AA190

Conceptual Design and Optimization of Clean

Hybrid Ultra Large Aircraft (CHULA)

Stanford University

Kyle Tsai

6/8/2011

This paper seeks to describe a rational approach to designing and optimizing the performance of a heavy-lift hybrid airship that is capable of transporting a 1,000,000 pound payload.

2

TABLE OF CONTENTS

INTRODUCTION ............................................................................................................................................................. 3

DESIGN OBJECTIVES ....................................................................................................................................................... 4

OPTIMIZATION STRATEGY ............................................................................................................................................. 5

NACA FOR-LOOP ........................................................................................................................................................ 6

K_LIFT FOR-LOOP ....................................................................................................................................................... 7

AR FOR-LOOP ............................................................................................................................................................. 7

DESIGN METHODOLOGY ................................................................................................................................................ 8

AIRSHIP SIZING ........................................................................................................................................................... 8

AIRSHIP WEIGHT ...................................................................................................................................................... 10

PAYLOAD .................................................................................................................................................................. 10

FLIGHT CREW ........................................................................................................................................................... 11

EMPTY WEIGHT ....................................................................................................................................................... 11

VOLUME OF ENVELOPE ........................................................................................................................................... 12

SHAPE ...................................................................................................................................................................... 12

LENGTH .................................................................................................................................................................... 13

REFERENCE AREA ..................................................................................................................................................... 15

DRAG BOOKKEEPING ................................................................................................................................................... 16

PARASITE DRAG ....................................................................................................................................................... 16

INDUCED DRAG ........................................................................................................................................................ 18

AERODYNAMIC LIFT ................................................................................................................................................. 18

TAKE-OFF FIELD LENGTH ............................................................................................................................................. 19

LANDING FIELD LENGTH .............................................................................................................................................. 20

RANGE ......................................................................................................................................................................... 21

STOL RANGE ............................................................................................................................................................. 21

VTOL RANGE ............................................................................................................................................................ 21

SUPERHEAT EFFECTS.................................................................................................................................................... 22

DISCUSSION OF RESULTS ............................................................................................................................................. 23

OPTIMIZATION OF GRID SIZE ................................................................................................................................... 23

DISCUSSION OF ERRORS .......................................................................................................................................... 24

12000 NM RANGE, 4500 FT. TOFL AND LFL CONSTRAINT ....................................................................................... 24

12000 NM RANGE, 1000 FT. TOFL AND LFL CONSTRAINT ....................................................................................... 26

CONCLUSIONS ............................................................................................................................................................. 27

FUTURE WORK ............................................................................................................................................................. 28

APPENDIX A: DERIVATION OF VOLUME OF ENVELOPE ............................................................................................... 29

APPENDIX B: AIRSHIP EMPTY WEIGHTS....................................................................................................................... 30

APPENDIX C: HOERNER’S FORMULA FOR DRAG OF REVOLVED BODY ........................................................................ 31

APPENDIX D: MATLAB MAIN FILE ................................................................................................................................ 32

APPENDIX E: VOLUME CALCULATION .......................................................................................................................... 35

APPENDIX F: DRAG COEFFICIENTS CALCULATION ....................................................................................................... 37

APPENDIX G: CALCULATES TOFL .................................................................................................................................. 38

APPENDIX H: CALCULATES LFL ..................................................................................................................................... 39

WORKS CITED .............................................................................................................................................................. 40

3

INTRODUCTION

The Clean Hybrid Ultra Large Aircraft (CHULA) is a heavy-lift freight airship. Hybrid airships

derive part of their overall lift from aerodynamic lifting forces, which necessitates taking off and

landing just like normal airplanes. In effect, hybrid airships, when operating at their max take-

off weight (MTOW), become short take-off and landing (STOL) aircraft. The CHULA is ‘clean’

because its cargo hold and crew compartments are completely enclosed within the airship

envelope, leaving the body aerodynamically pure.1 A clean exterior allows for the potential of

integrating boundary layer control (BLC) devices to increase the performance of the airship.

Cargo-carrying airships such as the CHULA could prove to be an integral part of the global

supply chain, which is currently dominated by three modes of conveyance: overland freight,

overseas shipping, and air transport. Airplanes provide the fastest means of shipping, but are the

costliest; thus, highly valuable and perishable goods like fresh seafood and flowers are typically

serviced by air cargo. Low value commodities with long shelf life are transported via a

combination of ships, rail, and trucks. Mid-range value and longer-lasting goods such as tropical

fruits and fresh meats currently do not have a mode of freight that is cheaper than planes and

faster than overland and marine shipping. However, recent studies have shown that airships can

fill this niche. Prentice, Beilock, and Phillips write that for airships with a cargo capacity of 220

tons2, operating costs would be comparable to that of trucking. At 1100 tons, freight rates

become competitive with marine and rail transport. When one also considers that airships do not

need runways, roads, or canals to operate, there is a compelling argument for the continuing

development of airship technologies.

Within the past two decades, there has been a revival in interest to develop such heavy-lift

freight airships. The recent Walrus program was a DARPA initiative to design a Hybrid Ultra

Large Aircraft (HULA) capable of carrying 1 to 2 million pounds of cargo up to 12,000 nautical

miles. The CargoLifter CL160 was a German program to build a 19 million cubic foot heavy

lifting cargo platform capable of carrying 160 metric tons or 350,000 pounds3. Two other

companies, the California-based Aeros and World SkyCat of the United Kingdom both offer

hybrid airship solutions, and are in varying stages of funding and development.

This paper presents a rational approach to design and optimize the performance of the CHULA.

1 The Hindenburg-class airship also attempted to integrate the crew, passenger, and payload sections within the

superstructure of the envelope. 2Throughout this paper, the unit ‘ton’ will be used to mean short ton, equal to 2000 pounds.

3 Its parent-company, CargoLifter AG, announced insolvency in 2002. The CL160’s hangar is now the site of

Tropical Islands, an artificial tropical resort in Brandenburg, Germany.

4

DESIGN OBJECTIVES

The main design point of the CHULA will be its full payload capacity of 1 million pounds. The

optimization of its design will involve three constraints: range, take-off field length, and landing

field length.

At full load, the CHULA will be operating as a short take-off and landing (STOL) aircraft When

it is carrying a sufficiently low fraction of its payload capacity, it will operate as a vertical take-

off and landing (VTOL) aircraft These two operating modes represent different performance

regimes, which affect the design constraints. Because performance conditions are most critical

when the CHULA is fully loaded, the constraints are evaluated at STOL mode.

The CHULA will be expected to meet the design range and payload capacity of the Walrus

HULA; thus, it will have a 1 million pound cargo capacity and must be able to fly 12,000

nautical miles at gross capacity. The CHULA will have a design take-off field length (TOFL)

and landing field length (LFL) of 4500 feet. The landing field length calculation will be

performed assuming two different ground surfaces: concrete and wet grass. Out of curiosity, an

attempt will also be made to assess if it can take-off and land within 1000 feet, allowing it to

operate out of any commercial airport in the world4. In practice, an airship of this size would

never be able to take-off or land safely in such a short space, but it is interesting to see how the

configuration would be affected.

