INTRODUCTORYAIRCRAFTPERFORMANCE

James F. Marchman, IIIProfessor of Aerospace & Ocean Engineering

Virginia Polytechnic Institute & State University

Copyright 1994 by James F. Marchman, III

INTRODUCTORY AIRCRAFT PERFORMANCE

James F. Marchman, IIIAssociate Dean of Engineering &

Professor of Aerospace and Ocean EngineeringViirginia Tech

PREFACE

Most aerospace engineering departments include some course in introductory aircraftperformance in their curricula; however, proper positioning of such a course within thecurriculum is sometimes a problem. At an introductory level, aircraft performance is arelatively elementary subject which can easily be taught to students in the sophomore year.Prerequisite material needed for such a course includes, for the most part, only basiccalculus, engineering statics and dynamics.

The only prerequisite subject matter beyond the above needed for a study ofintroductory level aircraft performance is some background in aerodynamics. This oftenpresents a problem to those who must decide on the proper place for the performancecourse within the aerospace curriculum since a full course in aerodynamics or basic fluiddynamics is, in reality, much more demanding for both the student and professor than theaircraft performance course which may follow.

It is the author’s opinion, based on over twenty years experience with curricula whichplaced a first aircraft performance course both before and after the first aerodynamics orfluids course, that the best place for the performance course is very early in the curriculum,immediately following the student’s introduction to engineering statics and dynamics. Inthis position the course can serve as a bridge between the generic, basic engineeringcourses of the freshman and sophomore years which the student often finds ratherunappealing, and the more difficult, more theoretical coursework of the junior year. Theapplied nature of the course has much appeal to the student who is, by this time, wonderingif he or she is ever going to get to a course which has anything to do with his or her chosencurriculum and early placement of the course in the curriculum can boost the morale of thestudent and prevent loss of good students to other curricula.

In order to give an aircraft performance course such early placement in an aerospacecurriculum, the course, and hence its text, must include proper introductory material on thesubject of aerodynamics. This need not include a complete theoretical background in suchareas as potential flow theory, vortex sheet theory or lifting line theory, but must includeenough introductory fluid flow theory to enable the student to understand the basic appliedaerodynamics which govern the performance of an aircraft in flight.

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This text is designed to give the student a “stand-alone” course in basic aircraftperformance. It contains sufficient introductory fluid mechanics to allow the student tounderstand the standard atmosphere in which the aircraft must perform and enough fluiddynamics and aerodynamics to enable the student to understand the origins of the fluidforces that govern the performance of the aircraft. The text can thus be used for an aircraftperformance course which preceeds a formal course in fluids or aerodynamics. The onlyprerequisites for the course are introductory courses in statics and dynamics such as thosenormally taught in the freshman and sophomore years of most engineering curricula, plusan understanding of basic calculus. Those students who take the course after completing acourse in aerodynamics or fluid dynamics should find the sections on those subjects in thistext a good review of the more practical aspects of those subjects.

It will be obvious to the reader that the present version of this text is relatively poorlyedited and needs much work to have consistent style and presentable graphics and to befree of error. A first cut at actually including the equations on the word processor file hasbeen made in the first three chapters, thanks to the patient, volunteer efforts of Mrs. BettyWilliams. She should not be blamed for any errors in the equations but only for her faultyjudgment in volunteering to try to learn to do equations on WordPerfect in her valiantattempts to help the author produce a readable text. The remainder of the equations in thetext are in the author’s poor handwriting and he must take all responsibility for both poorlegibility and technical error.

Writing style is seen to change throughout the text, depending on the mood andpurpose of the author at the times of writing. Chapter three was the first part of this textdeveloped. It was originally to be the sixth chapter in a longer, multiple authoredaerodynamics/aircraft performance text which died due to other priorities some 15 years ormore ago. The first two chapters were written when Virginia Tech converted from aquarter to a semester system and elements of a quarter based aerodynamics course werecombined with a quarter based performance course. The third chapter was tacked onto thefirst two and provided to the students as notes to supplement any of several “performance”texts used over the past few years. Finally, after tiring of changing from one text toanother year after year in an effort to find the “perfect” performance text, the author decidedthat his own meager attempts at constructing a satisfactory text for AOE 3104 would be asgood as anything on the market and, when duplicated and sold at no profit, would savethe students lots of money when compared to the usual $50 texts.

In future years the text will be further refined as needed and will continue to be madeavailable to students at cost. All suggestions of students and other faculty for improvingthis material are both encouraged and welcome.

ACKNOWLEDGMENTS

Thanks are due to Mrs. Betty Williams of the Virginia Tech Aerospace and OceanEngineering Department who retyped the first three chapters from my earlier typed notes.She said she wanted to do that work as a way to learn to use WordPerfect to typeequations.

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I hope it turned out to be worth the effort for her because it saved me a lot of work. Therest of the typing is my fault as are all the figures and handwritten equations.

Thanks are also due to Dr. Fred Lutze of the Virginia Tech Aerospace and OceanEngineering Department since the basic order of chapters 4 - 8 evolved from his teachingnotes provided me some 23 years ago when I first was asked to teach a section of AircraftPerformance. I owe much to those notes, which allowed me to learn much of the materialfor the first time myself, having had a rather meaningless encounter with these sameprinciples in my own undergraduate education. Dr. Lutze’s notes, in turn, are based onseveral classic source texts and his own experience. Anything I have added to those noteshas also come from my experience in subsonic aerodynamic research and teaching, fromtexts of others and from my experience as a private pilot and aircraft owner.

While being an aerospace engineer or even an aircraft performance specialist has little todo with being a pilot, it could easily be argued that one’s real world experience at thecontrols of an airplane does give an important perspective in teaching or understandingaircraft performance and aerodynamics. I owe my “hands on” flying experience to myfather who began flying in his teens and continues to own and fly an airplane as of thiswriting at age 78, and to Dr. David Manor who, as my Masters student, took on thechallenge of pushing me to get my pilot’s license. I will never live up to my father’s wishthat I share his intense love of being in an airplane whenever possible or to Dr. Manor’sdesire that all of his students become daredevil acrobatic pilots; however, even the relativelymundane flying in a straight line from point A to point B has its satisfying moments if oneis willing to put up with the hassles imposed by the FAA and the weather and theoutrageous expenses of flight which result primarily from government regulation and theunlimited greed of the American Bar Association.

Finally, thanks go to the hundreds of students who have been subjected to my methodsand demands in several versions of aircraft performance courses. While some of themhave sat in a stupor in the back of the classroom oblivious to everything, many haveresponded, questioned and even excelled, making the experience worthwhile for us all.

James F. Marchman, IIINovember 1991

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Since its first use in l 99 l, this text has served well as the text for several sections ofAOE 3 l04, Aircraft Performance. Thanks to student help (students delight in findingmistakes in work done by faculty) many errors have been found and corrected and somesmall sections have been rewritten. Also, a set of representative homework problems hasbeen added as an Appendix

Also, since 199l, I have become a victim of those high costs of flying which Imentioned on the previous page and have sold my airplane and gotten away from flying.Nonetheless the experience of being a pilot will always influence the way I teach any Aerocourse. My father also had to give up flying when he lost his medical certificate to LouGehrig’s disease (ALS) which later cost him his life

Perhaps it is my imagination, but I also have the perception that over the last few yearsengineering students have come to this course with an ever diminishing awareness of thephysical realities of how and why things work and with a decreased ability to recognizewhen their answers are not in the right “ball park”. In contrast to students of 25 years ago,today’s engineering student has probably never overhauled a car engine or built a stereoamplifier from a kit. Hence, a certain “feel” for physical reality which can only come fromgrease under one’s fingernails or through the smell of solder fumes is often lacking. He orshe has also never had to worry about the “magnitude” of an answer in a calculation, atalent which was forced on users of the slide rule, and thus often isn’t bothered whencalculations for the best rate of climb for a Cessna 150 give an answer of l2,000 ft/min.

Years of use of a “politically correct” but alien to everyday life unit system have createda numbness toward the magnitudes of numbers which allows the student to accept as OKanswers that would be immediately recognized as absurd were they in a familiar, everydayunit system. The same student who would demand his or her money back from a shoppingmall scale which gave his or her weight as 56 pounds would not be alarmed if a classroomcalculation showed that weight to be 250 Newtons! Having never learned that there mightbe physical relevance to the magnitude of answers and having 12 or more years ofreinforcement of the idea that if everything is done in SI units the answer will come outright, that same student has difficulty grasping the necessity of recognizing that a Reynoldsnumber of 600 or a Lift Coefficient of 9.3 for an airplane wing is a pretty good indicationof a blown calculation! And, it is a rare day indeed when one encounters a student whoknows how many feet are in a mile! These shortcomings of today’s student are often firstencountered in a course like Aircraft Performance and they can add a significant burden forthose teaching such a course, especially for one who finds it second nature to navigatesuccessfully through either the English or SI unit systems and whose vast experience inrecognizing ballpark answers makes it difficult to understand why others may find this anew and strange demand

J F. Marchman, III

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TABLE OF CONTENTS

Chapter 1: Introductory Concepts and the Standard Atmosphere 1Unit Systems and Their Use 2Fluid Forces - Hydrostatics 6Stratified Fluids and the Standard Atmosphere 12

Chapter 2: Introduction to Fluid Dynamics 22Steady Flow 24Conservation of Mass: The Continuity Equation 24Fluid Dynamics and Euler’s Equation 29Bernouli’s Equation: Conservation of Energy 34Velocity Measurement: The Pitot-Static Tube 42Momentum Theorem 46

Chapter 3: Airfoil Aerodynamics 56Forces and Moments 56Dimensional Analysis and Non-Dimensional Coefficients 58Force and Moment Coefficients 64Airfoil Geometry 66NACA Airfoil Designations 69Pitching Moment and Its Transfer 70Flaps and High Lift Devices 82Laminal Flow Airfoils 91Supercritical Airfoils 94Three-Dimensional Effects 95Summary 97

Chapter 4: Performance in Straight and Level Flight 98Static Balance of Forces 100Aerodynamic Stall 101Perspectives on Stall 103Drag and Thrust Required 104Minimum Drag 106Everything You Always Wanted to Know About Minimum Drag 109Review Minimum Drag Conditions for a Parabolic Drag Polar 112Flying at Minimum Drag 114Drag in Compressible Flow 114Review 115Thrust 116Minimum and Maximum Speeds 117Special Case of Constant Thrust 120Review for Constant Thrust 122Performance in Terms of Power 124Power Required 126Review 130

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Chapter 5: Altitude Change: Climb and Glide 132Gliding Flight 133Time to Descend 139Climbing Flight 142Time to Climb 149Power Variation with Altitude 150Ceiling Altitudes 151

Chapter 6: Range and Endurance 152Fuel Usage and Weight 154Range and Endurance: Jet 155Approximate Solutions 158Range and Endurance: Prop 163Approximate Solutions 165Wind Effects 166Let the Buyer Beware 168

Chapter 7: Accelerated Performance: Take-off and Landing 170Take-off Performance 171Take-off Without Rotation 178Thrust Augmented Take-off 180Ground Wind Effects 182Landing 184Effect of Wind 187

Chapter 8: Accelerated Performance: Turns 190The Two Minute Turn 196Instantaneous Versus Sustained Turns 197The V-n or V-g Diagram 199

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CHAPTER ONE

INTRODUCTORY CONCEPTS AND THE STANDARDATMOSPHERE

Aircraft Performance is a subject which examines the simpliest motions of an aircraft inthe atmosphere. It essentially treats the aircraft as a point mass which is subject to externalforces that cause the aircraft to move, accelerate and decelerate. As such, the subject ofaircraft performance is merely an extension or application of the material studied inintroductory courses in engineering statics and dynamics, the primary difference being thesource of the forces involved.

The primary forces which act on an aircraft in such a way as to cause or alter its motion::

Fluid dynamic forcesFluid static forcesGravitational forcesPropulsive forcesGround interaction forces

In this text the desire is to examine the way in which these forces determine the motionof an aircraft in the atmosphere; ie., the performance of the aircraft. Among the thingswhich will be examined are the following:

1. How fast can the aircraft fly?2. How high can it climb?3. How fast can it climb?4. How slowly can the plane fly?5. How far can the airplane glide?6. What is the time of descent in a glide?7. What is the range and endurance of the aircraft?8. What distances are needed for takeoff and landing?9. How quickly can the airplane make a turn?

10. What radius is required for a turn?

As stated above, the aircraft will be considered a point mass in modeling its motions.This means that rotational motions of the aircraft will not be considered. These motions,pitch, roll and yaw, are rotations of the aircraft about its center of gravity and are thesubject of texts dealing with aircraft stability and control. This text will thus consider thesimple translational motion of the center of gravity of the aircraft as it is subjected to theabove named forces.

The mathematics and mechanics background needed for the use of this text is relativelybasic. The equations derived in the text are fairly simple and can indeed be easily derived

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from Newton's Second Law, F = ma, and in most cases the right hand term is zero. Thedanger faced by the student in a course in elementary aircraft performance is that of beinglulled into complacency by the simplicity of the theoretical development of the course whilefailing to thoroughly comprehend the physical aspects of the problems. The result can becatastrophic when the student suddenly realizes that he or she is awash in a sea of simplebut similar theoretical relationships with no knowledge of the physics of the phenomenaunder investigation without which the correct selection of equations suitable to the solutionof the problems at hand is impossible. For this reason this text will emphasize the physicalaspects of the aircraft performance problems considered and will try to avoid the temptationto develop theoretical relationships in terms of anything other than the most basic physicalphenomena upon which the performance problem depends.

UNIT SYSTEMS AND THEIR USE

One of the most frequent sources of error in student solutions of engineering problemsis improper use of units. Many students, faculty and text authors seem to labor under themisconception that by using the International (SI) system of units one will never have toworry about units; however, in this author's experience such is not the case. Problemsseem to arise from a combination of two sources. The first is the student's lack ofunderstanding of the relationship among units within a given system of units and thesecond is the failure of the student to consider the physical reasonableness of the magnitudeof an answer.

The basic units used in this text are those of length (L), time (T) and mass (M). Allphysical properties considered will be defined in terms of one or more of these units.

These units are related through Newton's Second Law:

F = ma MLT −2( )This says that a force must have the units of mass multiplied by length and divided by

the square of time. This relationship is the key to dealing with the unit relationships in anysystem of units which may be used in aircraft performance problems.

There was a movement in the 1970's to purge all engineering texts of any unit systembut the SI system since this system was and still is being promoted as the desired standardworldwide unit system for use in all scientific and technical fields. Unfortunately, orfortunately, depending on one's point of view, this movement has not yet succeeded inmany engineering fields and, even if it had succeeded, one cannot now completely ignorethe vast store of technical literature which was written in terms of other unit systems. As aresult, most modern engineering texts employ both the SI unit system and the older,traditional unit system used in the subject under consideration.

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The unit system traditionally used in the Aerospace field is a variation of the standardEnglish system of units. This system uses the "slug" as the unit of mass whereas thestandard English system uses the pound-mass.

Let us examine the four unit systems most often encountered in engineering work.These are the gravitationally based English system which employs the pound-mass, the"slug" variation of the English system which is also gravitationally based, the "cgs" versionof the old metric system where the gram is the unit of mass, and the SI system which usesthe kilogram unit for mass. The latter two systems are not said to be gravitationally basedsince their unit relationship via Newton's Second Law does not depend on the definedgravitational acceleration at the Earth's surface. The standard English system istraditionally used by Mechanical Engineers, the "cgs" system by Electrical Engineers andthe slug based English system is commonly used in the Aerospace Engineering field.

UNIT STD ENGL. SLUG ENGL CGS SI

Mass lbmslug gm kg

Length foot foot cm meter

Time sec. sec. sec. sec.

Expressed in terms of Newton's Second Law these systems give the followingcombinations of units:

Standard English: 1 lb f = 1 lbm × 32.2 ft sec2

Modified English: 1 lb f = 1 slug × 1 ft sec2

CGS: 1 Dyne = 1 gm × 1 meter sec2

SI: 1 Newton = 1 kg × 1 meter sec2

The use of any of these systems is simple if one does not let his or her lack offamiliarity with the everyday use of the system intimidate. Unfortunately, most people areintimidated by one or more of these systems.

The SI system, which is supposed to be the "cure-all" for unit system woes, is aproblem for some because they have trouble remembering how many zeros go with thekilos, millis, centis, etc. Its biggest problem in reality resides in the lack of "feel" that mostpeople have for the physical magnitude of the units encountered in the system. Even thosewho have used the older metric system all their lives have such problems, having becomeaccustomed to the use of the kilogram-force as a unit of weight (force) for everyday useinstead of the

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Newton of the SI system. Only when the true SI system is used in common, everydaybusiness transactions in such places as food markets all over the world will it become atruly universal system. This author sees little indication that such is happening, either inthe United States or elsewhere.

Much of the engineering work in the aerospace field is still done in the English unitsystem. The modified English system which uses the slug as the unit of mass isformulated such that if mass is expressed in slugs all other units work out in terms ofpounds-force, feet and seconds if these same units are always used with the appropriateterms. This system has long been favored by Aerospace Engineers. Mechanical Engineersare more likely to use the basic English system in everyday work in the United States,using the pound-mass as the standard unit of mass. This results in the 32.2 feet persecond-squared term being used for acceleration in Newton's Law. While rememberingthat one poundforce is equal to one pound-mass times 32.2 feet per second squared (F =ma) would seem to be no greater a chore than remembering that one foot is equal to twelveinches or that one meter is equal to ten decimeters, it has been standard practice inMechanical Engineering texts to introduce the term g c as a means of dealing withcombinations of force and mass units. In this author's experience in teaching bothMechanical and Aerospace Engineering, courses the use of g c is both unnecessary andconfusing to the student who often tries to rely on memorizing where to put g c in everyequation rather than simply keeping up with all units as they are used and using F = ma tosimplify the units.

In using any unit system the best policy is to always write the appropriate units with allnumbers, combining the units as appropriate during the solution of the problem and usingthe unit relationship in Newton's Law to deal with combinations of mass and force unitswhenever and wherever they occur.

EXAMPLE

Suppose you are an astronaut-engineer assigned to the planet Zorg and are there toinstruct Zorgian engineers in an educational exchange program. In order to maintain gooddiplomatic relations you must instruct using the Zorgian Standard unit system (ZSI) inwhich mass is expressed in "lumps"", force is given in "whops", length in "leaps" and timeis in "zips". Fortunately, Newton's laws still hold anywhere in the universe, and theseunits are related through the Second Law:

1 whop = 1 lump × 827.6 leaps zip2

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You find that on a Zorgian scale you weigh 2134 whops at "crust-level standardconditions" where the gravitational acceleration is 92.77 leaps per zip squared. What is

your mass in ZSI units of lumps?

First you must use F = ma to calculate your mass:

2134 whops = mass × 92.77 leaps zip2

or mass = 23 whops × zip2 leaps

Now this must be converted to standard units of lumps. Using the second lawdefinition of the unit system you multiply the term on the right as follows:

23 whops × zip2

leap × 1 lump × 827.6 leaps zip2

The result gives your Zorgian mass:

mass = 19034.8 lumps.

As you calculate this you start planning to introduce the Zorgians to the SI unit systemand you begin to think that even the ancient English unit system wasn't so bad after all!

In this text, the unit systems used will be the modified or slug based English systemand the SI system since these are the systems most likely to be encountered by the engineerworking in the aircraft industry. Not only should the student be familiar with both systemsof units but he or she should also become familiar with the magnitudes of the variousperformance variables and fluid properties which will be commonly encountered inperformance problems.

It is not difficult for one to recognize an incorrect answer when dealing with familiarthings. For example, if calculating a person's height gives an answer of 42 feet, mostpeople would immediately recognize that an error had been made and would begin to try tofind the source of that error. Most Americans would, however, not be as quick torecognize an error if, when working in the metric system they calculated a person's heightto be 9752 mm and virtually no one would be able to recognize a meaningful quantity ifworking in the Zorgian units of the example above.

In an aircraft performance course it is necessary for the student to get a feel for the"normal" magnitudes of quantities found in the course. Is a rate of climb of 500 feet perminute normal or is 500 feet per second more normal? This is made more complicated bythe fact that the equations used will often give answers in units that are not commonly usedfor the quantity in concern such as aircraft endurance in seconds and range in feet whilehours and miles or kilometers are the more familiar terms. The ability to recognizereasonable magnitudes for such quantities will be an important asset to the student seeking

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success in an aircraft performance course and is certainly every bit as important as theability to choose the right equation for the solution of a particular problem.

FLUID FORCES-HYDROSTATlCS

Perhaps the most important source of forces which act on the aircraft in such a way asto cause it to move or change its motion is the fluid in which it is immersed. These effectsare found in the hydrostatic forces which determine the change of fluid properties throughthe atmosphere and in the aerodynamic forces which are used by the aircraft to fly andmaneuver. In order to understand these forces one needs to have at least a basicunderstanding of the fluid, in this case air, in which the aircraft operates.

Technically a fluid is any material which cannot resist a shear stress. A fluid movesand deforms continuously as long as a shear stress is applied. A fluid at rest is in a state ofzero shear stress.

Fluids are generally divided into two classes, liquids and gases. A liquid has close-packed molecules with strong cohesive forces and tends to retain its volume (isincompressible) and will form a free surface in a gravitational field. Gases have negligiblecohesive forces and are free to expand until they encounter confining walls. Gases cannotform a free surface. These are general definitions to which special case exceptions can befound.

In this text the fluid of primary concern is air, which is usually modeled as an "IdealGas" and is said to behave according to the Ideal Gas Equation of State,

P = RT

It is assumed that the student is already familiar with this relationship and with the fluidproperties involved. The Ideal Gas model is based on an assumption of no intermolecularinter-actions and, therefore, has limits in its application. The Ideal Gas model is, howevermore than adequate for use with air in the Earth's atmosphere in most aircraft performanceapplications. Typical magnitudes of these quantities and their units will be discussed later.It should be noted here, however, that pressure and temperature in this relationship, andindeed in all usage in this text must always be expressed in absolute values

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The first and most basic of the fluid principles which must be examined is Pascal'sLaw which states that the pressure acting at a point in a fluid at rest is the same in alldirections. This can be shown mathmatically by considering the triangular element of fluidshown below.

Figure 1.1

Pressure Px acts on the face dY, pressure Py acts on the face dX and pressure Pn actson the diagonal face dS. The weight of the element dW acts vertically downwards in thenegative Y direction.

dW = g (dX × dY 2)

where the term in parentheses is the area of the triangular element.

This element is static, ie., not in motion, and hence the forces on the element sum tozero. Summing these forces in the X and Y directions gives:

Fx = Px × dY( )∑ − Pn × dS( ) sin = 0

Fy Py × dX( )∑ − Pn × dS( ) cos − dW = 0

Realizing that:dY = dS sin

and that:dX = dS cos

the equations above become:

Fx = Px × dY( )∑ − Pn × dY( ) = 0

and Fy = Py × dX( )∑ − Pn × dX( ) − dW = 0

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From the first of these equations Fx = 0∑( ) it is found that:

Px = Pn,

and from the other equation:

Py = Pn + dW.

As the fluid element becomes small, the weight of the element goes to zero, giving:

Py = Pn.

Therefore,

Px = Py = Pn

and the pressure is shown to act equally in all directions at any given point in a fluid at rest.This pressure is known as the hydrostatic, or simply static, pressure.

The HYDROSTATIC EQUATION is probably the most basic relationship used influid mechanics . It is the basis for defining the model of the atmosphere which is used inaircraft performance predictions and provides the basis for the use of liquid manometers forthe measurement of pressures in the laboratory. It is also used to calculate the buoyancy oflighter than air vehicles in the atmosphere and for ships, submersibles and other vehiclesoperating in water.

Pascal's law dealt with the equality of pressures at a point within a static fluid. Thehydrostatic equation accounts for the changes in pressure within a fluid in a gravitationalfield resulting from the weight of one fluid element acting on the element below it. Again asimple fluid element is examined and the forces are summed over that element. Theelement below is a three dimensional block of fluid shown in the X - Z plane:

Figure 1.2

Note: the depth of the element is dY.

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In this derivation the element will not be reduced to a point in the limit and thus thevariation of pressure across the element must be considered. Pressure is assumed to varyvertically ( in the Z direction ) due to the force of one element stacked over another in thatdirection. The pressure at the center of the element is said to be P and to vary by an amount(dP dZ ) × (dZ 2) from the center of the element to the top or bottom surface.

Summing the forces on the element in the Z direction gives:

Fz∑ = − P +1

2

P

z

dxdy( )

+ P −

1

2

P

zdz

dxdy( )

− gdxdydz = 0

Dividing by the volume of the element ( dX In dY X dZ )gives:

−1

2

P

z−

1

2

P

z− g = 0

or

P

z+ g = 0

giving

P

z= − g.

Similarly summing forces in the X and Y directions will give:

P

x= 0,

P

y= 0

Note: there are no gravity forces in these directions.

Only the relation in the Z direction is meaningful in most cases and it is customarilywritten:

P

z=

P

h=− g

Note that this relation has two independent variables P and . Therefore, in order toto integrate this equation some relationship must he determined between two of thevariables. There are also situations where gravitational acceleration must be considered athird variable.

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There are many situations where both the density and the gravitational acceleration areconstant and this equation can be easily intergrated. One such case is when anincompressible liquid is the fluid. This results in the ability to use liquids as working fluidsin manometers to measure pressures in the laboratory. If and g are both constant

P = − g hor ∆P = − g∆h.

This form of the hydrostatic equation is the basis of pressure measurement using "U -tube" manometers where an incompressible liquid of known density is used to find thedifference in pressures between two points. One point is usually a convenient referencepoint of known pressure such as the atmosphere; thus allowing measurement of anunknown pressure at some other location. Such a case is shown below where a U - tubemanometer connects a point in a system with unknown pressure ( Ps ) to the surroundingatmosphere with a known atmospheric pressure ( Pa ). The fluid in the manometer has aknown density and, as shown, the higher pressure in the system causes the manometerfluid to rise in the atmospheric side of the U - tube. The difference in the heights of thefluid in the two sides of the U - tube manometer is the ∆ h term in the hydrostatic equationand the pressure in the system is found from the following relationship:

Ps = g ∆h( ) + Pa

Figure 1.3

It must be remembered when using this relationship that the density and the heightdifference used are for the manometer fluid and not for either the air in the atmosphere orfor the fluid in the system for which the pressure is being measured.

A common manometer fluid for use in wind tunnel experiments is water. Water isreadily available, non-toxic, and quite stable. Because of its common usage in wind tunnelmanometers where such U - tube manometers were once used as the primary means ofmeasuring ( via dynamic pressures ) the wind tunnel speed, it is common to hear windtunnel personnel refer to tunnel operating conditions in terms of "inches of water" and touse inches of water as a unit of pressure.

The only major problem with the use of water as a manometer fluid is its tendency toevaporate easily. For this reason other fluids such as mercury or oil are also commonlyused. The densities of water and mercury are given in metric and English units below:

water: =1.0 gm cc =1.94 sl cu ft.

mercury: =13.6 gm cc =0.002376 sl cu ft.

air: = 0.0012 gm cc =0.002376 sl cu ft.

(at sea level standard conditions)

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The density of air is shown for comparison purposes only. It is obviously not anacceptable manometer fluid for reasons such as its compressibility.

Oils ( usually with color added ) are commonly used as manometer fluids because theirdensity can be near that of water and they have little tendency to evaporate. Oils can beblended such that their density matches that of water and these special oils are commonlyfound in wind tunnel manometers.

It is interesting to note that although electronic manometers using electronic pressuretransducers have replaced fluid manometers in most modern wind tunnel facilities, theseinstruments are often calibrated to give readings in "inches of water".

EXAMPLE

A U - tube manometer is used to measure the pressure on the surface of a wing modelin a wind tunnel. The working fluid in the manometer is water and the reference end of themanometer is open to the atmosphere at sea level standard pressure of 2116 pounds-per-square-foot ( psf ). If the difference in the heights of the water on both sides of themanometer is three inches as shown below, find the pressure on the wing in units of psf.

Pa = 2116 psf

∆h =− 3"H2 O

Figure1.4

Px = g∆ h + Pa

H2 O = 1.94 sl ft3

g =32.17 ft sec2

Px = −3in ×1.94slft 3

× 32.17ft

sec2+ 2116

lb f

ft2

= − 312

ft × 62.4sl

ft2 .sec2+ 2116

lb f

ft2

= −15.6sl

ft.sec2×

1lb f

1sl ×1 ft

sec2

+ 2116lb f

ft2

Px = −15.6lb f

ft2+ 2116

lb f

ft2=2100.4

lb2

ft2

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Note: To convert from inches of water to psf one simply multiplies by 5.2.

g 12 = 62.4 12 = 5.2

Sometimes it is advantageous to "incline" or tilt a manometer in order to allow increasedaccuracy in reading its level as illustrated in the figure below.

Figure 1.5

The height difference across the manometer is the same for both the upright and theinclined manometers above. Increased precision is made possible using the inclinedmanometer by measuring the slant distance ∆ L rather than ∆ h and then calculating ∆ hfrom the measured ∆ L and the sine of the angle between the inclined tube and thehorizontal.

∆ h =∆ L sin

STRATIFIED FLUIDS AND THE STANDARD ATMOSPHERE

The operation of liquid manometers described above relies on the incompressibility orconstant density of the manometer fluid. In gases the density of the fluid usually changesas the forces on the gas increase or decrease and they are described as compressible. In theatmosphere, for example, the force on a particular element of the fluid (air) depends on itsaltitude, with that force, and hence the density, increasing as altitude decreases. Theelement at the base of the atmosphere near the Earth's surface experiences a large force dueto the weight of all the elements stacked above while the element high in the atmospheredoes not experience mush of a load. The density of the air thus varies with altitude and theatmosphere is said to be stratified.

It is intuitive that the performance of an aircraft will depend to some extent upon thedensity of the surrounding air as well as on the pressure and temperature of the air whichare also dependent on the air density through the ideal gas or other similar relationships. Itis therefore very important to have some model of the atmosphere and its pressure, densityand temperature variations in order to predict aircraft performance.

19

In order to model the atmosphere, two previously introduced relationships may beused. The first is the hydrostatic equation

dP dh = − g.

This relationship has at least two independent variables and cannot be solved by itself(density can no longer be considered a constant as it was in the liquid manometer case).

The second relationship which may be used is the Ideal Gas Equation of State:

P = RT

which also has two independent variables. Together, these two equations result in threeindependent variables P, h and T and another relationship is needed between two of thesethree variables or the definition of one of them as a constant.

The atmosphere is, of course, anything but static with pressures and temperatures (andhence, densities) varying with time of day, frontal movement and time of the year. Evenaverage conditions are different at every location on earth and on every day of the year.International agreement has been reached, however, on an International StandardAtmosphere (I.S.A) which defines sea level standard conditions and temperature variationswith altitude. This standard atmosphere is actually intended to approximate averageatmospheric conditions in the relatively temperate latitudes found in North America andEurope. Other standards have been defined for more extreme locations on the globe,giving standards such as the Arctic Minimum Atmosphere and the Tropical MaximumAtmosphere.

The definitions of these standard atmospheres are based on a knowledge of temperaturevariation within the atmosphere based on thousands of measurements from aircraft baloonsand sounding rockets. Through these measurements it has been determined that the bestway to characterize the property variations with altitude is through temperature.Temperature has been found to decrease very nearly linearly with increasing altitude up analtitude of 11,000 meters (36,100 feet) and to be essentially constant from that altitude upto those well above the maximum operating altitudes of most aircraft (there is somedisagreement among references regarding the upper limit of the constant temperatureregion). These temperature variations are shown graphically in Figure 1.6.

Temperature variation with altitude can best be defined by use of a term called the"lapse rate", L , which relates temperature to altitude.

dT dh = − L

Note that the minus sign is necessary due to the decrease in temperature with increasingaltitude.

20

If the lapse rate is assumed to be a constant, this integrates as

T = Tref + L href − h( ),a linear variation in temperature with altitude.

Figure 1.6

The reference values of temperature and altitude which are needed in the temperature,lapse rate relation above are chosen as the defined Sea Level Standard temperature and itsaccompanying zero altitude.

Sea level standard conditions are defined as follows in both SI and English units:

Pressure = 1.013 ×105 pascals or 2116lb ft2

Density = 1.23 kg m3 or 0.002378 sl ft3

Temperature = 288o K or 520 o R

Based on these sea level conditions, the defined lapse rate, the Ideal Gas assumptionand the hydrostatic equation conditions within the standard atmosphere can be calculated.The lower part of the atmosphere where the temperature is modeled as decreasing linearlywith altitude increase is called the TROPOSPHERE. In the troposphere the lapse rate is

21

defined as 6.5 degrees Kelvin per kilometer. This works out to 1.98 degrees Kelvin perthousand feet. Pilots learn to approximate this as two degrees per thousand feet, a figureboth useful and easy to remember since pilots in all parts of the world use feet as units ofaltitude.

In the Troposphere

T = Tsl − hL

and combining the hydrostatic equation with the ideal gas law,

dPdh =− g = −g P

RT

Rearranging gives

dPP = − g

RT( )dh

Substituting the lapse rate equation to reduce the number of variables for integration gives

dP

P=

− g

R Tsl − LH( ) dh

and integrating gives

ln P]P1

P2 =g

LRln Tsl − Lh( )[ ]h1

h2

or lnP2

P1

=

g

LRln T[ ]h1

h 2

=g

LRln

T2

T1

or finallyP2

P 1=

T2

T1

g

LR

22

Using the Ideal Gas law, this pressure temperature relationship can be rewritten as

or

2

1

=T2

T1

g − LR

LR

P2

P1

= 2

1

g

g− LR

The preceding equations are used to calculate the properties of the air at any altitudewithin the Troposphere. Such calculations will be examined in a later example. TheTroposphere is defined as extending up to 11,000 meters or 36,100 feet. Using thisaltitude in the Troposphere lapse rate relation for temperature results in a value of 216.5degrees Kelvin or 389.99 degrees Rankine at the upper limit of the Troposphere.

Above the Troposphere is a region where the temperature is modeled as a constant,invariant with altitude. This region is called the STRATOSPHERE and references arenot in agreement about the proper definition of the upper limit of this region It is commonfor modern jet aircraft to fly in the Stratosphere and it is necessary to extend the definitionof the standard atmosphere model through this region

In the Stratosphere the hydrostatic equation and the ideal gas law still hold, however thetemperature is now constant at 389.99˚ R or 2165˚ K making possible the integration of thehydrostatic equation over the single independent variable h.

dPP = − g

RTS

dh

where Ts is the constant temperature in the stratosphere.

Integrating gives: ln P]P1

P2 =− g

RTs

h2 − h1[ ]

orP2

P1

= e

g h1 − h2( )RT3

23

and since T is constant, the Ideal Gas Law gives:

P1

P2

= 1

2

and

2

1

= e

g h1 −h2( )RTs

These relations account for the variation of pressure and density in the stratosphere.

In using these relations it is convenient to use the conditions at the boundary betweenthe Troposphere and the Stratosphere as the reference or state one property values in theequations. This boundary between Troposphere and Stratosphere is known as theTROPOPAUSE.

EXAMPLE

Calculate the pressure and density at altitudes of 30,000 ft and 60,000 ft in the standardatmosphere.

(a) 30,000 feet is within the troposphere. The lapse rate equation gives:

T = 288˚K − (30,000 ft )1.98˚K 1000 ft.

or T = 228.6˚ K

This is now used in the pressure - temperature relationship for the troposphere:

PSL

P30,000

=TSL

T30,000

g / LR

24

Note that this example now contains a strange mixture of units with temperature in SI unitsand altitude in feet. Normal practice would be to convert everything to a common unitsystem; however, this is a rare case where such a combination of units is often seen in evenfairly recent literature and, as mentioned earlier, airplane pilots are still taught to use thislapse rate of about 2 degrees per thousand feet for calculations of atmospheric properties.It is therefore worthwhile to examine such a calculation. To make this work, the gasconstant R must be defined for air using this rather strange mixture of units. ( Asmentioned earlier, an engineer must be prepared to handle any system of units, even if itmeans working in leaps per whop!)

Thus, using:

R = 3092 ft − lbf sl °K

the pressure - temperature ratio becomes

PSL

P=

TSL

T

5.256

which, using sea level pressure as 2116 psf and sea level temperature as 288 degreesKelvin, finally gives

P30,000 = 630 psf .

