Conformal Transformations in Electrical Engineering

CONFORMAL TRANSFORMATIONS

W. J. GIBBS

CONFORMAL TRANSFORMATIONS IN ELECTRICAL ENGINEERING

by

W. J. GIBBS D.Sc., M.I.E.E.

CHAPM 1S 37 ESSE

44LL LTD EE

* C.2

First Published 1958 THE BRITISH THOMSON-HOUSTON CO.. I.TD. 1958

.Catalogue No. 60314 MADE AND PRINTED IN GREAT BRITAIN BY WILLIAM CLONES AND SONS, LIMITED, LONDON AND BECCLES

CONFORMAL TRANSFORMATIONS IN ELECTRICAL ENGINEERING

This Book has been written in the interests of those concerned with advanced theory and practice of engineering, and is one of a series recommended for publication by the Technical Papers Panel of The BRITISH Txonsov-HoujTox Oompany, ]Rugby i

PREFACE Although many young electrical engineers of today have a fair knowledge of the method of conformal transformations, very few attempt to use it in practice even when the problem they have to solve lends itself particularly to that method. The reason is that most solutions in practice involve elliptic functions and integrals, and engineers' knowledge of such functions is in general quite inadequate. There is no book written for engineers which deals only with conformal transformations and takes the reader up to the stage where he can follow the work of Carter and of Coe and Taylor. Miles Walker's Conjugate Functions for Engineers which is the classic engineering text on this subject and is now out of print does not go beyond problems involving elementary functions.

Other books also stop short at the very point where the real difficulties begin, that is, where more than two right angles are in the configuration to be transformed and elliptic functions enter the analysis. An attempt is made in this book to fill the deficiency.

In order to make it complete in itself the book covers in the early chapters the elements of field theory and of complex numbers both from a mathematical point of view. It is assumed that the reader is already to some extent familiar with these topics. The transformations start with the simplest and proceed to those of increasing difficulty until the final chapter. The second half of the book is entirely devoted to transformations involving elliptic functions. It is impossible to write such a book as this without being impressed by the immense debt owed by engineers and by electrical engineers in particular to F. W. Carter. Although others had already discussed conformal transformations in electromagnetism from the point of view of theoretical physics it was Carter who brought the method into the realm of practical work in electrical engineering. His work is described in Chapter 1 but 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 considered, at the time, to be beyond the ability of most engineers to understand. Thirty years have passed and most engineers still graduate without the mathematical knowledge that would enable them to follow papers of this standard. One aim of this book is to take the reader to the point where he can not only

follow and understand these papers but initiate and carry through similar work of his own. The final problem in this book is the first of the five dealt with in the paper by Coe and Taylor. A few years ago Professor G. W. Carter gave a lecture course at Rugby that included some lectures on conjugate functions.

The treatment given here in Chapters 3 to 9 is to some extent

based on his approach and I gratefully acknowledge my indebtedness to him. I would also thank him for his coinments on and criticism of the manuscript and for giving me the proofs 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 going over the manuscript, checked and corrected the mathematics. I am also grateful to Miss M. Oldfield, who checked much of

the numerical work and helped with the proof reading. Finally I must thank Mr. G. S. C. Lucas, Director arid Chief Electrical Engineer of the British Thomson-Houston Company, for his interest in the work and his permission to publish it. W. J. GIBBS RUGBY

May 1956

vi

CONTENTS Page V

PREFACE Chapter 1 INTRODUCTION 2

1

4

FIELD EQUATIONS

Field distribution problems. Inverse square fields-. Force equations. Divergence equations. Curl equations. Potential. Laplace's equation. 3

14

PROPERTIES OF COMPLEX VARIABLES

Complex numbers. Argand diagram. Polar form. Products. Division. 4

24

CONJUGATE FUNCTIONS

Potential function. Stream function. Orthogonal curves. Conjugate functions. 5

35

CONFORMAL TRANSFORMATIONS

Transformation of a curve. Transformation of a straight line. Transformation of shapes. Transformation of a quadrant. Conformal representation.

6

THE SCHWARZ-CHRISTOFFEL TRANSFORMATION

56

The Schwarz-Christoffel equation. 7

CONFIGURATIONS WITH No RIGHT ANGLES

.

60

Electric field at the edges of two capacitor plates. General determination of constants in the transformation. Finding

the origin and axes. Plotting equipotentials and stream lines. Density of charge. Increase of capacitance. Slotted plane surface. Density of flow. Point source. Collinear source and sink. 8

CONFIGURATIONS WITH ONE RIGHT ANGLE

.

87

Transformation of a quadrant. Field fringing from pole end to armature core. Distribution of flux density. 1)

CONFIGURATIONS WITH Two RIGHT ANGLES .

Flow through a slit. Single slot opening opposite a solid face. Amplitude of flux density ripple. The lost flux. Carter's coefficient.

vii

96

CONTENTS Page Chapter 10 CONFIGURATIONS WITH MORE THAN Two RIGHT 124 ANGLES 'T'ransformation of the inside of a rectangle. 11

ELLIPTIC INTEGRALS OF THE FIRST KIND Legendre's notation. Jacobi's notation.

129

Complete

integrals. Complementary modulus. Transformation of a rectangle. 12

ELLIPTIC FUNCTIONS

139

.

The function sn. More elliptic functions. Current density calculation. 13

DOUBLE TRANSFORMATIONS WITH ONE REQUIRING ELLIPTIC FUNCTIONS

148

Electrodes at the edge of a semi-infinite strip. 14

DOUBLE.

TRANSFORMATIONS

BOTH

REQUIRING

159

ELLIPTIC FUNCTIONS

Electrodes at one edge of a rectangle. Incomplete elliptic integrals. 15

ELLIPTIC INTEGRALS OF THE SECOND KIND

173

.

Incomplete and complete integrals of the second kind. Electrodes at the sides of a projection. Periodicity of elliptic functions. 16

AUXILIARY ELLIPTIC FUNCTIONS

.

.

185

Jacobi's Zeta function. Jacobi's Theta and J to functions. 17

ELLIPTIC INTEGRAL OF THE THIRD KIND Succession of equal slot openings. Elliptic integral of the

190

third kind. Jacobi's integral of the third kind. -Numerical example. Amplitude of the flux density ripple. Shape of the flux density curve. Carter's coefficient. Equations in terms 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 of

a bounded area in the plane of a complex variable by an area in the plane of another complex variable. Thus the method is a

branch of mathematics based on the theory of functions of a complex variable. Like most other branches of mathematics it has found applications in the sphere of theoretical physics, mainly through the powerful method of attack developed by Schwarz (1869)* and Christoffel (1867).* A chapter on the Schwarz-Christoffel transformation was included by J. J. Thomson in his book Recent Researches in Electricity and Magnetism (1893)*.

In 1900 F. W. Carter published a paper entitled `A Note on Airgap and Interpolar Induction'*; this was the first application of conformal mapping to an actual engineering problem. It was a major advance for it enabled engineers to find by direct calculation results hitherto only obtainable by rough graphical methods based on intuition. Carter was not only an engineer but also an able mathematician. He wrote of his own work : `Whatever of permanent value is to be found in my work lies 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 an Another paper of his, published in 1901,* gave the derivation of `Carter's coefficients' which are still in common use by electrical engineers, and it forms the principal subject of Chapter 9 of this book. In 1926 Carter published a more comprehensive paper on conformal mapping called `The Magnetic Field of the DynamoElectric Machine'.* In it twelve different configurations were example.

* See List of References, page 217. 1

1

CONFORMAL TRANSFORMATIONS

transformed some of which now required the use of elliptic functions. Hence the standard and range of mathematical knowledge needed to understand the paper was stepped up considerably.

After this other engineers contributed to the subject, but the most important paper was that written by R. T. Coe and H. W.

Taylor called `Some Problems in Electric Machine Design Involving Elliptic Functions'.* In it the authors analysed four different problems in electric machine design all of which required extensive use of elliptic functions. The paper is to some degree a follow up of Carter's 1926 paper, but it gives a more detailed analysis of each problem and takes the results to the point where numerical answers are given and graphs are plotted. In 1933 Miles Walker published the book Conjugate Functions for Engineers.* This work is confined to problems that need only elementary functions in their solution. Nevertheless the treatment is more extensive than that of any previous work and the book was long regarded as the standard text for engineers. An adequate knowledge of how to deal with fields should form an essential part of the engineer's training. Fortunately there are several ways of treating inverse square fields, some

based on drawing, some based on models or analogues, and some entirely mathematical. A good account of the various methods is given in the book Two-dimensional fields in Electrical Engineering by L. V. Bewley.*

To the question why, with so many techniques at his disposal, the engineer needs to know mathematical methods the answer is that he is not fully equipped unless he is familiar with all the methods available and can apply the appropriate one for the problem in hand. If he has little mathematical ability he must of necessity rely entirely on non-mathematical techniques. They may be quite adequate for his purpose, but

nevertheless there is considerable satisfaction to be obtained

from having these techniques supported by mathematical deductions.

A mathematical method never solves the actual problem posed 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 actual problem only gives immediate answers, and by reason of the fact that all the complications present are included in the set-up the analogue does not give the engineer a satisfactory basis for mental pictures of the factors at work. To take a very elementary example, the current locus of an electric motor not yet made may be predetermined from an analogue but the results obtained would not in themselves suggest anything helpful to the designer. On the other hand the current locus of the idealised motor might be part of a circle thereby providing a most helpful concept. Hence to the designer the idealised machine forms the standard from which the actual machine is a deviation. A similar comment applies to the more recondite problems of this book where a mathematical solution of the idealisation gives a better idea of the relative influences of the various ratios in the configurations. It is necessary to emphasise the remarkable identity between all the various kinds of fields met with in engineering. Although in this book a problem may be stated in terms of electric fields, the solution applies equally well to some other problem stated in terms of magnetic fields and vice versa, provided the fields

are static. The parallelism extends also to static fields in other 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 of elliptic functions and computation.

It is therefore useless for

the engineer to embark on any project connected with this analysis without being equipped with Tables of Integrals and Tables of Elliptic Functions as well as the usual Mathematical

Tables. Any special tables used in this book will be fully described when first used.

3

CHAPTER 2

Field Equations Field Distribution Problems

Engineers are frequently confronted with phenomena associated with irregularities of structure. For instance steady flow of a perfect fluid may be disrupted by a barrier as shown diagrammatically in Fig. 1, and the question arises: What are the actual characteristics of the flow past the barrier? For an

Fra. 1. Interrupted fluid flow.

example in electrical engineering consider the magnetic field between two pieces of magnetic material of opposite polarity. If there is a break in one face, as indicated diagrammatically in Fig. 2, how can the actual distribution of the lines of magnetic flux be determined?

Fia. 2. Example of a magnetic field. 4

FIELD EQUATIONS

Such problems are field distribution problems and many methods of dealing with them are available. Freehand sketching by guesswork as illustrated in Fig. 1 gives a crude idea of the field distribution. A much more accurate result is obtained by using the technique of flux plotting. Yet another method

is to devise a model or an analogue and then take readings of whatever is required. But very often the problem is suitable for mathematical solution by using the device of transformation. All field problems are really three-dimensional but a surprisinglylarge number can be treated as two-dimensional by assuming

that the field distribution does not vary in the third dimenWe can put it another way by saying that the structure is treated as though it extends an infinite distance each way in the direction of the dimension not being used. Although this is never quite true it is often a reasonable idealisation of the problem for that part of the structure for which a solution is required. For instance the width of the channel shown in Fig. 1 is persion.

pendicular to the plane of the paper. When this width is ten or more times the depth shown, it is clearly permissible to ignore the boundaries of the channel in that dimension for the general

solution of the problem; the conditions at the boundaries can subsequently be treated separately. By adopting this procedure we need only consider changes represented on the plane of the paper and the problem becomes two-dimensional. Such two-dimensional field problems are often readily and elegantly solved by the conformal transformations that form the subject of this book. Since the method cannot be applied to every kind of field that may appear in practice, it is necessary to

consider first what are the distinctive properties of the particular fields we have in mind. Inverse Square Fields

The fields we are interested in are fields of force called inverse square fields. They are so named because the magnitude of the

force at various points follows an inverse square law which is most easily conceived and explained in relation to one particular condition. The condition is that which obtains when the 5

CONFOR'AAL TRANSFORMATIONS

source of the field is concentrated at a point and is completely isolated. The field does not contain the point but with this exception pervades all space. The reason for using this concept of a single point source is that it makes the best starting point from which to develop the equations. With any point in the field there is associated a value of the force that would be exerted on any per unit test body placed at that point. The force on the test body must have direction as well as magnitude and hence such a field is called a vector field. The point source is the centre of the field and at every other point in it the force is directed either towards or away from this centre. Because it is an inverse square field the magnitude of the force is inversely proportional to the square of the distance from the centre. In order to describe the magnitude and direction of the force at any point in the field, a coordinate system is necessary not only to identify the position of the point in question but also to provide suitable components of the force at the point; these components considered together give both the magnitude and the direction of the force at the point selected. The coordinate system can be chosen to suit any particular problem and the

form of the resulting equations depends on the coordinate system chosen.

At this stage there is no reason why rectangular cartesian coordinates should not be used. Throughout this analysis either rectangular cartesian or polar coordinates are used ; both

kinds are orthogonal, i.e. the coordinate curves intersect at right angles. The Force Equations

For the present then, rectangular cartesian coordinates x and y as shown in Fig. 3 will be used. Since the field of force being considered is two-dimensional the component of force in the x

direction can be written Fz and the component of force in the y direction Fy. As shown in Fig. 3 this force is acting at the point 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 resultant force F is given by

F= T2-

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

Fz=Fxr kx ra

and by a similar process k3

Fy

r3

.

.

.

.

.

(2.2)

Now a field set up in this manner by a single point source can be handled readily by such simple equations as (2.1) and (2.2). But the fields encountered in practice are usually much more

complicated because there may be several sources, and the source or sources may be distributed instead of being concentrated at points. In dealing with these more complicated fields the analysis is helped considerably by the development of the differential 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 in terms of x and y by making the substitution x2+ y2

and this produces

F -

kx

(x2+y2)3j2

.

Fky

y = (x2+y2)3'2

(2.3)

(2.4)

Now find the partial derivative of Fx with respect to x and that of Fy with respect to y: aFz k{(x 2 +y 2 )- 3 /2 -2x 2(x 2+y 2)- 5 2 /} = OX and

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

ay Summing the two partial derivatives

OF, aFy 8x

+

ay

=

k{2(x 2 + y 2)-3/2

2 5/2 _ 2(x2 + y2 )(x + 2y )-}

= 0

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

aFx aFy

aFz

ax+ay+az

=0

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

1 a(rFr) 1 cFB aFz r & +r as + az

-0

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

divF=0 or

V.F=0 } S

(2.5)

FIELD EQUATIONS

Equation (2.5) which is called the divergence equation is introduced here only as a step to later developments. For the present it need only be noted that in the inverse square fields under consideration the divergence of the field at every point is zero. In two dimensions aFx aFy C"7=0

.

.

.

.

.

(2.6)

The Curl Equations

In a similar set of equations forming the next step we consider the rate of change of each component of force in one direction

with respect to the other direction. This leads to what are called the curl equations. From equations (2.3) and (2.4) form the partial derivative of Fx with respect toy and that of Fy with respect to x. Thus aFx vy

=

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

aFy ax

2xy(x2+y2)-5/2

The two derivatives give identical results and therefore aFy aFx = 0

-

ax

ay

(2

It It can also be shown that for a three-dimensional field using rectangular cartesian coordinates OF, ay

aFy

_

az

OF- aF az ax =

0

0

In cylindrical polar coordinates the equations are 1 aFZ _ aFa r 00

aFr aFZ or

0Z

1 a(rFe)

r

ar

_0

Oz

=0 0

OF,

ae 9

=

CONFORMAL TRANSFORMATIONS

In other coordinate systems the equations take other forms. As with the divergence equation, in order to express them in a

form that is brief and yet embraces all possible coordinate systems they are written

curl F = 0 or

(2.8)

VxF = 0 Notice that while V. means divergence V x means curl. All these equations apply only to the field itself; they do not hold at the points where sources are located. Similarly if the field 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 all inverse square fields the curl as well as the divergence at every point in the field is zero. The Potential

The fact that the curl of these fields is everywhere zero leads to an important result. Equations (2.8) state a necessary and sufficient condition that the force F at any point in the field can be 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 Fz =

.

.

.

.

.

(2.9)

a az

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

a20

ay

ax ay 10

FIELD EQUATIONS and

aFy

020

ax

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 by the first pair of equations (2.9) provided aFx _ aFy ay

_0

ax

and not otherwise. In general it can be proved that the equations curl F = 0 state a necessary and sufficient condition that 0 exists. The function 0 is called the potential of the force. It is simply a number and has no components corresponding to Fx and Fy. It is thus a scalar quantity. Hence the importance of the curl equations lies in the fact that we are released from the necessity of dealing with the vector components Fx and Fy and can deal instead with the scalar quantity ¢. If the curl of the field is not zero we cannot do this. It may be said here that if the source of the field is included in the field the curl is no longer zero, and the mathematics becomes more complicated. The concept of a vector field is difficult for some people. They have to make a mental effort, for instance, to develop the conception of the velocity field of a moving fluid. But provided there are no sources or sinks in the velocity field, the integration of the field produces a scalar field, viz. the field of flow. It is this scalar field associated with the potential that most people normally think of in connection with fluid motion. The scalar field is only a field in the mathematical sense ; it is really a map showing the potential at any point in a domain, and if lines are drawn through points of equal potential such lines are called equipotential lines. The difference of potential between two points in the field can be defined as the work done on a per unit

test body when transferring it from one point in the field to another. Laplace's Equation

Reverting now to the divergence equation, viz. (2.6) on page 9, by substituting for Fx and Fy from equations (2.9), 11

CONFORMAL TRANSFORMATIONS

page 10, we obtain what is called Laplace's Equation. The argument can be confined here to two dimensions. It follows from equations (2.9) that bFx

a2 a2o

aFy

axe + aye

ax + by

But from equation (2.6) the divergence equation

a Fx aFy _ dx

+

0

by

and hence 02d. +a2A

axeay2 =

0

.

(2.10)

This is Laplace's equation in two dimensions in rectangular cartesian coordinates. In polar coordinates it becomes a20

art

+

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

In other coordinate systems Laplace's equation takes yet other Therefore to express it briefly in a form that embraces all coordinate systems it is written V20 = 0

forms.

or

div grad = 0 It is as well not to be unduly disturbed or impressed by such words as divergence, curl and Laplace's equation. Nor should engineers allow themselves to be discouraged by the apparent clumsiness of some of the mathematical expressions. What it all amounts to is that the fields suitable for the treatment to be developed in this book are inverse square fields and that as a result they can be described by differential equations where the independent variable is a scalar denoted by which is a function of position in the plane of the field and is related to the components of the force acting at any point by the equations Fx

ax

Fy

ay 12

FIELD EQUATIONS

Laplace's equation does not yield easily to straightforward treatment: fortunately in the development of conformal transformations there is no need to seek a formal solution of the equation. It is only necessary to note first that all fields and functions to be considered in this book are those that satisfy the inverse square equation when emanating from a point source and therefore they also satisfy Laplace's equation. The second point is that the equation is necessary for the development of other important equations that govern these particular fields. Such fields are called Laplacian. When the field is not Laplacian, more recondite methods are necessary for determining its distribution. In vector analysis it is desirable to form physical conceptions of the meaning of gradient, divergence and curl. But for the purpose of conformal transformations it is sufficient to establish the equations for two dimensions in such a manner as to bring out certain characteristics of the fields under consideration. For a full development of the physical approach the reader is referred 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 a diversion from the main theme into that of complex numbers. But in fact the method of conformal transformations requires the use of complex numbers. Therefore it is desirable to review next some of the properties of complex variables. As a general symbol for a complex number the letter z can be used here be-

cause all problems discussed will be two-dimensional and therefore the z axis will never be required. The complex number z = x + jy consists of two parts : a real part x which is by itself a real number, and an imaginary part jy which consists of a real number y multiplied by the imaginary number '/ -1. The imaginary number,\/ --1 is represented by

the letter j ; this symbol is treated in accordance with all the normal laws of algebra except that whenever j2 appears it is replaced by - 1. With z=x+jy, the complex number x-jy is called the conjugate of z and is written z or z*. Now the product 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 is often written ¶z I or Ix + jy I . Therefore zz* = Iz12 The Argand Diagram

Real numbers can always be represented by points along a

straight line correctly spaced according to some scale and based on a given origin which corresponds to the number zero. This is indicated in Fig. 4. The statement includes negative numbers and irrational numbers. 14

PROPERTIES OF COMPLEX VARIABLES I

-5

-4

-3

-i

-2

FiG. 4.

O

2

I

3

I

4

5

Geometry of real numbers.

Similarly complex numbers can always be represented by points placed in a plane. The real part of the complex number can be represented along the horizontal line called the real axis,

and the imaginary part along the vertical line through the origin called the imaginary axis. Thus the number 1 + j2 is represented by the point A in Fig. 5. This representation of a complex number is called the Argand diagram. Imaginary Axis

4 3F

2

I

Real

l_ I

2

-2

-3

-4

FiG. 5.

The Argand diagram. 15

3 AxI

CONFORMAL TRANSFORMATIONS

All the algebraic operations using complex numbers can be represented geometrically on the Argand diagram. For instance 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+j6 is equivalent to the vector addition

OA+OB = OC shown in Fig. 6.

Fic. 6. Addition of complex numbers. Complex Numbers in Polar Form

Consider again the point A in Fig. 6 representing the number

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

16

PROPERTIES OP COMPLEX VARIABLES cos B1 =

real part of OA JOAJ

(22+42)1/2 9

x/(20)

imaginary part of OA

sill 61

IOAI

4

1/(20)

tan61 =

sin 61

cos01

-- y

Similarly if OB makes an angle with the real axis denoted by 02 and since OB = 4 + j 2 4

cos 02 =1/(20)

in 02 =

9

1/(20)

tan 02 = 0.5 The angles 01 and 02 are called the arguments of OA and OB

respectively. A complex number can be described by its modulus and its argument, e.g. 2+j4 = (20)1/2 tan-1 2

= r1 where ri -- OA = the modulus

01 = the argument. and Similarly OB can be described either by 4 + j 2 or by (20)1/2 tan-I 0i .e. by

x2102

Products of Complex Numbers

The advantage of expressing complex numbers in terms of 2

17

CONFORMAL TRANSFORMATIONS

their moduli and arguments-called their polar form-arises when such numbers are multiplied by or divided by other complex numbers. Complex numbers can be multiplied together by the ordinary

rules of algebra, treating j as if it were a real number except that 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 course another complex number. It is usually much easier to deal with these products when they are expressed in their polar form. For example if

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

a+jb = rl(cos 01+j sin01) = rieje, Similarly c + jd = r2ele2

Hence

(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 the moduli (a product of real numbers) and add the arguments. As an example, consider the two complex numbers shown as OA

and OB in Fig. 6. OA represents 2+j4 and OB represents 4+j2. Multiplied directly (2+j4)(4+j2) = (8-8)+j(16+4) = O+j20 Fig. 7 shows OA and OB again, together with their product

represented by OC. Now in polar form the numbers represented by OA and OB are OA = (20)1j2/tan-1 2 18

PROPERTIES OF COMPLEX VARIABLES imaginary Axis

20

C

~(OC

= OA- OB)

16

12

8

L Real 12 Axis

Fic. 7. Product of complex numbers.

as found on page 17.

The angle tan-' 2 from tables is 63° 26'

so that OA = (20)1/2/63- 26'

Similarly

OB = (20)1/2 tan-' 0.5 = (20)1/2 26° 34' Then OA.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 product of two complex numbers bears no relation to the scalar product 19

CONFORMAL TRANSFORMATIONS

of two plane vectors and it cannot be made to bear any relation

by taking the conjugate of one of the numbers, because, for instance

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 it becomes apparent when several complex numbers have to be multiplied together. The product of any quantity of complex numbers can be found by multiplying together all the moduli and adding up all the arguments, i.e. OA. OB.OC.OD ... = r1r2r3r4 ... Finding the product of a series of complex numbers in algebraic form by algebraic multiplication is very long and tiresome by comparison. Imaginary Axis

6

5

4 OC = OA-OB)

3

2

I

,

-3

-2

I

FIG. S.