In summary, the constraints used in this optimization are5:

Range: 12,000 n.mi.

TOFL: 4500 ft. and 1000 ft.

LFL: 4500 ft. and 1000 ft.

4 Juancho E. Yrausquin Airport (SAB), located on the island of Saba of the Netherlands Antilles, is home to the

world’s shortest commercial airport runway at 1300 ft. with only about 1000 ft. of useable runway. 5 The FAA Advisory Circular 150/5300-13 and the ICAO Annex 14 represent two common sources of geometric

specifications for airfield design. They establish that “the 80 meter box” is the maximum span and length that

commercial airports can accommodate. Originally, a span constraint of 80 meters was going to be implemented, but

it proved to be outside the design space of the airship.

5

OPTIMIZATION STRATEGY

The optimization of the CHULA’s performance will be conducted using a grid search. The

optimizer will compute every possible design point by running nested for-loops, each of which

vary one parameter within a certain range. These parameters are the airship’s cross-sectional

body profile NACA, the ratio of aerodynamic lift to total lift , and its aspect ratio AR. A

search function will be performed using the above constraints to winnow out the combinations

that do not meet the above constraints.

An example of the pseudo-code used to implement the optimization is shown below:

% MATLAB PSEUDOCODE

% DESIGN OBJECTIVES CARGO_WT = 1000000 % [lbs] FUEL_WT = 125000 % [lbs]

% VARYING PARAMETERS NACA K_LIFT AR

% OPTIMIZATION LOOPS for i = 1:length(NACA) for j = 1:length(K_LIFT) for k = 1:length(AR) evaluate TOFL(i,j,k)

evaluate LFL_CONCRETE(i,j,k) evaluate LFL_WETGRASS(i,j,k)

evaluate STOL_RANGE(i,j,k) end end end

% ESTABLISH FEASIBLE DESIGN SPACE Find TOFL <= 4500 & ... % [ft.] LFL_CONCRETE <= 4500 & ... % [ft.]

LFL_WETGRASS <= 4500 & ... % [ft.]

STOL_RANGE >= 12000 & ... % [n.mi]

maximize VTOL_RANGE

6

NACA FOR-LOOP

To expedite calculations, the airship is modeled as a buoyant wing section with the cross-

sectional geometry of a NACA 4-series airfoil. Because the cargo hold of the airship will most

likely be located along the lower surface of the airship for structural and roll stability reasons, an

airfoil with a relatively flat lower surface yields the least wasted space. Following visual

inspection, I concluded that 12 to 18 percent thick airfoils with 8 to 9 percent camber, the highest

point of which is located either 20 or 30 percent of the chord, provide suitable body profiles.

These airfoils and some of their 2-D performance information are listed below.6

NACA

Unit Chord Illustration

Ma

x C

l

Ma

x C

l A

ng

le

Ma

x L

/D

Ma

x L

/D

A

ng

le

Ma

x L

/D

C

l

Sta

ll

An

gle

Ze

ro-L

ift

An

gle

Ma

x C

l

8212

1.915 12.5 50.776 1.0 1.028 1.0 -7.0 1.915

8214

2.030 15.0 50.819 1.0 1.045 1.0 -7.0 2.030

8216

1.849 15.0 43.491 1.5 1.030 0.0 -7.0 1.849

8218

1.862 15.0 40.241 1.5 1.017 -0.5 -7.0 1.862

8312

1.973 12.5 54.702 0.5 1.027 7.0 -7.5 1.973

8314

2.093 15.0 50.086 0.5 1.043 1.0 -7.5 2.093

8316

1.960 15.0 48.526 0.0 0.996 0.0 -7.5 1.960

8318

2.000 15.0 41.458 3.5 1.260 -0.5 -7.5 2.000

6 Data taken from Airfoil Investigations Database (http://www.worldofkrauss.com)

7

9214

2.129 14.0 37.449 -0.5 0.980 -0.5 -8.0 2.129

9216

1.927 15.0 42.241 1.0 1.064 -0.5 -8.0 1.927

9218

1.941 15.0 39.719 1.0 1.058 -0.5 -8.0 1.941

9316

2.041 15.0 46.245 1.0 1.140 -0.5 -8.5 2.041

9318

2.089 15.0 43.340 0.5 1.082 -0.5 -8.5 2.089

K_LIFT FOR-LOOP

Generally, hybrid airships derive 30% of their lift from aerodynamic lift The range of values for

tested is from 0.26 through 0.45. Values below 0.26 and above 0.45 yield imaginary

results for the CHULA model.

AR FOR-LOOP

Aspect ratios range from 0.1 through 2. At an aspect ratio of 0, the airship ceases to have any

span and many equations that depend on this value result in a trivial value of 0 as well. I’ve

capped the analysis at an aspect ratio of 2 because if the airship is too much wider than it is long,

the chance of tip-strike while maneuvering low to the ground is high.

8

DESIGN METHODOLOGY

In the next sections, I will explain the aerodynamic equations and mathematical formulations by

which I approached the design and optimization of the CHULA.

AIRSHIP SIZING

The HULA’s volumetric displacement is written as . It is broken down into three main

terms:

The volume occupied by what would otherwise be a gondola or cockpit is assumed to be

negligible compared to the vast volumes necessary for lifting gases, ballast, and the cargo.

At cruise, the airship derives its lift from aerodynamic lift and buoyant force:

Accounting for the weight of the lifting gas and the air in the ballonets, we arrive at the equation

for the net lift, which is the amount of useful lift that can counteract the total weight of the

airship.

(1)

We can artificially set the amount of aerodynamic lift by introducing , the ratio of

aerodynamic lift to overall lift.

Substituting this relationship into equation 2 and rearranging, we arrive at a relationship for

as a function of and .

Substituting these relationships into the equation for total weight, we can rewrite the formulation

for or .

9

Using the Archimedes principle,

we can expand the equation for net lift to the following equation:

It’s a commonly held belief that as an airship ascends, its buoyancy diminishes as a function of

the decreasing density of air in which it floats. In actuality, the amount of buoyant lift is constant

because as the altitude increases, and the density of the surrounding air decreases, there is a

concomitant decrease in the density of the helium within the envelope caused by the enlargement

of the ballonets in which the gas is stored. Thus, an airship’s net lift is independent of the

altitude so long as the airship is below pressure altitude, the height at which the complete

fullness of lifting gas space is achieved (Burgess, 1927). The pressure altitude reflects the

altitude at which structural limit for the expansion of the helium ballonets and further ascent

necessitates an ejection of helium to relieve stress on the ballonet. The maximum fullness of the

helium gas space at sea level must not exceed the relative density of the air at the design pressure

altitude. The ratio of the volume of helium to the volume of the envelope is equal to that of the

relative density of the atmosphere at the pressure altitude. The volume of the helium gas cell can

be expressed:

The volume of the air ballast as solely a function of can be found by rearranging

equation 1 and substituting in the above equation:

Substituting and into the expanded form for net lift, yields:

A rearrangement of this equation produces an estimation of based on the total weight

of the airship and the atmospheric conditions of its cruise mode.

