙𝑎𝑛𝑑∗ 100 = 25.9636 %𝑝𝑒𝑟𝑐𝑒𝑛𝑡

Direct Operating Costs

𝑑𝑖𝑠𝑡 = 𝑅𝑎𝑛𝑔𝑒 ∗ 1.15 = 2875 (𝑠𝑡𝑎𝑡𝑢𝑡𝑒 𝑚𝑖𝑙𝑒𝑠)

𝑡𝑔𝑚 = .25 (𝑇𝑖𝑚𝑒 𝑓𝑜𝑟 𝐺𝑟𝑜𝑢𝑛𝑑 𝑀𝑎𝑛𝑢𝑒𝑣𝑒𝑟)

𝑡_𝑐𝑙 = .18 (𝑡𝑖𝑚𝑒 𝑡𝑜 𝑐𝑙𝑖𝑚𝑏)

𝑡_𝑑 = 0 (𝑡𝑖𝑚𝑒 𝑡𝑜 𝑑𝑒𝑠𝑐𝑒𝑛𝑑)

𝑡_𝑎𝑚 = .10 (𝑡𝑖𝑚𝑒 𝑓𝑜𝑟 𝑎𝑖𝑟 𝑚𝑎𝑛𝑢𝑒𝑣𝑒𝑟)

𝑑_𝑐 = 83(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑜 𝑐𝑙𝑖𝑚𝑏)

𝑑_𝑑 = 0 (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑜 𝑑𝑒𝑠𝑐𝑒𝑛𝑑)

𝑉𝑐𝑟 = 𝑉𝑐𝑟𝑢𝑖𝑠𝑒 ∗ 1.15 = 530.28 (𝑡𝑟𝑢𝑒 𝑎𝑖𝑟𝑠𝑝𝑒𝑒𝑑)

Ka = 1.02 (airway distance increment)

𝑡𝑐𝑟 =(𝑑𝑖𝑠𝑡 + 𝐾𝑎 + 20) − (𝑑𝑐 + 𝑑𝑑)

𝑉𝑐𝑟= 5.304702

𝑉𝑏 =𝑑𝑖𝑠𝑡

𝑡𝑔𝑚 + 𝑡𝑐𝑙 + 𝑡𝑑 + 𝑡𝑐𝑟 + 𝑡𝑎𝑚

= 492.74

𝑡𝑏 = 𝑡𝑔𝑚 + 𝑡𝑐𝑙 + 𝑡𝑑 + 𝑡𝑐𝑟 + 𝑡𝑎𝑚 = 5.834702

𝑓_𝑔𝑚 = 0;

𝑓_𝑐𝑙 = 9624;

𝑓𝑐𝑟𝑎𝑚= 𝑇𝑟𝑟𝑎𝑛𝑔𝑒

∗ (𝑐𝑟𝑎𝑛𝑔𝑒) ∗ (𝑡𝑐𝑟 + 𝑡𝑎𝑚) = 22371.68 (𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑇𝑒𝑐ℎ 𝐽𝑇9𝐷 𝑠𝑓𝑐)

𝑓_𝑑 = 0;

𝐹𝑏 = 𝑓𝑔𝑚 + 𝑓𝑐𝑙 + 𝑓𝑐𝑟𝑎𝑚+ 𝑓𝑑 = 31995.68

Flying Operations Cost

%Flight Crew

𝑃 =𝑊𝑃𝐿

2000= 12.8249 𝑇𝑜𝑛𝑠

𝑑𝑜𝑙𝑙𝑎𝑏𝑙𝑘ℎ𝑟 = 17.849 ∗ (𝑉𝑐𝑟𝑢𝑖𝑠𝑒 ∗ (𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓

105))

0.3

+ 40.83 = 161.68 (2 𝑚𝑎𝑛 𝑐𝑟𝑒𝑤)

𝐶𝑡𝑚𝐹𝐶 =𝑑𝑜𝑙𝑙𝑎𝑏𝑙𝑘ℎ𝑟

𝑉𝑏 ∗ 𝑃= 0.02558604

%Fuel&Oil

𝐶𝑓 = .40 ∗ (1

6.4) = 0.062500

𝐶𝑜𝑇 = 2.15

𝐶𝑡𝑚𝐹𝑂 = 1.02 ∗(𝐹𝑏 ∗ 𝐶𝑓) + (𝑛𝑢𝑚𝑒𝑛𝑔 ∗ 𝐶𝑜𝑇 ∗ .135)

𝑑𝑖𝑠𝑡 ∗ 𝑃= 0.05533532

%Hull Insurance

𝑊𝑎 = 𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓 ∗ (1 − 𝑊𝑓𝑊𝑡𝑜𝐽𝑇9𝐷𝑓𝑖𝑥) − 𝑊𝑃𝐿 − (𝑊𝑝𝑝 ∗ 𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓) = 54442.80

𝐶𝑎 = 2.4 ∗ 106 + (87.5 ∗ 𝑊𝑎) = 7163745.36

𝐶𝑒 = 590000 + 16 ∗ 𝑇𝑒 = 1068522.203

𝐶𝑇 = 𝐶𝑎 + (𝑛𝑢𝑚𝑒𝑛𝑔 ∗ 𝐶𝑒) = 9300789.77

𝐼𝑅𝑎 = .01

𝑈ℎ𝑢𝑙𝑙 = (630) +4000

1 + (1

𝑡𝑏 + .50)

= 4084.64

𝐶𝑡𝑚𝐻𝑢𝑙𝑙 =𝐼𝑅𝑎 ∗ 𝐶𝑇

𝑈ℎ𝑢𝑙𝑙 ∗ 𝑉𝑏 ∗ 𝑃= 0.003603

%Direct Maintenance

%Airframe Labor

𝐾𝐹𝐻𝐴 = (4.9169 ∗ (𝑙𝑜𝑔10 (𝑊𝑎

1000))) − 6.425 = 2.110445776328345

𝐾𝐹𝐶𝐴 = .21256 ∗ (𝑙𝑜𝑔10 (𝑊𝑎

1000))

3.7375

= 1.67010

𝑇𝐹 = 𝑡𝑏 − 𝑡𝑔𝑚 = 5.58470

𝑅_𝑙 = 8.6

𝐶𝑡𝑚𝐴𝐿 =𝑅𝑙 ∗ ((𝐾𝐹𝐻𝐴 ∗ 𝑇𝐹) + (𝐾𝐹𝐶𝐴))

