Aircraft performance

Aircraft Performance

Aircraft Performance. Maido Saarlas© 2007 John Wiley & Sons, Inc. ISBN: 978-0-470-04416-2

Aircraft Performance

Maido SaarlasProfessor

Department of Aerospace EngineeringU.S. Naval Academy

John Wiley & Sons, Inc.

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Library of Congress Cataloging-in-Publication Data:

Saarlas, Maido.Aircraft performance /by Maido Saarlas.

p. cm.Includes bibliographical references and index.ISBN-10: 0-470-04416-0 (cloth)ISBN-13: 978-0-470-04416-2 (cloth)

1. Airplanes—Performance. I. Title.

TL671.4.S228 2006629.132—dc22

2006043974

Printed in the United States of America

10 9 8 7 6 5 4 3 2 1

[To]the University of Illinois, where it all started

[and]my family who nurtured and supported me

vii

Preface

This is intended to be a textbook mainly for aeronautical engineeringstudents. Since college-level physics and integral calculus are the es-sential required background, the material in this book should also beaccessible to other engineering students or as reference material topractitioners in the aerospace industry.

The approach is to minimize aerodynamics and propulsion-drivenmethodologies that often lead to extensive (but necessary) approxi-mations and loss of overview of the problem. Thus, aerodynamics andpropulsion are considered as known inputs to the mechanics equations,which then permit exposing the salient features of the performanceproblem at hand without getting lost in some very detailed issues. Aminimal amount of aerodynamics and propulsion information, providedin the appendices, should provide (both a novice and an experiencedhand) a relatively safe path for estimating essential performance fea-tures of an aircraft. The methods described in Chapters 2 through 7—level flight, climb, range, take-off, and maneuvering—are applicablefor both nominal assessment as well as for extensive analysis neededfor mission evaluation and design. The difference lies in aerodynamicsand propulsion details (e.g., critical Mach number, high lift devices,etc.) and assumptions and approximations made to obtain practical re-sults. Approximations become a roadmap that defines validity of theresults and overlaps performance and design issues. Design, with rootsin performance requirements, means additional approximations and as-sumptions pursuant to requirements for a design; a small change in

viii PREFACE

requirements (e.g., from a 4g maneuver capability to 6g) may result inan entirely different aircraft design. Although this book is not aboutdesign, constraints for the performance-oriented design selection proc-ess is one of the topics in Chapter 8.

The central idea of this book is to maintain focus on basic aircraftperformance. The central theme is the energy method.

The seeds of this book can be found in early energy method devel-opments in the 1950s. Historically, the earliest works are by Lush,Kelly, and Rutowski (see bibliography), who developed and articulateda more fundamental approach to aircraft performance evaluation. Theirwork showed that use of the concept of total energy (kinetic � potentialenergies) and the rate of change of total energy as new independentvariables (rather that the usual altitude or speed) enhanced both theunderstanding and quality of aircraft performance evaluation.

This immediately spawned further work on optimization of perform-ance and flight paths, which provided results now regularly used inaircraft operations. Unfortunately, little of this approach could be foundin the textbooks of 1960s and 1970s, as the level of mathematics in-volved implied graduate level of involvement or the calculations wereviewed as excessively computer driven and cumbersome.

Thus, this set of notes was started with two very general goals inmind:

1. To include new and evolving approaches for performance analy-sis, starting with the basic energy concepts

2. To minimize the aerodynamic and propulsion-driven methodolo-gies (i.e., to treat aerodynamics and propulsion only as necessaryinputs for the mechanics equations)

Obviously, the second goal needs to be modified when high-speed aer-odynamics and supersonic inlets require more detailed and specializedtreatment. The basic methods still remain valid.

Energy method provides a somewhat unified approach, and it workswell over a practical subsonic-supersonic speed range. Performance isapproached from the point of energy production and balance culmi-nating in one fundamental performance equation. This provides a pointof departure for handling practical point performance problems. Butthe energy method can also handle the quasi-dynamic path performanceproblems arising from minimum-time and minimum-fuel consumptionrequirements. Rutowski’s work allows now evaluation of those basicoptimization problems with acceptable accuracy, as compared to com-plete and precise numerical optimization process.

PREFACE ix

Part of this success is due to evolution of (hand) calculators. Theavailable functions and solver processes can make short work now ofthe arctan function in range problems (see Chapter 5) and fourth-degreepolynomials regarding maximum velocity (see Chapter 3). Program-mable features reduce also the optimum path problems to a few ele-mentary steps. However, this does not necessarily obviate the classicalgraphical approach described in Chapter 3. It arose from necessity tocircumvent the cumbersome trial-and-error solution of the fourth-degree polynomial for maximum velocity. It still provides a quick over-view of the entire performance potential, and is also easily accessiblevia new graphing routines.

These notes have been used in late 1950s at UCLA, in the 1960s atthe University of Cincinnati, and since 1970 at the U.S. Naval Acad-emy. They have been continuously modified, set aside for every newtext that appeared, and picked up again to pursue the two goals pre-viously outlined. They have been updated, streamlined, and made morerelevant by the large number of military pilots and faculty who havepassed through this department. They are too numerous to mentionindividually, and lest a good friend will be omitted, they all shouldknow that the author owes them deep gratitude.

There is more material in these notes than can be covered comfort-ably in one semester. It has been found that the sections marked withan asterisk (*) can be omitted without loss of continuity and withoutsacrificing understanding of the topic at hand. Each chapter containsnumerical examples drawn from practice and numerous problems;some with answers. Appendices A through E contain altitude tablesand data on aircraft and propulsion.

xi

Contents

1 The General Performance Problem 1

1.1 Introduction / 1

1.2 Performance Characteristics / 21.2.1 Absolute Performance Characteristics / 31.2.2 Functional Performance Characteristics / 4

1.3 The Approach / 5

2 Equations of Motion 7

2.1 General Information / 7

2.2 The Energy Approach / 11

3 The Basics 16

3.1 Fundamental Performance Equation / 16

3.2 Stalling Speed / 19

3.3 Maximum Velocity and Ceiling / 253.3.1 General Considerations / 253.3.2 Drag and Drag Polar / 283.3.3 Flight Envelope: Vmax, Vmin / 34

xii CONTENTS

3.3.4 Power Required and Power Available / 473.3.5 Turboprop Engines / 52

3.4 Gliding Flight / 533.4.1 Glide Angle and Sinking Speed / 533.4.2 Glide Range and Endurance / 59

4 Climbing Flight 70

4.1 General / 70

4.2 Rate of Climb, Climb Angle / 71

4.3 Time to Climb / 74

4.4 Other Methods / 814.4.1 Shallow Flight Paths / 814.4.2 Load Factor n � 1* / 884.4.3 Partial Power and Excess Power

Considerations / 94

5 Range and Endurance 101


5.2 Approximate, But Most Used, Methods / 1035.2.1 Reciprocating Engine / 1055.2.2 Jet Aircraft / 111

5.3 Range Integration Method / 1175.3.1 Basic Methodology / 1185.3.2 An Operational Approach / 123

5.4 Other Considerations / 1265.4.1 Flight Speeds / 1265.4.2 Effect of Energy Change on Range / 128

5.5 Endurance / 1295.5.1 Reciprocating Engines / 1305.5.2 Turbojets / 1325.5.3 Endurance Integration Method* / 133

CONTENTS xiii

5.6 Additional Range and Endurance Topics / 1345.6.1 The Effect of Wind / 1355.6.2 Some Range and Endurance Comparisons / 142

6 Nonsteady Flight in the Vertical Plane 150

6.1 Take-off and Landing / 150

6.2 Take-off Analysis / 1516.2.1 Ground Run / 1536.2.2 Rotation Distance / 1606.2.3 Transition Distance* / 1626.2.4 Take-off Time* / 1666.2.5 Factors Influencing the Take-off / 166

6.3 Landing / 1726.3.1 Landing Phases / 1726.3.2 Landing Run / 1736.3.3 The Approach Distance* / 1756.3.4 The Flare Distance* / 176

6.4 Accelerating Flight* / 178

7 Maneuvering Flight 189


7.2 Turns in Vertical Plane: Pull-Ups or Push-Overs / 190

7.3 V–n Diagram / 192

7.4 Turning Flight in Horizontal Plane / 199

7.5 Maximum Sustained Turning Performance / 2087.5.1 Maximum Load Factor / 2097.5.2 Minimum Turn Radius / 2107.5.3 Maximum Turning Rate / 213

7.6 The Maneuvering Diagram / 220

7.7 Spiral Flight* / 224

xiv CONTENTS

8 Additional Topics 234

8.1 Constraint Plot / 2348.1.1 Take-off and Landing / 2368.1.2 Constraints Tied to Performance Equation / 238

8.2 Energy Methods / 245

A Properties of Standard Atmosphere 258

B On the Drag Coefficient 260

C Selected Aircraft Data 265

D Thrust Data for Performance Calculations 267

E Some Useful Conversion Factors 277

Index 280

1

1The General

Performance Problem

Wright Flyer

1.1 INTRODUCTION

An engineer-designer who decides on the configuration, size, arrange-ment, and the choice of power plant for an air vehicle must make thedecision on the basis of the performance that will be expected of thefinished product. For this reason, one must be familiar with the basicperformance characteristics and with the relationship of design factorsthat can influence these characteristics.

In addition to the direct design needs, an accurate knowledge ofaircraft performance is necessary for the operator of the aircraft. Theairlines need this information to determine how the aircraft can beoperated most efficiently and economically. Similarly, the armed ser-vices need to know what a proposed or given aircraft can do, and howit must be flown in order to gain the best possible advantage or providemost effective support.


2 THE GENERAL PERFORMANCE PROBLEM

Thus, an engineer needs a good understanding of aircraft perform-ance problems for use both in preconstruction design and in analysisor evaluation of the finished aircraft. Similarly, a knowledge of possibleperformance characteristics and limitations of various classes of aircraftis needed to establish sound strategy in both commercial and militaryoperations.

This book deals with the methods by which the performance of anaircraft can be determined from its aerodynamic and powerplant char-acteristics. For present purposes, it will be assumed that the aerody-namic and propulsive data are known and given, and for clarity themain topic will be performance. This implies that performance doesnot affect aerodynamics or propulsion. Strictly speaking, this is nottrue because different performance regimes (i.e., speed, range, altitude)may require different wing sweep, aspect ratio, or power plant—which,in turn, establishes different performance characteristics. However, sucha design iteration procedure is outside the scope of these notes. More-over, it is not intended to treat aerodynamics, propulsion, and perform-ance under the same cover, as each of these are independently worthytopics requiring a substantial amount of space for adequate coverage.In order to facilitate working example problems, and for otherwisehandy reference, special appendices cover some aerodynamic and pro-pulsive equations and data.

1.2 PERFORMANCE CHARACTERISTICS

This book does not attempt to solve the complete dynamic flight me-chanics problem where the translational and rotational motion of thevehicle is solved along a trajectory. Such a problem formulation re-quires a set of simultaneous differential equations describing the forces,moments, and the attitude of an aircraft as functions of position andtime. In general, only numerical solutions are then possible.

To preserve clarity and simplicity, and to provide analytical simpli-fications leading to a good overview of the physical problem involved,a quasi-steady approach will be used. The flight path will be confinedto either horizontal or vertical plane at a time over a ‘‘flat earth.’’ Asa result, cross-coupling and inertial terms can be neglected in the dy-namic equations of motion. Since the attitude of the vehicle is of sec-ondary interest in performance calculations, it will be assumed that theforces can be considered independently of the moments. This then per-mits ignoring the moments and the attitude of the vehicle, and the flight

1.2 PERFORMANCE CHARACTERISTICS 3

dynamics problem is reduced to considering the translational motionof a point mass with a variable mass due to propellant consumption.The performance problem then means studying various aspects oftranslational motion, such as how far the vehicle will fly, how fast, andso on.

From a mathematical point of view, one finds two basic problemsin aircraft performance: point performance and path performance (in-tegral performance). Point performance problems describe the partic-ular local performance characteristics at a given point on the pathindependent of the rest of the flight path. The path performance dealswith the study of the flight path as a whole, and involves integrationbetween given initial and final conditions along the flight path (i.e.,range, time to climb).

In general, the point performance problem represents a set of alge-braic equations obtained from the dynamic equations by neglecting thetime-dependent terms. Thus, one can find the local extrema by ele-mentary techniques of the differential calculus. The path performanceproblems usually require integration along the flight path and, withoutresorting to numerical methods, can be solved only with most restric-tive assumptions concerning either altitude or velocity, or both. Thedetermination of optimum flight conditions requires techniques of thecalculus of variations.

Although both point and path performance problems will be dis-cussed in the following chapters, another more functionally orienteddivision of flight performance problems transcends the two approaches.From the point of view of the dual purpose of performance calculationsoutlined in Section 1.1, it is natural to divide the performance char-acteristics of an aircraft into two groups as follows: absolute perform-ance characteristics and functional performance characteristics.

1.2.1 Absolute Performance Characteristics

The first group, called the absolute performance characteristics, are ofthe primary interest to the designer. They are called absolute becausethey are simply numbers representing the capabilities of the aircraft.The most important are:

• Maximum speed• Stalling speed• Best climbing speed• Best glide speed


• Rate of climb• Ceiling(s)• Maximum range and speed for maximum range• Maximum endurance and speed for maximum endurance• Take-off distance• Landing distance

1.2.2 Functional Performance Characteristics

The second group of performance characteristics will be called func-tional performance characteristics, as they are more important fromthe standpoint of the efficient operation of the aircraft. These charac-teristics are not expressed simply as numbers, but rather as functionsor curves (i.e., such as the variation of airspeed with altitude necessaryto accomplish a certain purpose). Typically, the characteristic functionsconstitute the answer to the following questions:

1. What is the program of speed and altitude that must be followedin order to go from a given altitude h1 to another altitude h2 inminimum time?

2. What is the program of airspeed and altitude to follow in orderto go from one flight condition (i.e., speed and altitude), to an-other in minimum time?

3. What is the program of altitude and speed such that the aircraftcan change from one flight condition (i.e., speed and altitude) toanother with minimum expenditure of fuel?

4. What variation in flight conditions will permit the aircraft to coverthe greatest distance over the ground?

It is easily seen that the so-called absolute characteristics in the firstgroup are design-compromise and specifications oriented, as there is adirect connection between these and the variables describing the aircraftgeometry, weight, and the power plant. The second group consists ofquestions of great importance to successful commercial or tactical useof the aircraft.

Last, and by no means least, it should be realized that the individualperformance characteristics, obtained by whatever means, must becombined into a unified description of the desired or specified flightprofile. The study of a flight profile and the requirements it places onthe performance characteristics, and vice versa, is called mission anal-

1.3 THE APPROACH 5

ysis. Although the mission analysis often precedes the detailed per-formance calculations, as it is used as a tool for generating the generalaircraft specifications, it also relies on the techniques used for perform-ance analysis. A full treatment of the mission analysis is beyond theintent and scope of these notes, however some discussion can be foundin Chapter 8, section 8.1.

1.3 THE APPROACH

Typically, a sound engineering approach to problem solving is to drawa free body diagram, consider all the forces and moments acting on it,and then apply all appropriate conservation laws. In a true dynamicenvironment, additional information must be provided concerning kin-ematics, aircraft attitude, and a reference system that helps to locatethe vehicle. As already pointed out, such an approach produces com-plete and accurate information about the behavior of the vehicle in theform of numerical solution. For many engineering purposes, analyticalsolutions—even if approximate—are desired, as they provide quickoverview, trends, and adequate numerical information.

Such an approach is followed here. Emphasis is on accessible anduseful solutions with reasonable and sensible accuracy. The generalthree-dimensional problem is divided into two separate problems lo-cated in horizontal and vertical planes. Equations of motion, based onthe free body diagram, are obtained for each separate plane. This stan-dard approach is provided for flexibility, familiarity, and ease of solu-tion of a number of point performance problems. It is also helpful forunderstanding and supporting the energy method used for most of top-ics at hand.

The main theme approach is to apply the concept of energy produc-tion and balance to aircraft performance evaluation. A single energyequation, fundamental performance equation (FPE), will be developedand related to existing equations of motion. This serves to clarify andamplify some standard point-performance methods. It also leads toquasi-steady maneuvering analysis and is a central function in solutionsof some path performance problems.


Shuttle

7

2Equations of Motion

F18

2.1 GENERAL INFORMATION

Before launching into performance analysis via the energy equation itmay be beneficial to review the standard dynamics approach that con-sists of a free body diagram of a point mass aircraft. They both shouldgive the same answers. When and why they diverge is the answersought in this book. Since the predominant portion of flight time is


8 EQUATIONS OF MOTION

h

x

W

ds

dx

dh

V

TL

D

A

t

n

γε

γ

ω

Figure 2.1 Flight Path in a Vertical Plane

spent either in almost rectilinear flight at constant altitude or in changefrom one altitude to another, it is appropriate to consider first the flightthat is confined entirely to a vertical plane defined by altitude h andhorizontal distance x. Combined with the mathematical simplificationsthe physical picture can then be maintained at such a level that practicaland significant results are easily obtainable. Flight in the horizontalplane—maneuvering flight and high-g turns—will be considered later.

The system of equations describing the motion is obtained fromNewton’s second law

→→ dVF � m (2.1)

dt

and the forces acting on the flight vehicle, as presented in Figure 2.1.The total force consists of the aerodynamic force , thrust , and

→ → →F A T

the gravitational term . , in turn, is given in terms of lift L and→→mg A

drag D. By convention, lift acts perpendicular to the velocity vectorand the drag is along and in the direction opposite to . Thus, Eq.

→ →V V2.1 becomes:

→→→ → dV→ ˙A � T � mg � m � mV (2.2)

dt

Since the flight path is in the vertical plane, Eq. 2.2 can be written inthe component form normal, , and tangent, , to the flight path. With

→→n tthe definition of the aerodynamic force

2.1 GENERAL INFORMATION 9

→ →→A � Ln � D t (2.3)

and the acceleration normal to the flight path

d�a � V � V� (2.4)

dt

one obtains the resolved equations in and directions, respectively:→ →t n

˙�D � T cos � � mg sin � � mV (2.5)

L � T sin � � mg cos � � mV� � mV� (2.6)

where � is the angle of the flight path with the horizontal plane. Thedot superscript signifies differentation with respect to time. � is therotation rate of the local radius vector of the path and is given by ,�the rate of change of inclination of the flight path to the horizontal.The velocity is tangent to the flight path.

→V

In addition, the following kinematic relationships hold along theflight path:

dx ds� cos � � V cos � (2.7)

dt dt

dh ds� sin � � V sin � (2.8)

dt dt

Eqs. 2.5 through 2.8 complete the set of equations of motion in avertical plane.

In general, the thrust T acts at a very small angle along the flightpath, and is a function of velocity and altitude. The lift L and drag Dare functions of altitude, velocity, and angle of attack �. Since the liftis directly a function of angle of attack, and since the latter does notappear explicitly in the formulation of Eqs. 2.5 to 2.8, it is convenientto replace � by the lift L. The equations of motion can then be writtenin the form with (W � mg):

h � V sin � (2.9)

x � V cos � (2.10)


V [T(h, V) cos � � D(h, V, L)]� � sin � (2.11)

g W

� [T(h, V) sin � � L(h, V)] cos �� (2.12)

g VW V

dWm � � (2.13)

dt

It is easily recognized that Eq. 2.9 represents the velocity in the verticaldirection—the rate of climb—and that Eq. 2.10 leads, by integration,to the range R � x. Eq. 2.13, an identity, states that the change of massis due to the change of the weight of the aircraft.

A simple inspection reveals that there are a total of six possiblevariables (m, h, V, �, L, x) and only four equations. Thus, additionalsimplifications must be made or the equations will lead to a two-parameter family of solutions. From the practical point of view, thissimultaneous set of differential equations is seldom solved for a para-metric solution. Sufficient approximations are made that then lead toan algebraic set of equations for various rates (dh /dt, etc.), and onlythe range equation remains to be integrated. Such approximations, al-though extremely practical and useful, can lead to serious errors, es-pecially in high-speed flight, and thus should be introduced with somecare.

The solution process falls into two general categories. For the directproblem, the aerodynamic and thrust terms are known and the solutionsare obtained for the kinematic variables (range, rate of climb, maximumvelocity, etc.). If the kinematic results are known, then the requiredaerodynamic or thrust data are sought to satisfy the specified condi-tions. This is the inverse of the design problem, and the solutions areobtained—often in parametric form.

This text will eventually introduce and follow most of the usualassumptions and methods, but the approach used here will be somewhatdifferent from those used in the standard textbooks. The process canbe broken down into three overall steps:

1. The basic performance will be formulated in terms of the aircrafttotal energy variation.

2. The standard steady-state problem will be deduced as a numberof special cases permitting a clear overview of all pertinent as-sumptions used, the range of applicability, and need for energyconsideration as the flight speed increases.

2.2 THE ENERGY APPROACH 11

3. The energy formulation will lead to a number of optimal per-formance problems that can provide engineering solutions with-out recourse to extensive computational equipment.

P51

2.2 THE ENERGY APPROACH

In formulation of a number of flight performance problems, it is con-venient, and perhaps necessary, to use the energy balance rather thanthe force equations. As the resulting equation represents a restatementof the force equations, it is illuminating to investigate the energy equa-tion and the conditions that must be satisfied for its use independentof the rest of the force equations.

The energy balance is essentially a statement about the source of theenergy and how the energy is spent. The source is the fuel used, whichis characterized by its energy content. The fuel/chemical energy isconverted into useful work by the aircraft powerplant at an overallefficiency �o. The energy content of the fuel Hf and fuel (mass) flowrate (dWf /g)/dt give the power available to the engine:

H dWf fP � � (2.14)a o g dt

This power is spent to overcome the drag and to provide energy forthe aircraft performance. Thus, the following energy balance must besatisfied:

Rate of � Power � Power � Rate of energychange of available dissipated change due to weight

aircraft energy from fuel by drag change of aircraft

The total energy of the aircraft E is given by the sum of its potentialand kinetic energies as


2VE � W h � � We (2.15)� �2g

where e is the specific energy, V is the flight speed, and h is the altitude:

2Ve � h � (2.16)

2g

Thus, the rate of change of aircraft specific energy may be written as

de dh V dV� � (2.17)

dt dt g dt

The energy balance may then be expressed as

dE edW� P � DV � (2.18)adt dt

Differentiating now Eq. 2.15

dE W de edW� � (2.19)

dt dt dt

and substituting into Eq. 2.18, one obtains the aircraft basic energyequation, which is used throughout this book:

P � DV dh V dV dea � � � � P (2.20)sW dt g dt dt

Eq. 2.20 shows that the rate of change of specific energy, e, is givenby excess (or lack of) power per pound of aircraft weight. More im-portant, the left-hand side may be viewed as the energy generationterm, where power available is balanced against power dissipated bydrag. The right-hand side may be considered as energy distributionbetween altitude and velocity changes. Either side may be zero. Theleft-hand side vanishes if all the power available is used to overcomedrag. de /dt may be zero for a variety of combination of altitude andvelocity variation. The commonly accepted symbol Ps, which may rep-

2.2 THE ENERGY APPROACH 13

resent either side of the energy equation (see Chapters 7 and 8), iscalled specific excess power.

Eq. 2.20 is a powerful tool in the study of efficient and comparativeaircraft maneuvering. Knowing the specific excess power determinesthe aircraft climb and acceleration capabilities. Conversely, knowing(or specifying) rate of climb and/or acceleration leads to knowledge(requirements) for aircraft power and drag levels (i.e., the excesspower).

Dimensionally, Eq. 2.20 represents a velocity. This is easily verifiedif one recalls the definitions of potential and kinetic energies. Potentialenergy is the work done by a force to raise the vehicle to height h, or

P.E. � Fh � mgh � Wh [lb -ft], [N-m], [J] (2.21)f

Kinetic energy is the work done by the resultant force to bring a vehiclefrom rest to motion with a speed V, or

1 1 12 2 2 2K.E. � Fx � (ma) at � ma t � mV (2.22)� �2 2 2

Thus, the units of energy E: [lb-ft] or [N-m] and for specific energy,e: [ft], [m] where e � E /W. Then, it follows that

de P � DV ft m� or (2.23)� � � �dt W sec sec

As examples of specific applications of Eq. 2.20, consider the un-accelerated level flight where h � constant, V � constant, and de /dt� 0. Then one obtains:

P � DV � P (2.24)a r

where Pr is the power required to sustain equilibrium level flight. Eq.2.24 is the basic static performance statement.

As a second example, if one assumes flight at constant velocity, thenit follows from Eq. 2.17 that de /dt � dh /dt, and Eq. 2.20 reduces tothe quasi-steady rate of climb equation:

dh P � DVa� (2.25)dt W


To place the energy equation, Eq. 2.20, into a different perspective,it can be derived also from the force equations, Eq. 2.11 and Eq. 2.9.To this end, Eq. 2.11 will be written

V T � D� sin � � (2.26)

g W

And using Eq. 2.9, we get:

VV (T � D)Vh � � (2.27)

g W

The left side can immediately be recognized as de /dt, and since TV �Pa, one recovers the energy equation, Eq. 2.20.

At this point it seems that the energy equation, after being first de-rived independently of the other path equations, could be used as anindependent equation. However, it is coupled to the other path equa-tions through lift and the flight path angle � (see Eqs. 2.9 to 2.12).Since these equations represent a set of simultaneous nonlinear differ-ential equations, certain assumptions must be made before individualequations can be treated as such.

There are three sets of assumptions, called energy assumptions, thatwill effectively decouple these simultaneous equations, Eqs. 2.9 to2.12:

1. Assuming that the flight path angle is constant, or that is very�small, with � � 0, the lift can be given independent of the pathangle and thrust, Eq. 2.12. Thus, Eq. 2.11 becomes quasi-independent.

2. Drag is only a function of altitude and velocity. Then sin � canbe eliminated as above, and one obtains the energy equation fromEq. 2.11.

3. Neglecting or both inertia terms: and , decouples the equa-˙ ˙V, V �tions.

The merits and disadvantages of these assumptions become evidentin those specific sections where individual performance problems aretreated. It is sufficient to point out at this time that these assumptions,although somewhat restrictive, are necessary in order to establish per-formance solutions for engineering expediency and accuracy. More-

PROBLEMS 15

over, a basic aspect of performance analysis is establishing (at least tofirst order) some optimum performance conditions. Without the use ofsome of the assumptions and the attendant simplification of the equa-tions, the optimizing process becomes extremely cumbersome and ame-nable only to computer methods.

PROBLEMS

2.1 What is the potential energy of a 40,000 lb aircraft at 40,000 ftabove sea level?

2.2 If the aircraft in Problem 2.1 is flying at 1,000 ft/sec, what is itskinetic energy? Its total energy?

2.3 If the aircraft in Problem 2.1 descends to 20,000 ft and continuesflying at 1,000 ft/sec, what is its energy now? What happened tothe energy difference?

2.4 The average heat content of JP-4 fuel or aviation gasoline is19,000 BTU/lb. Calculate the power available if the efficiency ofthe engine is 20 percent and the fuel flow rate is 750 lb/hr. As-sume a typical propeller efficiency to be 0.85.Ans: Pa � 952 HP.

16

3The Basics

F14

3.1 FUNDAMENTAL PERFORMANCE EQUATION

The general types of flight performance problems discussed in Chapter1 were divided into two basic groups: absolute and functional perform-ance characteristics. In addition, analytical development of the equationof motion and the nature of the motion indicates that most problems


3.1 FUNDAMENTAL PERFORMANCE EQUATION 17

can be divided into quasi-steady (static) or unsteady (dynamic) cate-gories. Some, like range and endurance, can be treated by either ap-proach, depending on the assumptions used. It turns out that mostabsolute performance characteristics can be treated mathematically assteady-state point-performance problems leading to simple algebraicexpressions. Local (path independent) optimum conditions are deter-mined by simple differential calculus. The dynamic problems, whicharise mainly from variation of velocity and altitude along the flightpath, deal with functional performance characteristics and the transientbehavior of the aircraft. Their solution requires the consideration ofmotion along the entire flight path, which leads to use of extensivenumerical methods. Some practical simplifications are found in Chap-ter 8.

The purpose of this chapter is to establish suitably simple engineer-ing expressions, while determining which data are required for evalu-ating the thrust T, drag D and power P. To that end, the main analysisvehicle is the basic energy equation Eq. 2.20, which is used to focuson those performance characteristics that establish the general boundsfor powered flight vehicle performance in steady nonmaneuveringflight: stalling speed, maximum and minimum speeds, and ceiling.These are often called the basic performance problems, and they definea theoretical region on an altitude-velocity plot where T � D and wheresteady-state flight is possible. This plot is called aircraft flight envelopeand will be established in subsequent sections.

For unpowered vehicles (gliders, space shuttles), glide performanceis important and will be discussed as the last section in this chapter.Stalling speed and glide problems, in their essential features, are in-dependent of the power plant and are essentially functions of aero-dynamics. However, power does influence both stall and glideperformance and may need to be considered in full aircraft performanceevaluation and design. As will be seen shortly, maximum and minimumspeeds and ceiling are obtained at full thrust (power) setting and thusare functions of the powerplant. Consequently, their solutions proceedaccording to whether the engine characteristics are thrust (jet engines)or power (propellers) oriented.

Before commencing discussion of the performance problems, theenergy equation will be recast in a different form suitable for directperformance calculations. The resulting equation is the FundamentalPerformance Equation (FPE), which will be used for most of the de-velopment in subsequent chapters.

18 THE BASICS

This FPE is obtained by rewriting the energy equation:

P � DV dh V dV dea � � � � P (3.1)sW dt g dt dt

Solving for the rate of climb yields

dh P � DV V dVa� � (3.2)� �dt W g dt

where Pa is the power available from the engine. Since Ps is just aconvenient shorthand symbol for either side, it plays no role in thisdevelopment. Its usefulness is demonstrated in Chapters 7 and 8.

The terms inside the square brackets constitute the classical funda-mental performance equation where the assumption has been used thatthe climb or descent velocity is constant, or, dV /dt � 0 (see also Eq.2.25). It is classical in the sense that when used in early low-speedflight analysis, ignoring the acceleration term did not introduce signif-icant errors and the bracketed term was easily derivable from simplethrust-drag equilibrium. With increasing engine sizes and availablesteep climb angles, the acceleration term became significant, as it rep-resented rapid energy change of the aircraft with attendant changes invelocity and/or altitude.

Considering Eq. 3.2 as representing excess (or lack) of power perpound of weight, several interpretations are possible (see Eqs. 2.24 to2.25), as a power-drag unbalance will lead to changes in specific en-ergy. The classical performance equation has its limitations, since onlypart of the energy balance is included. One way to improve accuracyfor high speed or steep climb problems has been to establish a correc-tion to the classical equation by rewriting Eq. 3.2 with

dV dV dh� (3.3)

dt dh dt

as

dh P � DV 1a� . (3.4)dt W V dV

1 �� g dh

3.2 STALLING SPEED 19

The term in square brackets is then evaluated as a kinetic energy cor-rection factor to the fundamental equation. For low speeds, the correc-tion factor is small, as expected, but can exceed 2 for high speeds. Asa result, the rate of climb dh /dt predicted by the classical formula maybe actually halved. Among many other parameters, the correction de-pends on the climb technique and the powerplant characteristics. Thisis discussed in some detail later in the chapter.

Spitfire

3.2 STALLING SPEED

The requirements for equilibrium flight are that the aircraft be flownat least at the speed that generates sufficient lift to counteract the air-craft weight and that the thrust be equal to drag. This is called thestalling speed. It occurs at the stall angle, or the maximum angle ofincidence, and it represents the low end of the steady-state flight spec-trum of the aircraft. For steady, level flight, equlibrium equations be-come (from Fig. 2.1 with � � 0):

20 THE BASICS

T � D

L � W

where, for simplicity, the thrust angle � has been set equal to zero.Lift equation, written in the usual form, is:

1 2L � �V C S � W (3.5)L2

where

slugs kg�—density ,� � � �3 3ft m

ft mV—velocity ,� � � �sec secC —lift coefficient, dimensionlessL

2 2S—wing reference area (ft ), (m )L—lift force (lb), (N)

This equation is also seen to be the simplest case of the equationsof motion in Chapter 2 and represents the force balance normal to theflight path (Eq. 2.12) in level, unaccelerated flight. The thrust is as-sumed to be parallel to the path with no normal components, and thelift is assumed to be independent of drag and determined only by at-titude, which also fixes the maximum value of lift. Thus, the velocityfor steady, level flight is obtained from Eq. 3.5:

2WV � � TAS (3.6)��SCL

It represents the relationship between the flight velocity and the re-quired lift coefficient. It is also called true airspeed TAS. Eq. 3.6 alsoestablishes the well-known relationship

2W2V C � � constL �S

for a constant altitude (and fixed W) flight.


The force balance along the flight path, Eq. 2.11, yields

W WT � D � � (3.7)

(L /D) (C /C )L D

where CD is the dimensionless drag coefficient defined by

1 2D � �V C S � T (3.8)D r2

Eq. 3.7 is the thrust required for level unaccelerated flight. Eq. 3.5shows that, for a given altitude, the lift depends only on the lift coef-ficient and the velocity. However, for a fixed weight and altitude, aflight attitude exists that gives a maximum value to the lift coefficient

(although the latter depends also on lift augmentation devices likeCL max

flaps and slats), which, in turn, indicates a minimum flight velocity inorder to satisfy the above equation. This minimum velocity is calledthe stall speed, and is given by

2WV � (3.9)s ��SCL max

Eq. 3.9 represents the minimum flight velocity at which steady sus-tained flight is possible. It depends on the altitude, maximum lift co-efficient, and, to some extent, on power at high angle of attack thatmust be determined experimentally. See also Problem 3.13.

It is convenient now to introduce the equivalent airspeed VE or EAS,which is defined as

1 12 2�V � � V (3.10)0 E2 2

where �0 is the sea-level density. The equivalent airspeed establishesequivalence of the dynamic pressure at sea level and at altitude, whichrenders lift and drag characteristics the same at sea level and at altitude.Moreover, the standard air-speed indicator is theoretically calibrated toread equivalent airspeed. In practical performance calculations, use ofVE eliminates the altitude effect from many equations. Introducing VE

in Eq. 3.9 gives

22 THE BASICS

2(W/S)V � (3.11)Es �� C0 L max

Eq. 3.11 states that, for a given wing loading W /S, the equivalentstall speed is a function of the maximum lift coefficient (attitude), forall altitudes. The significance of this result is that, for a given aircraft,

can be determined once and for all for any given combination ofCL max

flaps and/or slats, and can be calculated. Since, for a reasonablyVEs

good instrument and installation, indicated airspeed (IAS) is almostequal to EAS, the aircraft will always stall at approximately the sameIAS, which knowledge is of practical use to the pilot. Or, in otherwords, for a given aircraft, is then a function of aircraft weight WVEs

only.The indicated airspeed is the actual instrument indication in the air-

craft at any given flight condition and altitude. A number of factorscontribute to the difference between the figure shown by the IAS dialand the actual true airspeed (TAS): instrument error, installation error,position or instrument location error, compressibility at high speed, anddeviation of local density from that of the sea level value �0. Thecalibrated airspeed is

� instrument errorCAS � IAS � installation error

� position error

and, in turn,

V � CAS � compressibility correctionE

The standard airspeed indicator is calibrated to give a correct readingat standard sea-level conditions (14.7 psia, 59 F, 1 atm, 15 C). Sincemost flights take place in nonstandard conditions, corrections are madeto IAS such that EAS represents that flight speed at standard sea level(std sl) with the same free stream dynamic pressure as the actual flightcondition (Eq. 3.9). Finally, the true airspeed (TAS) is obtained whenthe EAS is corrected for density altitude (see Appendix A for atmo-spheric properties data):

EAS VEV � TAS � � (3.12)��

where � is the density ratio � /�0.


Since the density depends on both pressure and temperature, andsince CAS is corrected for pressure altitude, local sea-level pressureand temperature are needed to determine TAS from EAS (or CAS).

Theoretically, Eq. 3.11 also establishes the minimum velocity (atat which the aircraft can land. Practical flight safety rules, how-C )L max

ever, require that a landing speed above VE be used. In general, thefollowing requirements have been established:

V � 1.2 V � visual approachL s

V � 1.15 V � carrier landingL s

V � 1.3 V � instrument approachL s

These conditions then determine the required lift coefficients for re-spective landing conditions. The decreased lift coefficients and,therefore, the angle of attack, are reduced to provide required safetymargins. In practice, this set of rules effectively has two purposes. First,it sets a minimum flight speed for safe landing. Second, it establishesthe thrust requirements for safe landing (and take-off).

The following example explores the concepts of stall speed andCL � V relationship.

EXAMPLE 3.1

A Boeing 737 aircraft has the following characteristics:

W � 111,000 lbmax

W � 103,000 lbmaxland

2S � 980 ft

C � 2 at sea levelLland

V � 850 ft/sec at 40,000 ftmax

Calculate:

a. Vs

b. CL max

c. The relationship between the lift coefficient and the velocityover the usable speed range at the landing weight at 40,000 feet, ifthe safe landing speed is taken as 1.2 Vs

24 THE BASICS

a. Landing speed is calculated from

2W 2 � 103,000 ftV � � � 210L � �� SC 0.002377 � 980 � 2 sec0 L

Thus

210 ftV � � 175s 1.2 sec

b. The lift coefficient at stall, , can be calculated, because theCL max

weight and altitude are fixed.

2 2C � V � C � V � constL land L sland max

2V landC � 2 � 2 � 1.44 � 2.88L max 2V s

c. Since at sea level EAS � TAS, the stall speed at 40,000 feetcan be obtained from

ftV � 175Es sec

V 175 ftEsV � � � 352s 40,000 sec�� 0.247

Thus, the usable velocity range at 40,000 ft is

ft ft352 � V � 850

sec sec

The relationship between the lift coefficient and the flight velocityat 40,000 ft and at the maximum landing weight is given by

22W 2 � 103,000 ft2C V � � � 358,000L 2��S 0.002377 � 0.247 � 980 sec

and is shown in Table 3.1. The table shows that, theoretically, at40,000 ft altitude this aircraft may be able to maintain steady level

3.3 MAXIMUM VELOCITY AND CEILING 25

TABLE 3.1 Lift and Velocity Comparisons

V (ft / sec) 352 400 423 500 600 700 800 850CL 2.89 2.24 2.0 1.43 0.99 0.731 0.559 0.496

flight at 423 ft/sec at the point of stalling out the aircraft. In practice,for safety reasons, such low (stall) speed should never be attempted.Moreover, near-stall high lift coefficients may cause sufficiently highdrag that can exceed the powerplant capability.

F4

3.3 MAXIMUM VELOCITY AND CEILING

3.3.1 General Considerations

In the last section, the low end of flight spectrum was explored in somedetail. It was found that, for steady flight condition and in absence ofconcerns about thrust, there exists a minimum velocity—the stallspeed. Once the aircraft minimum velocity is established, the followingquestion naturally arises: What is the maximum speed for the aircraft,and how can it be determined? The answer to the first question isobviously affirmative, but how it is determined depends on the infor-mation available for the vehicle and is the subject of this section.Establishing the ceiling—how high can an aircraft fly, can be accom-

26 THE BASICS

plished in a number of ways and it is discussed in several chapters. Inthis section it appears as a byproduct of basic thrust—drag consider-ations.

Consider unaccelerated flight at constant altitude. The FPE (Eq. 3.2)becomes

P � DV � P (3.13)a r

Thus, the problem is reduced to finding the flight condition where theavailable power just balances the energy dissipated by drag, or thepower available is equal to the power required. The problem must nowbe further subdivided into two categories because there are basicallytwo types of powerplants used in aircraft:

• Power producing—reciprocating engines and turbine engines driv-ing a propeller

• Thrust producing—turbojets, ramjets, etc.

Power Producing The need for such a division becomes apparent ifone considers Eq. 3.13. Fundamentally, it is a power equation wherethe power is thrust horsepower and is given by

P � BHP � in HP (3.14)a p

where BHP is the total brake horsepower produced by the engines and�p is the propeller efficiency. Since the reciprocating and the turbopropengines are power-producing devices, the performance is described bythe product of the engine shaft horsepower and the efficiency of thepropeller. Thus, Eq. 3.13 may be written as

BHP� � DV (3.15)p

In general, both the horsepower and efficiency are functions of velocityand altitude and the power available is usually given either in graphicalor tabular format (see Ex. 3.4). Table 3.2 shows typical units in use.

Thrust Producing For jet engines, the output is thrust. Since the jetpower is a product of thrust and velocity, and since jet engine perform-ance curves are usually given directly in terms thrust, Eq. 3.13 simpli-fies with

P � TVa


TABLE 3.2 Units

P T V

P � TVft-lbsec

lbft

sec

P �TV550

HP lbft

sec

P � TV W Nm

sec

to

T � D (3.16)

A comparison of Eqs. 3.15 and 3.16 shows that reciprocating engineperformance should be considered from a power-available/power-required point of view and the jet engine case reduces to thrust-available/thrust-required formulation. The thrust of a jet engine canoften be described by analytical statements that can lead to simpleperformance calculations and analytical solutions (Appendix D). Forpropeller-powered aircraft, the power information is usually given ingraphical form, which implies that the propeller engine performanceproblems are best solved in graphical or numerical form. As in the casefor jet aircraft, simplified solutions can be found if the aircraft is op-erating essentially at constant velocity and at constant altitude. In thiscase, propeller efficiency remains practically constant and the poweravailable curve is also constant independent of velocity. Thus, the dif-ferent formulations of thrust and power data indicate that it is advan-tageous to pursue the questions concerning the maximum velocity andceiling separately for power- and thrust-oriented aircraft.

Both the Pa versus Pr and Ta versus Tr analyses presuppose a totalknowledge of the aircraft drag. Thus, it will be useful to review thedrag expressions in some detail. As a result, several specific relation-ships of general interest can be established that will be of use in sim-plifying the general performance calculations carried out in subsequentchapters.

747

28 THE BASICS

3.3.2 Drag and Drag Polar

Although the drag and the drag coefficient can be expressed in a num-ber of ways, for reasons of simplicity and clarity, the parabolic dragpolar will be used in all main analyses. For most of the existing (high-speed) aircraft, the drag cannot be adequately described by such asimplified expression. Exact calculations must be carried out using ex-tended equations or tabular data. However, the inclusion of more pre-cise expressions for drag at this stage will not greatly enhance basicunderstanding of performance, and thus, will be included only in somecalculated examples and exercises.