For a full derivation of this equation, please refer to Appendix A.

10

AIRSHIP WEIGHT

To solve for , an approximation of the total weight of the airship must be made. Its

weight can be broken down into the following components:

(12)

PAYLOAD7

There are three major types of air cargo containers: Unit Load Devices (ULD), pallets, and

intermodal containers. ULDs are most commonly used for cargo transport during commercial

airline operations because they are standardized for the airplane’s lower deck holds. These

containers directly lock into an aircraft’s restraint systems without the need for tie-downs or

exterior netting. Pallets are designed for use with the conveyor systems in terminals and in

aircraft. Cargo is secured to the pallet with cargo nets tightened with straps. The intermodal

container is the largest type of air container, which can directly interface with land and air

transportation without rehandling.

To facilitate analysis, I will assume that the entire airship’s payload is stored within enclosed 20’

by 8’ ISO intermodal containers. These containers have a tare weight of 4500 lbs and a payload

capacity of 40,000 lbs, making for a gross weight of 44500 lbs, assuming a full load (Bates,

2005).

The volume of the cargo hold can be estimated as:

where

.

The weight and volume of the cargo can be expressed:

(13)

(14)

7 In this section, I will use the terms payload and cargo interchangeably.

11

FLIGHT CREW

The augmented crew complement of the C-130H Hercules for an 18 hour crew duty is: three

pilots, two navigators, two flight engineers, and two loadmasters (GlobalSecurity.org). Because

the trans-oceanic cargo hauls of the CHULA will probably necessitate 24 hours shift.s, it would

probably require more personnel to operate. However, in this analysis, I will assume it has the

same crew complement as the C-130.

According to a 2004 FAA Advisory Circular on estimated aircraft loads, the average weight of

flight crewmembers can be considered to be 190 pounds, each with 50 pounds of baggage.

EMPTY WEIGHT

To perform a rapid approximation of the CHULA’s empty weight, I assumed a relationship

between existing airships’ size and their respective structural weights. Out of the 33 airships

sampled, 22 used hydrogen as a lifting gas and 11 used helium. Hydrogen is about 93% less

dense than helium per volume at sea level. To normalize the data, I assumed this ratio stays

constant through the operating altitudes of these airships, and divided the reported volumes of

hydrogen airships by 0.93 to determine how large they would have to have been in order to lift

the same gross weight had they been filled with helium instead. I plotted this normalized volume

against the Manufacturer’s Empty Weight (MEW) and performed a regression, which

determined that the following power curve fits the data with an R2

of 0.9823.

(16)

y = 0.0845x0.9326 R² = 0.9823

0

50,000

100,000

150,000

200,000

250,000

0 1,000,000 2,000,000 3,000,000 4,000,000 5,000,000 6,000,000 7,000,000 8,000,000

Emp

ty W

eig

ht

[lb

s]

Volume [ft3]

Empty Weight as a Function of Normalized Volume

12

One should be cautious in using this mathematical relationship because this study samples

airships produced across the entire gamut of powered flight, from airships built with WWI-era

manufacturing processes to those constructed with modern-day composite technology. The

relationship also does not take into account the differences in weight between rigid and semi-

rigid configurations, nor does it consider inequities in scaling. The specific data points used in

this regression can be found in Appendix B.

With this final relationship, we can express the total weight of the airship:

VOLUME OF ENVELOPE

Substituting the expression for total weight into the formulation for , we arrive at this

equation:

After some rearrangement, we can express it as a polynomial with which we can use polynomial

solvers to calculate .

SHAPE

The general shape of the airship has two distinct components: a cross-sectional geometry that is

an airfoil and sides that are revolved about the airfoil’s mean camber line. Mathematically, we

can treat the first as an extruded airfoil and the second as a revolved body. We can take the

length of the airship to be and the constant thickness span-wise portion to be . Because the

airship is so large, it has appreciable thickness, and a consideration must be made to the extra

span that the revolved “wingtips” add to the width. Its overall width can be defined as:

For modeling purposes, the airship’s cross-sectional geometry is that of a NACA 4-series airfoil.

The NACA four digit series family is a convenient way to shape the airship because it is

mathematically defined by the elements of its name. The first digit of a 4-series designation

indicates maximum camber ratio, multiplied by 100. The second is the chordwise location of

13

the maximum camber point, times 10. The third and fourth digits represent its thickness to

chord ratio, . The camberline of the airfoil is defined by the following two parabolas:

for

for

And the thickness distribution is given:

, in this equation, is the chord of the wing, which is equal to the length of the airship, . Since

the cargo hold will be located along the bottom of the airship, an airfoil with a relatively flat

lower surface would be preferable. 4-series airfoils with a maximum camber ratio of 8% or 9%,

a maximum camber point of 0.2 through 0.4 of the chord, and a thickness of 12% through 18%

yield such approximately flat-bottomed surfaces.

In order to complete the following calculations, it is useful to have a method of obtaining the

coordinates of the NACA 4-series airfoil for which the leading and trailing edges are

located on the x-axis. The following calculations assume such information is at hand and the

number of coordinates is known.8

LENGTH

The length of the airship can be calculated as a function of the airship’s aspect ratio and its

volume. To find this relationship, we break the airship into the extruded airfoil component and

the body of revolution.

VOLUME OF EXTRUDED AIRFOIL

The volume of the extruded airfoil is equal to the cross-sectional area of the airfoil times the

length of its extrusion, , which can be expressed as . Green’s theorem, which gives the

relationship between a line integral around a closed curve and a double integral of the plane

defined by that closed curve, can be applied to yield the area of the airfoil using its coordinates:

8 I used Professor Alonso’s Hess-Smith panel method MATLAB code to generate the airfoil coordinates used in this

analysis.

14

Usually, airfoil coordinates are non-dimensionalized to represent a unit airfoil. Therefore, it is

useful to represent the numerical implementation of Green’s theorem for a unit airfoil with

number coordinates as follows:

To scale the area of the unit airfoil up to the full size of the airship, we multiply it by . The

volume of the extruded component of the CHULA is merely the scaled area of the unit airfoil

times its span:

VOLUME OF REVOLVED WINGTIP

The volume of the revolved component can be found using the volume of a conical frustrum,

shown below:

The volume of a conical frustrum is expressed:

Because the airfoil may not be symmetrical, we must first average the -coordinates on the upper

side with their respective coordinates on the lower side to essentially make an equivalent

symmetrical airfoil. This process can be numerically implemented using the following process:

To scale to the full size of the airship, we must multiply by .

