AAE 451, SENIOR DESIGN - Purdue University...AAE 451, SENIOR DESIGN CONCEPTUAL DESIGN REVIEW TEAM 3: GOLDJET DIANE BARNEY DONALD BARRETT MICHAEL COFFEY JON COUGHLIN MARK GLOVER KEVIN

AAE 451, SENIOR DESIGN

CONCEPTUAL DESIGN REVIEW

TEAM 3: GOLDJET

DIANE BARNEY

DONALD BARRETT

MICHAEL COFFEY

JON COUGHLIN

MARK GLOVER

KEVIN LINCOLN

ANDREW MIZENER

JARED SCHEID

ERIC SMITH

Team GoldJet Conceptual Design Review

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Table of Contents I. Mission Statement 2

II. Outline of NASA Competition 2

III. Key Assumptions 3

IV. Executive Summary 4

V. System Requirements Definition 5

Market Research / Business Plan 5

Design and Economic Mission 6

Requirements Development 7

VI. System Definition 8

Constraint Analysis 8

Morphological Matrices 9

Pugh’s Method 11

Final Concept Refinement 12

VII. Conceptual Design 19

Aircraft Walk-around 19

Key Design Parameters 20

Aircraft Performance 20

Sizing 23

Aerodynamics 30

Propulsion 43

Weights and Balance 61

Structures 68

Stability and Control 76

Applications 79

VIII. Concept Summary 86

Requirements Compliance 86

Plausibility 88

Detailed Design Work Remaining 88

IX. References 89

X. Appendix 91


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I. Mission Statement

To design a profitable supersonic aircraft capable of Trans-Pacific travel to meet the

needs of airlines and their passengers around the world.

II. Outline of NASA Competition

The NASA Aeronautics Research Mission Directorate’s (ARMD) 2008-2009

University Competition calls for the design of an N+2 generation supersonic

aircraft which would have initial operational capability (IOC) in 2020. More

specific goals for the aircraft as outlined by the competition guidelines include:

Cruise speed of Mach 1.6 to 1.8

Design Range of 4000 nautical miles

Payload of 35-70 passengers, mixed class

Fuel Efficiency of 3 passenger-miles per pound of fuel

Takeoff field length < 10,000 feet for airport compatibility

Supersonic cruise efficiency

Low sonic boom (<70 PldB)

In addition, entries to the competition are to identify the possible market for a small

supersonic airliner, develop design and economic missions for the aircraft

(including likely routes), identify technologies that might enable the aircraft design,

and complete a conceptual sizing.

These guidelines proved to be a starting point for our aircraft. Deviations occurred

as a result of assumptions made, a market study, and sizing based on historical data.


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III. Key Assumptions

Our aircraft, the GoldJet FAST, is designed to have initial operational capability in

2020, with entry into service by 2023. In order for the FAST to be successful, there

are several precedents, both technological and political, that must be met.

1. There must be a change to the FAR regulations that prohibit supersonic

flight over land. Supersonic flight over land has been prohibited in the

United States since March 1972. Changes to these regulations must come in

the form of a complete repeal, or a modification to allow certain supersonic

corridors for flight over areas of low population density. According to a

statement released by Carl Burleson, Director of Environment and Energy

on October 16, 2008, it is anticipated that future regulations “would propose

any future supersonic airplane produce no greater noise impact on a

community than a subsonic airplane.”[1]

A design with a sonic boom

overpressure less than 0.3 lb/ft2 is a target of GoldJet in order to meet these

new anticipated regulations.

2. A number of Supersonic Business Jet (SSBJ) concepts are currently in

design and study phases of development. A change in supersonic flight

regulations would pave the way for the success of SSBJ’s over the next 10-

15 years, fueling technological innovation. Our aircraft will depend on this

field of research for products like more efficient supersonic engines, and

possibly composite materials with better temperature resistance.

IV. Executive Summary

The GoldJet FAST is the first trans-Pacific supersonic airliner capable of linking

the United States with Asia. Our aircraft’s design mission flies from Los Angeles

International Airport (LAX) to Narita International Airport (NRT) in Tokyo, Japan.

With a cruise Mach number of 1.8 the FAST cuts 46% off the 10 hour 45 minute

subsonic flight, dropping it to 5 hours and 45 minutes. The FAST has a double-delta

“cranked arrow” wing planform with 4 low-bypass turbofan engines podded under


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its wings. The FAST uses control canards for pitch control, a vertical tail and rudder

for lateral stability, and ailerons for roll control. Wing space has been budgeted for

high-lift devices, which allow for takeoff on runways smaller than 10,000 ft. The

maximum range for the FAST is 4,800 nm and an extra 240 nm has been factored in

for a missed approach and diversion to another airport.

Figure 1 Exterior aircraft walk-around

40-Passenger 2x2

luxury cabin

All fuel storage

in fuselage

Cranked-Arrow

wing planform

Vertical tail (no

horizontal tail)

4 Low-bypass

turbofans under wing

Under-Nose Cameras

for Landing Assistance

Canards for Pitch Control

and Low Boom

Two Emergency

Exits per side


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Figure 2 Requirements compliance matrix

Figures 1 and 2 show our aircraft walk-around chart and requirements compliance

matrix, respectively.

V. System Requirements Definition

Market Research / Business Plan

In determining the requirements our aircraft needs to fulfill, it is crucial to have an

accurate image of the market in order to ensure our aircraft’s profitability. Two

important factors for consideration resulted from market research:

Percentage of travelers to which our supersonic transport jet will appeal

Routes on which our supersonic transport jet will best be utilized (ie.

selection of most profitable city-pairs)

These two factors guide our decision for aircraft capacity, and allow a reasonable

estimate for total aircraft sold and profitability.

Requirement Units Target Threshold Design

TO Field Length ft 10,000 12,000 11,380

Boom Overpressure lb/ft2 0.3 0.5 0.36

TO Gross Weight lb 300,000 400,000 329,000

Landing Field Length ft 10,000 12,000 9,000

Range nmi 5,650 4,707 4,800

Turnaround Time min 30 45 27

Number of Pasengers pax 40 40 40

Crew # 3 3 3

Cruise Speed Mach 2.1 1.75 1.8

Max Speed Mach 2.2 1.8 1.9

Trip Time min 380 415 345

Cabin Volume ft3/pax 60 55 75.8

Cruise Altitude ft 50000 45000 45,000

Cruise Efficiency paxmi/lb fuel 3.5 3 1.2

Second Segment Climb Gradient % 5.0 3.0 3.0

Seat Pitch in 40 38 50

Seat Width in 26 22 26

Aisle Width in 22 20 22

Aisle Height in 76 72 76

Gold Jet Requirements Compliance Matrix


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We determined that the FAST will have a higher operating cost than subsonic

transport jets and, subsequently, a higher average ticket price. Our aircraft ticket

price was assumed to be comparable to a typical first class subsonic ticket price and

further research revealed approximately 2% of current airline travelers pay these

higher fares. We also made the assumption that an additional 1% of the total market

will be willing to purchase our higher priced-tickets for the added convenience of

shorter trip times and the novelty of flying in a supersonic jet. With these

assumptions we estimate that our supersonic transport has the potential of capturing

3% of the market share.

We next needed to determine routes that our aircraft would be most profitable to

operate on. City pairs were chosen based on the following criteria:

Sufficient distance so as to have significant reduction in trip time when

flying at supersonic speeds (approximately 1000 nmi and above)

Large volume of passengers to provide a market foothold

Transoceanic flight or easy access to supersonic corridors

Connection of large economic, social and cultural centers

Based on these criteria a collection of city-pairs were chosen and historical data on

traffic between them were used to estimate the volume of passengers our jet could

expect to carry on an average day.

The above market research yielded two significant conclusions:

1. The optimum capacity for our jet in order to minimize the number of empty

seats is 40 passengers

2. We project to sell at least 120 aircraft over its lifetime

Design and Economic Mission

Having determined the size of our market, we then selected specific missions to

define our aircraft requirements. The following section will highlight the details of

our economic and design missions.


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Our economic mission is the most profitable route for the FAST, combining both

high potential passenger volume and sufficient range for supersonic flight to

provide appreciable time savings. Based on market research, our economic mission

is John F. Kennedy International Airport (JFK, in New York) to Los Angeles

International Airport (LAX). This mission connects the two largest population

centers in the United States, both large and powerful economic and social centers.

This mission has a range of 2,520 nmi (great circle distance) with an estimated extra

400 nmi required for diversion to supersonic corridors, as outlined in the SRR. This

mission will be important during the design phase of our aircraft, as performance

will need to be such to maximize profitability.

Our design mission is the longest and furthest we would ever expect to travel,

which is crucial to sizing our aircraft. Our original design mission was from Los

Angeles International Airport (LAX) to Pudong International Airport (PVG) in

Shanghai, China. However, after completing more detailed sizing, and accounting

for headwinds, we decided that this range capability was not feasible at maximum

payload. In order to meet our threshold design range, we chose our design mission

to be from LAX to Narita International Airport (NRT) in Tokyo, Japan. This range

enables our aircraft to link Asia to North America, North America to Europe, and

the Middle East to most destinations. The design range is 4800 nmi. This mission

ultimately determines the requirements for our design as the range will determine

the required fuel and the gross takeoff weight of the aircraft.

Requirements Development

Our primary aircraft requirements are based on the range and capacity dictated by

the market research and design mission stated above. Our aircraft will need to carry

40 passengers a distance of 4800 nmi (plus necessary reserve fuel). Additional

requirements were developed to account for those customer needs and wants not

exclusively dependent on range.

A House of Quality (HoQ) was constructed in which we determined customer

wants and needs and ranked them according to their importance relative to each


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other. We found our most important customer needs to be profitable operations,

low trip time, long range, and marketability. The engineering requirements we

determined to be most important in satisfying our customers’ needs are cruise

speed, block time, and cruise efficiency. Hence it is clear that our cruise conditions

are crucial to meeting the customer needs. Based on this we decided to design the

FAST for a cruise Mach number of 1.8, allowing our aircraft to cruise at twice the

speed of most subsonic carriers.

Other important requirements to consider are takeoff and landing field lengths for

airport compatibility. All of the airports considered in our business plan have a

minimum field length of 10,000 feet or longer (many have field lengths greater than

12,000 feet). Hence, we have set a target field length of 10,000 feet with 12,000 feet

as our design threshold.

VI. System Definition

Constraint Analysis and Sizing

To begin analysis of our aircraft concepts, we begun with a constraint analysis,

using 5 main performance constraints:

Steady, level flight at Mach 1.8 and 45,000 ft

Subsonic 2g maneuver at 250 knots and 10,000 ft (such as would be executed

inside a landing pattern)

Takeoff ground roll of 6,000 ft at an altitude of 1,000ft on a +15° hot day

Landing ground roll of 6,000 ft at an altitude of 1,000ft on a +15° hot day

3% Second Segment Climb Gradient above an altitude of 1,000 ft on a +15°

hot day

The takeoff and landing distances come from a desired usable field length of 10,000

ft, and include an additional 2/3 of the takeoff distance (i.e. desiredTOTO sss 3

2 ).

Using the equations provided by Professor Crossley, and modifying the spreadsheet

he provided, we generate the constraint diagram shown in Figure 3. Each of these

constraints required assumptions about the aircraft, which were generated

independent of the aircraft configuration. Not having done any substantive analysis


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at this point, some assumptions were generic, and not based on any detailed

understanding of our aircraft. However, we had already determined that we would

have four engines. After investigating the available design space for two-, three-,

and four-engined designs, only four engines gave us practical values of TSL/W0 and

wing loading. Once we chose a concept and further developed our design (and

engine model), our constraints would become more defined and would give way to

carpet plots, which would allow us to choose a solid design point.

Figure 3: Constraint Diagram

Morphological Matrices

To begin defining our concepts, we used a two step approach. First, to better get an

idea of the options available to us, we broke down our design into eight categories:

wing planform, wingtips, engine location, canards, tail configuration, wing location,

fuselage type, and landing gear. We omitted low-boom technologies, high-lift

devices (flaps, slats, etc.) and spoilers, on the grounds that choosing one doesn’t

preclude the use of any of the others: flaps and slaps can be used on the same wing,

GoldJet Constraint Diagram

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

50 60 70 80 90 100 110 120 130 140 150

W/S [lb/ft2]

TS

L/W

0 Steady, Level Flight (1g), M = 1.8 @ h

= 45K ft

Subsonic 2g Manuever, 250kts @ h

=10K ft

Takeoff Ground Roll 5400 ft @ h = 1K

ft, +15° Hot Day

Landing Ground Roll 5400 ft @ h = 1K

ft, +15° Hot Day

Second Segment Climb Gradient

Above h = 1K ft, +15° Hot Day

Landing Ground Roll 5400 ft @ h = 1K

ft, +15° Hot Day, No TR


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but you cannot choose both a conventional tail and T-tail or low and high wing on

the same concept. We then listed all of the possible choices for each category in a

Morphological Matrix (Figure 4).

Figure 4: Morphological Matrix

Then, splitting up into four teams, we each chose a concept design – one choice

from each of the categories in the Morphological Matrix. In addition, we chose a

datum for Pugh’s Method: Concorde. Concorde was chosen because it is the only

commercial supersonic transport to achieve large-scale commercial success. The

concepts are summarized in Figure 5.

Figure 5: Initial Concepts

From these, we can note an immediate narrowing of the design choices – it can be

plainly seen that some choices are not feasible for a supersonic transport. Among

the four concepts, there are two wing platforms, two fuselage types, and one

landing gear style.


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Pugh’s Method

Following Pugh’s Method, we took a first round of comparisons, shown in Figure 6.

Note that these comparisons were based off the team’s engineering experience and

expertise, and had not been verified by any quantitative engineering analyses.

Figure 6: Pugh's Method, Round 1

At first, it seems that Concepts 1, 3, and 4 clearly distinguish themselves, but at a

second glance, none of the designs have very many negatives, and there is little to

distinguish them. We determined that this was due to our choice of datum:

Concorde. Attempting to compare a concept to come to market in 2020 to one

designed in the 1960s didn’t allow for proper analysis or meaningful comparisons.

To rectify this disparity, we switched datums – from Concorde to our Concept 2,

which was the concept most similar to the Concorde in the first round of Pugh’s

Method. This allowed us to compare our design on a level plane. In addition,

several categories were deemed either too difficult to judge or irrelevant, and were

eliminated in the second round.

The second round of Pugh’s Method compared our concepts 1, 3, and 4 against the

new datum, Concept 2 (Figure 7).


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Figure 7: Pugh's Method, Round 2

Final Concept Refinement

After the second round, we noticed that Concepts 1 and 3 were the most

outstanding, and more so, were fundamentally different: Concept 1 was based

around a double-delta wing, and Concept 3 a joined wing. We determined then to

finish with two concepts for further study: A double-delta concept, and a joined

wing concept (Figure 8). The joined wing concept is Concept 3, and the double

delta concept is a result of taking the best double delta concept (Concept 1), and

blending it with the other double delta concepts (2 and 4) to create the best possible

design.

Figure 8: Final Concepts


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These two concepts are illustrated here: Figures 9, 10, and 11 show our joined wing

design, and Figure 12 shows a sample double-delta configuration (please note that

this is simply an example, further analysis was done to refine it).

Figure 9: Joined Wing Concept, Front View

Figure 10: Joined Wing Concept, Side View


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Figure 11: Joined Wing Concept, Top View

Figure 12: Example Double-Delta Configuration


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These two concepts have very different needs and areas for further analysis, as well

as different benefits. The double-delta is relatively simple structurally and

aerodynamically, but is fundamentally a compromise, and requires further analysis

to choose the engine placement, canard choice, and tail configuration. The joined

wing is complicated structurally and aerodynamically, but has the potential for

some extremely attractive benefits (especially in aerodynamics). However,

choosing a joined wing planform would allow for fewer variations in layout and

configuration, simply by the nature of the planform. For example, the choices left to

make for the double delta design were engine placement and longitudinal control

surface (i.e. horizontal tail or canards. For a joined wing, mounting the engines on

the wing and forcing them to bear their weight would make the wing structure

enormously heavy and complex further exacerbating an already complex

optimization routine, mounting them on the fuselage precludes this. In addition, a

joined wing does not need an additional longitudinal control surface, as the wing

structure provides both pitch and roll control already. A canard would have been

employed to improve the aerodynamic properties of the design, as stated in

Wolkovitch (1985)[2]

.

Wing Planform

The first step in narrowing our concept choices down to one single design was to

select the wing planform – from this selection would flow many of our other design

decisions.

The first design presented was the joined wing planform. The benefits of this

planform were largely in the areas of aerodynamics, and the difficulties mostly in

structures (Wolkovitch, 1985 and Gallman and Kroo, 1993)[3]

. The double-delta, on

the other hand, is a compromise between high- and low-speed flight

aerodynamically, and has distinct structural benefits.

After some initial analysis of the joined wing design using lifting-line theory and a

much more detailed investigation on the magnitude of analysis required for a joined

wing, it was determined that the structural concerns of the joined wing outweighed


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the aerodynamic benefits, which proved to be difficult to quantify. In addition, the

double delta proved to be simple structurally, have good efficiency at high and low

speeds, have large internal volume for fuel, and provide weight savings. For these

reasons, the double-delta design was chosen.

After some further analysis and the discovery of a paper on the optimization of

cruise speed impact on supersonic aircraft planform (Hermann, 2004)[4]

, the final

design chosen is technically a “Cranked Arrow”, which is similar to a double-delta

in that the leading edge has two distinct sections – an area of high sweep closer to

the root, and an area of lower sweep closer to the tip, but differs from it in that the

wing root is not straight, but rather kinked towards the leading edge in planform

view, forming a rough arrowhead shape.

Figure 13 - Cranked Arrow Planform (Hermann, 2004)


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Engine Placement

Once the double delta design had been chosen, it remained to choose where the

engines would be placed: under the wings, in the manner of Concorde, or podded

and aft on the fuselage similar to the Embraer ERJ 145 (but with two engines per

pod).

To assist in making this decision, we evaluated the two options against each other

for eight design considerations: weight, noise, complexity, ease of maintenance,

airflow into the engine, dynamics, ground clearance, and crash safety. We indicated

a “+” if the design is superior and a “–“ if it we deemed it to be inferior. This

resulted in the following table:

Table 1: Engine Placement Comparison Table

The designs broke even at with 4 positives and 4 negatives apiece. To add an

additional level of distinction, we then determined that the most important

considerations to us were weight, noise, and ease of maintenance. The under-wing

design won two of those three considerations, and was therefore chosen. It is

important to note that these comparisons were all qualitative and not based on any

real calculations.

Longitudinal Control Surface


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The longitudinal control surface was the final major layout decision. As the design

choices hitherto made placed a large percentage of our weight at the back of the

aircraft, the double-delta wing and under-wing engines would both be far aft, we

would be required to have a large moment to trim the aircraft. To achieve this

moment with a horizontal tail would have resulted in a much longer aircraft to yield

the desired moment arm, or an excessively large control surface to attain the desired

moment magnitude. A canard would allow us to have the correct moment arm and

magnitude without increasing the length of the aircraft, and as a result was the

design we chose. In addition, canards give a second benefit of reducing sonic boom

(Yoshimoto, 2004)[5]

.

Figure 14: Canard Design for Low Boom (Yoshimoto, 2004)


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VII. Conceptual Design

Aircraft Walk-around

A brief overview of the GoldJet FAST is given in the following walk-around,

listing some key features of its exterior design and cabin layout:

Figure 15 Exterior aircraft walk-around

40-Passenger 2x2

luxury cabin

All fuel storage

in fuselage

Cranked-Arrow

wing planform

Vertical tail (no

horizontal tail)

4 Low-bypass

turbofans under wing

Under-Nose Cameras

for Landing Assistance

Canards for Pitch Control

and Low Boom

Galley Luxury Lavatory Overhead bins for

carryon storage

40-Passenger 2x2

Luxury Cabin

Two Emergency

Exits per side


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Figure 16 Interior and cabin layout

Key Design and Performance Parameters

A brief summary of aircraft’s key design and performance parameters follows:

General Performance

o Range: 4800 nautical miles

o Passengers: 40

o Length: 190 ft

o Takeoff Distance: 11,400 ft

o Landing Distance: 7,380 ft

Weight

o W0 = 329,000 lb

o We = 135,000 lb

o Wf = 185,000 lb

Engine

o Number: 4

o Diameter: 4.7 feet

o TSL: 35,400 lbs

o T/W: 0.425

Wing

o Wing Loading: 130 lb/ft2

o Area: 2541.5 ft2

o Sweep: 60° – 57°

o Span: 80.26 ft

o Aspect Ratio: 1.851

Canard

o Area: 500 ft2

o Sweep: 45°

o Span: 40 ft

o Aspect Ratio: 3.2

Sonic Boom

o Overpressure: 0.36 lb/ft2

Aircraft Performance

The V-n diagram for an aircraft shows the different load factors that the aircraft will

experience for a full envelope of equivalent velocities, which are scaled by

Two fuel tanks

under fuselage

Cargo Loading

Door

Canard

Box


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atmospheric conditions. According to Raymer, the highest positive load factor for a

transport should be three to four[6]

. This would cause the structural components of

the aircraft to experience three to four times the load they bear under normal

straight and level flight conditions. The highest acceptable negative loads on a

transport are between negative one to negative two.

The V-n diagram is made up of two different factors, maneuvering loads and gust

loads. Maneuvering loads are loads caused by the flight path of the aircraft. If the

plane were to suddenly pull up the load factor would increase, this also happens in

high banked and fast turns. Some of these maneuvering load factors are set by the

aerodynamic capabilities of the aircraft. Others are set by the structural capabilities.

Fig. 17 FAST maneuver loads V-n diagram

Figure 17 shows the maneuvering load factor as equivalent velocity is varied. The

curved portions beginning at a velocity of zero and continuing to the horizontal

lines at the top and bottom of the plot represent the maximum load factor our

aircraft can sustain at stall speed. These limits are set aerodynamically, as the

airplane cannot function any slower than is represented on the plot. The horizontal

lines at the top and bottom of the plot, positive three and negative 1, represent the

load factor our structure will be designed to. These numbers are within the range


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given by Raymer and will increase slightly when gust loads are factored in. The

vertical line at maximum equivalent velocity is dive speed, which is caped at

1.2Vcruise.

Gust loads account for times when there are strong wind forces that increase the

effective angle of attack of the aircraft. This creates sudden lift which causes the

aircraft to experience higher load factors, as if it were pulling up. The gust loads are

based on the aerodynamic properties of the aircraft such as Cl max , Cl alp ha , and the

mean chord length. These loads are also based on the atmospheric conditions like

gust velocity and air density.

Fig. 18 FAST gust loads V-n diagram

Figure 18 shows the maximum gust loads expected to be applied to our aircraft. The

peak gust load occurs at our aircraft’s cruise speed, and is only slightly higher than

the load experienced during our aircraft’s maximum allowable speed in turbulence.

This collection of gust loads is combined with the maneuvering loads calculated

earlier to form a total load envelope for all flight conditions. The highest

aerodynamically feasible load is recorded for each equivalent velocity to create the

envelope seen below in Figure 19.


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Fig. 19 FAST total loads V-n Diagram

The maximum load reached is 3.12 and the minimum reached is negative 1.12.

These are within the envelope set out by Raymer for transport aircraft.

Sizing

Summary

The purpose of the sizing code was to find the MGTOW of an aircraft meeting the

design range of 4800 nmi. The algorithm was based on Raymer Ch. 19 and

programmed using MATLAB. The algorithm is similar to that used for earlier

sizing (in the SRR and SDR), but uses more accurate equations based on actual

FAST performance, and smaller step sizes during mission segments, most notably

in cruise. The sizing code uses inputs from aerodynamics (for lift and drag), an

engine model (for thrust available and specific fuel consumption), and statistical

weight equations as described in Raymer Ch. 15.


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The gross weight is the sum of the payload weight, crew weight, empty weight, and

fuel weight. For a configuration of three crew, each at 200 pounds, the crew weight

is 600 pounds. Each passenger is estimated to have a total weight of 220 pounds,

based on an average person weighing 180 pounds with 40 pounds of baggage. For

40 passengers, the payload weight is 8800 pounds. Based on an initial guess for

gross weight, and fixed values for payload and crew weight, the fuel weight and

empty weight could be determined, and the sum could be compared to the guess for

gross weight. Iteration continued until the gross weight guess and calculated gross

weight converged.

AR 1.85

T/W 0.425

W0/S 130 lb/ft2

S 2541.5 ft2

Mcruise 1.8 Mach

number of Crew 3

number of Pax 40

weight per Crew 200 lb

weight per Pax 220 lb

Design Range 4800 nmi

cruise altitude 45000 ft

wind speed 100 kts

stall speed 168 kts

t/c 0.03

ΛLE 62 °

CLmax 1.2

Number of Engines 4

Table 2 Design Parameters

Finding Fuel Weight

In order to find fuel weight, we split the design mission into six segments: engine

start/taxi/takeoff, climb, cruise, descent, loiter, and reserve. Each segment,

including key equations and necessary assumptions, will be described.


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Engine start, taxi, takeoff

To account for the fuel used during this segment, Raymer suggests to calculate fuel

consumed for 14 min at ground idle and 1 min at takeoff thrust. The design

parameters chosen sized the engine at takeoff. The rubber engine model, discussed

in more detail later, was then used to find the specific fuel consumption at each

condition. Then we found the fuel fraction for each by the equation:

𝑊𝑖

𝑊𝑖−1= 1 − 𝐶𝑑

𝑇

𝑊 𝑖

Climb

The climb segment consists of two parts: a linearly increasing velocity climb to

10,000 ft, and an accelerating climb to cruise altitude, 45,000 ft. The climb to

10,000 ft was done in 1,000 ft increments, calculating lift, drag, and thrust required

at each point. We assumed a flight path angle of 14° based on suggestions from

Raymer, as well as on trial and error from engine performance. Due to difficulties

in plotting a trajectory and calculating drag through the transonic regime, an

approximate equation from Raymer Ch. 6 was used, which is:

where M is the cruise Mach number. Range credit was given to the climb segment

based on average horizontal velocity. Time credit was also given to the climb

segment based on average vertical velocity, obtained from climb rate.

Cruise

The equations of motion for steady, level, un-accelerated flight, as well as a

modified form of the Breguet range equation, were used to calculate the fuel

fraction for cruise. The Breguet range equation used was


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where R is the range, c is the specific fuel consumption, v is the velocity, and is

the lift-to-drag ratio. The range accounted for a headwind of 100 kts. During cruise,

the step size was 50 nmi, and L/D and SFC were calculated at each point to account

for weight loss due to fuel throughout the flight.

Descent/Landing

Raymer points out that the historical fraction for descent, used in Ch. 6, is a good

approximation even for detailed sizing. This is because very little fuel is used. No

range credit was given during descent. The result of this approximation is that fuel

use is slightly overestimated, but time of flight is slightly underestimated.

Loiter

The endurance equation of the form

where E is the time in hours, was used to calculate the fuel weight fraction. For the

purposes of sizing, 45 min of loiter time was used. L/D was estimated from an

approximation in Corke[7]

, but compared to the cruise L/D calculated to ensure it

was reasonable.