Writing now the drag force as

1 2D � �V C S (3.17)D2

and utilizing the parabolic drag polar (see Appendix B)

2 2C � C � C /(�ARe) � C � kC (3.18)D D L D L0 0

k � 1/(�ARe)where

One obtains

1 12 2 2D � �V C S � �V SkC (3.19)D L02 2

where

C is zero lift drag coefficientD0

2AR � b /S � aspect ratiob is wing span

Eliminating CL from the last equation by use of Eq. 3.5, and definingthe aircraft load factor n as

Ln � (3.20)

W

the drag equation becomes


D

V

D

V

Induced drag

Parasite drag

Total drag

min

Dmin

Increasingaltitude

Figure 3.1 Total Drag, Induced Drag, and Parasite Drag

2 21 kn W2D � �V C S � (3.21)D02 1 2�V S2

Eq. 3.21 is given by two terms, one proportional to V2 and the otherinversely proportional to V2. The first term, called parasite drag, rep-resents the aerodynamic cleanness with respect to frictional character-istics, and shape and protuberances such as cockpit, antennae, orexternal fuel tanks. It increases with the aircraft velocity and is themain factor in determining the aircraft maximum speed. The secondterm represents induced drag (drag due to lift). Its contribution is high-est at low velocities/high-g loading, and it decreases with increas-ing flight velocities/lower-g flight. For the rest of this chapter and formost of Chapters 4, 5, and 6, it is assumed that L � W, and thereforen � 1.

The relative significance of these two terms of Eq. 3.21 is shown inFig. 3.1, where a typical total drag curve is drawn for a parabolic dragpolar at sea-level altitude (� � �0). The effect of increase in altitude(decrease in �) is also shown, which shifts the total drag curves rightfor constant W, n, S, and Increasing W, n, S, and increasesC . CD D0 0

the drag and curves shift upward. Thus, the drag curve depends on fiveparameters, out of which only the reference area S will usually remainconstant during the flight. Although n is unity for steady level flight,as considered in most of these chapters, it will be included here ex-plicitly for basic development. This will permit an easy transfer to othercases in later chapters.

30 THE BASICS

For low-speed flight regime (where the Mach number effects can beignored), it is convenient to rewrite Eq. 3.21 by introducing the equiv-alent parasite area:

ƒ � C S (3.22)D0

2 2� W 2n2D � ƒV � (3.23)� � 22 b �e�V

The equivalent parasite area is often used to classify different aircraftand their component drags, and it leads to a logical way to a buildupof aircraft zero lift drag. For some tabulated data see Appendix C.

Since a graphical performance representation is often a convenientmeans toward solutions, the number of apparent parameters can bereduced by suppressing the altitude variation by introducing again theequivalent air speed from Eq. 3.10, which gives

2 21 2kn W B2 2D � � V C S � � AV � (3.24)0 E D E0 2 22 � V S V0 E E

where

1A � � C S0 D02

2 22kn WB �

� S0

and A and B depend on geometry, weight, and g-loading of the aircraft.Thus, for a given aircraft point performance problem at fixed weightand g-load, the drag can be given by a single curve for all altitudes,which helps to simplify graphical performance representation (compareFigures 3.1 and 3.2).

An item of considerable practical interest is the minimum point ofthe total drag curve, as it gives the minimum drag and the maximumlift/drag ratio and later serves to identify useful performance points.To find this point on the drag curve, any one of the drag expressionsgiven above, say Eq. 3.21, can be differentiated with respect to thevelocity to give

2 2dD 4kn W� �VC S � (3.25)D0 3dV �V S


Setting this expression equal to zero, and solving for velocity, one finds

1 /42Wn kV � (3.26)� �Dmin � �S CD0

Substituting Eq. 3.26 back into the drag equation Eq. 3.21, yields

D � Wn�kC � Wn�kC � 2Wn�kC (3.27)min D D D0 0 0

which states that the minimum drag occurs when the parasite and in-duced drags are equal and it is independent of altitude. From the dragcoefficient, Eq. 3.18, it follows then, since both of the drag terms areequal (this can be seen also from Fig. 3.1), that

2C � kC (3.28)D L0

and the total drag coefficient, at minimum drag, may be written as

2 2C � C � kC � 2C � 2kC (3.29)D D L D L0 0

Since the minimum drag occurs at the minimum value of D /L, thelatter ratio can be determined immediately by use of Eqs. 3.28 and 3.29as

D 1 1� � (3.30)�L EL mmin �D

max

CL C 1 1DL 0� � E � � (3.31)� � � �m �D C k 2C 2�kCD D0max max D0

NOTE

In keeping with the current practice in nomenclature usage, in the rest of thechapters the symbol Em will be used for the maximum lift /drag ratio L /D�max. It should be noted that, for the parabolic drag polar, the velocity forminimum drag occurs at Em; thus � and that the minimum drag,V VD Emin m

Eq. 3.27, is independent of altitude, which proves the assertion madeabove: the effect of altitude change is to shift the drag curve parallel tothe velocity axis. As with advanced nomencature, care should be exercisedin usage (e.g., using � may be ambiguous). Does E refer toV VD Emin m

32 THE BASICS

L /D ratio or to equivalent airspeed EAS? In this case it is best to use fullsubscript or simply The next example explores calculationV , V .(L / D)�max Dmin

of the drag parameters discussed above.

F86

EXAMPLE 3.2

An aircraft has a wing area of 255 ft2 and a weight of 10,000 lb anda clean drag polar (flaps and gear up) of

2C � 0.023 � 0.0735 C , AR � 5.07D L

Calculate:

a. (L /D)�max

b. at sea level and at 40,000 ftVDmin

c. Tmin for level flight

a. Eq. 3.31 gives Em directly

L 1 1� � � 12.16�D 2�kC 2�0.0735 � 0.023max D0

or, by Eq. 3.28 � we get:2C kC ,D L0

C 0.023D0C � � � 0.559LDmin � �k 0.0735

and by use of Eq. 3.31, one finds

C C 0.559L LE � � � � 12.16�m C 2C 2 � 0.023D D0max


b. The velocity (EAS) at minimum drag can be found from Eq.3.26 (n � 1)

1 /4 1 /42W k 2 � 10,000 0.0735V � �� EDmin � �� S C 0.002377 � 255 0.0230 D0

ft� 242.8

sec

This is also (TAS) at sea level. At 40,000 feet � 0.497)V (��Dmin

242.8 ftV � � 488.6Dmin 0.497 sec

c. The minimum thrust can also be found in a number of ways:

1. From Em, and for steady, level flight (W/T) � (L /D) onefinds from Eq. 3.7

W 10,000T � � � 822.4 lbmin E 12.16m

which is independent of altitude.2. Dmin can be obtained from Eq. 3.27, which gives

D � 2W�kC � 2 � 10,000 � �0.0735 � 0.023min D0

� 822.3 lb

3. From the basic drag expression, which can be written by useof Eq. 3.17

1 2 2D � � V C S � � V C Smin 0 E D 0 E DD min D 0min min2

1 2� 0.002377 � (242.8) � 2 � 0.023 � 2552

� 821.9 lb

C � 2C � 2 � .023where D Dmin 0

34 THE BASICS

Checking for L /D�max

L W 10,000� � � 12.17

D D 821.9min

a value in agreement with calculations above.These results are found also in Figure 3.4. The calculated rela-

tionships for Dmin and for can be read directly off the graph.VDmin

Also shown are induced drag

22kWD �i 2� V S0 E

and the parasite drag

1 2D � � V C Sp 0 E D02

for a weight of 10,000 lb. The influence of increased weight is alsoshown for 12,000 and 14,000 lb.

SR71

3.3.3 Flight Envelope: Vmax, Vmin

Returning to the original purpose of this chapter, to determine the max-imum and minimum velocities, it is necessary to consider the thrust-drag or power balance of the aircraft. For jet engine operation, thepower-available versus power-required statement reduces to thrust-required versus thrust-available, as already shown by Eq. 3.16 and Eq.3.24.

1 B2 2T � T � D � �V C S � AV � (3.32)a r D E 22 VE

Since, for steady, level flight, the thrust required is equal to the aircraftdrag, the developments in Subsection 3.3.2 can be applied here directly,


DDT

T

VV V

a

a

EES

S.L.

20,000 ft

30,000 ft

40,000 ft

50,000 ft

10,000 ft

Dmin

Figure 3.2 Aircraft Thrust Available and Drag

and Eqs. 3.18 to 3.31 give appropriate results for thrust required, min-imum thrust, and so on as shown in Example 3.2. If the thrust availableis known, then Eq. 3.32 can be solved directly for maximum or min-imum velocity at a given altitude and weight (at constant values of Aand B). In general, this means solving a fourth-degree algebraic equa-tion with two positive roots providing the maximum and minimumvelocities. Solution methods for such (polynomial) equations have beenknown for a long time, but the amount of labor and time involved hasresulted in preference to the graphical approach described in the nextsection (the exact method). Currently, any advanced hand-held calcu-lator can solve this problem in seconds (the Solver button), or one mayturn to a number of computer applications, such as MATLAB.

Unfortunately, jet engine performance data are not usually availablein a format that will permit a simple solution without recourse to sim-plifying assumptions. Thus, the method of solution of Eq. 3.32 dependson the type of engine data format—tabular, graphical, or equations—and willingness or need to use any of the appropriate simplifying equa-tions found in Appendix D.

Two practical methods are found in general use:

1. The exact or complete method, where detailed engine thrust dataare plotted over the drag curves on the T � V, (or VE) map withaltitude as a parameter (Figure 3.2). The more detailed curvesmay contain also fuel and air flow data (Appendix D). Exactnesshere depends on the completeness of drag data (i.e., inclusion of

36 THE BASICS

the effects of compressibility and high lift devices, stores, etc.)and the precision of drawing the curves and reading the resultsfrom the curves. The real advantage of this approach lies in theoverview it provides of the aircraft overall performance potential.Thus, it is also one way of presenting the aircraft flight envelope.

2. An approximate approach where the thrust expressions of the typeof Item D.1 or D.2 (Appendix D) are used to arrive at analyticalsolutions.

The Exact Method As shown in Figure 3.2, one finds typical sub-sonic jet engine data plotted over the drag data where the thrust de-creases with flight velocity (or Mach number) and shows a substantialdecrease with altitude. It should be noted that what is plotted is usuallythe maximum available thrust at a given engine rating. Any thrust be-tween that level and idle thrust is available and is a function of thethrottle setting. Moreover, standard engine curves are for uninstalledengines. For (subsonic) installed engies the data should be derated by5 to 10 percent, depending on the aircraft and installation. Since su-personic engine data is a strong function of the intake configuration,the data is best used in installed format.

The following information is immediately available from the curvesshown in Figure 3.2:

• Maximum and minimum velocities are found, at various differentaltitudes, at the intersection of thrust and drag curves. For a givenaltitude, the thrust curve crosses the drag curve at two widely dif-ferent velocities yielding the maximum and minimum level flightvelocities. The corresponding equivalent airspeed is read directlyoff the abcissa. Figure 3.3 shows typical EAS and TAS plottedagainst altitude. It is seen that, for jet aircraft, the maximum trueairspeed occurs at some intermediate altitude between sea leveland the ceiling. Since the overall Vmax occurs at an altitude higherthan sea level, the thrust required has also been reduced from thatat the sea level (see Figure 3.3). As the jet engine fuel consumptionis proportional to the thrust, it follows immediately that more ec-onomical flight occurs at altitudes higher than sea level.

It is seen that, for a given thrust level (altitude), there is a singleVmax. It should be noted that there are typically two minimumvelocities: the stall velocity Vs (Section 3.2) and a minimum valueVmin as determined by the available thrust level. If Vs is greater


V V

h

TASEAS

Ceiling

maxV

max

h

Figure 3.3 Airspeed as a Function of Altitude

than Vmin then, for steady level flight, the aircraft minimum velocityis determined by the maximum lift available at Vs. The relativemagnitude of these two velocities is determined by the aircraftmaximum lift coefficient and the characteristics of the particularpowerplant used.

• Ceiling also can be found directly from Figure 3.2. For steady,level flight, ceiling is defined by the condition where, at highestaltitude, Ta � D. This means, that the (absolute) ceiling is foundat the locus of the highest altitude Ta curve being tangent to a D(Treq) curve. The tangency condition also determines the velocityat which the (absolute) ceiling may be reached. For complete ceil-ing definitions, see the next chapter.

• The slight negative slope of the thrust curves in Figure 3.2 indi-cates that the velocity at the ceiling is at or near In this case,V .Dmin

the ceiling is shown to be at about 48,000 ft. If the thrust availablecurves were independent of altitude (a straight horizontal line),then the velocity at ceiling would occur at and then also it isVDmin

seen that at ceiling � Vmin � Vmax.VDmin

The results and graphs shown in Figures 3.2 and 3.4 require repetitiouscalculations for a given aircraft configuration and required data of thethrust as a function of velocity and altitude. However, the amount ofextra labor provides a general overview and realistic results with goodprecision. Example 3.3a shows the full calculation procedure to estab-lish the flight envelope for the aircraft discussed in Example 3.2.

38 THE BASICS

V E

W = 14,000 lb

W = 10,000 lb

W = 12,000 lb

T

T

ind

par

10,000 lb

10,000 lb

1000

2000

3000

4000

00 100 200 300 400 500 600 700 (ft/sec)

T (lb)

T Sea levela

10,000 ft

20,000 ft

30,000 ft

35,000 ft

40,000 ft

45,000 ft

Figure 3.4 Thrust Available and Required

EXAMPLE 3.3a

Suppose that the example aircraft treated in Example 3.2 is equippedwith two jet engines of P&W-60 class. The aircraft characteristicsare as follows:


2C � 0.023 � 0.0735 CD L

2S � 255 ft

W � 12,000 lb

C � 1.8L m

mT � T � per engineo

where To � 2,200 lb

m � 0.7, h � 36,000 ft

m � 1, h � 36,000 ft

The engine data is somewhat simplified (see Appendix D), and istaken to be independent of flight velocity. The drag has been cal-culated from Eq. 3.24 with

1A � .002377 � .023 � 255 � .00697

22 2B � 2 � .0735 � W / .002377/255 � .2425W

yielding

2 2 2D � .00696V � .2425W /VE E

The results, with engine data for twin engines, are shown in Figure3.4 for the aircraft weight ranging from 10,000 to 14,000 lb. Alsoshown are the individual curves for the induced and parasite dragsfor a weight of 10,000 lb. As expected, these drag terms are equalat � 242.8 ft/sec (see Example 3.2).VEDmin

The maximum speed at 30,000 ft altitude is read off as

V � V /�� 547/�.375 � 893 ft/secmax30 E 30max30

Also, the ceiling can be read off directly where the (imaginary) thrustline is tangent to 12,000 lb drag curve—at about 41,000 ft. Deter-mining the other results such as rate of climb, a more precise ceiling,and so on, will be discussed later in appropriate chapters.

40 THE BASICS

The Aproximate Approach This method provides quick results withless labor but also with a possible loss of accuracy. Here, drag data isgiven by the drag polar, and the (jet) engine data is approximated,usually by one of the analytical expressions found in Appendix D. Fordiscussion sake and to present the methodology jet engines, the dataleading to Figure 3.2 can be obtained approximately by one of thefollowing expressions:

mT � T � (3.33)o

2 mT � (A � BV )� (3.34)

where A, B, and m are constants obtained from curve-fitting themanufacturer-provided engine data and are not to be confused with thedrag representation with Eq. 3.24. Eqs. 3.33 and 3.34 lead to analyticexpressions for aircraft performance, with accuracy depending on thegoodness of fit to actual engine data.

To simplify calculations, and to generalize the results, it is practicalto use nondimensionalized equations, starting with Eq. 3.21 and intro-ducing as a reference condition at steady, level flight (n � 1), andVDmin

a nondimensional velocity as follows:V

VV �

VDmin

2 21 2W k kn W2D � �C V S � (3.35)� �D0 �2 S CD0 1 2W k 2�S V� � �2 �S CD0

Since (L /D)max � Em � 1/(2 this drag equation can be re-�kC ),D0

written as

2D 1 n2� V � (3.36)� �2W 2E Vm

By differentiating with respect to nondimensional velocity one canV,recover the results already obtained in Subsection 3.3.2. The minimumdrag occurs at

V � �n (3.37)

and is given by (W � constant)


D n� (3.38)�W L /D�maxmin

For steady, level flight, n � 1 and these results simplify further to

V � 1 (3.39)

D 1 1� � (3.40)�W L /D� Emax mmin

which is Eq. 3.30 and was entirely expected. The maximum velocitycan now be obtained by substituting a suitable thrust expression intoEq. 3.36.

Consider first Eq. 3.33—or for that matter, any thrust expressionwith T independent of velocity. Then Eq. 3.36 may be written, forsteady, level flight:

2TE 1m 2� V � � 0 (3.41)� �2W V

Introducing now a dimensionless thrust T

TEmT � (3.42)W

one obtains from Eq. 3.41

4 2V � 2TV � 1 � 0 (3.43)

Eq. 3.43 can be solved to give

2 2V � T � �T � 1 (3.44)

which, in turn, yields two more solutions. Only the two following pos-itive solutions are physically meaningful:

2V � �T � �T � 1 (3.45)1

2V � �T � �T � 1 (3.46)2

It is evident that represents the high-speed solution, � 1, andV V2 2

is the low-speed solution, � 1. This follows from the fact that,V V1 1

42 THE BASICS

for real solutions, must be larger than unity. If � 1, one seesT Timmediately that

V � V � 1 (3.47)1 2

which represents the condition where minimum and maximum veloc-ities coincide, and implies that V � Clearly, then, this gives aV .Dmin

solution for the ceiling where � 1 and the velocity at the ceilingTVceiling � Incidentally, Eqs. 3.45 and 3.46 yield, after multipli-V .Dmin

cation by each other, an interesting condition: � 1. This restatesV V1 2

the conclusion already obtained that one of the nondimensional veloc-ities must be larger than, and the other smaller than, unity. Thus, if onesolution is known, the other can be easily found from this condition.

The ceiling can now be calculated from the condition that � 1:T

TEmT � � 1 (3.48)W

Substituting for the thrust from Eq. 3.33, one gets the density ratio atthe ceiling:

1 /m 1 /m2W�kCDW 0� � � (3.49)� � � �c T E To m o

Caution should be exercised in applying Eq. 3.49, since the exponentm in the thrust equation may not apply over the entire altitude range(see Example 3.3a and Appendix D).

The maximum velocity can be calculated from . SubstitutingV T2

and recalling the definition of one obtains from Eq. 3.46V,

2V TE TEmax m m� � � 1 (3.50)� ��V W WDmin

Eq. 3.50 gives the maximum velocity, at a given altitude, for given T,W, S, k, CD. However, the altitude at which the maximum aircraft ve-locity occurs cannot be found easily from Eq. 3.50 (see Figure 3.3).To simplify calculations, nondimensional velocity shall be redefinedVin terms of the sea level � as follows by use of Eq. 3.26:V VD Emin D0 min


V VV � � (3.51)o 1 /4VDmin 2W ko � �� S C0 D0

Introducing this into Eq. 3.50 and using Eq. 3.33, one obtains, aftersome rearrangement (dividing both sides by ��),

2T E T E 1o m o mV � � � (3.52)� �omax 1�m 1�m 2��W� W� �

where To is the sea-level engine thrust and Em is given by Eq. 3.31. Itis seen that the altitude for Vmax increases as the thrust level is in-creased.

To round off the approximate method, a second approach to thevelocity solution is to set the thrust available equal to thrust required,Ta � Tr � D, by use of Eq. 3.21, as follows:

21 2kW2T � �V SC � (3.53)a D0 22 �V S

and to solve for the velocity V. After some laborious manipulations,one obtains

T 1aV � 1 � 1 � (3.54)� �2��SC (E (T /W))D m a0

where the positive sign is used for maximum velocity and negativesign for minimum velocity. It should be noted that Ta represents anyavailable thrust (at a given throttle setting) at an altitude and is assumedto be independent of velocity in Eqs. 3.53 and 3.54. Thus, Eq. 3.54avoids the curve-fitting process to determine the constants m, To, A,and B in Eqs. 3.33 or 3.34 and may seem to yield somewhat moreaccurate results. However, if thrust shows appreciable variation withvelocity as is more often the case, the calculation process may turn outto be rather laborious and iterative to determine the correct value of Ta

for use in Eq. 3.54.

44 THE BASICS

EXAMPLE 3.3b

(Example 3.3a continued) This example shows how the same per-formance data can be determined by the approximate approach.However, since simplified engine data is used, the agreement, asanticipated, is very good. In practice, such agreement should alwaysnot be expected.

In this example, Vmax and the ceiling are to be determined. SinceEm has been already calculated in Example 3.2 as 12.16, then

W 12,000T � � � 987 lbmin E 12.16m

is obtained from Eq. 3.26:VEDmin

1 /4 1 /42W k 2 � 12,000 0.0735V � �� EDmin � �� S C 0.002377 � 255 0.0230 D0

ft� 266

sec

The values for other altitudes are obtained from

VEDminV � (TAS)Dmin ��

The maximum velocity at altitude is found from Eq. 3.50 or 3.52.Using Eq. 3.42:

T E 4400 � 12.16o m � 4.45W 12,000

one finds that at 30,000 ft, � � 0.375 and

24.45 4.45 1V � � � � 3.36� �omax 0.3 0.3 2��0.375 0.375 0.375

Whereupon


TABLE 3.3 Velocity Comparisons, ft /sec; W � 12,000 lb

h (ft) 0 10,000 20,000 30,000 40,000 42,000

VDmin266 310 364 435 535 561

Vmax 790 826 862 891 684 561VEmax

790 710 629 546 334 266Vmin 90 100 155 210 430 561

ftV � 3.36 V � 3.36 � 266 � 894max E30,000 Dmin sec

which agrees well with 893 ft/sec found in Example 3.3a. Similarly,can be obtained by first calculating from Eq. 3.45VEmin

24.45 4.45 1V � � � � .7926� �omin .3 .3 2��.375 .375 .375

and then

ftV � .7926V � .7926 � 266 � 210min E30,000 Dmin sec

The maximum and minimum velocities at each altitude are shownin Table 3.3. To find the altitude at which the aircraft achieves itsabsolute maximum velocity, it is easiest to plot the maximum valuesas function of altitude, and to determine the maximum value fromthat curve. A rough interpolation from Table 3.3 indicates that themaximum velocity is about 900 ft/sec, and it occurs near 35,000 ftaltitude.

The density ratio at ceiling is obtained from Eq. 3.49:

1 /mW 1� � � � 0.225� �c T E 4.45o m

which gives, from the altitude table, h � 42,000 ft. The exponent,m has been assumed to be unity, as the ceiling was expected to occurat an altitude higher than 36,000 ft. In a later section in Chapter 4,where the rate of climb is studied, the ceiling can be determinedmore accurately since the ceiling can also be defined as the locationwhere the rate of climb is zero.

46 THE BASICS

Figure 3.5 Flight Envelope

For completeness, the stall velocity is obtained from

2W 2 � 12,000 ftV � � � 148Es � �� SC .002377 � 255 � 1.8 sec0 L m

The last line shows for direct comparison with the valuesVEmax

obtained from Figure 3.5, where Ta and D are plotted as functionsof VE.

The results obtained so far in this chapter and in Chapter 2 can besummarized and collected in Figure 3.5, which is commonly called theflight envelope. At this point, the graph of Vmin and Vmax represents, ata given thrust rating, only the theoretical steady, level flight boundaryof an aircraft. It has been calculated by use of a simple drag polarwithout any practical regards concerning the aircraft lifting capability(Table 3.3). Usually, Vs is larger than the drag polar calculated Vmin

over most of the lower speed range, thus shrinking the aircraft per-formance range. The resulting boundary shows, to a first approxima-tion, the possible velocity range at a given altitude and also the ceiling.In conjunction with Figure 3.4 it also determines the (excess) energyavailable (Ta � D) for steady-state maneuvering within that envelope.(More discussion follows in Chapter 4, and see also Figures 4.5 and


4.6.) Unsteady dynamic performance such as zooms and dives can takethe aircraft for a short time outside of this envelope.

Additional realistic constraints (compressibility and planform effectson drag polar, buffeting limits, and structural and q-load considerations)may considerably decrease the Vmax boundary. Inclusion of those effectsis a must for design and realistic performance analysis but is outsidethe intended scope of this text. However, the methodology already out-lined is applicable with any desired degree of precision and imagina-tion.

C130

3.3.4 Power Required and Power Available

It has been noted that it is more convenient to formulate propeller-driven aircraft performance in terms of power rather than via thrustexpressions. The power required Pr to fly an aircraft can be found fromthe drag expression by use of Eqs. 3.13 and 3.19:

3 2 2�V SCDV kn WD0P � � � (3.55)r 550 1100 275�VS

The first term gives the power required due to parasite drag. The secondterm is the induced power required. A typical power required curve isshown in Figure 3.6.

Due to proportionalities to V3 and V�1, the curve is skewed from thethrust required curve. The minimum power does not occur at the ve-locity for Em, and the minimum power shows a strong altitude effect.To determine these respective locations, Eq. 3.55 will be differentiatedwith respect to the velocity. Setting the result equal to zero, one obtainsfor the velocity at minimum power

1 /42Wn kV � (3.56)� �Pmin � �S 3CD0

By comparison with Eq. 3.26, it follows immediately that

48 THE BASICS

V

Pr

Parasite power

Induced power

Total P

Increasing altitude

VPrminV

max

r Line oftangency

(L/D)

Figure 3.6 Power Required

1 /4V � V � 3 V � 1.316V (3.57)D (L /D) P Pmin max min min

The minimum power required is found by substituting into Eq.VPmin

3.55, which gives for minimum power

3 /43 / 2 3 / 4 3 / 2 1 / 41 (2Wn) k k (2Wn) 31 /4 1 /4P � C � C� �min D D0 01100 3 1100��S ��S3 /2 3 /4 1 /42.48 (Wn) k CD0� (3.58)

550 ��S

An inspection of Eq. 3.58 reveals that, for minimum power required,the induced power required is three times the parasite power required.This implies that, at minimum power

2kC � 3C (3.59)L D0

2C � C � kC � C � 3C � 4C (3.60)D D L D D D0 0 0 0

The minimum power can be simplified as


CDV 1 1 D02 3P � � �V 4C SV � V S� (3.61)min D P0 min550 550 2 275

where is given by Eq. 3.56. Although these minimum expressionsVPmin

will not be used immediately, they are useful for determining the gen-eral shape of the power required curve and are essential later for es-tablishing other performance parameters.

Returning now to the basic problem at hand: Finding the maximumvelocity (a solution to Eq. 3.13) can be easily accomplished by plottingboth Pa and Pr curves versus V, as shown in Figure 3.7. The power-required curves are given for several altitudes at constant weight ornW. Power-available curves for engines as a function of altitude andRPM are supplied by the manufacturer as BHP (brake horsepower) ingraphical form (see Appendix D). Power available is calculated from

P � BHP � (3.62)a p

Since each propeller has its own different efficiency curves (dependingon its size, velocity, rpm, twist, etc.), families of Pa curves can bedeveloped with engine rpm as a parameter. It is impractical to derivethis information in analytical form. Thus, a graphical approach is usedfor general propeller aircraft performance calculations. For constant-altitude flight, the power-available curve may be approximated by astraight line (see Problem 3.11). Example 3.4 shows a typical approachto this problem.

EXAMPLE 3.4

Consider a twin-engine aircraft with the following characteristics:

2S � 300 ft

AR � 7

e � 0.852C � 0.024 � 0.0535 CD L

W � 9,600 lbT.O.

W � 1,450 lbfuel

50 THE BASICS

TABLE 3.4 Example 3.4 Engine Data

h (ft) 0 5,000 10,000 20,000

BHP 375 316 264 176

The engines are Lycoming I0-720 400 HP rated at 2,650 RPMfull throttle. The values for available power at 2,400 RPM are shownin Table 3.4 for altitude variation: The propellers are three-bladedHamilton Standard, 6 ft diameter, 100 activity factor, 0.3 integrateddesign lift coefficient. Table 3.5 shows propeller efficiency at thesealtitudes as a function of flight speed. The power required has beencalculated from Eq. 3.55 and is shown in Figure 3.7.

The power available was calculated by use of Eq. 3.62, BHP, andefficiency given in Tables 3.4 and 3.5. Using both Figure 3.7 andthe analytical equations, the following will be determined for aweight of 9,117 lb:

a. Pmin at sea levelb. at sea levelV(L /D)max

c. Vmax

d. ceiling

The answers follow in the same sequence.

a. Graphical solution is obtained from Figure 3.7 showing Pmin �203 HP at about 146 ft/sec. It can be found also by use of Eq. 3.58as

3 /2 3 /4 1 /42.4816 (9117) (0.0535) (0.024)P � � 203 HPmin 550 �0.002377 � 300

The velocity at minimum power follows from Eq. 3.56

1 /42 � 9117 0.0535 ftV � � 148� �Pmin �0.002377 � 300 3 � 0.024 sec

b. From Eq. 3.57 one finds

ftV � 1.316V � 1.316 � 148 � 195(L /D) Pmax min sec


TABLE 3.5 Propeller Efficiency, �

h (ft)V (mph)

0 10,000 20,000

50 0.29 0.30100 0.55 0.57 0.60150 0.74 0.75 0.78200 0.835 0.84 0.86250 0.881 0.89 0.88

V (ft/sec)

S.L.

10,000 ft

20,000 ft

r

aP

P

00

50 100 150 200 250 300

200

400

600

P (hp)

P

P

P

P

a

a

r

r

S.L.

S.L.

10k

10k

20k

20k

Figure 3.7 Power Required and Power Available

c. The values for the maximum velocity can be found from inter-sections of Pa and Pr curves, shown in Table 3.6.

d. The steady, level flight ceiling is given by the location wherethe Pr and the last Pa curves are tangent to each other. In this case,the last power available curve falls slightly below the Pr curve at20,000 ft altitude. Interpolation of calculated values indicates thatthe last Pa � Pr condition would occur near 19,500 ft and at about254 ft/sec. In a later section, where the rate of climb is studied, theceiling can be determined more accurately because the ceiling canalso be defined as the location where the rate of climb is zero.

52 THE BASICS

TABLE 3.6 Maximum Velocities

h (ft) 0 10,000 20,000

Vmax

ft� �sec330 300 255

Similar to the thrust-oriented approach, the intersections of thepower available and power-required curves represent (the velocity) so-lutions to Eq. 3.55, for a given W. The areas between the Pr and Pa

curves are called the steady flight envelope. In level flight, n � 1, andthe sequence of intersections shows that there are two possible speedsfor sustained level flight, Vmax and Vmin. From Example 3.4, the follow-ing conclusions can be drawn for propeller-driven aircraft:

1. The maximum velocity occurs at the sea level.2. The ceiling is found from the condition where Pa and Pr curves

are tangent to each other and it occurs at only one velocity—usually near for that altitude.V(L /D)max

E3C

3.3.5 Turboprop Engines

A turboprop engine develops both jet and propeller thrust. A typicalturboprop derives about 80 to 85 percent of the power from the shaft-propeller combination. The remaining fraction comes from the jet thrust(see Appendix D for T-56 engine data). Thus, according to the preced-ing sections, turboprop aircraft performance could be developed eitherin terms of power or in terms of thrust. Since the calculations centermainly around the propeller efficiency determination, it is convenientto analyze the turboprop performance in terms of the power expres-sions. The usual practice is to evaluate the thrust horsepower of a tur-boprop engine by means of

3.4 GLIDING FLIGHT 53

TVTHP � SHP � � (3.63)p 550

where SHP is the installed shaft horsepower. If the velocity is given inknots, then the thrust horsepower available becomes

TVTHP � SHP � � (3.64)p 326

Sometimes it is convenient (mainly for comparing different engines) toexpress the power in terms of equivalent shaft horsepower as

TVESHP � SHP � (3.65)

550�p

from which one can calculate the total power available as

THP � ESHP �p

The drag of a turboprop-driven aircraft must be stated in terms of powerrequired, and the basic performance analysis follows that of a typicalpropeller-driven aircraft.

3.4 GLIDING FLIGHT

3.4.1 Glide Angle and Sinking Speed

There are two basic approaches for the determination of the aircraftglide performance: power-off, and partial power-on. Since the power-on glide can be viewed as a special case of Section 3.3, with a certaindeficit in the thrust or power available, its discussion will be delayeduntil the next chapter where the excess power (or climb) problem willbe studied. The foundation of all glide problems is the FPE (Eq. 3.4):

dh P � DV 1a�dt W V dV

1 �� g dh

54 THE BASICS

with the implied assumption that � 0. This implies in Eq. 2.12 that�the change of the curvature of the flight path and the associated inertiaterm are negligible and the energy equation Eq. 2.20 can be used di-rectly. Since only the power-off case will be considered, Eq. 3.4 be-comes

dh DV 1� � (3.66)

dt W V dV1 �� g dh

As an energy equation, Eq. 3.66 states that the rate of change of altitudeis controlled by the drag power and by the rate of change of kineticenergy (the term in the denominator). As the aircraft loses altitude, thedrag power appears as the loss of potential energy, which is expendedin overcoming the drag. Ignoring the kinetic energy change, the lossof potential energy and also the rate of change of altitude (sinkingspeed) are minimum when the drag power DV is a minimum. In glideanalysis of typical slow-speed gliders and other aircraft operating inglide phase at modest altitudes, the kinetic energy change can be ig-nored because its contribution is small. In high-altitude operationswhere the flight speeds are very high (i.e., reentry vehicles), the kineticenergy change provides the major contribution to the glide range andendurance (see Example 3.6).

Following first the classical slow speed and shallow angle approach,the glide equation becomes

dh DV� � (3.67)

dt W

and the glide angle is obtained by substituting the kinematic relation-ship ([dh /dt] � V sin �) into Eq. 3.67, which gives

D CDsin � � � � � � (3.68)W CL

with the assumption that the equilibrium conditions hold for the glide(i.e., n � 1, or L � W). It follows that the minimum glide angle �occurs at Em, and one gets


1� � � � �2�k C (3.69)min D0Em

The flight speed for the minimum glide angle follows (see Eqs. 3.26and 3.31) from

1 /42W kV � V � � V (3.70)� �� (L /D) Dmin max min� �S CD0

The sinking speed v � �dh /dt can be developed from Eq. 3.67 withthe velocity

2WV � ��SCL

as follows:

C 2W 2W CD Dv � � (3.71)3 /2� �C �SC �S CL L L

To find the minimum sinking speed, or the lift coefficient that min-imizes the sinking speed, the above can be differentiated with respectto CL, and the result set equal to zero. Another approach yields thesame result if one recognizes, from Eq. 3.67, that v is minimum whenDV (the power required) is minimum. But this condition has alreadybeen developed for Eqs. 3.56 through 3.58. As a result, one finds thatthe minimum sink rate occurs atvcmin

V(L /D)maxV � � 0.76V (3.72)P (L /D)min max1 /43

where also

2C � 4C , k C � 3C (3.73)D D L D0 0

Evaluating one finds(L /D) ,Pmin

56 THE BASICS

�3 �33CL C 1DL 0� � � � E (3.74)� � � � m�D C k 4C 24�kCD D0P P Dmin min 0

With the aid of and the above expression, the equation for theVPmin

minimum sinking speed becomes

VDV WPmin 3 /4 1 /4v � � � 2.48 (k) (C ) (3.75)�c Dmin 0�W �SLPmin �DPmin

These results can be summarized as follows:

1. The horizontal velocity for minimum sinking speed isV vP cmin min

smaller than the velocity for the flattest glide V .(L /D)max

2. Both of these velocities are functions of the altitude, and theydecrease as the aircraft descends toward the sea level. This is truealso for the minimum sinking speed.

3. The L /D ratio is maximum for the flattest glide, but is /2 ��30.87 Em for the minimum sinking speed.

4. The minimum glide angle is independent of the altitude, and isonly a function of Em.

747 and Shuttle

EXAMPLE 3.5

Consider again the aircraft studied in Example 3.2. Assume that ithas a total power failure at 10,000 feet. Its clean drag polar and itsmaximum lift to drag ratio are as follows:

2C � 0.023 � 0.0735 CD L

E � 12.16m


Its minimum glide angle is

1 1� � � � 0.0823 rad � 4.71min E 12.16m

and occurs atV 243 ftEDminV � � � 283 (horizontal velocity)Dmin 0.859 sec��

Assuming that the pilot can maintain an indicated airspeed of 243ft/sec, the plane can glide a distance of

10,000x � � 121,500 ft � 23 miles

0.0823

The vertical velocity (sink speed) is Eq. 3.71 and is developed forCD and CL at V .Dmin

2CC 2W 2WDD 0v � � 3 /43 /2 � �C � �S � �SCL 0 0D0� �k

2 � 0.023 2 � 10,000 ft� � 23.23 /4 �0.002377 � 0.739 � 255 sec0.023� �0.0735

This completes the basic glide problem: calculation of the glide an-gle and the sinking speed for shallow paths (i.e., L � W and sin � �). For steeper angles of glide L � W, cos � and the glide anglebecomes

�1tan � � (3.76)

C /CL D

However, this is still not correct; for steep glide angles one must alsoconsider the kinetic energy contribution and use the energy equationEq. 3.66. The glide angle becomes

58 THE BASICS

C /CD Ltan � � � (3.77)V dV

1 �g dh

In order to obtain the glide angle, the velocity variation with altitudemust be specified, which implies that a glide schedule must be alsoknown. Otherwise, the problem becomes a path-integration problem. Itis possible, however, to reduce Eq. 3.77 to a point-performance prob-lem if the instantaneous velocity is known. This is achieved, for aconstant angle of attack, by differentiating the instantaneous path ve-locity:

2W cos �V � (3.78)� �SCL

with respect to altitude and determining dV /dh, since the angle ofattack is assumed to be constant. One obtains

dV 1 V d� V d� 1 V d�� tan � � �

dh 2 � dh 2 dh 2 � dh

(V/2) tan �(d�/dh) 1 V d�� (3.79)

dh /dt 2 � dh

because it was assumed that � 0. Thus, the glide angle becomes�

C /CD Ltan � � � (3.80)2V 1 d�1 �

2g � dh

Either V or CL can be eliminated from Eq. 3.80 by means of Eq. 3.78.Since CL is assumed to be constant, it is seen that the velocity decreasesas the altitude decreases (the density � increases), and the kinetic en-ergy correction in the denominator becomes increasingly smaller forlower altitudes. Since the density variation is given by different ex-pressions for troposphere and stratosphere, the problem needs to beconsidered in two steps.


Troposphere In the troposphere, h � 36,089 ft, the density variationis given by

�6 4.256� � (1 � 6.88 � 10 h)

and the glide angle, Eq. 3.80, becomes

C /CD Ltan � � � (3.81)�5 22.928 � 10 V

1 � .2352g�

This indicates that in the troposphere, where 1 � � � .297, the kineticenergy correction in the denominator is small for flight velocities lessthan 400 to 500 ft/sec.

Stratosphere In the stratosphere, � � .297, the density ratio is givenby

�5�4.8�10 (h�36089)� � .297e

which leads to the following expression for the glide angle:

C /CD Ltan � � � (3.82)�5 24.8 � 10 V

1 �2g

In the stratosphere, where the glide velocities must be high to sustainlift, the kinetic energy contribution is seen to be significant (see Ex-ample 3.6).

3.4.2 Glide Range and Endurance

The range and endurance can be calculated with some generality for aglide at constant angle of attack. The shallow glide path and the casewithout kinetic energy term will appear as a special case. The followingset of differential equations are obtained for the range and endurance.For range, the energy equation Eq. 3.2 can be written as

60 THE BASICS

2dh DV dV� � � (3.83)

dt W 2gdt

and if V is eliminated from the (DV)/W term by means of Eqs. 2.10and 2.12, x � R (range), 0 becomes�

2dR dV� �dh � (3.84)

L /D 2g

The equation for time is obtained from Eq. 2.11, (T � 0)

1 dV D� � � sin � (3.85)

g dt W

Eliminating sin � � (1/V)(dh /dt) (Eq. 2.9), and multiplying by dt, oneobtains (cos � 1)

dt dh dV� � � (3.86)

L /D V g

Since L /D � CL /CD and CD � CD(CL), it is possible to integrate Eq.3.84 and Eq. 3.86 if one assumes that the angle of attack is constant(CL � constant). With the following boundary conditions between anytwo points

R � 0, t � 01 1

one gets

2 2C V � VL 1 2R � h � h � (3.87)� �1 2C 2gD

To integrate Eq. 3.86 for endurance, the velocity will be eliminated byEq. 3.78, and one obtains for time:

2 2 2C C � S C V � VL L 0 L 1 2t � � ��dh � (3.88)� �� 1C 2W C 2gD D

As in the previous section, the integral containing the density ratio mustbe evaluated for two cases:


Troposphere

46467 C V � VL 1 2.74 .74t � (� � � ) � (3.89)� �trop 2 1v C gc Do

Stratosphere

41667 C V � VL 1 2t � (�� ) � (3.90)� �strat 2 1v C gc Do

The following results can be summarized:

1. The range equation is essentially an energy expression. The rangeis determined by the difference between the initial and final en-ergy heights h � V2/2g. The energy is converted into range bythe lift/drag ratio. Thus, the maximum range is obtained, forgiven initial and final conditions, at an angle of attack that yieldsEm.

2. The velocity for maximum range is given by Eq. 3.26:

V � V(L /D) Dmax min

3. The maximum endurance is essentially determined by CD / 3 /2C ,L

which implies that the path flight velocity is (see Eq. 3.56).VPmin

Due to the velocity term, where CL /CD appears, the angle ofattack for maximum endurance is such that the optimum flightvelocity is somewhere between andV V .(L /D) Pmax min

EXAMPLE 3.6

The glide range for Example 3.5 will now be reexamined using Eq.3.87:

2 2C V � VL 1 2R � h � h �� 1 2C 2gD

The potential energy contribution was calculated in Example 3.5,and it resulted in a range of 23 miles. The kinetic energy contributionis obtained from

62 THE BASICS

ftV � V � 2831 Dmin10,000 sec

ftV � V � 2432 Dmin0 sec

and the range increase is given by

2 2 2 2C V � V 283 � 243L 1 2 � 12.16 � � 4,973 ft� �C 2g 2 � 32.2D

However, the kinetic energy term becomes significant for a reentryvehicle even with a low CL /CD � 2.5, which is descending from80,000 ft at 17,400 ft/sec to a sea level landing at 300 ft/sec. Then,the total glide range is

2 22.5 17,400 � 300R � 80,000 � � 2,263 miles� �5,280 2 � 32.2

PROBLEMS

3.1 An aircraft weighing 18,000 lbs, with wing area of 350 ft2, fliesat a true air speed (TAS) of 250 knots. Calculate the aircraft liftcoefficient, CL, and the equivalent air speed (EAS) VE if theflight altitude is:a. sea levelb. 20,000 ft

3.2 After the aircraft in Problem 3.1 has consumed 3,000 lbs of fuel,what value of CL will be required to fly at the same TAS of 250knots at 20,000 ft altitude? The aircraft in Problem 3.1 is nowflying on its landing approach at its maximum lift coefficient

of 1.5. If the gross weight is 15,000 lbs, what is the cor-CL max

responding V ?Emin

Ans: 155 ft/sec.

3.3 It is much safer to fly at a speed marginally above (i.e.,VEmin

� 1.2 Calculate this speed and then calculate theV V ).land Emin

corresponding TAS for landing at an altitude of 6,000 ft.