𝑉𝑏 ∗ 𝑡𝑏 ∗ 𝑃= 0.0031385

%Airframe Material

𝐶𝐹𝐻𝐴 = 1.5994 ∗ (𝐶𝑎

106) + 3.4263 = 14.883994

𝐶𝐹𝐶𝐴 = 1.9229 ∗ (𝐶𝑎

106) + 2.2504 = 16.0255

𝐶𝑡𝑚𝐴𝑀 =𝐶𝐹𝐻𝐴 ∗ 𝑇𝐹 + 𝐶𝐹𝐶𝐴

𝑉𝑏 ∗ 𝑡𝑏 ∗ 𝑃= 0.00268899

%Engine-Labor

𝐾𝐹𝐻𝐸 =𝑛𝑢𝑚𝑒𝑛𝑔∗(

𝑇𝑒

1000)

.82715∗(𝑇𝑒

1000)+13.639

= 1.558618

𝐾𝐹𝐶𝐸 = .20 ∗ 𝑛𝑢𝑚𝑒𝑛𝑔 = .400

𝐶𝑡𝑚𝐸𝐿 =𝐾𝐹𝐻𝐸 ∗ 𝑇𝐹 + 𝐾𝐹𝐶𝐸

𝑉𝑏 ∗ 𝑡𝑏 ∗ 𝑃∗ 𝑅𝑙 = 0.00212351

%Engine Material

𝐶𝐹𝐻𝐸 = (28.2353 ∗ (𝐶𝑒

106) − 6.5176) ∗ 𝑛𝑢𝑚𝑒𝑛𝑔 = 47.30488

𝐶𝐹𝐶𝐸 = (3.6698 ∗ (𝐶𝑒

106) + 1.3685) ∗ 𝑛𝑢𝑚𝑒𝑛𝑔 = 10.57952

𝐶𝑡𝑚𝐸𝑀 =𝐶𝐹𝐻𝐸 ∗ 𝑇𝐹 + 𝐶𝐹𝐶𝐸

𝑉𝑏 ∗ 𝑡𝑏 ∗ 𝑃= 10.5795

%Total Maintenance Burden

𝐶𝑡𝑚𝑇𝑀 = (𝐶𝑡𝑚𝐴𝐿 + 𝐶𝑡𝑚𝐴𝑀 + 𝐶𝑡𝑚𝐸𝐿 + 𝐶𝑡𝑚𝐸𝑀) ∗ 2 = 0.02931543

%Depreciation

𝑎𝑑 = .06 ∗ (𝐶𝑇 − (𝑛𝑢𝑚𝑒𝑛𝑔 ∗ 𝐶𝑒)) = 4.2982472

𝑏𝑑 = .30 ∗ 𝑛𝑢𝑚𝑒𝑛𝑔 ∗ 𝐶𝑒 = 641113.32

𝐶𝑡𝑚𝑑𝑒𝑝 = (1

𝑉𝑏 ∗ 𝑃) ∗ (

𝐶𝑇 + 𝑎𝑑 + 𝑏𝑑

14 ∗ 𝑈ℎ𝑢𝑙𝑙

) = 0.028700

𝐷𝑂𝐶𝑡𝑜𝑛𝑚𝑖𝑙𝑒= 𝐶𝑡𝑚𝑑𝑒𝑝 + 𝐶𝑡𝑚𝑇𝑀 + 𝐶𝑡𝑚𝐻𝑢𝑙𝑙 + 𝐶𝑡𝑚𝐹𝑂 + 𝐶𝑡𝑚𝐹𝐶 = 0.14254 (

$

𝑡𝑜𝑛 𝑚𝑖𝑙𝑒)

𝐷𝑂𝐶𝑝𝑎𝑠𝑠𝑚𝑖𝑙𝑒= 𝐷𝑂𝐶𝑡𝑜𝑛𝑚𝑖𝑙𝑒

∗𝑃

𝑃𝐴𝑆𝑆= 0.01661894 (

$

𝑝𝑎𝑠𝑠𝑒𝑛𝑔𝑒𝑟 𝑚𝑖𝑙𝑒)

𝑊𝑒𝑖𝑔ℎ𝑡 = 𝑊𝑡𝑎𝑘𝑒𝑜𝑓𝑓 = 127369.71 (𝑙𝑏𝑠)

Center of Gravity Calculations (1970s Plane)

𝐹𝑢𝑠𝑒𝑙𝑎𝑔𝑒𝑐𝑜𝑒𝑓 = 1078.1

𝐹𝑢𝑠𝑒𝑙𝑎𝑔𝑒𝑤𝑒𝑖𝑔ℎ𝑡 = 𝐹𝑢𝑠𝑐𝑜𝑒𝑓 ∗ 𝑊𝑔𝑡𝑇𝑎𝑘𝑒𝑜𝑓𝑓0.235 = 18279 lb

𝑁𝑜𝑠𝑒𝑙𝑒𝑛𝑔𝑡ℎ =166.28

12= 13.8567 𝑓𝑡

𝐹𝑢𝑠𝑒𝑙𝑎𝑔𝑒𝑙𝑒𝑛𝑔𝑡ℎ = 127.51 𝑓𝑡

𝐹𝑢𝑠𝑒𝑙𝑎𝑔𝑒𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠 = 𝐹𝑢𝑠𝑙𝑒𝑛 ∗ 0.4 = 51.004 𝑓𝑡

𝑆 = 1737.8 𝑓𝑡2

𝑏 = 110.29 𝑓𝑡

𝑡𝑎𝑝𝑒𝑟 = 0.35

𝑤𝑖𝑛𝑔𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 = 0.007766

𝑊𝑖𝑛𝑔𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑤𝑖𝑛𝑔𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 ∗ 𝑊𝑒𝑖𝑔ℎ𝑡𝑇𝑎𝑘𝑒𝑜𝑓𝑓1.195 = 13843 𝑙𝑏