2

Real

L

3

4

5

Difference of two complex numbers. 20

Axis

PROPERTIES OF COMPLEX VARIABLES Subtraction and Division of Complex Numbers

The algebraic subtraction of one complex number from an-

other is equivalent to the vector subtraction of two plane vectors. For example if OA and OB are as before 2 + j4 and 4 + j 2 respectively, the difference is shown on the Argand diagram in Fig. 8 where OB is first reversed and then added to OA

giving the complex number OC which is seen to be - 2 + j 2. This is the result that would be obtained by subtracting the two numbers algebraically, and the algebraic method is by far the simplest. Division of complex numbers can also be accomplished in two ways. Here the more roundabout method is the algebraic where the denominator of the fraction representing the division

is made a real number by multiplying it by its conjugate and then the numerator is multiplied by the same conjugate. For instance OA OB

_ OA.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. Taking for example the two complex numbers used already 2+j4 _ (2+j4)(4-j2)

4+j2

42+22

16+j12 20

= 0.8+j0.6 But it is a much easier method to put the two numbers into their polar form and proceed thus : ri/Ct OA OB

r2/ rlej6i r2ei62 21

CONFOR'_14AL TRANSFORMATIONS

= r 11 e1(e1-e2) r2

ri

r2

/01-02

This result shows that to divide the complex number OA by the complex number OB we can first put both numbers into their polar form, divide the modulus of OA by that of OB and then subtract the argument of OB from that of OA. In the numerical example above the polar form of the numbers was found on page 19. Hence OA OB

(20)1/2163° 26'

(20)1/2/26- 34'

= 163° 26- 26° 34' = 1 36° 52

= 0.8+j0.6 Imaginary Axis

4r

A 1

3

2

I

C

3

Fiv. 9. Division of complex numbers. 22

4

PROPERTIES OF COMPLEX VARIABLES

The result is shown in Fig. 9. Notice that the division of complex numbers has no relation whatever with plane vectors where there is no such operation as division. The polar method is clearly most advantageous when there are several divisions combined with multiplications to be performed.

23

CHAPTER 4

Conjugate Functions The Potential Function

From the theory of complex numbers the conjugate functions

we require can be developed. Consider the two complex numbers

U = x+jy V = X-jy It is quite an elementary exercise to show by differentiation

that both u and v separately satisfy Laplace's equation. Furthermore x and y separately satisfy the same equation. But a far more important theorem will now be established, viz. that the real part and the imaginary part of any function of the complex 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 ax ax

Of(Z) az

az ax af(z)

Proceeding similarly 02

ax = f"(z) Again aX

Sf(z) az

ay

az ay = jf'(z) 24

.

.

.

.

.

(4.1)

CONJUGATE FUNCTIONS and 02X

_

3f'(z) az

ay az

8y

2 af'(z)

aw - f"(z)

.

.

.

.

(4.2)

Combining equations (4.1) and (4.2) gives 2X

02X

%x' + ?y-

=0

.

.

(4.3)

Now in terms of 0 and 0, since

X = 4+i/ G2X ax2

=

a2X

-

X20 + ax2

02` (4.4)

ax2

Similarly ayz

020

a2o

oy2 +J ay2

.

(4.5)

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

( 2-

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

a+ay20

.

(4.7)

and ax2+ay2=0

(4.8

Therefore the real part and the imaginary part of any function of the complex variable x+jy automatically satisfies Laplace's equation. These real and imaginary functions are called conjugate functions. To summarise, if X be any function of z, i.e.

where X = O + j o and z = x+jy

X=f(z) 25

CONFORMAL TRANSFORMATIONS

we know that and

.

.

.

.

.

(4.9)

0

V20

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

V20 0 is satisfied by a scalar function that is a function of position and 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, in equations (4.9), 0 is identified with potential and called the potential function, the conjugate function b is called the stream function or the flux function. The meaning of the stream function will become more apparent as the analysis is developed. The Stream Function

There is an important relation between the potential and stream functions expressed mathematically by what are called the Cauchy-Riemann equations. Writing as before

X = O+jo

z=x+jy

where X = F(z), then aX

ax

Furthermore

- aa4 ax + l ax

.

(4.10)

.

(4.11)

.

4.12)

.

4 . 13)

aX_aF(z)az ax

ax

az

OF(z)

az

From equations (4.10) and (4.11)

48oax _ eF(z)

TX

+l

az

Similarly it can be shown that a y+ j

_

y-

26

a J

ti

az )

CONJUGATE FUNCTIONS

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

0000 +j(LO+BO = 0 TX

ay

ax

ay)

Therefore since the real and imaginary parts of the left-hand side must separately be zero ao 00 TX - ay .

00

00

ax

ay

(4.14)

Equations (4.14) are the Cauchy-Riemann equations. They are important because they describe an important property of conjugate functions, a property now to be explained. If 0 is treated as a potential function the equation 0 = constant

gives a family of curves with x and y as coordinate axes, one curve for each value given to the constant. Each of these curves joins points of equal potential and the curves are therefore called equipotentials. To be quite clear in the argument that 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

Therefore equating the real parts 0 = sin x cosh y The family of curves given by 0 = constant 27

.

.

(4.15)

CONFORMAL TRANSFORMATIONS

can then be obtained from sin x cosh y = a or

cosh y = a cosec x where a can be assigned a series of values. Fig. 10 shows two curves, viz. : sin x cosh y = I and

sin x cosh y = 2 3

Y

2

a=2 f

a=

0

jr

1

2

Ex:. 10. Equipotential curves.

and both are curves joining points where 0 is constant, i.e. they are equipotential curves. It can be shown that they pass 28

CONJUGATE FUNCTIONS

through minimum values at x = it and then become asymtotic to the line x =7T. However, we are only concerned here with the early part of the curves shown in Fig. 10. Similarly a family of curves can be drawn from values of the constant given by the equation

0 = constant For instance in considering the function

x = sin z equation (4.15) showed by equating the imaginary parts that 0 = cos x sink y 3

b-2

Y

2

1

O

x

FiG. 11.

Stream lines. 29

CONFORMAL TRANSFORMATIONS

so that the family of curves given by

0 = constant can be obtained from cos x sinh y = b Fig. 11 shows two curves for two different values of b, i.e. the curves for cos x sinh y = 1 and cos x sinh y = 2

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

Comparison of these curves with those of Fig. 10 shows that

one family of curves intersects the other. In general the family of curves obtained from 0 = constant must intersect In Fig. 12 the two particular curves obtained from a = 2 and b = 2 are shown on the same diagram where they intersect. They happen to be the curves obtained from the specific function x = sin z but the argument is quite general. Let then therefore be called the curves the family of curves obtained from z& = constant.

0=a and 0=b

where 0 + j f is any function of x + j y.

The gradient at any point on either curve is given by dy/dx of 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) dy = 0

o dx + ay

Therefore on 0 =a

_

ao/hx 80/8y

dy _ dx

00/2x

dy dx Similarly on1, (= b

Way 30

CONJUGATE FUNCTIONS 3

)6-b

Y

2

. a

1

x 2

1

Fie. 12. Orthogonal curves.

and these relations apply at the point of intersection shown by the tangent lines in Fig. 12. The product of the two gradients is

(aolay)(aoiay)

and by using equation (4.14) this can be written o_l ay)

- -1

W/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 of curves 0 = constant intersect the family of curves 0 = constant orthogonally. This is an important result and it forms the basis of the flux-plotting techniques. It has been stated earlier that the function 0 is sometimes called the stream function and sometimes the flux function and that the curves given by 0 =constant show the distribution of flow when the curves = constant are regarded as equipotentials. But it is clear that since 0 was chosen quite arbitrarily

out of a pair of conjugate functions to be regarded as the potential function of the field, we could equally well have chosen 0 instead. Then the curves 4 = constant would have been the lines of equipotential and the curves O = constant would have been the lines of flow. The potential function and the stream function are mathematically interchangeable. These points can be brought out more clearly by considering a simple example of a pair of conjugate functions. The Function log z

Let X = log z where X and z have the same real and imaginary components as before. Then

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

x = r cos 0

y = rsin9 so that z

x+jy r(cos 9 + j sin 0) (4.16)

reie

Then 0 + j4 = log(r ell)

= logr+j9

.

.

.

.

(4.17)

Hence by equating real and imaginary parts

= log r 9 32

.

.

.

(4.18)

CONJUGATE FUNCTIONS

FiG. 13.

Conjugate functions.

Now plot the curves given by = constant on the Argand diagram for different values of the constant. The result is a family of concentric circles with their centre at the origin, as shown 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 a family of straight lines radiating out from the origin as shown in Fig. 13. Notice that all the intersections are orthogonal. For some problems 0 will be the potential function and then the circles will be lines of equipotential. This, for example, is the potential of an electrostatic field set up by a line charge through the plane of the paper and located at the origin. Then 3

33

CONFORMAL TRANSFORMATIONS

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

For other problems 0 will be the potential function and 0 the stream function, giving circular flux or flow lines and having the

radii as equipotentials. For example 0 is the potential of a magnetic field set up by a line current flowing through the plane

of the paper and located at the origin. It should be noted again that with either interpretation the field does not include the origin where the source of the field is located. The two functions can be expressed in rectangular cartesian coordinates by noting that x2+y2 = r2 and

y x

_ rsin8 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 Representation

This chapter deals with the central idea of this analysis, viz. conformal representation or mapping. Consider any complex number z. It has already been shown that it can be represented by a point on the Argand diagram as in Fig. 14 (a). This diagram will be called the z plane. Y

w

z 0

0

X

FIG. 14.

U

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

Suppose now that there is another complex number denoted by w which is some known function of z. Since w is a function of z there must be some relation between the point x + jy on the z 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 as the point x + jy because four quantities are involved; in other words we cannot plot on a plane a single locus that shows how w varies with z. For such a single curve a four-dimensional space would be necessary. Therefore two complex planes are used 35

CONFORMAL TRANSFORMATIONS

to depict the relation between w and z and there must be point to point correspondence between the two planes. Therefore the point u + jv is plotted on another Argand diagram shown here in Fig. 14 (b). This second diagram is called the w plane. In the w plane u is plotted along the real axis and v along the imaginary axis. Now clearly any point on the z plane can be related to a point on the w plane by two functions such as u = f (X, y)

v = 9(x, y) To each point x, y there are one or more points u, v although we shall confine our attention to functions that give one-to-one correspondence. Thus everything appearing on the z plane can be mapped on the w plane and vice versa. But although the functions written above are quite general this book will only be concerned with conjugate functions. This means that the transformations are confined to conformal transformations, and the distinctive features of such transformations are these :

(1) Adjacent points in one plane transform to adjacent points in the other plane. (2) In both planes the lines of flux and the equipotential lines intersect at right angles.

(3) Not only are these right angles conserved in the transformation but the sense of the angles is also conserved. The stage has now been reached when it is possible to explain the purpose of using conformal transformations. Suppose it is required to determine the distribution of the field between two equipotential boundaries that can be represented on a diagram and called the z plane. Suppose further that the two boundaries are 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 the

boundaries consist of magnetic material whose permeability could be considered infinite. In such a problem it may be possible to find and apply a transformation from this z plane to a

new plane called the 2v plane in which the shapes of the boundaries become something recognisable, something for which the distribution of the field is both regular and known. 36

CONFORMAL TRANSFORMATIONS

X

x

Fic. 15. Equipotontial boundaries. z plane.

For instance, a transformation may be found that gives the new boundaries shown in Fig. 16 and makes all the flux lines and equipotentials straight lines. With boundaries like this consisting of parallel straight lines and with all stream lines and

equipotentials straight, the field distribution is easily determined.

We can find for the new plane or we can find 0 instead, whichever is required for the problem in hand; then the equation of transformation is used to determine the corresponding field in the original plane. In other words the solution of the problem is readily found in the w plane, and this solution leads through the transformation equation to the solution in the z plane, i.e. to the solution of the actual problem. Notice that since is constant over the boundaries in the w plane it is also constant over the boundaries in the z plane and must be the required potential in the z plane. It is desirable at this 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 Curve

Consider the conformal transformation

z = 2cosw+jsinw

.

.

.

.

(5.1)

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

z = 2cosw+jsinw w

0

'7r f 6

z

2+j0

1.73+j0.50

w

77/1

z

0+j1.0

1

277/3 ---

-

-1.0+j0.87 38

7r/4

1r13

1.414-j0.71

1.00+j0.87

37r,14

5nj6

-1.41+j0.71

-1.73±j0.50

CONFORMAL TRANSFORMATIONS

N=o

(qi

z = --2 - Tr

z =2

1J=/- J l

0

-Tr

'/

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 w chosen in Table 1, the result is a straight line along the real axis as shown in Fig. 17 (b). On the other hand a line drawn in the z plane through the corresponding values of z in Table 1 forms the upper half of an ellipse. Further values of w from it to 27r give the lower half of the same ellipse ; this curve is shown in Fig. 17 (a). Thus an ellipse in the z plane is transformed by the equation to a straight line in the w plane. This is an example of the transformation of a curve into a straight line. However, it is instructive to go further than this to see what the same transformation means when all values of w real and 39

CONFORMAL TRANSFORMATIONS

complex are admitted. The procedure to adopt is to find x and y as functions of u and v by expanding equation (5.1) : z= 2 cos w+ j sin w Substituting 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 v

Equating 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 real values of w were considered, i.e. various values of it were chosen

with v zero throughout. Suppose now that v is made 0.5 throughout. Then cosh v = 1.1276 sinh v = 0.5211 and

2 cosh v - sinh v = 1.734 cosh v - 2 sinh v = 0854 Then from equations (5.2) x = 1.734 cos it

y = 0854 sin u and Table 2 can be drawn up as follows : TABLE 2

Transformation of an ellipse u a= cos u

x= 1.734a

'

0

7T/6

7r/3

7r/2

2ir/3

5ir/6

IT

1.000

0.866

0.500

0

-0-500

-0-866

-1-000

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

CONFORMAL TRANSFORMATIONS

On plotting these values in the z plane we find that the curve joining them is the upper half of another ellipse wholly inside the previous one shown in Fig. 17. The lower half of the new ellipse is obtained by taking values of u between 77 and 2ir. It will be shown presently that the new ellipse is confocal with the old one. When plotted in the w plane the values in Table 2 give an-

other straight line parallel with the real axis and passing through the point v = Consider now any general constant value of v. (5.2) can then be written with a = 2 cosh v -sinh v b = cosh v - 2 sinh v where a and b are constants. 't'hen

Equations

x = acosIt y = 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 v b 2 = cosh2 v+ 4 sinh2 v- 4 sinh v cosh v Therefore a2 - b2 = 3(cosh2 v - sinh2 v)

=3

Therefore whatever value is assigned to v the expression a2 - b2 is constant. Hence any change in the value of v which increases (or decreases) a2 by an amount h must also increase (or decrease) b2 by exactly the same amount ; hence the various

values assigned to v must give a family of curves that can be written x2

y2

a2+h+b2+h =

1

This is known to be the equation of a family of confocal ellipses.

Therefore the equation v = constant gives a family of 41

CONFORMAL TRANSFORMATIONS

(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 a family of confocal ellipses in the z plane. Both are shown in Fig. 18. Suppose now that u is held constant instead of v ; what happens when various values of v are selected for each value of the constant u? For instance consider the following values of u in turn :

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

At u = 0,

x = 2 cosh v - sink v

y=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 v

y = 0.500 cosh v-sinh v 42

CONFORMAL TRANSFORMATIONS

At u=7r/3,

x = cosh v - 0.500 sinh v y = 0.866 cosh v -1.732 sinh v From these three sets of equations Table 3 can be drawn up as follows : 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

1.500

1-662

2.240

3-380

5.380

8.73

0-578

0-772

1.176

1.881

3-066

5.04

-0.403

-0.953

-1-746

-2.484

-4.98

(u =7r/6)

e = 0.500a

y=e-c

(u=ir/6)

0

f = 0.500c

0-289

0.588

1-065

1-814

3-025

5.01

x=a-f

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

-0.700

-1.655

-3.030

(u=7r/3)

y=g-h (u =7r/3)

0

-5.180 -869

Notice that at u = jrr, x = 0 giving the imaginary axis of the z plane. When plotted on the z plane the two sets of values given in Table 3 of x and y give the lower halves of two hyperbolas. The upper halves can be found by giving negative values to v and positive values less than 0-55. The two hyperbolas are shown in Fig. 19. Note that negative values of x can and 37r/2. be found by selecting values of u between 43

2Tr

CONFORMAL TRANSFORMATIONS 8

4

U. 2

- 57 3

6

Y 1111

u' 6

2

u-O 4

6

8

-2

-4

6 -6 U_

Uz

2

3

-8

Confocal hyperbolas.

FIG. 19.

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

a

x = - (2 cosh v - sinh v) V3 (cosh v - 2 sinh v)

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

3

Then x2

y2

a2b2 44

CONFORMAL TRANSFORMATIONS

and this is the equation of a hyperbola. Now a2+b2 = 3(cos2 u+sin2 u)

=3

so that if any change in it increases (or decreases) a2 by an amount h it must also decrease (or increase) b2 by the same amount.

Thus for various values of It the following equation

holds : X2

y2

a2+h b2-h

(Q)

V

u

/6/ FiG. 20.

Curves of constant u. (a) z plane (6) w plane. 45

CONFORMAL TRANSFORMATIONS

/ol V

u

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

This is known to be the equation of a family of confocal hyperbolas. Therefore u = constant gives in the w plane a family of

straight lines parallel to the imaginary axis and in the z plane a family of confocal hyperbolas. Both planes are shown in Fig. 20 and again in Fig. 21, where the corresponding sets of straight lines and curves for v = constant are added. This then is an example of the transformation of a set of curves into a set of straight lines. The next thing to consider is the transformation of straight lines into straight lines. 46

CONFORMAL TRANSFORMATIONS

Transformation of a Straight Line

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

0

FIG. 22.

0

X

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

From equation (4.16), page 32,

z=rej° Therefore

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

the two points Qi and Q2 shown in Fig. 22 (b). These points are also very close together. Let ds2 be the length of the straight line joining Qi and Q2 and let this line make an angle a2 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 w

a relationship that is the basis of the transformation from one to the other w = f (z) and

dw = f(z) dz where f'(z) must be a complex number and can therefore be described another way by writing f'(z) = Aei$ where A is the modulus and fi the argument of f'(z). 47

CONFORMAL TRANSFORMATIONS

Then ds2eia2 = dw

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

ds2 = A ds1

and a2 = ai+N Thus, in transformation the line ds; is multiplied by A and also turned through the angle P. The change from the z plane to the w plane alters both the scale and the orientation. Note particularly that A = L82 =

dwl .

(5.4)

dz dsl Hence each small displacement from a point P in the z plane when transformed to the w plane is increased or decreased in the same ratio as the other displacements and is turned through the same angle. This result enables us to consider the trans-

formation of configurations or shapes consisting of straight lines and angles. It follows from the previous section that any small figure, say

the small triangle shown in the z plane Fig. 23 (a), remains a small triangle when transformed to the w plane Fig. 23 (b), but that it is either enlarged or diminished by the transformation and it is also turned through an angle. Fig. 23 shows the result of a typical transformation. The transformation fails at the points where A in equation (5.4) is zero or infinite. Y

D

4

U

(a)

Fm. 23. Transformation of a triangle. 48

14)

(a) z plane (b) w plane.

CONFORMAL TRANSFORMATIONS

The converse of this statement is equally true, viz. that the angles are bound to be conserved by the transformation unless A (i.e Idw/dzl) is zero or infinite. Transformation of a Definite Area

Consider again the transformation of equation (4.17) on page 32:

where q = log r

¢+j = logr+jO

z/r=B In the z plane of this transformation let the boundaries be

and

those of the shaded annular ring shown in Fig. 24 (a). We now

want to find what the corresponding boundaries are in the X

7;)

(Q)

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 in Fig. 24 (a). For numerical values let these points be

x = 4, 3, - 3 and - 4 respectively.

Consider first the point a where x = 4 and y = 0.

= logo = 1.386

8=0 Thus 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 b 4

49

CONFORMAL TRANSFORMATIONS

is 3 + j0, but in the X plane it becomes the point log 3 + j O, i.e. 1.099 +j0 as 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+j7r so that c becomes the point 1.099+j7r

Similarly d becomes the point 1.386+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 equations

show that in the X plane the points are connected by straight Therefore the shaded area in Fig. 24 (b) corresponds to the shaded annular area of Fig. 24 (a). In considering the transformations it is helpful to imagine lines.

that the area enclosed by the boundaries contains an ideal elastic substance or membrane corresponding to the equipotentials and the lines of flow. The material is ideal in the sense that it is infinitely expansible and infinitely compressible without its changing character and without there being constraints of any kind. The spaces enclosed by the equipotentials and lines of flow are regarded as cells of this ideal material, which possesses the property that the boundaries can be distorted in

any manner but that the lines forming the boundaries of the cells must always intersect at right angles. Thus the shape shown in Fig. 26 can be imagined picked up and compressed into the shape shown in Fig. 25 and vice versa. Transformation of a Quadrant

As a further example consider the transformation

w = z2

.

.

.

.

.

.

(5.5)

where for simplicity the restriction is imposed that z is confined 50

CONFORMAL TRANSFORMATIONS

F[G. 26.

Shape filled with hypothetical elastic substance.

FIG. 26.

Distorted shape.

wholly to the first quadrant. We wish to determine what this quadrant in the z plane becomes in the w plane under the transformation of equation (5.5). As before substitute for z and w thus:

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

Hence, separating real and imaginary parts u = x2 _ y2 11 v = 2xy 51

.

(5.6)

CON FORMAL TRANSFORMATIONS

V

c'

P

m

/''

;

C1

b

C

P

M u 6

0

b

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 confined to positive values. Draw the z plane Fig. 27 (a) and the w plane Fig. 27 (b). Note that only the first quadrant is required 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. In the z plane u = 0 corresponds to

x2-yz = 0 i.e. to

x=y

because only positive values can be admitted. This gives the line labelled `a' in both planes. The equation v = 0 gives the real axis of the w plane. In the z plane since v = 2xy

the values x = 0 and y = 0 both satisfy v = 0. Therefore the real axis of the w plane for both positive and negative values of u is

transformed to the x and y axes of the z plane giving the curve shown as m in both planes. Next find the curves in both planes for the value u = 2. In the w plane it is clearly the vertical line drawn two units away from the imaginary axis. But in the z plane x2 _ y2 = 2 so that

x2 = y2+2

and Table 4 can be drawn up as follows : 52

CONFORMAL TRANSFORMATIONS 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 in both planes Fig. 27. Similarly there must be curves b' in both planes obtained by putting u = - 2, i.e. y2 = x2 + 2. Then by putting u = 5 and u = - 5 the curves shown as c and c' in both planes are obtained. We can select as many values of u as we please and draw the corresponding curves in the two planes. Now consider various values of v in the equation v = constant. When v = 2

2xy = 2 and hence

x = l/y from which Table 5 can be drawn up : TABLE 5

Values of y and x for v = 2 y

0.25

0.50

1.00

x= l /y

4.00

2.00

1.00

2.00 4.00 __J ____ 0.50

0.25

Plotting the points obtained from Table 5 gives the curve labelled n in both planes.

Similarly by putting v = 5 we obtain

the curve labelled p in both planes. We can select as many values of v as we please and draw the corresponding curves in both planes. Notice that v cannot have negative values because if it had, either x or y would have to be negative, and negative values of x and y are inadmissible. 53

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

x2-y2 = a2 or x2

yz

a2

a2

and this is the equation of a rectangular hyperbola whose asymptote 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 whose asymptotes are the real and imaginary axes of the first quadrant. Clearly the two sets of curves intersect at right angles. As many curves as we please can be drawn in both planes by taking a sufficient number of values of the constants.

kio.'28. z plane.

Figs. 28 and 29 show a closer mesh in both planes. We can imagine the mesh in the z plane Fig. 28 to represent the cells of 54

CONFORMAL TRANSFORMATIONS

the ideal elastic material.

Then in imagination we can take the

y axis and force it round in such a way that it becomes the negative real axis. The hyperbolas forming the boundaries of the cells become straight lines and as a result the picture in the w plane shown in Fig. 29 appears.

s

4

-12

0

4

FIG. 29. w plane.

55

e

12

CHAPTER 6

The Schwarz-Christoffel Transformation IN ALL THE EXAMPLES discussed in previous chapters trans-

formation equations have been set down first and the corresponding configurations in both planes have been discovered afterwards. This is hardly a satisfactory procedure for actual problems because in practice the engineer knows the two configurations, the one he is concerned with and the simpler one possessing a known solution. From these known configurations he has to determine the appropriate equation of transformation.

If what has been done in previous chapters were the only possible procedure he would always have to look about for the cor-

rect equation of transformation. Naturally the preferred course from his point of view would be to specify a pair of configurations and from them find by a routine process the equation of transformation from one to the other. Now there is a known method of finding this equation when both of the equipotentials forming the source and sink are part of a closed polygon and the problem is entirely bounded by the polygon. At first sight this seems a severe limitation to the usefulness of the method, but in practice many field problems have as boundaries straight lines that can be treated as closed polygons. The method was first published by Schwarz and Christoffel * independently of each other. The Schwarz-Christoffel transformation opens out the

interior of a polygon in the z plane to the upper half of the w plane. Now the upper half of the w plane is not likely to be the final configuration that is required because the field might not be regular there, but by using a second Schwarz-Christoffel transformation we make another transformation from the con-

figuration wanted-say the X plane-to the same w plane as before.

Then the combination of the two transformations will * See List of References, page 217. 56

THE SCHWARZ-CHRISTOFFEL TRANSFORMATION

change the configuration in the z plane constituting the problem into the desired configuration in the X plane. Note that the interior of the polygon becomes the upper half of the w plane ; the exterior becomes the lower half. The sides of the polygon become the real axis of the w plane.