Density can now be calculated from the Ideal Gas Law

=P

RT

or

P30,000 = 0.00089sl

ft3

25

(b) An altitude of 60,000 feet is within the stratosphere which requires that the propertiesat the tropopause first be determined. These are found by using the troposphere relationswith an altitude of 36,100 feet, giving a pressure and temperature of 470 psf and 216.5

degrees Kelvin. Using thses as the base reference property values in the stratosphere, thepressure and density at 60,000 feet can be found.

PTROPOPAUSE

p60,000

= eg 60,000 − 36,100[ ]

RTS

P60,000 = 149 psf

and using the Ideal Gas Law

=P

RT= 0.000228 sl / ft3

The temperature is, of course, defined as constant in the stratosphere at the tropopausevalue of 216.5 degrees Kelvin.

It is a relatively simple matter to program the above equations in either BASIC orFORTRAN to readily calculate air properties anywhere in the troposphere or stratosphereand it is suggested that the student do this as a first step in building a library of computer

programs for elementary aircraft performance calculations. It is also convenient to tabulatestandard atmosphere conditions as is done in the following tables in both English and SI

units.

26

TABLE 1.1

PROPERTIES OF THE STANDARD ATMOSPHERE (ENGLISH UNITS)

h(ft)

T(˚F)

a(ft/sec)

p

lb/ft2( ) slugs/ft3( )× 107

(slugs/ ftsec)0 59.00 1117 2116.2 0.002378 3.719

1,000 57.44 1113 2040.9 .002310 3.6992,000 51.87 1109 1967.7 .002242 3.6793,000 48.31 1105 1896.7 .002177 3.6594.000 44.74 1102 1827.7 .002112 3.6395,000 41.18 1098 1760.8 .002049 3.6186,000 37.62 1094 1696.0 .001988 3.5987,000 34.05 1090 1633.0 .009128 3.5778,000 30.49 1086 1571.9 .001869 3.5579,000 26.92 1082 1512.9 .001812 3.536

10,000 23.36 1078 1455.4 .001756 3.51511,000 19.80 1074 1399.8 .001702 3.49512,000 16.23 1070 1345.9 .001649 3.47413,000 12.67 1066 1293.7 .001597 3.45314,000 9.10 1062 1243.2 .001546 3.43215,000 5.54 1058 1194.3 .001497 3.41116,000 1.98 1054 1147.0 .001448 3.39017,000 -1.59 1050 1101.1 .001401 3.36918,000 -5.15 1046 1056.9 .001355 3.34719,000 -8.72 1041 1014.0 .001311 3.32620,000 -12.28 1037 972.6 .001267 3.30521,000 -15.84 1033 932.5 .001225 3.28322,000 -19.41 1029 893.8 .001183 3.26223,000 -22.97 1025 856.4 .001143 3.24024,000 -26.54 1021 820.3 .001104 3.21825.000 -30.10 1017 785.3 .001066 3.19626,000 -33.66 1012 751.7 .001029 3.17427,000 -37.23 1008 719.2 .000993 3.15328,000 -40.79 1004 687.9 .000957 3.13029,000 -44.36 999 657.6 .000923 3.10830,000 -47.92 995 628.5 .000890 3.08631,000 -51.48 991 600.4 .000858 3.06432,000 -55.05 987 573.3 .000826 3.04133,000 -58.61 982 547.3 .000796 3.01934,000 -62.18 978 522.2 .000766 2.99735,000 -65.74 973 498.0 .000737 2.97440,000 -67.6 971 391.8 .0005857 2.96145,000 -67.6 971 308.0 .0004605 2.96150,000 -67.6 971 242.2 .0003622 2.96160,000 -67.6 971 150.9 .0002240 2.96170,000 -67.6 971 93.5 .0001389 2.96180,000 -67.6 971 58.0 .0000861 2.96190,000 -67.6 971 36.0 .0000535 2.961

100,000 -67.6 971 22.4 .0000331 2.961150,000 113.5 1174 3.003 .00000305 4.032200,000 159.4 1220 .6645 .00000062 4.277250,000 -8.2 1042 .1139 .00000015 3.333

Data taken from NACA TN 1428. Courtesy of the National Advisory Committee forAeronautics

27

TABLE 1.2

PROPERTIES OF THE STANDARD ATMOSPHERE (SI UNITS)

h

(km)

T

(˚ C)

a

(m / sec) p ×10− 4

N / m2( )(pascals)

kg / m3( )× 105

(kb / m sec)

0 15.0 340 10.132 1.226 1.7801 8.5 336 8.987 1.112 1.7492 2.0 332 7.948 1.007 1.7173 -4.5 329 7.010 0.909 1.6844 -11.0 325 6.163 0.820 1.6525 -17.5 320 5.400 0.737 1.6196 -24.0 316 4.717 0.660 1.5867 -30.5 312 4.104 0.589 1.5528 -37.0 308 3.558 0.526 1.5179 -43.5 304 3.073 0.467 1.48210 -50.0 299 2.642 0.413 1.44711 -56.5 295 2.261 0.364 1.41812 -56.5 295 1.932 0.311 1.41813 -56.5 295 1.650 0.265 1.41814 -56.5 295 1.409 0.227 1.41815 -56.5 295 1.203 0.194 1.41816 -56.5 295 1.027 0.163 1.41817 -56.5 295 0.785 0.141 1.41818 -56.5 295 0.749 0.121 1.41819 -56.5 295 0.640 0.103 1.41820 -56.5 295 0.546 0.088 1.41830 -56.5 295 0.117 0.019 1.41845 40.0 355 0.017 0.002 1.91260 70.8 372 0.003 3.9 ×10− 4 2.04775 -10.0 325 0.0006 8 ×10− 5 1.667

28

CHAPTER TWO

INTRODUCTION TO FLUID DYNAMICS

In this chapter the subject is the motion of a fluid and the changes in fluid propertiescaused by that motion. In order to calculate the performance of an aircraft one must havean understanding of the fluid forces which govern, in part, that performance, and to do thatone must have at least a fundamental understanding of the dynamics of the fluid itself.Numerous excellent fluid mechanics texts are available which will give the student a morethorough appreciation of this subject than is possible within the scope of this text; however,it is believed that the following coverage will be sufficient for a good understanding of thesubject of aircraft performance.

There are two fundamental approaches to the analysis of a fluid. The first approachstudies a fluid from the point of view of the individual fluid particle. This "fluid fixed"approach to the study of fluid dynamics is called the LAGRANGIAN approach and it isdesigned to describe directly the flow characteristics of individual particles. Usually,however, it is preferable to describe the behavior of a fluid through the overall flow patternthan by looking at individual particles.

The EULERIAN approach is a mathematical method of analysis which examines thetotal flowfield rather than the individual particle. This approach to fluid mechanics looks atfluid properties such as pressure, temperature and density as functions of time and spacethrough the fluid rather than looking at changes of these properties for a single particle as itmoves through the fluid. Properties of the fluid are described as functions of their spacialcoordinates and time. For example, the pressure in a flow field would be describedmathmatically as:

P = P (x,y,z,t).

In this study of fluid dynamics, as in most such studies, the Eulerian approach will beemployed. All fluid properties will be described as functions of x, y, z and t ( orappropriate cylindrical or spherical coordinates ) and the derivatives of the fluid propertiesmust include consideration of possible variation with respect to all of these. Thetemperature derivative would, for example, be written as follows:

dT =T

xdx +

T

ydy +

T

zdz +

T

tdt

A single particle can be followed in the Eulerian approach as well as in the Lagrangian.This is done by the use of the particle derivative or the "substantial derivative" of theproperty in concern. For example, to find the total change of temperature experienced by

29

a particle due to its motion over time dt, the substantial derivative DT / Dt is written asfollows:

DT

Dt=

T

x

dx

dt+

T

y

dy

dt+

T

z

dz

dt+

T

t

This equation recognizes that the changes in T with x and t, y and t, etc cannot be treatedindependently.

Knowing that three of the terms in the substantial derivative are velocities in the x, yand z directions,

dx dt = u, dy dt = v, dz dt = w

the substantial derivative can be written:

DT

Dt= u

T

x+ v

T

y+ w

T

z+

T

t

The first three terms on the right hand side of the above relation are called convective termsbecause they account for the change in the property in concern (in this case, temperature)due to its motion through the flowfield, while the last term accounts for the local rate ofchange with time.

The acceleration of a fluid particle in a flow field is one source of forces in a fluid. Particleacceleration is also written as a substantial derivative:

a =Dv

Dt=

v

t+ u

v

x+ v

v

y+ w

v

t,

where the velocity vector is

V = u ˆ i + v ˆ j + w ˆ k .

30

The x, y and z components of that acceleration are then written:

a x = ut

+ uux

+ vuy

+ wuz

a y =v

t+ u

v

x+ v

v

yw

v

z

az =w

t+ u

w

x+ v

w

y+ w

w

z

STEADY FLOW

In many elementary treatments of fluid mechanics it is convenient to simplify themathmatics by assuming steady flow. This means that the components of velocity are notfunctions of time; ie., at any time t the velocity of a particle in a flow field is dependent onlyon the position of that particle. This results in a simplification of the fluid flow equationsbecause the time derivatives of the velocity components are zero:

In most real world cases flows are actually unsteady; however, it is often possible toapproximate their behavior by assuming steady flow over short periods of time.

CONSERVATION OF MASS: THE CONTINUITY EQUATION

One of the most fundamental assumptions used in the development of the basicrelations governing the behavior of a fluid is that mass is conserved; ie., neither created ordestroyed within the fluid flow field under consideration. In order to develop amathmatical statement of this important principle, one need only examine the flow into andout of a simple fluid element in the flowfield.

Consider the cubical element of fluid shown below and assume that the fluid propertiesat the center of the element have values of p, , T. u, v, and w. Any of these properties ofthe fluid may change between the center and any of the element sides, for example the xdirected velocity has a value of u in the center of the element, a value of

u +u

x

dx

2

31

at the right face of the element and a value of

u −u

x

dx

2

at the left face.

Figure 2.1

The mass rate of flow of a fluid is given by the product of the fluid density, velocityand the area through which it flows:

mass rate of flow = ˙ m = AV .

Looking at the flow in the x direction, using the fluid density and velocity at the leftface of the element and the area of that face dy × dz( ) ), the mass rate of flow into theelement through the left face is:

32

−p

x

x

2

u −

u

x

dx

2

dydz

Similarly, the mass rate of flow out of the right face is:

−x

x

2

u −

u

x

dx

2

dydz

The net mass rate of flow in the positive x direction is therefore the difference betweenthese two flow rates, with the outflow being considered positive.

+p

x

dx

2

u +

u

x

d x

2

dydz − −

p

x

dx

2

u −

u

x

dx

2

dydz

Multiplying all terms as indicated gives:

1

2

u

xdxdydz( ) +

1

2u

p

xdxdydz( ) +

1

2u

p

xdxdydz( ) +

1

2u

p

xdxdydz( )

which reduces to: u −x

+u

x

dxdydz

Finally:x

u( )dxdydz

Similar mass flow terms are found by summing the flows in the y direction and in the zdirection.

33

yv( )dxdydz ,

zw( )dxdydz

The net mass flow balance into and out of the element must be balanced by any changein the mass within the element itself. The mass in the element is the product of the fluiddensity and the element volume and the change in that mass with time is:

tdxdydz( )

Equating this to the net mass flow through the element gives:

u( )x

+v( )

y+

w( )z

= −t

If the flow is steady and the fluid is incompressible (density is constant) the equationabove reduces to a simple form:

u

x+

v

y+

w

z= 0

These equations are known as the conservation of mass equations or continuity equations.The incompressible form of the equation is used repeatedly in elementary fluids coursesand students often loose sight of the fact that it is a special case with the compressible formof the equation being the most general.

In vector notation the incompressible continuity equation is expressed in terms of thedivergence if the velocity vector:

divV = 0

34

It is also often useful to know the continuity equation in cylindrical coordinates:

3 − D :ur

r+ ur

r+ 1

ru + w

z= 0

2 − D :ur

r+ 1

rUr + u

= 0

The continuity equation can be used to determine if a flow satisfies conservation ofmass. In fluid mechanics and aerodynamics mathmatics are used to model or describe flowfields. The student must realize; however, that it is possible to write equations for "flows"that are physically impossible because they do not satisfy conservation laws.

One way to write equations describing a flow is to develop relations for eachcomponent of velocity in terms of the coordinate system in which the flow exists. Suchrelations; ie., u = f(x,y,z), v = f(x,y,z) and w = f(x,y,z), should always be tested byapplying the continuity equation to see if mass is, indeed, conserved.

EXAMPLE

A two-dimensional flow field has been described by the following velocitycomponents:

u = 2 xy 2

v = 2 x2 y.

Is mass consemved in this flow?

If the continuity equation in two dimensions is satisfied then mass is conserved. Thepartial derivatives in the 2-D continuity equation become:

giving

ux

= 2y2 ,vy

= 2x2 ,

u

x+

v

y= 2y2 + 2x2 .

35

Since these terms do not add to zero mass is not conserved.

Note that if the X and Y components of velocity were reversed and the Y component ofvelocity above had been negative

u

x+

v

y= 4xy − 4yx = 0

mass would have been conserved.

Mass conservation is an important element in the mathmatical description of fluid flowsand the continuity equation is one of the fundamentals upon which theories ofaerodynamics and fluid dynamics are constructed.

FLUID DYNAMICS AND EULER'S EQUATIONS

The motion of a fluid can be described through Newton's Law, F = ma, whereacceleration refers to the acceleration of a particle or element of fluid. As mentioned earlier,in the Eulerian approach to the description of fluid motion, particle acceleration is given bythe substantial derivative:

a =DV

Dt

and, using the fluid density multiplied by the element or particle volume as the mass

F = ∆x ∆y∆ zDV

Dt

36

The substantial derivative of the velocity vector V has three components which may allbe functions of x, y, z and time. The u or x-directed velocity component of the substantialderivative is:

Du

Dt=

u

t+ u

u

x+ v

u

y+ w

u

z

The left hand side of F = ma must contain all the forces on the element of fluid. Thiscan include pressure forces, gravitational forces, friction forces, external forces, and evensuch things as electromagnetic forces for charged fluids. The following simpleexamination of the dynamics of a fluid element will only consider the pressure forces andthe gravitational forces. Friction forces will be ignored; ie. the fluid is assumed inviscid.Neglecting friction or viscous forces is valid as long as these forces are small compared tothe other forces. This is generally true outside of a normally thin region of flow next to thesurface of a body in a fluid. In this thin region, called the boundary layer, viscous forcesmust be considered and the equations of fluid motion become vastly more complicated.Texts on boundary layer theory are recommended for those wishing more discussion ofthese special but very important flow regions.

Again a simple cubical fluid element is used below to examine the forces in the fluid. Itshould be noted that it is popular in many fluid mechanics texts to use a more abstractshaped fluid element for derivation of these equations, with force vectors shown in genericnormal and tangential orientations to that element. While such "wounded amoeba"derivations may appeal to some, it is this author's experience that most students grasp thephysical reality of the mathmatics when a simple cubical element is used with force vectorsin the direction of the defined axes.

Figure 2.2

37

Pressure forces are shown acting on the left and right faces of the fluid element abovewhere the pressure on the faces may differ from that in the center of the element by theamount:

x

∆ x

2

Summing the pressure forces in the x direction gives:

P +P

x

∆ x

2

∆ y∆ z − P −

P

x

∆ x

2

∆ y∆ z = −

P

x∆ x ∆ y ∆ z

Similarly, in the y and z directions the sums of the pressure forces are:

−P

y∆ x∆y∆z, −

P

z∆ x∆ y∆z

Adding the pressure forces above gives the net pressure force on the element:

−P

x+

P

y+

P

z

∆ x∆ y∆ z = −grad P ∆ x∆ y∆z( ).

Gravitational acceleration also may represent a significant force on the element. This force,the weight of the element, is:

mg = g∆ x∆ y ∆z .

Adding this to the pressure force gives a total external force on the fluid element whichmust equal its mass multiplied by its acceleration.

g − grad P( )∆ x∆ y ∆z = mD V

Dt= ∆x ∆ y∆ z

D V

Dt

38

Dividing the volume of the fluid element out of the last relationship gives EULER'SEQUATION, which is normally written in one of the following three forms:

or

g − grad P = D V Dt

g −P

x+

P

y+

P

z

=

D V

Dt

g −1

grad P( ) =DV

Dt= a

Note that the simple looking equations above are vector relationships; ie., there is aseparate equation for each direction.

a x =u

t+ u

u

x+ v

u

y+ w

u

z= gx −

1 P

x

a y =v

t+ u

v

x+ v

v

y+ w

v

z= gy −

1 P

y

az = wt

+ u2x

+ vwy

+ 2wz

= gz − 1 Pz

These equations are really momentum equations.

EXAMPLE

Use the Euler equations to find the pressure at the bottom of a bucket of water which ison an elevator, accelerating downward at a rate of 10 feet per second-squared. The waterin the bucket is one foot deep.

The downward acceleration, az = 10 ft sec2

and the acceleration of gravity, also acting in the z direction is

g = 32.2 ft sec2

39

The pressure and the accelerations in this example all act in the z direction, thus, the az

relation is used.

az =w

t+ u

w

x+ v

w

y+ w

w

z= gz −

1 P

z

Since the term asked for in the problem is the pressure and the downward acceleration isknown, the solution need not deal with the velocity terms and only the following equationis needed:

or

az = −1 P

z− gz

−P

z= − az + gz

Entering the known quantities and the density of water:

−P

z= 1.94

sl

ft 3 10ft

Sec2 − 32.2ft

sec2

note: both az and gz act in negative z direction.

giving:

d P

dz=43

sl

ft 2 sec2

1 lb f

1sl + 1ft

sec2

40

or:

d P

dz= 43

lb f

ft3

Therefore, the water depth of one foot results in a pressure at the bottom of the bucket of

P = 43 lb f ft2 .

BERNOULLI'S EQUATION - CONSERVATION OF ENERGY

Euler's equation resulted from a force balance on a fluid element. As such it isessentially a momentum relationship since force is the rate of change of momentum.

A force acting over a distance produces work, a form of energy. That work isexpressed mathmatically as the dot product of the force vector and the distance vector.Therefore, a work or energy relationship will result if one takes the dot product of the Eulerequation and a distance over which the fluid element moves, ds.

g − grad P[ ]⋅ d s =Dv

Dt

⋅ d s

The result of this vector dot product is a scalar with each term being the product of thex, y or z components of Euler's equation with the like component of the distance vector:

gx − Px

dx + g y − P

y

d y + gz − P

z

dz

=Du

Dtdx +

Dv

Dtdy +

Dw

Dtdz.

41

This is divided by the fluid density to give:

Du

Dt+

1 P

x− gx

dx +

Dv

Dt+

1 P

y− gy

dy

+ DwDt

+ 1 Pz

− gz

dz = 0

Each of these three terms is similar. Considering only the x related term above andexpanding gives:

u

t+ u

u

x+ v

u

y+ w

u

z+

1 P

x− gx

dx

At this point an important assumption will be made, that of steady flow. Thiseliminates the time derivative term in the equation, making the equation simpler. Moreimportantly, it now means that any relationship developed after this point is only applicableto flows which are not time dependent. This assumption gives:

uu

xdx + v

u

ydx + w

u

zdx +

1 P

x− gx

dx

Another very basic assumption will now be made. It will be assumed that the flowunder consideration is on a STREAMLINE. A streamline is a path of the fluid flowalong which the velocity at any point is tangent to the streamline. By this definition thevelocity components along the streamline can be related to the directional derivatives of theflow:

v u = dy dx , w u = dz dx , w v = dz dy , etc.

Therefore:

vdx = udy, wdx = udz, etc.

It is therefore possible to rewrite the v and w terms in the x component of the dot productabove in terms of u:

42

vu

ydx = u

u

ydy, w

u

zdx = u

u

zdx

This gives:

uu

tdx + u

u

ydy + u

u

zdz

+

1 P

x− gx

dx

The first three terms in the above are simply udu. This part of the equation may thus bewritten:

udu +1 P

x− gx

dx

Similar terms will result for the y and z parts of the dot product. Combining theseterms into the full dot product gives a scalar equation for steady flow along astreamline:

udu + vdv + wdw( ) + 1 P

xdx + P

ydy + P

zdz

− gx dx + gy dy + gz dz[ ] = 0

The third group of terms above obviously deals with the effects of gravitationalacceleration. Gravitational acceleration is often expressed in terms of a gravitationalpotential where

g x =G

x, gy =

G

ygz =

G

z,

thus

g x dx + gy dy + gz dz = dG

43

The first group of terms can be rewritten as

udu + vdv + wdw = du2 + v2 + w2

2

and the second group as

1dP.

The full equation is now:

du2 + v2 + w2

2

+

1dP − dG = 0

This equation must now be integrated. In order to simplify that integration, the densityof the flow is assumed constant allowing the equation to be written as:

or

du2 + v2 + w2

2+

P+ G

= 0

dV2

2+ P − G

= 0

This is now an exact differential which integrates to:

V 2

2+

p− G = CONST .

44

This relationship states that the sum of the energies per unit mass of a fluid along astreamline in an incompressible, steady flow is constant. The first term in the equation isthe kinetic energy of the flow per unit mass, the second term is the internal energy per unitmass and the third term is the potential energy per unit mass.

Several more steps need to be taken to get this equation into the form usuallyrecognized as Bernoulli's equation. The first is to multiply by the fluid density, giving:

1

2V 2 + P − G = const .

Normally an axis system is chosen such that one axis is aligned with the direction ofgravitational acceleration. The selected axis is usually the z axis, thus:

gz = Gz = g

or

G = g z.

Integrating gives:

G = gz.

Bernoulli's equation is now written as:

1

2V 2 + P = const . + gz

It is worthwhile to pause at this point and examine the relative magnitudes of the termsin this equation. All the terms in the relationship will have units of pressure. The terms onthe left hand side of the equation are the DYNAMIC PRESSURE which results fromthe motion or kinetic energy of the flow and the STATIC PRESSURE or hydrostaticpressure of the fluid. At sea level conditions a velocity of 100 ft/sec would produce adynamic pressure of

45

1

2V 2 = 0.5 0.002378sl ft3( ) 100 ft sec( )2

or

1

2V 2 = 11.9psf ,

This is small compared to a sea level static pressure of

P = 2116 psf,

however the magnitude of the dynamic pressure term is sufficient to be very important inmost problems.

The gravity term on the right in the equation is, however, almost always quite small ifthe fluid is air. At sea level conditions in air this term becomes:

gz = 0.002378sl ft3( ) 32.2 ft sec2( )z

= 0.07657 psf × z

This is really a buoyancy term which says that at sea level a body displacing a one foothigh column of air will experience an upward buoyancy force of 0.07657 pounds persquare foot of surface area. This force, or more properly, pressure, is obviously quitenegligible when compared to either of the pressures on the left hand side of Bernoulli'sequation. For this reason it is common to neglect this term when using Bernoulli'sequation in air. Another approach which achieves the same result is to consider the termessentially constant and combine it with the constant of integration Either justification canbe used to obtain the following common form of BERNOULLI'S EQUATION for asteady, incompressible flow:

P +1

2v2 = P0

46

This is undoubtedly the single most important equation needed to understand the natureof a fluid flow. The student who understands the physics of this equations will have verylittle trouble in succeeding in studies of aerodynamics or aircraft performance. The studentwho, on the other hand, merely memorizes this relationship without understanding itsphysical consequences will have difficulty in these subjects.

Bernoulli's equation is an energy conservation statement. Its units arethose of pressure or energy per unit volume. The relationship can be thought of in twoways:

P + 1 2 V 2 = P0

STATIC PRESSURE + DYNAMIC PRESSURE = TOTAL PRESSURE

INTERNAL ENERGY + KINETIC ENERGY = TOTAL ENERGY

Bernoulli's equation can also be derived from the first law of thermodynamics(conservation of energy) by assuming an open system (control volume), steady flow andno heat transfer (adiabatic).

STEADY FLOW was assumed. This means that the relationship cannot be applied toa flow which has time dependent properties or behavior. There will be some exceptions tothis when flow changes are slow enough to consider the flow "quasi-steady".

INCOMPRESSIBLE FLOW was assumed, meaning that density was assumedconstant throughout the flow. A mathmatical analysis of this assumption would show thatit requires that the flow velocity in a gas must be lower than about one third the speed ofsound in the gas. In reality, for most problems encountered in aircraft performance andaerodynamics, the velocity can be as high as about 65 percent of the speed of sound, withthe limit being that free stream velocity at which the air speed over the wing has gonesupersonic and a local shock wave has formed. Liquids are almost always consideredincompressible at all speeds.

Compressible forms of Bernoulli's equation can be derived, either from Euler'sequation or from the first law of thermodynamics. Since there is little need to use these in acourse in elementary aircraft performance they are not derived here. A more advancedcourse in aircraft performance might require the use of such compressible relationshipswhich will undoubtedly be derived in any good aerodynamics text.

FLOW ALONG A STREAMLINE was assumed. The "constant" in Bernoulli'sequation, P0 may only be considered constant along the one streamline to which theequation is applied. P0 may be considered constant throughout the entire flow if it isknown that all streamlines emanate from a common source such as a quiescent atmosphereor a uniform flow. If the flow is known to have a velocity or pressure gradient,Bernoulli's equation must be used with care and only applied along streamlines for whichenough properties are known to define the total pressure.

47

NEGLIGIBLE BUOYANCY was assumed in neglecting the gravitationalacceleration term in the equation. It was seen that this is a good assumption for bodieswhich operate in the atmosphere and have a relatively small vertical displacement. Thereare vehicles, of course, which do not meet these restrictions. The obvious examples arelighter than air vehicles such as airships (blimps, derigibles) and baloons which depend onthis buoyancy force to remain aloft. A calculation of the magnitude of this term in waterwould reveal it to be significant due to the much higher density of water and the term mustalways be considered in liquids.

Several examples of the use of Bernoulli's equation are in order.

EXAMPLES:

1. An airfoil is moving through the air at a speed of 100 meters per second at analtitude of 1 km. Find the pressure on the leading edge of this airfoil.

Solution:

The pressure at the leading edge of the airfoil will be the stagnation pressure since theair will come to rest at that point.

The statement of the problem has the airfoil moving through stationary air. This is thesame as a stationary airfoil immersed in a uniform flowfield. The total pressure is the samethroughout the flowfield since it is a uniform field. The static pressure will change withinthe flowfield as the air accelerates or decelerates. In the freestream (at some distanceupstream of the airfoil) where the uniform flow velocity is 100 m/s the static pressure mustbe equal to the hydrostatic pressure in the atmosphere. This pressure is found, along withthe density of the air at that altitude, from the SI standard atmosphere table:

P = 8.987 ×104 pascals, = 1.112kg m3 .

The sketch below shows the flow streamline going from freestream conditions at infinity tothe stagnation point on the airfoil.

Figure 2.3

48

At the freestream location, Bernoulli's equation gives:

P0 = P∞ +1

2V∞

2 = 8.9870 pa +1.112

2

kg

m3 100m

s

2

= 95430 pa

Since this total pressure (stagnation pressure) is a constant all along the streamline, thestagnation pressure at the leading edge of the airfoil must also be 95430 Pascals.

2. For this same airfoil find the pressure over the airfoil at a point where the flowhas accelerated to 150 m/s.

Solution:

Since the same streamline examined above continues over the airfoil and the totalpressure along that streamline is constant, using that total pressure, the constant density andthe new velocity gives:

P = P0 −1

2V 2 = 95430 pa −

1.112

2

kg

m3 150m

s

2

= 82920 pa

VELOCITY MEASUREMENT: THE PITOT-STATIC TUBE

Bernoulli's equation provides an easy method for determining the speed of a fluid flow.Rearranging the equation to solve for the flow velocity gives:

V =2 P0 − P( )

In order to determine the speed, one need only find the difference in the total and staticpressures and the density of the fluid. It should, of course, be noted that, since theincompressible form of Bernoulli's equation is being used, the method of speedmeasurement being discussed is only valid for low speed flows in air.

The total or stagnation pressure in a flow is easy to measure if the flow direction isknown. One need only place a simple, open ended tube in the flow such that the tube isaligned with the local flow direction and the open end of the tube is pointing upstream. Theflow then stagnates or comes to rest at the tube's open end, giving a local pressure

49

equal to the total pressure. The tube can then be connected to any of a number of types ofpressure sensors to allow a determination of that pressure as shown below. This type oftube is called a pitot tube.

Figure 2.4. PITOT TUBE OPERATION

The static pressure is equally easy to measure using another type of tube with a closedend and pressure "taps" or openings around its circumference. This tube must also bealigned with the flow direction such that the flow is parallel to the pressure taps. Whenconnected to a sensor, the flow's static pressure will be indicated. It should be noted thatfor a static probe to obtain good readings the upstream end should be properly shaped toavoid flow separation and the pressure taps placed far enough from the end to avoid errorsthat might be produced by the local acceleration of the flow over the front of the probe.

Figure 2.5. STATIC TUBE OPERATION

Bernoulli's equation, as solved for the velocity, does not require the independentknowledge of the total and static pressures, but only a knowledge of their difference. Thismakes possible the use of a combination of the two types of instruments described above.This combined pressure probe is called a PITOT-STATIC TUBE.

50

The pitot-static tube is simply a static tube wrapped around a pitot tube as shown in thefigure below. The inner tube is connected to one side of a differential pressure sensor andthe outer tube, to the other side. The sensor used may be a simple U-tube manometer suchas those discussed in Chapter 1 where the pressure difference, P0 − P , is measured ininches of water.

Figure 2.6. PlTOT-STATIC TUBE

Many different styles of pitot-static probes exist with shapes and static port locationoptimized for particular applications. Static taps are usually placed at severalcircumferential locations around the probe, allowing accurate measurement of staticpressure even with slight probe/flow misalignment. The simple "Prandtl"-type probe,similar in shape to that shown in the above figure, is constructed such that with small andeven moderate flow/probe misalignment the errors in sensing of the two pressures tend tocounteract each other, allowing reasonably accurate pressure differential readings at anglesof ten to fifteen degrees from the flow direction.

Aircraft may also use pitot-static probes to measure speed but their probes are generallysturdier and incorporate heaters and drain systems for operation in rain and icingconditions. Most aircraft use separate pitot probes and static ports, still connected to acommon system to measure the pressure difference, with the pitot probe mounted on thewing or aircraft nose (away from the propellor airstream or jet intakes) and the static port ata carefully selected position on the side of the fuselage.

51

With the pitot-static tube to measure the difference in the total and static pressures, onlythe density is needed in order to calculate the flow velocity:

V =2 P0 − P( )

The density can be found using the ideal gas equation of state and measured values of staticpressure and temperature:

=P

RT

In aircraft performance work the velocity of interest to both the pilot and the engineer isthe sea level equivalent velocity which is defined as follows:

Vsl =2 P0 − P( )

sl

The use of sea level equivalent velocity allows the reference of all aircraft performanceparameters to a common atmospheric condition and eliminates the need to directly measurethe static pressure and temperature and calculate density in order to relate the performanceto a meaningful velocity. Fortunately, as will be seen in later chapters, most aircraftperformance capabilities such as rate of climb or take-off distance can be related directly tosea level standard conditions and the corresponding sea level equivalent velocity. Whenactual conditions at altitude are needed, the data is corrected using the ratio of actual airdensity to sea level standard density.

In most subsonic aircraft, the airspeed indicator actually shows the sea level equivalentvelocity rather than the actual velocity at altitude. To the pilot this is a better indication ofthe aircraft's actual performance under all operating conditions than the true velocity cangive. There are, of course, instrument calibration and other errors which must be includedin calculations when the precise airspeed is desired. Some of these will be discussed inlater chapters.

52

EXAMPLE:

The pitot-static tube on an aircraft measures a pressure differential of 300 psf. If theaircraft is flying at an altitude of 15,000 ft. Find the true airspeed and the sea levelequivalent airspeed.

Solution:

True airspeed: VT =

2 300 psf( )15,000

, 15,000 = 0.001497sl ft3

VT = 633 ft /sec

S. L. equivalent speed:

Vsl =2∆P

sl

=2 300 psf( )

0.002378sl / ft3 = 502 ft sec

Note: It is obvious that the true airspeed will always be greater than the sea level equivalentspeed unless the air density is higher than sea level standard density.

MOMENTUM THEOREM:

Conservation of mass and conservation of energy have been used to develop thecontinuity equation and Bernoulli's equation. These two equations are sufficient for thesolution of many basic problems in fluid mechanics; however, sometimes a thirdconservation law is useful. Euler's equation was developed earlier and it was pointed outthat it was a kind of momentum equation, but a more complete conservation of momentumrelation will prove more useful for many classes of problems.

As mentioned with Euler's equation a force is simply a rate of change of momentum:

F =dmv

dt=

d momentum( )dt

53

therefore, one way to determine forces developed on a body in a fluid flow is to find themomentum change in the fluid. An example is the determination of the drag on a wing bymeasurement of the change in momentum in the flow as it passes over the wing. The dragis equal to the momentum deficit in the flow downstream of the wing.

Figure 2.7

The thrust of a propellor or a jet engine can also be determined by finding the momentumadded to the flow.

Euler's equation can be used when a single particle or element of fluid is beingexamined; however when a large group of particles is involved a more general momentummethod is needed. Consider, then, a large group of particles located within the twodimensional area surrounded by the surface S in the figure below. Within a short time (astime goes from t to t1) this group of particles has moved to the right and is now surroundedby surface S1 at time + t1 .

Figure 2.8

During the time in which the defined group of particles has moved, individual particleswithin the group may have changed their relative positions or speeds, etc., such that thesurface around them may have changed shape as well as position. At time t1 the newsurface, S1 , surrounding the defined group of particles overlaps a portion of the areaoriginally enclosed by surface S. The areas within the original and final surfaces areshown crosshatched in different directions such that three regions, A, B and C, are definedas shown in the figure. These regions are defined as:

54

A = origional region minus overlap

B = overlapping region

C = new region minus overlap.

Consider now the momentum of the fluid in each of these three regions: MA , MB, MC ,where:

M A =A∫∫∫ V dA etc.

Note that this is a vector quantity.

At the original time t the fluid in the original region surrounded by S had a momentum:

M A t( ) + M B t( ),

and at the final time t1 the same particles now have momentum

M B t1( ) + M C t1( ).

The change in momentum of this group of fluid particles in the time observed must thenbe:

M B t1( ) + M C t1( ) − M A t( ) − M B t( )= MBt1 − M B t( )[ ]+ M C t1( ) − M A t( )[ ]

Dividing by the time t1 − t to get the rate of momentum change:

M B t1( ) − M B t( )t1 − t

+M C t1( ) − M A t( )

t1 − t1

The next step is to take the limit as t1 − t becomes small.In this limit S1 also approaches S.

55

In this limit the first term in the equation above represents the rate of change ofmomentum in the region which will now be called R which is enclosed by the surface S.Note that in the limit regions A and C go to zero. This term becomes:

d

dtV dR =

t∫∫∫R∫∫∫ V ( )dR

The second term represents, in the limit, the rate at which momentum leaves S minusthe rate at which it enters; ie., a momentum flux through S without outflow viewed aspositive. Defining the unit vector n as positive going out of S the second term in theequation becomes:

V V ⋅n ( )S∫∫ dS

The sum of these two terms must be equal to the force exerted on that defined group offluid particles during the time in question:

F =d

dtV dR+ V V ⋅ n( )dS .

S∫∫R∫∫∫

The force F may be the result of several things which are now defined as follows:

−Fe = force of body on fluid

− pn ds = forceof pressureon control surfaceS∫∫

− g dR = gravitational forces on body.R∫∫∫

56

Finally, the following relationship can be written:

d

dt R∫∫∫ V dR+ V V ⋅ ˆ n ( )S∫∫ dS

momentum changein fluid1 2 4 4 4 4 4 3 4 4 4 4 4

= − Fe − pS∫∫ ˆ n ds +

R∫∫∫ g dR

Forcescausin g momentumchange1 2 4 4 4 4 4 3 4 4 4 4 4

This relationship appears, at first, quite intimidating to most students, with all itsdouble and triple integrals over areas and surfaces. It can be said; however, that there aremany problems where the application of this theorem is fairly simple. The important thingabout the momentum theorem is that it can be used in cases where all that is known are theproperties of the flow at a system's boundaries and there is a desire to find out what ishappening within that system.

