conformal transformations in electrical engineering

CONFORMAL TRANSFORMATIONS

W. J. GIBBS

CONFORMAL

TRANSFORMATIONSIN ELECTRICAL ENGINEERING

by

W. J. GIBBSD.Sc., M.I.E.E.

CHAPM 1S 44LL LTD37 ESSE EE * C.2

First Published 1958

THE BRITISH THOMSON-HOUSTON CO.. I.TD.

1958

.Catalogue No. 60314

MADE AND PRINTED IN GREAT BRITAIN BYWILLIAM CLONES AND SONS, LIMITED, LONDON AND BECCLES

CONFORMAL TRANSFORMATIONSIN ELECTRICAL ENGINEERING

This Bookhas been written in the interests of those concernedwith advanced theory and practice of engineering, andis one of a series recommended for publication by the

Technical Papers Panel ofThe BRITISH Txonsov-HoujTox Oompany, ]Rugby

i

PREFACE

Although many young electrical engineers of today have afair knowledge of the method of conformal transformations,very few attempt to use it in practice even when the problemthey have to solve lends itself particularly to that method.The reason is that most solutions in practice involve ellipticfunctions and integrals, and engineers' knowledge of such func-tions is in general quite inadequate.

There is no book written for engineers which deals only withconformal transformations and takes the reader up to the stagewhere he can follow the work of Carter and of Coe and Taylor.Miles Walker's Conjugate Functions for Engineers which is theclassic engineering text on this subject and is now out of printdoes not go beyond problems involving elementary functions.Other books also stop short at the very point where the realdifficulties begin, that is, where more than two right angles arein the configuration to be transformed and elliptic functionsenter the analysis. An attempt is made in this book to fill thedeficiency.

In order to make it complete in itself the book covers in theearly chapters the elements of field theory and of complex num-bers both from a mathematical point of view. It is assumedthat the reader is already to some extent familiar with thesetopics. The transformations start with the simplest and pro-ceed to those of increasing difficulty until the final chapter. Thesecond half of the book is entirely devoted to transformationsinvolving elliptic functions.

It is impossible to write such a book as this without beingimpressed by the immense debt owed by engineers and by elec-trical engineers in particular to F. W. Carter. Although othershad already discussed conformal transformations in electro-magnetism from the point of view of theoretical physics it wasCarter who brought the method into the realm of practical workin electrical engineering. His work is described in Chapter 1but he also did some as yet unpublished work on this subject,

v

PREFACE

the results of which are still in constant use by design engineers.The 1928 paper of Coe and Taylor is also described in

Chapter 1. This paper and Carter's 1926 paper were con-sidered, at the time, to be beyond the ability of most engineersto understand. Thirty years have passed and most engineersstill graduate without the mathematical knowledge that wouldenable them to follow papers of this standard. One aim of thisbook is to take the reader to the point where he can not onlyfollow and understand these papers but initiate and carrythrough similar work of his own. The final problem in thisbook is the first of the five dealt with in the paper by Coe andTaylor.

A few years ago Professor G. W. Carter gave a lecture courseat Rugby that included some lectures on conjugate functions.The treatment given here in Chapters 3 to 9 is to some extentbased on his approach and I gratefully acknowledge myindebtedness to him. I would also thank him for his coin-ments on and criticism of the manuscript and for giving me theproofs of theorems duly acknowledged in the text.

I am also indebted for criticism and advice to _Mr. R. T. Coe(whose work, quoted above, has been an inspiration), to Mr.L. D. Anscombe and to Mr. N. Kerruish who, besides goingover the manuscript, checked and corrected the mathematics.I am also grateful to Miss M. Oldfield, who checked much ofthe numerical work and helped with the proof reading.Finally I must thank Mr. G. S. C. Lucas, Director arid ChiefElectrical Engineer of the British Thomson-Houston Company,for his interest in the work and his permission to publish it.

W. J. GIBBSRUGBYMay 1956

vi

CONTENTS

PREFACE

Chapter1 INTRODUCTION

PageV

1

2 FIELD EQUATIONS 4Field distribution problems. Inverse square fields-. Forceequations. Divergence equations. Curl equations. Poten-tial. Laplace's equation.

3 PROPERTIES OF COMPLEX VARIABLES 14Complex numbers. Argand diagram. Polar form. Pro-ducts. Division.

4 CONJUGATE FUNCTIONS 24Potential function. Stream function. Orthogonal curves.Conjugate functions.

5 CONFORMAL TRANSFORMATIONS 35Conformal representation. Transformation of a curve.Transformation of a straight line. Transformation ofshapes. Transformation of a quadrant.

6 THE SCHWARZ-CHRISTOFFEL TRANSFORMATION 56The Schwarz-Christoffel equation.

7 CONFIGURATIONS WITH No RIGHT ANGLES . 60Electric field at the edges of two capacitor plates. Generaldetermination of constants in the transformation. Findingthe origin and axes. Plotting equipotentials and streamlines. Density of charge. Increase of capacitance. Slottedplane surface. Density of flow. Point source. Collinearsource and sink.

8 CONFIGURATIONS WITH ONE RIGHT ANGLE . 87Transformation of a quadrant. Field fringing from pole endto armature core. Distribution of flux density.

1) CONFIGURATIONS WITH Two RIGHT ANGLES . 96Flow through a slit. Single slot opening opposite a solidface. Amplitude of flux density ripple. The lost flux.Carter's coefficient.

vii

CONTENTS

Chapter Page10 CONFIGURATIONS WITH MORE THAN Two RIGHT

ANGLES 124'T'ransformation of the inside of a rectangle.

11 ELLIPTIC INTEGRALS OF THE FIRST KIND 129Jacobi's notation. Legendre's notation. Completeintegrals. Complementary modulus. Transformation of arectangle.

12 ELLIPTIC FUNCTIONS . 139The function sn. More elliptic functions. Current densitycalculation.

13 DOUBLE TRANSFORMATIONS WITH ONE REQUIRINGELLIPTIC FUNCTIONS 148

Electrodes at the edge of a semi-infinite strip.

14 DOUBLE. TRANSFORMATIONS BOTH REQUIRINGELLIPTIC FUNCTIONS 159

Electrodes at one edge of a rectangle. Incomplete ellipticintegrals.

15 ELLIPTIC INTEGRALS OF THE SECOND KIND . 173Incomplete and complete integrals of the second kind.Electrodes at the sides of a projection. Periodicity ofelliptic functions.

16 AUXILIARY ELLIPTIC FUNCTIONS . . 185Jacobi's Zeta function. Jacobi's Theta and J to functions.

17 ELLIPTIC INTEGRAL OF THE THIRD KIND 190Succession of equal slot openings. Elliptic integral of thethird kind. Jacobi's integral of the third kind. -Numericalexample. Amplitude of the flux density ripple. Shape ofthe flux density curve. Carter's coefficient. Equations interms of modular and amplitude angles. Conclusion.

LIST OF REFERENCES 217

INDEX 218

V]11

CHAPTER 1

Introduction

CONFORMAL NAPPING as it is often called is the representation ofa bounded area in the plane of a complex variable by an area inthe plane of another complex variable. Thus the method is abranch of mathematics based on the theory of functions of acomplex variable. Like most other branches of mathematicsit has found applications in the sphere of theoretical physics,mainly through the powerful method of attack developed bySchwarz (1869)* and Christoffel (1867).* A chapter on theSchwarz-Christoffel transformation was included by J. J. Thom-son in his book Recent Researches in Electricity and Magnetism(1893)*.

In 1900 F. W. Carter published a paper entitled `A Note onAirgap and Interpolar Induction'*; this was the first applica-tion of conformal mapping to an actual engineering problem.It was a major advance for it enabled engineers to find bydirect calculation results hitherto only obtainable by roughgraphical methods based on intuition. Carter was not only anengineer but also an able mathematician. He wrote of his ownwork : `Whatever of permanent value is to be found in my worklies entirely in its attention to the correlation of two problems-the mathematical problem and the engineering problem'.In this correlation Carter was supremely gifted, and his applica-tion of conformal transformations to field problems is anexample. Another paper of his, published in 1901,* gave thederivation of `Carter's coefficients' which are still in commonuse by electrical engineers, and it forms the principal subject ofChapter 9 of this book.

In 1926 Carter published a more comprehensive paper onconformal mapping called `The Magnetic Field of the Dynamo-Electric Machine'.* In it twelve different configurations were

* See List of References, page 217.1 1


transformed some of which now required the use of ellipticfunctions. Hence the standard and range of mathematicalknowledge needed to understand the paper was stepped upconsiderably.

After this other engineers contributed to the subject, but themost important paper was that written by R. T. Coe and H. W.Taylor called `Some Problems in Electric Machine DesignInvolving Elliptic Functions'.* In it the authors analysedfour different problems in electric machine design all of whichrequired extensive use of elliptic functions. The paper is tosome degree a follow up of Carter's 1926 paper, but it gives amore detailed analysis of each problem and takes the results tothe point where numerical answers are given and graphs areplotted.

In 1933 Miles Walker published the book Conjugate Functionsfor Engineers.* This work is confined to problems that needonly elementary functions in their solution. Nevertheless thetreatment is more extensive than that of any previous work andthe book was long regarded as the standard text for engineers.

An adequate knowledge of how to deal with fields shouldform an essential part of the engineer's training. Fortunatelythere are several ways of treating inverse square fields, somebased on drawing, some based on models or analogues, andsome entirely mathematical. A good account of the variousmethods is given in the book Two-dimensional fields in Elec-trical Engineering by L. V. Bewley.*

To the question why, with so many techniques at his dis-posal, the engineer needs to know mathematical methods theanswer is that he is not fully equipped unless he is familiar withall the methods available and can apply the appropriate onefor the problem in hand. If he has little mathematical abilityhe must of necessity rely entirely on non-mathematical tech-niques. They may be quite adequate for his purpose, butnevertheless there is considerable satisfaction to be obtainedfrom having these techniques supported by mathematicaldeductions.

A mathematical method never solves the actual problemposed but only an idealisation of it. This appears at first a

* See List of References, page 217.2

INTRODUCTION

disadvantage. But an analogue that simulates the actualproblem only gives immediate answers, and by reason of thefact that all the complications present are included in the set-upthe analogue does not give the engineer a satisfactory basis formental pictures of the factors at work. To take a very elemen-tary example, the current locus of an electric motor not yetmade may be predetermined from an analogue but the resultsobtained would not in themselves suggest anything helpful tothe designer. On the other hand the current locus of theidealised motor might be part of a circle thereby providing amost helpful concept. Hence to the designer the idealisedmachine forms the standard from which the actual machine isa deviation. A similar comment applies to the more reconditeproblems of this book where a mathematical solution of theidealisation gives a better idea of the relative influences of thevarious ratios in the configurations.

It is necessary to emphasise the remarkable identity betweenall the various kinds of fields met with in engineering. Althoughin this book a problem may be stated in terms of electric fields,the solution applies equally well to some other problem statedin terms of magnetic fields and vice versa, provided the fieldsare static. The parallelism extends also to static fields inother branches of physics and engineering.

Finally it should be said that any practical work on this sub-ject entails a fair amount of integration, manipulation ofelliptic functions and computation. It is therefore useless forthe engineer to embark on any project connected with thisanalysis without being equipped with Tables of Integrals andTables of Elliptic Functions as well as the usual MathematicalTables. Any special tables used in this book will be fullydescribed when first used.

3

CHAPTER 2

Field Equations

Field Distribution Problems

Engineers are frequently confronted with phenomena asso-ciated with irregularities of structure. For instance steadyflow of a perfect fluid may be disrupted by a barrier as showndiagrammatically in Fig. 1, and the question arises: What arethe actual characteristics of the flow past the barrier? For an

Fra. 1. Interrupted fluid flow.

example in electrical engineering consider the magnetic fieldbetween two pieces of magnetic material of opposite polarity. Ifthere is a break in one face, as indicated diagrammatically inFig. 2, how can the actual distribution of the lines of magneticflux be determined?

Fia. 2. Example of a magnetic field.4

FIELD EQUATIONS

Such problems are field distribution problems and manymethods of dealing with them are available. Freehand sketch-ing by guesswork as illustrated in Fig. 1 gives a crude idea of thefield distribution. A much more accurate result is obtainedby using the technique of flux plotting. Yet another methodis to devise a model or an analogue and then take readingsof whatever is required. But very often the problem issuitable for mathematical solution by using the device oftransformation.

All field problems are really three-dimensional but a surpris-inglylarge number can be treated as two-dimensional by assumingthat the field distribution does not vary in the third dimen-sion. We can put it another way by saying that the structure istreated as though it extends an infinite distance each way in thedirection of the dimension not being used. Although this isnever quite true it is often a reasonable idealisation of the prob-lem for that part of the structure for which a solution is required.For instance the width of the channel shown in Fig. 1 is per-pendicular to the plane of the paper. When this width is tenor more times the depth shown, it is clearly permissible to ignorethe boundaries of the channel in that dimension for the generalsolution of the problem; the conditions at the boundaries cansubsequently be treated separately. By adopting this pro-cedure we need only consider changes represented on the planeof the paper and the problem becomes two-dimensional.

Such two-dimensional field problems are often readily andelegantly solved by the conformal transformations that formthe subject of this book. Since the method cannot be applied toevery kind of field that may appear in practice, it is necessary toconsider first what are the distinctive properties of the parti-cular fields we have in mind.

Inverse Square FieldsThe fields we are interested in are fields of force called inverse

square fields. They are so named because the magnitude of theforce at various points follows an inverse square law which ismost easily conceived and explained in relation to one parti-cular condition. The condition is that which obtains when the

5

CONFOR'AAL TRANSFORMATIONS

source of the field is concentrated at a point and is completelyisolated. The field does not contain the point but with thisexception pervades all space. The reason for using this conceptof a single point source is that it makes the best starting pointfrom which to develop the equations.

With any point in the field there is associated a value of theforce that would be exerted on any per unit test body placed atthat point. The force on the test body must have direction aswell as magnitude and hence such a field is called a vector field.The point source is the centre of the field and at every otherpoint in it the force is directed either towards or away from thiscentre. Because it is an inverse square field the magnitude ofthe force is inversely proportional to the square of the distancefrom the centre.

In order to describe the magnitude and direction of the forceat any point in the field, a coordinate system is necessary notonly to identify the position of the point in question but also toprovide suitable components of the force at the point; thesecomponents considered together give both the magnitude andthe direction of the force at the point selected. The coordinatesystem can be chosen to suit any particular problem and theform of the resulting equations depends on the coordinatesystem chosen.

At this stage there is no reason why rectangular cartesiancoordinates should not be used. Throughout this analysiseither rectangular cartesian or polar coordinates are used ; bothkinds are orthogonal, i.e. the coordinate curves intersect atright angles.

The Force EquationsFor the present then, rectangular cartesian coordinates x and

y as shown in Fig. 3 will be used. Since the field of force beingconsidered is two-dimensional the component of force in the xdirection can be written Fz and the component of force in they direction Fy. As shown in Fig. 3 this force is acting at thepoint P(x, y) in the plane.

Let r be the distance of the point P from the origin.Then

r2 = x2 + y2

6

FIELD EQUATIONS

FY#

Source

FIG. 3. Components of a force.

and since the field follows the inverse square law the resultantforce F is given by

F=T2-

where k is a constant of proportionality. Then the componentsof force F. and F,, are given by

Fz=Fxrkx

ra

and by a similar process

Fyk3

. . . . . (2.2)r3

Now a field set up in this manner by a single point source canbe handled readily by such simple equations as (2.1) and (2.2).But the fields encountered in practice are usually much morecomplicated because there may be several sources, and thesource or sources may be distributed instead of being concen-trated at points. In dealing with these more complicated fieldsthe analysis is helped considerably by the development of thedifferential equations of the field.

7

CON-FORMAL TRANSFORMATIONS

Divergence Equations

The first step is to set up a differential equation called the.divergence equation. Equations (2.1) and (2.2) can be put interms of x and y by making the substitution

x2+ y2and this produces

F - kx(x2+y2)3j2 . (2.3)

Fky (2.4)y = (x2+y2)3'2

Now find the partial derivative of Fx with respect to x andthat of Fy with respect to y:

aFz2 2 2 2 22 2 3 5

andOX

+y )-= k{(x / -2x (x +y )- / }

aFy _ k{(x2+y2)-3/2 -2y2(x2+y2)-5/2}ay

Summing the two partial derivatives

OF, aFy=

2 2 3/2 _ 2 2 2 2 5/28x

+ay

k{2(x + y )- 2(x + y )(x + y )-}

= 0

It can also be shown that in a three-dimensional field usingrectangular cartesian coordinates

aFx aFy aFzax+ay+az =0

This equation expressed in cylindrical polar coordinates(r, 0, z) is

1 a(rFr) 1 cFB aFzr & +r as + az -0

In other coordinate systems the equation takes other forms.Therefore to express it briefly in a form that embraces all pos-sible coordinate systems it is written

or

divF=0

V.F=0S

} (2.5)

FIELD EQUATIONS

Equation (2.5) which is called the divergence equation isintroduced here only as a step to later developments. For thepresent it need only be noted that in the inverse square fieldsunder consideration the divergence of the field at every point iszero. In two dimensions

aFx aFyC"7- =0 . . . . . (2.6)

The Curl EquationsIn a similar set of equations forming the next step we consider

the rate of change of each component of force in one directionwith respect to the other direction. This leads to what arecalled the curl equations.

From equations (2.3) and (2.4) form the partial derivative ofFx with respect toy and that of Fy with respect to x. Thus

aFx

The two derivatives

(2ax ay

It It can also be shown that for a three-dimensional field usingrectangular cartesian coordinates

OF, aFy _ay az

OF- aF

0

az ax = 0

In cylindrical polar coordinates the equations are1 aFZ _ aFa _ 0r 00 Oz

aFr aFZ = 00Z or

1 a(rFe) OF, 0r ar ae =

vy

aFy

= 2jx(x2+?12)-5/2

2xy(x2+y2)-5/2ax

give identical results and thereforeaFy aFx- = 0

9


In other coordinate systems the equations take other forms.As with the divergence equation, in order to express them in aform that is brief and yet embraces all possible coordinatesystems they are written

curl F = 0or (2.8)

VxF = 0Notice that while V. means divergence V x means curl.

All these equations apply only to the field itself; they do nothold at the points where sources are located. Similarly if thefield terminates not at infinity but at a point known as a sink,the sink itself must not be considered included in the field.

The differential equations developed above show that in allinverse square fields the curl as well as the divergence at everypoint in the field is zero.

The PotentialThe fact that the curl of these fields is everywhere zero leads

to an important result. Equations (2.8) state a necessary andsufficient condition that the force F at any point in the field canbe completely described in terms of another function of position 0,where 0 is related to the components of F by the set of equations

F2= a0 )

Fy = 0y . . . . . (2.9)

Fz = aaz

In practice it is convenient in most fields to define 0 as inequations (2.9) but with negative signs before the derivatives,but this convention need not be used here. Considering now atwo-dimensional field, assume that there is such a function 0related to the force F by the first two of equations (2.9). Thenby partial differentiation of Fz with respect to y and of Fy withrespect to x.

aFx a20

ay ax ay10

FIELD EQUATIONS

andaFy 020ax ay ax

It is clear now that if the two partial derivatives are not equal,the function cannot exist.

Therefore F can be described in terms of the function 0 bythe first pair of equations (2.9) provided

aFx _ aFy _ 0ay ax

and not otherwise. In general it can be proved that the equa-tions curl F = 0 state a necessary and sufficient condition that 0exists. The function 0 is called the potential of the force. It issimply a number and has no components corresponding to Fxand Fy. It is thus a scalar quantity. Hence the importanceof the curl equations lies in the fact that we are released from thenecessity of dealing with the vector components Fx and Fy andcan deal instead with the scalar quantity ¢. If the curl of thefield is not zero we cannot do this. It may be said here that ifthe source of the field is included in the field the curl is no longerzero, and the mathematics becomes more complicated.

The concept of a vector field is difficult for some people. Theyhave to make a mental effort, for instance, to develop the con-ception of the velocity field of a moving fluid. But providedthere are no sources or sinks in the velocity field, the integrationof the field produces a scalar field, viz. the field of flow. It isthis scalar field associated with the potential that most peoplenormally think of in connection with fluid motion. The scalarfield is only a field in the mathematical sense ; it is really a mapshowing the potential at any point in a domain, and if lines aredrawn through points of equal potential such lines are calledequipotential lines. The difference of potential between twopoints in the field can be defined as the work done on a per unittest body when transferring it from one point in the field toanother.

Laplace's EquationReverting now to the divergence equation, viz. (2.6) on

page 9, by substituting for Fx and Fy from equations (2.9),11


page 10, we obtain what is called Laplace's Equation. Theargument can be confined here to two dimensions. It followsfrom equations (2.9) that

bFx aFy a2 a2oax + by axe + aye

But from equation (2.6) the divergence equation

a Fx aFy _0

dx + by

and hence02d. +a2A

0 (2.10)axeay2 = .

This is Laplace's equation in two dimensions in rectangularcartesian coordinates. In polar coordinates it becomes

a20 100 1 a20art + r ar '{- r2 a92 = 0

In other coordinate systems Laplace's equation takes yet otherforms. Therefore to express it briefly in a form that embracesall coordinate systems it is written

V20 = 0or

div grad = 0It is as well not to be unduly disturbed or impressed by such

words as divergence, curl and Laplace's equation. Nor shouldengineers allow themselves to be discouraged by the apparentclumsiness of some of the mathematical expressions. What itall amounts to is that the fields suitable for the treatment to bedeveloped in this book are inverse square fields and that as aresult they can be described by differential equations where theindependent variable is a scalar denoted by which is a functionof position in the plane of the field and is related to the com-ponents of the force acting at any point by the equations

Fx

Fy

12

ax

ay

FIELD EQUATIONS

Laplace's equation does not yield easily to straightforwardtreatment: fortunately in the development of conformal trans-formations there is no need to seek a formal solution of theequation. It is only necessary to note first that all fields andfunctions to be considered in this book are those that satisfy theinverse square equation when emanating from a point sourceand therefore they also satisfy Laplace's equation. The secondpoint is that the equation is necessary for the development ofother important equations that govern these particular fields.Such fields are called Laplacian. When the field is not Lap-lacian, more recondite methods are necessary for determiningits distribution.

In vector analysis it is desirable to form physical conceptionsof the meaning of gradient, divergence and curl. But for thepurpose of conformal transformations it is sufficient to establishthe equations for two dimensions in such a manner as to bringout certain characteristics of the fields under consideration.For a full development of the physical approach the reader isreferred to standard works on vector analysis.

13

CHAPTER 3

Properties of Complex Variables

Complex Numbers

It is necessary at this point to make what might seem to be adiversion from the main theme into that of complex numbers.But in fact the method of conformal transformations requiresthe use of complex numbers. Therefore it is desirable to reviewnext some of the properties of complex variables. As a generalsymbol for a complex number the letter z can be used here be-cause all problems discussed will be two-dimensional andtherefore the z axis will never be required.

The complex number z = x + jy consists of two parts : a realpart x which is by itself a real number, and an imaginary partjy which consists of a real number y multiplied by the imaginarynumber '/ -1. The imaginary number,\/ --1 is represented bythe letter j ; this symbol is treated in accordance with all thenormal laws of algebra except that whenever j2 appears it isreplaced by - 1. With z=x+jy, the complex number x-jy iscalled the conjugate of z and is written z or z*. Now theproduct zz* is a real number because

zz* = (x+jy)(x-jy)= x2+y2

The real number (x2 + y2)1/2 is called the modulus of z and isoften written ¶z I or Ix + jy I . Therefore

zz* = Iz12

The Argand DiagramReal numbers can always be represented by points along a

straight line correctly spaced according to some scale andbased on a given origin which corresponds to the number zero.This is indicated in Fig. 4. The statement includes negativenumbers and irrational numbers.

14

PROPERTIES OF COMPLEX VARIABLES

I I

-5 -4 -3 -2 -i O I 2 3 4 5

FiG. 4. Geometry of real numbers.

Similarly complex numbers can always be represented bypoints placed in a plane. The real part of the complex numbercan be represented along the horizontal line called the real axis,and the imaginary part along the vertical line through theorigin called the imaginary axis. Thus the number 1 + j2 isrepresented by the point A in Fig. 5. This representation of acomplex number is called the Argand diagram.

Imaginary Axis

4

3F

2

I

-2

-3

-4

l_ RealI 2 3 AxI

FiG. 5. The Argand diagram.15


All the algebraic operations using complex numbers can berepresented geometrically on the Argand diagram. Forinstance in Fig. 6 the point A is 2 + j 4 and the point B is 4 + j 2.The algebraic addition

(2+j4)+(4+j2) = 6+j6is equivalent to the vector addition

shown in Fig. 6.OA+OB = OC

Fic. 6. Addition of complex numbers.

Complex Numbers in Polar FormConsider again the point A in Fig. 6 representing the number

2 + j 4. On the Argand diagram the line OA makes an anglewith the real axis that can be denoted by B1. Then

16

PROPERTIES OP COMPLEX VARIABLES

real part of OAcos B1 =

JOAJ

sill 61

(22+42)1/29

x/(20)

imaginary part of OA

IOAI4

1/(20)

sin 61 - ytan61 = -cos01

Similarly if OB makes an angle with the real axis denoted by02 and since OB = 4 + j 2

cos 02 =1/(20)

in 0 =

4

92 1/(20)

tan 02 = 0.5

The angles 01 and 02 are called the arguments of OA and OBrespectively. A complex number can be described by itsmodulus and its argument, e.g.

2+j4 = (20)1/2 tan-1 2

= r1where ri -- OA = the modulusand 01 = the argument.

Similarly OB can be described either by 4 + j 2 or by

(20)1/2 tan-I 0-

i .e. by x2102

Products of Complex Numbers

The advantage of expressing complex numbers in terms of2 17


their moduli and arguments-called their polar form-ariseswhen such numbers are multiplied by or divided by other com-plex numbers.

Complex numbers can be multiplied together by the ordinaryrules of algebra, treating j as if it were a real number exceptthat j2 whenever it appears must be replaced by -1. Thus

(a+jb)(c+jd) = (ac+j2bd)+j(bc+ad)= (ac-bd)+j(bc+ad)

The product of two complex numbers is in general of courseanother complex number.

It is usually much easier to deal with these products whenthey are expressed in their polar form. For example if

a+jb = ri/01then since a = ri cos 01and b = ri sin 01

a+jb = rl(cos 01+j sin01)= rieje,

Similarly

Hencec + jd = r2ele2

(a+jb)(c+jd) = rlr2001+e2)= rlr2 01 + 02

Therefore a convenient way of finding the product of two com-plex numbers is to put them into polar form, multiply themoduli (a product of real numbers) and add the arguments. Asan example, consider the two complex numbers shown as OAand OB in Fig. 6. OA represents 2+j4 and OB represents4+j2. Multiplied directly

(2+j4)(4+j2) = (8-8)+j(16+4)= O+j20

Fig. 7 shows OA and OB again, together with their productrepresented by OC. Now in polar form the numbers repre-sented by OA and OB are

OA = (20)1j2/tan-1 2

18


imaginary Axis

20

16

12

8

C ~(OC = OA- OB)

L Real12 Axis

Fic. 7. Product of complex numbers.

as found on page 17. The angle tan-' 2 from tables is 63° 26'so that

SimilarlyOA = (20)1/2/63- 26'

OB = (20)1/2 tan-' 0.5= (20)1/2 26° 34'

ThenOA.OB = (20)1/2(20)1/2/63° 26'+26° 34'

= 20290°= 20(cos 90°+j sin 90°)= 20(0+jl)= O+j20

as obtained by direct multiplication. Notice that the productof two complex numbers bears no relation to the scalar product

19


of two plane vectors and it cannot be made to bear any relationby taking the conjugate of one of the numbers, because, forinstance

OA.OB* = (a- jb)(c-jd)= (ac+bd)+j(bc-ad)

and the product is still a complex number.The advantage of using the polar forms for multiplication is

not too clear when two numbers only are concerned, but itbecomes apparent when several complex numbers have to bemultiplied together. The product of any quantity of complexnumbers can be found by multiplying together all the moduliand adding up all the arguments, i.e.

OA. OB.OC.OD ... = r1r2r3r4 ...

Finding the product of a series of complex numbers in algebraicform by algebraic multiplication is very long and tiresome bycomparison.

Imaginary Axis

6

5

4

OC = OA-OB) 3

- 3 -2

2

I

, L RealI 2 3 4 5 Axis

FIG. S. Difference of two complex numbers.20


Subtraction and Division of Complex NumbersThe algebraic subtraction of one complex number from an-

other is equivalent to the vector subtraction of two planevectors. For example if OA and OB are as before 2 + j4 and4 + j 2 respectively, the difference is shown on the Argand dia-gram in Fig. 8 where OB is first reversed and then added to OAgiving the complex number OC which is seen to be - 2 + j 2.This is the result that would be obtained by subtracting the twonumbers algebraically, and the algebraic method is by far thesimplest.

Division of complex numbers can also be accomplished in twoways. Here the more roundabout method is the algebraicwhere the denominator of the fraction representing the divisionis made a real number by multiplying it by its conjugate andthen the numerator is multiplied by the same conjugate. Forinstance

OA _ OA.OB*OB OB.OB*

or

a+jb _ (a+jb)(c-jd)c+jd (c+jd)(c-jd)

(ac+bd)+j(bc-ad)c2+d2

which is easily split into its real and imaginary parts.for example the two complex numbers used already

2+j4 _ (2+j4)(4-j2)4+j2 42+22

16+j1220

= 0.8+j0.6

Taking

But it is a much easier method to put the two numbers intotheir polar form and proceed thus :

OA ri/CtOB r2/

rlej6ir2ei62

21

CONFOR'_14AL TRANSFORMATIONS

= r11 e1(e1-e2)

r2ri /01-02r2

This result shows that to divide the complex number OA by thecomplex number OB we can first put both numbers into theirpolar form, divide the modulus of OA by that of OB and thensubtract the argument of OB from that of OA.

In the numerical example above the polar form of the num-bers was found on page 19. Hence

OA (20)1/2163° 26'

Imaginary Axis4r

3

2

I

C

OB (20)1/2/26- 34'= 163° 26- 26° 34'= 1 36° 52= 0.8+j0.6

A1

3 4

Fiv. 9. Division of complex numbers.22


The result is shown in Fig. 9. Notice that the division ofcomplex numbers has no relation whatever with plane vectorswhere there is no such operation as division. The polar methodis clearly most advantageous when there are several divisionscombined with multiplications to be performed.