𝐸𝑛𝑔𝑖𝑛𝑒𝑤𝑒𝑖𝑔ℎ𝑡 = 3218 𝑙𝑏

𝑇ℎ𝑟𝑢𝑠𝑡𝑒𝑥𝑖𝑡 = 14500 𝑙𝑏

𝑇ℎ𝑟𝑢𝑠𝑡𝑛𝑒𝑒𝑑𝑒𝑑_1970_𝑒𝑛𝑔𝑖𝑛𝑒 = 20652.5 lb

𝐸𝑛𝑔𝑖𝑛𝑒𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑒𝑥𝑖𝑠𝑡= 95.6 𝑖𝑛

𝐸𝑛𝑔𝑖𝑛𝑒𝑑𝑛𝑒𝑒𝑑𝑒𝑑= 𝐸𝑛𝑔𝑖𝑛𝑒𝑑𝑒𝑥𝑖𝑠𝑡

∗ 𝑠𝑞𝑟𝑡 (𝑡ℎ𝑟𝑢𝑠𝑡𝑛𝑒𝑒𝑑𝑒𝑑

𝑡ℎ𝑟𝑢𝑠𝑡𝑒𝑥𝑖𝑠𝑡

) = 114.093 𝑖𝑛

𝐸𝑛𝑔𝑖𝑛𝑒𝑙𝑒𝑛𝑔𝑡ℎ = 128.2 𝑖𝑛

𝐸𝑛𝑔𝑖𝑛𝑒𝑤𝑒𝑖𝑔ℎ𝑡_𝐽𝑇8𝐷 = 𝐸𝑛𝑔𝑖𝑛𝑒𝑤𝑒𝑖𝑔ℎ𝑡 ∗𝑇ℎ𝑟𝑢𝑠𝑡𝑛𝑒𝑒𝑑𝑒𝑑

𝑇ℎ𝑟𝑢𝑠𝑡𝑒𝑥𝑖𝑡

= 4583.4

𝑁𝑎𝑐𝑒𝑙𝑙𝑒𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 1.1 ∗ 𝐸𝑛𝑔𝑖𝑛𝑒𝑑𝑛𝑒𝑒𝑑𝑒𝑑= 125.5 𝑖𝑛

𝑁𝑎𝑐𝑒𝑙𝑙𝑒𝑙𝑒𝑛𝑔𝑡ℎ = = 1.1 ∗ (0.7 ∗ 𝑒𝑛𝑔𝑖𝑛𝑒𝑑𝑛𝑒𝑒𝑑𝑒𝑑+ 𝑒𝑛𝑔𝑖𝑛𝑒𝑙𝑒𝑛) = 228.87 𝑖𝑛

𝑁𝑎𝑐𝑒𝑙𝑙𝑒𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠 = 𝑛𝑎𝑐𝑒𝑙𝑙𝑒𝑙𝑒𝑛𝑔𝑡ℎ ∗0.4

12= 7.6291 𝑖𝑛

𝑃𝑜𝑤𝑒𝑟𝑃𝑙𝑎𝑛𝑡𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑁𝑎𝑐𝑒𝑙𝑙𝑒𝑤𝑒𝑖𝑔ℎ𝑡 + 𝐸𝑛𝑔𝑖𝑛𝑒𝑤𝑒𝑖𝑔ℎ𝑡 ∗ 2 = 12248 𝑙𝑏

Payload

𝑃𝑎𝑦𝑙𝑜𝑎𝑑𝑤𝑒𝑖𝑔ℎ𝑡 = 25650 𝑙𝑏 (𝐴𝑡 𝑃𝑙𝑎𝑛𝑒′𝑠 𝐶𝑒𝑛𝑡𝑒𝑟 𝑀𝑎𝑠𝑠 − 𝐴𝑠𝑠𝑢𝑚𝑒𝑑)

𝐶𝑎𝑟𝑔𝑜𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟_𝑎𝑟𝑒𝑎 = 3400 𝑖𝑛2

𝐶𝑎𝑟𝑔𝑜𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟_𝑙𝑒𝑛𝑔𝑡ℎ = 42.3𝑖𝑛

𝐶𝑎𝑟𝑔𝑜𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟_𝑣𝑜𝑙𝑢𝑚𝑒 = 𝐶𝑎𝑟𝑔𝑜𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟_𝑎𝑟𝑒𝑎 ∗𝐶𝑎𝑟𝑔𝑜𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟𝑙𝑒𝑛𝑔𝑡ℎ

123= 83.2292

𝐶𝑎𝑟𝑔𝑜𝑤𝑒𝑖𝑔ℎ𝑡 = 200 𝑙𝑏

𝐹𝑟𝑜𝑛𝑡𝐵𝑜𝑥𝑒𝑠 = 3

𝐹𝑟𝑜𝑛𝑡𝐵𝑜𝑥𝑒𝑠𝑙𝑒𝑛𝑔𝑡ℎ = 152 in

𝐵𝑎𝑐𝑘𝐵𝑜𝑥𝑒𝑠 = 7

𝐵𝑎𝑐𝑘𝐵𝑜𝑥𝑒𝑠𝑙𝑒𝑛𝑔𝑡ℎ = 356 𝑖𝑛

𝐹𝑟𝑜𝑛𝑡𝐵𝑜𝑥𝑙𝑒𝑛𝑔𝑡ℎ_𝑡𝑜_𝑓𝑢𝑠𝑒𝑙𝑎𝑔𝑒 = 73.55 𝑖𝑛

𝐵𝑎𝑐𝑘𝐵𝑜𝑥𝑙𝑒𝑛𝑔𝑡ℎ_𝑡𝑜_𝑓𝑢𝑠𝑒𝑙𝑎𝑔𝑒 = 640.8 𝑖𝑛

𝐹𝑟𝑜𝑛𝑡𝐵𝑜𝑥𝑒𝑠𝑤𝑒𝑖𝑔ℎ𝑡 = 𝐶𝑎𝑟𝑔𝑜𝑤𝑒𝑖𝑔ℎ𝑡 ∗ 𝐹𝑟𝑜𝑛𝑡𝐵𝑜𝑥𝑒𝑠 + 40 ∗ 110 ∗ (𝐹𝑟𝑜𝑛𝑡𝐵𝑜𝑥𝑒𝑠

9) = 2066.7 𝑙𝑏

𝐹𝑟𝑜𝑛𝑡𝐵𝑜𝑥𝑒𝑠𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠 = 26.31 𝑓𝑡

𝐵𝑎𝑐𝑘𝐵𝑜𝑥𝑒𝑠𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠 = 82 𝑓𝑡

Class Seating

𝑓𝑖𝑟𝑠𝑡𝑐𝑙𝑎𝑠𝑠𝑤𝑒𝑖𝑔ℎ𝑡 = 2100 𝑙𝑏

𝑓𝑖𝑟𝑠𝑡𝑐𝑙𝑎𝑠𝑠𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠 = 𝑁𝑜𝑠𝑒𝑙𝑒𝑛𝑔𝑡ℎ +20