15

Expanding the equation using the equations for

and , we find:

Because span can be expressed as , we can substitute this relationship into the above

equation and rearrange to find the length of the airship.

REFERENCE AREA

With the length of the airship calculated, we can calculate the reference area of the airship using

its aspect ratio.

16

DRAG BOOKKEEPING

Drag bookkeeping involves the decomposition of the total drag of an aircraft into its

components: parasite drag and induced drag. Parasite drag is the component of drag independent

of lift which includes friction and pressure drag. Induced drag or vortex drag is the lift-

dependent component that quantifies the drag force that occurs whenever a moving object

redirects flow to produce lift In the application of this method to the CHULA, wave drag need

not be considered as the airship’s cruise speed is nowhere near the transonic regime. Thus, total

drag can be expressed in non-dimensionalized form:

PARASITE DRAG

Typically, the parasite drag of conventional cigar-shaped airships is calculated using Hoerner’s

formula for the volumetric drag coefficient of revolved bodies.9 However, because the CHULA

is essentially a buoyant wing section, I use equations for calculating the drag of wings instead.

The calculation of is described in Kroo’s aircraft design text.

is the form factor, which is a reflection of the increase in drag due to the higher surface

velocity of air caused by the wing’s thickness, . According to Kroo, a value of agrees

well with empirical data.

If there is no sweep, the equation simplifies to:

is the skin friction coefficient for a flat plate. In the case of the CHULA, I assume that it is

fully turbulent.

9 Hoerner’s formula for the volumetric drag coefficient of revolved bodies can be found in Appendix C.

17

is the wetted area, which Shevelle approximates as:

More accuracy can be achieved using numerical calculation. Again, we can think of the airship

as the summation of two parts: an extrusion and a revolution.

The surface area of the extrusion is the arc length of the airfoil multiplied by its span. With

airfoil coordinates that describe a unit airfoil, one can calculate the unit arc length and then scale

it by the length of the airship.

The surface area of the revolved airfoil can be computed by discretizing it into a series of conical

frustrums and summating their respective lateral surface areas. The lateral surface area of a

conical frustrum is given by the following expression:

The radii of the frustrum can be thought of as the y-coordinates of the airfoil, and the height can

be the x-coordinates.

To calculate the actual surface area of the revolution, one must scale this result by .

18

INDUCED DRAG

The equation for induced drag has two components: , the variation of with due to

viscous effects; and , the uneven distribution downwash across the span of the wing.

where is the correction for fuselage interference and is the planform efficiency factor, which

we will assume to be 0.99. Because there is no fuselage interference, is simply 1.

Raymer provides a general relationship that can be derived from Shevelle’s quoted values of

:

AERODYNAMIC LIFT

The equation to calculate the 3-D coefficient of lift for an elliptical wing is as follows:

Because the CHULA is untwisted, the angle of attack does not vary along the wing and thus remains constant. Therefore,

19

TAKE-OFF FIELD LENGTH

Kroo, in Aircraft Design: Synthesis and Analysis, provides a statistical method for determining

the take-off field length using a parameter defined as:

where the lift off speed is taken as 1.2 times the stall speed. Although Kroo does not provide an

equation of line for a 4 propeller plane, I was able to extrapolate the data points and create my

own fit.

y = 0.0275x2 + 6.4521x + 564.43 R² = 0.9997

0

1000

2000

3000

4000

5000

6000

7000

8000

50 100 150 200 250 300 350 400

Take

Off

Fie

ld L

en

gth

[ft

]

Parameter

4-engine Prop TOFL vs Parameter

20

LANDING FIELD LENGTH

Kroo also describes a method to determine landing field length, which approximates it as the

summation of an air run and a ground deceleration distance.

The air distance models a steady state glide with deceleration at constant altitude:

where is the effective drag, is 1.3 times the stall speed, and is taken at

approximately 1.25 times the stall speed. The ground distance is expressed:

where is the final weight and is the total stopping force. is calculated:

Here, is the braking coefficient of friction, and D is the total drag of the aircraft including

flaps, slats, and spoilers. To be conservative, my analysis will not take into account these

devices and so will be the same as .

Currey provides measured friction coefficients in his book, Aircraft Landing Gear Design:

Principles and Practices.

Dry Wet Surface 5 MPH 40 MPH 5 MPH 40 MPH Asphalt 0.95 0.75 0.95 0.65

Concrete, rough 0.92 0.73 0.70 0.40

Concrete, smooth - - 0.58 0.45

Gravel 0.65 0.72 0.70 0.71 Snow, dry, packed - - 0.45 0.45

Snow, moist, packed - - 0.50 0.52

0.5 in. Snow over ice - - - 0.3 Grass - 0.4 - 0.2

21

RANGE

As I wrote earlier, the CHULA will have two operating modes depending on its load situation:

vertical take-off and landing mode and short take-off and landing mode. When its payload is

small enough for the airship to achieve flight through gas buoyancy alone, it will operate in

VTOL mode. When it is loaded past that point, it will have to take off and land like an airplane,

so it will be in STOL mode. Presumably, the airship will fly farther flying under VTOL mode

because no energy need be expended to overcome gravity as it is neutrally buoyant.

STOL RANGE

For airplanes, the chemical energy in fuel is ultimately used to overcome drag and gravity. The

Breguet range equation is as follows:

According to Shevelle, a typical markup on manufacturer’s engine specification is about 3%.

When applied to airships, the initial and final weights can be considered as the portion of the

total weight with and without fuel that exceeds the amount of lift gas buoyancy provides.

VTOL RANGE

The VTOL range is calculated as:

Here, , in units of horsepower, is the power required to overcome the drag force at cruise

velocity. Assuming that drag and velocity are in SI units, it is calculated:

22

SUPERHEAT EFFECTS

Caused by the radiative heat of the sun, superheating can cause significant changes in the

buoyancy of airships. According to Buckley, for light colored ships, superheat can reach 15

degrees Fahrenheit, and 40 for dark airships. When calculating the volume of the ballonets as a

function of the pressure height of the airship, margin must be included in order to account for the

effect of superheating.

The ideal gas law is as follows:

Substituting for

and rearranging, we see that the volume of the envelope is also a function of

the temperature of the gas inside it.

The previously derived equation for airship envelope volume assumes standard atmospheric

conditions and does not account for the effect of superheating. In order to incorporate a margin

for these temperature variations, there should be a distinction between the airship’s maximum

safe operating ceiling and the pressure altitude.