Since the transformation is conformal the equation has the form w = f(z)

Let the vertices of the polygon ABCD in Fig. 30 (a) transform to

the points A', B', C' and D' on the real axis of the w plane shown in Fig. 30 (b). Open the polygon at some arbitrary point chosen to suit the problem ; in Fig. 30 (a) it has been opened between A and D. This is not at one of the corners of the polygon. If, however, for convenience the opening is made at a corner, that corner takes no part in the transformation. y w=b

A x

/a)

V

-s

w-+ CN. A

C

B

D

U

/61'

FIG. 30. The Schwarz-Christoffel transformation.

(a) z plane (b) w plane. 57

CONFORMAL TRANSFORMATIONS

Let one side of the opening become + oo in the w plane and the other side become - oo, as shown in Fig, 30 (a). In effect we now force open the polygon at the selected point, bend it into a straight line, stretch it to infinity at each end, and then place it along the real axis of the w plane. Consider the line AB in the z plane. Along this line the small length dz becomes in the w plane another small length dw. The angle between dz and dw must remain constant between A and B in the z plane, but at the point B it then changes abruptly to a new value because dz changes in direction at the point B whereas dw does not change at the corresponding point B'. On page

48 it is shown that when the transformation is performed the element of line is multiplied by A, the modulus of dw/dz, and is at the same time turned through an angle fi which is the argument of dw/dz. Therefore the change in angle when transforming from z to w is the argument of dw/dz. This argument

has one constant value between the points A and B, another constant value between the points B and C, and so on. The transformation has to be such that these conditions are fulfilled and they are in fact fulfilled by dz dw

A (w - a )(-In)-i (w-b)(alir)-1(w-c)(rl+r)-1 ,

(6.1)

where the number of factors in addition to A is the number of sides in the polygon. Note that if the polygon is opened at a corner there must be one less factor in equation (6.1) because that corner is not transformed. In this transformation a, 9, y, etc., are the internal angles of the polygon while the numbers a, b, c, etc., are the coordinates of the points A, B, C, etc., in the w plane, i.e. the points corresponding to A, B, C, etc., on the real axis to which the vertices of the polygon are transformed. Notice that in the Schwarz-Christoffel transformation dw/dz must be either zero or infinite at each of the corners. Equation (6.1) satisfies that condition at every corner. The constant A fixes the size of the z picture and since it can be a complex number it also fixes the orientation of the z picture. The other constants a, b, c, etc., must be allocated values that make the lengths of the sides of the polygon have the correct ratios. 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 others have to be left undetermined until the end of the solution. The use of the Schwarz-Christoffel transformation always entails performing an integration. Sometimes the integration is reasonably elementary but not always. For engineers it is eminently desirable to have at hand, and be familiar with, a book of Tables of Integrals ; they should avoid as far as possible wasting time on doing actual integrations. In the present book no integrations will be performed if they can be taken from, or derived from, a book of this nature. Wherever possible we shall refer here to Dwight's Tables of Integrals and Other Mathematical Data,* and give a numbered reference to any integration it is found necessary to use from it. For example the result of an integration will be followed by a reference such as 'Dwight 194.11'.

The details of the procedure followed in Schwarz-Christoffel transformations are best explained in an example. The trans-

formation is usually applied to configurations in which the angles 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 be transformed. For this reason it is desirable to consider transformations of each type in turn, beginning with the simplest,

viz. the transformation of configurations containing no right angles whatever. * See List of References, page 217.

59

CHAPTER 7

Configurations With No Right Angles Electric Field at the Edges of Two Capacitor Plates

Let us examine as a first example of a configuration with no right angles the electric field between the edges of two capacitor plates. Remember, however, that the same configuration and transformation could apply to other problems. Let A and B in Fig. 31 be the edges of two capacitor plates that are parallel and distant d apart. They are assumed to extend an infinite distance in the direction to the right as indicated in Fig. 31 and

to a sufficient distance in the direction perpendicular to the

plane of the paper to make the problem in effect twodimensional. A

B

Fic. 31. Edges of two capacitor plates.

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

Ac

D

d E

B

Fia. 32.

F

Representation of two capacitor plates. 60

CONFIGURATIONS WITH NO RIGHT ANGLES

not regular the second step is to find the equation that transforms two parallel plates of infinite length in all directions to the upper half of the w plane. The combination of the two equations of transformation gives the required solution. The regular field 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 of

Fig. 32 to the w plane, the configuration in Fig. 32 does not appear to be a closed polygon. But it is possible to imagine that the points D and E are joined together and that the points C and F are also joined together; the figure then becomes a degenerate 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 open the polygon at the corner CF. Then, as explained on page 57, this point will not be transformed. Note that the interior angle at A and B is 360° while the interior angle at the corner DE is zero. We now have to allocate values to w in the z plane and the chosen values are shown in Fig. 34. Since the polygon is to be opened at the corner CF, the point C is made w = oo and the point F made w = - oo. Then the symmetry of the figure suggests that w = 0 should be the value at the corner DE. w=+00 C

W-1

A

D

W=0 E F

Fic. 34.

Values of w as points in the z plane. 61

CONFORMAL TRANSFORMATIONS

With regard to the values of w at the corners A and B, it happens that they can be chosen arbitrarily in this example. The rule is that there must be one unspecified value of w for every dimensional value in the z plane more than one. For example, suppose there are two independent dimensions in the z plane giving one ratio ; then there must be one unspecified value of w which has to be called, say, w = a. In general when

the values zero and infinity have been assigned, additional values of w can be specified at two corners only. In this particular example there is only the distance between the plates specified and hence no ratios. Therefore there need be no unspecified values of w. But we have to ensure that the value 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 lie between the values at E and F ; hence w must be negative at B. The expressions obtained will be simplest if the values chosen are w =1 at A and w = -1 at B as shown in Fig. 34. Then the w plane will appear as shown in Fig. 35. Notice that although

the origin in the iv plane has now been determined, it is not necessary to fix the origin of the z plane yet. One reason for choosing the corner DE as the origin in the w plane is that it is a

point of potential discontinuity ; it is nearly always advantageous to choose such a point for the origin. Y

F

B

D E

A

0

-1

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

Since there are now three corners in the z plane to be opened

out there must be three factors besides the constant in the Schwarz-Christoffel equation of transformation. Thus the equation is dz

dw -

A (w - a )

(al-)-I

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

62

(7.1)

CONFIGURATIONS WITH NO RIGHT ANGLES

In equation (7.1) a is the point B where w= - 1, and a is the angle 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 also zero ; hence in the equation b = f = 0. Finally, c is the point A where w =1 and the angle is again 360 °, giving c = I and y = 21T. Therefore equation (7.1) becomes dw

= A(w-1)w-I(w+ I) A(w2-1) w

Therefore z

Ai (w- l)dw w

.

.

.

.

(7.2)

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 B can only affect the position of the origin in the z plane, and since

this is not yet fixed we can make B zero now and find the position of the origin later. Determination of Constants

The next step is to find the value of the constant A. In this particular example there are two ways of establishing the value of A, and since either may be used in subsequent examples both methods will be described now. One very effective method can be used where there is a corner or are corners in the z plane at plus or minus infinity. In the example under consideration one of the corners is at z = + oo where w = 0. Now when crossing from one plate to the other in a direction perpendicular to the plates the distance traversed is d in the z plane. In the to plane the corresponding path traversed is a semicircle centred on the origin. The nearer we cross to the point z = cc the smaller is the radius of the semicircle in the w plane although the distance traversed in the z plane remains equal to d. Therefore as the point of crossing approaches the corner, z-- oe and r-,0 where r is the radius of the small semicircle. 63

CONFORMAL TRANSFORMATIONS

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

dz = -jd

.

.

.

.

.

(7.3)

.

.

(7.4)

In terms of w, from equation (7.2)

Al (w-w) dw= -jd

where the path L is a small semicircle of radius r. In polar coordinates w = reie

and

dw = jrei° dO

Hence equation (7.4) can be written

A fv (reJe -1 a-JO) jrele dO = - id r

J

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

AJ' (-j) dO

= -jd

Performing the integration jA7r = jd whence

A=d/rr

However, in this example the other method is quicker. Starting from

z = A(w2-log w)

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

w = 1, z=JA w = -1, z = ZA{1- 2log (-1)} = JA (1- 2j,r) Since we have not yet found the origin in the z plane we do not know the value of z at these points, but we do know that the distance between the points in the z plane is jd. The difference 64

CONFIGURATIONS WITH NO RIGHT ANGLES

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 used where, when a corner of the polygon is at z = ± oo, the value of w there is not zero but is itself plus or minus infinity. When

this happens, as the path from one side to the other in the z plane approaches the corner at infinity, so the radius of the semi-

circle in the w plane gets larger and larger and in the limit becomes infinite.

If this large radius is called R we can write

z = f(w) w = Reje dw = jReie dO

Hence

B

f(Rei9)j Reie d6 = - id

J Then the constant B can be found by letting R--co and then performing the integration. An example of the procedure will be found on page 98. The Second Transformation

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

w in the t pla.ue.

5

65

CONFORMAL TRANSFORMATIONS

between the plates. Here again there are no ratios of dimensions. The point of potential discontinuity is at either end ; it is therefore convenient to put w = 0 at one end and w = oo at the 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 dw

= A(w-a)

(Jw)-i

where now a = 0 and a = 0 so that

dt=`4dw w Hence by integration (Dwight 82.1)

t = Alogw+B where A and B are new constants. The value of B once again only affects the position of the origin in the t plane and since that 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 of A which corresponds to a distance d between the plates in the t plane (thus making the distance the same as in the z plane), or we can make A unity and find what distance between plates in the t plane corresponds to the distance d in the z plane. For the present we shall take the second course and put A equal to unity so that the equation of transformation becomes t = log w or

w = et . . . . (7.6) The direct transformation from the z plane to the t plane can now 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. 34 to that of Fig. 36. 66

CONFIGURATIONS WITH NO RIGHT ANGLES The Axes in Both Planes

Since equation 7.7 gives a direct relation between the z and t planes, the intermediate w plane can now be ignored. To fix the origins and axes in the two planes proceed as follows. In the t plane, Fig. 36, the origin can be set anywhere. The

imaginary axis must clearly be perpendicular to the plates through the origin and it is convenient to place the origin on the 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 position of the axes in the z plane it is necessary to find the minimum value of z in terms of t by differentiation. From equation (7.7) dz dt

_ d (201-2) 27r

This is zero when e2t is unity, i.e. when t is zero. Differentiating again gives d2z dt2

_

d (4e2t) 27r

and this is positive for all values of t. Therefore the value of z at t = 0 is a minimum value. Hence the minimum value of z is given by equation (7.7) when t=0, i.e. when z = d/27r. Thus the 67

CONFORMAL TRANSFORMATIONS

edge of one of the plates in the z plane lies a distance d,f2rr along

the real axis from the origin. The next thing to find out is whether that plate from which the origin is fixed lies along the real axis of the z plane. Let the real and imaginary parts of t be p + jq as marked on Fig. 37. When t is wholly real q is zero and t = p. Hence from equation (7.7) d

z=

2Tr

(ezp- 2p)

and therefore z is also wholly real. As p tends to infinity so does z ; furthermore as p tends to minus infinity z again tends to plus infinity. This establishes that the plate whose edge in the z plane is distant d,12ir from the origin lies along the positive real

axis as shown in Fig. 38. In that picture the upper plate is shown as the plate concerned but we have now to prove that the upper plate is the one lying along the real axis. Y

d 2n

A

1

M

d

-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

CONFIGURATIONS WITH NO RIGHT ANGLES

We already know that when q is zero z gives the edge of the plate 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 a distance d below the real axis of the z plane and therefore it is the 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, and

the corresponding point B in the t plane lies on the imaginary axis (p = 0) and at a distance 7r from the origin. This is shown in Fig. 37. In the t plane the upper plate lies along the line

t = p+j7r From equation (7.7) this corresponds in the z plane to

z=

dd (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 d below it as shown in Fig. 38. As p tends either to plus infinity or 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 of Fig. 38. The origins and axes in both planes have thus been determined and further problems can now be considered. Equipotentials and Stream Lines

Usually there is a specific problem to solve that does not en-

tail drawing the equipotentials and the stream lines. For instance, in the present example it may be required to find what increase in capacitance is caused by the fringing of the field at the ends of the plates. This increase can be calculated from the 69

CONFORMAL TRANSFORMATIONS

equation of transformation without going to the trouble of calculating and mapping the equipotentials and flow lines. But if for other reasons a picture of the field distribution is required the procedure is as follows. Equation (7.7) gave (e2t - 2t)

z=

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

=2

(et - t)

by putting in real and imaginary parts

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

d 21T

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

Separating real and imaginary parts d

x = r (eP cos q-p) 27

y _=

d (epsinq-q)

Notice that in changing from 2t to t we have made the distance between 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 on in steps of 0.5 to as high a value of p as is convenient. Then for each value of p (i.e. treating p as constant at that value) give q the 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 a straight line parallel to the imaginary axis. Similarly, for each value of q the values of p give a straight line parallel to the real axis. Therefore in the t plane the field is perfectly regular. 70

CONFIGURATIONS WITH NO RIGHT ANGLES

In the z plane there is for each value of p and q a value of x and y which can be calculated from equations (7.8). For

example take the value p =1 and let the distance d be 27r. Then equations (7.8) become

x = 2.718 cos q- I y = 2.718 sin q-q From these equations Table 6 can be drawn up. TABLE 6

Field outside two capacitor plates

42

0

46

43

a=cosq

1

0.866

0.500

0

b= sin q

0

0.500

0.866

1

1.354

0.359, -1

0.830

1.307

q

x=2.718a-1

1.718

y=2.718b-q

0

7a/6

q

47x/3

342

1.147

543

27r/3

546

-0.500 -0.866 0.866

0.500

0

0260 -1.264 -3.142 114r)'6

21r

-0.866 -0.500

0

b=sinq

-0.500 -0.866

-1

x=2.718a-1

-3.354 -2.359 -1

y=2.718b-q

-5.024 -6.543 -7.4311 -7.590 -7.1196.283

01866

1

-0866 -0.500

0

0.359

-1

-2.359 -3.354 -3.718

a=cos q

0.500

7r

1.354

1.718

These values give the field outside the plates. The field inside the plates can be calculated by starting with negative Table 6 gives the points for one curve of p = constant. The regular field is as shown in Fig. 39. The actual field is shown drawn in Fig. 40. values of p.

Density of Charge

The strength of the field depends on the difference of potential between the plates and not on the absolute values. Hence the solution is not restricted in any way if one of the plates is considered to be at zero potential. Considering now the t plane, if 71

CONFORMAL TRANSFORMATIONS

0 Fm. 39. The regular field.

P

t plane.

Y

x

FiG. 40. The actual field.

z plane.

the potential of plate A is called zero and the potential of plate B is called V, the potential at any point is given by V i2

CONFIGURATIONS WITH NO RIGHT ANGLES

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

¢+Jo =

X=

And

V

-- (p+jq) Vt

Let the density of charge be denoted by v. On the plate A, a is equal to the normal component of the electric field, and this component is proportional to the rate of change of ti in a

direction perpendicular to the plate with a negative sign. Therefore o =

in M.K.S. rationalised units, where aojan is the rate of change

mentioned and co is the primary electric constant equal to 8.854 x 10-12.

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

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

at I Eo

an az

But 00

00

On

aq V 21r

73

CONFORMAL TRANSFORMATIONS

Furthermore since z=

_

dz

dt

d ?7r

d

(et - t)

2iT

let-11

=

1epelq-

d

_ _IeP(cos q+ j sin q) -1 1eP-1I

because on the surface q is zero. Hence V 21r Cr =

1

27r d IeP - l l E0 VE0

d(eP-1) when p is positive and VE°

d(1 -eP)

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

V

Eo = ao

(say)

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

The charge on the outside of the plates tends to infinity making it impossible to calculate the increase of capacitance directly.

However, an approximation to the effect of the 74

CONFIGURATIONS WITH NO RIGHT ANGLES ad F-o

3.0

2.5

20

1.5

Inside

1.0

0.5

Outside 0

0.25

0.5

FiG. 41.

1 25

1-0

0 75

15

175

20

x d

Distribution of charge density for V = 1.

charge on the outside has been made by Prof. G. W. Carter who obtains the total effect of the fringing as follows. From above, the charge density on a plate is given by a°

a = IeP-1I where ao is the density far into the plates. The extra charge due to fringing on the inner surface is therefore w

(a - a°) dx

Distances x along the surface of the plate are given by d

x=

so that

2 (ev --p)

dx=2

(eP-1)dp

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

f

°

°°

I Ad

a0(1-ep217(eP-I)dp =

75

a°d 27r

CONFORMAL TRANSFORMATIONS

Thus the extra charge due to fringing on the inner surface is equivalent to widening the distance between the plates by the amount d/21r. The extra charge on the outer surface is

r0 adx Jd/2n

00

(To

d

(ell -1)dp = 1, o eP-1 2ir

and r0°

= 27rfo dp=o0 However, the form of the integral shows that the charge is so distributed that each addition of unity to p brings in a charge corresponding with a plate width of d/27r in the uniform part of the field. From the formula for x in terms of p we obtain these values : TABLE 7

Charge on the outer surfaces of two capacitor plates

Extra charge, (a0d/27r) times:

1

2

3

4

Width of plate back, d times:

0.27

0.86

2.72

8.06

5

22.8

If, therefore, the plate is not of infinite width making a capacitance correction impossible to calculate but has a finite width of (say) 16.12d, we can approximate closely to the additional capacitance by saying that the fringing adds an effective width of d/27r for the charges inside the plates and four times d/27r for the charges outside the plates. Distance between Plates in the t Plane

In the transformation from the t plane to the w plane given on page 66 it was stated that the constant A could be determined in such a way as to make the distance between the plates the same in the t plane as it was in the original z plane. That

distance was called d and, putting in the constant, the transformation is

t = A log w Now the point B in Fig. 38, page 68, is given by

z=

dr-j 76

CONFIGURATIONS WITH NO RIGHT ANGLES

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

jd = A log (-1) = jA7r

Therefore A=d,7r and the transformation becomes

t=

d

log IV

.

.

.

.

.

(7.9)

W

Slotted Plane Surface

For another example of a transformation with no right angles

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

2a

Fir.. 42.

31.

Slotted plane surface.

slot be denoted by 2a. Some of the field must penetrate the slot and therefore the configuration to be transformed is that shown

in Fig. 43. This figure can be made the z plane. 77

CONFORMAL TRANSFORMATIONS y

w=-A

E

B

W,-I

-a

Fic. 43.

F

0

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 in Fig. 42 become rectangular squares in the w plane. Considera-

tion of the rough picture of Fig. 42 suggests the values of w We wish in effect to bend up the portion BE through 180° so that it lies along the axis, bend up the portion CF so that it too lies along the real axis, and then separate the two portions until the point B meets the point C. The whole configuration in the z plane then becomes the real axis of the w plane, and the field in the new plane is perfectly regular. Having thus assigned values to w at A, B, C and D we have now to give values to w at the edges of the slot, points E and F in Fig. 43. Since there is only the one dimension a and hence no ratios in the z plane the values of w at the points E and F are at our disposal provided the value at E is negative and the value at F is positive. It is therefore desirable to make w unity at these points, i.e. w =1 at F and w = -1 at E. The position of these points in the w plane is shown in Fig. 44. shown in Fig. 43.

V

E

B

FIG. 44.

C

0

-1

1

The unbroken surface. w plane. 78

D

F

U

cc

CONFIGURATIONS WITH NO RIGHT ANGLES

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

dz_ dw - A(wb)cs/-)--1(wc)(Y1n)-1(w-d)csin>-1 where now b = -1, c=0, d=1,

P = 2ir

y= -17 8=2ir

Therefore dz

dw = 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 necessary to find what value of B matches those origins. At w = 1, z = a,

and putting these values into the equation of transformation gives

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

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

z = 2(w+w)

.

.

.

.

.

(7.10)

Density of Now across the Slot

Suppose that in the original problem the horizontal lines of Fig. 42 represented stream lines and it is required to find what the distribution of density of flow is across the line of symmetry through the slot. In other words what the distribution of

stream line density is on the imaginary axis of the z plane Fig. 43. Call this density E and let the distance a be 2 units. 79

CONFORMAL TRANSFORMATIONS

Now the density is given by Idwl

multiplied by the appropriate primary constant. Since we are not concerned with particular units here nor even with the kind of field, this constant is made unity. From equation (7.10) by putting a = 2

w=

1/2

Z2

Z

2+I4-1

Therefore

dwl

E_

dz

2I1+ (z2 z4)1/2I But since the values of the density are those along the imaginary axis of the z plane, put x = 0.

Then

z = jy 12

Ey =

jy 1+j(y2+4)1/2

(7.11)

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

Now as y-oo, E1. Therefore unity is the value of the undisturbed flow.

Again as y-> - oc, Z y-->0 as it should E 1.0

09

0.6

0.2

4

3

2

1

0

1

2

3

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

4

5y

CONFIGURATIONS WITH -NO RIGHT ANGLES

By taking various values of y-positive, negative and zerowe can determine from equation (7.11) how the density of flow .E

changes along the y axis. A graph such as that shown in Fig. 45 can then be plotted. Plotting the Stream Lines and Equipotentials

To plot the stream lines and equipotentials start from equation (7.10) and replace z and w by their real and imaginary parts thus : 1 x+]y = acu+jv+ u+jv, a u-jV

- 2(u+jz+u,2+v2

Therefore

u a x=,,+ 2 u2 +v2

y=

v u2 + V2

2(V_

(7.12) }

V , , ,

4 t ,

I 1

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 in the w plane plotted from equations (7.12) for a=2. Fig. 47 shows the lines obtained for the same values of u = constant and v = constant in the z plane. 6

81

CONFORMAL TRANSFORMATIONS

FIG. 47.

The actual field.

Transformation of a Point Source

It may happen sometimes that the source of the field can be idealised to a single point source, i.e. a line source in the threedimensional field. Such sources are relatively easy to transform : in fact a point source has already been dealt with on

page 32 when the transformation X=log z was discussed. Consider first a z plane consisting of two parallel straight lines

representing a strip of conducting material carrying current. The strip extends indefinitely in both directions as shown in

waO

W. +a*

FIG. 48.

/el (a) Transformation of a point source. (a) z plane (b) to plane. 82

CONFIGURATIONS WITH NO RIGHT ANGLES

Fig. 48 (a).

Let there be a uniform flow of current in the direc-

tion shown by the arrow so that there are equipotential lines perpendicular to the lines shown. There is clearly a current source an infinite distance away in the negative direction. By choosing values for w as shown in Fig. 48 (a) the configuration can be opened out into the w plane as shown in Fig. 48 (b) by using the transformation z = A log w . . . . . (7.13) The stream lines have become radii in the w plane and the flow of current is outwards from a single point, viz. the centre of the system. The equipotentials have become semicircles centred on the origin in the w plane. Now the current flow from a line source whose cross-section is located at w = 0 must extend in all directions, with the result

that the current flow in the upper half-plane is only half the total. Allowance must be made for this. Hence the transformation of equation (7.13) transforms the field due to a point source into a uniform field where twice the current in the uniform field has to be taken and the constant A has to be determined from data given in the actual problem. Assume for the present that A is unity. Then

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

x=log y

y=B

and these are the equations required. Collinear Source and Sink

Equation (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 point source located at the point w= - a. Furthermore

z = -log (w - a) gives the transformation of a point sink (owing to the negative 83

CONFORMAL TRANSFORMATIONS

sign) located at the point w = a. posed so that the equation

The two fields can be super-

log w-a w+a gives the transformation of a point source and a point sink z=

which 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-Christoffel transformation from the z plane of Fig. 49 (a) to the w plane of Fig. 49 (b). Let the values assigned to w in the z plane be those marked in Fig. 49 (a) and let the configuration be opened up from the break shown in the diagram. If the current is flowing from right to left there is in the w plane a point source at w = a

and a point sink at w = -a. By the Schwarz-Christoffel theorem the transformation is given by dz

dw =

A(w-a)-'(w+a)-1 A W2-a2

Therefore (from Dwight 140.1)

z=

A

-a

w-a

log w+a+B 84

(7.14)

CONFIGURATIONS WITH NO RIGHT ANGLES

Polar coordinates can be used by taking radii from to = a and

w= -a so that .w = r1e,Bi +a and

w = r2ejea - a Suppose that in equation (7.14) the origin is chosen so as to make B zero and the potential is such as to make A unity. Sub-

stitution for w in that equation gives 1

z=

(log rlei°' -log r2ejel)

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

-

1 rl x = 2a log r2 y = 01-02 In the z plane the field is regular and the lines of flow are

given by y = constant, i.e. by 01-02 = constant

.

.

.

7.15)

in the w plane. Fig. 50 shows that 01- B2 is the angle APB where 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 geometry

that if the point P is moved in such a manner that the angle subtended by AB at P is kept constant., the locus of P must be P

0 FiG. 50.

B

Collinear source and sink. 85

CONFORMAL TRANSFORMATIONS

a circle through the points A and B. Therefore the various values of the constants in equations (7.15) give various circles

all passing through the points - a and +a as shown in 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 Fig. 49 (b).

rl

- = constant r2

Referring back to Fig. 50, it is shown in books on coordinate geometry that when A and B are fixed points and P moves in such a manner that the ratio PA/PB is kept constant, the locus of P must be a circle whose centre is on an extension of the line AB or of BA. Different values of the constant give different circles as shown in Fig. 49 (b). These circles form the equi-

potentials due to the source and the sink. Thus the stream lines and equipotential lines are intersecting circles whose centres lie along the u and v axes in the w plane.