23

CHAPTER 4

Conjugate Functions

The Potential Function

From the theory of complex numbers the conjugate functionswe require can be developed. Consider the two complexnumbers

U = x+jyV = X-jy

It is quite an elementary exercise to show by differentiationthat both u and v separately satisfy Laplace's equation.Furthermore x and y separately satisfy the same equation. Buta far more important theorem will now be established, viz. thatthe real part and the imaginary part of any function of the com-plex variable x + jy satisfies Laplace's equation.

Let X = 0 + ji be any function of z = x + jy, i.e.

X = f(x+jy)= f(z)

Then

Proceeding similarly

Again

ax Of(Z) az

ax az axaf(z)

02ax = f"(z) . . . . . (4.1)

aX Sf(z) azay az ay

= jf'(z)24

CONJUGATE FUNCTIONS

and02X _ 3f'(z) azay az 8y

2 af'(z)

aw

- f"(z) . . . . (4.2)

Combining equations (4.1) and (4.2) gives

2X 02X = 0 . .+

(4.3)?y-%x'

Now in terms of 0 and 0, since

X = 4+i/02`

X20G2X = + (4.4)ax2 ax2 ax2

Similarlya2o

020a2X -.+J (4.5)ay2ayz oy2

combining equations (4.3), (4.4) and (4.5) gives,920

( 2-

Therefore since the real and imaginary parts in this equationmust separately be zero

a+ay20and

. (4.7)

ax2+ay2=0 (4.8

Therefore the real part and the imaginary part of any functionof the complex variable x+jy automatically satisfies Laplace'sequation. These real and imaginary functions are called con-jugate functions.

To summarise, if X be any function of z, i.e.

X=f(z)where X = O + j oand z = x+jy

25


we know that

and . . . . . (4.9)V20 0

The functions 0 and 0 are conjugate functions.Now it has already been shown that the equation

V20 0is satisfied by a scalar function that is a function of positionand is also the potential of a vector field such as a field of force;the relations were given in equations (2.9), page 10. If, inequations (4.9), 0 is identified with potential and called thepotential function, the conjugate function b is called the streamfunction or the flux function. The meaning of the stream func-tion will become more apparent as the analysis is developed.

The Stream FunctionThere is an important relation between the potential and

stream functions expressed mathematically by what are calledthe Cauchy-Riemann equations. Writing as before

X = O+joz=x+jy

where X = F(z), thenaX - aa4

.+ l

(4.10)ax ax ax

FurthermoreaX_aF(z)azax az ax

OF(z). (4.11)

az

From equations (4.10) and (4.11)

48o _ eF(z).+ l 4.12)

ax azTX

Similarly it can be shown that

tiaaj

_)

4 13)y+

.y - J az.

26

CONJUGATE FUNCTIONS

Now add j times equation (4.13) to equation (4.12) and collectreal and imaginary terms. This produces

0000 +j(LO+BO = 0TX ay ax ay)

Therefore since the real and imaginary parts of the left-handside must separately be zero

ao 00

TX - ay

00 00

ax ay

. (4.14)

Equations (4.14) are the Cauchy-Riemann equations. Theyare important because they describe an important property ofconjugate functions, a property now to be explained. If 0 istreated as a potential function the equation

0 = constant

gives a family of curves with x and y as coordinate axes, onecurve for each value given to the constant. Each of thesecurves joins points of equal potential and the curves are there-fore called equipotentials. To be quite clear in the argumentthat follows it is best to deal with an actual function of z.

Typical Conjugate Functions

For instance suppose that the function is

X = sin z

where X and z are as specified on page 26. Then

+jo = sin (x+jy)= sin x cos jy + cos x sin jy= sin x cosh y + j cos x sink y . . (4.15)

Therefore equating the real parts

0 = sin x cosh y

The family of curves given by

0 = constant

27


can then be obtained from

sin x cosh y = aor

cosh y = a cosec x

where a can be assigned a series of values. Fig. 10 shows twocurves, viz. :

and

3

Y

2

a=2

sin x cosh y = I

sin x cosh y = 2

f

0

a=

1

Ex:. 10. Equipotential curves.2jr

and both are curves joining points where 0 is constant, i.e. theyare equipotential curves. It can be shown that they pass

28

CONJUGATE FUNCTIONS

through minimum values at x = it and then become asymtoticto the line x =7T. However, we are only concerned here withthe early part of the curves shown in Fig. 10.

Similarly a family of curves can be drawn from values of theconstant given by the equation

0 = constantFor instance in considering the function

x = sin z

equation (4.15) showed by equating the imaginary parts that0 = cos x sink y

3

Y

2

1

b-2

O x

FiG. 11. Stream lines.29


so that the family of curves given by

0 = constantcan be obtained from

cos x sinh y = b

Fig. 11 shows two curves for two different values of b, i.e.the curves for

andcos x sinh y = 1

cos x sinh y = 2

Since 0 is called the stream function or flux function, thelines are called stream lines or lines of flux or flow.

Orthogonal Curves

Comparison of these curves with those of Fig. 10 shows thatone family of curves intersects the other. In general thefamily of curves obtained from 0 = constant must intersectthe family of curves obtained from z& = constant. In Fig. 12 thetwo particular curves obtained from a = 2 and b = 2 are shownon the same diagram where they intersect. They happen to bethe curves obtained from the specific function x = sin z but theargument is quite general. Let then therefore be called thecurves

0=a and 0=bwhere 0 + j f is any function of x + j y.

The gradient at any point on either curve is given by dy/dxof the curve at the point chosen. Consider the curve 0 = a.Since 0 is constant on the curve its total derivative is zero.Hence from

= f(x+jy)

o dx +ay

dy = 0

Therefore on 0 =a

Similarly on1, (= b

dy _ ao/hxdx 80/8y

dy _ 00/2xdx Way

30

CONJUGATE FUNCTIONS

3

Y

2

)6-b

. a

1

1

Fie. 12. Orthogonal curves.

2x

and these relations apply at the point of intersection shown bythe tangent lines in Fig. 12. The product of the two gradientsis

(aolay)(aoiay)and by using equation (4.14) this can be written

o_l ay) - -1W/ay)(ao/ay)

Since the product of the gradients is - 1 at the point of inter-section the tangents there cut at right angles. Hence the

31

CONFORMAL TRANSFORITATIONS

Cauchy-Riemarni equations are a statement that the family ofcurves 0 = constant intersect the family of curves 0 = constantorthogonally. This is an important result and it forms thebasis of the flux-plotting techniques.

It has been stated earlier that the function 0 is sometimescalled the stream function and sometimes the flux function andthat the curves given by 0 =constant show the distribution offlow when the curves = constant are regarded as equipoten-tials. But it is clear that since 0 was chosen quite arbitrarilyout of a pair of conjugate functions to be regarded as thepotential function of the field, we could equally well havechosen 0 instead. Then the curves 4 = constant would havebeen the lines of equipotential and the curves O = constantwould have been the lines of flow. The potential function andthe stream function are mathematically interchangeable. Thesepoints can be brought out more clearly by considering a simpleexample of a pair of conjugate functions.

The Function log z

LetX = log z

where X and z have the same real and imaginary componentsas before. Then

O+jo = log (x+jy)Now in this example it is advantageous to put z into its polarcoordinates, i.e. to put

x = r cos 0y = rsin9

so that

Then

z x+jyr(cos 9 + j sin 0)reie (4.16)

0 + j4 = log(r ell)= logr+j9 . . . . (4.17)

Hence by equating real and imaginary parts

= log r (4.18)9

. . .

32

CONJUGATE FUNCTIONS

FiG. 13. Conjugate functions.

Now plot the curves given by = constant on the Arganddiagram for different values of the constant. The result is afamily of concentric circles with their centre at the origin, asshown in Fig. 13.

Now plot the curves given by 0 = constant on the same dia-gram for different values of the constant. The result is afamily of straight lines radiating out from the origin as shownin Fig. 13. Notice that all the intersections are orthogonal.

For some problems 0 will be the potential function and thenthe circles will be lines of equipotential. This, for example, isthe potential of an electrostatic field set up by a line chargethrough the plane of the paper and located at the origin. Then

3 33


//i in this problem is the stream function and the radii are linesof flow or flux.

For other problems 0 will be the potential function and 0 thestream function, giving circular flux or flow lines and having theradii as equipotentials. For example 0 is the potential of amagnetic field set up by a line current flowing through the planeof the paper and located at the origin. It should be notedagain that with either interpretation the field does not includethe origin where the source of the field is located.

The two functions can be expressed in rectangular cartesiancoordinates by noting that

x2+y2 = r2and

y _ rsin8x r cos 8

= tan 0

Hence, substituting for r and 0 in equations (4.18),

= log (x2 + y2)1/2

>/i = tan-1 (y/x)

34

CHAPTER 5

Conformal Transformations

Conformal RepresentationThis chapter deals with the central idea of this analysis, viz.

conformal representation or mapping. Consider any complexnumber z. It has already been shown that it can be representedby a point on the Argand diagram as in Fig. 14 (a). This dia-gram will be called the z plane.

Y

w

z

0 X 0 U

FIG. 14. Transformation of a point. (a) z plane (b) w plane.

Suppose now that there is another complex number denotedby w which is some known function of z. Since w is a functionof z there must be some relation between the point x + jy on thez plane and the point a+jv on the w plane shown in Fig. 14 (b).Now the point u + jv cannot be plotted on the same plane asthe point x + jy because four quantities are involved; in otherwords we cannot plot on a plane a single locus that shows how wvaries with z. For such a single curve a four-dimensional spacewould be necessary. Therefore two complex planes are used

35


to depict the relation between w and z and there must be pointto point correspondence between the two planes.

Therefore the point u + jv is plotted on another Argand dia-gram shown here in Fig. 14 (b). This second diagram is calledthe w plane. In the w plane u is plotted along the real axis andv along the imaginary axis. Now clearly any point on thez plane can be related to a point on the w plane by two functionssuch as

u = f (X, y)v = 9(x, y)

To each point x, y there are one or more points u, v althoughwe shall confine our attention to functions that give one-to-onecorrespondence. Thus everything appearing on the z plane canbe mapped on the w plane and vice versa. But although thefunctions written above are quite general this book will only beconcerned with conjugate functions. This means that thetransformations are confined to conformal transformations, andthe distinctive features of such transformations are these :

(1) Adjacent points in one plane transform to adjacentpoints in the other plane.

(2) In both planes the lines of flux and the equipotential linesintersect at right angles.

(3) Not only are these right angles conserved in the trans-formation but the sense of the angles is also conserved.

The stage has now been reached when it is possible to explainthe purpose of using conformal transformations. Suppose it isrequired to determine the distribution of the field between twoequipotential boundaries that can be represented on a diagramand called the z plane. Suppose further that the two boundariesare of awkward shape such as, for instance, the surfaces repre-sented by the heavy lines `a' and `b' in Fig. 15.

Such boundaries could be those of a magnetic field where theboundaries consist of magnetic material whose permeabilitycould be considered infinite. In such a problem it may be pos-sible to find and apply a transformation from this z plane to anew plane called the 2v plane in which the shapes of theboundaries become something recognisable, something forwhich the distribution of the field is both regular and known.

36


X

x

Fic. 15. Equipotontial boundaries. z plane.

For instance, a transformation may be found that gives the newboundaries shown in Fig. 16 and makes all the flux lines andequipotentials straight lines. With boundaries like this con-sisting of parallel straight lines and with all stream lines andequipotentials straight, the field distribution is easily deter-mined.

We can find for the new plane or we can find 0 instead,whichever is required for the problem in hand; then the equa-tion of transformation is used to determine the correspondingfield in the original plane. In other words the solution of theproblem is readily found in the w plane, and this solution leadsthrough the transformation equation to the solution in the zplane, i.e. to the solution of the actual problem.

Notice that since is constant over the boundaries in the wplane it is also constant over the boundaries in the z plane andmust be the required potential in the z plane. It is desirable atthis stage to illustrate the transformation by an example.

37

CONFO'RMAL TRANSFORMATIONS

V

a'

b'

u

Fir,. 16. Equipotential boundaries. w plane.

Transformation of a CurveConsider the conformal transformation

z = 2cosw+jsinw . . . . (5.1)

For the purpose of this example let us consider first real valuesonly of to. Table 1 shows a series of real values of w and thecorresponding values of z obtained from equation (5.1).

TABLE 1

z = 2cosw+jsinw

w 0 '7r f 61

7r/4 1r13

z 2+j0 1.73+j0.50 1.414-j0.71

-1.00+j0.87

w 77/1 277/3 --- 37r,14 5nj6

z 0+j1.0 -1.0+j0.87 -1.41+j0.71 -1.73±j0.50

38


N=o

(qi

z = --2 1J=/- J l

- Tr -Tr 0

z =2 '/z=l1 I/ tm -2

2

/6J

Fro. 17. Ellipse transformed to straight line.(a) z plane (b) w plane

If in the w plane a line is drawn through all the values of wchosen in Table 1, the result is a straight line along the real axisas shown in Fig. 17 (b). On the other hand a line drawn in thez plane through the corresponding values of z in Table 1 formsthe upper half of an ellipse. Further values of w from it to 27rgive the lower half of the same ellipse ; this curve is shown inFig. 17 (a). Thus an ellipse in the z plane is transformed by theequation to a straight line in the w plane. This is an example ofthe transformation of a curve into a straight line.

However, it is instructive to go further than this to see whatthe same transformation means when all values of w real and

39


complex are admitted. The procedure to adopt is to find x andy as functions of u and v by expanding equation (5.1) :

z= 2 cos w+ j sin wSubstituting for z and w

x+jy = 2 cos (u+jv)+j sin (u+jv)= 2 cos u cos j v- 2 sin u sin j v

+ j sin u cos jv + j cos it sin jv= 2 cos u cosh v - 2j sin it sinh v

+ j sin It cosh v - cos it sinh vEquating real and imaginary parts

x = cos u(2 cosh v - sink v)y = sin u(cosh v - 2 sinh v) (5.2)

In the earlier treatment of this transformation only realvalues of w were considered, i.e. various values of it were chosenwith v zero throughout. Suppose now that v is made 0.5throughout. Then

cosh v = 1.1276sinh v = 0.5211

and2 cosh v - sinh v = 1.734cosh v - 2 sinh v = 0854

Then from equations (5.2)x = 1.734 cos ity = 0854 sin u

and Table 2 can be drawn up as follows :

TABLE 2

Transformation of an ellipse

u 0 7T/6 7r/3 7r/2 2ir/3 5ir/6 IT

a= cos u ' 1.000 0.866 0.500 0 -0-500 -0-866 -1-000

x= 1.734a 1.734 1.502 0.867 0 -0-8 7 -1-502 -1-734

b =sin u 0 0.500 0.866 1.000 0.866 0.500 0

y 0 0427 0740 0854 0740 0427 0

40


On plotting these values in the z plane we find that the curvejoining them is the upper half of another ellipse wholly insidethe previous one shown in Fig. 17. The lower half of the newellipse is obtained by taking values of u between 77 and 2ir. Itwill be shown presently that the new ellipse is confocal with theold one.

When plotted in the w plane the values in Table 2 give an-other straight line parallel with the real axis and passingthrough the point v =

Consider now any general constant value of v. Equations(5.2) can then be written with

a = 2 cosh v -sinh vb = cosh v - 2 sinh v

where a and b are constants. 't'hen

x = acosIty = bsin It

Therefore

cos 2 It + sine It

1

This is the equation of an ellipse since a2 and b2 are both positive.Reverting to equations (5.3)

a2 = 4 cosh2 v + sinh2 v - 4 sinh v cosh vb 2 = cosh2 v+ 4 sinh2 v- 4 sinh v cosh v

Thereforea2 - b2 = 3(cosh2 v - sinh2 v)

=3Therefore whatever value is assigned to v the expression

a2 - b2 is constant. Hence any change in the value of v whichincreases (or decreases) a2 by an amount h must also increase(or decrease) b2 by exactly the same amount ; hence the variousvalues assigned to v must give a family of curves that can bewritten

x2 y2a2+h+b2+h = 1

This is known to be the equation of a family of confocalellipses. Therefore the equation v = constant gives a family of

41


(a>

V

0 U

(6)

FIG. 18. Curves of constant v. (a) z plane (b) w plane.

straight lines parallel to the real axis in the w plane and afamily of confocal ellipses in the z plane. Both are shown inFig. 18.

Suppose now that u is held constant instead of v ; what hap-pens when various values of v are selected for each value of theconstant u? For instance consider the following values of u inturn :

At u = 0,u=0; u=rr/6; u=ir/3

x = 2 cosh v - sink vy = 0

Therefore this value of u gives the real axis in the z plane.At u=7T/6,

x = 1.732 cosh v - 0.866 sinh vy = 0.500 cosh v-sinh v

42


At u=7r/3,x = cosh v - 0.500 sinh vy = 0.866 cosh v -1.732 sinh v

From these three sets of equations Table 3 can be drawn up asfollows :

TABLE 3

Transformation of hyperbolas

v 0-55 1.0 1.5 2.0 2.5 3.0

a = cosh v 1.155 1.543 2.352 3-762 6.132 10.07

b = 1.732a 2.000 2.680 4.080 6.520 10.620 17.40

C = sink v 0.578 I 1.175 2.129 3-627 6.050 10-02

d = 0.866c 0.500 1.018 1.840 3-140 5.240 8.67

x=b-d(u =7r/6) 1.500 1-662 2.240 3-380 5.380 8.73

e = 0.500a 0-578 0-772 1.176 1.881 3-066 5.04

y=e-c(u=ir/6)

0 -0.403 -0.953 -1-746 -2.484 -4.98

f = 0.500c 0-289 0.588 1-065 1-814 3-025 5.01

x=a-f(u=7r/3) 0-866 0-955 1-287 1-948 3-107 5-06

g=0-866a 1.000 1.340 2-035 3-260 5-310 8-71

h= 1.732c 1-000 2-040 3-690 6-290 10-490 17-40

y=g-h(u =7r/3) 0 -0.700 -1.655 -3.030 -5.180 -869

Notice that at u = jrr, x = 0 giving the imaginary axis of thez plane. When plotted on the z plane the two sets of valuesgiven in Table 3 of x and y give the lower halves of two hyper-bolas. The upper halves can be found by giving negativevalues to v and positive values less than 0-55. The two hyper-bolas are shown in Fig. 19. Note that negative values of x canbe found by selecting values of u between

2Trand 37r/2.

43


8

6

Y

2

-2

-4

-6

-8

574U. 2 -3

1111

u' 6

u-O

4 6 8

6

U_2

U z3

FIG. 19. Confocal hyperbolas.

It will now be shown that all the curves given by u = constantare confocal hyperbolas. For a constant value of u equations(5.2) can be written

x = a- (2 cosh v - sinh v)

where a = -%/3 cos ub = -\/3 sin u

Then

V3

3(cosh v - 2 sinh v)

x2 y2

a2b244


and this is the equation of a hyperbola. Nowa2+b2 = 3(cos2 u+sin2 u)

=3so that if any change in it increases (or decreases) a2 by anamount h it must also decrease (or increase) b2 by the sameamount. Thus for various values of It the following equationholds :

X2y2

a2+h b2-h

(Q)

V

u

/6/FiG. 20. Curves of constant u. (a) z plane (6) w plane.

45


/ol

V

u

/al

FIG. 21. (a) z plane (b) w plane.

This is known to be the equation of a family of confocal hyper-bolas. Therefore u = constant gives in the w plane a family ofstraight lines parallel to the imaginary axis and in the z plane afamily of confocal hyperbolas. Both planes are shown inFig. 20 and again in Fig. 21, where the corresponding sets ofstraight lines and curves for v = constant are added.

This then is an example of the transformation of a set of curvesinto a set of straight lines. The next thing to consider is thetransformation of straight lines into straight lines.

46


Transformation of a Straight Line

In Fig. 22 (a) there are shown two points Pi and P2 very closetogether. Let dsi be the distance between them and let ai bethe angle between the straight line PIP2 and the real axis, asshown on the diagram.

0 X 0

FIG. 22. Transformation of a straight line. (a) z plane (b) w plane

From equation (4.16), page 32,z=rej°

Therefore

dz = dsieialThe transformation of these two points to the w plane produces

the two points Qi and Q2 shown in Fig. 22 (b). These pointsare also very close together. Let ds2 be the length of thestraight line joining Qi and Q2 and let this line make an anglea2 with the real axis as shown in Fig. 22 (b). Then as before

dw = dsreia2

Now since there is a fundamental relationship between z and wa relationship that is the basis of the transformation from oneto the other

andw = f (z)

dw = f(z) dzwhere f'(z) must be a complex number and can therefore bedescribed another way by writing

f'(z) = Aei$where A is the modulus and fi the argument of f'(z).

47


Then

Therefore

and

ds2eia2 = dw= Aeia dsl eial= A ds1 ei (i+a)

ds2 = A ds1

a2 = ai+NThus, in transformation the line ds; is multiplied by A and

also turned through the angle P. The change from the z planeto the w plane alters both the scale and the orientation. Noteparticularly that

= L82Adsl =

dwl

dz. (5.4)

Hence each small displacement from a point P in the z planewhen transformed to the w plane is increased or decreased inthe same ratio as the other displacements and is turned throughthe same angle. This result enables us to consider the trans-formation of configurations or shapes consisting of straight linesand angles.

It follows from the previous section that any small figure, saythe small triangle shown in the z plane Fig. 23 (a), remains asmall triangle when transformed to the w plane Fig. 23 (b), butthat it is either enlarged or diminished by the transformationand it is also turned through an angle. Fig. 23 shows theresult of a typical transformation. The transformation fails atthe points where A in equation (5.4) is zero or infinite.

Y

D 4

U

(a) 14)

Fm. 23. Transformation of a triangle. (a) z plane (b) w plane.48


The converse of this statement is equally true, viz. that theangles are bound to be conserved by the transformation unlessA (i.e Idw/dzl) is zero or infinite.

Transformation of a Definite AreaConsider again the transformation of equation (4.17) on

page 32:¢+j = logr+jO

where q = log rand z/r=B

In the z plane of this transformation let the boundaries bethose of the shaded annular ring shown in Fig. 24 (a). We nowwant to find what the corresponding boundaries are in the X

(Q) 7;)

Fia. 24. Transformation of an area. (a) z plane (b) X plane.

plane. Label the corners of the ring a, b, c and d as shown inFig. 24 (a). For numerical values let these points be

x = 4, 3, - 3 and - 4respectively. Consider first the point a where x = 4 and y = 0.

= logo= 1.386

8=0Thus in the z plane the point a becomes the point

1.386+j0

as shown in Fig. 24 (b). Similarly in the z plane the point b4 49


is 3 + j0, but in the X plane it becomes the point log 3 + j O, i.e.

1.099 +j0as shown in Fig. 24 (b).

Now the point c in the z plane is given by

(r, 0) = (3, Ir)

Hence in the X plane it is given by

¢+jo = log 3+j7rso that c becomes the point

1.099+j7r

Similarly d becomes the point1.386+j7r

as shown in Fig. 24 (b).Thus the four boundary points in the z plane become the four

boundary points shown in Fig. 24 (b). Moreover the equationsshow that in the X plane the points are connected by straightlines. Therefore the shaded area in Fig. 24 (b) corresponds tothe shaded annular area of Fig. 24 (a).

In considering the transformations it is helpful to imaginethat the area enclosed by the boundaries contains an idealelastic substance or membrane corresponding to the equipoten-tials and the lines of flow. The material is ideal in the sensethat it is infinitely expansible and infinitely compressible with-out its changing character and without there being constraintsof any kind. The spaces enclosed by the equipotentials andlines of flow are regarded as cells of this ideal material, whichpossesses the property that the boundaries can be distorted inany manner but that the lines forming the boundaries of thecells must always intersect at right angles.

Thus the shape shown in Fig. 26 can be imagined picked upand compressed into the shape shown in Fig. 25 and vice versa.

Transformation of a QuadrantAs a further example consider the transformation

w = z2 . . . . . . (5.5)

where for simplicity the restriction is imposed that z is confined50


F[G. 26. Shape filled with hypothetical elastic substance.

FIG. 26. Distorted shape.

wholly to the first quadrant. We wish to determine what thisquadrant in the z plane becomes in the w plane under the trans-formation of equation (5.5). As before substitute for z and wthus:

u+jv = (x+jy)2= x2 _ y2 + 2jxy

Hence, separating real and imaginary parts

u = x2 _ y2v = 2xy

11

51

. (5.6)

CON FORMAL TRANSFORMATIONS

P

m

c' ;

M6

V

C1 b C

0 b

u

P

c

/'' /6)

Fro. 27. Transformation of a quadrant. (a) z plane (b) w plane.

Since z is to be confined to the first quadrant, x and y are con-fined to positive values. Draw the z plane Fig. 27 (a) and thew plane Fig. 27 (b). Note that only the first quadrant is re-quired in Fig. 27 (a) and only the upper half plane in Fig. 27 (b).The equation u = 0 gives the imaginary axis of the w plane. Inthe z plane u = 0 corresponds to

x2-yz = 0i.e. to

x=ybecause only positive values can be admitted. This gives theline labelled `a' in both planes.

The equation v = 0 gives the real axis of the w plane. In thez plane since

v = 2xy

the values x = 0 and y = 0 both satisfy v = 0. Therefore the realaxis of the w plane for both positive and negative values of u istransformed to the x and y axes of the z plane giving the curveshown as m in both planes.

Next find the curves in both planes for the value u = 2. Inthe w plane it is clearly the vertical line drawn two units awayfrom the imaginary axis. But in the z plane

x2 _ y2 = 2so that

x2 = y2+2and Table 4 can be drawn up as follows :

52


TABLE 4

Values of y and x for u = 2

y2 0 1 4 9 16 25

x2=y2+2 2 3 6 11 18 27

x 1.41 1.73 2.45 13.32 4.24 5.20

Plotting the points obtained from Table 4 gives curve b inboth planes Fig. 27. Similarly there must be curves b' in bothplanes obtained by putting u = - 2, i.e. y2 = x2 + 2. Then byputting u = 5 and u = - 5 the curves shown as c and c' in bothplanes are obtained. We can select as many values of u as weplease and draw the corresponding curves in the two planes.

Now consider various values of v in the equation v = constant.When v = 2

and hence

2xy = 2

x = l/yfrom which Table 5 can be drawn up :

TABLE 5

Values of y and x for v = 2

y 0.25 0.50 1.00 2.00__J 4.00____x= l /y 4.00 2.00 1.00 0.50 0.25

Plotting the points obtained from Table 5 gives the curvelabelled n in both planes. Similarly by putting v = 5 we obtainthe curve labelled p in both planes. We can select as manyvalues of v as we please and draw the corresponding curves inboth planes. Notice that v cannot have negative values be-cause if it had, either x or y would have to be negative, andnegative values of x and y are inadmissible.

53

CONFORMALTRANSFORMATION$

The squares shaded in Fig. 27 are corresponding regions in thetwo planes. From equations (5.6) if u = a2

x2-y2 = a2or

x2 yz

a2 a2

and this is the equation of a rectangular hyperbola whoseasymptote in the first quadrant is the line given by the equation

x= y

Again from equation (5.6) if v=c

xy = 2c

and this is the equation of a rectangular hyperbola whoseasymptotes are the real and imaginary axes of the first quad-rant. Clearly the two sets of curves intersect at right angles.

As many curves as we please can be drawn in both planes bytaking a sufficient number of values of the constants.

kio.'28. z plane.

Figs. 28 and 29 show a closer mesh in both planes. We canimagine the mesh in the z plane Fig. 28 to represent the cells of

54


the ideal elastic material. Then in imagination we can take they axis and force it round in such a way that it becomes thenegative real axis. The hyperbolas forming the boundaries ofthe cells become straight lines and as a result the picture in thew plane shown in Fig. 29 appears.

s

4

-12 0 4 e 12

FIG. 29. w plane.

55

CHAPTER 6

The Schwarz-Christoffel Transformation

IN ALL THE EXAMPLES discussed in previous chapters trans-formation equations have been set down first and the corres-ponding configurations in both planes have been discoveredafterwards. This is hardly a satisfactory procedure for actualproblems because in practice the engineer knows the two con-figurations, the one he is concerned with and the simpler onepossessing a known solution. From these known configurationshe has to determine the appropriate equation of transformation.If what has been done in previous chapters were the only pos-sible procedure he would always have to look about for the cor-rect equation of transformation. Naturally the preferredcourse from his point of view would be to specify a pair of con-figurations and from them find by a routine process the equationof transformation from one to the other.

Now there is a known method of finding this equation whenboth of the equipotentials forming the source and sink are partof a closed polygon and the problem is entirely bounded by thepolygon. At first sight this seems a severe limitation to theusefulness of the method, but in practice many field problemshave as boundaries straight lines that can be treated as closedpolygons. The method was first published by Schwarz andChristoffel * independently of each other.

The Schwarz-Christoffel transformation opens out theinterior of a polygon in the z plane to the upper half of the wplane. Now the upper half of the w plane is not likely to be thefinal configuration that is required because the field might notbe regular there, but by using a second Schwarz-Christoffeltransformation we make another transformation from the con-figuration wanted-say the X plane-to the same w plane asbefore. Then the combination of the two transformations will


THE SCHWARZ-CHRISTOFFEL TRANSFORMATION

change the configuration in the z plane constituting the probleminto the desired configuration in the X plane. Note that theinterior of the polygon becomes the upper half of the w plane ;the exterior becomes the lower half. The sides of the polygonbecome the real axis of the w plane.

Since the transformation is conformal the equation has theform

w = f(z)

Let the vertices of the polygon ABCD in Fig. 30 (a) transform tothe points A', B', C' and D' on the real axis of the w planeshown in Fig. 30 (b). Open the polygon at some arbitrarypoint chosen to suit the problem ; in Fig. 30 (a) it has beenopened between A and D. This is not at one of the corners ofthe polygon. If, however, for convenience the opening is madeat a corner, that corner takes no part in the transformation.

y

w=b

/a)

A B

/61'

A

x

V

w-+ CN. -sC D U

FIG. 30. The Schwarz-Christoffel transformation.(a) z plane (b) w plane.