It is often useful to consider the individual x, y and z components of these forces withina cartesian coordinate system and gravitational forces are often negligible since their changewith time is usually small. The equation can now be written as three equations.

ddt R∫∫∫ udR + u V ⋅ ˆ n ( )

S∫∫ ds = −Fx − pS∫∫ cos ˆ n , ˆ i ( )dS

d

dt R∫∫∫ vdR + v V ⋅ ˆ n ( )S∫∫ dS = −Fy − p

S∫∫ cos ˆ n , ˆ j ( )dS

d

dt R∫∫∫ wdR + w V ⋅ ˆ n ( )S∫∫ dS = −Fz − p

S∫∫ cos ˆ n , ˆ k ( )dS

In each of the above equations is a term such as cos (n,i). This term is the cosinebetween the normal unit vector and the x, y, or z direction unit vector.

At this point it will be helpful to look at a few practical examples of the use of themomentum theorem.

As mentioned earlier, one use of the momentum theorem is in finding the drag on abody from the momentum deficit in the flow. Consider the flow control volume below:

Figure 2.9

57

The control volume is bounded by streamlines far enough from the body that the velocityon those streamlines is undisturbed and the static pressure is constant. The end stations arealso far enough away that the pressure is constant over them. Note that the height of thedownstream surface is not necessarily the same as that upstream.

By making the boundaries large enough to have an undisturbed pressure around thoseboundaries the pressure integral becomes zero:

p ˆ n ds = 0S∫∫

Thus, the pressure force on the control volume is zero. Steady flow is also assumed.

d

dt= 0

Since the velocities at the boundaries are all in the x direction, only the single x equationis needed.

F e = − V V ⋅n( )S∫∫ ds

= −ˆ i V22∫ V2dy2( ) + ˆ i V1

1∫ V1dy1( )

The force resulting is then the x directed force on the body; ie., its DRAG.

In solving, mass conservation must also be considered to account for the fact that a"stream tube" going between an element dy at the upstream boundary may become larger orsmaller before it reaches the downstream boundary to form a downstream dy. (A streamtube is a "tube" of flow in which the mass rate of flow is constant. The mass coming intothe upstream boundary must equal that leaving downstream or the momentum carried byany mass entering or leaving through the sides of the volume must be included in thecalculations)

˙ m = V1dy1 = V2 dy2

58

The momentum equation now becomes:

D = − V2

2∫ dy2 + V11∫ V2dy2( )

or simply:

D = V22∫ V1 − V2( )dy2

This approach is used in wind tunnels to measure drag without using a force balance.In this derivation only linear momentum has been considered. If high angles of attack areinvolved or there are other sources of rotational momentum change are present, morecomplicated relations are needed.

EXAMPLE

Another common example of the application of the momentum theorem is for thecalculation of the force exerted on a pipe by flow passing through a bend in the pipe. Thefollowing figure shows a flow entering a pipe bend at station one and exiting at station twowith possible changes in area, density and pressure between the two stations.

Figure 2.10

In this problem the direction of the resulting force is unknown and must be found from thevector form of the solution. For solution a control surface is indicated by the dashed line.Writing the general momentum equation and assuming steady flow and neglectinggravitational forces gives:

V V ⋅n ( )S∫∫ ds = − F e − pn ds

S∫∫

59

The pressure integral becomes:

pn ds = ˆ i P1 − Pa( )S∫∫ A1 + ˆ j P2 − Pa( )A2

where Pa is the ambient pressure acting around the entire control volume.

The momentum flux term is:

V V ⋅ n ( )S∫∫ ds = − 2

ˆ j V2 − ˆ j V2( ). − ˆ j ( )[ ]A2 + 1ˆ i V1

ˆ i V1( ). − ˆ i ( )[ ]A1

= − 2V22 A2

ˆ j − 1V12 A1

ˆ i

The momentum equation now becomes:

− 2V22 A2

ˆ j − 1V12 A1

ˆ i =− F e + ˆ i P1 − Pa( )A1 + ˆ j P2 − Pa( )A2

and the force has both x and y components:

F e = ˆ i 1 A1V12 + P1 − Pa( )A1[ ] + ˆ j 2 A2V2

2 + P2 − Pa( )A2[ ]

Now, since mass is conserved,

˙ m = 1V1A1 = 2V2 A2

Thus,

F e = ˆ i ˙ m V1 + P1 − Pa( )A1[ ]+ ˆ j ˙ m V2 + P2 − Pa( )A2[ ]

60

This allows the force to be determined from a knowledge of the mass rate of flow in thepipe, the velocities going into and out of the bend, the pressures into and out of the bendand the external pressure.

EXAMPLE

As a final example of applications of the momentum theorem, consider flow throughthe set of blades or vanes shown in the figure below. These could represent compressor orturbine blades in a jet engine or turning vanes at the corner of a wind tunnel.

Figure 2.11

The control surface is made up of streamlines AB and CD and the end surfaces AC andBD. Steady flow will be assumed as well as uniform flow over AC and BD. This lastassumption says that any momentum loss is due to turning of the flow and not to any dragof the vanes ( not any worse than the frictionless bearings and massless points assumed inother texts! ).

Conservation of mass gives:

˙ m AC = u1h = ˙ m BD = U2h, ThusU1 = U2

Neglecting gravity again, the resulting momentum equation is:

V V ⋅n ( )S∫∫ ds = Fe − pn ds

S∫∫

where

V V ⋅ n ( )ds = ˆ i 0( ) + ˆ j U2h v2 − v1( ) since U1 = U2

61

The pressure term becomes:

pn ds = ˆ i P2 − P1( )S∫∫ h

In this example it is convenient to use Bernoulli's equation to eliminate the pressureterms from the momentum equation.

P1 +1

2V1

2 = P2 +1

2V2

2

V22 = u2

2 + v22, V12

= u12 + v1

2

V22 − V1 2

= u22 − u1

2

01 2 3 + v2

2 − v12 or P1 − P2 = 1

2v2

2 − v12( )

Finally

F e = ˆ i 1

2v2

2 − v12( )h

+

ˆ j ˙ m v1 − v2( )

or

Fx =1

2h v2

2 − v12( ) = p1 − p2( )h

Fy = ˙ m v1 − v2( )

62

CHAPTER 3AIRFOIL AERODYNAMICS

3.1 FORCES AND MOMENTS

It is conventional to separate aerodynamic forces and moments into three forcecomponents (lift, drag, sideforce) and three moments (pitch, yaw, roll). Thesecomponents may be defined relative to the wind direction (wind axis system) or relative tothe vehicle centerline (body axis system) or a combination of the two. One must be carefulin the computation or use of force and moment data to use the proper axis system and to beconsistent in its use. The most commonly used system is the wind axis system where theforces are defined either along the free stream velocity vector or perpendicular to it asshown in Figure 3.1.

Figure 3.1. Wind Axis System

To many people this axis system appears inverted and somewhat unnatural. It was chosenprimarily because it is a standard right hand system. It is often more intuitive to invert partof the system to make the z axis point "ups and the x axis go with the wind; however, inthat arrangement the moments do not follow the right hand rule. Either system can be usedif one is careful in its use.

It is important to note that the axis system is aligned with the wind, rather than thehorizon or the vehicle axis. This is an easy source of confusion since it is common tovisualize the wind vector concurrent with the horizon or along the aircraft axis. Indeed, ina straight and level flight situation for an aircraft the free stream wind vector might coincidewith the vehicle axis and the horizon; however it is best not to think in terms of that specialcase. Some people prefer to think of the x axis shown in Figure 3.1 as lying along the pathof flight of the vehicle. Figure 3.2 illustrates the problem by showing a typical glidesituation for an aircraft.

63

Figure 3.2. Aircraft in Glide

The three orthogonal forces are lift, L, drag, D, and side force, Y. Lift is defined as theforce along the negative z axis (normally "upward") and acting perpendicular to the freestream direction. Note that this is not necessarily upward with respect to the aircraft axis orthe horizon as indicated in Figure 3.2.

Drag is defined as the force in the direction of the relative wind or along the negative xaxis. Drag can always be thought of as the force which resists the motion of the vehicle.

Side force, given the symbol Y, is defined as mutually perpendicular to both lift anddrag and is positive outhe the right hand or starboard side of the vehicle.

The three moments, pitch (M), roll (LR) and yaw (N) are the moments which tend toresult in a rotation of the vehicle about the y, x or z axes respectively.

Pitching moment, M, is by far the most widely discussed of the three moments since itmust be considered in two dimensional (x, z plane) problems as well as in 3-D cases.Pitching moment is defined as positive when it tends to raise the nose of the vehicle. It isthis basic definition of the sense of pitch that requires the use of an "inverted" (zdownward) axis system in order to have a right had coordinate system. The pitchingmoment acts about the positive y axis.

The rolling moment, LR, is the moment which causes rotation about the x axis orcauses an aircraft to roll one wing up and the other down. Note that this definition of rollmay not coincide with that of an airplane pilot who thinks of roll as occurring about theplane's body axis rather than the wind axis.

Yawing moment, N. rotates the vehicle around the lift direction and is defined aspositive when it is clockwise or results in a nose right motion. Again the use of a bodyaxis rather than a wind axis may result in a different value for yawing moment.

64

3.2 DIMENSIONAL ANALYSIS AND NON-DIMENSIONALCOEFFICIENTS

It is convenient in engineering work to deal with non-dimensional terms or unitlessnumbers rather than everyday dimensional terms. The resulting non-dimensionalparameters not only remain unchanged from one unit system to another but they are usuallymore meaningful in terms of the physics of a problem than conventional dimensionalnumbers.

It is possible to develop the concept of a non-dimensional force coefficient and toexamine the important physical parameter groupings on which aerodynamic forces dependby using a simple process known as dimensional analysis. Dimensional analysis, as usedhere, is simply a process of first identifying the parameters on which fluid forces dependand then grouping these parameters in such a way that the units or dimensions balance.Assume then that it is know that the forces one body in a fluid depend on the following:

1. The properties of the fluid itself, pressure, P, density, , viscosity, , and thespeed of sound (fluid's elastic properties), a. Note that one need not includetemperature since P and are considered.

2. The speed of the body relative to the fluid, V.

3. The acceleration of gravity, g.

4. The characteristic size or dimension of the body, l, or the distance of a bodyfrom a fluid boundary, also designated l.

Hence it can be said that the force on a body in a fluid is a function of all of the above,

F = f( ,V,l, ,g,P,a)

[3.1]

or to be completely general, the force is a function of each variable to some power,

F = f( A,VB ,lC , D ,g E ,P G,aH )

[3.2]

Since the left hand side of this equation, [3.1] has the dimensions of force, the righthand side must also have the same dimensions. Force has units of mass multiplied byacceleration (kilograms meters/seconds2 or slugs feet/seconds2). To be general then it canbe said that force has dimensions of (mass)x(length)/(time)2 and letting M, L, and Trepresent these physical dependencies one can write the dimensions for force as MLT-2.Likewise, one can write the dimensions for all the other terms in equation 3.2 in terms ofmass, length, and time:

65

Parameter Dimensionvelocity (V) LT-1

length (L) Ldensity ( ) ML-3

viscosity ( ) ML-1T-1

pressure (P) ML-1T-2

gravitationalacceleration (g) LT-2

speed of sound (a) LT-1

force (F) MLT-2

Now, substituting these into equation [3.2] the result is a dimensional equation of theform:

MLT2 = ML−3( )ALT −1( )B

L( )CML− 1T− 1( )D

LT −2( )EML− 1T− 2( )G

LT −1( )H

[3.3]

The task is now to balance the above equation dimensionally; i.e.., the sum of the massexponents on the left side of the equation must equal those on the right, etc. Equatingexponents of mass, length and time respectively leads to three equations:

(Mass) 1 = A + D + G

(Length) 1 = -3A + B + C - D + E - G + H

(Time) 2 = B + D + 2E + 2G + H

[3.4]

Now, equations [Figure 3.4] give a set of three equations and seven unkowns. Theseequations may be solved for the values of any three of the unknowns in terms of theremaining four. Solving then for A, B, and C in terms of the remaining terms gives:

A = 1 - D - G

B = 2 - D - 2E - 2G - H

C = 2 - D + E

[3.5]

Substituting these solutions [Figure 3.2.5] into the original relationship [3.2.2] andgrouping all terms of like exponents gives:

F = V 2l2 Vl

− Dgl

V 2

EP

V 2

GVa

− H

[3.6]

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It is noted that each of the terms with an unknown exponent is a dimensionless term;

i.e., the term

Vl is unitless as are

gl

V2 , p

V 2 and V a . If the equation is divided

by V2 l2

, both side become unitless.

F V2l2 = f Vl

glV 2

P

V2

Va

[3.7]

Two conclusions can be drawn from equation [3.7]. The first is that the proper way tonondimensionalize a fluid force is to divide it by the fluid density, the square of the velocityand the square of the characteristic dimension of the body or the body's representative area.

Since the term 1

2V 2 represents the dynamic pressure of the fluid as found in Bernoulli's

equation, a factor of 1

2 is introduced into the relationship and a "force coefficient" is

defined as

CF = F1

2 V 2S(unitless)

[3.8]

where S is the representative area of the body. In two dimensional problems acharacteristic length (2-D area) is used in the denominator instead of S.

The second conclusion is that the nondimensional fluid force is dependent on thegroups of parameters on the right of equation [3.7]. These groups are known as "similarityparameters" and are important in relating nondimensional force coefficients found on onebody in a fluid to those on a geometrically similar body of different size. Technically,equation [3.7] says that for force coefficients on two geometrically similar bodies to beequal, each of the grouped terms or "similarity parameters" must be identical for the flowsaround the two bodies. It is obvious that it would be quite a task to make all of thesesimilarity parameters equal for tests on bodies of two different sizes or in different fluids.Fortunately, it is seldom necessary to match all four of these similarity parameters at thesame time as an examination of the meaning of each term will show.

The first term on the right of equation [3.7] is known as Reynolds number, Re.

Re = Vl

[3.9]

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Reynolds number is a parameter relating inertial effects in a fluid to viscous effects. This isan important parameter for flow similarity because it is found that laminar-turbulenttransition in a boundary layer is a function of Re and a body at low Re values may have asignificantly different behavior from one at high Reynolds number. The classic example ofReynolds number effects is found in the flow around a sphere or cylinder, where at low Reboundary layer separation occurs early, resulting in a high wake drag and at high Reseparation is delayed by turbulence and the wake drag is reduced.

Reynolds number is an important similarity parameter which must be considered inevery flow. However, there are some cases where it may be ignored. These are generallywhere flow separation occurs at a sharp corner on a body and the separation point will notbe influenced by the laminar or turbulent character of the flow. For this reason Re may notbe a factor when considering flows around some non-streamlined shapes. However, thesesituations are rare and Reynolds number is almost always the most important factor inconsiderations of flow scaling and similarity.

The second most important similarity parameter in most aerospace problems is Machnumber, M, where

M = Va

[3.10]

This parameter indicates the relevance of compressibility effects. Compressibility effectsoccur due to elastic compression and expansion of a fluid as it passes over a body. Theseare important only when compression or mach waves begin to form in a fluid as it flowsaround a body. Hence, Mach number similarity need not be considered at speeds givingMach numbers less than 0.5 or so. In water Mach number need not be considered sincewater is an incompressible fluid.

In some aerospace problems both Reynolds number and Mach number are importantbut it is impossible to satisfy both types of similarity at one time. Here, tests are usuallydone to examine separately the effects of each parameter and the resulting scaling of datamust be done using engineering judgement based on past experience and a thoroughunderstanding of the problem at hand.

The remaining two parameter groupings on the right side of equation [3.7] areencountered primarily in the fields of naval architecture or ocean engineering. The first ofthese is the inverse square root of a widely used similarity parameter known as Froudenumber (F), where

F = V gL

[3.11]

Froude number is the ratio of inertial forces to gravitational forces and it is essentially ameasure of the importance of the effects of a fluid boundary or interface on the forces onthe body. Here the term L refers to a distance which may be the distance of the vehicleabove or below the ground or air-water interface or the height of waves generated by a

68

ship. The importance of Froude number is perhaps most easily understood whenconsidering the motion of a submarine below the surface. When a submarine is sufficientlyfar below the surface it can move without disturbing the surface; however, if it is close tothe surface, waves are generated. The energy present in these waves represents an energyloss by the submarine and consequently must be treated as part of the vehicle's drag. Inlike manner any vehicle moving over, under or through the air-sea interface which causessuch surface waves develops a wave drag and for proper simulation of this drag in testingFroude number for aerodynamic similarity, hydrodynamic problems often involve a needfor both Re and F similarity at the same time, a condition which may not be easilyachievable.

The last of the four similarity parameters developed in equation [3-7] is Euler number,

Euler number = PV 2

[3.12]

which is a measure of the ratio of inertial forces and pressure forces. This term assumesimportance when cavitation is a problem on ship hulls or propellers. Basically, cavitationoccurs when the pressures caused by motion of water around a body become low enoughto result in boiling of the water. Cavitation can result in loss of lift on hydrofoils, loss ofthrust on propellers, and high drag on hulls and is thus a very important phenomenon.Proper scaling of flows where cavitation may occur therefore nessitates the use of Eulernumber to insure similarity. In practice Ocean Engineers define a slightly different numbercalled the cavitation number, , where

= P − Pv

V 2

and Pv is the vapor pressure of the water. This is used in place of Euler number as thecavitation similarity parameter.

Example 3.1. An aircraft is designed to fly at 250 mph at an altitude of 25,000 feetwhere the pressure, temperature and density are standard. We wish to test a one-tenth scalemodel of this plane in a wind tunnel at sea level standard conditions. What problems mightwe have in achieving flow similarity?

The Reynolds number for the full scale aircraft would be

Re f = Vlf

and, at standard conditions for 25,000 feet and 250 mph we have

= 0.001066sl ft3

= 3.196x10-7 sl ft

V = 250mph = 367.5 fps

69

This gives a Reynolds Number per foot of

Ref lf = 1.226x10 6 ft -1

Note: It is common practice in wind tunnel testing to speak of Reynolds Numberper foot.

To achieve this Reynolds Number on a one-tenth scale model in the wind tunnel atsea level conditions we need to make

Re f = Rem = SLVm1m SL = Re f 1 f = 1.223 ×1061b

Thus, the speed in the wind tunnel test section must be

Vm = 1.226x106 l f SL( ) SLlm

Using SL = 0.002376 sl ft3 , SL = 3.719 x 10−7 sl ftsec and l f lm = 10, weget:

Vm = 1919ft sec = 1305mph

But this supersonic speed quite obviously violates the Mach Number similarityrequirement! Does this mean that it is impossible to properly test for full scaleaerodynamics effects by using small scale models in a wind tunnel? Fortunately not.

One solution is to use a sealed, variable density wind tunnel. Most of the earlywing aerodynamics tests of the National Advisory Committee for Aeronautics ( NACA ),NASA's predecessor, were done in such a tunnel at Langley Field using 1/20th scalemodels tested in a tunnel pressurized to 20 atmospheres, giving a density twenty timesnormal and matching full scale Reynolds Numbers.

Another commonly used method, provided that the model scale is not too small, isto "fool" the flow into behaving like it would at a higher Reynolds Number. The primaryflow influence of Reynolds Number on a streamlined shape is to determine where on theshape the flow in the thin "boundary layer" next to the body changes from a smooth"laminar"" behavior to a turbulent behavior. This, in turn, partly determines where flowseparation from the surface might occur. We can "fool" the flow by forcing or tripping"boundary layer transition from laminar to turbulent flow by using a "trip strip" which canbe as simple as a line of fine sand grains glued to the surface.

70

3.3 FORCE AND MOMENT COEFFICIENTS

In the preceeding section it was shown that thenondimensional force coefficient tookthe form

CF = Force12

V 2S

Hence, force coefficients can be defined for the aerodynamic forces of lift, drag andsideforce:

lift coefficient, CL = lift1

2V 2S

drag coefficient, CD = drag12

V 2S

side force coefficient, C Y = side force12

V 2S

As with all force coefficients these are unitless (nondimensional) and therefore are invariantfrom one unit system to another.

One problem which appears when using force coefficients is the area, S, to be used inthe denominator. This is merely a representative area which is characteristic of the body orcase being considered. For example, when using lift coefficient the area S used is almostalways the planform or projected area of the wing. This is logical since the wing is the liftproducing element of the vehicle. The best area to use for drag coefficient is not as obviousand any one of several areas will be found in common usage; wing planform area whenconsidering drag wings, fuselage or hull cross sectional when considering such shapes, or"wetted" (total surface) area when skin friction drag is being treated. Similar variations willbe found when defining characteristic areas for side force coefficients. Hence, it isimportant that one be extremely careful in interpreting data in coefficient form sincethemagnitude of the resulting coefficient may vary greatly depending on the choice of arepresentative area. For the same reason extreme care must be exercised in combining

71

coefficients. The drag coefficients of various parts of a vehicle cannot merely be added toget a total drag coefficient unless all the coefficients are based on a common area

Force coefficients can also be similarly defined in two dimensions where only a twodimensional section of a body such as an airfoil section is under consideration. In thesecases only lift and drag are considered and the forces are two dimensional forces; that is,they are expressed in units of force-per-unit-length such as force-per-foot or meter of wingspan. Therefore, to obtain a nondimensional coefficient only a characteristic length isneeded in the denominator.

CL2− D= L

1

2V 21

CD2 − D= D

12

V 21

3.13

The length commonly used for two dimensional lift and drag coefficients is the wing chord(distance from leading-to-trailing edge) when dealing with airfoils.

Moment coefficients are similarly defined except that an additional length factor isneeded in the coefficient denominator. The length most commonly used is the wing's meanchord c or the length of the body.

CM = M12

V 2S c

3.14

In like manner coefficients for rolling and yawing moment may be defined with appropriatecharacteristic lengths.

In two dimensions the pitching moment coefficient becomes

CM2 −D= M

12

V 2c2

3.15

72

3.4 AIRFOIL GEOMETRY

In order to discuss airfoil aerodynamics it is necessary to have a grasp of theterminology commonly used to define the geometry of a wing. This terminology will bediscussed in reference to Figure 3.3 which shows the "planform" view of a wing andFigure 3.4 which shows a typical airfoil section.

Figure 3.3. Wing Planform Geometry

Figure 3.4. Wing Section Geometry

73

The wing span, b, is the tip-to-tip dimension perpendicular to the fuselage centerlineindlucing the width of the fuselage.

Several chords may be defined for a 3-D wing. The root chord, Co , is the distancefrom the airfoil leading edge to trailing edge taken parallel to the fuselage centerline at hewing-fuselage junction. The tip chord CT is the same dimension taken at the wing tip.Two

different definitions of a mean or average chord are in common usage. The mean chord cis defined as

c = ∫ ob 2 cdy∫ o

b 2 dy

[3.16]

where y is measured from fuselage centerline along the span to the wing tip. Also definedis the "aerodynamic mean chord", cA ,

c A = ∫ob 2 c2dy∫o

b 2 cdy

[3.17]

Wing area, S, is usually thought of as the planform area rather than the actual surfacearea. The planform area includes the imaginary area between the wings in the fuselage.When this fuselage area is to be ignored, ie., the planform area of the exposed wing only isto be used, this is referred to as the net wing area SN .

A term which is of some importance in airfoil fluid dynamics is the "aspect ratio"AR . Aspect ratio is a measure of the "narrowness" of the wing planform and is the ratio ofthe wing span to the mean chord

AR = b c

[3.18a]

which is often written as

AR = b2

bc= b2

S

[3.18b]

a form which is easier to use. Aspect ratio will be found to be a measure of the 3-Defficiency of a wing.

74

While some airfoils have a rectangular planform, most are swept and/or tapered to somedegree. Wing sweep can be measured at the leading or trailing edge or at some other pointsuch as a line along the quarter chord (C/4 behind the leading edge) and the sweep angle isgiven the symbol Λ . Taper is defined in terms of a taper ratio where = C r Co . Thewing angle of attack ( ) shown in Fig. 3.4 is defined as the angle between the chord lineand the fluid velocity vector.

Wing camber is usually expressed as percent camber of an airfoil section where percentcamber is the percentage of the airfoil chord represented by the maximum perpendiculardistance between the chord and camber lines. Wing thickness is also defined in terms ofpercent chord.

Figure 3.5 illustrates wing dihedral which is defined as a dihdral angle due to theinclination of the wings from a common plane. Dihedral is built into wings for rollstabaility purposes and the dihedral angle is defined as 2 , the sum of the inclination anglesof both wings. A negative dihedral is called "anhedral".

Figure 3.5. Wing Dihedral

75

3.5 NACA AIRFOIL DESIGNATIONS

In the early days of wing development wing shapes were often given names andnumbers in a very nonsystematic manner. Sections were names after researchers such asClark or Eiffel or after laboratories or research groups (RAF, Gottingen) and each groupdeveloped its own series of airfoil shapes. In an attempt to systematize airfoil research anddesignation, the National Advisary Committee for Aeronautics or NACA (NASA'sforerunner), devised a systematic scheme of testing and classifying airfoil sections.Hundreds of airfoil sections were thoroughly tested and catalogues. These airfoils weredesignated by four, five, and six digit numbers according to basic shape and performance.The data from these tests is reported in numerous NACA publications and some of it willbe referenced in later sections. To properly use this data one should understand themeaning of the NACA airfoil designations.

Most of the NACA airfoils fall into the four, five or six digit airfoil series as explainedin examples below.

NACA 24122 - The maximum camber of the mean line is 0.02c4 - the position maximum camber is 0.4c12 - the maximum thickness is 0.12c

NACA 230212 - the maximum camber of the mean line is approximately 0.02c (also the

design lift coefficient is 0.15 times the first digit for this series)30 - the position of the maximum camber is at 0.30/2 = 0.15c21- the maximum thickness is 0.21c

NACA 632 -215 (laminar flow series)6 - series designation3 - the maximum pressure is at 0.3c2 - the drag coefficient is near its minimum value over of lift coefficients of

0.2 above and below the design CL2 - the design lift coefficient is 0.215 - the maximum thickness is 0.15c

There are other series of airfoil sections besides the ones given above, however, theseare most common. There is a new class of airfoil which will be discussed later that is beingdeveloped out of the "super critical" design of wing. NASA has now began tosystematically number these shapes in categories of low, medium and high speed airfoilswith shapes given designations such as LS(1)-0417 and MS (1)-0313 etc. Hence, theprocess of systematically defining and designating airfoil shapes still continues.

76

3 .6 . PITCHING MOMENT AND ITS TRANSFER

Before a discussion of airfoil characteristics can be meaningful a better understandingof pitching moment is helpful. Since most of the later discussion of airfoils will centeraround the dimensional case the only moment of concern will be the pitching moment. It isobvious that pitching moment can be defined as acting about any chosen point on theairfoil, however, its value will vary depending on the point of definition. It is thereforeconvenient to choose some sort of standard reference points for definition of pitchingmoment which will be meaningful physically. One then needs to be able to transfer apitching moment which has been measured or calculated at a given point on the airfoil toone of the chosen reference points or any other location.

In order to transfer the pitching moment all forces and moments acting on the airfoilmust be known. In two dimensions this means that one must know the lift, drag andpitching moment at some point. Suppose, for example, that wing has been mounted in awind tunnel and the lift, drag and pitching moment measured at the mounting point (a), adistance, a, behind the airfoil's leading edge, as shown in Figure 3.6 (a) and that one needsto know the pitching moment about a different point x where a structural member is to belocated.

Figure 3.6. Transfer of Pitching Moment

The forces are invariant in the transfer, however, their effects on the moment must beconsidered.

La = Lx

Da = Dx

77

To examine the effect of the moment transfer take the moments in each case relative to acommon reference point, the leading edge. In case a, the lift and drag produce momentsaround the leading edge of

a( ) Lcos( )and a( ) Dsin( )

Since both moments are counterclockwise they are considered negative.

Thus, the entire moment about the leading edge is

M LE = Ma - Lacos - Dasin

Likewise for wing b

MLE = M x -Lxcos -Dxsin

Since these two moments must be identical, equating [3.19] and [3.20] gives

M x = Ma - Lcos = Dsin( ) a - x( )

converting to coefficient form by dividing by 1

2V 2c2 gives

CMx = CMa − CL cos + CD sin( ) ac

− xc

As mentioned previously, there are some cases of special interest regarding placementof pitching moment. One such case is the point where the pitching moment becomes zeroand another is the point where the moment becomes a constant over a wide range of angleof attack or lift. Both of these may be important both aerodynamically and structurally inthe design of a wing. These two points are called the center of pressure and theaerodynamic center, respectively.

The aerodynamic center is probably the most commonly used reference point forforces and moment on an airfoil. If one were to measure the forces and pitching moment onan airfoil over a wide range of angles of attack or lift coefficient and then for each value ofCL look at the values of CM at various chordwise positions on the wing, one special pointwould be found where CM was virtually constant for all values of CL This point is theaerodynamic center.

78

The aerodynamic center is defined as the point along the chord where the pitchingmoment coefficient is constant and independent of the lift coefficient. Note that CM isconstant at the aerodynamic center and not necessarily zero. There are some limits to thisdefinition since at high angles of attack where CL does not change linearly with theaerodynamic center may shift; however, the concept of an aerodynamic center is veryuseful and valid over the normal operating range of CL 's for an airfoil.

It is therefore useful to find an equation which will locate the aerodynamic center oncethe forces and pitching moment about some point on the airfoil have been found.Returning to equation [3.22] and assuming that the unknown position x is the aerodynamiccenter

CMxac= CMa − CL cos + CD sin( ) a

c− xac

c

[3.23]

Several assumptions can be used to simplify this equation based on the alreadymentioned fact that the definition for the aerodynamic center is only meaningful formoderate angles of attack. At these angles cos is approximately ten times the magnitudeof sin ,

cos ≈ 10sin

Also at moderate one normally finds that CL is about twenty times the magnitude ofCD ,

CL ≈ 20CD

therefore,

CLcos ≈ 200C Dsin

and the latter term can be neglected. Using this and assuming Cos is approximatelyunity gives

CMxac= CMa

− CLac

− xac

c

[3.24]

Now the definition for aerodynamic center can be utilized. This definition states that atthe aerodynamic center CM does not vary as CL is changed, or

dCMac

dCL

= 0

79

To use this the derivative of equation [3.23] is taken with respect to CL ,

dCMac

dCL

= dCMa

dCL

− dCL

dCL

ac

− xac

c

By definition then the term on the left becomes zero and the derivative of CL with respectto itself is unity; thus, the equation can be rearranged as

xac

c= a

c− dCMa

dCL

[3.24]

Using the equation 3.24 the position of the aerodynamic center can be found as adistance xac from the leading edge by finding the value of the derivative of the knownmoment coefficient with respect to the lift coefficient. This can be easily obtained byplotting a graph of CM versus CL and measuring the slope of the curve. Such a curve willbe essential linear over a normal range of angle of attack for most common airfoils.

EXAMPLE 3.2 For a particular airfoil section the pitching moment coefficient about apoint 1/3 chord behind the leading edge varies with the lift coefficient in the followingmanner:

CLK 0.2 0.4 0.6 0.8

CM K − 0.02 0.00 + 0.02 + 0.04

Find the aerodynamic center and the value of CMo .

It is seen that CM varies linearly with CL , value of dCM dCL being

0.04 − −0.02( )0.80 − 0.20

= + 0.060.60

= +0.10

Therefore, from equation 3.12, with a/c = 1/3

xac

c= 1

3− 0.10 = 0.233

The aerodynamic centre is therefore at 23.3% chord behind the leading edge.

Plotting CM against CL gives the value of CMo , the value of CM when CL = 0, as-0.04.

80

Theoretically, on a flat plate or circular arc airfoil in subsonic flow the aerodynamiccenter is exactly one forth of the chord from the leading edge or at a point called the"quarter chord". In practice it is usually from 23 to 25% of the chord from the leading edgeat incompressible speeds. For this reason it is popular to define aerodynamic forces andpitching moments about the quarter chord as is the practice in may reports of tabulatedairfoil aerodynamic data.

When speeds are high enough for compressibility to become a factor (M ≥ 0.5) theaerodynamic center will start to move rearward along the airfoil. The theoretical position ofthe aerodynamic center in supersonic flow is at 50% chord. This shift will be examined in alater section where compressibility effects on airfoils are discussed.

The center of pressure, while not necessarily a fixed point over a range of orCL , is important since this is the point where the moment disappears. It is the point, notnecessarily on the chord, where the pitching moment is zero for a particular value of lift.Returning to equation [3.20] the following relation can be written relating the moment at theleading edge, MLE to the moment at the center of pressure MMP .

M LE = MMCP − Lcos +Dsin( )kCPC

Now, recognizing that by definition MCP = O and writing XCP as kCPC where kCP is thefraction of the chord to the center of pressure, the result is

MLE = MCP − L cos a + Dsin a( )kCPC .

Writing the same equation with the moment about the aerodynamic center in the equationgives

MLE = Mac − Lcos + Dsin( )Xac

Now equating the two equations [3.25] and [3.56] and dividing by 1

2V 2c2 to get the

relation in coefficient form gives

CMac − CL cos + CD sin( ) Xacc

= − CL cos + CD sin( )kcp

Solving for kcp, the position of the center of pressure as a fraction of the chord,

kcp = Xacc

− CMac

CL cos + CD sin

81

Here again, since is relatively small and CL > > CC , the term CD sin can beneglected and cos is assumed to be unity, giving

kcp = Xacc

− CMac

CL

Hence a relation has been developed which will determine the position of the center ofpressure for any value of CL once the pitching moment at the aerodynamic center and thelocation of the aerodynamic center known.

Since CMac is almost always negative it can be seen from equation [3.28] that

kcp −Xac

c will be positive, indicating that the center of pressure is almost always behind

the aerodynamic center.

Example 3.3 For the airfoil section of Example 3.2, plot a curve showing theapproximate variation of center of pressure position with lift coefficient, for lift coefficientsbetween zero and unity.

For this case, kcp ≅ 0.233 − −0.04CL

≅ 0.233 + 0.04CL

82

The corresponding curve is shown above. It shows that kcp tends asymptotically to Xac asCL increases, and tends to infinity behind the airfoil as CL tends to zero. For values ofCL less that 0.05 the center of pressure is actually behind the airfoil.

For a symmetrical section (zero camber) and for some special camber lines, the pitchingmoment coefficient about the aerodynamic center is zero. It then follows, that kcp = xac ;i.e., the center of pressure and the aerodynamic center coincide, and that for moderateincidences the center of pressure is therefore stationary at about the quarterchord point.

3.7 AIRFOIL AERODYNAMIC PERFORMANCE

Now that the meaning of forces and moments on airfoils have been discussed and theterminology of airfoil geometry is understood, the effects of changes in airfoil geometryand fluid flow behavior can be explored. The discussion that follows will be very generaland many exceptions can undoubtedly be found to the examples given; however, the airfoilbehavior discussed will be that typical of most airfoils.

Any discussion of airfoils should begin with the simple symmetrical airfoil whichtypically exhibits an aerodynamic behavior similar to that shown in Figure 3.7.Theoretically the lift coefficient for a two dimensional airfoil increases linearly with a curveslope of 2 per radian when plotted against . When the CL curve begins to becomenon-linear, flow separation is beginning along the upper surface of the airfoil. Asseparation or stall progresses the curve slope decreases until a peak is reached and beyondthis angle of attack CL will decrease. At this peak the value of lift coefficient is termedCLmax The nature of the stall region will vary with leading edge radius, Reynolds numberand other factors which will be discussed later.

Fig. 3.7. Symmetrical airfoil characteristics.

83

The pitching moment coefficient shown is for the quarter chord position and since thisis roughly the position for the aerodynamic center for an airfoil in incompressible flow,CMc 4 is constant over the range of where dCL d is constant.

The drag coefficient for the symmetrical airfoil is seen to be at a minimum at a CL ofzero which corresponds to a zero angle of attack. Drag coefficient rises rapidly in the stallregion as wake drag increases.

The effects of camber on an airfoil are illustrated in Figure 3.8.

Fig. 3.8. Effects of Camber

The primary effect of camber on an airfoil is to shift the CL vs curve to the left whichmeans that the cambered airfoil produces a finite lift at a zero nominal angle of attack. Thecambered airfoil experiences a zero lift at some negative angle of attack designated LO .