3.4 The landing speed of an airplane is 10 mph greater than itsstalling speed. � 1.5, � 1.8. Find the stalling speed.C CL Ll s

PROBLEMS 63

3.5 An airplane has the following characteristics:

C � 1.9L max

� deg �2L�0

Lift curve slope a � 0.075 per degV � 60 mph at 3,000 ftsmax

At what velocity will it fly at sea level if the angle of attack is � 1/deg?Ans: 245 ft/sec.

3.6 The aircraft described in Example 3.2 is a T-2B jet trainerequipped with J60-P-6 turbojets, and its empty weight is 10,000lb. The drag polar given there is for a clean configuration. If theaircraft is loaded with two racks of rockets (rockets � racks �487 lbs), its parasite area is now 6.25 ft2 and the span efficiencydrops to 0.82. If it carries 2,500 lb fuel, calculate:a. The new drag polarb. (L /D)�max

c. Tr at minimum drag and at 20,000 ft

3.7 The aircraft in the last problem is equipped with 2 J60-P-6 jetengines and weighs 13,000 lb. The thrust can be approximatedby

mT � 2600 �m � 0.72 h � 36,000 ftm � 1 h � 36,000 ft

Find (a) maximum velocity at 25,000 ft (b) its ceiling

3.8 The following data are given for P-3C Orion ASW aircraft:

2S � 1300 ft2f � 29.1 ft

AR � 7.5W � 128,000 lbb � 99 ft (wing span)span efficiency � 0.948 (low speed)

Calculate the thrust horsepower required for 15,000 ft at a speedof 200 knots.Ans: 5,588 HP.

64 THE BASICS

3.9 An aircraft has the following characteristics:

2W/S � 40 lb/ft2S � 400 ft

2C � 0.018 � 0.062 CD L

Calculate:a. Minimum drag speed, EAS (ft/sec)b. Minimum power required speed, EAS (ft/sec)c. Minimum power requiredd. Minimum thrust required for steady, level flight

3.10 A transport aircraft cruise weight is about 650,000 lb. Its bestcruise speed is 625 mph at 42,000 ft. It is equipped with fourJTD9-3 engines. Calculate its drag coefficient and estimate thethrust-specific fuel consumption. Its wing area is 5,500 ft2.Ans: CD � .0227.

3.11 Show that

1 /44 1 1 WV � � �Pmin �3�ƒe � � b0

where b is the wing span, and

WV1.15 PminP �min 550 Em

3.12 The clean drag polar of a fighter is

2C � 0.04 � 0.09CD L

Its landing drag polar (flaps, slats and gear down) is

2C � 0.2 � 0.32CD L

It is equipped with one engine whose thrust can be approximatedby

mT � 15,000 � m � 1 h � 36,000 ft

m � 0.8 h � 36,000 ft

PROBLEMS 65

Its maximum lift coefficient is 2.16. Its take-off speed is thesame as its landing speed. Assume that the landing speed is 1.2Vs. Calculate:a. T /W at ceilingb. T /W at takeoff

Ans: a: 0.12; b: 0.613

3.13 Show that for level flight, with the engine thrust inclined at angleT to the flight path, the power-on stall velocity is given by

2W TV � 1 � sin � �s T��SC WL max

Hint: Consider Eqs. 2.9 to 2.13. Derive the full equation.

3.14 A wing is gliding from a height h into a horizontal head windof Vw mph. If the wing loading is lw (psf), find an equation forhorizontal distance d (ft) that the wing travels before strikingground. Show that

C � CL Ld � h 1 � V� �w�C 2 lD w

Assume small angles.

3.15 A jet aircraft is at 20,000 ft with the following characteristics:

W � 13,000 lbT � 5,200 lb

2S � 255 ftAR � 5.07

2C � 0.0245 � 0.0765 CD L

There is a flame-out and the engine cannot be started. Calculate:a. Potential maximum glide rangeb. Path anglec. Maximum range sinking speedd. Horizontal velocitye. The glide angle at the minimum sinking speedf. Minimum sinking speedg. Range at minimum sinking speed

66 THE BASICS

3.16 Calculate the minimum sinking speed and resulting glide rangefrom the data given in Example 3.5.Ans: vcm � 20.4 ft/sec; R � 19.9 mi.

3.17 A lifting reentry body with CL /CD � 2.5 is gliding at 80,000 ftwhen its flight speed is 17,400 ft/sec. Calculate its flight pathangle without and with the kinetic energy term.Ans: �21.8, �.1 deg, resp.

3.18 The drag polar for a glider is given by

2C � 0.017 � 0.021CD L

AR � 20.4b � 102 ftW � 840 lbh � 35,000 ft

Calculate its range and time to glide to 5,000 ft at constant angleof attack.

3.19 A twin-engine, propeller-driven airplane has the following char-acteristics:

E � 12m2f � 17.45 ft

2S � 500 ftW � 67,000 lb

The pilot is told to hold at minimum fuel flow speed at 5,000ft. Calculate the holding speed.Ans: � 300 ft/sec.VPmin

3.20 A 15,000 lb airplane is coming in for a landing on instrumentapproach. Its landing drag polar is CD � 0.042 � 0.06 and2CL

is 2.9. Calculate the thrust required at the landing speed.CL max

Ans: T � 1,913 lb.

3.21 An Airbus A330-200, with a crew of 13 and 293 passengers,runs out of fuel (a bad leak) at an altitude of 39,000 ft. Emptyweight of the aircraft is 266,000 lb. Determine:a. Whether this aircraft can glide to the nearest airport, 85 nmi

away

PROBLEMS 67

b. The flight speed for the maximum range, at altitude, at sealevel (Normal landing speed is 250 km/hr, flight manual rec-ommends a flight speed of 170kt for a deadstick landing.)

c. The sinking speed corresponding to b. at sea leveld. How much range would be available if the energy term would

be included in the calculations.Ans: a. 129 mi.

3.22 A propeller-driven airplane weighs 4,000 lb, has a wing area of200 ft with an aspect ratio of 6.5, an efficiency factor of .7, anda zero-lift drag coefficient of .0277. It has two engines, eachproducing a maximum of 200 HP at propeller efficiency of .82.Estimate the maximum sea-level speed attainable by this aircraft.

3.23 Calculate minimum required power for the aircraft in problem3.22. Hint: Problem 3.11.Ans: 91.7 HP.

3.24 Verify Eq. 3.81.

3.25 The following is known about a multiengined aircraft:

2S � 1,300 ft2ƒ � 29.1 ft

AR � 7.5span � 99 ftspan efficiency � .948C � 2.6L max

W � 128,000 lb

Determine:a. How much horsepower is required to fly this aircraft at sea

level at 200 knotsb. The landing speedc. The maximum weight to land at 100 knots


2C � .025�.057CD L

W � 30,000 lb2S � 550 ft

AR � 5e � .895span � 52.44 ft

68 THE BASICS

The thrust at 10,000 ft altitude is constant at 8,000 lb. DetermineVmax at that altitude.Ans: 805 ft/sec.

3.27 At sea-level testing, a twin-engine jet aircraft was observed tohave a maximum velocity of 746 ft/sec. Its weight was 14,000lb with a wing area of 232 ft2. Engine data for a JT15 engineindicated that the thrust would be, at that altitude and speed,about 1,575 lb/engine. Estimate the parasite drag coefficient.Ans: .0205

3.28 A twin-engine jet utility aircraft, having been in the holdingpattern (at minimum drag speed to conserve fuel) at 5,000 ft,was instructed to reduce velocity by 90 ft/sec prior to com-mencing landing approach. Aircraft data:

W � 10,254 lb2S � 260 ft

AR � 5.4e � .87k � .0678C � 2.7, C � 1.7L Lmax,flaps max,clean

C � .017D0

Determine:a. The amount of thrust change (added or reduced)b. Whether the resulting flight speed is a safe flying speed


2C � .025 � .057CD L2W � 30,000 lb, S � 550 ft

AR � 5, e � .895span � 52.44 ft, V � 320 ft/sec

Thrust and chord are along fuselage centerline, 0L � 0, wingquarter chord sweep is 12o, twin tail.a. Obtain the equations of motion when flying up in a perfectly

vertical direction (Hint: use Eqs. 2.9 to 2.12).b. What is the thrust required at sea level when this aircraft is

in steady vertical flight (i.e., apply the equations obtained inpart a)?

Ans: 31,670 lb

PROBLEMS 69

3.30 An A7E has flame-out at 15,000 ft and cannot be started. It hasthe following characteristics:

2C � .015 � .1055CD L

C � 1.6L mclean

C � 2.5L mfl,slats

2W � 25,000 lb, S � 375 ft

Determine to sea level:a. Range at minimum sink rateb. Rate of sink at touchdownc. Touchdown velocityd. Estimate time to fly

3.31 It is considered that a single-engine jet aircraft, with the follow-ing data, be uprated by adding another identical engine. Assum-ing that the second engine does not affect the aerodynamics,how much can the aircraft weight be increased to reach the sameceiling? The aircraft has the following parameters: equivalentparasite area 4.84 ft2, full span flaps, AR � 8, W � 11,000 lb, S� 242 ft2, T � 1,000 lb, e � .663, concrete runway, hot standardday.

3.32 Using Eq. B.2 in Appendix B, show that the minimum dragcoefficient is now given byCDmin

2C � (k � k )C � kCD 1 L Lmin o

and L /D�max becomes

L 1�� 2 2D 2�(C � kC )(k � k ) � 2kCmax D L 1 L0 o o

70

4Climbing Flight

F15

4.1 GENERAL

Climbing performance of an aircraft can be viewed as an energyexchange between the powerplant and the kinetic and potential energiesof the aircraft. If the aircraft is in steady, level flight and the power isincreased to exceed the amount needed for sustained level flight, thenmore work is being done on the aircraft than is required to overcome


4.2 RATE OF CLIMB, CLIMB ANGLE 71

the drag. As a result, the kinetic or potential energy, or both, mustincrease. Which one increases depends on how the aircraft is beingoperated.

Keeping the angle of attack constant will tend to increase V until anew T � D balance is achieved. The new increased V, at constant �(and CL), will produce excess lift, which in turn leads to an increasein potential energy. Moving the control stick to decrease the angle ofattack at constant power will decrease CL and CD, which at constantpower will lead to an increase in V. Since the total lift will remainapproximately constant, most of the energy change will go into anincrease of the kinetic energy. If the velocity is kept at some constantvalue, then the excess power will be converted into an increase inpotential energy and the aircraft will climb.

The fundamental performance equation, Eq. 3.4,

dh P � DV 1a�dt W V dV

1 �� g dh

is again appropriate as the basic energy equation for the aircraft. As-suming that Pa � DV (excess of power), dh /dt is positive and is takenas rate of climb (usually given in feet per minute, fpm). If Pa � DV,then dh /dt is negative, and the problem is an extension of the glidingflight with partial power on. Normally it is assumed that climb takesplace at essentially constant velocity, which justifies then omitting thekinetic energy correction factor 1 � (V /g) (dV /dh) in the denominator.This works well for most propeller-driven aircraft or aircraft with mod-est thrust-to-weight ratio. For high-performance aircraft (T/W � O(1)),this assumption needs to be reevaluated. For example, at high speedsand shallow climb angles, dV /dh may be small but the entire term maystill be significant.

4.2 RATE OF CLIMB, CLIMB ANGLE

For the time being, it will be assumed that the climb will take place atconstant velocity, or at such a rate that the term dV /dh is small. Thus,one obtains immediately the following equations.

The rate of climb is given by the following equations (see also Figure4.1):

72 CLIMBING FLIGHT

R/C

VV VV max

max

maxmin

γ

γ

Figure 4.1 Aircraft Climb Performance

TABLE 4.1 Climbing Conversion Factors

dh /dt P V

P � DVa 33000� �WHP ft /sec

T � D60V� �W

HP ft /sec

T � D88V� �W

HP mph

T � D101.4V� �W

HP knots

dh P � DVa� v � (4.1)�dt W prop

dh (T � D)V� v � (4.2)�dt W jet

and the climb angle is established by means of Eq. 2.9, which gives

v P � DVa� sin � � (4.3)�V VW prop

v T � D� sin � � (4.4)�V W jet

Since the rate of climb is usually given in units of ft/min (fpm), Table4.1 will give the appropriate conversion factors: dh /dt in fpm, T, D, Win lb.

4.2 RATE OF CLIMB, CLIMB ANGLE 73

For any instantaneous values of power, drag, and velocity, Eqs. 4.1to 4.4 give the aircraft rate of climb, v, and the local inclination of theflight path to the horizon, sin � (see Figure 2.1). For practical problems,similarly to calculating the maximum velocities in Chapter 3, climbcalculations can also be divided into exact and approximate approaches.The exact method depends on having information available in the formof Figures 3.3 or 3.7 and calculations proceed straightforward accord-ing to Eqs. 4.1 to 4.4. This will be the focus of the current section.Approximate methods will be taken up later.

Now further subdividing:

A. Jets For the climb angle, T � D can be calculated or readdirectly for a range of values of V from Figures 3.2 or 3.4 andthen divided by the weight W. The maximum value and the cor-responding flight velocity V are found from the resulting tableof values or from a graph of sin � vs V (see Example 4.1). Ifthrust is independent of velocity, then the flight velocity for thesteepest climb angle occurs at and Em. The rate of climb isVDmin

obtained from the same table by multiplying the values of (T �D) by corresponding flight velocity V.

B. Propeller aircraft For rate of climb, (Pa � Pr) can be read di-rectly for a range of values of V from Figure 3.7 and then dividedby weight W. The maximum value and corresponding flight ve-locity V are found again from the resulting table or a graph ofdh /dt vs V. The climb angle is obtained from the same table bydividing the rate of climb values by corresponding flight velocityV. See also Example 4.1 and Table 4.2 for more details.

This approach to either type of aircraft is straightforward and simplebut requires that Figures 3.3 or 3.7 or equivalent data (T � D) in tabularform be already available.

Figure 4.1 shows, for constant altitude, essential features of the climbperformance. In addition, the maximum rate of climb can be plottedas a function of altitude, as shown in Figure 4.2, providing anotherway to determine aircraft ceiling.

The absolute ceiling is defined by the altitude where the rate of climbapproaches zero and it occurs in the stratosphere (above 36,000 ft) forjet-powered aircraft, and usually in the tropopause (below 36,000 ft)for propeller-driven aircraft. From an operational point of view, anotherceiling is defined: the service ceiling. This occurs when the maximumrate of climb is 100 fpm. Although the ceiling of the propeller-drivenaircraft tends to occur below tropopause, its flight path usually exhibits

74 CLIMBING FLIGHT

R/C

h

Absoluteceiling

Serviceceiling

Prop

Jet

Tropopause

100 fpm max

Figure 4.2 Maximum Climb Rate and Ceiling

a steeper climb angle up to almost the service ceiling. Some notableexceptions are found for very-high-performance aircraft with a thrustweight ratio of near unity, or higher. In that case, extremely high climbangles approaching 90 degrees can be obtained.

NOTE

The calculations described on immediately preceeding pages are simple andstraightforward but are entirely dependent on the type and quality of dragand thrust data. So are the results. Chapter 3 with all the appendix infor-mation is a reminder that much work is required to prepare accurate drag(i.e., ) and thrust data for predicting reliable performance information.CD0

Chapter 3 and the appendices will provide sufficient aerodynamics andpropulsion information only for basic simple cases. Rather, the basics arereviewed in Chapter 3, which should be of help in deciding the necessarycourse of action and provide a warning that all material beyond Chapter 3is dependent on Chapter 3.

4.3 TIME TO CLIMB

An important aspect of the aircraft climb problem is the amount oftime needed to climb from one given altitude h1 to another altitude h2.The aircraft rate of climb, or the vertical velocity is defined by

4.3 TIME TO CLIMB 75

1R/C

hh h1 2

Area = time to climb

Figure 4.3 Calculation of Time to Climb

dhv � � R/C

dt

and the time to climb can be determined from

h2 dht � t � � (4.5)2 1

h v1

The burden of the problem lies in finding a suitable expression forthe rate of climb and integrating this between any two desired altitudes.The last section and the developments in Section 3.3.4 have indicatedthat the rate of climb for a propeller-driven aircraft must be obtainedby numerical means, implying also that the time to climb must beevaluated by numerical integration techniques. As will be seen, for jetaircraft, some analytical expressions are available but the resulting in-tegral expression in Eq. 4.5 turns out to be unduly cumbersome and isbest evaluated by numerical techniques.

There are two commonly used methods for calculating the time toclimb to altitude. The more precise method for any type of aircraftconsists of evaluating the rate of climb v for several altitudes. The areaunder the 1/v curve then gives the required time for climbing betweenany two altitudes. See Figure 4.3 and Example 4.1. However, as de-veloped by approximate methods in Chapter 3, some simplificationsare possible for rate of climb of the jet aircraft (see also the nextsection).

76 CLIMBING FLIGHT

The other method is an approximate one since it assumes the rateof climb versus altitude curve is (at least piecewise) linear:

v � v2 1v � v � (h � h ) (4.6)1 1h � h2 1

When Eq. 4.6 is substituted into Eq. 4.5, the following elementaryintegral is obtained, which yields

h2 (h � h ) dh h � h v2 1 2 1 2t � � � ln (4.7)h v h � v h � h (v � v ) v � v v1 1 2 2 1 2 1 2 1 1

where it has been assumed, without any loss of generality, that t1 � 0.Eq. 4.7 can be further simplified if the following expansion is used forthe logarithm:

3x � 1 1 x � 1ln x � 2 � � . . . , x � 0 (4.8)� � � �x � 2 3 x � 2

which gives for the time

2(h � h )2 1t � (4.9)v � v1 2

Either of Eqs. 4.7 or 4.9 can be applied in a stepwise manner tocalculate the rate of climb over linear portions of the v � h curve ifthe altitude range is subdivided into suitable segments over which v islinear with altitude. The sum of the times over the individual segmentsis then the total time to climb.

Since the aircraft can consume considerable quantities of fuel duringthe climb to high altitudes, the effect of the weight change can beaccounted for in an approximate manner in the stepwise calculation ifthe fuel consumption rate is multiplied by the calculated climb timeand the total weight is then reduced accordingly for the following al-titude increment. Moreover, such a stepwise calculation also permitsallowance for thrust reduction due to the altitude (e.g., Eq. 3.33), whichfurther improves the approximations made above and also yields anestimate for the fuel used in the climb.

The basic methodology for climb problems just discussed is de-scribed in Example 4.1. Since the calculations are extensive, the details


are shown only for the sea-level altitude. The results are shown inFigure 4.5 for a full spectrum of altitude and velocity. Although theexample has been carried out for the propeller-driven aircraft, themethod is fully valid for a jet aircraft.

EXAMPLE 4.1

For the propeller-driven aircraft discussed in Example 3.4, determinethe climb performance and ceiling for the weight W � 9,117 lb.

2C � .024 � .0535CD L

AR � 7

e � .85

The rate of climb is determined by use of Eq. 4.1:

P � Pa rv � R/C �W

and the climb angle from

R/Csin � �

V

To find the desired performance over the pertinent range of ve-locities and to establish the maximum values, the calculations arebest carried out in a tabular format as shown in Table 4.2. The Pa

and Pr values are obtained from Example 3.4 with the suitable rangeof velocities taken from Figure 3.7. Such a table of values needs tobe calculated for a number of altitudes in order to estimate the air-craft ceiling. Table 4.2 shows the values calculated at sea level con-ditions.

Figure 4.4 shows the rate of climb plotted against the aircraft flightvelocity. The maximum rate of climb can be read from the top ofthe graph to be 17.5 ft/sec (1050 fpm) or estimated from the tableto be about 1040 fpm at 220 ft/sec flight speed. A tangent drawnfrom the origin to the rate of climb curve gives for the maximumpath angle 5.3� at about 170 ft/sec.

78 CLIMBING FLIGHT

TABLE 4.2 Rate of Climb Calculation Procedure

Vmph

Vft / sec

Pa

HPPr

HPExcess PPa � Pr

R /Cfpm

�deg

50 73 217 315 �9875 110 330 227 103 372 3.2

100 147 413 204 208 753 4.9125 183 495 220 275 995 5.2150 220 555 269 286 1035 4.5175 256 593 352 240 869 3.2200 293 626 471 155 561 1.8225 330 648 629 19 69 0.2350 367 661 831 �170

Although Figure 4.4 provides a quick overview of the climb per-formance at one altitude, it is limited and impractical because it isdifficult to read the values with sufficient precision from the figure.In any case, the information is already contained in Table 4.2 inadequate detail.

A much more useful representation is found in Figure 4.5, whichshows the excess power as a function of velocity for several selectedaltitudes. Since the flight path angle is usually of no great interest,and since only the maximum rate of climb is of general value, theexcess power plot saves extra calculations. Moreover, where the ex-cess power vanishes, one finds either Vmax or Vmin for that particularaltitude.

The maximum rate of climb values for several altitudes are shownplotted in Figure 4.5. The altitude where the maximum rate of climbapproaches zero is the aircraft absolute ceiling, in this case, at about19,500 ft altitude. The curve of maximum ratio of climb is usuallydetermined for several altitudes below the ceiling and then is ex-trapolated to R /C � 0. The extrapolation is assisted by plotting theVmax and Vmin curves and also by determining the locus of R /Cmax

from the excess power curves. Thus, Figure 4.5, which is based onthe same calculation as Figure 4.4 but conveys more information ina more precise manner, is to be preferred for practical calculations.

NOTE

Multiplying the excess power curves in Figure 4.5 by V /W yields curvesfor the specific excess power Ps. It is easily seen that, at a given altitude,the maximum rate of climb is determined by either the maximum valueof the excess power curve, which is


maxγ

0 100 200 300 V (ft/sec)

10

20

R/C(ft/sec)

0

Figure 4.4 Rate of Climb

�[P � P ] � 0a r h�c�V

or

� (P � P )Va r � 0� ��V W h�c

This fully anticipates the minimum-time-to-climb process discussed inChapter 8 under ‘‘Energy Methods’’ (see Eqs. 8.7 and 8.8).

F106

As the last item in this example, the time to climb from sea levelto 15,000 ft will be determined. Assume that the weight remainsconstant at 9,117 lb.

The time to climb can be calculated, with best accuracy, from Eq.4.5, by first plotting 1/vc � 1/(R /C) from Figure 4.5 or Table 4.2the original data, as shown in Figure 4.3. Numerical evaluation ofthe area under the 1/vc curve gives

80 CLIMBING FLIGHT

0

0

250

50

500

100

750

150

1000

0 50 100 150 200 250

10,000

20,000

0

0

100

200

300

200

R/C (fpm)

V (mph)

h (ft)

Excesspower (hp)

V (mph)

S.L.

10,000 ft

15,000 ft

R/C Vmax

Vmin

Absoluteceiling

Serviceceiling

20,000 ft

Figure 4.5 Climb Performance

15 dht � � � 1,802 sec15

0 vc

For comparison, the time is also evaluated from Eqs. 4.7 and 4.9,respectively:

h � h v 15,000 17.52 1 2t � ln � � ln � 1,689 secv � v v 17.5 � 3.7 3.72 1 1

4.4 OTHER METHODS 81

and

2(h � h ) 2 � 15,0002 1t � � � 1,415 secv � v 3.7 � 17.51 2

It is seen that Eq. 4.7 gives a fair approximation, but Eq. 4.9 wouldmiss the answer by about 27 percent. It should be pointed out thatassuming the aircraft weight to be constant introduces an error inthe answers that can be removed by evaluating, in stepwise manner,the fuel flow during the climb and recalculating the values in Table4.2. It should be noticed also that there is an implicit assumptionmade in this procedure that 1/vc is evaluated from a sequence ofR/C �max values. This, in turn, leads to a climb with velocity increas-ing toward ceiling (see Figure 4.5). If climb time at constant velocityis desired, the consistent values of R /C must be chosen from Figure4.5. All these will not be R /C �max.

4.4 OTHER METHODS

In case there is a need for a fast and closed form of solution, theapproximate methods described in this section are useful. It should berecognized that the approximate methods work well for jet aircraft butare not always suitable for propeller-driven aircraft, as explained inChapter 3.

4.4.1 Shallow Flight Paths

For a jet aircraft, the climb angle can be developed from Eq. 4.4 andthe nondimensional form of the drag, Eq. 3.36, as follows (with n �1):

T D T 1 12sin � � � � � V � (4.10)� �2W W W 2E Vm

where again

V VV � �

1 /4VDmin 2W k� � �S CD0

82 CLIMBING FLIGHT

Multiplying the above by V, one obtains the rate of climb, or verticalvelocity

dhv � � V sin � � R/C (4.11)

dt

which becomes, in nondimensional form, after dividing by VDmin

v TV 1 13v � � � V � (4.12)� �V W 2E VD mmin

The maximum values may be obtained by means of simple differen-tiation, but it must be kept in mind that the expression for the thrustmay be a function of velocity. As an example for this procedure usingthe thrust expression given by Eq. 3.33, which is independent of ve-locity, one gets

mT � 1 1o 2sin � � � V � (4.13)� �2W 2E Vm

and

mT � V 1 1o 2v � � V � (4.14)� �2W 2E Vm

The maximum values can be obtained in a straightforward manner bydifferentiating these two equations with respect to V

d sin � 1 1� � V � � 0 (4.15)� �3dV E Vm

or � 1, which states that the steepest climb occurs atV

V � V (4.16)Dmin

or that

mT � 1osin � � � (4.17)max W Em


Sukhoi27

EXAMPLE 4.2

For a high-performance aircraft with a thrust-to-weight ratio of 0.9and a L /Dmax of 10, calculate the maximum climb angle. A straight-forward substitution into Eq. 4.17 with

mT �o � 0.9W

and

E � 10m

gives

1sin � � 0.9 � � 0.8max 10

Therefore

� � 53.1�max

which is a rather steep climb angle and exceeds the concept of ashallow path angle.

The condition for fastest climb is obtained from Eq. 4.12:

84 CLIMBING FLIGHT

dv T 1 12� � 3V � � 0 (4.18)� �2dV W 2E Vm

which gives, after some rearrangement,

4 23V � 2TV � 1 � 0 (4.19)

with

TEmT � .W

This biquadratic equation can be easily solved (see Eqs. 3.45 and3.46) to give one physically possible solution

2T � T � 3V � (4.20)vmax 3

Substituting back into Eq. 4.12 gives, by use of Eq. 4.20, for the max-imum rate of climb

1 13v � TV � V �� max 2E Vm

0.3849 2 2� T � T � 3 2T � T � 3 (4.21)� �Em

The result given by Eq. 4.16 should have been anticipated, as theclimb angle for jet aircraft is determined by the excess power andoccurs, for the constant thrust case, at (see also Example 3.3). IfVDmin

the thrust varies with velocity, the steepest climb location decreases orincreases beyond depending on whether the T � V curve slope isVDmin

negative or positive, respectively. This is also seen if Eq. 3.34 is usedfor the thrust in Eqs. 4.10 and 4.12. Only the results are given below;the details are left as an exercise. Upon substituting Eq. 3.34 into 4.10,the climb angle is obtained as


2 m(A � BV )� 1 12sin � � � V �� 2W 2E Vm

m 2 m1 2A� E 2V B� 1m 2� � � V �� 22E W 2W Vkm �SCD0�S CD0

1 12� 2A � V (B � 1) � (4.22)� �22E Vm

where

mA� EmA �W

m2B�B �

�SCD0

Similarly, the rate of climb is obtained from Eq. 4.12:

1 13v � 2AV � V (B � 1) � (4.23)� �2E Vm

The maximum values of the climb angle and climb velocity are

1 12 3 /2sin � � 2A � (1 � B) � (4.24)� �max 2Em 1 � B1 / 4V � (1 � B) (4.25)

The maximum rate of climb is found to be

1 1 2V 13v � 2AV � V (1 � B) � � A �� max 22E V 3E Vm m

22A � A � 3(1 � B)2� 2A � A � 3(1 � B)� �

33(1 � B)Em

(4.26)

Eqs. 4.22 and 4.26 reduce for the case of B � 0 (thrust is independentof the velocity) to Eqs. 4.17 and 4.21, respectively.

86 CLIMBING FLIGHT

EXAMPLE 4.3

An aircraft has the following characteristics:

W � 36,000 lb2S � 450 ft

T � 6,000 lb0

2C � 0.014 � 0.05 CD L

If its thrust is assumed to be constant and it climbs at a speed of300 ft/sec at a shallow angle, calculate:

a. Angle of climbb. Rate of climbc. Maximum angle of climbd. Maximum rate of climb

a. The problem is most easily solved if the climb angle is calculatedfirst from Eq. 4.10 by establishing

T 6,000 1� �

W 36,000 6

1 / 4 1 / 42W k 2 � 36,000 0.05V � �� Dmin �S C 0.002377 � 450 0.014D0

ft� 356.6

sec

V 300v � � � 0.841

V 356.6Dmin

1 1E � � � 18.89m 2kC 20.014 � 0.05D0

Therefore

1 1 12sin � � � 0.841 � � 0.1105� �26 2 � 18.89 0.841


whence

� � 6.34�

b. The rate of the climb can be calculated from Eq. 4.12, or, froma simpler expression,

ftv � V sin � � 300 � 0.1105 � 33.15 � 1989 fpm

sec

It should be noted that the climb does not take place at (L /D)max

(it is only a convenient means for establishing Eq. 4.10), but at

2W 72,000C � � � 0.748L 2 2� V S 0.002377 � 300 � 450o

which gives an L /D of

C 0.748L � � 17.82C 0.014 � 0.05 � 0.748D

The maximum angle of climb is obtained from Eqs. 4.16 and 4.17as follows:

T 1 1�1 �1� � sin � � sin 0.1667 � � 6.53�� max W E 18.89m

and occurs at V � � 356.6 ft/sec and at (L /D)max.VDmin

Maximum rate of climb is obtained from Eq. 4.21 with

T 6,000T � E � � 18.89 � 3.15mW 36,000

0.3849 2v � 3.15 � 3.15 � 3max 18.892� (2 � 3.15 � 3.15 � 3) � 0.1432

where

ftv � 0.1432V � 0.1432 � 356.6 � 51.05 � 3,063 fpmmax Dmin sec

88 CLIMBING FLIGHT

It is obtained at the following flight speed (Eq. 4.20):

2 2T � T � 3 3.15 � 3.15 � 3V � � � 1.4994vmax 3 3

or

ftV � 1.4994V � 1.4994 � 356.6 � 534.7Dmin sec

and at

72,000C � � 0.235L 20.002377 � 534.7 � 450

and corresponding

L� 14.03

D

This concludes the discussion of shallow flight paths where n � 1,or L � W. From a practical engineering point of view, many of theclimb problems can be analyzed by use of the techniques presented inthis section. In cases where the thrust-weight approaches unity, or theflight path angle is very steep, or both, the above approach fails to takeinto account the variation of the velocity with the changing altitude.Even this omission will not cause great errors in the flight path angle,but the errors become significant in the calculation of the optimumvelocity.

In the next section, a technique will be discussed for evaluatingperformance at steep angles where the load factor n � 1.

4.4.2 Load Factor n � 1*

For small climb angles it has been assumed that n � L /W � 1, or inother words, cos � � 1. At steeper climb angles, this approximationceases to be accurate. An improved solution is obtained if one considersEq. 2.12 with � � 0 and � 0. This gives�


L � W cos � (4.27)

n � cos � � 0 (4.28)

Eq. 2.11 supplies the equation tangent to the flight path, with � 0,Vwhich can be immediately restated in nondimensional form, similar toEq. 4.10, as follows

2T 1 n2sin � � � V � (4.29)� �2W 2E Vm

Eliminating n2, and using cos2 � � sin2 � � 1, one obtains

T4 2 2V � 2V E � sin � � 1 � sin � � 0 (4.30)� �m W

This equation can be solved for either or sin �. In the first case, theVsolution depends on the type of the thrust expression available, i.e. isit a function of velocity (Eq. 3.34). The case for sin � can be solvedwithout recourse to an explicit thrust statement.

To illustrate the method, it will be assumed that the thrust is inde-pendent of the velocity and Eq. 3.33 holds. Use of Eq. 3.34 will beleft as an exercise. Solving for the velocity, one gets

2T T2 2V � E � sin � � E � sin � � (1 � sin �) (4.31)� � � �m mW W

Eq. 4.31 gives, for a fixed thrust-weight ratio and flight path angle, therequired flight path velocity. Both signs give physically possible results.The positive sign gives the higher-speed solution (see Section 3.3.3).To obtain � and v, Eq. 4.30 can be rewritten as follows:

T2 2 2 4sin � � 2sin �E V � 2E V � 1 � V � 0 (4.32)m m W

Solving for sin � gives

2 2 4 2 4sin � � V E � E V � (2TV � 1 � V ) (4.33)m m

where

90 CLIMBING FLIGHT

TT � E (4.34)mW

(for any expression of thrust), from which the rate of climb is obtainedas

v 3 4 2 4v � � V sin � � V E � VE V � (2TV � 1 � V )m mVDmin

(4.35)

Again, both signs yield possible solutions. The negative sign is usedfor � 1. Once the flight velocity is found from Eq. 4.31, the rate ofTclimb can then be obtained from Eq. 4.35. Of course, this process canalso be reversed (i.e., the path angle can be determined) for a given

and T /W, from Eq. 4.33, whereupon the rate of climb is found fromVEq. 4.35.

It should be emphasized again that Eqs. 4.33 and 4.35 representgeneral results for any value or expression of thrust. Eq. 4.31, however,is the result of using the simple thrust formula Eq. 3.33, which isindependent of velocity.

EXAMPLE 4.4

The thrust of the aircraft in Example 4.3 is increased to 16,000 lband it is climbing at a speed corresponding to Em. Calculate the pathangle and the rate of climb.

In this case

V � 1

T L 16,000T � � � 18.89 � 8.396�W D 36,000max

Eq. 4.33 is used:

2 2 4 2 4sin � � V E � E V � (2TV � 1 � V )m m

2� 18.89 � 18.89 � (2 � 8.396 � 3) � 0.3956


Therefore

� � 23.3�

The rate of climb is

ftv � V sin � � V sin � � 356.6 � 0.3956 � 141Dmin sec

� 8466 fpm

NOTE

The climb angle solution, and consequently the rate of climb, can be simpli-fied and expressed in terms of level flight conditions for any angle and forunrestricted thrust expressions. To this end, the ratio of climb thrust tolevel flight thrust is formed by dividing Eq. 4.29 by Eq. 3.41, as follows:

21 n2sin � � V �� 22(L /D) VmaxT� (4.36)

T 1 1L 2V �� L 22(L /D) Vmax L

where the subscript L refers to the level flight values at which the climbstarts. It is also assumed that both climb and level flight occur at the sameL /D. Recognizing that the denominator can also be expressed as 1/(L /D)(see Eq. 3.41), Eq. 4.36 becomes

2n2V � 2T L V� sin � � (4.37)

T D 1L 2V �L 2VL

And since

n � cos �

2 2V � V cos �L

L � L cos �L

92 CLIMBING FLIGHT

Eq. 4.37 becomes

T L� sin � � cos � (4.38)

T DL

Eq. 4.38 determines, for a given L /D and thrust, the climb angle �. Con-versely, for a given L /D and desired climb angle, the increase of thrustover the level flight value, T /TL, can be established.

EXAMPLE 4.5

Solve Example 4.3 by using Eq. 4.38.

W � 36,000 lb

T � 16,000 lb2C � .014 � .05CD L

The aircraft in Example 4.3 is climbing at a speed corresponding toEm or at . This is assumed to be also the original level flightVDmin

speed.At level flight TL � D, and since at Em

C � 2C � 2 � 0.014 � 0.028D D0

and

2C � C � kCD D L0 i

then

0.014C � � 0.529L 0.05

The drag can be found from

C 0.028DD � W � 36,000 � � 1905 lb � TLC 0.529L

Then


T 16,000� � 8.4

T 1,905L

and with

L� E � 18.89mD

one finds from Eq. 4.38 that

� � 23�

It is of interest to obtain two limiting cases from Eq. 4.10. For � �0 (level flight), Eq. 4.10 or 4.31 reduces to the basic situation of Eq.3.48. For � � 90� (vertical flight), one obtains from Eq. 4.31, with thenegative sign, � 0. The positive sign givesV

L T2V � 2 � 1 (4.39)� � �D Wmax

Introducing the definition of (see Eq. 4.10), and remembering thatVDmin

Em � one finds after some rearrangement that1/2kC ,D0

2C �SVD0T � W � (4.40)2

or

T � W � D (4.41)L�0

which gives the condition for powered vertical climb if the thrust isindependent of velocity. Eq. 4.41 states that, in vertical flight, the thrustmust be at least equal to the aircraft weight, plus the zero lift drag.The implication here is that lift is zero for vertical flight and it is thethrust that contributes a balancing force to counter the weight W.

As a matter of completeness, it is possible to calculate from Eqs.4.33 and 4.35 the maximum climb angle and the maximum rate ofclimb. However, considering the fact that the simplified thrust expres-sion (Eq. 3.33) needs to be used for practical results and that the ac-

94 CLIMBING FLIGHT

celeration effect has been neglected ( term in Eq. 2.11), or the termVin the denominator of Eq. 3.4, the slight improvement of results overthose for the shallow flight path angle would hardly justify the extralabor involved.

4.4.3 Partial Power and Excess Power Considerations

In Section 4.1 it was shown that climb is the direct result of excesspower available above and beyond the amount required to overcomeaircraft drag. Both rate of climb and climb angle are determined byapplied excess power. Since the climb starts usually from some steady-state level flight condition, it is of interest to reconsider what happensif steady-state power is increased, or how much power increase is re-quired to achieve certain desired climb conditions. Moreover, what hap-pens if the power is reduced?

In steady-state flight, each specific angle of attack gives a specificCL with the corresponding EAS for a particular aircraft weight. Anincrease in angle of attack will produce a reduction in velocity, and adecrease in angle of attack will lead to an increase in velocity. Duringthe change in angle of attack and while a new steady-state flight con-dition is established the aircraft may climb or descend if there is nochange in power setting but the fundamental cause-effect relationshipis between the angle of attack and the airspeed. Thus, one can concludethat the angle of attack is the primary control of the aircraft velocityin steady flight.

If the aircraft power is increased during steady-state flight conditions,the resulting excess power would cause a positive climb rate dh /dt, asalready established in Section 4.2. An increase of power while main-taining the same angle of attack will result in aircraft climbing to ahigher altitude. A decrease in power results in descent. Therefore,power setting is the primary control of altitude during steady-flight con-ditions.

The previous discussion has tacitly implied that typically ‘‘normal’’relationships obtain between power applications and resulting flightspeeds (i.e., a higher power setting gives a higher steady state flightspeed, and vice versa). This is true if the flight speed is sufficientlyhigh (higher than , or the speed for maximum endurance), but atVPmin

speeds lower than an increase in power is required as the flightVPmin

speed decreases. This is seen in Figures 3.2 and 3.7. Since the increasein required power with a decrease in velocity is contrary to normal


12

Powerdeficiency

P

aP

r

P

VPmin V

Region ofreverse command

Excess power

3

Figure 4.6 Power versus Velocity

command of flight, the region between and the stall speed VsVPmin

(minimum control speed) is called the region of reversed command,back side of power curve, or the slow flight region.

Consider an aircraft flying at point 1 (Figure 4.6) in the normalcommand (high speed) region. If the angle of attack is decreased with-out changing the power setting, a speed increased results with a defi-ciency in power and the aircraft will descend. If the angle of attack isincreased, without change in power, the speed will decrease to point 2and a climb will result (between points 1 and 2) due to an excess ofpower.

Flight in the reversed command means that a lower power setting isrequired at higher speeds, and vice versa, to maintain altitude. Thus, ifaircraft angle of attack is increased at point 2 to produce an airspeedcorresponding to point 3, power setting must be increased or descentwould follow. Similarly, if the angle of attack is continuously decreasedfrom point 3 to, say, the increasing flight velocity permits a re-V ,Pmin

duction in power if altitude is to be maintained. Otherwise, the aircraftwould climb. Since the back side of the power curve is characterizedby high lift coefficients, the reversed command operations occur duringlanding and take-off phases of flight. Most of the airplane flight takesplace in the region of normal command.

96 CLIMBING FLIGHT

PROBLEMS

4.1 Calculate the total power required (ft-lb)/sec for the rate of de-scent of 1,200 fpm and for the rate of climb of 1,200 fpm, foran aircraft in flight under the following conditions:

V � 300 ft/sectrue

C � 0.05D2S � 400 ft

h � 10,000 ftW � 5400 lb/hrf

W � 18,000 lb

4.2 An aircraft weighing 20,000 lb begins a climb from sea level to30,000 ft at an equivalent airspeed of 250 knots (423 ft/sec).This EAS will be maintained constant on the pilot’s airspeedindicator throughout the climb. Assume a constant power avail-able of 3,030 HP, and a constant average drag of 2,000 lb force.Calculate:a. The TAS at the end of the climb (30,000 ft) and a mean TAS

for the climb (TAS30 � TASS.L.) /2b. The power consumed in overcoming drag at the mean TAS,

and the excess power available for the climb.c. The resulting rate of climb (fpm), assuming constant TAS at

the mean value (1,650 fpm).d. The correction factor arising from acceleration during the

climb, and the corrected mean rate of climb (fpm)

4.3 A P-3 has the following characteristics:

2C � 0.0225 � 0.0448 C (clean)D L

W � 110,000 lb2S � 1300 ft

h � 500 ft� � 0.82p

It is cruising at maximum endurance speed (i.e., V )Pmin

It is equipped with four Allison T56-A-14 turboprop engines.

What is its potential rate of climb if full power is applied?Ans: 3910 fpm.

PROBLEMS 97

4.4 A jet aircraft is in steady, level flight at 20,000 ft altitude. Itreduces thrust and descends to 15,000 ft altitude in 3.25 minutes.For the given aircraft data: W � 24,000 lb, W /S � 60, V � 500mph (assume to be constant), CD � .025 � determine:2.06C ,L

a. The resulting path angleb. The rate of sinkc. Horizontal distance coveredd. Maximum power-off glide range over the same altitude

changee. Flight speed corresponding to d.

4.5 For the turbojet aircraft with the characteristics given below,determine how much thrust must be developed at 300 knots tohold the rate of sink to 1,000 fpm at 10,000 ft. Consider thrustline inclination effects if applicable. The incidence angle of thewing is 2.5 with respect to the axis of the fuselage. The enginethrust line is along the axis. The �L�0 of the wing is along theaxis of the fuselage.

2C � 0.015 � 0.06 CD L

W � 30,000 lb2S � 750 ft

a � 0.071 1/deg

4.6 A homemade glider has the following parameters: W � 3,200lb, CD � .014 � S � 200ft2. It is easily established that2.05C .L

the maximum glide range from 10,000 ft is 36 miles. How muchthrust is needed from a retrofit jet engine to double this mileage?Ans: about 250 lb.