57


Let one side of the opening become + oo in the w plane andthe other side become - oo, as shown in Fig, 30 (a). In effectwe now force open the polygon at the selected point, bend itinto a straight line, stretch it to infinity at each end, and thenplace it along the real axis of the w plane.

Consider the line AB in the z plane. Along this line the smalllength dz becomes in the w plane another small length dw. Theangle between dz and dw must remain constant between A and Bin the z plane, but at the point B it then changes abruptly to anew value because dz changes in direction at the point B whereasdw does not change at the corresponding point B'. On page48 it is shown that when the transformation is performed theelement of line is multiplied by A, the modulus of dw/dz, and isat the same time turned through an angle fi which is the argu-ment of dw/dz. Therefore the change in angle when trans-forming from z to w is the argument of dw/dz. This argumenthas one constant value between the points A and B, anotherconstant value between the points B and C, and so on.

The transformation has to be such that these conditions arefulfilled and they are in fact fulfilled by

dzA w - a (-In)-i

dw ( ) (w-b)(alir)-1(w-c)(rl+r)-1 , (6.1)

where the number of factors in addition to A is the number ofsides in the polygon. Note that if the polygon is opened at acorner there must be one less factor in equation (6.1) becausethat corner is not transformed. In this transformation a, 9, y,etc., are the internal angles of the polygon while the numbersa, b, c, etc., are the coordinates of the points A, B, C, etc., in thew plane, i.e. the points corresponding to A, B, C, etc., on thereal axis to which the vertices of the polygon are transformed.

Notice that in the Schwarz-Christoffel transformationdw/dz must be either zero or infinite at each of the corners.Equation (6.1) satisfies that condition at every corner. Theconstant A fixes the size of the z picture and since it can be acomplex number it also fixes the orientation of the z picture.The other constants a, b, c, etc., must be allocated values thatmake the lengths of the sides of the polygon have the correctratios. It will be shown soon that some of them may be given

58

THE SCHWARZ-CHRISTOFFEL TRANSFORMATION

arbitrary values to facilitate the calculation, but that the othershave to be left undetermined until the end of the solution.

The use of the Schwarz-Christoffel transformation always en-tails performing an integration. Sometimes the integration isreasonably elementary but not always. For engineers it iseminently desirable to have at hand, and be familiar with, abook of Tables of Integrals ; they should avoid as far as possiblewasting time on doing actual integrations. In the present bookno integrations will be performed if they can be taken from, orderived from, a book of this nature. Wherever possible weshall refer here to Dwight's Tables of Integrals and Other Mathe-matical Data,* and give a numbered reference to any integrationit is found necessary to use from it. For example the result ofan integration will be followed by a reference such as 'Dwight194.11'.

The details of the procedure followed in Schwarz-Christoffeltransformations are best explained in an example. The trans-formation is usually applied to configurations in which theangles are either right angles (7T/2 or 37r/2) or are zero angles(0 or 27r). The complexity of the resulting equation of trans-formation increases with the number of right angles to betransformed. For this reason it is desirable to consider trans-formations of each type in turn, beginning with the simplest,viz. the transformation of configurations containing no rightangles whatever.

* See List of References, page 217.

59

CHAPTER 7

Configurations With No Right Angles

Electric Field at the Edges of Two Capacitor PlatesLet us examine as a first example of a configuration with no

right angles the electric field between the edges of two capacitorplates. Remember, however, that the same configuration andtransformation could apply to other problems. Let A and Bin Fig. 31 be the edges of two capacitor plates that are paralleland distant d apart. They are assumed to extend an infinitedistance in the direction to the right as indicated in Fig. 31 andto a sufficient distance in the direction perpendicular to theplane of the paper to make the problem in effect two-dimensional.

A

B

Fic. 31. Edges of two capacitor plates.

Since each side of each plate is an equipotential surface thetrue representation of the system to be transformed is shown inFig. 32. This figure will be called the z plane and the first stepin finding a solution is to transform this configuration into theupper half of the w plane. Since the field in this w plane is still

AcC

D

d

EB

F

Fia. 32. Representation of two capacitor plates.60

CONFIGURATIONS WITH NO RIGHT ANGLES

not regular the second step is to find the equation that trans-forms two parallel plates of infinite length in all directions to theupper half of the w plane. The combination of the two equa-tions of transformation gives the required solution. The regularfield plane is shown in Fig. 33 and it will be called the t plane.

Fic. 33. Regular field plane. t plane.

Reverting now to the first transformation from the z plane ofFig. 32 to the w plane, the configuration in Fig. 32 does notappear to be a closed polygon. But it is possible to imaginethat the points D and E are joined together and that the pointsC and F are also joined together; the figure then becomes adegenerate quadrilateral with corners at A, B, CF and DE.Therefore it can be treated as a polygon.

Examination of Fig. 32 suggests that it is desirable to openthe polygon at the corner CF. Then, as explained on page 57,this point will not be transformed. Note that the interiorangle at A and B is 360° while the interior angle at the cornerDE is zero. We now have to allocate values to w in the z planeand the chosen values are shown in Fig. 34. Since the polygonis to be opened at the corner CF, the point C is made w = oo andthe point F made w = - oo. Then the symmetry of the figuresuggests that w = 0 should be the value at the corner DE.

w=+00W-1 C

AD

W=0

E

F

Fic. 34. Values of w as points in the z plane.61


With regard to the values of w at the corners A and B, ithappens that they can be chosen arbitrarily in this example.The rule is that there must be one unspecified value of w forevery dimensional value in the z plane more than one. Forexample, suppose there are two independent dimensions in thez plane giving one ratio ; then there must be one unspecifiedvalue of w which has to be called, say, w = a. In general whenthe values zero and infinity have been assigned, additionalvalues of w can be specified at two corners only.

In this particular example there is only the distance betweenthe plates specified and hence no ratios. Therefore there needbe no unspecified values of w. But we have to ensure that thevalue of w at the point A lies between the values at C and D,i.e. w must be positive at A. Similarly the value at B must liebetween the values at E and F ; hence w must be negative at B.The expressions obtained will be simplest if the values chosenare w =1 at A and w = -1 at B as shown in Fig. 34. Then thew plane will appear as shown in Fig. 35. Notice that althoughthe origin in the iv plane has now been determined, it is notnecessary to fix the origin of the z plane yet. One reason forchoosing the corner DE as the origin in the w plane is that it is apoint of potential discontinuity ; it is nearly always advan-tageous to choose such a point for the origin.

Y

F B D E A

-1 0

Fir,. 36. The intermediate plane. w plane.

Since there are now three corners in the z plane to be openedout there must be three factors besides the constant in theSchwarz-Christoffel equation of transformation. Thus theequation is

dz A w - a (al-)-I

dw - ( )(w-b)(alir)-1(w-c)cYln>-a (7.1)

62


In equation (7.1) a is the point B where w= - 1, and a is theangle there, viz. 360'; hence in the equation a = -1 and a = 2ir.Similarly b is the point DE, where w is zero and the angle is alsozero ; hence in the equation b = f = 0. Finally, c is the point Awhere w =1 and the angle is again 360 °, giving c = I and y = 21T.Therefore equation (7.1) becomes

= A(w-1)w-I(w+ I)dw

A(w2-1)w

Therefore

z Ai (w- l)dw . . . . (7.2)w

By direct integration (Dwight 81.1 and 82.1)

z = A(2lw2-log w)+B

where B is the constant of integration. Now the value of Bcan only affect the position of the origin in the z plane, and sincethis is not yet fixed we can make B zero now and find theposition of the origin later.

Determination of ConstantsThe next step is to find the value of the constant A. In this

particular example there are two ways of establishing the valueof A, and since either may be used in subsequent examples bothmethods will be described now. One very effective method canbe used where there is a corner or are corners in the z plane atplus or minus infinity. In the example under consideration oneof the corners is at z = + oo where w = 0. Now when crossingfrom one plate to the other in a direction perpendicular to theplates the distance traversed is d in the z plane. In the to planethe corresponding path traversed is a semicircle centred on theorigin. The nearer we cross to the point z = cc the smaller isthe radius of the semicircle in the w plane although the distancetraversed in the z plane remains equal to d. Therefore as thepoint of crossing approaches the corner, z-- oe and r-,0 wherer is the radius of the small semicircle.

63


Now taking account of direction the distance from D to Eis - jd, i.e.

Edz = -jd . . . . . (7.3)

D

In terms of w, from equation (7.2)

Al (w-w) dw= -jd . . (7.4)

where the path L is a small semicircle of radius r. In polarcoordinates

w = reie

anddw = jrei° dO

Hence equation (7.4) can be written

A fv (reJe -1 a-JO) jrele dO = - idr J

As we approach the corner at z = oo, r-->0, but the right-handside remains unchanged. Hence in the limit

AJ' (-j) dO = -jdPerforming the integration

jA7r = jdwhence

A=d/rrHowever, in this example the other method is quicker.

Starting fromz = A(w2-log w)

put in known values of w. From this equation, when

w = 1, z=JAw = -1, z = ZA{1- 2log (-1)}

= JA (1- 2j,r)

Since we have not yet found the origin in the z plane we do notknow the value of z at these points, but we do know that thedistance between the points in the z plane is jd. The difference

64


between the two values of z given by the equation is jATr.Hence

A = d/ir

as before and the equation of transformation becomes

z = d(w2 - 2 log w) . . . . (7.5)

The method of integrating round a semicircle can also be usedwhere, when a corner of the polygon is at z = ± oo, the value ofw there is not zero but is itself plus or minus infinity. Whenthis happens, as the path from one side to the other in the zplane approaches the corner at infinity, so the radius of the semi-circle in the w plane gets larger and larger and in the limitbecomes infinite. If this large radius is called R we can write

z = f(w)w = Reje

dw = jReie dOHence

BJ

f(Rei9)j Reie d6 = - id

Then the constant B can be found by letting R--co and thenperforming the integration. An example of the procedure willbe found on page 98.

The Second Transformation

The second part of the problem can now be tackled, i.e. todetermine the equation giving the transformation from the tplane shown in Fig. 36 where the field is regular to the w planeof Fig. 35. The t plane represents two parallel plates extendingan infinite distance in all directions thus giving a regular field

w in the t pla.ue.655


between the plates. Here again there are no ratios of dimen-sions. The point of potential discontinuity is at either end ; itis therefore convenient to put w = 0 at one end and w = oo atthe other end, + oo for one plate and - oo for the other plate.These values are shown in Fig. 36.

From the Schwarz-Christoffel equation

dt(Jw)-i= A(w-a)

dw

where now a = 0 and a = 0 so that

dt=`4dww

Hence by integration (Dwight 82.1)

t = Alogw+Bwhere A and B are new constants. The value of B once againonly affects the position of the origin in the t plane and sincethat position can be anywhere without affecting the problem,B can be made zero.

With regard to the constant A, we can either find the value ofA which corresponds to a distance d between the plates in thet plane (thus making the distance the same as in the z plane),or we can make A unity and find what distance between platesin the t plane corresponds to the distance d in the z plane. Forthe present we shall take the second course and put A equal tounity so that the equation of transformation becomes

t = log wor

w = et . . . . . . (7.6)

The direct transformation from the z plane to the t plane cannow be found by combining the two transformations of equa-tions 7.5 and 7.6 and eliminating w, giving

z = 2 (e2t - 2t) . . . . (7.7)

This is the equation that changes the configuration of Fig. 34to that of Fig. 36.

66


The Axes in Both PlanesSince equation 7.7 gives a direct relation between the z and t

planes, the intermediate w plane can now be ignored. To fixthe origins and axes in the two planes proceed as follows.

In the t plane, Fig. 36, the origin can be set anywhere. Theimaginary axis must clearly be perpendicular to the platesthrough the origin and it is convenient to place the origin onthe lower plate. In Fig. 37 it is the point marked A and the

q

B

P

FIG. 37. Fixing axes in the t plane.

lower plate becomes the real axis. In order to fix the positionof the axes in the z plane it is necessary to find the minimumvalue of z in terms of t by differentiation. From equation (7.7)

dz _ d (201-2)dt 27r

This is zero when e2t is unity, i.e. when t is zero. Differentiat-ing again gives

d2z _ d (4e2t)dt2 27r

and this is positive for all values of t. Therefore the value of zat t = 0 is a minimum value. Hence the minimum value of z isgiven by equation (7.7) when t=0, i.e. when z = d/27r. Thus the

67


edge of one of the plates in the z plane lies a distance d,f2rr alongthe real axis from the origin. The next thing to find out iswhether that plate from which the origin is fixed lies along thereal axis of the z plane. Let the real and imaginary parts of tbe p + jq as marked on Fig. 37. When t is wholly real q is zeroand t = p. Hence from equation (7.7)

z =d

(ezp- 2p)2Tr

and therefore z is also wholly real. As p tends to infinity sodoes z ; furthermore as p tends to minus infinity z again tends toplus infinity. This establishes that the plate whose edge in thez plane is distant d,12ir from the origin lies along the positive realaxis as shown in Fig. 38. In that picture the upper plate isshown as the plate concerned but we have now to prove that theupper plate is the one lying along the real axis.

Y

-d

d2n

A 1 M

d

FI(G. 38. Axes in the z plane.

When t is wholly imaginary p is zero and t=jq. Hence

z = -(cos 2q+j sin 2q-j2q)

68


We already know that when q is zero z gives the edge of theplate lying along the real axis. But when q = 7r,

z = d{1+j(0-21r)}

d= 27- -id

Now the negative sign in this result shows that this plate lies adistance d below the real axis of the z plane and therefore it isthe upper plate as shown in Fig. 38 that lies along the real axis.Thus when q = 7r, we have the point B in the z plane, Fig. 34, andthe corresponding point B in the t plane lies on the imaginaryaxis (p = 0) and at a distance 7r from the origin. This is shownin Fig. 37.

In the t plane the upper plate lies along the line

t = p+j7rFrom equation (7.7) this corresponds in the z plane to

dz = d (e'-Pe2j--2p-j27r)

(e2P-2p-j27r)

=d (e2P-2p)-jd

This is a line in the z plane parallel to the x axis and distant dbelow it as shown in Fig. 38. As p tends either to plus infinityor to minus infinity z tends to plus infinity.

This confirms that the transformation given by equation (7.7)does in fact transform the boundaries of Fig. 37 to those ofFig. 38. The origins and axes in both planes have thus beendetermined and further problems can now be considered.

Equipotentials and Stream LinesUsually there is a specific problem to solve that does not en-

tail drawing the equipotentials and the stream lines. Forinstance, in the present example it may be required to find whatincrease in capacitance is caused by the fringing of the field atthe ends of the plates. This increase can be calculated from the

69


equation of transformation without going to the trouble of cal-culating and mapping the equipotentials and flow lines. But iffor other reasons a picture of the field distribution is requiredthe procedure is as follows.

Equation (7.7) gave

z = (e2t - 2t)

but the expression can be simplified by writing t for 2t sincesuch a change only affects the relative size of the two picturesbut at the same time makes the mathematics much clearer.Then from

z (et - t)= 2by putting in real and imaginary parts

dx+jy = {ep(cos q+j sin q)-p-jq}

=d

21T{(ep cos q-p) +j(ep sin q-q)}

Separating real and imaginary parts

x = dr (eP cos q-p)

27

=d

y _ (epsinq-q)

Notice that in changing from 2t to t we have made the distancebetween plates in the t plane 2ir instead of 7r. In equations (7.8)put p equal to zero, then to 0.5, 1.0, 1.5 in succession and so onin steps of 0.5 to as high a value of p as is convenient. Then foreach value of p (i.e. treating p as constant at that value) give qthe following values :

0, 7T/6, 7r/3, 1r/2, 2ir/3, 5ir/6, 7r,

and so on up to q = 29r.

In the t plane, for each value of p the values of q give astraight line parallel to the imaginary axis. Similarly, for eachvalue of q the values of p give a straight line parallel to the realaxis. Therefore in the t plane the field is perfectly regular.

70


In the z plane there is for each value of p and q a value of xand y which can be calculated from equations (7.8). Forexample take the value p =1 and let the distance d be 27r.Then equations (7.8) become

x = 2.718 cos q- Iy = 2.718 sin q-q

From these equations Table 6 can be drawn up.

TABLE 6

Field outside two capacitor plates

q 0 46 43 42 27r/3 546 7r

a=cosq 1 0.866 0.500 0 -0.500 -0.866 -1b= sin q 0 0.500 0.866 1 0.866 0.500 0

x=2.718a-1 1.718 1.354 0.359, -1 -2.359 -3.354 -3.718

y=2.718b-q 0 0.830 1.307 1.147 0260 -1.264 -3.142

q 7a/6 47x/3 342 543 114r)'6 21r

a=cos q -0.866 -0.500 0 0.500 01866 1

b=sinq -0.500 -0.866 -1 -0866 -0.500 0

x=2.718a-1 -3.354 -2.359 -1 0.359 1.354 1.718

y=2.718b-q -5.024 -6.543 -7.4311 -7.590 -7.1196.283

These values give the field outside the plates. The fieldinside the plates can be calculated by starting with negativevalues of p. Table 6 gives the points for one curve of p = con-stant. The regular field is as shown in Fig. 39. The actualfield is shown drawn in Fig. 40.

Density of ChargeThe strength of the field depends on the difference of potential

between the plates and not on the absolute values. Hence thesolution is not restricted in any way if one of the plates is con-sidered to be at zero potential. Considering now the t plane, if

71


0Fm. 39. The regular field. t plane.

Y

P

x

FiG. 40. The actual field. z plane.

the potential of plate A is called zero and the potential of plateB is called V, the potential at any point is given by

V

i2


since 27r is the distance between the plates in the t plane. Andsince by equations (4.14), page 27,

¢+Jo = --V

(p+jq)

X = V t

Let the density of charge be denoted by v. On the plateA, a is equal to the normal component of the electric field, andthis component is proportional to the rate of change of ti in adirection perpendicular to the plate with a negative sign.Therefore

o =

in M.K.S. rationalised units, where aojan is the rate of changementioned and co is the primary electric constant equal to8.854 x 10-12.

To translate this into the z plane it is necessary to multiply by

ataz

an operation that takes care of the scale as described on page 48.In the z plane therefore

a

an

at IEo

az

But00 00On aq

V

21r

73


Furthermore since

z = d(et - t)

2iT

dz _ dlet-11

dt ?7r

= 1epelq-

_ _IeP(cos q+ j sin q) -1d

1eP-1I

because on the surface q is zero.Hence

Cr =

when p is positive and

V 21r 1

27r d IeP - l l E0

VE0

d(eP-1)

VE°

d(1 -eP)

when p is negative. Thus as p- - oo,

Va ->-

dEo = ao (say)

Therefore the equation for negative values of p applies to theinside of the plates, confirming the statement made on page 71.A graph showing a both inside and outside the plate for unitpotential is shown in Fig. 41.

Increase of CapacitanceThe charge on the outside of the plates tends to infinity

making it impossible to calculate the increase of capacitancedirectly. However, an approximation to the effect of the

74

CONFIGURATIONS WITH NO RIGHT ANGLESadF-o

3.0

2.5

20

1.5

1.0

0.5

Inside

Outside

0 0.25 0.5 0 75 1-0 1 25 15 175 20 xd

FiG. 41. Distribution of charge density for V = 1.

charge on the outside has been made by Prof. G. W. Carter whoobtains the total effect of the fringing as follows.

From above, the charge density on a plate is given bya°

a = IeP-1Iwhere ao is the density far into the plates. The extra chargedue to fringing on the inner surface is therefore

w(a - a°) dx

Distances x along the surface of the plate are given byd

x = 2 (ev --p)

so that

dx=2 (eP-1)dp

Now substitute for dx and a and put in the new limits ofintegration, noting that when p is zero x=d/27r, and that whenp = - oo, x= + oo. The extra charge is therefore

f°° I Ad a°d

°a0(1-ep217(eP-I)dp =

27r

75


Thus the extra charge due to fringing on the inner surface isequivalent to widening the distance between the plates by theamount d/21r.

The extra charge on the outer surface is

r0 adx = 1,00 (To d

(ell -1)dpJd/2n o eP-1 2ir

and r0°= 27rfo dp=o0

However, the form of the integral shows that the charge is sodistributed that each addition of unity to p brings in a chargecorresponding with a plate width of d/27r in the uniform part ofthe field. From the formula for x in terms of p we obtain thesevalues :

TABLE 7

Charge on the outer surfaces of two capacitor plates

Extra charge, (a0d/27r) times: 1 2 3 4 5

Width of plate back, d times: 0.27 0.86 2.72 8.06 22.8

If, therefore, the plate is not of infinite width making acapacitance correction impossible to calculate but has a finitewidth of (say) 16.12d, we can approximate closely to the addi-tional capacitance by saying that the fringing adds an effectivewidth of d/27r for the charges inside the plates and four timesd/27r for the charges outside the plates.

Distance between Plates in the t PlaneIn the transformation from the t plane to the w plane given

on page 66 it was stated that the constant A could be deter-mined in such a way as to make the distance between the platesthe same in the t plane as it was in the original z plane. Thatdistance was called d and, putting in the constant, the trans-formation is

t = A log w

Now the point B in Fig. 38, page 68, is given by

z = dr-j

76


and this point corresponds to w = - 1. In the t plane we nowwant this point to be t = jd in order to make the distance betweenplates the same in both planes. Hence

jd = A log (-1)= jA7r

Therefore A=d,7r and the transformation becomes

t = dlog IV . . . . . (7.9)

W

Slotted Plane SurfaceFor another example of a transformation with no right angles

in the configuration, consider a field meeting a slotted planesurface. One electrode (or line source) is assumed to be aninfinite distance above another electrode (or line sink) whichhas the form of a slotted plane surface. The expected fielddistribution is roughly drawn in Fig. 42. Let the width of the

2a 31.

Fir.. 42. Slotted plane surface.

slot be denoted by 2a. Some of the field must penetrate the slotand therefore the configuration to be transformed is that shownin Fig. 43. This figure can be made the z plane.

77


B

w=--A E

W,-I 0-a

y

F

Fic. 43. Representation of boundaries.We want to open this out into a continuous line along the real

axis of the w plane in such a way that the curvilinear squares inFig. 42 become rectangular squares in the w plane. Considera-tion of the rough picture of Fig. 42 suggests the values of wshown in Fig. 43. We wish in effect to bend up the portion BEthrough 180° so that it lies along the axis, bend up the portionCF so that it too lies along the real axis, and then separate thetwo portions until the point B meets the point C. The wholeconfiguration in the z plane then becomes the real axis of the wplane, and the field in the new plane is perfectly regular.

Having thus assigned values to w at A, B, C and D we havenow to give values to w at the edges of the slot, points E and Fin Fig. 43. Since there is only the one dimension a and henceno ratios in the z plane the values of w at the points E and F areat our disposal provided the value at E is negative and the valueat F is positive. It is therefore desirable to make w unity atthese points, i.e. w =1 at F and w = -1 at E. The position ofthese points in the w plane is shown in Fig. 44.

V

E B

-1

C F D0 1 U cc

FIG. 44. The unbroken surface. w plane.78


There are three `corners' in the `polygon' to be transformedand therefore

dz_dw - A(wb)cs/-)--1(wc)(Y1n)-1(w-d)csin>-1

where now b = -1, P = 2irc=0, y= -17

Therefore

d=1, 8=2ir

dzdw = A(w+1)w-2(w-1)

= A(1-w2)

Therefore (Dwight 80 and 82.2)

z=Aw+1+B

Since the origins have already been fixed it is now necessaryto find what value of B matches those origins. At w = 1, z = a,and putting these values into the equation of transformationgives

a = 2A+BAt w = -1, z = - a, and putting these values in gives

-a= -2A+BHence B is zero and a = 2A so that the transformation becomes

z = 2(w+w) . . . . . (7.10)

Density of Now across the SlotSuppose that in the original problem the horizontal lines of

Fig. 42 represented stream lines and it is required to find whatthe distribution of density of flow is across the line of symmetrythrough the slot. In other words what the distribution ofstream line density is on the imaginary axis of the z planeFig. 43. Call this density E and let the distance a be 2 units.

79


Now the density is given byIdwl

multiplied by the appropriate primary constant. Since we arenot concerned with particular units here nor even with the kindof field, this constant is made unity. From equation (7.10) byputting a = 2

Z Z2 1/2w = 2+I4-1

Therefore

E_ dwl

dz

2I1+ (z2 z4)1/2IBut since the values of the density are those along the

imaginary axis of the z plane, put x = 0. Thenz = jy

jyEy =12

1+j(y2+4)1/2

2 E'+ (y2+4)1/2J(7.11)

Now as y-oo, E1. Therefore unity is the value of the un-disturbed flow. Again as y-> - oc, Z y-->0 as it should

E1.0

09

0.6

0.2

4 3 2 1 0 1 2 3 4 5 y

Fie.. 45. Density of flow across a slot.so

CONFIGURATIONS WITH -NO RIGHT ANGLES

By taking various values of y-positive, negative and zero-we can determine from equation (7.11) how the density of flow .Echanges along the y axis. A graph such as that shown inFig. 45 can then be plotted.

Plotting the Stream Lines and EquipotentialsTo plot the stream lines and equipotentials start from equa-

tion (7.10) and replace z and w by their real and imaginary partsthus :

x+]y = acu+jv+ 1u+jv,a u-jV

- 2(u+jz+u,2+v2Therefore

a ux=,,+2 u2 +v2

y=2(V_ v

u2 + V2

V

}(7.12)

,

,

,

4

t

I1 3

,

2

1

-4 -3 -2 O I 2 3 4 u

Fic. 46. The regular field.

Fig. 46 shows the pattern of stream lines and equipotentials inthe w plane plotted from equations (7.12) for a=2. Fig. 47shows the lines obtained for the same values of u = constant andv = constant in the z plane.

6 81


FIG. 47. The actual field.

Transformation of a Point SourceIt may happen sometimes that the source of the field can be

idealised to a single point source, i.e. a line source in the three-dimensional field. Such sources are relatively easy to trans-form : in fact a point source has already been dealt with onpage 32 when the transformation X=log z was discussed.

Consider first a z plane consisting of two parallel straight linesrepresenting a strip of conducting material carrying current.The strip extends indefinitely in both directions as shown in

waO

W. +a*

(a) /elFIG. 48. Transformation of a point source. (a) z plane (b) to plane.

82


Fig. 48 (a). Let there be a uniform flow of current in the direc-tion shown by the arrow so that there are equipotential linesperpendicular to the lines shown. There is clearly a currentsource an infinite distance away in the negative direction.

By choosing values for w as shown in Fig. 48 (a) the con-figuration can be opened out into the w plane as shown inFig. 48 (b) by using the transformation

z = A log w . . . . . (7.13)The stream lines have become radii in the w plane and the flowof current is outwards from a single point, viz. the centre of thesystem. The equipotentials have become semicircles centredon the origin in the w plane.

Now the current flow from a line source whose cross-sectionis located at w = 0 must extend in all directions, with the resultthat the current flow in the upper half-plane is only half thetotal. Allowance must be made for this. Hence the trans-formation of equation (7.13) transforms the field due to a pointsource into a uniform field where twice the current in the uni-form field has to be taken and the constant A has to be deter-mined from data given in the actual problem. Assume for thepresent that A is unity. Then

x+jy = log (re")= log r+j8

Hencex=log yy=B

and these are the equations required.

Collinear Source and SinkEquation (7.13) gives the transformation of a point source

located at the origin. Similarly

z = log (w + a)

when the constant A is unity is the transformation of a pointsource located at the point w= - a. Furthermore

z = -log (w - a)gives the transformation of a point sink (owing to the negative

83


sign) located at the point w = a. The two fields can be super-posed so that the equation

z = log w-aw+a

gives the transformation of a point source and a point sinkwhich are collinear and of equal but opposite potential.

w. . w.+e

w.-4 ... W.e

w.o

/a) (4)

Fie.. 49. Collinear point source and sink. (a) z plane (b) w plane.

This result can be shown by making a Schwarz-Christoffeltransformation from the z plane of Fig. 49 (a) to the w plane ofFig. 49 (b). Let the values assigned to w in the z plane be thosemarked in Fig. 49 (a) and let the configuration be opened upfrom the break shown in the diagram. If the current is flowingfrom right to left there is in the w plane a point source at w = aand a point sink at w = -a. By the Schwarz-Christoffeltheorem the transformation is given by

dzdw = A(w-a)-'(w+a)-1

AW2-a2

Therefore (from Dwight 140.1)

A w-az = -a log w+a+B (7.14)

84


Polar coordinates can be used by taking radii from to = a andw= -a so that

and.w = r1e,Bi +a

w = r2ejea - a

Suppose that in equation (7.14) the origin is chosen so as to makeB zero and the potential is such as to make A unity. Sub-stitution for w in that equation gives

z = 1 (log rlei°' -log r2ejel)

{log r2+j(01-e2)}Hence

x = 1log

rl-2a r2

y = 01-02

In the z plane the field is regular and the lines of flow aregiven by y = constant, i.e. by

01-02 = constant . . . 7.15)

in the w plane. Fig. 50 shows that 01- B2 is the angle APBwhere P is a point on one of the stream lines, A is the point - a,and B is the point +a.

Now AB is a fixed line and it is known in elementary geometrythat if the point P is moved in such a manner that the anglesubtended by AB at P is kept constant., the locus of P must be

P

0 B

FiG. 50. Collinear source and sink.85


a circle through the points A and B. Therefore the variousvalues of the constants in equations (7.15) give various circlesall passing through the points - a and +a as shown inFig. 49 (b). These are the stream lines from source to sink.

The equipotentials in the z plane are given by x = constant ;therefore in the w plane they are given by

rl- = constantr2

Referring back to Fig. 50, it is shown in books on coordinategeometry that when A and B are fixed points and P moves insuch a manner that the ratio PA/PB is kept constant, the locusof P must be a circle whose centre is on an extension of the lineAB or of BA. Different values of the constant give differentcircles as shown in Fig. 49 (b). These circles form the equi-potentials due to the source and the sink. Thus the streamlines and equipotential lines are intersecting circles whosecentres lie along the u and v axes in the w plane.

86

CHAPTER 8

Configurations with One Right Angle

Transformation of a QuadrantOn page 50 the transformation w= z2 was considered with no

knowledge of what such a transformation might represent.After analysing the transformation it was seen that, providedreal and imaginary parts of z were confined to positive values,it represented a transformation from a quadrant to the upperhalf of a plane.