Stall usually occurs at a lower nominal angle of attack for the cambered airfoil, howeverCLmax will usually be somewhat increased over that for a symmetrical airfoil of comparablethickness and leading edge radius.

Increasing camber is seen to produce a larger negative CMc 4 and to cause shift in the

drag polar to correspond to the CL shift.

The effects of changes in airfoil thickness depend largely on where the maximumthickness of the airfoil occurs. It is somewhat intuitive that a thicker section will produce ahigher drag coefficient, however, the primary effect of thickness is on the wings stallbehavior and the lift curve slope. The effect of thickness on the lift curve slope varies withthe distribution of the thickness along the chord as shown in Figure 3.9

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Fig. 3.9 Effect of Thickness on Lift Curve Slope

For the NACA 4 and 5 digit series airfoils the slope of the lift curve is seen to decreaseslightly as thickness increases (the theoretical slope should be 2 radian or0.10966/degree)/ whereas, for the 63 series airfoil it shows an increase.

The thickness can also either increase or decrease CLmax as illustrated in Figure 3.10.An increase in thickness up to a point can lead to a higher CLmax because of its effect onimproving the leading edge radius, however further increases increase the likelihood ofearlier separation over the rear of the airfoil.

Figure 3.10. Thickness effects on CLmax

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The most significant effect of thickness is to be seen near the airfoil's leading edge interms of leading edge radius. Too small a leading edge radius can result in an abrupt stallwith a sudden decrease in lift coefficient. A larger leading edge radius can smooth the stalland give an increase in CLMax as shown in Figure 3.11.

Figure 3.11

The effects of Reynolds number on airfoil performance are due to the influence ofReynolds number on the airfoil boundary layer and on flow separation. These effects havebeen discussed to some extent earlier. At low values of Re a laminar boundary layer existsover a large distance from the leading edge of the airfoil. A laminar boundary layer is verypoor at resisting separation over the region of the airfoil where the flow is slowing downand the pressure is increasing, and early flow separation may result, giving larger drag andearly stall. At high Re values the boundary layer goes turbulent at an early point on theairfoil and because of the ability of a turbulent boundary layer to resist flow separation,wake drag may be reduced and stall delayed. These effects are illustrated in Figure 3.12for a typical airfoil.

Figure 3.12. Reynolds Number Effects on Lift and Drag

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One of the more important effects of Reynolds number is its influence on the near stallregion at high angles of attack. Increasing Reynolds number almost always increasesCLMax and it can also have a significant effect on the nature of stall as shown in Figure3.13. For a particular airfoil it is seen that at low Re a fairly smooth stall exists with arelatively low CLMax . As Re is increased a higher CLMax is obtained but the stall is asharp one. Here the high Re flow is able to resist separation but when it finally occurs itsuddenly covers a large portion of the airfoil. At even higher Re the turbulent boundarylayer is able to resist a sudden leading edge separation and an ever higher CLMax isachieved with a somewhat improved stall behavior.

Figure 3.13. Effect of Re on Stall

The effect shown in Figure 3.13 is also a function of the airfoil thickness andparticularly the leading edge radius even small increases in Re may increase CLMax whileon a thin airfoil with small leading edge radius it may be necessary to go to very highvalues of Re before significant CLMax increases occur, as shown in Figure 3.14. A smallleading edge radius produces very large flow accelerations at high angles of attack resultingin large pressure deficits. The subsequent pressure rise as the flow slows down again overthe airfoil's upper surface is also very rapid and this large "adverse" pressure gradientcauses flow separation. Very high Reynolds numbers are required to resist the separationinducing effects of this large pressure gradient. on a thicker airfoil with a larger leadingedge radius the pressure gradient is not as large and it is not necessary to reach as large avalue of Re to control separation.

Fig. 3.14. Effects of thickness and Re on CLMax

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Compressibility effects on an airfoil must be considered at Mach numbers above 0.5 orso. Even at relatively modest subsonic speeds the local velocities of air over a wing mayapproach the speed of sound. When the free stream velocity is great enough that the localflow at some point on the upper surface of the airfoil reaches the speed of sound the airfoilis said to have reached its "critical Mach number, Mcrit ." For example, the NACA 4412airfoil flying in air at sea level standard conditions will begin to experience locallysupersonic flow at about 20 - 30% of the chord behind the leading edge at a speed of 592fps or 180m/s. Since the speed of sound is 1117 fps at sea level the wing is flying at aMach number of 0.53. Hence its critical Mach number Mcrit = 0.53.

Once a region of supersonic flow beings to form on the airfoil a shock wave will formas the flow "shocks" down to subsonic flow. In a shock wave the flow deceleration occursvery suddenly resulting in a sharp, essentially instantaneous, pressure increase. It is thispressure increase or compression wave which results in the sonic boom when the wave isstrong enough to reach the ground. The sudden pressure increase in the shock wave actsas an almost infinite adverse pressure gradient on the wing's boundary layer, resulting inalmost certain flow separation. This process is illustrated in Figure 3.15.

Figure 3.15. Transonic Flow Patterns Around an Airfoil

As this supersonic flow region and accompanying shock wave grow the separatedwake also grows, increasing drag and changing the manner in which lift is produced on thewing. This causes the aerodynamic center to shift and large changes in the pitchingmoment about the quarter chord. These changes with Mach number are shown in Figure3.16 as changes in the slope of the lift curve, drag coefficient and moment coefficient aboutthe quarter chord.

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Figure 3.16. Compressibility Effects on Lift Curve Slope, Drag and Pitching Moment

It is these large changes which created the myth of the "sound barrier" in the early daysof transonic flight. For an aircraft designed for subsonic flight these changes could easilylead to disaster when the wing reached and passed its critical Mach number in a high speeddive. When these speeds were reached there was a sudden sharp rise in drag whichrequired thrust beyond that available in the engine technology of the day, and worse, thesudden aerodynamic center shift and flow separation caused severe stress in the structureand loss of control. As shown in Figure 3.16-c the value of CMc 4 may go from a negativeto a positive value, resulting in a complete reversal of the stability behavior of the aircraftand control behavior. The result was often a disaster where the aircraft went out of controland broke up in mid-air. Once powerplants were large enough to handle the drag rise andaircraft were designed to handle the moment shifts it became possible to fly in and throughthe transonic regime.

3.8 FLAPS AND HIGH LIFT DEVICES

In order to fly, an aircraft must produce enough lift to counteract its weight. From theequation for lift coefficient [3.12] it can be seen that the lift generated by a wing is afunction of its area, (S), the square of its velocity (V∞ ), the air density ( ), and the liftcoefficient (CL ).

Lift = 12

V∞2SCL

Hence, lift can be increased by increasing any one of these four parameters. This presentslittle problem in high speed flight since for a given wing and angle of attack the liftincreases as the square of the speed. There are however, problems at lower speeds.

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The lower limit for the airspeed of a wing is its stall speed VSTALL . This is the speedat which a wing of given area will stall at a given altitude (or density). Stall occurs whenthe lift coefficient has reached its maximum (CLMax). Thus for a given wing area, densityand weight of aircraft, the minimum flying speed is given by:

V∞min =L

1

2SCLmax

=2W

SCL max

This is obviously then the speed which will result in the shortest takeoff or landing roll.Naturally, it would not be safe to land or takeoff an aircraft at CLMax conditions since it ison the verge of stall; thus, a slightly higher speed is used to give a safety factor. Thisspeed is usually figured to be 1.3VSTALL . However, the fact still remains that theminimum safe flying speed and, hence, the minimum landing or takeoff distance isdetermined by CLMax for a wing at a given altitude, weight and wing area.

In the early days of aviation, when more lift was needed to keep landing and takeoffdistances within limits, a larger wing area was used. However, this also increased the dragof the aircraft and limited its cruise speed. A more highly cambered wing or one whichwas relatively thick to give a larger leading edge radius could also be used to increaseCLMax but these also increased drag and limited cruise speed. Therefore, the cruise speedof an aircraft was effectively limited by the length of runway available for the takeoff andlanding. What was needed was a way to use a minimal drag wing for high speed cruiseand then increase wing area and/or camber for takeoff and landing to reduce the requiredtakeoff and landing speeds.

The answer to this requirement was the flap. The basic idea behind the use of the flapis that of developing a variable camber airfoil which will normally have a low camber, lowdrag shape but can have its camber increased when a higher maximum lift coefficient isneeded for low speed flight. Adjustable camber airfoils were built and flown as early as1910 and the hinged flap can be traced to as early as 1914 in England.

Theory can show that a change in camber near the trailing edge of an airfoil has a muchgreater effect on CL than changes at any other position. Hence flaps are, in their simplestform, merely hinged portions of the airfoil's trailing edge as shown in Figure 3.17. Thetypical effects of flap deflection are also shown in the figure to be essentially the same asadding camber.

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Figure 3.17. Aerodynarnic Effects of the Flap .

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There are many different designs for flaps, some more effective than others. Theeffectiveness of a variety of flaps is illustrated in Figure 3.18.

Designation Diagram CLMaxat CLMax(degrees)

L Dat CLMax

ReferenceNACA

Basic airfoilClark Y

1.29 15 7.5 TN 459

.30cPlain flap

deflected 45o

1.95 12 4.0 TR 427

.30cSlotted flap

deflected 45o

1.98 12 4.0 TR 427

.30cSlit flap

deflected 45o

2.16 14 4.9 TN 422

.30c hinged at .80cSplit flap (zap)deflected 45o

2.26 13 4.43 TN 422

.30c hinged at .90cSplit flap (zap)deflected 45o

2.32 12.5 4.45 TN 422

.30cFowler flap

deflected 40o 2.82 13 4.55 TR 534

.40cFowler flap

deflected 40o

3.09 14 4.1 TR 534

Fixed slot 1.77 24 5.35 TR 427

Handley Pageautomatic slot

1.04 28 4.1 TN 459

Fixed slot and.30c plain flapdeflected 45o

2.18 19 3.7 TR 427

Fixed slot and.30c slotted flapdeflected 45o

2.26 18 3.77 TR 427

Handley Page slot and.40c Fowler flapdeflected 40o

3.36 16 3.7 TN 459

Data taken form NACA 7 x 10 ft tunnel, wing AR=6, Re=609,000

Fig. 3.18. Effectiveness of Flaps and Slots on a Clark Y airfoil.

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Some of the flaps have slots between the flap and the main wing and some both deflect andextend. The Fowler flap, which deflects and extends to open a slot is seen to be the mosteffective of the single flap systems, increasing CLMax by a factor of 2.4.

The leading edge slot and flap is also illustrated in Figure 3.18. It is seen that theleading edge flap and slot can be very effective in increasing CLMax by itself and its usewith trailing edge flaps improves the performance of the wing even further.

The leading edge flap, slot of "slat" does not achieve its effectiveness through a changein camber, because a camber change at the leading edge has very little effect; but works byreducing the likelihood of flow separation and stall at high angles, much like an increasedleading edge radius or increased Reynolds number would. Figure 3.19 illustrates the basicinfluence of the slot on an airfoils performance.

Figure 3.19. Effect of Leading Edge Slot

Figure 3.20 shows several types of leading edge devices which have been used onwings. The first four do not incorporate slots and achieve their effect essentially byproviding an easier path for the flow to follow over the leading edge. This reduces the lowpressure peak experienced by the upper surface of the airfoil at high angles of attack sincethe flow does not have to accelerate as rapidly to get around the leading edge.Subsequently the following adverse pressure gradient is not as strong and separation is lesslikely.

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Figure 3.20. Leading Edge Devices

The other two devices shown in Figure 3.20 incorporate a slot of some type. Theleading edge flap with a slot is much more effective than one without it. The action of theleading edge slot has often been incorrectly explained as that of a nozzle which directs highspeed air into the wing's boundary layer, "energizing" it so that it can continue on aroundthe wing without separating. However, an examination of the flow through the slot willshow that it is, in fact, a very low speed flow and that, instead of adding high speed air tothe flow above the wing, it slows the flow.

The slot effect is best understood if the main wing and the leading edge flap or slat areviewed as two separate wings. If these two wings (A and B in Figure 3.21) were testedalone at the orientation to the free stream shown, a pressure distribution similar to thatshown by the dashed lines in the plot would result. When the two are placed in proximityto each other however, the blockage of flow through the slot slows the flow over thebottom of wing A and the leading edge of wing B. This reduction in the velocity of theflow has two effects. The most important effect is that by slowing the flow over theleading edge of wing B, the low pressure peak is reduced in magnitude, thus reducing theseverity of the adverse pressure gradient which follows and delaying separation. Thismeans the wing can go to higher angles of attack before stall. As is always the case,however, this benefit is not free. The solid line on the CP graph shows that the newpressure distribution reduces the lift generated by the wing.

The loss of lift on wing B is more than counteracted by the increase i lift on wing A dueto the reduced velocity and hence, higher pressure below wing A. In actuality, the flowaround wing B results in an effective large positive angle of attack on wing A. Thepressure distribution on wing A thus changes as shown in Figure 3.21 and the lift on wingA changes from negative to a large positive value.

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The net effect of the flow over the two wings is a slotted airfoil which can go to highangles of attack before separation and stall. In actuality, the total flow is a bit morecomplicated than that just described and includes boundary layer interaction anddownstream influences. However, the essentials of the flow fit the above description.

Figure 3.21. Action of a Leading Edge Slot.

Slotted trailing edge flaps work on the same principle as that just described for theleading edge slot. In this case the slots allow the trailing edge flaps to be deflected to verylarge angles without flap stall. Modern transport aircraft which need to be able to cruise athigh speeds and land in reasonable distances often use multiple flap and slot systems suchas the one shown in Figure 3.22.

Fig. 3.22. Geometry of Leading Edge Slat and Triple Slotted Flap

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A number of other devices exist which are intended to increase the maximum liftcoefficient of an airfoil. Most of these devices are in some way dependent on power fromthe aircraft and are essentially boundary layer control devices. With non-power augmenteddevices such as mechanical flaps and slats it is possible to achieve lift coefficients in therange of 3.5 to 4.0. According to inviscid theory the upper limit on CL is 4 or about12.5. Of course, this theory does not account for the effects of viscosity in retarding theflow in the boundary layer and subsequent flow separation. By attempting to control theboundary layer to prevent separation it is possible to achieve higher maximum liftcoefficients.

The two primary types of power augmented boundary layer control high lift devicesused have been suction and blowing. Boundary layer suction can be effective indelaying separation on an airfoil. Suction is applied through a slot in the wing surface inthe region where the adverse pressure gradient would be likely to lead to separation. Thesuction pulls away the "stale" boundary layer and, essentially a new boundary layer begins.It is possible to roughly double CLmax for an airfoil by using suction properly. Theimprovement in CLmax increases as the suction increases.

A certain amount of power is required to create the suction used to control the boundarylayer and this power usage must be accounted for. Since power used in a vehicle is usuallythat used for propulsion to overcome drag and the use of suction for boundary layer controlmust be considered as part of the total vehicle power requirement, this power is usuallytreated as a drag penalty. Hence the question must always be asked whether or not the gainin lift is worth the price paid.

Boundary layer blowing may be accomplished in several different ways and is morecommon than suction as a boundary layer control device. Boundary layer control byblowing is usually accomplished by introducing a jet through a slot such that the jet istangential to the boundary layer (Figure 3.23). This jet "energizes" the boundary layer byintroducing a high speed stream and by entraining fluid from outside the boundary layer.Like suction similarly placed, boundary layer blowing on the wing's upper surface canessentially double the CLmax of the airfoil. Again, the power required for blowing must beconsidered as a penalty and is usually treated as drag.

Blowing is often used to control the flow over the flaps where fluid is blown over theflap either from a slot in the aft portion of the main airfoil or from a slot in the leading edgeof the flap itself. This can be extremely effective in increasing CLmax for a wing bypreventing flow separation over the flaps at high wing angles of attack and high flapdeflection angles (Figure 3.24).

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Fig. 3.23. Boundary Layer

Figure 3.24. Internally Blown Flaps

97

A type of blowıng with sinıilar effect is found in the externally flow flap (Figure 3.25)where the jet engine or fan exhaust flows over the airfoil enhancing the normal flap, slot,slat action to increase CLmax . Engine placement is critical in this system and the exhaustmust be designed to cover as much of the wing as possible.

Many other types of boundary layer control, high lift devices exist or have beeninvestigated. These are too numerous to mention within the scope of this text. Theseinclude the augmentor wing upper surface blowing, the Coanda effect, jet flaps and otherdevices.

Figure 3.25. Externally Blown Flaps.

3.9 LAMINAR FLOW AIRFOILS

An examination of boundary layer behavior would show that a laminar boundary layercauses less skin friction drag than a turbulent boundary layer; however, the turbulentboundary layer is much better at resisting flow separation. Knowing this, NACAresearchers in the late 1930's and 1940's designed a low drag series of airfoils calledlaminar flow airfoils. These airfoils, know as the NACA 6-series airfoils, were designedto take advantage of the low skin friction of a laminar boundary layer by encouraging alaminar flow over the first 30 to 50 percent of the airfoil surface.

In order to maintain laminar flow over a large portion of an airfoil something must bedone to prevent laminar-turbulent transition and to prevent flow separation, which occursrather easily in a laminar boundary layer. In the laminar flow airfoil this can be achieved atlow to moderate angles of attack by shaping the airfoil to produce a favorable pressuregradient over the area where laminar flow is required. A favorable pressure gradient, thatis one where pressure is decreasing and flow is accelerating, has the effect of suppressingthe development of turbulence in the boundary layer. Laminar-turbulent transition in aboundary layer is caused by the growth and merger of small scale turbulence disturbancesin the flow. The likelihood of the growth of these disturbances is spoken of in terms ofboundary layer stability and a favorable pressure gradient is a stabilizing one while anadverse gradient is de-stabilizing.

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The most straightforward way of creating a favorable pressure gradient on an airfoil isto shape the wing in such a way that the flow accelerates relatively slowly over a largeportion of the wing. This is done primarily by reducing the leading edge radius to cutdown on large accelerations there, and by moving the maximum thickness of the airfoilrearward. This will result in easier separation at high angles of attack and a lower CLmaxbut will give a lower drag at more moderate angles of attack hence, saving fuel in a longdistance cruise condition. Figure 3 26 shows the difference in wing shape and pressuregradient between the conventional NACA 0015 airfoil and its laminar flow counterpart, theNACA 653 -015.

Figure 3.26. Conventional and Laminar Flow Airfoils

Figure 3.27 shows the result of this change in comparing an NACA 2415 airfoil with aNACA 642 -415 wing. Both airfoils are 15% thickness and have the same camber,however the latter has a region of reduced drag coefficient over a range of CL from 0 to0.5. This area of reduced CD is known as the "drag bucket" and it is this "bucket" that isthe primary characteristic of laminar flow airfoils. It should be noted that at CLs above thebucket, the drag of the laminar flow airfoil is actually higher than its conventionalcounterpart because of the effect of the sharper leading edge at high , however thereduction of drag coefficient by a factor of two in the drag bucket region may be moreimportant in a given wing design than the effects at higher angles of attack.

The position of the drag bucket can be changed as shown in Figure 3.28 with the firstdigit in the second series of three numbers in the NACA designation referring to the valueof CL at the center of the drag bucket. This gives the designer an important tool, allowingher or him to select a laminar flow airfoil which is best for his or her design. If an aircraftis to be designed for long distance cruise and a CL of 0.2 is needed in cruise, the designercan select the laminar flow airfoil which will position the drag bucket around the CL = 0.2range, giving a low drag cruise and a high fuel economy in cruise. Likewise if a highperformance aircraft requires low drag during maneuvers and climbs where CL is high, thedrag bucket can be placed there.

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Fig. 3.27: Drag Bucket for a Laminar Airfoil

Figure 3.28. Placement of Drag Bucket

100

Laminar flow airfoils have been used on almost all transport aircraft and on manygeneral aviation aircraft designed since about 1940. They can greatly reduce drag in cruiseand with proper use of flaps and other high lift devices it is still possible to achieve highCLmax values when needed for landing and takeoff. It should be noted however, that it ispossible to loose the drag bucket if sufficient dirt or roughness accumulates on the forwardpart of the airfoil and, hence, such wings must be kept clean to work at their best.

3.10 SUPERCRITICAL AIRFOILS

The problem of drag rise, moment shift and lift loss during transonic flow wasdiscussed earlier. When the critical Mach number is reached, a shock wave forms on theairfoil's upper surface and the flow is very likely to separate due to the large adversepressure gradient imposed by the shock. Anything which can be done to increase thecritical Mach number; i.e., delay the onset of the shock wave to a higher speed, will allowhigher speed subsonic flight with low drag. Hence a higher Mcrit will allow flight athigher speed on a given amount of engine power or flight at the same speed as wings withlow Mcrit at reduced power and fuel levels.

The most obvious way to delay the onset of an upper wing surface shock is to reducethe curvature of the upper surface in such a way that the flow does not accelerate to as higha speed as it would with a conventional or even a laminar flow airfoil. With a "flatter"upper surface the region of supersonic flow may be spread over a larger portion of theairfoil and the flow does not accelerate to as high a supersonic speed. This accomplishestwo things, it delays the onset of the shock until a more aft position where any resultingseparation will affect less of the airfoil and it gives a weaker shock which is less likely tocause separation. Because of the lower supersonic speeds over the airfoil the pressuresdeveloped are not as low and lift due to the flow over the upper surface may be reducedfrom that on a conventional airfoil. Hence, the new airfoil is designed with a large cambercreated by a cusp in the lower surface at the trailing edge which creates enough lift to makeup for that lost on the upper surface. The resulting design is shown in Figure 3.29.

Figure 3.29. Supercritical Airfoils

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The supercritical airfoil has proved to be very effective at lower speeds as well as in thetransonic regime. As of the mid 1970's supercritical airfoils were appealing on transportdesigns for production in the 1980's and beyond. Low speed versions of the airfoil havebeen tested on general aviation aircraft and are appearing on new models of such airplanes.the airfoil design has an advantage of giving low drag with a relatively thick wing section(13 to 21% thickness) and thus provides room for improved structure or fuel capacity. Thelarge thickness and large leading edge radius generally result in a higher CLmax in lowspeed applications.

The primary drawback for the supercritical airfoil design has been the large negativepitching moment caused by the large magnitude of lift generated near its trailing edge.Another problem has been in the manufacture of the sharp trailing edges which result fromthe trailing edge cusp. It is, however, possible to design around these problems and thesupercritical airfoil is expected to be the wing used on most aircraft for the foreseeablefuture.

3.11 THREE DIMENSIONAL EFFECTS

Most of the discussion in the previous sections of this chapter has dealt with twodimensional airfoil behavior. A two dimensional airfoil is, of course, only a section orslice of a 3-D airfoil and two dimensional airfoil aerodynamics must be considered the idealcase. Three dimensional effects can often be thought of as simply factors which limit thenormal 2D performance of an airfoil. The two primary three dimensional factors whichaffect wing performance are aspect ratio, AR, and wing sweep.

As mentioned in section 3.4 aspect ratio is a measure of the ratio of the span to themean chord. Aspect ratio has a significant effect on the performance of the total wingbecause of wing tip losses. Ideally, the lift generated by an untapered wing would beconstant at every point along the span. However, due to flow around the wing tip there arelift losses near the tip as shown in Figure 4.30. Since there is a lower pressure on top ofthe wing than below the fluid will flow around the wingtip to the area of lower pressure.This, therefore reduces the lift generated near the wingtip and must be considered as a 3-Dloss. This flow around the wingtip is also the source of the trailing vortex, a swirling flowcoming off of each wingtip.

The loss of lift at the tip of a 3-D wing goes inboard over some percentage of thespan. Hence, for a short stubby wing with low AR a greater percentage of the total wingarea experiences some tip loss than in a high aspect ratio wing. Therefore a high aspectwing will produce more lift than a low aspect ratio wing of the same area and airfoilsection. The 3-D wing has a higher drag coefficient than the 2D airfoil and this additionaldrag is inversely proportional to aspect ratio. The net result is that a high aspect ratio airfoilhas higher lift and lower drag than a low aspect ratio wing of the same area and airfoilshape.

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Figure 3.30. 3-D Airflow Effects

Wing sweep primarily affects the performance of a 3-D airfoil in the transonic flowregime. The primary effect of sweeping the wing is to raise the critical Mach number of theairfoil. The shock wave, which develops on an airfoil in transonic flow, is developed inresponse to the component of the free stream to the line down the quarter chord of thewing. Hence, for a given free stream velocity, the greater the wing sweep the lower will bethe component of velocity normal to the quarter chord line. It is not until this normalcomponent of the flow reaches the critical Mach number that the drag rise will occur andeven when it does occur the drag rise is not as great as in the unswept case. Figure 3.31shows the effect of varymg degrees of sweep on one senes of wings.

Figure 3.31. Sweepback Effect on Drag Rise vs. Mach No.

As is the case with almost any change that is of benefit in some way a price must bepaid for the favorable effects of sweep. This price is sweep-induced cross flow tip stall.The span-wise flow which results from the sweep can cause severe adverse pressuregradients at the wing tips and tip stall. Severe structural as well as aerodynamic problemscan result.

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Summary

A review of this chapter will show that almost all phenomena which occur on airfoilscan be understood in terms of simple pressure-velocity behavior and their effects on thebehavior and their effects on the behavior of the boundary layer. All that is needed for aphysical understanding of airfoil aerodynamics is an appreciation of the meaning of thepressure-velocity relationship called Bernoullis equation. This, combined with a physicalfeel for the meaning of laminar, turbulent and separated boundary layers can explain theaerodynamic behavior of flow about any shape. Other texts will show how to take thissimple physical understanding and build it into a useful mathematical description of fluidflows.

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CHAPTER 4

PERFORMANCE IN STRAIGHT AND LEVEL FLIGHT

Now that we have examined the origins of the forces which act on an aircraft in theatmosphere, we need to begin to examine the way these forces interact to determine theperformance of the vehicle. We know that the forces are dependent on things likeatmospheric pressure, density, temperature and viscosity in combinations that become"similarity parameters" such as Reynolds number and Mach number. We also know thatthese parameters will vary as functions of altitude within the atmosphere and we have amodel of a standard atmosphere to describe those variations. It is also obvious that theforces on an aircraft will be functions of speed and that this is part of both Reynoldsnumber and Mach number.

Many of the questions we will have about aircraft performance are related to speed.How fast can the plane fly or how slow can it go? How quickly can the aircraft climb?What speed is necessary for lift-off from the runway?

In the previous section on dimensional analysis and flow similarity we found that theforces on an aircraft are not functions of speed alone but of a combination of velocity anddensity which acts as a pressure that we called dynamic pressure. This combinationappears as one of the three terms in Bernoulli's equation

P +1

2V 2 = P0

which can be rearranged to solve for velocity

V = 2 P0 − P( )

In chapter two we learned how a Pitot-static tube can be used to measure the differencebetween the static and total pressure to find the airspeed if the density is either known orassumed. We discussed both the sea level equivalent airspeed which assumes sea levelstandard density in finding velocity and the true airspeed which uses the actual atmosphericdensity. In dealing with aircraft it is customary to refer to the sea level equivalent airspeedas the indicated airspeed if any instrument calibration or placement error can beneglected. In this text we will assume that such errors can indeed be neglected and the termindicated airspeed will be used interchangeably with sea level equivalent airspeed.

VIND = Ve = VSL =2 P0 − P( )

SL

105

It should be noted that the equations above assume incompressible flow and are notaccurate at speeds where compressibility effects are significant. In theory, compressibilityeffects must be considered at Mach numbers above 0.3; however, in reality, the aboveequations can be used without significant error to Mach numbers of 0.6 to 0.7.

The airspeed indication system of high speed aircraft must be calibrated on a morecomplicated basis which includes the speed of sound:

VIND =2aSL

2

− 1

P0 − P

SL

+1

−1

−1

where as l = speed of sound at sea level and PSL = pressure at sea level.Gamma is the ratio of specific heats Cp Cv( ) for air.

Very high speed aircraft will also be equipped with a Mach indicator since Machnumber is a more relevant measure of aircraft speed at and above the speed of sound.

In the rest of this text it will be assumed that compressibility effects are negligible andthe incompressible form of the equations can be used for all speed related calculations.Indicated airspeed (the speed which would be read by the aircraft pilot from the airspeedindicator) will be assumed equal to the sea level equivalent airspeed. Thus the true airspeedcan be found by correcting for the difference in sea level and actual density. The correctionis based on the knowledge that the relevant dynamic pressure at altitude will be equal to thedynamic pressure at sea level as found from the sea level equivalent airspeed:

1

2V 2

alt

≡1

2 SLVe2

Ve =SL

Valt = Valt

An important result of this equivalency is that, since the forces on the aircraft depend ondynamic pressure rather than airspeed, if we know the sea level equivalent conditions offlight and calculate the forces from those conditions, those forces (and hence theperformance of the airplane) will be correctly predicted based on indicated airspeed and sealevel conditions. This also means that the airplane pilot need not continually convert theindicated airspeed readings to true airspeeds in order to gauge the performance of theaircraft. The aircraft will always behave in the same manner at the same indicated airspeedregardless of altitude (within the assumption of incompressible flow). This is especiallynice to know in take-off and landing situations!

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STATIC BALANCE OF FORCES

Many of the important performance parameters of an aircraft can be determined usingonly statics; ie., assuming flight in an equilibrium condition such that there are noaccelerations. This means that the flight is at constant altitude with no acceleration ordeceleration. This gives the general arrangement of forces shown below.

Figure 4.1 Static force balance in straight and level flight.

In this text we will consider the very simplest case where the thrust is aligned with theaircraft's velocity vector. We will also normally assume that the velocity vector is alignedwith the direction of flight or flight path. For this most basic case the equations of motionbecome:

T - D = 0

L - W = 0

Note that this is consistent with the definition of lift and drag as being perpendicular andparallel to the velocity vector or relative wind.

Now we make a simple but very basic assumption that in straight and level flight lift isequal to weight,

L = W

We will use this so often that it will be easy to forget that it does assume that flight isindeed straight and level. Later we will cheat a little and use this in shallow climbs andglides, covering ourselves by assuming "quasi-straight and level" flight. In the final part ofthis text we will finally go beyond this assumption when we consider turning flight.

Using the definition of the lift coefficient

CL =L

1

2V∞

2S

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and the assumption that lift equals weight, the speed in straight and level flight becomes:

V =2W

SCL

The thrust needed to maintain this speed in straight and level flight is also a function ofthe aircraft weight. Since T = D and L = W we can write

D L = T W

or

T =D

LW =

CD

CL

W

Therefore, for straight and level flight we find this relation between thrust and weight:

The above equations for thrust and velocity become our first very basic relations whichcan be used to ascertain the performance of an aircraft

AERODYNAMIC STALL

Earlier we discussed aerodynamic stall. For an airfoil (2-D) or wing (3-D), as the angleof attack is increased a point is reached where the increase in lift coefficient, whichaccompanies the increase in angle of attack, diminishes. When this occurs the liftcoefficient versus angle of attack curve becomes non-linear as the flow over the uppersurface of the wing begins to break away from the surface. This separation of flow may begradual usually progressing from the aft edge of the airfoil or wing and moving forward;sudden, as flow breaks away from large portions of the wing at the same time; or somecombination of the two. The actual nature of stall will depend on the shape of the airfoilsection, the wing planform and the Reynolds number of the flow.

Figure 4.2

108

We define the stall angle of attack as the angle where the lift coefficient reaches amaximum, CLmax, and use this value of lift coefficient to calculate a stall speed for straightand level flight.

VSTALL =2W

SCL max

Note that the stall speed will depend on a number of factors including altitude. If we lookat a sea level equivalent stall speed we have

Ve STALL=

2W

SLSCL max

It should be emphasized that stall speed as defined above is based on lift equal toweight or straight and level flight. This is the stall speed quoted in all aircraft operatingmanuals and used as a reference by pilots. It must be remembered that stall is only afunction of angle of attack and can occur at any speed. The definition of stallspeed used above results from limiting the flight to straight and level conditions where liftequals weight. This stall speed is not applicable for other flight conditions. For example,in a turn lift will normally exceed weight and stall will occur at a higher flight speed. Thesame is true in accelerated flight conditions such as climb. For this reason pilots are taughtto handle stall in climbing and turning flight as well as in straight and level flight.

For most of this text we will deal with flight which is assumed straight and level andtherefore will assume that the straight and level stall speed shown above is relevant. Thisspeed usually represents the lowest practical straight and level flight speed for an aircraftand is thus an important aircraft performance parameter.

We will normally define the stall speed for an aircraft in terms of the maximum grosstakeoff weight but it should be noted that the weight of any aircraft will change in flight asfuel is used. For a given altitude, as weight changes the stall speed variation with weightcan be found as follows:

VSTALL 2 = VSTALL 1

W2

W1

It is obvious that as a flight progresses and the aircraft weight decreases, the stall speedalso decreases. Since stall speed represents a lower limit of straight and level flight speed itis an indication that an aircraft can usually land at a lower speed than the minimum takeoffspeed.

109

For many large transport aircraft the stall speed of the fully loaded aircraft is too high toallow a safe landing within the same distance as needed for takeoff. In cases where anaircraft must return to its takeoff field for landing due to some emergency situation (such asfailure of the landing gear to retract), it must dump or burn off fuel before landing in orderto reduce its weight, stall speed and landing speed. Takeoff and landing will be discussedin a later chapter in much more detail.

PERSPECTIVES ON STALL

While discussing stall it is worthwhile to consider some of the physical aspects of stalland the many misconceptions that both pilots and the public have concerning stall.

To the aerospace engineer, stall is CLmax, the highest possible lifting capability of theaircraft; but, to most pilots and the public, stall is where the airplane looses all lift! Howcan it be both? And, if one of these views is wrong, why?

The key to understanding both perspectives of stall is understanding the differencebetween lift and lift coefficient. Lift is the product of the lift coefficient, the dynamicpressure and the wing planform area. For a given altitude and airplane (wing area) lift thendepends on lift coefficient and velocity. It is possible to have a very high lift coefficient CL

and a very low lift if velocity is low.

When an airplane is at an angle of attack such that CLmax is reached, the high angle ofattack also results in high drag coefficient. The resulting high drag normally leads to areduction in airspeed which then results in a loss of lift. In a conventionally designedairplane this will be followed by a drop of the nose of the aircraft into a nose down attitudeand a loss of altitude as speed is recovered and lift regained. If the pilot tries to hold thenose of the plane up, the airplane will merely drop in a nose up attitude. Pilots are taught tolet the nose drop as soon as they sense stall so lift and altitude recovery can begin as rapidlyas possible. A good flight instructor will teach a pilot to sense stall at its onset such thatrecovery can begin before altitude and lift is lost.

It should be noted that if an aircraft has sufficient power or thrust and the high dragpresent at CLmax can be matched by thrust, flight can be continued into the stall and post-region. This is possible on many fighter aircraft and the post-stall flight realm offers manyinteresting possibilities for maneuver in a fight.

The general public tends to think of stall as when the airplane drops out of the sly. Thiscan be seen in almost any newspaper report of an airplane accident where the story line willread "the airplane stalled and fell from the sly, nosediving into the ground after the enginefailed". This kind of report has several errors. Stall has nothing to do with engines and anengine loss does not cause stall. Sailplanes can stall without having an engine and everypilot is taught how to fly an airplane to a safe landing when an engine is lost. Stall alsodoesn't cause a plane to go into a dive. It is, however, possible for a pilot to panic at theloss of an engine, inadvertently enter a stall, fail to take proper stall recovery actions andperhaps "nosedive" into the ground.

110

DRAG AND THRUST REQUIRED

As seen above, for straight and level flight, thrust must be equal to drag. Drag is afunction of the drag coefficient CD which is, in turn, a function of a base drag and aninduced drag.

CD = CD0 + CDi

We assume that this relationship has a parabolic form and that the induced drag coefficienthas the form

CDi = KL2.

We therefore write

CD = CD0 + KL2 .

K is found from inviscid aerodynamic theory to be a function of the aspect ratio andplanform shape of the wing

where e is unity for an ideal parabolic form of the lift distribution along the wing's spanand less than one for non-ideal spanwise lift distributions.

The drag coefficient relationship shown above is termed a parabolic drag "polar"because of its mathematical form and is actually only valid for inviscid wing theory. Inthis text we will use this equation as a first approximation to the drag behavior of an entireairplane. While this is only an approximation, it is a fairly good one for an introductorylevel performance course. It can, however, result in some unrealistic performanceestimates when used with some real aircraft data.