Reserve

To account for diversions to alternate airport, as well as trapped fuel, 6% was added

to the fuel from the previous segments. Any diversion to an alternate airport would

be done at subsonic speeds.

W1/W0 0.973 Takeoff

W2/W1 0.952 Climb

W3/W2 0.55 Cruise

W4/W3 0.995 Land


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W5/W4 0.926 Loiter

Wf/W0 0.562 Total Fuel Fraction

Table 3 Weight fractions

Finding Empty Weight

After one iteration of the fuel fraction, key parameters (such as engine weight,

engine thrust, fuel weight, and cruise L/D) are passed to a statistical component

weights function. Other inputs to the component weights function include wing and

control surface area, fuselage data, and landing gear. The method for calculating

component weights from these data will be discussed in a later section.

A summary of the results of the sizing code are presented in Table 4.

W0 (approx) 329,000 lb

We (approx) 135,000 lb

Wf (approx) 185,000 lb

Range 4800 nmi

L/Dcruise 8.85

Engine Diameter 4.7 ft

Engine Thrust 35,400 lb

Total Thrust 141,600 lb

Cruise time 5:03

Flight time (no diversions or loiter, estimate) 5:45

Flight time (includes diversions and loiter, estimate) 7:16

Table 4 Sizing code results

As mentioned before, the range of 4800 nmi represents a flight from LAX-NRT

(Tokyo, Japan). The estimated flight time for subsonic transports is 10 hr, 44 min.

The FAST’s estimated flight time of 5:45 then gives a savings of about 5 hours, or

46%.

Carpet Plots


28

For the conceptual sizing, the design parameters chosen were based on a generated

constraint diagram (featured in the SDR report). The values were then refined

through structural and aerodynamic constraints, and actual engine performance.

This constraint diagram, however, was based on a more general supersonic aircraft,

not specifically for the FAST. The next stage of design would use the carpet plots to

re-center the baseline around the optimal thrust-to-weight and wing loading values

for our aircraft.

Performance Parameters

The sizing code, as it stands, ensures that the FAST will be able to meet the

required mission range. After sizing the engine, the code also ensures that the

aircraft is capable of sustaining a supersonic cruise by comparing drag to available

thrust. To further test our aircraft, the carpet plots introduce five other parameters:

fuel efficiency, takeoff distance, landing distance, subsonic 2g maneuver, and

second segment climb gradient. The equations listed below are from Raymer,

Chapters 5 and 17.

Fuel Efficiency:

𝐹𝐸 =𝑛𝑝𝑎𝑥 𝑅

𝑊𝑓

Takeoff Distance:

𝐵𝐹𝐿 =0.863

1 + 2.3𝐺

𝑊 𝑆

𝜌𝑔𝐶𝐿𝑐𝑙𝑖𝑚𝑏+ 𝑕𝑜𝑏𝑠𝑡𝑎𝑐𝑙𝑒

1

𝑇𝑎𝑣 𝑊 − 𝑈+ 2.7 +

655

𝜌 𝜌𝑆𝐿

Landing Distance:

𝑆𝐺 =1

2𝑔

𝑑(𝑉2)

𝐾𝑇 + 𝐾𝐴𝑉2

𝑉𝑓

𝑉𝑖

2g Maneuver:


29

𝑃𝑠 = 𝑉 𝑇

𝑊−

𝑞𝐶𝐷0

𝑊 𝑆 − 𝑛2

𝐾

𝑞

𝑊

𝑆

Second segment climb gradient:

𝑇𝑆𝐿𝑊0

=𝛽

𝛼

𝑁

𝑁 − 1 𝐶𝐺𝑅 +

1

𝐿 𝐷

After finding the constraints on thrust-to-weight and wing loading using these

equations, we determined that the fuel efficiency requirement set by NASA of 3

pax-mi/lb of fuel was not feasible for our aircraft. Achieving this fuel efficiency

would require extensive aerodynamics work beyond the scope of this project.

Therefore, fuel efficiency was removed from the carpet plot constraints. Figure 20

shows the resulting carpet plot.

Figure 20 Carpet plot


30

The plot shows that our current design point is very close to the optimal, and meets

all of the required constraints. Takeoff distance and required climb gradient appear

to be the constraining performance parameters for the FAST. Changing our baseline

design point to the lowest weight (T/W = 0.417, W0/S = 132 lb/ft2) would achieve a

weight savings of approximately 10,000 lb, according to rough estimates.

Aerodynamics

In order for us to compute the thrust required at cruise, we performed an analysis of

the drag polar for free stream conditions (Figure 21) of standard atmosphere at

45,000 feet and Mach 1.8. The sizing code provided the weight at various points

during cruise. By equating lift to weight, we could find drag and therefore thrust

required.

Figure 21 Drag polar at cruise


31

The initial and final weights for the cruise mission segment provide upper and

lower limits for the lift required for level flight. This gives the range for cruise

angle of attack, which is at its maximum at the beginning of cruise and diminishes

as fuel is burned off and less lift is needed as shown in Figure 22.

Figure 22 Lift vs angle of attack at cruise

The maximum and minimum angles of attack show the maximum and minimum

drag demonstrated in Figure 23.


32

Figure 23 Drag vs angle of attack at cruise

This lift and drag data was then input to the cruise mission segment of the sizing

code.

Also of importance were the takeoff conditions. We studied the drag polars of take-

off conditions (Figure 24) to ensure the aircraft could hypothetically lift off at the

initial weight of 329,000 lbf.


33

Figure 24 Takeoff drag polar

For standard sea level conditions with 60% effective plain flaps (Raymer), Figure

25 shows that the aircraft requires an eight degree angle of attack at 150 knots to lift

off the ground, which is reasonable.


34

Figure 25 Lift vs angle of attack at takeoff

The approach configuration varies from the take-off configuration in speed, weight

and flap effectiveness. Figure 26 gives the drag polar on normal approach with no

high-lift devices employed. In Figure 27, the flaps are assumed to be 100%

deployed, as suggested in Raymer. We computed the effect of the flaps on

coefficient of lift using estimated values of CL and the flapped area as given in

Raymer.


35

Figure 26 Drag polar on approach with no high-lift devices

Figure 27 Drag polar on approach with high-lift devices


36

Wing Planform

Our wing planform was chosen from Herrmann, “CISAP: Cruise Speed impact on

Supersonic Aircraft Planform; a Project Overview.”[4]

In the CISAP project, Airbus

created three datum wing body configurations, chosen for three Mach numbers, and

they were then optimized by multiple organizations. The key performance

parameters for each aircraft design were mission range, the lift to drag ratio, and the

mass of the wing. They also made sure that the wings would perform at low

speeds, to guarantee a feasible planform. We chose the wing optimized by DLR for

Mach 1.6 because that is closest to our design space and it has the best performance.

Their optimal design, improved range by 25.4%, lift to drag ratio by 13.2% and

reduced weight by 11%. These improvements in performance made it the logical

choice for our aircraft. They also presented the optimized structural layout for the

wing that facilitated the weight reduction.

Figure 28 The wing planform optimized for Mach 1.6 by DLR.


37

Drag prediction

We predicted the drag of the aircraft using a MATLAB function as a superposition

of three components – induced, wave, and parasite - with an addition of three

percent of this total to account for miscellaneous factors. However, after receiving

feedback during the CoDR presentation to Lockheed Martin, we discovered that a

higher percentage, approximately fifteen percent, would have been more reasonable

to account for miscellaneous factors.

The method for evaluating induced drag used is a function of lift and angle of attack

of the lifting surfaces completely independent of Mach number[8]

. All calculations

were performed assuming the canard was fixed at the same angle of attack as the

wing.

Wave drag is only non-zero at supersonic conditions. The method for finding wave

drag we used is based on an approximate area profile of the aircraft found with

basic geometric representations of the nose, engines, fuselage, wings, tail, and

canards. From this area profile, a code generated points on the surface of an

equivalent body of revolution. The code then found the boundary of the intersection

of the body of revolution with a plane rotated from the vertical by the Mach angle

of the free stream at many positions along the length of the body.


38

Figure 29 Equivalent body of Revolution

Figure 29 illustrates the body of revolution (blue) with an example of a Mach cut

(green) at Mach 1.8. The area of each boundary projected onto the plane

perpendicular to the longitudinal axis was then used to form a new area profile. The

second derivative of this new area profile, dependent on Mach number, was used in

a formula along with a reference area to find the coefficient of wave drag[9]

.

Parasite drag was found by summing the contributions of the nose, engines,

fuselage, wings, tail, and canards. For supersonic conditions, only the wetted area

and coefficient of friction under turbulent conditions were relevant for finding

parasite drag. Subsonic conditions required the estimation of an interference factor,

the wetted area, the friction factor and coefficient of friction with estimates for

percent length of turbulent flow along each component (Raymer).


39

Lift Prediction

Whole wing lift prediction model is given in Lee, “An Analytical Representation of

Delta Wing Aerodynamics.”[8]

The model is designed to be accurate in the range of

aspect ratios between 1 and 3, where lifting line theory and thin wing theory are

inaccurate.

Figure 30 The regions of aspect ratio where lifting line theory and thin wing

theory are accurate. The model in Lee fills in the area around an Aspect Ratio

of 2 for delta wings.[10]

To accurately predict the lift, Lee derived the equations for the normal force on the

delta wing with no approximation assumptions. The equations were simplified to

get the coefficient of lift in terms of a linear and non-linear factor. When

simplified, these functions are factors of only angle of attack, leading edge sweep,

and aspect ratio. With this model, airfoil section and thickness are irrelevant.


40

Figure 31 Theory compared to experiments on a wing with an aspect

ratio of 2. While not completely accurate, it is much closer than lifting line or

thin with theory.

Due to this theory’s accuracy for the aspect ratio of our wing, we approximated the

wing of the Gold Jet FAST as a delta wing of identical aspect ratio, using the

average leading edge sweep. The resulting lift curve for the planform of the Gold

Jet FAST can be seen in Figure 32.

Figure 32 The lift curve slope for the Gold Jet FAST planform


41

Based upon some of the more complete experimental data presented in Lee, a

CLMAX of 1.2 was chosen at an angle of attack of 27 degrees. While approximate,

experimental of computational validation is required. For further work a higher

fidelity, most likely CFD based model should be used, or at the very least

experimental verification that the assumptions made to use this model are accurate.

Sonic Boom Prediction

Sonic-boom prediction was carried out using a simple “N-wave” prediction. The

method is based on Carlson’s “Simple Sonic-Boom Prediction” method.[11]

It uses

a series of non-linear factors to estimate the effects of shape, and altitude on the

propagation of the sonic boom. The non-linear factors were determined from charts

and graphs in the paper. To improve the generality of the code and simplicity of the

input file, the non-linear factors were interpolated in Microsoft Excel. This change

made it so the input file only requires data about the aircraft shape, Mach number,

and altitude. The shape can be accounted for, either from the effective area

distribution or based on the weight, Mach number, altitude, curve fit to historical

data. Test cases were run against the examples given in the paper to verify

accuracy.

Initial prediction of the sonic-boom signature was calculated based upon the cross

sectional area, Mach number and altitude. Fidelity is improved over the last

estimate by calculating shape factor based on the cross sectional area instead of

basing it off of Concorde[11]

. The resulting sonic boom signature is lower but not

substantially.


42

Figure 33 The predicted sonic-boom signature, modified in both

Experimentally changing the area distribution had little effect upon the resulting

sonic boom signature. The baseline signature had an overpressure of 1.5 lb/ft2

which is far above the acceptable level of 0.5 lb/ft2 and nowhere near the target of

0.3 lb/ft2. To improve the overpressure, shaping of the aircraft is required.

Aircraft shaping reduces the overpressure by making the shockwaves parallel to

prevent them from coalescing into a stronger shockwave. Since doing so requires

quite a bit of CFD and experimentation, it is infeasible for this level of design. To

simulate the effects of aircraft shaping, reduction factors were created based upon

the NASA F-5 Shaped Sonic Boom Demonstrator(SSBD), and the SAI Quiet

Supersonic Transport. NASA successfully demonstrated that shaping of the aircraft

body can reduce the sonic boom overpressure at the ground. They did this by

modifying the nose of an F-5 based upon CFD optimization. They managed to

reduce the sonic boom overpressure when compared to an unmodified F-5 by

almost 40 percent[12]

. Comparing their experimental results to the ones predicted


43

using Carlson’s method, a shaping factor of 0.63 was used. This resulted in a

reduction of boom overpressure to 0.9 lb/ft2. This is still well above the threshold.

To see the effects of shaping the whole aircraft, another shaping factor was created

based upon the SAI QSST. The QSST boasts an overpressure of 0.3 lb/ft2 which is

our target value[13]

. This is accomplished by shaping of the nose, use of canards at a

high dihedral angle, and a gull wing[14]

. Comparing the predicted overpressure with

the number given on the website resulted in a shaping factor of 0.37. When this is

applied to the Gold Jet FAST, the overpressure is reduced to 0.36 lb/ft2 with a

duration of 0.14 seconds. This is within our threshold and means that the Gold Jet

FAST will require substantial CFD and shaping to reduce the overpressure to an

acceptable amount. This will be a large focus of the aerodynamics team during

preliminary design.

Propulsion

Engine Configuration and Performance Summary

The propulsion system on FAST has been designed with 4 Low bypass turbofan

engines contained in two nacelles. It may have been possible to achieve the

necessary thrust with fewer engines, however the size of such engines would have

required an unreasonable increase in noise and drag. Each nacelle is designed to

hold two engines (as depicted in the Figure below) with one nacelle placed

underneath each wing.


44

Figure 34: Engine Nacelles

Since there is currently no other supersonic commercial jet, and considering the

Olympus engines of Concorde were designed with 1960s hardware, it seemed

unreasonable to expect any currently operating engine to be optimal for FAST. For

this reason we are proposing a new engine program which will be more suited for

FAST’s design goals and optimized to achieve the best possible performance. The

Table below details the primary design parameters for this engine:

Inlet Type 2D Ramp inlet with variable geometry

Nozzle Type Converging-Diverging, Variable exit area

Fan Face Diameter 4.7 ft2

Engine Length 17.5 ft

Inlet and Duct Length 22.2 ft

Bypass Ratio 0.5

Overall Compressor Pressure Ratio 20

Table 5: Engine Specifications


45

From this chosen engine configuration and with the use of a rubber engine model,

which will be discussed in the following section, we generated the following

performance curves.

Figure 35: Max Power Thrust TSFC at multiple flight conditions


46

Figure 36: Max Power Thrust at multiple flight conditions

To summarize the results depicted in the charts shown above, each single engine

has the following important performance characteristics:

Max Sea Level, Static Thrust 35,361 lbf

Max Cruise (45k, Mach 1.8) Thrust 9,102 lbf

Cruise TSFC 0.9492 (lbm/hr)/lbf

Table 6: Performance characteristics for each engine

To ensure that these engines are producing the necessary thrust at various flight

conditions, the total thrust of all 4 engines is plotted against the required thrust (ie.

Drag) as functions of Mach number at important altitudes.


47

Figure 37: Thrust available, required as function of Mach # varying altitude

The above chart shows that there is more than sufficient thrust at low altitudes for

slow speeds. From this chart it is clear that the crucial condition is in fact the cruise

point. At 45,000 ft the thrust available drops below the thrust required at Mach

1.85. Hence there is sufficient power to operate at the desired cruise point of 1.8,

however nearly full power will be required.

It should also be noted that there is a minimum speed at which FAST can operate at

45,000 ft as the required thrust (Drag) increases when slowing down to the

transonic region. This has been taken into account when considering the climbing

and acceleration pattern used to reach cruise.

Engine Model

As already discussed, we at Gold Jet are proposing a new engine that will be

optimized specifically for FAST’s design point. For this reason, during the design

of FAST a “rubber engine model” was needed in order to perform optimization

0

20000

40000

60000

80000

100000

120000

140000

0 0.5 1 1.5 2 2.5

Thru

st (

lbs)

Mach #

Thrust Required (Drag) and Thrust Available vs Mach # for Important Altitudes

Assuming Steady level flight at Max Cruise Weight (307,000 lb)

T req - 0 ftTreq - 10k ftTreq - 45k ftTavail - 0 ftTavail - 10k ftTavail - 45k ft


48

studies as well as to create a final engine model which could be implemented along

with our sizing code to determine engine size and compute engine performance.

We created a thermodynamic model based on Hill & Peterson’s text for a turbofan

engine[15]

. The engine model accounts for varying flow conditions depending on the

operating conditions of the aircraft (ie. flow through aircraft mach cone,

compression shocks across inlet etc.).

Inlet Design

For FAST we have decided to use a 2D ramp inlet with variable geometry to allow

for optimal capture area in both subsonic and supersonic flight conditions. The

figure below shows the inlet design in both the supersonic and the subsonic

configuration. Standard relations for normal and oblique shocks were used to

calculate inlet conditions.


49

Figure 38: 2D – Ramp variable inlet geometry

The geometry depicted above was designed specifically so that the shown capture

areas would be met for two conditions: (1) The supersonic capture area sized for

required engine airflow at design cruise conditions (Mach 1.8 at 45000 ft), (2) The

subsonic capture area sized for required engine flow at takeoff (Mach 0.32 at Sea

Level). These two capture areas were computed from (10.19) in Raymer’s text:

𝐴𝑐𝑎𝑝𝑡𝑢𝑟𝑒 = 𝑚 𝑒 1+

𝑚 𝑠𝑚 𝑒

𝜌∞𝑢∞ 1 +

𝐴𝑏

𝐴𝑐𝑎𝑝𝑡𝑢𝑟𝑒

Where, 𝑚 𝑒 is the airflow required by the engine, 𝑚 𝑠 is the secondary flow

requirement, 𝐴𝑏 is the boundary layer bleed port area, and 𝜌∞ and 𝑢∞ represent the


50

ambient density and flow velocity (note: for supersonic flight, 𝑢∞ represents the

flow velocity inside the aircraft’s Mach cone).

Based on typical values given in Raymer’s text, the secondary bleed fraction was

assumed to be 𝑚 𝑠

𝑚 𝑒 = 0.20 (Table 10.2 in Raymer for fighter) and the boundary

layer bleed area was assumed to be 𝐴𝑏

𝐴𝑐𝑎𝑝𝑡𝑢𝑟𝑒= 0.04 (Fig. 10.19 in Raymer, Slot

Bleed at M=1.8).

Having arrived at an inlet geometry design, this inlet then needed to be included in

the engine model. The shock angles depicted in the Figure above are computed for

the design condition of Mach 1.8 at 45,000 feet. However, the engine model must

be capable of computing Inlet performance at virtually any given flight condition.

Figure 39: Shock pattern across inlet

For supersonic flow, assuming the conditions are known at Station 1 (𝑀1, 𝑃1, 𝑇1)

and for the known turning angle, 𝛿1, the shock angle, 𝜎1, and the conditions at 1a

can be computed using the oblique shock relations. Likewise from the conditions at

1a, the conditions at 1b can be computed and then from which the conditions at 1c

can be computed using the Normal Shock relations. This flow in 1c is then passed

into a subsonic diffuser, which continues to decelerate the flow to a Mach number

of 0.4 or less at the fan face (Station 2).


51

From this setup the stagnation pressure loss across all the shocks through the inlet

can be computed:

𝑃01

𝑃01𝑐=

𝑃01

𝑃01𝑎

𝑃01𝑎

𝑃01𝑏

𝑃01𝑏

𝑃01𝑐

Nozzle Geometry

We have designed the exit nozzle to be a converging-diverging nozzle to create

supersonic flow out the exit in order to maximize the thrust output of the engine.

Furthermore we have chosen a variable exit area design to allow for perfect

expansion to ambient conditions. The Figure below shows the conceptual design for

the nozzle.

Figure 40: Converging-Diverging Variable Nozzle Design

We recognize that a variable exit geometry adds significant complexity to the

engine design, increasing production and maintenance cost. However, considering

that FAST will have to perform well over such a wide range of conditions, we

believe the performance benefit of a variable nozzle design will outweigh the extra

complexity of the design.

Engine Required Airflow


52

Total thrust and fuel consumption of a jet engine are dependent on airflow through

the engine, which is a function of altitude and mach number. For the purpose of this

analysis, the total airflow required by the engine will be assumed based on data

from Raymer’s text in Appendix E.1 for an afterburning turbofan. While our engine

does not have an afterburner, this approximation is considered close enough for the

conceptual design stage.

Figure 41: Total Required Airflow from Engine in Raymer’s Appendix E.1

Data from the Raymer’s plots for required engine airflow (𝑚 𝑒𝐸1) were tabulated as

functions of Mach number and altitude, however it is recognized that this airflow

value is for an engine of a specific size (Fan-face diameter, 𝐷𝑓𝐸1 44 inches). Hence

this airflow needs to be scaled to match our engine diameter, 𝐷𝑓 . Assuming that

mass flow is proportional to fan-face area, the total airflow required by our engine

is then:

𝑚 𝑒 = 𝑚 𝑒𝐸1 𝐷𝑓

2

𝐷𝑓𝐸 12

0

100

200

300

400

500

600

700

0 0.5 1 1.5 2 2.5 3

Re

qu

ire

d A

irfl

ow

(lb

m/h

r)

Mach Number

Total Required Airflow

SL

10k

20k

30k

36k

40k

50k

60k


53

This total engine airflow will then be split into two airflows, bypass flow, 𝑚 𝑏 , and

the core flow, 𝑚 𝑎 . Hence for a given Bypass ratio, 𝐵𝑝𝑟 =𝑚 𝑏

𝑚 𝑎 then the core and

bypass flows can be computed:

𝑚 𝑎 =𝑚 𝑒

𝐵𝑝𝑟 +1 , 𝑚 𝑏 = 𝑚 𝑒 − 𝑚 𝑎

Thermodynamic Modeling

Figure 42: Thermodynamic Model and Stations of Turbofan engine[citation e-2]

Let the numbers depicted in the Figure above be the station definitions throughout

the engine. Station 1c being the flow entering the subsonic diffuser (after external

compression from the shocks). Station 2 is then the condition at the fan face, Station

3 is the exit of the compressor and the inlet of the combustor/burner, Station 4 is the

combustor exit and the inlet to the turbine, Station 5 is the exit of the turbine and

the inlet of the afterburner, Station 6 is the entrance to the Nozzle and Station 7 is

the core nozzle exit position. For the bypass flow, Station 8 is the flow exiting the

fan in the bypass duct, and station 9 is the bypass duct exit.

1c 2 3 4 5 6 7

8 9


54

At this point, in order to thermodynamically model the engine, several assumptions

must be made. The Table below lists these assumptions.

Diffuser Efficiency 𝜂𝑑 = 0.97

Fan Efficiency 𝜂𝑓 = 0.85

Fan Pressure Ratio 𝑃08𝑃02

= 𝑃𝑟𝑓 = 1.6

Compressor Efficiency 𝜂𝑐 = 0.85

Burner Pressure Drop 𝑃04𝑃03

= 𝑃𝑟𝑏 = 0.99

Max Turbine Inlet Temperature 𝑇04𝑚𝑎𝑥 = 1500 𝐾

Turbine Efficiency 𝜂𝑐 = 0.90

Core Nozzle Efficiency 𝜂𝑛𝑐 = 0.98

Fan Nozzle Efficiency 𝜂𝑛𝑓 = 0.98

Fuel Heating Value 𝑄𝑟 = 45,000 𝑘𝐽/𝑘𝑔

Specific Heat Ratio prior to combustion 𝛾1 = 1.4

Constant Press Spec heat prior to combustion 𝐶𝑝1 = 1005 𝐽/𝑘𝑔𝐾

Specific Heat Ratio after combustion 𝛾2 = 1.31

Constant Press Spec heat after combustion 𝐶𝑝2 = 1240 𝐽/𝑘𝑔𝐾

Table 7: Engine Assumptions

1) Conditions inside Mach cone

Let the ambient conditions be designated as Station a. If the aircraft is flying at

supersonic speeds, then there will be a Mach cone generated with an angle:

𝜇 = 𝑠𝑖𝑛−1 1

𝑀

If the flow conditions inside the Mach cone are designated as Station 1, then Mach

number, Temperature and Pressure (𝑀1, 𝑇1, 𝑃1) can be computed using the oblique

shock equations for a shock angle 𝜇.

If the aircraft is flying at subsonic speeds then the conditions at Station 1 are

equivalent to the ambient conditions.

2) External Compression across inlet


55

For the computed flow conditions at Station 1 (𝑀1, 𝑇1, 𝑃1) and for the given inlet

geometry, the flow through the inlet can be computed up to Station 1c as described

in the Inlet Design section.

If 𝑀1 < 1 then condition at 1c are assumed equivalent to conditions at Station 1.

3) Compressor Inlet Conditions

For inlet conditions, 𝑇1𝑐 and 𝑀1𝑐 and assuming adiabatic flow through the subsonic

diffuser, the stagnation temperature and Pressure are computed:

𝑇02 = 𝑇1𝑐 1 +𝛾1−1

2𝑀1𝑐

2

𝑃02 = 𝑃1𝑐 1 + 𝜂𝑑 𝑇02

𝑇1𝑐− 1

𝛾1 𝛾1−1

4) Compressor outlet conditions

For a known compressor pressure ratio, 𝑃𝑟𝑐 , the conditions at the compressor exit

can be computed:

𝑇03 = 𝑇02 1 +1

𝜂𝑐 𝑃𝑟𝑐

𝛾1−1 𝛾1

− 1

𝑃03 = 𝑃02𝑃𝑟𝑐

5) Fan outlet conditions

Similar to the compressor, for an assumed fan pressure ratio, 𝑃𝑟𝑓 , the fan outlet

conditions can be computed:

𝑇08 = 𝑇02 1 +1

𝜂𝑓 𝑃𝑟𝑓

𝛾1−1 𝛾1

− 1

𝑃08 = 𝑃02𝑃𝑟𝑓

6) Burner fuel-air ratio


56

For a specified Power setting, Pwr, given in percent (ie Max throttle: Pwr = 100)

the turbine inlet temperature (burner outlet temperature) can be defined as follows:

𝑇04 = 𝑇04𝑚𝑎𝑥 −𝑇03

100 𝑃𝑤𝑟 + 𝑇03

Hence, the fuel-air ratio which will achieve this turbine inlet temperature is:

𝑚 𝑓

𝑚 𝑎= 𝑓 =

𝑇04𝑇03

−1

𝑄𝑟𝐶𝑝1𝑇03

−𝑇04

𝑇03

Where 𝑚 𝑓 is the total fuel flow and 𝑚 𝑎 is the total core air flow.