23

DISCUSSION OF RESULTS

To summarize, the three parameters I varied were the body profile shape, ; the ratio of

aerodynamic lift to total lift, ; and the aspect ratio of the airship, . I varied them

accordingly:

Parameter Range : 8212, 8214, 8216, 8218, 8312, 8314, 8316, 9214, 9216, 9218, 9316, 9318 : 0.26 through 0.45

: 0.1 through 2

After obtaining an i × j × k - sized grid of results for take-off field length, landing field length,

and STOL range, a search was performed to filter out only results that fulfilled the following

criteria:

Range: 12,000 n.mi.

TOFL: 4500 ft. /1000 ft.

LFL: 4500 ft. /1000 ft.

OPTIMIZATION OF GRID SIZE

The NACA for-loop has a fixed number of elements, but the k_lift and AR for-loops are free to be

any size. Too small and the model risks not being thorough enough. Too large and computation

time might be too long to be practical. I tested differently sized loops and noted the ratio of

successful configurations to total grid size as well as the computation time.10

NACA K_LIFT AR Grid Size Good Config. Success Rate Calc Time

13 50 50 32,500 1,081 3.33% 7.1338 13 100 100 130,000 3,708 2.85% 17.4926

13 50 150 97500 3,244 3.33% 12.9219 13 200 200 520,000 13,681 2.63% 68.6598

13 300 300 1,1700,000 29,893 2.55% 110.0709

I determined that 100 elements in both the and aspect ratio for-loops would be sufficient.

10

I performed the calculations on a HP Pavilion dm3 laptop with an Intel Core2 Duo P9300 running at 2.26 GHz

and 4 gigabytes of RAM.

24

DISCUSSION OF ERRORS

One of the main shortcomings of my code is that it does not reliably calculate drag. The

methods I use assume a conventionally sized airplane with a wing of aspect ratio between 3 and

14. Prandtl-Lanchester Lifting Line theory does not hold well below an AR of 3, which is the

design space of the CHULA. I believe the inaccuracy in drag significantly affected my VTOL

range calculation because they fell short of my expectations. As I wrote earlier, the VTOL range

values should exceed those for STOL range, but my MATLAB code indicates that in the best

scenario, it was 650 nautical miles, which is far too small.

Another source of error was that my propulsion parameters do not change with altitude or

velocity. The code uses constant values for engine SFC and thrust, which affect range and take-

off field length calculations.

12000 NM RANGE, 4500 FT. TOFL AND LFL CONSTRAINT

After evenly spacing the and AR arrays to have 100 elements, a feasible design space

comprising of 3708 different possible airship configurations were found, taking 17.5 seconds to

compete the search. An airship with the above constraints yields a success rate of 2.85%. The

main distinguishing parameters were the landing field lengths for concrete and wet grass, and the

STOL range. The configurations that produce the top 20 ranges of the airship are shown below:

NACA K_LIFT AR TOFL LFL WETGRASS LFL CONCRETE STOL RANGE

8312 0.26 2.00 564.6024 2096.206678 2100.678352 32048.46

8212 0.26 2.00 564.6103 2095.328475 2099.731583 31997.31

8312 0.26 1.99 564.603 2089.655019 2094.174283 31948.46

8212 0.26 1.99 564.6109 2088.782198 2093.231648 31897.5

8312 0.26 1.97 564.6036 2083.077504 2087.645355 31848.09

8212 0.26 1.97 564.6115 2082.210121 2086.706876 31797.31

8312 0.26 1.96 564.6042 2076.473846 2081.091316 31747.34

8212 0.26 1.96 564.6122 2075.611959 2080.157011 31696.75

8312 0.26 1.95 564.6048 2069.843756 2074.511907 31646.21

8212 0.26 1.95 564.6128 2068.987424 2073.581798 31595.8

8312 0.26 1.94 564.6054 2063.186937 2067.906865 31544.69

8212 0.26 1.94 564.6134 2062.336221 2066.980972 31494.46

8312 0.26 1.92 564.606 2056.503085 2061.275924 31442.77

8212 0.26 1.92 564.6141 2055.658049 2060.354267 31392.73

8312 0.26 1.91 564.6067 2049.791892 2054.61881 31340.46

8212 0.26 1.91 564.6148 2048.952602 2053.701411 31290.61

8312 0.26 1.90 564.6073 2043.053044 2047.935247 31237.75

8212 0.26 1.90 564.6154 2042.219566 2047.022128 31188.09

8312 0.26 1.89 564.6079 2036.286218 2041.224954 31134.64

25

At a of 0.26 the CHULA would need to have a volume of 28,323,000 cubic feet, four times

as voluminous as the LZ-129 Hindenburg. For aspect ratios between 2 and 1.89, its length

would be between 555 ft. and 565 ft.—about the length of the British R-23 (1917).

Under the 4500 ft. take-off field length and landing field length constraint, the results indicate

that they were never critical, even when operating on wet grass. The largest ranges were

achieved when the aspect ratio was as large as possible and when the amount of aerodynamic lift

was as small as possible—both of which work towards minimizing induced drag. It also

suggests that the airship velocity, which is the lowest at a of 0.26, was not critical in

influencing maximum STOL range as a faster velocity would have made the range greater.

It’s interesting to note that two similar airfoils had a monopoly over these mega-ranges: the

NACA 8212 and the NACA 8312. Their sectional max L/D’s are 50.776 and 54.702, respectively.

Though high, these values are on par with some of the other airfoils such as the NACA 8214 and 8314.

One must look at the overall L/D of the airship to differentiate the NACA 8212 and 8312 from the 8214

and 8314. The lift component of the overall L/D is only the aerodynamic lift of the airship, which is kept

constant by fixing . Since the aerodynamic lift of the airship should be more or less constant

across the board, it is the drag that we must look at. Airships with profiles of the NACA 8212

and 8312 are thinner than those with the NACA 8214 and 8312, and so have less parasite drag.

At these ranges, however, the landing field lengths are also as high as they can be because the air

run portion of the landing distance is directly proportional to L/D as well. This reveals a tradeoff

between landing field length and range—with range, in this case, driving the design.

26

12000 NM RANGE, 1000 FT. TOFL AND LFL CONSTRAINT

In trying to satisfy the somewhat artificial constraint of landing within 1000 ft., we arrive at a

design space that is significantly smaller. Using the same and AR arrays of 100 elements,

only 85 possible configurations exist—a 0.06% success rate.