The drag of the aircraft is found from the drag coefficient, the dynamic pressure and thewing planform area:

D = CD

1

2V∞

2S

Therefore,

D = CD0 + KCL2( ) 1

2V∞

2 S

111

Realizing that for straight and level flight, lift is equal to weight and lift is a function ofthe wing's lift coefficient, we can write:

CL =L

1

2V∞

2S=

W1

2V∞

2S

giving:

D = CD0

1

2V∞

2S +KW2

1

2V∞

2S

The above equation is only valid for straight and level flight for an aircraftin incompressible flow with a parabolic drag polar.

Let's look at the form of this equation and examine its physical meaning. For a givenaircraft at a given altitude most of the terms in the equation are constants and we can write

where

D = AV2 + BV2

A =1

2SCD0

B =KW2

1

2S

The first term in the equation shows that part of the drag increases with the square ofthe velocity. This is the base drag term and it is logical that for the basic airplane shape thedrag will increase as the dynamic pressure increases. To most observers this is somewhatintuitive.

Figure 4.3

112

The second term represents a drag which decreases as the square of the velocityincreases. It gives an infinite drag at zero speed, however, this is an unreachable limit fornormally defined, fixed wing (as opposed to vertical lift) aircraft. It should be noted thatthis term includes the influence of lift or lift coefficient on drag. The faster an aircraft flies,the lower the value of lift coefficient needed to give a lift equal to weight. Lift coefficient, itis recalled, is a liner function of angle of attack (until stall). If an aircraft is flying straightand level and the pilot maintains level flight while decreasing the speed of the plane, thewing angle of attack must increase in order to provide the lift coefficient and lift needed toequal the weight. As angle of attack increases it is somewhat intuitive that the drag of thewing will increase. As speed is decreased in straight and level flight, this part of the dragwill continue to increase exponentially unto the stall speed is reached

Figure 4.4

Adding the two drag terms together gives the following figure which shows the completedrag variation with velocity for an aircraft with a parabolic drag polar in straight and levelflight.

Figure 4.5

113

MINIMUM DRAG

One obvious point of interest on the previous drag plot is the velocity for minimumdrag. This can, of course, be found graphically from the plot. We can also take a simplelook at the equations to find some other information about conditions for minimum drag.

The requirements for minimum drag are intuitively of interest because it seems that theyought to relate to economy of flight in some way. Later we will find that there are certainperformance optima which do depend directly on flight at minimum drag conditions.

At this point we are talking about finding the velocity at which the airplane is flying atminimum drag conditions in straight and level flight. It is important to keep thisassumption in mind. We will later find that certain climb and glide optima occur at thesesame conditions and we will stretch our straight and level assumption to one of"quasi"-level flight.

We can begin with a very simple look at what our lift, drag, thrust and weight balancesfor straight and level flight tells us about minimum drag conditions and then we will moveon to a more sophisticated look at how the wing shape dependent terms in the drag polarequation (CD0 and K) are related at the minimum drag condition. Ultimately, the mostimportant thing is the speed for flight at minimum drag because this is the quantity whichwill enable the pilot to achieve such flight.

Let's look at our simple static force relationships:

L = W, T = D

to write

D = W × D L,

which says that minimum drag occurs when the drag divided by lift is a rninimum or,inversely, when lift dinded by drag is a maximum.

This combination of parameters, L D , occurs often in looking at aircraft performance.In general, it is usually intuitive that the higher the lift and the lower the drag, the better anairplane. It is not as intuitive that the maximum lift-to drag ratio occurs at the same flightconditions as minimum drag. This simple analysis, however, shows that

MINIMUM DRAG OCCURS WHEN L/D IS MAXIMUM.

Note that since CL CD = L D we can also say that minimum drag occurs when CL CD ismaximum. It is important to note that minimum drag does not connoteminimum drag coefficient.

Minimum drag occurs at a single value of angle of attack where the lift coefficientdivided by the drag coefficient is a maximum:

Dmin when CL CD( )max

114

As noted above, this is not at the same angle of attack at which CD is at a minimum. It isalso not the same angle of attack where lift coefficient is maximum. This should be ratherobvious since CLmax occurs at stall and drag is very high at stall.

CL

CD

max

≠CL max

CD min

Since minimum drag is a function only of the ratio of the lift and drag coefficients andnot of altitude (density), the actual value of the minimum drag for a given aircraft at a givenweight will be invariant with altitude. The actual velocity at which minimum drag occurs isa function of altitude and will generally increase as altitude increases.

If we assume a parabolic drag polar and plot the drag equation

D = CD0

1

2V∞

2S + 2 KW SV∞2

for drag versus velocity at different altitudes the resulting curves will look somewhat likethe following:

Figure 4.6

Note that the minimum drag will be the same at every altitude as mentioned earlier and thevelocity for minimum drag will increase with altitude.

We discussed in an earlier section the fact that because of the relationship betweendynamic pressure at sea level with that at altitude, the aircraft would always perform thesame at the same indicated or sea level equivalent airspeed. Indeed, if one writes the dragequation as a function of sea level density and sea level equivalent velocity a single curvewill result.

115

Figure 4.7

To find the drag versus velocity behavior of an aircraft it is then only necessary to docalculations or plots at sea level conditions and then convert to the true airspeeds for flightat any altitude by using the velocity relationship below.

VALT = Ve

EVERYTHING YOU ALWAYS WANTED TO KNOW ABOUT MINIMUMDRAG!

We know that minimum drag occurs when the lift to drag ratio is at a maximum, butwhen does that occur; at what value of CL or CD or at what speed?

One way to find CL and CD at minimum drag is to plot one versus the other as shownbelow. The maximum value of the ratio of lift coefficient to drag coefficient will be wherea line from the origin just tangent to the curve touches the curve. At this point are thevalues of CL and CD for minimum drag. This graphical method of finding the minimumdrag parameters works for any aircraft even if it does not have a parabolic drag polar.

116

Figure 4.8

Once CLmd and CDmd are found, the velocity for minimum drag is found from theequation below, provided the aircraft is in straight and level flight

VMD =2W

SCLMD

As we already know, the velocity for minimum drag can be found for sea level conditions(the sea level equivalent velocity) and from that it is easy to find the minimum drag speed ataltitude.

Ve MD=

2W

SLSCLMD

It should also be noted that when the lift and drag coefficients for minimum drag are knownand the weight of the aircraft is known the minimum drag itself can be found from

Dmin =W

LD( )

max

For most of the examples considered in this text a parabolic drag polar will be assumed.In such cases where we know the equation for drag, the minimum drag parameters can befound analytically. For the parabolic drag polar

CD = CDO + KCL2

117

it is easy to take the derivative with respect to the lift coefficient and set it equal to zero todetermine the conditions for the minimum ratio of drag coefficient to lift coefficient, whichwas a condition for minimum drag.

CD

CL

=CDO + KCL

2

CL

Hence,

dCD

CL

dCL

=CL 2KCL( ) − CDO − KCL

2

CL2 = 0

This gives

2K − K − CDO

CL2 = 0

or

CDO

CL2 = K

and

CDO = KCLMD

2

The above is the condition required for minimum drag to occur.

Now, we return to the drag polar

CD = CDO + KCL2

and for minimum drag we can write

CDMD= CDO + KCLMD

2

which, with the above, gives

118

or

CDMD= 2CDO = 2KCL MD

2

CLMD=

CDO

K

From this we can find the value of the maximum lift-to-drag ratio in terms of basicdrag parameters

L D( )max

=CLMD

CDMD

=CDO K

2CDO

LD( )

max=

1

2 CDOK

And the speed at which this occurs in straight and level flight is

VMD =2W

SCLMD

=2W

S CDO KSo we can write the minimum drag velocity as

VMD =2W

SCLMD

K

CDO

4

or the sea level equivalent minimum drag speed

Ve MD=

2W

SLS

K

CDO

4

REVIEW: MINIMUM DRAG CONDITIONS FOR A PARABOLIC DRAGPOLAR

At this point we know a lot about minimum drag conditions for an aircraft with aparabolic drag polar in straight and level flight. The following equations may beuseful in the solution of many different performance problems to be considered later in thistext. There will be several flight conditions which will be found to be optimized whenflown at minimum drag conditions. It is therefore suggested that the student write thefollowing equations on a separate page in her or his class notes for easy reference.

119

Dmin = WL D( )max

= 2W CDOK

CDMD= 2CDO = 2KCL MD

2

CLMD= CDO K

VMD =2W

S

K

CDO

4

L D( )max

= 12 CDOK

EXAMPLE

An aircraft which weighs 3000 pounds has a wing area of 175 square feet and an aspectratio of seven with a wing aerodynamic efficiency factor (e) of 0.95. If the base dragcoefficient, CDO, is 0.028, find the minimum drag at sea level and at 10,000 feet altitude,the maximum lift-to-drag ratio and the values of lift and drag coefficient for minimum drag.Also find the velocities for minimum drag in straight and level flight at both sea level and10,000 feet.

We need to first find the term K in the drag equation.

Now we can find

Dmin = 2W CDOK = 220lb

CDMD= 2CDO = 0.056

CLMD= CDO

K= 0.764

L D( )min

= CLMDCDMD

=0.764

0.0506=13.64

We can check this with

L D( )MAX

=2

2 CDOK=

1

0.0733= 13.64

The velocity for minimum drag is the first of these that depends on altitude.

120

VMD =2W

S

K

CDO

4

At sea level

VMDS. L .= 14,430

ft 2

sec2

0.048

0.0284 = 137.5 ft sec.

To find the velocity for minimum drag at 10,000 feet we an recalculate using the density atthat altitude or we can use

VMDALT= VMDS. L.

VMD10 K= SL

10 K

VMDSL= 0.002376

0.001756137.5( ) =160 ft sec.

It is suggested that at this point the student use the drag equation

D = CDO

1

2V∞

2S

+

2KW 2

V∞2 S

and make graphs of drag versus velocity for both sea level and 10,000 foot altitudeconditions, plotting drag values at 20 fps increments. The plots would confirm the abovevalues of minimum drag velocity and minimum drag.

FLYING AT MINIMUM DRAG

One question which should be asked at this point but is usually not answered in a texton aircraft performance is "Just how the heck does the pilot make that airplanefly at minimum drag conditions anyway?"

The answer, quite simply, is to fly at the sea level equivalent speed for minimum dragconditions. The pilot sets up or "trims" the aircraft to fly at constant altitude (straight andlevel) at the indicated airspeed (sea level equivalent speed) for minimum drag as given inthe aircraft operations manual. All the pilot need do is hold the speed andaltitude constant.

DRAG IN COMPRESSIBLE FLOW

For the purposes of an introductory course in aircraft performance we have limitedourselves to the discussion of lower speed aircraft; ie, airplanes operating in incompressibleflow. As discussed earlier, analytically, this would restrict us to consideration of flightspeeds of Mach 0.3 or less (less than 300 fps at sea level), however, physical realities of

121

the onset of drag rise due to compressibility effects allow us to extend our use of theincompressible theory to Mach numbers of around 0.6 to 0.7. This is the range of Machnumber where supersonic flow over places such as the upper surface of the wing hasreached the magnitude that shock waves may occur in flow deceleration resulting in energylosses through the shock and in drag rises due to shock-induced flow separation over thewing surface. This drag rise was discussed in Chapter 2.

As speeds rise to the region where compressiblility effects must be considered we musttake into account the speed of sound a and the ratio of specific heats of air, gamma.

a2 = RT = P , = CP CV , P = RT

Gamma for air at normal lower atmospheric temperatures has a value of 1.4.

Starting again with the relation for a parabolic drag polar, we can multiply and divideby the speed of sound to rewrite the relation in terms of Mach number.

D = CDO

1

2V 2S + 2KW2 SV2

P

P

where M2 = V 2 a2 = V2 P

or D =2

SPCDOM2 + 2W2 K SPM2

The resulting equation above is very similar in form to the original drag polar relation andcan be worked with mathmatically in a similar fashion. For example, to find the Machnumber for minimum drag in straight and level flight we would take the derivative withrespect to Mach number and set the result equal to zero. The complication is that someterms which we considered constant under incompressible conditions such as K and CDOmay now be functions of Mach number and must be so evaluated.

dD

dM= SPCDOM +

SPM2

2

dCDO

dM−

4W2K

SPM 2 +2W2

SPM 2

dK

dM= 0

Often the equation above must be solved itteratively.

REVIEW

To this point we have examined the drag of an aircraft based primarily on a simplemodel using a parabolic drag representation in incompressible flow. We have furtherrestricted our analysis to straight and level flight where lift is equal to weight and thrustequals drag.

122

The aircraft can fly straight and level at a wide range of speeds, provided there issufficient power or thrust to equal or overcome the drag at those speeds. The student needsto understand the physical aspects of this flight.

We looked at the speed for straight and level flight at minimum drag conditions. Onecould, of course, always cruise at that speed and it might, in fact, be a very economicalway to fly (we will examine this later in a discussion of rang and endurance). However,since "time is money" there may be reason to cruise at higher speeds. It also might just bemore fun to fly faster. Flight at higher than minimum drag speeds will require less angle ofattack to produce the needed lift (to equal weight) and the upper speed limit will bedetermined by the maximum thrust or power available from the engine.

Cruise at lower than minimum drag speeds may be desired when flying approaches tolanding or when flying in holding patterns or when flying other special purpose missions.This will require a higher than minimum drag angle of attack and the use of more thrust orpower to overcome the resulting increase in drag. The lower limit in speed could also bethe result of the drag reaching the magnitude of the power or the thrust available from theengine; however, it will normally result from the angle of attack reaching the stall angle.Hence, stall speed normally represents the lower limit on straight and level cruise speed.

It must be remembered that all of the preceding is based on an assumption of straightand level flight. If an aircraft is flying straight and level at a given speed and power orthrust is added, the plane will initially both accelerate and climb until a new straight andlevel equilibrium is reached at a higher altitude. The pilot can control this addition ofenergy by changing the plane's attitude (angle of attack) to direct the added energy into thedesired combination of speed increase and/or altitude increase. If the engine output isdecreased, one would normally expect a decrease in altitude and/or speed, depending onpilot control input.

We must now add the factor of engine output, either thrust or power, to ourconsideration of performance. It is normal to refer to the output of a jet engine as thrustand of a propeller engine as power. We will first consider the simpler of the two cases,thrust.

THRUST

We have said that for an aircraft in straight and level flight, thrust must equal drag. Ifthe thrust of the aircraft's engine exceeds the drag for straight and level flight at a givenspeed, the airplane will either climb or accelerate or do both. It could also be used to maketurns or other maneuvers. The drag encountered in straight and level flight could thereforebe called the thrust required (for straight and level flight). The thrust produced actuallyby the engine will be referred to as the thrust available.

Although we can speak of the output of any aircraft engine in terms of thrust, it isconventional to refer to the thrust of jet engines and the power of prop engines. Apropeller, of course, produces thrust just as does the flow from a jet engine; however, foran engine powering a propeller (either piston or turbine), the output of the engine itself ispower to a shaft. Thus when speaking of such a propulsion system most references are toits power. When speaking of the propeller itself, thrust terminology may be used.

123

The units employed for discussions of thrust are Newtons in the SI system and poundsin the English system. Since the English units of pounds are still almost universally usedwhen speaking of thrust, they will normally be used here.

Thrust is a function of many variables including efficiencies in various parts of theengine, throttle setting, altitude, Mach number and velocity. A complete study of enginethrust will be left to a later propulsion course. For our purposes very simple models ofthrust will suffice with assumptions that thrust varies with density (altitude) and throttlesetting and possibly, velocity. In fact, we will often assume that thrust is invariant withvelocity for a simple jet engine.

If we know the thrust variation with velocity and altitude for a given aircraft we can addthe engine thrust curves to the drag curves for straight and level flight for that aircraft asshown below. We will normally assume that since we are interested in the limits ofperformance for the aircraft we are only interested in the case of 100% throttle setting. It isobvious that other throttle settings will give thrusts at any point below the 100% curves forthrust.

Figure 4.9

In the figure above it should be noted that, although the terminology used is thrust anddrag , it may be more meaningful to call these curves thrust available and thrustrequired when referring to the engine output and the aircraft drag, respectively.

MINIMUM AND MAXIMUM SPEEDS

The intersections of the thrust and drag curves in the figure above obviously representthe minimum and maximum flight speeds in straight and level flight. Above the maximumspeed there is insufficient thrust available from the engine to overcome the drag (thrustrequired) of the aircraft at those speeds. The same is true below the lower speedintersection of the two curves.

The true lower speed limitation for the aircraft is usually imposed by stall rather than theintersection of the thrust and drag curves. Stall speed may be added to the graph as shownbelow:

124

Figure 4.10

The area between the thrust available and the drag or thrust required curves can becalled the flight envelope. The aircraft can fly straight and level at any speed between theseupper and lower speed intersection points. Between these speed limits there is excessthrust available which can be used for flight other than straight and level flight. This excessthrust can be used to climb or turn or maneuver in other ways. We will look at some ofthese maneuvers in a later chapter. For now we will limit our investigation to the realm ofstraight and level flight.

Note that at the higher altitude, the decrease in thrust available has reduced the "flightenvelope", bringing the upper and lower speed limits closer together and reducing theexcess thrust between the curves. As thrust is continually reduced with increasing altitude,the flight envelope will continue to shrink until the upper and lower speeds become equaland the two curves just touch. This can be seen more clearly in the figure below where alldata is plotted in terms of sea level equivalent velocity. In the example shown, the thrustavailable at 22,000 feet falls entirely below the drag or thrust required curve. This meansthat the aircraft can not fly straight and level at that altitude. That altitude is said to be abovethe "ceiling" for the aircraft. At some altitude between 20,000 and 22,000 feet there willbe a thrust available curve which will just touch the drag curve. That altitude will be theceiling altitude of the airplane, the altitude at which the plane can only fly at a single speed.We will have more to say about ceiling definitions in a later section.

125

Figure 4.11

Another way to look at these same speed and altitude limits is to plot the intersections of thethrust and drag curves on the above figure against altitude as shown below. This showsanother version of a flight envelope in terms of altitude and velocity. This type of plot ismore meaningful to the pilot and to the flight test engineer since speed and altitude are twoparameters shown on the standard aircraft instruments and thrust is not.

Figure 4.12

126

It may also be meaningful to add to the figure above a plot of the same data usingactual airspeed rather than the indicated or sea level equivalent airspeeds. This can be donerather simply by using the square root of the density ratio (sea level to altitude) as discussedearlier to convert the equivalent speeds to actual speeds. This is shown on the graphbelow. Note that at sea level V = Ve and also there will be some altitude where there is amaximum true airspeed.

Figure 4.13

SPECIAL CASE OF CONSTANT THRUST

A very simple model is often employed for thrust from a jet engine. The assumption ismade that thrust is constant at a given altitude. We will use this assumption as our standardmodel for all jet aircraft unless otherwise noted in examples or problems. Later we willdiscuss models for variation of thrust with altitude.

The above model (constant thrust at altitude) obviously makes it possible to find arather simple analytical solution for the intersections of the thrust available and drag (thrustrequired) curves. We will let thrust equal a constant

T = T0

therefore, in straight and level flight where thrust equals drag, we can write

T0 = D = CD

1

2V∞

2 S

= CDqS

where q is a commonly used abbreviation for the dynamic pressure.

127

T0 = CDOqS + KW2 qSor

T0qS = CDO qS( )2 + KW2

and rearranging as a quadratic equation

CDO qS( )2 − T0 qS( ) + KW2 = 0

Solving the above equation gives

or

qS = 12

V∞2S = T0

2CDO

± 12

T02

CDO2

− 4KW2

CDO

V 2 =T0

CDO S±

T0

CDO S

2

−4KW2

CDO2S2

In terms of the sea level equivalent speed

Ve2 =

T0

CDO SLS±

T0

CDO SLS

2

−4KW2

CDO SL2 S2

These solutions are, of course, double valued. The higher velocity is the maximumstraight and level flight speed at the altitude under consideration and the lower solution isthe nominal minimum straight and level flight speed (the stall speed will probably be ahigher speed, representing the true minimum flight speed).

There are, of course, other ways to solve for the intersection of the thrust and dragcurves. Sometimes it is convenient to solve the equations for the lift coefficients at theminimum and maximum speeds. To set up such a solution we first return to the basicstraight and level flight equations T = T0 = D and L = W.

T0

W=

D

L=

CD

CL

=CDO + KCL

2

CL

TW

CL = CDO + KCL2

or

CL2 −

T0

W

CL

K+

CDO

K= 0

128

solving for CL

CL =T0

2KW±

1

2

T0

KW

2

−4CDO

K

This solution will give two values of the lift coefficient. The larger of the two valuesrepresents the minimum flight speed for straight and level flight while the smaller CLis for the maximum flight speed. The matching speed is found from the relation

V =2W

SCL

REVIEW FOR CONSTANT THRUST

The figure below shows graphically the case discussed above. From the solution of thethrust equals drag relation we obtain two values of either lift coefficient or speed, one forthe maximum straight and level flight speed at the chosen altitude and the other for theminimum flight speed. The stall speed will probably exceed the minimum straight and levelflight speed found from the thrust equals drag solution, making it the true minimum flightspeed.

Figure 4.14

As altitude increases T0 will normally decrease and VMIN and VMAX will move togetheruntil at a ceiling altitude they merge to become a single point.

It is normally assumed that the thrust of a jet engine will vary withaltitude in direct proportion to the variation in density. This assumption issupported by the thrust equations for a jet engine as they are derived fromthe momentum equations introduced in chapter one of this text. We cantherefore write

129

TALT = TSL = ALT

SL

TSL

EXAMPLE

Earlier in this chapter we looked at a 3000 pound aircraft with a 175 square foot wingarea, aspect ratio of seven and CDO of 0.028 with e = 0.95. Let us say that the aircraft isfitted with a small jet engine which has a constant thrust at sea level of 400 pounds. Findthe maximum and minimum straight and level flight speeds for this aircraft at sea level andat 10,000 feet assuming that thrust available varies proportionally to density.

If, as earlier suggested, the student, (yes you!) plotted the drag curves for this aircraft,a graphical solution is simple. One need only add a straight line representing 400 poundsto the sea level plot and the intersections of this line with the sea level drag curve give theanswer. The same can be done with the 10,000 foot altitude data, using a constant thrustreduced in proportion to the density.

Given a standard atmosphere density of 0.001756 sl ft3 , the thrust at 10,000 feet willbe 0.739 times the sea level thrust or 296 pounds. Using the two values of thrust availablewe can solve for the velocity limits at sea level and at l0,000 ft.

VSL2 = T0

CDO SLS± T0

CDO SLS

2

− 4KW2

CDO SL2 S2

= 34357 ± 8.2346 × 108

= 63053 or 5661

VSL = 251 ft sec max( )or = 75 ft sec min( )

Thus the equation gives maximum and minimum straight and level flight speeds as 251 and75 feet per second respectively.

It is suggested that the student do similar calculations for the 10,000 foot altitude case.Note that one cannot simply take the sea level velocity solutions above and convert them tovelocities at altitude by using the square root of the density ratio. The equations must besolved again using the new thrust at altitude. The student should also compare theanalytical solution results with the graphical results.

As mentioned earlier, the stall speed is usually the actual minimum flight speed. If themaximum lift coefficient has a value of 1.2, find the stall speeds at sea level and add themto your graphs.

130

PERFORMANCE IN TERMS OF POWER

The engine output of all propeller powered aircraft is expressed in terms of power.Power is really energy per unit time. While the propeller output itself may be expressed asthrust if desired, it is common to also express it in terms of power.

While at first glance it may seem that power and thrust are very different parameters,they are related in a very simple manner through velocity. Power is thrust multiplied byvelocity. The units for power are Newton-meters per second or watts in the SI systemand horsepower in the English system. As before, we will use primarily the Englishsystem. The reason is rather obvious. The author challenges anyone to find any pilot,mechanic or even any automobile driver anywhere in the world who can state the powerrating for their engine in watts! Watts are for light bulbs: horsepower is for engines!Actually, our equations will result in English system power units of foot-pounds persecond. The conversion is

one HP = 550 foot-pounds/second.

We will speak of two types of power; power available and power required.Power required is the power needed to overcome the drag of the aircraft

PREQ = D × V .

Power available is equal to the thrust multiplied by the velocity.

PAV = T × V .

It should be noted that we can start with power and find thrust by dividing by velocity,or we can multiply thrust by velocity to find power. There is no reason for not talkingabout the thrust of a propeller propulsion system or about the power of a jet engine. Theuse of power for propeller systems and thrust for jets merely follows convention and alsorecognizes that for a jet, thrust is relatively constant with speed and for a prop, power isrelatively invariant with speed.

Power available is the power which can be obtained from the propeller.Recognizing that there are losses between the engine and propeller we will distinguishbetween power available and shaft horsepower. Shaft horsepower is the powertransmitted through the crank or drive shaft to the propeller from the engine. The enginemay be piston or turbine or even electric or steam. The propeller turns this shaft power(Ps) into propulsive power with a certain propulsive efficiency, P.

PAV = PPS

131

The propulsive efficiency is a function of propeller speed, flight speed, propeller designand other factors.

It is obvious that both power available and power required are functions of speed, bothbecause of the velocity term in the relation and from the variation of both drag and thrustwith speed. For the ideal jet engine which we assume to have a constant thrust, thevariation in power available is simply a linear increase with speed.

Figure 4.15

It is interesting that if we are working with a jet where thrust is constant with respect tospeed, the equations above give zero power at zero speed. This is not intuitive but isnonetheless true and will have interesting consequences when we later examine rates ofclimb.

Another consequence of this relationship between thrust and power is that if poweris assumed constant with respect to speed (as we will do for prop aircraft) thrustbecomes infinite as speed approaches zero. This means that a Cessna 152 whenstanding still with the engine running has infinitely more thrust than a Boeing 747 withengines running full blast. It also has more power! What an ego boost for theprivate pilot!

In using the concept of power to examine aircraft performance we will do much thesame thing as we did using thrust. We will speak of the intersection of the power requiredand power available curves determining the maximum and minimum speeds. We will findthe speed for minimum power required. We will look at the variation of these with altitude.The graphs we plot will look like that below.

Figure 4.16

132

While the maximum and minimum straight and level flight speeds we determine fromthe power curves will be identical to those found from the thrust data, there will be somedifferences. One difference can be noted from the figure above. Unlike minimum drag,which was the same magnitude at every altitude, minimum power will be different at everyaltitude. This means it will be more complicated to collapse the data at all altitudes into asingle curve.

POWER REQUIRED

The power required plot will look very similar to that seen earlier for thrust required(drag). It is simply the drag multiplied by the velocity. If we continue to assume aparabolic drag polar with constant values of CDO and K we have the following relationshipfor power required:

P = DV = CD

1

2V∞

3S

= CDO

1

2V∞

3S +2KW2

V∞S

We can plot this for given values of CDO, K, W and S (for a given aircraft) for variousaltitudes as shown in the following example.

Figure 4.17

We will note that the minimum values of power will not be the same at each altitude.Recalling that the minimum values of drag were the same at all altitudes and that powerrequired is drag times velocity, it is logical that the minimum value of power increaseslinearly with velocity. We should be able to draw a straight line from the origin through theminimum power required points at each altitude.

The minimum power required in straight and level flight can, of course be takenfrom plots like the one above. We would also like to determine the values of lift and dragcoefficient which result in minimum power required just as we did for minimum drag.

133

One might assume at first that minimum power for a given aircraft occurs at the sameconditions as those for minimum drag. This is, of course, not true because of the addeddependency of power on velocity. We can begin to understand the parameters whichinfluence minimum required power by again returning to our simple force balanceequations for straight and level flight

D = DW

L= W

CD

CL

, V =2W

SCL

PR = DV = WCD

CL

V = WCD

CL

2W

SCL

PR = 2W3

SCD

CL3 2

Thus, for a given aircraft (weight and wing area) and altitude (density) the minimumrequired power for straight and level flight occurs when the drag coefficient divided by thelift coefficient to the two-thirds power is at a minimum.

PRminwhen CD CL

3 2( )min

Assuming a parabolic drag polar, we can write an equation for the above ratio ofcoefficients and take its derivative with respect to the lift coefficient (since CL is linear withangle of attack this is the same as looking for a maximum over the range of angle of attack)and set it equal to zero to find a maximum.

d

dCL

CD CL3 2( ) = d

dCL

CDO + KCL2

CL3 2

= 0

CL32 2KCL( ) − CDO + KCL

2( ) 3

2CL

1 2 = 0

3CDO = KCL2

CLmin P=

3CDO

K

Note that

CLMP= 3CLMD

134

The lift coefficient for minimum required power is higher (1.732 times) than that forminimum drag conditions.

Knowing the lift coefficient for minimum required power it is easy to find the speed atwhich this will occur.

VMP =2W

SCL MP

=2W

S

K

3CDO

4

Note that the velocity for minimum required power is lower than that for minimum drag.

VMP =1

3

1 4

VMD = 0.76 VMD

The minimum power required and minimum drag velocities can both be foundgraphically from the power required plot. Minimum power is obviously at the bottom ofthe curve. Realizing that drag is power divided by velocity and that a line drawn from theorigin to any point on the power curve is at an angle to the velocity axis whose tangent ispower divided by velocity, then the line which touches the curve with the smallest anglemust touch it at the minimum drag condition. From this we can graphically determine thepower and velocity at minimum drag and then divide the former by the latter to get theminimum drag. Note that this graphical method works even for nonparabolic drag cases.Since we know that all altitudes give the same minimum drag, all power required curves forthe various altitudes will be tangent to this same line with the point of tangency being theminimum drag point.

Figure 4.18

135

One further item to consider in looking at the graphical representation of power requiredis the condition needed to collapse the data for all altitudes to a single curve. In the case ofthe thrust required or drag this was accomplished by merely plotting the drag in terms ofsea level equivalent velocity. That will not work in this case since the power required curvefor each altitude has a different minimum. Plotting all data in terms of Ve would compressthe curves with respect to velocity but not with respect to power. The result would be aplot like the following:

Figure 4.19

Knowing that power required is drag times velocity we can relate the power required atsea level to that at any altitude.

or

DVe = DVVe

V= DV

P = CDO

12 SLSVe

3 + 2KW 2

SLSVe

The result is that in order to collapse all power required data to a single curve we must plotpower multiplied by the square root of sigma versus sea level equivalent velocity. This,therefore, will be our convention in plotting power data.

Figure 4.20

136

REVIEW

In the preceding we found the following equations for the determination of minimumpower required conditions:

PRminwhen CD CL

3 2( )min

, when 3CDO = KCL2

CLMP=

3CDO

K= 3CL MD

VMP = 2WS

K3CDO

4 = 0.76 VMD

We can also write

CDMP = CDO + KCL2 = 4CDO

CDMP= 2CD MD

Thus, the drag coefficient for minimum power required conditions is twice that forminimum drag. We also can write

L D( )MP

=CLMP

CD MP

=3

16KCDO

= 0.866 L D( )max

Since minimum power required conditions are important and will be used later to findother performance parameters it is suggested that the student write the above relationshipson a special page in his or her notes for easy reference.

Later we will take a complete look at dealing with the power available. If we know thepower available we can, of course, write an equation with power required equated to poweravailable and solve for the maximum and minimum straight and level flight speeds much aswe did with the thrust equations. The power equations are, however not as simple as thethrust equations because of their dependence on the cube of the velocity. Often the bestsolution is an itterative one.

If the power available from an engine is constant (as is usually assumed for a propengine) the relation equating power available and power required is

P = CONST =1

2CDO V 3S +

2KW2

VS

137

For a jet engine where the thrust is modeled as a constant the equation reduces to that usedwhen thrust was discussed.

EXAMPLE

For the same 3000 lb airplane used in earlier examples calculate the velocity for minimumpower.

VMPSL=

2W

SL S

K

CDO

4 = 14430 0.57144

VMPSL= 104.44 ft sec

It is suggested that the student make plots of the power required for straight and levelflight at sea level and at 10,000 feet altitude and graphically verify the above calculatedvalues. It is also suggested that from these plots the student find the speeds for minimumdrag and compare them with those found earlier.

138

CHAPTER 5

ALTITUDE CHANGE: CLIMB AND GLIDE

Through the basic power and thrust performance curves considered in the last chapterwe have been able to investigate the straight and level flight performance of an aircraft. Wemust now add another dimension to our study of performance, that of changes in altitude.We know that from the straight and level data we can determine the theoretical maximumaltitude, or ceiling, for a given aircraft. The question to be answered now is how do we getthe aircraft from one altitude to the other? This discussion must include the investigation ofpossible rates of climb and descent, the distance over the ground needed to climb a givenaltitude and the range of the aircraft in a glide. How fast can I get from altitude A to altitudeB? How far can I glide after my engine fails? If I take off 600 feet from the end of therunway, can I clear the trees ahead?

To look at altitude changes we need to think in terms of energy changes. In climb weare turning kinetic and internal (engine) energy into an increase in potential energy. In aglide we are converting potential energy into velocity (kinetic energy) which will give usneeded lift for flight.

One of the questions above involved the rate of climb. In climbing, the aircraft isincreasing its potential energy. Rate of climb then involves the change of potential energyin a given time. The engine provides the needed energy for climb and the engine energyoutput per unit time is power (work per unit time). We are aware that a certain amount ofpower is required for straight and level flight at a given speed. To climb at that same speedthen requires extra power and the amount of that extra power will determine the rate atwhich climb will occur. The maximum rate of climb at a given speed will then depend onthe difference between the power available from the engine at that speed and the powerrequired for straight and level flight. This can be determined from the power performanceinformation studied in the last chapter.

The concept of adding power to increase altitude (climb) is usually not intuitive. Mostof us are conditioned by experience with cars, boats and bicycles to think of speed increaseas a consequence of adding power. These, of course, are vehicles limited to the altitude ofthe road or water surface. If we think about a car going over a hill, however, the process isnot hard to understand. If a car is traveling at, say, 55 mph (since none of us would thinkof driving at speeds over the limit!) and we start up a hill holding the accelerator (throttle)steady, the car will decelerate as it climbs the hill. To maintain our 55 mph as we move upthe hill we must add power. The same is true in an aircraft.

One of the most difficult things for a flight instructor to teach a new pilot is that thethrottle controls the altitude and the control stick or yoke controls the speed. This is, ofcourse, not entirely true since the two controls are used simultaneously; however, this isthe analogy that will best serve the pilot in a difficult situation. For example, on anapproach to landing the pilot is attempting to hold a steady descent toward the runway. If asudden downdraft causes a loss of altitude the pilot must take immediate action to regain the

139

lost altitude or run the risk of an unplanned encounter with the ground short of the runway!Pulling back on the control to bring the nose of the aircraft up is the most commoninstinctive response since the aircraft is descending with the nose down. This will,however, merely increase the angle of attack and result in a reduction in speed, possiblyleading to stall and certainly leading to further loss of lift and altitude. The properresponse, adding power, will result in a climb to recover from the altitude loss. Theultimate control of the aircraft in such a circumstance will require the coordinated use ofboth controls to regulate both speed and altitude during this most difficult phase of flight.

The pilot in the above situation is not going to stop think about her or his aircraft'spower available or power required performance curves. This is the job of the engineer whodesigns the airplane to be able to meet the pilot's needs in such a situation. This is our jobin the sections that follow. In this study we must add an angle to our previous illustrationof the balance of forces on the airplane. This will be the angle of climb, , which will beconsidered positive in a climb and negative in a glide or descent.

Figure 5.1

GLIDING FLIGHT

The first case we will consider will be the simple case of non-powered descent, orglide. This is a very important performance situation for an aircraft since all aircraft aresusceptible to engine failure. One of the first things a student pilot is taught to do is toproperly handle an "engine-out" in his or her aircraft; how to set up the best speed tooptimize the rate of descent in order to allow maximum time to call for help, restart theengine, prepare for an emergency landing, etc. For some aircraft of course, the unpoweredglide is normal. Sailplanes and hang gliders come to mind immediately but one should alsoconsider that the Space Shuttle is nothing but an airplane with an "engine-out" in its descentfrom orbit to landing!

In an unpowered glide there are only three forces acting on the aircraft, lift, drag andweight. These forces must reach an equilibrium state in the glide. It is up to the pilot tomake sure that the equilibrium reached is optimum for survival and, in most aircraft, it is upto the aircraft designer to make the airplane so that it will seek a reasonable equilibriumposition on its own. The airplane which stalls and goes into a spin upon loss of an enginewill not be very popular with most pilots! We must now determine what those optimumconditions are.