86

CHAPTER 8

Configurations with One Right Angle Transformation of a Quadrant

On 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, provided real and imaginary parts of z were confined to positive values, it represented a transformation from a quadrant to the upper half of a plane.

We can now approach this matter from the other end and suppose that it is required to transform a quadrant in the z plane

to the upper half of the w plane. We can use the SchwarzChristoffel method to show that w = 22 is the transformation required. The z plane is shown in Fig. 51 (a) and since the transformation is equivalent to bending the imaginary axis back

until it lies along the negative end of the real axis, it is convenient to make the values of w those shown in Fig. 51 (a). Note that the origins in the two planes are at corresponding points. W-00 Y

V

(b)

(a)

Z.0

O

j.+m U -00

w.+co

W.01

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

Since there is only one corner to be straightened out dz

dw

= A(w-a)(a/') 1 87

CONFORMAL TRANSFORMATIONS

and a = zr. Hence dz = Aw-1j2 dw and by direct integration (Dwight 181.1) where a = 0

z = 2Aw1/2+B

But since when z is zero w is zero, B must also be zero. Therefore w = 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 this example because the complete analysis was done in Chapter 5 where a was taken as unity. Fringing of the Field at a Pole End

Another problem, this time a more practical one, requiring the transformation of only one right angle is that of the fringing of the magnetic field between the end of a pole of an electric machine and the end of the armature core. The problem is illustrated by the section shown in Fig. 52 where the length of the airgap between pole and armature is shown as g. Pole

I

-----

-- -- -

9 ----

------ --

L Armature core

Fic. 52. Fringing problem.

At first sight this looks like a problem involving the transformation of two right angles, but the broken line in Fig. 52 indicates a line of symmetry between pole and core and this means 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 right angles from two to one. 88

CONFIGURATIONS WITH ONE RIGHT ANGLE w.-Cc iy

(a)

U 9/

W.0

w=+OD

O

X

V

(b)

z=j ou

Z. Go -1

0

a

Z

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 convenient points for w = 0, w = oo and w = - co are readily chosen at the points shown in the figure. Clearly we want to straighten

out the configuration by making the line of symmetry the positive real axis and the pole end the negative real axis in the w plane as shown in Fig. 53 (b). Since again there are no ratios to consider but just the single dimension g, the value of w at the corner in the z plane can be given any arbitrary value between

zero and minus infinity. The equation of transformation appears in its simplest form when the magnitude of w is made unity at the corner, i.e. w = -1 there. The stream lines in the z plane should therefore become semicircles in the w plane, centred on the origin. In pulling out the configuration to a straight line two angles have to be straightened and hence dz

dw

= A(w-a)(arn>

where a=-I, a= 37r/2 b=0, P=O. 89

1

CONFORMAL TRANSFORMATIONS

Notice particularly that the configuration was straightened out by making the outside of the corner become the upper side of 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)

z = A 2(w+1)1/2+Iog (w+1)

12-

11 +B

(w+ 1)1/2+ 1

But since the origin in the z plane can be fixed quite arbitrarily B can be made zero. Furthermore, to suit later developments the equation can be re-written with the log term inverted and subtracted, 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 = -jao

W = + oo,

z = + co

=A(2-logoc) _ -co

w= 0,

These are the values of w shown in Fig. 53 (a) and the equation of transformation (8.2) is thereby checked. But the value of the constant A has still to be determined. For this the method of page 63 can be used. In crossing the air gap near the point

w = 0 and z = - co, the distance traversed in the z plane is jjg while in the w plane it is a semicircle of small diameter centred on the origin. Thus if w = reie

jjg =

fiu

dz

0

= A fo 1e-ie(rei° r = jA

+

1)1/2jrei8 dO

o (reie+ 1)1/2 d8 J0 90

CONFIGURATIONS WITH ONE RIGHT ANGLE and as r--->o

?j9=jA

fodO 0

= jATr

Hence

A = g/2Tr

.

.

.

.

(8.3)

.

Putting this value of A into equation (8.2) gives z = 2TT

= 9 =

2(w+ 1)1/2-log (w+ 1)1/2+1]

(w+1)1/2-1

E2-

2 -Tr

1)1/2-log

{(w+

1)0/2+

1}21

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

(8.4)

Note that when iv= -1, z .=

=

log (-1)

2n 29r

07T)

ij9 Thus the origin in the z plane is on the line of symmetry and in line with the pole edge. Hence the pole edge lies along the y axis. The stream lines become semicircles in the w plane. In the X plane shown in Fig. 54 the field is regular, and equation (7.9) q,

0

W. 4 o

Fia. 54. Regular field plane. 91

CONFORMAL TRANSFORMATIONS

on page 77 shows that for unit distance between the plates in the X plane the transformation is given by X = I log w and hence w=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 the original z plane to the X plane where the field is regular. Distribution of Flux Density

This particular problem is typical of others in which the boundaries in the original plane extend to infinity in various directions. Because of this the total flux passing from the pole end to the end of the core cannot be determined because it must turn out to be infinite. Of course in practice the configuration

cannot extend to infinity in any direction whatever but must be bounded in all directions. If finite lengths are defined for the boundary surfaces (or the electrodes) in the problem there can no longer be a line of symmetry in the airgap and both right angles will need to be straightened out. Transformations involving 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 obtained that can be used as an aid to judgment. For instance stream lines and equipotentials can be drawn. Write x + j y as before for z and O + j i for X. Then by putting O = a constant and 0 a constant in turn, values of x and y are determined that enable us to draw the stream lines and equipotentials in the z plane. 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 separating real and imaginary parts, values of x and y are obtained that 92

CONFIGURATIONS WITH ONE RIGHT ANGLE

provide points in the z plane. When these points are joined, the line formed is a stream line. Other stream lines are obtained by taking other constant values of . Similarly by putting 0 = a constant and taking various values of 0 for each value of 0, a series of points can be obtained in the

z plane which, when connected up, provide a family of equipotential lines. Another quantity that can readily be found is the density of flow, called the flux density, at any point along the axis of symmetry. Thus a curve can be drawn showing how rapidly the flux density falls away as we recede from the pole end. The procedure is as follows. The magnitude of the flux density at

any point is given by

B=

dX dz

µo

.

(8.7)

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

B=

dx dw dw T z1

2irw

7rwg(w+7)I12

µo

(using equations (8.1) and (8.5))

_

I(w+ 1)-1/2 /Jo

.

.

.

.

.

.

(8.8)

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

and if this uniform value is called unity the flux density at any other point is found from B = I(w+ 1)-1/'-I . . . (8.9) .

If only positive values of w are now accepted z must be wholly real because all values of w between zero and plus 93

I

2.19

2.098

2.050

1

0.976

0.953

-0-522

-0-750

x = d/7r

B.. =1/a

-1-640

-2.357

2a-2b+c=d 0.913

-0-286

-0.90

-1-61

-2-303

-2-996

log w = c

2a

-1-48

0.740

-1-435

0.7173

1.095

0.20

-1-411

0.7055

log (a+ 1) =b

1.049

0.10

- 2b

1.025

0.05

(w+1)112=a

w

0.881

1.414

1.00

0.817

0.050

0.157

2.450

- 0.693

0.707

0.339

1-066

2.828

0

- 1-600J -1-762

0.800

1.225

0.50

Distribution of Flux Density

TABLE 8

0.500

0.923

2.901

4.000

1.099

-2-198

1.099

2.000

3.00

0.333

1.689

5307

6.000

2.079

- 2.772

1.386

3.000

8.00

0.250

2.384

7.490

8.000

2.708

- 3.218

1.609

4.000

15.00

0.111

5.658

17.776

18.000

4.382

-4-606

2.303

9.000

80.00

CONFIGURATIONS WITH ONE RIGHT ANGLE

infinity lie along the x axis as shown in Fig. 53 (a). The x axis is the line of symmetry bisecting the airgap and it is therefore the 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 the curve shown in Fig. 55 can be drawn.

It shows the magnitude of the distribution of flux density along the x axis, i.e. along the line half-way across the airgap.

0.3

0.6

O.4

O.2

-1

Fro.. 55.

0

1

3

2

4

s

r

Distribution of flux density along the axis of symmetry.

From examination of Fig. 55 some idea can be formed of what

approximations can be made. For instance it is possible to judge at what distance from the armature core it is justifiable to ignore the fringing of the field. However, if it is essential to obtain precise answers to specific questions the field has to be precisely delineated. The transformation to the plane is then usually a transformation to a rectangular configuration. Since this configuration contains four right angles the analysis is considerably more difficult. Such problems are therefore left for the later chapters.

95

CHAPTER 9

Configurations with Two Right Angles Flow through a Slit

Transformations to the upper half of the w plane of configurations containing two right angles still lead to elementary functions, usually circular or hyperbolic functions. Consider two plates, each containing a right-angle bend, separated only by a small aperture symmetrically placed as shown in Fig. 56.

k

d

Slit

Fic. 56. Flow through a slit.

The two plates form a trough with a slit in the bottom. Any liquid in the trough will have lines of flow as it pours through the slit. There will also be lines of equipotential intersecting the

lines of flow at right angles. The equipotentials are lines of equal velocity potential. Let the distance between the parallel portions 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 thing to 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 as shown.

Then the values w = oo and w = - oo should be at

z = j oo the negative sign being attached to the left side plate. Since there are no ratios but only the distance d in the configuration 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 whole of the inside of the trough opens out into the upper half of the w plane as shown in Fig. 58.

FIG. 58. w plane.

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

-

dw

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

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

dw = A(w2- 1)-1/2

.

.

.

.

.

.

(9.1)

Hence (Dwight 260.01) z = A cosh-1 w + B 7

97

.

.

(9.2)

CONFORMAL TRANSFORMATIONS

It is convenient to make the constant B zero and accept the origin so determined in the z plane. To find the constant A the method of page 65 can be used whereby an integration is made

from one plate to the other near to z =joo. In going across from right to left the distance traversed is - d. In the w plane the path of integration is a semicircle of large radius R, and as z- co in the z plane R---oo in the w plane. Hence from equation (9.1), since w = Rej° dw = j Rei° dO

-d - A

f

"

o

j Rei8 dO

(R2e2J6- 1)1/2

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

-d=A fojd9 0

=jA,r Hence

A = jd/nr and equation (9.2) becomes z =

w =cosh `` =COs

L cosh-1

w

(9.3)

l d) J d

(_j) (Dwight 654.1)

(9.4)

Now when z= -d, w = cos 7r = -1. Therefore the point C in Fig. 57 where w = -1 is the point x = - d. Furthermore when z is zero, w is unity (cos 0). Therefore the point D in Fig. 57 where 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. The stream lines in the z plane become radii in the w plane while the equipotentials become semicircles in the w plane. We now require the transformation between the w plane and

the X plane where the stream lines and equipotentials are 98

CONFIGURATIONS WITH TWO RIGHT ANGLES

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

X=dlogw IT

.

.

.

.

.

(9.5)

provided the distance between plates in the X plane is also d. This is shown in equation (7.9), page 77. The direct transformation from the z plane to the X plane is found by combining equations (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 that d/ir is unity. The problem as stated on page 96 is to draw the equipotentials and lines of flow in the z plane. Therefore substitute 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

0+1o = log (m+jn)

.

.

.

.

(9.7)

and if

m+jn = re3° r = (m2+n2)1"2 0 = tan-' (nlm.) Then

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) 99

(9.8)

CONFORMAL TRANSFORMATIONS and

sG = tan-'(n/m)

tan-' (- tan x tanh y)

.

.

.

(9.9)

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

cost x cosh2 y+sin2 xsink' y = constant

The expressions can be simplified by using the following relations.

cos2x = (1+cos2x)

(9,10)

cosh2 y = ? (I+ cosh 2y)

sin2x = -(I-cos2x) sinh2 y = (- I + cosh 2y)

l

(9,11)

2

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 found to satisfy equation (9.12). These values plotted in the z plane when connected together give an equipotential line. A whole family of equipotentials can be obtained by taking sufficient values of the constant. Table 9 shows a typical calculation for three values of the constant.

Similarly the stream lines can be determined by giving various values to the constant in 0 = constant. But if 0 is constant 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 a sufficient number of values of the constant. Table 10 shows a typical calculation for three values of the constant. 100

y

cosh 2y = 4 - cos 2x

y

0.882

3

0.659

2

0

y

cosh 2y = 3 - cos 2x

0

1

cosh 2y = 2 -cos 2x

2y

1

0

2x

cos 2x

0

x

0.931

3.293

0.736

2.293

0.374

0.748

1.293

1.032

4

0.882

3

0.659

1.317

2

0

-7T/2

-7r/4 0.707

-7r/4

-7r/8 --

1.115

4-707

0.992

3 7 07

0.827

1.654

!

1.146

5

1.032

4

0.882

1.763

3

1

1.115

4-707

0.992

3.707

0.827

1.654

2.707

-0-707

-1

-0-707 2.707

- 57x/4

-77

-57x/8

--

- 37x/4

-37x/8

- -JI -7r`2

Flow through a slit Calculation of cquipotentials

TABLE 9

1.032

4

0.882

3

0.6559

1.317

2

0

- 37x(2

1

0.931

3.293

0.736

2.293

0.374

0.748

1.293

0.707

- 77x/4

-37x/4 - 77x/8

0.882

3

0.659

2

0

0

1

1

- 21r

-;r

CONFORMAL TRANSFORMATIONS TABLE 10

Flow through a slit Calculation of stream lines -7r/8

x

cot x

- 2.414

tanh y = + 0.4 cot x y

tanh y=

±0.2 cot x y

tank y = 0.1 cot x

y

-Tr/4

- 37r/8

-1.000 - 0.414

- 5Tr/8

- 344 -7Tr/8

0.414

1.000

2.414

0.966

0.400

0.166

0.166

0.400

0.966

2.029

0.424

0.168

0.168

0.424

2.029

0.483

0.200

0.083

0.083

0.200

0.483

0.527

0.203

0.083

0.083

0.203

0.527

0.242

0.100

0.042

0.042

0.100

0.242

0.247

0.100

0.042

0.042

0.100

0 247

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

At x=0,

y = co

x = j Tr,

y=0

x = - TT,

y= 00

From calculations such as those shown in Tables 9 and 10 the

field can be mapped. Some of the stream lines and equipotentials are shown in Fig. 59. Note that this transformation applies equally well to the magnetic field between two plates shaped as in Fig. 56, the potential of one plate being zero and the potential of the other being, say, unity. In that application of the transformation 4 and 0 change roles. The curves

0 = constant give the equipotentials and the curves

¢ = constant give the lines of flow. 102

CONFIGURATIONS WITH TWO RIGHT ANGLES

Fia. 59. Flow through a slit. Equipotential and stream lines. Slot Opening Opposite a Solid Face

There is a much more practical and more celebrated transformation of a configuration containing two right angles. It arises in problems associated with two magnetic faces separated by an airgap where one of the faces is slotted. The configuration is shown diagrammatically in Fig. 60 where g is the length of the airgap, this length being considered to be uniform, and 8 is the width of the slot opening. This transformation is a well-

known piece of work first undertaken by F. W. Carter and published in 1901.* Its purpose is to take account of the e$ects

of having slot openings at the airgap surfaces in electric machines. The results of this transformation have been so successful and are in such widespread use today that it is worth while to treat it rather exhaustively. For the purpose of the transformation it is assumed that the

curvature of the surfaces at the airgap can be neglected and that the depth of the slot has so little influence on the results that it can be considered infinite. In actual electric machines the length of the airgap and the width of the slot opening are both so small compared with the diameter at the airgap that it * See List of References, page 217. 103

CONFORMAL TRANSFORMATIONS

is quite reasonable to ignore the curvature for purely local problems such as those to be dealt with here. More exact methods such as those used later in this book show that, for the depths of slots found in practice, it is justifiable to treat the slot as if it had infinite depth. The usual assumption is also made that 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 the slotted surface, i.e. the lower line, can be treated as another equipotential at zero potential. But for the presence of the slot opening the magnetic field would be uniformly distributed along the airgap. In practice the problem is never really that of a single slot but that of a succession of equal slot openings. Now the relative dimensions of slot width, tooth width and airgap length are sometimes, though not often, such that it is unwise to simplify the problem to that of a single slot. The problem of a succession of equal slot openings will be dealt with fully in Chapter 17 and it will then be interesting and instructive to compare the results with those obtained in this chapter and determine which aspects of the problem can justifiably be treated as the problem of a single slot opening and which can not. The question of a slot of finite depth, although not dealt with in Chapter 17, can readily be analysed by the methods of that chapter and it is found that the field does not penetrate very far into the slot opening. In fact the penetration is so slight that with normal depths 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 may 104

CONFIGURATIONS WITH TWO RIGHT ANGLES I

w=O

A

B W= b

0

of

w= -00

9 T

S

C w.1

FiG. 61. z plane.

seem rather drastic, it is not so in fact, and the configuration shown there can be made the basis of the z plane shown with certain values of w in Fig. 61. But there are several ways of opening out this configuration into the upper half of the w plane.

However, since it will be advantageous to have the resulting stream lines in the w plane semicircles it is best to make the origin of the w plane where the real part of z is at minus infinity.

Then naturally where z = + oo, at the upper surface make w = - oo and at the lower surface make w = + oo. These are the values of w shown in Fig. 61. There are three more corners, shown as A, B and C, where values of w have to be fixed. Only two of these three corners can have arbitrary values assigned to them, so that one value of 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 of wv at each of the corners A, B and C must be positive, for the points lie between the points w = 0 and w = oo. The following choice of values can be made : at A let w = a, at C let w = 1, and at B let w = b. Note that the value b is not independent of a ; only one value is independent because there is only one definite

ratio in the z plane. Therefore there is a relation between a and b that must be determined later on. Notice also that a lies between w = 0 and w =1 and is therefore less than unity. From the symmetry of the configuration it might be expected that b is 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 point where the imaginary part of z is minus infinity. 105

CONFORMAL TRANSFORMATIONS

By the Schwarz-Christoffel equation the transformation from the z plane to the upper half of the w plane is given by dz _ A(w-a)112(w-b)'12 w(w -1) dw

(9.13)

V

z.w+jq

z. -co

z;-jco

0

-CO

a

1

z-0 b

tz.*ao 4

Fm. 62. w plane.

It is sometimes an advantage to determine some or all of the constants from the equation of transformation before the integration is attempted. This will be the procedure here. The w plane is shown in Fig. 62.

Determination of Constants

The value of A can be found by using the method of page 65 of integrating along a large semicircle in the w plane. In the z plane, Fig. 61, the distance between surfaces as z approaches

infinity is the constant value g so that, taking into account direction, the value of the integration across the airgap is jg at all points after w = b. In the w plane the path of integration is along a large semicircle radius R where B can be made as large as we please. Then, as before, w = Reie

and dw = jReje dO

Substitution into equation (9.13) gives f dz = fIT A(ReiO-a)1/2(BeJB-b)"2 jReia d9 o

.Reie(ReJO-1) 106

CONFIGURATIONS WITH TWO RIGHT ANGLES

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

-

f

=

I

A(ReiORei8)1/2 Reia d8 ReJeReie

jA dO

I

0'ff

=jiA But in the z plane f dz is jg and therefore A = g/,;T

.

.

(9.14)

.

A similar investigation at the other end of the configuration in the z plane where w = 0 gives the relation between a and b.

This is the method described on page 63.

Here again the

integration in the z plane across the airgap remains constant as z approaches - oo and the value is again jg. In the w plane the path of integration is a small semicircle centred on the origin and of radius r where r can be made as small as we please. Then as before w = reJ0 dw = jreJ° dO

f dz = f o

A(reJ°-a)1/2(reie-b)1/2 . reje dO reie(reJe - 1)

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

f dz =

rv A(ab)1/2 f oo

.

r &B(-1)

J0

-

(ab)1/217.

-jg(ab)1/2

But f dz in the z plane is jg and hence (ab)1/2 = - 1

ab=1 b = 1/a 107

.

.

.

.

.

(9.15)

CONFORMAL TRANSFORMATIONS

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 the separate symbols. Then substituting for A in equation (9.13) we obtain dz = g (w-a)1'2(w-b)112 dw IT

w(w-1)

Hence dw z=.y f (w-a)1/2(w-b)1/2 w(w-1) 7r

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

w-b

yw-a

so that

w= ape-b p2-1

.

Then

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

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

dw

w(w-1)

_

2p(b -a) dp

(ape-b){(a-1)p2-(b-1)} 2(b -a)

a(a-1)

p/ dp

-11

(p2--b2)I p2-ba-1/J

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

since from a= I lb, b-1 a-1 = -b 108

(9.16)

CONFIGURATIONS WITH TWO RIGHT ANGLES

Furthermore from equation (9.16)

w-a = ape-b-ape+a p2-1

a-b p2 - 1 and

vl - h = ape-b-bp2+b

pZ-1

(a- b)p2

p2-1 Hence

(w-a)(w-b)

(a-b) (a-b)p2

p2-1 p2-1

and

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

p2-

Combining these results z

_ 2g (' (b-a)2p2 dp = it J a(1-a)(p2-1)(p2-b2)(b+p2)

and substituting 1/b for a

z = 2g (' (b+1)2(b-1)p2dp (9.17) 7T J (1-p2)(b2-p2)(b+p2) Clearly, when equation (9.17) has the integrand split into partial fractions the resulting forms will be found in Dwight's Tables.

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)

-A- B+C = 0 p°: b3A+bB+b'-C = 0 i.e.: b2A+B+bC = 0

Equating coefficients of p4 :

109

.

.

(a)

.

.

(b)

CONFORMAL TRANSFORMATIONS

Adding (a) and (b) :

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

C = (1-b)A

Substitute this value of C in equation (a,):

-B-bA = 0 Hence

B= -b.4

Now 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 divide through by the factor (1- b) : bA+bA+(1+b2)A = (b+1)2 Hence, A =1, B = - b, C = (1- b). Therefore z=

2g

7

(' dp _f bdp _f (b -1) dp b+p2 J 1-p2 J b2-pa

Using Dwight 140, 140.1 and 160.01,

z = g[log I i +p -log

p

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

Ib-p

b

(9.18)

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

z=

g

E

log 1-log 1- 2(b

- 1

tan-10

b

= 0

Thus the origin in the z plane occurs when p is zero.

But, from

equation (9.16), when p is zero w=b; this means that the origin in the z plane is at the point B in Fig. 61. Finding the Value of b in terms of s/g

The value of b depends on the ratio s/g. Since

p2 _ w-b w-a 110

CONFIGURATIONS WITH TWO RIGHT ANGLES

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

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

gb-1 IT

g(b - 1)

,/b But in the z plane, Fig. 61, w = a at the point A where z Hence g(b-1)

s.

,/b

from which b-1

/b

= 9s

(9.19)

In any specific problem it is best to proceed with the quantities a and b and convert them after the solution into terms of 8/g by equation (9.19). The Distribution of Flux Density

In the X plane, Fig. 63, the field is uniform. It was shown on

page 77 that if the distance between plates in the x plane is unity and the potential difference is V, the transformation from the w plane to the X plane can be written

x=

Ylog w 7T

W= - oo

W. 0

0 Fio. 63. X plane. 111

0

W:}o0

CONFORMAL TRANSFORMATIONS

As before the flux density at any point is given by µo Therefore for a potential difference across the airgap of V the flux density B is IdX/dzI.

B=V

dX dw Uo

dwdz

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

V

(w-1)

9

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

µo

.

(9.20)

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

(9.21)

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

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

dB dw

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

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

This is zero when

(w-1)

1

1

w-bMw-a

i.e. when

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

2

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

(a+b+2)1JY 112

.

.

(9.22)

CONFIGURATIONS WITH TWO RIGHT ANGLES

This expression for the ratio of minimum to maximum flux density enables us to determine the amplitude of the ripple in the flux density wave caused by a slot opening opposite a pole face. The ripple in the wave sets up in the pole face eddy current losses that are proportional to the square of the amplitude of the ripple. Hence in order to calculate these losses the ratio Burin/Borax is required. The Flux Density Curve

Sometimes it is necessary to know the shape of the flux density distribution curve because this is also the shape of the curve

of variation in time of the density of the eddy currents at any point on the pole face. Usually there are pronounced harmonics in this curve, harmonics that make a large contribution to the total eddy current loss. The extra loss caused by the eddy currents can be calculated provided the amplitude of the various harmonics are known and these amplitudes are found by making an analysis of the flux density distribution curve.

The problem is even more important in such apparatus as inductor-type eddy current brakes and couplings, for they have a similar magnetic structure and operate entirely by virtue of

the ripple set up in the flux density wave by slot openings. Since the air gap in such machines is relatively very short the amplitudes of the harmonics are correspondingly large. The harmonic currents so caused assist in producing torque and therefore have to be taken into account. To draw the flux density curve take various values of w and

from equation (9.20) determine B for each value chosen, choosing real values only. Then from equation (9.16) determine 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 be drawn. The values of w chosen need only lie between zero and -1 for this reason : It was shown on page 112 that the minimum

value of B occurs when w = -1. Now at this point from equation (9.16) 1+b p

1 + 1/b

b 8

113

CONFORMAL TRANSFORMATIONS

Hence from equation (9.18), when w= -- 1, 2(b - 1)

z=

7b

tall-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 of minimum flux density but also on the axis of symmetry of the Therefore it is only necessary to take real values of w between zero and ---1; this gives the curve of one half-period and owing to the symmetry about the point w= -1 the curve for the whole of the period can then be drawn in. For a numerical example consider a ratio 81g= 1-5. Then from equation (9.19), page 111, wave.