NACA K_LIFT AR TOFL LFL WETGRASS LFL CONCRETE STOL RANGE

9214 0.26 0.41 564.8965 992.8518 933.4777 14232.9

9318 0.26 0.46 565.0206 995.5886 928.1874 14142.11

8312 0.26 0.37 564.8964 992.744 928.8786 14120.37

8212 0.26 0.37 564.9178 997.8247 928.5336 14099.01

9316 0.26 0.43 564.9941 988.3507 923.9187 14063.82

8314 0.26 0.39 564.9246 968.6228 916.1769 13958.98

8214 0.26 0.39 564.9478 972.0796 916.038 13940.4

9214 0.26 0.39 564.9121 961.0008 912.3261 13909.69

8318 0.26 0.45 565.0794 976.9758 912.9047 13884.66

8218 0.26 0.45 565.1529 992.0363 914.5548 13870.71

9318 0.26 0.45 565.0383 966.3103 909.7924 13861.13

9218 0.26 0.45 565.1092 978.4999 911.2651 13843.55

8316 0.26 0.41 565.0489 966.4229 906.987 13783.44

8216 0.26 0.41 565.1055 976.519 907.9983 13767.43

8312 0.26 0.35 564.9135 954.9023 905.5984 13765.73

9316 0.26 0.41 565.0125 956.8943 903.9315 13758.87

8212 0.26 0.35 564.9357 957.5765 905.2617 13744.89

9216 0.26 0.41 565.0649 964.618 904.6396 13738.8

8314 0.26 0.37 564.9422 937.7723 894.3566 13625.84

At a of 0.26, the CHULA would still need to have a volume of 28,323,000 cubic feet.

However, for aspect ratios between 0.41 and 0.37, its length would be between 905 ft. and 932

ft., a little longer than the Titanic.

These results indicate a trend of low and low aspect ratio. Whereas the 4500 ft. constraint

allows the STOL range to be maximized at the expense of field length, the 1000 ft. constraint

caps the STOL range at a little more than 14,000 nautical miles.

27

CONCLUSIONS

There were a few pieces of information gleaned from this research:

The design of 1,000,000 pound payload airships will require that they be the largest man-

made objects ever created.

These airships will never be able to land on conventional runways because of their size

and will almost always require a large field or harbor.

When airfoils are used for the cross-sectional profile of a HULA, only its thickness

relative to the other airfoils used in the analysis is important.

There is a tradeoff between landing field length and STOL range.

For maximum range, the airship will want to have low lift and high aspect ratio.

For minimum landing field length, the airship will also want to have both low

aerodynamic lift and low aspect ratio.

28

FUTURE WORK

Though computationally inexpensive, the drag bookkeeping methods used in this analysis were

not accurate enough to provide a good performance model. Computational fluid dynamics or a

3-D panel code would go a long ways towards improving it.

29

APPENDIX A: DERIVATION OF VOLUME OF ENVELOPE

Equation (5) is as follows:

Expanding this equation,

Rearranging and collecting the terms,

And, finally, isolating , we arrive at the final relationship:

30

APPENDIX B: AIRSHIP EMPTY WEIGHTS

Total Volume [ft.3] Lifting Gas Normalized

Empty Weight [lbs]

Lotte 2/3 3,848 Helium 3,848 216 American Blimp MZ-3 170,000 Helium 170,000 6,300 LZ N 07 289,580 Helium 289,580 7,255 LZ N 17 600,349 Helium 600,349 12,831 Skycat 20 1,099,416 Helium 1,099,416 45,292 RA-180 Holland Navigator 2,754,544 Helium 2,754,544 94,799 ZR-1 USS Shenandoah 2,290,000 Helium 2,290,000 80,200 ZR-2/R38 2,724,000 Helium 2,724,000 66,000 LZ-126/ZR-3 USS Los Angeles 2,760,000 Helium 2,760,000 91000 ZRS-4 USS Akron 6,500,000 Helium 6,500,000 221000 ZRS-5 USS Macon 6,500,000 Helium 6,500,000 216,000 S.S.P. 70,000 Hydrogen 75,524 2,880 N.S. 360,000 Hydrogen 388,407 13,900 C 180,000 Hydrogen 194,203 7,900 Zodiac 328,000 Hydrogen 353,882 11,900 M 441000 Hydrogen 475,798 18,100 O 127,000 Hydrogen 137,021 5,200 P.V 176,000 Hydrogen 189,888 8,300 Roma 1,250,000 Hydrogen 1,348,634 44500 N-1 700,000 Hydrogen 755,235 24,200 R-9 930,000 Hydrogen 1,003,384 42,100 R-23 1,040,000 Hydrogen 1,122,064 39,900 R-31 1,610,000 Hydrogen 1,737,041 68,300 R-33 2,100,000 Hydrogen 2,265,706 81,700 L-33 2,100,000 Hydrogen 2,265,706 67,200 L-49 2,100,000 Hydrogen 2,265,706 58,200 L-70 2,340,000 Hydrogen 2,524,643 61,800 Bodensee 875,000 Hydrogen 944,044 28,700 L-100 4,075,000 Hydrogen 4,396,548 78,600 LZ-127 Graf Zeppelin 3,707,550 Hydrogen 4,000,103 122,000 LZ-129 Hindenberg 7,063,000 Hydrogen 7,620,323 206,350 R100 5,156,000 Hydrogen 5,562,847 204,000

31

APPENDIX C: HOERNER’S FORMULA FOR DRAG OF REVOLVED BODY

Hoerner’s formula for the volumetric drag coefficient is:

where is the fineness ratio of the airship; calculated:

The volumetric drag force using Hoerner’s formula is as follows:

32

APPENDIX D: MATLAB MAIN FILE

% START CLOCK tic; ticID = tic;

% MISSION OBJECTIVE Wpayload_US = 1000000; % [lbs] Wfuel_US = 125000; % [lbs]

% SET ALTITUDES ALT_SL = 0; % Sea evel altitude [ft.] ALT_to = 0; % Take off altitude [ft.] ALT_cruise = 5000; % Cruise altitude [ft.] ALT_land = 0; % Landing altitude [ft.]

% ATMOSPHERIC CONDITIONS (OUTPUTS IN SI) [rho_SL,nu_SL,Vsonic_SL] = getAtmosphere(ALT_SL*0.3048); [rho_to,nu_to,Vsonic_to] = getAtmosphere(ALT_to*0.3048); [rho_cruise,nu_cruise,Vsonic_cruise] = getAtmosphere(ALT_cruise*0.3048); [rho_land,nu_land,Vsonic_land] = getAtmosphere(ALT_land*0.3048);

% TENTATIVE CHULA CHARACTERISTICS Nengines = 4; Npanels = 100; engineParameters = getEngineParameters;

% VARIABLES NACA_list = [8212,8214,8216,8218,8312,8314,8316,8318,9214,9216,9218, ... 9316,9318]; % creates NACA_list vector k_lift = (linspace(0.26,0.45,100))'; % creates k_lift vector AR = linspace(0.1,2,100); % creates AR vector