140

Using the figure above and using a thrust of zero we can write the following twosimple force balance relations in the lift and drag directions:

L − W cos = 0

−D − W sin = 0.

Dividing the second equation by the first gives

Tan = − D L =−1L

D( ) .

There's that term again, L D !

This tells us a very simple and very important fact: the glide angle depends onlyon the lift-to-drag ratio.

g = − andTan g =1

LD( ) .

Something seems wrong here. Does this mean that glide angle has nothing to do withthe weight of the aircraft? It sure seems lice a heavy airplane wouldn't glide like a lightone. Will a Boeing 747 glide just like a Cessna 152? What about the Space Shuttle?

Yes, the equation doesn't contain the weight of the aircraft even though it was in theoriginal force balance equations. The glide angle depends only on the lift-to-drag ratio andthat ratio depends on parameters such as CD 0 , K and e as discussed in the last chapter.

But doesn't the fact that there must be sufficient lift to support the weight (or at leastmost of it) mean that weight really is a factor? Not really, since drag is also seen to be afunction of weight and in the ratio of lift-to-drag, the weight "divides outs" of therelationship. A Boeing 747 can indeed glide as well as a Cessna 152.

So, what are our concerns in a glide? Essentially we want to know how far the aircraftcan glide (range) and how long will it take to reach the ground (endurance).

We will look at the range assuming an absence of any natural wind. This is, of course,rarely the case in real life but is the easiest case for us to examine. We will also assume asteady glide, meaning that the pilot has set (or trimmed) the aircraft such that it will hold theselected angle of glide during the entire descent. The geometry of the situation is rathersimple as shown below.

141

Figure 5.2

From the figure it is clear that the glide angle is the arctangent of the change in altitudedivided by the range.

tan g=h2 − h1

R

This gives a range of:

R =h2 − h1

tan g

and since the tangent of the glide angle is simply the lift-to-drag ratio we have

R =L

Dh2 − h1( )

Maximum range in a glide occurs at the maximum-to-drag ratio; ie, atminimum drag conditions! We already know how to find anything we may wish toknow about minimum drag conditions so we know how to determine conditions formaximum range in an unpowered glide.

To the pilot this means that he or she must, upon loss of engine power, trim the aircraftto glide at the indicated (sea level equivalent) airspeed for minimum drag, a speed which theengineer has provided in the aircraft owner's handbook, if maximum range isdesired. The pilot would then fly the aircraft to hold the desired speed in the glide.

Usually, maximum range is not the most desirable goal in an "engine-out" situation.The best solution is usually to optimize the time before "ground encounter" (hopefully alanding!). This means going for minimum rate of descent rather than maximum range.This is, again, one of those things which may not be intuitive to most people, even topilots, and there are many cases where planes have crashed as pilots tried unsuccessfully tostretch their range after loosing an engine. Wind, which wasn't included in the abovecalculations, can cut range to zero or can enhance it. Distances are hard to judge from theair. Student pilots are taught that in an "engine-out" situation it is time that should be

142

optimized rather than range. The pilot needs to know the airspeed for minimum rate ofdescent rather than for maximum range in order to trim the aircraft for a descent which willallow the maximum time to try to restart the engine, to prepare for an emergency landing, toradio for help, etc. This means we are interested in the rate of descent.

Looking at rate of descent is a little more complicated than looking at range. We willconsider two cases, the small glide angle case where we can make some simplifyingassumptions and the general or large angle case. In looking at the small angle case we willuse the usual mathmatical assumption that the cosine of an angle is close enough to unitythat we can approximate it as one. The usual limit of this assumption is about 5 degreessince a check on our calculator will show that cos 5˚ = 0.099619. However, our angle ofinterest is the glide angle which we already know is equal to the arc-tangent of D L . Wewould also like to assume that the sine of that angle is approximately equal to its tangent.Because of this we will stretch the applicability of the small angle rationalization to includeglide angles up to about 15 degrees.

(At fifteen degrees the cosine is 0.9659, so we are still within 5% to of our goal ofcosine = 1.0. This is usually pretty good in the real world. Also the tangent of fifteendegrees is 0.2679 while the sine is 0.2588, making our sine = tangent assumption goodwith less than 4% error.)

A useful result of the small angle assumption is that it will allow us to further assumethat lift is approximately equal to weight. Since we had

L − W cos = 0

and if

cos ˜ = 1 then L ˜ = W .

This might be referred to as "quasi-level" flight. The main advantage of thisassumption is that it allows us to continue to relate velocity to weight through

V ˜ = 2W

SCL

even though flight isn't really straight and level.

Now we want to begin to look at rate of change of altitude, dh dt where dh

dt= ˙ h .

This is the rate of climb when defined in terms of a positive change ofaltitude.

143

Figure 5.3

From the figure above we see that the rate of climb is equal to the plane's airspeedmultiplied by the sine of the angle of climb. Referring to our earlier force balance equationsfor the glide case (no thrust) we can write

or−D − W sin = 0

sin = −D W

and using the small angle assumption that weight is approximately equal to lift gives

˙ h = V sin = −VD W ˜ = −V D L

Changing to a form which uses the force coefficients

˙ h ˜ = −VCD

CL

.

Now use the other small ande assumption for velocity

V ˜ = 2W

SCL

,

we have

˙ h ≅ −2W

SCL

CD

CL

or finally

˙ h ˜ = −2W

S

CD

CL3 2

144

Note that this is a negative rate of climb since we are looking at the case of glide or descent(we assumed no thrust).

From the above it is obvious that for the minimum rate of descent for a given aircraftand altitude will occur when CD CL

3 2 is at a minimum. Looking back at our study ofpower in the previous chapter we find that this is the same condition found forminimum power required.

In review, we have found the conditions needed for flight in an unpowered glide fortwo optimum cases, minimum rate of descent and maximum range with no wind. Theseare found to occur when the descending aircraft is trimmed to hold an indicated airspeed forminimum power required conditions and for minimum drag, respectively. We knoweverything about both of these conditions from the previous chapter's discussion.

We have found that for any glide, the range with no wind is

R =L

Dh2 − h1( ) =

CL

CD

h2 − h1( ),

and for glide angles of fifteen degrees or less the rate of descent is

˙ h ˜ = −2W

S

CD

CL3 2

These can be used to find the range and rate of descent for any glide condition wherewe know the appropriate lift and drag coefficients (angle of attack) and are not limited to theoptimum cases. In addition, we know that to optimize range we need to fly at minimumdrag conditions while for a minimum rate of descent, we need to fly at the conditions forminimum power required.

Most aircraft in a glide will satisfy the fifteen degree small angle assumption used in theabove. A few, such as the Space Shuttle, will not. It is therefore worthwhile to back upand briefly consider the case of steep glide angles. This is, of course, the general casewithout the small angle assumption. We must use the force balance equations as developedwithout the approximations. These become

andL = W cos ,

D = −W sin .

The velocity equation cannot assume straight and level flight and the first of the abovetwo equations must be used to insert aircraft weight into the relationship.

V = 2L SCL , W cos = L

or

V = 2W SCL cos( )145

The glide angle definition is unchanged

tan = −D L =− CD CL

and we can use this relation with some simple trigonometry to find a relationship betweenthe cosine of the glide or climb angle and the lift and drag coefficients.

sin =−CD

CL2 + CD

2, cos =

CL

CL2 + CD

2

The rate of climb (rate of descent) equation now becomes

˙ h = V sin = 2WSCL

cos( )sin

˙ h = −2W

S

CD

CL2 + CD

2( )3 4

This is a relation which will determine the rate of descent for any glide angle. It is notedthat this equation is not really any more complicated mathmatically than that found using thesmall glide angle approbation. The difference is that there is now no correlation betweenthe minimum rate of descent and the condition for minimum power required.

TIME TO DESCEND

Using the rate of descent and the altitude change it is possible to determine the timerequired for that descent.

dt = dh ˙ h

If the rate of descent is constant this can become

t = ∆h ˙ h In reality we have already shown that for both the general and the small angle cases the

rate of descent is not constant but depends on altitude since it is a function of density. Thecomplete equation would therefore be

146

t = − dh2W

S

CD

CL2 + CD

2( )3 4

h1

h 2

∫

and by using the equations for density variation in the standard atmosphere one could insertdensity as a function of h to give a general equation for time of descent. However, to get asimpler picture of the time to descend problem we will assume that an incremental approachcan be used where the density, and thus rate of descent, can be assumed constant overreasonably small increments of altitude during descent. For example, over an increment ofaltitude of 1000 feet we can base our calculations on the density (rate of descent) half-waybetween the upper and lower altitude without introducing much error. This can be repeatedincrementally to find the time of descent over larger altitude changes. A few simpleexamples might help illustrate this process.

Example 1:A sailplane weighs one-thousand pounds and has a wing loading (W S) of 12.5 pounds

per square foot with a drag polar given by CD = 0.010 + 0.022CL2 . Find the time to glide

from 1000 feet to sea level at minimum rate of descent (minimum sink rate).

Solution: Minimum sink rate occurs at conditions for minimum power required

CLMP=

3CD0

K= 1.17, CDMP

= 4CD0 = 0.04

We can check the resulting lift-to-drag ratio to determine if the small angle approximationsare valid

L D( )MP

=1.17

0.04= 29.2, = tan−1 1

LD

= 1.96°.

Thus we can find the velocity from the "quasi-level" equation

VMP =2W

SCL MP

,

and using the density for a 500 foot altitude we have

500 = 0.002343sl ft3 , VMP = 95.5 fps,

and the rate of descent becomes˙ h = V sin = −3.27 fps,

147

giving a time to descend the 1000 feett = ∆h ˙ h = 306sec.

EXAMPLE 2Consider descent of the same sailplane from a much higher altitude. We can use a descentfrom 20,000 feet to investigate the inaccuracies of using the incremental approach to thetime to descend problem. Suppose that in order to get a first guess for the time to descendwe assumed a single increment using the density at 10,000 feet. We will first find anairspeed

V10K =2W

SCL

= 110.27 fps,

then a rate of descent

˙ h = V sin = −3.771 fps,

giving a time for descent of

t =−20,000 ft

−3.771fps= 5303sec = 88.4min.

We should expect improved accuracy if we use four increments of 5000 feet each,calculating velocities and rates of descent at 17,500; 12,500; 7,500; and 2,500 foot

h(ft) h( mean)( ft) V(fps) ˙ h = V sin fps( )

20,00015,000

17,500 0.5793 124.49 4.258

15,00010,000

12,500 0.6820 114.74 3.924

10,0005,000

7,500 0.7982 106.05 3.627

5,0000SEA LEVEL

2,500 0.9288 98.32 3.363

VSL = 94.752 fps

altitudes as shown in the following table.

The total time to descend is found by summing the incremental times from each of the 5000foot descents.

148

t = ∆h˙ h

20 −15

+ ∆h˙ h

15−10

+ ∆h˙ h

10 − 5

+ ∆h˙ h

5 − 0

t = 5313.8sec = 88.6min

This gives a time to descend from 20,000 feet of 88.6 minutes, a difference of only 0.2minutes or 10.8 seconds from the gross, single increment solution.

Does the above show that there is little point in breaking the glide into increments tofind the time of descent or simply that the increments chosen were too large to make muchdifference? A solution of the "exact" integral equation for the 20,000 foot descent willresult in a time of descent of 5426.5 seconds or 90.4 minutes. There is only a two minutedifference between the "exact" solution and the worst possible approximation; a 2% error!

CLIMBING FLIGHT

As discussed earlier, the addition of power above that required for straight and levelflight at a given speed will make possible either an increase in altitude or a change in speedor both. If speed is held constant while power (or thrust) is added the result will be aclimb. Since climb is best thought of as an increase in potential energy we can best analyseit on an energy usage basis as reflected in power or energy addition per unit time. To beginour look at climb we can return to the figure used earlier and again write force balanceequations in the lift and drag directions, this time adding the thrust vector.

L − W cos = 0

T − D − W sin = 0

It should be emphasized that we are assuming that climb occurs at constant speed. Thismeans physically that climb is a straight exchange of energy from the engine for a gain inpotential energy. It also means that our force balance equations sum to zero; ie, are staticequations with no acceleration. We will not, however, restrict ourselves too much. Asevery good engineer should we will fudge a little by saying that we are flying at "quasi-steady: conditions and tolerate very small accelerations which are inevitable in real flight.

The rate of climb relation is still

˙ h = V sin

From the Thrust/Drag force balance above we can write the angle of climb

sin = T − D( ) W

The rate of climb is then

˙ h =T − D

W

V =

PAV − PREQ

W.

149

Note from the above that the angle of climb depends on the amount of excess thrustwhile the rate of climb depends on the amount of excess power. Not surprisingly, this isthe same kind of dependence we found in the gliding case except there we spoke of draginstead of thrust.

Since the angle of climb and rate of climb both can be directly related to previouslydiscussed performance curves for an aircraft, we can take a look at these parameters as theyrelate to these graphs. A typical plot of thrust and drag (thrust required) is shown below.At any given velocity the difference between the thrust and drag curves can be divided bythe aircraft weight to determine the maximum possible angle of climb at that speed using therelationship defined earlier. Of course, at any given speed, not all of the excess availablethrust need be used for climb if a lower angle of climb is desired. As the thrust and dragcurves move together to the left and right, the possible angle of climb narrows toward zeroat the velocities where thrust equals drag.

Figure 5.4

The velocity where the maximum possible angle of climb occurs is that for which thevertical distance between the thrust and drag curves is maximum. This could be foundfrom an actual data plot by simply using a ruler or a pair of dividers to find this maximum.It could also be found analytically if functional relationships are known for the thrust anddrag curves by taking the derivative of the difference in thrust and drag with respect to thevelocity and setting that equal to zero to determine the maximum.

A simple case occurs when it can be assumed that the thrust available from an engine isconstant, an assumption often made for jet engines. If the thrust is constant the maximumdifference between thrust and drag and, hence, the maximum angle of climb, must occurwhen the drag is minimum. Once again minimum drag conditions become the optimum fora performance parameter. It should also be obvious that when thrust is not a constant,minimum drag is probably not the condition needed for maximum angle of climb.

The reader should note that no reference has been made in the above to a parabolic dragpolar and the conclusions reached are not restricted to such a case. In the case of the

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parabolic drag polar we know how to determine the lift and drag coefficients and the speedfor minimum drag from our previous study.

A typical plot of power versus velocity is shown below. We know from above that therate of climb is equal to the difference in the power available and that required, at a givenspeed, divided by the aircraft weight. Thus the power available / power required graph canbe used to graphically determine the rate of climb at any speed in the same manner as thethrust curves were used above. In cases where the power available is assumed constant, asis often the case in a simple representation of a propeller powered aircraft, the maximumrate of climb will occur at the speed where power required is a minimum. We know fromthe previous chapter how to determine the conditions for minimum power required. Ifpower available is not constant, maximum rate of climb will not necessarily occur at thespeed for minimum power required.

Note that the graph shown below plots P versus Ve since this allows the powerrequired data at all altitudes to collapse to a single curve as derived in Chapter 4.

Figure 5.5

It should also be noted that maximum rate of climb and maximum angle of climb do notoccur at the same speed.

It is interesting to compare the power performance curves, and hence the rate of climb,for the two simple models we have chosen for jet and prop aircraft. In the plot whichfollows, the prop aircraft is assumed to have constant power available and the jet to haveconstant thrust. Since power available equals thrust multiplied by velocity, the jet poweravailable data lies in a diagonal line starting at the origin. The power required curveassumes a common aircraft. In other words this is a comparison of the same airplane withtwo different types of engine. It is obvious that at lower speeds the rate of climb for theprop exceeds that for the jet while at higher speeds the jet can outclimb the prop. Thiscomparison, while fictional, is typical of the differences between similar jet and prop

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aircraft. It shows one reason why one would not design a jet powered crop duster sincesuch an aircraft needs a high rate of climb at very low speeds.

Figure 5.6

Special Case: Constant Thrust

In the case mentioned above as a simple model for a jet aircraft, finding the maximumangle of climb is easy since it must occur at the speed for minimum drag or maximum lift-to-drag ratio. The conditions for maximum rate of climb are not as simple. Looking at rateof climb again we recall

˙ h = V sin ,

and assuming quasi-level flight we can write

V =2W

SCL

.

Thus, we have a relationship which has the lift coefficient as the variable.

˙ h =2W

SCL

1 2T − D

W

= CL

−1 2 2W

S

1 2T − D

W

.

152

Again rising the quasi-level assumption which assumes that lift is essentially equal toweight

D W = D L = CD CL .

We now have a relation which includes both lift and drag coefficients as variables.However, we know that drag coefficient depends on the lift coefficient in the drag polar.This gives

or

˙ h =2W

S

T

WCL

−1 2 −CD

CL3 2

˙ h =2W

S

T

WCL

−1 2 −CD0 + KCL

2

CL3 2

.

The above equation is for the constant thrust case and shows the rate of climb as a functionof only one variable, the lift coefficient. To determine the optimum rate of climb it is thennecessary to take the derivative of this equation with respect to the lift coefficient. Only theterms in the brackets need be included in the derivative since it will be set equal to zero.

d

dCL

T

WCL

−1 2 −CDO + KCL

2

CL3 2

= 0.

This gives

KCL2 +

T

WCL − 3CD0 = 0

which can be solved via the quadratic equation to find the value of the lift coefficient whichwill give the highest rate of climb for this special case of constant thrust.

CL ˙ h max=

T

W± T W( )2 +12CD0K

2K, T = CONST.

EXAMPLE

A given aircraft has CDO = 0.013, K = 0.157, W = 35,000 lb, S = 530 sqft, T W = 0.429and thrust is constant with speed. Find the best rate of climb and the associated angle ofclimb.

Before starting our solution we should make sure we understand what is being asked.Note that the best angle of climb was not requested. The angle of climb sought was that for

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the best rate of climb case. Students sometimes assume that the answer sought is alwaysfor some optimum case.

To find the maximum rate of climb we use the relation found above to solve for the liftcoefficient.

CL =−0.429 ± 0.429( )2 +12 0.013( ) 0.157( )

2 0.151( ) = 0.088.

This can then be used to find the associated speed of flight for maximum rate of climb.

Ve =2W

SL SCL

= 794.1 fps.

The angle of climb for maximum rate of climb (not maximum angle of climb) can then befound as follows:

sin = T − D

W= T

W− CD CL

= 0.429 −0.013 + 0.157 0.088( )2

0.088= 0.267

= 15.51°

Finally, these are used together to find the rate of climb itself.

˙ h = V sin = 794.1fps 0.267( )= 212.4 fps = 12,743 ft min .

(units of feet per minute are traditional)

Now let's look at the other optimum, that of maximum angle of climb for this same aircraft.Maximum angle of climb occurs at conditions for minimum drag or maximum L D.

sin( )max

=I

W−

CD

CL

mindrag

=T

W−

1

L D( )max

L D( )max= 1

2 KCD 0

= 11.067

sin = 0.429 − 1

11.067= 0.3386

max = 19.79°

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We can then find the lift coefficient associated with maximum angle of climb and theairspeed at which that occurs.

CLMD= CD0

K= 0.288

Ve =2W

SCL

= 438.96 fps.

Finally, the rate of climb for maximum angle of climb

˙ h = V sin = 148.63fps

= 8918 ft min.

Lets look at the answers above and make sure they are logical.

The maximum rate of climb should be higher than the rate of climb for maximum angleof climb. Is that true?

The climb angle for the maximum rate of climb case should be less than the maximumangle of climb. Is that true?

Maximum angle of climb should occur at a lower airspeed than that for maximum rateof climb. Is that the case?

In all cases the above questions are satisfied. These are some of the questions that thestudent should ask in reviewing the solutions to a problem. Often asking questions such asthese can catch errors which might otherwise be ignored.

One situation in which all pilots are interested in both rate of climb and angle of climb ison takeoff. In a normal takeoff the pilot wants to initially climb at the speed which willgive the maximum rate of climb. This will allow the aircraft to gain altitude in as short atime as possible, an important goal as a precaution against engine or other problems intakeoff. Should an engine fail on takeoff, maximum altitude is desired to allow time torecover and make an emergency landing. There are, however, some situations in which itis in the pilot's best interest to forgoe best rate of climb and go for best angle of climb. Anobvious case is when the aircraft must clear an obstacle at the end of the runway such as atree or tower. The figure below illustrates both cases.

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Figure 5.7

The airplane which flew at maximum rate of climb would have reached the desiredaltitude faster than the plane which flew at maximum angle of climb if that darn tree hadn'tbeen in the way!

TIME TO CLIMB

To find the time to climb from one altitude to another we must integrate over the timedifferential

dtt1

t 2

∫ =dh˙ h h1

h2

∫ =dh

V sinh1

h2

∫

To integrate this expression we must know how V sin varies as a function of altitude.We are usually going to be interested in the minimum time to climb as a limiting case. Thiswill, of course, occur at the speed for maximum rate of climb. This speed will be afunction of altitude.

If we can find the rate of climb at each altitude we can plot rate of climb versus altitudeas shown below. The area under the curve between the two desired altitudes represents thetime to climb between those two altitudes.

156

Figure 5.8

Either of the above methods can be used to find the time to climb. In reality they are thesame. The analytical method may not be as simple as it appears at first since the equationsmust account for the velocity and climb angle variation with altitude, necessitating theincorporation of the standard altitude density equations into the integral. The equationscould be simplified by the assumption of a constant velocity climb or a constant angleclimb.

Power Variation with Altitude

We dealt earlier with the variation of power required (to overcome drag) with altitudeand how the power required curves could be merged into one by plotting power multipliedby the square root of the density ratio. The power available must also be multiplied bythe square root of the density ratio to be included on the same performance plot. Inaddition to this we must be aware of how the power available actually varies with altitude.

For both jet engines (turbojet, fan-jet and turbo-prop) and piston engines the powerproduced by the engine drops in proportion to the decrease in density with increasedaltitude.

PAV ∝ , PALT = ALT

SL

PSL = Psl

For a turbocharged piston engine the turbocharger is designed to maintain sea level intakeconditions up to some design altitude. A simple model of power variation with altitude fora turbocharged engine will have power constant at its sea level value up to about 20,000feet and dropping in direct proportion to decreasing density at higher altitudes.

PAV = CONST]SL

20,000, Ph > 20 K = P20 K

ALT

20 K

.

More complicated situations are possible with multiple stages of turbocharging.157

It must be remembered that in plotting power data versus the sea level equivalentvelocity we must both account for the real variation in power available as just discussedand multiply that result by the square root of the density ratio to make the power availablecurves compatible with the power required curves. This is not redundant. The first changeis made to account for the real altitude effects and the second for a plotting scheme neededto collapse all power required data to a single curve.

CEILING ALTITUDES

In earlier discussion we spoke of the ceiling altitude as that at which climb was nolonger possible. This would be the altitude where the power available curve just touchesthe power required curve, indicating that the aircraft can fly straight and level at only onespeed at that altitude. Here the maximum rate of climb is zero. We define this altitude asthe absolute ceiling. This definition is, however, somewhat misleading.

Theoretically, based on our previous study, it would take an infinite amount of time toreach the ceiling altitude. One could look at the rate of climb possible for an aircraft whichis, say, 500 feet below its absolute ceiling. A very low rate of climb would be found,resulting in a very large amount of time required to climb that last 500 feet to reach theabsolute ceiling. Because of this we define a more practical ceiling called the serviceceiling. The definition of the service ceiling is based on the rate of climb; ie, at whataltitude is the maximum rate of climb so low as to make further climb impractical. This isdifferent for jet and piston powered aircraft. For the piston aircraft, the service ceilingis the altitude at which the rate of climb is 100 feet per minute (or 0.5 meters per second).For the jet aircraft, the service ceiling is the altitude at which the rate of climb is 500 feetper minute (or 2.5 meters per second).

It should be noted that many fighter and high performance aircraft may, in reality, beable to exceed even their absolute ceiling through the use of energy managementapproaches. An aircraft may climb to its service ceiling, for example, and then go into adive, building up excess kinetic energy, and then resume a climb, using both the excesspower and the excess kinetic energy to climb to altitudes higher than that found as"absolute". Also, at very high altitude it may be necessary to include orbital dynamics in theconsideration for climb and ceiling capabilities.

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CHAPTER 6

RANGE AND ENDURANCE

In the earliest days of powered flight the primary concern was getting the aircraft intothe air and back down safely (with safely meaning the ability to limp away after the"landing"). The Wright's famous first flight was shorter than a football field and even for acouple of years after December of 1903 they were content to circle around the family farmThe Wright's home built engines couldn't run for long periods of time and they simplydidn't envision the need or desire for flights over distances of over a few miles. In 1908,however, Scientific American magazine challenged the fledgling aviation community of thetime to produce an aircraft or "aero-plane"* which could fly, in public view, over a distanceof one mile! While the Wright's claimed to be able to make such a flight, their obsessionwith secrecy as they sought military sales and their egotistical belief that no one else couldapproach their expertise in aviation led them to ignore the prize offered by ScientificAmerican for the one mile public flight.

It was Glenn Curtiss, a builder of motorcycle engines and holder of numerous worldspeed records in motorcycle racing, who, in July of 1908, made the first public one-mileflight. Curtiss, who had worked with Alexander Bell and others to develop their ownairplanes, made the flight with newspaper reporters watching and with movie camerasrecording the flight. Curtiss became the top aviator in America and the Wrights werefurious, leading to numerous legal suits as Wilbur and Orville sought to prove in the courtsthat Curtiss and Bell had infringed on their patents. Curtiss went on to outperform theWrights and others in aviation meets in America and Europe. The Wright's subsequentpatent suits aimed at reserving for themselves the sole rights to design and build airplanesin the United States, stagnated aircraft development in America and shifted the scene ofaeronautical progress to Europe where it remained until after World War I.

As aircraft and aviation continued to develop, range and endurance became the primaryobjective in aircraft design. In war, bombers needed long ranges to reach enemy targetsbeyond the front lines and by the end of World War I huge bombers had been developed inseveral countries. After the war, European governments subsidized the conversion of thesegiants into passenger aircraft. Some larger planes had even been built as passenger carryingvehicles before the conflict. Sikorski's early designs are good examples In the UnitedStates, however, with little government interest in promoting air travel for the public untilthe late 1920's, long range aircraft development was more fantasy than fact. In 1927Lindberg's trans-Atlantic flight captured the public's imagination and interest in long rangeflight increased. Lindberg's

* The term "aero-plane" originally referred to a wing, a geometrically planar surface meantto support a vehicle in flight through the air. By the time of the 1903 Wright flight the termhad become associated with the entire flying vehicle. Aviation publications of the first twodecades of the 20th century continued to write of "aeroplanes" then, by the 1920's,"airplanes".

159

flight, like that of Curtiss, was prompted by a prize from the printed media, illustrating therole of newspapers and magazines in spurring technological progress.

By the late 1930's the public began to see flight as a way to travel long distances inshort times, national and international airline routes had developed, and planes like the"China Clipper" set standards for range and endurance. World War II forced people andgovernments to think in global terms leading to wartime development of bombers capableof non-stop flight over thousands of miles and to post-war trans-continental andtrans-ocean aircraft. Since 1903 we have seen aircraft ranges go from feet to non-stopcircling the globe and endurances go from minutes to days!

Fuel Usage and Weight

In studying range and endurance we must, for the first time in this course, consider fuelusage. In the aircraft of Curtiss and the Wrights, it was not uncommon for the engine toquit from mechanical problems or overheating before the fuel ran out. In today's aircraft,range and endurance depend on the amount of fuel on board. When the last drop of fuel isgone the plane has reached its limit for range and endurance. One could, of course, includethe glide range and endurance after the aircraft runs out of fuel, but an airline that operatedthat way would attract few passengers!

Fuel usage depends on engine design, throttle settings, altitude and a number of otherfactors. It is, however, not the purpose of this text to study engine fuel efficiency or thepilot's use of the throttle. We will assume that we are given an engine with certainspecifications for efficiency and fuel use and that the throttle setting is that specified in theaircraft handbook or manual for optimum range or endurance at the chosen altitude. It isassumed that the student will take a separate course on propulsion to study the origins ofthe figures used here for these parameters.

Our primary concern in fuel usage will be the change in the weight of the aircraft withtime. Many of our performance equations used in previous chapters include the aircraftweight. In those chapters we treated weight as a constant. Weight is, in reality, constantonly for the glider or sailplane. For other aircraft the weight is always changing, alwaysdecreasing as the fuel is burned. This means that the aerodynamic performance of theairplane changes during the flight. This does not, however, negate the value of themethods used earlier to study cruise and climb. Those calculations will normally be doneusing the maximum gross weight of the airplane which will lead to a conservative or "worstcase" analysis of those performance parameters. We can also use the methods developedearlier to look at the "instantaneous" capabilities of the aircraft at a given weight, realizingthat at a later time in flight and at a lower weight, the performance may be different.

In considering range and endurance it is imperative that we consider weight as avariable, changing from maximum gross weight at take-off to an empty fuel tank weight atthe end of the flight. To do this we will deal with fuel usage in terms of the weight of thefuel (as opposed to fuel volume, in gallons, which we normally use for automobiles).When our concern is endurance we are interested in the change in weight of the fuel perunit time

dWf

dt

160

and, when range is the concern we want to know how the weight of fuel decreases withdistance traveled.

dWf

dS

Aircraft engine manufacturers like to specify engine fuel usage in terms of "specificfuel consumption". For jet engines this becomes a thrust specific fuel consumptionand for prop aircraft, a power specific fuel consumption. Since thrust and power bringdifferent units into the equations we must consider the two cases separately.

RANGE AND ENDURANCE: JET

We speak of the engine output of a jet engine in terms of thrust; therefore, we speak ofthe fuel usage of the jet engine in terms of a thrust specific fuel consumption,Ct . Ct , is the mass of fuel consumed per unit time per unit thrust. The unit oftime should be seconds and the unit for thrust should be in pounds or Newtons of thrust.

Ct[ ] =sl sec

lb − thrustor

kg sec

N − thrust.

The above is the proper definition of thrust specific fuel consumption, however, it isnot really exactly what we need for our calculations. We would prefer a definition basedon weight of fuel consumed instead of the mass. We will thus define a weightspecific fuel consumption, T , as the weight of fuel used per unit time per unit thrust.This gives units of one-over-time.

T = g Ct( ) inlb − fuel sec

lb − thrustor

N − fuel sec

N − thrust.

The reader should be aware that many aircraft performance texts and propulsion textsare very vague regarding the units of specific fuel consumption. Some even define it interms of mass and give it units of 1 sec onds , making it dimensionally incorrect. Part ofthe confusion, particularly in older propulsion texts, lies in the use of the pound-mass asthe unit of mass. This gives a combination of pounds-mass divided by pounds-force,which, in reality gives sec2 ft . The situation is then further complicated by the authorseemingly throwing in a term called gc which is supposed to resolve the lbm lbf issue. Atany rate it is very important that the engineer using specific fuel consumption carefullyconsider the units involved before beginning the solution of a range or endurance problem.A correctly specified weight specific fuel consumption will have units of sec−1 andwill do so without the use of anything called gc.

Some sources of specific fuel consumption data use units of hours−1 since the hour is amore convenient unit of time. The use of seconds is, however, correct in any standard

161

unit system and the student may be well advised to convert hours to seconds beforebeginning calculation even though this will ultimately result in the calculation of endurancein seconds, giving rather large numbers for answers.

To find endurance we want the rate of fuel weight change per unit time which can bewritten in terms of the thrust specific fuel consumption

dWf

dt= T T.

And, in straight and level flight where thrust equals drag

dWf

dt= T D.

For maximum endurance we want to minimize the above term. This clearly shows thatfor maximum endurance the jet plane must be flown at minimum drag conditions.We will look at how to find that endurance after taking a brief look at range.

To find the flight range we must look at the rate of change of fuel weight with distanceof flight. We might pause a little at this point to realize that this may be more complicatedthan endurance because range will depend on more than simply the aerodynamicperformance of the airplane. It will also require consideration of the wind speed. Anairplane can fly forever at a speed of 100 mph into a 100 mph head-wind and still have arange of zero! For now, however, we will put those worries aside and look at the simplemathmatics with which we begin consideration of the problem. We looked above at therate of weight change with time. We can combine this with the change of distance withtime (speed) to get the rate of change of weight with distance.

dWf

dS=

dWf dt

dS dt= T T

V= T

D

V.

Note that we still assume D = T or straight and level flight.

From the above it is obvious that maximum range will occur when the drag divided byvelocity ( D V ) is a minimum. This is not a condition which we have studied earlier but wecan get some idea of where this occurs by looking at the plot of drag versus velocity for anaircraft.

162

Figure 6.4

On this plot a line drawn from the origin to intersect the drag curve at any point has atangent equal to the drag at the point of intersection divided by the velocity at that point.The minimum possible value of D V for the aircraft represented by the drag curve mustthen be found when the line is just tangent to the drag curve. This point will give thevelocity for maximum range. Note that it is a higher speed than that for minimum drag(which, in turn was higher than the speed for minimum power).

In the above we have found the conditions needed to achieve maximum range andendurance for a jet aircraft. We have not yet found equations for the actual range orendurance. To find these we need to return to the time and distance differentials andintegrate them. For time we have

dt =dWfuel

TT.

We now wish to put the equations in a form which includes the weight of the aircraftinstead of the weight of the fuel. Since the change in weight of the aircraft in flight is equaland opposite the weight of fuel consumed

dW = −dWfuel ,

we have

dt = −dW

TT.

163

Finally, integrating over time to find the endurance gives

E = −dW

TTW1

W2

∫

In a similar manner the range is found from the distance differential

dS =−VdW

TT,

or

R = −Vdw

TTW1

W2

∫ .

In the above equations we must know how the aircraft velocity, thrust and specific fuelconsumption vary with aircraft weight. At this point we need to make some assumptionsabout the way the flight is to be conducted. This is sometimes called the "flight schedule".

Approximate solutions for range and endurance for a jet

The first assumption to be made in finding range and endurance equations is that theflight will be essentially straight and level. In order to give ourselves some leeway we willcall this "quasi-level" flight, our desire being merely to use the L = W. T = D relations.

T = D = WD

L= W

CD

CL

,

and we can also use

V =2W

SCL

Substituting these into the range and endurance relationships above give

E = −dW

TD=−

W1

W 2

∫1

T

CL

CD

dW

WW1

W2

∫ , R = −1

TW1

W2

∫2

s

1 2CL

1 2

CD

dW

W1 2 .

At this point we need to make some further assumptions about the flight schedule inorder to simplify the integration of these equations. For example, in the enduranceequation, if we assume that the flight is made at constant angle of attack, we are assuming

164

that the lift and drag coefficients are constant for the entire flight. If we also assume thatthe specific fuel consumption is constant for the flight the only variable left in the integral isweight itself and the integral becomes:

E = −1

T

CL

CD

dW

WW1

W2

∫ =1

T

CL

CD

lnW1

W2

const( )

The range integral contains an additional variable, the density of the atmosphere. It isstill possible to make a couple of combinations of assumptions which will result in simpleintegration and realistic flight conditions. The first case will be to assume cruise at bothconstant altitude and constant angle of attack giving both density and the lift and dragcoefficients as constants in the integration.

R =2 2

T S

CL1 2

CD

W1 − W2( ) constconstalt .

.

A second simple case combines the assumptions of constant angle of attack andconstant speed, which can be used with the earlier form of the range equation.

R = −VdW

TT= −

V

TW1

W2

∫W 1

W2

∫CL

CD

dW

W

giving

R =V

T

CL

CD

lnW1

W2

const

constV

.

Note that the last equation above is simply the endurance equation multiplied by thevelocity. This should not be surprising since this is the case where velocity is constant.

In the final equations above for range and endurance we should note that if standardunits are used with specific fuel consumptions in sec−1, range will be given in feet ormeters and endurance in seconds. We may find it easier to ascertain the degree to whichour answers are realistic if we convert these answers to miles or kilometers and hours.

165

In finding the above equations for range and endurance we have looked only at specialcases which would result in simple integrations. If we know more complicated flightschedules we can determine the functional relationships between the lift and dragcoefficients, velocity, density, etc. and weight loss during flight and insert them into theoriginal integrals to solve for range and endurance. The above cases are, however, verydose to actual operational cruise conditions for long range aircraft and will probably sufficefor an introductory study of aircraft performance. Let's take a look at those simple cases.