7) Turbine inlet pressure

Assuming a value for the burner pressure drop, 𝑃𝑟𝑏 , the stagnation pressure at the

turbine inlet can be computed:

𝑃04 = 𝑃03𝑃𝑟𝑏

8) Turbine outlet conditions

For the power balance in the engine, the total power output by the turbine must

equal the power required for the compressor and the fan. Hence in this case the

power balance is:

𝑚 𝑎 + 𝑚 𝑓 𝐶𝑝2 𝑇04 − 𝑇05 = 𝑚 𝑎𝐶𝑝1 𝑇03 − 𝑇02 + 𝐵𝑝𝑟𝑚 𝑎𝐶𝑝1 𝑇08 − 𝑇02

Where 𝐵𝑝𝑟 is the Bypass ratio. Hence, solving this balance for 𝑇05 gives:

𝑇05 = 𝑇04 −𝑚 𝑎𝐶𝑝1 𝑇03 − 𝑇02 + 𝐵𝑝𝑟𝑚 𝑎𝐶𝑝1 𝑇08 − 𝑇02

𝑚 𝑎 + 𝑚 𝑓 𝐶𝑝2

And the turbine exit stagnation pressure can then be computed:


57

𝑃05 = 𝑃04 1 −1

𝜂𝑡 1 −

𝑇05

𝑇04

𝛾2 𝛾2−1

9) Core Nozzle Inlet conditions

Assuming no afterburner, then

𝑇06 = 𝑇05 , 𝑃06 = 𝑃05

10) Core Nozzle Exit Velocity

For variable nozzle geometry which perfectly expands the exit gas to ambient

pressure (𝑃1) the core nozzle exit velocity can be computed:

𝑢𝑒𝑐 = 2𝜂𝑛𝑐

𝛾2

𝛾2 − 1𝑅2𝑇06 1 −

𝑃1

𝑃06

𝛾2−1𝛾2

Where, 𝑅2 = 𝐶𝑝2 1 −1

𝛾2 .

11) Fan Nozzle Exit Velocity

Similarly to the core nozzle, if the fan nozzle perfectly expands the gas to ambient

pressure then:

𝑢𝑒𝑓 = 2𝜂𝑛𝑓

𝛾1

𝛾1 − 1𝑅1𝑇06 1 −

𝑃1

𝑃08

𝛾1−1𝛾1

Where, 𝑅1 = 𝐶𝑝1 1 −1

𝛾1 .

12) Thrust and TSFC


58

Hence, having calculated the nozzle exit velocities, and knowing the mass flows in

the engine, the total installed thrust of the engine can be computed:

ℱ = 𝑚 𝑎 + 𝑚 𝑓 𝑢𝑒𝑐 + 𝑚 𝑏𝑢𝑒𝑓 − 𝜌1𝑢1𝐴𝑐𝑎𝑝𝑡𝑢𝑟𝑒 𝑢1

The Thrust Specific Fuel Consumption (TSFC) is then simply the total fuel flow,

divided by this thrust value:

𝑇𝑆𝐹𝐶 =𝑚 𝑓

ℱ

With the above equations, an engine model was created (in MATLAB) which

computes Thrust and TSFC with flight Mach number, altitude, power setting,

bypass ratio and compressor ratio as inputs. This “rubber engine” model then

allowed for multiple combinations of bypass ratio and overall compressor pressure

ratio to be checked during the engine optimization.

Engine Optimization results

Two parameters which can have a significant impact on engine performance are

bypass ratio, and overall compressor pressure ratio. For this reason, after

developing an engine model, we performed an optimization analysis to determine

what values would best suite the design goals for FAST.


59

Figure 43: TSFC as it depends on Compressor Pressure Ratio

The Figure above shows how TSFC depends on compressor pressure ratio for

different flight conditions, the most important of which is the design cruise point

shown by the green line. This plot clearly shows the TSFC is minimized for a very

high pressure ratio of nearly 70.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 20 40 60 80 100

TS

FC

[(l

bm

/hr)

/lb

f]

Compressor Pressure Ratio

TSFC vs Compressor Pressure RatioT04 = 1500 K, BPR = 0

M = 0, Alt = 0

M = 0.8, Alt = 30k

M = 1.8, Alt = 45k

0

0.005

0.01

0.015

0.02

0.025

0.03

0 20 40 60 80 100

Sp

ecif

ic T

hru

st

[lb

f/(l

bm

/s)]

Compressor Pressure Ratio

Specific Thrust vs Compressor Pressure RatioT04 = 1500 K, BPR = 0

M = 0, Alt = 0

M = 0.8, Alt = 30k

M = 1.8, Alt = 45k


60

Figure 44: Specific Thrust as it depends on Compressor Pressure Ratio

While a high value of pressure ratio optimize TSFC, this plot shows that specific

thrust (ℱ/𝑚 𝑎 ) is optimized for a relatively low compressor pressure ratio of

approximately 5. Some tradeoff must be made between optimization of fuel

consumption, and optimization of thrust. Based on these two plots, a compressor

pressure ratio of 20 appears to be the optimum compressor pressure ratio.

Figure 45 TSFC vs. bypass ratio

This plot shows how TSFC depends on Bypass Ratio. High bypass ratios have a

beneficial impact on fuel consumption. Based on this chart alone, a fairly high

bypass ratio would seem optimum. However, a high bypass ratio, while it would

produce a lower TSFC would require a larger fan area in order to meet thrust

demands. This would increase drag which would hurt overall aircraft performance.

A high bypass ratio engine, while initially temping, would not be beneficial in the

end for a supersonic transport.

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5

TS

FC

[(l

bm

/hr)

/lb

f]

Bypass Ratio

TSFC vs Bypass RatioT04 = 1500 K, CPR = 20, FPR = 1.6

M = 0, Alt = 0

M = 0.8, Alt = 30k

M = 1.8, Alt = 45k


61

There is, however, some benefits to having a relatively small bypass ratio. A bypass

ratio of 0.5 would give a benefit of 0.04 (lbm/hr)/lbf in TSFC with relatively little

increase in engine size needed. While this number seems small, for 4 engines

producing a total of approximately 36,000 lbs of thrust at cruise during a six hour

flight, this improvement in TSFC leads to a total fuel reduction of 10,000 lbs.

Our optimization study lead us to an overall compressor pressure ratio of 20 and a

bypass ratio of 0.5.

Weights and Balance

Component Weights

As discussed earlier, our sizing code calculates the empty weight of FAST by

estimating the weight of each individual component and then summing the weights

to create a total composite weight. Raymer gives statistical models of each

component based on the type of aircraft being modeled. There are three different

types of component equations: fighter, transport and general aviation. Based on

mission profile, FAST does not fit entirely in any of the groups of equations. To

solve this problem, a logical, intuitive approach was used to model each component

with the most appropriate equation. Because of the high cruise speed, the wings and

vertical tail have to be built stronger than ordinary transport aircraft in order to

withstand increased stresses at supersonic speed. This includes the higher drag, as

well as the effects of increased temperature due to skin friction. We felt that the

fighter equations would more accurately reflect the weight of these components.

The canard was modeled using the transport equation because it would never have

to achieve the maneuverability required of fighter aircraft, and thus would not have

to be built as tough as the fighter canard. FAST's engines are low bypass turbofans

that are mounted flush with the wings, and utilize converging/diverging nozzles and

a ramp inlet. This configuration is the most common one for fighter aircraft, so the

fighter equations were used for the entire engine group. The remainder of the

component equations used were from the transport section, because they are not

affected by the increased speed of the aircraft. Use of advanced materials on the


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FAST is also reflected in the component weight codes. Because of the use of

GLARE panels for the skin of FAST, a materials weight factor of .95 was used for

the fuselage. This number was not higher because the interior structure extensively

used traditional aluminum alloys. For the wings, canard and vertical tail, a factor of

0.90 was used, due to the combination of GLARE for the skin and Al-Li alloy 2090

for the leading edges. To account for the total increase of weight during the final

design phases, the entire empty weight was multiplied by a factor of 1.02. Raymer

describes this 2% increase as reasonable for modern computer aided design

techniques. The weight statements of the individual components are shown in

Tables 8-10.

Table 11 shows the various payloads that are carried. For the purposes of this early

design, the passengers are assumed to be 180 pounds, including carry-on luggage.

Checked baggage is represented by the cargo weight.

The overall FAST weight statements are compiled in Table 12. The empty weight is the

weight of the components that cannot be removed from the aircraft, including structures

and systems. OWE is the empty weight plus the weight of the crew and their baggage

and equipment. MZFW is the zero fuel weight, and includes the crew and the maximum

possible payload. MFW is the zero fuel weight, plus the reserve fuel. Finally, MTOGW

is the maximum takeoff weight for the aircraft. In this case, it was assumed that taxiing

the aircraft resulted in negligible fuel loss. Figures 46-48 show the statistical

breakdown of the weight empty, at MZFW and at MTOGW, respectively.

Figure 46 MTOGW Figure 47 Zero Fuel Weight


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Figure 48 Empty Weight

Landing Gear

When sizing the landing gear, we used Raymer’s equations for transports and bombers.

These equations give the diameter and the width of the landing gear tires based on the

amount of weight each carries. The equation for both diameter and width is 𝐴 ∗ 𝑊𝑤𝐵,

where 𝑊𝑤 is the weight on each wheel, and the coefficients A and B depend on the

parameter being calculated. For the diameter equation, 𝐴 = 1.63 and 𝐵 = .315, giving

a tire diameter of 44.35 inches for the main landing gear wheels. This also gives the

nose wheel diameter of 39.74 inches. For the width equation the coefficients change to

𝐴 = .1043 and 𝐵 = .48, according to Raymer’s sizing for transports. This gives the

widths of the tires to be 16.01 inches and 13.54 inches for the main and nose wheels,

respectively. Since there are no real-world tires that fit those specifications exactly, we

used a database of 100 actual tires and passed into the tire sizing code. The program

then selected the lightest tire that met both the diameter and width criteria. These tires

also have properties that lead to the calculation of tire pressure and the stroke distance

of the oleo cylinder. We deemed those calculations to be too detailed at this point in the

design process, so they were not used.

The main landing gear had to be placed first on the structure in order to maintain static

stability. The main landing gear must be aft of the center of gravity so that the aircraft

does not tip on its tail. Since the FAST’s engines are mounted under the wings, the


65

main landing gear had to be behind the nozzle inlets. This is to avoid debris being

kicked into the engine, and also to eliminate turbulent air from being ingested by the

engine. After placing the main landing gear, the nose landing gear was then placed to

take 15 percent of the static load of the aircraft. According to Raymer, the nose gear is

supposed to be smaller than the main gear, but it must take at least 8 percent of the load

to allow for steering. This dictates that the main landing gear be much closer to the CG

than the nose gear.

Once the main and nose landing gear were placed, their length was calculated. It is

important to make the main landing gear long enough so that the tail does not hit the

ground during rotation for takeoff.

Figure 49 FAST dimensions

The angle between the main landing gear and the tip of the tail must be greater than the

takeoff angle. We designed the FAST for a maximum takeoff angle of 15°, which

means that the length of the gear from ground to bottom of fuselage would need to be at

least 6.5 feet.

Center of Gravity

Another important reason for using component based weight prediction is center of

gravity (CG) predictions. CG prediction and management is an important part of the

design process, as it is necessary for ensuring the aircraft is stable. To calculate the CG,

we simply treated the aircraft as a static beam and found the location where the

moments balance. This is shown in the following simple equation.


66

𝑊𝐴𝑖𝑟𝑐𝑟𝑎𝑓𝑡 𝑥𝐶𝐺 = 𝑊𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑥𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑜 𝑛𝑜𝑠𝑒

𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠

The weight of each component was found as described earlier using the component

weight equations. The weight of the aircraft fluctuates in flight, based on the fuel

consumption. This results in a shift in the CG during flight, which affects the static

margin. Stability is compromised when the static margin is no longer positive, and

would require an expensive stability augmentation control system to correct.

The goal of CG management is to ensure that the aircraft CG remains between forward

and aft limits. To do this, components are placed so that the CG is maintained in the

proper location at all times. The majority of the components must be located based on

other factors, such as the wing for aerodynamics and the vertical tail and canard for

pitch and yaw control. The cockpit is at the tip of the fuselage, and the cabin is

constrained by operating restrictions. The cargo and fuel tank placement took potential

CG impact into consideration, but were treated as fixed at this point in the design. The

remainder of the components were placed in order to achieve the necessary balance.

The main segments that were moved were the avionics and APU. The final location for

the APU is directly behind the aft fuel tank, and above the wing box. This is shown in

blue in Figure 60. The avionics were kept in the nose, as shown later in Figure 62.

After final placement, the CG is calculated for all points of the mission. These include

the design weights shown in Table 12, as well as the different stages where the fuel

tanks are emptied. From this information, the most forward and aft CGs are found.

These numbers are 111 and 128 feet from the nose, respectively, and this is shown

graphically in Figure 50.


67

Figure 50 Fore and aft CG

The CG for each point are also plotted as a function of the gross weight in Figure 51,

and this was used heavily to fine tune the locations of components. Important to note is

the different forward CG limits. This occurs because the aerodynamic center moves

back during supersonic flight. Based on fuel consumption calculations from the sizing

code, FAST will weigh approximately 310,000 pounds when it enters supersonic flight,

which is represented in the graph by the horizontal dashed red line.


68

Figure 51 CG travel

During the first stages of design, the forward fuel tanks were modeled as one, but

during CG travel calculation it was found that this produced a CG that was ahead of the

supersonic forward limit when the aircraft weighed 310,000 pounds. To correct this,

the forward fuel tanks were split in two, and the forward-most tank would be emptied

first. This served to move the CG back fast enough so that the CG remained in between

the limits at all times.

Structures

At this stage of design, the structure is approached at a very basic level. The goal of the

team is to ensure that the concept chosen is structurally feasible, and that there is

adequate space allocated for all necessary structural components. Figure 52 highlights

the initial structural design of FAST modeled in CATIA.

Figure 52 Structural layout

One of the most important structural considerations is the load path. FAST utilizes a

keel beam that is located directly underneath the cabin, as shown in Figure 53. This

nonstandard location was chosen to simplify the structure in the fuselage. Locating the

keel beam directly beneath the cabin reduces the distance between the deck and the

load path, eliminating the need for intervening structure and reducing weight. The


69

position of the keel beam also allows the landing gear and canard box to integrate

directly to the load path, further simplifying the structure.

Figure 53 Keel beam connection

The keel beam is connected to the fuselage via the deck and reinforced deck stringers,

shown in blue in Figure 54. This figure also shows the forward fuel tanks suspended

from the keel beam, shaded in brown.

Figure 54 Keel beam and fuselage


70

The fuselage itself is designed using traditional stringers and frames. The actual sizing

of structural members would be accomplished after preliminary design, but a few

important concepts are highlighted here. First, both the stringers and frames are spaced

such that they reinforce the areas around the windows. The window edges form areas

of stress concentration due to the joining of dissimilar materials, so the structure must

be reinforced at these locations. This is also true for the door, which is larger and has

increased weight from opening and closing mechanisms. Another important design

choice was the reinforcement of the stringers that attach the deck to the fuselage. As

stated earlier, these stringers serve to connect the main load path (the keel beam), with

the fuselage, and do so along the length of the fuselage. This means that they are larger

than the typical stringer.

Figure 55 shows the area under the fuselage where the nose landing gear attaches to the

keel beam (highlighted in green).

Figure 55 Nose gear and keel beam

The location of the wings presented special structural challenges. FAST's wings are

located aft and vertically central, and not on the keel's axis. This means that we had to

account for a mechanism to transfer the loads from the keel beam to the wing box. This

is shown in Figure 56.


71

Figure 56 Wing structure

The junction box then transfers the load to a pair of longerons that form the main load

path in the wing. These are significant because they are responsible for carrying a large

portion of FAST's loads, including the weight and thrust of the engines, the main

landing gear loads, and the lift generated by the wings.

Figure 57 Landing gear and engine nacelle

Figure 57 shows a cutaway of the main landing gear and engine nacelle locations.

Because of the size of the main landing gear trucks, they could not fit inside the tapered

fuselage or the thin wing, necessitating the outer structure. The landing gear box,

shown in green, is designed to act as a fairing to reduce its impact on drag. The final

shape of these landing gear housings would be dependent on more in-depth

aerodynamic design.

The next major layout decision was the location of the cargo section. In most

conventional airliners, cargo is located under the cabin. This works well with cabins

with large diameters, but FAST was designed specifically to reduce cabin diameter.


72

Consequently, the area underneath the cabin is only 28 inches tall at its largest point.

We decided that this would make it difficult to store cargo in this space, as this is

approximately the same size as some checked luggage. The small size would also make

it difficult to access areas that were not directly adjacent to a cargo door, increasing the

amount of time needed to load and unload the aircraft. Based on the total number of

bags, and the maximum volume of a typical checked bag (approximately 6 ft3), we

determined that adding a 9 ft section of cabin aft of the main cabin would account for

all of the necessary cargo, with a 150% margin of tolerance. The added length was

deemed acceptable due to the minimal structural increase and the improved fineness

ratio. This cargo section is shown in Figure 58, and is shown separate from the main

cabin for clarity. Note the large cargo door that allows easy and rapid loading. The

door next to the cargo section is an emergency exit.

Figure 58 Fuselage structure

The next large complication for the FAST was finding room for the fuel. Due to the

long range required for transpacific flight, FAST carries over 180,000 pounds of fuel.

This works out to be over 3,600 cubic feet. In many typical transports fuel is carried in

the wings. However, for FAST this represented problems in balance and volume. In

most conventional aircraft, the wings are located at or near the longitudinal center of

the aircraft. This means that any fuel carried in this location does not significantly

affect the center of gravity. However, FAST has wings located at the most aft section of

the fuselage, like a typical supersonic fighter. This also dictated that the engines and

landing gear were located extremely far aft. This combination resulted in a very

rearward empty weight CG. Adding large amounts of fuel in the wings would serve to


73

further move the CG aft, resulting in control issues. The FAST’s wings also have a t/c

of 3%, making them very thin, and reducing potential volume available for fuel. As a

result of these complications, we decided to forgo placing fuel tanks in the wings.

Instead of putting fuel in the wings, we decided to place them under the fuselage and in

the aft fuselage section. As previously stated, the area under the fuselage was

eliminated as a location for cargo. However, it is an ideal location for fuel tanks, as they

could be molded around structural elements such as the canard box, landing gear and

keel beam. The two forward fuel tanks are shown in Figure 59.

Figure 59 Fuselage and canard box

Volume was made available in the aft fuselage section due to the aerodynamic shaping

of the fuselage, which needed a rapidly tapering shape to approximate a Sears-Haack

body and reduce wave drag. The increasingly small diameter proved to be of little use

for cargo and passengers, but is adequate for fuel storage. Its location ahead of the

aerodynamic center also helped with balance. One complicated factor for this location

is the internal structure running through this section of fuselage. We would prefer not to

encase the load path structures in fuel, as this can cause corrosion and other fatigue

related issues. However, for preliminary structural design, we are not making detailed

plans for this fuel tank. As an approximation, we modeled the fuel tank occupying the

entire available volume as shown in light brown in Figure 60, and then applied a factor

of 0.80 to the volume to account for losses due to structural losses resulting from

further design work.


74

Figure 60 Aft fuel tank structure

The last major structural component considered in our preliminary design was the

cockpit and nose. The exterior nose fuselage is shown in Figure 61, and shows the

continuation of the frame and stringer system used in the main fuselage.

Figure 61 Nose structure

The nose was shaped primarily for aerodynamic reasons. This works well to reduce

overpressure and drag, but also reduces visibility for the cockpit. To increase visibility,


75

we decided to make a raised cockpit section. This is shown in Figure 62. This section

also includes an extra seat for the third crewmember.

Figure 62 Cockpit interior

With all of the major structural components placed, the next step was to choose

materials. The primary requirements for material choice are set by the high

temperatures created during supersonic flight. FAST cruises at Mach 1.8, but to be safe

we chose materials based on conditions at Mach 2.2. Based on Raymer Figures 14.18

and 14.19, the leading edge temperatures will reach 270 ºF, while the nose will reach

300 ºF. This presents problems for most conventional Al alloys, which cannot operate

above 250 ºF. In aerospace, materials ranging from titanium to advanced alloys of rare

materials are used for operation at very high temperatures. FAST employs Al-Li 2090,

an advanced alloy that that combines aluminum with lithium and copper to produce

sheets that are lighter and stronger than traditional alloys. Alcoa claims that 2090 has

8% less density and 10% higher elastic modulus than 7075 Al, along with operating

temperatures that allow supersonic flight. This material is used in the leading edges of

the Eurofighter Typhoon, a fighter that according to the German Luftwaffe, can cruise

at Mach 1.5 and reach top speeds of Mach 2+. FAST uses Al-Li 2090 in the wing

leading edges as well as the nose cone.


76

The skin panels of FAST are made with GLARE (GLAss REinforced fiber metal

laminate). GLARE is made up of bonded layers of aluminum and glass fibers, a

combination that gives excellent strength and durability. GLARE panels are 10% less

dense than conventional Al, and as a result give excellent weight savings. Also,

GLARE has shown excellent fatigue properties, with a large critical crack length.

GLARE is currently utilized extensively on the Airbus A380, and as such will be fully

certified for use prior to FAST.

Despite the advantages of composites and advanced alloys, most of FAST's structure is

made up of traditional aerospace aluminum. This is done to reduce cost, in terms of

purchase cost as well as certification cost. The purchase cost is easy to quantify, as

Alloy 2090 (for example) costs approximately five times as much as traditional Al.

FAST reduces cost by balancing the light weight of the advanced materials in critical

components with the cheaper price of traditional materials. Traditional materials also

save on maintenance cost. Airlines already have maintenance procedures in place for

conventional aluminum. This includes specific inspection procedures, tools and training

to handle problems. Introducing new materials can require new techniques. For

example, composites do not respond well to magnetic inspection, as most composites

are not magnetic, and they also are not receptive to dye penetrant, because their flaws

are usually subsurface. Titanium requires special tools to repair, including harder drill

bits and special coatings that do not etch the material. Using materials that are familiar

to the airline maintenance systems allows for more seamless integration of FAST into

existing airline procedures.

Stability and Control

The canard and vertical tail provide the longitudinal and lateral stability necessary for

controlled flight. The size of the canard determines the aircraft’s available pitching

moments as well as the aircraft’s ability to trim in pitch. The vertical tail provides

lateral stability and the rudder, attached to the rear of the tail, provides the aircraft’s

yaw moment and lateral trim ability. Vertical tail and rudder size are constrained by the

requirement to land in a strong crosswind and fly with one engine inoperative.


77

Estimates for vertical tail and canard sizes (Table 13) were obtained with parameters

from the sizing code, wing data, and estimates of aircraft center of gravity.

Planform Area Estimates

Canard 500 ft2

Vertical Tail 282 ft2

Table 13 Planform Area Estimates

These estimates are fed into the canard and vertical tail functions, which iterate along

with the sizing code outputs, to converge on the minimum allowable control surface

sizes.

Canard

Canard size is constrained by supersonic and subsonic trim conditions, as well as

takeoff performance. The supersonic trim constraint considers the aft-moving

aerodynamic center (AC) during supersonic cruise. This aft AC creates a larger moment

to be countered by the canard, which must remain at relatively low deflection to reduce

trim drag. The takeoff constraint requires the canard to be sized to provide enough

moment to rotate the plane to the required takeoff angle of attack. Each constraint will

require a different planform area, and the canard function selects the more demanding

of the two for final size.

The aircraft CG position moves during the flight, changing the moment arm of the

canard and affecting its ability to trim. With our canard being in the front of the aircraft,

its shortest moment arm occurs with the CG at its forward-most point. The canard

function converges on canard size while using the forward-most, and most demanding,

CG for the aircraft. The remaining function inputs were determined from preliminary

aircraft layout. Within the canard function, assumptions were made for canard sweep

and coefficients of aerodynamic performance said to be typical of control canards in

Roskam.[16]


78

The canard function iterates canard size for both subsonic and supersonic trim

conditions with takeoff condition as a minimum constraint. The canard function outputs

a planform size for each trim condition once all of the moments are appropriately

balanced. The subsonic trim condition was determined to be the limiting factor in

canard size for our final aircraft weight and a final planform area of approximately 500

ft2.

Vertical Tail

Vertical tail and rudder size are constrained by the aircraft’s ability to land in a

crosswind and fly stable with one engine inoperative. The ability to trim in each of

these conditions is largely determined by rudder size, which is measured in percent

chord of vertical tail. Figure 2.23 in Roskam shows a plot of rudder effectiveness vs.

rudder percent chord. We decided to use a rudder that was 30% chord in order to

achieve a rudder effectiveness of 50%. For the landing in a crosswind condition, we

used maximum crosswind velocity of 20% of the takeoff speed. Using equations from

Roskam, the code finds the various yawing moment coefficients applied to the aircraft

while in sideslip. With a maximum rudder deflection of 20o, the code finds the yawing

moment coefficient due to rudder deflection. When the size of the vertical tail becomes

large enough, the loop stops and the tail area is saved. The next step is to find the tail

area required to trim with one engine out. The same steps are taken as in the crosswind

condition, except the actual moments are used, not the coefficients. The loop iterates

until the vertical tail and rudder are large enough to provide lateral stability with one

engine inoperative. The one engine inoperative condition is the limiting factor in our

design, and it corresponds to a tail area of approximately 282 ft2.

Aircraft Trim

Aircraft trim diagrams (Figure 63) were created to show longitudinal trim capability for

subsonic and supersonic flight. These diagrams were created using methods found in

Roskam chapter 4. The diagrams show that the aircraft can be trimmed at any required

flight condition. The black lines on the left and right of the diagrams show the center of

gravity limits, fore and aft, respectively. The upper black line shows the limit on angle


79

of attack, and the colored lines represent the corresponding angles of incidence of the

canard.

Figure 63 Trim Diagrams

The lateral trim conditions require a rudder deflection of 20o for one engine inoperative

and 10.5o for a crosswind landing.

Applications

Gate Compatibility

Airport compatibility was a major consideration when looking at design tradeoffs for

the FAST. Our aircraft’s canard and wing positions were the limiting factors when

pulling up to the gate and placing the service equipment required for turnaround. Our

aircraft is expected to require one passenger bridge, two fuel trucks, one potable

water/lavatory service truck, one galley/cabin service truck, and a maximum of two

bulk cargo carriers. Figure 64 shows an example arrangement of aircraft servicing

equipment during a typical turnaround.


80

Figure 64 Gate compatibility diagram

Our wing and canard placement allow for maximum service equipment access to the

aircraft. The leading edge of our canard is located 10 feet aft of our main passenger

door, which allows for plenty of space to maneuver a passenger bridge without any

significant threat of damaging the canard. Had the canard been placed forward of the

main door there would be an increased concern with the bridge operator’s ability to

maneuver in and around the canard without causing damage. The FAST’s wing is

located aft of the cabin and cargo service doors, allowing easy access by ground crews

for loading and unloading cargo, cleaning the cabin, and refreshing the galley. The two

forward fuel tanks for the fast are located in the belly of the fuselage and are expected

to be serviced by the same fuel truck, while the rear fuel compartment will be serviced

by the second truck.