4.7 A high-performance aircraft cruises at 10,000 ft under minimumthrust conditions. The aircraft has the following characteristics:

2W � 17,000 lb, S � 380 ft.8T � 14,000 �

2C � .022 � .0505CD L

Full available thrust is applied and the plane starts to climb atmaximum climb angle. Determine:a. The resulting climb angle at 10,000 ft altitudeb. The time to climb to 20,000 ft

Ans: 35.4�, about 66 sec.

98 CLIMBING FLIGHT

4.8 Show that, for the same CL, the power in climbing flight is givenin terms of level flight power by

P L3 /2� cos � 1 � tan �� P DL

Hint: First show that V /VL � cos1 / 2�.

4.9 Since maximum rate of climb is almost linear with altitude showthat absolute ceiling hc can be calculated from

h v1 oh �c v � vo 1

where vo and v1 are maximum rates of climb at sea level andaltitude h1, respectively.

4.10 Using the results in Problem 4.9, calculate the absolute ceilingfrom the data tabulated in Example 4.5.Ans: 19,020 ft.

4.11 If the maximum rate of climb varies linearly with the altitudeshow that the time to climb to altitude h is given by

h hc ct � lnv h � ho c

4.12 Calculate the time to climb in Example 4.6 from the expressionobtained in Problem 4.11.Ans: 1,689 sec.

4.13 A propeller aircraft is climbing at the rate of 3,500 fpm withthe following data:

V � 190 mph h � sea levelW � 6,000 lb b � 44 ft

2S � 240 ft ƒ � 5.76e � 0.91 � � 0.85p

Determine the angle of climb and the Pr for the climb.

PROBLEMS 99

4.14 A test aircraft at 20,000 ft is flying straight and level at itsminimum drag speed of 280 KTAS. Pilot then adds full powerand commences a level acceleration run. Five seconds into therun, the airspeed has increased to 295 KTAS. Given the follow-ing aircraft parameters: W � 40,000 lb, S � 400 ft2, � .025,CD0

Ta is independent of V, determine:a. Thrust available, Ta (Hint: For FPE first calculate average

acceleration.)b. Vmax

c. Potential rate of climb at VDmin

d. Maximum climb angleAns: Ta � 9,046 lb, Vmax � 1,195 ft/sec, R /C � 4,380 fpm,8.9�

4.15 A 24,000 lb jet aircraft in steady, level flight at a speed of 500mph cuts its thrust back by 600 lb. W /S � 60, CD � .02 �

Determine:2.0535C .L

a. Resulting flight path angleb. Its rate of sinkc. Horizontal distance covered when it reaches 15,000 ft

Ans: �1.43�, 18.33 ft/sec, 37.8 mi

4.16 A propeller aircraft establishes a steady-state climb at a constantequivalent airspeed of 140 knots. Passing through 5,000 ft thevertical speed indicator reads 600 fpm. Given: SHPrated � 300HP, �p � .85, W /S � 10 lb/ft2, W � 3,000 lb, k � .0497 SHPavail

� (1.132� � .132)SHP.Determine: Preq, f.

4.17 A jet with S � 200ft2, CD � .014 � is in a steady-state2.05CL

climb at 10,000 ft altitude. Determine:a. The percentage increase in level flight thrust TL required to

achieve a climb angle of 30� if V � .VDmin

b. If Tmax � 3Tmin, what is the climb angle, at the same airspeed?c. If V � 300 ft/sec, T /TL � 7, and the climb angle is 45�, how

much does the aircraft weigh (this leads to a quadratic equa-tion to solve)?Ans: 930%, about 6�, 6,200 lb.

100 CLIMBING FLIGHT

Concorde

101

5Range and Endurance

747

5.1 INTRODUCTION

Range is the distance covered on the ground while the aircraft is flyingfrom one point in the space to another point. Endurance, at its basiclevel, may be defined as the time spent to cover that distance. However,an aircraft may spend four hours flying (theoretically) at 150 mph intoa 150-mph headwind and not cover any distance at all. Thus, enduranceneeds to be defined in terms of flight time available, for a given amountof fuel, at a given flight condition.

Range calculation is subject to many conditions—in other words, itdepends on the mission or flight profile. Different aircraft (or even thesame aircraft) can cover the same distance in various ways resulting indifferent fuel consumption and endurance figures. The simplest missionwould consist of take-off, cruise, and landing segments.


102 RANGE AND ENDURANCE

This chapter is concerned with the basic methods for calculating theaircraft range and endurance that would be applicable to any of theindividual mission segments. The cruise range will be studied first insome detail, as the cruise conditions usually represent the aircraft de-sign requirements and since most of the flying time is spent in thecruise configuration. The approach taken here is to use the same basicenergy balance relationship, Eq. 2.20, as used for the majority of theproblems treated in previous chapters. Thus, the change in energy levelarising during the flight and its influence on the range can be correctlyidentified, and the influence of fuel and powerplant characteristics onthe range can be studied separately. Eq. 2.20 can be rewritten by usingEq. 2.14:

� Ho fW de � P dt � DV dt � dW � DV dt (5.1)a fg

Making use of the kinematic relationship Eq. 2.10, for level flight dx� dR � V dt, one obtains from Eq. 5.1

� H dW W deo f fdR � � (5.2)gD D

Since also dWf � �dW, L � W, and D � W /(L /D) one gets

�� H (L /D)dW Lo fdR � � de (5.3)gW D

Eq. 5.3 is the general form for calculating the aircraft range. The basicrequirements are that lift, drag, and weight variations be given as afunction of two other flight parameters; velocity and altitude (see Sec-tion 2.1), or that some specific assumptions be introduced for approx-imate solutions. In particular, this leads to two different approaches:exact range integration—a numerical technique (Section 5.3), or anumber of approximate methods (Section 5.2). The essential differenceis that the range integration method uses a minimal number of as-sumptions but requires detailed thrust and drag data over applicablealtitude and velocity ranges. This results in extensive numerical com-putations but gives highly accurate and reliable results. In practicalapplications, however, a number of simplifying assumptions are usedto obtain usable but somewhat approximate equations and results. Sev-

5.2 APPROXIMATE, BUT MOST USED, METHODS 103

eral averaging or stepwise sequential calculations are then introducedto find adequate solutions.

Both approximate and integral methods are discussed in some detail.Initially, the term containing specific energy variation is neglected. Itseffect on range (normally very small) is discussed in a later section.As already discussed in Chapter 3, separate solutions are obtained forpropeller and jet aircraft.

5.2 APPROXIMATE, BUT MOST USED, METHODS

Before specializing Eq. 5.3 to particular flight paths or specific aircraft,it is possible to carry out a simple generalized integration if it is as-sumed that the lift-drag ratio and the overall power efficiency are con-stant. Then one obtains

H L W Lf 1R � � ln � (e � e ) (5.4)� �� o 1 2g D W D2

which represents the aircraft range between points 1 and 2, providedthat the energy assumptions discussed in Section 2.2 and L � W aresatisfied. It is easily recognized that the first term is due to change ofweight (burning of the fuel) and the second term arises due to changein the specific energy level (h � V2/2g). For constant altitude andvelocity cruise the second term vanishes. Its effect on range will beconsidered later in Section 5.4.2. Retaining, for the time being, onlythe first term in Eq. 5.4, one gets

H L Wf 1R � � ln (5.5)� �� o g D W2

which is a convenient form to observe the role of the fuel heat (energy)content in the range calculation process. The energy content Hf is aproperty of the fuel used, and �o describes the ability of a particularengine to convert this energy into useful work. The lift/drag ratio L /D represents aircraft efficiency in conversion of the work into range.The role of the weight ratio W1/W2 is discussed further in Section 5.3.For hydrocarbons such as aviation gasoline a typical value for Hf is18,800 BTU/lb (42,900 kJ/kg). If the heat capacity in BTU’s is con-verted entirely by using the mechanical equivalent of heat, 778 ft-lb/


BTU (i.e., �o � 1), it is seen that the remaining dimension is that oflength

BTU ft-lb18,800 � 778� � � �lb BTUH gf c� � � 2770 miles (5.6)o g g ft

5,280 � �mi

Hf can be expressed in suitable units for calculation of the range. Atfirst glance, the order of magnitude of the range can be estimated sim-ply from the heat content of the fuel (implied here per lb of fuel). Theinfluence of the other factors that determine the range is seen by as-suming some typical values for a propeller driven aircraft as follows:

E � 12m

� � 0.3o

A 1 lb weight change in a 10,000 lb aircraft (i.e., ln(W1/W2) � 0.0001)and assuming that fuel weighs 6 lb/gallon, then

R mi� 0.3 � 2,770 � 6 � 12 � 0.0001 � 6

gal gal

which is a typical value for reciprocating engines.Although the last equation is not particularly useful for range cal-

culation of specific aircraft it is suitable for comparative representationof aircraft performance. Individual aircraft range depends on particularpower plant characteristics that are given in terms of specific fuel con-sumption. Thus, the previous equations must be further specialized forreciprocating and jet aircraft operations. In this chapter, following his-torical development, reciprocating engine will be treated first.

KingAir 200


5.2.1 Reciprocating Engine

In keeping with the energy approach used in Chapter 2, Eq. 2.14 canbe cast in several different forms:

� H dW � � H dWo f f p t f fP � � � � BHP (5.7)a pg dt g dt

where

�t � engine thermal efficiency�p � propulsive (propeller) efficiency

BHP � Brake horsepower

Since the specific fuel consumption can be expressed as

(dW /dt) lbfC � (5.8)� �BHP HP � hr

one obtains from Eqs. 5.7 and 5.8

� H � BHP � BHP �o f p p p� � � (5.9)g (dW /dt) C BHP Cf

Eq. 5.9 forms the basis for calculating the reciprocating aircraft range.Owing partly to historical developments and partly to calculational dif-ficulties, two distinctive methods have emerged for calculating the re-ciprocating aircraft range. Breguet method was developed in thebeginning of the twentieth century and is best known for its simplicity.The second one, having no specific label, could be called constantvelocity method.

Breguet Equation The range, R, can be found directly upon substi-tuting Eq. 5.9 and the conversion factor

ft � lb sec550 � 3,600� � � �HP � sec hr lb � mi

� 375 � �HP � hrft5,280 � �mi


VV VRmax (L/D)max

Pr

θ

=

Figure 5.1 Maximum Range

into Eq. 5.5 yielding the well known Breguet range equation for recip-rocating engines. It should be remembered that the assumptions leadingto Eq. 5.5 are �p, C, and L /D are constant:

� L Wp 1R � 375 ln [miles] (5.10)C D W2

For a typical heat content of Hf � 18,500 BTU/lb for aviation gas-oline, the specific fuel consumption is

0.138 lbC � � 0.45 � 0.55 � �� HP � hrt

It is seen that, for constant � and C, the maximum range is obtainedat Em and at The velocity for maximum range can be obtainedV .Em

also from the power required velocity curve, Figures 3.4 or 5.1, bydrawing a tangent to the curve from the origin. This is seen easily ifone forms the tangent from Pr and V, whence

3P �V C Sr Dtan � � � (5.11)V 2V

By eliminating V2 � 2W /�SCL one gets


C 1Dtan � � W � (5.12)C C /CL L D

Maximum value of CL /CD, or Em, gives the minimum angle that is justtangent to the curve. With CL � constant and with the flight velocitygiven by

1 /42(W/S kV � (5.13)� �(L /D)max � � DD0

the flight velocity needs to be continuously changed as the aircraftweight decreases. If the velocity is held constant, then W/� must alsoremain constant, which implies that as W decreases, the density mustdecrease. This implies that the aircraft will continuously drift up duringthe weight decrease. This is called cruise-climb schedule.

There are two problems with such a scheme. First, the air trafficcontrol will have difficulties with the flights drifting all over the sky atvarying altitudes and speeds. Thus, the Federal Aviation Administration(FAA), through the Federal Aviation Regulations (FAR) 21, 23, and25, issues regulations for civil aviation concerning climb speeds, var-iation of cruise velocity, altitude variation, and so on. For example,even-numbered altitudes are assigned to traffic going in one directionand odd-numbered altitudes flying in opposite direction with altitudechanges restricted to discrete steps. Military aviation is regulated bysimilar regulations through specific MILSPECS. Second, such a cruise-climb schedule may not be the best flight schedule for propeller aircraft,as the best propeller efficiency and the best fuel consumption cannotalways be found at Em.

Although the propeller efficiency and the specific fuel consumptionC vary somewhat with power and rpm, they can be kept almost constantover the duration of flight. Since the Breguet equation tends to over-estimate the range (the longer the range, the larger the discrepancy)there are two different approaches to Breguet equation for more ac-curate range prediction:

1. Mean, or average, quantities are used to establish both V andL /D.

2. Range is calculated from Eq. 5.10 in several short segments. Thismethod is preferred in flight operations, as it yields a simulta-neous fuel consumption check with the range schedule.


Constant Velocity Method Since the propeller aircraft tend to beflown at constant altitude and at constant throttle setting, a better rangeresult can be found if one assumes a constant airspeed at constantaltitude. From Eq. 5.3 and 5.9 one obtains

� dWR � �375 � (5.14)

C D

Expressing the drag as

2kWD � qSC � (5.15)D0 qS

it follows that

� 1 �dW k2R � 375 � with a �� 2 2 2 2C qSC 1 � a W q S CD D0 0

� 1� 375 [arctan aW � arctan aW ] (5.16)� �1 2C qSC aD0

Eq. 5.16 can be simplified to read:

�E Wm 2R � 750 arctan 2C E k � arctan 2C E k (5.17)� �L m L m1 1C W1

Eq. 5.17 can also be expressed, as found in literature, as

�E E (W /W )m 1 2 1R � 750 arctan (5.18)� �C 2E (1 � kC E W /W )m L 1 2 11

where

LE � and C�1 L1D 1

are determined at the beginning of the flight. Although this approachis better suited for general propeller aircraft analysis, the reason for thelack of its wide usage has been the difficulty of evaluation of the arc-tangent format with the tools prior to 1980s.


EXAMPLE 5.1

Determine the range of the following propeller-driven aircraft, withthe following data, at a constant airspeed of 180 mph at 8,000 ftaltitude:

W � 18,500 lb1

W � 6,000 lbfuel

2S � 939 ft

� � 0.85p

C � 0.45 lb/HP-hr2C � 0.0192 � 0.047CD L

For use with Eq. 5.15 the following items need to be calculated

1E � � 16.64m 20.0192 � 0.047

2WC �L1 2�V S

2 � (18,500)� � 0.3032(0.002377) � (0.786) � (264) � (939)

then

2 � C � E � k � 2 � .303 � 16.64 � .047 � .4739L m1

and the range becomes

750 � 16.64 � .85R � [arctan .4739

.45

� arctan(.4739 � (12,500/18.500)] � 3,123 miles

For comparison, the range will also be evaluated by means of Bre-guet equation, Eq. 5.9. To this end, L /D will be evaluated at 180mph:

2W 2 � 15,500C � � � 0.254L 2 2�V S (0.002377) � (0.786) � (264) � 939


where an average weight of 15,500 lb has been used. Then,

L 0.254� � 11.422D 0.0192 � 0.047 � (0.254)

The range is obtained from Eq. 5.9 as

� L W 0.85 18,5001R � 375 ln � 375 � � 11.42 � lnC D W 0.45 12,5002

� 3,171 miles

Which shows unusually good agreement since the Breguet equation,in general, tends to overestimate the range by 10 to 15 percent. Itshould be noticed that an average weight and an L /Dmean were usedin the last calculation. If, as usually assumed, one uses Em in theBreguet equation, then the range becomes

16.64R � 3,171 � 4,620 miles

11.42

Thus, extreme care should be excercised in evaluating the commonparameters in use of the Breguet equation.

Lear Jet


5.2.2 Jet Aircraft

For a jet engine the available energy from the fuel can be expressed as

P TV Va� H � � � (5.19)o f (dW /dt) TC Cf

with the specific fuel consumption C now defined as

WfC � [lb /hr/lb ] (5.20)m fT

Substituting now into Eq. 5.3, one obtains the following range integral:

V dW L V L dW LR � �� de � �� de (5.21)

C D D C D W D

Similarly to propeller aircraft, simple analytical expressions can befound if a number of simplifying assumptions are made (i.e., aircraftcruise flight program will be prescribed). Three commonly used pro-grams (sets of assumptions) will be discussed below.

Breguet Cruise—Climbing Flight The most commonly used ex-pression for the jet engine range calculation is the Breguet analog forthe jet aircraft. It is assumed that V � const, CL � const, (L /D) �const (if the simple drag polar is used) and one obtains (neglectingagain the second term in Eq. 5.21)

V L W1R � ln [miles] (5.22)C D W2

where velocity is in mph. It is easily seen that, for maximum rangewith a constant C and a given weight ratio, V(L /D) must be maximum.Thus, one finds that

C2W C 2W L1 L 1V(L /D) � � (5.23)� ��SC C �S CL D D

must be a maximum, or CD / must be a minimum, and one obtainsCL

with the parabolic drag polar the following minimum condition


2C � kCd D L0 � 0 (5.24)� �dC CL L

After carrying out the differentiation and simplifying, one finds

2C � 3kC (5.25)D L0

and

C CD L0 DminC � �LRmax � 3k 3

It also follows that

1 /4V � 3 V � 1.316V � 1.316V (5.26)R D D Emax min min m

One can also show that for maximum range

3L� EmD 2

which permits writing the maximum range expression from Eq. 5.22as

V E WD mmin 1R � 1.14 ln (5.27)max C W2

Since Eq. 5.25 gives the lift coefficient for maximum range, the (equiv-alent) airspeed for the initial weight W1 can be calculated. The appro-priate cruise altitude can be obtained approximately from the thrustavailable data at a point where the thrust-altitude curve at cruisingvelocity crosses the drag curve at EAS (see Figure 3.5). SimilarlyVRmax

to the propeller-driven aircraft, the velocity for maximum range can beobtained from the thrust-velocity curve by drawing a tangent from theorigin to the thrust-required curve.

The effect of weight change on the range can be assessed as follows.In the integration of Eq. 5.21 it was assumed that L /D � constant. Iftwo weights are compared, at the same airspeed and constant CL,


1 2W � � � C V S (5.28)1 0 1 L2

1 2W � � � C V S (5.29)2 0 2 L2

which gives

W �2 2� (5.30)W �1 1

This shows that, as the aircraft weight decreases due to fuel beingburned, the density ratio also must decrease. Thus, similar to the de-velopments for propeller aircraft, as the fuel is expended the aircraftwill gain altitude or will drift up and air traffic control issues are en-countered again.

EXAMPLE 5.2

An F-86 aircraft has the following characteristics:

2C � 0.0159 � 0.075 C (clean)D L

W � 14,200 lb1

2q � 192 lb/ft2S � 290 ft

W � 3,000 lbf

W � 1,000 lb/hrf

Calculate:

a. The range at constant velocity starting at 25,000 ftb. The final altitude at the end of this range

a. For use with Eq. 5.22 CL and CD will be evaluated as follows:


W 14,200C � � � 0.225L qS 192 � 290

2C � 0.0159 � 0.075 � (0.255) � 0.0208D

which gives

L 0.255� � 12.26

D 0.0208

TSFC is obtained by first determining drag

D � C qS � 0.0208 � 192 � 290 � 1,158 lbD

and then calculating

W 1,000 lbf mC � � � 0.863T 1,158 lb � hrf

The range is obtained as

V L W 1 2q L W1 1R � ln � ln�C D W C � D W2 2

0.6818 2 � 192 14,200� (12.26) ln � 1380 miles�0.863 0.002377 � 0.448 11,200

b. Final altitude is obtained by calculating the density ratio:

W 11,2002� � � � 0.448 � 0.353� � � �2 1 W 14,2001

which corresponds approximately to 32,000 ft.

Constant Altitude Cruise In addition to the constant altitude as-sumption, this program requires that either CL � const and a parabolicdrag polar exists or CL � const and CD � const. Eq. 5.21 can be writtenas


c(L /D) dW (L /D) 2 dWR � � � V � � �� 1C W C �SC WL

(L /D) 2 (L /D) W2� 2 (W � W ) � 2 V 1 � miles� �1 2 1� �C �SC C WL 1

(5.31)

where

2W1V � mph1 ��SCL

is calculated at the beginning of the cruise. For calculation purposes,Eq. 5.31 is similar to Eq. 5.22, but in actual flight situations this flightprogram is not very practical since, as the weight changes, the flightvelocity must continuously be reduced. As in the previous section, themaximum range is obtained at /CD)max, which reduces to(CL

2C � 3kC (5.32)D L0

as in Eq. 5.25.

EXAMPLE 5.3

Calculate the range for the aircraft in Example 5.2 for the constantaltitude flight at 25,000 ft.

2W 2 � 14,2001V � �1 � ��SC 0.002377 � 0.498 � 290 � 0.255L

ft� 570 � 388 mph

sec

The range is found from

2 � 12.26 11,200R � � 388 � 1 � � 1,234 miles� ��0.863 14,200


Constant Altitude—Constant Velocity Cruise Since much of thecruise flight takes place at constant altitude and velocity, it is practicalto consider flight with the assumptions that h and V are constant. Inthis case the range equation is written from Eq. 5.20:

V dWR � � �

C D

which is similar to the constant speed propeller range calculations, Eq.5.18. Without going into the details, the range becomes

V 1 �dWR � � (5.33)� �2 2C qSC 1 � a WD0

with

k2a � 2 2q S CD0

Upon integration, the range can be expressed in two fully equivalentforms:

2V E Wm 2R � arctan 2kC E � arctan 2kC E (5.34)� �L m L m1 1C W1

or

2V E E (W /W )m 1 f 1R � arctan (5.35)� �C 2E (1 � kC E W /W )m L 1 f 11

where again

LE � and C�1 L1D 1

are evaluated at the beginning of flight.This flight schedule is more practical, as it is able to conform with

the general flight traffic control regulations. To maintain constant ve-locity, the thrust must be reduced as the aircraft loses weight due tothe fuel usage. This can be carried out stepwise with the increase inthe flight attitude.

5.3 RANGE INTEGRATION METHOD 117

5.3 RANGE INTEGRATION METHOD

A more accurate method for range calculation is a numerical-graphicalintegration technique. The advantage of this approach lies in the factthat it is independent of the type of power plant used. That is, it worksequally well for reciprocating, turbojet, or turboprop engines. The onlyrequirements are that the fuel flow rate, weight variation, and the trueairspeed be given in a manner consistent with thrust or horsepowerused. Thus, during the calculations airspeed and altitude may be variedcontinuously, which, in turn, determine the variations in thrust (orhorsepower) required and the fuel used. This permits the calculation ofthe range under actual flight conditions, including the climb, cruise,and landing operations. Such calculations may be laborious but willprovide highly reliable results, if so desired.

The starting point of the range calculation is the integral form of therange equation Eq. 5.21, which can be expressed (with the specificenergy term neglected) as

W W2 2V VW dWR � � dW � � (5.36)˙ ˙W WW W W1 1f f

where

dWfW �f dt

is the fuel flow rate. The quantity V / is called the specific range RsWf

and is a direct measure of the distance flown per unit fuel consumed.It is expressed as mi/lb, km/kg, or nautical air miles/lb (nam/lb). Itis evident that the use of maximum values V / in Eq. 5.36 will alsoWf

maximize the range. Since the fuel flow or the thrust specific fuelW ,fconsumption (TSFC), is a function of altitude, velocity, and thrust, andsince fuel flow rate is usually given in graphical form, determinationof maximum values of specific range requires an optimization processthat is best carried out by numerical-graphical means. Several variationsof the specific range integration method can be found, depending onthe user’s purpose, approximations made concerning the altitude range,and the choice of dependent variable (h, V, M, or W). Three are dis-cussed in the next section.

The first one contains the full method, as practiced in industry. Thesecond version is contained in the example following the first one,which is a somewhat abbreviated version leading to practical results


with a modicum of an effort. The basic idea here is to indicate howelementary path information can be extracted by user choice simplifi-cation. The third approach (an operational approach) simplifies themethodology as it would be applicable to mission analysis taking intoaccount dropping the stores and refueling.

5.3.1 Basic Methodology

Version One The essential features, which are common to all vari-ants, are given in the following step-by-step technique:

1. Select an altitude.2. Select at least four flight gross weights to cover the expected

range weight variations for each weight.3. For each weight select at least four cruise velocities. At each

cruise velocity determine the vehicle drag corresponding to se-lected weight, altitude, and speed conditions. Assuming that T �D.

4. Determine the net thrust per engine, Tn.5. Determine for the required Tn, at selected velocity and altitude,

the fuel flow rate from the engine data. The total fuel flowWf

rate is obtained from

˙ ˙W � W � # of enginesf fn

From engine data find TSFC, then

W � TSFC � Tf

6. Calculate the specific range Rs � V / for each of the cruiseWf

velocities selected in Step 3.7. Repeat steps 2 through 6 and 1 through 6 for all desired weights

and altitudes, respectively. The calculation process is facilitatedby setting up Table 5.1.

The calculated results are plotted as shown in Figure 5.2.The maximum range is obtained at the maximum values of Rs, for

a chosen weight, that determine the velocity (or Mach number), whichmust be maintained to achieve the best range. The effect of cruise atconstant velocity, or more important, at constant Mach number, isshown as a vertical line. The line identified as holding speed is com-


TABLE 5.1 Range Integration Method

h � constant

Step 1 2 3 4 5 6

Variable W V or M T � D Tn Wfn Rs �VWf

Action Select Select Calculate fromdrag polar

# of enginesengine data

Determine from# of engines Step

25

Rs

V

RmaxHoldingspeed LRC

h = constant

Wincr

Constantvelocity

Figure 5.2 Specific Range at Constant Altitude

monly established at a velocity corresponding to that at minimum drag(or slightly higher), and it occurs at a fuel flow slightly higher than theminimum flow.

The long-range cruise (LRC) line defines 99 percent of maximumrange to account for about 1 percent reduction of Rs due to requiredaltitude increase as the weight decreases during the flight. In practice,this altitude adjustment cannot be made continuously because it takesa finite time to burn up fuel and to achieve a new weight that requiresthat the thrust be increased to maintain flight at the proper altitude fora high Rs. Experience shows that the (stepwise) thrust increase leadsto about 1 percent reduction in Rs.

The effect of weight change on altitude is shown in Figure 5.3, whichrepresents the data calculated above and presented in Figure 5.2 forseveral altitudes when cross-plotted at constant W.

It is clearly seen that as W decreases, the altitude for maximum Rs

increases with an increase in the optimum flight velocity. Although theeffect of increasing altitude is to increase the specific range Rs, the


R S

h

W

W

W

1

2

3V

V

V

1

2

3

Wdecr

opt

opt

opt

Figure 5.3 Effect of Weight and Altitude on Specific Range

RF

W

Wdecr

h = 40,000 ft h = 35,000 ft h = 30,000 ft

Figure 5.4 Optimum Altitude and Weight

thrust requirements (for increased altitude and velocity) lead to a de-crease of Rs at higher altitudes. Thus, for each W there exists an opti-mum altitude, up to a point where engine thrust limitations or decreasein Rs restrict further altitude. The optimum altitude for each weight isdetermined if a quantity known as the range factor Rf, also called cruiseconstant, is plotted as a function of weight for a number of altitudes,as shown in Figure 5.4.

The range factor is defined as

VW V LR � � W R � (5.37)f sW C Df

which permits writing the range expression as


W1R � R lnf W2

(see Eqs. 5.36 and 5.22), and it is obtained from the data in Figure 5.2.It can be determined for maximum range, long-range cruise, or anyother desired flight schedule. If plotted against weight W, the rangefactor often remains almost constant for maximum range (also LRC)for a climbing cruise schedule. Thus, the assumptions made for ap-proximate integration of Eq. 5.21 are justified for long-range climbingcruise operation. Assuming the range factor to be a constant may notbe valid for other range flight schedules.

Version Two This example illustrates a simplified version of therange integration method, which permits a quick evaluation of bestcruise Mach number and cruise altitude. Obviously, these results canbe sharpened by improved drag and thrust data.

EXAMPLE 5.3

For an eight-passenger corporate jet, determine the cruise Machnumber, altitude, and fuel used for a range of 2,400 miles. The air-craft has the following characteristics:

W � 22,000 lb, W � 22,500 lb, W � 16,000 lb (zero fuel)cruise TO e

2S � 300 ft , AR � 7.7

T � 2 � 2,100 � � 55, h � 20,000 ft

The drag data are given in tabular form for the simple drag polar:

M .6 .7 .8 .9C .017 .017 .0175 .0195D0

k .05 .05 .052 .055

Although the specific fuel consumption C varies with both altitudeand Mach number, for simplicity only the variation with altitude isincluded here as C � .86, .88, .89 [lbm/hr/lbf], at 30,000, 40,000,and 50,000 ft altitude, respectively.


For calculations, it is assumed that the drag polar CD � �CD0

holds, the weight is constant at 22,000 lb, and the range is2kCL

calculated from Eq. 5.37. Thus, calculating only the range factor forselected Mach numbers (similar to Figure 5.2 will be

M qS CL CD D (lb) Ta (lb) Clbm/hr/lbf Rf mi

.6

.7

.8

.9

29700450005280066900

.741

.543

.417

.329

.0444

.0318

.0265

.0255

1320128614011702

1942194219421942

.88

.88

.88

.88

7510896594428708

sufficient for providing a selection process for the best Mach numberand cruise altitude. The calculations are carried out in tabular formatand are shown only for 40,000 ft altitude.

The range factor Rf is plotted in Figure 5.5 as a function of Machnumber for 30,000, 40,000, 45,000, and 50,000 ft altitude. Best rangeis indicated by the highest value of the range factor. For simplicity,it will be assumed that the cruise Mach number corresponding tobest range factor is the same as already given by the tabulated data.This permits a simple evaluation of the long-range cruise values at.99Rf. The following table, with liberal round-offs, summarizes theresults.

h,kft MRmaxRfmi .99Rfmi MLRC

30404550

.7

.8.82.89

823094409850

10250

815093509750

10000

.74

.83

.85

.84

It may be concluded that the cruise Mach number should be inthe .81 to .83 range. With the simplified drag polar and thrust values,ceiling is reached at about 50,000 ft and the realistic drag rise maybe somewhat higher than used here. Thus, the long-range cruiseMach number is selected at a value less than .9. Also the specificfuel consumption C may be larger at Mach numbers higher than .8,thus further reducing the range factor. A compensating factor maybe the use of constant weight W. In any case, this example showsthe methodology, which may be sharpened by use of more detailedand complete thrust, drag, and weight data. The cruise fuel may becalculated from


Figure 5.5 Range Factor

W 22,000 2,4001 � � exp � 1.279� �W W 9,7502 2

and

1W � 22,000 1 � � 4,800 lb� �f 1.279

5.3.2 An Operational Approach

From an operations point of view, range can also be evaluated if thespecific range has been calculated for a particular cruise flight scheduleby plotting Rs versus the weight, Figure 5.6. Since the data in the tableabove have been obtained for various weights, altitudes, and speeds,Figure 5.6 may represent any desired cruise schedule and is not limitedto, say, constant altitude or velocity schedule.

The area under the curve, obtained by numerical integration betweeninitial and final weights, is the desired range. W1 is the initial weight(W2 � Wfuel), and W2 is the empty weight, (empty weight � fuel re-serves).

The weight ratio W1/W2 has a strong influence on the range (seealso Eqs. 5.9, or 5.21. On one hand, the lighter the aircraft (small W2),the more range can be obtained for a given amount of fuel. On theother hand, increasing the dry weight W2 requires a larger increase in


RS(mi/lb)

R

W2

W2

R

W1

W1 W

∆

∆

∆

∆

Figure 5.6 Range Solution

fuel weight if the range is to remain constant. In Figure 5.6 an emptyweight increase �W2 (due to equipment, stores, etc.) indicates that anarea �R has been lost. For same desired range an equal �R must bemade available by an increase of W1 by �W1. The primary reason forthis disproportionate increase in fuel weight is the much higher specificrange at the end of the flight (near W2) where lower aircraft weightrequires less thrust power. For convenience, the results of integrationof the Figure 5.6 can be displayed as a plot of range against the in-stantaneous weight, Figure 5.7, which permits a quick evaluation ofthe available range at any given weight and is a valuable tool in missionanalysis.

For practical calculations and mission analysis, the range integrationprocess just described can be simplified to a few essential steps:

1. Fix initial aircraft weight W and altitude (�).2. Choose V, or calculate V .Rmax

Jet Aircraft

1 /42W kV � 1.316V � 1.316V � 1.316 � �R D (L /D)max min max � �S CD0

(5.38)

Propeller Aircraft

1 /42W kV � V � (5.39)� �R (L /D)max max � �S CD0


W1

W

Wf

Wstoresdropped

W2

RLineardistance

Range

Rmission

Refuel

Wf

Figure 5.7 Mission Range (Radius)

or, the maximum range velocity can be obtained by drawing atangent to the Pr or Tr curves (at a suitable W) as discussed inSection 5.1.

3. Calculate Tr or Pr from Eq. 5.15 or 3.55, or determine from Tr orPr curves.

4. From manufacturer’s data, determine the fuel flow rate (lb/Wf

hr).5. Calculate the specific range Rs:

V mi/hr miR � �� s W # of engines � lb/hr lbf

6. Plot Rs vs. W.7. Repeat steps 1 through 6 by varying W and � as desired.8. Integrate the Rs � W curve to get instantaneous distance as a

function of aircraft instantaneous weight.

Figure 5.7 shows the instantaneous weight as a function of down-stream distance. At W1 the distance covered is zero; full range is foundat W2. Such a graph may be prepared for any flight schedule (varyingaltitude, velocity, etc.). For most of the staightforward flight schedules,


the curve is almost a linear one, and the radius (flight out and return)can be established simply by reversing the curve at the same slope toend at the empty weight W2, as shown in Figure 5.7. Also shown arethe effects of dropping the stores (bombs, fire retardant fluid, agricul-tural sprays, etc.) and refueling. The latter case is shown with the as-sumption that the aircraft is refueled back up to its original full weight,W1. This may or may not be possible, depending on the configurationof volume available for fuel in case of dropped stores.

E3

5.4 OTHER CONSIDERATIONS

5.4.1 Flight Speeds

The range equations developed in Sections 5.1 and 5.2 are, at best,good approximations of what can be expected under actual flight con-ditions from take-off to landing. In addition to assumptions necessaryfor reaching useful solutions, air traffic control, weather, and variationsin flight speed contribute to deviations from expected results. Similarly,the corresponding flight velocities (e.g., Eqs. 5.38 and 5.39), deducedwhile establishing the range expressions, also represent good guidelinesand have long been subject to discussions as to what is the right orbest speed to achieve a desired/maximum range.

For jet aircraft, it is intuitive to assume that a high flight speed iscalled for a long range. Indeed, Eq. 5.38 shows that high flight veloc-ities are to be expected. For high wing loading (W /S) and high altitude(low �), the calculated maximum range velocity can approach andVRmax

even exceed the sonic speed. This invalidates the assumptions of con-stant and k, which can double and be reduced by a factor of twoCD0

to three, respectively, beyond the critical Mach number range (see alsoAppendix B). Introduction of the local compressible flight regime val-ues for and k will help solve this problem and yield Eq. 5.38 as aCD0

5.4 OTHER CONSIDERATIONS 127

reasonable estimate of the flight velocity. It should be noted that Eq.5.38 also implies that the flight take place at /CDmax.CL

For propeller aircraft, where at lower wing loadings and at loweraltitude and k remain constant, Eq. 5.39 tends to predict flightCD0

velocities that are too low for practical flight purpose. The same con-clusions can be drawn from Eq. 5.16, which predicts optimum flightspeeds in the region of reversed command (V � Obviously, thisV ).Em

contradicts the reasons for flight (high speeds at reasonable economy),as now the flight is supposed to take place at low speed and higherpower (higher fuel consumption).

To reconcile the desire for higher speed and better economy, Carson(see bibliography) proposed, by reviewing the small aircraft perform-ance, that minimizing the weight of fuel used per unit of velocity leadsto a higher and more economical flight speed. Consequently, the equa-tion for preferred flight speed turns out to be that given by Eq. 5.38.Thus, the expression for speed for desired economy for small aircraftand the best range speed for jet aircraft coincide. This does not meanthat a small propeller-driven aircraft be flown at the jet speed. It meansthat Eq. 5.38 be used, with the values appropriate for a small aircraft,for determining the flight speed. As a result, the flight velocity turnsout to be higher than given by Eq. 5.39 but appreciably lower than thespeed obtained for a jet aircraft from Eq. 5.38.

This somewhat surprising conclusion is also available from theenergy-based range equation, Eq. 5.2. For jet aircraft, Eq. 5.21 provides

V 1 VdR � � dW � � dW

W C T

and maximizing

CV 2L� �T C �WSD

maximizes the range with the best range speed given by Eq. 5.38 (seealso Problem 5.7).

For propeller aircraft (see Eq. 5.9)

� BHP � �� H dW L dWp p po fdR � � � � dW � � dW � �˙g D DW CT C D W

which clearly implies that the best range is obtained at (Eq. 5.39).VEm

Rewriting this now as


dW C T� � dR

V � Vp

it is easily seen that the Carson economy is obtained at maximumV/T, implying that the best speed then is at V � 1.316V .Em

5.4.2 Effect of Energy Change on Range

In Section 5.2 it was assumed that the specific energy (h � V2/2g)remains constant during the cruise portion of the flight. Consequently,the second term in the range equation vanishes. However, in derivingthe range equation (e.g., Eq. 5.4), it was assumed that L /D is constant,which, in turn, implies that aircraft attitude is constant. Subsequentsections have shown that either the velocity or altitude must thenchange, with a resulting change in energy level. Usually, velocity iskept constant, and altitude variation brings about the resulting changes.

In order to estimate the magnitude of the omitted term, Rutowski(see bibliography) replaced specific energy variation de with altitudedh and developed from the hydrostatic equation

dW RT dp RT dWdh � � � � � �

g� g p g W

where

dW dp�

W p

is obtained from logarithmic differentiation of the lift equation

� 2L � W � pM SCL2

Thus, the energy equation may be written for a jet aircraft as

V RT L W1R � � ln� �C g D W2

For a typical jet aircraft in stratosphere (T � 390R) at 800 ft/sec andwith a specific fuel consumption of C � 1.0 lbm/lbf/hr, the secondterm amounts to about .8 percent correction.

5.5 ENDURANCE 129

Since weight change is the basic cause of change(s) in energy, anapproximate assessment of this effect can be made by rewriting Eq.5.4 and Eq. 5.21 as

L V W E E2R � ln � �� D C W W W1 2 1

Assuming now that E � constant, which implies a cruise-climb oper-ation at constant Mach number in the troposphere, then the energy termcan be expressed as

1 1 E ƒE � �� W W W 1 � ƒ2 1 1

where ƒ is the fuel fraction ƒ � Wf /W1. Then the range equation be-comes

L V W ƒ2R � ln � e� �1D C W 1 � ƒ1

For an aircraft traveling at a speed of 800 ft/sec at 40,000 ft, with aweight of 400,000 lb and a specific fuel consumption of C � .6 lbm/lbf/hr, the relative magnitude of the second term is about 1.2 percentof the Breguet range—the first term.

5.5 ENDURANCE

Endurance is the time an airplane is able to remain in flight withoutlanding. In the simplified analysis to follow this means also withoutrefueling. Endurance with refueling can be included along the ideasoutlined in the previous sections. Endurance can be obtained from therange equation by replacing dR � Vdt. Thus, Eq. 5.3 becomes

H L dW (L /D)fdt � � � de� � � �o g D VW V

H L dW (L /D (L /D)f� � � dh � dV (5.40)� � � �o g D VW V g

where the specific energy has been used to expand the last term. Formost ordinary applications, the last two terms can be neglected. Again,


the nature of useful work done �o Hf leads to different expressions forthe reciprocating and jet engine endurance calculations.

5.5.1 Reciprocating Engines

Using Eq. 5.8 one obtains

� L dWdt � � (5.41)

C D VW

Since the weight changes, with L /D � constant, also the aircraft speedchanges. Substituting the expression for the velocity

2WV � ��SCL

Eq. 5.41 becomes

3 / 2� �S C dWLdt � � (5.42)3 /2�C 2 C WD

Assuming now that the flight occurs at constant altitude, and that �,C, /CD, and L /D are constant, then the integration can be carried3 /2CL

out to give

3 /22� �S C 1 1LE � t � t � �� 2 1 �C 2 C W WD 2 1

2� �SC C WL L 1� � 1� �� C 2W C W1 D 2

2� C /C WL D 1� � 1 (5.43)� ��C V W1 2

Maximum endurance is obtained if the multiplier

5.5 ENDURANCE 131

3 /2� 2�S CL�C W CD

is maximum. For constant-altitude flight, the maximum occurs when/CD is maximum. This condition can easily be found for the par-3 /2CL

abolic drag polar if the following differentiation is carried out:

3 /2d CL � 0� �2dC C � kCL D L0

Upon differentiation and simplification, one obtains

23C � kC or C � 4CD L D D0 0

But this is the condition for minimum power, as already established inSection 3.3.3. It occurs at

1 /42W kV � � �Pmin � �S 3CD0

Then also

33CC 1 DL 0� ��C 4C k 4kCD D0 D0

is constant, and /CD is given by3 /2CL

3 /2 3 /4C 3L � 3 /4 1 /4C 4k CD D0

As the aircraft weight W changes during the flight, the best speed formaximum endurance decreases (see Figure 3.7 with Example 3.4),VPmin

thus requiring that the aircraft speed be continuously changed duringthe flight. Since V2 (speed at the end of flight) will be less than V1, thespecific energy term indicates a slight improvement in the endurance.In practice, however, such continuous speed change is impractical tomaintain. Also, the best propeller efficiency � and the best fuel con-sumption C cannot always be maintained. Eq. 5.43 shows that the pro-


peller airplane endurance is maximum at sea level. It can be written ina consistent set of units as follows:

3 /2� �S C W � C 1 WL 1 L 1E � 1,100 � 1 � 1,100 � 1� � � �� C 2W C W C C V W1 D 2 D 1 2

(5.44)

For the case of a parabolic drag polar the maximum propeller aircraftendurance can be expressed as

628�� S W1E � � 1 (5.45)� �3 /4 1 /4 � �Ck C 2W WD 1 20

In Eqs. 5.44 and 5.45

E—Hours �— Propeller efficiencyW—lb V—mph

lb3�—slugs/ft C—BHP � hr

Subscripts 1 and 2 indicate quantities at the beginning and end of flight,respectively.

5.5.2 Turbojets

The endurance of a turbojet aircraft can be obtained also from Eq. 5.40with the fuel energy expression used in Section 5.2.2.