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

B

w-4 w-0.25

w-1

(9.23)

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

using equation (9.21). Substituting these values in equation (9.18), page 110, and taking values for which z is wholly real 4+p x = g log 1+p - log -3 tan-1 (1p)} 1-p 4-p

7r

.

(9.24)

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

_

2 (6.25)112

= 0.8

Since we are choosing values of w lying between zero and minus one, equation (9.23) is best rewritten I- W

B=

(0.25-w)1"2(4-w)112

B(9.25) max

Now choose various values of to between the limits mentioned 114

CONFIGURATIONS WITH TWO RIGHT ANGLES

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

4-w

2

0.25-w Then for each value of w calculate B from equation (9.25), making Bmax unity. Finally, for each value of p determined above 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 the smooth surface from the point opposite the centre of the slot, i.e. where z = - zs + jg can be drawn. As before it is best to draw up a table, and Table 11 shows a typical calculation. TABLE 11

Calculation of flux density distribution -0.01

0.10 i-0.20 -0.40

-0.05

-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

0-989

0-953

0.918

0.873

0-824

0-809

0-800

3-927

3-674

3-422

3-056

2-602

2-326

2-000

1-683

1-747

1.826

1.973

2 248

2 508

3 000

7.475

4-131

3.779

3-000

B

1-w

Bm,X- (3) p=(2)=/2/(1)1%2

4) =p i i

±

(5)=4 p

108-6

23-54

12-84

(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

tan-1(p)

(9) =Stan-1(kp) I

1Tx=(8)-(9)

1.100

1.072

3-300

3-216,

0

1-042

0-991

0.915

0.861

0.785

3.126

2-973

2-745

2-583

2-355

i

-7.469 -5-817 -5-077 -4.305 -3.354 -2.994 -2.355 ,

9

9

-2-377 -1-852 -1-616 -1.370 -1.068 - 0.953 -0,750 115

CONFORMAL TRANSFORMATIONS B Bmaz. 1.0

09

08

07

9

06

0

4

9 =1.5;

_

0.612

3

X

Fin,. 64.

4

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 in Fig. 64 is obtained. It shows that the ratio B%Bmax reaches its uniform value of unity at apparently x =1.22. This means

that if the tooth is so narrow that the edge of the next slot is nearer than at x = 2.44 this analysis for a single slot is invalid for a succession of equal slots. The transformation that takes care 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 tooth ratio of s/t = 0.612. The curve shows the presence of harmonics

including second, fourth and higher even harmonics of quite large amplitudes. The machine dimensions giving the curve of Fig. 64 are typical of salient-pole synchronous machines with open armature slots. In eddy-current inductor brakes and couplings the airgap lengths are by comparison very small and the ratio s/g may be as high as 80. Then the amplitudes of the harmonics are even higher. Fig. 65 shows the flux density distribution curve calculated in the same manner for a ratio s/g = 12 and for g = 0.1. This curve is continued for a tooth width equal to the slot width s. 116

CONFIGURATIONS WITH TWO RIGHT ANGLES a

1

Bmax.

n

10

0.75

O.5

025 O 2

0

1

S

9=12

1

X 2

t=10 S

;

i

I

S

FiG. 65.

Waveform of B with small gap.

It is clear that for these values the analysis is valid for any practical width of tooth because the flux density becomes uniform 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 amplitude of the variation in flux density depends on the ratio s/g. The Lost Flux

By the lost flux is meant the diminution in flux crossing the airgap caused by the presence of the slot opening. This can always be obtained from the flux density distribution curve by

integrating graphically the area between that curve and the straight line B/Borax = 1

But it is just as easy to integrate the equations and obtain a general expression.

The lost flux is

f

00

i.e. the difference between

m(1--B)dx

dx and f. B w

117

CONFORMAL TRANSFORMATIONS

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

B=

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

if Bma, is unity, and

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

Tr

Fig. 62 shows that when x = -00,

w = 0

X = oo,

w = ± oo,

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

(1-B) dx

0a = f-".O

g f-00

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

0°

f

o

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

The first part of the integrand has already been put in terms of p in equation (9.17), page 109. Considering now the second part and putting it in terms of p, from equation (9.16) p2-1

ape - b

w

b(p2-1) p2-b2 It is shown on page 108 that

dw =

2p(b - a) dp (p2 - 1)2

Hence dw

2p(b2-1) dp

w

(p2 - b2)(p2 - 1)

With regard to limits of integration when changing from w to p, since p2 =

w-b

w-a

118

CONFIGURATIONS WITH TWO RIGHT ANGLES

when w = 0, p = (b/a)1 J2 = b, and when w = - oo, p=1. fore fb 2p(b2 -1) dp dw _

There-

-Ji (p2-b2)(p2-1) Adding now the first part of the integrand as given in Jo

w

equation (9.17) and putting in the limits of integration

i--

b

2g(b2 _ 1) ar J1

(b + 1)p2

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

(p2- l)(p2-b2) dp

- 2g'7

_

2

(b+l)p2-p(p2+b)

-1)

1 Sb

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

1

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

-p{p2-(b+l)p+b}

_ _ 29 ( b2 _ 1 ) 7r

=

29 (b2-1) 7r

=

2g

n

(b2-1)

-

p(p-1)(p-b) dp

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

p dp

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

C B1 p2+b+p+b+p+1 2gJb(A

it

1

dp

where

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 = 1 therefore

2g 01

= =

2g

b

b-1

1

1

p2+b+p+b p+l dp log b11 b

tan_1

V/b 29

Lb

b

p+

!tan-1 /b -tan-1 fib} 119

(Dwight 120.01, 90.1) 1)z}

-log (b 4b

CONFORMAL TRANSFORMATIONS

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

A = tan-1 Vb

B = tan-1 i.e.

tan A = V b and

I

tan B=1/i/b. Then

tan (A - B) =

tan A - tan B 1 + tanA tan B

_ \/b-1/Vb _

s

2

2g

from equation (9.19), page 111. Therefore 1 A - B = tan-' \/b-tan-1 Vb

tan-1

S

2g

With regard now to the second term, (b+1)2 4b

=

4b+(b-1)2 4b

1+4(ti/b- 1 2 \Ib)

Therefore I

(s

=9 _s

s tan 1 2g r -log (1 9 tan-1 2g

log

S2

+ 4g2 2

1+

1

l (9.26)

= so

where 1+42

or =

tan-1 9-s log I(

(9.27) 92/

The coefficient a in O = sa has been called by some authors 120

CONFIGURATIONS WITH TWO RIGHT ANGLES

Carter's coefficient. However, it is not the coefficient in use by practising engineers. Their coefficient is derived from a as follows. Carter's Coefficient

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

0 = t+s Now when there is a tooth width dimension sufficient to give substantially uniform flux density before x reaches the value it we know that the actual flux crossing the airgap for one slot

pitch must be 0 as above, less the lost flux calculated on page 120, i.e.

Oa=0-01 = t+s-scr The ratio of the undisturbed flux gS to this actual flux C

is

t+s t+s-sa

0 Oa

_

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 procedure 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 from equation (9.28) for various values of s/t. A family of curves can 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/g between 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 if the ordinates and the abscissae are scaled on a basis that makes the family of curves into a family of straight lines. Furthermore the spacing of the lines can be greatly improved. The 121

CONFORMAL TRANSFORMATIONS 10

9 3

$

i

8

9 7

s

C I

r7

rs 1.A

13

12

1 05 f

2

3

4

6

1

8

9

-

11

t FIG. 66. Carter's coefficients.

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 difficult to devise one. The Equivalent Airgap Length

There is another way of expressing Carter's coefficient. Let the actual physical airgap length be g giving a flux across the slotted gap of 0. the quantity we have called the actual flux. Let g' be the equivalent airgap length that would give exactly the same flux 0a with the same excitation but with both airgap surfaces unslotted. Considering now both surfaces unslotted the flux obtained for an airgap length g would be 0, the quantity we have called the undisturbed flux. For the greater airgap length g' the flux obtained 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

r

s

3

2

1.5

1.0 (>I r

,

0.6

p J

-

4 r.O?

05

T

'Ojs

0.4

I

03

0.2 0.15

0'2

03

0{

05

06

08

b5

2

S 3

1c

F

FiG. 67.

Carter's coefficients with special scales.

excitation held constant) is inversely proportional to the airgap length it follows that Therefore y 9

_ 0 = C (equation 9.28)

a

123

CHAPTER, 10

Configurations with more than Two Right Angles The Inside of a Rectangle

In all the problems considered in previous chapters the field has extended in the w plane to plus and minus infinity along the real axis so that in the x plane the two plates that give a uniform field have also been of infinite length. In practice there are many problems, in fact a majority of them, in which the equipotentials forming the boundaries of the problem do not take up the whole of the real axis when transformed to the w plane. The equipotentials are already finitely bounded in the z plane and the Schwarz-Christoffel transformation produces defined lengths along the real axis of the w plane to correspond to these equipotentials. A typical w plane is shown in Fig. 68 where AB V

A

C

B

D

u

FiG. 68.

represents one equipotential boundary and CD the other. The field'in the w plane extends from AB to CD but ranges all over the upper half of the plane ; it is only bounded by the real axis. Notice that this is an entirely different transformation from

that of a definite portion of the w plane in which the equipotentials are radii and the lines of flow are semicircles. The field in the w plane is then bounded by semicircles as shown in Fig. 69. The area bounded by EFGH in the w plane Fig. 69 transforms 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 on page 76 and given by equation 7.9:

X=dlogw 7T

where d is the distance between plates in the X plane.

The equation applies to the field between the values of w at the four corners E, F, G and H.

I

E

F

I

d

G

I

0

H

FIG. 70. Bounded configuration in the X plane.

In most transformations that follow in this book the only boundaries in the upper half of the w plane will be the lengths AB and CD on the real axis ; these lengths can be called for convenience `electrodes', since in some problems the equipotential boundaries will actually be electrodes. It is comparatively easy to transform the w plane of Fig. 68 to the t plane of Fig. 71 where the field extends indefinitely between plates both above and below. But this is not what is sometimes required. For instance if we are concerned with current flow from one electrode to the other we shall want the 125

CONFORMAL TRANSFORMATIONS

Fr- (G FIG. 71.

Field in the t plane.

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

A

W.! 00

D

8

w=O

C

0

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

are completely bounded, the transformation from the regular field in the X plane, Fig. 72, to the w plane, Fig. 68, is a transformation from the inside of a rectangle to the upper half of the w plane. This transformation requires the straightening out of four right angles. When more than two right angles have to be straightened out

the integration required in the Schwarz-Christoffel transformation cannot be performed in terms of elementary functions. The integral is either itself an elliptic integral or an integral that can be broken down into one or more elliptic integrals and one or more elementary integrals. Hence the integrations 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 SchwarzChristoffel transformation of the inside of a rectangle to the w plane, the transformation now under consideration. It is therefore the one most suitable for opening the subject. To transform the configuration of Fig. 72 to that of Fig. 68 it is clear that the values w = 0 and w = ± oo must be placed at the points shown in Fig. 72. Values of iv must now be assigned at the four points A, B, C and D where, however, by symmetry the values at A and B must be the negatives of the values at D and C respectively. There is only one ratio in the X plane, viz. the ratio of width to height of the rectangle, i.e. BC/AB. Therefore one of the values of w must be left unknown and denoted by

a 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

A[

w=a

0

I

0

8

0

w=-1

w=1

W= 0

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

W=1

at C

w=a

at D

w = -1 at B

w=-a atA

From this choice it is clear that a exceeds unity. By the Schwarz-Christoffel transformation A(w-1)-1/2(w-a)-1/2(w+1)-1/2(w+a)-1/2

dw = Hence

_ X

dw

A f (u'2- 1)1/2(w2-a2)1/2 127

.

(10.1)

CONFORMAL TRANSFORMATIONS

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

= B

-

dw

(1-w2)'/2(1- k2w2)li2

(10.2)

where B = - kA, and the integrand is now recognisable as that of an elliptic integral. Before further consideration can be given to this equation which transforms the interior of a rectangle to the upper half of

a plane, it is necessary to digress and consider the general method of handling elliptic integrals.

128

CHAPTER 11

Elliptic Integrals of the First Kind Jacobi's Form

Most of the difficulties and complexities confronting the engineer when he tries to tackle elliptic integrals are more apparent than real. Quite often the difficulties are largely associated with nomenclature and terminology, and for this reason it is well worth while to proceed slowly in the initial stages. Consider again the integrand of equation (10.2).

The constant B can be made unity for the present since its value only affects the scale change in the transformation. Now an indefinite integral can always be expressed by a definite integral because

f(w) = f(0)+

f'(x) dx J`0

Therefore we can investigate the definite integral in

_

-

fWdw

(11.1)

o (1-w2)I/2(l-k2w2)I12 Notice that X is a function not only of w but also of k. The constant k is called the modulus of X. Normally it is quite un-

necessary to mention the modulus explicitly throughout the analysis provided the same modulus applies always and there is no danger of ambiguity.

Equation (11.1) can be evaluated numerically for various values of the amplitude w ; the results can then be tabulated It is clear from the original problem that since a exceeds unity the value of k cannot be greater than unity. In engineering problems k is always a real number. Moreover, since it only appears in the integral as k2 negative values of k give the same results as the corresponding positive values. Therefore the only values of k to be considered are for use and indeed have been.

9

129

CONFORMAL TRANSFORMATIONS

those between zero and unity. For this reason k can be, and often is, designated by sin 0 where 0 is called the modular angle and in engineering problems is always real. The integral in equation (11.1) is called the elliptic integral of the first kind. There are other kinds as will be seen later. Such integrals were called elliptic because they were first encountered in the determination of the length of are of the ellipse. The form of the integral in equation (11.1) is known as Jacobi's form ; it is said to be expressed in Jacobi's notation. Legendre's Notation

Another notation is that of Legendre ; it can be obtained from

that of Jacobi by putting w = sin o

.

.

.

.

.

(11.2)

where 0 is called the amplitude angle. Distinguish this angle carefully from 0 the modular angle. From equation (11.2)

dw = cos 0 d

.

.

.

.

.

(11.3)

.

.

.

(11.4)

.

.

.

(11.5)

and hence from equations (11.1) to (11.3) X

_

do

-

p (1-k2Sj2 0)1/2 The integral in equation (11.4) is the elliptic integral of the first kind in Legendre's notation. It is usual to write

0-amX

.

.

signifying that 0 is the amplitude of X. Tables of elliptic integrals are often given in terms of 0 and 0 instead of in terms of w and k. Complete Integrals

Refer back to equation (11.1) . If the upper limit of the integral is made unity the integral is said to be complete. The complete elliptic integral of the first kind is always denoted by K and hence

K

dw

- fO (1 -w2)1/2(1 -k2w2)1/2 130

(11.6)

ELLIPTIC INTEGRALS OF THE FIRST KIND

Equation (11.6) gives K in Jacobi's notation. In Legendre's notation, since w = sin q, the limits of integration are from zero to it/2. Hence in Legendre's notation fn/2

-

c

(11. 7)

(1 - 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 elliptic integral of the first kind, modulus k. Normally, however, the modulus can be understood without fear of ambiguity. Notice

that although w may be a complex number the complete elliptic integral K is always a real number. The Complementary Modulus

We now need to define a complementary modulus k' related to the modulus k by the equation

k2+k'2 = I or k' = (I - k2)1/2

.

From equation (11.8), when k is written sin 0, k' must be written cos 0.

Now the complete elliptic integral of the first kind to

modulus k' must be entirely different from K which has modulus k. This new complete integral is naturally enough denoted by K'. Thus

K'(k) = K(k')

.

.

.

.

.

(11.9)

Hence K and K' are related through their moduli and K' must also be a real number. By this definition K' is given by the integral K'-

fl

dw

(11.10)

(1-w2)'12(1-k'2w2)1/2

The next thing to find is the equation defining Kin terms of the normal modulus k. In equation (11.10) let

t=

I (I _. k'2w2)1/2 131

.

(11.11)

CONFORMAL TRANSFORMATIONS

Then w2 =

k12(1-12

t2-1 k''212

Therefore k'2t2 - t2 + 1

1-w2 =

k'2t2

1-t2(1-k'2) k'2t2

1-k2t2 k'2t2

and hence (1-k212)1/2

_ z (1-w)1/2 =

k't

.

(11.13)

Furthermore from equation (11.12) (t2 _ 1)1/2

w=

k't

Therefore 1)-1/2 _ (t2 - 1)1/2

dw =

12(t2 _

dt

k't2

t2-(t2- 1)

k't2(t2 - 1)1/2 dt dt k't2(t2 - 1)1/2

Substitution into equation (11.10) from equations (11.11), (11.13) and (11.14) gives

K

dt

-

k't Jk't2(t2-1)1j2(1-k212)1/2

t

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/k 132

ELLIPTIC INTEGRALS OF THE FIRST KIND

Therefore simplifying the integrand and adding the new limits dt

1/k

K' _

(t2 - 1)1/2(1 -k2t2)1/2 1/k

dt j(1 -t2)1/2(1 - k2t2)1/2

fl

and this is the required relation. Notice that the integral is the

same as that for K but that (a) the limits are changed and (b) the integral is imaginary. Combining the integrands of K and K' gives the total integration from the lower limit zero to the upper limit l /k and it is the complex number % dw K+jb' = fo,( 1-w2)1/2(1-k2w2)1/2

Transformation of a Rectangle

at this point it is possible to return to the problem of transforming the inside of a rectangle to the upper half of the w plane. On page 127 values of w at the corners of the rectangle were assigned and later on the substitution a = 1/k was made. Y_

D

C

0

0

W.1

wz0

Flea. 74.

X plane.

The X plane is therefore as shown in Fig. 74. The transformation to the w plane is given by equation (10.2), page 128: w X

YI0

dw (1- w2)1/2(1- k2w2)1/2

.

(11.16)

The constant of integration can be made zero by letting the origin in the X plane be at the point w = 0. 133

The constant B only

CONFORMAL TRANSFORMATIONS

affects the scale of the transformation and it is therefore convenient for the present to make it unity. It may be necessary to reintroduce it later on. By dropping these constants equation (11.16) becomes identical with equation (11.1).

-1

1.

O

1

L

u

k FmG. 75. w plane in the transformation of a rectangle. k

The w plane is shown in Fig. 75. Starting at the origin in Fig. 74, as we go along the real axis from 0 to C we go from w = 0 to w = 1.

Therefore the value of X at the point C is given

by dw

1

X(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 move along the imaginary direction in the X plane from C to D, so that the whole path from 0 to D in the X plane is given by X(at D) =

f

1/k

o

dw (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 and the length CD by K'.

Similarly by using negative values in each case it can be shown that X(at B) _ _K X(atA) _

-K+jK'

Therefore in the X plane the width of the rectangle is given by

2K and the height by P. 134

ELLIPTIC INTEGRALS OF THE FIRST KIND

Before proceeding further let us consider a very simple example that requires this transformation only. Electrodes and a Semi-infinite Conductor

In this example, suppose that there is a conductor of unit thickness but extending so far in the other two directions that for the purpose of the problem the conductor can be considered semi-infinite. Two electrodes are applied to the edge of the sheet and a known voltage exists between one electrode and the other. The electrodes are, of course, collinear; let them each be one inch long and spaced two inches apart. Suppose that the problem is to find the effective resistance of the current path per inch thickness of sheet. V

A

B

-Y

-1

X -K+ jK'

X.' -K

FIG. 76.

C

D

0

1

X0

X-K

Yk

u

X=K+ jK'

Values of X in the w plane.

The w plane can be made to represent the sheet and the electrodes as shown in Fig. 76. Here the dimensions chosen

give the values of w at A, B, C and D as - 2, -1, 1 and 2 respectively, and these are the values written into Fig. 76. Y_

(w`41'.

w (w.%K.2)

-2)

* -K+j K'

X-K+jK'

A

D

B

X -K w=-1

FIG. 77.

C

0 W=O

XK W.1

Values of X in the X plane. 135

0

The

CONFORMAL TRANSFORMATIONS

X plane is shown in Fig. 77 and, if we make the point D correspond as before to Ilk where k is the modulus in the transformation, it is clear that k = 0.5. This value of k satisfies the requirements for w at the points A and B. In the X plane it is known that the resistance between CD and BA for unit thickness is

R=pBDohms where p is the resistivity of the material of the conductor in ohm-inches. We now have to show that the effective resistance between electrodes is invariant under a conformal transformation. Prof. G. W. Carter's proof is as follows. Let perfectly conducting electrodes in the z plane form the equipotentials 0 = 01 and 0 = 02 in a conducting sheet having a resistance of p8 units per square. Let 0 + jo be known all over the sheet. On an electrode EnI

an

and the current per unit width is

la _

ps an

1a:/i ps as

Hence the current flowing to or from either electrode is [fi]/ps where [0] is the change in 0 when one makes a complete circuit of the electrode. The resistance is thus given by 0 R-p8 1-2 [0]

1

If the electrodes are transformed by conformal transformations into others that are still supposed to be at potentials 01 and 02 in a sheet having a resistance per square of ps, all the quantities in the formula for R are unchanged. Thus the resistance in the transformed system is the same. In the w plane the lengths BC and CD are 2K and K' respectively, and therefore in the w plane

R=p

2K 136

ohms

ELLIPTIC INTEGRALS OF THE FIRST KIND

where K and K' are to modulus k, i.e. to modulus 0.5. From tables of elliptic functions, at k = 0.5 or 0 = 30 °,

K = 1-686 K' = 2.157 Therefore

R = 1.560p ohms

for unit thickness. Any change in the ratio of width of electrodes to their spacing must bring about a change in the value

of the modulus k and therefore a change in the effective resistance.

In this particular example suppose that the material of the conductor 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 applied across the electrodes is therefore

_

10-3

1.287 x 10-6

amps

= 777 amps

Notice that in determining the effective resistance the constant B in equation (11.16) has no effect on the result, because the resistance depends only on the ratio of width to height of the rectangle and not on its size. If instead of being made

unity the constant B had been carried through the analysis, the final result would have been

2BK R=pBK' 2K p as before.

137

CONFORMAL TRANSFORMATIONS

There is more work to be done on this transformation but it requires familiarity with elliptic functions as well as elliptic integrals. A knowledge of elliptic functions is also necessary for plotting the fields in the z plane. Therefore further consideration of this particular problem will be deferred until after the next chapter, in which elliptic functions are discussed.

138

CHAPTER 12

Elliptic Functions The Function sn

The elliptic functions of interest to engineers are Jacobi's elliptic functions, so called because Jacobi first developed them although the credit for the original idea lies with Abel. Referring back to equation (11.1), page 129, it is clear that when k is given its minimum value of zero fw X

dw

= 0 (1- w2)1/2

= sin-1 w . Thus the effect of getting rid of k is to make X a function only of w and moreover an elementary function of w. Now k = 0 can

be regarded as giving a special case of the elliptic integral. Notice that in equation (12.1) both X and w can be complex numbers. Now in handling trigonometric functions of a complex variable, such as sin z, it is usually preferable by far to avoid the inverse functions because they are less manageable. Wherever

possible the relation

w = sinX

.

(12.2)

.

is used rather than that of equation (12.1) or the relation given by the integral. Suppose that in dealing with the elliptic integral it happened that the case of k zero was being considered ; anyone would then naturally prefer to use equation (12.2) rather than the integral. In the same way, when k is not zero it is generally easier to deal with the elliptic function than with the elliptic integral. Using the 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 integral can be obtained by putting . (12.3) . to = sin 0 .

139

CONFORMAL TRANSFORMATIONS

Immediately afterwards, in equation (11.5), was described as the amplitude angle of X and written . . (12.4) 0 = am X From equations (12.3) and (12.4) (12.5) w = sin am X . . . 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 nowadays written sn X and called either `sine am' or `ess en' X. This notation was invented by Jacobi and later modified and added to 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 implied. But if the modulus has to be brought in explicitly the equation can be written w = sn (X, k) To sum up then, it is clear that when k is not zero w is a function of X analogous to sin X and it actually becomes sin X when k is zero. Expressed mathematically, sn (X, 0) = sin X .

.

.

.

Transformation of a Rectangle in Terms of Elliptic Functions

The transformation of the inside of a rectangle to the upper half of the w plane was given in equation (11.I6), page 133, in terms of the elliptic integral. By normal inversion this can be written sn-1 w X=

.

.

.

.

.

(12.7)

Equation (12.7) expresses the transformation from the X plane to the w plane in terms of an inverse elliptic function. Clearly it is more convenient to express the transformation by equation (12.6).

Now when the integral is complete, w in equation (12.7) is unity ; furthermore X in the same equation becomes K. Therefore

K = sn-1 (1) and by inversion of this snK = 1 140

.

.