Both range cases included our endurance assumption of constant angle of attack andspecific fuel consumption. The first case combined these assumptions with specification ofconstant altitude. This appears to be the simplest case to actually fly but to see what itactually means we need to go back to the straight and level flight velocity relation

V =2W

SCL

, V ∝ W ]CL =CONST

=CONST.

If altitude (density) and angle of attack (lift coefficient) are both constant it is obvious thatthe velocity must change as the weight changes. In other words, for this flight schedule asfuel is burned and the weight of the aircraft decreases, the flight speed must decrease inproportion to the square root of the weight.

The other case, constant speed combined with constant angle of attack, is seen from thevelocity relation above to require that density decrease in proportion to the weight.

W = const]CL =const

V = const.

This means that as the aircraft burns off fuel, the aircraft will slowly move to higheraltitudes where the density is lower. This is commonly known as the drift-up flightschedule. This is actually very similar to the way that commercial airliners fly long distanceroutes. Those of you who have been on such flights will recall the pilot announcing that"we are now cruising at 35,000 feet and will climb to 39,000 feet after crossing theMississippi" or some such plan. While the FAA will not allow aircraft to simply "drift-up"as they fly from coast-to-coast, they will allow schedules which incrementally approximatethe drift-up technique.

It must be noted that the two range equations above will give two different answers forthe same amount of fuel. Also note that the equations are based on only the cruise portionof the flight. An actual flight will include take-off, climb to cruise altitude, descent andlanding in addition to cruise. Allowance also must be made for reserve fuel to handleemergency situations and "holds" imposed by air traffic controllers.

The biggest assumption used in all the integrations above is that of constant angle ofattack. While this fits our conditions for optimum cases such as maximum endurance

166

occurring at maximum lift-to drag ratio (minimum drag), it may not fit real flight very well.While the pilot can easily monitor his or her airspeed and altitude, the airplane's angle ofattack is not as easily monitored and directly controlled.

The equations above for range and endurance are valid for any flight condition whichfalls within the assumptions made in their derivation. If we have a Boeing 747 flying at anangle of attack of eight degrees and a speed of 250 miles per hour these equations can beused to find the range and endurance even though this is obviously not an optimum speedand angle of attack. Should we wish to determine the optimum range or endurance we usethe values of lift and drag coefficient and the velocity which we found earlier to be neededfor these optimums.

Earlier we found that for maximum endurance the aircraft needs to fly at minimum dragconditions. Our actual endurance equation confirms this, showing endurance as a functionof the lift-to-drag coefficient ratio which will be a maximum if drag is a minimum.

We also found that range would be optimum if the drag divided by velocity was aminimum. The correlation between this condition and the range equations derived is not asobvious as that of minimum drag with the endurance equation. Using the straight and levelflight force relations which can be manipulated to show

D = WD

L

= W

CD

CL

,

the quantity V D can be written

V D =V

D

CL

CD

,

Now using the velocity relation for straight and level flight

V =2W

SCL

we find

V

D=

2W

S

1

W

CL1 2

CD

.

Therefore, we find that the maximum range occurs when, for a given weight and altitude

CL1 2

CD

is a maximum.

167

If we assume a parabolic drag polar with constant CD 0 and K we can write

CL1 2

CD

=CL

12

CD0 + KCL2

To find when this combination of terms is at a maximum we can take its derivative withrespect to its variable (CL ) and set it equal to zero.

d

dCL

CL1 2

CD

=

CD0 + KCL2( ) 1

2CL

−1 2 − CL1 2 2KCL( )

CD2 = 0

Solving this gives

1

2CD 0 + KCL

2( )CL−12 − 2KCL

2( )CL−1 2 = 0

orCD 0 + KCL

2 − 4KCL2 = 0

then

CD 0 = 3KCL2

and, finally

CL = CD0 3K

Thus, for maximum range

CLMAXR=

CD 0

3K.

Using this in the drag polar gives the value of drag coefficient for maximum range

CDMAXR= CD 0 + KCL MR

2 = CD 0 + KCD

3K=

4

3CD0 .

168

These are referred to as the conditions for "instantaneous" maximum range. The terminstantaneous is used because the calculations are for a given weight and we know thatweight is changing during the flight. In other words, at any point during the flight, at theweight and altitude at that point, the lift and drag coefficients found above will give the bestrange.

RANGE AND ENDURANCE: PROP

An examination of range and endurance for aircraft which have performance measuredin terms of power (propeller aircraft) by defining a power specific fuel consumptionsimilar to the thrust specific fuel consumption used for jets. The power specific fuelconsumpfion is defined as the mass of fuel consumed per unit time per unit shaftpower. The units are slugs per unit power per second in the English systemor kilogram per unit power per second in SI units.

CP[ ] = sl power ⋅ sec or Kg power ⋅ sec.

The power units used are horsepower in the English system and watts in SI units.

Just as in the jet (thrust) case we will often find an alternate definition of specific fuelconsumption given in terms of the weight of fuel consumed instead of the mass.

While the proper time unit is seconds we will often find such data given for an engine interms of hours. We will develop our equations in terms of the fundamental units (secondsfor time) and, as in the jet case, assume "quasi-level" flight which has

PAVAIL = PREQ = DV .

In dealing with prop engines we must consider the propulsive efficiency, P , whichrelates the shaft power coming from the engine itself to the power effectively used by theprop to transfer momentum to the air.

P = DV P

As for the jet, to find endurance we must consider

dWf

dt= PP = P

DV

P

,

169

and for range are interested in

dWf

dS=

dWf dt

dS dt= PD

P

.

From the above equations it is obvious that, for a given specific fuel consumption andefficiency, the rate of fuel use is a minimum (instantaneous endurance is a maximum) whenthe power required (DV) is a minimum. It is also obvious that the fuel use per amount ofdistance traveled is a minimum (instantaneous range is a maximum) when the drag is aminimum.

So we again run into our old friends minimum power required and minimumdrag as conditions needed for optimum flight. We already know how to find thesegraphically from power versus velocity plots as shown below. This graphicaldetermination of minimum power and minimum drag speeds is valid for any drag polar,even if not parabolic.

Figure 6.5

At this point we should pause and say: "Hey, wait just a minute! It was only acouple of pages back that you said that maximum endurance occurred atminimum drag conditions. Now you say it is maximum range that I get atminimum drag conditions. Make up your mind, for Pete's sake!"

The problem is that in one case we are talking about jets and the other, prop aircraft. Thismeans that we must be very careful to see which type of plane we aredealing with before starting any calculations. It is very easy to get into a big rushand get the two cases mixed up (especially in the heat of battle on a test or exam!).

170

Now, as for the jet, we can develop integrals to determine the range or endurance forany flight situation. For endurance we have

E = dtt1

t 2

∫ =− Pdw

P DV,

W1

W2

∫

and for range

R = dS = − PdW

PD.

W1

W2

∫S 1

S 2

∫

Approximate solutions for range and endurance for a prop aircraft

Once more we will assume "quasi-level" flight and manipulate the terms in our forcebalance relations to give

D = WD

L= W

CD

CL

.

This makes the endurance integral

E = − P

PW1

W2

∫1

V

CL

CD

dW

W3

2.

Using the straight and level velocity relation

E = − P

P

S

2

CL3 2

CD

dW

W.

W1

W2

∫

The range integral can be written in a similar fashion as

R = − P

P

CL

CD

dW

WW1

W2

∫

Now we need to consider the same flight schedules examined in the jet case. Constantangle of attack flight will give constant lift and drag coefficients and constant altitude willgive constant density. We will also assume constant specific fuel consumption.

171

For range we need only to use the constant angle of attack assumption to give a simpleintegral. The resulting range is

R = P

P

CL

CD

lnW1

W2

, const∝( ).

For endurance we will consider tvo cases. The first holds both altitude and angle ofattack constant, giving

E = − P

P

S

2

CL3 2

CD

dW

W 3 2W1

W2

∫

which integrates to

E = P

P

2 SCL

3 2

CD

1

W2

−1

W1

,

const ∝const

.

The second case has angle of attack and velocity constant

E = − P

P

1

V

CL

CD

dW

WW1

W2

∫

or

E = P

P

1

V

CL

CD

lnW1

W2

,

const ∝constV

.

This is the "drift up" flight schedule.

WIND EFFECTS

The above range and endurance equations for both jet or prop aircraft were derivedassuming no atmospheric winds. The speeds in the equations are the airspeeds, not speedsover the ground. If there is a wind the airspeed is, of course, not equal to the speed overthe ground.

172

Endurance calculations are not altered by the presence of an atmospheric wind. If ourconcern is how long the aircraft can stay in the air at a given airspeed and altitude and wedon't particularly care if it is making progress over the ground we need not worry aboutwinds. We are doing endurance calculations based only on the aerodynamic behavior ofthe airplane at a given speed and altitude in a mass of air.

Range is related to speed across the ground rather than the airspeed; thus, if there is awind our range equation results need to be re-evaluated to account for the wind. The logicof this is simple: a headwind will slow progress over the ground and reduce range while atailwind will increase range. What is not so obvious is how to correct the calculations toaccount for this wind. Since our usual concern is to find the maximum range, we willexamine the correction for wind effects only for this optimum situation.

Maximum range for a jet was found to occur when D V was a minimum while, for aprop, maximum range occurred at minimum drag conditions. The velocities for both casescan be determined graphically by finding the point of tangency for a line drawn from thezero velocity origin on either the drag versus velocity curve in the jet case or the powerrequired versus velocity curve for the prop plane. We can use an extension of thisgraphical approach to find the speed for best range with either a head wind or a tail wind.

The important first step in determining optimum range in the presence of anatmospheric wind is to find a new airspeed for best range with a wind. This newspeed will then be used to calculate a new value of the optimum range. The new value ofbest range airspeed is found as illustrated in the figures below. The first task is to draw aconventional drag versus velocity (for a jet) or power required versus velocity (for a prop)plot. To this plot is added a new origin, displaced to the left by the value of atailwind or to the right by the magnitude of the tailwind. A line is then drawnfrom the displaced origin, tangent to the drag or power curve and the point of tangencylocates the new velocity for optimum range with a wind. The magnitude of this newoptimum range velocity is read with respect to the original origin (not thedisplaced origin). This speed is an airspeed, not a ground speed.

Figure 6.6

173

This new optimum range velocity is then used to find a new range value from the sameequations developed previously. Using the new velocity, new values of lift and dragcoefficients are first calculated and these new coefficients and velocity are used to find theoptimum range with the wind. To this new range must be added another range whichresults purely from the aircraft's time of exposure (endurance) to the wind. This enduranceis also found using the newly found optimum range velocity and associated coefficients.The final corrected range for maximum range in a wind is

RW WIND = RCORRECTED + VWECORRECTED tailwind

or

RW WIND = RCORRECTED − VWECORRECTED headwind

LET THE BUYER BEWARE

Airplane manufacturers, like those of automobiles and other products, like to doanything they can to make their product look good and sometimes they hope that the buyerdoesn't look too closely at the contradictions in their specifications and advertising. A carmay be advertised as having seating for five, an EPA fuel economy rating of 38 mpg, theability to go 542 miles on a single tank of gas and a top speed of 120 miles per hour. Mostof us, however, know not to expect that car to go 542 miles on a single tank of gas whilecarrying 5 people at a speed of 120 mph! Those who believe it will would also probably bedumb enough to pay sticker price.

What about airplanes? Is this product of an industry which is regulated at every step bythe FAA just as subject to contradiction in specifications as a car?

Let's look at a few simple examples taken from the 1976 GAMA Aircraft FleetDirectory (a little old, but things haven't really changed much). A Cessna 150, the mostwidely used two place aircraft in the country, quotes a range of 815 nautical miles on 32gallons (210 pounds) of fuel. The plane has an empty weight (no pilot, passenger,baggage or fuel) of 1104 pounds and a maximum gross takeoff weight of 1600 pounds.This means that with the full fuel tanks needed for maximum range there is only a 286pound allowance for both pilot and passenger, hardly enough for two adults and luggage!This is why one of the favorite questions of flight examiners who are preparing for aprivate pilot check-ride in a Cessna 150 involves weight and balance of the aircraft and whysometimes pilots may have to actually pump fuel out of an airplane before takeoff.

A Cessna 172, the most popular four place aircraft in the world, is a little better than the150 cited above. It has an empty weight of 1387 pounds and to reach its advertised rangeof 742 miles it has a fuel tank which holds 288 pounds of gas. This gives a total weightfor airplane and fuel of 1675 pounds. The maximum gross takeoff weight of the 172 is2300 pounds, leaving 625 pounds allowance for four passengers and their stuff; an averageof 156 pounds each! It is beginning to look like airplanes are designed like those "four-place" cars which have a rear seat about large enough to seat two small poodles!

174

With another Cessna product, the all around best of their 4 seat line , the Skylane,things are a little better. Its listed empty weight of 1707 pounds, range of 979 nauticalmiles on 474 pounds of fuel and max gross weight of 2950 pounds leave 769 pounds forpilot, passengers and accessories (192 pounds each). Finally an airplane for real people!

Lest the naive get the idea that this is only a problem for small single engine airplanes,let's look at one more example, one of my favorites, the Learjet 25C. It claims a range of2472 miles, just the ticket for the rich young business tycoon to fill with seven of herclosest friends for a transcontinental weekend jaunt**. That fuel load, weighing 7393pounds, adds to the "zero fuel" weight of 11,400 pounds to give a 18,793 pound airplane.So how much is left for those 8 passengers? A max gross weight of 20,000 pounds wouldallow about 150 pounds for each, perhaps enough if everyone has been on the latest OprahWinfrey diet. The listed max gross weight of the Learjet 25C is 15,000pounds! With a full tank of gas the airplane is over its maximum allowabletakeoff weight! With a 160 pound pilot and no other passengers or load this airplanecan carry enough fuel for a real range of about 1150 miles, less than half that advertised.Why claim a range of almost 2500 miles? Well, the fuel tanks are big enough to carry theneeded fuel. If only the airplane could get off the ground!

* Obviously a biased opinion since the author once owned a Cessna 182!

* * Did everyone catch the "politically correct" reference to the historicallyunderrepresented female of the species?

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CHAPTER 7

ACCELERATED PERFORMANCE:TAKE-OFF AND LANDING

To this point, all of our discussion has related to static or unaccelerated flight where F =ma = 0. Even in climb and descent we assumed "quasi-level" conditions where the forceson the aircraft summed to zero. If we are to look at the performance of an airplane duringtake-off and landing we must, for the first time, consider acceleration (during takeoff) anddeceleration (during landing). We will also have a couple of new forces to consider in theground reaction force and ground friction.

In take-off, the airplane accelerates from zero groundspeed (but not necessarilyairspeed!) to a speed at which it can lift itself from the ground. The thrust must exceeddrag for acceleration to take place and the lift won't equal weight until the moment of lift-off.

In landing, deceleration must be provided through braking and aerodynamic drag toslow the plane to zero speed; hopefully before it reaches the end of the runway!

Any pilot will tell you that take-off and landing are what flight is all about. The thrill offull throttle and maximum acceleration as the plane roars down the runway, followed by thefreeing of the soul which comes from cheating gravity and breaking the bond with the earthis incomparable. Of course the pilot hopes this occurs before the end of the runway isreached and in such a way as to allow clearance of the water tower at the end of the strip!

Landing is the ultimate challenge of man* against nature as the pilot once again attemptsto remain in control of a planned encounter with the ground in a vehicle moving at speedswhich can result in instant mutilation and death if there is the slightest miscalculation ofcrosswind or downdraft. Of course, all of this must be done in such a manner as to assurethe passenger that every move is as safe and natural and controlled as a Sunday afternoondrive to the golf course.

Wind will be a factor in take-off and landing and one would think it would be obviousthat the pilot should position the aircraft at the end of the runway which will result inoperation into the wind. This will result in a reduction in the length of the ground roll ineither take-off or landing. To some, however this may not be obvious.

The author once sat on a graduate committee of a student in Transportation Engineeringwho had taken several courses in airport design. When asked what role the prevailingwinds played in the design of airports the student appeared puzzled. Given a hint that ithad something to do with the way the runways were aligned, he still drew a blank.

* The term "man" is here used in its pure and innocent generic sense without prejudice orpreference with regard to sex, race, creed or political affiliation!

176

Finally, when asked to draw a runway and show an airplane getting ready to take-off at oneend and to explain which way the wind would be blowing, the student's eyes lit up in anapparent revelation of truth. He drew the runway horizontal across the center of theblackboard with the airplane at the right end, ready to begin a take-off roll toward the left.Then he triumphantly drew an arrow to indicate a wind moving from right-to-left, the samedirection as the motion of the aircraft!

As dispair and gloom settled over the faculty in the room I, rather reluctantly, askedhim why the airplane would take-off in the same direction as the wind blew. He repliedthat the answer was obvious, "So the wind will carry the pollution away with theairplane!" Watch out for environmentalists who design airports!

To study aircraft performance in take-off and landing we must make sure we haveproper definitions of what these phases of flight entail. Then we must consider the forcesacting on the airplane. We will begin this study by looking at take-off.

TAKE OFF PERFORMANCE

The definition used by the Federal Aviation Administration for take-off includes theground run from zero ground speed to the point where the wheels leave the ground, plusthe distance required to clear a 50 foot obstacle. The distance over the ground for all of theabove is computed at maximum gross weight at sea level standard conditions. The "worstcase" condition is often also calculated for a hot day at high altitude (100˚F in Denver).

We will concern ourselves only with the ground run portion of the take-off run,knowing that we can find the distance to clear a 50 foot obstacle from our climb equations.That climb would be calculated for maximum angle of climb conditions.

The first step in the calculation of the ground run needed for take-off is an examinationof the forces on the aircraft. In addition to the lift, drag, thrust and weight, we must nowconsider the ground friction and the "Resultant" force of the ground in supporting all orpart of the weight of the aircraft. These are shown in the figure below.

Figure 7.1

177

A summation of the vertical forces in the preceding figure gives

L + R - W = 0

or

R = W-L

Summing the horizontal forces gives

T − D − R = mdV dt.

Note that in the above relation we have, for the first time, an acceleration. These forceschange as the aircraft accelerates from rest to take-off speed.

Combining the two equations above we have a single relation

T − D − W − L( ) =W

g

dV

dt,

which can be rearranged to give

gT

W−

−

g

WD − L( ) =

dV

dt.

Our desire is to integrate this or a related equation to get the time and distance neededfor the take-off ground run. To do this we must first account for the dependence of bothlift and drag on velocity. This gives

dV

dt= g

T

W−

−

g

W

1

2V 2S CD − CLg( ).

The above equation still contains thrust and weight, both of which may well changeduring the take-off ground run. Thrust is known to be a function of velocity, however,weight will be a function of the rate of fuel use (specific fuel consumption) and will be afunction of time rather than speed. In order to keep our analysis relatively simple we willconsider the weight change during the take-off roll to be negligible and treat weight as aconstant in the equation. We can assume a rather simple model for thrust variation withspeed.

T = TO − aV2 .

In this equation TO is the thrust at zero velocity or the "static thrust", a is a constant (whichcould be zero) and T is the thrust at any speed. Substituting this model for thrust into ouracceleration equation gives:

178

dV

dt= g

TO

W−

−

g

W

1

2S CD − CLg( ) + a

V

2

It should be noted that the velocity in this equation is the airspeed and not the speedrelative to the ground. When we look at the take-off distance we will have to be concernedwith both the ground speed and the airspeed. The simplest case will be when there is noground wind; ie, when the airspeed and ground speed are equal.

In the above relation all of the terms in brackets and parentheses are essentially constantfor a given aircraft at a given runway altitude and for a given runway surface. The liftcoefficient is given the special designation of CLg to denote that it is the value for theground run only. In a normal take-off roll the airplane accelerates to a pre-determinedspeed and then "rotates" to a higher angle of attack which will produce enough lift to resultin lift-off at that speed. Hence, the ground run lift coefficient will probably not be the sameas the take-off lift coefficient. The drag coefficient could be similarly subscripted;however, since CD is a function of CL and will subsequently be written in that manner, thiswill not be done at this point.

Since most of the terms in the equation can be treated as constants the equation can besimplified as follows:

dV dt = A − BV 2

where

A = gTO

W−

, B =

g

W

1

2S CD − CLg( ) + a

This can be integrated to obtain the time for the ground run of take-off.

dtt1

t 2

∫ =dV

A − BV2V1

V2

∫

Assuming that the airplane starts the take-off run from rest and that there is no ground windand that the other limit is the take-off velocity VTO, we have

t =1

ABtan h−1 VTO

B

A

= time for take-off.

NOTE: This may be the first time that the student has ever seen an inverse hyperbolictangent. What should follow is a frantic search of the student's calculator to see if thereis any such key or combination of keys combined by an equally harried check of the indicesof high school and college trig and calculus texts to see just what the heck this thing is. Inreality, the student will probably shrug his or her shoulders at this time in the assumption

179

that this will never appear on a test or homework. The author will not be available atmidnight the night before the homework is due to answer these questions!

A question which should be considered here is "What is a good value for the take-offspeed?"

The very lowest speed at which the airplane can possibly lift off of the ground is thestall speed for straight and level flight at the runway altitude. It is, however, not safe toattempt takeoff at this minimum speed with the airplane right on the verge of stall. Asomewhat higher than stall speed will give a margin of safety which will allow take-off at afairly low speed without risk of stall due to unexpected gusts or similar problems. Acommonly used value for take-off speed is a speed 20 percent higher than straight and levelstall speed.

VTO = 1.2VSTALL

Far more important than the time required for the take-off ground run is the distancerequited. It is always nice to know that the pilot can get the airplane into the air before itreaches the end of the runway! To find the take-off distance we must integrate overdistance instead of time.

dV

dS=

dV dt

dS dt=

A − BV 2

V.

Rearranging this gives

dS =VdV

A − BV2,

which is integrated to get

S2 − S1 = −1

2Bln A − BV2( )]

V1

V2

Finally

S2 − S1 =1

2Bln

A − BV12

A − BV22

.

Now, assuming that the airplane starts from rest, no wind and lift off at VTO we have

STO =1

2Bln

A

A − BVTO2

.

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We will later investigate the case of take-off in a wind.

Before going further with an analytical analysis of the takeoff ground run it isworthwhile to pause and examine the physical aspects of the problem. These are too oftenlost in the equations, especially when we have hidden a lot of terms behind convenientterms like A and B. Let's first write the last equation for take-off distance in its full glory.

STO =W

g S CD − CLg( ) + a[ ] ln

TO

W−

TO

W−

− 1

W

1

2S CD − CLg( ) + a

VTO

2

It is obvious that a great number of factors influence the take-off distance.

It is, for example, intuitive that the ground friction will retard take-off. The retardingforce due to friction will decrease as the lift increases during the take-off run. So, itappears that it might be to our advantage to move down the runway at a high angle of attacksuch that high lift is generated which will result in a reduction in the friction force andenhance the airplane's acceleration to take-off speed. On the other hand, a high angle ofattack will also give a high drag coefficient, retarding acceleration. At some point in thetake-off run the drag force will exceed the friction force. Does this mean the pilot shouldbegin the take-off run at a high angle of attack and then lower it to reduce drag so as to holdsome friction/drag ratio at an optimal value?

What about the value of the friction coefficient? Do we use one type of ground run on aconcrete runway and another on a grass strip? What about soft dirt? Typical values offriction coefficient are:

Concrete, asphalt 0.02Hard Turf 0.04Normal turf, short grass 0.05Normal turf, long grass 0.10Soft ground 0.10 to 0.30

Indeed, for a "soft field" take-off such as on long grass or soft ground, pilots are taught todo several things to reduce the role of ground friction on the take-off roll. Usually, the useof flaps is recommended to increase the lift coefficient and, if the airplane has a tricycle typeof landing gear (nose wheel and two main wheels), the pilot is taught to keep the nose up,which will both reduce the friction on that wheel and give a higher angle of attack and liftcoefficient. One reason for the popularity of the "tail dragger" style of aircraft in the earlydays of aviation was it's natural superiority in soft field takeoffs, which were common atthe airfields of the day.

In a normal take-off, as mentioned earlier, the aircraft accelerates along the runway at afairly constant angle of attack until the desired take-off speed is reached. The plane is thenrotated to give an increased angle of attack and lift coefficient such that lift equals orexceeds the weight, allowing lift-off. The angle of attack during that ground roll

181

and, hence the lift and drag coefficients, is largely determined by the relative lengths of thelanding gear and the angle at which the wing is attached to the fuselage.

Many factors influence the size and placement of the landing gear. It is nice if the gearstruts are long enough to keep the propeller from hitting the runway (this can be a realproblem with a tail mounted prop) and it is also good if the center of gravity of the aircraftis between the main and auxillary gear. The main gear should be close to the CG to allowease of rotation but far enough away to prevent inadvertent rotation. There is also thequestion of where the gear are stored in a retractable system

The wing angle of placement on the fuselage will primarily be a function of optimalcruise considerations such that things like the fuselage drag is minimized and pilot visibilityis satisfactory when the wing is at the best combination of lift and drag coefficient forcruise as determined by using relations of previous chapters.

An important task for the designer (perhaps you in a few years) is to find the wingangle of attack which will minimize the take-off ground run and then to design the landinggear such that under normal conditions the plane sits on its gear with the wing at that angle.Let's try to find that angle or, more precisely, the lift and drag coefficients at that angle ofattack.

We first return to the equation for acceleration in the ground run.

dV

dt= g

TO

W−

−

g

W

1

2S CD − CLg( ) + a

V

2

Our desire is to maximize this acceleration at all times during the run. Assuming that theonly variable is angle of attack and thus CL and CD and assuming that we have a parabolicdrag polar and further assuming that the take-off speed VTO is independent of CLg, we canfind the maximum acceleration by taking the derivative with respect to CLg and equating theresult to zero. The assumption that VTO is independent of CLg means that the plane will berotated at VTO to achieve lift off rather than allowed to continue to accelerate until lift of foccurs at CLg.

d

dCL

dV

dt

=

d

dCL

CD − CLg( ) =d

dCL

CDO + KCLg2 − CLg( ) = 0

or

2KCLg − = 0

This gives the best value of the ground run lift coefficient for minimum ground run length.

CLg = 2k

182

A factor not previously noted in this discussion is that we have accounted for the outputof the aircraft's propulsion system in terms of thrust and not power. This was naturalbecause we were dealing with force equations. What do we do when we have an aircraftwhich has a power based propulsion system (propeller)? We know that thrust is equal topower divided by velocity but how do we use that in the equations? Perhaps an examplewill provide an answer

EXAMPLE

For an aircraft with the following properties find the minimum ground run distance atsea level standard conditions.

W = 56,000 lb VTO = 1.15 VSTALL P = 75%

S = 1000 sq.ft. CD = 0.024 + 0.04CL2 TO = 13000 lb

CL max = 2.2 = 0.025 Pshaft = 4800hp

Let's first find the stall speed and then the take-off speed.

Vstall =2W

SCL max

=146 fps.

VTO = 1.15Vstall = 168 fps.

Now we must face the problem of having power information and equations thatdemand thrust data. We have been given the static thrust and we can assume that thepower available which was given will be the power in use at the moment of take-off. Wethen have to determine how thrust varies and how to fit it to our assumed thrust versusvelocity relation used in the take-off acceleration equation.

At take-off speed

PAVAIL = S = 0.75 4800hp( ) = 3600hp,

so, the thrust at take off is

TTO =PAV

VTO

=3600hp

168 fps= 11800lb

Our thrust versus velocity relationship is

T = TO − aV2

so, substituting the takeoff speed and thrust and the static thrust we can find the value for a.

183

11800lb =13000 lb − a 168 ft sec( )2

a = 0.0422lb s ft( )2

Our thrust relationship to be used in the take-off equations is then

T = 13000 − 0.0422V 2

Now we need to determine the lift coefficient for minimum ground run.

CLg =2K

=0.025

0.080= 0.3125.

The drag coefficient at the minimum ground run lift coefficient is:

CD = CD0 + KCLg2 = 0.024 + 0.04 0.3125( )2 = 0.0279.

Finally we can use all of the above to determine the take-off ground run.

S =1

2Bln

A

A − BVTO2

A = gTO

W−

= 32.2 ft sec2 13000

56000− 0.025

= 6.65 ft sec2

B =g

W

1

2S CD − CLg( ) + a

= 3.80 ×10−5 ft−1

S = 12 3.8 × 10−5( ) ft × ln

6.656.65 − 3.8 ×10−5( ) 168( )2

S = 2314 ft.

Takeout without rotation

As described previously, a conventional take-off run would be made at the angle ofattack dictated by the airplane configuration and the landing gear geometry, all of which hasprobably been designed to give near optimal ground run acceleration. When apredetermined take-off speed is reached the pilot raises the nose of the aircraft to increasethe angle of attack and give the lift needed for lift-off. But, what would happen if, insteadof rotating, the airplane was allowed to simply continue to accelerate until it gained enoughspeed to lift off without rotation?

184

Continued ground run acceleration to take-off without rotation is not an optimum wayto achieve flight. It will always require more runway than a conventional take-off. Thereare, however, a limited number of aircraft which are designed for this type of lift-off. Onewell know example is the B-52 bomber. This aircraft has what might be described as"bicycle" type landing gear with the gear located entirely in the long fuselage and placedwell fore and aft the center of gravity. This placement and the long, low fuselage makerotation virtually impossible. The result is the need for very long runways and very long,shallow approaches to landing.

Optimizing the takeoff run for an aircraft like the B-52 is different from the maximumacceleration optimum for a conventional take-off. Since the plane cannot rotate, the geardesign and wing placement on the fuselage must be arranged such that the wing angle ofattack is that desired for a safe and efficient lift-off. Too high an angle of attack mightresult in take-off at conditions too near stall and too low an angle might require too muchrunway. In the following example we look at such an aircraft where the design is such thatthe ground run angle of attack of the wing is set to give take-off at a speed 20% above stallspeed.

EXAMPLE

The aircraft defined below is designed for take-off with no rotation, thus the groundrun angle of attack (and, therefore, CL and CD) is the same as that at take-off. Find thetake-off distance at sea level standard conditions.

W = 75,000 lb CD = 0.02 + 0.05CL2 = 0.02

S = 2500 sq.ft. CL max =1.5 T = TO = 12,000 lbVTO = 1.2 VSTALL

We must first find the stall speed, the take-off speed and the related take-off (and, thus,ground run) lift coefficient.

Vstall =2W

SCL max

=129.7 fps

VTO = 1.2Vstall = 155.7 fps

We can find the lift coefficient for this take-off speed.

CLg = CLTO =W

1

2VTO

2 S=1.042.

Actually we could have skipped some of the above is we realized the following.:

CLg = CLTO =Vstall

VTO

2

CL max =1

1.2

2

1.5( ) = 1.042.

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Using the above lift coefficient we find the drag coefficient.

CD = CD0 + KCL2 = 0.02 + 0.051.042( )2 = 0.0742.

We are now ready to find the takeoff distance.

S =1

2Bln

A

A − BVTO2

A = gTO

W−

= 4.54 ft sec2

B =g

W

1

2S CD − CLg( ) + a

= 6.85 ×10−5 ft−1

S =10

2 6.85( ) ln4.54

4.54 − 6.85 ×10− 5( ) 155.7( )2

S = 3324 ft.

Thrust Augmented Take-off

Although not commonly seen today, a technique once regularly used by military cargoaircraft and bombers like the B-52 to reduce the take-off distance involved the augmentationof ground run thrust through the use of strap-on or built in solid rockets. This system wasoften referred to as JATO for jet assisted take-off even though it used rockets and not jets.Calculation of ground runs for this type of take-off require breaking the ground rundistance integral into two parts to account for the two different levels of thrust used in therun. The resulting equation is as follows:

STO = 12B

lnA

A − BVTO

2

2

+ 12B

lnA1 − B

VTO

2

2

A1 − BVTO2

,

A1 = gTO + TR

W−

.

186

Let us return to the last example and see what happens if we try to shorten the groundrun of this airplane by the use of 15,000 pounds of extra thrust obtained from JATO unitswhich are fired for the first ten seconds of the ground run to boost the plane's initialacceleration.

The total thrust during the first ten seconds of the ground run will be 27,000 pounds.Thus, for that portion of the run the A term in the ground run distance equation will be

A = gTO + TR

W−

= 32.2

2700

7500− 0.02

= 11.0 ft sec2

The B term will not be changed.

Now we must determine the velocity of the aircraft at the end of this first ten seconds ofacceleration since the limits on the distance equation are velocities. To find this we go tothe relationship for take-off ground run time

t1 − t0 =1

ABtanh−1 V1

B

A

− tanh−1 V0

B

A

.

Since the initial velocity is zero and t1 − t0 =10sec. sec we have

10 AB sec = tanh− 1 V1

B

A

.

Solving gives the speed at the end of the augmented thrust portion of the take-off run.

V1 = 107 fps

(The student should verify this answer by using the values of A and B found above andworking through the units.)

The entire distance for take-off can now be found as follows:

S1 − S0 =1

2Bln

A

A − BV12

= 540 ft

S2 − S1 =1

2Bln

A − BV12

A − BV22

= 1939 ft

STOTAL = 2480 ft.

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The JATO boost in this example gave a 25% reduction in the ground run needed fortakeoff. This could be important for such an aircraft if it is operating out of short, remoteairfields often found in "third world" countries or in military operations.

Ground Wind Effects

Earlier we mentioned the importance of ground wind in the take-off of aircraft. It israre that a ground wind does not exist, thus, our "no-wind" equations are, hopefully, worstcase predictions since taking off into the wind will reduce the distance for the ground run.Finding the distance required for take-off into a ground wind (assuming the pilot has thegood sense to fly into the wind and not attempt a "downwind" take-off) requires anotherlook at the equations. [There are conditions such as "downhill" runways or end-of-runwayobstacles which sometimes necessitate a downwind takeoff.]

In the take-off equations it is important to realize that, as noted when first presented, thedistance and acceleration are measured relative to the ground; however, the aerodynamicforces in the equations are obviously dependent on airspeed and not ground speed. Wemust consider this in our equations. In doing so we will use the following designations fordifferent speeds:

VG = GROUND SPEED

VA = AIRSPEED

VW = WIND SPEED (PARALLEL TO RUNWAY)

This gives

VA = VG ± VW, + if a head wind, - if a tail wind.

Returning to the basic equation of motion we have

dVG

dt= A − BVA

2.

However,

VG = VA m VW

Thus

dVG

dt=

dVA

dt= A − BVA

2

So, to determine the time for take-off we use

188

dt =dVA

A − BVA2 .

To find take-off distance we use

dVG

dS=

dVG dt

dS dt=

dVA dt

dS dt=

1

VG

dVA

dt

or

VG

dVG

dS=

dVA

dt= VG

dVA

dS.

This becomes

VA m VW( ) dVA

dS=

dVA

dt= A − BVA

2.

Finally we have a differential which includes wind effects. We will write it only for thecase of the headwind since this would be the normal situation.

dS =VAdVA

A − BVA2 −VW

dVA

A − BVA2

Now, we must also note that the take-off speed of the aircraft is airspeed and notground speed. The time and distance equations above may be integrated above to give

t2 − t1 =1

ABtanh−1 VA2

B

A

− tanh−1 VA1

B

A

S2 − S1 = 12B

lnA − BVA1

2

A − BVA2

2

− VW t2 − t1( )

Finally, realizing that take-off usually starts from rest at zero ground speed (at t = 0), weobtain

189

t =1

ABtan h−1 VATO

B

A

− tanh−1 VW

B

A

S = 12B

lnA− BVW

2

A − BVATO

2

− Vwt.

Note that the take-off speed in these equations is the airspeed for takeoff and not theground speed.

LANDING

Landing, like take-off, is properly defined as having two parts; the "terminal glide"over a 50 foot obstacle to touchdown and the landing ground run. We have alreadyconsidered gliding flight and should be able to deal with this portion of flight. This"terminal glide" will usually not actually be a non-powered glide as studied earlier. Thenormal approach to landing for most aircraft is a powered descent. The FAA definition ofthe landing terminal glide over an obstacle is, however based on an unpowered glide as thelimiting case. A real descent can be the most interesting portion of the flight for a pilot ashe of she corrects for side-winds, updrafts and downdrafts while aiming for a hoped-fortouchdown point on the runway. All of this is done at a descent rate of about 500 feet perminute (about 8 mph).