H Vf� �� o g C

Thus, one obtains, with the energy term omitted, and assuming thatL /D is approximately constant

L /D dWdt � � (5.46)

C W

and upon integrating

5.5 ENDURANCE 133

L /D W1E � ln (5.47)C W2

It should be noticed that, with the assumptions used, jet aircraft en-durance could be obtained directly from the range expression, Eq. 5.22,by dividing the range by V. Eq. 5.47 shows that the jet aircraft maxi-mum endurance occurs at Em (with the fuel consumption C being con-stant). For a parabolic polar this condition was shown in Section 3.3.2to give Em � 1/ and it occurs at the velocity for minimum(2kC )D0

drag

1 /42W kV � (5.48)� �Dmin � �S CD0

Thus, the endurance can be rewritten for parabolic polar:

1 W E W1 m 1E � ln � ln (5.49)W C W2CkC 2 2D0

shows that the best endurance speed is dependent on the aircraftVDmin

weight and decreases as the flight progresses (see Figure 3.4 with Ex-ample 3.3). According to Eq. 5.49, the endurance does not seem to bedependent on the altitude. However, the specific fuel consumption Ctends to increase with increasing altitude and increasing thrust. Thus,it is expected that improved endurance can be expected at lower alti-tudes at modest thrust levels.

P3

5.5.3 Endurance Integration Method*

Similarly to the range integration method discussed in Section 5.2,aircraft endurance can also be evaluated for any flight path with goodaccuracy from


Area = Endurance

1Wf

W2 W1 W

.

Figure 5.8 Endurance Calculation

W1 dWE � � ˙W W2 f

For jet aircraft

1 Rs�W Vf

can be obtained directly from tabulated range calculations in Section5.5. The propeller aircraft fuel flow rate is evaluated as

CDVW �f ��

as seen from Eq. 5.41. The calculation and graphing procedure for1/ as a function of altitude and weight is the same as for the range,Wf

Figure 5.8, and needs not be repeated here again.

5.6 ADDITIONAL RANGE AND ENDURANCE TOPICS

The previous sections of this chapter have clearly shown that cruiserange and endurance evaluation is simple and straightforward if com-monly accepted special conditions can be shown to be valid. The nu-merical (range) integration scheme hinted of the complexities involvedif more exact results or noncruise flight schedule results are desired.Indeed, accurate range calculations under a variety of conditions and

5.6 ADDITIONAL RANGE AND ENDURANCE TOPICS 135

V (tail wind) –V VRm tail wind

Tr

VWVRm

VRmhead wind

W

Figure 5.9 Effect of Wind on Range Speeds

factors involved are important for any successful commercial operation.For military and special-purpose airplanes, range or endurance calcu-lations depend on type of mission flown, refueling, stores dropped dur-ing flight, and so on. In the following sections the effect of wind onrange and endurance will be considered as it influences all other results,regardless of the flight schedules. Also, some general comparisons ofrange and endurance results will be offered.

5.6.1 The Effect of Wind

Up to this point, the basic performance analyses were based on theassumption that the flight path was determined by lift and thrust andnot subject to wind disturbances. In practical operations, such a con-dition would be the exception rather than the rule. The wind can havea strong influence, both adverse and beneficial, on aircraft performance:range, take-off, landing, and so on. However, wind has no practicalinfluence on endurance. To illustrate this, consider the Tr versus V curvein Figure 5.9.

The same argument holds for the power-required curve of propeller-driven airplanes. The presence of wind does not change the shape ofthe curve. Only the locus of the range velocity on the curve (the amountof drag) is modified by wind. If the airplane is flying, say, into the


wind, its relative airspeed, Vw, is increased but its ground speed, Vg, isdecreased:

V � V � V (head wind) (5.50)g w

V � V � V (tail wind) (5.51)g w

As the range is defined by the ground speed, it is decreased by thehead wind. In Figure 5.9 this amounts to moving the origin to the right.Drawing a tangent to the thrust-required curve locates the maximumrange speed for the flight with the head wind. The tail wind shifts theorigin to the left. If the plane flies into the head wind with the airspeedequal to the wind speed Vw, the ground speed and the range becomezero. This flight, at zero range, can be continued (theoretically) untilthe aircraft runs out of fuel.

The zero range finite endurance flight is consistent with the defini-tions of endurance and range. Range is dependent on ground speedwhile endurance is determined by the fuel available for propulsion. Itshould be pointed out that the specific fuel consumption C is affectedby flight speed and the thrust level (drag). However, once the lowestdrag is established on the thrust curve, then also the best endurancerelative air speed is determined and the endurance depends only on theamount of fuel available. Rather than work with the elusive tangentsin Figure 5.9 it is possible to formulate the previous concepts in ana-lytical form. Propeller and jet powered aircraft will be treated sepa-rately.

Propeller Aircraft To properly formulate the range expression withthe wind effect, included it is best to return to the basic range energyequation, Eq. 5.1, and to identify the velocities as follows:

W de � � H dW � DV dt (5.52)o f f a

and

dR � V dt (5.53)g

where

Va � airspeedVg � ground speed


Combining Eq. 5.52 and Eq. 5.53 gives

V V� dW Wg gdR � � � de (5.54)C V D V Da a

With the assumptions of de � 0, CL � const and V � const, the rangeequation becomes

� L V Ww 1R � 375 1 � ln (5.55)� �C D V Wa 2

If the velocity is nondimensionalized by use of (Eq. 3.26), andVDmin

Eq. 3.36 is introduced in the following form:

2L 2V� E� � m4D V � 1

then the range equation with the wind can be written as

� 2V W1R � 375 E (V � V ) ln (5.56)� �m w 4C WV � 1 2

where

VV �

VDmin

VwV �w VDmin

The aircraft does not necessarily fly at Em (where � 1) but at L /DVconsistent with the actual airspeed Em appears only as a conse-V .quence of introducing Eq. 3.41. The maximum range, in the presenceof wind, is not obtained at Em but at a different L /D, as indicated inFigure 5.9. In general, it is rather cumbersome and difficult to read offtangents from power, or thrust, curves. Rather, the tangent to the Pr

curve can be related to the ground speed if a simple drag polar isavailable. In nondimensional form, and again by use of Eq. 3.36, thetangency condition is given by


d(P/W) d 1 1 (P /W)r3� V � �� 2EdV dV V V � Vm w

1 1 13� V � (5.57)� �2E V V � Vm w

Carrying out the differentiation and clearing the fractions, one finds

4 43V � 1 V � 1� (5.58)

V V � Vw

which gives the following inverse relationship:

42V (1 � V )V � (5.59)w 43V � 1

It is inverse in the sense that a relationship of the form isV � ƒ(V )w

desired rather than � However, Eq. 5.59 can be solved onceV ƒ(V).w

and for all and is shown plotted on Figure 5.10 as the prop curve.Thus, if the wind fraction is known, the best range speed�V Vw

can be found from the curve and the range calculated from Eq. 5.56.A useful approximation for the solution of Eq. 5.59 is given by

4 4 Vw1 /43 V � � V � (5.60)� �w3 9 3

which can be used to explicitly evaluate the best range speed if isVw

known.

Jet Aircraft Similar to the developments leading to Eq. 5.57, therange for jet aircraft is given by

V L Wg 1R � ln (5.61)C D W2

where CL � const and V � const. Using the same nondimensionalparameters as for the propeller-driven aircraft, the range expressionbecomes


Jet

Prop

Tail windHead wind-1.0 1.0 VW

V

0

1.0

1.5

2.0

Figure 5.10 Maximum Range Velocity with the Wind

V L WDmin 1R � (V � V ) lnwC D W2

2V V � V WDmin w 12� V E ln (5.62)� � m4C WV � 1 2


Again, the aircraft does not necessarily fly at Em and These ap-V .Dmin

pear as a consequence of this particular nondimensional procedure. Tofind the airspeed that maximizes the range in the presence of wind, thetangency condition to the Tr curve needs to be established. Thus,

d(T /W) d 1 1 (T /W)r r2� V � �� 22EdV dV V V � Vm w

1 1 12� V � (5.63)� � � �22E V V � Vm w

Upon simplification, one gets the following implicit equation for �Vƒ(V ):w

4V V � 3V � � (5.64)� �w 42 1 � V

This has been plotted on Figure 5.10 as the jet curve and can be givenapproximately as

4 4 VwV � � V � (5.65)� �w3 9 3

The propeller and jet optimum flight velocity expressions, Eqs. 5.60and 5.65, respectively, differ by the factor 31 / 4 (see also Section 5.2.2),which makes these equations convenient to use. Thus, only one curveis sufficient in Figure 5.10.

EXAMPLE 5.5

Calculate the range and endurance of a propeller aircraft with 50mph head wind at sea level. The aircraft has the following charac-teristics:

W � 15 percent of total weightf

� � 0.8

lbC � 0.5

HP-hr


W/S � 342C � 0.022 � 0.0606CD L

Calculate the range for CL � const and V � const.The range, from Eq. 5.56, can be written as

� 2V W1R � 375 E (V � V ) ln� �m m 4C WV � 1 2

2V� (V � V ) R� �w no wind4V � 1

The no-wind range is determined at Em and Calculating these:V .Dmin

1 1E � � � 13.7m 2kC 20.022 � 0.0606D0

1 / 4 1 / 42W/S k 2 � 34 0.0606V � �� Dmin � �� C 0.002377 0.022D0

ft� 217 � 148 mph

sec

provides

� W 0.8 11R � 375 E ln � 375 13.7 ln � 1,336 milesno wind mC W 0.5 0.852

The wind fraction becomes Vw � 50/148 � 0.338, which gives, formaximum range from Figure 5.10, � 1.12. The range, with theVoriginal no-wind range, can be calculated as follows:

2 � 1.12R � (1.12 � 0.338) 1336 � 909.3 miles� �41.12 � 1

Thus, the original no-wind maximum range of 1,336 miles has beenreduced by the 50 mph head wind to 909.3 miles. The flight takesplace at

V � 1.12


or

V � 1.12V � 1.12 � 148 � 166 mphDmin

and at 166 � 50 � 116 mph ground speed. The actual flight lift-drag ratio is now

2L 2V 2 � 1.12� E � 13.7 � 11.9m 44D 1.12 � 1V � 1

As the flight speed is constant, the time to fly, or the endurance is

909.3E � � 7.83 hours

116

As a first-order approximation that sometimes gives surprisinglygood results, one can use

R � R � V Ew no wind w

For this case

1336R � 1336 � 50 � � 1336 � 451 � 885 milesw 148

The preceding development and the example were carried out onlyfor the cruise-climb flight schedule. Similar analysis applies for otherflight paths (e.g., constant altitude, constant velocity cruise, as givenby Eqs. 5.16 and 5.34). The constant ground speed needs to be replacedby its proper wind formulation, and Figure 5.10 can be used to findthe correct flight (air) speed. As pointed out in Section 5.2, those rathercumbersome calculations are often replaced (or rather approximated)by the Breguet equations. Thus, that particular development will beomitted here.

5.6.2 Some Range and Endurance Comparisons

Maximum range and maximum endurance were calculated in the pre-ceding sections for several different flight schedules. It is evident by


TABLE 5.2 Comparison of Maximum Range andEndurance

Prop Jet

Emax

ERmax

1.14 1.155

Rmax

REmax

1.155 1.14

inspection that these conditions do not occur simultaneously. For ex-ample, maximum range and maximum endurance for a jet aircraft(Breguet expressions) occur at and at 1.316 respectively.V V ,D Dmin min

Moreover, jet aircraft endurance is independent of altitude. Thus, a jetflying at at 20,000 ft and at sea level will have the same maximumVDmin

endurance as (L /D)max (with C being approximately constant). How-ever, the plane flying at higher altitude will have more range due tohigher airspeed. Similar arguments hold also for ranges flown at dif-ferent schedules.

From Eq. 5.9, and assuming that the specific fuel consumption C isconstant,

R (L /D)max max�R (L /D)E Emax max

at (L /D)max

L 1�

D 2kCD0

at (L /D)Emax

L 3�

D 4kCD0

and

4kCDR 20max � � � 1.155R 23kC 3Emax D0

144

TAB

LE5.

3S

umm

ary

of

Ran

ge

and

End

uran

ceE

qua

tio

ns

Pro

pV

max

R,E

C

DV

0m

ax

Ran

ge(m

i)C

L�

cons

t�

LW

p1

375

lnC

DW

2

VE

m2

C�

kCD

L0

h�

cons

t,V

�co

nst

�E

Wp

m2

�1

�1

750

tan

[2kC

E]

�ta

n2k

CE

��

��L

mL

m1

1C

W1

VE

m2

C�

kCD

L0

End

uran

ce(h

r)�

�S

CL

WL

111

00�

1�

��

�C

2WD

W1

2

VP

min

23C

�kC

DL

0

C�

SFC

(lb/h

r/H

P)

Jet

Ran

ge(m

i)C

L�

cons

t,V

�co

nst,

crui

se�

clim

bV

LW

1ln

CD

W2

1.31

6VE

m2

C�

3kC

DL

0

CL

�co

nst,

h�

cons

t2

2WL

W1

21

� ��

��

C�S

CD

WL

1

1.31

6VE

m2

C�

3kC

DL

0

h�

cons

t,V

�co

nst

2VE

Wm

2�

1�

1ta

n[2

kCE

]�

tan

2kC

E�

��

Lm

Lm

11

CW

1

VE

m2

C�

kCD

L0

End

uran

ce(h

r)1

LW

1ln

CD

W2

VE

m2

C�

kCD

L0

C�

SFC

(lb/h

r/T)


Tr

EmaxRmax

Dmin

Pr Emax

Pmin

Rmax

VPmin VDmin 1.316 VDminV

T, P

Figure 5.11 Range and Endurance

which states that a propeller aircraft maximum range is always 15.5percent higher than the range flown at the best endurance condition.Similarly, a comparison of endurance for a propeller aircraft yields

3 /2E C /C )max L D max� 3 /2E (C /C )R L D (L /D)max max

In Section 5.5.1 it was shown that for maximum endurance

3 /2 3 /4C 3L � 3 /4 1 /4C 4k CD D0

Since, for (L /D)max, CL � CD � thenC /k, 2C ,D D0 0

3 / 4 3 / 4 3 /43 k 2CE 3Dmax 0� � � 1.143 /4 1 /4 3 /4E 4k C C 2R D Dmax 0 0

The maximum endurance of a propeller airplane is 14 percent longerthan the endurance at maximum range. Table 5.2 summarizes the re-


sults. Table 5.3 summarizes the general results for propeller and jetaircraft range and endurance. Figure 5.11 gives a graphical interpreta-tion of the results found in Table 5.3.

PROBLEMS

5.1 Derive the jet aircraft range expression with the assumption thatthe cruise velocity is constant.

5.2 Calculate the range of the aircraft in Example 3.4 at 10,000 ft,assuming the specific fuel consumption of 0.5 lb/hr-HP.Ans: 1,455 mi.

5.3 Calculate the maximum range of the aircraft in Example 3.3, as-suming that it carries 2,000 lb fuel and is equipped with two J60engines. What is its final flight altitude with 10 percent fuel re-maining for descent and landing if the cruise climb schedule be-gins at 25,000 ft?Ans: 797 and 723 mi.

5.4 Determine the ranges of the aircraft in Problem 5.3 with a 40knot head wind and 40 knot tail wind.Ans: 652 mi, head wind.

5.5 A jet aircraft has the following characteristics:

W � 70,000 lbS � 1500 ft2

h � 30,000 ftCD � 0.018 � 0.04 2CL

Wpayload � 9000 lb (dropped at destination)Wfinal � 42,000 lb (including 5 percent fuel reserve)

TSFC � 0.6 (constant)

A long-range cruise drift-up technique is used. If an instanta-neous altitude change takes place upon releasing the payload,calculate the distance at which the payload was dropped and thefinal altitude when reaching home base.

5.6 For the aircraft in Problem 5.5, calculate the distance at whichthe refueling should take place in order to increase the totalrange by 50 percent, 100 percent.

PROBLEMS 147

5.7 Show that

CV 2L� �T C �W SD

which leads to Eq. 5.38 (see Section 5.2.2).

5.8 Boeing 707-320B, Intercontinental Series, weighs 320,000 lb atthe beginning of its long-range cruise at 40,000 ft. Its drag polar,up to M � .85, is

2C � .0185 � .036CD L

Other pertinent data:

AR � 7.3S � 2892 ft2

Four JT3-1 engines

Calculate the amount of fuel needed to fly from London to NewYork, 3,500 mi.Ans: about 88,000 lb.

5.9 Gates Learjet 56 Longhorn is a 13-place executive transport withtwo 3,700 lb turbofan engines. It has a maximum cruise weightof 20,000 lb, S � 265 ft2, AR � 7.2, e � .85, � .018, cruiseCD0

airspeed 508 mph (.77M). With a normal fuel load of 5,500 lb,it has a range of 3,000 mi (sfc � .75), more than sufficientcapacity to fly from Los Angeles to New York (2,450 mi). Dur-ing the last flight, the pilot, planning to capitalize on a favorablejet stream, took on an extra 1,500 lb of commercial cargo andonly 4,000 lb of fuel, which normally would be insufficient toreach New York without an appreciable tail wind. Determine thefollowing:a. What would be the minimum jet stream speed that the pilot

needed to make good the distance?b. At what airspeed would the pilot fly to get the best range?

Ans: a. 77 mph, b. 486 mph.


5.10 A jet aircraft weighs 29,950 lb at the beginning of a constantaltitude cruise at 20,000 ft altitude. S � 640 ft2, sfc � .87. Fromthe thrust setting it can be estimated that the thrust is initially2,650 lb. Calculate the range if the cruise fuel weighs 8,150 lb.The drag polar is

2C � .0255 � .0616CD L

Ans: 1117 mi.

5.11 An A-4 aircraft is being tested at 5,000 ft altitude. The aircraftis equipped with one J-52 engine capable of delivering a thrustof 8,500 lb up to M � .8. Fuel flow rate is given in terms ofthe thrust and Mach number by wf � .9T(1�.36M) lb/hr. Aspectratio is 2.91, S � 260 ft2. Weight of the aircraft, at the beginningof the flight, is 16,000 lb, and the available fuel weighs 1,700lb. CD � .022 � The ground radar measures the speed2.19C .L

to be .6M. Assuming that the simple drag polar holds, estimatethe aircraft range under these (constant) flight conditions.Ans: about 240 mi.

5.12 A four-engine turbojet has the following characteristics: �CD0

.018, Mcr � .77, S � 1,500 ft2, AR � 5.5, e � .82, W /STO � 70� 2.0, T � 1,600 � lb/engine, C � .87 lbm/lbf/hr.CLTO

The aircraft uses 3,000 lb fuel to take off and to climb to 30,000ft. Determine:a. Rmax for constant altitude flight at 30,000 ft for 24,000 lb

available cruise fuelb. The velocity at the destination

5.13 The following information is known about 727-200:

W � 135,000 lbS � 1,700 ft2

AR � 6.9e � .83T � 15000��8/engine, 3 engines

TSFC � .95 lbm/hr/lbT

�CD0.019, � .092 (including air brakes)CD0 land

�CLmax.96, � 2.6, CLTO � 2.1CLmax flaps

PROBLEMS 149

Determine:a. How much fuel is needed to fly from Baltimore to Chicago

(700 miles) using the maximum range schedule at constantaltitude

b. Upon arriving in Chicago, how much thrust is required tomaintain a glide slope of 2 degreesAns: 13,265 lb, 3739 lb.

150

6Nonsteady Flight in the

Vertical Plane

DC10

6.1 TAKE-OFF AND LANDING

The quasi-steady analysis presented in previous chapters determinesthe general bounds of the aircraft performance: stall, maximum veloc-ity, ceiling, climb, range, and endurance. Although corrections wereincluded for speed variations, the average steady-state nature of thephenomena permitted solutions without recourse to path-dependent in-tegration and without the need for direct consideration of accelerations.Take-off and landing are accelerated operations involving velocity var-iations from zero to safe flight speed, thus requiring a more exactinganalysis of equations of motions. Although take-off and landing rep-resent only a small portion of the total operation of an aircraft, per-formance of these two phases is considered very important due to two


6.2 TAKE-OFF ANALYSIS 151

entirely different reasons. First, a great majority of accidents (mostlyattributed to pilot error) occur during landing or take-off. Second, it isthe take-off portion that establishes the engine sizing (in conjunctionwith air worthiness requirements) for design of civil aircraft. For mul-tiengine design, the critical condition is provided by the failure of oneor more engines at the least favorable moment.

Thus, owing to the variability of the pilot techniques human element,it is not practical to introduce too many exacting mathematical com-plications in the analysis of take-off or landing performance. On onehand, experience indicates that it is practical to introduce those sim-plifying assumptions without destroying the fundamental aspects of thephenomena, which lead to analytical solutions permitting inclusion ofall important factors affecting the performance. On the other hand,design studies may require that extensive computer calculations be car-ried out to provide systematic studies concerning the effect of variousparameters on aircraft performance. The wealth of resulting numericaldetail can lead to useful design results, but there is also a real dangerof obscuring the basic features of the problem. To support the detailednumerical studies an analysis is needed that is sufficiently simple yetpermits a good overview of all pertinent features of the problem; thisis the purpose of the current chapter.

6.2 TAKE-OFF ANALYSIS

The take-off analysis can be conveniently divided into two phases: theground run and the air phase. The air phase begins when the aircraftbecomes airborne and lasts until the aircraft has reached a safe flightcondition. The safe flight condition is defined as the condition wherethe aircraft has cleared a specified height and commences to climb ata steady rate. This height is called obstacle height or screen height—FAA specifies 50 ft. For more details, see FAR23 and FAR25 series orapplicable MILSPEC’s.

Depending on the type of aircraft and the take-off performance ca-pabilities, the airborne phase may be divided into two parts: a transitionphase and a steady climb phase. If the 50-ft height is reached beforesteady flight is achieved, the airborne phase is approached as one con-tinuous operation. Due to operational safety aspects FAA uses also theconcept of balance field length (BFL). At its simplest, it can be definedas a total length when the distance required for take-off is equal to adistance needed to come to a complete stop on the ground (aborting

152 NONSTEADY FLIGHT IN THE VERTICAL PLANE

Ground run, sg Transition distance, st c

Climb distance s

Stall safety

Vs

VCR

VMCG

Acceration at full power Take-off possible with one engine

Continue take-off if enginefails after this point

Stop take-off if enginefails before this point

hc

Total take-off distance

VT

cγ

Figure 6.1 Take-off Distances

the take-off). Its real intent is to assure safe landing or take-off dis-tances for multiengined aircraft experiencing engine(s) failure at take-off. Very little generalization is possible here. In practice, the largenumber of possible cases arising from different aircraft subject to dif-ferent FAR rules at a variety of flight conditions require highly indi-vidual and detailed treatment. Although the necessary methodology isdisplayed below, it is not practical to pursue these details within theintent and given scope of the topic treatment here.

The total take-off distance is calculated as the ground distance toclear the screen height as shown in Figure 6.1, and is the sum of thefollowing four distances:

s —ground rung

s —rotation distancer

s —transition distancet

s —climb distance to reach screen heightc

In the following sections these distances will be evaluated separately,as they represent individual portions of the take-off operation. However,they are each subject to some of the following factors:


�—runway slope or gradientV —wind speed, engine failure during take-offw

These factors will be considered inasmuch as practical results can beobtained.

6.2.1 Ground Run

Ground roll can be defined as the distance between the points wherethe brakes are released and where the take-off velocity, VT is reached.During the ground roll the velocity increases progressively through anumber of (operationally) important velocities, as shown in Figure 6.1,and they are stated in the order in which they occur on a multiengineaircraft.

1. VMCG, minimum control speed on the ground. If an engine failureoccurs above this speed the pilot is able to maintain control anda straight path in the original direction without retrim or reductionof thrust.

2. VCR, critical speed. One engine failure at this speed permits thepilot to continue take-off to screen height or stop, in the samedistance. Engine failure at lower speed requires stopping the air-craft, while a failure at higher speed means that the take-off becontinued.

3. Vs, stall speed (decision speed). This is calculated in the take-offconfiguration.

4. VT, take-off speed. After leaving the ground, the airspeed shouldnot be allowed to fall below this speed. Also called take-off safetyspeed, it is usually chosen as $1.2 Vs. The aircraft can be liftedoff the ground (unstick, or getaway point) at any speed above Vs,but no appreciable height can be gained until VT has been reached.It is questionable whether completing ground roll with wheels offthe ground will shorten the ground phase at all since, while inground contact, above Vs there will be very little load on thewheels anyway and the induced drag is reduced due to favorableground effect. Lifting the aircraft out of the ground effect earlymay increase the drag, and thus also increase the distance inwhich VT is reached.

While an aircraft is in airborne motion, the governing equations arethose given by Eqs. 2.9 through 2.13. Two additional forces appear in


L

DT

V

W

R

R

µ

Figure 6.2 Forces on an Aircraft During Take-off

the ground phase: the reaction force R of the runway on the aircraftand the tangential force due to rolling friction �R; see Figure 6.2.

Average values for the friction coefficient � are given in Table 6.1.They can be used for either take-off or landing. Thus, if it is assumedthat the runway is horizontal (� � 0) and that the weight variation dueto fuel consumption can be neglected, the equations of motion can berewritten as

x � V (6.1)

VW� T � D � �R (6.2)

g

0 � L � W � R (6.3)

If the reaction R is now eliminated and the velocity V is selected asa new independent variable, one obtains the following differential equa-tions for the distance and time (s � x):

ds VW/g� (6.4)

dV T � D � �(W � L)

dt W/g� (6.5)

dV T � D � �(W � L)

In general, since T, D, and L are functions of velocity, exact solutionsare possible only by carrying out numerical integration.

Several schemes are available in literature for integrating these ex-pressions. The two most common assumptions have made use of either


TABLE 6.1 Average Coefficient of Friction Values

Type of SurfaceAverage Coefficient

No BrakesAverage Coefficient

Full Brakes

Concrete 0.02 to 0.05 0.4 to 0.6Hard Turf 0.05 0.4Soft Turf 0.07Wet Concrete 0.05 0.3Wet Grass 0.10 0.2Snow/ Ice Covered Field 0.02 0.7 to 1.0

a constant average force idea (the denominator is the acceleration force)or an average acceleration concept. Although such methods requireadditional assumptions concerning individual parameters, they do pro-vide practical engineering approximations. However, this results in hid-ing the trends and basic features of the problem, thus preventing a goodoverview of the role of the significant parameters involved.

A fully equivalent and more flexible solution becomes available ifone assumes that CL and CD are constants during the ground phase andthe take-off portion of the thrust can be approximated (by simple curve-fitting procedure) by an equation of the form

2T � T � BV � CV (6.6)o

Then the expressions for distance and time become

V dVds � (6.7)2aV � bV � c

dVdt � (6.8)2aV � bV � c

where

�gS Cga � � (C � �C ) �D L2W W

Bgb �

WToc � g � �� W


Since the solutions reappear in landing calculations and due to theneed to evaluate the integral between any two speeds as the thrust mayvary substantially when the take-off assisting rockets are fired or anengine fails, the general solutions are given by (see any good table ofintegrals) between two arbitrary limits V1 and V2 as

21 aV � bV � c2 2s � ln � �g 22a aV � bV � c1 1

b /a 2aV � b 2aV � b2 1�1 �1� tanh � tanh� �2 2 2�b � 4ac �b � 4ac �b � 4ac(6.9)

or

21 aV � bV � c b /2a (1 � a )(1 � a )2 2 2 1s � ln � ln � �g 2 22a aV � bV � c (1 � a )(1 � a )�b � 4ac1 1 2 1

(6.10)

where

(2aV � b)2a �2 2�b � 4ac(2aV � b)1a �1 2�b � 4ac

The time for take-off becomes

2 2aV � b 2aV � b1 2�1 �1t � tanh � tanh (6.11)� �2 2 2�b � 4ac �b � 4ac �b � 4ac

1 (1 � a )(1 � a )1 2� ln (6.12)� �2 (1 � a )(1 � a )�b � 4ac 1 2

Simple calculations will show that, in most cases, the quantity (b2 �4ac) is positive. Thus, the choice of inverse hyperbolic tangent solution.In case 4ac � b2 (this may happen if large reverse thrust is applied,To � O). Then the inverse hyperbolic tangent must be replaced byinverse tangent function, and the radical in the denominators will bereplaced by Furthermore, Eqs. 6.10 and 6.12 cease to be2�4ac � b .


valid for this case, as they represent the equivalent functions only forthe case of inverse hyperbolic tangent function.

To bring out the factors that are essential for establishing the groundrun distance, consider the following (rather general) case: normal un-assisted, level runway, no wind, starting at rest, constant thrust take-off. Thus (see Eq. 6.6):

V � 0 B � 01

T � T C � 0o

and one obtains from either Eqs. 6.11 or 6.12

21 aV � c W/S 12s � ln � lng 2a c �g (C � �C ) C � �CD L D L� �1 �(T /W � �)Co LT

(6.13)

The take-off lift coefficient is obtained from the condition of theCLT

take-off speed when the aircraft just leaves the ground:

2W/SC � , V � V (6.14)L T 2T 2�VT

The coefficients CD and CL are average values evaluated during theground run. Eq. 6.13 can be further simplified by expanding the log-arithm as

2x � 1 1 x � 1 1ln x � � � � � � x � (6.15)� �x 2 x 2

Letting now x � 1/(1 � Z), with Z being the term in the squarebrackets in Eq. 6.13, then

1 Zln � Z � � � � � (6.16)

1 � Z 2

and Eq. 6.13 can be written as


W/S 1 C � �CD Ls � 1 � � � � � (6.17)g 2T C To LT o� ��g � � � � C� � � � LTW W

where the significant remaining term in the brackets is usually less than0.1, and the discarded terms are entirely negligible.

NOTE

For rapid calculations and comparative estimates, Eq. 6.17 is simplified toread

W /Ss �g �gC (T /W)L oT

1.44(W /S)� (6.18)

�gC (T /W)L om

Eq. 6.18 brings out all the salient features of the take-off ground run. Itshows that the ground run is directly proportional to the aircraft wingloading W /S and the altitude of the runway. It varies inversely with thetake-off lift coefficient and To /W (� being about 0.02, and thus may beignored). Increase in ambient temperature works through density (� � 1/T) and through degradation of engine thrust to increase the take-off dis-tance.

Calculation of the take-off ground run is not very sensitive to aircraftattitude and, thus, to evaluation of CL, especially since it appears incombination with rolling friction coefficient �. However, an optimumlift coefficient can be determined that gives theoretically a minimumground distance and provides some guidance for selection of the groundrun lift coefficient. When differentiating Eq. 6.13 with respect to CL, itleads to the requirement that

da d 2� (C � kC � �C ) � 0 (6.19)D L L0dC dCL L

and

2kC � � � 0 (6.20)L


It then follows that

� ��AReC � � (6.21)Lg 2k 2

for the ground run lift coefficient that tends to give the shortest dis-tance.

The drag coefficient is evaluated by means of Eq. 6.21, and CD0

should include the additional drag due to landing gear, flaps, and theground effect. The effect of ground, or water, is to reduce the aircraftdownwash and the strength of tip vortices that makes the aircraft be-have as if its aspect ratio were increased. The spanwise lift distributionis then altered in a favorable manner to produce more lift and lessinduced drag and a smaller angle of attack is required to produce thesame lift, which is also significant during the landing operations. Thus,to account for the ground run, the induced drag may be reduced by afactor of 0.6:

2C � C � 0.6kCD D L0

In transition and climb-out phases, the ground effect is generally ig-nored.

For approximate preliminary design estimates the following expres-sion can be obtained from Eq. 6.18 by ignoring the terms in the brack-ets and the rolling coefficient in comparison with T /W, and byintroducing the take-off velocity:

2W2V � (6.22)T �SCLT

as

2WVTs � (6.23)g 2gTo

where W is the take-off weight. For use with Eq. 6.22, the thrust isevaluated at (0.7 � 0.8) VT.

Summarizing: Eqs. 6.23, 6.18, 6.17, 6.13, and 6.9 calculate theground run in an increasing order of accuracy. Eqs. 6.23 and 6.22 arepractically the same first-order approximations. Eq. 6.13 marginallyimproves the accuracy over the simplified version of Eq. 6.17.


F16A

6.2.2 Rotation Distance

At the end of ground roll and just prior to going into transition phase,most aircraft are rotated to achieve an angle of attack to obtain thedesired take-off lift coefficient CL. Since the rotation consumes a finiteamount of time (1–4 seconds), the distance traveled during rotation sr,must be accounted for by using

s � V �t (6.24)r T

where �t is usually taken as 3 seconds.

EXAMPLE 6.1

A twin-engine jet aircraft has the following characteristics at sealevel:

W � 50,000 lbT � 5000 � 3.28V per engine

2S � 1500 ftC � 1.5L max

2C � 0.02 � 0.05CD L

Calculate the take-off distance on a smooth, dry concrete runway (�� .02) at sea level for no-wind conditions.

Ground run is calculated from Eq. 6.9. Assuming VT � 1.2Vs, oneobtains

C 1.5L maxC � � � 1.04LT 1.44 1.44


and

22W 2 � 50,000 ft2V � 1.44 � 1.44 � 26,925T 2�SC 0.002377 � 1500 � 1.5 secL max

Ground run lift coefficient is obtained from Eq. 6.21:

� 0.02C � � � 0.2L 2k 2 � 0.05

which gives for the drag coefficient:

2C � 0.02 � 0.05(0.2) � 0.022D

For evaluating the ground run from Eq. 6.20 the coefficients a, b,and c are

��gS �0.002377 � 32.2 � 1,500a � (C � �C ) �D L2W 2 � 50,000

�5� (0.022 � 0.02 � 0.2) � �2.02 � 10

Bg 6.56 � 32.2 �3b � � � � �4.225 � 10W 50,000

where B � 2 � 3.28

T 10,000oc � g � � � 32.2 � 0.02 � 5.796� � � �W 50,000

The ground run is then from Eq. 6.9, V1 � 0, � 0.0222�b � 4ac

5�10s � ln

2.02 � 2�5 2 �3�2.02 � 10 � (164) � 4.225 � 10 � (164) � 5.796� �5.796

�3�4.225 � 10�

�5�2.02 � 10 � 0.022�5 �32 � (�2.02 � 10 ) � 164 � 4.225 � 10�1� tanh� 0.022

�3�4.225 � 10�1�tanh � 5,938 � 3,288 � 2,650 ft�0.022


Two comparisons are now of great interest. First, if it is assumed,as it is often the case, that the take-off thrust is constant (B � 0, b� 0), then the second term in Eq. 6.9 drops out and the ground runbecomes

5 �5 2�10 �2.02 � 10 � (164) � 5.796s � ln � 2,436 ft

2.02 � 2 5.796

Second, using the approximate result, Eq. 6.16, one gets

1 1 C � �CD Ls � 1 ��gS 2T � � To oC � � CLT � � LT� � �W W W

1 0.5 � 0.0176� 1 �� 0.002296 � 1.04 � 0.18 0.18 � 1.04

� 2,326 � 1.04 � 2,435 ft

Rotation distance is evaluated from Eq. 6.23:

s � 3V � 3 � 164 � 492 ftr T

Thus, the total distance on the ground is sg � 2,650 � 492 � 3,142ft.

6.2.3 Transition Distance*

Evaluation of the transition distance is cumbersome and results tend tobe unreliable due to various factors like landing gear retraction, varyingground effect, different piloting techniques, and so on. In addition, thenumber of parameters tends to be large or parameters appear in theanalysis that can be evaluated only by experimental means. It becomessignificant only for modestly powered and/or extremely loaded aircraft.Thus, a method that is convenient and relatively simple to apply withacceptable accuracy is obtained from Eq. 6.8 by recognizing that � �0 for the airborne phase, and by assuming that T �D � constant. Thus,by direct integration of Eq. 6.8 one gets


2 2W V � Va Ts � (6.25)t 2g T � D

where Va is the airspeed at the end of transition, and T and D are tobe evaluated at the average velocity (Va � VT)/2 during transition. Theinherent drawback of this method is the fact that the airborne velocityVa, at the end of transition is not known and must be either assumedor iteratively determined. Furthermore, the process is complicated bythe fact that the transition distance is not the distance in which thesteady-state climb path is achieved. Rather, it is the distance from thelift-off point to the location where the climb path slope intersects theground (Figure 6.1). However, it should be recognized that high-performance aircraft with high thrust-weight ratios will reach the screenheight before transition is actually completed.

The following steps can be used to facilitate calculation procedurefor determining st:

1. VT � 1.2Vs is assumed.2. Va � (1.2 � d)Vs is chosen, where d may be taken anywhere

between 0.05 to 0.15.3. Evaluate T and D at (Va � VT)/2.4. Calculate st from Eq. 6.25.5. Calculate the average transition �t from

T � Dsin � � (6.26)t W jet

P � DVa avgsin � � (6.27)t WVavg prop

6. Check the height ht reached when achieving the (assumed) air-speed Va from

h � s tan � (6.28)t t

Then:a. If ht � hscreen, the screen height has been passed during the

transition and st calculated in step 4 above is too large. Thus,a new and smaller Va must be assumed and the procedurerepeated until ht � hscreen.


b. If ht � hscreen, it is assumed that the transition is completedand the climb distance can be calculated for the resultingsteady climb phase to the screen height.

The climb distance is evaluated from the following (see Figure 6.1):

h hc cs � � (6.29)c tan � �c c

For small angles of climb, with L � W, �c can be written

T D T CD0� � � � � � kC (6.30)c L cW L W CL c

whence

hcs � (6.31)c T /W � C /C � kCD L L0 c c

Strictly speaking, the climb-lift coefficient should be evaluated at aclimb speed higher than Va since the aircraft will continuously pick upspeed. For practical purposes, however, evaluating at Va will pro-CL c

vide satisfactory results.Thus, summarizing, the basic take-off distance can be calculated

from Eqs. 6.11, 6.24, 6.25, and 6.31, or

S � s � s � s � s (6.32)T.O. g r t c

EXAMPLE 6.2

For Example 6.1, determine the transition distance and the total take-off distance.

Transition distance is calculated according to procedures given inSection 6.2.2.

1. VT � 1.2Vs � 1.2 � 136.7 � 164 ft/sec2. Assume d � 0.05, then Va � 1.25Vs � 1.25 � 136.7 � 170.9

ft/sec


3.

V � Va T � 167.5 ft/sec2

C 1.5L maxC � � � 0.96L 2 2(1.25) (1.25)2C � 0.02 � 0.05(0.96) � 0.066D

2D � 1/2(0.002377)(167.5) (1500)(0.066) � 3,301 lb

T � 2(5,000 � 3.28(167.5)) � 8901 lb

4.

2 2 2 2W V � V 50,000 170.9 � 164a Ts � � � 320 ftt 2g T � D 2 � 32.2 8,901 � 3,301

5.

T � D 8,901 � 3,301sin � � � � 0.112

W 50,000

� � 6.4�

6. ht � st tan � � 320 � tan 6.4� � 36 ft, which is less than thescreen height of 50 ft; thus, one can proceed to calculation ofthe climb phase distance.

The climb distance is evaluated as follows:

hss �c �c

T D 8,901 0.066� � � � � � 0.1093c W L 50,000 0.96

50s � � 458 ftc 0.1093

Finally, then the total take-off distance is

s � s � s � s � s � 2650 � 492 � 320 � 458 � 3,920 ftT.O. g r t c


6.2.4 Take-off Time*

The time to take off consists of the same individual segments as shownfor the ground run in Eq. 6.32. The duration of the ground run followsdirectly from Eq. 6.11 with V1 � 0 as

2 b 2aV � bT�1 �1t � tanh � tanh (6.33)� �2 2 2�b � 4ac �b � 4ac �b � 4ac

with the constants a, b, c given with Eq. 6.6. If the thrust is constant,then b � 0, and the above can be simplified to give

V2t � (6.34)g g(T /W � �)o

For the rotation it is assumed that tr � const. Calculation of the tran-sition phase starts with Eq. (6-2) as now, � � 0. Then, assuming thatT � D � const,

VaW dV W V � Va Tt � � � (6.35)tVg T� D g T � DT

The airspeed Va can be calculated by the procedure given in the lastsection.

The time during the climb-out phase is given by

sct � (6.36)c Va

where sc is given by Eq. 6.31. The total time to reach height is obtainedfrom

t � t � t � t � t (6.37)g r t c

6.2.5 Factors Influencing the Take-off

For the take-off portion, it remains now to evaluate the various effectsthat influence the take-off distance.

Runway Gradient The effect of runway gradient, �, is to modifythe effective thrust in Eqs. 6.7 and 6.8 by a factor of �W. Thus, thecoefficient c in Eq. 6.7 becomes


Toc � 2g � � � � (6.38)� �W

where � and � are used for downhill or uphill slope, respectively.

Engine Failure In multiengine design, one of the critical cases oc-curs when an engine fails at take-off. Then it is needed to know thedistance to complete take-off run to the screen height, to bring theaircraft to a stop, or to evaluate VCR for a given runway length.

The effect of engine failure can be accounted for by derating thethrust-weight ratio by a factor ƒ, which stands for the fraction of thrustavailable after one (or) more engine failures. Thus, wherever T /W orTo/W appears it is to be replaced by ƒ(T /W) after the engine failure.

Although an engine failure may occur anywhere during the take-off,for simplicity it is assumed that the failure will occur during the groundrun up to the lift-off at VT. This will require dividing the take-off por-tion into two sections and applying Eq. 6.11:

a. V1 � 0, V2 � Vf, during the starting portion, where Vf is the speedat which engine failure occurs.

b. V1 � Vf, V2 � VT, for the section up to the lift off if the take-offis to be continued. If take-off is aborted then the procedures forlanding run apply (see next section). For this portion of take-offT and D will be evaluated at an average velocity Vm � (Vf � VT)/2.

This concept can be extended to cover engine failure in any take-offphase.

Thrust Augmentation There are several ways for shortening theground run or achieving satisfactory take-off performance under unfa-vorable conditions. Ground run can be reduced by lifting off the run-way at a lower speed since the ground distance is directly proportionalto (note the group in Eq. 6.23). VT can be reduced if2V W/�SCT LT

can be increased, by means of flaps, slats, boundary layer control,CLT

and so on. A point of diminishing return is reached quickly, however,since flap deflection causes drag to increase, which may, in turn, in-crease the take-off distance. Thus, a careful choice and use of flapsdeflection is necessary during the take-off.

Thrust augmentation by various means (water injection to reduceeffective inlet temperature, after-burner, auxiliary jets, or rockets) isvery effective in reducing the ground run, as it is inversely proportional


to thrust; see Eq. 6.18. If thrust augmentation is used throughout theentire ground run, the calculations proceed in the same manner as inthe preceding sections, with the thrust given at the new augmentedlevel. Often the augmentation time is shorter than the ground run (es-pecially in case of rocket firing) and the calculations must be carriedout in several steps. Moreover, it is necessary then to determine whenthe augmenting unit should be fired.

The augmenting units usually produce constant thrust, which is ap-plied at a varying speed. It is most effective when the power produced,or work done, is maximum. The available work is

W � �T � V � �t (6.39)

and since �T and �t are fixed by the engine characteristics, the workdone increases with increasing velocity. Clearly, then, the thrust aug-mentation should be used during the latter part of ground run with theburnout to coincide with the lift-off.

Assuming that the unit is fired at the time a ground speed of V1 isreached, the calculations can be carried out as follows.