2= 23.85 𝑓𝑡

𝐸𝑐𝑜𝑛𝑤𝑒𝑖𝑔ℎ𝑡 = 17150 𝑙𝑏

𝐸𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠 = 74.63 𝑓𝑡

𝑃𝑎𝑦𝑙𝑜𝑎𝑑𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠 = 69.41

𝐹𝑢𝑒𝑙𝑤𝑒𝑖𝑔ℎ𝑡 = 46173𝑙𝑏

𝐹𝑢𝑒𝑙𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 50.41𝑙𝑏

𝑓𝑡3

𝐹𝑢𝑒𝑙𝑣𝑜𝑙𝑢𝑚𝑒𝑤𝑖𝑛𝑔=

𝐹𝑢𝑒𝑙𝑤𝑒𝑖𝑔ℎ𝑡

𝐹𝑢𝑒𝑙𝑑𝑒𝑛𝑠𝑖𝑡𝑦

2− 𝐶𝑎𝑟𝑔𝑜𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟𝑣𝑜𝑙𝑢𝑚𝑒

∗ 2 = 291.51 𝑓𝑡3

𝐹𝑢𝑒𝑙𝑤𝑖𝑛𝑔𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠= 13.5 𝑓𝑡

𝐹𝑢𝑒𝑙𝑓𝑢𝑠𝑒𝑙𝑎𝑔𝑒𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠= 43 𝑓𝑡

𝐹𝑢𝑒𝑙𝑐𝑔 =𝑓𝑢𝑒𝑙𝑣𝑜𝑙𝑢𝑚𝑒𝑤𝑖𝑛𝑔

∗ 𝐹𝑢𝑒𝑙𝑑𝑒𝑛𝑠𝑖𝑡𝑦 ∗ 2 ∗ 𝑓𝑢𝑒𝑙𝑤𝑖𝑛𝑔𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠+ 𝑐𝑎𝑟𝑔𝑜𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟𝑣𝑜𝑙𝑢𝑚𝑒

∗ 𝐹𝑢𝑒𝑙𝑑𝑒𝑛𝑠𝑖𝑡𝑦 ∗ 4 ∗ 𝑓𝑢𝑒𝑙𝑓𝑢𝑠𝑒𝑙𝑎𝑔𝑒𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠

𝐹𝑢𝑒𝑙𝑤𝑒𝑖𝑔ℎ𝑡

𝐹𝑢𝑒𝑙𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠(𝑐𝑔) = 24.22 𝑓𝑡

𝐹𝑖𝑥𝑒𝑑𝐸𝑞𝑢𝑖𝑝𝑊𝑒𝑖𝑔ℎ𝑡 = 14641 + 0.035 ∗ 𝑊𝑒𝑖𝑔ℎ𝑡𝑇𝑎𝑘𝑒𝑜𝑓𝑓 = 20599 𝑙𝑏

𝐹𝑖𝑥𝑒𝑑𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠 = 0

𝐿𝑎𝑛𝑑𝑖𝑛𝑔𝐺𝑒𝑎𝑟𝑤𝑒𝑖𝑔ℎ𝑡 = 0.04 ∗ 𝑊𝑒𝑖𝑔ℎ𝑡𝑇𝑎𝑘𝑒𝑜𝑓𝑓 = 6808.6 𝑙𝑏

Iterate

𝑋𝑤𝑖𝑛𝑔 = 51.0040

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑇𝑎𝑖𝑙𝑐𝑒𝑛𝑡𝑒𝑟𝑚𝑎𝑠𝑠 = 125.004 𝑓𝑡

Iterate Wing Position

full fuel, full payload

full fuel, no payload

no fuel, full payload

no fuel, no payload

𝑃𝑙𝑎𝑛𝑒𝑐𝑒𝑛𝑡𝑒𝑟𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 68.8328𝑓𝑡

𝑋𝑤𝑖𝑛𝑔 = 51.0040𝑓𝑡

Wing Sizing Equations

halfspan =b

2

𝑐𝑒𝑛𝑡𝑒𝑟 =𝑆

𝑏∗

2

1 + 𝑡𝑎𝑝𝑒𝑟

𝑡𝑖𝑝 = 𝑐𝑒𝑛𝑡𝑒𝑟 ∗ 𝑡𝑎𝑝𝑒𝑟

𝑐𝑒𝑛𝑡𝑒𝑟𝑞𝑢𝑎𝑟𝑡𝑒𝑟 =𝑐𝑒𝑛𝑡𝑒𝑟

4

𝑡𝑖𝑝_𝑞𝑢𝑎𝑟𝑡𝑒𝑟 = 𝑡𝑖𝑝/4

𝑚𝑎𝑐 = (2

3) ∗ ((𝑐𝑒𝑛𝑡𝑒𝑟 + 𝑡𝑖𝑝) − 𝑐𝑒𝑛𝑡𝑒𝑟 ∗

𝑡𝑖𝑝

𝑐𝑒𝑛𝑡𝑒𝑟 + 𝑡𝑖𝑝)

𝑦𝑚𝑎𝑐 = (1

3) ∗

𝑐𝑒𝑛𝑡𝑒𝑟 + 2 ∗ 𝑡𝑖𝑝

𝑐𝑒𝑛𝑡𝑒𝑟 + 𝑡𝑖𝑝∗ (

𝑏

2)

𝐶𝑜𝑀 = 𝑐𝑒𝑛𝑡𝑒𝑟𝑞𝑢𝑎𝑟𝑡𝑒𝑟 + 𝑦𝑚𝑎𝑐 ∗ 𝑡𝑎𝑛𝑑(𝑙𝑎𝑚𝑏𝑑𝑎)

Matlab Code

%MAE 159 %Sample Calculation clc clear format short tic M=0.80; H=35000; %ft W_cargo=2000; Range=2500; %nautical miles TOFL=6000; %ft LandV=130; %knots PASS=110; Abreast=4; num_eng=2; Aisle=1; %Sea Level Initial Conditions gamma=1.4; R=1718; %English Ps=2116.2; %lb/ft^2 Rhos=.002377; %Density Sea Level = slug/ft^3 Ts=543.67; %rankine from 84 F - hot day Rho=Ps/(R*Ts); DensR= Rho/Rhos; %Normal 35,000 ft Conditions Temperature=394.09; %Rankine Pressure=498.8667; %lb/ft^2 Density=0.00073751; %slug/ft^3 PR=0.23574; %Pressure Ratio DR=0.31027; %Density Ratio Visc= 0.00073751; %lb-s/ft^2 KVisc= 0.00040495; %ft^2/s %Engine location factor in weight calculation % engineplacement=1; % if engineplacement == 1 %engines on wing % Kw = 1.00; % Kts = 0.17; % dragfactor = 0.35; % else %engines on fuselage % Kw = 1.03; % Kts = 0.25; % drag_factor = 0.45; % end % % % if num_eng==4; % treq1=0.5; % treq2=3.0; % treq3=1.7; % treqAP=2.7; % treqLNG=.2; % elseif num_eng==3 % treq1=0.3; % treq2=2.7;