Turnaround Analysis

We modeled the FAST’s turnaround procedures after a combination of the Embraer

ERJ 145 and the Boeing 777 procedures. The ERJ 145 is a short to mid-range

commercial aircraft with a cabin size of 50 passengers. The passenger load and unload

times are expected to remain similar between the two aircraft, however the FAST’s

trans-pacific range along with its luxury market require significantly larger galleys for


81

full meals and beverages. The ERJ 145 was also used for modeling the cabin cleaning,

potable water, and lavatory servicing times. The FAST’s long range requires a large

fuel volume and high refueling rates, which were modeled after the Boeing 777. The

gate servicing procedures for both the 777 and ERJ 145 were found in the airport

planning manuals supplied on their manufacturers’ websites.[17][18]

The planning

manuals contain the technical specifications for baggage, fuel, and passenger loads, as

well as the total time required for ground crews to perform each service. Table 14

shows the specifications for each plane, as well as a passengers/minute (pax/min) bulk

rate for each service performed.

Table 14 Fast Turn Time

We used conservative estimates for every service rate with the exception of fueling

rate, which we set equal to the Boeing 777 with 303 gallons/minute. The fueling rate

assumes 2,985 gallons remaining in the fuel tanks for our missed approach reserves.

Measurement Units FAST ERJ 145 777

Passenger Capacity pax 40 50 375

Pax Bridges # 1 1 2

Pax Unload Time min 5.0 4.0 7.5

Pax Unload Rate pax/min 8.0 12.5 50.0

Pax Load Time min 9.0 5.0 12.5

Pax Load Rate pax/min 4.4 10.0 30.0

Baggage Doors # 1 1 3

Bag Load Time (Total) min 8.0 8.0 52.5

Bag Unload Time (Total) min 8.0 6.0 52.5

Baggage rate pax/min 5.0 6.3 7.1

Cabin Cleaning Time min 8.0 6.0 26.5

Cabin Cleaning Rate pax/min 5.0 8.3 14.2Fuel Capacity gal 27,612 1,690 31800

Fuel Trucks # 2.0 2.0 2.0

Fuel Ports # 4.0 2.0 4.0

Fuel Time min 20.3 14.0 23.0

Fuel Rate gal/min 303.0 125.0 303.0

Lav Time min 10.0 5.0 15.0

Water Time min 10.0 5.0 17.0

Galley Time (Total) min 12.0 7.0 56.5

Galley Rate pax/min 3.3 7.1 6.6

Turn Time Stat Table


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We added a considerable amount to galley, water, and lavatory service times to account

for the additional capacity of our long-range flights. Figure 65 shows the complete

FAST turnaround procedure for a full load of 40 passengers.

Figure 65 FAST turnaround procedure

The dark blue bars represent actual service time, while the light blue bars represent

equipment placement time. With these assumptions, the total turnaround time for our

aircraft is 27 minutes.

Cost Model

Cost estimation and analysis for the GoldJet FAST was broken down into two parts:

Research, Development, Tooling, and Engineering (RDT&E) and Production Costs


83

Operating Costs

The RDT&E Costs and Production Costs, divided by the number of aircraft to be

produced and multiplied by a profit factor gives the Civil Purchase Price. We estimated

these using the the modified DAPCA IV Cost Model from Raymer which he dervices

from a 1987 RAND Corporation publication by Hess and Romanoff,[19]

and includes an

engine cost model from Birkler, Garfinkle, and Marks (1982), also from the RAND

Corporation.[20]

The equation used to calculate the RDT&E and Production costs (and

the purchase price) was:

Q

FFFFCCnCCCCRHFRHRHRHC

PovAdvISAvengengMFDQQMMMRTEE int

Where:

Engineering Hours = 163.0894.0777.086.4 QVWH eE

Tooling Hours = 263.0696.0777.099.5 QVWH eT

Manufacturing Hours = 641.0484.082.037.7 QVWH eM

Quality Control Hours = MQ HH 133.0

Manufacturing Hours = 641.0484.082.037.7 QVWH eM

Development Support Cost = 3.1630.00.66 VWC eD

Flight Test Cost = 21.1882.0325.01.1807 FTAVWC eF

Manufacturing Materials Cost = 799.0621.0921.016 QVWC eM

Engine Production Cost

o TurbTurbInleteng FTMTC 2228969.025.24343.00.2251 maxmax

Avionics Cost: eav WC 4000

Interior Cost: PassnC 2500int

The constants used in these equations are:

We = Empty Weight: 329,000 lb


84

V = Maximum Velocity: 1,032.5 knots

Q = Lesser of production quantity or number to be produced in five years: 120

FTA = Number of flight-test aircraft, 6 (Raymer says typically 2-6)

neng = Number of engines: 4

Tmax = Engine maximum sea-level static thrust: 35,361 lb

Mmax = engine maximum mach number: 1.8

TTurbInlet = Maximum Turbine Inlet Temperature: 2700°R

nPass = Number of passengers: 40

And the wrap rates are:

Engineering: 86$ER

Tooling: 88$TR

Quality Control: 81$QR

Manufacturing: 73$MR

In addition, several adjustment factors were used, as follows:

Advanced Materials Manufacturing Cost Factor: 25.1MF

o Settled upon after consultation with structures and materials group

Turbofan Cost Factor: 2.1TurbF

o Raymer says to multiply the engine cost model by 1.15-1.2 for a turbofan

engine

Initial Investment (“Profit”) Factor: 25.1PF

o Raymer: Roughly estimated as 1.1-1.4

Initial Spares: 1.1ISF

o Raymer: Initial spares add roughly 10%-15% to an aircraft’s purchase price

Overestimation Factor: 9.0OVF

o Raymer: DAPCA tends to over-predict commercial aircraft development costs

by roughly 10%

Note, however, that Raymer’s equations give costs in 1999 dollars. To adjust them to

2009 dollars, we used the Bureau of Labor Statistics’ Consumer Price Index, which


85

models the average change over time in prices paid by consumers for a market basket

of consumer goods and services.[21]

This gave a CPI adjustment factor of 1.2705,

yielding a purchase price in 2009 dollars of $236.78 million.

To calculate the operating cost model, we again used Raymer. He indicates that the

three factors which make up the operating cost are:

Crew Salaries

Fuel Costs

Maintenance (Labor and Materials)

In addition, we included insurance, which he states is an additional 1%-3% of the cost

of operations (because of our aircraft’s high speed and revolutionary design and

technology, we used the high end of 3%).

The crew salaries were calculated based on the three-man crew cost from Raymer:

172

1068

3.0

5

0WVBHCCrew

Where BH is the block hours per year (from the design mission), V is the cruise

velocity in knots, and W0 is the takeoff gross weight in pounds

Maintenance costs are estimated from Raymer as well: the labor costs using the

estimated maintenance man hours per flight hour (which Raymer gives as 5-15, and we

chose as 15), the number of flight hours per year (3500, inside Raymer’s range of 2400-

4500 for a civil transport) and the manufacturing wrap rate from the RDT&E model.

Maintenance material costs are estimated using Raymer’s equation:

19

10582.10

103.3

66

eeng

aMat

Cn

CFHC

Where:


86

FH = flight hours per year, 3500

Ca = Cost of aircraft (not including profit) minus engine

Ce = Cost of engine

neng = number of engines, 4

These numbers also needed adjusting to 2009 dollars, as Raymer also gives them in

1999 dollars.

The fuel costs are calculated using the weight of fuel burned during the design mission

(minus the missed approach), and the price of jet fuel (Jet A-1) as of April 3, 2009,

from the International Air Transport Association’s Jet Fuel Price Monitor.[22]

Combined, these yield an approximate operating cost per year of $36.9 million per

year, per aircraft.

VIII. Concept Summary

The GoldJet FAST is the first trans-Pacific supersonic airliner capable of linking the

United States with Asia. Our aircraft’s design mission flies from Los Angeles

International Airport (LAX) to Narita International Airport (NRT) in Tokyo, Japan.

With a cruise Mach number of 1.8 the FAST cuts 46% off the 10 hour 45 minute

subsonic flight, dropping it to 5 hours and 45 minutes. The FAST has a double-delta

“cranked arrow” wing planform with 4 low-bypass turbofan engines podded under its

wings. The FAST uses control canards for pitch control, a vertical tail and rudder for

lateral stability, and ailerons for roll control. Wing space has been budgeted for high-lift

devices, which allow for takeoff on runways smaller than 10,000 ft. The maximum

range for the FAST is 4,800 nm and an extra 240 nm has been factored in for a missed

approach and diversion to another airport.

Requirements Compliance

Our requirements compliance matrix is a collection of technical specifications

necessary to meet our customer needs determined early on in our market research.


87

Figure 66 Requirements compliance matrix

The GoldJet FAST meets all of our initial requirement thresholds with the exception of

cruise efficiency. We found that regardless of our chosen design tradeoffs, we were not

able to design a supersonic transport with a cruise efficiency higher than 1.6

paxmi/lbfuel. Flying fast is expensive and without strong advancements in lowering

wave drag during supersonic flight we feel it is unreasonable to expect a cruise

efficiency much higher than 2 paxmi/lbfuel. Another notable number is our boom

overpressure, which we were unable to reduce below 0.36 lb/in2. We had assumed at

the beginning of our design phase that an aircraft with a boom overpressure below 0.3

lb/in2 will be allowed to fly in supersonic corridors above land by the year 2020. Our

overpressure does not make this target value, however we did not completely optimize

our design for boom overpressure and it is reasonable to assume that with more time an

airliner could be designed to fly supersonic with an overpressure in the acceptable

range. The GoldJet FAST excels in the areas of cabin size and trip time savings, where

it is seated well within typical first class range and has nearly 50% time savings,

respectively.

Requirement Units Target Threshold Design

TO Field Length ft 10,000 12,000 11,380

Boom Overpressure lb/ft2 0.3 0.5 0.36

TO Gross Weight lb 300,000 400,000 329,000

Landing Field Length ft 10,000 12,000 9,000

Range nmi 5,650 4,707 4,800

Turnaround Time min 30 45 27

Number of Pasengers pax 40 40 40

Crew # 3 3 3

Cruise Speed Mach 2.1 1.75 1.8

Max Speed Mach 2.2 1.8 1.9

Trip Time min 380 415 345

Cabin Volume ft3/pax 60 55 75.8

Cruise Altitude ft 50000 45000 45,000

Cruise Efficiency paxmi/lb fuel 3.5 3 1.2

Second Segment Climb Gradient % 5.0 3.0 3.0

Seat Pitch in 40 38 50

Seat Width in 26 22 26

Aisle Width in 22 20 22

Aisle Height in 76 72 76

Gold Jet Requirements Compliance Matrix


88

Considering that our top four customer needs in order were profitable operations, low

trip time, long range, and marketability, the GoldJet FAST meets its requirements with

a high degree of success. The FAST’s largest obstacle is its cruise efficiency which,

being as low as it is, may make the concept implausible for mass production.

Plausibility

With the work done thus far the GoldJet team has come to the conclusion that the

GoldJet FAST, as designed, could be built economically, but not flown profitably.

GoldJet’s main plausibility concerns include the following:

sonic boom shaping has never been tested on a large passenger aircraft

the FAST uses four large under-wing engines with little noise shielding, which is

assumed to be considerable on takeoff and landing

the FAST’s awful cruise efficiency compared to long-range subsonic passenger

airliners, which may drive its ticket price too high for any reasonable market

large uncertainty in predicting 2020 market size as new technologies like video

conferencing emerge

Detailed Design Work Remaining

The FAST is currently in the preliminary design stages. The detailed design work

expected includes:

significant CFD to avoid canard vortex ingestion by wing-mounted engines

significant boom shaping to reduce boom overpressure to predicted level

engine R&D to replace the “rubber engine” model in propulsion calculations

complex avionics for fly-by-wire landing

structural member sizing and testing

manufacturing/assembly process design and contractor assignments/logistics


89

IX. References

[1]"Civil Supersonic Airplane Noise Type Certification Standards and Operating Rules,"

FR Doc E8-25052, Federal Aviation Administration, Washington, D.C., October 22,

2008. [http://edocket.access.gpo.gov/2008/E8-25052.htm. Accessed 2/03/09.]

[2]Wolkovitch, J., " The Joined Wing: An Overview," AIAA 23

rd Aerospace Sciences

Meeting, American Institute of Aeronautics and Astronautics, Reno, NV, 1985.

[3]Gallman, J.W. and Kroo, I.M., " Structural Optimization for Joined-Wing

Synthesis," Journal of Aircraft, Vol. 33, No. 1, 1993, pp. 214-223.

[4]Herrmann, U., “CISAP: Cruise Speed Impact on Supersonic Aircraft Planform; a

Project Overview,” 10th AIAA/ISSMO Multidisciplinary Analysis and Optimization

Conference, American Institute of Aeronautics and Astronautics, Albany, NY, 2004.

[5]Yoshimoto, M. and Uchiyama, N., " Optimization of Canard Surface Positioning of

Supersonic Business Jet for Low Boom and Low Drag Design," 33rd AIAA Fluid

Dynamics Conference and Exhibit, American Institute of Aeronautics and Astronautics,

Orlando, FL, 2003.

[6]Raymer, D., “Aircraft Design: A Conceptual Approach,” 4

th ed., AIAA, Reston, VA,

2006.

[7]Corke, T., “Design of Aircraft,” Prentice Hall, Upper Saddle River, NJ, 2002.

[8]Lee, Shiang-yu, "An Analytical Representation of Delta Wing Aerodynamics," 4th

AIAA Theoretical Fluid Mechanics Meeting, Toronto, Ontario, June 6-9, 2005.

[9]Kroo., I., “Volume Wave Drag,” Aircraft Design, Synthesis, and Analysis,

http://adg.stanford.edu/aa241/drag/volumedrag.html [accessed 06 Mar 2009].

[10]Lyrintis, A. (2009). AAE 514 Intermediate Aerodynamcis. West Lafayette, Indiana:

Purdue University.

[11]Carlson, H. W. (1978). Simplified Sonic-Boom Prediction. Hampton, VA: NASA

Scientific and Technical Information Office.


90

[12]

Plotkin, K., Page, J., Graham, D., Pawlowski, J., Schein, D., Coen, P., et al. (2003).

Ground Measurements of a Shaped Sonic Boom. AIAA/CEA8 Aeronautics Conference.

Manchester: American Institute of Aeronautics and Astronautics.

[13]Supersonic Aerospace International. (n.d.). QSST Quiet Supersonic Transport.

Retrieved April 21, 2009, from QSST Quiet Supersonic Transport:

http://www.saiqsst.com/popup.htm?images/drawing2.jpg

[14]Morgenstern, J. (2004). Patent No. 6729577. United States of America.

[15]Hill, P. and Peterson, C., “Mechanics and Thermodynamics of Propulsion,” 2

nd ed.

Addison-Wesley Publishing Company, Reading, MA, 1992.

[16]Roskam, J., Airplane Flight Dynamics and Automatic Flight Controls,

DARcorporation, Lawrence, KS, 1995.

[17]The Boeing Company. (December 2007). 777-200LR/-300ER/ FreighterAirplane

Characteristics for Airport Planning. Revision C. Seattle, WA:

http://www.boeing.com/

[18]Embraer. (May 30, 1997). Embraer EMB 145 Airport Planning Manual. Revision C.

San Jose Dos Campos, Brazil: http://www.embraer.com.

[19]Hess, R. W., and Romanoff, H. P., “Airframe Cost Estimating Relationships,” Rand

Corp., Rept. R-3255-AF, Santa Monica, CA, 1987.

[20]Birkler, J. L., Garfinkle, J. B., and Marks, K. E.., “Development and Production Cost

Estimating Relationships for Aircraft Turbine Engines,” Rand Corp., Rept. N-1882-AF,

Santa Monica, CA, 1982.

[21]"Consumer Price Index," Bureau of Labor Statistics, Washington DC, March 2009.

[ftp://ftp.bls.gov/pub/special.requests/cpi/cpiai.txt Accessed 3/22/09.]

[22]"Jet Fuel Price Monitor," International Air Transport Association, Montreal, Canada,

3 April 2009. [http://www.iata.org/whatwedo/economics/fuel_monitor/index.htm

Accessed 4/6/09.]


91

X. Appendix

Sizing Code

constants_FAST.m

% Constants file % Design Fixed Parameters specs = zeros(1,18); % initialize specs(1) = 328180.867719279; % W0 guess [lb] specs(2) = 1.85; % Aspect Ratio [ND] specs(3) = .425; % T/W ratio specs(4) = 2541.5; % Wing Area [ft^2] specs(5) = 1.8; % Cruise Mach [ND] specs(6) = 2; % Maximum Mach [ND] specs(7) = 3; % Number of Crew specs(8) = 40; % Number of Pax specs(9) = 200; % Weight Per Crew Member specs(10) = 220; % Weight Per Pax specs(11) = 4800; % Design Mission Range [nmi] specs(12) = 0.8; % estimate for cruise SFC [1/hr] - UNUSED specs(13) = 1; % estimate for loiter SFC [1/hr] specs(14) = 45000; % cruise altitude [ft] specs(15) = 100; % design mission headwind [kts] specs(16) = 280; % stall speed [ft/sec] specs(17) = 62; % wing leading edge sweep [deg] specs(18) = 14; % climb flight path angle gamma [deg]

refinedSizing.m

% Refined Sizing % Follows Raymer 19.3 pg 580+ % Main file - run this % In order to use the files in subdirectories, you must go to File -> Set % Path..., click on "Folder with sub-directories" and choose the current % folder. To check whether it has worked correctly, type "which Engine" % into the MATLAB command window. If it finds the file, then you are now % ready to run this code % clc clear constants_FAST; % call the constants file

% Give more readable variable names to constants W0 = specs(1) % W0 initial estimate [lb] - "as-drawn" aircraft takeoff

weight AR = specs(2) % AR T_W = specs(3) % Thrust to Weight S = specs(4) % Wing Area Mcruise = specs(5) % Cruise Mach Number Mmax = specs(6) % Max Mach Number (typically .2 higher than Mcruise) Wcrew = specs(7) * specs(9) % Wcrew [lb] Wpayload = specs(8) * specs(10) % Wpayload [lb] Range = specs(11) % Range [nmi]


92

SFCcruise = specs(12) % SFC at cruise [1/hr] SFCloiter = specs(13) % SFC at loiter [1/hr] hcruise = specs(14) % cruise altitude [ft] wind = specs(15) % max headwind [kt] Vstall = specs(16) % Stall speed [kt] WingSweep = specs(17) % Wing Sweep [°]

n = 0; % number of iterations completed (for tracking purposes) -

initialization

while (1<2) n = n+1;

% Find fuel weight fraction [Wf Wland cruiseLD WeightEngine Tengine Dengine dTO time Wfi] =

findFuelWeight(specs);

% Find empty weight from component weight equations We = call_compweight(specs, Wland, Wf*specs(1), Vstall, cruiseLD,

WeightEngine, Tengine, Dengine); % finds We based on component weight

equations

W0 = Wcrew + Wpayload + Wf*specs(1) + We; % Wcrew + Wpayload + Wf + We if abs((W0 - specs(1))) < .000001 % Checks if solution has converged break end specs(1) = (3*W0 + specs(1)) / 4; % Chooses intermediate value for

next iteration end

fprintf('Final W0 is %5.0f lb\n', W0) fprintf('Final Fuel weight is %5.0f lb\n', Wf*W0) fprintf('Final Empty weight is %5.0f lb\n', We) disp('Fuel fraction segments:') Wfi fprintf('Total iterations: %i\n', n)

% Call cost model GJCost(W0, We, Wf*W0, Tengine)

findFuelWeight.m

% Fuel Weight Function % Uses Raymer Ch. 3,6,19 and historical data

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Goes through each mission segment to find fuel weight fraction % There are a total of 10 mission segments: Taxi/Takeoff, Climb, Cruise, % Land, re-Takeoff, partial climb, divert to alternate, loiter for 45 min, % land, Taxi to Gate. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


93

function [Wf Wland cruiseLD WeightEngine Tengine Dengine dTO time Wfi] =

findFuelWeight(specs)

% called values used W0 = specs(1); % W0 guess [lb] AR = specs(2); % Aspect Ratio T_W = specs(3); % Thrust to Weight ratio S = specs(4); % Wing Area [ft^2] Mc = specs(5); % Cruise Mach R = specs(11); % Range [nmi] Cruiseh = specs(14); % cruise altitude [ft] wind = specs(15); % wind speed [kts] W_S = W0/S; % wing loading [lb/ft^2] Vstall = specs(16); gamma = specs(18); % flight path angle for climb

% hardcoded values for missions segments Ralt = .05*R; % Extra range for diversion to alternate airport - 5% of

design range E = .75; % Loiter time [hr] - nighttime or IFR - Raymer pg. 19

% need to fix these SFC's - not sure what to do for loiter - talk to Kevin % C = specs(12); % SFC [1/hr] Cloiter = specs(13); % SFC for loiter [1/hr]

% These will be refined through the following functions % LDmax = 11 / sqrt(specs(5)); % Supersonic L/Dmax - Estimate from Corke % LDcruise = 0.866*LDmax; % most efficient L/D for cruise - from Raymer LDloiter = 1.4*AR+7.1; % subsonic L/D max for loiter - estimate from

Corke

W = zeros(1,5); % 5 distinct segments

% Second segment - climb and accelerate [W(2) R_end Dengine WeightEngine Tengine dTO time1] =

findClimbFraction(Cruiseh, AR, W0, W_S, S, T_W, Vstall, gamma, Mc, 0, 0);

% split into small segments and using Raymer 19.8 and 19.9

% First segment - engine start, taxi, takeoff % W(1) = (1-C_idle*14*(T_idle/W0))*(1-C*1*(Tengine/W0)); % W1/W0, 14 min

at % ground idle plus 1 min at takeoff thrust - Raymer pg. 582 [T_lbs C_taxi Deng Leng Weng Lin] = Engine(0, 0, 20, Dengine); [T_lbs C_TO Deng Leng Weng Lin] = Engine(0, 0, 100, Dengine); W(1) = (1-C_taxi/60*14*4*Tengine/W0)*(1 - C_TO/60*2*1*Tengine/W0); %W(1) = 0.97; % historical value (for verification)

% Third segment - cruise [W(3) cruiseLD time2] = findCruiseWeightFraction(Mc, R-R_end, S, Cruiseh,

wind, Dengine, W0, W(1), W(2)); % Derived from Breguet Range equation -

(Raymer 6.11)

% Fourth segment - land W(4) = 0.995; % based on historical data - values 0.990-0.995 - (Raymer

6.22)


94

% Raymer says this is a good estimate even for detailed sizing

% Fifth segment - loiter W(5) = exp(-E*Cloiter/LDloiter); % Endurance Equation, with max L/D %

W_tot = 1; % initialize for loop

for m = 1:length(W) W_tot = W_tot*W(m); % Fuel fraction for total mission end

Wf = 1.06*(1-W_tot); % Fuel weight fraction - add 1% for trapped fuel Wfi = W;

time = 15 + time1/60 + time2/60 + 60+45; % total time of flight in minutes Wland = W0 * W_tot; % taxi/takeoff (estimate), climb time, cruise time, loiter + land time

(estimate) end

findClimbFraction.m

% Climb Fuel Weight Function % Uses Raymer 19.8, 19.9

function [Wi R_end Dengine WeightEngine Tengine dTO time] =

findClimbFraction(hMax, AR, W, W_S, S, T_W, Vstall, gamma, Mc, flag,

Dengine)

% V is in ft/sec for this segment

WeightEngine = 0; Wi = 1; % initialize g = 32.174; % ft/sec^2

T = T_W * W; % find max thrust % Call Mizener's code for takeoff distance

dTO = TODist2(W, W_S, T_W*W); % This TakeOffDistance is for OEI - one engine inoperative

% find speed at beginning of climb V = 1.2*Vstall; % [ft/sec] [t p rho a_prev] = Atmosphere(35,1); M = V/a_prev; % using a_prev for loop to follow Tengine = .25*T; % since there are 4 engines % call Kevin's Engine Size function to find single engine required size if flag == 0 [Dengine Lengine WeightEngine Linlet] = EngineSize(Tengine, 0, M); end % We will be using Dengine (engine diameter) in the remainder of function


95

time = 0; % initialize time required hstep = 1000; % altitude step in for loop hstart = 0; % initial altitude - should this be 35 ft? W_init = W; % store initial Weight

R = 0; % initialize range credit during climb

% step for each 1000 ft % first step : climb & accelerate to 10000 ft, Mach .58ish for h = hstart:hstep:10000 [t, p, rho, a] = Atmosphere(h,1); % find properties of std. atm. % use steady climb V = V+2; % linearly increase velocity for first 10000 ft L = W*cosd(gamma); % using fixed flight path angle - Raymer 17.36 M = V/a; if M >= Mc V = a*Mc; % don't expect this to happen, but just in case end % prevents acceleration beyond that required C_L = L/(.5*rho*V^2*S)*g; % Find C_L, converting rho from lbm/ft^3 to

slugs/ft^3 % uses previous value of V to compute C_L C_D=Drag(C_L,3,M, 0); % call drag function - assumed alpha %C_D=Drag(C_L,alpha, M, 0); % Call Drag function D = .5*rho * V^2*S*C_D/g; % Compute total Drag and fix units

T = D + W*sind(gamma); % find thrust required

[Thrust TSFC FuelRate Pwr] = EnginePower(0.25*T, h, M, Dengine);

Vv = V*sind(gamma); % vertical velocity dt = hstep / Vv; % time to climb this amount - seconds R = R + (V*cosd(gamma)*dt)*0.000164578834 ; % find range credit and

convert ft to nautical miles C = TSFC; dW = -C/3600*T*dt; % Weight loss from fuel burn W = W + dW; % Calculate new aircraft weight time = time + dt; % Calculate time of segment end

% Use Raymer 6.10 for the remainder of flight Wi1 = 0.991-0.007*M-0.01*M^2; Wi2 = 0.991-0.007*Mc-0.01*Mc^2; Wi12 = Wi2/Wi1;

[t p rho a1] = Atmosphere(10000,1); [t p rho a2] = Atmosphere(hMax,1); Vavg2 = (3*M*a1 + Mc*a2)/4; % average velocity for second part of climb gamma2 = 4; % assume average climb angle Vv2 = Vavg2*sind(gamma2); % average vertical velocity dt = (hMax-10000)/Vv2; time = time + dt; % total time


96

R_end = R + (Vavg2*cosd(gamma2)*dt)*0.000164578834; % distance traveled,

convert to nmi Wi = W/W_init * Wi12; % fuel fraction

findCruiseWeightFraction.m

% Cruise Fuel Weight Function

function [Wi cruiseLD time] = findCruiseWeightFraction(Mach, R, S, h,

wind, Dengine, W0, W1, W2)

% input R is actually the Range remaining - Range credit from ascent

% Choose range step Rstep = 50; % [nmi]

% Convert M to V (in kts) [t, p, rho, a] = Atmosphere(h, 1); % call std atm function V_in_ft_per_sec = a * Mach; V = 0.592483801 * V_in_ft_per_sec; % V in knots % Correct Range for Headwind R = R*V/(V-wind); W = W0*W1*W2; % W1 and W2 are fuel fractions, W0 is gross takeoff weight cruiseLD = 0;