1. The ground distance, s1 can be evaluated from Eq. 6.11 or 6.17as outlined in Section 6.2.1 with velocity V1.

2. The ground distance s2 from V1 to VT is obtained from Eq. 6.11,with the augmented thrust-weight ratio included in

T To augc � 2g � � �� W W

3. The time spent during the augmentation between V1 and VT canbe evaluated from Eq. 6.12, with a and b determined in step 2above, or by use of the average value of the time integral

W V � VT 1t � � �2 g T � �W � D � �L

where D and L are obtained at Vaug and T is the total augmentedthrust.

4. The time t2, obtained in step 3, is now compared with the firingtime of the augmenting unit. If t2 � taug then a new and smaller


V1 must be chosen and the steps 1 through 3 repeated. If t2 � taug

then a larger V1 is used for the next iteration step. The iterationcan be repeated until t2 and taug agree within satisfactory limits.

Effect of Wind The effect of wind is to reduce the ground run dis-tance and time if the take-off is into a head wind. With take-off intodownwind, both distance and time are appreciably increased. To estab-lish the effect of wind it is necessary to calculate the distance traveledwith respect to ground, as the distance traveled through air is not thesame due to relative motion of air with respect to ground.

The ground speed V is then

V � V � V (6.40)a w

where

Va is airspeedVw is wind velocity [� head wind, � tail wind] (assumed to be

constant)

The ground distance can be obtained by first establishing the acceler-ation and then integrating.

Now for the head wind, the acceleration ag is

2dV dV dV dVa a aa � V � (V � V ) � � V (6.41)g a w wds ds 2ds ds

and

2dV Va wds � � dV (6.42)a2a ag g

Since ag �T � D � �(W � L)

W/g

one obtains

V VT TV dV dVs � � � V � (6.43)w

�V �Va aw wg g


and by use of Eq. 6.9

21 aV � bV � c b /a � 2VT T ws � ln �g 2 22a aV � bV � c �b � 4acw w

2aV � b �2aV � bT w�1 �1� tanh � tanh (6.44)� �2 2�b � 4ac �b � 4ac

where the top sign is used for the head wind, and the bottom sign fordownwind take-off.

NOTE

A convenient, but rather approximate, expression for the ground distance withthe wind can be obtained from

dVad � (V � V ) (6.45)s a w ag

which integrates into

2(V � V )ws � (6.46)2ag

For transition and climb portions, the effect of headwind may be ob-tained approximately by multiplying the distance obtained in still air (Sec-tion 6.2.2) by the factor

(V � V )f w (6.47)Vf

where the signs stand for

� headwind� tailwind

andV � VT aV � , for the transition phasef 2

V � V , for the climb phasef a


L1011

EXAMPLE 6.3

For Example 6.1, determine the take-off distance with 15 knot headwind. With head wind, the ground run is calculated from Eq. 6.44,the coefficients a and b remaining the same. Thus, Vw � 25.35 ft/sec, � 643 ft2/sec22Vw

5�10s � ln

2 � 2.02�5 2 �3�2.02 � 10 � (164) � 4.225 � 10 � 164 � 5.796

�5 �3�2.02 � 10 � 643 � 4.225 � 10 � 25 � 5.796

2 � 25.35 �1� 9500 � �0.5391 � tanh� � �0.022�5 �3�2.02 � 10 � 2 � 25.35 � 4.225 � 10 �0.022

� 5426 � 3492 � 1934 ft

The effect of head wind on transition distances can be estimated bydetermining

V � V ftT aV � � 167.5f 2 sec

and calculating

V � V 167.5 � 23.35f ws � � � 320 � 272 ftt V 167.5f

Climb phase distance is estimated by letting Vf � Va � 170.9 ft/secand from the last part of the example:


Transition

Ground run, sgr

Total landing distance

Airborne phase

hf

Flare, sGlide, s Float

hs

γ

g f

Figure 6.3 Total Landing Distance

170.9 � 25.35s � � 458 � 390 ft� �c 170.9

by summing up, the total take-off distance is found to be, with thehead wind,

s � s � s � s � s � 1934 � 492 � 272 � 390 � 3,088 ftT.O. g r t cw

6.3 LANDING

6.3.1 Landing Phases

Similarly to take-off operations, landing of a conventional aircraftshould be divided into separate phases, as indicated in Figure 6.3:

1. The final approach, or gliding flight, at a relatively steady speedand rate of descent.

2. The flare, or flattening out transition phase. Here the pilot at-tempts (ideally) to reduce the rate of sink to zero and the forward

6.3 LANDING 173

speed to a minimum, which must be larger than Vs. To assuresufficiently high landing speed, Federal Air Regulations requirethat the velocity at screen height must be at least 1.3 Vs and 1.15Vs at touchdown.

3. The floating phase, which is a result if the flare of zero rate ofdescent is achieved at the end of flare and the velocity must yetbe reduced. The float is often eliminated for aircraft with nose-wheel type of landing gear, which, due to a relatively low groundangle of attack, are able to make a touchdown at speeds wellabove the stall speed. In general, the float occurs when the aircraftbecomes subject to ground effect, which requires speed reductionfor touchdown as the lift is now somewhat increased.

4. The ground run. In many cases, the ground run should be cal-culated in two sections. First, right after touchdown, the free roll-ing distance needs to be calculated while the pilot gets the nosewheel on the ground and starts applying brakes and reverse thrust.Second, the relatively longer deceleration distance must be de-termined that is required for the aircraft to come to rest or toreach a speed sufficiently low to be able to turn off the runway.

In practice, similar to take-off, piloting skill and varying techniques,different aerodynamic characteristics, and other factors such as differ-ences of height and path angle all reduce the reliability of the calculatedlanding results. Inasmuch as take-off occurs under full power and atincreasing speed, it is more repeatable. Landing, however, takes placeunder partial power and on the back side of the power curve. Thus, itis inherently less stable and pilots tend to find safety in speed. Tenpilots, in the same aircraft under same conditions, will produce tendifferent landings. Thus, the calculated final landing distance is mul-tiplied by 1.67 to obtain a safe landing length. In case of wet runway,this distance is increased by another 15 percent.

Owing to the uncertainties just discussed, airborne phase of landingdoes not lend itself to a practical and reliable analysis. The groundphase, however, is quite significant for aircraft-airport compatibility(i.e., is there a sufficiently long runway to be able to land this particularaircraft?). Thus, more attention is devoted here to the landing run, andthe airborne phase is included later only as a matter of completeness.

6.3.2 Landing Run

The landing ground run analysis is identical to the take-off run, andEq. 6.9 can be used with b � 0, C � 0; see Eq. 6.6 and 6.8:


2�1 c � aVTs � ln (6.48)� �gr 22a c � aV2

Where VT is now the touchdown velocity. V2, in general, will be zero,and

Toc � g � � (6.49)� �W

�S�ga � (C � �C ) (6.50)D L2W

The essential differences are found in the coefficient c being possiblynegative due to reverse thrust, braking parachute, or due to the fact that� is now much larger than for take-off when brakes are used for aportion of the landing run. When reverse thrust is applied, then thethrust term To/W becomes negative in the constant c. Since it is un-likely that a pilot will apply brakes at the beginning of the ground run,for fear of getting into skid, the ground run may have to be dividedinto two or more portions to account for initial rolling friction, andthen for the full brake section (usually at speeds below 80 knots). Whenthe aircraft has been brought to rest, the landing run is determined by

1 a 2s � ln 1 � V (6.51)� �gr T2a c

where CD (especially must be evaluated with all the aerodynamicC )D0

braking devices and landing gear effects taken into account. isCLT

obtained from touchdown speed VT, which is between 1.15 and 1.2 Vs.To account for the free rolling distance during which the pilot brings

the nose wheel to the ground, one estimates similarly to take-off dis-tance:

s � V �t (6.52)r T

where �t is taken to be about 2 to 3 seconds.The total calculated landing distance is now obtained from

s � s � s � s � s (6.53)land gl f r gr

which is given by Eqs. 6.57, 6.62, 6.51, and 6.52, or equivalent. Aspointed out, the safe landing distance on a dry runway is obtained from

6.3 LANDING 175

s � 1.67s (6.54)l

NOTE

Some newer jets have a system called Autobrake, where the pilot can enterthe desired deceleration during the ground run (after the nose wheel hasbeen brought to the ground at a velocity V1). Since the sensors establish aconstant deceleration, call it D1, the ground run can easily be determinedfrom Eq. 6.4 as follows. The aircraft acceleration (a � F /m) is

T � D � �(W � L)a � � D (6.55)1W /g

Thus, Eq. 6.4 can be integrated to give

2 2V dV V1s � � � (6.56)r1 D 2D1 1

Thus, if an aircraft makes contact with the ground at 160 ft /sec and theautobrake is set at 13 ft /sec2, the expected ground distance is

2160s � � 985 ftgr 2 � 13

6.3.3 The Approach Distance*

The approach, or glide, distance can be estimated by assuming astraight glide path which gives

h � hs fs � (6.57)gl �

where � is a small angle and hs is the screen height. Then

D � T 1 Ttan � � � � (6.58)

L L /D L

where L /D is usually evaluated at a mean speed between 1.3 Vs, and1.15 Vs, or simply assumed to be 8. For a first approximation, the thrustterm is often neglected; however, a better procedure is to use the idlethrust value. The glide velocity is assumed to be constant at 1.3 Vs.


6.3.4 The Flare Distance*

Estimation of the flare distance is the least reliable of the landing dis-tance calculation procedures, as it requires a number of assumptionsconcerning the shape of the path, velocities, start and end points, andso on.

Thus, there are many methods and variations that define the arc usedto calculate the flare distance. They all give similar results based on avariety of reasonable physical assumptions that are judiciously tem-pered by experimental observations. In what follows, two approachesare given. The first one, Method A, is included as a typical exampleto indicate the basic methodology. A thoughtful observer has no diffi-culty raising more questions than are answered here. Method B is basedon the energy conservation and seems to yield results comparable toother more elaborate methods. Its main advantages lie in simplicity andthe individual number of assumptions involved.

Method A The following steps and assumptions are usually involved:

1. The flare path is a circular arc of such a radius that the normalacceleration is 0.1 g.

2. The flare is carried out at a constant speed:

1.3V � 1.15Vs sV � � 1.23Vf s2

Thus, the radius of the circle is (see Figure 6.3)

2 2 2(1.23) V Vs 2r � � 15.1f 0.1g g

Furthermore, it is assumed that the flight angle is sufficiently small thatthe horizontal distance is equal to the arc length. Then, if the flareheight hf is known, or assumed:

2V �ss � 15.1f g

where � can be evaluated from the following (see Figure 6.4):

6.3 LANDING 177

hf

Flare, s

γ

f

r = 15.1Vs

2

gγ

Figure 6.4 Transition Geometry

215.1V g � hs fcos � � 215.1V gs

Otherwise, it may be assumed that � is determined by L /D (� 8).If the flare starts at a height above the screen height hs, then

215.1V h � hs s fs � � � (6.59)f g tan �

Method B Considering now the uncertainties involved in evaluatingthe glide and flare distances, one might as well calculate the entireairborne distance by considering the energy equation derived in Chapter2. For simplicity, the energy equation can be applied between the screenheight and the touchdown point. Thus, one obtains with Eq. 2.15

We � We � D � T (6.60)s T S Sa a

Upon substituting kinetic and potential energies at screen height andat landing (h � 0), one obtains:

2 2V Vsc TW h � � � s (D � T)� �s a2g 2g

or


2 2 2 2(1.3) V (1.15) Vs sW h � � � s (D � T) (6.61)� �s a2g 2g

where

s is the airborne distancea

e is the specific energy at screen heights

e is the specific energy at touchdown, (h � 0)T

V is the velocity at screen heightsc

V is the touchdown velocityT

V is the stall speeds

Then the airborne distance is obtained as

2W 0.3675V ss � h � (6.62)� �a sD � T 2g

where D must be evaluated at an average speed Vf corresponding to alift that yields an L /D � 8. Often, thrust is assumed to be zero. Amuch better approach is to use engine idle thrust value, which amountsto about 6 to 9 percent of rated thrust.

6.4 ACCELERATING FLIGHT*

In level flight, an aircraft accelerates or decelerates every time its thrust(or speed) is changed. Since most of the flight takes place at relativelyconstant velocity, the problems involving accelerations are of interestnear beginning or end of the flight or arise due to special maneuvers—for example, in combat. In addition, many aircraft are equipped withair brakes to provide speed control in dive or combat, or to assist inflight path control on landing approach. The essential problems arefinding the time and distance to accomplish the speed change and toevaluate the effectiveness of the air brakes.

Consider an aircraft in steady, level flight with velocity V1 at analtitude h (Figure 6.5).

At this flight condition, the drag is exactly balanced by the thrust:

6.4 ACCELERATING FLIGHT* 179

Tamaxat h

Vmax

D

T , Da

VsT2 at h

T1 at h

Excess energyavailable

Deficit in energyavailable

V1

V2

Tidle at h

V

Figure 6.5 Energy for Acceleration

D � T � T1 1 maxavail

An increase in the thrust, at the given altitude h, accelerates the aircraftfrom V1 up to any desired velocity V2 Vmax depending on the levelof thrust T2. Excess energy Ta � D will be available to accelerate theaircraft until the maximum velocity Vmax is reached where the availablethrust exactly balances the drag force. Acceleration beyond Vmax maybe possible if it is accompanied by a change in altitude. Conversely, ifthe aircraft is in steady level flight at V2 and the thrust is reduced toT1, the excess of drag energy (deficit in energy available) will serve todecelerate the aircraft to V1. Suppose now that the thrust is further cutto its idle value Ti, then the aircraft will continue decelerating to reachstall velocity Vs if constant altitude can be maintained according to L� W, or CLV2 � constant. In general, it cannot, as shown in Figure6.5.

The general solution of accelerating flight requires integration of thedynamic Eqs. 2.9 through 2.13 and requires a complete description ofthe flight path, as discussed in Chapter 1. However, if the problem isapproached from the point performance point of view, or the equationsof motion are integrated by the use of simplifying assumptions, somepractical solutions can be obtained.

Eqs. 2.9 through 2.12 reduce, for level flight or for flight with smallpath angle change, to


dx� V (6.63)

dt

1 dV T� D� (6.64)

g dt W

L � W (6.65)

It is expedient now to change the independent variable to velocityV, as carried out in Section 6.2.1, and these equations become

W V dV V dVdx � � (6.66)

g T � D a

W dV dVdt � � (6.67)

g T � D a

Eq. 6.65 reduces to stating that, for level flight, CLV2 � constant, whichimplies that the angle of attack must be changed whenever the velocityis changed.

In point performance approach, the following numerical integrationprocedure can be used (for either acceleration or deceleration):

1. Initial and final velocities are assumed.2. At selected velocity intervals calculate

CL �const

2VCD � ƒ(CL), from a given drag polarT, from engine data at selected V or h (usually assumed to beconstant)

3. Evaluate, at selected V

1 w V� and

a (T � D)g a

4. Plot 1/a and V /a vs. V5. The areas under the curves in Figure 6.6 give the time or distance

in accelerated flight.

At small flight path angles, the above process will provide usefulapproximations to climbing or gliding flight, provided a consistentaltitude/velocity program is chosen.


Area = Time todecelerate from

V to V

V V1

21

2

1/a

Area = Distanceto accelerate from

V to V1 2

1 2V V

V/a

Figure 6.6 Time and Distance Calculations

The effect of speed brakes is evaluated by including the additionaldrag due to speed brakes in calculating the drag coefficient CD.

EXAMPLE 6.4

An aircraft weighs 105,000 N, has a wing area of 75 m2 and istraveling at 100 m/s at 6100 m altitude. Estimate the time to accel-erate this aircraft to a speed of 250 m/sec if its thrust is increasedto a constant value of 40,000 N.

2C � 0.018 � 0.055CD L

The calculations are carried out for initial and final velocities of100 and 250 m/s, respectively, at 50 m/s intervals. The lift coeffi-cient is evaluated from

2W 2 � 105,000 4,244C � � �L 2 2 2�SV 1.2256 � 0.5383 � 75 � V V

The drag is calculated by D � W(CL /CD), whence

T � D 40,000 � D 40,000 � Da � � �

W/g 105,000/9.807 10,707

Calculations at 50 m/s intervals produce Table 6.2.


TABLE 6.2 Example 6.4 Data

V 100 150 200 250CL 0.424 0.189 0.106 0.068CD 0.0279 0.02 0.0186 0.0183D 6909 11,091 18,442 28,187a 3.09 2.7 2.01 1.11/a 0.324 0.37 0.497 0.906

A straightforward numerical integration gives the result that ittakes about 75 seconds to accelerate from 100 to 250 m/sec. (Seealso Problem 6.9).

Eqs 6.66 and 6.67 can be integrated directly to give time and distanceif it is assumed that

1. The thrust T is independent of velocity (is essentially assumed tobe constant).

2. The simple drag polar

2C � C � kCD D L0

is applicable.

Integration is facilitated by introducing the following nondimensionalvariables:

xgx � 2V ED mmin

tgt � 2V ED mmin

VV �

VDmin

TT � EmW

Eqs. 6.66 and 6.67 become


32V dVdx � � (6.68)

4 2V � 2TV � 122V dV

dt � � (6.69)4 2V � 2TV � 1

Integration now leads to

2V E 1D mmin 4 2x � � ln (V � TV � 1) � TQ � constant (6.70)� �g 2

where

2 2V � T � �T � 11 2Q � ln , if T � 1 � 02 2 22�T � 1 V � T � �T � 1

21 V � T�1 2Q � tan , if T � 1 � 02 22�1 � T �1 � T

1 2Q � � , if T � 1 � 02V � T

and the expression for time becomes

V E V � V V � VD mmin 2 1t � V ln � V ln� �2 12 2g(V � V ) V � V V � V2 1 2 1

2� constant, if T � 1 � 0 (6.71)

where

2V � V � �T � �T � 11 max

2V � V � �T � �T � 12 min

and


2V E sin � V � 1 � 2V cos �D mmint � � ln� 22g V � 1 � 2V cos �2�1 � T

cos � V � cos � V � cos ��1 �1� tan � tan� ��sin � sin �2�1 � t2� constant, if T � 1 � 0 (6.72)

where

2�1 � T1�1� � tan .

2 T

It should be pointed out that the tan�1 terms should not be added,as is often a customary procedure. There is a hidden singularity at

� 1.V

EXAMPLE 6.5

Calculate the distance to decelerate for an aircraft, which is decel-erating at sea level from initial velocity Vi � 600 ft/sec to finalvelocity Vf � 300 ft/sec at idle thrust of 250 lb. The aircraft weighs18,000 lb and has a clean drag polar for a wing area of 300 ft2.

2C � 0.014 � 0.06CD L

Upon calculating

1E � � 17.25m 2�0.014 � 0.06

T 250 � 17.25T � E � � 0.24mW 18,000

and since

2T � 1 � 0

one finds the proper form of Eq. 6.70 to be

PROBLEMS 185

2 4 2V E 1 V � 2TV � 1 TD mmin i ix � ln �� 4 2 2g 2 V � 2TV � 1 �1 � Tf f

2 2V � T V � Ti f�1 �1tan � tan� ��2 2�1 � T �1 � T

Evaluating now

1 / 4 1 / 42W k 2 � 18,000 0.06V � �� Dmin �S C 0.002377 � 300 0.014D0

ft� 323

sec

V 600iV � � � 1.86i V 323Dmin

V 300fV � � � 0.929f V 323Dmin

The solution for the deceleration distance is found to be

2(323) � 17.25 1 11.31 0.24x � ln ��32.2 2 1.33 0.971

�1 �1(tan 3.316 � tan 0.641) � 69,589 ft�

PROBLEMS

6.1 For Example 6.1, evaluate ground run from Eq. 6.22 with 0.8VT.Ans: 2,285 ft.

6.2 For Example 6.1, evaluate ground run when the thrust reversalis used from 0.8VL to 60 mph. Assume that the thrust reversesproduce 50 percent of sea level rated thrust, and that lift anddrag coefficients during the landing run are CL � 0.2 and CD �0.1. Assume that no brakes are applied and that the idle thrust


is 5 percent of the rated sea level value.Ans: 8,570 ft.

6.3 A twin jet aircraft is taking off from a concrete runway (� �0.04). The take-off polar is

2C � 0.075 � 0.04CD L

with

2S � 54.8 mAR � 6W � 266,880 NC � 2.7L max

T/engine � 80,064 N � constT � 0.07Tidle

Calculate:a. The ground runb. The ground run with one engine failing at 100 mph.c. The head wind required to reduce the ground run in b to that

obtained in a—that is, what head wind is required to com-pensate for the loss of one engine?

d. Time for a.e. Time for b.

Ans: a. 400 m; b. 672 m; c. 24.4 m/s; d. 11.8 s; e. 17 s.

6.4 For Example 6.1, calculate the ground run with a 15 knot tailwind.Ans: 2,994 ft

6.5 Evaluate the effect of W /S, and T /W on ground run distanceC ,LT

for two aircraft, at W /S � 50 and 100. Assuming that the take-off polar for each is CD � 0.03 � and that � � 0.02,20.05C ,L

plot the ground run distance vs. with T /W and W /S as par-CLT

ameters. What conclusions can be drawn regarding increase ofC ?LT

6.6 A transport aircraft weighs 290,000 lbs and is equipped withfour JT3D jet engines. It has the following characteristics:

PROBLEMS 187

C � 2.2L max

AR � 7.3462S � 2,892 ft

C � 0.021D0

Calculatea. Ground run, standard day, in Los Angelesb. Ground run, 100�F, in Los Angelesc. Ground run, standard day, Mexico City (7,400 ft elevation)

6.7 For Example 6.5, calculate the time to decelerate from 600 ft/sec to 300 ft/sec.Ans: 2.7 min.

6.8 For Example 6.5, calculate the distance and time to deceleratefrom 600 ft/sec to 300 ft/sec if speed brakes are activated thathave a drag coefficient of 1.0, based on the brakes area of 20ft2.Ans: 2.5 mi, 32 s.

6.9 Verify the answer obtained in Example 6.4 by using the appro-priate integrated result for time.

6.10 A twin-engine cargo plane is being loaded for takeoff. The air-craft has the following characteristics:

2 2C � .019 � .048C , S � 880 ftD Lclean

2C � 2.1, C � .03 � .06CLTO DTO L

W � 59,000 lb (incl fuel and crew).empty

Sea-level rated thrust is 10,000 lb/engine. Ignore �.

Airport runway is rather short, translating into a ground run ofabout 2,000 ft, before rotation. It is a hot day with the air densitybeing only 92 percent of that of the standard day, and the thrustavailable has thus dropped to 92 percent of the rated value.

The charts show that the maximum weight allowable is about10,000 lb cargo. The crew chief insists that 13,000 lb is fine.Who is right?Ans: The charts are right.


6.11 A large transport aircraft has the following characteristics:

W � 605,000 lbTO2S � 3,650 ft

3 engines, 61,500 lb each, Tmax2C � .016 � .042C , C � 2.4D L L max

Determine:a. Lift-off velocityb. The take-off rollc. The time to lift off

6.12 An A7E has just landed. It takes 3 seconds to bring the nosegear down and to stabilize for the ground run. It has the follow-ing characteristics:

T � 11,000 lba2C � .015 � .1055CD L

C � 1.6Lmclean

C � 2.5Lmfl,slats2W � 25,000 lb, S � 375 ft

T � .09Tidle a

Calculate the total distance covered on the ground if:a. No brakes are applied at all.b. Brakes are applied after the end of first three seconds (use �

� .04).Ans: about 190,000 ft; 2,100 ft.

6.13 The following is given for a twin-engine F18 aircraft:

2C � .0245 � .13C , clean; add �C � .075 for landing gearD L D0

2S � 400 ft , W � 40,000 lbC � 2.8; C � .2L Lmax TO

T � 9,300 lb/engine

Calculate the take-off ground run (� � .03) for:a. The data given aboveb. Take-off with full afterburners, 13,000 lb/engine

Ans: Using Eq. 6.13, 1,642 ft; 1,130 ft

189

7Maneuvering Flight

MIRAGE 2000

7.1 INTRODUCTION

In a steady, unaccelerated flight all the forces acting on an aircraft arein equilibrium. For about 90 percent of flight, the simple equilibriumcondition CLV2 � constant holds, which governs simple and minoradjustments to the flight path. In point performance approach, this maybe assumed even for modest rectilinear acceleration. This chapter con-siders flight where the path direction changes sufficiently rapidly sothat the above condition may not be valid.


190 MANEUVERING FLIGHT

In order to accelerate or curve the flight path in any plane—that is,to pull up or to turn—the force equilibrium acting on the aircraft mustbecome unbalanced. This can be accomplished essentially in threeways: change lift, change velocity, or bank the aircraft. The lift coef-ficient will be changed so rapidly that the velocity cannot keep up withthe change and the equilibrium condition (CLV2 � const) will not besatisfied. The unbalanced force then acts to create an acceleration nor-mal to the flight path. If a velocity change is the cause, it must beproduced sufficiently rapidly so that CL cannot compensate. The resultis again a normal acceleration. In vertical plane (pull-ups, push-overs)rapid lift and velocity changes are the principal mechanisms for causingaccelerations.

Accelerations out of the vertical plane (not necessarily normal to it,but normal to the flight path) can be caused by rolling the aircraft intoa turn. The lift vector is then rotated out of the vertical plane by rollingthe aircraft (banking the wings). This action produces a lift componentto act as an accelerating force for a turn and leaves a portion of liftforce to balance the weight vector.

If the vertical component of the lift force, usually by getting someassistance from a vertical component of thrust, does not exactly balancethe weight force, then climb or descent will accompany the turningmotion. The analysis of resulting three-dimensional turning perform-ance is rather complicated, and the governing set of differential equa-tions admits only a few solutions, which provide scant insight into thephenomena at hand.

Thus, in keeping with the general point performance approach takenabove, the motion will be restricted to either vertical or horizontal planeand analytic solutions will be considered at distinct points of the flightpath. The resulting equations should provide adequate tools for ana-lyzing the essential turning performance of an aircraft.

7.2 TURNS IN VERTICAL PLANE: PULL-UPS ORPUSH-OVERS

To determine what turn rate may be obtained, or more important, whataccelerations could be produced in a pull-up maneuver, consider Figure7.1, where the lift generated now can be expressed as

L � nW (7.1)

7.2 TURNS IN VERTICAL PLANE: PULL-UPS OR PUSH-OVERS 191

T

L

V

WD

εγ

h

x

Figure 7.1 Pull-up Maneuver

The load factor n, usually expressed as g’s of acceleration, is definedby Eq. 7.1 and represents the excess of lift generated over the weightto produce the required maneuver.

Since the flight path is in the vertical plane, Eqs. 2.5 and 2.6 applydirectly. As the maneuver is performed very rapidly, the flight speedremains instantaneously unchanged and those equations uncouple.Thus, only Eq. 2.6 needs to be considered.

L � T sin � � mg cos � � mV�

which becomes with L � nW and mg � W

g(n � cos �) Tg sin �� (7.2)

V WV

and is usually simplified to give

g(n � 1)� � (7.3)

V

Eq. 7.3 gives the rotation rate for a modest thrust-to-weight ratioaircraft at the beginning of pullout from level flight. Since T /W ratiofor high-performance aircraft approaches, or even exceeds, unity, Eq.7.2 shows that thrust contribution may be significant and should notalways be ignored.

Centrifugal acceleration may now be determined easily by recallingthat a � V, whence�


Tg sin �a � g(n � cos �) � � g(n � 1) (7.4)

W

Eq. 7.4 indicates the acceleration on the pilot and aircraft in an n ‘‘g’’pull-out. It should be recognized that acceleration has the same effecton the pilot and the structure: to change the direction of the flight path.The pilot, however, senses the load factor only in excess of unity. Atn � 1 (level flight) the pilot feels no load in excess to that felt whensitting at home on the couch. At n � 1, the wings already feel the load,weight W, of the aircraft, and at n � 2 must be able to generate liftcapable of supporting 2W. Thus, as the load factor increases, so doesthe load acting on the aircraft. But same load factors do not necessarilyproduce same loads as the weight of the aircraft changes during flightand its missions. A load factor of 5 may be just within safe operatingload limits of an aircraft carrying only a part load of fuel. When fullyfueled and with external stores attached, the total weight could increaseby 50 percent. Now a load factor of 5 may require wing loads thatcould have disastrous structural consequences.

F14

7.3 V–n DIAGRAM

The operation of an aircraft is subject to rather specific operatingstrength limitations, usually presented on what is known as the V–n orV–g diagram. The basic static strength requirement is stated as thepositive limit load factor nl. This means that an aircraft, in its designor mission configuration, is expected to meet a peak load correspondingto limit load factor nl and to withstand this load without permanentdeformation of the structure. To allow for situations where mission oremergency situations may lead to exceeding the limit load factor, theultimate load factor 1.5 nl is provided. At this load, the primary load-carrying structure may exhibit some permanent deformations but noactual failures should take place. Some typical values for the limit loadfactors are listed below. These values are only typical, as they maychange with specific mission requirements.

7.3 V–n DIAGRAM 193

Type of Aircraft Limit Load Factor

Fighter, Trainer, Attack 7–9Transport, Patrol 3Passenger 3–4

The above load factors provide only a bound for loads that must beobserved carefully during aircraft operation and may easily be exceededif due regard is not given to the operating limits that must be estab-lished from additional considerations.

The loads on an aircraft are produced in two ways. The intentionalloads are due to maneuvers and the unintentional ones are due to airdisturbances—gusts. In attack and fighter aircraft design and perform-ance analysis, the maneuver loads are more important. In passengeraircraft operations, where operating load factors seldom exceed 1.5, thegust loads are more significant. Although this chapter is essentiallyconcerned with maneuvers, the gust-originated load factors will also beconsidered due to their impact on high-speed flight at lower altitudes.

The load factor can be related to velocity by dividing the instanta-neous values of lift and weight. Of importance is the maximum loadfactor, which is obtained from maximum lift at any airspeed:

1 2L � C �V S (7.5)max L max2

At stall speed, maximum lift is equal to the weight:

1 2W � C �V S (7.6)L smax2

From definition of the load factor, then

2L Vmaxn � � (7.7)� �max W Vs

Eq. 7.7 shows that the load factor varies as the square of the aircraftvelocity (ratio of the stall speed). Thus, if the aircraft is flying at thestall speed, load factor is unity. At twice the stall speed, a load factorof 4 can be obtained if the angle of attack is rapidly increased toproduce maximum lift. At four times the stall speed, a load factor of16 will result, which exceeds the structural capabilities of all aircraft,


-5

-4

-3

-1

-2

0

2

1

3

5

4

6

8

7

9

10

100 200 300 600400 500

Positive limit load factor, n

Positive ultimate load factor, 1.5 n

Negative limit load factor

Negative ultimate load factor

-3

-4.5

6

9

V, (mph)

n

A

4.63

5�10

V

2.70�

10V

–62

–52

l

l

1+0.00779V

1 – 0.00779V

S+

S-

B

Vc Vmax

Figure 7.2 V–n Diagram

with the possible exception of special-purpose research planes. It fol-lows, then, that high-speed aircraft, which fly at several times the stallspeed, are capable of g loads that will exceed all structural limits. Toprevent exceeding these limits, safe flight boundaries are presented ona V–n diagram, Figure 7.2.

The horizontal boundaries represent the limit load factors as definedfor this particular aircraft. Also shown are the ultimate load factors.


The curved boundaries S–A and S–B are obtained by means of Eq.7.7; VS � 100 mph where n � 1. S–A and S–B can also be obtainedfrom Eq. 7.5, as follows:

Consider an aircraft with the following sea level characteristics.

2S � 650 ftW � 20,000 lb�C � 1.2L max

�C � 0.7L max

a � 0.08 per deg (both positive and negative lift)V � 570 mph, at sea levelmax

�n � 6lmax

�n � 3lmax

V � 100 mph (�C )s L max

V � 131 mph (�C )s L max

Then the load factor can be expressed as

1 S 2n � C �VL max2 W

1 650 2� � 0.002377 � 1.2 � V� �2 20,000�5 2� 4.635 � 10 V for positive lift

�5 2� �2.704 � 10 V for negative lift (7.8)

When evaluating n from Eq. 7.8, care should be exercised in eval-uating . It may not be sufficient to use the static value asC CL Lmax max

presented in standard tables or charts. First, under dynamic loadingconditions when a wing is rotated rapidly to a high angle of attackhigher maximum lift conditions than for the static loading may beobtained for brief periods. Thus, higher load factors can be developedthan predicted by static stall speed. Second, the lift-generating capa-bility of the tail of some aircraft may be of sufficiently high magnitude(10 to 20 percent) that it should be included in evaluating forCL max

calculating the load factor n.Eq. 7.8 shows that as the altitude and weight change, a new diagram

must be constructed. In addition, and S may also change depend-CL max

ing on configuration, external stores, and so on. Eq. 7.7 remains valid


�

�Resultant velocity

Static airspeed

Gust velocity of KU

static

�

Figure 7.3 Effect of Gust

in any case, since the altitude and weight changes can be representedas a change in Vs, which then only requires shifting the velocity scale.

The upper, positive, n scale implies that a pull-up maneuver is per-formed. Some aspects of turning maneuvers will later be superimposedon that portion of the diagram. The lower, negative scale is used fornegative accelerations in a push-over maneuver. The n values aresmaller because the values of negative are, in general, smaller (inCL max

this case �0.7) than the normal positive . Since a pilot’s abilityCL max

to withstand negative max g’s is more limited than in the case of pos-itive values, aircraft normally are designed for smaller values of neg-ative load factors, as shown in Figure 7.2. The vertical limit definingthe maximum speed is calculated from the terminal dive speed or isdetermined by compressibility considerations or from buffeting bound-ary.

The superimposed dashed lines are determined by calculating theaircraft response from anticipated gust loads. Gusts are horizontal andvertical velocity disturbances found in the atmosphere. In general, gustsare rarely found in stratosphere but occur frequently in the lower at-mosphere in the vicinity of thunderstorms, mountains, shorelines, andin areas where thermal gradients exist. Crossing the wake of anotheraircraft also has the effect of a gust.

A horizontal gust causes a change in dynamic pressure over thewing, which does result in change of lift, but the load factors producedare relatively small and they can be ignored. A horizontal gust doesnot change the angle of attack. A vertical gust causes a change in angleof attack, as shown in Figure 7.3, and a change in load factor.

A gust consists of a wave-shaped velocity variation with low veloc-ities at extremes and a maximum near the center. For analytical pur-poses, it is customary to assume that a gust behaves as a sharp-edgedimpulse and an equivalent velocity value of KU � 30 ft/sec is assumedto represent the strongest gust likely to be encountered in a normaloperation. K is the airplane response factor (0.4–0.7) and U is themaximum gust velocity of 50 to 75 ft/sec. Thus, a normal operationcombination of KU � 30 is assumed.


The effect on load factor can now be calculated by evaluating thechanges in angle of attack and in the lift coefficient. For small ��

KU�� (7.9)

V

From definition of the lift curve slope

C �CL La � � (7.10)� ��

KU�C � a�� a (7.11)L V

The change in lift is obtained as

1 KU 2�L � �aV S (7.12)2 V

and the change in load factor becomes

�L � KUV�n � � a (7.13)

W 2 W/S

Since the aircraft is flying, before encountering the gust, at n � 1,the load factor can be written as

�aKUVn � 1 � (7.14)

2W/S

For negative gusts the second term becomes negative as KU � 0. Sim-ilarly to maneuver load induced load factors it is seen from Eq. 7.13that altitude, velocity, and the physical characteristics of the aircraftdetermine the gust load factors. At higher altitudes the load factor de-creases rapidly as the density is decreased. This effect is offset some-what due to higher airspeeds required to sustain flight at high altitudes.But, as already pointed out, gusts occur very infrequently at high al-titudes.

The effect of airspeed on gust load factor is important for flightperformance and operations. Since the gust load factor increases lin-early with the velocity the effect is more pronounced at higher speeds.Thus, an aircraft operating at high speeds is expected to avoid turbulent


regions or to reduce airspeed. Otherwise, structural damage may result.For the example aircraft in Figure 7.2, the gust load factors can bedetermined from Eq. 7.14. One obtains

0.002377 � 0.08 � 57.3 � 30 � 1.467 � Vn � 1 �

2 � 20,000/650

� 1 � 0.00779V V mph, positive lift, sea level

� 1 � 0.00779V V mph, negative lift, sea level

The results are shown in Fig. 7.2 as 30 fps gust lines. For thisparticular aircraft at sea level, the gust produced load factors are ex-pected to be less than the limit load factor of 6. The same aircraft withan improved powerplant and an increased Vmax � 750 mph should notbe operated at a speed higher than 642 mph when turbulence is antic-ipated. Higher speeds will produce a gust load factor higher than 6,which points to a probability of structural damage. At 20,000 ft altitude(� � 0.5328), this aircraft can increase its speed to 1,204 mph and stillstay within safe load-factor limits.

The lift curve slope, a, is another factor that indicates aircraft re-sponse to gust loads. A low-aspect-ratio, swept-wing aircraft has a lowlift curve slope and will experience lower gust load factors. An aircraftwith a high lift curve slope, and straight, high-aspect-ratio wings,would be more sensitive to gusts.

At the first glance it seems that an aircraft with higher wing loading,W/S, will be less sensitive to gusts. This is quite true when comparingtwo different aircraft, but can be somewhat misleading when one par-ticular airplane is considered at two different weights. At low weightsthe gust-induced accelerations are higher, and at full loads the reverseis true. For given atmospheric conditions same lift acts on differentmasses and produces different accelerations as sensed by the pilot. Theaircraft, however, is subjected to same loads.

The V–n diagram in Figure 7.2 shows the (safe) flight boundariesin level flight or in vertical plane maneuvers. The boundaries have beenestablished from maximum lift capabilities or from structural strengthlimitations. Additional boundaries may be imposed by thrust limita-tions. The ceiling of an aircraft, which is thrust limited, has alreadybeen discussed in Chapter 3. Thrust-limited turns will be considered inthe next section. From the point of view of aircraft operations, the lift-related boundaries are the most significant ones and will be consideredagain briefly below.

7.4 TURNING FLIGHT IN HORIZONTAL PLANE 199

The lines of maximum lift capability indicate that an aircraft cannotproduce higher steady-state load factors aerodynamically and is con-fined to fly below this line. This is true at both positive and negativelift coefficients except for the flight speeds at corresponding load fac-tors, which are higher at negative lift coefficients due to a smallernegative maximum lift coefficient. At increasing flight speeds, evenbelow the lift limit line, an aircraft can generate very large load factorsthat can exceed the limit load factor or even the ultimate load factor.If flight speeds are restricted to values below point A (intersection ofthe positive limit load factor and the line of maximum positive lift)then the aerodynamically generated load factors remain in the saferange below the limit load factor. At speeds greater than point A theaircraft is capable of producing positive lift and load factors that cancause damage to its structure. The same argument holds for negativelift and the corresponding point B.

The velocity corresponding to point A is the minimum airspeed atwhich the limit load factor can be produced aerodynamically. It iscalled maneuver speed or the corner speed, as it will give minimumturn radius and is the maximum velocity for alleviating stall due togust. Combined aerodynamic and predictable gust loads cannot damageaircraft structure if the aircraft is flying below the corner speed. Thus,the corner speed is a useful reference point since below this speed onedoes not anticipate damaging flight loads. At speeds higher than thecorner speed careful attention must be paid to the airspeed and the gload to be attempted. The corner speed is evaluated from

V � V �n (7.15)c s l

and fully reflects an aircraft’s physical properties, aerodynamic liftingcapability, and the altitude at which it is flying.

Corsair and Stuka

7.4 TURNING FLIGHT IN HORIZONTAL PLANE

For normal turning flight the aircraft is rolled to produce an unbalancein static equilibrium by rotating the lift vector out of the vertical plane.


Figure 7.4 Turning Flight

The resulting force unbalance and acceleration then cause the aircraftto turn. A turn is primarily described by turn rate and turn radius, whichare controlled by aircraft lifting, structural and thrust characteristics.

In a steady, coordinated turn the inclined lift vector produces a hor-izontal force component to equal the centrifugal force and a verticalcomponent to balance the weight of the aircraft, Figure 7.4; W � Lcos .

The bank angle is such that the above forces are exactly balancedand no sideways motion (yaw, sideslip) occurs. If a side force is gen-erated, mainly for path correction, it can cause lateral acceleration lead-ing to a turn at zero bank angle. This is called a flat turn. For acoordinated level turn, a simple force balance permits evaluation of theload factor from Figure 7.4 as

L 1n � � (7.16)

W cos

which shows that each bank angle produces a specific load factor n.For example, a bank angle of 60 requires a load factor of 2 for asteady coordinated turn and therefore a corresponding lift of L � 2W.If a different lift is generated at the same bank angle (through changeof T and V), then acceleration components would exist (in both hori-zontal and vertical planes) and the turn would not be steady. As alreadypointed out, there should not be any aerodynamic forces in the span-wise direction and the aircraft should be pointed along the velocityvector. Strictly speaking, no yaw is allowed but the aircraft may bepitched in the lift vector plane (plane of aircraft symmetry).


Plan

e of a

/c sy

mm

etry

Flig

ht p

ath

L

D

W

Vertical plane

Horizontal plane

V

T

�

�

Figure 7.5 Turn in a Horizontal Plane

To obtain expressions for turning velocity, radius, bank angle, andtime, and to better define the coordinated turn, consider the trajectoryand the forces acting on an aircraft when turning in a horizontal plane,Figure 7.5.

As shown in the Figure 7.5, T, L, D, and V are all acting in the planeof symmetry which is inclined to the vertical at the bank angle . Thethrust T may be inclined at angle � with respect to the horizontal plane.Weight W is in the vertical plane. While the aircraft is turning the flightpath and the velocity vector remain in the horizontal plane and planeof symmetry and are oriented at angle from a (reference) verticalplane.

The governing equations are easily obtained by referring to Figure2.1 and to the horizontal plane and plane of symmetry in Figure 7.5.


The equation tangent to the flight path is

˙T cos � � D � mV, (7.17)

In horizontal plane,

2V˙(T sin � � L) sin � mV� � m (7.18)R

In the vertical plane,

(T sin � � L) cos � mg (7.19)

Some basic results can now be obtained without further assumptions.Dividing Eq. 7.18 by Eq. 7.19 one gets

2Vtan � (7.20)

gR

which shows that the bank angle is proportional to the velocity squaredand inversely proportional to the radius of turn. The turn rate may beexpressed as

V� � (7.21)

R

which becomes with Eq. 7.20, upon eliminating R:

g� � tan (7.22)

V

Since the bank angle was obtained as a function of the load factor,the turn rate and radius of turn may also be expressed as a function ofthe load factor. Thus, from

1n �

cos

and


12sin � �1 � cos � 1 � 2� n

one obtains

1 2tan � n 1 � � �n � 12� n

Then, Eqs. 7.20 and 7.22 become

2 2V VR � � (7.23)

2g tan g�n � 1

and

g 2� � �n � 1 (7.24)V

To estimate the time to turn through an angle �, it will be assumedthat the flight path is a circle which then yields for time to turn through� radians

2 R � �Rt � � (sec) (7.25)

V 2 V

Eliminating now R by means of Eq. 7.23, one finds

V�t � (7.26)

g tan

For unaccelerated level flight � 0, and Eqs. 7.17 and 7.19 can beVcombined to give the turning velocity

2Wn 1V � (7.27)� �S C � C tan �L D

Combining Eqs. 7.26 and 7.27 gives


� 2W nt � (7.28)2� �g �S (C � C tan �) n � 1L D

and the turn radius becomes

2Wn 2R � �n � 1 (7.29)g�S (C � C tan �)L D

Furthermore, the rate of turn can now be written

2n � 1 �S (C � C tan �)L D� � g (7.30)� �n 2W

A10

A number of conclusions can be drawn concerning R, t, and . First�of all, a practical assumption needs to be made concerning the loadfactor n (or the bank angle ). Since a bank angle 90 implies a loadfactor of infinity, and since from previous discussions concerning theV–n diagram an aircraft is structurally limited to a specified value ofn � nl (see Figure 7.2), both the bank angle and the load factor arethen limited by the limit load factor nl. However, a more serious lim-itation may be imposed by the available thrust, as is seen shortly.