% treq3=1.5; % treqAP=2.4; % treqLNG=3.2; % elseif num_eng==2 % treq1=0; % treq2=2.4; % treq3=1.2; % treqAP=2.1; % treqLNG=3.2; % end %Conventional Code %Supercritical Code % filename=sprintf('SCrit%i.txt',swp); % [X,Y,X_fit,Y_fit,Y_des] = dataReadAndFit(filename,2,Mdiv); %2 is for polynomial 2nd power --> Can change

higher % tc = Y_des; % swp=35*(pi/180); %deg to rad ARmain= 6:1:12; SW= 10:5:40; SW=SW*(pi/180); qc = 0; Cl=0.50; i=1; Wf_Wto_JT9D_fix=.390; j=1; R=1; SWPmain=35*(pi/180); AR=7; % for jj=1:numel(ARmain) % AR=ARmain(jj) % for ii=1:numel(SW) % SWPmain=SW(ii); % Cl=0.50; i=1; % Wf_Wto_JT9D_fix=.390; j=1; R=1; while j while i deltaM=-0.2088*(Cl)^2 - 0.1048*Cl + 0.1206; %Supercritical Mdiv=M+.004-deltaM; if SWPmain==40*(pi/180) tc = -390.07*(Mdiv)^3 + 1027.3*(Mdiv)^2 - 903.44*(Mdiv) + 265.44; %40 degree sweep elseif SWPmain==35*(pi/180) tc = -36.767*(Mdiv)^3 + 98.73*(Mdiv)^2 - 89.063*(Mdiv) + 27.07; %35 degree sweep elseif SWPmain==30*(pi/180) tc = -13.822*(Mdiv)^3 + 37.829*(Mdiv)^2 - 34.956*(Mdiv) + 10.967; %30 elseif SWPmain==25*(pi/180) tc = -19.926*(Mdiv)^3 + 51.111*(Mdiv)^2 - 44.216*(Mdiv) + 12.978; %25 elseif SWPmain==20*(pi/180) tc = -18.32*(Mdiv)^3 + 46.069*(Mdiv)^2 - 39.129*(Mdiv) + 11.301; %20 elseif SWPmain==15*(pi/180) tc = -21.623*(Mdiv)^3 + 53.138*(Mdiv)^2 - 44.06*(Mdiv) + 12.402; %15 elseif SWPmain==10*(pi/180) tc = -20.061*(Mdiv)^3 + 49.022*(Mdiv)^2 - 40.478*(Mdiv) + 11.361; %10 end

Q=((cos(SWPmain))^2)*((tc)^2)*AR; Cl_max_to= -30.631*(Q)^2 + 12.95*(Q) + 1.1749; Cl_max_ldg= -24.489*(Q)^2 + 11.544*(Q) + 2.1581; Wload_ldg= (LandV/1.3)^2*((DensR*Cl_max_ldg)/296); %Wing load landing lb/ft^2 Vcruise=M*576.4; %0.592484 to convert ft/s to knots R_ao=Range+200+(0.75*Vcruise); % Wf_Wto_JT8D=(3E-24*(R_ao)^6)-(5E-20*(R_ao)^5)+(8E-17*(R_ao)^4)+(3E-12*(R_ao)^3)-(3E-

08*(R_ao)^2)+(0.0001*(R_ao))-0.0028; % sfc8=.78; % sfc9=.61; % Wf_Wto_JT9D=Wf_Wto_JT8D*(sfc9/sfc8); Chi=0; % plus 100% Max Fuel Wload_to=Wload_ldg/(1-Chi*Wf_Wto_JT9D_fix);%Wing load takeoff lb/ft^2 Wload_IC=(.965*Wload_to);%Wing Load Initial Cruise %.965 included climb and taxing Cl_IC= Wload_IC/(1481*PR*(M)^2); Clerror=abs(Cl-Cl_IC)/Cl_IC; if Clerror > .00000000001 Cl = (Cl_IC + Cl)/2; else i = 0; end end %3 Engine tofl=TOFL*10E-4; if num_eng==2 K =-0.3417*tofl^2+39.332*tofl-103.81; elseif num_eng==3 K=-0.3963*tofl^2+43.794*tofl-103.53; elseif num_eng==4 K= -0.4414*tofl^2+45.553*tofl-94.24; end W_T_7Vlo= (K/Wload_to)*Cl_max_to*.953; V_lo=1.2*((296*Wload_to)/(.953*Cl_max_to))^0.5; %Kts M_lo=V_lo/(661)/sqrt(.953); M_7lo=0.7*M_lo; %From JT9D Chart @ Sea Level TSLST=45000; T_m=37200; %At M=0.21 W_T=W_T_7Vlo*(T_m/TSLST); %% %Weight lambda=0.35; W_w=((.00945*AR^0.8*((1+lambda)^0.25)*1.01*(3.75^(.5))))/(((tc+.03)^0.4)*(cos(SWPmain))*(Wload_to^.695)); l_fus=(3.76*(PASS/Abreast)+33.2)*1.10; d_fus=(1.75*Abreast+1.58*Aisle+1.0)*1.10; W_fus=.6727*11.5*(l_fus)^.6*((d_fus)^.72)*(3.75)^.3; W_lg=.04; W_n_p=(.0555/W_T); W_n_p=W_n_p*0.8; %Composite W_ts=(.17+(.08/num_eng))*W_w; W_w_ts=W_w+(W_ts); W_w_ts=W_w+(W_ts)*0.70; %Composite W_pp=(1/(3.58*W_T)); W_f=1.0275*(Wf_Wto_JT9D_fix); W_f=W_f*0.85; %Composite N_FC=2; %Number of Flight Crew