We can now approach this matter from the other end andsuppose that it is required to transform a quadrant in the z planeto the upper half of the w plane. We can use the Schwarz-Christoffel method to show that w = 22 is the transformationrequired. The z plane is shown in Fig. 51 (a) and since thetransformation is equivalent to bending the imaginary axis backuntil it lies along the negative end of the real axis, it is con-venient to make the values of w those shown in Fig. 51 (a).Note that the origins in the two planes are at correspondingpoints.

W-00Y

Z.0O

(a) (b)V

W.01 w.+co

0 X

j.+mU -00

Fic. 51. Transformation of a quadrant. (a) z plane (b) w plane.

Since there is only one corner to be straightened out

dz = A(w-a)(a/') 1dw

87


where a = 0 and a = zr. Hence

dz = Aw-1j2 dw

and by direct integration (Dwight 181.1)

z = 2Aw1/2+BBut since when z is zero w is zero, B must also be zero.

Thereforew = az2

where 4/a =112A.The constant a only affects the relative scales of measurement

in the two planes. There is no need to proceed with thisexample because the complete analysis was done in Chapter 5where a was taken as unity.

Fringing of the Field at a Pole EndAnother problem, this time a more practical one, requiring

the transformation of only one right angle is that of the fringingof the magnetic field between the end of a pole of an electricmachine and the end of the armature core. The problem isillustrated by the section shown in Fig. 52 where the length ofthe airgap between pole and armature is shown as g.

Pole

I----- 9 ---- -- -- - ------ --

LArmature

core

Fic. 52. Fringing problem.

At first sight this looks like a problem involving the trans-formation of two right angles, but the broken line in Fig. 52indicates a line of symmetry between pole and core and thismeans that the problem can be confined to that of Fig. 53 (a)where the line of svmmetrv is taken as one of the boundaries.This modification of the problem reduces the number of rightangles from two to one.

88

CONFIGURATIONS WITH ONE RIGHT ANGLE

w.-Cciy

(a)

W.0U9/

w=+OD

O X

z=j ou

V

(b)

Z. Go

-1 0 aZ

FiG. 53. Pole end to a line of symmetry. (a) z plane (b) w plane.

Fig. 53 (a) can be considered the z plane ; the most con-venient points for w = 0, w = oo and w = - co are readily chosenat the points shown in the figure. Clearly we want to straightenout the configuration by making the line of symmetry thepositive real axis and the pole end the negative real axis in thew plane as shown in Fig. 53 (b). Since again there are no ratiosto consider but just the single dimension g, the value of w at thecorner in the z plane can be given any arbitrary value betweenzero and minus infinity. The equation of transformationappears in its simplest form when the magnitude of w is madeunity at the corner, i.e. w = -1 there. The stream lines in thez plane should therefore become semicircles in the w plane,centred on the origin.

In pulling out the configuration to a straight line two angleshave to be straightened and hence

dz= A(w-a)(arn> 1

dw

where a=-I, a= 37r/2b=0, P=O.

89


Notice particularly that the configuration was straightenedout by making the outside of the corner become the upper sideof the real axis in the w plane ; therefore in this case a is not 90 °but 2700. Inserting these values

dz = w(w+ 1) 1/2 dw . . . . (8.1)

The integration gives (Dwight 194.11)

12-z = A 2(w+1)1/2+Iog (w+1) 11 +B

(w+ 1)1/2+ 1

But since the origin in the z plane can be fixed quite arbitrarilyB can be made zero. Furthermore, to suit later developmentsthe equation can be re-written with the log term inverted andsubtracted, so that

z = A 2(w+1)1/2-log (w+1)1/2+1(w+ 1)1/2- 1

(8.2)

From equation (8.2) when

w = -co, z = -jaoW = + oo, z = + cow= 0, =A(2-logoc) _ -co

These are the values of w shown in Fig. 53 (a) and the equationof transformation (8.2) is thereby checked. But the value ofthe constant A has still to be determined. For this the methodof page 63 can be used. In crossing the air gap near the pointw = 0 and z = - co, the distance traversed in the z plane isjjg while in the w plane it is a semicircle of small diametercentred on the origin. Thus if

w = reiefiu

jjg = dz0

= A fo 1e-ie(rei°+ 1)1/2jrei8 dOr

= jAJ

o (reie+ 1)1/2 d80

90


and as r--->o

?j9=jA fodO0

= jATrHence

A = g/2Tr . . . . . (8.3)

Putting this value of A into equation (8.2) gives

z = 2(w+ 1)1/2-log (w+ 1)1/2+1](w+1)1/2-12TT

= 9

E2- 1)1/2-log{(w+

1)0/2+1}21

=2

[2(w+ 1)1/2-2 log {(w+1)1/2+1}+log w] (8.4)-Tr

Note that when iv= -1,

z .=2n

log (-1)

=29r

07T)

ij9Thus the origin in the z plane is on the line of symmetry and inline with the pole edge. Hence the pole edge lies along the yaxis.

The stream lines become semicircles in the w plane. In theX plane shown in Fig. 54 the field is regular, and equation (7.9)

q,

Fia. 54. Regular field plane.91

0W. 4 o


on page 77 shows that for unit distance between the plates inthe X plane the transformation is given by

X = I log w

and hencew=e'TX . . (8.5)

Substituting this value of w in equation (8.4) gives

z = [2(enx+1)1/2-2log{(e"X+1)1/2+1}+7TX] (8.6)

Equation (8.6) gives the complete transformation from theoriginal z plane to the X plane where the field is regular.

Distribution of Flux DensityThis particular problem is typical of others in which the

boundaries in the original plane extend to infinity in variousdirections. Because of this the total flux passing from the poleend to the end of the core cannot be determined because it mustturn out to be infinite. Of course in practice the configurationcannot extend to infinity in any direction whatever but mustbe bounded in all directions. If finite lengths are defined forthe boundary surfaces (or the electrodes) in the problem therecan no longer be a line of symmetry in the airgap and both rightangles will need to be straightened out. Transformations in-volving two right angles are dealt with in Chapter 9.

Although direct answers cannot be obtained to all the ques-tions that may be asked in such problems, results can be ob-tained that can be used as an aid to judgment. For instancestream lines and equipotentials can be drawn. Write x + j y asbefore for z and O + j i for X. Then by putting O = a constantand 0 a constant in turn, values of x and y are determined thatenable us to draw the stream lines and equipotentials in the zplane. For example, when is zero,

x+jy = 2 2(ejn4c+ 1)11e-2 log l}+jirb]

Now put various values of 0 into this equation ; by separatingreal and imaginary parts, values of x and y are obtained that

92


provide points in the z plane. When these points are joined,the line formed is a stream line. Other stream lines are obtainedby taking other constant values of .

Similarly by putting 0 = a constant and taking various valuesof 0 for each value of 0, a series of points can be obtained in thez plane which, when connected up, provide a family of equi-potential lines.

Another quantity that can readily be found is the density offlow, called the flux density, at any point along the axis ofsymmetry. Thus a curve can be drawn showing how rapidlythe flux density falls away as we recede from the pole end. Theprocedure is as follows. The magnitude of the flux density atany point is given by

B= dX

dzµo . (8.7)

where µo = 1.257 x 10-6 is the primary magnetic constant.Therefore

B= dx dwdw T z-

(using equations (8.1) and (8.5))

7rwg(w+7)I121 2irw

µo

_ I(w+ 1)-1/2 /Jo . . . . . . (8.8)

Now as w--0 and z---> - oo the flux density tends to its uni-form value in the middle of the core and the uniform value isalso the maximum value. It is usual to call the maximumvalue unity. From equation (8.8) as w tends to zero

dX

dz

and if this uniform value is called unity the flux density at anyother point is found from

B = I(w+ 1)-1/'-I . . . . (8.9)

If only positive values of w are now accepted z must bewholly real because all values of w between zero and plus

93

TA

BL

E 8

Dis

trib

utio

n of

Flu

x D

ensi

ty

w0.

050.

100.

200.

501.

003.

008.

0015

.00

80.0

0

(w+

1)11

2=a

1.02

51.

049

1.09

51.

225

1.41

42.

000

3.00

04.

000

9.00

0

log

(a+

1)

=b

0.70

550.

7173

0.74

00.

800

0.88

11.

099

1.38

61.

609

2.30

3

- 2b

-1-4

11-1

-435

-1-4

8-

1-60

0J -

1-76

2-2

-198

- 2.

772

- 3.

218

-4-6

06

log

w =

c-2

-996

-2-3

03-1

-61

- 0.

693

01.

099

2.07

92.

708

4.38

2

2a2.

050

2.09

82.

192.

450

2.82

84.

000

6.00

08.

000

18.0

00

2a-2

b+c=

d-2

.357

-1-6

40-0

.90

0.15

71-

066

2.90

153

077.

490

17.7

76

x =

d/7

rI

-0-7

50-0

-522

-0-2

860.

050

0.33

90.

923

1.68

92.

384

5.65

8

B..

=1/

a1

0.97

60.

953

0.91

30.

817

0.70

70.

500

0.33

30.

250

0.11

1


infinity lie along the x axis as shown in Fig. 53 (a). The x axisis the line of symmetry bisecting the airgap and it is thereforethe line along which the distribution of flux density is required.Therefore Table 8 can be drawn up by using equations (8.4) and(8.9).

From the values given in the last two rows of Table 8 thecurve shown in Fig. 55 can be drawn. It shows the magnitudeof the distribution of flux density along the x axis, i.e. alongthe line half-way across the airgap.

30.

0.6

O.4

O.2

-1 0 1 2 3 4 s r

Fro.. 55. Distribution of flux density along the axis of symmetry.

From examination of Fig. 55 some idea can be formed of whatapproximations can be made. For instance it is possible tojudge at what distance from the armature core it is justifiableto ignore the fringing of the field. However, if it is essential toobtain precise answers to specific questions the field has to beprecisely delineated. The transformation to the plane is thenusually a transformation to a rectangular configuration. Sincethis configuration contains four right angles the analysis is con-siderably more difficult. Such problems are therefore left forthe later chapters.

95

CHAPTER 9

Configurations with Two Right Angles

Flow through a SlitTransformations to the upper half of the w plane of con-

figurations containing two right angles still lead to elementaryfunctions, usually circular or hyperbolic functions. Considertwo plates, each containing a right-angle bend, separated onlyby a small aperture symmetrically placed as shown in Fig. 56.

k d

SlitFic. 56. Flow through a slit.

The two plates form a trough with a slit in the bottom. Anyliquid in the trough will have lines of flow as it pours through theslit. There will also be lines of equipotential intersecting thelines of flow at right angles. The equipotentials are lines ofequal velocity potential. Let the distance between the parallelportions of the plates be d. For this problem we are required,say, to draw the stream lines and the equipotentials.

Fig. 57 shows the z plane of the problem and the first thingto do is to open out the ends to form the real axis of the w plane.Clearly the origin in the w plane should be at the slit itself asshown. Then the values w = oo and w = - oo should be atz = j oo the negative sign being attached to the left side plate.Since there are no ratios but only the distance d in the con-figuration it is again permissible to make the values of w at the

96

CONFIGURATIONS WITH TWO RIGHT ANGLES

W=-oo W = o0

,y

E

+-- - - d -----I

w._1 w.o w.1

FIG. 57. z plane.

corners C and D equal to - 1 and + 1 respectively. The wholeof the inside of the trough opens out into the upper half of thew plane as shown in Fig. 58.

FIG. 58. w plane.

Since there are two corners in the z plane there must be twoterms in the equation of transformation which is therefore

- = A(w-a)(d/n>-1(w-b)csln>-1dw

where now a= -1, b=1, and a = fl = jir. Therefore

dzdw = A(w2- 1)-1/2 . . . . (9.1)

Hence (Dwight 260.01)

z = A cosh-1 w + B . . . . (9.2)

7 97


It is convenient to make the constant B zero and accept theorigin so determined in the z plane. To find the constant A themethod of page 65 can be used whereby an integration is madefrom one plate to the other near to z =joo. In going acrossfrom right to left the distance traversed is - d. In the w planethe path of integration is a semicircle of large radius R, and asz- co in the z plane R---oo in the w plane. Hence from equa-tion (9.1), since

w = Rej°dw = j Rei° dO

-d - A " j Rei8 dO

fo(R2e2J6- 1)1/2

and by making R-->oo, in the limit

-d=A fojd90

=jA,rHence

A = jd/nr

and equation (9.2) becomes

z = L cosh-1 w (9.3)

lw =cosh ``

d)d

J

=COs (_j) (Dwight 654.1) (9.4)

Now when z= -d, w = cos 7r = -1. Therefore the point C inFig. 57 where w = -1 is the point x = - d. Furthermore whenz is zero, w is unity (cos 0). Therefore the point D in Fig. 57where w is unity is the point z = 0 and CD lies along the x axis.Furthermore the line DE forms the y axis in the z plane. Thestream lines in the z plane become radii in the w plane while theequipotentials become semicircles in the w plane.

We now require the transformation between the w plane andthe X plane where the stream lines and equipotentials are

98


straight lines. From previous results the transformation to theplane where the field is regular is given by

X=dlogw . . . . . (9.5)IT

provided the distance between plates in the X plane is also d.This is shown in equation (7.9), page 77. The direct trans-formation from the z plane to the X plane is found by combiningequations (9.4) and (9.5) giving

X = d log cos d z . . (9.6)

since cos ( -A)= cos A.In order to simplify what follows and to avoid obscuring the

essential points let the distance between the plates be ir, so thatd/ir is unity. The problem as stated on page 96 is to draw theequipotentials and lines of flow in the z plane. Therefore sub-stitute real and imaginary parts for z and X in equation (9.6) :

+j = log cos (x+jy)= log (cos x cos jy - sin x sin jy)

log (cos x cosh y - j sin x sinh ,y)

Let m = cos x cosh y,n = -sin x sinh y.

Then

and if

Then

0+1o = log (m+jn) . . . . (9.7)

m+jn = re3°r = (m2+n2)1"20 = tan-' (nlm.)

0+j' = log (m+jn)= log r+j8= lz log (m2+n2)+j tan-1 (n!m)

By separating real and imaginary parts

= log (m2+n2)log (cost x cosh2 y+sin2 x sinh2 y) (9.8)

99


andsG = tan-'(n/m)

tan-' (- tan x tanh y) . . . (9.9)

Equipotential lines are obtained by giving various values tothe constant in the equation 0 = constant, i.e.

cost x cosh2 y+sin2 xsink' y = constant

The expressions can be simplified by using the followingrelations.

cos2x = (1+cos2x) (9,10)cosh2 y = ? (I+ cosh 2y)

sin2x = -(I-cos2x) l (9,11)sinh2 y =

2(- I + cosh 2y)

Therefore, when = constant,(1 +cos 2x) (1 + cosh 2y) + (I - cos 2x)(- I + cosh 2y) = constant

By multiplying out and collecting terms it is seen that = con-stant is equivalent to

cos 2x + cosh 2y = constant . . . (9.12)

For each value of the constant, values of x and y can be foundto satisfy equation (9.12). These values plotted in the z planewhen connected together give an equipotential line. A wholefamily of equipotentials can be obtained by taking sufficientvalues of the constant. Table 9 shows a typical calculation forthree values of the constant.

Similarly the stream lines can be determined by givingvarious values to the constant in 0 = constant. But if 0 isconstant so is tan 0 and from equation (9.9)

tan 0 = - tan x tanh y

Therefore the stream lines are given by

tan x tanh y = constant

The whole family of stream lines can be calculated by taking asufficient number of values of the constant. Table 10 shows atypical calculation for three values of the constant.

100

TA

BL

E 9

Flow

thro

ugh

a sl

itC

alcu

latio

n of

cqu

ipot

entia

ls

x0

-7r/

8-7

r/4

-37x

/8--

JI -

7r`2

-57x

/8--

-37x

/4 -

77x

/8-;

r2x

0-7

r/4

-7T

/2-

37x/

4-

--7

7-

57x/

4-

37x(

2-

77x/

4-

21r

cos

2x1

0.70

70

-0-7

07-1

-0-7

070

0.70

71

cosh

2y

= 2

-co

s 2x

11.

293

22.

707

32.

707

21.

293

1

2y0

0.74

81.

317

1.65

41.

763

1.65

41.

317

0.74

80

y0

0.37

40.

659

0.82

70.

882

0.82

70.

6559

0.37

40

cosh

2y

= 3

- c

os 2

x2

2.29

33

3 7

07!

41

3.70

73

2.29

32

y0.

659

0.73

60.

882

0.99

21.

032

0.99

20.

882

0.73

60.

659

cosh

2y

= 4

- c

os 2

x3

3.29

34

4-70

75

4-70

74

3.29

33

y0.

882

0.93

11.

032

1.11

51.

146

1.11

51.

032

10.

931

0.88

2


TABLE 10

Flow through a slitCalculation of stream lines

x -7r/8 -Tr/4 - 37r/8 - 5Tr/8 - 344 -7Tr/8

cot x - 2.414 -1.000 - 0.414 0.414 1.000 2.414

tanh y =+ 0.4 cot x 0.966 0.400 0.166 0.166 0.400 0.966

y 2.029 0.424 0.168 0.168 0.424 2.029

tanh y=±0.2 cot x 0.483 0.200 0.083 0.083 0.200 0.483

y 0.527 0.203 0.083 0.083 0.203 0.527

tank y =0.1 cot x

0.242 0.100 0.042 0.042 0.100 0.242

y 0.247 0.100 0.042 0.042 0.100 0 247

In this table those values of x are omitted for which the valueof y is unchanged whatever the value of the constant chosen.These values are :

At x=0,x = j Tr,

x = - TT,

y = co

y=0y= 00

From calculations such as those shown in Tables 9 and 10 thefield can be mapped. Some of the stream lines and equi-potentials are shown in Fig. 59. Note that this transformationapplies equally well to the magnetic field between two platesshaped as in Fig. 56, the potential of one plate being zero andthe potential of the other being, say, unity. In that applicationof the transformation 4 and 0 change roles. The curves

0 = constant

give the equipotentials and the curves

¢ = constantgive the lines of flow.

102


Fia. 59. Flow through a slit. Equipotential and stream lines.

Slot Opening Opposite a Solid FaceThere is a much more practical and more celebrated trans-

formation of a configuration containing two right angles. Itarises in problems associated with two magnetic faces separatedby an airgap where one of the faces is slotted. The configura-tion is shown diagrammatically in Fig. 60 where g is the lengthof the airgap, this length being considered to be uniform, and 8is the width of the slot opening. This transformation is a well-known piece of work first undertaken by F. W. Carter andpublished in 1901.* Its purpose is to take account of the e$ectsof having slot openings at the airgap surfaces in electricmachines. The results of this transformation have been sosuccessful and are in such widespread use today that it is worthwhile to treat it rather exhaustively.

For the purpose of the transformation it is assumed that thecurvature of the surfaces at the airgap can be neglected andthat the depth of the slot has so little influence on the resultsthat it can be considered infinite. In actual electric machinesthe length of the airgap and the width of the slot opening areboth so small compared with the diameter at the airgap that it



is quite reasonable to ignore the curvature for purely localproblems such as those to be dealt with here. More exactmethods such as those used later in this book show that, for thedepths of slots found in practice, it is justifiable to treat the slotas if it had infinite depth. The usual assumption is also madethat the problem is entirely two-dimensional.

{9

S-

Fir.. 60. Single slot opening.

The smooth surface at the airgap, i.e. the upper line in Fig. 60,can be treated as an equipotential at, say, potential V, and theslotted surface, i.e. the lower line, can be treated as anotherequipotential at zero potential. But for the presence of the slotopening the magnetic field would be uniformly distributed alongthe airgap.

In practice the problem is never really that of a single slotbut that of a succession of equal slot openings. Now the rela-tive dimensions of slot width, tooth width and airgap length aresometimes, though not often, such that it is unwise to simplifythe problem to that of a single slot. The problem of a succes-sion of equal slot openings will be dealt with fully in Chapter 17and it will then be interesting and instructive to compare theresults with those obtained in this chapter and determine whichaspects of the problem can justifiably be treated as the problemof a single slot opening and which can not. The question of aslot of finite depth, although not dealt with in Chapter 17, canreadily be analysed by the methods of that chapter and it isfound that the field does not penetrate very far into the slotopening. In fact the penetration is so slight that with normaldepths of slot the depth might as well be considered infinite ;the effect on the results is negligible.

Thus although the idealisation indicated by Fig. 60 may104


Iw= -00

w=O A BW= b

0of S

9

T

Cw.1

FiG. 61. z plane.

seem rather drastic, it is not so in fact, and the configurationshown there can be made the basis of the z plane shown withcertain values of w in Fig. 61. But there are several ways ofopening out this configuration into the upper half of the w plane.However, since it will be advantageous to have the resultingstream lines in the w plane semicircles it is best to make theorigin of the w plane where the real part of z is at minus infinity.Then naturally where z = + oo, at the upper surface makew = - oo and at the lower surface make w = + oo. These arethe values of w shown in Fig. 61.

There are three more corners, shown as A, B and C, wherevalues of w have to be fixed. Only two of these three cornerscan have arbitrary values assigned to them, so that one valueof w is left unspecified to be found later in terms of the ratio s/g.Let this unspecified value be denoted by a. Now the value ofwv at each of the corners A, B and C must be positive, for thepoints lie between the points w = 0 and w = oo. The followingchoice of values can be made : at A let w = a, at C let w = 1, andat B let w = b. Note that the value b is not independent of a ;only one value is independent because there is only one definiteratio in the z plane. Therefore there is a relation between aand b that must be determined later on. Notice also that a liesbetween w = 0 and w =1 and is therefore less than unity. Fromthe symmetry of the configuration it might be expected that bis the reciprocal of a ; it will be shown presently that this is so.Finally note that the `corner' C (where w= 1) is at the pointwhere the imaginary part of z is minus infinity.

105


By the Schwarz-Christoffel equation the transformation fromthe z plane to the upper half of the w plane is given by

dz _ A(w-a)112(w-b)'12dw w(w -1)

V

(9.13)

z.w+jq z. -co z;-jco z-0 z.*ao-CO 0 a 1 b t 4

Fm. 62. w plane.

The w plane is shown in Fig. 62. It is sometimes an advantageto determine some or all of the constants from the equation oftransformation before the integration is attempted. This willbe the procedure here.

Determination of ConstantsThe value of A can be found by using the method of page 65

of integrating along a large semicircle in the w plane. In thez plane, Fig. 61, the distance between surfaces as z approachesinfinity is the constant value g so that, taking into accountdirection, the value of the integration across the airgap is jg atall points after w = b. In the w plane the path of integration isalong a large semicircle radius R where B can be made as largeas we please. Then, as before,

w = Reieand

dw = jReje dO

Substitution into equation (9.13) gives

fITf dz = A(ReiO-a)1/2(BeJB-b)"2 jReia d9o .Reie(ReJO-1)

106


But as R-* oo it greatly exceeds a, b and unity. Hence forvalues of R approaching infinity

If- A(ReiORei8)1/2 Reia d8

ReJeReie

= I jA dO0'ff

=jiABut in the z plane f dz is jg and therefore

A = g/,;T . . . (9.14)

A similar investigation at the other end of the configurationin the z plane where w = 0 gives the relation between a and b.This is the method described on page 63. Here again theintegration in the z plane across the airgap remains constant asz approaches - oo and the value is again jg. In the w plane thepath of integration is a small semicircle centred on the originand of radius r where r can be made as small as we please. Thenas before

w = reJ0dw = jreJ° dO

A(reJ°-a)1/2(reie-b)1/2 .f dz = f reje dOo reie(reJe - 1)

But now as r approaches zero it is very much smaller than eithera, b or unity, and hence for values of r approaching zero

rv A(ab)1/2 .

f dz = f oo r &B(-1)

0J

- (ab)1/217.

-jg(ab)1/2

But f dz in the z plane is jg and hence

(ab)1/2 = - 1ab=1b = 1/a . . . . . (9.15)

107


This result was anticipated on page 105.Although equation (9.15) has established a relationship be-

tween a and b it is convenient for the present to retain theseparate symbols. Then substituting for A in equation (9.13)we obtain

dz = g (w-a)1'2(w-b)112

HenceIT w(w-1)

dw

y f (w-a)1/2(w-b)1/2 dwz=.7r w(w-1)

The integrand is not in a form in which Dwight's Tables canbe applied and some preliminary treatment is necessary. Put

w-byw-a

so that

Then

and

ape-bp2-1w=

w-1 = (a-1)p2-(b-1)p2- 1

dw = (p2-1)2ap-(ape-b)2p dp(p2-1)2

2p(b -a) dp

Therefore

dw _ 2p(b -a) dpw(w-1) (ape-b){(a-1)p2-(b-1)}

2(b -a) p/ dpa(a-1)

(p2--b2)I p2-b -11a-1/J

2(b-a)p dpa(a -1)(p2 - b2)(p2+b)

since from a= I lb,b-1

= -ba-1108

. (9.16)


Furthermore from equation (9.16)

and

Hence

and

ape-b-ape+ap2-1

a-b

w-a =

vl - h =

p2 - 1

ape-b-bp2+bpZ-1

(a- b)p2p2-1

(w-a)(w-b) (a-b) (a-b)p2p2-1 p2-1

(w-a)1/2(w-b)1/2 = (a b)pp2-

Combining these results

_ 2g (' (b-a)2p2 dpz

= it J a(1-a)(p2-1)(p2-b2)(b+p2)

and substituting 1/b for a

z = 2g (' (b+1)2(b-1)p2dp7T J (1-p2)(b2-p2)(b+p2)

(9.17)

Clearly, when equation (9.17) has the integrand split intopartial fractions the resulting forms will be found in Dwight'sTables. Break up the integrand as follows.

(b+ 1)2(b-1)p2 _ A B C

(1 -p2)(b2 -p2)(b +p2) - 1-p 2 + b2 -p2 + b +p2

From this

A{b3-b(1-b)p2-p4}+B{b+(1-b)p2-p4}+C{b2 - (1 + b2)p2 +p4} = p2(b + 1)2(b - 1)

Equating coefficients of p4 : -A- B+C = 0 . . (a)p°: b3A+bB+b'-C = 0

i.e.: b2A+B+bC = 0 . . (b)

109


Adding (a) and (b) :

Hence(b2-1)A+(b+1)C = 0

C = (1-b)ASubstitute this value of C in equation (a,):

-B-bA = 0Hence

B= -b.4Now equate coefficients of p2:

-b(1-b)A+(1-b)B-(1+b2)C = (b+1)2(b-1)In this equation first substitute for B and C and then dividethrough by the factor (1- b) :

bA+bA+(1+b2)A = (b+1)2Hence, A =1, B = - b, C = (1- b). Therefore

z = 2g (' dp _f bdp _f (b -1) dp7 J 1-p2 J b2-pa b+p2

Using Dwight 140, 140.1 and 160.01,

z = g[log I i +pp

-log Ib-pb-p

- 2(b -1) tan (9.18)/b -1b

In equation (9.18) the constants of integration have been madezero since the origin in the z plane is not yet fixed. To deter-mine this origin it can be noted that when p is zero

g 2(b - 1z =E

log 1-log 1-b

tan-10

= 0

Thus the origin in the z plane occurs when p is zero. But, fromequation (9.16), when p is zero w=b; this means that theorigin in the z plane is at the point B in Fig. 61.

Finding the Value of b in terms of s/gThe value of b depends on the ratio s/g. Since

p2 _ w-bw-a

110


as w- tea, p--- oo. Therefore near the point w = a the value of pis very large and in the limit where w = a,

z = {log (-1)-log (-1)-2(N/b1) tan-' 00

gb-1=

IT

g(b - 1)

,/bBut in the z plane, Fig. 61, w = a at the point A where z s.Hence

g(b-1)

from which,/b

b-1 s

/b = 9 (9.19)

In any specific problem it is best to proceed with the quan-tities a and b and convert them after the solution into terms of8/g by equation (9.19).

The Distribution of Flux Density

In the X plane, Fig. 63, the field is uniform. It was shown onpage 77 that if the distance between plates in the x plane isunity and the potential difference is V, the transformation fromthe w plane to the X plane can be written

x= Ylog w7T

W. 0

0Fio. 63. X plane.

111

W= - oo

0W:}o0


As before the flux density at any point is given by µo IdX/dzI.

Therefore for a potential difference across the airgap of V theflux density B is

B = V dX dwdwdz Uo

9

V -7r w(w -1)7rw g (w-a)1/2(w-b)112 µo

V (w-1)(w-a)1/2(w-b)1/2 µo . (9.20)

It is clear from the z plane, Fig. 61, that the flux densityreaches a maximum value at w = 0, an `infinite' distance awayfrom the slot opening. In equation (9.20) as w--->0, B->Vµolg,and this must be Bmax Therefore

B (w-a)1/2(w-b)1/2 Bmax (9.21)

for all real values of w. The minimum flux density can now befound in terms of Bmax by putting dB/dwv equal to zero andfinding the value of w at this point.

dB= (iv-a)-112(W- b) "1/2-

z(w-1)(w-a)-1/2(w-b)-1/2

dw{(w - b)-1 + (w -a)-1}

This is zero when

(w-1)

i.e. when

1 1

w-bMw-a

Hence Brn in is obtained by putting w = - 1 in equation (9.21) ;therefore

Bmin 2

Bmax (1 +a)1)2(1 +b)1/2

2

(a+b+2)1JY112

. . (9.22)


This expression for the ratio of minimum to maximum fluxdensity enables us to determine the amplitude of the ripple inthe flux density wave caused by a slot opening opposite a poleface. The ripple in the wave sets up in the pole face eddy cur-rent losses that are proportional to the square of the amplitudeof the ripple. Hence in order to calculate these losses the ratioBurin/Borax is required.

The Flux Density CurveSometimes it is necessary to know the shape of the flux den-

sity distribution curve because this is also the shape of the curveof variation in time of the density of the eddy currents at anypoint on the pole face. Usually there are pronounced harmonicsin this curve, harmonics that make a large contribution to thetotal eddy current loss. The extra loss caused by the eddycurrents can be calculated provided the amplitude of the variousharmonics are known and these amplitudes are found by makingan analysis of the flux density distribution curve.

The problem is even more important in such apparatus asinductor-type eddy current brakes and couplings, for they havea similar magnetic structure and operate entirely by virtue ofthe ripple set up in the flux density wave by slot openings.Since the air gap in such machines is relatively very short theamplitudes of the harmonics are correspondingly large. Theharmonic currents so caused assist in producing torque andtherefore have to be taken into account.