From Eqs. 7.23 and 7.29, it is seen that, for a given maximum nl,the minimum turn radius is obtained at lowest airspeed or at maximumlift coefficient . As altitude increases, turn radius also increasesCL max

due to decreased density. It follows, then, that the shortest turn is madewhen the aircraft is on the point of stalling. The same conclusions applyto the fastest turn (minimum time turn) when one observes Eqs. 7.26and 7.28. Then is the desired condition for the maximum turnCL max

rate . This follows easily from Eq. 7.30 or by considering that is˙ ˙� �inversely proportional to R for which minimum conditions were beingconsidered.

Therefore, it seems that the best turning performance is governed byaerodynamic and structural limitations and that minimum turn radius


n

n1

Vs Vc V

CLmax

Rincr

Vs Vc V

CLmax

incr

•�

1

n

n1

1

Figure 7.6 Constant R and on V–n Diagram�

and maximum turn rate are obtained at a point where and max-CL max

imum load factor nl are obtained at the same time. This can also beseen from Figure 7.6, where lines of constant and R are superimposed�on the V–n diagram. This condition was already established in Section7.3 as the corner velocity (maneuver speed), which is defined also byEq. 7.27:

2WnlV � � V �n (7.31)c s l��S (C � C tan �)L Dmax

and then also a relationship between stall-in-turn and stall velocitiesand Vs, respectivelyVst

V � V �ns st

Thus, in absence of any other restrictions, one obtains

2VcR � (7.32)min 2g�n � 1l

2g�n � 1l� � (7.33)max Vc


� nlt � V (7.34)min c 2�g n � 1l

V � V � V (7.35)�max R cmin

It should be noticed that Eqs. 7.21, 7.23, and 7.24 are generic kinematicexpressions, whereas Eqs. 7.32 and 7.33 are aircraft-specific via Vs

( ) and the limit load factor nl.CL max

Summarizing: Aircraft turning performance, as governed only byaerodynamic and structural limitations, often referred to as instanta-neous performance, consists of turns made at an instant of time withoutany consideration of sustaining that performance for any period of time.Of all the theoretical values of and R, only the values inside the�maneuver bounds (defined by , nl curves, and Vmax, see Figure 7.2)CL max

are available. At speeds below the corner speed Vc turning performanceis aerodynamically limited by . Above Vc, maximum aerodynamicCL max

load factor can be larger than nl and then the turning performance isstructurally limited. Eqs. 7.32 and 7.33 do not assure that such maximacan be achieved. Rather, they establish potential limits subject to otherconstraints (thrust). It is also evident that the (instantaneous) turn rateand radius are basically determined, through Vc, by the maximum liftcoefficient and wing loading W /S. These two parameters play aCL max

significant role in sustained maneuvering performance (see also Prob-lem 7.11).

In practice, a much more severe restriction is placed on the turningperformance by the fact that the thrust, or power, is limited. For high-performance aircraft, where T /W � 1, this will not be a problem atlow altitudes. But degradation of engine thrust at high altitudes willimpose turning restrictions on all aircraft.

As a result of lift increase required to produce a turn while banking,the induced drag is increased above the level flight value. This dragincrease can be observed, for steady flight, from Eqs. 7.17 and 3.22 as

21 2kW2T � D � �V SC � (7.36)r D0 2 22 �V S cos

When Eq. 7.36 is plotted for constant � and W, as a function ofvelocity and bank angle with the thrust available curves also present,the increase in induced drag and the role of the bank angle are clearlyevident. A bank angle of � 0 represents the steady, level flightresults. Figure 7.7 shows that as the bank angle (load factor) increases,


T

Tc

Vc VE

Tr

Ta h

Ta SL

= 0

45

50

30

c

CLmax

Figure 7.7 Effect of Bank Angle on Thrust Required

the drag may exceed the thrust available. To maintain constant altitude,the bank angle must be decreased to the point where T � D. Thus, theturning performance becomes thrust limited. Turns at higher bank angle(higher load factor) may be accomplished, but the altitude and velocitycannot be maintained at the same time.

Thrust limitation also influences the minimum radius and maximumturn rate performance. Figure 7.8 shows constant thrust curves super-imposed on the V–n diagram. The significance of these curves lies inthe fact that they show the amount of thrust required to maintain cornervelocity. This, of course, can be calculated by use of Eqs. 7.31 and7.36. If there is sufficient thrust available to maintain Vc (Ta � Tr), theTr curve must pass above Vc–nl intersection and the maximum turningperformance Eqs. 7.32 through 7.35 hold.

For the configuration shown in Figure 7.7 one can conclude that atTa shown for the sea level turning performance is not thrust limited.At altitude h, however, curve falls below Tc point, as required byTah

Vc, and there will be an increase in Rmin and a reduction of from�max

the values predicted by Eqs. 7.32 and 7.33, respectively. Those equa-tions, or rather Eqs. 7.23 and 7.24, can still be used to evaluate R and


Vs

n

n1

1

Vc V

Tr

incr

CLmax

Figure 7.8 Constant Tr Lines

but a new load factor must be determined from Eq. 3.21 (or any�,similar expression) by use of the actual thrust available at that particularaltitude.

At thrust levels where Tah � Tc, turns can be made at n � nl, thestructural limit, but altitude and/or velocity may change (decrease),allowing the recovered energy to be converted into a temporary turn.What actually happens is governed by the energy equation Eq. 2.20. Anumber of options are available, depending on operational needs. Suchproblems fall in the area of energy maneuverability, some of which canbe approached via the point-performance techniques and will be dis-cussed in Section 7.6.

7.5 MAXIMUM SUSTAINED TURNING PERFORMANCE

In the previous section, aircraft maneuvering performance was foundto be constrained primarily by its aerodynamic and structural charac-teristics leading to expressions describing the instantaneous perform-ance potential. It was found also that the thrust was the deciding factorwhether this potential can be achieved and then maintained.

In practical flight-performance analysis, the items of great interestare the maximum sustained turn rate and the minimum sustained turnradius. Implicit in such a development is the assumption that the motionand aircraft attitude be steady and could be maintained for some prac-tical, but undetermined, time. Returning to Eq. 2.20 where aircraft per-

7.5 MAXIMUM SUSTAINED TURNING PERFORMANCE 209

formance was formulated in terms of the energy balance, it is seen thatsteady and maintained would have to mean that the altitude and ve-locity must remain constant and therefore the energy rate must be iden-tically zero (Ps � 0, see Eq. 3.1). It follows also that T � D, and theproblem falls clearly into point-performance category. This is calledsustained performance. Its meaning and definition have found wide-spread use in aircraft performance description, comparison, and evenacquisition.

7.5.1 Maximum Load Factor

First of all, it is helpful to develop more understanding and an expres-sion for the maximum load factor. This is obtained by rewriting thethrust required (T � D) expression Eq. 3.21 as:

q T CD02n � � q (7.37)� � ��kW/S W (W/S

or

q 1 T /Sn � � C (7.38)� �D0�W/S k q

Eqs. 7.37 and 7.38 give the load factor for a given thrust loading T /Wand velocity (q � 1/2�V2). For each value of q the local maximumvalue of the load factor is determined by the maximum value of T /W(see Figure 7.8). The velocity that maximizes the load factor is foundfrom Eq. 7.38 by differentiation with respect to q and then setting theresult equal to zero. One obtains for the dynamic pressure at maximumload factor

T /Sq � (7.39)nm 2CD0

which yields for the velocity

T /SV � (7.40)nm ��CD0

Substituting now Eq. 7.39 into Eq. 7.37 gives


Tn � E (7.41)max mW

Eq. 7.41 shows that the maximum load factor is obtained at maximumlift-drag ratio, Em at a given T /W ratio (any T /W). The true maximumavailable load factor is obtained also at Em but at the maximum avail-able thrust-weight ratio (T /W)max. In other words, this occurs when theaircraft is operating simultaneously at both maxima:

Tn � E (7.42)� �m mW m

The corresponding flight speed can be brought in sharper focus if Eq.7.40 is nondimensionalized by , obtainingVDmin

V T /S � C Tn Dm 0V � � � E (7.43)n mm � � ��V �C 2W/S k WD Dmin 0

Eqs. 7.40 and 7.43 show that the velocity for maximum available loadfactor is proportional to the available thrust. It decreases as the altitudeincreases with the decrease of thrust with altitude.

F14

7.5.2 Minimum Turn Radius

The minimum turn radius is found by differentiating Eq. 7.23:

2VR �

2g�n � 1

with respect to V and setting the result equal to zero. Since dn /dV isequal to �V dn /dq, it is convenient to change variables, and one findsthe well-known result


dn2n � 1 � qn � 0 (7.44)dq

Substituting now dn /dq from Eq. 7.37, after some manipulation, oneobtains the following results for the condition of minimum turn: ve-locity of the turn (assumed to be constant):

4k(W/S)V � , (7.45)Rmin � �(T /W)

applicable load factor, from Eqs. 7.37 and 7.45:

1n � 2 � , (7.46)R 2min T 2E� � � mW

minimum turn radius:

2V rminR � , (7.47)min 2g�n � 1rmin

the turning rate for minimum radius turn follow form Eq. 7.33:

2g�n � 1Rmin� � , (7.48)Rmin VRmin

or

g tan Rmin� � , (7.49)Rmin VRmin

where

1 � . (7.50)Rmin nRmin

Eqs. 7.45 to 7.50 provide necessary tools for evaluating the minimumturn radius and, at the first glance, show that high thrust-weight ratio


and a high Em with a low wing loading provide the desired results.However, a comparison of the with the stall velocity in showsVRmin

that if the available thrust-weight ratio T /W is higher than, obtained by equating Vs2kC /nL Rmax min

V �n � V (7.51)s Rmin

in Eq. 7.45, then � Vs and the turn cannot be accomplished aero-VRmin

dynamically. In other words, the locus of ( , n) is on the T /W curveVRmin

but to the left of the stall line. In this case, the minimum turn radius,for that given T /W, is on the stall line where the T /W curve crossesthe stall line. That intersection can be found by equating the load fac-tors determined from the equation of the stall line:

qCq L maxn � � (7.52)q (W/S)s

to the load factor from Eq. 7.37:

1 /2C qqC q T DL 0m � � (7.53)� � ��W/S k(W/S) W W/S

This gives the stall speed in turn as

(T /W)(W/S)q � (7.54)Rmin CDmin

where � �2C kC C .D L Dm m 0

The corresponding stall load factor is obtained by substituting Eq.7.54 back into Eq. 7.52, giving

(T /W)C TL maxn � � E (7.55)r 2min C WDm

where

CL maxE � (7.56)2 CDmax


Thus, a turn can be made according to Eqs. 7.54 and 7.55 but at higherturning radius, which can then be calculated again from Eq. 7.47 butwith values of V and n at , or, the turn is made at edge of stall.CL max

NOTE

Since the stall line velocity, determined from Eq. 7.53, must be equal (on thestall line) to the velocity obtained from Eq. 7.45, a simpler check for theturn can be found by equating these equations. Thus, one finds that if

2T2kC � � �Dm W

the turn can be made. Then � and the locus of (VRmin, n) is onV VR Rminmin s

the T /W curve but on, or to the right of, the stall line.

7.5.3 Maximum Turning Rate

The maximum steady-state turning rate, also called the fastest turn, isfound through a process similar to that used in the previous section forthe minimum turn radius. The equation for the turn rate, Eq. 7.33, isdifferentiated with respect to q. Setting the result equal to zero givesthe basic condition for the maximum turning rate as

dn2n � 1 � 2nq � 0 (7.57)dq

When combining with Eq. 7.38, one obtains for the fastest turningvelocity

W kq � (7.58)t �S CD0

whence

.252(W/S) kV � � V (7.59)� �t Dmin� � CD0

is the velocity in level flight at that particular altitude. RecombiningEq. 7.58 with Eq. 7.38 gives


Tn � 2 E � 1t m� W

and

Tn � 2 E � 1 � �2n � 1. (7.60)� �t m maxmax � W

m

Recalling that L � nW, Eq. 7.58 also yields a limiting condition

CD0C � n � n C � � C (7.61)L t t L E Lt M max� k

The maximum turning rate can be calculated simply from Eq. 7.33 as

g 2� � �n � 1 (7.62)tVt

Using Eq. 7.58 again, one obtains a more explicit equation:

1 /2� T

� � g � 2�kC (7.63)� � ��D02(W/S)k W

It is seen that a high turning rate requires a large T /W but low valuesof the wing loading W /S, k, and . Small k means large aspect ratioCD0

AR. Small k and imply also a high Em. Thus, conflicts and neces-CD0

sary compromises arise in the design stage.Transport and small noncombatant aircraft rely on high Em and rel-

atively low T /W values for economic operation. High turning ratesbecome significant only for acrobatic categories. Fighter aircraft needlow aspect ratio for high speed and structural considerations. Thus, highT /W and are then usual trade-offs for lower aspect ratio and EmCL max

with attendant range penalties. Eqs. 7.59 and 7.63 also show why air-to-air combat takes place at subsonic speeds ( ) and on deck dueVDmin

to maximum air density at the sea level.Eqs. 7.60 and 7.61 highlight another hidden constraint. A high-

performance aircraft with a high T /W may be able to produce theoret-ically a sufficiently high nmax to generate, in turn, high lift requirement( in Eq. 7.61), which may exceed the aircraft maximum lifting ca-CLt


pability . Then, similarly to minimum turn radius case, the turnCL max

may have to be performed (at reduced T /W) at lift limit curve at (near)stall conditions.

In summary, all three limitations due to aerodynamics, structuralconsiderations, or thrust may affect at the same time aircraft turningperformance. In general, aerodynamic and structural limitations are thesignificant ones at low altitudes. At high altitudes, thrust considerationspredominate. At high airspeeds may also be limited due to MachCL max

number effects.

EXAMPLE 7.1

An aircraft has the following characteristics:

C � 1.5L max

2S � 125 ftW � 2,000 lbn � 4l

2C � 0.02 � 0.05CD L

T � 300 lb for all flight speed at sea level (turbojet)� � 0

Calculate the minimum time added to perform a constant altitude180 turn. Assuming that there will be adequate thrust to performthe turn, Eqs. 7.31 and 7.34 can be used. The drag coefficient is

2C � 0.02 � 0.05(1.5) � 0.1325D

and the stall speed is

2WV �s ��SCL max

2 � 2,000� �0.002377 � 125 � 1.5

� 94.7 ft/sec

The corner speed becomes


ftV � V �n � 95�4 � 190c s l sec

The time for minimum turn follows from

�V � 190ct � � � 4.79 secmin 2g�n � 1 32.2 � �15l

Checking now the validity of adequate thrust assumption

1 12 2T � �V SC � � 0.002377 � (190)r D2 2� 125 � 0.1325 � 710 lb

which clearly exceeds the thrust available and indicates that a 4gturn at the corner speed is not possible. What is actually possiblecan be calculated as follows: the load factor at the available thrustof 300 lb is evaluated from

T 300n � 2 E � 1 � 2 � � 1 � 1.934� �t m� �W 2,000

In order to calculate the flight velocity, the turn lift coefficient isobtained from Eq. 7.61

C .02D0C � n � 1.934 � 1.22L tt � �k .05

Turn velocity is

2Wn 2 � 2000 � 1.934tV � � � 146 ft/sect � ��SC .002377(125)1.22Lt

Bank angle is calculated from

2 2tan � �(n ) � 1 � �1.934 � 1 � 1.655t

which indicated a bank angle of 59. The time to turn is obtainedfrom Eq. 7.26


V� 146 t � � � 8.59 sec

g tan 32.2 � 1.655

This problem can be solved also in an indirect manner from Eqs.7.27 and 7.28, which can be written as

tg V�

2� �n � 1

The problem now resolves into one of finding the combination of Vand n that minimizes tg/�. This can be determined by solving theTa � D equation for n, for selected values of V, and then calculatingV/ . The drag equation becomes2�n � 1

2 21 2kW n2T � �V SC �a D0 22 �SV

or

1 2300 � � 0.002377 � V � 125 � 0.022

2 22 � 0.05 � (2,000) � n� 20.002377 � 125 � V

Which simplifies to

2 4300V � 0.002971V 2� n1346235

Constructing Table 7.1 gives a minimum value:

tg� 88.07

�

which occurs at a flight velocity of 146 ft/sec. The time to turn canbe calculated from


TABLE 7.1 Data for Example 7.1

V n V/ (n2 � 1)

140 1.876 88.19145 1.926 88.09146 1.936 88.07147 1.946 88.08150 1.974 88.13

� 88.07t � � 8.59 sec

32.2

The turn radius is calculated from

2 2V 146R � � � 399 ft

2 2g�n � 1 32.2 � �1.936 � 1

This process amounts to assuming that the lift coefficient CL re-mains constant throughout the turn. In other words, in level flightprior to turning or when returning to level flight after the turn, thelift coefficient is the same as in the turn:

2Wn 2 � 2000 � 1.936C � � � 1.22L 2 2�SV 0.002377 � 125 � (146)

Thus, when the aircraft returns to (or starts from) level flight, itsflight speed is reduced to

2W 2 � 2000 ftV � � � 105L � ��SC 0.002377 � 125 � 1.22 secL

The same result can also be calculated from VL � V / . Then the�nthrust for level flight can be reduced from 300 lb to [with CD � 0.02� 0.05(1.22)2 � 0.0944]

1 2T � � 0.002377 � 0.0944 � 125 � (105) � 155 lb2


EXAMPLE 7.2

The aircraft in Example 7.1 is in level flight at 125 ft/sec and setsout to perform the same 180 turn but keeps the flight velocity con-stant while turning. This, however, requires that the lift coefficientbe increased to maintain altitude and to produce turning acceleration.Thus, the drag coefficient also will increase with an increase in thrust(over the level flight value of 155 lb). Eq. 7.26 permits calculationof the required load factor n from

V�t �

2g�n � 1

whence

2 2V� 125 � n � � 1 � � 1 � 1.74� � � �� tg 8.59 � 32.2

The required CL, and CD, are obtained by means of Eq. 7.27, whichgives

2Wn 2 � 2,000 � 1.74C � � � 1.5L 2 2�SV 0.002377 � 125 � 125

and the drag coefficient as

C � 0.1325D

The thrust required to accomplish this turn is

1 2T � � 0.002377 � 0.1325 � 125 � 125 � 307 lb2

and the resulting turn radius is

2125R � � 341 ft

232.2 � �1.74 � 1


7.6 THE MANEUVERING DIAGRAM

The last two sections established the important parameters required foranalyzing aircraft maneuvering performance in horizontal plane. Usefulequations were obtained predicting minimum turn radius and the max-imum turn rate. In addition, it was found that the maximum turn ca-pability is limited in an intricate fashion by three factors:

1. Structural strength limit2. Lifting capability3. Thrust limit

All this information can be combined in a common presentationdiagram called the maneuvering diagram or the energy maneuverabilitydiagram. It has the advantage of presenting the aircraft turning capa-bility in a very compact and complete manner. However, each altitudeand gross weight require separate diagrams. The diagram is also veryuseful for presenting and comparing several tactical aircraft turningperformances. As is seen in Figure 7.9, the horizontal axis representsthe velocity and the vertical axis gives the turn rate. The basic graphconsists of a carpet plot of the load factor and the turn radius. Figure7.9, generated by Eqs. 7.21, 7.23, and 7.24, and by its purely kinematiccontent, is generic and valid for any aircraft. The aircraft-specific datais entered through the limiting factors.

The structural limit nl and the lift limit (also called accelerated stallboundary) are entered via V � Vs and superposed on the graph;�nl

see Figure 7.10. This introduces the aircraft-specific informationthrough chosen altitude and weight data. Due to its shape, the resultinggraph is also known as the doghouse plot. Figure 7.10 shows the curvesof A-4 Skyhawk for the following data:

2 2W � 18,000 lb, C � .0177 � .156C , S � 260 ft , T � 10,000 lbD L

For illustration purposes it will be assumed that the limit load factornl � 5. Then from Fig. 7.10, at the point A, where the stall line crossesthe locus of nl � 5, one finds that

• The maximum (instantaneous) turn rate is about 21.5 degrees.• The minimum (instantaneous) turn radius is about 1,200 ft.• The corner speed is about 255 KTAS.

7.6 THE MANEUVERING DIAGRAM 221

Figure 7.9 The Maneuvering Diagram

For evaluating sustained performance capabilities, it is necessary toconsider also the thrust limit. This is accomplished via the energy equa-tion (Eq. 2.20), which is used in the form

de dh V dV T � DP � � � � V� �s dt dt g dt w

C qT W/SD0 2� Ma � � kn� � ��W WS q

where the drag has been introduced from Eq. 3.21. Next, a parametriccurve calculation is established for an arbitrary range of �500 � Ps �250 and a practical range of velocities with the load factor n as anotherparameter to assist in superimposing the Ps curves in Fig. 7.6. Thecurve for Ps � 0 signifies T � D, which provides the locus for sustained(steady) performance evaluation.

Following the line for load factor nl � 5 to the Ps � 0 curve (pointB) gives the following:


Figure 7.10 A-4 Skyhawk Energy Maneuverability Diagram, Sea Level, W � 18,000 lb

• The minimum sustained turn radius is little less than 4,000 ft atabout 470 KTAS, with a turn rate of about 12 degrees (point B).

• The maximum sustained turn rate is found at the top of Ps � 0curve at about 13 degrees/sec. This takes place at a load factor oflittle more than 3 and a turn radius of about 1,800 ft.

However, this is not the complete picture. To better understand therole of Ps it is useful to return to the energy rate equation Eq. 2.20 andbriefly consider how energy can be gained and then redistributed invarious maneuvers.

The energy production side of Eq. 2.20

T � DP � Vs W

clearly shows that the aircraft specific energy is increased if T � D.Conversely, if Ps � 0 the aircraft has no capability to increase its

7.6 THE MANEUVERING DIAGRAM 223

specific energy and it represents the upper limit of sustainable maneu-vering (at T � Tmax). If Ps is positive, then the aircraft can (must)accelerate, climb, or maneuver. If Ps is negative, it can still performsome of these functions but now it also has to decelerate or lose alti-tude, as it is losing energy. Thus, the Ps � 0 curve is a useful boundaryfor separating sustained maneuvers from those that lead to energy loss.

For some specific examples, consider again the A-4 Skyhawk. Atsea level 1-g loading and at 500 KTAS, its Ps is

(T � D)V 10,000 � 4,140P � � 845 � 275 ft/secs W 18,000

From the energy conservation side of Eq. 2.20

de dh V dVP � � �s dt dt g dt

one can conclude that

• At constant KTAS climb (dV /dt � 0) the aircraft can climb at therate of 275 ft/sec; or

• If the altitude h is constant (dh /dt � 0), the aircraft can accelerateat

dV 275g 2� � 275g /845 � 10.5 ft/sec ; ordt V

• If the aircraft is set into a vertical climb where dh /dt � V � 275ft/sec, then

dh V dV V dVP � � � V �s dt g dt g dt

whence

dV 275 2� � 1 g � �21.7 ft/sec� �dt 845

and the aircraft is decelerating at about .67g.

Returning now to the example concerning instantaneous perform-ance, if the pilot wants to perform a 5g turn at the corner speed (about


t

VT

L

D

n

�

��

�

Figure 7.11 Spiral Flight

250 KTAS), the aircraft would lose altitude at the rate of about 330 ft/sec.

In its essence, the maneuvering diagram represents point-performance approach since it is tied to a given altitude and weight.Change of altitude has no effect on the limit load factor, as it is onlya function of structural design. For a given load factor, altitude gaindoes increase the stall velocity. An increase in altitude will decreasethe available thrust. The combined effect is to shrink the positive Ps

region with attendant reduction of the turn rate and an increase ofturning radius. Effect of the altitude change may be estimated by pre-paring the diagrams similar to the one in Fig. 7.10 at different altitudesand weight, but such procedure cannot really predict interaltitude be-havior because of time dependent behavior of the aircraft velocity. Ac-tual (time-dependent) problems are the topic of energy maneuverabilityapplied to flight paths.

7.7 SPIRAL FLIGHT*

An aircraft that is rolled into a (steady) coordinated turn at a constantangle of bank , and flies at a constant angle or descent � moves ona helical path on an imaginary circular cylinder with a radius of R, (seeFigure 7.11).

7.7 SPIRAL FLIGHT* 225

This spiral path can represent also instantaneous (point performance)conditions of a transient flight path where only centripetal and tangen-tial accelerations are included. Thus, as it was assumed in Section 7.4,the forces are exactly balanced without any yaw or sideslip takingplace. Referring to Figures 7.3 and 3.2, the equation tangent to theflight path is

˙T cos � � D � W sin � � mV (7.64)

the equation normal to the flight path (along the radius R) is

2sin mV˙(T sin � � L) � mV� � cos � (7.65)cos � R

and in the vertical plane the equation is

(T sin � � L) cos � W cos � (7.66)

Here is the turn rate about a vertical axis. The angle �, in general,�is not small for a hard banked turn, as the thrust sector may have asizable inclination to the velocity sector. This is due to the thrust in-stallation angle �, and the aircraft angle of attack �, which will nowapproach the stall angle (usually exceeding 20).

� � � � � (7.67)

In previous chapters the thrust angle � was ignored by assuming eitherthat the angle of attack was small or that the thrust contribution to thelifting effort was small (see Eq. 2.12). In a highly banked turn withrequired large amount of thrust, the angle � will become important, asthe thrust will contribute significantly to the turning lift force. Similarto the developments in Section 7.4, the load factor comes from Eq.7.66 as

T sin � � L cos �� n (7.68)

W cos

Also

2V cos �tan � (7.69)

gR


Figure 7.12 Climbing, Turning Flight

and

V cos � g tan � � � (7.70)

R V

Eqs. 7.68 through 7.70 are equivalent to Eqs. 7.20 through 7.24 ofsection 7.4 and are valid for climbing or descending flight (� � 0). Forthe case of flight without power (thrust), this set of equations can bedeveloped further to yield some practical and interesting results. Inearlier developments and examples it was shown that more power isneeded in turns than in level flight. It will be shown now that as anairplane is descending in glide, the angle of glide will be increased ifthe plane is turning. In glide, as the plane turns through a heading of, the height lost is

2 2WV cos ��h � �R tan � � � tan � (7.71)� gL sin

This can be simplified by

2W/S2V ��C cos L

and


1tan � �

L /D

to give

24(W/S) cos ��h � � (7.72)� 2g�(C /C ) sin 2L D

To find the criteria for minimum height loss it can be assumed that theflight path is shallow; thus cos2 � � 1. Also � can be replaced by 2 (one full turn). The height loss is clearly then minimized if the bankangle � 45, and CL � which provides also the maximumC ,L max

condition for Eq. 7.72 shows also that an aircraft with a higher2C /C .L D

wing loading (W /S) loses weight more rapidly. As an aircraft descends,the density increases and the height loss is also decreased. A compar-ison can be made between the angle of glide � in pure glide, where

Lcot � � (7.73)

D

and in turning glide where

Lcot � � cos (7.74) D

It is easy to see that � � �(cot � � cot �) as cos � 0.7, and for pureglide L /D � (L /D)max. For turning glide L /D corresponds to C ,L max

which yields a smaller L /D.

EXAMPLE 7.3

Consider a light aircraft with the following characteristics:

W � 3,000 lb2S � 225 ft

C � 1.2L max

2C � 0.025 � 0.06CD L

h � 5,000 ft

The aircraft is in a 45 steady, banked, gliding turn


Calculate the glide performance for one full turn. In a pure glidemode, the glide angle would be

1 1�1 �1� � tan � tan � � 4.43E 12.9m

In a banked turn, assume CL � � 1.2, which gives an L /D �CL max

8.8. The velocity in the glide can be calculated from Eq. 7.37 with

2C � 0.025 � 0.06(1.2) � 0.1364D

cot � � 8.8 � 0.707 � 6.22

� � 9.13

2W sin �V � � �SCD

2 � 3,000 � sin 9.13� �0.002377 � 0.8617 � 225 � 0.1364

ft� 123

sec

The radius of turn in a 45 bank is given by Eq. 7.70:

2V cos �R � � 464 ft

g tan

The loss of altitude in one complete turn is

S � 2 R tan � � 2 � � 464 � tan 9.13 � 468 ft

EXAMPLE 7.4

A high-performance aircraft has the following characteristics:

W � 30,000 lbT � 20,000 lbmax

2S � 460 ft


C � 0.06/degL�

C � 1.2L max

� � 3 (thrust angle)2C � 0.0185 � 0.06CD L

M � 0.8

is in a steady 10 climbing turn under full thrust at 30,000 ft altitude.Calculate the load factor and the turn radius. The load factor n canbe calculated from Eq. 7.68. Since � is not known, the proceduremust be carried out iteratively. Assuming first that � � 0, Eq. 7.64gives

D � T � W sin � � 20,000 � 30,000 sin 10 � 14,790 lb

The drag coefficient follows with

D DC � �D 2�/2pM S qS

14,790� � 0.114

0.7 � 2,117 � 0.2970 � 0.64 � 460

The lift coefficient can be found from

C � C 0.114 � 0.0185D D0C � � � 1.26L � �k 0.06

which requirement exceeds . For successive approximation pur-CL max

poses, one can proceed and calculate the following:

C 1.26L� � � � 21a 0.06

Thus

� � 21 � 3 � 24

D � T cos � � W sin � � 20,000 cos 24 � 30,000 sin 10

� 13,061 lb

Then


C � 0.1D

C � 1.17L

� � 19.4� � 22.4

These values are sufficiently close. Calculating now n, from Eq.7.68:

5T sin � � C qS 20,000 sin 22.4 � 1.17 � 1.3 � 10Ln � �W 30,000

� 5.32

The bank angle comes from Eq. 7.68:

cos � cos 10�1 �1 � cos � cos � 79.3n 5.32

The radius can be calculated from (V � .8a � 800 ft/sec):

2 2 2 2V cos � 800 cos 10R � � � 3,687 ft2 2 2 2� �g(n � cos �) 32.2(5.32 � cos 10)

PROBLEMS

7.1 An airplane has the following characteristics:

V � 120 m.p.h., EASh � 10,000 ftW � 3,400 lb

2W/S � 24 lb/ftL /D � 10T � 6,200 lba

It makes a 90 turn in 18 seconds maintaining altitude and in-cidence angle

Calculate the load factor, bank angle, radius of turn, and thethrust horsepower required.Ans. � 31.3; R � 2,520 ft; 160 HP.

PROBLEMS 231

7.2 An aircraft is flying at 20,000 ft at M � 0.5. The aircraft isbanked into a 180 turn, which is supposed to take 29.24 secondsto complete.

Assume C � constantL

T � 6,200 lba

W � 26,000 lbn � 7.5l

2S � 375 ft2C � 0.015 � 0.1055CD L

C � 1.6L max

Calculate the turn radius and show why this turn (can/cannot)be made.Ans: R � 8,763 ft; cannot, since D � Ta.

7.3 Show that the turn radius in a coordinated climbing maneuvercan be given by

2 2V cos �R �

2 2g�n � cos �

7.4 Show that the turn rate in a coordinated climbing maneuver canbe given by

2 2�n � cos ��

V cos �

7.5 Calculate the F-18A turn rates for the following conditions: S� 400 ft2, W � 38,000 lb, T /W � .486, CD � .0245 � ,2.13CL

� 2.8, nl � 7.0CLmax

a. The turn rate at its minimum turn radiusb. The maximum turn ratec. The maximum instantaneous turn rated. The minimum velocity and the turn rate at which the instan-

taneous turn rate can be sustained, (i.e. the values at the edgeof stall)Ans: 8.7 deg/sec; 11 deg/sec; 28.6 deg/sec; 193 ft/sec; 7.94deg/sec


7.6 If the aircraft in the Example 7.4 is climbing at a 10 angle butis maintaining a 60 bank angle, what is the resulting accelera-tion at full thrust?

7.7 Solve Example 7.4 if the aircraft is in a 10 descending flight.

7.8 The aircraft in Problem 7.2 loses all power and it will startgliding. In order to attempt to land at a nearby airport, the pilotmakes a 180 turn. How much height is lost in that maneuver?Ans: 4498 ft.

7.9 The aircraft in Example 7.4 is cruising at M � 0.8 at 30,000 ft.The aircraft is then banked to 50 angle and full power is ap-plied. How much height is gained if the aircraft keeps flying atthe same speed through a 270 turn?

7.10 A 727-200 is traveling at an altitude of 4,000 ft with an approachspeed of 300 ft/sec TAS. Heading is due north, wind is fromthe south at 20 mph.a. What is the minimum separation for the flight path from (say,

the edge of) the runway so that the aircraft can perform a180 turn to be conveniently aligned with the runway for thefinal approach?

b. With the assumptions and data outlined below, how muchaltitude will the aircraft lose?

At the beginning of the turn, the pilot will maintain the thrustand attitude and put the aircraft into a 30 bank angle. It isreasonable to assume that the airspeed in the turn will remainconstant. Wheels are up and the aircraft weighs 130,000 lb. Forcalculation purposes, use sea-level air properties. The rest ofknown data:With partial flaps CD � .021 � � 2.2; �2.076C , C CL L Lmax max,land

2.6, S � 1,700 ft2. Three JT80-11 engines at 15,000 lb ea.Ans: a. separation � 2R � 9,840 ft, b. �h � �270 ft.

7.11 Two aircraft are engaged in air-to-air combat. Minimum (sus-tained) turn radius at sea level determines the winner.Aircraft A. CD � .0225 � � 1.6, T /W � .5, S �2.11C , CL L max

375 ft2, W � 40,000 lbAircraft B. CD � .0245 � � 2.8, T /W � .486, S2.13C , CL L max

� 400 ft2, W � 38,000 lbAns: A � 1,882 ft, B � 1,368 ft.

PROBLEMS 233

7.12 A jet transport aircraft at 300 mph is vectored into a holdingpattern at a turn rate of 1.5 deg/sec. Calculate the turn radiusand the required T /W. Aircraft data at 20,000 ft altitude:

2 2W/S � 100 lb/ft , S � 2,000 ft2C � .015 � .06C , C � 1.8D L L max

L /D� � 16.67max

Ans: R � 16,800 ft, T /W � .074

7.13 Construct the energy maneuverability diagram for F15C with thefollowing characteristics at sea level:

2W � 60,000 lb, S � 608 ft

C � 1.785, C � .023, k � .133L Dmax 0

T � 50,000 lb, n � 7.33l

It can be determined from the diagram (or, also calculated moreprecisely) that: Vc � 340 KTAS, maximum instantaneous turnrate is 23 deg/sec at 345 KTAS, maximum instantaneous turnradius is 1450 ft at 345 KTAS, maximum sustained turn rate is15 deg/sec at 250 KTAS, minimum sustained turn radius is1,510 ft at 250 KTAS. If 6 g’s are maintained at the corner speed(Ps � �250 ft/sec), then the altitude loss amounts to 260 ft/sec.

234

8Additional Topics

F5A

8.1 CONSTRAINT PLOT

One of the more useful and productive tools in mission analysis andpreliminary design is the so-called constraint plot. Usually it is foundin the format where take-off thrust loading To/Wo is plotted against thetake-off wing loading Wo/S for mission requirements of interest. For agiven set of mission specifications, called constraints, calculation of therange of typical To/Wo versus Wo/S values defines the available solutionspace for that particular set of missions (see Figure 8.1). A plot of To

/Wo against Wo/S with all the applicable mission specifications as par-


8.1 CONSTRAINT PLOT 235

Figure 8.1 Jet Trainer Constraint Plot

ameters provides a map indicating a range of solutions available forthe aircraft designer.

This provides one basis for choosing the design point of the aircraft.Since the design point seldom consists of a unique set of parameters,the constraint plot permits easy change or modification of constraintsto establish trends and, via crossplotting unto carpet plots, render it-eration accessible. Of course, the set of constraints (a wish list) maybe established to be so overwhelming that, on one hand, they leave noroom for choice and compromise of parameters—the essence of design.On the other hand, they may lead to physical conflicts (ultrashort run-way) or exceeding existing technological limits (T /W exceeding, say,5). In any case, a negative outcome here may yield also useful infor-mation.

Next, the most used mission constraints with their characteristic fea-tures will be listed in Table 8.1. Then the applicable equations usedfor evaluating thrust and wing loadings for each mission constraint willbe reviewed. It is assumed that the general lift, drag, and thrust char-acteristics of the proposed aircraft are available. Finally, a numericalexample will be used to demonstrate the constraint plot methodology.

236 ADDITIONAL TOPICS

TABLE 8.1 Typical Mission Constraints

Mission Specs Characteristics Parameters

Take-off Ground run, rotation distance VT, T, Tr

Landing Free roll, ground roll, braking VT, T, �, �CD0

Air superiority Energy rate Ps M, T, altitudeRate of climb Rate of climb M, T, altitudeMaximum speed V or M specified T, altitudeAcceleration �M or g’s specified T, altitudeSustained turning (a) g-load (n) M, altitude

(b) turn rate M, altitude

The equations used to calculate the thrust loading as a function ofthe wing loading (or the reverse) are the same ones as used to evaluateperformance of the individual missions in the point performance sense.However, since some of the expressions and procedures (e.g., take-offrun) may be rather lengthy and involved and in order to gain quickoverview, usually simplified equations are used. This also includesmany other simplifying assumptions concerning the variation of themain parameters with altitude and velocity. The emphasis in this meth-odology is on trends and relative behavior rather than on precision.

8.1.1 Take-off and Landing

Take-off Take-off distance is usually specified as

s � s � sg r T

which means that the ground run and the distance covered while theaircraft is in the rotation phase must be less that some prescribed totaldistance sT. Ground run and rotation distances are given by

1.44W/Ss �g �gC T /WLm

s � 1.2t Vr r s

2W/S� 1.2tr��CLm

Rotation time tr must be specified. Combining the above equationsyields


1.44.W/S 2W/S� 1.2t � sr T��gC T /W �CLm Lm

This can be considered as a quadratic equation in as a function�W/Sof T /W, or, which is much simpler, solve the equation directly for thethrust loading T /W as

1.44W/ST �gC aW/SLm� �W s � b�W/S2W/S T

s � 1.2T ��CLm

Landing Landing distance is usually specified as

s � s � sfr b L

with the total distance sL and free roll time tr specified. Free roll dis-tance sfr may be expressed as

s � 1.1t V .fr r s

Landing distance with braking and retarding devices may be expressedas

1.21W/Ss �b C qSD0�gC � �� Lm W

During landing, V (in q) is evaluated at .7VT, as q � .5�(.7VT)2 �.245�(1.1Vs)2. and � specifications must reflect expected brakingCD0

configuration (e.g., brakes, chutes, spoilers, and runway conditions).Normally, summation of free roll and braking distance yields a quad-

ratic equation in independent of T /W:�W/S

2W/S 1.21W/S1.1t �r��C C qSLm D0�gC � �� Lm W


8.1.2 Constraints Tied to Performance Equation

The following six constraints will be evaluated by use of the funda-mental performance equation (for the excess energy (rate)), which isthen specialized for each individual requirement:

de dh V dV T � DP � � � � V� �s dt dt g dt W

C qT W/SD0 2� Ma � � kn� � ��W W/S q

In the sequence of constraints 3 through 7, usually a full thrust is used.Thrust may be given by an equation of the form T � Toƒ(M)�, whereƒ(M) is normally curve-fitted to provide an expression like (1 � aM),(1 � aM2), etc.

Air Superiority For air superiority, Ps, h, M are specified, n � 1, andthe following equation results:

T b� a � � cW/S

W W/S

Rate of Climb For rate of climb, set Ps � dh /dt and dV/dt � 0, withh and M specified, n � 1. One obtains then again

T b� a � � cW/S

W W/S

as in air superiority, only the constants a, b, c will take on differentvalues.

Acceleration For acceleration, Ps � dV /dt, dh /dt � 0 and M and hare specified. One obtains the same type of equations as in the twoprevious constraints.

Maximum Speed For maximum speed conditions, set Ps � 0 (dV/dt and dh /dt are both zero), specify V � Vmax and h, n � 1. Thefollowing equation results:


T b� � cW/S

W W/S

Again, the constants b and c will take on different values from previousconstraints settings with:

b � C qD0

kc �

q

Sustained Turning

1. Given g-load (n).Here, set Ps � 0 with n, M, and h specified. Again, the followingequation is obtained:

T b� � cW/S

W W/S

with the constants b and c:

b � C qD0

2knc �

q

2. Turn rate given.Here, set Ps � 0 with M and h specified. Turn rate is given by

and the same equation as given g-load applies. n is now deter-�mined from

g 2� � �n � 1V

This completes a brief survey of the typical specifications used forestablishing an aircraft-performance constraint plot. The list is not ex-haustive. To explore the two primary design parameters, thrust-weightratio and the wing loading, just about any aircraft-performance param-eter may be turned into a requirement specification—turn time, range,


endurance, and so on. Example 8.1 serves to illustrate intricacies andpossibilities of this technique to focus on what is possible, what isneeded, or what may be excessive.

B58

EXAMPLE 8.1

In this example, the requirements for a new Navy jet trainer areexplored. It is anticipated (from existing aircraft data) that the fol-lowing aerodynamic properties should be applicable:

2C � .018 � .15C , C � 1.7D L Lmax

The following constraints are of interest:

1. Take-off distance at sea level not to exceed 1,500 ft at atemperature of 103�F. Include ground roll plus 1 second rota-tion time.

2. Maximum flight speed at 30,000 ft is .9M. Assume W � .8Wo,T � To(1�M /2)�

3. Maintain a minimum rate-of-climb ROC � 100 fpm (serviceceiling) at 40,000 ft, .8M, W � .8Wo, T � To(1 � M /2)�.

4. Maintain a minimum ROC � 500 fpm under tropical day con-ditions (90�F) and wave-off condition, one engine out, at max-imum landing � .1, W � .7Wo, T � To�.CD0


5. Landing distance not to exceed 1,500 ft at 103�F. W � .7Wo,� .1, brakes only � � .3, L � T � 0, 2 sec free roll.CD0

The constraints will be considered in the same order as given in therequirements statement above. Note that only for simplicity, five areused from Table 8.1.