% CREATE I x J x K GRID for i = 1:length(NACA_list); % RETRIEVE (X,Y) COORD AND PLACES INTO MATRIX [NACA_coordinates(:,2*i-1),NACA_coordinates(:,2*i)] = ... naca4(NACA_list(i),Npanels); % CALCULATE UNIT PARAMETERS [unitConicSurfaceArea(:,i),unitConicVolume(:,i), ... unitArcLength(:,i),unitCrossSectArea(:,i)] = ... getUnitParameters(NACA_coordinates(:,2*i-1), ... NACA_coordinates(:,2*i)); % OBTAIN AIRFOIL PARAMETERS [NACA(i),CLmax(i),CLmax_angle(i),LDmax(i),LDmax_angle(i), ... LDmax_CL(i),stallAngle(i),zeroLiftAngle(i),thickness(i)] = ... getAirfoilParameters(NACA_list(i));

for j = 1:length(k_lift); % CALCULATE VOLUME AND WEIGHT [Volume(j,:),Wi_US(j,:),Wf_US(j,:)] = ... getVolume(Wpayload_US,Wfuel_US,k_lift(j),rho_cruise, ... rho_SL,Nengines); Wi(j,:) = Wi_US(j,:)*4.44822162; % Takeoff Wt in [lbs]

33

Wf(j,:) = Wf_US(j,:)*4.44822162; Wavg(j,:) = (Wi(j,:) + Wf(j,:))/2;

for k = 1:length(AR); Length(j,i,k) = (Volume(j)/(unitCrossSectArea(i)*AR(k) ... + unitConicVolume(i))).^(1/3); % CALCULATE Sref Sref(j,i,k) = AR(k)*Length(j,i,k)^2; % [m^2] Sref_US(j,i,k) = Sref(j,i,k)*10.7639104; % [ft.^2] % CALCULATE Cruise Coefficients CL_assumed = 0.025; Vcruise(j,i,k) = ((2*Wavg(j))/(CL_assumed*rho_cruise* ... Sref(j,i,k)))^0.5; % [m/s] [CLcruise(j,i,k),CDp_cruise(j,i,k),CDi_cruise(j,i,k), ... CDcruise(j,i,k)] = getCoefficients(Vcruise(j,i,k), ... ALT_cruise, AR(k),Length(j,i,k),Sref(j,i,k), ... unitArcLength(i), unitConicSurfaceArea(i),thickness(i)); LDcruise(j,i,k) = CLcruise(j,i,k)/CDcruise(j,i,k); Dcruise_STOL(j,i,k) = CDcruise(j,i,k)*0.5*rho_cruise* ... Vcruise(j,i,k)^2*Sref(j,i,k); % [N]

% CALCULATE RANGE for STOL SFC = engineParameters(2); Range_STOL(j,i,k) = 325*(0.8/SFC)*(LDcruise(j,i,k))* ... log(Wi_US(j)/Wf_US(j)); % [n.mi]

% CALCULATE VTOL RANGE Dcruise_VTOL(j,i,k) = CDp_cruise(j,i,k)*0.5*rho_cruise* ... Vcruise(j,i,k)^2*Sref(j,i,k); Vcruise_kts(j,i,k) = Vcruise(j,i,k)*1.94384449; % [knots] Pcruise_VTOL(j,i,k) = Vcruise(j,i,k)*Dcruise_VTOL(j,i,k); Pcruise_VTOL_HP(j,i,k) = Pcruise_VTOL(j,i,k)*0.00134102209; Range_VTOL(j,i,k) = Wfuel_US*(Vcruise_kts(j,i,k)/ ... (SFC*Pcruise_VTOL_HP(j,i,k))); % [n.mi]

Pcruise_STOL_HP(j,i,k) =

Vcruise(j,i,k)*Dcruise_STOL(j,i,k)*0.00134102209; fuelFlowRate(j,i,k) = Pcruise_STOL_HP(j,i,k)*SFC; % [lb/hr] t_endurance(j,i,k) = Wfuel_US/fuelFlowRate(j,i,k); % [hr] Range_VTOL2(j,i,k) = Vcruise_kts(j,i,k)*t_endurance(j,i,k);

Range_VTOL3(j,i,k) =

Wfuel_US*325*(0.8/SFC)*(1/Dcruise_VTOL(j,i,k));

% CALCULATE TAKE-OFF FIELD LENGTH Vstall(j,i,k) = (2*Wi(j)/(CLmax(i)*rho_to*Sref(j,i,k)))^0.5; Vto = 1.2*Vstall(j,i,k); % [m/s] SHPto = Nengines*engineParameters(3); % [SHP] Tto = SHPto*745.699872/Vto; % [N] Tto_US = Tto*0.224808943; % [lbf] RDto = rho_to/rho_SL; % relative density at TO TOFL(j,i,k) = getTOFL(Wi_US(j),RDto,CLmax(i), ... Sref_US(j,i,k),Tto_US); % [ft.]

% CALCULATE LANDING FIELD LENGTH

34

V_50(j,i,k) = 1.3*Vstall(j,i,k); % [m/s] Vland(j,i,k) = 1.25*Vstall(j,i,k); % [m/s] [CLland(j,i,k),CDp_land(j,i,k),CDi_land(j,i,k), ... CDland(j,i,k)] = getCoefficients(Vland(j,i,k), ... ALT_land, AR(k),Length(j,i,k),Sref(j,i,k), ... unitArcLength(i), unitConicSurfaceArea(i),thickness(i)); Lland(j,i,k) = CLland(j,i,k)*0.5*rho_land*Vland(j,i,k)^2* ... Sref(j,i,k); Dland(j,i,k) = CDland(j,i,k)*0.5*rho_land*Vland(j,i,k)^2* ... Sref(j,i,k); mu_wetgrass = 0.2; % Braking coef. of fric. for wet grass mu_concrete = 0.75; % Braking coef. of fric. for asphalt LFL_wetgrass_US(j,i,k) = getLFL(Lland(j,i,k),Dland(j,i,k), ... V_50(j,i,k),Vland(j,i,k),mu_wetgrass,Wf(j)); LFL_concrete_US(j,i,k) = getLFL(Lland(j,i,k),Dland(j,i,k), ... V_50(j,i,k),Vland(j,i,k),mu_concrete,Wf(j));

% CALCULATES MAX WIDTH Width(j,i,k) = Length(j,i,k)*AR(k) + thickness(i)* ... Length(j,i,k); Span(j,i,k) = Length(j,i,k)*AR(k);

end

end

end

% GRID SEARCH % RETURNS SUCCESSFUL ROW, COLUMN, AND PAGE INDICES [r,c,p] = ind2sub(size(Range_STOL),find(Range_STOL > 12000 & ... TOFL < 4500 & LFL_wetgrass_US < 1000 & LFL_concrete_US < 1000));

r_c_p = horzcat(r,c,p); % r = j; c = i; p = k

for l = 1:length(r) Range_STOL_success(l) = Range_STOL(r(l),c(l),p(l)); TOFL_success(l) = TOFL(r(l),c(l),p(l)); LFL_wetgrass_success(l) = LFL_wetgrass_US(r(l),c(l),p(l)); LFL_concrete_success(l) = LFL_concrete_US(r(l),c(l),p(l)); end