To draw the flux density curve take various values of w andfrom equation (9.20) determine B for each value chosen,choosing real values only. Then from equation (9.16) deter-mine p for each value of w and hence find x from equation (9.18)for the same values. A curve connecting B and x can then bedrawn. The values of w chosen need only lie between zero and-1 for this reason : It was shown on page 112 that the minimumvalue of B occurs when w = -1. Now at this point fromequation (9.16)

1+b

8

p1 + 1/b

b

113


Hence from equation (9.18), when w= -- 1,

z =2(b - 1)

7btall-1 (1)

= 1) - 2}9

S_ -9+Jg

This value of z is the point on the smooth face exactly oppo-site the centre of the slot. Thus it is not only the point ofminimum flux density but also on the axis of symmetry of thewave. Therefore it is only necessary to take real values of wbetween zero and ---1; this gives the curve of one half-periodand owing to the symmetry about the point w= -1 the curvefor the whole of the period can then be drawn in.

For a numerical example consider a ratio 81g= 1-5. Thenfrom equation (9.19), page 111,

b=4 and a=0-25,n2 = w-4

w-0.25w-1

B (w- 0.25)112(w- 4)1/2 Bmax (9.23)

using equation (9.21). Substituting these values in equation(9.18), page 110, and taking values for which z is wholly real

x = g log7r

1+p1-p

- log4+p4-p

-3 tan-1 (1p)} . (9.24)

Furthermore, for these values of a and b, from equation (9.22)

Bmin _ 2 = 0.8Bmax (6.25)112

Since we are choosing values of w lying between zero andminus one, equation (9.23) is best rewritten

B= I- W B(9.25)(0.25-w)1"2(4-w)112 max

Now choose various values of to between the limits mentioned114


and for each value of w find the corresponding value of p from

24-w

0.25-wThen for each value of w calculate B from equation (9.25),making Bmax unity. Finally, for each value of p determinedabove calculate the corresponding value of x from equation(9.24); if g is unknown the value found will be x,lg.

Hence a curve showing the value of B at any point along thesmooth surface from the point opposite the centre of the slot,i.e. where z = - zs + jg can be drawn. As before it is best todraw up a table, and Table 11 shows a typical calculation.

TABLE 11Calculation of flux density distribution

-0.01 -0.05 0.10 i-0.20 -0.40 -0-60 -1.0

(1)=0-25-w 0.260 0.300 0-350 0-450 1 0.650 0.850 1.250

(2) = 4 - w 4-010 4-050 4-100 4.200 4-400 4-600 5.000

(3) _ (1)1/2(2)1/2 1-021 1.102 1-198 1-375 1.700 1.977 2-500

B 1-wBm,X- (3)

0-989 0-953 0.918 0.873 0-824 0-809 0-800

p=(2)=/2/(1)1%2 3-927 3-674 3-422 3-056 2-602 2-326 2-000

4) =p i i 1-683 1-747 1.826 1.973 2 248 2 508 3 000

±p(5)=4 108-6 23-54 12-84 7.475 4-131 3.779 3-000

(6) =log (4) 0-521 0.558 0-602 0-680 0-810 0.919 1.099

(7) =log (5) 4-688 3.1591 2-553 2-012 1-419 1.330 1-099

(8)=(6)-(7) -4-169 -2.601 -1.951 -1.332 -0-609 -0-411 0

tan-1(p) 1.100 1.072 1-042 0-991 0.915 0.861 0.785

(9) =Stan-1(kp)I

3-300 3-216,i

3.126 2-973 2-745 2-583 2-355

1Tx=(8)-(9) -7.469 -5-817 -5-077 -3.354-4.305 -2.994 -2.3559

,

--2-377 -1-852 -1-616 -1.370 -1.068 - 0.953 -0,750

9

115


BBmaz.

1.0

09

08

07

06

0

9 9 =1.5; _ 0.612

4 3 4X

Fin,. 64. Waveform of B with large gap.

From Table 11 the curve can be drawn for any value of g.By taking g unity and therefore s =1.5, the curve shown inFig. 64 is obtained. It shows that the ratio B%Bmax reaches itsuniform value of unity at apparently x =1.22. This meansthat if the tooth is so narrow that the edge of the next slot isnearer than at x = 2.44 this analysis for a single slot is invalidfor a succession of equal slots. The transformation that takescare of a succession of equal slots will be discussed in Chapter 17.

In Fig. 64 a tooth width of 2.45 is shown giving a slot to toothratio of s/t = 0.612. The curve shows the presence of harmonicsincluding second, fourth and higher even harmonics of quitelarge amplitudes. The machine dimensions giving the curveof Fig. 64 are typical of salient-pole synchronous machines withopen armature slots. In eddy-current inductor brakes andcouplings the airgap lengths are by comparison very small andthe ratio s/g may be as high as 80. Then the amplitudes of theharmonics are even higher. Fig. 65 shows the flux densitydistribution curve calculated in the same manner for a ratios/g = 12 and for g = 0.1. This curve is continued for a toothwidth equal to the slot width s.

116


aBmax.

10

0.75

O.5

n1

025

O1 0

S S

12 X 2

9=12 ; t=10Ii

S

FiG. 65. Waveform of B with small gap.

It is clear that for these values the analysis is valid for anypractical width of tooth because the flux density becomes uni-form at a distance of only s/6 from the edge of the slot opening.The curves of Figs. 64 and 65 show how greatly the amplitudeof the variation in flux density depends on the ratio s/g.

The Lost FluxBy the lost flux is meant the diminution in flux crossing the

airgap caused by the presence of the slot opening. This canalways be obtained from the flux density distribution curve byintegrating graphically the area between that curve and thestraight line

B/Borax = 1But it is just as easy to integrate the equations and obtain ageneral expression. The lost flux is

f00

m(1--B)dx

i.e. the difference between

dx and f. Bw

117


where the scale is so arranged that Bmax is unity. Now fromequations (9.15) and (9.21)

B=(w-a)1/2(w-b)1/2

if Bma, is unity, and

dx = g (w - a)1/2(w - b)1/2dw

Tr w(w- 1)Fig. 62 shows that when

x = -00, w = 0

X = oo, w = ± oo,

and since we are integrating along the smooth surface, the valuew = - oo should be taken. Therefore

0a =

The first part of the integrand has already been put in termsof p in equation (9.17), page 109. Considering now the secondpart and putting it in terms of p, from equation (9.16)

p2-1

w ape - bb(p2-1)p2-b2

(1-B) dxf-".O

g f-00 W-1 }(w_a)l/2(w_b)1;2_ a 1-(w-a)1/2(w-b)1/2 w(w-1)dv)

o w(w -1) wTr

g 0° (w-a)1/2(w-b)1/2 1dwf

w-1

It is shown on page 108 that

Hence

dw = (p2 - 1)2

dw 2p(b2-1) dp

2p(b - a) dp

w (p2 - b2)(p2 - 1)

With regard to limits of integration when changing from w to p,since

p2 = w-bw-a

118


when w = 0, p = (b/a)1 J2 = b, and when w = - oo, p=1. There-fore

dw _ fb 2p(b2 -1) dpJo w -Ji (p2-b2)(p2-1)

Adding now the first part of the integrand as given inequation (9.17) and putting in the limits of integration

i - - 2g(b2 _ 1) b (b + 1)p2ar J1 (p2-1)(p2-b2)(p2+b)

p(p2- l)(p2-b2) dp

_2g 2

Sb

(b+l)p2-p(p2+b)- -1) d'7 p1 (p2-1)(p2-b2)(p2+b)

29 b2 1 -p{p2-(b+l)p+b}d

_ ( _ )_7r 1

(p2 - 1)(p2-b2)(p2+b) p

= 29(b2-1) p(p-1)(p-b) dp7r f1 (p2-1)(p2-b2)(p2+b)

=2g(b2-1)

b

p dpn J1 (p+1)(p+b)(p2+b)

B 1- 2gJb(A C

dpit 1 p2+b+p+b+p+1where

A{p2+(b+1)p+b)+B{p3+bp2+bp+b2}+C{p3+p2+bp+b} _ (b2-1)p

Equating coefficients of each index of p gives

A=(b-1), B=-1, C = 1therefore

01

b2g b-1 1 1

p2+b+p+b p+l dpb

=2g

V/btan_1

b-

log

p+

b11 (Dwight 120.01, 90.1)

=29

L b !tan-1 /b -tan-1 fib} -log (b 4b

1)z}

119


The next step is to get this result into a form that containss and g instead of b. For the first term put

A = tan-1 Vb

B = tan-1I

i.e. tan A = V b and tan B=1/i/b. Then

tan (A - B) = tan A - tan B1 + tanA tan B

_ \/b-1/Vb _ s

2 2g

from equation (9.19), page 111. Therefore

A - B = tan-' \/b-tan-1 1

VbS

tan-12g

With regard now to the second term,

(b+1)2 4b+(b-1)24b - 4b

=1+4(ti/b- 1 2

\Ib)

Therefore

where

(s sr S2

I = 9 tan 1 2g -log (1+ 4g2

2

_ s tan-1 2g9

log1+

1l= so (9.26)

(or = tan-1 9-s log I1+42

92/(9.27)

The coefficient a in O = sa has been called by some authors120


Carter's coefficient. However, it is not the coefficient in use bypractising engineers. Their coefficient is derived from a asfollows.

Carter's Coefficient

The flux crossing the airgap for one slot pitch when there is asuccession of equal slot openings is the flux across a width oft + s where t is the tooth width. If the slots were not presentand if again units are chosen that make Bmax unity, the flux forone slot pitch would be simply

0 = t+sNow when there is a tooth width dimension sufficient to givesubstantially uniform flux density before x reaches the valueit we know that the actual flux crossing the airgap for one slotpitch must be 0 as above, less the lost flux calculated onpage 120, i.e.

Oa=0-01= t+s-scr

The ratio of the undisturbed flux gS to this actual flux is

0 t+sC

Oa t+s-sa_ t/s + 1

t/s+1-o (9.28)

and this C is what most authors call Carter's coefficient.The factor a obtained from equation (9.27) is independent of

the ratio t/s and depends only on s/g. Therefore the best pro-cedure is to calculate a for various values of s/g from equation(9.27) and for each value of a so found to calculate C fromequation (9.28) for various values of s/t. A family of curvescan then be drawn giving C for any combination of s/g and s/t.Such a family of curves is shown in Fig. 66 for all values of s/gbetween 0.2 and 10, and for all values of s/t between 0.15 and 10.

However, it is easier to interpolate for intermediate values ifthe ordinates and the abscissae are scaled on a basis that makesthe family of curves into a family of straight lines. Further-more the spacing of the lines can be greatly improved. The

121


10

9

$ 8

9

7

s

3 i

I

1 05

f 2 3 4 6 1 8 9 - 11

tFIG. 66. Carter's coefficients.

C

r7

rs

1.A

13

12

curves of Fig. 66 redrawn in this manner are shown in Fig. 67.For practical use a nomogram is even better and it is not diffi-cult to devise one.

The Equivalent Airgap LengthThere is another way of expressing Carter's coefficient. Let

the actual physical airgap length be g giving a flux across theslotted gap of 0. the quantity we have called the actual flux.Let g' be the equivalent airgap length that would give exactlythe same flux 0a with the same excitation but with both airgapsurfaces unslotted.

Considering now both surfaces unslotted the flux obtained foran airgap length g would be 0, the quantity we have called theundisturbed flux. For the greater airgap length g' the fluxobtained would be 0. because g' was chosen to give this result.Since the flux obtained (when the surfaces are unslotted and the

122

CONFIGtiRATIONS WITH TWO RIGHT ANGLES

3

2

1.5

1.0

(>I

0.6

05

0.4

03

0.2

r s

rp J

4 r.O?, T-'Ojs I

0.15 0'2 03 0{ 05 06 08 b5 2 3S

F

FiG. 67. Carter's coefficients with special scales.

1c

excitation held constant) is inversely proportional to the airgaplength it follows that

Thereforey 0 = C (equation 9.28)9

_ a

123

CHAPTER, 10

Configurations with more than Two Right Angles

The Inside of a RectangleIn all the problems considered in previous chapters the field

has extended in the w plane to plus and minus infinity along thereal axis so that in the x plane the two plates that give a uniformfield have also been of infinite length. In practice there aremany problems, in fact a majority of them, in which the equi-potentials forming the boundaries of the problem do not takeup the whole of the real axis when transformed to the w plane.The equipotentials are already finitely bounded in the z planeand the Schwarz-Christoffel transformation produces definedlengths along the real axis of the w plane to correspond to theseequipotentials. A typical w plane is shown in Fig. 68 where AB

V

A B

FiG. 68.

C D u

represents one equipotential boundary and CD the other. Thefield'in the w plane extends from AB to CD but ranges all overthe upper half of the plane ; it is only bounded by the real axis.

Notice that this is an entirely different transformation fromthat of a definite portion of the w plane in which the equi-potentials are radii and the lines of flow are semicircles. Thefield in the w plane is then bounded by semicircles as shown inFig. 69. The area bounded by EFGH in the w plane Fig. 69transforms to the corresponding rectangle in the X plane shown

124

CONFIGURATIONS WITH OVER TWO RIGHT ANGLES

X- id X.o

Fic. 69. Bounded configuration in the w plane.

in Fig. 70. This transformation is the one developed onpage 76 and given by equation 7.9:

X=dlogw7T

where d is the distance between plates in the X plane. Theequation applies to the field between the values of w at the fourcorners E, F, G and H.

I

F E

Id

IG 0 H

FIG. 70. Bounded configuration in the X plane.

In most transformations that follow in this book the onlyboundaries in the upper half of the w plane will be the lengthsAB and CD on the real axis ; these lengths can be called for con-venience `electrodes', since in some problems the equipotentialboundaries will actually be electrodes.

It is comparatively easy to transform the w plane of Fig. 68to the t plane of Fig. 71 where the field extends indefinitelybetween plates both above and below. But this is not what issometimes required. For instance if we are concerned withcurrent flow from one electrode to the other we shall want the

125


Fr- (GFIG. 71. Field in the t plane.

current in the X plane to be bounded by a rectangular area suchas that shown in Fig. 72. Inside the rectangle the equipoten-tials are parallel to the sides AB and CD, while the lines of floware straight lines intersecting the equipotentials at right angles.Therefore, in problems in which the equipotentials in the z plane

Y,

A

8

W.! 00

w=O

D

C 0Fir,. 72. Regular field in the X plane.

are completely bounded, the transformation from the regularfield in the X plane, Fig. 72, to the w plane, Fig. 68, is a trans-formation from the inside of a rectangle to the upper half of thew plane. This transformation requires the straightening out offour right angles.

When more than two right angles have to be straightened outthe integration required in the Schwarz-Christoffel trans-formation cannot be performed in terms of elementary func-tions. The integral is either itself an elliptic integral or anintegral that can be broken down into one or more ellipticintegrals and one or more elementary integrals. Hence theintegrations are considerably more difficult. It so happens

126

CONFIGURATIONS WITH OVER TWO RIGHT ANGLES

that the simplest of them is the integral given by a Schwarz-Christoffel transformation of the inside of a rectangle to the wplane, the transformation now under consideration. It is there-fore the one most suitable for opening the subject.

To transform the configuration of Fig. 72 to that of Fig. 68 itis clear that the values w = 0 and w = ± oo must be placed at thepoints shown in Fig. 72. Values of iv must now be assigned atthe four points A, B, C and D where, however, by symmetry thevalues at A and B must be the negatives of the values at D andC respectively. There is only one ratio in the X plane, viz. theratio of width to height of the rectangle, i.e. BC/AB. There-fore one of the values of w must be left unknown and denoted bya while the other can be given an arbitrary value, say unity.Hence the values of w chosen are those shown in Fig. 73, viz.

Y_

w=-a w=-ooJW=+oo w=aA[ I 0

w=-1 0W= 0

w=108

Fio. 73. Values of w in the X plane.

From this choice it is clear that a exceeds unity. By theSchwarz-Christoffel transformation

dw =A(w-1)-1/2(w-a)-1/2(w+1)-1/2(w+a)-1/2

Hence_ dw

X A f (u'2- 1)1/2(w2-a2)1/2

W = 1 at Cw = -1 at Bw = a at Dw=-a atA

. (10.1)

127


Now the integral in equation (10.1) cannot be evaluated interms of elementary functions. But by putting k=1 Ja wemake the equation assume the better-known form

dwX

=- B(1-w2)'/2(1- k2w2)li2 (10.2)

where B = - kA, and the integrand is now recognisable as thatof an elliptic integral.

Before further consideration can be given to this equationwhich transforms the interior of a rectangle to the upper half ofa plane, it is necessary to digress and consider the generalmethod of handling elliptic integrals.

128

CHAPTER 11

Elliptic Integrals of the First Kind

Jacobi's FormMost of the difficulties and complexities confronting the

engineer when he tries to tackle elliptic integrals are moreapparent than real. Quite often the difficulties are largelyassociated with nomenclature and terminology, and for thisreason it is well worth while to proceed slowly in the initialstages.

Consider again the integrand of equation (10.2). The con-stant B can be made unity for the present since its value onlyaffects the scale change in the transformation. Now an inde-finite integral can always be expressed by a definite integralbecause

f(w) = f(0)+J`0

f'(x) dx

Therefore we can investigate the definite integral in

_-fWdwo (1-w2)I/2(l-k2w2)I12

(11.1)

Notice that X is a function not only of w but also of k. The con-stant k is called the modulus of X. Normally it is quite un-necessary to mention the modulus explicitly throughout theanalysis provided the same modulus applies always and thereis no danger of ambiguity.

Equation (11.1) can be evaluated numerically for variousvalues of the amplitude w ; the results can then be tabulatedfor use and indeed have been. It is clear from the original prob-lem that since a exceeds unity the value of k cannot be greaterthan unity. In engineering problems k is always a real number.Moreover, since it only appears in the integral as k2 negativevalues of k give the same results as the corresponding positivevalues. Therefore the only values of k to be considered are

9 129


those between zero and unity. For this reason k can be, andoften is, designated by sin 0 where 0 is called the modular angleand in engineering problems is always real.

The integral in equation (11.1) is called the elliptic integral ofthe first kind. There are other kinds as will be seen later. Suchintegrals were called elliptic because they were first encounteredin the determination of the length of are of the ellipse. Theform of the integral in equation (11.1) is known as Jacobi'sform ; it is said to be expressed in Jacobi's notation.

Legendre's NotationAnother notation is that of Legendre ; it can be obtained from

that of Jacobi by puttingw = sin o . . . . . (11.2)

where 0 is called the amplitude angle. Distinguish this anglecarefully from 0 the modular angle. From equation (11.2)

dw = cos 0 d . . . . . (11.3)

and hence from equations (11.1) to (11.3)

_ doX - . . . (11.4)p (1-k2Sj2 0)1/2

The integral in equation (11.4) is the elliptic integral of thefirst kind in Legendre's notation. It is usual to write

0-amX . . . . . (11.5)

signifying that 0 is the amplitude of X. Tables of ellipticintegrals are often given in terms of 0 and 0 instead of in termsof w and k.

Complete IntegralsRefer back to equation (11.1) . If the upper limit of the

integral is made unity the integral is said to be complete. Thecomplete elliptic integral of the first kind is always denoted byK and hence

dwK - fO (1 -w2)1/2(1 -k2w2)1/2(11.6)

130

ELLIPTIC INTEGRALS OF THE FIRST KIND

Equation (11.6) gives K in Jacobi's notation. In Legendre'snotation, since w = sin q, the limits of integration are from zeroto it/2. Hence in Legendre's notation

fn/2

(1 - (11. 7)- c k2 sin2 0)1/2

It may sometimes be necessary to state the modulus of K.It is then necessary to write .Ii (k) for the complete ellipticintegral of the first kind, modulus k. Normally, however, themodulus can be understood without fear of ambiguity. Noticethat although w may be a complex number the completeelliptic integral K is always a real number.

The Complementary Modulus

We now need to define a complementary modulus k' relatedto the modulus k by the equation

k2+k'2 = Ior

k' = (I - k2)1/2 .

From equation (11.8), when k is written sin 0, k' must be writtencos 0. Now the complete elliptic integral of the first kind tomodulus k' must be entirely different from K which has modulusk. This new complete integral is naturally enough denoted byK'. Thus

K'(k) = K(k') . . . . . (11.9)

Hence K and K' are related through their moduli and K' mustalso be a real number.

By this definition K' is given by the integral

flK'- dw(1-w2)'12(1-k'2w2)1/2

(11.10)

The next thing to find is the equation defining Kin terms of thenormal modulus k. In equation (11.10) let

t=I

(I _. k'2w2)1/2

131

. (11.11)


Then

w2 = k12(1-12

Therefore

1-w2 =

t2-1k''212

k'2t2 - t2 + 1

k'2t2

1-t2(1-k'2)k'2t2

1-k2t2

and hencek'2t2

z _(1-k212)1/2

(1-w)1/2 = k't

Furthermore from equation (11.12)

Therefore

w=(t2 _ 1)1/2

k't

dw = 12(t2 _1)-1/2 _ (t2 - 1)1/2

dtk't2

. (11.13)

t2-(t2- 1)dtk't2(t2 - 1)1/2

dt

k't2(t2 - 1)1/2

Substitution into equation (11.10) from equations (11.11),(11.13) and (11.14) gives

dt k'tK - tJk't2(t2-1)1j2(1-k212)1/2

with new limits. To find the new limits note that when w = 0,t=1 and when w = 1

t=1

(1 -k'2)1/2

= 1/k132


Therefore simplifying the integrand and adding the new limits1/k dtK' _ (t2 - 1)1/2(1 -k2t2)1/2

1/k dtfl j(1 -t2)1/2(1 - k2t2)1/2

and this is the required relation. Notice that the integral is thesame as that for K but that (a) the limits are changed and (b)the integral is imaginary. Combining the integrands of K andK' gives the total integration from the lower limit zero to theupper limit l /k and it is the complex number

% d wK+jb' = fo,( 1-w2)1/2(1-k2w2)1/2

Transformation of a Rectangleat this point it is possible to return to the problem of trans-

forming the inside of a rectangle to the upper half of the wplane. On page 127 values of w at the corners of the rectanglewere assigned and later on the substitution a = 1/k was made.

Y_

0wz0

D

CW.1

0

Flea. 74. X plane.

The X plane is therefore as shown in Fig. 74. The transforma-tion to the w plane is given by equation (10.2), page 128:

w dwX

YI0(1- w2)1/2(1- k2w2)1/2

. (11.16)

The constant of integration can be made zero by letting theorigin in the X plane be at the point w = 0. The constant B only

133


affects the scale of the transformation and it is therefore con-venient for the present to make it unity. It may be necessaryto reintroduce it later on. By dropping these constants equa-tion (11.16) becomes identical with equation (11.1).

1. -1 O 1 L uk k

FmG. 75. w plane in the transformation of a rectangle.

The w plane is shown in Fig. 75. Starting at the origin inFig. 74, as we go along the real axis from 0 to C we go fromw = 0 to w = 1. Therefore the value of X at the point C is givenby

1 dwX(at C) = Jo (1-w2)1/2(1 -k2w2)1/2

= K (from equation (11.6), page 130)

As we proceed further in the w plane from unity to Ilk we movealong the imaginary direction in the X plane from C to D, so thatthe whole path from 0 to D in the X plane is given by

1/k dwX(at D) = fo (1-w2)1/2(1-k2w2)1/2

= K+jK' (from equation (11.15), page 133)Therefore the length OC in the X plane is represented by K andthe length CD by K'.

Similarly by using negative values in each case it can beshown that

X(at B) _ _KX(atA) _ -K+jK'

Therefore in the X plane the width of the rectangle is given by2K and the height by P.

134


Before proceeding further let us consider a very simpleexample that requires this transformation only.

Electrodes and a Semi-infinite ConductorIn this example, suppose that there is a conductor of unit

thickness but extending so far in the other two directions thatfor the purpose of the problem the conductor can be consideredsemi-infinite. Two electrodes are applied to the edge of thesheet and a known voltage exists between one electrode and theother. The electrodes are, of course, collinear; let them eachbe one inch long and spaced two inches apart. Suppose thatthe problem is to find the effective resistance of the current pathper inch thickness of sheet.

V

A B C D-Y -1 0 1

Yku

X -K+ jK' X.' -K X 0 X-K X=K+ jK'

FIG. 76. Values of X in the w plane.

The w plane can be made to represent the sheet and theelectrodes as shown in Fig. 76. Here the dimensions chosengive the values of w at A, B, C and D as - 2, -1, 1 and 2respectively, and these are the values written into Fig. 76. The

Y_

(w`41'.-2)

* -K+j K'A

B

X -Kw=-1

0W=O

w (w.%K.2)

X-K+jK'D

C

XKW.1

0FIG. 77. Values of X in the X plane.

135


X plane is shown in Fig. 77 and, if we make the point Dcorrespond as before to Ilk where k is the modulus in thetransformation, it is clear that k = 0.5. This value of k satisfiesthe requirements for w at the points A and B.

In the X plane it is known that the resistance between CD andBA for unit thickness is

R=pBDohms

where p is the resistivity of the material of the conductor inohm-inches. We now have to show that the effective resistancebetween electrodes is invariant under a conformal transforma-tion. Prof. G. W. Carter's proof is as follows.

Let perfectly conducting electrodes in the z plane form theequipotentials 0 = 01 and 0 = 02 in a conducting sheet having aresistance of p8 units per square. Let 0 + jo be known all overthe sheet. On an electrode

EnIan

and the current per unit width is

la 1a:/ips an

_ps as

Hence the current flowing to or from either electrode is [fi]/pswhere [0] is the change in 0 when one makes a complete circuitof the electrode. The resistance is thus given by

1-2R-p8[0]

01

If the electrodes are transformed by conformal transformationsinto others that are still supposed to be at potentials 01 and 02in a sheet having a resistance per square of ps, all the quantitiesin the formula for R are unchanged. Thus the resistance inthe transformed system is the same.

In the w plane the lengths BC and CD are 2K and K' respec-tively, and therefore in the w plane

R = p 2Kohms

136


where K and K' are to modulus k, i.e. to modulus 0.5. Fromtables of elliptic functions, at k = 0.5 or 0 = 30 °,

K = 1-686K' = 2.157

Therefore

R = 1.560p ohms

for unit thickness. Any change in the ratio of width of elec-trodes to their spacing must bring about a change in the valueof the modulus k and therefore a change in the effectiveresistance.

In this particular example suppose that the material of theconductor is copper with a resistivity

p = 0.825 x 10-6 ohm-inches

and that the thickness of the sheet is one inch. Then

R = 1.560 x 0.825 x 10-6 ohms= 1.287 x 10-6 ohms

The total current in the sheet when one millivolt is appliedacross the electrodes is therefore

_ 10-31.287 x 10-6

amps

= 777 amps

Notice that in determining the effective resistance the con-stant B in equation (11.16) has no effect on the result, becausethe resistance depends only on the ratio of width to height ofthe rectangle and not on its size. If instead of being madeunity the constant B had been carried through the analysis,the final result would have been

2BKR=pBK'2K

pas before.

137


There is more work to be done on this transformation but itrequires familiarity with elliptic functions as well as ellipticintegrals. A knowledge of elliptic functions is also necessaryfor plotting the fields in the z plane. Therefore further con-sideration of this particular problem will be deferred until afterthe next chapter, in which elliptic functions are discussed.

138

CHAPTER 12

Elliptic Functions

The Function snThe elliptic functions of interest to engineers are Jacobi's

elliptic functions, so called because Jacobi first developed themalthough the credit for the original idea lies with Abel. Refer-ring back to equation (11.1), page 129, it is clear that when k isgiven its minimum value of zero

X =fw

0 (1-dww2)1/2

= sin-1 w .

Thus the effect of getting rid of k is to make X a function only ofw and moreover an elementary function of w. Now k = 0 canbe regarded as giving a special case of the elliptic integral.Notice that in equation (12.1) both X and w can be complex num-bers. Now in handling trigonometric functions of a complexvariable, such as sin z, it is usually preferable by far to avoid theinverse functions because they are less manageable. Whereverpossible the relation

w = sinX . . (12.2)

is used rather than that of equation (12.1) or the relation givenby the integral.

Suppose that in dealing with the elliptic integral it happenedthat the case of k zero was being considered ; anyone would thennaturally prefer to use equation (12.2) rather than the integral.In the same way, when k is not zero it is generally easier to dealwith the elliptic function than with the elliptic integral. Usingthe integral is equivalent to using the inverse function and it is,indeed, actually using it when k is zero.

It was shown on page 130 that Legendre's form of the integralcan be obtained by putting

to = sin 0 . . . (12.3)

139


Immediately afterwards, in equation (11.5), was described asthe amplitude angle of X and written

0 = am X . . . . . (12.4)

From equations (12.3) and (12.4)w = sin am X . . . . (12.5)

to modulus k throughout.Now sin am X is the elliptic function needed to correspond

with sin X ; indeed, when the modulus is zero it becomes sin X.But in order to shorten the nomenclature sill am X is nowadayswritten sn X and called either `sine am' or `ess en' X. Thisnotation was invented by Jacobi and later modified and addedto by Glaisher. Equation (12.5) is therefore written

w = sn X to modulus k . . . (12.6)

Normally it is unnecessary to mention the modulus ; it is im-plied. But if the modulus has to be brought in explicitly theequation can be written

w = sn (X, k)To sum up then, it is clear that when k is not zero w is a functionof X analogous to sin X and it actually becomes sin X when k iszero. Expressed mathematically,

sn (X, 0) = sin X

Transformation of a Rectangle in Terms of Elliptic FunctionsThe transformation of the inside of a rectangle to the upper

half of the w plane was given in equation (11.I6), page 133, interms of the elliptic integral. By normal inversion this can bewritten

X = sn-1 w . . . . . (12.7)

Equation (12.7) expresses the transformation from the X planeto the w plane in terms of an inverse elliptic function. Clearlyit is more convenient to express the transformation by equation(12.6).