1. Sea-level take-off The only other item is to establish the airdensity at 103�F as

3� � .002377(520/563) � .0022 sl/ft

Thus, the landing equation will be used as given above:

W1.44�gCLmaxT So �

Wo W/Ss � 1.2tT r��CLmax

1.44W/S

.0022(32.2)1.7�

21500 � 1.2 �W/S�.0022(1.7)

11.96W/S�

1500 � 27.75�W/S

This permits setting up a table of values, W � Wo, T � To

2. Level flight maximum speed, .9M, 30,000 ft At 30,000 ft � �.375, V � Ma � .9(995) � 896 ft/sec.

n � 1

1 .00089072 2 2q � �V � 896 � 357 lb/ft2 2

T � T (1.45).375 � .544To o

Thus, the maximum speed constraint becomes

C qT kD0� � W/SW W/S q


.544T .018(375) .15(.8)W /So o� �.8W .8W /S 357o o

T 11.81o � � .000494(W /S),oW W /So o

which then yields

W /S 20 40 60 80 100o

T /W .6 .315 .226 .187 .168o o

3. Maintain a minimum ROC � 100 fpm at 40,000ft At 40,0000ft � � .246, V � Ma � .8(968) � 774 ft/sec.

n � 1

.002377 2 2q � (.246)774 � 176 lb/ft2

T � T (1.4).246 � .344To o

dh 100P � � � 1.667 ft/secs dt 60

The general constraint equation becomes

C qT kD0P � V � � W/S� �s W W/S q

.344T .018(176) .15o1.6677 � 774 � � .8W /S� �o.8W .8W /S 176o o

T 9.209o � � .00159W /S � .005oW W /So o

which yields the following set of values for the thrust and wingloading:

W /S 20 40 60 80 100o

T /W .5 .3 .254 .247 .256o o

4. Maintain an ROC of 500 fpm at wave-off, one engine, sealevel At landing conditions � .1, W � .7Wo, � � 520/CD0

550 � .945. Speed of sound a � 1,116/.945 � 1,148 ft/sec.


n � 1

ToT � � � .4725To2

2(.7)V � 1.1V � 1.1 �W/S � 21.06�W/Ss �.002377(.945)1.7

.002377 2q � .945(21.06) W /S � .5W /So o2

dh 500P � � � 8.33 ft/secs dt 60

Using again the general constraint equation:

C qT kD0P � V � � W/S� �s W W/S q

8.33 � 21.06�W/S

.475T .1(.5)W /S .15(.7)W /So o o� � � , or� �.7W .7W /S .5W /So o o

T .582o � � .4137W �W /So o

The corresponding set of thrust-wing loading values are

W /S 20 40 60 80 100o

T /W .544 .506 .489 .479 .472o o

5. Landing distance, sT � 1,500 ft At landing conditions: CD0

� .1, � � .3, L � T � 0, tr � 2 sec. Hot, 103�F day, � ��o(520/563)�.924 �o.

W � .7Wo

2(.7)W /SoV � 1.1V � 1.1 � 21.3�W /SL s o�.002377(.924)1.72q � .245(.002377).924(21.3) W /S � .244W /So o

The landing constraint equation, from 2. above, becomes


1.21W/S1,500 � t V �r L C qD0�gC � �� Lm W/S

1.21(.7)W /So1,500 � 2(21.3)�W /S �o .1(.244)W /So.002377g1.7 � .3� �.7W /So

1,500 � 42.6�W /S � 19.44W /So o

Solving the quadratic in yields Wo/S � 60.2 a result that is�W /So

independent of the thrust loading T /W. The five limiting curves re-sulting from the means part calculations have been plotted on Figure8.1.

These curves, arising from the requirements set above, show that aminimum sea-level thrust–weight ratio of .5, at Wo/S � 44, is neededto satisfy the initial constraints. If now the wave-off constraint wereeliminated, then the thrust–weight ratio would be reduced to about .34and the wing loading would drop to about 35. Such a move wouldconsiderably alleviate the initial thrust requirements and, in turn, yielda lighter aircraft design.

Whether totally realistic or not, this example serves to demonstrateusefulness of the constraint plot in establishing practical specificationsfor new aircraft and then to evaluate the implications on design andtechnology.

F15

8.2 ENERGY METHODS 245

8.2 ENERGY METHODS

The problem of determining aircraft performance optimum conditions(fastest climb, highest altitude, minimum fuel consumption, etc.) hasbeen around since the advent of flight. For earlier low-performanceaircraft, the standard point-performance techniques (discussed and usedextensively in preceeding chapters) were sufficient to estimate the air-craft performance potential by using methods of single variable ordi-nary calculus. Faster and higher flying aircraft with more complex andsupersonic missions have brought along the need to optimize the entireflight path. Thus, there are more variables to consider with attendantneed for increased accuracy, and the point-performance (also calledquasi-steady-state) methods must yield to an integral performance ap-proach requiring use of calculus of variations or variational theory. Thisusually results in an increased level of calculational complexity and aloss of generality due to numerical computer solutions of nonlineardifferential equations.

A simplified method, based on calculus of variations, is the energy-state (also called the energy-height) approximation, which has beenshown to agree well with exact solutions. Here the aircraft is consideredto be a point mass and its state is given in terms of its total energy,consisting of the sum of potential and kinetic energy. Now altitude andvelocity are included in a single variable, energy, which then allowsthe use of the point-performance methods to build up approximate op-timal trajectories.

The total energy concept was already introduced in Chapter 2 andused in Chapters 4 and 7 with the energy-maneuverability diagram.Aircraft specific energy is defined as

2E Ve � � h � � h , (energy height) (8.1)eW 2g

and has the dimensions of length. Since h represents altitude, the unitsof he are given in feet or meters, and the specific energy is often calledthe energy height. It represents the theoretical altitude h that the aircraftcould reach if all of the kinetic energy (V2/2g) were converted (by azoom) into potential energy. Figure 8.2 shows constant specific excesspower and energy height curves on the altitude-airspeed map. Any air-craft at point A, at an altitude of 30,000 ft and at 400 knots (676 ft/sec) has an energy height of 37,088 ft because it is moving with avelocity of 676 ft/sec. Ideally, it could zoom to an altitude of 37,088ft (with a final zero speed), or it could dive to sea level with a final


Figure 8.2 he and Ps Constant Lines; Conventional and Optimum Paths

(theoretical) velocity of 1,545 ft/sec (915 knots). In either case, itstotal specific energy remains at 37,088 ft.

Consider now points B and C on the he � 45,000 ft line. From pointA to either points B or C there are many possible paths to follow butno matter which path is taken there will be a change of 7912 ft (45000-37088) in the energy height. Both B and C have the same specificenergy state, both of which are at a higher level than point A. Theability of an aircraft to change its energy state is called energy maneu-verability, as already seen in Chapter 7. The fundamental reason forthe energy maneuverability method is to determine how to move fromone energy state to another in some optimal manner—that is, the fastesttime from A to B, minimum fuel flight from A to C, and so on. Aminimum time transfer will be considered first, with other options in-troduced later.

The usefulness and need for the energy method will be demonstratedby considering flight from point A to point C, which requires an in-crease in the energy state. This can be accomplished by simply increas-ing the thrust to accelerate, at constant altitude, to a higher velocity. Itwill be seen later that it would be faster to dive first to a lower altitudewhere energy can be increased quickly and then zoom climb to reachpoint C.

The ability to change aircraft energy state comes from the fuel en-ergy and is commonly expressed by the specific excess power

P � DV (T � D)VaP � � (8.2)s W W

which is equal to the rate of change of specific energy dhe /dt. FromChapter 2, Eq. 2.20:


Figure 8.3 Minimum Time Path by Energy Method

dh dE/W dh V dV (T � D)Ve � � � � � P (8.3)sdt dt dt g dt W

The left-hand side (kinematics) represents the sum of accelerationand the rate of climb and describes its maneuvering behavior. The right-hand side (aircraft specific power generation) defines the maneuveringcapability (i.e., the energy needed to accelerate and/or climb, or movefrom, say, point A to point B or C). It should be noted that Eq. 8.3 issimply a restatement of the fundamental performance equation used inprevious chapters.

Since Ps plays an important role in determining aircraft climb-acceleration capabilities, it is useful to briefly review the origin andmeaning of the numerical values of the specific power. Specific excesspower can be developed by use of Eq. 3.24 to yield

2qCT kn W/SD0P � V � � (8.4)� �s W W/S q

which shows that the specific power is also a function of the load factor.At a load factor of unity, if Ps � 0, thrust is equal to drag and a levelflight is described. When plotted and superimposed for a typical aircrafton Figure 8.3, the result is a one-g level flight envelope. As the loadfactor increases, the values of Ps (specific excess power) will decreasewith the attendant shrinking of the flight envelope. This is depicted inFigure 8.2 for the subsonic flight where the constants C1, C2, . . .represent either an increased g-level or a reduced value of Ps, or both.At the same time, a Ps � 0 contour defines, at a particular load factor,a constant-speed level altitude turning performance already studied


with the doghouse plot in Chapter 7. When Ps is negative, the aircraftwill lose altitude or decelerate. Local climb (zoom) may be accom-plished, but at a loss of airspeed.

Consider now an F-18 aircraft flying at 30,000 ft with a velocity of726 ft/sec, .7M (q � 334 lb/ft2). Its weight is 38,000 lb, wing area is400 ft2, CD � .0245 � This gives a lift coefficient CL � .2852.13C .L

and a drag coefficient CD � .035. If the local thrust/weight ratio is .4,the excess power is (from Eq. 8.2):

T qC 334 � .035DP � V � � 726 .4 � � 201 ft/sec (8.5)� � � �s W W/S 95

The positive value of Ps means that the aircraft can accelerate orclimb or a combination of both as long as the sum (the total specificenergy rate) does not exceed 201 ft/sec. In this case, the instantaneousclimb rate is

dh� 201 ft/sec�dt V�c

at one g and .7M. A level acceleration can be carried out at

dV P g 201 � 32.2s 2� � � 8.9 ft/sec�dt V 726h�c

A level flight cruise at .7M requires a reduction of thrust until Ps �0. This clearly shows the usefulness of the specific energy in consoli-dating variables (h,V) and in representing aircraft performance capa-bilities.

Returning now to the problem of determining optimal paths, specif-ically to find the minimum time to change from one speed and altitudeto another, the time can be expressed as (similar to Eq. 4.5)

h he2 e21 1t � dh � dh (8.6)e e

h hdh Pe1 e1e s

dt

where

dh (T � D)Ve � � ƒ(V, h) � Psdt W


The problem now becomes one of minimizing the above integral. Thecalculus of variations provides a simple condition for minimizing thisintegral (Rutowski):

� dh � dhe e� 0 or � 0� � � ��V dt �h dth �c h �ce e

or

� (T � D)V � (T � D)V� 0 or � 0 (8.7)� � � ��V W �h Wh �c h �ce e

Equation 8.7, in any of the equivalent forms, is a requirement that theminimum time path satisfy this condition at every point. An analyticalexpression for the path cannot be defined in general, but the conditionitself is sufficient to provide a solid basis for determining a minimumtime path using the energy state concept.

This is easiest seen if one returns to the conventional method (Chap-ter 4) where the minimum time (maximum rate of climb) conditionsatisfies the following condition:

� (T � D)V)� 0 (8.8)� ��V W h�c

Equation 8.8, simply stated, implies that, at a given altitude, maximumspecific excess power (Ps) defines the best climb rate. See also Figure8.2, where the meaning of Eq. 8.8 is that the lines of constant altitudeare tangent to Ps � const curves (see also Figure 4.5). The implicationsfrom conditions given by Eq. 8.7 are similar. The locations of thetangent point of constant he curves to the Ps � const curves define theminimum time path—that is, the optimum velocity at a given altitude,which also amounts to operating at maximum excess energy per poundof aircraft.

At lower altitudes, the path given by the energy state method shownin Figure 8.2 differs little from the conventional one but starts deviatingat higher speeds and altitudes. The difference is seen by comparing thetangency conditions of the conventional and optimal paths. At super-sonic speeds, the differences become more pronounced. Indeed, theconventional process, as given by Eq. 8.8 (see also Figure 4.5), is notappropriate because of the large speed changes involved, as indicatedin Figure 8.3. The energy method is uniquely suited for this case.


Following Figure 8.3, a supersonic aircraft initially climbs at sub-sonic speed until it reaches the intersection of an he � constant line,which is tangent to two equal-valued Ps lines (the dashed line goesthrough two A quick dive follows at constant energy (he) untilCs).2

the high-speed portion of C2 is reached. Then a climb at increasingMach number is accomplished according to Eq. 8.7 following the he

� Ps tangency points to a desired altitude and speed, or as defined bythe dynamic pressure-operating envelope limit. Since the aircraft is thenwithin the operating envelope (Ps � 0), higher altitudes to, say, pointM at Ps � 0 can be reached by a (constant he) zoom; losing some speedand gaining altitude.

A shortcoming of the energy optimization method, for the aircraftshown in Fig. 8.3, is that the conversion of energy (from potentialenergy to kinetic in dive, reverse in zoom) is theoretically accomplishedin zero time due to �he � 0 in Eq. 8.9 along constant energy lines.The near-sonic dive portion to the Ps bubble appears mainly in aircraftwith marginal sonic and supersonic excess thrust, and some judgmentmust be exercised in establishing the dive path. Then time can be es-timated by elementary kinematic techniques.

For modern high-performance aircraft with more powerful andlighter engines, the subsonic-supersonic conversion via a dive may notappear, since the excess energy (Ps) is given by a rather regular layerof curves ascending to higher altitudes and toward Ps � 0, Figure 8.4(see Example 8.2). The optimum path then simply follows the he � Ps

tangency points (the dashed lines) according to Eq. 8.7. However, azoom technique may still be necessary to reach higher altitudes abovethe curve defining the optimum flight path.

The overall path time from one energy level (h1, V1) to another (h2,V2) can be evaluated by one of the following two methods. The firstone is a simple approximation of Eq. 8.6 by a summation process

he2 1 1 �het � dh � �h � (8.9) e eh dh dh /dt Pe1 e e savg

dt

which is a process analogous to the one used with the standard climbtechnique given by Eq. 4.5. Now the integration limits are set by he2

and rather than altitudes h2 and h1.he1

The second approach arises from the integral side of Eq. 8.9. Figure8.5 shows the integrand (1/Ps) plotted as a series of curves for constantaltitude between the limits and Minimizing the integrand ish h .e e1 2


Figure 8.4 Excess Energy Curves

accomplished by selecting the points on the constant altitude curvesaccording to Eq. 8.7 (i.e., at tangency points to the constant altitudecurves). A sequence of such points establishes the optimum path en-velope (shown by the short dashed lines), and the area under the en-velope represents the minimum value of the time integral.

The standard climb curve, according to Eq. 8.8 (longer dashed lines),is also shown as the locus of the minimum points at constant altitudecurves. The area under this curve is somewhat larger that the one cor-responding to the optimum climb. Thus, the aircraft flying the optimumpath reaches the same energy height before the aircraft using the stan-dard climb technique. As previously pointed out, the difference in areas(climb time) is small at lower altitudes and speeds, but it shows asubstantial increase at higher altitudes. The running time and the dif-ference in the time for the energy path and the conventional path isshown in the upper part of Figure 8.5.

For commercial and general aviation aircraft, flying at more modestspeeds and altitudes, the energy advantage is of no importance. Thequestion of how to get from point A to point B, or to climb from take-off to cruise altitude, with a minimum fuel consumption, is an item ofmuch more consequence.


Figure 8.5 Minimum Time Paths

The amount of fuel required to fly from one energy level toh he e1 2

is given by

h he2 e2dw dw 1f fw � dh � dh (8.10)f e eh hdh dt dhe1 e1e e

dt

Since the rate of fuel consumption dwf /dt is usually expressed as

dwf � CTdt

where C is the thrust specific fuel consumption and is usually a func-tion of speed and altitude, then Eq. 8.10 can be written as

he2 CTw � dh (8.11)f e

h Pe1 s


Figure 8.6 Minimum Fuel Path

This integral is minimized (the condition of minimum fuel path) when

� P � Ps s� 0 or � 0 (8.12)� � � ��h CT �V CTh he�c e�c

which is similar to Eq. 8.7.A typical minimum fuel path solution is shown in Figure 8.6 with

some of the computational details given in Example 8.2. The sameaircraft is used for both minimum time and minimum fuel problems.The amount of fuel can be calculated by the same technique as usedin Figure 8.5 but one plots now CT /PC � 1/ƒs against he for constantaltitude as a parameter. The area under the envelope curve representsthe amount of fuel used.

P3C


EXAMPLE 8.2

Consider a hypothetical supersonic capable aircraft with the follow-ing characteristics:

2C � C � kC , C � .016D D L D0 0

k � 1(ARe), e � .8M/62S � 480 ft , AR � 5

mT � T (1 � .5M)� , T � 20,000 lbo o

m � .7 h � 40,000 ft

m � .8 h � 40,000 ft

C � C ��, C � .85(1 � M/6) lb /lb /hro o m f

The purpose of this excercise is to determine:

a. Minimum time flight pathb. The time along this flight pathc. Minimum fuel flight path

a. Minimum time path is obtained in three steps.

1. Calculate, for a number of altitudes, Ps as a function of Machnumber from Eq. 8.3:

(T � D)VP �s W

and plot Ps against Mach number for a number of altitudes.2. Obtain constant Ps curves by crossplotting the Ps values, at

constant altitude, against Mach number, as shown in Figure8.4.

3. Obtain the minimum time energy path from the he � constanttangency points to the Ps � constant curves.

Several such points are shown. One (B), where he � 80,000 andthe other (C), for he � 118,000 constant curves are tangents to PS

� 400 and 300 curves, respectively. The others are found in Tables8.1 and 8.2.


TABLE 8.2 Standard Flight Path, Time

Point Ps he Alt. M �t

A1 469 54900 23k 1.45B1 410 69300 30k 1.6C1 300 98000 43.5k 1.94D1 100 117000 61k 1.94

TABLE 8.3 Optimal Path Points, Time

Point Ps he Alt. M �t

A 450 62800 20k 1.6B 400 80000 30k 1.8C 300 118000 40k 2.3D 100 131000 43.5k 2.44

b. The time along the flight path, from one altitude to another, isobtained by plotting 1/Ps curves, at constant altitudes against he.The numerical values have been computed already in a. The calcu-lations can be carried out by use of Eq. 8.9 in the difference formor the integral form. Here, both have been used. The results areshown in Figure 8.5.

The area under the short dashed lines gives the time for the op-timum path between any desired energy height he. The longer dashedlines give the time computed under the conventional method. Theshort dashed lines represent the envelope drawn to the constant al-titude curves which represents time for the optimum climb path.

To assist in obtaining and verifying the results for the optimumand standard fight paths, the pertinent data is given in the tablesbelow. In both tables columns are shown for Ps, he, altitude, andMach number at the four points, also shown in Figure 8.4. The timecolumn is left open to be filled out by the reader.

Table 8.2 contains the data for the path to be calculated by thestandard techniques (Chapter 4). The calculation procedure is givenfor both the standard method and the optimal path by Eq. 8.9. Forthe standard method one uses the value of he as found by the constantaltitude tangent line to a Ps constant line (i.e., at the points A1 . . .D1. The optimal path is calculated from the data found at points A. . . D.

Table 8.3 lists the data for the optimal path points.


TABLE 8.4 Optimal Path Points, Fuel

Alt. 1 /ƒs he �lb

10k .01295 30,00020k .01295 42,00030k .01095 57,50040k .0106 72,00050k .01024 90,500

The results obtainable from both tables, as compared to Figures8.4 and 8.5, may not agree precisely due to roundoffs in the tablesand due to plotting of lines with high curvature to obtain tangentpoints.

When calculating path time from Tables 8.1 and 8.2 (or any otherdata source), extreme care should be taken in interpreting the results.The energy method delivers the results, shows advantage, betweenspecific energy values—but not between two altitudes. For example,the times between points A1 and C1 (standard) and between A andC (optimal) turn out to be (about) 115 and 146 sec, respectively.Even if the optimal path aircraft spends only, say 4 seconds to climbto same 43,500 ft altitude, it is still 30 seconds behind. However, ithas a much higher specific energy due to higher Mach number. Thismay be of tactical advantage.

When comparing same specific energy end points, D1 and C, thenthe standard-path aircraft needs another 100 sec to get to point D1

and the tables are turned in favor of the optimal path approach.c. Minimum fuel path can be calculated by plotting 1/ƒs � CT /

Ps against the energy height he, as shown in Figure 8.6. Area underthe envelope drawn to the constant altitude curves gives the fuel usedbetween any energy heights. It turns out that the envelope curvefollows very closely the minimum points on the curves. Since it ismuch easier to locate the minimum points during calculations, it ispractical to carry out the minimum fuel calculations on the curvedrawn through the minimum points, rather than through the tangentpoints.

c. To facilitate following through with some numerical results,Table 8.4 gives the values for 1/ƒs for a number of altitudes shownin Figure 8.6. � is the local atmospheric temperature ratio.

PROBLEMS 257

PROBLEMS

8.1 In Example 8.2, show the energy path from sea level at M � 1.0to Ps � 0 at M � 2.0.

8.2 In Example 8.2, calculate and compare the flight times for theenergy path and the one obtained by conventional method be-tween 20,000 ft and 40,000 ft. Compare your results with thevalue obtainable from Figure 8.5.

8.3 In Example 8.2, calculate the amount of fuel used by a climbfrom 20,000 ft to 50,000 ft.Ans: 660 lb.

258

A

Properties ofStandard Atmosphere

U.S. Standard Atmosphere, 1962

hft

ToR

plb / ft2 � �

aft / sec

� � E7slug/ ft /sec

� � E4ft2 /sec

�2000 526 2274 1.074 1.060 1124 3.75 1.49�1000 522 2193 1.037 1.030 1120 3.73 1.52

0 519 2116 1.000 1.000 1116 3.72 1.561000 515 2041 .964 .971 1113 3.70 1.602000 512 1968 .930 .943 1109 3.70 1.643000 508 1897 .896 .915 1105 3.66 1.684000 504 1828 .864 .888 1101 3.64 1.725000 501 1761 .832 .862 1097 3.62 1.776000 497 1696 .801 .836 1093 3.60 1.817000 494 1633 .772 .811 1089 3.58 1.868000 490 1572 .743 .786 1085 3.56 1.909000 487 1513 .715 .762 1081 3.54 1.95

10000 483 1455 .688 .739 1077 3.52 2.0011000 480 1400 .661 .716 1073 3.50 2.0512000 476 1346 .636 .693 1069 3.47 2.1113000 471 1294 .611 .671 1065 3.45 2.1614000 469 1243 .588 .650 1061 3.43 2.2215000 465 1194 .564 .629 1057 3.41 2.2816000 462 1147 .542 .609 1053 3.39 2.3417000 458 1101 .520 .589 1049 3.37 2.4018000 454 1057 .499 .570 1045 3.35 2.5419000 451 1014 .479 .511 1041 3.33 2.6120000 447 973 .460 .533 1037 3.31 2.6121000 444 933 .441 .515 1033 3.28 2.68


PROPERTIES OF STANDARD ATMOSPHERE 259

hft

ToR

plb / ft2 � �

aft / sec

� � E7slug/ ft /sec

� � E4ft2 /sec

22000 440 894 .422 .500 1029 3.26 2.7623000 437 856 .405 .481 1024 3.24 2.8324000 433 820 .388 .464 1020 3.22 2.9225000 430 785 .371 .448 1016 3.20 3.0026000 426 752 .355 .433 1012 3.17 3.0927000 422 719 .340 .417 1008 3.15 3.1828000 419 688 .325 .417 1003 3.13 3.2729000 415 658 .311 .388 989 3.11 3.3730000 412 629 .297 .374 995 3.09 3.4531000 408 600 .284 .361 990 3.06 3.5732000 405 573 .271 .347 987 3.04 3.6833000 401 547 .259 .335 982 3.02 3.8034000 397 522 .247 .322 977 3.00 3.9335000 394 498 .235 .310 973 2.97 4.0436000 390 475 .224 .298 969 2.95 4.1837000 390 453 .214 .284 968 2.96 4.3838000 390 432 .204 .271 968 2.96 4.5939000 390 411 .194 .258 968 2.96 4.8240000 390 392 .185 .246 968 2.96 5.0641000 390 373 .176 .235 968 2.96 5.3142000 390 356 .168 .224 968 2.96 5.5743000 390 339 .160 .213 968 2.96 5.8444000 390 323 .153 .203 968 2.96 6.1345000 390 308 .146 .194 968 2.96 6.4350000 390 242 .115 .152 968 2.96 8.1855000 390 190 .090 .120 968 2.96 10.4160000 390 151 .071 .094 968 2.96 13.2265000 390 118 .056 .074 968 2.96 16.8370000 392 93 .0430 .058 971 2.96 21.55175000 395 73 .0345 .045 974 2.96 27.6780000 398 58 .027 .036 978 2.96 35.0585000 401 46 .022 .028 981 2.96 44.6390000 403 36 .017 .022 985 2.96 56.7195000 406 29 .014 .017 988 2.96 71.98

100000 409 23 .011 .014 991 2.96 91.22

h altitude T temperature a speed of soundp pressure � � p / po pressure ratio � � � / �o density ratio� coefficient of

viscosity� � � / � coefficient of kinematic

viscositypo � 14.7 psia�o � .002377slugs / ft3

� � T / T0 temperature ratio T0 � 519�R

260

B

On the Drag Coefficient

Aircraft drag can be expressed as

1 2D � �V C SD2

where CD is the aircraft total drag coefficient. A typical set of curvesof the drag coefficient is shown in Figure B.1 as a function of the Machnumber for selected lift coefficients. It is seen that the drag coefficientremains nearly constant in the subsonic region up to the critical Machnumber and starts increasing rapidly in the transonic range after reach-ing the drag divergence Mach number. In the supersonic range the dragcoefficient tends to decrease rapidly.

The level of depends on the type of aircraft (glider, fighter,CD0

transport, etc.). For some typical data of see Appendix C.CD0

The drag coefficient CD plays a crucial role in aircraft performanceevaluation because it represents most of the aircraft aerodynamics inits variation with the flight parameters.

In its accurate and complete representation, the drag coefficient maybe given in a graphical (see Figure B.1) or piecewise polynomial formto accommodate both the required lift and Mach number ranges. Forpractical engineering calculations where analytical results are desirable,the following expression is used for the drag coefficient of an airplane:


ON THE DRAG COEFFICIENT 261

Figure B.1 A-4M Drag Rise Characteristics

C � C � CD D D0 L

2� C � kC (B.1)D L0

where is the drag due to lift and is the zero lift drag coefficientC CD DL 0

due to parasite (viscous � form) drag. Due to its parabolic shape anddue to its early representation in polar form, Eq. B.1 has acquired thelabel drag polar.

Eq. B.1 is a direct consequence of finite wing theory results andworks best for uncambered aircraft in low subsonic range. The coeffi-cient k is given by

1k �

�ARe

The Oswald span efficiency factor e is a function of the aspect ratio,increasing with increasing aspect ratio. Typical values range from .6 to.9, but it becomes a function of the Mach number and CL near and

262 ON THE DRAG COEFFICIENT

Figure B.2 A-4M Efficiency Factor

above the critical Mach number where it exhibits a continuing decreaseto about 50 percent of its low speed value, Figure B.2.

Although the efficiency factor e depends on a number of parameters,the following correlation for wing-body combinations should serve asa first order of approximation:

e � 4.61(1 � .045AR � 68)cos � � 15 � 3.1

where � should be determined by the locus of t /c�max for the wing. Inpractice, leading-edge value �LE is used for quick estimation.

Typical drag polar at subsonic speeds for A-7E aircraft is shown inFigure B.3. At supersonic speeds the form of Eq. B.1 is still valid butthe wave drag effects need to be included in both k and due toCD0

their lift and nonlift contributions, respectively.Supersonic drag can be estimated from

Total drag � Parasite drag � Wave drag � Induced drag

or

ON THE DRAG COEFFICIENT 263

Figure B.3 A-7E Drag Polar

Figure B.4 Cambered Aircraft Polar

264 ON THE DRAG COEFFICIENT

C � C � Zero lift wave drag � Wave drag due to liftD Dp

� Induced drag

At its simplest, it can be written as

22 �M � 149(t /c) 2 2C � C � � C � kCD D L Lp 2 4�M � 12� C � C � (k � k)CD D 1 L0 0w

2� C � KCD L0

An improved version of k1 in supersonic flight is

2AR(M � 1)k � cos �1 24AR�M � 1 � 2

In transonic region, K � k transitions slowly from k to k � k1. Intransonic region, beyond the critical Mach number, more advanced cor-relations must be used to estimate the zero lift drag rise to low super-sonic Mach numbers.

For an aircraft with camber (or with flaps extended), the drag polarwill still be a parabola, but it will be shifted up along the lift axis, asshown in Figure B.4. Thus, the drag polar is given by a simple trans-lation:

2 2C � C � k(C � C ) � k C (B.2)D D L L 1 Lmin o

For many aircraft with relatively low camber, the difference betweenand is small and may be neglected. Thus, is usually usedC C CD D D0 min 0

in practice. For clarity and simplicity, Eq. B.1 is used throughout mostof the development in this book.

265

C

Selected Aircraft Data


266

Airc

raft

Wm

We

SA

RC

D0

ke

L/D

mR

CL m

Wf

767-

200

320,

000

177,

000

3,08

07.

9.0

18.8

5,20

011

2,00

074

7-20

083

0,00

038

0,00

05,

500

7.0

148

.85

17.7

6,80

02.

531

0,00

073

7-20

013

9,00

073

,700

1,14

07.

9.0

1917

.62,

700

3.2

34,5

00D

C10

-30

555,

000

241,

000

4,60

07.

6.0

162

17.7

4,70

02.

5A

330-

200

507,

000

266,

000

3,90

010

.018

17.5

5,60

024

7,00

0D

C3

25,0

0017

,700

987

9.14

.024

9.7

514

.7D

HC

834

,500

23,0

0058

512

.4.0

2.0

322

.81.

55,

700

A-4

M24

,500

10,5

0026

02.

91.0

213

.132

.82

1,90

05,

440

A-7

42,0

0019

,000

375

4.0

15.1

055

.87

2,25

01.

610

,000

F4E

62,0

0029

,000

530

2.8

.022

48.

61,

900

F5A

21,0

008,

100

170

3.8

2F1

5C68

,000

28,0

0060

83

.023

.133

1.78

5F1

6A35

,400

14,6

0030

03

.018

1.65

F18A

51,0

0024

,000

400

3.5

.024

3.1

32.

8F1

4A74

,000

41,0

0056

57.

3/1

.9.0

255

Raf

ale

40,0

0037

5.0

225

.11

1.6

Mira

geIII

30,0

0016

,000

375

1.9

.013

1P

iper

2,45

01,

400

170

6.0

358

.76

101.

7C

hero

kee

No

tes.

Mos

tof

the

dat

ain

this

tab

lesh

ould

be

cons

ider

edas

app

roxi

mat

e,or

aver

age,

and

may

cros

sth

esp

ecifi

catio

nsof

seve

ral

confi

gura

tions

.Fo

rex

amp

le,

the

max

imum

wei

ght

Wm

isa

func

tion

ofth

eai

rcra

ftlo

adin

gan

dth

eam

ount

offu

elon

boa

rd.

Thus

,it

may

vary

sub

stan

tially

and

also

have

anim

pac

ton

the

valu

esin

the

rang

e(R

)an

dfu

elw

eigh

t(W

f)co

lum

ns.

Sim

ilarly

,th

ed

ata

inth

ed

rag-

due

-to-

lift

(k)

colu

mn,

the

effic

ienc

y(e

),an

dth

eva

lues

for

Wm

–the

max

imum

wei

ght

and

We–t

heem

pty

wei

ght

rep

rese

ntso

me

avai

lab

leav

erag

eva

lues

and

shou

ldb

eus

edw

ithso

me

caut

ion.

267

D

Thrust Data forPerformanceCalculations

TURBOJETS

Engine data, for several sample engines, are given for J-60, J52, JT9D-3, JT8D-9, TF-30, TFE731-2, GE F404-400, FJ-44, Allison T-56 turbo-prop, and AVCO Lycoming I0-540 reciprocating engines. Although allthese engines are somewhat dated and all have been superceded to givehigher thrust and better fuel consumption, they do represent typicalperformance trends. As is clearly seen, the thrust and specific fuelconsumption (SFC) curves vary widely with speed and altitude. Thus,there are no general-duty expressions available that would permit car-rying out easy and simple performance calculations. However, severalapproximate expressions, shown below, have been used with some en-gineering success. As a first-order rough approximation, engine thrustcan be scaled linearly (for similar engines) and the fuel consumptioncan be decreased by at least 5 percent.

The process consists of curve-fitting the thrust equation as a functionof altitude, velocity, or both. Each engine may require its own specialcurve-fitting expression and may have to be accomplished in a piece-wise fashion over the velocity and/or altitude range.

Jet Engines

For subsonic flight the simplest correlation is


268 THRUST DATA FOR PERFORMANCE CALCULATIONS

Figure D.1 Pratt and Whitney J-60 Turbojet Engine

Figure D.2 AVCO Lycoming IO-540 Reciprocating Engine

T � T � (D.1)ref

where Tref may be taken as To, the sea level thrust value.A somewhat improved expression is

TURBOJETS 269

Figure D.3 AVCO Lycoming IO-540 Reciprocating Engine

nT � T � (D.2)ref

which is often written as

T n� � h � 36,089 ftTo

� � h � 36,089 ft

A better correlation, but more cumbersome to curve fit and to use, is

T 2� (A � BV )� (D.3)To

The advantage of Eq. D.3 lies in taking into account the realisticthrust variation with velocity. Usually this is not very significant athigher altitudes and velocities but may be 10 percent or more during


Figure D.4 Pratt and Whitney TF-30 Turbofan Engine

the take-off portion of flight (see TF-30 and JT9D data). At higherspeeds, another correlation that has been used is

T� (1 � cM)� (D.4)

To

where, typically, .25 � c � .5.Specific fuel consumption varies with both altitude and velocity and

defies generalization with both of those parameters. It has been foundthat the velocity effect can be correlated for some engines, very ap-proximately, by

TSFC� 1 � .5M (D.5)

TSFCref

TURBOJETS 271

Figure D.5 Williams/Rolls FJ-44 Turbofan Engine

Reciprocating Engines

Reciprocating engines admit more generalizations:

• BHP is independent of velocity V.• SFC tends to be independent of both velocity and altitude.

For engine brake horsepower, the commonly accepted altitude vari-ation is


Figure D.6 Garrett TFE-731-2 Turbofan Engine

BHP� 1.132� � .132 (D.6)

BHPo

where subscript o refers to the sea-level value.For supercharged engines, it is assumed that BHP remains constant

to at least 25,000 ft altitude. Correlations used for higher altitude su-percharged engines are:

BHP .765� � , h � 36,089 ftBHPo

� 1.331�, h � 36,089 ft (D.7)

TURBOJETS 273

Figure D.7 GE F404-400 Installed Performance


Figure D.8 Pratt and Whitney JT8D-9 Turbofan Engine

Figure D.9 Pratt and Whitney JT9D-3 Turbofan Engine

TURBOJETS 275

Figure D.10 Pratt and Whitney J52 Turbojet Engine

Figure D.11 Allison T-56-A Turboprop Engine, Horsepower


Figure D.12 Allison T-56-A Turboprop Engine, Thrust

Figure D.13 Allison T-56-A Turboprop Engine, Specific Fuel Consumption

277

E

Some UsefulConversion Factors

Multiply by to obtain

meters 3.281 feetmeters2 10.76 ft2

Newton, N .224 lb.102 kgf

miles 5280 feet1.609 km

mph 1.467 ft /sec1.609 km/hr.869 knots

km/hr .9113 ft /sec.2778 m/sec.6214 mph.54 knots

m/sec 3.281 ft /sec3.6 km/hr1.944 knots2.237 mph

knots 1.689 ft /sec1.852 km/hr.5144 m/sec

1.151 mph


278

Bibliography

In Historical Order

1. Diehl, W. S. Engineering Aerodynamics. New York: Ronald Press, 1928.2. Collection. Handbook of Aeronautics. London: H.M. Stationary Office,

1931.3. Jones, B. Elements of Practical Aerodynamics. New York: Wiley, 1936.4. Millikan, C. B. Aerodynamics of the Airplane. New York: Wiley, 1941.5. Mises, R. Theory of Flight. Dover: McGraw-Hill, 1945.6. Hemke, P. E. Elementary Applied Aerodynamics. New York: Prentice-

Hall, 1946.7. Wood, K. D. Technical Aerodynamics. Dover: McGraw-Hill, 1947.8. Dwinnell, J. H. Principles of Aerodynamics. Dover: McGraw-Hill, 1949.9. Perkins, C. D. Airplane Performance, Stability and Control. New York:

Wiley, 1949. Hage, R. E.10. Dommasch, D. O., Sherby, S. S., and Connolly, T. F. Airplane Aerody-

namics. New York: Pitman, 1951–1967, 4 editions.11. Miele, A. Flight Mechanics. Reading, MA: Addison-Wesley, 1960.12. Houghton, E. L. Aerodynamics for Engineering Students. Arnold, 1960

and Brock, A. E.13. McCormick, B. Aerodynamics, Aeronautics, and Flight Mechanics. New

York: Wiley, 1979.14. Lan, E., and Roskam, J. Airplane Aerodynamics and Performance.

Roskam Aviation, 1980.


BIBLIOGRAPHY 279

15. Hale, F. J. Intro to Aircraft Performance, Selection, and Design. NewYork: Wiley, 1984.

16. Asselin, M. An Intro to Aircraft Performance. AIAA, 1997.17. Anderson, J. D. Aircraft Performance and Design. McGraw-Hill, 1999.18. Eshelby, M. Aircraft Performance, Theory and Practice. AIAA, 2000.19. Phillips, W. F. Mechanics of Flight. Hoboken: Wiley, 2004.

Two special references:

20. Rutowski, E. S. ‘‘Energy Approach to the General Aircraft PerformanceProblem.’’ Journal of the Aeronautical Sciences, 21 (March 1954).

21. Carson, B. H. Fuel Efficiency of Small Aircraft. AIAA Paper No. 80-1847, American Institute of Aeronautics and Astronautics, Washington,D.C., 1980.

280

Index

Absolute ceiling, 73, 80Accelerating flight, 178Accelerated stall line, 220Angle of attack, 9Aerodynamic force, 8Altitude, 20, 21, 22, 158Aspect ratio, 28, 198, 214Atmosphere, 58, 258

Balanced field length, 151Breguet range equation

jet, 111propeller, 106

Calibrated airspeed, 22Carson’ speed, 127Ceiling

absolute, 37, 42, 73, 80service, 73, 80

Climb angle, 72, 91Climbing flight, see Rate of climbCompressibility, 126Constraint plot, 235Corner (maneuvering) speed, 199Cruise-climb, 107, 113

Doghouse plot, 220Downwash, 159

Dragforce, 8, 21coefficient, 21due to lift, 29, 262parasite, 29, 262polar, 28, 264

Dynamic pressure, 21

Efficiencyoverall, 11, 103propeller, 15, 49, 105thermal, 105

Endurancedefinition, 101jet, 132propeller, 130table of equations, 144

Energyassumptions, 14balance, 11content of fuel, 103equation, 12generation, 12height, 245maneuverability diagram, 220specific, 12state, 246

Equations of motion, 9


INDEX 281

Equivalent airspeed (EAS), 21Equivalent parasite area, 30Equivalent shaft horsepower, 53

Fastest turn, 213Flat turn, 200Flattest glide, 55Flight envelope, 17, 46Flight path angle, 9, 191Forces

general, 8take-off, 154turning, 191, 201

Fuel consumption, 105Fuel energy, 11, 105Fundamental Performance Equation

(FPE), 5, 17, 71

Glide angle, 54Gliding flight

engine offeffect of altitude, 58maximum range, 61

minimum sink rate, 56sink rate for maximum range, 55,

57sink rate for minimum glide

angle, 55partial power, 94

Ground effect, 159, 173Ground run

take-off, 152landing, 173

Ground speed, 136Gust loading, 196

Instantaneous maneuvering, 206, 220

Landingapproach speed, 172flare distance, 176ground run, 173speed, 23

Level flight, 19Lift

coefficient, 20force, 8, 20

Lift curve slope, 197Lift-drag-ratio, 31

Liftoff speed, 23, 152Load factor

general, 28, 190, 192, 193, 195, 200gust, 197limit, 192, 194maximum, 209stall, 212ultimate, 192, 194

Maneuver (corner) speed, 199Maneuvering diagram, 220Maximum climb rate, 78, 84Maximum endurance and range

comparisons, 143Maximum lift-drag ratio, 31Minimum drag

general, 31airspeed, 31

Minimum powergeneral, 48airspeed, 47

Moments, 2

Optimizationpaths, 248

minimum fuel, 252minimum time, 250

Oswald efficiency factor, 28, 261

Plane of symmetry, 201Point performance, 3, 17, 189, 245Power

altitude effect, 272available, 11, 26, 49excess, 78, 80minimum, 48required, 47

Propeller efficiency, 15, 49, 105Pull-up, 191

Rangebasic equation, 102Breguet

jets, 111reciprocating, 105

constant altitude, jets, 114constant velocity

jets, 116reciprocating, 108

282 INDEX

Range (continued )cruise-climb, 107effect of energy change, 128effect of wind, 135factor, 120integration method, 117

simplified, 124table of equations, 144

Rate of climbjets, 72, 73maximum climb angle, 72, 91maximum value, 78, 84quasi-steady, 13reciprocating, 72, 73

Screen height, 151Service ceiling, 73, 80Specific excess power, 12, 18, 221,

246, 247Specific fuel consumption, 105Specific range, 117Stall, 19Stall speed, 21Standard atmosphere, 258

stratosphere, 59troposphere, 59

Sustained maneuvering, 208, 223

Take-offbalanced field length, 151critical speed, 153effect of engine failure, 167effect of wind, 169equations, 159

friction coefficient, 155ground run, 152lift-off (take-off) speed, 23, 152minimum control speed, 153minimum distance, 158thrust augmentation, 167

Thrustavailable, 26, 40equations, 40, 269required, 27, 34, 38

Thrust to weight ratio, 206, 212Tightest turn, 210Time to climb, 75True airspeed (TAS), 22Turn radius, 210Turn rate, 213

Ultimate load factor, 194Unpowered flight, see Gliding flight

Velocityequivalent airspeed (EAS), 21landing, 23maximum, 35, 36, 42, 78maximum glide range, 55, 61minimum, 35, 36, 42, 78minimum glide angle, 55minimum glide sink rate, 55true airspeed (TAS), 22

V-n diagram, 192

Wave drag, 192Wind fraction, 138Wing loading, 158, 234

Technology

Aircraft performance