N_CA=6; %Number of Cabin Attendants W_PL=215*PASS+W_cargo; W_FE=(132*PASS)+((300*num_eng))+(260*N_FC)+(170*6); %JT9D % W_FE=(132*PASS)+(0.9*(300*num_eng))+(260*N_FC)+(170*6); %JT8D Lighter W_FE=W_FE*0.9; %Composite syms f(x) f(x)=(W_w_ts*x^1.195)+(W_fus*x^.235)+((W_lg+W_n_p+W_pp+W_f+.035-1)*x)+(W_FE+W_PL); sol=vpasolve(f,x,[0,10E5]); if isempty(sol) qc = qc+1; break end W_takeoff=double(sol); S=W_takeoff/Wload_to; b=(AR*S)^(1/2); mac=S/b; T=W_takeoff/W_T; Te=T/num_eng; %Thrust Per Engine RNK=2.852E+6*.5; %/ft RN_wing=RNK*mac; Cf_wing=.0027; %Gotta Double Check RN_fus=RNK*l_fus; Cf_fus=.0018; %Wing Z=((2-M^2)*cos(SWPmain))/sqrt((1-M^2*(cos(SWPmain)^2))); K_w=1+(Z*tc)+(100*tc^4); Swet_wing= 2*(S-(20*30))*1.02; K_w=31.592*(tc)^3 - 2.4733*(tc)^2 + 1.6691*tc + 1.01; f_wing=Swet_wing*Cf_wing*K_w; %Fuselage Swet_fus=0.9*pi*d_fus*l_fus; L_D=l_fus/d_fus; K_fus= -0.0015*(L_D)^3 + 0.0394*(L_D)^2 - 0.3641*(L_D) + 2.3273; f_fus=Swet_fus*Cf_fus*K_fus; %Tail f_ts=.38*f_wing; %Nacelle Swet_Nac=2.1*sqrt(Te)*num_eng; K_Nac=1.25; f_Nac=Cf_wing*Swet_Nac*K_Nac; %Pylons f_pyl=.20*f_Nac; %Total f f_tot=(f_wing+f_fus+f_ts+f_Nac+f_pyl)*1.06; C_Do=f_tot/S; eff=1/(1.035+(0.38*C_Do*pi*AR)); Sigma=.5702; %Sigma = Density Ratio?? W_climb=((1+.965)/2)*W_takeoff; V_cl=1.3*(12.9/(f_tot*eff)^(1/4))*(W_climb/(Sigma*b))^(.5); a_climb=576.54; %speed of sound M_climb=V_cl/a_climb; T_r_cl=((Sigma*f_tot*(V_cl^2))/296)+(94.1/(Sigma*eff))*((W_climb/b)^2*(1/V_cl)^2); T15= 17.413*M_climb^4 - 44.292*M_climb^3 + 45.427*M_climb^2 - 30.119*M_climb + 27.441; T25 = -14.58919*M_climb^4 + 32.77303*M_climb^3 - 23.49468*M_climb^2 + 3.82968*M_climb + 15.36777; T20= (((20000-15000)*( T25-T15))/(25000-15000)+T15)*1000; c15 = (-2.645E-04*T15^3 + 1.991E-02*T15^2 - 5.188E-01*T15 + 5.022); %JT9D

c25 = (-0.01471*T25^3 + 0.77220*T25^2 - 13.33893*T25 + 76.43786); %JT9D % c15 = (-2.645E-04*T15^3 + 1.991E-02*T15^2 - 5.188E-01*T15 + 5.022)*1.1; %JT8D % c25 = (-0.01471*T25^3 + 0.77220*T25^2 - 13.33893*T25 + 76.43786)*1.1; %JT8D c20 = ((20000-15000)*(c25-c15))/(25000-15000)+c15; T_a=(Te/TSLST)*T20; RC=(101*((num_eng*T_a)-T_r_cl)*V_cl)/W_climb; Time_cl=H/RC; %Min Range_cl=V_cl*(Time_cl/60); Wf_cl=(num_eng*T_a)*c20*(Time_cl/60); W0=W_takeoff-Wf_cl; W1=(1-Wf_Wto_JT9D_fix)*W_takeoff; CL_avg=((W0+W1)/(2*S))/(1481*PR*M^2); CDi=(CL_avg)^2/(pi*AR*eff); deltaCD_C=.0010; CD=C_Do+CDi+deltaCD_C; LD=CL_avg/CD; T_r_range=((W0+W1)/2)/LD; T_r_range_JT9D=T_r_range*(TSLST/Te); T_r_range_per=T_r_range_JT9D/num_eng; tsfc=[50:5:70]; for i = 1:length(tsfc); fig = sprintf('JT9D_35000_%i.txt',tsfc(i)); [X,Y,X_fit,Y_fit,X_des] = dataReadAndFitInverted(fig,2,T_r_range_per/1000); mach(i) = X_des; end LB = find(mach<M); i1 = LB(end); UB = find(mach>M); i2 = UB(1); diff = tsfc(i2)-tsfc(i1); c_range = (diff)/100*(M-mach(i1))/(mach(i2)-mach(i1))+tsfc(i1)/100; Rcruise=(V_cl/(c_range))*LD*log(W0/W1); %JT9D % Rcruise=(V_cl/(c_range*1.1))*LD*log(W0/W1); %JT8D R=Range_cl+Rcruise; clear x clear f R_error=abs(R-R_ao)/R_ao; if R_error < .01 j = 0; elseif R > R_ao Wf_Wto_JT9D_fix = Wf_Wto_JT9D_fix-.001; else Wf_Wto_JT9D_fix = Wf_Wto_JT9D_fix+.001; end end % if qc == 1 %continue check % qc = 0; % Weight(jj,ii) = 0; % DOC_ton_mile(jj,ii) = 0; % continue % end %PAGE 9 CLic=(W0/S)/(1481*PR*M^2); Cdi= (CLic^2)/(pi*AR*eff); C_D=C_Do+Cdi+.0010;

L_D=CLic/C_D; T_req=W0/L_D; T_req_eng=T_req/num_eng; T_req_JT9D=T_req_eng*(TSLST/Te); % %Climb Gradients %1st Segment CL_to=Cl_max_to/(1.2)^2; CL_to_maxto=1/(1.2)^2; deltaCDo= 0.16*(CL_to_maxto)^4 - 0.1489*(CL_to_maxto)^3 + 0.0894*(CL_to_maxto)^2 - 0.0707*(CL_to_maxto) +