Now when the integral is complete, w in equation (12.7) isunity ; furthermore X in the same equation becomes K. There-fore

K = sn-1 (1)and by inversion of this

snK = 1 . . (12.8)

140

ELLIPTIC FUNCTIONS

Since from equation (11.15), page 133,

A+JA' =

by inversion

l;x dw

fo (1-w2)1/2(1-k2w2)1/2= sn-1 (1/k)

sn (K+jK') = l /k . . . (12.9)

The results of equations (12.8) and (12.9) can be used tointerpret the meaning of equation (12.6). Thus when

w ±1, X=±KW = 0, x = 0

W = + Ilk, X = ±K+jK'These results, obtained with the help of equations (12.8) and

(12.9), establish the values of X and w at the four corners of therectangle more concisely and more elegantly than they wereestablished in Chapter 11, but the method requires familiaritywith the elliptic function sn.

More Elliptic FunctionsIt is not the purpose of this book to present an exhaustive

treatment of elliptic functions. This can be found in mathe-matical works. Engineers need to take advantage of whateveraids are available in dealing with such functions. A necessaryaid for each engineer working in this field is a book of data onelliptic functions that can be used for reference. Such a bookis the one entitled Handbook of Elliptic Integrals for Engineersand Physicists,* and this book will be referred to freely in whatfollows by the letters E.I.E.P. followed by the figures givingthe reference in detail. In E.I.E.P. we can readily find manyrelations between the various elliptic functions, their derivativesand their integrals.

On page 140 sn X was obtained by using the relation w = sinwhere q became the amplitude angle of X. From this

(1- w2)1/2 = cos0= cos am X

* By Byrd and Friedman. See List of References, page 217.141


and like the shortening of sin am X this is written

on X to modulus kor

on (X, k)

Since sine ¢ + cost it can be deduced that

sn2 X+cn2 X = 1

This result is given in E.I.E.P. 121.00; many other such rela-tions are given.

Of equal importance to sn X and on X is the function do Xwhich is defined as follows.

do X = (1 - k2w2)1/2(1-k2sln2c/,)1/2(I - k2 sn2 X)1/2

The three functions sn X. on X and do X are the principalJacobian elliptic functions. There are others derived from twoor more of these three such as

sn Xto X =

on X

and just as sn-1 is written for the inverse of sn, there are theinverse functions en-1, do-1, to-1 and so on.

The functions need not be discussed exhaustively at this stage,if indeed at any stage, in this book but they will be consideredin rather more detail wherever necessary as they arise in thetransformations. Most of the information required will befound in E.I.E.P. and the references therein will be given.E.I.E.P. 120.02 gives the nomenclature for such functions as1/sn = ns, and this nomenclature will be used throughout thepresent work.

Current Density in Semi-infinite SheetIt is now possible to return to the problem of the two elec-

trodes placed at the edge of a semi-infinite conductor. At theend of Chapter 11 the effective resistance of the sheet had beendetermined, together with the total current produced by a vol-

142

ELLIPTIC FUNCTIONS

tage difference between electrodes of one millivolt. Now thecurrent density anywhere is given by

_ IdXJ -dw

where J is the magnitude of the current density.Hence from equation (11.16), page 133,

J= B(1 - w2)1/2(1 - k2w2)1/2 1

where the scale constant B has had to be reintroduced and isyet to be determined.

Along the imaginary axis of w, u is zero. Therefore by put-ting w = jv we find that the current density for unit thickness ofsheet along the v axis, i.e. the axis of symmetry between theelectrodes, is given by

_ B (12)(1 + v2)1/2(1 + k2v2)1/2

.10

noting that v is wholly real. In the problem being consideredhere, that of page 135, k was found to be 0.5 and therefore

2B (12.11)(1 + v2)1/2(4 + v2)1/2

To determine the value of B it is necessary to find the totalcurrent along the sheet from equation (12.11) and equate it to777 amps, the value found on page 137.

The total current is found by integrating J along the v axisfrom zero to infinity. Thus

1= B dv(1 + v2)1/2(1 + v2)1/2f000

where I is the total current. This integral is given in E.I.E.P.130.05:

I = B tn-1 (oo, k')where

k'_(1-k2)"2= (1-0.25)1'2= 0866

143


HenceI = B to-1 (co, 0.866)

= BK(0.866) (from E.I.E.P. 132.03)= 2.156B (from tables of K)= 777 (from page 137)

Therefore B = 777/2.156= 360

Hence from equation (12.11)720

(1+v2)112(4+v2)1(z ampsfsq. in. (12.12)

Equation (12.12) gives the distribution of current densityalong the v axis of the w plane, Fig. 75. Values assigned to vgive different points along the axis of symmetry between thetwo electrodes. For each value of v a value of J to correspondcan be calculated from equation (12.12). The results can beplotted on a graph and such a graph is shown in Fig. 78.

400

300

c200

100

2 4

I

V 8

Inches

FIG. 7 8. Current density distribution along the axis of symmetry be-tween two electrodes at the edge of a semi-infinite conductor.

144

ELLIPTIC TUNCTI0 \ S

Plotting the Stream Lines and Equipotentials

In this type of problem it is normally unnecessary to ploteither the stream lines or the equipotentials ; it is not difficultto imagine the pattern of lines in all the planes. But if forsome reason an accurate plot is required the procedure is thesame as that described in earlier examples ; since, however,elliptic functions now appear in the transformations the diffi-culties tend to increase considerably.

As an example, suppose that the stream lines and equipoten-tials are required to be drawn on Fig. 76, page 135. The trans-formation is given by equation (12.6) page 140:

w = snXHence

u+jv = sn (0+jt)

and from E.I.E. P. 125.01

sn(0+jo) = Ocn

where all the functions of 0 are to modulus k and all the functions ofare to modulus V.Separating real and imaginary parts

91, =

v=(12.13)

1-sn2 dn20 J

each argument with different moduli as before.At this stage, tables of elliptic functions are required, prefer-

ably a set containing values of sn u, en u and do u.The tables used in this book are the Smithsonian Elliptic

Function Tables of 1947 by G. W. and R. M. Spenceley.*Another very suitable book for this purpose is Jacobian EllipticFunction Tables by Milne-Thomson.*

In this example k= 0.5 so that for functions of 0, 6 = 30°, and

I-sn2grdnil ¢sn0cn0cn0dn

* See List of References, page 217.10 145


for functions of ,, 0 = 60 °. The calculation of one set of pointsis given in Table 12. In this set of points 0 = 30'; there arevarious values of Vr put in terms of r° where r is defined by

90°r = K

For Table 12, the following values are first calculated :

0 = 30°

sn di = 0.527 to modulus :30°en q = 0.850do = 0.965

All the functions of shown in Table 12 are to modulus 60°.All values are taken from the Smithsonian Tables referred toabove. In these Tables the quantity r defined above is used andfor this reason intervals of r are used to start the calculations.

U

Fic. 79. Equipotentials and streamlines from two electrodes at theedge of a semi-infinite conductor.

Similar calculations can be made for other values of 0 between0° and 90°. The resulting lines of flow and equipotentials areshown in Fig. 79.

146

ELLIPTIC FUNCTIONS

TABLE 12

Calculation of stream lines and equipotentials, = 30°

r 0 150 30° 450 60° 75° 900

sn 0.346 0.627 0.816 0.928 0.983 1

en 1 0.938 1 0.779 0.577 0.373 0.182 0dn 1 0.954 j 0.840 0.707 0.595 0.524 0.500

a=sn2 w dn2 0 10.111 0.366 0.620 0.802 0.900 0.931b =1- a 1 1 0.889 0.634 0.380 0.198 0.100 0.069c =sn 0 (in 0 0.527 0.503 0.443 0.373 0.314 0.276 0.264za = clb 0.327 0.566 0.699 0.982 1.586 2.760 3.826

d = sn 0 en 0 en do 0 0.266 0.401 0.386 0.284 0.147 0v = d/b 0 0.299

I

0.632 1.016 1.434 1.470 0

147

CHAPTER 13

Double Transformations with One RequiringElliptic Functions

IN THE LAST TWO CHAPTERS a single transformation from theinterior of a rectangle to the upper half of the w plane was con-sidered. A slightly more difficult problem arises in the trans-formation of a configuration that makes the w plane just anintermediate plane and requires an initial transformation to thew plane from a configuration containing only two right angles.

Electrodes at the Edge of a StripFor a simple example of this, suppose that the two electrodes,

shown in Fig. 76, page 135, at the edge of a semi-infinite plane,were placed at the edge of a strip of conductor of finite widthbut of infinite length in one direction, as shown in Fig. 80. In

A B 0 C DIFic. 80. Electrodes at the edge of a strip.

other words the conductor is bounded in all directions exceptone ; therefore the resistance between the electrodes must begreater than that obtained on page 137.

148

DOUBLE TRANSFORMATIONS : ELLIPTICAL FUNCTIONS

The configuration is shown in Fig. 80 which can be made thez plane. What is required is the effective resistance for unitthickness of plate of the current path between the electrodes.The configuration and transformation apply, of course, to otherproblems ; for instance, they apply equally to an idealisation ofa problem concerning the end effects of eddy currents in a solidconductor.

As before the electrodes are labelled AB and CD, and for ageneral solution let the distance OC (or OB) be g/2 and let theratio OD/OC be r. We know that, after the transformationfrom the inside of the strip to the upper half of the w plane, asecond transformation will be required from the inside of arectangle in the plane to the upper half of the w plane. Theeffective resistance of the current path in the X plane is easilyfound and by combining the two transformations we can arriveat the effective resistance in the z plane.

From previous work it is clear that we want the values of wto be Ilk at D, unity at C, -1 at B and -1/k at A. Further-more the origin in the w plane should preferably correspond tothe point in the z plane midway between the electrodes. Wethus arrive at the z plane and the w plane shown in Fig. 81.

The First TransformationNow in the first transformation there are only two right

angles to straighten, and therefore the integration to be per-formed must lead to elementary functions. By the Schwarz-Christoffel theorem the transformation is given by

1 2 -1/2d = A w-k) l (w+k)whence

A dw=z

(w2 - 1/k2)1/2

Using this equation the value of A can be found by themethod of page 65, i.e. by integrating along a large semicirclein the w plane. In the z plane, as we cross the strip at pointswhere w is very large

fdz= -rgthe negative sign appearing because the path of integration is

149


w=-m w=.m

A B OI C D

W. -% w.-1 w.0 w.1 w. %k

(a)

(b)

A B 0

V

C D U

I--Go I=-r2 zz-2 2.0 Z.I Z- L92 2 2 2

FIG. 81. (a) z plane (b) w plane.

from right to left in Fig. 81 (a). But in the w plane the path ofintegration is along the semicircle of radius R where w = ROB.As before

dw = jRele dO

so that the equation of transformation gives

f dz = foj ReJBA d8

(R2e21e- 1/k2)1/2

where f dz = - ry and as R--* co

- ry = f r jA dO

jA-a

Hence

150

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTIONS

Using this value of Arg

fdw

Z =IT

J J(w2- l/k2)112

r_g

(1.10dw

IT j -w2)1/2

-T sin- r kw + C (Dwight 320.01)7T

For convenience let C = rg. Thenrz = - (7T- sin-1 kw)IT

L sin-1 kwaNotice that when w is zero

rgz = - sin-' 0IT

= 0

and this result checks that the zeros in the two planes given byequation (13.1) are identical as shown in Figs. 81 (a) and 81 (b).

From equation (13.1) the value of k corresponding to theunknown ratio r can be determined as follows. Since atz=,q/2, w=1, equation (13.1) gives

2

= rg sin-' k

Hencek = sin (7r/2r) . . . . . (13.2)

This result means that the first transformation provides thevalue of the modulus k for the second transformation. Thusthe modulus depends entirely on the ratio r and not on theactual dimensions, and this is what we should expect to find.In the example on page 135 the modulus was the directreciprocal of the ratio r whereas here it is the function of r givenby equation (13.2).

The Second TransformationThe first part of the problem has now been completed. The

second part has already been done in Chapter 12 and the151

CONFORMAL TRANTSFORMATIONS

required transformation is given by equation (12.6) on page 140:

iv = sn X to modulus k

where the modulus k is the k of equation (13.2).The X plane where the field is regular is shown in Fig. 82 from

which it is clear that the effective resistance of the current pathbetween the electrodes is

R = pK, (k)

Y,

-K+jK'

-K 0K -0

Fu:. 82. Transformation of electrodes at the edge of a strip (X plane).

In the previous example OC was one inch and OD two inches.If these dimensions are repeated r = 2 and

k = sin (Ir/4)= 0.707

From tables of elliptic integrals, when k = 0.707,

K = K' = 1.854

Therefore the effective resistance for unit thickness is

R = 2.000p ohms

as against 1.560p ohms in the previous example.If as before the thickness of sheet is one inch and the resis-

tivity of the material is 0.825 x 10-6 ohm-inches, the resistanceis

R = 2.000 x 0.825 x 10-6 ohms1.6510 x 10-6 ohms

152

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTION -S

Then the total current flowing in the sheet when one millivoltis applied at the electrodes is

_ 10-31.650 x 10-6

amps

= 606 amps . . . (13.3)

Current Density in the w PlaneThe distribution of current density magnitude across the axis

of symmetry of the strip can now be calculated. The totalcurrent in the w plane can be found as before from equation(12.10), page 143:

Jv

I - B dz

0 (1 +v2)1/2(1 +I-v2)1/2

= B to-1 (oo, k') (from E.I.E.P. 130.05)

where now k=0-707 as found above. Hence now

BV (1 -+2)1/2(1 +v2)1/2

and the total current is given by

where

Hence

B(1 +v2)1/2(1 +k2v,2)1/2

k' = (1-k2)1/2= 0.707

I = B to-1 (oo, 0-707)= BK(0.707) (from E.I.E.P. 132.03)= 1.854B (from tables of K) . . . (13.4)

Then from equations (13.3) and (13.4)

B = 327 . . . . . (13.5)

Current Density Distribution in the z PlaneIn the z plane the magnitude of the current density anywhere

153


is given by

TrB

4K

dX

dz

dX dwdw dz

B IT

-k2w2)1J24.(1-k2w2)1/2

(1-w2)1/2(ll

(1-q()2)-1(2

But here k = 0.707 and (from equation (13.5)) B = 327. Hence

Jz =36

I

z

1/2 . . (13.6)(1-2v)The next step is to put equation (13.6) in terms of z instead of w.Equation (13..1) shows that

1 . TTzw_- sin-9

Inserting the values g = r = 2 and k = 0.707 gives

is = 1.414 sin (7rz/4)Hence

_ 36311Z

i {1 - 2 sln2 (irz/4)}1/2

Along the y axis where the current density distribution isrequired x = 0 and hence z = jy. Thus

J -363

Y {1- 2 sln2 (7rjy/4)}1/2

_ 363

{1+2 sinh2 (iry/4)11/2(13.7)

since sin jy = j sinh y. Equation (13.7) shows that at the originJ(y0) = 363 amps/sq. in. Similarly at the point y=2

Try/4 = 1.571

and from tables of hyperbolic functions

sinh 1571 = 2.302154


Hence

400

300

100

0

363J(y=2) _(1 + 2 x 2.302 x 2.302)1,2

= 106.5 amps/sq. in.

2 4 6 V6Inches

Fic. 83. Current density distribution alon, the axis of ynirnetry be-tween two electrodes at the edge of a strip.

Similarly by taking various other values of y the graph ofFig. 83 can be calculated and drawn. This graph shows thedistribution of current density magnitudes across the axis ofsymmetry through the strip and between the electrodes.

Plotting the Field in the z PlaneThe procedure for plotting the stream lines and the equi-

potentials in the z plane follows the previous method but israther more complicated. The transformation from the X planeto the w plane is given by

w=snXso that the relations between u, v and are again those of

155


equation (12.13), page 145. But it is now necessary to con-sider in addition the transformation from the z plane to the wplane given by equation (13.1) :

z = -- sin-1 kwL7r

From this

x+jy = gsin-1k(u+jv)7T

Using Dwight 507.10 to expand the right-hand side

x+jy = 9{n,r+(-1)nsin-1p+q+j(-1)-cosh-1p2q'IT

p = {(1+ku)2+k2v2}1/2q = {(1-k2L)2+k2v2}1/2 t

(13.8)

the positive values being taken in each case. In these expres-sions the principal values found when n is zero can be used sothat

rg 2kux = - sin-1IT p+q

y= 9 cosh-1 p2 q

(13.9)

Thus the procedure is to put in the numerical values for theparameters in the equation and determine the value of k andhence of 0 from the first transformation. Then choose valuesfor 0 and 0. For each pair of values so chosen, u and v can becalculated from equations (12.13), page 145.

Then calculate values of p and q corresponding to these valuesof it and v from equations (13.8). Finally equations (13.9) givethe values of x and y corresponding to the values of 0 andchosen originally.

For the numerical example already used on page 152

rg = 4k = 0.7070=450

and k' = k156

I

Fic.

84.

Stre

am li

ne, a

nd e

quip

oten

tials

fro

m tw

o el

ectr

odes

at th

eed

ge o

f a

stri

p.


Hence 45° is the modular angle for the functions of as well asfor the functions of 0 in equations 12.13. A typical calculationfor 0 = 30 ° is given in Table 13 below.

Similar calculations can be made for other values of 0 between0° and 90°. The resulting equipotentials and lines of flow areshown in Fig. 84.

TABLE 13

Calcidation of the field in the z planeSince d = 30°, cn 0 = 0.565; en 0 = 0.825; (in = 0.917

r 0 15' 30° 45' 60° 75° 90°

sn ¢ I 0 0.302 0.565 0.765 0.900 0.976 1

en I 0.953 0.825 0.644 0.435 0.218 0do 1 0.977 0.917 0.841 0.771 0.724 0.707

a=sn2 dn2 0 0.077 0.268 0.492 0.681 0.801 0.841b = 1- a 1 0.923 0.732 0.508 0.319 0.199 0.159c=sn 0 do 0 0.565 0.552 0.518 0.475 0.436 0.409 0.399u=c/b 0.565 0.598 0.708 0.935 1.367 2.055 2.509

d = sn 0 en 0 en do 0 0.218 0.353 0.373 0.296 0.161 0v=d/b 0 0.236 0.482 0.734 0.928 0.809 0

(k=0.707) ku 0.399 0.423 0.501 0.661 0.966 1.453 1.774kr 0 0.167 0.341 r 0.519 0.656 0.572 0

f=(1+ku)2 1.956 2.025 2.253 2.759 3.865 6.018g=k2v2 0 0.028 0.116 0.269 0.430 0.3271) =(f+g)'12 1.399 1.433 1.539 1.740 2.072 2.519 2.774

h=(1-ku)2 0.360 0.333 0.249 0.115 0.001 0.205 0.5994=(h+g)112 0.600 0.601 0.604 0.620 0.657 0.729 0.774

rn=,i(p+4) 1.000 1.017 1.072 1.180 1.365 1.624 1.774n=cosh- 1 rra 0 0.184 0.377 0.591 0.830 1.066 1.175J=4n/a 0 0.234 0.480 0-752 1-057 1.357 1.496

r=ku/m 0.416 0.467 0.560 0.708 0.895 1.0008=sin-'r 0.410 0.429 0.486 0.594 0.787 1.108 1.571x=48/or 0.522 0.546 0.619 0.756 1.002 1.411 2.000

158

CHAPTER 14

Double Transformations Both RequiringElliptic Functions

A YET MORE DIFFICULT PROBLEM occurs when both trans-formations to the intermediate plane require the use of ellipticfunctions ; the simplest example is that where a rectangle isconcerned in both transformations.

Electrodes at One Edge of a RectangleConsider two electrodes again but this time let them be

applied to a conductor bounded in all directions. In parti-cular let the two electrodes be applied to one edge of a rectangleas shown in Fig. 85. Suppose that the problem is again to find

y

W k,E

A B 0 C 0W.-1 w=0 W=1

X

FiG. 85. Electrodes at one edge of it rectangle.

the effective resistance of the current path between the elec-trodes. This is an idealisation of a problem concerning endeffects in an eddy current machine. It is also applicable tothe determination of the permeance of the leakage path for theflux of a machine from one tooth tip to an adjacent one.

The configuration of Fig. 85 can be called the z plane and forcomparison with the two previous problems the electrodes arelabelled AB and CD, with the length OD called g/2 and theratio OD/OC called r. The problem is similar to that given by

159


Fig. 80 the difference being that the conductor is now limited inthe y direction. This gives an additional dimension and there-fore an additional ratio, viz. DE/10D: let this ratio belabelled s.

The First TransformationFrom previous work we know that the transformation to the

w plane must introduce the elliptic integral of the first kind, andhence the values of w for the various points in the z plane arechosen as shown in Fig. 85. Any other arbitrary values wouldentail later substitutions to bring the integrals into forms thatare recognisable as elliptic integrals. Thus the value w =1/k1at the point E is the reciprocal of the modulus in the integral ;kl has been chosen as the label for this modulus because k willbe required in the second transformation.

V

A B

1 -1

k'

10

C

Fia. 86. w plane.

The w plane is shown in Fig. 86, and with the values of wchosen we already know that the equation of transformationfrom the z plane to the w plane is

w = sn Bz to modulus k l . . . (14.1)where the constant B is retained because its value must dependon the relative ratios in the rectangle. The first thing to deter-mine is the value of the constant B and then the value of themodulus kn in terms of the ratio s which is known. Now when

z = rg/2, w = 1But when w is unity, from equation (14.1)

sn Bz = 1But equation (12.8), page 140, shows that sn K = 1 and hence

Bz = K(kn)160


andB = K(kl)Iz

2K(ki)rg

since z = rg/2 at the point in question.Therefore from equation (14.1)

w = sn 2K(kl)zrg

From equation (14.3) when w=1/k1

iSn

2K(ki)z=

ki rg

But equation (12.9), page 141, shows that

1/k, = sn (K + jK') to modulus klHence

and

2Kzrg

= K + jK' to modulus kl

z2 +J r2K (kl)

(14.2)

(14.3)

Now this is the value of z at the point E. Separating realand imaginary parts

OD = rg/2

DE =2ryKK' (ki)Therefore

s = DE/OD = K'/K(kl) . . . . (14.4)

Thus for any given value of the ratio s, it can readily be foundfrom Tables of complete integrals for what value of kl the ratioK'/K is equal to s. This determines kl which is the quantityreally required from the first transformation. To take anumerical example at this point, suppose that the height of therectangle is one quarter of the width. Then

s=DE/OD= 0.511 161


For this example, from equation (14.4)K'/K(kl) = 0.5

and henceK' = 2K (both to modulus k1)

From tables of complete integrals, it is seen that K= 2Kwhen k1= 0.985. This completes the first part of the problem.

The Second TransformationThe second transformation is exactly the same as in the

example of Chapter 13 and is given by equation (12.6) page 140.

w=-1 w.IA D

B O k-

W--k w.p w=k

Fiu. 87. X plane.

The X plane is shown in Fig. 87 where the values of w are thoseshown. Notice in Fig. 86 that at the point C, w is less thanunity, while at the point D it has already been made unity.Therefore it is convenient to make w = k at the point C, where kis now the modulus of the elliptic functions in the secondtransformation.

Since in Fig. 87 the scale is k times that of equation (12.6) theequation of transformation is

w = k sn (X, k) . . . . . (14.5)

This can be checked at the known points as follows :When

w = ±k, X = ±Kw=0, X 0w= ±1, X= ±K+jK'

and we know that these are correct.162


As before the effective resistance of the current path is givenby

but the new modulus k is not yet known. A relation between kand the known k1 must next be established.

In all three planes the ration OC/OD corresponds to the ratiok in the w plane, i.e. C in the z plane corresponds to w = k.Therefore since w = k when z = g/2, from equation (14.3)

k = sn (k1)rg

= snr

(k1) . (14.6)

Now when r happens to be an integer of low value it is possibleto evaluate k directly from equation (14.6). For example if,as in the previous example of Chapter 13, the electrodes are twoinches apart and are each one inch in length, OC =DE =1 andOD = 2. Then s = 0.5 and, as found on page 162, k1= 0.985.Furthermore r = 2 as before. Hence putting these values intoequation (14.6) we find

k=snK (k1)

= (1+k1')-I/2 (from E.I.E.Y. 122.10)where

k1' _ (1-k12)1 2= 0.173

Hencek = (1.173)-1/2

0.923

Therefore the required effective resistance is

R = p K to modulus 0.923

and from tables of complete elliptic integrals, at this modulus

K = 2.395K' = 1.634

163


and thereforeR = 2.93p ohms

for unit thickness of conductor.

Incomplete Elliptic IntegralsHowever, in general the ratio r is not a simple integer. When

it is not, the equation (14.6) needs to be interpreted in a dif-ferent way, a way that requires the use of tables of incompleteelliptic integrals. Equation (11.7) on page 131 gives the com-plete elliptic integral of the first kind as

_ /2 dK - J0 (1-k2 sin2 0)1/2

But if the upper limit of the integral is some value of 0 less thanir/2, the integral is incomplete and is then written k). Thus

k)= fo(1_k2sin2)uf2 . (14.7)

The integral is frequently written in terms of the amplitudeangle and the modular angle as F(0, 0) and rather less frequentlyin terms of the amplitude and modulus as F(w, k). All theseare different descriptions of the same integral. Clearly when

=7r/2F(w, k) = F(7r/2, k) - K(k) - K

Comparison of equation (14.7) with equation (11.4) page 130shows that

F(w, k) = F(o, k) = X

But from equation (12.6), page 140,

X = sn-1 (w, k)and hence

sn-1 (w, k) = F(w, k) . . . (14.8)

General Solution for kReturning now to equation (14.6) page 163, we have to find a

general solution for

k = sn (k1)

164


where r can have any positive value greater than unity. Fromthe inversion of this equation

sn-1 (k, k1) = K/r= F(k, k1) (from equation (14.8)) (14.9)

Since K, r and k1 are all known, by looking up tables of incom-plete elliptic integrals k can be determined from equation (14.9).

The procedure can be summarised thus : given the ratio s,use tables of complete elliptic integrals to find for what value ofk1 the ratio K'/K is equal to s. This finds not only k1 but alsoK(k1). Hence (K/r)(kl) can be calculated. Now look up tablesof incomplete elliptic integrals to find what value of k satisfies

K(k1) = F(k,

k1)

When k is thus determined tables of complete elliptic integralsgive K and K' to modulus k, and the effective resistance is thengiven by

2KR = K, (k)p

To clarify the procedure outlined above by a numericalexample, suppose that s = 0.5 as before but that now r =1.6.Note that the value of r does not affect k1 which remainsunchanged at 0.985. Therefore

k = sn1.6'

0.985)

But from tables, K to modulus 0.985 is /3.156. Therefore

Hencek = sn (1.973, 0.985)

F(k, 0.985) = 1.973Using tables of incomplete elliptic integrals

sin-1 k = 75.5°whence

k = 0.968

Then from tables of complete elliptic integrals it is found thatat k=0.968

K = 2.80K' = 1.59

165


and therefore_

R5.60

1.59 p

= 3.52p ohms

for unit thickness of sheet.

Rectangle of Infinite Height

As DE in Fig. 85 tends to an infinite length s->oo. Thereforefrom equation (14.4), page 161,

K'/K(kl) --* ooHence

kl = 0

Let r have the value given in the problem of Chapter 13, viz.r = 2. Then

k=Sn12,01and

F(k, 0) = K/2 to modulus zero.

But from E.I.E.P. 111.02, K(0) _ -r/2 and hence

F(k, 0) = rr/4

We then find from tables of F(o, 0) that k = 0.707 and fromtables of complete integrals, at this modulus

K = K'Hence

R = 2.00p ohms

for unit thickness, as obtained on page 152 for electrodes at theedge of a strip of infinite length.

Variation of Effective Resistance with Height of Rectangle

At what height of rectangle is the effective resistance so nearto that of the electrodes at the edge of a strip that increase of

166


height has little effect? This can be determined by takingvarious values of s and calculating .P for each value. The cal-culation is shown in Table 14.

TABLr. 14

CnlcvlatiTg the change in effective resistance

s I 0.289II

0-426 0-502 1.000 1-440 2-000

kj 0.9998 0.995 0985 0-707 0.400 0-173

r2-718 1-848 1.578 0-927 0-820 0.791

82-3 72-5 67.4 49-9 46-2 45.2

89-0 84-3 80-1 45-0 23-6 10-0

k 0.991 0-954 0-923 0-765 0-722 0.710

K(k) 3-408 2.628 2.395 1-934 1-873 1-857

K'(k) 1-577 1.608

--1-634 1-788 1-836 1.851

R 2K7V V + 4-32

-3-27 2-93 I 2-16 2-04 2-01

Fig. 88 is plotted from the results shown in Table 14 and itshows how R/p varies with s. For low values of s it is prefer-able to plot the curve of the reciprocal, i.e. p/R against s. Thecurves indicate that when the height of the rectangle exceedsone half of the width, the rectangle can be considered to have aninfinite height as far as the effective resistance is concerned.