(12.8)

ELLIPTIC FUNCTIONS

Since from equation (11.15), page 133,

A+JA' =

l;x

fo

dw (1-w2)1/2(1-k2w2)1/2

= sn-1 (1/k)

by inversion

sn (K+jK') = l /k . . (12.9) The results of equations (12.8) and (12.9) can be used to .

interpret the meaning of equation (12.6). Thus when

±1,

w

X=±K

W = 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 the

rectangle more concisely and more elegantly than they were established in Chapter 11, but the method requires familiarity with the elliptic function sn. More Elliptic Functions

It is not the purpose of this book to present an exhaustive treatment of elliptic functions. This can be found in mathematical works. Engineers need to take advantage of whatever aids are available in dealing with such functions. A necessary aid for each engineer working in this field is a book of data on elliptic functions that can be used for reference. Such a book is the one entitled Handbook of Elliptic Integrals for Engineers and Physicists,* and this book will be referred to freely in what follows by the letters E.I.E.P. followed by the figures giving the reference in detail. In E.I.E.P. we can readily find many relations between the various elliptic functions, their derivatives and their integrals. On page 140 sn X was obtained by using the relation w = sin where 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

CONFORMAL TRANSFORMATIONS

and like the shortening of sin am X this is written on X

to modulus k

or

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 relations are given. Of equal importance to sn X and on X is the function do X which 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 principal Jacobian elliptic functions. There are others derived from two or more of these three such as to X =

sn X on X

and just as sn-1 is written for the inverse of sn, there are the inverse 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 considered in rather more detail wherever necessary as they arise in the transformations. Most of the information required will be

found in E.I.E.P. and the references therein will be given. E.I.E.P. 120.02 gives the nomenclature for such functions as 1/sn = ns, and this nomenclature will be used throughout the present work. Current Density in Semi-infinite Sheet

It is now possible to return to the problem of the two electrodes placed at the edge of a semi-infinite conductor. At the end of Chapter 11 the effective resistance of the sheet had been determined, together with the total current produced by a vol142

ELLIPTIC FUNCTIONS

tage difference between electrodes of one millivolt. Now the current density anywhere is given by

J _- IdX dw where J is the magnitude of the current density. Hence from equation (11.16), page 133, B

J=

(1 - w2)1/2(1 - k2w2)1/2 1

where the scale constant B has had to be reintroduced and is yet to be determined. Along the imaginary axis of w, u is zero. Therefore by putting w = jv we find that the current density for unit thickness of

sheet along the v axis, i.e. the axis of symmetry between the electrodes, is given by

_

B (1 + v2)1/2(1 + k2v2)1/2

(12) .10

noting that v is wholly real. In the problem being considered here, 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 total current along the sheet from equation (12.11) and equate it to 777 amps, the value found on page 137. The total current is found by integrating J along the v axis from zero to infinity. Thus B dv

1= f000

(1 + v2)1/2(1 + v2)1/2

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

CONFORMAL TRANSFORMATIONS

Hence

I = 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 density along the v axis of the w plane, Fig. 75. Values assigned to v give different points along the axis of symmetry between the two electrodes. For each value of v a value of J to correspond can be calculated from equation (12.12). The results can be plotted on a graph and such a graph is shown in Fig. 78. 400 I

300

c

200

100

2

4

V

8

Inches

Current density distribution along the axis of symmetry between two electrodes at the edge of a semi-infinite conductor.

FIG. 7 8.

144

ELLIPTIC TUNCTI0 \ S Plotting the Stream Lines and Equipotentials

In this type of problem it is normally unnecessary to plot either the stream lines or the equipotentials ; it is not difficult to imagine the pattern of lines in all the planes. But if for some reason an accurate plot is required the procedure is the same as that described in earlier examples ; since, however, elliptic functions now appear in the transformations the difficulties tend to increase considerably. As an example, suppose that the stream lines and equipotentials are required to be drawn on Fig. 76, page 135. The transformation is given by equation (12.6) page 140: w = snX Hence

u+jv = sn (0+jt) and from E.I.E. P. 125.01

Ocn

sn(0+jo) =

where all the functions of 0 are to modulus k and all the functions of are to modulus V.

Separating real and imaginary parts 91, =

I-sn2grdnil ¢

sn0cn0cn0dn v= 1-sn2 dn20

(12.13) J

each argument with different moduli as before. At this stage, tables of elliptic functions are required, preferably 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 Elliptic Function Tables by Milne-Thomson.* In this example k= 0.5 so that for functions of 0, 6 = 30°, and * See List of References, page 217. 10

145

CONFORMAL TRANSFORMATIONS

for functions of ,, 0 = 60 °. The calculation of one set of points

is given in Table 12. In this set of points 0 = 30'; there are various values of Vr put in terms of r° where r is defined by

r = 90° K For Table 12, the following values are first calculated : 0 = 30°

sn di = 0.527 to modulus :30° en q = 0.850 do

= 0.965

All the functions of shown in Table 12 are to modulus 60°. All values are taken from the Smithsonian Tables referred to above.

In these Tables the quantity r defined above is used and

for this reason intervals of r are used to start the calculations.

U

Fic. 79. Equipotentials and streamlines from two electrodes at the edge of a semi-infinite conductor.

Similar calculations can be made for other values of 0 between 0° and 90°. The resulting lines of flow and equipotentials are shown in Fig. 79. 146

ELLIPTIC FUNCTIONS TABLE 12

Calculation of stream lines and equipotentials, 0

r

sn en

1

dn

1

0

10.111

1

1 0.889

c =sn 0 (in 0

0.527 0.327 0 0

d = sn 0 en 0 en

v = d/b

do

450

0.346 0.627 0.816 0.938 1 0.779 0.577 0.954 j 0.840 0.707

a=sn2 w dn2 b =1- a za = clb

30°

150

= 30°

60°

75°

900

0.928 0.373 0.595

0.983 0.182 0.524

1

0.500

0

0.503 0.566

0.366 0.634 0.443 0.699

0.620 0.380 0.373 0.982

0.802 0.198 0.314 1.586

0.900 0.100 0.276 2.760

0.931 0.069 0.264 3.826

0.266 0.299

0.401 0.632

0.386 1.016

0.284 1.434

0.147 1.470

0 0

I

147

CHAPTER 13

Double Transformations with One Requiring Elliptic Functions IN THE LAST TWO CHAPTERS a single transformation from the

interior of a rectangle to the upper half of the w plane was considered. A slightly more difficult problem arises in the trans-

formation of a configuration that makes the w plane just an intermediate plane and requires an initial transformation to the w plane from a configuration containing only two right angles. Electrodes at the Edge of a Strip

For 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 width but of infinite length in one direction, as shown in Fig. 80. In

I A

B

0

C

D

Fic. 80. Electrodes at the edge of a strip.

other words the conductor is bounded in all directions except one ; therefore the resistance between the electrodes must be greater than that obtained on page 137. 148

DOUBLE TRANSFORMATIONS : ELLIPTICAL FUNCTIONS

The configuration is shown in Fig. 80 which can be made the z plane. What is required is the effective resistance for unit thickness of plate of the current path between the electrodes. The configuration and transformation apply, of course, to other problems ; for instance, they apply equally to an idealisation of a problem concerning the end effects of eddy currents in a solid conductor. As before the electrodes are labelled AB and CD, and for a general solution let the distance OC (or OB) be g/2 and let the

ratio OD/OC be r. We know that, after the transformation from the inside of the strip to the upper half of the w plane, a second transformation will be required from the inside of a rectangle in the plane to the upper half of the w plane. The effective resistance of the current path in the X plane is easily found and by combining the two transformations we can arrive at the effective resistance in the z plane. From previous work it is clear that we want the values of w to be Ilk at D, unity at C, -1 at B and -1/k at A. Furthermore the origin in the w plane should preferably correspond to the point in the z plane midway between the electrodes. We thus arrive at the z plane and the w plane shown in Fig. 81. The First Transformation

Now in the first transformation there are only two right angles to straighten, and therefore the integration to be performed must lead to elementary functions. By the SchwarzChristoffel theorem the transformation is given by

d = A w-k)

1 2

l (w+k)

-1/2

whence z

=

A dw (w2 - 1/k2)1/2

Using this equation the value of A can be found by the method of page 65, i.e. by integrating along a large semicircle in the w plane. In the z plane, as we cross the strip at points where w is very large

fdz= -rg

the negative sign appearing because the path of integration is 149

CONFORMAL TRANSFORMATIONS w=.m

w=-m

A W. -%

C

OI

B

w.-1

w.0

D

w.1

w. %k

(a)

V

(b)

A

I--Go

I=-r2 2

FIG. 81.

0

B

zz-2

2.0

2

C

Z.I 2

D

U

Z- L9 2

(a) z plane (b) w plane.

from right to left in Fig. 81 (a). But in the w plane the path of integration is along the semicircle of radius R where w = ROB. As before dw = jRele dO

so that the equation of transformation gives

f dz = fo

j 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 A rg

Z =

IT

dw

f J

J(w2- l/k2)112

dw IT j (1.10 -w2)1/2

r_g

-T sin- r kw + C (Dwight 320.01) 7T

For convenience let C = rg. Then

r z = - (7T- sin-1 kw) IT

L sin-1 kw

a

Notice that when w is zero

z = r-ITgsin-' 0 = 0

and this result checks that the zeros in the two planes given by equation (13.1) are identical as shown in Figs. 81 (a) and 81 (b).

From equation (13.1) the value of k corresponding to the unknown ratio r can be determined as follows. Since at z=,q/2, w=1, equation (13.1) gives = rg sin-' k Hence

2 k = sin (7r/2r)

.

.

.

.

.

(13.2)

This result means that the first transformation provides the value of the modulus k for the second transformation. Thus the modulus depends entirely on the ratio r and not on the actual dimensions, and this is what we should expect to find.

In the example on page 135 the modulus was the direct reciprocal of the ratio r whereas here it is the function of r given by equation (13.2). The Second Transformation

The first part of the problem has now been completed.

The

second part has already been done in Chapter 12 and the 151

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 path between the electrodes is

R = pK, (k) Y,

-K+jK'

-K

K

0

-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 resistivity of the material is 0.825 x 10-6 ohm-inches, the resistance is

R = 2.000 x 0.825 x 10-6 ohms 1.6510 x 10-6 ohms 152

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTION -S

Then the total current flowing in the sheet when one millivolt is applied at the electrodes is _

10-3 amps 1.650 x 10-6

= 606 amps

.

.

.

(13.3)

Current Density in the w Plane

The distribution of current density magnitude across the axis

of symmetry of the strip can now be calculated. The total current in the w plane can be found as before from equation (12.10), page 143: Jv

B (1 +v2)1/2(1 +k2v,2)1/2

where now k=0-707 as found above. Hence now B V

(1 -+2)1/2(1 +v2)1/2

and the total current is given by I

0

B dz

(1 +v2)1/2(1

= B to-1 (oo, k')

+I-v2)1/2

(from E.I.E.P. 130.05)

where

k' = (1-k2)1/2 = 0.707 Hence

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)

.

(13.5)

Then from equations (13.3) and (13.4) B = 327

.

.

.

.

Current Density Distribution in the z Plane

In the z plane the magnitude of the current density anywhere 153

CONFORMAL TRANSFORMATIONS

is given by dX dz

dX dw

dw dz B

IT

(1-w2)1/2(ll -k2w2)1J24.

(1-k2w2)1/2

TrB (1-q()2)-1(2

4K

But here k = 0.707 and (from equation (13.5)) B = 327. Hence 36 z

Jz = (1-2v) 1/2

.

.

(13.6)

I

The next step is to put equation (13.6) in terms of z instead of w. Equation (13..1) shows that

w_- sin-9 1

.

TTz

Inserting the values g = r = 2 and k = 0.707 gives is = 1.414 sin (7rz/4) Hence

_ 11Z i

363 {1 - 2 sln2 (irz/4)}1/2

Along the y axis where the current density distribution is required x = 0 and hence z = jy.

JY _

Thus 363

{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 origin J(y0) = 363 amps/sq. in. Similarly at the point y=2 Try/4 = 1.571

and from tables of hyperbolic functions

sinh 1571 = 2.302 154

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTIONS

Hence J(y=2) _

363 (1 + 2 x 2.302 x 2.302)1,2

= 106.5 amps/sq. in. 400

300

100

0

4

2

V6

6

Inches

Fic. 83. Current density distribution alon, the axis of ynirnetry between two electrodes at the edge of a strip.

Similarly by taking various other values of y the graph of Fig. 83 can be calculated and drawn. This graph shows the distribution of current density magnitudes across the axis of symmetry through the strip and between the electrodes. Plotting the Field in the z Plane

The procedure for plotting the stream lines and the equipotentials in the z plane follows the previous method but is rather more complicated. The transformation from the X plane to the w plane is given by

w=snX so that the relations between u, v and 155

are again those of

CONFORMAL TRANSFORMATIONS

equation (12.13), page 145. But it is now necessary to consider in addition the transformation from the z plane to the w plane given by equation (13.1) : z = --Lsin-1 kw 7r

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}1/2 q = {(1-k2L)2+k2v2}1/2

(13.8) t

the positive values being taken in each case. In these expressions the principal values found when n is zero can be used so

that

x = -rgsin-1 IT

2ku

p+q

y= 9 cosh-1 p

(13.9)

2 q

Thus the procedure is to put in the numerical values for the parameters in the equation and determine the value of k and hence of 0 from the first transformation. Then choose values for 0 and 0. For each pair of values so chosen, u and v can be calculated from equations (12.13), page 145. Then calculate values of p and q corresponding to these values of it and v from equations (13.8). Finally equations (13.9) give the values of x and y corresponding to the values of 0 and chosen originally. For the numerical example already used on page 152

rg = 4 k = 0.707

0=450

and

k' = k

156

Fic. 84.

Stream line, and equipotentials from two electrodes at the edge of a strip.

I

CONFORMAL TRANSFORMATIONS

Hence 45° is the modular angle for the functions of as well as for the functions of 0 in equations 12.13. A typical calculation for 0 = 30 ° is given in Table 13 below. Similar calculations can be made for other values of 0 between 0° and 90°. The resulting equipotentials and lines of flow are shown in Fig. 84. TABLE 13

Calcidation of the field in the z plane

Since d = 30°, cn 0 = 0.565; en 0 = 0.825; (in r

sn ¢ en

I

0

15'

30°

45'

60°

75°

90°

0 I

0.302 0.953 0.977

0.565 0.825 0.917

0.765 0.644 0.841

0.900 0.435 0.771

0.976 0.218 0.724

1

0.707

0.077 0.923 0.552

0.268 0.492 0.732 0.508 0.518 0.475 0.708 0.935

0.681 0.319 0.436 1.367

0.801 0.199 0.409 2.055

0.841 0.159 0.399 2.509

0.373 0.734

0.296 0.928

0.161 0.809

0 0

0.399 0.423 0.167 0 1.956 2.025 0 0.028 1.399 1.433

0.501 0.661 0.341 r 0.519 2.253 2.759 0.116 0.269 1.539 1.740

0.966 0.656 3.865 0.430 2.072

1.453 0.572 6.018 0.327 2.519

1.774

2.774

0.001 0.657

0.205 0.729

0.599 0.774

do

1

a=sn2 dn2 0 b = 1- a c=sn 0 do 0

1

0.565

u=c/b

d = sn 0 en 0 en

v=d/b

(k=0.707) ku kr

f=(1+ku)2

g=k2v2 1) =(f+g)'12

h=(1-ku)2

0.565 0.598 do

= 0.917

0

0

0.218 0.236

0.353 0.482

0

0

4=(h+g)112

0.360 0.600

0.333 0.601

0.249 0.115 0.604 0.620

rn=,i(p+4)

1.000

1.017 0.184 0.234

1.072 0.377 0.480

1.180 1.365 0.591 0.830 0-752 1-057

1.624 1.066 1.357

1.774 1.175 1.496

0.416 0.429 0.546

0.467 0.486 0.619

0.560 0.594 0.756

0.895 1.108

1.000 1.571 2.000

n=cosh- 1 rra

0

J=4n/a

0

r=ku/m

8=sin-'r x=48/or

0.410

0.522

158

0.708 0.787 1.002

1.411

CHAPTER 14

Double Transformations Both Requiring Elliptic Functions A YET MORE DIFFICULT PROBLEM occurs when both trans-

formations to the intermediate plane require the use of elliptic functions ; the simplest example is that where a rectangle is concerned in both transformations. Electrodes at One Edge of a Rectangle

Consider two electrodes again but this time let them be applied to a conductor bounded in all directions. In particular let the two electrodes be applied to one edge of a rectangle as shown in Fig. 85. Suppose that the problem is again to find y W k, E

A

B

W.-1

0

C

w=0

FiG. 85.

0

X

W=1

Electrodes at one edge of it rectangle.

the effective resistance of the current path between the electrodes. This is an idealisation of a problem concerning end effects in an eddy current machine. It is also applicable to the determination of the permeance of the leakage path for the flux of a machine from one tooth tip to an adjacent one. The configuration of Fig. 85 can be called the z plane and for comparison with the two previous problems the electrodes are labelled AB and CD, with the length OD called g/2 and the ratio OD/OC called r. The problem is similar to that given by 159

CONFORMAL TRANSFORMATIONS

Fig. 80 the difference being that the conductor is now limited in the y direction. This gives an additional dimension and there-

fore an additional ratio,

viz.

DE/10D: let this ratio be

labelled s. The First Transformation

From previous work we know that the transformation to the w plane must introduce the elliptic integral of the first kind, and hence the values of w for the various points in the z plane are chosen as shown in Fig. 85. Any other arbitrary values would

entail later substitutions to bring the integrals into forms that are recognisable as elliptic integrals. Thus the value w =1/k1 at the point E is the reciprocal of the modulus in the integral ; kl has been chosen as the label for this modulus because k will be required in the second transformation. V

A 1

-1

B

10

C

k'

Fia. 86. w plane.

The w plane is shown in Fig. 86, and with the values of w chosen we already know that the equation of transformation from 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 depend on the relative ratios in the rectangle. The first thing to determine is the value of the constant B and then the value of the modulus kn in terms of the ratio s which is known. Now when z = rg/2, w=1 But when w is unity, from equation (14.1) sn Bz = 1 But equation (12.8), page 140, shows that sn K = 1 and hence Bz = K(kn) .

160

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTIONS

and

B = K(kl)Iz 2K(ki)

(14.2)

rg

since z = rg/2 at the point in question. Therefore from equation (14.1)

w = sn

2K(kl)z

(14.3)

rg

From equation (14.3) when w=1/k1

i ki

= Sn

2K(ki)z rg

But equation (12.9), page 141, shows that 1/k, = sn (K + jK') to modulus kl Hence 2Kz = K + jK' rg

to modulus kl

and z

2

+J r2K (kl)

Now this is the value of z at the point E. and imaginary parts OD = rg/2

Separating real

DE = 2ryK K' (ki) Therefore

s = DE/OD = K'/K(kl)

.

.

.

.

(14.4)

Thus for any given value of the ratio s, it can readily be found from Tables of complete integrals for what value of kl the ratio K'/K is equal to s. This determines kl which is the quantity

really required from the first transformation. To take a numerical example at this point, suppose that the height of the rectangle is one quarter of the width. Then

s=DE/OD= 0.5 11

161

CONFORMAL TRANSFORMATIONS

For this example, from equation (14.4)

K'/K(kl) = 0.5 and hence

K' = 2K (both to modulus k1) From tables of complete integrals, it is seen that K= 2K This completes the first part of the problem.

when k1= 0.985.

The Second Transformation

The second transformation is exactly the same as in the example of Chapter 13 and is given by equation (12.6) page 140. w.I

w=-1

A

D

B W--k

k-

O

w=k

w.p Fiu. 87.

X plane.

The X plane is shown in Fig. 87 where the values of w are those

Notice in Fig. 86 that at the point C, w is less than unity, while at the point D it has already been made unity. shown.

Therefore it is convenient to make w = k at the point C, where k

is now the modulus of the elliptic functions in the second transformation. Since in Fig. 87 the scale is k times that of equation (12.6) the equation of transformation is w = k sn (X, k) . . . . . (14.5) This can be checked at the known points as follows : When

w = ±k,

w=0, w= ±1,

X = ±K X

0

X= ±K+jK'

and we know that these are correct. 162

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTIONS

As before the effective resistance of the current path is given by

but the new modulus k is not yet known. A relation between k and the known k1 must next be established. In all three planes the ration OC/OD corresponds to the ratio k 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 = sn

rg (k1)

r

(k1)

.

(14.6)

Now when r happens to be an integer of low value it is possible

to evaluate k directly from equation (14.6).

For example if, as in the previous example of Chapter 13, the electrodes are two inches apart and are each one inch in length, OC =DE =1 and

Then s = 0.5 and, as found on page 162, k1= 0.985. Furthermore r = 2 as before. Hence putting these values into equation (14.6) we find OD = 2.

k=snK (k1) =

(1+k1')-I/2

(from E.I.E.Y. 122.10)

where

k1' _ (1-k12)1 2 = 0.173

Hence k = (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 = 2.395 K' = 1.634 163

CONFORMAL TRANSFORMATIONS

and therefore

R = 2.93p ohms for unit thickness of conductor. Incomplete Elliptic Integrals

However, in general the ratio r is not a simple integer. When

it is not, the equation (14.6) needs to be interpreted in a different way, a way that requires the use of tables of incomplete elliptic integrals. Equation (11.7) on page 131 gives the complete elliptic integral of the first kind as

d

/2

K_

- J0

(1-k2 sin2 0)1/2

But if the upper limit of the integral is some value of 0 less than ir/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 amplitude angle and the modular angle as F(0, 0) and rather less frequently in terms of the amplitude and modulus as F(w, k). All these are different descriptions of the same integral. Clearly when =7r/2

F(w, k) = F(7r/2, k) - K(k) - K Comparison of equation (14.7) with equation (11.4) page 130 shows 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 k

Returning now to equation (14.6) page 163, we have to find a general solution for

k = sn 164

(k1)

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTIONS

where r can have any positive value greater than unity. From the inversion of this equation sn-1 (k, k1) = K/r (14.9) = F(k, k1) (from equation (14.8)) Since K, r and k1 are all known, by looking up tables of incomplete 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 of k1 the ratio K'/K is equal to s. This finds not only k1 but also K(k1). Hence (K/r)(kl) can be calculated. Now look up tables of incomplete elliptic integrals to find what value of k satisfies K k1) (k1) = F(k, When k is thus determined tables of complete elliptic integrals give K and K' to modulus k, and the effective resistance is then given by 2K

R = K, (k)p

To clarify the procedure outlined above by a numerical example, suppose that s = 0.5 as before but that now r =1.6.

Note that the value of r does not affect k1 which remains unchanged at 0.985.

Therefore

k = sn

1.6'

0.985)

But from tables, K to modulus 0.985 is /3.156. Therefore k = sn (1.973, 0.985)

Hence

F(k, 0.985) = 1.973 Using tables of incomplete elliptic integrals sin-1 k = 75.5° whence k = 0.968 Then from tables of complete elliptic integrals it is found that

at k=0.968

K = 2.80 K' = 1.59 165

CONFORMAL TRANSFORMATIONS

and therefore R

_

5.60

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. from equation (14.4), page 161,

Therefore

K'/K(kl) --* oo

Hence kl = 0

Let r have the value given in the problem of Chapter 13, viz. r = 2.

Then

k=Sn12,01 and

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 from tables of complete integrals, at this modulus

K = K' Hence

R = 2.00p ohms for unit thickness, as obtained on page 152 for electrodes at the edge of a strip of infinite length. Variation of Effective Resistance with Height of Rectangle

At what height of rectangle is the effective resistance so near

to that of the electrodes at the edge of a strip that increase of 166

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTIONS

height has little effect? This can be determined by taking various values of s and calculating .P for each value. The calculation is shown in Table 14. TABLr. 14

CnlcvlatiTg the change in effective resistance s

r

I

0.289

0-426

0-502

1.000

1-440

2-000

0.9998

0.995

0985

0-707

0.400

0-173

2-718

1-848

1.578

0-927

0-820

0.791

II

kj

k

82-3

72-5

67.4

49-9

46-2

45.2

89-0

84-3

80-1

45-0

23-6

10-0

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

4-32

3-27

2-93

2-16

2-04

2-01

R 2K 7V V

+

-- -

I

Fig. 88 is plotted from the results shown in Table 14 and it shows how R/p varies with s. For low values of s it is preferable to plot the curve of the reciprocal, i.e. p/R against s. The curves indicate that when the height of the rectangle exceeds one half of the width, the rectangle can be considered to have an infinite height as far as the effective resistance is concerned. Total Current

To find the total current flowing consider a particular example.

Let us use the same numerical values as those of

page 163 where the effective resistance was found to be 2.93p

ohms for unit thickness of sheet. If as in Chapter 13 the 167

CONFORMAL TRANSFORMATIONS 0-5

P

R 0.4

O.3

0.2

01

O5

2.0

15

110

S

Fic. 88. Effective resistance for varying heights of rectangle.

thickness of sheet is one inch and the resistivity of the material 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 total current is 10-3

2417 x

10-6

amps

= 413.7 amps

.

.

.

(14.10)

Determination of Constant

From equation (12.10), page 143, the magnitude of the current density along the imaginary axis of the a plane is

J = V

B (1 +v2)1/2(1 + k12v2)1/2 168

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTIONS

where B is a constant yet to be evaluated and k1(from page 162) is 0.985. Hence the total current is B dv

of°D

(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 amps Hence

B=261.3

.