0.0328; CDgear=C_Do; CD1=C_Do+deltaCDo+CDgear+((CL_to)^2/(pi*AR*eff)); LD_to=CL_to/CD1; Treq1=W_takeoff/LD_to; Ta_eng1=(Te/TSLST)*34500; %34500 from the graph Grad1=((num_eng-1)*Ta_eng1-Treq1)/W_takeoff*100; % if Grad1 <Treq1 % DOC_ton_mile=0; % continue % end %Second Segment CD2=C_Do+deltaCDo+((CL_to)^2/(pi*AR*eff)); LD_2=CL_to/CD2; Treq2=W_takeoff/LD_2; Grad2=((num_eng-1)*Ta_eng1-Treq2)/W_takeoff*100; % if Grad2 <Treq2 % DOC_ton_mile=0; % continue % end %Third Segment Cl_35clean=-303.8*(tc)^3 + 79.898*(tc)^2 - 2.522*(tc) + 0.859; V3=1.2*sqrt((296*Wload_to)/(.925* 1.10)); %Knots %.925 is at 1000ft on a hot day M_3=V3/659; Cl_3=Cl_35clean/(1.2)^2; CD3=C_Do+((Cl_3)^2/(pi*AR*eff)); LD3=Cl_3/CD3; Treq3=W0/LD3; Ta_eng3=(Te/TSLST)*26000; Grad3=((num_eng-1)*Ta_eng3-Treq3)/W_takeoff*100; %percent % if Grad3 <Treq3 % DOC_ton_mile=0; % continue % end %Approach Cl_Ap=Cl_max_to/(1.3)^2; Cl_Ap_max=1/(1.3)^2; deltaCDoAp= 0.16*(Cl_Ap_max)^4 - 0.1489*(Cl_Ap_max)^3 + 0.0894*(Cl_Ap_max)^2 - 0.0707*(Cl_Ap_max) +

0.0328; CD_Ap=C_Do+deltaCDoAp+((Cl_Ap)^2/(pi*AR*eff)); LD_Ap=Cl_Ap/CD_Ap; Wland=Wload_ldg*S; Treq_Ap=Wland/LD_Ap; V_Ap=sqrt((296*Wload_ldg)/(.953*Cl_Ap)); Ta_Ap=(Te/TSLST)*29500; GradAp=((num_eng-1)*Ta_Ap-Treq_Ap)/Wland*100; %percent

% if GradAp <TreqAP % DOC_ton_mile=0; % continue % end %Landing Cl_land=Cl_max_ldg/(1.3)^2; Cl_Clmax=1/(1.3)^2; deltaCDoLand= 0.16*(Cl_Clmax)^4 - 0.1489*(Cl_Clmax)^3 + 0.0894*(Cl_Clmax)^2 - 0.0707*(Cl_Clmax) + 0.0328; CDland=C_Do+deltaCDoLand+CDgear+((Cl_land)^2/(pi*AR*eff)); LD_land=Cl_land/CDland; Treq_land=Wland/LD_land; Ta_Land=(Te/TSLST)*37200; GradLand=(num_eng*Ta_Land-Treq_land)/Wland*100; %percent % if GradLand <TreqLNG % DOC_ton_mile=0; % continue % end %% Direct Operating Costs dist=Range*1.15; %statute miles t_gm=.25; %Time for Ground Manuever t_cl=.18; %time to climb t_d=0; %time to descend t_am=.10; %time for air manuever d_c=83; %distance to climb d_d=0; %distance to descend Vcr=Vcruise*1.15; %true airspeed Ka=1.02; %airwaydistance increment t_cr=((dist+Ka+20)-(d_c+d_d))/Vcr; %block speed Vb=dist/(t_gm+t_cl+t_d+t_cr+t_am) ; %block time hour t_b=t_gm+t_cl+t_d+t_cr+t_am; %block fuel f_gm=0; f_cl=9624; f_cr_am=T_r_range*(c_range)*(t_cr+t_am); %Current Tech JT9D sfc f_d=0; Fb=f_gm+f_cl+f_cr_am+f_d; %Flying Operations Cost %Flight Crew P=W_PL/2000; %Tons dolla_blkhr=17.849*(Vcruise*(W_takeoff/(10^5)))^0.3+40.83; %2 man crew Ctm_FC=dolla_blkhr/(Vb*P); %Fuel&Oil Cf=.40*(1/6.4); %#/lb CoT=2.15; %#/lb Ctm_FO=1.02*((Fb*Cf)+(num_eng*CoT*.135))/(dist*P); %Hull Insurance Wa=W_takeoff*(1-Wf_Wto_JT9D_fix)-W_PL-(W_pp*W_takeoff); Ca=2.4*10^6+(87.5*Wa); Ce=590000+16*Te; CT=Ca+(num_eng*Ce); IRa=.01; U_hull=(630)+4000/(1+(1/(t_b+.50))); Ctm_Hull=(IRa*CT)/(U_hull*Vb*P);

%Direct Maintenance %Airframe Labor KFHA=(4.9169*(log10(Wa/1000)))-6.425; KFCA=.21256*(log10(Wa/1000))^3.7375; TF=t_b-t_gm; R_l=8.6; Ctm_AL=(R_l*((KFHA*TF)+(KFCA)))/(Vb*t_b*P); %Airframe Material CFHA=1.5994*(Ca/10^6)+3.4263; CFCA=1.9229*(Ca/10^6)+2.2504; Ctm_AM=(CFHA*TF+CFCA)/(Vb*t_b*P); %Engine-Labor KFHE=(num_eng*(Te/1000))/(.82715*(Te/1000)+13.639); KFCE=.20*num_eng; Ctm_EL=(KFHE*TF+KFCE)/(Vb*t_b*P)*R_l; %Engine Material CFHE=(28.2353*(Ce/10^6)-6.5176)*num_eng; CFCE=(3.6698*(Ce/10^6)+1.3685)*num_eng; Ctm_EM=(CFHE*TF+CFCE)/(Vb*t_b*P); Ctm_EM=Ctm_EM; %JT9D Ctm_EM=Ctm_EM*0.9; %JT8Df %Total Maintenance Burden Ctm_TM=(Ctm_AL+Ctm_AM+Ctm_EL+Ctm_EM)*2; %Depreciation ad=.06*(CT-(num_eng*Ce)); bd=.30*num_eng*Ce; Ctm_dep=(1/(Vb*P))*((CT+ad+bd)/(14*U_hull)); DOC_ton_mile=Ctm_dep+Ctm_TM+Ctm_Hull+Ctm_FO+Ctm_FC DOC_pass_mile=DOC_ton_mile*P/PASS Weight= W_takeoff % end % end clear, clc Weight_Takeoff = 170215.55; %Origin to Fuselage Fuselage_coefficient = 1078.1; Fuselage_weight = Fuselage_coefficient*Weight_Takeoff^0.235; %lb Nose_length = 166.28/12; %ft Fuselage_length = 127.51; %ft Fuselage_cg = Fuselage_length*0.4; %ft S = 1737.8; %ft^2 b = 110.29; %ft taper = 0.35; wing_c = 0.007766; Wing_weight = wing_c*Weight_Takeoff^1.195; %lb engine_weight = 3218; %lb thrust_given_exit = 14500; %lb thrust_needed = 22433.58; %lb engine_diam = 95.6; %in engine_diameter_scaled = engine_diam*sqrt(thrust_needed/thrust_given_exit); %in engine_length = 128.2; %in engine_weight = engine_weight*thrust_needed/thrust_given_exit; %lb nacellediam = 1.1*engine_diameter_scaled; %in nacellelength = 1.1*(0.7*engine_diameter_scaled + engine_length); %in nacellecg = nacellelength*0.4/12; %ft nacelle_inner = 0.7*engine_diameter_scaled/12; %in nacelleweightt = 0.0181*Wgt_Takeoff; %ft powerplantweight = nacelleweightt+engine_weight*2; %lb PayloadWeight = 25650; %lb %Assumed to be at center of mass