Wi = 1; % initialize time = 0; % time initialization % start step for m = Rstep:Rstep:R % L = W, T = D - use these concepts to find SFC L = W; C_L = L/(.5*rho*V_in_ft_per_sec^2*S)*32.174; % Find C_L, converting

rho from lbm/ft^3 to slugs/ft^3 alpha=Cruise_AOA(L); C_D=Drag(C_L,alpha,Mach,1); % Call Drag function

D = .5*rho * V_in_ft_per_sec^2*S*C_D/32.174; % Compute total Drag and %fix units [Thrust TSFC FuelRate Pwr] = EnginePower(0.25*D, h, Mach, Dengine); %

Since at cruise, T = D, find C (for one engine) C = TSFC; % specific fuel consumption % find fuel consumed and duration dt = Rstep / V * 3600; time = time + dt; dW = -C/3600*D*dt; % Weight change during segment W = W + dW; Wi = Wi * exp(-Rstep*C/(V*L/D)); % from Breguet Range Equation if m == Rstep cruiseLD = L/D; end end


97

Performance

TODist2.m

% GoldJet Aerospace % AAE 451, Spring 2009 % Senior Design %%%%% % GJ FAST TO Distance % v 2.0 Andrew Mizener %%%%%

function [BFL] = TODist2(W_0, WL, T_sl)

% clc % clear format short eng

% Inputs: CL_alpha = 1.2632; % Lift-Curve Slope (1/rad OR 1/deg) CL_max = 1.2; % Maximum Lift Coefficient % W_0 = 270000; % Takeoff Weight % WL = 125; % Wing Loading, lbf/ft^2 % T_sl = 35000; % Sea-Level Static Thrust

% Takeoff and Clearing 35 Foot Obstacle

% Choose desired Takeoff Conditions % rho = 0.002193; % Altitude = 1000 ft, +15 degree C day,

slugs/ft^3 rho = 0.0023769; % Sea-Level, Normal Day, slugs/ft^3

rho_sl = 0.0023769; % Standard day sea-level density, slugs/ft^3 g = 32.2; % Acceleration Due To Gravity, ft/s h_obj = 35; % Height of object to clear, ft BPR = 0.1; % Bypass Ratio

gamma_c_deg = 12; % Takeoff angle, degrees gamma_c_rad = deg2rad(gamma_c_deg); % Takeoff angle, radians gamma_climb = 1*tan(gamma_c_rad); % Takeoff rate, % gamma_min = 0.030; % Minimum rate of climb: 3%

CL_climb = CL_alpha*gamma_c_rad; % Climb Lift Coefficient

G = gamma_climb - gamma_min; U = 0.01*CL_max + 0.02; T_av = 0.75*T_sl*((5+BPR)/(4+BPR)); T_av/W_0;

BFL = (0.863/(1+2.3*G))*(WL/(rho*g*CL_climb) + h_obj)*(1/(T_av/W_0-

U))+(655/sqrt(rho/rho_sl));


98

disp(['Balanced Field Length: ', num2str(BFL,5),' ft (including clearing

35 foot obstacle)']); end

LDist.m

% GoldJet Aerospace % AAE 451, Spring 2009 % Senior Design %%%%% % GJ FAST Landing Distance % v 0.1 Andrew Mizener %%%%%

function [S_L, S_L_Field] = LDist(W_0, T_sl)

% % Inputs: % W_0 = 270000; % lb % T_sl = 4*3500; % lb

rho_sl = 0.0023769; % Standard day sea-level density, slugs/ft^3 g = 32.2; % Acceleration Due To Gravity, ft/s h_obj = 50; % Height of object to clear, ft n = 1.2; % Raymer: Most aircraft n = 1.2 t_delay = 2; % Delay in hitting the brakes, seconds (Raymer: 1-

3 s) mu = 0.2; % Braking Coefficient

A_W = 2541.5; % Wing Area, ft^2 AR = 1.85; % Aspect Ratio CL_0 = 0.0; % Zero-angle Lift-Coefficient CD_0 = 0.0045517; % Parasite Drag Coefficient WL = W_0/A_W; % Wing Loading L2D = 9; % Lift-to-Drag Ratio on Approach

T2W = T_sl/W_0; % Thrust-Weight Ratio Lamb_LE = deg2rad(62); % Average Leading-Edge Sweep

V_stall = 283; % ft/s V_a = 1.3*V_stall; % Approach Velocity V_F = 1.23*V_stall; % Average Velocity during flare V_TD = 1.15*V_stall; % Touchdown Velocity V_f = 0; % Final Velocity on Land

gamma_a = -asin(T2W-(1/(L2D))); % Approach Angle, radians (ALSO:

gamma_a = (T-D)/W R = V_F^2/(g*n-2); % Flare Arc Radius h_F = R*(1-cos(gamma_a)); % Height of Flare F_Trev = 0.25; % Reverse Thrust Factor T_rev = -F_Trev*T_sl; % Reverse Thrust


99

e = 4.61*(1-0.045*AR^0.68)*(cos(Lamb_LE))^0.15-3.1; % Span Efficency

Factor K = 1/(pi*AR*e); % Drag-Due-to-Lift

Factor

K_T = (T_rev/W_0)-mu; K_A = rho_sl/(2*WL)*(mu*CL_0-CD_0-K*CL_0^2);

S_a = (h_obj-h_F)/tan(gamma_a); S_F = R*sin(gamma_a); S_FR = V_TD*t_delay; 1/(2*g*K_A); log((K_T + K_A*V_f^2)/(K_T + K_A*V_TD^2)); S_B = 1/(2*g*K_A)*log((K_T + K_A*V_f^2)/(K_T + K_A*V_TD^2));

S_L = S_a + S_F + S_FR + S_B; S_L_Field = S_F + S_FR + S_B;

disp(['Landing Distance: ', num2str(S_L,5),' ft (including clearing 50

foot obstacle on approach)']); disp(['Landing Field Distance: ', num2str(S_L_Field,5),' ft']); end

V_n.m clc

clear all

close all

%% Variable Definition

M_cruise = 1.8; % Mach number at cruise

a_cruise = 968.1; % Speed of sound at cruise altitude in ft/s

v_cruise = M_cruise*a_cruise; % Velocity of cruise in ft/sec

rho_SL = .0023769; % Density of air at sea level in slug/ft^3

rho_cruise = 0.0004623; % Density of air at cruise altitude in slug/ft^3

v_e_cruise = v_cruise*sqrt(rho_cruise/rho_SL) % Equivalent flight speed

at cruise in ft/sec

M_dive = M_cruise+.2; % Dive mach number according to Raymer

v_dive = M_dive*a_cruise; % Velocity of dive in ft/sec

v_e_dive = v_dive*sqrt(rho_cruise/rho_SL) % Equivalent flight speed for

dive in ft/sec

v_turb = .75*v_cruise; % Turbulence flight velocity

v_e_turb = 501 % Equivalent flight speed velocity for turbulence

%http://www.flightsimaviation.com/data/FARS/part_25-335.html

cl_max = 1.2;

%http://books.google.com/books?id=ltx92YbEMWMC&pg=PA244&lpg=PA244&dq=v-

n+diagram+calculations&source=bl&ots=6lIH10PZhX&sig=1CqONNfPq5sy9hGFMMYJvu

J9v0o&hl=en&ei=kt7tSfOsNJHIMrCuhfcP&sa=X&oi=book_result&ct=result&resnum=4

#PPA260,M1

AR = 1.8508; % Aspect ratio of the wing

cl_alpha= 7/9*pi*AR*tand(31);

g = 32.2; % Gravitational constant in ft/sec

weight_total = 270000; % Total aircraft weight in lbs

s_wing = 2400; % Planform area of wings


100

%% Maneuver array build and plot

v_e_stall(2,100) = 0;

top(2,100) = 0;

bottom(2,100) = 0;

neglim(2,100) = 0;

back(2,100) = 0;

% (n,v)

v_e_stall(1,:) = linspace(-1,3,100);

v_e_stall(2,:) =

sqrt(2*abs(v_e_stall(1,:))*weight_total/rho_SL/s_wing/cl_max);

top(1,:) = 3;

top(2,:) = linspace(max(v_e_stall(2,:)), v_e_dive, 100);

bottom(1,:) = -1;

bottom(2,:) = linspace(v_e_stall(2,1), v_e_cruise, 100);

neglim(1,:) = linspace(-1, 0, 100);

neglim(2,:) = linspace(v_e_cruise, v_e_dive, 100);

back(1,:) = linspace(0, 3, 100);

back(2,:) = v_e_dive;

plot(v_e_stall(2,:),v_e_stall(1,:))

hold on

plot(top(2,:),top(1,:))

plot(bottom(2,:),bottom(1,:))

plot(neglim(2,:),neglim(1,:))

plot(back(2,:),back(1,:))

axis([0 1000 -1.5 4])

xlabel('V_e (ft/s)')

ylabel('Load Factor')

title('V-n Gust Loads')

%% gust array build and plot

u_de_stall = 30; % Gust velocity for slow speeds in ft/sec

u_de_cruise = 30; % Gust velocity for high speeds in ft/sec

u_de_dive = 15; % Gust velocity for high speeds in ft/sec

u_de_turb = 45; % Gust velocity for high speeds in ft/sec

c_bar = 41.62 ; % Mean geometric chord of wing

mu_sub = 2*(weight_total/s_wing)/(rho_SL*g*c_bar*cl_alpha);

mu_super = 2*(weight_total/s_wing)/(rho_cruise*g*c_bar*cl_alpha);

k_sub = .88*mu_sub/(5.3+mu_sub);

k_super = mu_super^1.03/(6.95+mu_super^1.03);

u_gust_stall = k_sub*u_de_stall;

u_gust_cruise = k_super*u_de_cruise;

u_gust_dive = k_super*u_de_dive;

u_gust_turb = k_super*u_de_turb;

delta_n_zero = 1;

delta_n_turb =

rho_cruise*u_gust_turb*v_e_turb*cl_alpha/(2*weight_total/s_wing);

delta_n_cruise =

rho_cruise*u_gust_cruise*v_e_cruise*cl_alpha/(2*weight_total/s_wing);

delta_n_dive =

rho_cruise*u_gust_dive*v_e_dive*cl_alpha/(2*weight_total/s_wing);

for test = 1:100


101

if top(2,test) >= v_e_turb

turb = test;

break

end

end

for test2 = 1:100

if top(2,test2) >= v_e_cruise

cruise = test2;

break

end

end

n_zero = delta_n_zero

n_turb = top(1,turb) + delta_n_turb

n_cruise = top(1,cruise) + delta_n_cruise

n_dive = top(1,100) + delta_n_dive

n_gust_1(1,:) = linspace(n_zero,n_turb,100);

n_gust_1(2,:) = linspace(0,v_e_turb,100);

n_gust_2(1,:) = linspace(n_turb,n_cruise,100);

n_gust_2(2,:) = linspace(v_e_turb,v_e_cruise,100);

n_gust_3(1,:) = linspace(n_cruise,n_dive,100);

n_gust_3(2,:) = linspace(v_e_cruise,v_e_dive,100);

n_gust_4(1,:) = linspace(min(n_gust_3(1,:)), 2 - min(n_gust_3(1,:)),100)

n_gust_4(2,:) = v_e_dive;

plot(n_gust_1(2,:),n_gust_1(1,:),'r')



plot(n_gust_1(2,:),2-n_gust_1(1,:),'r')




Atmosphere.m

This was an open source function from the MathWorks. Code written by Richard Rieber.

% function [t, p, rho, a] = atmosphere(height, unit) % Richard Rieber % [email protected] % Updated 3/17/2006 % % Function to determine temperature, pressure, density, and speed of sound % as a function of height. Function based on US Standard Atmosphere of % 1976. All calculations performed in metric units, but converted to % english if the user chooses. Assuming constant gravitational % acceleration. % % Input o height (feet or meters), can be a 1-dimensional vector % o unit - optional; default is metric; % boolean, True for english units, False for metric % Output o t - Temperature (K (default) or R) % o p - Pressure (pa (default) or psi) % o rho - Density (kg/m^3 (default) or lbm/ft^3)


102

% o a - speed of sound (m/s (default) or ft/s)

function [t, p, rho, a] = Atmosphere(height,unit)

if nargin < 1 error('Error: Not enough inputs: see help atmosphere.m') elseif nargin > 2 error('Error: Too many inputs: see help atmosphere.m') elseif nargin == 2 unit = logical(unit); elseif nargin == 1 unit = 0; end

for j = 1:length(height) if height(j) < 0 error('Height should be greater than 0') end

m2ft = 3.2808; K2R = 1.8; pa2psi = 1.450377377302092e-004; kg2lbm = 2.20462262184878;

if unit height(j) = height(j)/m2ft; end

g = 9.81; %Acceleration of gravity (m/s/s) gamma = 1.4; %Ratio of specific heats R = 287; %Gas constant for air (J/kg-K)

%Altitudes (m) Start = 0; H1 = 11000; H2 = 20000; H3 = 32000; H4 = 47000; H5 = 51000; H6 = 71000; H7 = 84852;

%Lapse Rates (K/m) L1 = -0.0065; L2 = 0; L3 = .001; L4 = .0028; L5 = 0; L6 = -.0028; L7 = -.002;

%Initial Values T0 = 288.16; %(k) P0 = 1.01325e5; %(pa) Rho0 = 1.225; %(kg/m^3)


103

if height(j) <= H1 [TNew, PNew, RhoNew] = Gradient(Start, height(j), T0, P0, Rho0,

L1);

elseif height(j) > H1 & height(j) <= H2 [TNew, PNew, RhoNew] = Gradient(Start, H1, T0, P0, Rho0, L1); [TNew, PNew, RhoNew] = IsoThermal(H1, height(j), TNew, PNew,

RhoNew);

elseif height(j) > H2 & height(j) <= H3 [TNew, PNew, RhoNew] = Gradient(Start, H1, T0, P0, Rho0, L1); [TNew, PNew, RhoNew] = IsoThermal(H1, H2, TNew, PNew, RhoNew); [TNew, PNew, RhoNew] = Gradient(H2, height(j), TNew, PNew, RhoNew,

L3);

elseif height(j) > H3 & height(j) <= H4 [TNew, PNew, RhoNew] = Gradient(Start, H1, T0, P0, Rho0, L1); [TNew, PNew, RhoNew] = IsoThermal(H1, H2, TNew, PNew, RhoNew); [TNew, PNew, RhoNew] = Gradient(H2, H3, TNew, PNew, RhoNew, L3); [TNew, PNew, RhoNew] = Gradient(H3, height(j), TNew, PNew, RhoNew,

L4);

elseif height(j) > H4 & height(j) <= H5 [TNew, PNew, RhoNew] = Gradient(Start, H1, T0, P0, Rho0, L1); [TNew, PNew, RhoNew] = IsoThermal(H1, H2, TNew, PNew, RhoNew); [TNew, PNew, RhoNew] = Gradient(H2, H3, TNew, PNew, RhoNew, L3); [TNew, PNew, RhoNew] = Gradient(H3, H4, TNew, PNew, RhoNew, L4); [TNew, PNew, RhoNew] = IsoThermal(H4, height(j), TNew, PNew,

RhoNew);

elseif height(j) > H5 & height(j) <= H6 [TNew, PNew, RhoNew] = Gradient(Start, H1, T0, P0, Rho0, L1); [TNew, PNew, RhoNew] = IsoThermal(H1, H2, TNew, PNew, RhoNew); [TNew, PNew, RhoNew] = Gradient(H2, H3, TNew, PNew, RhoNew, L3); [TNew, PNew, RhoNew] = Gradient(H3, H4, TNew, PNew, RhoNew, L4); [TNew, PNew, RhoNew] = IsoThermal(H4, H5, TNew, PNew, RhoNew); [TNew, PNew, RhoNew] = Gradient(H5, height(j), TNew, PNew, RhoNew,

L6);

elseif height(j) > H6 & height(j) <= H7 [TNew, PNew, RhoNew] = Gradient(Start, H1, T0, P0, Rho0, L1); [TNew, PNew, RhoNew] = IsoThermal(H1, H2, TNew, PNew, RhoNew); [TNew, PNew, RhoNew] = Gradient(H2, H3, TNew, PNew, RhoNew, L3); [TNew, PNew, RhoNew] = Gradient(H3, H4, TNew, PNew, RhoNew, L4); [TNew, PNew, RhoNew] = IsoThermal(H4, H5, TNew, PNew, RhoNew); [TNew, PNew, RhoNew] = Gradient(H5, H6, TNew, PNew, RhoNew, L6); [TNew, PNew, RhoNew] = Gradient(H6, height(j), TNew, PNew, RhoNew,

L7); else warning('Height is out of range') end

t(j) = TNew; p(j) = PNew;


104

rho(j) = RhoNew; a(j) = (R*gamma*t(j))^.5;

if unit t(j) = t(j)*K2R; p(j) = p(j)*pa2psi; rho(j) = rho(j)*kg2lbm/m2ft^3; a(j) = a(j)*m2ft; end end

function [TNew, PNew, RhoNew] = Gradient(Z0, Z1, T0, P0, Rho0, Lapse) g = 9.81; %Acceleration of gravity (m/s/s) gamma = 1.4; %Ratio of specific heats R = 287; %Gas constant for air (J/kg-K)

TNew = T0 + Lapse*(Z1 - Z0); PNew = P0*(TNew/T0)^(-g/(Lapse*R)); % RhoNew = Rho0*(TNew/T0)^(-(g/(Lapse*R))+1); RhoNew = PNew/(R*TNew);

function [TNew, PNew, RhoNew] = IsoThermal(Z0, Z1, T0, P0, Rho0) g = 9.81; %Acceleration of gravity (m/s/s) gamma = 1.4; %Ratio of specific heats R = 287; %Gas constant for air (J/kg-K)

TNew = T0; PNew = P0*exp(-(g/(R*TNew))*(Z1-Z0)); % RhoNew = Rho0*exp(-(g/(R*TNew))*(Z1-Z0)); RhoNew = PNew/(R*TNew);

Aerodynamics

Drag.m

function [C_D,C_Di,C_Dw,C_D0]=Drag(C_L,alpha,M_inf,cruise)

%cruise 1 for it is, 0 for it isn't % C_L=0; % alpha=0; %deg % M_inf=1.3;

fast=0; %Fast uses the max diameter of the equivelent body of revolution,

0 uses Mach cuts

S_wing=2168+36.8*10.15; %ft^2. Planform area

load('Area_Profile_Appx.mat') S_size=size(S_ac); n=0; for i=1:8:max(S_size) n=n+1;


105

S_ac_use(n)=S_ac(i); %ft^2. x-z plane slices corresponding to the

positions specified by x_ac x_ac_use(n)=x_ac(i); end

S_ac=S_ac_use; x_ac=x_ac_use;

t_c_wing=.03; %Thickness to chord ratio x_max_c_wing=.4; %The percent of chord back from the leading edge that

the max thickness is located lam_max_wing=60; %The sweep of the max thickness c_root_wing=67.39; %ft. Root chord %C_L=.4; AR=1.85;

S_wet_tail=262*2; %ft x_max_c_tail=x_max_c_wing; t_c_tail=t_c_wing; lam_max_tail=45; %45 deg c_root_tail=25; %25 ft

S_wet_fuse=5607.4; %ft^2 l_fuse=180; %ft d_fuse=11.67; %ft

S_wet_eng=524.91; %ft l_eng=36.8; %ft d_eng=10.15; %ft

S_can=500; %ft (total planform of canards) t_c_can=t_c_wing; x_max_c_can=x_max_c_wing; lam_max_can=45; %ft c_root_can=20; %ft

P_L_wing=.3; P_L_tail=.3; P_L_fuse=.2; P_L_eng=.2; P_L_can=.3;

mu_inf=asind(1/M_inf);

%Re=rho_inf*v_inf*c_root/mu_inf;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% %Induced %Dependant on C_L, AR C_Di=C_L*tand(alpha);


106

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% %Wave %Dependent on Area Distribution as a function of x (S_ac, x_ac, M_inf)

%Generate Equivelent Body of Revolution %This will most certainly work if the initial and final areas are zero if M_inf<1 C_Dw=0; elseif cruise C_Dw=0.0071474; else C_Dw = 0; res=60; S_ref=2*S_wing;

x_ac_size=size(x_ac); for i=1:x_ac_size(2) x_eq(:,i)=x_ac(i).*ones(res,1); theta_eq(:,i)=rot90(fliplr(linspace(0,360-360/res,res))); r_eq(:,i)=sqrt(S_ac(i)/pi).*ones(res,1); end

if fast C_Dw=4*pi^2*max(max(r_eq))^2/max(x_ac)^2;

else for j=1:x_ac_size(2) for i=1:res [Y_eq(i,j), Z_eq(i,j),

X_eq(i,j)]=pol2cart(theta_eq(i,j)*pi/180,r_eq(i,j),x_eq(i,j)); end end

S_ac=[0,S_ac,0]; S_size=size(S_ac); x_ac=[min(x_ac)-.001*min(x_ac),x_ac,max(x_ac)+.001*max(x_ac)];

%Sums Areas of Vertical Projections of the Mach cuts for i=1:(x_ac_size(2)) %for a mach cut along x n=0; x_mid=x_ac(i); %The x coordinate where the Mach slice

intersects the x axis x_int=0; y_int=0; z_int=0; for j=1:res for k=1:x_ac_size(2) vec(1)=X_eq(j,k); vec(2)=Y_eq(j,k); vec(3)=Z_eq(j,k); if sqrt((vec(1)-x_mid)^2+vec(3)^2)==0 vec_angle(k)=acosd(dot([vec(1),vec(3)]-

[x_mid,0],[1,0]));


107

else vec_angle(k)=acosd(dot([vec(1),vec(3)]-

[x_mid,0],[1,0])/sqrt((vec(1)-x_mid)^2+vec(3)^2)); end

if k>1 if

(X_eq(j,k)==x_mid)&((theta_eq(j,k)==0)|(theta_eq(j,k)==180)) n=n+1; x_int(n)=x_mid; y_int(n)=Y_eq(j,k); z_int(n)=Z_eq(j,k); elseif (theta_eq(j,k)<180)&((vec_angle(k-

1)>mu_inf)&(vec_angle(k)<mu_inf)) n=n+1; x_int(n)=(X_eq(j,k)+X_eq(j,k-1))/2; y_int(n)=(Y_eq(j,k)+Y_eq(j,k-1))/2; z_int(n)=(Z_eq(j,k)+Z_eq(j,k-1))/2; elseif (theta_eq(j,k)>180)&((vec_angle(k-

1)>(mu_inf+90))&(vec_angle(k)<(mu_inf+90))) n=n+1; x_int(n)=(X_eq(j,k)+X_eq(j,k-1))/2; y_int(n)=(Y_eq(j,k)+Y_eq(j,k-1))/2; z_int(n)=(Z_eq(j,k)+Z_eq(j,k-1))/2; end end end end S_proj(i)=polyarea(y_int,z_int); end

S_ac=S_ac(1,2:(S_size(2)-1)); x_ac=x_ac(1,2:(S_size(2)-1)); S_size=size(S_ac);

S_proj_size=size(S_proj);

for i=2:(S_size(2)-1) dS_proj(i-1)=(S_proj(i-1)-S_proj(i+1))/(x_ac(i-1)-x_ac(i+1)); end

for i=3:(S_size(2)-2) d2S_proj(i-2)=(dS_proj(i-2)-dS_proj(i))/(x_ac(i-1)-x_ac(i+1)); end

%Computes C_Dw x_ac_for_d2=x_ac(1,3:(S_size(2)-2));%/x_ac(1,S_size(2)-1); C_Dw=0; for x1=1:(S_proj_size(2)-4) for x2=1:(S_proj_size(2)-4) if x2~=x1 if (x1>1)&(x1<(S_proj_size(2)-4)) if (x2>1)&(x2<(S_proj_size(2)-4)) C_Dw=C_Dw-

1/(2*pi*S_ref)*d2S_proj(x1)*d2S_proj(x2)*log(abs(x_ac_for_d2(x1)-


108

x_ac_for_d2(x2)))* abs(x_ac_for_d2(x1-1)-x_ac_for_d2(x1+1))/2 *

abs(x_ac_for_d2(x2-1)-x_ac_for_d2(x2+1))/2; elseif x2==1 C_Dw=C_Dw-



abs(x_ac_for_d2(x2)-x_ac_for_d2(x2+1))/2; elseif x2==(S_proj_size(2)-4) C_Dw=C_Dw-



abs(x_ac_for_d2(x2-1)-x_ac_for_d2(x2))/2; end elseif x1==1 if (x2>1)&(x2<(S_proj_size(2)-4)) C_Dw=C_Dw-


x_ac_for_d2(x2)))* abs(x_ac_for_d2(x1)-x_ac_for_d2(x1+1))/2 *







abs(x_ac_for_d2(x2-1)-x_ac_for_d2(x2))/2; end elseif x1==(S_proj_size(2)-4) if (x2>1)&(x2<(S_proj_size(2)-4)) C_Dw=C_Dw-


x_ac_for_d2(x2)))* abs(x_ac_for_d2(x1-1)-x_ac_for_d2(x1))/2 *







abs(x_ac_for_d2(x2-1)-x_ac_for_d2(x2))/2; end end end end end end end


109

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% %Parasite %Dependent on (thickness over chord) t_c, (position of max thickness as %%chord) x_max_c, lam_max, M_inf S_ref=S_wing;

if M_inf<1 %Components: wing, tail, fuselage, 4 engines, canards %C_f, FF, Q, S_wet %FIND BETTER ESTIMATES FOR PERCENT LAMINAR

%Wing Q_wing=1; S_wet_wing=2*S_wing;

FF_wing=(1+.6/x_max_c_wing*t_c_wing+100*t_c_wing^4)*(1.34*M_inf^.18*cosd(l

am_max_wing)^.28); C_f_wing=Friction_Coefficient(P_L_wing,c_root_wing/l_fuse,M_inf);

C_D0_wing=(C_f_wing*FF_wing*Q_wing*S_wet_wing)/S_ref;

%Tail Q_tail=1; S_wet_tail=S_wet_tail;

FF_tail=(1+.6/x_max_c_tail*t_c_tail+100*t_c_tail^4)*(1.34*M_inf^.18*cosd(l

am_max_tail)^.28); C_f_tail=Friction_Coefficient(P_L_tail,c_root_tail/l_fuse,M_inf);

C_D0_tail=(C_f_tail*FF_tail*Q_tail*S_wet_tail)/S_ref;

%Fuselage Q_fuse=1; S_wet_fuse=S_wet_fuse; f_fuse=l_fuse/d_fuse; FF_fuse=1+60/f_fuse^3+f_fuse/400; C_f_fuse=Friction_Coefficient(P_L_fuse,l_fuse/l_fuse,M_inf);

C_D0_fuse=(C_f_fuse*FF_fuse*Q_fuse*S_wet_fuse)/S_ref;

%Engines Q_eng=1.5; S_wet_eng=S_wet_eng; f_eng=l_eng/d_eng; FF_eng=1+.35/f_eng; C_f_eng=Friction_Coefficient(P_L_eng,l_eng/l_fuse,M_inf);

C_D0_eng=(C_f_eng*FF_eng*Q_eng*S_wet_eng)/S_ref; C_D0_eng=2*C_D0_eng;

%Canards Q_can=1; S_wet_can=2*S_can;


110

FF_can=(1+.6/x_max_c_can*t_c_can+100*t_c_can^4)*(1.34*M_inf^.18*cosd(lam_m

ax_can)^.28); C_f_can=Friction_Coefficient(P_L_can,c_root_can/l_fuse,M_inf);

C_D0_can=(C_f_can*FF_can*Q_can*S_wet_can)/S_ref;

%Total

C_D0=(C_D0_tail+C_D0_tail+C_D0_fuse+C_D0_eng+C_D0_can)+.03/.95*(C_D0_tail+

C_D0_tail+C_D0_fuse+C_D0_eng+C_D0_can); C_D0=C_D0;

else %Wing S_wet_wing=2*S_wing; C_f_wing=Friction_Coefficient(0,c_root_wing/l_fuse,M_inf);

C_D0_wing=(C_f_wing*S_wet_wing)/S_ref;

%Tail S_wet_tail=S_wet_tail; C_f_tail=Friction_Coefficient(0,c_root_tail/l_fuse,M_inf);

C_D0_tail=(C_f_tail*S_wet_tail)/S_ref;

%Fuselage S_wet_fuse=S_wet_fuse; C_f_fuse=Friction_Coefficient(0,l_fuse/l_fuse,M_inf);

C_D0_fuse=(C_f_fuse*S_wet_fuse)/S_ref;

%Engines S_wet_eng=2*S_wet_eng; C_f_eng=Friction_Coefficient(0,l_eng/l_fuse,M_inf);

C_D0_eng=(C_f_eng*S_wet_eng)/S_ref;

%Canards S_wet_can=2*S_can; C_f_can=Friction_Coefficient(0,c_root_can/l_fuse,M_inf);

C_D0_can=(C_f_can*S_wet_can)/S_ref;

%Total

C_D0=(C_D0_wing+C_D0_tail+C_D0_fuse+C_D0_eng+C_D0_can)+.03/.95*(C_D0_wing+

C_D0_tail+C_D0_fuse+C_D0_eng+C_D0_can); C_D0=C_D0;

end

C_D=C_Di+C_Dw+C_D0;


111

function C_f=Friction_Coefficient(P_L,l,M_inf)

l=l*180; C_fL=1.328/sqrt(38.21*(l/(1.33*10^-5))^1.053); P_T=1-P_L; C_fT=.455/(log10(44.62*(l/(1.33*10^-

5))^1.053*M_inf^1.6))^2.58*(1+.144*M_inf^2)^.65;

C_f=(P_L*C_fL+P_T*C_fT)/2;

function alpha_cruise=Cruise_AOA(lift)

alpha = [-5;0;0.5;1;1.5;2;2.5;3;3.5;4;4.5;5;10]; L = [-

606386.857523051;0;63125.1033443721;126627.805134608;190488.489867199;2546

87.251878909;319203.901703487;384017.972631856;... 449108.727473489;514455.16551657;580036.029684362;645829.813885109;1310584

.52067415]; alpha_cruise=interp1(L,alpha,lift); end

function [AR, S_wing, c_root, lam_ave, x_ac_wing]=

Wing_Sizing(M_inf,M_max,S_wing,w_fuse,x_cabin,L_cabin) %w_fuse is fuselage width %x_cabin is the distance from the nose to the begining of the cabin %L_cabin is the length of the cabin %I LIKE SI AND DEGREES!