.

(14.11)

Distribution of Current Density

It is now possible to determine what the distribution of current density is along the axis of symmetry between the two electrodes, 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 is given by dX dw, J= dwdz

Now since, from the inversion of equation (14.3), z

dz

_

rg

rg

2K sn-1 (w, k1) 1

2K(kl)

dw

=

(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

dw

(1 __ w2/k2)1!2(1 -.w2)1/2

(from E.I.E.P. 732.01)

Note that the constant B has had to be reintroduced. Its 169

CONFORMAL TRANSFORMATIONS

value has already been found and given in equation (14.11). Therefore

.I =

2K(k1)B I (1 -w2)1/2(1 - k12w2)1/2 rg (1-w2)1/2(1-w2/'k2)1/2

2K(k1)B (1 -k12w2)1/2 rg 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 now required.

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) rg

z = F(jv, ki)

.

.

.

(14.13)

An interpretation of the elliptic integrals with imaginary arguments 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 0 Here jq is the amplitude angle so that in the present example jv =sin j0 = j sinh 0. Hence

tan fi = sinh 0 = v and 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)

out

the real and

Equation (14.15) enables us to separate

imaginary parts of equation (14.13). Selecting the imaginary parts of both sides and rearranging gives y

_

79

2K(kl)

F

v

(1+V2)1/2 '

170

k1, .

.

(14.16)

DOUBLE TRANSFORMATIONS: ELLIPTICAL FUNCTIONS

Then by putting w = jv into equation (14.12)

()

2K(kl)B (1+k12v2)1/2 (1+v2/k2)1/2 rg

14.17

Equations (14.16) and (14.17) give, for different values of v, corresponding values of current density and of y that can be plotted. The parameter v can have any real values from zero to infinity. In the example being considered, that of page 163, r = g = 2, and we already know that k1' = 0.173,

k1 = 0.985,

k = 0.923

and from equation (14.11), B=261-3. Hence from tables K(ki) = 3.156 Then from equation (14.17) 2 x 3.156 x 261.3 (1 +0.97v2)1/2 (1 + 1.175v2)1/2

4

(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)

(14.1/11.18)

from equation (14.16). From equations Table 15 can be calculated.

and (14.19)

TABLE 15 Calculation of variation in current density

0

0.5

1

2

0

0.447

0.707

0.894

0.949

0.981

0.995

1.000

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 2.175

4.880 5.700

9.730 11.58 0.917

25.25 30.380 0.912

98.00 118.5

378

376

375

v

a

(1

+V

2

d=1+1.175v2

1

P,(e'd)1/2

1

1.294 0.980

412

404

J = 412e

N

0.952 0925 392

I

381

171

3

5

10

00

0909 0908 374

CONFORMAL TRANSFORMATIONS 1

J 4

c 300

EE 200

tOO

1

O

FIG. 89.

02

06

O'4

Os

Inches

Y10

Current density distribution along the axis of symmetry between two electrodes on one edge of a rectangle.

Fig. 89 shows the variation in current density across the axis of symmetry between the electrodes plotted from the values calculated in Table 15.

172

CHAPTER 15

Elliptic Integrals of the Second Kind Incomplete Elliptic Integrals of the Second Kind

Elliptic integrals of the second kind are best introduced in the course of an example. Consider a semi-infinite strip with a rectangular projection as shown in Fig. 90 where the projection is labelled PQRS. Suppose, for instance, that the strip V

P

L_

Q w=-1

a

0

W.0

R

x

w.1

Fia. 90. Rectangular projection. z plane.

is an electrical conductor and has electrodes of opposite polarity placed at PQ and SR. The effective resistance between PQ and SR is required. Fig. 90 can itself be regarded as the z plane. There is only one ratio of dimensions, viz. RS/OR and there must therefore be one undetermined value of w in the z plane. Since there are four right angles in the configuration to be transformed there must be an elliptic integral contained in the Schwarz-Christoffel transformation. Therefore it is best to allot the values of w at the corners shown in Fig. 90. The transformation to the w plane shown in Fig. 91 is given by the Schwarz-Christoffel equation dz

dw =

1k') 1/2 A(w-1)-1/2(w+1)-1/2 w-

173

1

w+ 1)

1;2

CONFORMAL TRANSFORMATIONS V

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 S are each 3,r/2. Therefore

z=A

(w2- 1/k2)1/2 dw /

J

(w2- 1)1;2

1. (1-k2w2)1/2

Ak2 ff

(1-w2)12 dw

.

.

.

(15.1)

Consider now the integral (1 - k2w2)1/2

fw p

dw

(1 - 202)1/2

This is Jacobi's form of the elliptic integral of the second kind and it is denoted by E(w, k) where, as before, w is the argument

or amplitude and k is the modulus. As with the elliptic integral of the first kind it can be put into Legendre's form by making 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 where 0 is the amplitude of u. Then

sin 0 = sn u and by differentiating both sides cos 0 do = cn u do u du (E.I.E.P. 731.01) Therefore en u do u du do

(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)1/2dnudu 0

('u JI

dn2 u du

(15.4)

a

Equation (15.4) defines the elliptic integral of the second kind in terms of elliptic functions. Extensive Tables are available for numerical values of E(0, 0) ; they are usually found with tables of F(j, 0). Complete Elliptic Integrals of the Second Kind

If 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 the

modulus has to be named E(k). When w is made unity 0 becomes 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 complete integral K' with modulus k', the complete integral E has an associated 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 Projection

Return now to the transformation from the z plane of Fig. 90 to the w plane of Fig. 91 given by equation (15.1). The transformation can now be written

z = B E(w, k) +C where B = Ak2.

The origins in the two planes have already been chosen and from that choice the value of C is fixed and has now to be determined. Since, when z = 0, w = 0

0 = B E(0, k) +C = 0+C (from E.I.E. P. 111.00) 175

CONFORMAL TRANSFORMATIONS

Hence C is zero and

z = B E(w, k)

.

.

.

.

.

(15.6)

Let the corner R in Fig. 90 be the point z =a and let the corner B be the point z =a+ j b. At these points, w = 1 and w =1 /k respectively. In equation (15.6) first put z = a and w=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' .

E

a

(15.8)

The ratio b/a is known for any given configuration. Hence for any value of this ratio there is a corresponding value of Ic y P

r- a Q

z.-a FIG. 92.

p

(K!E') R

z.a

x

Values in the z plane.

satisfying equation (15.8). In Fig. 92 the z plane is shown again with values of z given both in numbers and in terms of elliptic integrals. 176

ELLIPTICAL INTEGRALS OF THE SECOND KIND

For a numerical example let a= 1-25b. Then

K'-El B

= 0. 800

It is now necessary to choose or estimate a value for the modulus

k and check whether for this modulus the equation holds. In this example choose k=0-400. Then from tables of complete elliptic integrals K'=2-359, E=1-151 and E=1.506. From these values K'-El

E

= 0. 802

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 is necessary to transform from the w plane to the X plane where

WR -}

AEr-

FL W-1

FiG. 93.

W. I

W-M

1BH

C

0

W=0

0

W-1

Rectangular projection. X plane.

the field is regular. The X plane is shown in Fig. 93. From this 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 plane of Fig. 91 is given by equation (12.6), page 140: w = sn (X, k)

12

177

.

.

.

.

.

(15.9)

CONFORMAL TRANSFORMATIONS

Hence as in previous results the effective resistance is

R = p K to modulus k where k has been determined by the first transformation. In the present example where a= 1.25b the modulus has been determined by equation (15.8) to be 0.400. Hence from tables of complete elliptic integrals K' = 2.359 as before and K =1.640. Hence R = 1.390p ohms

for unit thickness of conductor. Total Current

To find the distribution of current density it is necessary to find first the total current flowing in the material. The effective resistance when the conductor is one inch thick and the resistivity 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 total current is _ 10-3 1.147 x 10-6

amps

= 872 amps

.

.

.

(15.10)

Determination of Current of Integration

On page 168 the current density distribution along the imaginary axis of the uw plane is given by B V

(1+v2)1/2(1+k2v2)1112

where in the present example k = 0.400.

Then

k' = (I-k2)1/2 = 0.917 Therefore, 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

ELLIPTICAL INTEGRALS OF THE SECOND KIND

Hence, from equation (15.10), B = 370

.

.

.

.

.

(15.11)

Distribution of Current Density

The current density at any point in the z plane is, as already shown, given by

J=

dX dw

dw dz

Now from equation (15.7), page 176,

z = E E(w, k) and from equation (15.2), page 174, a (1 - k2w2)1/2 dw E (1 -2'2)1/2 dz

Furthermore, from equation (15.9), X = B sn-1 (w, k)

.

.

.

(15.12)

where the constant B has had to be reintroduced to enable us to obtain the actual current densities. The value of B is given by equation (15.11). From equation (15.12) dX

dw

=

B

(1-w2)112(1-k2w2)1%2

(E.I.E.P. 732.01).

The product of the two differential coefficients gives

j _I _

(1-w2)1/2 BE a(1-w2)1/2(1-k2w2)1/2 (1-k2w2)1;2 l

BE a(1- k2w2)

(15.13) I

Equation (15.13) gives the magnitude of the current density at any point in the z plane in terms of w ; it is now necessary therefore to determine corresponding values of y also in terms of w. Equation (15.7) gives z = E E(w, k) 179

CONFORMAL TRANSFORMATIONS

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 given in E.I.E.P. 161.02 thus :

z=j

a 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 sin P =

V

(1 +v2)1/2

Equating imaginary parts of both sides '2 ,2

Y=

{F((1+x2)1/2'

k')-E((1+x2)1/2, k')+v(1-i+x2) /

1,2

\

(15.14)

The corresponding values of current density are found by putting w = jv into equation (15.13) thus :

all

BE 2)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) +v

(1+0.16'02 1+v2

11/2

(15.16)

and

I=

.

446

1 + 0.16v2 180

(15.17)

ELLIPTICAL INTEGRALS OF THE SECOND KIND

By taking various values of v from zero to infinity, corresponding values of current density and of y can be calculated from equations (15.16) and (15.17). Typical values and calculations are shown in Table 16. TABLE 16

Calculation of current density distribution 0

0.5

1

a-(1+v2)112

0

0.447

b=F(a, 0-917)

0

c = F(a, 0-917) b-c

0 0

0 + 0.16v2) d= (1+ 0.16v2)1/2

1

e=ad

2

3

5

10

0.707

0-894

0-949

0.981

0.995

0-477 0-449 0-028

0-861 0.721

1.345 0.936

1-647 1.014

0-140 0409 0.633

1.896 1-071 0-825

2-124 1-097 1-027

0

1-040 1-020 0-456

1.160 1-077 0-761

1-640 1.281 1-145

2-440 1.562 1.482

5-000 2.236 2-194

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

429

384

272

183

89-20

26-24

V

1

446

1+0. 6

1

+

17-00

4.123 4-102

Fig. 94 shows the variation in current density across the axis of symmetry between the electrodes plotted from the values of J and y obtained from Table 16. Periods of Elliptic Functions

It was shown in Chapter 12 that sn X = sin 0. Now since sin 0 is periodic sn X must also be periodic, although the length of the period is not necessarily the same. While the period of sin 0 is 2ir that of sn X can be determined as follows. Starting from do X

- Jo

(1 -k2 sin2 #)1/2

the increase in X over a complete period of 0 is f +2A 0

do do _ f 2, (1 - V g1I12.)1/2 (1 -k2 sln2 0)1/2 Jo 181

CONFORMAL TRANSFORMATIONS

00

300

200

100

2

3

S

Inches

Fm. 94. Current density distribution along the axis of symmetry between the sides of a projection.

But owing to the symmetry possessed by the function the increase in x over the whole cycle of sin is four times that over a quarter cycle. Thus the increase in X over the cycle (which gives the period of sn x) is 4

d

,,0

117/2

(1 -k2 Sln2 )1/2

4K

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. They

are 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 of sn X, en X and do X plotted against the argument for a modulus of 0.8 taking real values of the argument. 182

ELLIPTICAL INTEGRALS OF THE SECOND KIND k.o e f(X

Fla. 95. Curves of sn, en and do X. Periods of Functions with Complex Arguments

In the previous section real values of the argument only were considered. But it is usual in conformal transformations to

find that the argument of the functions is complex. It then appears that elliptic functions have an additional period, and it can be shown that the ratio of the two periods must be imaginThis means that compared with trigonometrical functions which have real periods and with hyperbolic functions which have imaginary periods elliptic functions have two periods, one real and the other imaginary. Elliptic functions are therefore called doubly periodic. To find the imaginary period of sn u we note that ary.

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 also en u and do u, E.I.E.P. shows that the periods of these principal elliptic functions are sn u : en u : do u :

4K and 2jK' 4K and 2K + 2jK' 2K and 4j K'

However, the elliptic integrals, as distinct from the elliptic functions are not periodic. Fig. 96 shows curves of the incomplete elliptic integrals of the first and second kinds plotted against the arguments with modulus 0.5. 183

CONFORMAL TRANSFORMATIONS K I-$ F(O) £(O)

0 2

Fic. 96. Curves of

0

and

Because the integrals are not periodic it is desirable to be able to use certain auxiliary functions to be developed in the next chapter. These auxiliary functions are required for the problems of Chapter 17.

184

CHAPTER 16

Auxiliary Elliptic Functions WHEN THE INTEGRALS RESULTING from a Schwarz-Christoffel

transformation either are or contain elliptic integrals but yet not in any standard form, it is often necessary to have recourse to one or more of the auxiliary functions described in this chapter. They arise particularly in the expansion and the development of elliptic integrals of the second and third kinds. Jacobi's Zeta Function

One of those most commonly met is Jacobi's Zeta function denoted by Z. This Z function must not be confused with the better-known zeta function of Riemann denoted by . Jacobi's function is similar to the function E(c, k) but it is periodic with only 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 Function

More important still are Jacobi's Theta functions. His original Theta function is denoted by O. The functions are 185

CONFORMAL TRANSFORMATIONS 4. sin 750 0 Z(0)

0.

O

2K 4

2

5

V,

-0.2

-0 FIG. 97. Curve of Jacobi's Zeta function.

widely used and they are indeed almost indispensable when numerical 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 imaginary 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 by equation (16.2) is the original Jacobi's Theta function and is still the principal one. In this section therefore it has been called the Theta function. Series for the Theta Function

Jacobian elliptic functions can all be expanded into various kinds of series and E.I.E.P. 900 to 908 gives the different series for the principal functions. The series for the Theta function is 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 m=1

K

.

.

(16.4)

where q = e-nK'/K

(16.5)

The quantity q defined by equation (16.5) is a basic number in the numerical evaluation of elliptic functions and it appears in most tables of elliptic integrals and functions. It is called the `nome'. Note that when K 00

0(K) = 1 +2 = 1+2q+2q4+2q9+ ... and that when ¢ is zero .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 expression with the complete elliptic integrals to modulus k' is denoted by the primed nome q'. 187

CONFORMAL TRANSFORMATIONS

The Function 91

Another Theta function is defined in terms of the principal function by

01(0) = 0(4+K)

.

.

.

.

(16.6)

This function has the series qm-2

01(9c) = 1 + 2

cos

m7TO

K

1

Note from the series that when = K 01(K) = 1+ 2

(-1)mgm2

= 0(0)

and that when 0 is zero 01(0) =

1+2

q

2

= 0(K)

Jacobi's Eta Functions

Closely related to Jacobi's Theta functions are his Eta functions 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)nrr 2K

Note that by putting zero we find H(0) = 0 Putting 0 = K gives 00

H(K) = 2

(-1)m-1q(m-1/2)2 sin

2m-

1

V

2

=281/4+289/4+ ...

The period of H is shown in E.I.E.P. 1051.03 and 1051.07 and it is seen that the function is simply periodic with the period 4K. The other Eta function denoted H 1 is defined in terms of H

by the relation

H1(0) = H(¢+K) 188

.

.

.

.

(16.8)

AUXILIARY ELLIPTIC FUNCTIONS

and it has the series 00

q(m-1,2) a cos

HI(o) = 2

(2m-1)07r

1

2K

Note that H1(0) = 2

q(m-1/2)2

= 281/4+289/4+

= H(K)

and that H1(K) = 2

q(m-1/2)2 cos

2m- I

_7r

2

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 by The relations between the old and the new functions are given in E.I.E.P. 1050.01. While the ao(v),1&1(V), 82(v) and klr3(v).

newer forms of Theta functions are those encountered in mathematical treatises the older forms are found to be the more useful in conformal transformation calculations.

189

CHAPTER 17

Elliptic Integral of the Third Kind THE ELLIPTIC INTEGRAL of the third kind is another integral best

introduced in connexion with an example. Any manipulations of this integral usually lead to the introduction of one or more of the functions described in the previous chapter. Succession of Equal Slot Openings

The most suitable example for this purpose is that of a succession of equal slot openings opposite a solid face, particularly since the results are well worth comparing with those obtained from the problem of a single slot opening covered in Chapter 9. In that chapter the effect of a single slot opening on the distribution of the magnetic field between two plane magnetic surfaces was discussed. On page 104 it was explained that the actual problem encountered is that of a succession of equal slots and that sometimes it should not be simplified to the problem of one slot. Let us now consider therefore the more difficult but more practical condition of a number of equal slot openings opposite a solid face and across an airgap as shown in Fig. 98. This problem is one of those discussed by Carter in his 1926 paper * and much more fully by Coe and Taylor in 1928.* The solution requires the use not only of the functions described in Chapters 12 and 16 but also of the elliptic integral of the third kind.

Fic. 98. Succession of equal slot openings. * See List of References, page 217. 190

ELLIPTICAL INTEGRAL OF THE THIRD BIND

It is unnecessary to use the whole of Fig. 98 to find a solution because the pattern of field distribution keeps repeating itself. It is only necessary to take the smallest repeating section of the field 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 ; therefore the portion of airgap space to be considered is that portion

bounded by the straight line AB and the path EFGH. C

A E

Y

n

G

D

B

Image pattern.

FIG. 99.

In this problem there are three parameters fixed by the design, viz. the slot opening s, the airgap length g, (both as found in Chapter 9) and in addition the tooth width t. Thus there are two ratios instead of one and it is convenient to take either g,s and t;s or their reciprocals. The z plane is therefore the configuration shown in Fig. 100

where the portion HG can be considered to have a potential V above zero while the image portion EF has the potential - V. The lines AB and FG are boundary lines of flow. The potential is zero along the y axis. y w=-W

}

W.- oo

W.-I 1h 2f C

1

F W= 1

0

G

W=0

w=1

FIG. 100. z plane. 191

D

CONFORMAL TRANSFORMATIONS

Since there are two ratios in the z plane there must be two values of w at the corners left undetermined, but owing to the symmetry in the figure one set of two can be the negative of the

other set. In the configuration to be transformed there are

four right angles and this means that the integral in the Schwarz-Christoffel transformation must either be, or must contain, an elliptic integral. Therefore it is desirable to make the undetermined values of w equal to 1/k and 1/k,, so that k and k1 become 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

z=A

(w-1)-1/2(w+1)-1/2

w-kl (+) 1

1/2

1

1/2

x

(w-k1)-1(w+k11-1 dw (1-k2w2)1/2 dw

= C J (1-k12w2)(1-w2)1/2

(17.1)

where the new constant C in terms of A is C = - k12A/k Now the form of the integral in equation (17.1) and the presence of the two moduli suggest that an elliptic integral of the third kind is present ; therefore the best procedure is to try to manipulate the integrand to find this integral. Hence the next step is to examine the elliptic integral of the third kind so that it is readily recognised. Elliptic Integrals of the Third Kind

Legendre's form of the elliptic integral of the third kind is defined by the equation w

11 (w, k1, k)

=

dw

o (1-k12w2)(1-k2w2)1/2(1-w2)1/2

.

(

17.2 )

Notice the presence of two moduli in the integral. As in the elliptic integral of the first kind another form is found by putting w = sin 0 giving

ki, k)

fo (1--k12 sing 0)(1-k2 &n2,6)1/2 192

(17.3)

ELLIPTICAL INTEGRAL OR THE THIRD KIND

There is yet another form of Legendre's integral derived from that of equation (17.3) as follows. If the integral is denoted by it, 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) =

du

(17.4)

.

01-k12sn2u

IU

All these three forms of Legendre's elliptic integral of the third kind are given in E.I.E.P. 400.01. Jacobi's Elliptic Integral of the Third Kind

Jacobi adopted not merely a different form but an entirely different definition of the elliptic integral of the third kind and it so happens that his definition gives us an integral rather more. suitable for the kind of analysis required in conformal transformations. Equation (17.4) is modified by introducing a new parameter 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) becomes du 1- k2 sn2 asn2u .

Jacobi used this integrand in his definition of the elliptic integral of the third kind and for the full definition wrote fI (u, a) = k2 sn a en a do a

f

sn2 u du

u

0

2 1-ksnza snz u

(17.6)

This is the definition quoted in the footnote to E.I.E. P. 400.01. Jacobi's definition of the integral can be expressed in terms of his Zeta function and also in terms of his Theta functions as follows.

2snuenadna

sn (u+a)+sn (u-a) _ 1- k2 sn2 u sn2 a 13

193

(E.I .

E.P. 123.02)

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. 142.02

k2 sn a sn u sn (u+a) = -Z(u+a)+Z(u)+Z(a) and

k2 sn a sn u sn (u -a) = Z (u - a) - Z(u) + Z(a) Hence

u {Z(u-a)-Z(u+a)+2Z(a)} du

N(u, a) =

(17.7)

J0

Equation (17.7) gives a definition of Jacobi's elliptic integral of the third kind in terms of his Zeta function. From equation (16.3), page 186, f('u

Z(u) du = log O(u)+C

.

.

.

(17.8)

Jo

Hence from equations (17.7) and (17.8)

II(u, a) =

rr

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 elliptic integral of the third kind this time in terms of his Theta and Zeta functions. Transformation to the w Plane

It is now possible to return to the transformation of equation 17.1 and manipulate the expressions to yield results in familiar forms. Recall first what we expect to obtain from this first transformation. As in all previous examples involving elliptic functions the second transformation gives the answers we are seeking to the various problems ; but the answers are in terms of elliptic functions whose modulus or moduli we have to determine. These moduli are determined from the first transformation in terms of the known ratios of the configuration in the z plane. Thus for the problem now in hand the first transformation should give values of k and kl in terms of the two ratios g/s and t/s. 194

ELLIPTICAL INTEGRAL OF THE THIRD KIND V

AE -1K,

0

YK

z.-g+jit

z. -w-jZt

BH

G

F

z--'9

1

1/K

z-g

YK,

z-m+j't

z=g+jlt

Fio. 101. w plane.

Equation (17.1) gives the first transformation, that from the z plane of Fig. 100 to the w plane of Fig. 101. One familiar form we are seeking is the integrand of equation (17.6). First put

w = snp

.

.

.

.

.

(17.10)

Then

dw = on p do p dp (E.I.E.P. 731.01) But do p = (1- k2 sn2 p)1/2

(E.I.E.P. 121.00)

Hence

dw = cnp(1-k2sn2p)1/2dp Equation (17.1) in terms of p becomes z

p (1 - k2 sn2 p)'/2(l - k2 sn2 p)112 on p dp

(1-k12sn2p)enp p 1-k2sn2p = C fo 1-k12sn2pdp Cfo

C

1-=

( \1+1-k2sn2p-1+k12sn2p\

dp

dp (17.11) } l fp0 I -k12 sn p)) Now reintroduce the parameter a as in equation (17.5) :

C{p+(kl2_k2)

k1 = ksna Then

k12-k2 = -k2(1 -sn2 a) = - k2 cn2 a (E.I.E.P. 121.00) 195

CONFORMAL TRANSFORMATIONS

Equation (17.11) then becomes

z = C(p - k2 cn2 a

sn2 dp Po 1-k2sn2asn2p p

1

``

= C,}p-

ell a

snadnu rt(p, a)}

(17.12)

from equation (17.6), page 193, and all the functions are to modulus k. Determination of Constant

The constant C in equation (17.12) can be determined by choosing a point where the values of both z and w are known and then inserting those known values into equation (17.1). Referring to Fig. 101 we see that the value of w as we move along the positive real axis from the origin passes through the point 1/k1. Fig. 100 shows that the corresponding value of z changes abruptly at the point corresponding to w = 1 /k1 by the amount 'js. Hence from equation 17.1 1CJs

C1

w (1- k2/k12)1/2 dw 1/k1 E fO I -k1w}1/kl--e (1 + k12(k12)(1 - 1/k12)1/2

Notice that w =1 /k1 is not substituted into the remainder of the integrand shown above because doing so would render the result indeterminate. Simplifying the expressions and performing the integration gives (k2 - k12)1/2

2Js = C 2(1

1/k1+E

1 -k12)1/2(-k1)

log (kiu -1)}141-E

Now as w passes from zero through the value 1/k1 the expression (kiw - 1) changes from negative to positive. Reference to Dwight 604.1 shows that log (k,w-1) increases by the amount

- jir if the principal value is taken because the logarithm of a negative number differs by jw from the logarithm of the corresponding positive number. Therefore the change in value from Ilk, -E to 1/k1+E of -(1/k1) log (kiw-1) is j7r/k1 so that (k2 - k12)1/2

2j`S = j

2k,(I - k12)1/2 196

ELLIPTICAL INTEGRAL OF THE THIRD KIND Therefore

C _ s k1(1 -k,2)1/2 .r (k2-kl2)1/2

Now substitute k sn a for k1:

C, - s k sn a(l - k2 8112 ken a T

a)1;'2

ssnadna

(17.13) (E.I.E.P. 121.00) cn a Now k and k1 are real numbers since they are points on the real axis of the w plane, and since from Fig. 100 k is greater than k1, a must also be real. Hence all the other functions in equation (17.13) are real. Therefore C is a real number. Substitution of C from equation (17.13) into equation (17.12) ar

gives z

l _ s{psnadna-1(p a)} cna J

(17.14)

Equation (17.14) gives values of z virtually in terms of w because from equation (17.10) p = sn--1 w.

