D-1992 EXACT SOLUTION FOR THE ELECTRORAGNETIC FIELDS OF A UNIFORM LINE--ETC(U)APR 82 E A LE WIS , J L HECKACHER

UNCLASSIFIED RACTR-0292 N

RADC-TR-82-92In-House ReportApril 1982

EXACT SOLUTION FOR THE ELECTRO-MAGNETIC FIELDS OF A UNIFORMLINE-CURRENT PARALLEL TO AFLAT HOMOGENEOUS EARTH

Edward A. LewisJohn L. Heckscher

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EXACT SOLUTION FOR THE ELECTROMAG- In-HouseNETIC FIELDS OF A UNIFORM LINE CURRENT__________PARALLEL TO A FLAT HOMOGENEOUS EART1 ' PER"FORMING 040. REPORT NUMBER

7. AUTIIOR(e) 11 CONTRACT ON GRANT NUMBER(.)

Edward A. Lewis*John L. Heckscher

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* Consultant to Megapulse, Inc. , 8 Preston Ct. , Bedford, MA 01730

It KEY WORDS (C-on.n. 00 I--hO side if n e,.y end idenity by block n-~b*,)

EM boundary-value problemInduced resistivityHorizontal LF antennaLine -current excitation

20 ABSTRACT (Coninue on r-orea ide it ncesay ed identify by block -~obe')

%%A detailed step-by-step description is given of the exact solution of theboundary-value problem of calculating the fields of a uniform, sinusoidallyvarying, straight-line current flowing at a fixed height above a flat homo-geneous earth of arbitrary dielectric constant and conductivity. The solutionsatisfies Maxwell's equations and boundary conditions, reduces to Ampere'sLaw at the current itself, and hence includes the radiation fields. The resulticontain integrals with infinite limits, but these can be evaluated numerically

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20. Abstract (Continued)

in specific cases of interest As an example, the resistivity induced in a wire10 m above the earth of VOj mho/m conductivity is calculated for a frequencyof 35 kHz.

UnclassifiedSE-UPITY CL ASSIFICATION OF PAGIE'Whof Dora E-od

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2

Contents

1. INTRODUCTION 5

2. APPROACH 5

3. FREE-SPACE FIELDS 6

4. FIELDS OVER A PERFECTLY CONDUCTING EARTH 9

5. SUPPLEMENTARY FIELDS AND FORMAL SOLUTION 13

6. PLAUSIBLE APPROXIMATIONS 17

7. NUMERICAL EXAMPLE 19

REFiRENCES 27

Illustrations

1. Coordinates in a Plane Transverse to the Z-axis, Which isDirected Toward the Reader

2. Radiation Resistivity of a Wire Carrying a Uniform Currentas a Function of Height Above a Perfectly Conducting FlatEarth 12

3. Graphs of the Functions Qr and Qi i

4. Graph of the Function J1(p)/p 22

3

5. A Portion of the Function 6x23

6. Graph of -Q 1 F 1 25

4

Exact Solution For The Electromagnetic Fieldsof a Uniform Line Current Parallel

to a Flat Homogeneous Earth

1. INTROD CTION

In hi. pioneer paper, Carson1 obtained approximate solutions for essentiaUy

the same problem addressed here, but the magnitude of the errors caused by sim-

plifying assumptions, such as neglecting displacement currents, is unclear,

especially in cases of poor conductivity and high frequencies. The aiternate ap-

proach adopted here is to develop the exact solutions which, as might be expected,are more cumbersome. They do, however, offer the possibility of numerical

evaliition to a desired degree of accuracy in specific cases.

2. A'ROACH

Figure I represents a rectangular coordinate system XYZ, witi. the XGZ

plane coincident with the flat earth-surface. A line-current of I rms amperes

flows in the positive Z-direction at height h, and passes through the point x = 0,

y = h, z = 0. The material of the uniform earth has a dielectric constant E F/m

and a conductivity of a mho/m. The region y > 0 is taken to be free space of

(]Received for publication 5 April 1982)

1. Carson, J.R. (1926) Wave propagation in overhead wires with grounti return,Bell System Technical Journal 5:539-554.

5

Y

P( x,y)

-It

Figure 1. Coordinates in a Plane Transverse tothe Z-axjs, Which is Directed Toward the Reader

dielectric constant o and both media have the magnetic permeability j o of free

space. MKS units are used throughout.

The field at an arbitrary point P(x, y, z) above the earth is regarded as consist-

ing of three parts: (1) the fields which the current would produce if it were in free

space, the earth being absent; (2) the fields of a perfect image of the current at

x = 0. y = -h, as if the ep:th were perfectly conducting; and (3) "supplementary'

fields which account for the earth not being a perfect conductor. The fields inside

the earth have no source-singularity and are described as just one field-complex.

The free-space field is discussed in Section 3, the effects of the image are obtained

in Section 4. and the supplementary fields are developed in Section 5.