%Required input Examples M_inf=1.6; %Freestream Mach number S_wing=201.4; %m^2 Planform area of both wings M_max=1.6; %Maximum operating Mach number (for placing wings within Mach

cone) w_fuse=2; %m Width of fuselage x_cabin=5; %m Axial position of the start of the cabin (negative

distance from nose) L_cabin=10; %m Length of cabin

%NOT NEEDED %For alt = 14km (45,000 ft), weight 270,000lbs, area 2168 ft^2 %gamma=1.4; %R=287; %KJ/Kg-K %T_inf=216; %K %P_inf=14090; %Pa %v_inf=M_inf*sqrt(gamma*R*T_inf); %m/s %W_ac=122470*9.81; %N


112

%Geometry (DLR) show=1; %If you want a plot of the planform, show=1 otherwise, show=0

%Find c_root, wing_span, lam_ave, AR %Dependent on S_wing, w_fuse geo_tol_err=.0001; %% geo_inc=1; %m low=1; high=100;

err=1; c_1last=0; while err>geo_tol_err n=0; err_1last=1; err=1; for c=low:geo_inc:high err_2last=err_1last; err_1last=err;

n=n+1; a=.28*c; b=.47*c; t=.1*c; A=2*( .5*a^2/tand(25) + (b-a)*tand(57)*(a+.5*(b-a)) + b*(t-(b-

a)*tand(10)) + .5*(b-a)^2*tand(10) + .5*a^2/tand(75) );

err=abs((S_wing-A)/S_wing);

if (n>3)&(err_2last>err_1last)&(err>err_1last) high=c; low=c_2last; end

if err<=geo_tol_err c_root=c; wing_span=2*b+w_fuse; end

c_2last=c_1last; c_1last=c; end geo_inc=geo_inc/2; end

AR=(2*b)^2/S_wing; lam_ave=atand((c_root-t)/((wing_span-w_fuse)/2));

%Find x_root (position closest to the nose the wing may be and not %interfere with the bow shock %Dependent on M_max, x_cabin, L_cabin, previous geometry

mu=asind(1/M_max); theta=atand(b/(a/tand(25)+(b-a)*tand(57)));


113

if mu>theta %Keeps wing behind shocks x_root=x_cabin-L_cabin; else %Keeps wing behind the cabin if -(b/tand(mu)-c_root)>(x_cabin-L_cabin) x_root=x_cabin-L_cabin; else x_root=-(b/tand(mu)-c_root); end end

%Generate LE & TE lines res=100; y_LE=linspace(w_fuse/2,b+w_fuse/2,res); y_TE=y_LE;

x_LE(1)=x_root; for i=1:res if y_LE(i)<(a+w_fuse/2) x_LE(i)=-1/tand(25)*(y_LE(i)-w_fuse/2)+x_root; x_TE(i)=1/tand(75)*(y_LE(i)-w_fuse/2)+x_root-c_root; else x_LE(i)=-1/tand(33)*(y_LE(i)-a-w_fuse/2)+x_root-a/tand(25); x_TE(i)=-1/tand(80)*(y_LE(i)-a-w_fuse/2)+x_root-c_root+a/tand(75); end end

%Find x_ac_wing %Dependent on M_inf, previous geometry c=x_LE-x_TE; if M_inf<1 for i=2:(res) x_ac(i-1)=(x_LE(i)+x_LE(i-1))/2-.25*(c(i-1)+c(i))/2; y_ac(i-1)=(y_LE(i)+y_LE(i-1))/2; weighted_ac(i-1)=((x_LE(i)+x_LE(i-1))/2-.25*(c(i-

1)+c(i))/2)*(y_LE(i)-y_LE(i-1))*(c(i-1)+c(i))/2; end else for i=2:(res) x_ac(i-1)=(x_LE(i)+x_LE(i-1))/2-.4*(c(i-1)+c(i))/2; y_ac(i-1)=(y_LE(i)+y_LE(i-1))/2; weighted_ac(i-1)=((x_LE(i)+x_LE(i-1))/2-.4*(c(i-

1)+c(i))/2)*(y_LE(i)-y_LE(i-1))*(c(i-1)+c(i))/2; end end

x_ac_wing=sum(weighted_ac)/(.5*S_wing);

if show plot(x_LE,y_LE) axis equal hold on


114

plot(x_TE,y_TE) plot([x_LE(res), x_TE(res)],[y_LE(res), y_TE(res)])

plot(x_LE,-y_LE) plot(x_TE,-y_TE) plot([x_LE(res), x_TE(res)],[-y_LE(res), -y_TE(res)])

plot(x_ac_wing,0,'.g') plot(x_ac,y_ac,'g') plot(x_ac,-y_ac,'g') title('Wooooosh') end

Lift.m

function C_L=Lift(alpha,config) %alpha=2; %config=0; %0 for normal config, 1 for take-off, 2 for landing

S_can=500; %ft (total planform of canards) lam_max_can=45; %deg lam_min_can=atand(12/20); lam_ave_can=(lam_max_can+lam_min_can)/2; w_fuse=11.67; %ft Width of fuselage can_span=20; %ft

lam_ave_wing=62; AR_wing=1.85; AR_can=(2*can_span-w_fuse)^2/S_can; S_wing=2168+36.8*10.15; %ft^2. Planform area S_ref=S_wing; %alpha=-5:.01:20;

n=0; %for ALPHA=alpha ALPHA=alpha; n=n+1; CNLa=(7/9)*pi*AR_wing*tand(lam_ave_wing/2); CNN=2/(1+AR_wing); CL_wing(n)=CNLa*sind(ALPHA)*cosd(ALPHA)+CNN*sind(ALPHA)^2*cosd(ALPHA); alpha0=0; CL_max=1.2; CL_alpha=CNLa;

CNLa=(7/9)*pi*AR_can*tand(lam_ave_can/2); CNN=2/(1+AR_can); CL_can(n)=CNLa*sind(ALPHA)*cosd(ALPHA)+CNN*sind(ALPHA)^2*cosd(ALPHA); %end

S_flaped=polyarea([-173,-167.9,-140.6,-100],[6,24.76,24.76,6]); %ft^2 if config==0 dC_L_flap=0; dC_L_slat=0; elseif config==1


115

dC_l_flap=1.3*.6; dC_L_flap=.9*dC_l_flap*(S_flaped/S_ref)*cosd(75);

dC_l_slat=.4*.6*1; dC_L_slat=.9*dC_l_slat*(S_flaped/S_ref)*cosd(60) elseif config==2 dC_l_flap=1.3*.8; dC_L_flap=.9*dC_l_flap*(S_flaped/S_ref)*cosd(75);

dC_l_slat=.4*.8*1; dC_L_slat=.9*dC_l_slat*(S_flaped/S_ref)*cosd(60)

end

C_L=CL_wing+CL_can+dC_L_flap+dC_L_slat;

Cost Model GJCost.m % GoldJet Aerospace % AAE 451, Spring 2009 % Senior Design %%%%% % GJ FAST Cost Model % v 1.0 Andrew Mizener %%%%%

function [] = GJCost(W_0, W_e, W_f, Thrust_max) % Inputs Required: W_e, Thrust_max, W_0, W_f

% % Inputs for Testing % clc % clear % W_0 = 270000; % Gross Takeoff Weight, lbs % W_e = 103000; % Empty Weight, lbs % W_f = 155000; % Fuel Weight, lbs % Thrust_max = 35000; % Engine Maximum Sea-Level Static Thrust, lb

%%% Constants %%% M_cruise = 1.8; % Cruise Mach Number a_cruise = 968.1; % Speed of Sound (ft/s) at cruise altitude:

45,000 ft V_fps = M_cruise*a_cruise; % Maximum Velocity, feet/second V = V_fps*0.592483801; % Maximum Velocity, knots Q = 120; % Less of Production Quantity or # to be

produced in 5 years n_FTA = 6; % # of Flight-Test Aircraft n_eng = 4; % # of Engines n_pass = 40; % # of passengers M_max = 2.0; % Maximum Flight Envelope Mach Number T_Turb_inlet = 2700; % Turbine Inlet Temperature, degrees R


116

%%% Factors %%% F_M = 1.25; % Advanced Materials Manufacturing Cost Factor F_Turb = 1.2; % Turbofan Cost Factor F_P = 1.5; % Investment Cost ("Profit") Factor F_IS = 1.1; % Inital Spares F_adv = 1; % Advanced Design Factor F_ov = 0.9; % Overestimation Factor CPI_Adjust = 1.2705; % CPI Adjustment Factor: 1999 - 2009

%%% Hours %%% H_E = 4.86*W_e^0.777*V^0.894*Q^0.163; % Engineering H_T = 5.99*W_e^0.777*V^0.696*Q^0.263; % Tooling H_M = 7.37*W_e^0.820*V^0.484*Q^0.641; % Manufacturing H_Q = 0.133*H_M; % Quality Control

%%% Wrap Rates %%% R_E = 86; % Engineering R_T = 88; % Tooling R_Q = 81; % Quality Control R_M = 73; % Manufacturing

%%% Costs %%% % Devel Support Cost C_D = 66.0*W_e^0.630*V^1.3; % Flight Test Cost C_F = 1807.1*W_e^0.325*V^0.822*n_FTA^1.21; % Manufacturing & Materials Cost C_M = 16*W_e^0.921*V^0.621*Q^0.799; % Engine Production Cost C_eng = 2251.0*(0.043*Thrust_max+243.25*M_max+0.969*T_Turb_inlet-

2228)*F_Turb; % Avionics Cost C_av = 4000*W_e; % Interior Cost C_int = 2500*n_pass;

%%% RDT&E + Flyaway Cost in 1999 Dollars %%% C = (H_E*R_E + H_T*R_T + H_M*R_M*F_M + H_Q*R_Q + C_D + C_F + C_M +

C_eng*n_eng + C_av + C_int)*F_IS*F_adv*F_ov/Q;

% Cost per Aircraft, excluding engine C_a = (H_E*R_E + H_T*R_T + H_M*R_M*F_M + H_Q*R_Q + C_D + C_F + C_M + C_av

+ C_int)*F_IS*F_adv*F_ov/Q;

% Cost per Engine C_e = C_eng*F_IS*F_ov;

%%% Purchase Price in 1999 Dollars %%% P_1999 = C*F_P;

%%% Purchase Price in 2009 Dollars %%% P_2009 = P_1999*CPI_Adjust; P_2009_Millions = P_2009/1e6;


117

disp(['Gold Jet FAST Purchase Price, in millions of 2009 dollars: $',

num2str(P_2009_Millions)]);

%%%% Operating and Maintenance Costs %%%%

% Block and Flight Time FH = 2500; % Flight Hours Per Year Per Aircraft Raymer =

2500-4500 BHR = 1.041204795; % Block Hour Ratio - Block Hours/Flight Hours,

based on design mission BH = BHR*FH; % Block Hours Per Year (Per AC)

% Crew Cost CCpBH = 68*(V*W_0/(1e5))^0.3 + 172; % 3-man crew cost per block hour

(2 Pilots, one Flight Attendant) C_C = BH*CCpBH; % Crew Cost (incl. salary and

benefits) per year

% Maintenance Cost MMHpFH = 15; % Maintenance Hours per Flight Hour, Raymer =

5-15 MMHpY = MMHpFH*FH; % Maintenance Hours per Year C_MaintLab = MMHpY*R_M; % Maintenance Labor Costs, approximated using

Manufacturing Wrap Rate

C_MatpFH = 3.3*(C_a/1e6)+10.2+(58*(C_e/1e6)-19)*n_eng; % Materials

Cost of Maintenance per Flight Hour C_MaintMat = C_MatpFH*FH; % Materials

Cost per Year

C_Maint = C_MaintLab + C_MaintMat; % Total Maintenance Cost

% Fuel Cost W_fb = 0.9*W_f; % Weight of Fuel Burned per flight, lb T_DesMiss = 5.663; % Design Mission Flight Time, hours W_fbph = W_fb/T_DesMiss; % Weight of Fuel Burned per Flight Hour rho_fuel = 6.7389141; % Density of fuel, lb/gallon C_Fpg = 1.46; % Fuel Price, in $/gallon, as of April 3,

2009 (from http://www.iata.org/whatwedo/economics/fuel_monitor/index.htm) C_Fplb = C_Fpg/rho_fuel; % Fuel Price, in $/lb C_Fuel = W_fbph*C_Fplb*FH; % Fuel Cost, in $ 2009

C_OM = (C_C + C_Maint)*CPI_Adjust + C_Fuel; C_OM_Millions = C_OM/1e6;

disp(['Gold Jet FAST Operating and Maintenance Costs (Per year), in

millions of 2009 dollars: $', num2str(C_OM_Millions)]);

Control Surface Sizing and Stability

vertical_tail.m


118

function [s_vt, b_vt, AR_vt, taper_vt, sweep_vt] = vertical_tail(b_w, s_w,

l_vt, T, y_e, W, fuse_volume)

%AAE 451

%Spring 2009

%Team 3

%

%This function uses the size of the wing and locations of the engines to

%find a preliminary size of the vertical tail.

%inputs:

%b_w = wingspan (ft)

%s_w = planform area of wing (ft^2)

%l_vt = distance from A/C c.g. to vertical tail aerodynamic center(ft)

%T = thrust of one engine(lbf SLS)

%y_e = moment arm of engine out(ft)

%W = weight of A/C (lbs)

%fuse_volume (ft^3)

%

%outputs

%s_vt = planform area of vertical tail (ft^2)

%b_vt = height of v.t. (ft)

%AR_vt= aspect ratio of v.t.

%taper_vt = taper ratio for v.t.

%sweep_vt = sweep angle for v.t.

%constants (some can change)

rho = .002378; % slugs/ft^3 (sea level)

V = 368; %ft/s (takeoff speed)

q_bar = .5*rho*V^2; %lb/ft^2

AR_vt = 2;

taper_vt =.7;

sweep_vt = 45; %degrees

alpha_rudder = .5; %effectiveness of rudder (30% chord is rudder, see

roskam fig 2.23)

C_L_alpha_v = 2.1; %estimate from Roskam fig 2.7

c_vt = .07; %vertical tail volume coefficient (assume = .07 for

jet fighter)

s_vt1 = c_vt*b_w*s_w/l_vt; %initial estimate of tail size (Raymer)

%crosswind =20% TO speed

beta = .197; %rad

delta_rudder = -20*pi/180; %rad

Cn_wind = -1; %intial value to trigger loop

s_vt2 = 1; %initial estimate can be larger

while (Cn_wind<0)||(Cn_wind>.01); %loops until yawing moment coefficient

is close to zero

s_vt2 = s_vt2+1;

Cn_beta_fuse = -1.3*fuse_volume/(s_w*b_w); %yawing moment due to

fuselage

Cn_beta_vt = C_L_alpha_v*1*1*s_vt2*l_vt/(s_w*b_w); %yawing moment due

to V.T.

Cn_beta = Cn_beta_fuse + Cn_beta_vt; %total yawing moment due to

sideslip

Cn_deltarudder = -C_L_alpha_v*alpha_rudder*1*s_vt2*l_vt/(s_w*b_w);

%yawing moment due to rudder deflection


119

Cn_wind = Cn_beta*beta + Cn_deltarudder*delta_rudder; %total yawing

moment coefficient

end

%one engine inoperative (OEI)

%assume no slideslip, drag of dead engine is assumed small compared to the

thrust of the opposite engine

N_oei = -T*y_e; %yawing moment caused by engine out

N = N_oei;

s_vt3 = 1;

delta_rudder = -20*pi/180; %rad

while N<0; %loops until the rudder moment outdoes the oei moment

s_vt3 = s_vt3+1; %initial estimate can be larger

Cn_deltarudder = -C_L_alpha_v*alpha_rudder*1*s_vt3*l_vt/(s_w*b_w);

%yawing moment due to rudder deflection

b_vt = sqrt(AR_vt*s_vt3);

N_rudder = Cn_deltarudder*delta_rudder*q_bar*s_vt3*b_vt;

N = N_oei + N_rudder;

end

s_vt = [s_vt1, s_vt2, s_vt3];

b_vt = sqrt(AR_vt*s_vt3); %corresponding tail height

echo off

canard.m

function [s_c, b_c, AR_c, taper_c, sweep_c] = canard(b_w, s_w, l_c, W, cg)

%AAE 451

%Spring 2009

%Team GoldJet

%

%This function sizes the canard

%inputs:

%b_w = wingspan (ft)

%s_w = planform area of wing (ft^2)

%l_c = distance from A/C c.g. to canard aerodynamic center

%W = weight of A/C (lbs)

%cg = location of A/C center of gravity aft of MAC leading edge

% (Roskam fig 3.24)

%

%outputs

%s_c = planform area of canard (ft^2) [Raymer estimate, Takeoff,

supersonic]

%b_c = wingspan of canard (ft) [Raymer estimate, Takeoff, supersonic]

%AR_c= aspect ratio of v.t.

%taper_c = taper ratio for canard

%sweep_c = sweep angle for canard

%disp('------------------------CANARD------------------------');disp(' ')

%constants (some can change)

rho = .002378; % slugs/ft^3 (sea level)

V = 368; %ft/s (approach speed)


120


AR_c = 2; %assumed value of AR, used for calculating C_L_alpha_c

taper_c =.5;

sweep_c = 30; %degrees

alpha_e = 1; %effectiveness of elevator (all moving surface, see

roskam fig 2.23)

C_L_alpha_c = 2.5; %estimate from Roskam fig 2.7 (assume higher AR than

wing)

C_L_0_c = .3; %zero angle of attack lift coefficient of canard

c_c = .1; %canard volume coefficient (assume = .1 for canard -

Raymer)

a_bar_ac_wf = .25; %assume at 1/4 MAC for subsonic

C_L_alpha_wf = 2.1; %estimate from Roskam fig 2.7

eta_c =1;

MAC1 = 2/3*(82.4+18 - 82.4*21.85/(82.4+18));

MAC2 = 2/3*(4.2+18 - 4.2*18/(4.2+18));

MAC = (MAC1+MAC2)/2;

c_bar = MAC; %mean aerodynamic chord

x_bar_cg = cg/c_bar; %non dimensional cg location

x_bar_ac_c = l_c/c_bar; %nondimensional moment arm of canard

alpha_max_c = 15*pi/180; %max angle of attack for canard during TO(rad)

s_c1 = c_c*b_w*s_w/l_c; %initial estimate of tail size (Raymer)

%parameters to start loop

s_c = 1;

M_diff = -1;

%subsonic loop begin

while M_diff < 0;

s_c = s_c+1;

%find aerodynamic center (Roskam eqn 3.40)

x_bar_ac_a = (a_bar_ac_wf -

C_L_alpha_c/C_L_alpha_wf*eta_c*s_c/s_w*x_bar_ac_c)/(1 +

C_L_alpha_c/C_L_alpha_wf*eta_c*s_c/s_w*x_bar_ac_c);

%find static margin

SM = -(x_bar_cg-x_bar_ac_a);

%find trim force needed from canard

M_weight = -SM*c_bar*W; %ft-lbf

M_wing_lift = -(a_bar_ac_wf-x_bar_ac_a)*.95*W;%assume wing handles

most of the lift

M_total = M_weight+M_wing_lift;%ft-lbf

%find max trim force possible

C_L_c_max = C_L_0_c + C_L_alpha_c*alpha_max_c;

L_c_max = C_L_c_max*q_bar*s_c;

M_c_max = L_c_max*l_c;

%check moments

M_diff = M_total + M_c_max;

end

s_c2 = s_c;


121

%supersonic loop

a_bar_ac_wf = .4; %assume at .4 chord for supersonic

alpha_max_c = 2*pi/180; %want low canard deflection for low cruise drag

rho = 0.0004622; % slugs/ft^3 (sea level)

V = 1500; %ft/s (cruise speed M= 1.8)


s_c = 1;

M_diff = -1;

%subsonic loop begin

while M_diff < 0;

s_c = s_c+1;

%find aerodynamic center (Roskam eqn 3.40)

x_bar_ac_a = (a_bar_ac_wf -

C_L_alpha_c/C_L_alpha_wf*eta_c*s_c/s_w*x_bar_ac_c)/(1 +

C_L_alpha_c/C_L_alpha_wf*eta_c*s_c/s_w*x_bar_ac_c);

%find static margin

SM = -(x_bar_cg-x_bar_ac_a);

%find trim force needed from canard

M_weight = -SM*c_bar*W; %ft-lbf

M_wing_lift = -(a_bar_ac_wf-x_bar_ac_a)*.9*W;%assume wing handles most

of the lift

M_total = M_weight+M_wing_lift;%ft-lbf

%find max trim force possible

C_L_c_max = C_L_0_c + C_L_alpha_c*alpha_max_c;

L_c_max = C_L_c_max*q_bar*s_c;

M_c_max = L_c_max*l_c;

%check moments

M_diff = M_total + M_c_max;

end

s_c3 = s_c;

%calculate canard dimensions

s_c = [s_c1 s_c2 s_c3];

b_c = (AR_c*s_c).^.5;

%find aerodynamic centers at subsonic and supersonic conditions to

estimate C.G. limits

a_bar_ac_wf = .25;

x_bar_ac_a2 = (a_bar_ac_wf -

C_L_alpha_c/C_L_alpha_wf*eta_c*s_c2/s_w*x_bar_ac_c)/(1 +

C_L_alpha_c/C_L_alpha_wf*eta_c*s_c2/s_w*x_bar_ac_c);

a_bar_ac_wf = .4;

x_bar_ac_a3 = (a_bar_ac_wf -

C_L_alpha_c/C_L_alpha_wf*eta_c*s_c3/s_w*x_bar_ac_c)/(1 +

C_L_alpha_c/C_L_alpha_wf*eta_c*s_c3/s_w*x_bar_ac_c);

Trimdiagram_goldjet.m


122

% Aircraft trim diagrams

% AAE 451 - Team GoldJet

%SUBSONIC trim diagram

CL0=0;

CLalpha=2.35*pi/180; % Per degree

CLdeltaE=.43*pi/180; % Per degree

CLiH=2.5*s_c2/s_w*pi/180; %roskam pg 79

CM0=.04;

CMalpha=-.05*pi/180; % Per degree

CMdeltaE=0; %Per degree

CMiH = 2.5*s_c2/s_w*(l_c/MAC-x_bar_cg)*pi/180; %use roskam pg 85

deltaE=0;

iH=[0 4 8 12 16]; % Degree

mom_ref_pt=x_bar_ac_a2 ; % Moment reference point in % chord

forward_cg=-29.83/MAC ; % forward cg limit in % chord

aft_cg=-12/MAC ; % aft cg limit in % chord

alpha_stall=30 ; % angle of attack (deg) for stall

Cl_PlotMax=1.5; % maximum Cl for plots in figure 2 (the aircraft trim

diagram)

alpha_PlotMax=30; % maximum angle of attack (deg) for figure 1

% End of data input section

% Plotting information

color=['-bo-gx-r+-c*-md-yv-k^'];

s1=['iH=',num2str(iH(1)),' deg.'];





% End plotting information

alpha=0:1:alpha_PlotMax;

dCMdCL=CMalpha/CLalpha;

CM0bar=CM0-dCMdCL*CL0;

CMiHbar=CMiH-dCMdCL*CLiH;

CMdeltaEbar=CMdeltaE-dCMdCL*CLdeltaE;

%%

% Plot aircraft trim diagram

figure(1) % <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<

clf

for i=1:length(iH)

iHx=iH(i);

CL=CL0+CLalpha*alpha+CLiH*iHx+CLdeltaE*deltaE;