The next step is to obtain some equations relating the parameters k and a to the ratios g/s and t/s. This means eliminating

the variable p from equation (17.14), and it can be done as follows. Elimination of p

The quantity p must have a definite value depending on the dimensions in the z configuration. That value can be determined by considering another pair of corresponding values of z and w. At the point where w = 1j k, z has the value g + s j t as shown in Figs. 100 and 101. Inverting equation (17.10) p = sn-1 w u

=

dw ( 1 - w2)1/2(l - k2z(. ,2)1/2

o

(E.I.E.P. 130.02)

and therefore at w =1 j k 1 /k

P

fo

dw (1-w2)1j2(l -k22C2)1/2

= K+jK' (from equation (11.15), page 133) 197

CONFORMAL TRANSFORMATIONS

Substituting these values for p and z into equation (17.14) gives 7Tj(K+jK')sn

9+zjt =

a-II(K+jK')a)) }

on a

(17.15)

All that remains is to equate real and imaginary parts of equation (17.15) and thus produce the direct relation between the ratios ,g js and t/s and the parameters a and k. But first it is necessary to find out what are the real and imaginary parts of

the elliptic integral of third kind which has the argument K+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 equation (17.8), page 194. The integration of cs u do u can be found thus :

resudnudu J

=

fen udnudu (E.I.E.P. 120.02) snu

f dsnu snu

(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

ELLIPTICAL INTEGRAL OF THE THIRD KIND

By subtraction

log O(K-a+jK')-log O(K+a+jK') O(K-a) sn (K-a) = log O(K+a)+1og sn

j 'r (K+a)+2K(2a)

From E.I.E.P. 1051.03 l0g

O(K-a) - to O1(-a O(K+«) - g 01(a) (E.I.E.P. 10:51.02)

= log 1

=0 Similarly log

sn (K - a)

sn (K+a)

= log 1

(E.I.E.P. 122.03)

= 0

Hence

log @(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 imaginary

part is K+K' Z(a)

With this result, equation (17.15) can now be separated into real and imaginary parts yielding the two equations g

_ Krsnadna-Z(a)} en a

s

it

t

2K sn a do a a cna - Z(a) I-W

s

(17.18)

}

The two equations (17.18) give the relation between the known ratios g/s and t/s and the parameters a and k. Numerical Example

Before 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

CONFORMAL TRANSFORMATIONS

Tables quoted on page 145 will be used again. From these tables

K = K' = 1.845 sn a = 0.765 on a = 0.644 do a = 0.841 and from the tables in E.I.E.P. K Z(a) = 0.2715 Hence from equations (17.18)

-

1.845

g

s

5 I- 0271 1.845

= 0.501

t s

_

2 x 1-845 IT

j, 1

0-2715 1.845 )

0.927 1.845

= 1.002 - 0.502 = 0.500

Hence when the slot width is twice the tooth width and twice the airgap length we know that in the problems we wish to solve k is 0.707 and a is 0.927. Note that in some tables of elliptic

functions the values are listed in terms of modular angles. These angles are given by 0 = sin-1 k

k1=ksna

= ksin0

Hence

= 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 as K Z(P) where fi is written for 0. In this numerical example we started with arbitrary values of L-. and a and found the corresponding values of g/s and t/s. In practice this is, of course, starting from the wrong end. The 200

ELLIPTICAL INTEGRAL OF THE THIRD KIND

ratios in the z plane are what we know for they are fixed by the

Hence we have to start with them and find k and a. This necessitates first finding a whole series of results starting design.

with the parameters and then using the method of crossplotting, so that values of k and a can be read off for any given combination of g/s and t/s. This will be discussed again later in the chapter. Transformation to the X Plane

What we have obtained so far is the transformation from the z plane to the w plane in terms of the parameters k and a and

hence the relations between the parameters and the actual dimensions in the configuration, The second part of the transformation is that between the w plane and the X plane where the field is regular.

P

-1p

W=1

FiG. 102. X plane.

The X plane is shown in Fig. 102 and the equation of transformation 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 appropriate value of A can be found in the usual way as follows. Fig. 100 shows that when w is unity the value of X must be the potential between HG and the origin, viz. V. Thus V = A sn-1 1 = AK(k1) (from (12.8), page 140) Hence

A = V/K(kl) 201

CON FORMAL TRANSFORMATIONS and

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 a single slot opening was assumed. Then a comparison can be made between the two sets of results. In Chapter 9, after the transformation is established the first problem considered is that of the distribution of flux density. Distribution of Flux Density

As before the magnitude of the flux density at any point in the z plane is given by,ao IdX/dz I. Therefore denoting the flux

density by B B =

dX dw µo

dw dz

From equation (17.19) and using E.I.E.P. 732.01 V

dX

dw

1

K(ki) (1 - w2)1/2(1 - k12w2)1/2

and from equation (17.1) dw

dz ~

(I -k12w2)(1 -w2)1/2 C(1 -k2w2)1/2

where C is the constant determined by equation (17.13), page 197.

Hence (1 -k12w2)1/2I

V

B

CK(k1) (1-k2w2)1/2

H`°

I

and substituting for C from equation (17.13)

B

_

V

cn a

sK(k1)snadna

(1 -k12w2)1/2 µo (1 - k2w2)1/2

(17.20)

Maximum and Minimum Values of Flux Density

The maximum value of flux density occurs at the point on the solid face which is exactly opposite the centre of a tooth, 202

ELLIPTICAL INTEGRAL OF THE THIRD KIND

i.e. at the point shown in Fig. 100 where w is zero. Hence, by putting w=0 into equation (17.20), Bmax =

V ena sK(kl)snadnat`c

(11.21)

Therefore equation (17.20) can be written B=

(1-k12w2)' 2 (1-k2W2)1'2 (Bmax

(17.22)

The minimum value of flux density occurs at the point on the solid face which is exactly opposite the centre of a slot opening, i.e. at the point shown in Fig. 100 where w is infinite. Hence by 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 slot openings is given by the equation

B.I. - sn a Bmax

.

.

.

.

.

(17.23)

In the numerical example of p. 200, where g/s is 0.501 and t/s is 0.500, it was found that sn a = 0.765. Hence the magnitude of the tooth ripple for these ratios is given by the relation Bmin Bmax

= 0.765

The Mean Flux Density

Referring to Fig. 10O the total flux between AB and CD divided by the area of the airgap enclosed by these lines gives the value of the mean flux density, Bmean. The area is clearly I (s + t) while the total flux is given by the increase in X as w

increases from unity to Ilk,. From equation (17.19) this change in X is {sn_1 k - sn-1 (1)}1

203

all to modulus kl

CONFORMAL TRANSFORMATIONS

This expression interpreted by equations (12.8) and (12.9), page 141, becomes

K(K+jK'-K) VK' K

o modulus k1

Hence

B mean =

2 VK'µo

K(t+s) V µ02K (g/s)

g K(1+t/s)

to modulus k1

.

(17.24)

If the solid face were opposite an unslotted surface (i.e. opposite another solid face) with the same airgap length g and the same potential difference V, the flux density would have the uniform value of .oVi`g. Hence it is reasonable to call this value of density unity. Then Bmean = 2K'(gls)

K(l + (s)

to modulus k1

.

(17.25)

From equation (17.21) for the same value of unit density ng ell Ix Bmax = sK(ki) sn a do a

.

(17.26)

Numerical Example

It is desirable now to consider a numerical example iii order to compare the results with those obtained in Chapter 9 for a single slot opening.

Take the following values. k = 0.675 0 = sin-1 0675 = 42.4° a = 1.071

From tables sn a = 0.8396.

Then

= sin-1 0.8396 = 57-l' 204

ELLIPTICAL INTEGRAL OF THE THIRD KIND

Also from tables

en a = 0.5426 do a = 0.8234 K(k) = 1-819 K'(k) = 1-893 KZ(a) = 0.2265 Therefore

Z(a) = 0.1245

kl=ksna=0.566 K(k1) = 1-726

K'(kl) = 2.047 From the above values and equation (17.18) 1.819 (1.275-0.1245) s

g

7r

= 0.667

Notice that this is the value of g/s used on page 114 where, however, it is written 8/g =1.5. Again from equations (17.18) t

s

=

2 x 1.897

1.071

x1 1505-1819

it = 1.389 - 0.589 = 0-800

According to the equations derived in this Chapter, for these values of g/s and tls the various values of flux density are : Bmean

2 x 2-047 x 0-667 1.726 x 1.800

(from equation (17.25))

= 0.878 Bmax =

0-6677T 1.726 x 1.275 )

(from equation (17.26))

= 0-952

Bmi = 0-952 x 0.8396 = 0.800

(from equation (17.23))

Now consider a single slot opening for the same ratio of g/s. 205

CONFORMAL TRANSFORMATIONS

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 is the same for a succession of equal slot openings as it is for a single slot opening, but the value of Bmax is less, the difference depending on the width of the teeth. The Flux Density Curve

Then in

The term µoV/g has been called unit flux density. terms of unit flux density equation (17.20) becomes _ 7rg/s cn a (1 - kl2w2)1/2 B K(ki) sn a do a (1- k2w2)1i2

(17.27)

From this we want to find the variation in value of the flux density along the solid face, i.e. the distribution of flux density

along the lines z=jy and w=jv. Substituting jv for w in equation (17.27) gives

B_

cn a (1- k12v2)1'2 K(k1) sn a do a (1 +k2,r2)1/2 7rg/8

(

17.28) -

Equation (17.28) gives B along the imaginary axis of the w plane and it is now necessary to find a correspondence between y in the z plane and v in the w plane. From equation (17.14); page 197,

s[psnadna

rl cn a (p' By inversion of equation 17.10 page 195 z

7r l

a)

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)

.

.

.

.

.

206

.

.

.

.

.

.

(17.29)

ELLIPTICAL INTEGRAL OF THE THIRD KIND

where fi can have values between zero and K'. Substituting for p in equation (17.14) gives

S Snadlla Jg- II(jg, a)

Z

IT

I

on a

and from equation (17.9), page 194, this becomes z

=

s snachi a,Jf3-j(i Z(«}- 2 log Ei(a-jp}

()(a+jp) To break this equation up into real and imaginary parts we en a

7,1

have to find the imaginary part of the log term. The following well-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,

O(a+jfl) = 1+2

(-1)mgm2cos

irm

(a+jfl)

(-1)mgm2 x

=1+2 1

(cos K a cosh K fi- J sin K a sink K Therefore the imaginary part of log ®(a + jfl) is given by

-2

(-1)mgm2

sin Ka sinh

xP

tan-1 1+2

(-1) mgm2 cos

Ka cosh

Kfl

Similarly the imaginary part of log 0(a - jB) is given by

+2

(-1)mgm2

tan 1+2

sin Ka sinh

K_

(-1)mgm2cosKacosh -_

Now tan-1 A = - tan-I (- A) and therefore tan-1 (alb) - tan-1 (- a/b) = 2 tan-1 (a/b) 207

CONFORMAL TRANSFORMATIONS

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

(- 1) mg m2 s i n

_K sink -

= tan-1 1

1 ) mg m2

+2

cos

7rm a

cos h

irn 8

K Therefore equating real and imaginary parts of both sides of equation (17.30) gives, for the imaginary parts,

y= 7r II

K

sn a do on a

(- 1)mgnz 2 sin Ka sinh -r

-2

tan'

1 + 2 , (- 1)mgm2 cos

ka cosh 'rh-

(17.31)

'

1

Thus the procedure is to take a series of values of v. Then for each value of v the flux density is given by equation (17.28). Next from the tables find

F and call this fl.

(1

v

+22)1/2

k

,

Using that value of fi find y/s from equation

(17.31).

A numerical example follows. Numerical Example

To keep continuity in the numerical examples use the values

of g/s and tJs found on page 205 that were obtained from k=0.675(0=42-4-) and a=1-071. The values of the various elliptic integrals and functions required are given on page 205 and need not be repeated here. From equation (17.28) and using these values of the functions B

-

0.952(1 + 0.321v2)1/2

(1+0.456v2)1/2 208

(17.32)

ELLIPTICAL INTEGRAL OF THE THIRD KIND The next thing to find is the value of the nome q. is defined by equation (16.5), page 187, as q = e-nK'/K e-3.27 = 0.038

.

This quantity

.

.

(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 are quite negligible. Hence in equation (17.31) put ?n unity and q=0.038 and use the values of the functions found on page 205. Then y 8

-

2(0.038 sin 0.5897r sinh 7r,8/K) 1- 2(0.038 cos 0'5891r cosh 7rr/K) 0.073 sink 1.729P } (17.34) 1.159+tan-1 _ 11 1 + 0.021 cosh 1.729P j 1

1

where by equation (17.28) r8 is calculated from

P = r'

v

(1 +v2)1/2' k

,

Now k' = (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 v and the corresponding calculations of y are shown in Table 17. The curve showing the wave form of the flux density distribution as calculated in Table 17 is shown in Fig. 103. It is interesting to compare this curve and these results with those obtained by assuming a single slot only. They were in fact calculated for the same value of g;s in Table 11, page 115. The resulting curve is shown in Fig. 64. The tooth width chosen then was such as to allow B/Bmax to reach a value as close to unity as the eye could detect in the drawing. But if the tooth width is reduced until the ratio t/s is 0.8 (the value used for Fig. 103) and the abscissa scale changed to suit Fig. 103, the

resulting 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

B Flux Density

up)

9

9S .1.5

0

Fic. 103.

i-1.25

2

1

Wave-forui of flux density with a succession of equal slot openings.

10

09F

0$

B Flux Density

(per u nit)

a .

9

9

s

f=

t

1 2s

1 -V

I

1

FiG. 104.

,

O

2

3

Comparison of waveforms.

It is clear that for narrow tooth widths the more complicated equations (17.28) and (17.31) must be used when the amplitudes of the harmonics in the flux density wave are required. The discrepancy is marked and raises a doubt now to be investigated as to whether equation (9.28), page 121, is sufficiently valid to give Carter's coefficient for ratios of g/s and t/s such as we have just used. In the following sections we shall see that the simple equation, surprisingly enough in view of Fig. 104, does give the coefficient with remarkable accuracy. 210

(

1

V

v

0.689 0.721 1.311 1.025 1.423

0.321

0.321 1.149 0.456 1-206

0.0288 1.014 0.041 1.020

0

-

-

B=0-952p

0.994 0-946

1

0-952

1

n=(1 +m)1"2

p_hfn

0

1

,n = 0.456v2

h=(1+g)1'2

g=0.321v2

0.907

0.953

0.955

0-877

0.921

1-234

0-210

13

0.180

0-338

0.138 0-137

1.047

0.145

2-219

0

3.118 3-274 0-228 1.069

1.981

(for s=1.5) y=1 5 (e+f)1n

1-855

1.435

0

c1d

d=1+0.021 cosh b

1

0

f =1.15)3

56.3 1-073

0.832

45.0 0.830

0.707

3-25

1.5

0-0378 0.038

-

-1

0 0

16.7 0-294

0-287

--- -

e=tan-1 (c1d)

0

0

0

0

1.09

0-3

0 1,021

i

I

1

0

0-508 0.530 1-132 0-0387 1.024

b =1.729P sink b cosh b c = 0.073 sinh b

+ v2) 11Z 4=sin-' a =F(¢, k)

a

1 +vZ

TABLE 17

-

-

I

5

2

0 .836

0.899

1-282 1.511 1.823 1.680

0.798

1.405

0-273 0.267

2-113 4.076 4.197 0-297 1-088

63.4 1-222

0.894

--

Calculation of flux density distribution

0.831

0-873

0.812

0.853

3.520

3-002 1 1.30

8.01

1-107

1.853

0-503 0.466

1-171

2-785 8.069 8.131 0.589

1.611

4.10 2-258

f

0.981

26

5

78.8

-- -

1-971

2-882

0.965

1-653

0-386 0.368

2-485 5.959 6-042 0.435 1-127

1-137

71.6

0.949

10

3

i

---

0.800

0.840

0.675v

oc

0-566v

] 347

2-177

0-753 0.645

0-962 1.277

3-273 13-18 13.21

90.0 1.893

1

Oc

m

CONFORMAL TRANSFORMATIONS Carter's Coefficient

Carter's coefficient was defined and developed in the final problem of Chapter 9. On page 123 it was shown that the coefficient 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 of airgap that would give with unslotted surfaces the same mean flux density as actually appears. Hence Bmean =

V

7

Po

Therefore V ito

g'Bmean

Hence from equation (17.24), page 204,

C - K(I+t/s)

(17.35)

2K`g/s

where the elliptic integrals are to modulus k1. Numerical Examples

For actual figures consider the example for which Fig. 103 was calculated and drawn. The values of the relevant elliptic functions and integrals were given on page 205. From those values and the subsequent calculation K =1.726, K'= 2.047, t/s = 0.800 and gjs = 0.667. Substituting these values into equation (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 the equation governing a single slot opening, i.e. equation (9.27), page 120,

v=

l

tan-1

2g

-91og (1+s2/4g2)}

= 2 (tan-1 0.75 - 0.667 log 1.5625) IT

= 0.221 212

ELLIPTICAL INTEGRAL OF THE THIRD KIND

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+ 1

C

tj8+1-v 1.3

1-8-0-221 = 1.14

exactly as was obtained in equation (17.36) by using the more complicated equations governing a succession of equal slot openings.

Although these results show excellent agreement by the two methods it is not an isolated agreement. To show this consider as a contrast an example where C is large, i.e. where, as in an induction type machine, the airgap length is small while at the 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 = 0.334 on a = 0.942 do a = 1.000

K Z(a) = 0.000678 K(k) = 1.572 K'(k) = 4.340

k1 = ksna = 0.1745 K(kl) = 1.571 K'(ki) = 5.435 These are all the values needed for the calculation. From equations (17.18), page 199, 1.572 s

s

Tr

(0-361-0-0004)

= 0.180

t

8.680 (0.3606) -0-216

8

= 0.780 14*

213

CONFORMIAL TRANSFORMATION'S

Hence

C=

1.571 x 1.780

2x5.435x0.180 = 1.43 . . .

(from equation (17.3:5)) .

.

.

.

.

.

(17.37)

again to three significant figures. Now obtain Carter's sigma from equation (9.27), page 120, using the values of g/$ and t1:8 obtained above :

n

(tan-1 2.78 - ()-IS log 8.73)

= 2 (1-2255-0-3892) = 0.533 IT

Then equation (9.28), page 121, gives Carter's coefficient for a single slot opening C=

1.780 = 1.43 1.247

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 a single slot opening although the problem in practice is that of a succession of slot openings. Equations in Terms of Angles

Throughout this chapter whenever numerical examples have been quoted to illustrate the procedure, the starting point has been to choose values of k and a. It was pointed out on page 200

that this procedure starts at the wrong end of the practical problem where values of the two ratios g/s and t/s are specified

and k and a have to be determined from them. But in an investigation of this kind, when the equations have been developed they are seldom used merely to obtain a specific answer to a single question. The equations are usually employed to evaluate a whole series of results to be embodied in tables or curves for future use. For instance Carter's coefficient is never worked out for a single pair of ratios s/g and s/t but

a complete family of curves is calculated and drawn such as those shown in Figs. 66 and 67, page 122. Therefore it is quite a reasonable procedure to select a large series of values of k and a 214

ELLIPTICAL INTEGRAL OF THE THIRD KIND

and take all possible pairs of the selected values. Then g/s and t/s can readily be calculated for each pair from equations (17.18). It is then necessary to resort to the method of cross-plotting so

that it becomes possible to start from the specified ratios and read off the corresponding values of k and a. Some books of tables of elliptic functions do not list values of

sn a, do a and on a. It is then desirable to work with the modular and amplitude angles 8 and 0, and this in fact was the

method used by Coe and Taylor in their paper.* It is best explained in connexion with a specific example and for this purpose equations (17.18) on page 199 will be expressed in terns of 0 and 0. Now

k=sill 0

and

k1 = ksna = ksin0 from equations (12.3) and (12.6), pages 139 and 140. Now let 01 = sin-1 k'1

Then (1-k12)1/2

cos 91

(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

snadna on a

= tan ¢ cos 01

In tables of elliptic functions r' is defined by the equation r'/90' = a/K. Substituting all these expressions into equations (17.18), page 199, gives 9

s

=

h {tan

cos 01- Z(a)}

7T

= 2K' {tan 0 cos 01 - Z(a)} S

90°

* See List of References, page 217. 215

00-FORMAL TRANSFORMATIONS

By using these equations and the method of cross-plotting Coe and Taylor found values of e1 and 0 for many combinations

of 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 equations will also be found in the paper by Coe and Taylor. Conclusion

The reader who has been able to follow and understand the problems worked through in this chapter will now be able to

attack most of the field distribution problems that arise in For further examples of the same standard of difficulty he is referred to the papers of F. W. Carter, and more practice.

particularly to the paper of R. T. Coe and H. W. Taylor, already extensively quoted. In their paper there are four problems very fully discussed and solved, as well as the one used in this chapter, and there should be no difficulty in following them.

216

REFERENCES (1) Papers CARTER, F. W., `A Note on Airgap and Interpolar Induction', J. Instn elect. Engrs, 29 (1900), p. 925. SCHWARZ, H. A., `Uber einige Abbildungsaufgaben', Crelle's J. Math., 70 (1869), p. 105. CHRISTOFFEL, E. B., Ann. Mat. pura appl., 1 (1867), p. 95. CARTER, F. W., `Airgap 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) Books THOMSON, 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 (Macmillan, 1948).

(3) Books of Tables, etc. DWIGHT, H. B., Tables of Integrals and Other Mathematical Data (Macmillan, 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 Publications).

217

Index Abel, 139

FLOW LINES, 30, 32

Amplitude angle, 130 Annular ring, 49 Argand diagram, 14 Argument, 17

Flow through a slit, 96 Flux density distribution, 92, 111 Flux function, 30, 32 Force equations, 6 Fringing of field, 88

Bewley, L. V., 2

IMAGINARY NUMBERS, 15

CAPACITOR PLATES, 60

Incomplete elliptic integrals, 164 Increase of capacitance, 74 Inverse square fields, 5

Carter, F. W., v, vi, 103, 190, 212, 216

Carter, G. W., vi, 75, 136 Carter's coefficient, 1, 121 Cauchy-Riemann equations, 27 Christoffel, E. B., 1, 56 Coe, R. T., v, vi, 2, 190, 215, 216 Collinear electrodes, 135 Collinear source and sink, 83 Complementary modulus, 131 Complete elliptic integrals, 130 Complex numbers, 14 Complex planes, 35-6 Concentric circles, 33 Confocal ellipses, 41-2 Confocal hyperbolas, 44-6 Conformal transformations, 35-6 Conjugate functions, 25-30, 36 Conservation of angles, 49 Constants of integration, 63 Curl, 9 Current density, 142, 153-4, 169172, 179-181 DENSITY OF CHARGE, 71

Density of flow, 79 Distance between capacitor plates, 76

Distribution of flux density, 92, 111

Jacobi, 129, 130, 139, 185-9, 193

Laplace, 11-13, 24-5 Legendre, 130, 139, 192, 193 Line charge, 33 Lost flux, 117 MODULAR ANGLE, 130

Modulus, 14, 129 NOME, 187 ORTHOGONAL CURVES, 30

PERIODS OF ELLIPTIC FUNCTIONS, 181

Point source, 6, 7, 82 Polar form of complex numbers, 16 Polygon, 57 Potential, 10, 11, 24 Primary electric constant, 7:3 Primary magnetic constant, 9:1 Product of complex numbers, 17 QUADRANT, 50-5, 87-8

Divergence, 8

Division of complex numbers, 21 E@UIPOTENTIALS, 11, 27, 32, 69, 81, 145

Equivalent airgap, 122 Eta functions, 188

Schwartz, H. A., 1, 56 Single slot opening, 103 Sinks, 10 Slotted plane surface, 77 Stream function, 26, 32

218

INDEX Stream lines, 69, 81. 148 Succession of slot openings, 190

Taylor, H. TV., v, vi, 2, 190, 215, 216

Theta functions, 185 Thomaon, J. J., 1 Transformation of a curve, 38-46 definite area, 49-50

point source, 82 quadrant, 50-5, 87-8 rectangle, 124, 133, 140 straight line, 47 triangle, 48 Walker, Milea, 2 ZETA FUNCTION, 185

219