We will concern ourselves with only the touchdown through full stop portion of thelanding. Again our primary concern will be ground run distance with the hope that fullstop occurs before the end of the runway.

The equations of motion for the landing ground run are identical to those for takeoff,however, the terms in the equations can assume very different magnitudes from those intake-off. To slow the aircraft in its landing ground run high drag is desirable, negative or"reverse" thrust may be used and certainly brakes will be used during much of the run togreatly increase the friction term. The boundary conditions on the integrals are essentiallyreversed with the initial speed being the touchdown or "contact" speed and the final groundspeed being zero.

Before we look at the equations let's look at a typical landing as seen by a small plane,general aviation pilot. The approach-to-landing descent will probably be made using fullflaps, at least in its final "glide" (this will be true for almost any aircraft). This will lowerthe stall speed and allow approach and touchdown at a lower flight speed. It will alsosteeper the approach glide and, on the ground, add to the drag to help slow the aircraft.

190

As soon as the pilot feels that the aircraft is under full control after touchdown he or shewill probably raise the flaps. While this reduces the drag and contributes to a longerground roll, it also reduces the lift, increasing ground friction forces and allowing betterdirectional control of the aircraft in a crosswind. After this is done the brakes will beapplied to further slow the aircraft to a stop. Larger, jet aircraft may apply reverse thrustvery soon after touchdown and before use of brakes to improve deceleration.

Now, let's look again at the equations of motion for an aircraft on the ground. We canstill use

dS = VdV A− BV2( ).

And, if we define VC as the speed of initial ground contact on landing at some point definedas S1 and conditions at any other point in the ground roll as S2 and V2 we have thefollowing integrated equation:

S2 − S1 = −1

2Bln A − BV2

2( ) − ln A − BVC2( )[ ]

or

S2 − S1 =1

2Bln

A − BVC2

A − BV22

.

If the total ground run distance is our interest we have a final speed V2 = 0, giving

S =1

2Bln 1 −

B

AVC

2

.

The time for this landing ground roll is found from

dV

dt= A − BV 2

or

t2 − t1 = dt∫ =dV

A − BV 2V1

V2

∫

191

Integration of this equation can take several different forms depending on the relativemagnitudes and signs of A and B. Looking again at these terms

A = gTO

W−

B =g

W

1

2S CD − CLg( ) + a

,

Note that A will almost always be negative since thrust will always be zero or negative, ifnot at touchdown, then very quickly thereafter. Braking forces could also be large enoughto make B negative, depending on the relative magnitudes of the lift and drag coefficients.In the previously considered take-off case, both A and B would both be positive undervirtually any condition. In various landing situations it may be possible to have anycombination of negative or positive terms and this affects the form of the integral. Thedifficulty arises in the fact that integration gives a square root of the product of A and B aswell as other terms with square roots of A and B individually or ratios of A and B. Theresult can be an imaginary answer if the correct solution is not chosen.

The time of landing ground roll solution is given for the four possible combinations ofA and B below.

1. A > 0, B > 0 : t2 − t1 = 1

2 ABln

A + V B

A − V B

2. A > 0, B < 0: t2 − t1 =1

− ABtan−1 V

−B

A

3. A < 0, B > 0: t2 − t1 = 1− AB

tan−1 VB

−A

4. A < 0, B < 0: t2 − t1 =1

2 ABln

V − B − −A

V −B + −A

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Effect of Wind on Landing Ground Roll

As in the case of taking off, all landings should be made into the wind (with the sameexceptions noted for take-off). The equations must then be written to account for thedifferent velocity terms. This is done exactly as it was for the take-off case.

dVG

dS=

dVG dt

dS dt=

A − BVA2

VG

or

VG

dVG

dS= A − BVA

2 = VG

dVA

dS

For the headwind case this gives:

VA − VW( ) dVA

dS= A − BVA

2

and

dS =VAdVA

A − BVA2 −VW

dVA

A − BVA2 .

Integrating and noting that when the aircraft has come to rest on the ground the velocity willequal that of the wind component along the runway VW,

S =1

2Bln

A − BVC2

A − BVW2

− VW t2 − t1( ).

The last term is evaluated using the time equation already discussed.

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EXAMPLE

The following aircraft touches down in landing at a speed 30% above its stall speed. Thepilot applies the brakes when the plane has slowed to 80% of its touchdown speed. If thereis no wind, find the distance required for the aircraft to come to a complete stop on therunway.

W = 30,000 lb B = 0.5 = 0.02S = 750 sq.ft. CL max = 2.2 with flaps( )

Assume that the lift-to drag ratio at 1.3 times the stall speed has a value of eight and isconstant throughout the ground roll and that thrust is zero at touchdown and throughout theground roll.

Since everything is related to the stall speed we will first find its value.

Vstall =2W

SCL max

=123.6 fps

giving a touchdown speed of

VC = 1.3Vstall = 160.7 fps.

This speed gives a lift coefficient of

CLg = CLVC=

2W

SVC2 =1.30.

We will assume this lift coefficient is constant through the ground run.

We were not given a drag polar equation or its constants but we do know the liftto-dragratio and can find the drag and drag coefficient as follows:

DVe=

W

L D= 3750lb, CDg

=D

1

2VC

2S= 0.1627

Now we can find the A and B terms for the distance solution. We must solve for thedistance in two parts, the distance between touchdown and application of the brakes and theremaining distance to full stop.

194

Before braking = 0.02( )

A1 = gTW

−

= −0.6434 ft sec2

B1 =g

W

1

2S CD − CLg( ) + a

= 1.3085 ×10−4 ft−1

giving a distance of

S1 =1

2B1

lnA1 − B1VC

1

A1 − B1VB2

= 1376 ft

After braking = 0.5( )

A2 = gTW

− B

= −16.085 ft sec2

B2 =g

W

1

2S CD − BCLg( )

= 4.663× 10− 4 ft−1

giving the rest of the ground roll distance as:

S2 =1

2B2

lnA2 − B2VB

2

A2

= 699.4 ft

The total ground roll in landing is the sum of the two distances above:

S = S1 + S2 = 2075.4 ft

195

FAA AND OTHER DEFINITIONS OF TAKEOFF ANDLANDING PARAMETERS

Takeoff

As discussed earlier, there are many components which may be included in thecalculations of takeoff and landing distances. In the previous calculations only the actualground run distances were considered and these, especially during landing, may becomposed of multiple segments where different values of friction coefficient and thrustapply. A complete look at takeoff must also include the distance between the initiation ofrotation and the establishment of a constant rate of climb and the distance needed to clear adefined obstacle height as shown in the figure below.

Figure 7.2

Several different terms may be used in a complete discussion of takeoff. These include thefollowing:

Ground Roll: The distance from the start of the ground run or release ofbrakes until the point where the wheels leave the ground. This includes thedistance needed to achieve the needed lift to equal the weight during rotation.The takeoff velocity must be at least 1.1 times the stall speed and is normallyspecified as between 1.1 and 1.2 times that speed.

Obstacle Clearance Distance: The distance between the point of brakerelease and that where a specified altitude is reached. This altitude is usuallydefined as 50 feet for military or smaller civil aviation aircraft and 35 feet forcommercial aircraft.

Balanced Field Length: The length of the field required for safecompletion of takeoff should one engine on a multi-engine aircraft fail at theworst possible time during takeoff ground run. This distance includes theobstacle clearance distance. The balanced field length is sometimes also calledthe FAR Takeoff Field Length because it is a requirement for FAAcertification in FAR 25 for commercial aircraft and includes the 35 foot obstacleclearance minimum. In the early part of the takeoff ground run the loss of oneengine would usually lead to a decision to abort the takeoff, apply brakes andcome to a stop. The "worst possible time" for engine failure would be when it is

196

no longer possible to stop the aircraft before reaching the end of the runway andthe decision must be made to continue the takeoff with one engine out.

Decision Speed (V1 ): The speed at which the distance to stop after thefailure of one engine exactly equals the distance to continue takeoff on theremaining engines and to clear the FAA defined obstacles. In calculating thisspeed one cannot assume the possibility of using reverse thrust as part of thebraking process.

Landing

As in takeoff, landing actually includes several possible segments as shown in thefigure below. Our previous calculations included only the actual ground roll distance but acomplete definition may also include the portion of the approach needed to clear a definedobstacle and that needed to transition from a steady approach glide to touchdown (the "flaredistance"). Note that the landing ground run could also include portions with reversedthrust used alone or with the brakes.

The weight of the aircraft at landing is normally less than that at takeoff due to the useof fuel during the flight, however it is common to calculate the landing distance of traineraircraft and of most propeller driven aircraft at takeoff weight. For non-trainer jets, landingweight is normally assumed to be 85% of the takeoff weight. Military requirements usuallyassume landing with a full payload and about half of the fuel.

Figure 7.3

As in takeoff, there are several definitions associated with landing which should befamiliar to the performance engineer:

FAR 23 Landing Field Length: This distance includes that needed toclear a 50 foot obstacle at approach speed flying down a defined approachglidepath (normally about 3 degrees). Touchdown is usually at about 1.15times the stall speed. This total distance is usually about twice that of thecalculated ground roll distance. This distance is normally about the same as thatspecified in requests for proposals for military aircraft.

FAR25 Landing Field Length: This distance adds to that of FAR 23above an arbitrary twothirds as a safety margin.

197

CHAPTER 8

ACCELERATED PERFORMANCE:TURNS

Thus far all of our performance study has involved straight line flight. Unfortunately,unless our airplane is flying from a runway that is exactly in line with our destinationrunway and there is no wind on the route, straight line flight isn't very practical! We needto be able to turn.

While the need to be able to turn is fairly obvious to us, a look at early aviation willshow that it often was the last thing on the mind of many aviation pioneers. Theuniqueness of the Wright Flyer was not its ability to fly a few feet in a straight line over thesand at Kitty Hawk. It was unique in its ability to turn and maneuver. There are claims forseveral other early aviators in this country and abroad who may indeed have made short,uncontrolled "hops" or even legitimate straight line "flights" in powered vehicles beforeDecember 17, 1903 but there are no claims for "controlled" flight of a powered, heavierthan air, man (or woman) carrying vehicle prior to this date.

There are several ways to turn a vehicle in flight. Early experimenters such as OttoLilienthal in Germany and Octave Chanute in this country knew that shifting the weight ofthe "pilot" suspended beneath their early "hang gliders" would tilt or "bank the wings toallow turns. Others such as Samuel P. Langley, the turn of the century director of theSmithsonian who had government funding to build and fly the first airplane, designed theircraft to be steered with a rudder like a ship. Neither method of turning was very efficient.Langley's heavier than air powered models, for example, flew very well but couldn't adjustfor winds and flew in long circles instead of a straight line as they had been designed to do.

Banking the wings (called the aero-planes in the 1890's) tilts the lift force to the sideand the sideward component of the lift results in a turn. Using a rudder alone results in aside force on the fuselage of the aircraft and, hence, a turning force. Neither method,employed alone, provides a very satisfactory means of turning and the result is usually avery large radius turn.

The Wright brothers designed a complex mechanism involving coordinated rudders andtwisting of wings to combine both roll and yaw in a "coordinated", efficient turn. Whenthe Wrights took their aircraft to Europe in 1908 they amazed European aviators with theircraft's ability to turn and maneuver. French airplanes, which were the most sophisticatedin Europe, used only rudders to turn. The Wright Flyer, with its "wing warping" systemand coordinated rudder, was literally able to fly circles around the French aircraft.

The Wrights had made this system of ropes and pulleys which connected rudder totwisting wing tips to a cradle under the pilot's body, the central focus of their patent on theairplane. When world motorcycle high speed record holder and engine designer GlennCurtiss, with funding from Alexander Graham Bell and others, built and flew an airplane

198

with performance as good as or better than the Wright Flyer, the Wrights sued for patentviolation. The Curtiss planes, which used either small, separate wings near the wing tipsor wingtip mounted triangular flaps (later to be called ailerons) and which relied on pilotoperation of separate controls like today's stick and rudder system, were able to achieve thesame turning performance as the Wrights Flyer. Curtiss, a much more flamboyant andpublic figure than either of the Wrights, quickly captured the attention and imagination ofthe American public, infuriating the Wrights who had shunned public attention whileconvincing themselves that no one else was capable of duplicating their aerial feats.

The decade long court battle between Curtiss and the Wright family over patent rights todevices capable of efficiently turning an airplane is credited by most historians as allowingEuropean aviators and designers to forge far ahead of Americans. The Wrights were soabsorbed with protecting their patent that they made no further efforts to improve theairplane and the threat of a Wright lawsuit kept all but Curtiss out of the business. Curtiss,whose lack of respect for caution had earlier enabled him to set the world motorized speedrecord on a motorcycle with a V-8 engine, with moral and financial support from Bell andHenry Ford and others, kept the patent suit in court through appeal after appeal andcontinued to build and sell airplanes. To get around the Wright patent, Curtiss, at one timebuilt his aircraft without ailerons or other roll controls and then shipped them to nearbyCanada where one of Bell's companies added the ailerons before the planes were shippedto customers in Europe! Meanwhile, Curtiss continued to experiment and innovate and it isno accident that when the first World War drew American participation it was Curtiss andnot Wright aircraft that went to war. After the war it was the famed Curtiss "Jenny" thatbrought the "barnstorming" age of aviation to all America.

I hope the reader will pardon the above slip into historical fascination. By now some ofyou are asking what the heck all of this has to do with aircraft performance in turns? Thefacts are, however, that the first ten to fifteen years of American flight really weredominated by the airplane's ability to turn.

To keep the physics of our discussion as simple as possible, lets consider only turns atconstant radius in a horizontal plane. This is the ideal turn with no loss or gain ofaltitude which every student pilot practices by flying in circles around some farmer's silo orother prominent landmark.

Our objectives in looking at turning performance will be to find things like themaximum rate of turn and the minimum turning radius and to determine the power or thrustneeded to maintain such turns. We will begin by looking at two types of turns.

Today's airplanes, in general, make turns using the same techniques pioneered by theWrights and improved by Curtiss; coordinated turns using rudder and aileron controls tocombine roll and yaw. The primary exception would be found in some evasive turningmaneuvers made by military aircraft and in the everyday turns of most student pilots!

199

Non-winged vehicles such as missiles, airships, and submarines still make turns likethose of early French aviators and of Langleys "Aerodrome", using rudder and body orfuselage sideforce to generate a "skiding" turn. We will examine this technique beforelooking at the more sophisticated coordinated turn.

The acceleration in any turn of radius R is given by the following relation:

ar = V2 R.

This acceleration is directed radially inward toward the center of the circle.

We can also consider the acceleration from the perspective of the rate of change of the"heading angle", as shown in the figure below.

ar = V ˙

Figure 8.1

The "skid to turn" technique is illustrated below for a constant radius, horizontalturn. A rudder (or even vectored thrust) is used to angle the vehicle and the sideforcecreated by the flow over the yawed body creates the desired acceleration.

Figure 8.2

The equations of motion become

L − W = 0

Y = mV ˙ =mV 2

R.

200

For the skid turn examined above, the turning rate and radius depend on the amount ofside force which can be generated on the body of the vehicle. Note that the lift (orbuoyancy in the cases of submarines and airships) does not enter into the problem.

R =WV2

gY, ˙ = gY

WV .

Now let's look at the coordinated turn. In the ideal coordinated turn as illustrated inthe figure below, the aerodynamic lift is used to both balance the weight such thathorizontal flight is maintained and to provide a side force which produces the desiredturning acceleration. No actual side force is generated on the fuselage of the aircraft.

Figure 8.3 Coordinated Turn

In reality the pilot uses both aileron (roll) and rudder (yaw) to enter such a turn. If theturn is properly coordinated the resulting combined acceleration and gravitational force feltby both airplane and pilot will be directed "down" along the vertical axis of the aircraft andwill be felt by the pilot as an increased force into the seat. The improperly coordinated willbe felt as including a side force pushing the pilot left or right in the seat. These same forcesact on the "ball" in the aircraft's "turn-slip" indicator, moving the ball off center in anuncoordinated turn.

If a turn is not coordinated several results may occur. The turning radius will not beconstant with the airplane either "skidding" outward to a larger radius turn or "slipping"inward to a smaller radius. There could also be a gain or loss of altitude.

In the coordinated turn, part of the lift produced by the wing is used to create theturning acceleration. The remainder of the lift must still counteract the weight to maintainhorizontal flight. Now we look at our only situation where lift is not assumed equal toweight. Lift must be greater than the weight.

201

We must now define a load factor, n, where

L = nW .

This load factor can then be related to the bank angle used in the turn, to the turn radiusand to the rate of turn. Returning to the vertical force balance equation we have

Lcos − W = nW cos − W = 0or

cos =1 n .

Using the other equation of motion we can find the turn radius

R =mV2

Lsin=

mV2

nW sin=

V 2

n

1

sin

Knowing that the cosine of the bank angle is equal to 1 n we can find the value of thesine of the bank angle by constructing a right triangle

hence,

sin =n2 −1

n2 , tan = n2 −1

and the turning radius becomes

R =V 2

g

1

n2 − 1.

In dealing with turns we must remember that lift is no longer equal to weight. The liftcoefficient is then

CL =L

1

2V 2S

=2nW

V2S,

therefore202

V 2 =2nW

SCL

.

The above allows us to write the turning radius in another manner,

R =2W

gSCL

n

n2 − 1

.

It should be noted here that if a small turning radius is desired a high load factor and liftcoefficient are needed and low altitude will help. High wing loading W S( ) will also allowa tighter turn.

The rate of turn in a coordinated turn is

˙ =Lsin

mV=

nmg

mVsin =

ng

V

n2 − 1

n

or

˙ =g

Vn2 − 1.

Alternatively,

˙ =g SCL

2W

n2 − 1

n.

The same factors which contribute to small turning radii give high rates of turn.

Load factor (n)

From the above equations it is obvious that the load factor plays an important role inturns. In straight and level flight the load factor, n, is 1. In maneuvers of any kind theload factor will be different than 1. In a turn such as those described it is obvious that n willexceed 1. The same is true in maneuvers such as "pull ups".

The load factor is simply a function of the amount of lift needed to perform a givenmaneuver. If the required bank angle for a coordinated turn is 60° the load factor mustequal 2. This means that the lift is equal to twice the weight of the aircraft and that thestructure of the aircraft must be sufficient to carry that load. It also means that the pilot andpassengers must be able to tolerate the loading imposed on them by this turn, a load whichis forcing their body into their seat with an effect twice that of normal gravity. This "2g"load or acceleration is also forcing their blood from their heads to their feet and havingother interesting effects on the human body.

203

If we look at the lift relation

Lmax = nmaxW = CL max

1

2V 2S

we see that the maximum lift and therefore the maximum load factor that may be generatedaerodynamically is a function of the maximum lift coefficient (stall conditions).

One must realize that the aircraft, or, more precisely, its wings, may be capable ofgenerating far higher load factors than either the pilot and passengers or the aircraftstructure may be able to tolerate. It is not hard to design aircraft which can tolerate farhigher "g-loads" than the human body, even when the body is in a prone position in aspecially designed seat and uniform. Engineers in the industry will tell you that they coulddesign far more agile fighters at much lower cost if the military didn't insist on havingpilots in the cockpit!

All aircraft, from a Cessna 152 to the X-31, are designed to tolerate certain load factors.The aerobatic version of the Cessna 152 is certified to tolerate a higher load factor than the"commuter" version of that aircraft.

The FAA also imposes certain flight restrictions on commercial aircraft based onpassenger comfort. It is possible to do aerobatics in a Boeing 767 but most of thepassengers wouldn't like it. Passenger carrying commercial flight is therefore normallyrestricted to "g-loads" of 1.5 or less even though the aircraft themselves are capable ofmuch more.

The two-minute turn

General aviation pilots are usually familiar with the "standard rate" or "two-minute"turn. This turn, at a rate of three-degrees per second (0.05236 rad/sec) is used inmaneuvers under controlled instrument flight conditions. To make such a turn the pilotuses an instrument called a "turn-slip" indicator. This instrument, illustrated below,consists of a gyroscope which is partially restrained and attached to a needle indicator, anda curved tube containing a ball in a fluid. As the airplane turns, the gyroscope deflects theindicator needle as it attempts to remain fixed in orientation. The force of the gyroscopeand the resulting needle displacement is proportional to the turn rate. The accuracy of thisindication is not dependent on the degree to which the turn is coordinated. The ball in thecurved tube will stay centered if the turn is coordinated while it will move to the side (rightor left) if it is not coordinated. There is a mark on the face of the instrument which denotesthe needle position for the two-minute turn rate.

204

Figure 8.4 Turn-Slip Indicator

To make a two-minute turn the pilot need only place the aircraft in a turn such that theneedle is at the standard turn indication in the desired direction. To turn 90˚ the turn rate ismaintained for 30 seconds, one minute for 180˚, etc. The vertical speed indicator (rate ofclimb) is used to maintain altitude and the ball is kept centered to coordinate the turn.

Many pilots are taught, incorrectly, that the two-minute turn mark on the turn-slipindicator is an indication of a 15 degree bank angle, with the next mark being 30˚ and soon. Some pilots even refer to the turn-slip indicator as the "turn-bank" indicator when theinstrument has absolutely no way to detect bank. It is possible, using a "cross control"technique, to turn the aircraft via yaw with no bank (much like a missile turns) and see thatthe instrument indicates the correct rate of turn even though there is no bank and, similarly,the aircraft may be placed in roll without turning and the indicator will remain centered.

Why would this error in flight instruction occur? The answer lies partly in the difficultyin erradicating longstanding lore and partly in the fact that, for a small general aviationtrainer, a coordinated two-rninute turn does occur at about a 15 degree bank angle. Let'slook at the numbers.

From our previous equations we have

tan = V˙

g.

Inserting fifteen degrees as the bank angle and a two minute turn rate (0.05236 rad/sec)gives a velocity of 165 ft/sec or 112 mph.

This is indeed close to the speed at which such an airplane would fly in a turn. If we,however, look at a faster aircraft, lets say one that is operating at 350 miles per hour, anduse the two-minute rate of turn we get a very different bank angle of 30 degrees!

Suppose you are a passenger in a Boeing 737 traveling at 600 mph and the pilot set upa two minute turn. This would give a bank angle of 55 degrees. It would also give a loadfactor of 1.75! This is higher than the FAA allows for airline operations. For this reasonairliners use turn rates slower than the two minute turn.

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Instantaneous versus sustained turn conditions

The previously derived relations will give the instantaneous turn rate and radius for agiven set of flight conditions. In other words, for a given set of flight conditions we candetermine the turn rate and radius, etc. Another question which must be asked is "Can theairplane sustain that turn rate?" The pilot may be able to, for example, place the plane in a60 degree bank at 250 mph but may find that there is not enough engine thrust to hold thatspeed, bank angle and maintain altitude.

EXAMPLE

For the airplane with the specifications below find the main turn rate and minimumradius of turn and the speeds at which they occur. Also determine if this turn can besustained at sea level standard conditions.

W / S = 59.88lb ft2 S =167sq ft CL max = 1.5

nmax = 6 CD = 0.018 + 0.064CL2

Tmax = 5000 pounds

The maximum turning rate is

˙ max = g

SCL max

2W

nmax2 − 1

nmax

˙ max = 0.424rad sec = 24.29° sec.

The velocity for this turn rate is

V =2nW

SCL

= 448.6 ft sec.

The minimum turning radius is

Rmin =2W

gSCL

n

n2 − 1=1058 ft = V

˙ .

Now we must see if the plane has enough thrust to operate at these conditions. The dragcoefficient at maximum lift coefficient is

CD = 0.018 + 0.064 CL max( )2= 0.162.

At the speed found above the drag is then

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D = CD

1

2V 2S = 6479lb.

This drag exceeds the thrust available from the aircraft engine!

If the above aircraft enters a coordinated turn at the maximum turn rate it will quicklyslow to a lower speed and turning rate with a larger turn radius or it will loose altitude.

The V-n or V-g Diagram

A Plot which is sometimes used to examine the combination of aircraft structural andaerodynamic limitations related to load factor is the V-n or V-g diagram. This is a plot ofload factor n versus velocity.

We know that when lift exceeds weight

L = nW = CL

1

2V 2S.

We know that one limit is imposed by stall

L = nW = CL max

1

2V2 S.

Rearranging this we can write

n =1

2V 2 CL max

W S,

and we can rearrange this as

V =2n

CL max

W

S.

Plotting n versus V will then give a curve like that shown below.

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Figure 8.5

We can also consider negative load factors which will relate to "inverted" stall; ie, stall atnegative angle of attack. At negative angle of attack, unless the wing is untwisted andconstructed of symmetrical airfoil sections, will have a CLmax different from that atpositive angle of attack. This will give a different but similar curve below the axis.Combining this with the plot above gives the following plot.

Figure 8.6

To the left of this curve is the post-stall flight region which, with the exception of highperformance military aircraft, represents a out-of-bounds area for flight.

Other limits must also be considered. There will obviously be an upper speed limitsuch as that found earlier for straight and level flight. There will also be limits imposed bythe structural design of the aircraft. Depending on the aircraft's structural category (utility,

208

aerobatic, etc.) it will be designed to structurally absorb load factors up to a given limit atpositive angle of attack and another limit at negative angle of attack. Once these aredefined, the complete V-n diagram denotes an operating envelope in terms of load factorlimits.

Figure 8.7

The point where the structural limit line and the stall limit intersect is termed a "cornerpoint". The velocity at this point is limited by both maximum structural load factor andCLmax. The velocity at that point is

Vcorner =2nmax

CL max

W

S.

At speeds below the "corner velocity" it is impossible to structurally damage the airplaneaerodynamically because the plane will stall before damage can occur. At speeds above thisvalue it is possible to place the aircraft in a maneuver which will result in structural damage,provided the plane has sufficient thrust to reach that speed and load.

It is possible for a wind "gust" to cause loads which exceed the above limits. Suchgusts may be part of what is referred to as wind shear and are common aroundthunderstorms or mountain ridges. Gusts can be in either the vertical or horizontaldirection. The primary effect of a horizontal gust is to increase or decrease the likelihood ofstall due to the change in speed relative to the wing. This is often the cause of wind shearaccidents around airports where the aircraft is operating at near-stall conditions.

If a gust is vertical, we can look at its effect in terms of change of angle of attack.Suppose we have a vertical gust of magnitude wg. Its effect on the angle of attack and CLis seen below. If, for example, an aircraft in straight and level flight encounters a verticalgust of magnitude wg the new load factor is

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∆ ∝≅wg

V∞

∆CL = dCL

d∆ ∝

∆CL = a∆ ∝≅ awg

V∞

Figure 8.8

so the change in lift is

∆L =∆ CL

1

2V∞

2S

≅a

2SV∞wg .

Thus

∆n =∆L

W=

a

2

SV∞

Wwg.

If, for example, an aircraft in straight and level flight encounters a vertical gust ofmagnitude wg the new load factor is

n1 = n + ∆n = 1+a V∞wg

2 WS

.

(for straight and level flight n = 1)

The effect of the gust on the load factor is therefore amplified by the flight speed V. Thiseffect can be plotted on the V-n diagram to see if it results in stall or structural failure.

Figure 8.9

210

For the case illustrated above, the gust will cause stall if it occurs at a flight speedbelow Va and can cause structural failure if it occurs at speeds above Vb.

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APPENDIX A

HOMEWORKPROBLEMS

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AOE 3104 ASSIGNMENT 1

Read Chapter 1 in the text

1. Write a computer program in either BASIC or FORTRAN to calculate standardatmosphere conditions (pressure, temperature and density) for any altitude in thetroposphere and stratosphere in both SI and English units. Turn in a listing of theprogram and a print-out for conditions every 1,000 meters (SI units) and every 1000feet (English units) up to 100,000 feet or 30,000 meters.

2. A compressed air tank is fitted with a window of 150 mm diameter. A U-tubemanometer using mercury as its operating fluid is connected between the tank and theatmosphere and reads 1.80 meters. What is the total load acting on the bolts securingthe window? The relative density of mercury is 13.6.

3. On a certain day the sea level pressure and temperature are 101,500 N m2 and 25°C,respectively. The temperature is found to fall linearly with altitude to -55°C at 11,300meters and be constant above that altitude.

An aircraft with no instrument errors and with an altimeter calibrated to ISAspecifications has an altimeter reading of 5000 meters. What is the actual altitude ofthe aircraft? What altitude would the altimeter show when the plane lands at sealevel?

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AOE 3104 HOMEWORK 2

1. Which of the following flows satisfy conservation of mass for an incompressiblefluid?

(a)u = − x3sin y

v = −3x 2 cosy

(b)u = x3 sin y

v =− 3x2 cosy

(c)Ur = 2r sin cos

U = −2r sin2

2. A model is being tested in a wind tunnel at a speed of 100 mph.

(a) If the flow in the test section is at sea level standard conditions, what is thepressure at the model's stagnation point?

(b) The tunnel speed is being measured by a pitot-static tube connected to a U-tube manometer. What is the reading on that manometer in inches of water.

(c) At one point on the model a pressure of 2058 psf is measured. What is thelocal airspeed at that point?

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AOE 3104 HOMEWORK 3

The following information pertains to the flow of air and fuel through a jet engine:

inlet velocity 300 fpsinlet flow density 0.0023 sl ft3

inlet area 4 ft2

exit flow velocity 1800 fpsexit flow density unknownexit area 2 ft2

fuel flow rate 5lbm sec

Use the momentum theorem to find the thrust from this engine. The solution shouldinclude finding the entrance and exit mass flow rates and the exit flow density.

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AOE 3104 HOMEWORK 4

1. To tow a certain body through the atmosphere at sea level requires a power PO

Power required is equal to the drag multiplied by the velocity. Calculate thepower needed to tow the same body under aerodynamically similar conditions(same Reynolds Number) at the base of the stratosphere where the density is0.0007103sl ft3 and viscosity is 2.97 ×10− 7 sl ft.sec.

2. A certain aircraft is designed so that when on the runway during takeoff itswing will be at an angle of attack of five degrees. If the lift curve slope is 0.08per degree and the angle of attack for zero lift is minus one degree, what speedwould the aircraft have to obtain before it lifts off the runway without "rotation"when its "wing loading" (W/S) is 75 lb ft2 . Solve the problem at sea levelstandard conditions and at a standard altitude of 5000 ft.

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AOE 3104 HOMEWORK 5

An aircraft weighs 3000 lb and has a wing area of 175 ft2 , AR = 7, e = 0.95. If CD 0 is0.028, plot drag versus velocity for sea level and 10,000 feet altitudes, plotting drag in 20fps intervals. Confirm that Dmin and VMD are the same as those calculated in the text.Also, using TSL = 400 lb (constant) and the proportional value of thrust at 10,000 ft, findthe maximum and minimum speeds at sea level and 10,000 ft and compare those with thevalues calculated in the text examples.

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AOE 3104 HOMEWORK 6

1. An aircraft weighs 56,000 pounds and has a wing area of 900 ft2 . Its dragequation is given by

CD = 0.016 + 0.04 CL2 .

The airplane is powered by a turbojet engine whose thrust is constant at altitudeas follows:

altitude (ft 0 5000 10,000 15,000 20,000 25,000 30,000thrust (lb) 6420 5810 5200 4590 4000 3360 2700

a. Calculate the minimum thrust required for straight and level flight and thecorresponding true airspeeds at sea level and at 30,000 ft.

b. Calculate the minimum power required and the corresponding trueairspeeds at sea level and 30,000 ft.

2. For the above aircraft:

a. plot thrust and drag vs Ve for straight and level flight.

b. plot altitude vs Ve max and Vmax for straight and level flight.

c. find the altitude for maximum true airspeed.

d. find the maximum obtainable altitude.

e. compare V at minimum drag from the plot with that calculated above.

f. calculate L D( )max

.

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AOE 3104 HOMEWORK 7

1. An aircraft has the following specifications:

W = 24,000 lbS = 600 p,2CD 0 = 0.02K=0056

This aircraft has run out of fuel at an altitude of 30,000 ft. Find the initial andfinal values of its airspeed for best range, the glide angle for best range, itsrate of descent at this speed and find the time taken to descend to sea level at thisspeed.

2. For the aircraft above, assume a sea level thrust of 6000 pounds and assumethat thrust at altitude is equal to the sea level thrust times the density ratio(sigma). Find the true airspeeds for best rate of climb at sea level, at 20,000 ft,30,000 ft and 40,000 ft. Also find the ceiling altitude.

3. For an aircraft where:

W = 10,000 lbW S = 50 psfCD 0 = 0.015K=0.02

find the best rate of climb and the velocity for best rate of climb at sea levelwhere T = constant = 4000 lb and at an altitude of 40,000 ft where T = 2000 lb.

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AOE 3104 HOMEWORK 8

We wish to compare the performance of two different types of "General Aviation"aircraft; the popular Cessna Citation III business jet and the best all-around, four place,single engine, piston plane in the business, the Cessna 182. Approximate aerodynamic andperformance characteristics are given in the table below:

CITATION III CESSNA 182

Wingspan 53.3 ft 35.8 ft

Wing area 318 ft2 174 ft2

Normal gross weight 19,815 lb 2,950 lb

Total thrust at sea level 7300 lb ----------

Usable power at sea level --------- 230 hp

CD 0 0.02 0.025

Oswald Efficiency Factor (e) 0.81 0.80

1. Calculate and tabulate the thrust required (drag) versus Ve data for both aircraftand plot the results on the same graph. Plot the sea level thrust available curvesfor both aircraft on the same graph.

2. Calculate the maximum velocity at sea level for both aircraft and compare withthat indicated on the graph.

3. Calculate and tabulate the power required versus Ve data for both aircraft andplot each on a separate graph. Plot the sea level power available on the samegraphs.

4. Calculate and tabulate the rate of climb (in ft min) versus velocity data at sealevel for both aircraft for normal gross weight and plot the data on the samegraph.

5. Calculate and tabulate the maximum rate of climb versus altitude data for bothaircraft and plot it on the same graph. Determine the absolute ceilings of bothaircraft.

6. Calculate the time required to climb from sea level to 20.000 ft for both aircraft.Assume that the curves in (5) are close enough to linear to use a linearapproximation for the calculation.

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AOE 3104 HOMEWORK 9

1. A jet aircraft with a max gross weight of 18,000 pounds can carry one third ofits weight in fuel (fuel weight = 6000 lbs.) and has a 400 ft2 wing area. If thespecific fuel consumption at an altitude of 30,000 ft is 1.3 lb per pound thrustper hour and if CD 0 = 0.02 and K = 0.06, what is the plane's maximum rangeand maximum endurance at 30,000 feet? Check to make sure the plane is notexceeding its critical Mach number of 0.85 at its CL for best range.

2. Determine the maximum range, maximum endurance and their related speeds atan altitude of 10,000 feet and a max gross weight of 10,000 lb. The aircraft hasa wing area of 200 ft2 , a fuel and oil load of 4000 pounds, an oil consumptionof one gallon for every 25 gallons of fuel consumed, a specific fuelconsumption of 0.5lb hp − hr and a propeller efficiency of 90%. The dragpolar is not necessarily parabolic but the power required characteristics aretabulated as follows for flight at max gross weight at an altitude of 10,000 ft.:

V (mph) Preq hp( )

403 1350350 925300 600250 400200 250175 215150 200140 205130 220125 240

Obtain the speeds for best range and endurance from the power required curveand then find the lift and drag coefficients from the velocity and power required.

Use fuel = 6 lb gal and oil = 7.5 lb gal .

Note: A very large portion of the aircraft's weight is fuel and theresult will be a rather incredible range and indurance.

3. For the airplane in problem 2 find the maximum range with a 30 knot tailwind.

1 Knot = 1.15 mph

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AOE 3104 HOMEWORK 10

1. An aircraft has a drag polar where CD 0 = 0.02 and K = 0.04. The wing'smaximum lift coefficient is 1.5. The plane weighs 115,000 pounds, has a 59psf wing loading and has four jet engines with 6900 pounds of thrust from eachengine. Find the takeoff ground run for this aircraft with a takeoff speed of 1.2times the stall speed and a runway friction coefficient of 0.02 for the followingtwo cases:

(a) The takeoff angle of attack is used during the entire ground run

(b) Optimum ground run conditions.

2. Find the ground run time for the two cases in the above problem.

3. For commercial aircraft the load factor is limited to a value of 1.5. For theaircraft described in problem 1 find:

(a) the maximum radius for a coordinated turn.

(b) the speed for this turn.

(c) the time required for a 360˚ turn.

222