CM=CM0bar+dCMdCL*CL+CMiHbar*iHx+CMdeltaEbar*deltaE;

plot(alpha,CL); hold on

end

% End temporary plot


123

% Extend plots

z1=axis;

clf

CLexpended=z1(3):.1:z1(4);

for i=1:length(iH)

iHx=iH(i);


hold on

CMexpanded=CM0bar+dCMdCL*CLexpended+CMiHbar*iHx+CMdeltaEbar*deltaE;

hold on; plot(CMexpanded,CLexpended,color(3*(i-1)+1:3*(i-1)+3));

end

LOC='SouthEast';

legend(s1,s2,s3,s4,s5,'Location',LOC)

% end of expanded plotting

plot([0 0],[0 z1(4)],'k') %plot zero line on CM plot

hold off

z1=axis;axis([z1(1) z1(2) 0 Cl_PlotMax])

% plot the forward cg line of the trim triangle

delta_cg1=mom_ref_pt-forward_cg;

Cm_forward=+Cl_PlotMax*delta_cg1;

hold on; plot([0 Cm_forward],[0 Cl_PlotMax],'k') %plot zero line on CM

plot

% plot the aft cg line of the trim triangle

delta_cg2=mom_ref_pt-aft_cg;

Cm_aft=+Cl_PlotMax*delta_cg2;

hold on; plot([0 Cm_aft],[0 Cl_PlotMax],'k') %plot zero line on CM plot

str1=['CM about ',num2str(mom_ref_pt),' c'];

xlabel(str1)

ylabel('CL')

title('Fig 4.9(right), Fig 4.10(right),Fig 4.11b(right) with De=-30')

grid on

%text(.18,.75,['iH= ',num2str(iH),' deg.'])

strXF=['fwd cg xbar=',num2str(forward_cg)];

strXR=['aft cg xbar=',num2str(aft_cg)];

%text(.16,1.18,strXF)

%text(-.02,1.18,strXR)

stralpha=['alpha stall=',num2str(alpha_stall), ' deg.'];

% text(.09,.95,stralpha)

ylabel('CL')

% xlabel('alpha (deg)')

title('Subsonic Trim Diagram')

grid on

%text2(.1,.95,['iH= ',num2str(iH),' deg.'])

hold off

% Plot alpha_stall line in figure 2

CL=CL0+CLalpha*alpha_stall+CLiH*iH+CLdeltaE*deltaE;

CM=CM0+CMalpha*alpha_stall+CMiH*iH+CMdeltaE*deltaE;

hold on; plot(CM,CL,'k');

axis([0 .7 0 1.45])

set(gca, 'XDir', 'reverse'); % reverse the plotting direction on the x

axis


124

% End alpha_stall line plotting

%%

%SUPERSONIC

CL0=0;

CLalpha=2.35*pi/180; % Per degree

CLdeltaE=.43*pi/180; % Per degree

%CLiH=.25*pi/180;% Per degree

CLiH=2.5*s_c2/s_w*pi/180; %roskam pg 79

CM0=.04;

CMalpha=-.05*pi/180; % Per degree

CMdeltaE=0; %Per degree

% CMiH=1.5*pi/180 ;%Per degree

CMiH = 2.5*s_c2/s_w*(l_c/MAC-x_bar_cg)*pi/180; %use roskam pg 85

deltaE=0;

iH=[0 4 8 12 16]; % Degree

mom_ref_pt=x_bar_ac_a3 ; % Moment reference point in % chord

forward_cg=-29.83/MAC ; % forward cg limit in % chord

aft_cg=-12/MAC ; % aft cg limit in % chord

alpha_stall=30 ; % angle of attack (deg) for stall

Cl_PlotMax=1.5; % maximum Cl for plots in figure 2 (the aircraft trim

diagram)

alpha_PlotMax=30; % maximum angle of attack (deg) for figure 1

% End of data input section

% Plotting information

color=['-bo-gx-r+-c*-md-yv-k^'];






% End plotting information

alpha=0:1:alpha_PlotMax;

dCMdCL=CMalpha/CLalpha;

CM0bar=CM0-dCMdCL*CL0;

CMiHbar=CMiH-dCMdCL*CLiH;

CMdeltaEbar=CMdeltaE-dCMdCL*CLdeltaE;

%%

% supersonic trim diagram

figure(2)

clf

for i=1:length(iH)

iHx=iH(i);


CM=CM0bar+dCMdCL*CL+CMiHbar*iHx+CMdeltaEbar*deltaE;

plot(alpha,CL); hold on


125

end

% End temporary plot

% Extend plots

z1=axis;

clf

CLexpended=z1(3):.1:z1(4);

for i=1:length(iH)

iHx=iH(i);


hold on

CMexpanded=CM0bar+dCMdCL*CLexpended+CMiHbar*iHx+CMdeltaEbar*deltaE;

hold on; plot(CMexpanded,CLexpended,color(3*(i-1)+1:3*(i-1)+3));

end

LOC='SouthEast';

legend(s1,s2,s3,s4,s5,'Location',LOC)

% end of expanded plotting

plot([0 0],[0 z1(4)],'k') %plot zero line on CM plot

hold off

z1=axis;axis([z1(1) z1(2) 0 Cl_PlotMax])

% plot the forward cg line of the trim triangle

delta_cg1=mom_ref_pt-forward_cg;

Cm_forward=+Cl_PlotMax*delta_cg1;

hold on; plot([0 Cm_forward],[0 Cl_PlotMax],'k') %plot zero line on CM

plot

% plot the aft cg line of the trim triangle

delta_cg2=mom_ref_pt-aft_cg;

Cm_aft=+Cl_PlotMax*delta_cg2;

hold on; plot([0 Cm_aft],[0 Cl_PlotMax],'k') %plot zero line on CM plot

str1=['CM about ',num2str(mom_ref_pt),' c'];

xlabel(str1)

ylabel('CL')

title('Fig 4.9(right), Fig 4.10(right),Fig 4.11b(right) with De=-30')

grid on

%text(.18,.75,['iH= ',num2str(iH),' deg.'])

strXF=['fwd cg xbar=',num2str(forward_cg)];

strXR=['aft cg xbar=',num2str(aft_cg)];

%text(.16,1.18,strXF)

%text(-.02,1.18,strXR)

stralpha=['alpha stall=',num2str(alpha_stall), ' deg.'];

% text(.09,.95,stralpha)

ylabel('CL')

% xlabel('alpha (deg)')

title('Supersonic Trim Diagram')

grid on

%text2(.1,.95,['iH= ',num2str(iH),' deg.'])

hold off

% Plot alpha_stall line in figure 2

CL=CL0+CLalpha*alpha_stall+CLiH*iH+CLdeltaE*deltaE;

CM=CM0+CMalpha*alpha_stall+CMiH*iH+CMdeltaE*deltaE;

hold on; plot(CM,CL,'k');


126

axis([0 .7 0 1.45])

set(gca, 'XDir', 'reverse'); % reverse the plotting direction on the x

axis

% End alpha_stall line plotting

Engine Model

Engine.m

function [T_lbs TSFC_lbs Dengine Lengine WeightEngine Linlet] =

Engine(Alt, M, Pwr, Df)

%*************************************************************************

% Engine Definition

%*************************************************************************

Bp = 0.5; % Bypass Ratio (0 for TurboJet)

Prf = 1.6; % Fan Pressure Ratio (1 for TurboJet)

Prc = 20; % Compressor Pressure Ratio

% Component Efficiencies

nd = 0.97; % Diffuser Eff

nf = 0.85; % Fan Eff

nc = 0.85; % Compressor Eff

nb = 1.00; % Burner Eff

nt = 0.90; % Turbine Eff

nn = 0.98; % core Nozzle Eff

nnf = 0.97; % Fan Nozzle Eff

DP = 0.99; % Ratio for Burner Pressure Drop

% Material Limit (For Turbine Inlet Temperature)

Tmax = 1500; % Max Temp - (K)

%*************************************************************************

*

% CALCULATIONS

%*************************************************************************

*

% Air properties

gam = 1.4;

cp = 1.005*10^3; % J/kgK

R = 287.06; % J/kgK

% Compute Fan Face Area

Df = Df*0.3048; % Convert Feet into Meters

Af = pi*(Df/2)^2;

% Compute Ambient Conditions

[Ta,Pa,rho_a,Hp_a] = atmosphere4(Alt,1);

% Convert Ambient Condition units (From English to SI)

Ta = Ta/1.8; %From R to K

Pa = Pa*(1/144)*6894.75729; %From psf to Pa

rho_a = rho_a*515.378818; %From slug/ft^3 to kg/m^3


127

T0a = Ta*(1 + (gam-1)/2*M^2); % Stagnation Temp

P0a = Pa*(1 + (gam-1)/2*M^2)^(gam/(gam-1)); % Stagnation

Pressure

% Engine Air flows

[m_dot_e Dengine Lengine WeightEngine] = EngineFlow(M, Alt, Df); %

Engine Required flow

m_dot_a = m_dot_e/(Bp + 1); % Core flow

m_dot_b = m_dot_e - m_dot_a; % Bypass Flow

% Inlet Conditions - (Station 1)

% Flow properties inside Mach Cone

if M >= 1

sig = asin(1/M);

[M1 T1_Ta P1_Pa P01_P0a delt] = ObliqueShock(M, sig, gam);

T1 = Ta*T1_Ta;

P1 = Pa*P1_Pa;

rho_1 = P1/(R*T1);

else

M1 = M;

T1 = Ta;

P1 = Pa;

rho_1 = rho_a;

end

u = sqrt(gam*R*T1)*M1; % Flow Velocity

T01 = T1*(1 + (gam-1)/2*M1^2); % Stagnation Temp

P01 = P1*(1 + (gam-1)/2*M1^2)^(gam/(gam-1)); % Stagnation Pressure

% Inlet (Diffuser) - (Station 2)

[T02 P02 M2 m_dot_cap Acap AthroatSup AthroatSub Linlet] =

inlet1(M1, gam, T1, P1, T01, P01, nd, Af, m_dot_e, R, u);

% Bypass

% Fan Exit - (Station 8)

P08 = P02*Prf;

T08 = T02*(1 + (1/nf)*(Prf^((gam-1)/gam) - 1));

% Fan Nozzle Exit - (Station 9)

uef = sqrt(2*nnf*(gam/(gam-1))*R*T08*(1 - (P1/P08)^((gam-

1)/gam)));

e_f = (((gam+1)/2)^(1/(gam-

1))*(P1/P08)^(1/gam)*sqrt(((gam+1)/(gam-1))*(1-(P1/P08)^((gam-1)/gam))))^-

1;

% Engine Core

% Compressor Exit - (Station 3)

P03 = P02*Prc;

T03 = T02*(1 + (1/nc)*(Prc^((gam-1)/gam) - 1));

% Burner (Combustor) Exit - (Station 4)

[T04 P04 f m_dot_f m_dot_t Qr gam R Cpaf] = ...

burner2(T03, P03, Pwr, nb, DP, m_dot_a, Tmax, gam, R, cp);

% Turbine Exit - (Station 5)

T05 = T04 - (m_dot_a*cp*(T03 - T02) - Bp*m_dot_a*cp*(T08 -

T02))/(m_dot_t*Cpaf);

P05 = P04*(1 - (1/nt)*(1 - T05/T04))^(gam/(gam-1));


128

% After Burner Exit (Currently Assumes no Afterburner) - (Station

6)

T06 = T05;

P06 = P05;

% Core Nozzle Exit - (Station 7)

ue = sqrt(2*nn*(gam/(gam-1))*R*T06*(1 - (P1/P06)^((gam-

1)/gam)));

e_c = (((gam+1)/2)^(1/(gam-

1))*(P1/P06)^(1/gam)*sqrt(((gam+1)/(gam-1))*(1-(P1/P06)^((gam-1)/gam))))^-

1;

% Thrust

% Thrust

T = ((m_dot_a + m_dot_f)*ue + m_dot_b*uef - m_dot_cap*u);

T_lbs = T*0.224808943; % Convert from Newtons to lbs

% Specific Thrust (T/m_dot_a)

Tsp = T/m_dot_a;

m_dot_a_lbs = m_dot_a * 7936.64144; %lbm/hr

Tsp_lbs = T_lbs/m_dot_a_lbs;

% Fuel Consumption

% TSFC

m_dot_f_lbs = m_dot_f*7936.64144; % Convert from kg/s into

lbm/hr

TSFC = m_dot_f/T;

TSFC_lbs = m_dot_f_lbs/T_lbs;

% Convert Engine Dimensions to English Units

Dengine = Dengine * 3.2808399; % ft

Lengine = Lengine * 3.2808399; % ft

WeightEngine = WeightEngine * 2.20462262; % lbs

Linlet = Linlet * 3.2808399; % ft

end

%*************************************************************************

% INLET 1 - 2D Two ramps (8, 8) - Variable geom for subsonic perf

% Designed for Fan Face Mach = 0.4

%*************************************************************************

function [T02 P02 M2 m_dot_cap Acap Athroatdes AthroatdesSub Linlet] =

inlet1(M1, gam, T1, P1, T01, P01, nd, Af, m_dot_e, R, u1)

% Design Variables

Mdes = 1.8; %Super sonic Designed Inlet Mach #

M2des = 0.4; %Designed Fan Face Mach # for given Designed Inlet

Mach #

Altdes = 45000; % Design Altitude

delT1 = [8 8]; % Ramp Angles (Degrees)

% Secondary Flow fractions

sec_m_ratio = 0.20; %Raymer Table 10.2 - Fighter

bleed_A_ratio = 0.04; % Raymer Fig 10.19 - Slot Bleed

% Compute Design Flow Properties

[T1des,P1des,rho_1des,Hp_1des] = atmosphere4(Altdes,1);


T1des = T1des/1.8; %From R to K


129

P1des = P1des*(1/144)*6894.75729; %From psf to Pa

rho_1des = rho_1des*515.378818; %From slug/ft^3 to kg/m^3

u1des = Mdes*sqrt(gam*R*T1des);

[m_dot_edes Ddes Ldes Wdes] = EngineFlow(Mdes, Altdes,

sqrt(Af*4/pi));

% Compute Takeoff Flow Properties

MTO = 0.3;

AltTO = 0;

[T1TO,P1TO,rho_1TO,Hp_1TO] = atmosphere4(AltTO,1);


T1TO = T1TO/1.8; %From R to K

P1TO = P1TO*(1/144)*6894.75729; %From psf to Pa

rho_1TO = rho_1TO*515.378818; %From slug/ft^3 to kg/m^3

u1TO = MTO*sqrt(gam*R*T1TO);

[m_dot_eTO DTO LTO WTO] = EngineFlow(MTO, AltTO, sqrt(Af*4/pi));

% Design Inlet Capture Area

Acapdes = (m_dot_edes*(1 + sec_m_ratio)/(rho_1des*u1des))*(1 +

bleed_A_ratio);

AcapTO = (m_dot_eTO*(1 + sec_m_ratio)/(rho_1TO*u1TO))*(1 +

bleed_A_ratio);

if Acapdes > AcapTO

Acap = Acapdes;

else

Acap = AcapTO;

end

% Design Supersonic Inlet Geometry

[Athroatdes Linlet] = rampinletdes(Mdes, M2des, gam, T1des, P1des,

nd, Af, m_dot_edes, R, delT1, sec_m_ratio, bleed_A_ratio, rho_1des, u1des,

Acap);

% Design Subsonic Throat Area

Df = sqrt(4*Af/pi);

AthroatdesSub = Acap; % Assume variable inlet geometry which

allows full capture area open for subsonic flow

% Total Mass flows

m_dot_s = sec_m_ratio*m_dot_e;

m_dot_BL = m_dot_e*(bleed_A_ratio);

m_dot_cap = m_dot_e + m_dot_s + m_dot_BL;

if M1 >= 1

% Oblique Shocks from Ramp

for j = 1:length(delT1)

if j == 1

[M2r(j) T2r_T1(j) P2r_P1(j) P02r_P01(j) check(j) sig(j)] =

ObliqueShockd(M1, delT1(j)*pi/180, gam);

if check(j) == 1

T2r(j) = T1*T2r_T1(j);

P2r(j) = P1*P2r_P1(j);

P02r(j) = P01*P02r_P01(j);

T02r(j) = T01;

else

break


130

end

else


ObliqueShockd(M2r(j-1), delT1(j)*pi/180, gam);

if check(j) == 1

T2r(j) = T2r(j-1)*T2r_T1(j);

P2r(j) = P2r(j-1)*P2r_P1(j);

P02r(j) = P02r(j-1)*P02r_P01(j);

T02r(j) = T02r(j-1);

else

break

end

end

end

% If Local M # below min for given delta, assume Normal Shock to

% subsonic Flow

if j > 1 && check(j) == 2

M2o = M2r(j-1);

T2r(j) = T2r(j-1);

P2r(j) = P2r(j-1);

P02r(j) = P02r(j-1);

elseif j == 1 && check(j) == 2

M2o = M1;

T2r = T1;

P2r = P1;

P02r = P01;

T02r = T01;

else

M2o = M2r(j);

end

% Normal Shock Into Engine

[M2n T2n_T2r P2n_P2r P02n_P02r] = NormalShock(M2o, gam);

T2n = T2r(j)*T2n_T2r;

P2n = P2r(j)*P2n_P2r;

P02n = P02r(j)*P02n_P02r;

rho_2n = P2n/(R*T2n);

u2n = M2n*sqrt(gam*R*T2n);

% Diffuser To Fan face

M2 = Area_Mach(Athroatdes, Af, M2n, gam);

T02 = T2n*(1 + (gam-1)/2*M2n^2);

P02 = P2n*(1 + nd*(T02/T2n-1))^(gam/(gam-1));

T2 = T02/(1 + (gam-1)/2*M2^2);

P2 = P02/(1 + (gam-1)/2*M2^2)^(gam/(gam-1));

else

T02 = T1*(1 + (gam-1)/2*M1^2);

rho_1 = P1/(R*T1);

m_dot_1 = rho_1*u1*AthroatdesSub;

if m_dot_1 < m_dot_e

p = (m_dot_e-m_dot_1)/m_dot_e;

P02 = P01*(0.97 - 0.11*p);

else

P02 = P1*(1 + nd*(T02/T1-1))^(gam/(gam-1));


131

end

M2 = Area_Mach(AthroatdesSub, Af, M1, gam);

T2 = T02/(1 + (gam-1)/2*M2^2);

P2 = P02/(1 + (gam-1)/2*M2^2)^(gam/(gam-1));

end

end

%**********************************************************************

% function rampinletdes - Computes throat area for "Design" ramp inlet

%**********************************************************************

function [Athroat Linlet] = rampinletdes(M1, M2, gam, T1, P1, nd, Af,

m_dot_e, R, delT1, sec_m_ratio, bleed_A_ratio, rho_1, u1, Acap)

Df = sqrt(4*Af/pi);

T01 = T1*(1 + (gam-1)/2*M1^2);

P01 = P1*(1 + (gam-1)/2*M1^2)^(gam/(gam-1));

% Oblique Shocks from Ramp

for j = 1:length(delT1)

if j == 1


ObliqueShockd(M1, delT1(j)*pi/180, gam);

T2r(j) = T1*T2r_T1(j);

P2r(j) = P1*P2r_P1(j);

P02r(j) = P01*P02r_P01(j);

T02r(j) = T01;

else


ObliqueShockd(M2r(j-1), delT1(j)*pi/180, gam);

T2r(j) = T2r(j-1)*T2r_T1(j);

P2r(j) = P2r(j-1)*P2r_P1(j);

P02r(j) = P02r(j-1)*P02r_P01(j);

T02r(j) = T02r(j-1);

end

end

% Normal Shock Into Engine

[M2n T2n_T2r P2n_P2r P02n_P02r] = NormalShock(M2r(j), gam);

T2n = T2r(j)*T2n_T2r;

P2n = P2r(j)*P2n_P2r;

P02n = P02r(j)*P02n_P02r;

% Diffuser To Fan face

Athroat = Af*(M2/M2n)*((1 + (gam-1)/2*M2n^2)/(1 + (gam-

1)/2*M2^2))^((gam+1)/(2*(gam-1)));

T02 = T2n*(1 + (gam-1)/2*M2n^2);

P02 = P2n*(1 + nd*(T02/T2n-1))^(gam/(gam-1));

T2 = T02/(1 + (gam-1)/2*M2^2);

P2 = P02/(1 + (gam-1)/2*M2^2)^(gam/(gam-1));

% Compute Inlet Geometry

th1 = delT1(1) * pi/180;

th2 = (delT1(1) + delT1(2)) * pi/180;

sig1 = sig(1);

sig2 = sig(2) + delT1(1)*pi/180;

sigma = sig*180/pi;

Ht = Athroat/Acap;

L1 = 1;


132

B = [1 0 0; 0 1 -tan(th2); 1 1 0];

X = [L1*tan(th1); 0; (1-Ht)];

H = inv(B)*X;

L2 = H(3); H1 = H(1); H2 = H(2);

hcap = Acap/Df;

l1 = L1*hcap;

l2 = L2*hcap;

h1 = H1*hcap;

h2 = H2*hcap;

ht = Ht*hcap;

Lramps = l1 + l2;

Lduct = Lramps/2; %Assumption

Linlet = Lramps + Lduct;

end

%*************************************************************************

% function burner2 - Computes Property Changes through Combustor

%*************************************************************************

function [T04 P04 f m_dot_f m_dot_t Qr gam R Cpaf] = ...

burner2(T03, P03, Pwr, nb, DP, m_dot_a, Tmax, gam, R, cp)

T04 = ((Tmax - T03)/100)*Pwr + T03;

P04 = P03*DP;

Qr = 45000000;

% Mass Flow Properties with Added Fuel mass

f = ((T04/T03) - 1) / (Qr/(cp*T03) - T04/T03);

m_dot_f = f*m_dot_a;

m_dot_t = m_dot_a + m_dot_f;

% Redefine gam equal to gam of exhaust gasses and R - Assumptions

based

% on CEA code runs for combustion of Jet-A with Air

gam = 1.31;

Cpaf = 1240;

R = Cpaf*(1 - 1/gam);

end

%*************************************************************************

*

% FUNCTION Engine Flow - Computes Engine Required Airflow

%*************************************************************************

*

function [m_dot_a D L W] = EngineFlow(M, Alt, D)

% Table for Engine Required Airflow adapted from Raymer pg. 796

Mach = [0; 0.25; 0.5; 0.75; 1.0; 1.25; 1.5; 1.75; 2.0; 2.25; 2.5;

2.75];

Altitude = [0 10000 20000 30000 36000 40000 50000 60000];

% Alt SL 10k 20k 30k 36k 40k 50k 60k % Mach #

flow = [246 175 120 85 55 45 30 25 ; %

0.0

260 180 125 90 60 50 35 25 ; %

0.25

285 200 140 95 65 55 35 25 ; %

0.5

305 240 160 110 80 60 40 30 ; %

0.75


133

340 270 200 140 110 90 50 40 ; %

1.0

385 300 230 180 130 120 80 50 ; %

1.25

450 360 280 205 160 140 85 55 ; %

1.5

510 395 330 225 180 160 100 75 ; %

1.75

580 470 370 280 235 215 120 90 ; %

2.0

0 0 410 310 260 220 140 100 ; %

2.25

0 0 0 350 300 260 160 120 ; %

2.5

0 0 0 0 340 280 180 140 ]; %

2.75

% Sample Engine Fan Face Diameter

Ds = 1.02; % meters

Ls = 4.07; % meters

Ws = 1361; % lb

% Scale Engine Based on Equations from Raymer

SF = (D/Ds)^2; % Eq 10.2

L = Ls*SF^0.4; % Eq 10.1

W = Ws*SF^1.1; % Eq 10.3

% Interpolate on Mach #

flow_Alt = [interpo(M, Mach, flow(:,1)) ...

interpo(M, Mach, flow(:,2)) ...






interpo(M, Mach, flow(:,8))];

% Interpolate on Altutude

m_dot_a = interpo(Alt, Altitude, flow_Alt);

% Convert lbm/s into kg/s

m_dot_a = m_dot_a*0.45359237;

% Scale mass flow for Actual Engine Diameter

m_dot_a = m_dot_a*(D^2/Ds^2);

end

%*************************************************************************

*

% FUNCTION NormalShock - Computes Property Changes accross Normal Shock

%*************************************************************************

*

function [M2 T_ratio P_ratio P_Stag_ratio] = NormalShock(M1, gam)

% Propertry Changes for Normal Shock

T_ratio = (1 + (gam-1)/2*M1^2)*(2*gam/(gam-1)*M1^2 - 1)/(

(gam+1)^2/(2*(gam-1)) * M1^2 );


134

P_ratio = 2*gam/(gam+1)*M1^2 - (gam-1)/(gam+1);

M2 = sqrt((M1^2 + 2/(gam-1))/(2*gam/(gam-1)*M1^2 - 1));

% Stagnation Pressure Ratio

P_Stag_ratio = ((1 + (gam-1)/2*M2^2)/(1 + (gam-1)/2*M1^2))^(gam/(gam-

1))* P_ratio;

end

%*************************************************************************

*

% FUNCTION ObliqueShock - Computes Property Changes accross Oblique Shock

% for given shock angle sig

%*************************************************************************

*

function [M2 T_ratio P_ratio P_Stag_ratio delta] = ObliqueShock(M1, sig,

gam)

% Determine Deflection Angle delta

delta = sig - atan(2*(1 + ((gam -

1)/2)*(M1^2)*(sin(sig)^2))/((gam+1)*(M1^2)*sin(sig)*cos(sig)));

% Normal and Parallel Upstream Mach

M1n = M1*sin(sig);

M1t = M1*cos(sig);


T_ratio = (1 + (gam-1)/2*M1n^2)*(2*gam/(gam-1)*M1n^2 - 1)/(

(gam+1)^2/(2*(gam-1)) * M1n^2 );

P_ratio = 2*gam/(gam+1)*M1n^2 - (gam-1)/(gam+1);

M2n = sqrt((M1n^2 + 2/(gam-1))/(2*gam/(gam-1)*M1n^2 - 1));

M2t = M1t*sqrt(1/T_ratio);

% Downstream Mach

M2 = sqrt(M2n^2 + M2t^2);



1))* P_ratio;

end

%*************************************************************************

*

% FUNCTION ObliqueShockd - Computes Property Changes accross Oblique Shock

% for given turning/ramp angle delta

%*************************************************************************

*

function [M2 T_ratio P_ratio P_Stag_ratio check sig] = ObliqueShockd(M1,

delta, gam)

% Determine Shock Angle Sigma

sig = 45*pi/180; %Guess

check = 0;

counter = 0;

while check == 0

counter = counter + 1;

sig2 = atan(2*(1 + ((gam -

1)/2)*(M1^2)*(sin(sig)^2))/((gam+1)*(M1^2)*sin(sig)*cos(sig))) + delta;

if abs(sig - sig2) > 0.000001


135

sig = (sig + sig2)/2;

else

check = 1;

end

if counter > 10000

check = 2;

%disp('Warning - Shock solution not converged')

end

end

% Normal and Parallel Upstream Mach

M1n = M1*sin(sig);

M1t = M1*cos(sig);


T_ratio = (1 + (gam-1)/2*M1n^2)*(2*gam/(gam-1)*M1n^2 - 1)/(

(gam+1)^2/(2*(gam-1)) * M1n^2 );

P_ratio = 2*gam/(gam+1)*M1n^2 - (gam-1)/(gam+1);

M2n = sqrt((M1n^2 + 2/(gam-1))/(2*gam/(gam-1)*M1n^2 - 1));

M2t = M1t*sqrt(1/T_ratio);

% Downstream Mach

M2 = sqrt(M2n^2 + M2t^2);



1))* P_ratio;

end

%*************************************************************************

% function Area_Mach - Computes downstram Mach # from Area change Only

% valid for the subsonic case

%*************************************************************************

function M2 = Area_Mach(A1, A2, M1, gam)

M2_v = [1:-0.001:0];

A2_A1 = (M1./M2_v).*sqrt(((1+(gam-1)/2*M2_v.^2)/((1+(gam-

1)/2*M1^2))).^((gam+1)/(gam-1)));

M2 = interpo(A2/A1, A2_A1, M2_v);

end

%*************************************************************************

*

% FUNCTION atmosphere4 - Standard Atmosphere

%*************************************************************************

*

function [temp,press,rho,Hgeopvector]=atmosphere4(Hvector,GeometricFlag)

%function

[temp,press,rho,Hgeopvector]=atmosphere4(Hvector,GeometricFlag)

% Standard Atmospheric data based on the 1976 NASA Standard Atmoshere.