Total Current

To find the total current flowing consider a particularexample. Let us use the same numerical values as those ofpage 163 where the effective resistance was found to be 2.93pohms for unit thickness of sheet. If as in Chapter 13 the

167


0-5

PR

0.4

O.3

0.2

01

O5 110 15 2.0S

Fic. 88. Effective resistance for varying heights of rectangle.

thickness of sheet is one inch and the resistivity of the mat-erial 0.825 x 10-6 ohm-inches, the effective resistance is

R = 2.93 x 0.825 x 10-6 ohms= 2.417 microhms

When one millivolt is applied across the electrodes the totalcurrent is

10-3

2417 x 10-6 amps

= 413.7 amps . . . (14.10)

Determination of ConstantFrom equation (12.10), page 143, the magnitude of the cur-

rent density along the imaginary axis of the a plane is

J =B

V (1 +v2)1/2(1 + k12v2)1/2

168


where B is a constant yet to be evaluated and k1(from page 162)is 0.985. Hence the total current is

of°D B dv(1 +v2)1/2(1 + 0.9722)1/2

= B to-1 (oc, k1) (from E.I.E.P. 130.05)B to-1 (oe, 0.173) (since k1' = (1- k12)1/2)

= BK(0.173) (from E.I.E.P. 132.03)= 1.583B (from tables of K)

But from equation (14.10)

I = 413.7 ampsHence

B=261.3 . . (14.11)

Distribution of Current Density

It is now possible to determine what the distribution of cur-rent density is along the axis of symmetry between the twoelectrodes, i.e. along the y axis in the z plane shown in Fig. 85,page 159. The current density at any point in this plane isgiven by

J= dX dw,

dwdz

Now since, from the inversion of equation (14.3),

g=

rz 2K sn-1 (w, k1)

dz _ rg 1

dw 2K(kl) (1-w2)1/2(1-k12w2)1/2(from E.I.E.P. 732,011

Furthermore, since from equation (14.5) also inverted

x = B sn-1 w(k)k

dX B (from E.I.E.P. 732.01)dw (1 __ w2/k2)1!2(1 -.w2)1/2

Note that the constant B has had to be reintroduced. Its169


value has already been found and given in equation (14.11).Therefore

.I = 2K(k1)B I (1 -w2)1/2(1 - k12w2)1/2rg (1-w2)1/2(1-w2/'k2)1/2

2K(k1)B (1 -k12w2)1/2rg I (1-w2/k2)1/2I

(14.12)

Equation (14.12) gives the current density in terms of w;therefore corresponding values of y in terms of w are nowrequired. Again, from equation (14.3)

2K(kl)z = sn_1 (w, k1)

rg= F(w, ki) (from equation (14.9))

Along the imaginary axis of w, since w = jv there,

2K(k1)z = F(jv, ki) . . . (14.13)

rg

An interpretation of the elliptic integrals with imaginary argu-ments is given in E.I.E.P. from which E.I.E.P. 161.02 gives

F(jo, k) = j F(P, k') . . . (14.14)where

tan g = sinh 0Here jq is the amplitude angle so that in the present examplejv =sin j0 = j sinh 0. Hence

tan fi = sinh 0 = vand

sin R =V

(1+v2)1/2

Then using amplitudes instead of amplitude angles equation(14.14) can be written

I'(jv, k1) = j F`(1 +v2)1/2'k1 l (14.15)

Equation (14.15) enables us to separateout

the real andimaginary parts of equation (14.13). Selecting the imaginaryparts of both sides and rearranging gives

79 Fv

k1,. . (14.16)

_'y 2K(kl) (1+V2)1/2

170


Then by putting w = jv into equation (14.12)

2K(kl)B (1+k12v2)1/214.17

rg (1+v2/k2)1/2()

Equations (14.16) and (14.17) give, for different values of v,corresponding values of current density and of y that can beplotted. The parameter v can have any real values from zeroto infinity.

In the example being considered, that of page 163, r = g = 2,and we already know that

k1 = 0.985, k1' = 0.173, k = 0.923

and from equation (14.11), B=261-3. Hence from tablesK(ki) = 3.156

Then from equation (14.17)

2 x 3.156 x 261.3 (1 +0.97v2)1/24 (1 + 1.175v2)1/2

(1 + 0.97v2)1/2= 412x(1+1.175v2)1/2 (14.18)

and

y = 0.634 x F((1 +V2)1/2' 0.173) . . (14.19)

from equation (14.16). From equations(14.1/11.18)

and (14.19)Table 15 can be calculated.

TABLE 15

Calculation of variation in current density

v 0 0.5 1 2 3 5 10 00

aV

0 0.447 0.707 0.894 0.949 0.981 0.995 1.000(1+ 2

b=F(a, 0!173) 0 0.463 0.788 1.114 1264 1.388 1.485 1.577

J=0.6346 0 0.294 0-5)00 0.706 I 0.801 0.880 0.941 1.000

c= 1+0.97v2 1 1.970 4.880 9.730 25.25 98.00d=1+1.175v2 1 1.294 2.175 5.700 11.58 30.380 118.5P,(e'd)1/2 1 0.980 N 0.952 0925 0.917 0.912 0909 0908

J = 412e 412 404 392 I 381 378 376 375 374

171


J

4

c300

EE

200

tOO

O 02 O'4

1

061

OsInches

Y10

FIG. 89. Current density distribution along the axis of symmetry be-tween two electrodes on one edge of a rectangle.

Fig. 89 shows the variation in current density across the axisof symmetry between the electrodes plotted from the valuescalculated in Table 15.

172

CHAPTER 15

Elliptic Integrals of the Second Kind

Incomplete Elliptic Integrals of the Second KindElliptic integrals of the second kind are best introduced in

the course of an example. Consider a semi-infinite strip witha rectangular projection as shown in Fig. 90 where the projec-tion is labelled PQRS. Suppose, for instance, that the strip

V

P

L_ a

Q 0 Rw=-1 W.0 w.1

x

Fia. 90. Rectangular projection. z plane.

is an electrical conductor and has electrodes of opposite polarityplaced at PQ and SR. The effective resistance between PQand SR is required.

Fig. 90 can itself be regarded as the z plane. There is onlyone ratio of dimensions, viz. RS/OR and there must thereforebe one undetermined value of w in the z plane. Since there arefour right angles in the configuration to be transformed theremust be an elliptic integral contained in the Schwarz-Christoffeltransformation. Therefore it is best to allot the values of w atthe corners shown in Fig. 90.

The transformation to the w plane shown in Fig. 91 is givenby the Schwarz-Christoffel equation

dzdw =

A(w-1)-1/2(w+1)-1/2 w-

1k') 1/2

w+1

1)

1;2

173


P Q R S

FIG. 91. Rectangular projection. w plane.

since the angles at Q and R are each 1r,/2 and those at P and Sare each 3,r/2. Therefore

z = AJ

(w2- 1/k2)1/2 dw/ (w2- 1)1;2

V

1. (1-k2w2)1/2Ak2 ff (1-w2)12 dw . . . (15.1)

Consider now the integral

f w

(1 - k2w2)1/2dw

p (1 - 202)1/2

This is Jacobi's form of the elliptic integral of the second kindand it is denoted by E(w, k) where, as before, w is the argumentor amplitude and k is the modulus. As with the ellipticintegral of the first kind it can be put into Legendre's form bymaking w = sin 0, thus giving

E(o, k) = fo (1- k2 sin2 0)1;2 do . . (15.3)

Another form is obtained by introducing the variable u where0 is the amplitude of u. Then

sin 0 = sn uand by differentiating both sides

cos 0 do = cn u do u du (E.I.E.P. 731.01)Therefore

doen u do u du(1-sn2 u)112

= do u du (E.I.E. P. 121.00)174

ELLIPTICAL INTEGRALS OF THE SECOND KIND

Hence equation (15.3) can be written

E(u, k) _ f(l_kssn2u)1/2dnudu0

('uJI dn2 u du

a(15.4)

Equation (15.4) defines the elliptic integral of the second kindin terms of elliptic functions. Extensive Tables are availablefor numerical values of E(0, 0) ; they are usually found withtables of F(j, 0).

Complete Elliptic Integrals of the Second KindIf the upper limit of the integral of (15.2) is made unity the

integral is said to be complete. It is then written E, or if themodulus has to be named E(k). When w is made unity 0becomes rr/2, so that the upper limit in the integral of equation(15.3) is z/2 when the integral is complete. Hence

E(1, k) = E(1 7T, k) = E(k) =_ E . . (15.5)

Just as the complete integral K has an associated completeintegral K' with modulus k', the complete integral E has anassociated complete integral E' also to modulus k'. Therefore

E'(k) = E(k') = E'where as before

k' _ (1-k2)1/2

Electrodes at the Sides of the ProjectionReturn now to the transformation from the z plane of Fig. 90

to the w plane of Fig. 91 given by equation (15.1). The trans-formation can now be written

z = B E(w, k) +Cwhere B = Ak2.

The origins in the two planes have already been chosen andfrom that choice the value of C is fixed and has now to be deter-mined. Since, when z = 0, w = 0

0 = B E(0, k) +C= 0+C (from E.I.E. P. 111.00)

175


Hence C is zero andz = B E(w, k) . . . . . (15.6)

Let the corner R in Fig. 90 be the point z =a and let thecorner B be the point z =a+ j b. At these points, w = 1 andw =1 /k respectively. In equation (15.6) first put z = a andw=1:

a = B E(1, k)= BE (from equation (15.5))

Hence B = alE and

z - jE(w, k) . . . . . (15.7)

Now put z=a+jb and w=1/k in equation (15.7):

a+jb=EE(kk)

= j{E +j(K' - E')} (from E.I.E.P. 111.09)

and by equating imaginary parts

b_ K'-E'a E

. (15.8)

The ratio b/a is known for any given configuration. Hencefor any value of this ratio there is a corresponding value of Ic

P

y

r- a (K!E')

Q p R xz.-a z.a

FIG. 92. Values in the z plane.

satisfying equation (15.8). In Fig. 92 the z plane is shownagain with values of z given both in numbers and in terms ofelliptic integrals.

176


For a numerical example let a= 1-25b. Then

K'-El= 0. 800

B

It is now necessary to choose or estimate a value for the modulusk and check whether for this modulus the equation holds. Inthis example choose k=0-400. Then from tables of completeelliptic integrals K'=2-359, E=1-151 and E=1.506. Fromthese values

K'-El802= 0.

E

Thus for a= 1-25b it is sufficiently close to take k = 0.400.

The Effective Resistance between Electrodes

To obtain the effective resistance between electrodes it isnecessary to transform from the w plane to the X plane where

WR -}

AEr-

FLW-1

W. I

1BHW-M

0W=0

C

W-1

FiG. 93. Rectangular projection. X plane.

0

the field is regular. The X plane is shown in Fig. 93. Fromthis it is clear that the required resistance in the X plane is

R = QR x resistivity

The transformation from the X plane of Fig. 93 to the w planeof Fig. 91 is given by equation (12.6), page 140:

w = sn (X, k) . . . . . (15.9)

12 177


Hence as in previous results the effective resistance is

R = p K to modulus k

where k has been determined by the first transformation. Inthe present example where a= 1.25b the modulus has beendetermined by equation (15.8) to be 0.400. Hence from tablesof complete elliptic integrals K' = 2.359 as before and K =1.640.Hence

R = 1.390p ohms

for unit thickness of conductor.

Total CurrentTo find the distribution of current density it is necessary to

find first the total current flowing in the material. The effectiveresistance when the conductor is one inch thick and the resis-tivity of the material is 0.825 x 10-6 ohm-inches is

R = 0.825 x 10-6 x 1-390 ohms= 1.147 x 10-6 ohms

When one millivolt is applied across the electrodes the totalcurrent is

_ 10-31.147 x 10-6

amps

= 872 amps . . . (15.10)

Determination of Current of IntegrationOn page 168 the current density distribution along the

imaginary axis of the uw plane is given by

BV (1+v2)1/2(1+k2v2)1112

where in the present example k = 0.400. Then

k' = (I-k2)1/2 = 0.917Therefore, as on page 169, the total current is now

I = B to-1 (oo, 0.917) (E.I.E.P. 130.05)= BK(0.917) (E.I.E.P. 132.03)= 2.359B (from tables of K).

178


Hence, from equation (15.10),

B = 370 . . . . . (15.11)

Distribution of Current DensityThe current density at any point in the z plane is, as already

shown, given by

J= dX dwdw dz

Now from equation (15.7), page 176,

z = E E(w, k)

and from equation (15.2), page 174,

dz a (1 - k2w2)1/2dw

-E (1 -2'2)1/2

Furthermore, from equation (15.9),

X = B sn-1 (w, k) . . . (15.12)

where the constant B has had to be reintroduced to enable usto obtain the actual current densities. The value of B is givenby equation (15.11). From equation (15.12)

dX = B (E.I.E.P. 732.01).dw (1-w2)112(1-k2w2)1%2

The product of the two differential coefficients gives

j_ I

BE (1-w2)1/2

la(1-w2)1/2(1-k2w2)1/2 (1-k2w2)1;2

_ BEI

(15.13)a(1- k2w2)

Equation (15.13) gives the magnitude of the current densityat any point in the z plane in terms of w ; it is now necessarytherefore to determine corresponding values of y also in termsof w. Equation (15.7) gives

z = E E(w, k)

179


Along the imaginary axis of the w plane w = j v, so that

z = E E(jv, k)

The interpretation of E with an imaginary argument is givenin E.I.E.P. 161.02 thus :

az = j E{F(8, k') - E(,8, k') + tan,8(1- k'2 sin2 l)1/2}

where as before tan fi = sinh . Here again sinh c = v and hence

Vsin P =

(1 +v2)1/2

Equating imaginary parts of both sides'2 ,2 1,2

Y = {F((1+x2)1/2' k')-E((1+x2)1/2, k')+v(1-i+x2)/ \

(15.14)

The corresponding values of current density are found by put-ting w = jv into equation (15.13) thus :

all BE2)1/2 . . . (15.15)

In the present example

B = 370 (from equation (15.10))a = 1-25 (for unit value of b)k = 0.400 (from equation (15.8))

E(0.4) = 1.506 (from tables of E)

These values` put into equations (15.14) and (15.15) give

y = 0.83{FI (1 +x2)1/2' 0.917) -E((1 +v2)1'2 0.917)

(1+0.16'02 11/2

+v (15.16)1+v2

and446

.1 + 0.16v2

180

I= (15.17)


By taking various values of v from zero to infinity, cor-responding values of current density and of y can be calculatedfrom equations (15.16) and (15.17). Typical values and cal-culations are shown in Table 16.

TABLE 16

Calculation of current density distribution

V 0 0.5 1 1 2 3 5 10

a-(1+v2)112 0 0.447 0.707 0-894 0-949 0.981 0.995

b=F(a, 0-917) 0 0-477 0-861 1.345 1-647 1.896 2-124c = F(a, 0-917) 0 0-449 0.721 0.936 1.014 1-071 1-097b - c 0 0-028 0-140 0409 0.633 0-825 1-027

0 + 0.16v2) 1 1-040 1.160 1-640 2-440 5-000 17-00d= (1+ 0.16v2)1/2 1 1-020 1-077 1.281 1.562 2.236 4.123e=ad 0 0-456 0-761 1-145 1.482 2-194 4-102

f=b-c+e 0 0-484 0901 1-554 2.115 3-019 5.129

y=0.83f 0 0-402 0-748 1-290 1.755 2.506 4.257

446

61+0. 446 429 384 272 + 183 89-20 26-24

Fig. 94 shows the variation in current density across the axisof symmetry between the electrodes plotted from the values ofJ and y obtained from Table 16.

Periods of Elliptic FunctionsIt was shown in Chapter 12 that sn X = sin 0. Now since

sin 0 is periodic sn X must also be periodic, although the lengthof the period is not necessarily the same. While the period ofsin 0 is 2ir that of sn X can be determined as follows. Startingfrom

doX - Jo (1 -k2 sin2 #)1/2

the increase in X over a complete period of 0 isf +2A do _ f 2, do

0 (1 - V g1I12.)1/2 Jo (1 -k2 sln2 0)1/2181


00

300

200

100

2 3 S

Inches

Fm. 94. Current density distribution along the axis of symmetry be-tween the sides of a projection.

But owing to the symmetry possessed by the function theincrease in x over the whole cycle of sin is four times that overa quarter cycle. Thus the increase in X over the cycle (whichgives the period of sn x) is

4K4117/2

d,,0 (1 -k2 Sln2 )1/2

Hence the period of sn X is 4K.Similarly the functions en X and do X can be shown to be

periodic and their periods can be similarly determined. Theyare shown in E.I.E.P.:

do (u + 2K) = do u (E.I.E.P. 122.04)and to check that 2K is not twice the period,

do (u+K) = k' nd u (E.I.E.P. 122.03)Hence the period of do u is 2K. In the same way E.I.E.P.shows that the period of cn u is 4K. Fig. 95 shows graphs ofsn X, en X and do X plotted against the argument for a modulusof 0.8 taking real values of the argument.

182


f(Xk.o e

Fla. 95. Curves of sn, en and do X.

Periods of Functions with Complex ArgumentsIn the previous section real values of the argument only were

considered. But it is usual in conformal transformations tofind that the argument of the functions is complex. It thenappears that elliptic functions have an additional period, and itcan be shown that the ratio of the two periods must be imagin-ary. This means that compared with trigonometrical functionswhich have real periods and with hyperbolic functions whichhave imaginary periods elliptic functions have two periods, onereal and the other imaginary. Elliptic functions are thereforecalled doubly periodic.

To find the imaginary period of sn u we note that

sn (u+2jK') = sn u (E.I.E.P. 122.19)

and this is not twice the period because

sn (u + jK') _ (ns u)/k (E.I.E.P. 122.07)

Hence 2jK' is the imaginary period of sn u. Considering alsoen u and do u, E.I.E.P. shows that the periods of these prin-cipal elliptic functions are

sn u : 4K and 2jK'en u : 4K and 2K + 2jK'do u : 2K and 4j K'

However, the elliptic integrals, as distinct from the ellipticfunctions are not periodic. Fig. 96 shows curves of the incom-plete elliptic integrals of the first and second kinds plottedagainst the arguments with modulus 0.5.

183


K

I-$

F(O)

£(O)

0 02

Fic. 96. Curves of and

Because the integrals are not periodic it is desirable to beable to use certain auxiliary functions to be developed in thenext chapter. These auxiliary functions are required for theproblems of Chapter 17.

184

CHAPTER 16

Auxiliary Elliptic Functions

WHEN THE INTEGRALS RESULTING from a Schwarz-Christoffeltransformation either are or contain elliptic integrals but yetnot in any standard form, it is often necessary to have recourseto one or more of the auxiliary functions described in thischapter. They arise particularly in the expansion and thedevelopment of elliptic integrals of the second and third kinds.

Jacobi's Zeta FunctionOne of those most commonly met is Jacobi's Zeta function

denoted by Z. This Z function must not be confused with thebetter-known zeta function of Riemann denoted by . Jacobi'sfunction is similar to the function E(c, k) but it is periodic withonly a single period. It can be defined by

Z(0) = E(0)-F(@) K . . . (16.1)

all to modulus k. From equation (16.1)

K Z(0) = K E(0) -E .F(')

With regard to the period of Z(6)

Z(0+2K) = Z(0) (E.I.E.P. 141.01)and

Z(4+njK') # Z(0)Thus Z(¢) has a real period 2K but no imaginary period.Tables of K Z(0) are given in E.I.E.P.

A graph of Z(0) plotted against 0 for a modular angle of 75°is shown in Fig. 97.

Jacobi's Theta FunctionMore important still are Jacobi's Theta functions. His

original Theta function is denoted by O. The functions are185


0

Z(0)

0.

O

-0.2

-0

4. sin 750

2 4 5V,

2K

FIG. 97. Curve of Jacobi's Zeta function.

widely used and they are indeed almost indispensable whennumerical values of elliptic functions have to be calculated.Jacobi's Theta function is related to his Zeta function by

% )(16 2))= .

a relation given by E.I.E.P. 144.01. Hence

ZW = d{logg (4)}

and

rZ()d o = log O(ff) +C . . (16.3).0

With regard to the periods of the Theta functions

O(¢+2K) = O(c) (E.I.E.P. 1051.04)and

O(0+K) 0 O(o) (E.I.E.P. 1051.03)

so that the real period of O is 2K. With regard to the imagin-ary period

0(0+njK') 0 O(0)186

AUXILIARY ELLIPTIC FUNCTIONS

and therefore, like the Zeta function, O is simply periodic.Although there are other Theta functions the one defined byequation (16.2) is the original Jacobi's Theta function and isstill the principal one. In this section therefore it has beencalled the Theta function.

Series for the Theta Function

Jacobian elliptic functions can all be expanded into variouskinds of series and E.I.E.P. 900 to 908 gives the different seriesfor the principal functions. The series for the Theta functionis very useful and necessary and it is given in E.I.E.P. 1050.01.It is a Fourier series frequently quoted as the definition of O :

U(¢) = 1+2 > (-1)mgm cos K . . (16.4)m=1

whereq = e-nK'/K (16.5)

The quantity q defined by equation (16.5) is a basic number inthe numerical evaluation of elliptic functions and it appears inmost tables of elliptic integrals and functions. It is called the`nome'.

Note that when K00

0(K) = 1 +2

and that when ¢ is zero

= 1+2q+2q4+2q9+ ...

.r,

0(0) = 1+2 (-1)mgm2

= 1-2q+2q4-2q9+ ...

The Primed Nome

In the definition of the nome q (equation (16.5)) the two com-plete integrals appearing in the equation are to modulus k.When q is defined to this particular modulus the same expres-sion with the complete elliptic integrals to modulus k' is denotedby the primed nome q'.

187


The Function 91Another Theta function is defined in terms of the principal

function by01(0) = 0(4+K) . . . . (16.6)

This function has the series

01(9c) = 1 + 2 qm-2 cosm7TO

1 KNote from the series that when = K

01(K) = 1+ 2 (-1)mgm2

=and that when 0 is zero

0(0)

01(0) =1 + 2 q 2

= 0(K)

Jacobi's Eta FunctionsClosely related to Jacobi's Theta functions are his Eta func-

tions denoted by H and H1. The H function is defined by

H(0) _ -j exp (2jiro-1irK'/K) times 0(o+jK') (16.7)

where exp (x) is written for ex. This function has the series

H (q) = 2 (-1)m-1q(m-1/2)2 sin (2m - 1)nrr2K

Note that by putting zero we find

H(0) = 0Putting 0 = K gives

00 12H(K) = 2 m-(-1)m-1q(m-1/2)2 sin V

2

=281/4+289/4+ ...

The period of H is shown in E.I.E.P. 1051.03 and 1051.07 andit is seen that the function is simply periodic with the period 4K.

The other Eta function denoted H 1 is defined in terms of Hby the relation

H1(0) = H(¢+K) . . . . (16.8)

188

AUXILIARY ELLIPTIC FUNCTIONS

and it has the series

Note that

00 (2m-1)07rHI(o) = 2 q(m-1,2) a cos

2K1

H1(0) = 2 q(m-1/2)2

= 281/4+289/4+= H(K)

and that2m- IH1(K) = 2 q(m-1/2)2 cos

2_7r

1

= 0

= H(0)

In his later work Jacobi replaced the four functions O(ff),01(x), H(0) and HI(o) by four other functions denoted byao(v),1&1(V), 82(v) and klr3(v). The relations between the old andthe new functions are given in E.I.E.P. 1050.01. While thenewer forms of Theta functions are those encountered in mathe-matical treatises the older forms are found to be the more usefulin conformal transformation calculations.

189

CHAPTER 17

Elliptic Integral of the Third Kind

THE ELLIPTIC INTEGRAL of the third kind is another integral bestintroduced in connexion with an example. Any manipulationsof this integral usually lead to the introduction of one or moreof the functions described in the previous chapter.

Succession of Equal Slot OpeningsThe most suitable example for this purpose is that of a suc-

cession of equal slot openings opposite a solid face, particularlysince the results are well worth comparing with those obtainedfrom the problem of a single slot opening covered in Chapter 9.In that chapter the effect of a single slot opening on the dis-tribution of the magnetic field between two plane magnetic sur-faces was discussed. On page 104 it was explained that theactual problem encountered is that of a succession of equal slotsand that sometimes it should not be simplified to the problemof one slot.

Let us now consider therefore the more difficult but morepractical condition of a number of equal slot openings oppositea solid face and across an airgap as shown in Fig. 98. Thisproblem is one of those discussed by Carter in his 1926 paper *and much more fully by Coe and Taylor in 1928.* The solutionrequires the use not only of the functions described in Chapters12 and 16 but also of the elliptic integral of the third kind.

Fic. 98. Succession of equal slot openings.


ELLIPTICAL INTEGRAL OF THE THIRD BIND

It is unnecessary to use the whole of Fig. 98 to find a solutionbecause the pattern of field distribution keeps repeating itself.It is only necessary to take the smallest repeating section of thefield together with its image in the solid face as shown in Fig. 99.The lines AB and CD are axes of symmetry in the field ; there-fore the portion of airgap space to be considered is that portionbounded by the straight line AB and the path EFGH.

C A

E

Y

n G

D B

FIG. 99. Image pattern.In this problem there are three parameters fixed by the

design, viz. the slot opening s, the airgap length g, (both asfound in Chapter 9) and in addition the tooth width t. Thusthere are two ratios instead of one and it is convenient to takeeither g,s and t;s or their reciprocals.

The z plane is therefore the configuration shown in Fig. 100where the portion HG can be considered to have a potentialV above zero while the image portion EF has the potential- V. The lines AB and FG are boundary lines of flow. Thepotential is zero along the y axis.

y

w=-W

C

W.-IW.- oo }

1

F 0 GW= 1 W=0 w=1

FIG. 100. z plane.191

1h2fD


Since there are two ratios in the z plane there must be twovalues of w at the corners left undetermined, but owing to thesymmetry in the figure one set of two can be the negative of theother set. In the configuration to be transformed there arefour right angles and this means that the integral in theSchwarz-Christoffel transformation must either be, or must con-tain, an elliptic integral. Therefore it is desirable to make theundetermined values of w equal to 1/k and 1/k,, so that k and k1become the moduli of the elliptic functions that must appear.Hence values of w are assigned as shown in Fig. 100.

Then the Schwarz-Christoffel transformation to the w plane(shown in Fig. 101) is given by

1 1/2 1 1/2z = A (w-1)-1/2(w+1)-1/2 w-kl (+) x

(w-k1)-1(w+k11-1 dw

= C J(1-k2w2)1/2 dw

(1-k12w2)(1-w2)1/2(17.1)

where the new constant C in terms of A isC = - k12A/k

Now the form of the integral in equation (17.1) and the pre-sence of the two moduli suggest that an elliptic integral of thethird kind is present ; therefore the best procedure is to try tomanipulate the integrand to find this integral. Hence the nextstep is to examine the elliptic integral of the third kind so thatit is readily recognised.

Elliptic Integrals of the Third KindLegendre's form of the elliptic integral of the third kind is

defined by the equation

11 (w, k1, k) =w dw

17.2o (1-k12w2)(1-k2w2)1/2(1-w2)1/2 . ( )

Notice the presence of two moduli in the integral. As in theelliptic integral of the first kind another form is found by puttingw = sin 0 giving

ki, k)fo (1--k12 sing 0)(1-k2 &n2,6)1/2 (17.3)

192

ELLIPTICAL INTEGRAL OR THE THIRD KIND

There is yet another form of Legendre's integral derived fromthat of equation (17.3) as follows. If the integral is denoted byit, then 0 is the amplitude of it. Then from 0 = am it,

do = do u du (E.I.E.P. 731.00)sin 0 = sin am u = sn it

Furthermore(1-k2sin20)1/2 = (1- k2 sn2u)1/2

= do u (E.I.E.P. 121.00)Collecting these results and inserting them into equation

(17.3) gives

ll(u, kl k) =I U

du . (17.4)01-k12sn2u

All these three forms of Legendre's elliptic integral of the thirdkind are given in E.I.E.P. 400.01.

Jacobi's Elliptic Integral of the Third KindJacobi adopted not merely a different form but an entirely

different definition of the elliptic integral of the third kind andit so happens that his definition gives us an integral rather more.suitable for the kind of analysis required in conformal trans-formations. Equation (17.4) is modified by introducing a newparameter denoted by a to take the place of k1 by putting

kl = k sn a to modulus k . . .

Then the integrand of equation (17.4) becomesdu

1- k2 sn2 asn2uJacobi used this integrand in his definition of the ellipticintegral of the third kind and for the full definition wrote

fI (u, a) = k2 sn a en a do au sn2 u duf 2 z (17.6)

0 1-ksnza sn uThis is the definition quoted in the footnote to E.I.E. P. 400.01.

Jacobi's definition of the integral can be expressed in termsof his Zeta function and also in terms of his Theta functions asfollows.

sn (u+a)+sn (u-a) _ 2snuenadna E.P. 123.02)(E.I .1- k2 sn2 u sn2 a

13 193

CON-rORNTAL TRANSFOR1TATTONS

and hence substituting into equation (17.6)

11(u, a) = k2 sn a J0 sn a{sn (u+a)+sn (u-a)} du

But from E.I.E. P. 142.02k2 sn a sn u sn (u+a) = -Z(u+a)+Z(u)+Z(a)

and

Hencek2 sn a sn u sn (u -a) = Z (u - a) - Z(u) + Z(a)

N(u, a) =J

u {Z(u-a)-Z(u+a)+2Z(a)} du (17.7)0

Equation (17.7) gives a definition of Jacobi's elliptic integral ofthe third kind in terms of his Zeta function.

From equation (16.3), page 186,f('u

Jo

Z(u) du = log O(u)+C . . . (17.8)

Hence from equations (17.7) and (17.8)rr

II(u, a) = z{log O(u-a) -log O(u+a)}+J0 Z(a) du

0(u-a}= log

O(u+a)+u Z(a) . . . . (17.9)

Equation (17.9) gives yet another definition of Jacobi's ellipticintegral of the third kind this time in terms of his Theta andZeta functions.

Transformation to the w PlaneIt is now possible to return to the transformation of equation

17.1 and manipulate the expressions to yield results in familiarforms. Recall first what we expect to obtain from this firsttransformation. As in all previous examples involving ellipticfunctions the second transformation gives the answers we areseeking to the various problems ; but the answers are in termsof elliptic functions whose modulus or moduli we have to deter-mine. These moduli are determined from the first transforma-tion in terms of the known ratios of the configuration in the zplane. Thus for the problem now in hand the first transforma-tion should give values of k and kl in terms of the two ratios g/sand t/s.

194

ELLIPTICAL INTEGRAL OF THE THIRD KIND

AE F

-1K, YK

z. -w-jZt z.-g+jit z--'9

V

G BH

0 1 1/K

z-g z=g+jltYK,

z-m+j't

Fio. 101. w plane.

Equation (17.1) gives the first transformation, that from thez plane of Fig. 100 to the w plane of Fig. 101. One familiarform we are seeking is the integrand of equation (17.6). Firstput

Then

But

Hence

w = snp . . . . . (17.10)

dw = on p do p dp (E.I.E.P. 731.01)

do p = (1- k2 sn2 p)1/2 (E.I.E.P. 121.00)

dw = cnp(1-k2sn2p)1/2dp

Equation (17.1) in terms of p becomes

p (1 - k2 sn2 p)'/2(l - k2 sn2 p)112 on p dpz

Cfo (1-k12sn2p)enpp 1-k2sn2p

= C fo 1-k12sn2pdp

C( \1+1-k2sn2p-1+k12sn2p\

d p

fp

1-=C{p+(kl2_k2)

dp} (17.11)

l 0 I -k12 sn p))

Now reintroduce the parameter a as in equation (17.5) :

k1 = ksnaThen

k12-k2 = -k2(1 -sn2 a)= - k2 cn2 a (E.I.E.P. 121.00)

195


Equation (17.11) then becomes

z = C(p - k2 cn2 a P sn2 p dp1`` o 1-k2sn2asn2p

= C,}p- ell art(p, a)} (17.12)snadnu

from equation (17.6), page 193, and all the functions are tomodulus k.