% RETURNS SUCCESSFUL AIRSHIP CHARACTERISTICS % [NACA,AR,K_LIFT,TOFL,LFL_WETGRASS,LFL_CONCRETE,RANGE_STOL] AirshipResults = horzcat(NACA(c)',k_lift(r),AR(p)', .... TOFL_success',LFL_wetgrass_success',LFL_concrete_success', ... Range_STOL_success');

% STOP CLOCK toc(ticID); elapsedTime = toc

35

APPENDIX E: VOLUME CALCULATION

% CALCULATE VOLUME OF AIRSHIP ENVELOPE

function [Venvelope,Wi_US,Wf_US] = getVolume(Wpayload_US,Wfuel_US, ... k_lift,rho_cruise,rho_SL,Nengines)

% CONTAINER PARAMETERS Vcontainer_US = 20*8*8; % [ft.^3] Wcontainer_US = 40000; % [lbf] Wtare_US = 4500; % [lbf] Ncontainer = ceil(Wpayload_US/Wcontainer_US);

% CALCULATION OF Wcargo and Vcargo Vcargo_US = Ncontainer*Vcontainer_US; % [ft.^3] Wcargo_US = Ncontainer*(Wcontainer_US + Wtare_US); % [lbf] Vcargo = Vcargo_US*(0.3048)^3; % Conversion to [m^3] Wcargo = Wcargo_US*4.44822162; % Conversion to [N]

% CALCULATION OF Wfuel % [lbf] Wfuel = Wfuel_US * 4.44822162; % Conversion to [N]

% CALCULATION OF Wengines engineParameters = getEngineParameters; Wengines_US = Nengines*engineParameters(1); % [lbf] Wengines = Wengines_US*4.44822162; % Conversion to [N]

% CALCULATION OF Wcrew Ncrew = 9; Wperson_US = 190; Wbaggage_US = 50; Wcrew_US = Ncrew*(Wperson_US + Wbaggage_US); % [lbf] Wcrew = Wcrew_US*4.44822162; % Conversion to [N]

% CONSTANTS g = 9.8065; % [m/s^2] rho_helium = 0.1786; % [kg/m^3]

% CALCULATION OF RDair RDair = rho_cruise/rho_SL;

function F = myfun(Venvelope) F = ((rho_cruise*g/(1-k_lift)) - (rho_cruise*g) + ... (RDair*rho_cruise*g) - (RDair*rho_helium*g))*Venvelope - ... 10.443*Venvelope^0.9326 + ... rho_cruise*g*Vcargo - (Wcargo+Wfuel+Wengines+Wcrew); end

Venvelope = fsolve(@myfun,600000); % [m^3] Venvelope_US = Venvelope*(3.2808399)^3; % [ft.^3] Wempty_US = 0.0845*Venvelope_US^0.9326; % [lbs] Wempty = 10.443*Venvelope^0.9326;

36

Lbuoyant_US = rho_cruise*g*Venvelope*0.224808943; % [lbf]

Wi_US = Wcargo_US + Wengines_US + Wcrew_US + Wempty_US - Lbuoyant_US + ... Wfuel_US; Wf_US = Wcargo_US + Wengines_US + Wcrew_US + Wempty_US - Lbuoyant_US;

end

37

APPENDIX F: DRAG COEFFICIENTS CALCULATION

% CALCULATES DRAG COEFFICIENTS

function [CLmaxrange,CDp,CDi,CD] = ... getCoefficients(V,ALT,AR,L,Sref,UnitArcLength,UnitConicSurfaceArea,t);

% GET AIRFOIL PARAMETERS [rho,nu,Vsonic] = getAtmosphere(ALT);

% CALCULATE Length Re Re_l = V*L/nu;

% CALCULATE Mach M = V/Vsonic;

% FORM FACTOR, K C = 1.1; k = 1 + 2*C.*t./(1 - M.^2).^0.5 + C^2*t^2.*(1 + 5)./(2*(1 - M.^2));

if AR < 1 % AR = 1; end

% WETTED AREA, Swet Swet = L.^2*(UnitArcLength.*AR + UnitConicSurfaceArea);

% COEFFICIENT OF FRICTION, Cf Cf = 0.455./(log10(Re_l)).^2.58;

% COEFFICIENT OF DRAG (PARASITE), CDp CDp = k*Cf*Swet./Sref;

% INTERIM CALCULATIONS sweep = 0; k_viscous = CDp.*(0.38 + 57*10^-6 * sweep^2); u = 0.99; % Planform efficiency factor s = 1; % Fuselage interference correction e = 1./((pi.*AR.*k_viscous) + (1/(u*s)));

% CL FOR MAX RANGE CLmaxrange = ((CDp.*pi.*AR.*e)/3).^0.5; % Shevelle pg. 279

% INDUCED DRAG CALCULATIONS CDi = k_viscous.*CLmaxrange.^2 + CLmaxrange.^2/(pi*AR*u*s);

% TOTAL DRAG CD = CDp + CDi;

end

38

APPENDIX G: CALCULATES TOFL

% CALCULATES TAKE-OFF FIELD LENGTH

function TOFL_4PROP = getTOFL(TOW_US,sigma,CLmax,Sref_US,Tto_US)

Index = TOW_US^2/(sigma*CLmax*Sref_US*Tto_US);

%TOFL_2 = 857.4 + 28.43*Index + .0185*Index^2; %TOFL_3 = 667.9 + 26.91*Index + .0123*Index^2; %TOFL_4 = 486.7 + 26.20*Index + .0093*Index^2 TOFL_4PROP = 0.0275*Index^2 + 6.4521*Index + 564.43;

end

39

APPENDIX H: CALCULATES LFL

% CALCULATES LANDING FIELD LENGTH

% Aircraft landing gear design: principles and practices By Norman S. % Currey pg. 130

function LFL_US = getLFL(Leff,Deff,V_50,Vland,mu,Wf)

LDeff = Leff/Deff; % L/D g = 9.8065; % [m/s^2]

R = mu*(Wf - Leff) + Deff; % Overall braking force

d_air = 50*LDeff + (LDeff/(2*g))*(V_50^2 - Vland^2); % [m] d_g = Vland^2*Wf/(2*g*R); % [m] LFL = d_air + d_g; % [m] LFL_US = LFL*3.2808399;

end

40

WORKS CITED

Anderson, J. D. (1991). Fundamentas of Aerodynamics 2nd Edition. McGraw-Hill.

Bain, A., & Schmidtchen, U. (2000). Afterglow of a Myth: Why and how the "Hindenburg" burnt. Berlin: Deutscher

Wasserstoff- und Brennstoffzellen-Verband (DWV).

Bates, J. C. (2005, March-April). Containerizing the Joint Force. Retrieved April 24, 2011, from Army Logistician:

http://www.almc.army.mil/alog/issues/MarApr05/joint.html

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