% Hvector is a vector of altitudes.

% If Hvector is Geometric altitude set GeometricFlag=1.

% If Hvector is Geopotential altitude set GeometricFlag=0.

% Temp, press, and rho are temperature, pressure and density

% output vectors the same size as Hgeomvector.

% Output vector Hgeopvector is a vector of corresponding geopotential

altitudes (ft).

% This atmospheric model is good for altitudes up to 295,000

geopotential ft.


136

% Ref: Intoduction to Flight Test Engineering by Donald T. Ward and

Thomas W. Strganac

% index Lapse rate Base Temp Base Geopo Alt Base

Pressure Base Density

% i Ki(degR/ft) Ti(degR) Hi(ft) P, lbf/ft^2

RHO, slug/ft^3

format long g

D= [1 -.00356616 518.67 0

2116.22 0.00237691267925741

2 0 389.97 36089.239

472.675801650081 0.000706115448911997

3 .00054864 389.97 65616.798

114.343050672041 0.000170813471460564

4 .00153619 411.57 104986.878

18.1283133205764 2.56600341257735e-05

5 0 487.17 154199.475

2.31620845720195 2.76975106424479e-06

6 -.00109728 487.17 170603.675

1.23219156244977 1.47347009326248e-06

7 -.00219456 454.17 200131.234

0.38030066501701 4.87168173794687e-07

8 0 325.17 259186.352

0.0215739175227548 3.86714900013768e-08];

R=1716.55; %ft^2/(sec^2degR)

gamma=1.4;

g0=32.17405; %ft/sec^2

RE=20926476; % Radius of the Earth, ft

K=D(:,2); %degR/ft

T=D(:,3); %degR

H=D(:,4); %ft

P=D(:,5); %lbf/ft^2

RHO=D(:,6); %slug/ft^3

temp=zeros(size(Hvector));

press=zeros(size(Hvector));

rho=zeros(size(Hvector));

Hgeopvector=zeros(size(Hvector));

% Convert from geometric altitude to geopotental altitude, if

necessary.

if GeometricFlag

Hgeopvector=(RE*Hvector)./(RE+Hvector);

%disp('Convert from geometric altitude to geopotential altitude in

feet')

else

Hgeopvector=Hvector;

%disp('Input data is geopotential altitude in feet')

end

ih=length(Hgeopvector);

n1=find(Hgeopvector<=H(2));

n2=find(Hgeopvector<=H(3) & Hgeopvector>H(2));





137



n8=find(Hgeopvector<=295000 & Hgeopvector>H(8));

icorrect=length(n1)+length(n2)+length(n3)+length(n4)+length(n5)+length(n6)

+length(n7)+length(n8);

if icorrect<ih

disp('One or more altitutes is above the maximum for this

atmospheric model')

icorrect

ih

end

% Index 1, Troposphere, K1= -.00356616

if length(n1)>0

i=1;

h=Hgeopvector(n1);

TonTi=1+K(i)*(h-H(i))/T(i);

temp(n1)=TonTi*T(i);

PonPi=TonTi.^(-g0/(K(i)*R));

press(n1)=P(i)*PonPi;

RonRi=TonTi.^(-g0/(K(i)*R)-1);

rho(n1)=RHO(i)*RonRi;

end

% Index 2, K2= 0

if length(n2)>0

i=2;

h=Hgeopvector(n2);

temp(n2)=T(i);

PonPi=exp(-g0*(h-H(i))/(T(i)*R));


RonRi=PonPi;


end

% Index 3, K3= .00054864

if length(n3)>0

i=3;

h=Hgeopvector(n3);







end

% Index 4, K4= .00153619

if length(n4)>0

i=4;

h=Hgeopvector(n4);







138


end

% Index 5, K5= 0

if length(n5)>0

i=5;

h=Hgeopvector(n5);

temp(n5)=T(i);



RonRi=PonPi;


end

% Index 6, K6= -.00109728

if length(n6)>0

i=6;

h=Hgeopvector(n6);







end

% Index 7, K7= -.00219456

if length(n7)>0

i=7;

h=Hgeopvector(n7);







end

% Index 8, K8= 0

if length(n8)>0

i=8;

h=Hgeopvector(n8);

temp(n8)=T(i);



RonRi=PonPi;


end

end

%*************************************************************************

*

% FUNCTION interpo - Linear Interpolation Function

%*************************************************************************

*

function valuedat2 = interpo(valuedat1, data1, data2)


139

if valuedat1 < data1(1)

valuedat2 = data2(1);

elseif valuedat1 > data1(length(data1))

valuedat2 = data2(length(data2));

else

check = 0;

count = 0;

while check == 0

count = count + 1;

if valuedat1 == data1(count)

valuedat2 = data2(count);

check = 1;

elseif (valuedat1 > data1(count)) && (valuedat1 < data1(count

+ 1))

valuedat2 = ((data2(count+1)-

data2(count))/(data1(count+1)-data1(count)))*(valuedat1 - data1(count)) +

data2(count);

check = 1;

elseif (count) == length(data1)

valuedat2 = -999;

check = 1;

end

end

end

end

EnginePower.m

%************************************************************************* % EnginePower - Determines Required Power Setting for Engine to % achieve a given Thrust (T) at a Given Altitude and Mach # (Alt, M) for % use in Sizing Code: % % Returns Engine Thrust(lbf), TSFC(lbm/hr/lbf), FuelRate(lbm/hr), % Power Level(% - note this is based on T04, not Thrust) % %*************************************************************************

*

function [Thrust TSFC FuelRate Pwr] = EnginePower(T, Alt, M, D)

Pwrguess = 100; tol = 0.1; check = 0; counter = 0; while check == 0 counter = counter + 1; [Thrust TSFC De Le We Li] = Engine(Alt, M, Pwrguess, D); if abs(T - Thrust) <= tol check = 1; else Pwrguess = Pwrguess + (T - Thrust)/T*100; end if counter > 1000


140

check = 2; disp('WARNING!! EnginePower solution did not converge') disp(T) end end Pwr = Pwrguess; if Pwr > 100 disp('WARNING!! Engine is operating at greater than 100%. Larger

Engine is Needed') disp(T) disp(Pwr) end FuelRate = TSFC*Thrust; end

EngineSize.m

%************************************************************************* % EngineSize - Determines Required Engine Size to % achieve a given Thrust (Treq) at a Given Altitude and Mach # (Alt, M) % when at Full Throttle (100% Pwr), for use in Sizing Code: % % Returns Engine Dimensions: Engine Diameter Dengine (ft), % Engine Length Lengine(ft), Engine Weight WeightEngine(lbs) % Inlet & duct length Linlet (ft) %*************************************************************************

*

function [Dengine Lengine WeightEngine Linlet] = EngineSize(Treq, Alt, M) for i = 1:length(Treq) Diamguess = 5; %feet tol = 0.1; check = 0; counter = 0; while check == 0 counter = counter + 1; [Thrusti(i) TSFCi(i) Di(i) Li(i) Wi(i) Linleti(i)] =

Engine(Alt(i), M(i), 100, Diamguess); if abs(Treq(i) - Thrusti(i)) <= tol check = 1; else Diamguess = Diamguess + (Treq(i) -

Thrusti(i))/Treq(i)*(Diamguess/2); end if counter > 1000 check = 2; disp('WARNING!! EngineSize solution did not converge') end end

FuelRatei(i) = TSFCi(i)*Thrusti(i); end Dengine = max(Di); Lengine = max(Li);


141

WeightEngine = max(Wi); Linlet = max(Linleti); end

Component Weights

call_compweight.m

function [weight] = call_compweight(specs, Wland, Wf, Vstall, cruiseLD,

WeightEngine, Tengine, Dengine) %Global G(1) = specs(1); %Flight design gross weight (lb) G(2) = Wland; %Landing design gross weight (lb) G(3) = Wf; %Fuel Weight (lb)

%Flight Characteristics F(1) = Vstall*0.592483801; % Stall Speed (kts) F(2) = specs(5); %Max mach number F(3) = cruiseLD; %Lift/Drag

%Structure S(1) = 6; %Ultimate landing load factor S(2) = 4.75; %Ultimate load factor

%Wing w(1) = specs(2); %Aspect Ratio w(2) = 80; %Wing Span (ft) w(3) = 500; %Control surface area on wing (ft^2) w(4) = w(2)^2 ./ w(1); %trapezoidal wing area (ft^2) | S =

b^2/AR w(5) = .03; %T/c w(6) = .1; %taper ratio of wing w(7) = 56*pi/180; %wing sweep at .25 chord

%Canard C(1) = 1.25; %Aspect ratio of horizontal tail C(2) = 35; %horizontal tail span (ft) C(3) = 11.375; %fuselage width at horizontal tail intersection

(ft) C(4) = 95.65; %tail length; wing quarter MAC to tail MAC (ft) C(5) = 500; %elevator area (ft^2) C(6) = 500; %horizontal tail area (ft^2) C(7) = 45*pi/180; %tail sweep at .25 chord

%Vertical Tail V(1) = 1.125; %Aspect ratio of vertical tail V(2) = 0; %Horizontal tail height (0 for canard) (ft) V(3) = 15; %vertical tail heigh above fuselage (ft) V(4) = 200; %vertical tail area (ft^2) V(5) = 67; %Rudder area (ft^2) V(6) = .4; %Taper ratio of tail V(7) = 35*pi/180; %sweep ratio of vertical tail at .25 chord


142

%Main Gear L(1) = 10; %extended length of main gear (ft) L(2) = 8; %number of main wheels L(3) = 2; %number of main shock struts

%Nose Gear L(4) = 4.25; %extended nose gear length (ft) L(5) = 2; %Number of nose wheels

%Fuselage f(1) = 190; %Fuselage Structural Length (ft) f(2) = 5607.4; %fuselage wetted area (ft^2)

%Engine E(1) = WeightEngine; %Engine weight (lb) E(2) = Tengine; %Thrust of each engine E(3) = 18; %duct length (ft) E(4) = Dengine; %Diameter of engine (ft) E(5) = 34; %Length of engine shroud (ft) E(6) = 4; %Length of tailpipe (ft)

W = comp_weight(G,F,S,w,C,V,L,f,E)'; weight = sum(W); end

comp_weight.m

function W = comp_weight(G,F,S,w,C,V,L,f,E)

%GoldJet %Weight Code %Donald Barrett

%%%%%%%%%%%%%%%%%%%%%%%% %Equation Constants %%%%%%%%%%%%%%%%%%%%%%%%

%Global %G(1) %Flight design gross weight (lb) %G(2) %Landing design gross weight (lb) %G(3) %Fuel Weight (lb)

%Flight Characteristics %F(1) %Stall Speed (kts) %F(2) %Max mach number %F(3) %Lift/Drag

%Structure %S(1) %Ultimate landing load factor %S(2) %Ultimate load factor


143

%Wing %w(1) %Aspect Ratio %w(2) %Wing Span (ft) %w(3) %Control surface area on wing (ft^2) %w(4) %trapezoidal wing area (ft^2) %w(5) %T/c %w(6) %taper ratio of wing %w(7) %wing sweap at .25 chord

%Canard %C(1) %Aspect ratio of horizontal tail %C(2) %horizontal tail span (ft) %C(3) %fuselage width at horizontal tail intersection (ft) %C(4) %tail length; wing quarter MAC to tail MAC (ft) %C(5) %elevator area (ft^2) %C(6) %horizontal tail area (ft^2) %C(7) %tail sweap at .25 chord

%Vertical Tail %V(1) %Aspect ratio of vertical tail %V(2) %Horizontal tail height (0 for canard) (ft) %V(3) %vertical tail heigh above fuselage (ft) %V(4) %vertical tail area (ft^2) %V(5) %Rudder area (ft^2) %V(6) %Taper ratio of tail %V(7) %sweep ratio of vertical tail at .25 chord

%Main Gear %L(1) %extended length of main gear (ft) %L(2) %number of main wheels %L(3) %number of main shock struts

%Nose Gear %L(4) %extended nose gear length (ft) %L(5) %Number of nose wheels

%Fuselage %f(1) %Fuselage Structural Length (ft) %f(2) %fuselage wetted area (ft^2)

%Engine %E(1) %Engine weight (lb) %E(2) %Thrust of each engine %E(3) %duct length (ft) %E(4) %Diameter of engine (ft) %E(5) %Length of engine shroud (ft) %E(6) %Length of tailpipe (ft)

%Passenger and Crew N_c = 3; %Number of crew N_pass = 40; %Number of passengers


144

W_p_c = 100; %Weight of checked baggage per passenger

(lb) W_c = N_pass*W_p_c; %Maximum cargo weight (lb)

%%%%%%%%%%%%%%%%%%%%%%%% %Structure Group %%%%%%%%%%%%%%%%%%%%%%%%

%Wing (Modeled from the Raymer, fighter (eq 15.1))

%Constant K_dw = .768; %0.768 for delta wing, 1.0 other K_vs = 1; %1.19 for variable sweep wing, 1.0 for other K_wing = 1.85; %correction factor for larger size than wing

%Weight W(1) =

K_wing*0.0103*K_dw*K_vs*((G(1)*S(2))^.5)*(w(4)^0.622)*(w(1)^.785)*(w(5)^(-

0.4))*((1+w(6))^0.05)*((cos(w(7)))^(-1))*(w(3)^0.04);

%Horizontal Tail/Candard (Raymer Cargo eq 15.26)

%Constants K_uht = 1; %1.143 for all moving tail, 1.0 other K_y = 0.3*C(4); %aircraft pitching radius of gyration (ft)

(~=0.3L_t)

%Weight W(2) = 0.0379*K_uht*((1+(C(3)/C(2)))^-

.25)*(G(1)^.639)*(S(2)^.1)*(C(6)^0.75)*(C(4)^(-

1))*(K_y^.704)*((cos(C(7)))^(-1))*(C(1)^.166)*((1+(C(5)/C(6)))^.1);

%Vertical Tail (Raymer Fighter eq 15.3)

%Constants K_rht = 1; %1.047 for rolling horizontal tail, 1.0 other

%Weight W(3) =

0.452*K_rht*((1+(V(2)/V(3)))^.5)*((G(1)*S(2))^.488)*(V(4)^.718)*(F(2)^.341

)*(C(4)^(-

1))*((1+(V(5)/V(4)))^.348)*(V(1)^.223)*((1+V(6))^.25)*((cos(V(7)))^(-

.323));

%Main Gear (Raymer Cargo 15.29)

%Constants K_mp = 1; %1.126 for kneeling gear, 1.0 other

%Weight


145

W(4) =

0.0106*K_mp*(G(2)^.888)*(S(1)^.25)*(L(1)^0.4)*(L(2)^0.321)*(L(3)^(-

.5))*(F(1)^.1);

%Nose Gear (Raymer Cargo 15.30)

%Constants K_np = 1; %1.15 for kneeling gear, 1.0 other

%Weight W(5) = 0.032*K_np*(G(2)^.646)*(S(1)^.2)*(L(4)^0.5)*(L(5)^.45);

%Fuselage (Raymer Cargo 15.28)

%Constants K_door = 1.06; %1.0 no cargo door, 1.06 one side dorr, 1.12 two side,

1.12 one aft, 1.25 two side one aft K_ws = 0.75*((1+2*w(6))/(1+w(6)))*(w(2)*(tan(w(7)/f(1)))); K_lg = 1; %1.12 if fuselage mounted gear, 1.0 other

%Weight W(6) =

0.3280*K_door*K_lg*((G(1)*S(2))^.5)*(f(1)^.25)*(f(2)^.302)*((1+K_ws)^.04)*

(F(3)^.1);

%%%%%%%%%%%%%%%%%%%%%%%% %Propulsion Group %%%%%%%%%%%%%%%%%%%%%%%%

%Engine

%Constants N_en = 4; %Number of engines

%Weight W(7) = E(1)*N_en;

%Engine Section (Raymer fighter 15.9)

%Constants %None

%Weight W(8) = 0.01*(E(1)^.717)*N_en*S(2);

%Air Induction System (Raymer fighter 15.10)

%Constants K_vg = 1; %1.62 for variable geometry, 1.0 other K_d = 2.75; %duct constant (see raymer fig 15.2)

%Weight W(9) = 13.29*K_vg*(E(3)^.643)*(K_d^.182)*(N_en^1.498)*E(4);


146

%Engine Mounts (Raymer fighter 15.7)

%Constants T = E(2)*N_en; %Total thrust (lbf) K_mount = 10; %Correction factor for different mounting factor

(fighter nacelles are integral)

%Weight W(10) = K_mount*0.013*(N_en^.795)*(T^.579)*S(2);

%Fuel System (Raymer cargo 15.34)

%Constants V_t = G(3)*6.7; %Total volume of fuel tanks (gal) V_i = V_t; %Volume of integral fuel tanks (gal) V_p = 0; %Volume of self sealing fuel tanks (gal) N_t = 4; %Number of fuel tanks

%weight W(11) = 2.405*(V_t^.606)*((1+(V_i/V_t))^(-1))*(1+(V_p/V_t))*(N_t^.5);

%Oil Cooling (raymer fighter 15.13)

%Constants %None

%Weight W(12) = 37.82*(N_en^1.023);

%Engine Cooling (Raymer fighter 15.12)

%Constant

%Weight W(13) = 4.55*E(4)*E(5)*N_en;

%Tailpipe (Raymer fighter 15.11)

%Constants

%Weight W(14) = 3.5*E(4)*E(6)*N_en;

%%%%%%%%%%%%%%%%%%%%%%%% %Equipment Group %%%%%%%%%%%%%%%%%%%%%%%%

%Furnishings (Raymer cargo 15.41, note, doesnt not include seats)

%Constants

%Weight


147

W(15) = 0.0577*((N_c+N_pass)^0.1)*(W_c^0.393)*(f(2)^0.75);

%APU Installed (Raymer cargo 15.36)

%Constants W_apu_un = 576; %Weight of APU uninstalled (lb) (Garrett GTCP 331-

400)

%Weight W(16) = 2.2*W_apu_un;

%Avionics (Raymer cargo 15.40)

%Constants W_uav = 800; %Uninstalled avioncis weight (lb) (typically 800-1400

lb)

%Weight W(17) = 1.73*(W_uav^0.983);

%Instruments (Raymer cargo 15.37)

%Constants K_r = 1; %1.133 if reciprocating engine, 1.0 other

%Weight W(18) = 4.509*K_r*(N_c^.541)*N_en*((f(1) + w(2))^.5);

%Electrical (Raymer cargo 15.39)

%Constants R_kva = 90; %System electrical rating (kV) (typically 40-60 for

transports, 110-160 for fighter/bomber) L_a = 100; %electrical routing distance (ft), from generator

to cockpit N_gen = N_en; %number of generators

%Weight W(19) = 7.291*(R_kva^.0782)*(L_a^0.346)*(N_gen^0.10);

%Hydraulics (Raymer cargo 15.38)

%Consants N_f = 3; %Number of functions performed by controls

%Weight W(20) = 0.2673*N_f*((f(1) + w(2))^0.937);

%Air Conditioning (Raymer Cargo 15.42)

%Constants N_p = N_c+N_pass; %Number of passengers and crew V_pr = 500; %Volume of pressurized section (ft^3)


148

%Weight W(21) = 62.36*(N_p^0.25)*((V_pr/1000)^0.604)*(W_uav^0.10);

%Anti-ice (Raymer cargo 15.43)

%Constants %None

%Weight W(22) = 0.002*G(1);

%Seats (Barrett)

%Constants W_seat = 40; %Weight of seat (lb)

%Weight W(23) = W_seat*N_p;

%Lavs (Barrett)

%Constants W_lav = 5*(N_pass^1.33); %Weight of lav (lb) N_lav = 2; %Number of lavs

%Weight W(24) = W_lav*N_lav;

End

Landing Gear % Landing Gear Code

% Michael Coffey

% 4-6-09

clc

%% Tire sizing code

per = 0.15; % Percentage of weight on nose gear

weight_tot= 270000; % Total Weight of aircraft

weight_nose= per*weight_tot; % Weight on nose gear assuming 8% minimum

for steering

weight_main= (1-per)*weight_tot; % Weight on the main gear

N_nose= 2; % number of nose gear tires

N_main=8; % number of main gear tires

g_f=1.25; % growth factor for weight of aircraft

A_diam=1.63; % Coefficient based on Raymer's Landing gear sizing

B_diam=.315; % Coefficient based on Raymer's Landing gear sizing

A_width=.1043; % Coefficient based on Raymer's Landing gear sizing

B_width=.480; % Coefficient based on Raymer's Landing gear sizing


149

W_w_nose=weight_nose/N_nose*g_f; % weight per tire of the nose gear

adjusted for growth

W_w_main=weight_main/N_main*g_f; % weight per tire of the main gear

adjusted for growth

Diam_nose=A_diam*W_w_nose^B_diam; % Diameter of the nose gear tires

based on Raymer's landing gear sizing

Width_nose= A_width*W_w_nose^B_width; % Width of the nose gear tires


Diam_main=A_diam*W_w_main^B_diam; % Diameter of the main gear tires


Width_main= A_width*W_w_main^B_width; % Width of the main gear tires


%.043 lb/ft^2%

%% Tire Selection code

data = xlsread('w:\Personal\AAE 451\tires');

sortd_main = sortrows(data,4);

boold_main = sortd_main(:,4)>=Diam_main;

indexd=find(boold_main,1,'first');

subd_main = sortd_main(indexd:end,:);

sortwid_main=sortrows(subd_main,5);

boolw_main=sortwid_main(:,5)>=Width_main;

indexd_main=find(boolw_main,1,'first');

subwid_main=sortwid_main(indexd_main:end,:);

sortwgh_main=sortrows(subwid_main,3);

selection_main=sortwgh_main(1,:);

diam_tire_main=selection_main(4)

width_tire_main=selection_main(5)

r_radius_main=selection_main(6);

sortd_nose = sortrows(data,4);

boold_nose = sortd_nose(:,4)>=Diam_nose;

indexd_nose=find(boold_nose,1,'first');

subd_nose = sortd_nose(indexd_nose:end,:);

sortwid_nose=sortrows(subd_nose,5);

boolw_nose=sortwid_nose(:,5)>=Width_nose;

indexd_nose=find(boolw_nose,1,'first');

subwid_nose=sortwid_nose(indexd_nose:end,:);

sortwgh_nose=sortrows(subwid_nose,3);

selection_nose=sortwgh_nose(1,:);

diam_tire_nose=selection_nose(4)

width_tire_nose=selection_nose(5)

r_radius_nose=selection_nose(6);

% Weight of tires

W_tires_main=N_main*selection_main(3)

W_tires_nose=N_nose*selection_nose(3)

W_tires_tot=W_tires_main+W_tires_nose

% Tire pressure


150

A_p_main=2.3*(sqrt(diam_tire_main*width_tire_main)*(diam_tire_main/2-

r_radius_main));

tire_pressure_main=W_w_main/A_p_main

A_p_nose=2.3*(sqrt(diam_tire_nose*width_tire_nose)*(diam_tire_nose/2-

r_radius_nose));

tire_pressure_nose=W_w_nose/A_p_nose

%% Landing gear placement code

h_gear=0:.25:10;

x_cg= 132; % location of cg in x direction measured from nose

h_cg= 4+h_gear ; % height of the center of gravity off the ground

x_tail= 190 ; % location of the back end of the plane measured from the

nose

h_tail= 5.5+h_gear ; % tail clearance

a_stall= pi/12; % wing stall angle

x_main= x_tail-(h_tail/tan(a_stall)) % location of the main gear along

the length of the aircraft

B= (x_main-x_cg)/per; % distance between the nose gear and main gear

x_nose=x_main-B %location of the nose gear along the length of the

aircraft

% Dynamic Braking load

g=32.2; % gravitational constand in ft/s/s

dbl_nose= 10*h_cg*weight_tot/g/B; % maximum loading on the nose gear

%% Strut and damper code

V_vert = 10; % vertical velocity in ft/s

weight_land = weight_tot; % maximum weight at landing (assumes emergency)

KE_vert = .5*(weight_land/g)*V_vert^2; % vertical kinetic energy needing

to be dissipated at landing

eta_oleo = .8; % oleo efficiency from Raymer

l_oleo = weight_land/2; % average load on the oleo during landing

pressure_oleo = 1800; % internal oleo pressure

stroke_tire_main = diam_tire_main/2-r_radius_main; % amount the tire will

deform upon loading

stroke_tire_nose = diam_tire_nose/2-r_radius_nose; % amount the tire will

deform upon loading

eta_tire = .47; % tire efficiency from Raymer

n_gear = 3; % gear load factor from Raymer

stroke_oleo_main = V_vert^2/(2*g*eta_oleo*n_gear)-

eta_tire/eta_oleo*stroke_tire_main;

stroke_oleo_nose = V_vert^2/(2*g*eta_oleo*n_gear)-

eta_tire/eta_oleo*stroke_tire_nose;

l_oleo_main = 2.5*stroke_oleo_main; % the length of the oleo shock

absorber

diameter_oleo_main = sqrt(4*l_oleo/(pressure_oleo*pi)); % inside diameter

of oleo


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out_diameter_oleo_main = 1.3*diameter_oleo_main; % external diameter of

the oleo

l_oleo_nose = l_oleo_main; % the stroke and length of the oleo on the

nose and the main gear is equal for taxi comfort

diameter_oleo_nose = sqrt(4*(dbl_nose+weight_nose)/(pressure_oleo*pi)); %

internal diameter of the nose oleo

out_diameter_oleo_nose = 1.3*diameter_oleo_nose; % external diameter of

the nose strut

Documents

AAE 451, SENIOR DESIGN - Purdue University...AAE 451, SENIOR DESIGN CONCEPTUAL DESIGN REVIEW TEAM 3: GOLDJET DIANE BARNEY DONALD BARRETT MICHAEL COFFEY JON COUGHLIN MARK GLOVER KEVIN