Determination of Constant

The constant C in equation (17.12) can be determined bychoosing a point where the values of both z and w are knownand then inserting those known values into equation (17.1).Referring to Fig. 101 we see that the value of w as we movealong the positive real axis from the origin passes through thepoint 1/k1. Fig. 100 shows that the corresponding value of zchanges abruptly at the point corresponding to w = 1 /k1 by theamount 'js. Hence from equation 17.1

1 1C(1- k2/k12)1/2 w dw 1/k1 E

CJs(1 + k12(k12)(1 - 1/k12)1/2 fO I -k1w}1/kl--e

Notice that w =1 /k1 is not substituted into the remainder ofthe integrand shown above because doing so would render theresult indeterminate. Simplifying the expressions and per-forming the integration gives

(k2 - k12)1/2 1 1/k1+E

2Js = C 2(1-k12)1/2(-k1)

log (kiu -1)}141-E

Now as w passes from zero through the value 1/k1 the expres-sion (kiw - 1) changes from negative to positive. Reference toDwight 604.1 shows that log (k,w-1) increases by the amount- jir if the principal value is taken because the logarithm of anegative number differs by jw from the logarithm of the cor-responding positive number. Therefore the change in valuefrom Ilk, -E to 1/k1+E of -(1/k1) log (kiw-1) is j7r/k1 so that

(k2 - k12)1/22j`S = j 2k,(I - k12)1/2

196


Therefore

s k1(1 -k,2)1/2C _.r (k2-kl2)1/2

Now substitute k sn a for k1:

C, - s k sn a(l - k2 8112 a)1;'2

T ken assnadna

(E.I.E.P. 121.00) (17.13)ar cn a

Now k and k1 are real numbers since they are points on thereal axis of the w plane, and since from Fig. 100 k is greater thank1, a must also be real. Hence all the other functions in equa-tion (17.13) are real. Therefore C is a real number.

Substitution of C from equation (17.13) into equation (17.12)gives

lz _ s{psnadna-1(pa)} (17.14)cna J

Equation (17.14) gives values of z virtually in terms of wbecause from equation (17.10) p = sn--1 w.

The next step is to obtain some equations relating the para-meters k and a to the ratios g/s and t/s. This means eliminatingthe variable p from equation (17.14), and it can be done asfollows.

Elimination of pThe quantity p must have a definite value depending on the

dimensions in the z configuration. That value can be deter-mined by considering another pair of corresponding values of zand w. At the point where w = 1j k, z has the value g + s j t asshown in Figs. 100 and 101. Inverting equation (17.10)

p = sn-1 wu dw

=o ( 1 - w2)1/2(l - k2z(. ,2)1/2

(E.I.E.P. 130.02)

and therefore at w =1 j k1 /k dw

Pfo (1-w2)1j2(l -k22C2)1/2

= K+jK' (from equation (11.15), page 133)197


Substituting these values for p and z into equation (17.14) gives

9+zjt =7Tj(K+jK')sn

on aa-II(K+jK')a)) } (17.15)

All that remains is to equate real and imaginary parts ofequation (17.15) and thus produce the direct relation betweenthe ratios ,g js and t/s and the parameters a and k. But first itis necessary to find out what are the real and imaginary parts ofthe elliptic integral of third kind which has the argumentK+jK'.

Real and Imaginary Parts of n(K+jK', a)From equation (17.9), page 194,

II(K+jK', a) = 'flog 0(K-a+jK')-log 0(K+a+jK')}+(K+jK') Z(a) (17.16)

From E.I.E.P. 141.01

Z(K - a + jK') = Z(K-a)+cs (K - a) do (K-a)-jrr/2K

Integrate both sides of this equation with respect to (K - a).The integration of the Zeta functions can be taken from equa-tion (17.8), page 194. The integration of cs u do u can befound thus :

resudnudu = fen udnudu(E.I.E.P. 120.02)

J snu

f dsnusnu (E.I.E.P. 731.01)

= log sn u

The complete integration therefore produces

log 0(K-a+jK')+Cl= log 0(K-a)+C2+log sn (K-a)-j,r(K-a)l2K

Similarly

log 0(K+a+jK')+C1= log 0(K+a)+C2+Iog sn (K+a)-jir(K+a)/2K

198


By subtraction

log O(K-a+jK')-log O(K+a+jK')O(K-a) sn (K-a) j 'r= log O(K+a)+1og

sn(K+a)+2K(2a)

From E.I.E.P. 1051.03

l0O(K-a) - to O1(-a

g O(K+«) - g 01(a)

= log 1 (E.I.E.P. 10:51.02)= 0

Similarly

logsn (K - a)

= log 1 (E.I.E.P. 122.03)sn (K+a)= 0

Hencelog @(K-a+jK')-log O(K+a+jK') = j7ra/K (17.17)

From equations (17.16) and (17.17)

11(K+jK', a) = j7ra/2K+(K+jK') Z(a)

The real part of the right-hand side is K Z(a), and the imaginarypart is

K+K' Z(a)

With this result, equation (17.15) can now be separated intoreal and imaginary parts yielding the two equations

g _ Krsnadna-Z(a)}s it en at 2K sn a do a as cna - Z(a)I-W

}(17.18)

The two equations (17.18) give the relation between the knownratios g/s and t/s and the parameters a and k.

Numerical ExampleBefore proceeding further it is desirable to work through a

numerical example. Take the following typical values :k=0-707 and (x=0.927. In this example the Smithsonian

199


Tables quoted on page 145 will be used again. From thesetables

K = K' = 1.845sn a = 0.765on a = 0.644do a = 0.841

and from the tables in E.I.E.P.

K Z(a) = 0.2715

Hence from equations (17.18)

g 1.845 02715I-s 1.845

= 0.501

t _ 2 x 1-845- j, 0-27151 )

0.927s IT 1.845 1.845

= 1.002 - 0.502= 0.500

Hence when the slot width is twice the tooth width and twicethe airgap length we know that in the problems we wish to solvek is 0.707 and a is 0.927. Note that in some tables of ellipticfunctions the values are listed in terms of modular angles.These angles are given by

0 = sin-1 kk1=ksna

= ksin0Hence

= sin-' (ki/k)

In the numerical example given above

0 = sin-' 0.707 = 45°0 = sin-' 0.765 = 50°

Notice also that in E.I.E.P. the Zeta function is tabulated asK Z(P) where fi is written for 0.

In this numerical example we started with arbitrary values ofL-. and a and found the corresponding values of g/s and t/s. Inpractice this is, of course, starting from the wrong end. The

200


ratios in the z plane are what we know for they are fixed by thedesign. Hence we have to start with them and find k and a.This necessitates first finding a whole series of results startingwith the parameters and then using the method of cross-plotting, so that values of k and a can be read off for any givencombination of g/s and t/s. This will be discussed again laterin the chapter.

Transformation to the X PlaneWhat we have obtained so far is the transformation from the

z plane to the w plane in terms of the parameters k and a andhence the relations between the parameters and the actualdimensions in the configuration, The second part of the trans-formation is that between the w plane and the X plane where thefield is regular.

-1pP

W=1

FiG. 102. X plane.

The X plane is shown in Fig. 102 and the equation of trans-formation is exactly that of equation (12.6), page 140, viz. :

x = A sn-1 w to modulus kI

where the constant A is the scale factor. The appropriatevalue of A can be found in the usual way as follows. Fig. 100shows that when w is unity the value of X must be the potentialbetween HG and the origin, viz. V. Thus

V = A sn-1 1= AK(k1) (from (12.8), page 140)

HenceA = V/K(kl)

201

and

CON FORMAL TRANSFORMATIONS

V sn-1 (w, k1)17.19X =

K(kl). . . ()

With the two transformations of equations (17.14) and (17.19)we can solve the same problems as those of Chapter 9 where asingle slot opening was assumed. Then a comparison can bemade between the two sets of results. In Chapter 9, after thetransformation is established the first problem considered isthat of the distribution of flux density.

Distribution of Flux DensityAs before the magnitude of the flux density at any point in

the z plane is given by,ao IdX/dz I. Therefore denoting the fluxdensity by B

B = µodX dwdw dz

From equation (17.19) and using E.I.E.P. 732.01

dX V 1

dw K(ki) (1 - w2)1/2(1 - k12w2)1/2

and from equation (17.1)

dw (I -k12w2)(1 -w2)1/2dz ~ C(1 -k2w2)1/2

where C is the constant determined by equation (17.13),page 197. Hence

BV

I

(1 -k12w2)1/2I

CK(k1) (1-k2w2)1/2 H`°

and substituting for C from equation (17.13)

_ V cn aB sK(k1)snadna

(1 -k12w2)1/2(1 - k2w2)1/2 µo (17.20)

Maximum and Minimum Values of Flux DensityThe maximum value of flux density occurs at the point on

the solid face which is exactly opposite the centre of a tooth,202


i.e. at the point shown in Fig. 100 where w is zero. Hence,by putting w=0 into equation (17.20),

V enaBmax = sK(kl)snadnat`c (11.21)

Therefore equation (17.20) can be written

=(1-k12w2)' 2

B(1-k2W2)1'2 (Bmax (17.22)

The minimum value of flux density occurs at the point on thesolid face which is exactly opposite the centre of a slot opening,i.e. at the point shown in Fig. 100 where w is infinite. Henceby letting w tend to infinity in equation (17.22)

Bmin = k Bmax

= Bmax sn a

Thus the amplitude of the flux density ripple set up by the slotopenings is given by the equation

B.I. - sn a . . . . . (17.23)Bmax

In the numerical example of p. 200, where g/s is 0.501 andt/s is 0.500, it was found that sn a = 0.765. Hence the magni-tude of the tooth ripple for these ratios is given by the relation

Bmin = 0.765Bmax

The Mean Flux DensityReferring to Fig. 10O the total flux between AB and CD

divided by the area of the airgap enclosed by these lines givesthe value of the mean flux density, Bmean. The area is clearlyI (s + t) while the total flux is given by the increase in X as wincreases from unity to Ilk,. From equation (17.19) thischange in X is

{sn_1 k - sn-1 (1)}1 all to modulus kl

203


This expression interpreted by equations (12.8) and (12.9),page 141, becomes

K(K+jK'-K)

Hence

VK'K o modulus k1

2 VK'µomean K(t+s)

V µ02K (g/s)to modulus k1 . (17.24)g K(1+t/s)

B =

If the solid face were opposite an unslotted surface (i.e.opposite another solid face) with the same airgap length g andthe same potential difference V, the flux density would have theuniform value of .oVi`g. Hence it is reasonable to call thisvalue of density unity. Then

Bmean = 2K'(gls) to modulus k1 . (17.25)K(l + (s)

From equation (17.21) for the same value of unit density

ng ell IxBmax = sK(ki) sn a do a

. (17.26)

Numerical ExampleIt is desirable now to consider a numerical example iii order

to compare the results with those obtained in Chapter 9 for asingle slot opening. Take the following values.

k = 0.6750 = sin-1 0675 = 42.4°a = 1.071

From tables sn a = 0.8396. Then

= sin-1 0.8396 = 57-l'204


Also from tables

Therefore

en a = 0.5426do a = 0.8234K(k) = 1-819K'(k) = 1-893

KZ(a) = 0.2265

Z(a) = 0.1245kl=ksna=0.566

K(k1) = 1-726K'(kl) = 2.047

From the above values and equation (17.18)

g - 1.819(1.275-0.1245)

s 7r

= 0.667

Notice that this is the value of g/s used on page 114 where, how-ever, it is written 8/g =1.5.

Again from equations (17.18)

t 2 x 1.897 1.071

s=

itx1 1505-1819

= 1.389 - 0.589= 0-800

According to the equations derived in this Chapter, for thesevalues of g/s and tls the various values of flux density are :

Bmean2 x 2-047 x 0-667

(from equation (17.25))1.726 x 1.800

= 0.878

0-6677TBmax = (from equation (17.26))

1.726 x 1.275 )

= 0-952

Bmi = 0-952 x 0.8396 (from equation (17.23))= 0.800

Now consider a single slot opening for the same ratio of g/s.205


On page 112 it is shown that

Borax = µoV /g = Unity

as arranged on page 204. On page 114 it is seen that

Bmin = 0.8

Hence for this particular value of g/8, the value of Bmin isthe same for a succession of equal slot openings as it is for asingle slot opening, but the value of Bmax is less, the differencedepending on the width of the teeth.

The Flux Density CurveThe term µoV/g has been called unit flux density. Then in

terms of unit flux density equation (17.20) becomes

_B

7rg/s cn aK(ki) sn a do a

(1 - kl2w2)1/2(1- k2w2)1i2

(17.27)

From this we want to find the variation in value of the fluxdensity along the solid face, i.e. the distribution of flux densityalong the lines z=jy and w=jv. Substituting jv for w inequation (17.27) gives

B _ 7rg/8 cn a (1- k12v2)1'217.28)

K(k1) sn a do a (1 +k2,r2)1/2 ( -

Equation (17.28) gives B along the imaginary axis of the w planeand it is now necessary to find a correspondence between y inthe z plane and v in the w plane. From equation (17.14);page 197,

zs[psnadna

rl a)7r l cn a (p'By inversion of equation 17.10 page 195

p = sn-1 w to modulus k-= F(w, k) (by equation (14.8), page 164)

Thus when jv is written for w

p = F(jv, k)

jF{(1+v2)"2' k'} (from equation (14.15), page 170)

= jR (say) . . . . . . . . . . . (17.29)206


where fi can have values between zero and K'. Substitutingfor p in equation (17.14) gives

S Snadllaon a

Jg- II(jg, a)ZITI

and from equation (17.9), page 194, this becomes

= s snachi a, Ei(a-jp}z

7,1 en a Jf3-j(i Z(«}- 2 log ()(a+jp)

To break this equation up into real and imaginary parts wehave to find the imaginary part of the log term. The followingwell-known relation is in Dwight 604:

Im log (x + jy) = tan-1 (y!x)

where Im stands for the `imaginary part'. From equation(16.4), page 187,

irmO(a+jfl) = 1+2 (-1)mgm2cos (a+jfl)

= 1 + 2 (-1)mgm2 x1

(cos K a cosh K fi- J sin sink KK a

Therefore the imaginary part of log ®(a + jfl) is given by

- 2 (-1)mgm2 sin Ka sinh xP

tan-11 + 2 (-1) mgm2 cos Ka cosh

KflSimilarly the imaginary part of log 0(a - jB) is given by

+2 (-1)mgm2 sin Ka sinhK_tan

1+2 (-1)mgm2cosKacosh -_

Now tan-1 A = - tan-I (- A) and therefore

tan-1 (alb) - tan-1 (- a/b) = 2 tan-1 (a/b)207


Hence collecting the results and considering only the imaginary -parts throughout

2 log 0(a+jf3) = 2{log 0((x +j/3)-log 0(a-jg)}

2 _K sink -(- 1) m m2 i- g s n= tan-1

1 2irn 87rm am m21 h+ g) cos

Kcos K

Therefore equating real and imaginary parts of both sides ofequation (17.30) gives, for the imaginary parts,

y=I I

sn a do7r on a

(-- 2 1)mgnz 2 sin Ka sinh -rtan' (17.31)

1 + 2 , (- 1)mgm2 coska

cosh 'rh-1 '

Thus the procedure is to take a series of values of v. Thenfor each value of v the flux density is given by equation (17.28).Next from the tables find

v ,F (1+22)1/2

k

and call this fl. Using that value of fi find y/s from equation(17.31).

A numerical example follows.

Numerical ExampleTo keep continuity in the numerical examples use the values

of g/s and tJs found on page 205 that were obtained fromk=0.675(0=42-4-) and a=1-071. The values of the variouselliptic integrals and functions required are given on page 205and need not be repeated here. From equation (17.28) andusing these values of the functions

B - 0.952(1 + 0.321v2)1/2(1+0.456v2)1/2

(17.32)

208


The next thing to find is the value of the nome q. This quantityis defined by equation (16.5), page 187, as

q = e-nK'/K

e-3.27 = 0.038 . . . (17.33)

Now q is so small in this example that all terms in equation(17.31) except those obtained by putting m equal to unity arequite negligible. Hence in equation (17.31) put ?n unity andq=0.038 and use the values of the functions found on page 205.Then

y - 1 2(0.038 sin 0.5897r sinh 7r,8/K)8

11- 2(0.038 cos 0'5891r cosh 7rr/K)

_ 11 1.159+tan-1 0.073 sink 1.729P } (17.34)1 + 0.021 cosh 1.729P j

where by equation (17.28) r8 is calculated from

v ,P = r' (1 +v2)1/2' k

Nowk' = (1- k2)1/2

and here k = 0.675, from which k'= 0.7375 and

Fj (j +,2)1/2' 0.73751

A set of calculations of flux density for various values of vand the corresponding calculations of y are shown in Table 17.The curve showing the wave form of the flux density distribu-tion as calculated in Table 17 is shown in Fig. 103.

It is interesting to compare this curve and these results withthose obtained by assuming a single slot only. They were infact calculated for the same value of g;s in Table 11, page 115.The resulting curve is shown in Fig. 64. The tooth widthchosen then was such as to allow B/Bmax to reach a value asclose to unity as the eye could detect in the drawing. But if thetooth width is reduced until the ratio t/s is 0.8 (the value usedfor Fig. 103) and the abscissa scale changed to suit Fig. 103, theresulting curve is that shown in broken line in Fig. 104.The full line is the result calculated in Table 17.

14+ 209

CONTORbf AI, TRANSFORMATIONS

0.9

o-4

BFlux

Density

up) 9 9S .1.5 i-1.25

0 1 2

Fic. 103. Wave-forui of flux density with a succession of equal slotopenings.

10

09F

0$

B

FluxDensity

(peru nit) 9 a . s f= 1 2s

9

t1

I -V1 O , 2 3

FiG. 104. Comparison of waveforms.

It is clear that for narrow tooth widths the more complicatedequations (17.28) and (17.31) must be used when the amplitudesof the harmonics in the flux density wave are required. Thediscrepancy is marked and raises a doubt now to be investigatedas to whether equation (9.28), page 121, is sufficiently valid togive Carter's coefficient for ratios of g/s and t/s such as we havejust used. In the following sections we shall see that the simpleequation, surprisingly enough in view of Fig. 104, does give thecoefficient with remarkable accuracy.

210

TA

BL

E 1

7C

alcu

latio

n of

flu

x de

nsity

dis

trib

utio

n

v0

0-3

1

---

-1.5

23

5m

1 +

vZ1

1.09

- -

-3-

25--

510

---

26--

-O

c

VI

0a

0-287

0.707

0.832

0.894

0.949

0.981

11

v2)11Z

+(

4=si

n-' a

016

.745

.056

.363

.471

.678

.890

.0=

F(¢,

k)

00-

294

0.83

01-

073

1-22

21-

137

1.61

11.

893

b =

1.72

9P0

0-50

81.

435

1-85

52-

113

2-48

52-

785

3-27

3si

nk b

00.

530

1.98

13.

118

4.07

65.

959

f

8.06

913

-18

cosh

b1

1-13

22-

219

3-27

44.

197

6-04

28.

131

13.2

1c

= 0

.073

sin

h b

00-

0387

0.14

50-

228

0-29

70.

435

0.58

90-

962

d=1+

0.02

1 co

sh b

1,02

11.

024

1.04

71.

069

1-08

81-

127

1-17

11.

277

c1d

00-

0378

0.13

813

0-27

30-

386

0-50

30-

753

e=ta

n-1

(c1d

)0

0.03

80-

137

0-21

00.

267

0.36

80.

466

0.64

5

f =1.

15)3

00-

338

0.95

51-

234

1.40

51-

653

1.85

32-

177

(for

s=

1.5)

i0

y=1

5 (e

+f)

1n0.

180

0.32

10.

689

0.79

80.

965

1-10

7] 3

47

g=0.

321v

20

0.02

880.

321

0.72

11-

282

2-88

28.

01h=

(1+

g)1'

21

1.01

41.

149

1.31

11.

511

1-97

13-

002

0-56

6v,n

= 0

.456

v20

0.04

10.

456

1.02

51.

823

4.10

1 1.

30oc

n=(1

+m

)1"2

1

--

-

1.02

01-

206

1.42

3

-

1.68

02-

258

3.52

0i

0.67

5v

p_hf

n1

0.99

40.

953

0.92

1I

0.89

90-

873

0.85

30.

840

B=

0-95

2p0-

952

0-94

60.

907

0-87

70

.836

0.83

10.

812

0.80

0


Carter's Coefficient

Carter's coefficient was defined and developed in the finalproblem of Chapter 9. On page 123 it was shown that thecoefficient could be defined by C = g'lg where g is the actual air-gap length and g' is the equivalent length, i.e. the length ofairgap that would give with unslotted surfaces the same meanflux density as actually appears. Hence

VPoBmean = 7

Therefore

V itog'Bmean

Hence from equation (17.24), page 204,

C - K(I+t/s)2K`g/s

where the elliptic integrals are to modulus k1.

(17.35)

Numerical ExamplesFor actual figures consider the example for which Fig. 103

was calculated and drawn. The values of the relevant ellipticfunctions and integrals were given on page 205. From thosevalues and the subsequent calculation K =1.726, K'= 2.047,t/s = 0.800 and gjs = 0.667. Substituting these values intoequation (17.35) gives

C= 1.726 x 1.80

2x2.047x0.667= 1.14 . . (17.36)

to three significant figures. Using the same value of g/s in theequation governing a single slot opening, i.e. equation (9.27),page 120,

v = tan-1 -91og (1+s2/4g2)}l 2g

= 2 (tan-1 0.75 - 0.667 log 1.5625)IT

= 0.221212


Then using the same value of t/s, viz. 0.800 in the equation(9.28), page 121, governing a single slot opening

t/s+ 1C tj8+1-v

1.3

1-8-0-221= 1.14

exactly as was obtained in equation (17.36) by using the morecomplicated equations governing a succession of equal slotopenings.

Although these results show excellent agreement by the twomethods it is not an isolated agreement. To show this con-sider as a contrast an example where C is large, i.e. where, as inan induction type machine, the airgap length is small while atthe same time open slots are used giving a large slot opening.Let us take for example

k = 0.0523 (or 0 = 3°) and a .= 0.340

Then from tables of elliptic integrals and functions

sn a = 0.334on a = 0.942do a = 1.000

K Z(a) = 0.000678K(k) = 1.572K'(k) = 4.340

k1 = ksna = 0.1745K(kl) = 1.571

K'(ki) = 5.435These are all the values needed for the calculation.From equations (17.18), page 199,

s s1.572

Tr(0-361-0-0004)

= 0.180t 8.680 (0.3606) -0-2168

= 0.78014* 213

CONFORMIAL TRANSFORMATION'S

Hence

C= 1.571 x 1.780(from equation (17.3:5))

2x5.435x0.180= 1.43 . . . . . . . . . (17.37)

again to three significant figures. Now obtain Carter's sigmafrom equation (9.27), page 120, using the values of g/$ and t1:8obtained above :

(tan-1 2.78 - ()-IS log 8.73)n

= 2 (1-2255-0-3892) = 0.533IT

Then equation (9.28), page 121, gives Carter's coefficient for asingle slot opening

1.780C =

1.247= 1.43

exactly as obtained in equation (17.37).These results justify the usual procedure of using a set of

Carter's coefficients calculated from the equations governing asingle slot opening although the problem in practice is that of asuccession of slot openings.

Equations in Terms of AnglesThroughout this chapter whenever numerical examples have

been quoted to illustrate the procedure, the starting point hasbeen to choose values of k and a. It was pointed out on page 200that this procedure starts at the wrong end of the practicalproblem where values of the two ratios g/s and t/s are specifiedand k and a have to be determined from them. But in aninvestigation of this kind, when the equations have beendeveloped they are seldom used merely to obtain a specificanswer to a single question. The equations are usually em-ployed to evaluate a whole series of results to be embodied intables or curves for future use. For instance Carter's coeffi-cient is never worked out for a single pair of ratios s/g and s/t buta complete family of curves is calculated and drawn such asthose shown in Figs. 66 and 67, page 122. Therefore it is quitea reasonable procedure to select a large series of values of k and a

214


and take all possible pairs of the selected values. Then g/s andt/s can readily be calculated for each pair from equations (17.18).It is then necessary to resort to the method of cross-plotting sothat it becomes possible to start from the specified ratios andread off the corresponding values of k and a.

Some books of tables of elliptic functions do not list values ofsn a, do a and on a. It is then desirable to work with themodular and amplitude angles 8 and 0, and this in fact was themethod used by Coe and Taylor in their paper.* It is bestexplained in connexion with a specific example and for this pur-pose equations (17.18) on page 199 will be expressed in terns of0 and 0. Now

andk=sill 0

k1 = ksna = ksin0from equations (12.3) and (12.6), pages 139 and 140. Now let

01 = sin-1 k'1Then

cos 91 (1-k12)1/2(1 - k2 8I12 a)1/2

= do a (E.I.E.P. 121.00)Furthermore

on a = (1-8112 a)1/2 (E.I.E.P. 121.00)= (I-sine= cos 0

Hence

on a= tan ¢ cos 01

In tables of elliptic functions r' is defined by the equationr'/90' = a/K. Substituting all these expressions into equations(17.18), page 199, gives

9 h= {tan cos 01- Z(a)}s 7T

S= 2K'

{tan 0 cos 01 - Z(a)} -90°

snadna


00-FORMAL TRANSFORMATIONS

By using these equations and the method of cross-plottingCoe and Taylor found values of e1 and 0 for many combinationsof g/s and t/8 and a large selection is given in their paper.Similar changes can be made by the same method to equations(17.20), (17.21) and (17.31), and the new form of these equationswill also be found in the paper by Coe and Taylor.

Conclusion

The reader who has been able to follow and understand theproblems worked through in this chapter will now be able toattack most of the field distribution problems that arise inpractice. For further examples of the same standard of diffi-culty he is referred to the papers of F. W. Carter, and moreparticularly to the paper of R. T. Coe and H. W. Taylor,already extensively quoted. In their paper there are fourproblems very fully discussed and solved, as well as the oneused in this chapter, and there should be no difficulty in follow-ing them.

216

REFERENCES(1) PapersCARTER, F. W., À Note on Airgap and Interpolar Induction', J. Instn

elect. Engrs, 29 (1900), p. 925.SCHWARZ, H. A., Ùber einige Abbildungsaufgaben', Crelle's J. Math.,

70 (1869), p. 105.CHRISTOFFEL, E. B., Ann. Mat. pura appl., 1 (1867), p. 95.CARTER, F. W., Àirgap Induction', Elect. World, N.Y., 38 (1901), p. 884.CARTER, F. W., `The Magnetic Field of the Dynamo-Electric Machine',

J. Instn elect. Engrs, 64 (1926), p. 1115.COE, R. T., and TAYLOR, H. W., `Some Problems in Electrical Machine

Design involving Elliptic Functions', Phil. Mag., 6 (1928), p. 100.

(2) BooksTHOMSON, J. J., Recent Researches in Electricity and Magnetism (Oxford

University Press, 1893, Chapter 3).WALKER, MILES, Conjugate Functions for Engineers (Oxford University

Press, 1933).BEWLEY, L. V., Two-dimensional Fields in Electrical Engineering (Mac-

millan, 1948).

(3) Books of Tables, etc.DWIGHT, H. B., Tables of Integrals and Other Mathematical Data (Mac-

millan, 1947).BYRD, P. F., and FRIEDMAN, M. D., Handbook of Elliptic Integrals for

Engineers and Physicists (Springer, 1954).SPENCELEY, G. W., and R. M., Smithsonian Elliptic Function Tables

(Smithsonian Institution, Washington, 1947).MILNE-THOMSON, Jacobian Elliptic Function Tables (Dover Publica-

tions).

217

Index

Abel, 139Amplitude angle, 130Annular ring, 49Argand diagram, 14Argument, 17

Bewley, L. V., 2

CAPACITOR PLATES, 60Carter, F. W., v, vi, 103, 190, 212,

216Carter, G. W., vi, 75, 136Carter's coefficient, 1, 121Cauchy-Riemann equations, 27Christoffel, E. B., 1, 56Coe, R. T., v, vi, 2, 190, 215, 216Collinear electrodes, 135Collinear source and sink, 83Complementary modulus, 131Complete elliptic integrals, 130Complex numbers, 14Complex planes, 35-6Concentric circles, 33Confocal ellipses, 41-2Confocal hyperbolas, 44-6Conformal transformations, 35-6Conjugate functions, 25-30, 36Conservation of angles, 49Constants of integration, 63Curl, 9Current density, 142, 153-4, 169-

172, 179-181

DENSITY OF CHARGE, 71Density of flow, 79Distance between capacitor plates,

76Distribution of flux density, 92,

111Divergence, 8Division of complex numbers, 21

E@UIPOTENTIALS, 11, 27, 32, 69,81, 145

Equivalent airgap, 122Eta functions, 188

FLOW LINES, 30, 32Flow through a slit, 96Flux density distribution, 92, 111Flux function, 30, 32Force equations, 6Fringing of field, 88

IMAGINARY NUMBERS, 15Incomplete elliptic integrals, 164Increase of capacitance, 74Inverse square fields, 5

Jacobi, 129, 130, 139, 185-9, 193

Laplace, 11-13, 24-5Legendre, 130, 139, 192, 193Line charge, 33Lost flux, 117

MODULAR ANGLE, 130Modulus, 14, 129

NOME, 187

ORTHOGONAL CURVES, 30

PERIODS OF ELLIPTIC FUNCTIONS,181

Point source, 6, 7, 82Polar form of complex numbers, 16Polygon, 57Potential, 10, 11, 24Primary electric constant, 7:3Primary magnetic constant, 9:1Product of complex numbers, 17

QUADRANT, 50-5, 87-8

Schwartz, H. A., 1, 56Single slot opening, 103Sinks, 10Slotted plane surface, 77Stream function, 26, 32

218

INDEX

Stream lines, 69, 81. 148Succession of slot openings, 190

Taylor, H. TV., v, vi, 2, 190, 215,216

Theta functions, 185Thomaon, J. J., 1Transformation of a curve, 38-46

definite area, 49-50

point source, 82quadrant, 50-5, 87-8rectangle, 124, 133, 140straight line, 47triangle, 48

Walker, Milea, 2

ZETA FUNCTION, 185

219

conformal transformations in electrical engineering

Documents