cargo_contarea = 3400; %in^2 cargo_contlength = 42.3; %in cargo_contvolume = cargo_contarea*cargo_contlength/12^3; %ft^3 cargo_b = ceil(750/cargo_contvolume); cargoweight = 2000/cargo_b; %lb FrontLoc = 73.55; %in BackLoc = 640.8; %in Frontb = 3; Frontlength = 51*(Frontb-1)+50; %in Backb = 7; Backb_len = 51*(Backb-1)+50; %in FrontWeight = cargoweight*Frontb+40*110*(Frontb/9); %lb FrontCG = Nose_length+(FrontLoc+Frontlength/2)/12; %ft BackWeight = cargoweight*Backb+40*110*(Backb/9); %lb BackCG = Nose_length+(BackLoc+Backb_len/2)/12; %ft firstclasswt = 175*12; %lb firstclassCG = Nose_length + 20/2; %ft econclasswt = 175*98; %lb econclassCG = Nose_length+20+81.56/2; %ft PayloadCG = (firstclasswt*firstclassCG+econclasswt*econclassCG+FrontWeight*FrontCG+BackWeight*BackCG)/PayloadWeight;

%ft FuelWT= 46173; %lb Jetfuel_dens = 50.41061; %in lb/ft^3 fuelv_wing = FuelWT/Jetfuel_dens/2 - cargo_contvolume*2; %ft^3 fuelwingCG = 13.5; %ft fuelfusCG = (43*2)/2; %ft FuelCG = (fuelv_wing*Jetfuel_dens*2*fuelwingCG + cargo_contvolume*Jetfuel_dens*4*fuelfusCG)/FuelWT; %ft FixEqWT = 14641 + 0.035*Wgt_Takeoff; %lb FixECG = 0; %ft lg_wgt = 0.04*Wgt_Takeoff; %lb X_WING = Fuselage_length*0.40; %ft horizontaltailCG = 117.66+24.77/2-5; %ft verticaltailCG = horizontaltailCG - 5; %ft TailSurWT = 0.17*Wing_weight; while 1 %%%% Horizontal Tail len_hortail = horizontaltailCG-X_WING-wing_CoM; S_hortain = TailSizing(S,mac,b,len_hortail,0); %%%% Vertial Tail len_vertail = verticaltailCG-X_WING-wing_CoM; S_vertail = TailSizing(S,mac,b,len_vertail,1); hortailWT = TailSurWT*S_hortain/(S_hortain+S_vertail); vertailWT = TailSurWT-hortailWT; PP_CoM = X_WING+nacellecg; check = 0; n = 0.001; WingCG = X_WING + wing_cg; Fuel_CoM = X_WING + FuelCG; for i = 1:4 if i == 1 %all fuel all payload lane_CoM(i) =

CoM(Fuselage_weight,Fuselage_cg,Wing_weight,WingCG,FuelWT,Fuel_CoM,powerplantweight,PP_CoM,PayloadWeight,Payload

CG,hortailWT,horizontaltailCG,vertailWT,verticaltailCG); diff(i) = Plane_CoM(i) - WingCG; if Plane_CoM(i) > (WingCG+mac*0.1) X_WING = X_WING + n; check = 0; disp('xa') break elseif Plane_CoM(i) < (WingCG-mac*0.1) X_WING = X_WING - n; check = 0;

disp('xb') break else check = check + 1; end elseif i == 2 %all fuel no payload Plane_CoM(i) =

CoM(Fuselage_weight,Fuselage_cg,Wing_weight,WingCG,FuelWT,Fuel_CoM,powerplantweight,PP_CoM,0,PayloadCG,hortailWT,

horizontaltailCG,vertailWT,verticaltailCG); diff(i) = Plane_CoM(i) - WingCG; if Plane_CoM(i) > (WingCG+mac*0.1) X_WING = X_WING + n; check = 0; disp('ya') break elseif Plane_CoM(i) < (WingCG-mac*0.1) X_WING = X_WING - n; check = 0; disp('yb') break else check = check + 1; end elseif i == 3 %no fuel all payload Plane_CoM(i) =


CG,hortailWT,horizontaltailCG,vertailWT,verticaltailCG); diff(i) = Plane_CoM(i) - WingCG; if Plane_CoM(i) > (WingCG+mac*0.1) X_WING = X_WING + n; check = 0; disp('za') break elseif Plane_CoM(i) < (WingCG-mac*0.1) X_WING = X_WING - n; check = 0; disp('zb') break else check = check + 1; end elseif i == 4 %no fuel no payload Plane_CoM(i) =


CG,hortailWT,horizontaltailCG,vertailWT,verticaltailCG); diff(i) = Plane_CoM(i) - WingCG; if Plane_CoM(i) > (WingCG+mac*0.1) X_WING = X_WING + n; check = 0; disp('4a') break elseif Plane_CoM(i) < (WingCG-mac*0.1) X_WING = X_WING - n; check = 0; disp('4b') break else check = check + 1; end end end if check == 4

break end disp(' ')

halfspan = b/2; center = S/b * 2/(1+taper); tip = center*taper; center_quarter = center/4; tip_quarter = tip/4; mac = (2/3)*((center+tip) - center*tip/(center+tip)); y_mac = (1/3)*(center+2*tip)/(center+tip)*(b/2); CoM = center_quarter+y_mac*tand(lambda); end

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Justin Oyas MAE 159 Final Report 1 - WordPress.com · Justin Oyas – MAE 159 Final Report 5 1. Introduction The objective of the MAE 159 final report is to study the cost and performance