3. FREE SPACE FIELDS

In discussing the free-space fields it is convenient to use an auxiliary cylindri-

cal coordinate system (r_, ,z) whose z-axis coincides with the line-current, as

shown in Figure 1. The current I and all its fields vary with time t according to

the factor e" J (which is not explicitly written Out in the expressions to follow)

r 1X Figue 1.Coodinaes n a laneTrasvere tthe -axi, Whch s Diecte Toard he Rade

dielectric........ constant.. F_-51 0- ,"''; and-othmeda hve he-agneic.-rmabiityP 0of r ee ': "-"

where w is the angular frequency corresponding to the current frequency f. Infree apace, Maxwell's equations for the electric field 2 and the magnetic fieldH

are

VXHi = W -t 0 E (2)

Because of symmetry and the absence of free charges, only the field componentsE. and Ho exist, and these do not vary with z. In cylindrical coordinates theMaxwell equations become

BE5rZ iWAoH(3)

r 00Z

TheHanelfuntins ad H ar related by the recurrence relations2

dH ()-- 1

I(d)0)(5)

1U To 1) ( u) H (" (6)

Using these it is readily shown that Eqs. (3) and (4) are satisfied by

E H H1) (kr..) (7)

H H l(") (8)r* 4 1 (r

where

k= c

2. Abramowitz, A.,* and Ste gun, .LA. (1964) Handbook of Mathematical Functions.National Bureau of Standards, Applied Mathematics S-eries 55, U.S. Gover-n--nient Printing Office, Washington, D. C.

.7

where c is the velocity of electromagnetic waves in free space, and X is the

wavelength.Other properties of the Hankel and Bessel functions of the first kind (J), and

the second kindt (Y) which will be needed are:

H(1) = J + iY

H(l) = J1 + Y

and for u -0,

H()(u) I- 1- In 2 (10)

(l)( U 2i iU U

H () U + -In -

where 'Y = 1. 78107.

Also, for v -o,

(1)(} 2 i ,, e ( u - r/4)

0 Iro

(1) 2 i(u-3r/4)H (U) -e

Close to the line current, kr_ - 0 and

JAWI 2i 2

z 4 7- kr-j

H - (12)

The constants in Eqs. (7) and (8) were pre-chosen so that Eq. (12) agrees with

Ampere's Law.

tSome writers represent this function by N.

8

At large distances from the line current, kr_ -oo and

E 0 r i (kr- - /4)

H -. 1 ei(kr_ -w/4) (14)

Since the time factor is e - i t and k is a positive real number these fields have the

form of outgoing waves, as they should. Thus the fields given by Eqs. (7) and (8)

meet all the requirements for an exact solution for the free-space case. Also, forkr.. co,

E (15)

E L

which is the impedance relation for plane waves in free-space. The (radiated)

power flowing through a 1 m section of a large cylinder is then

2rr_ l 1H I = '4 (

This power loss is as if the line-current encountered a resistivity 0o!'/4 ./m, andit may be seen from Eq. (11) that the in-phase, or real part of Ez at the line cur-

rent is negative and opposes the current flow. The power required to drive the

current against this field is p o 2/4 W/m and is equal to the radiated power The0

imaginary part of E has a logarithmic singularity at r- = 0, as can be seen fromzEq. (11). This is a consequence of having a current of infinitely small cross sec-

tion; if the current were distributed over a finite cross section the field would

remain finite.

4. FIELDS OVER A PERFECTLY CONDUCTING EARTH

The fields inside a perfectly conducting earth are of course zero, as is the

tangential component of the electric field at the surface. The boundary conditions

are met by placing a 180 degree out-of-phase image at x = 0, y = -h. The fields

at a point P above the surface are then a superposition of fields of the type discussed

in Section 3. Thus

re~n m=nnm I .x: HI I _J I - - I 9

= - ... ... L l r~ / n I "1" I i'l .... .I l ... .. ... -- .. . ... ..... . .

E JA0 [H (1(kr-)-H (1) (k r+)] (17)

where the subscript oo is used to denote the case of the perfectly conducting earth,

and

r xt + (y~h) (18)

Since the magnetic fields are in the &.and 0+ directions, as shown in Figure 1

and since

cos~ h-y snsi 0- F:

and (9

Cos~ h+y sin 0+

resolving the magnetic fields into x- and y-components gives

H =ikl [h- H (1)(kr) + r~y H 1(k+ (20)

H = M-i[ H,51(kr) - AH 1 '(k r+ (21)

Just above the earth, as YO0, E Z 0and H,. vanish while

H =ikhl I )k,2 x+ h2 y (22),/2 + h

Fromn Eq. (17) the electric field at the current itself (x 0, v h) is

E (,h) 0 W H [(1)(0) H H(1)(2kh)] (23)

10

Since the real part of the Hankel function H 0 l ) is J0 and since J (0) 1. the corn-

ponent of field opposing the current flow is

j0-Re[EzQ] - 4 11 - J 0 (2kh)] (24)

The power required to drive the current is I times this, and the effective resis-

tivity is (Po /4)[ I - J (2kh)J SI/m. A portion of this function is illustrated in

Figure 2, which was plotted from tabulated values of J and from the appropriate

series. The physical reason for the oscillatory behavior becomes clear from con-

sidering the fields at a great distance from the current and its image. Then

r x = x 2 +y +h 2 ± 2hy v/r 2 ± 2hy = r ±h sin a (25)

where r = x + y and a = sin (y/r) is the elevation angle of the point P. On

using Eq. (13), Eq. (17) becomes

EoA i (k - 7r/4) [e-ikh sin a ikh sin a

,/ -8- i-kr-n

0 (11 ei(kr - ir/4) sin(kh sin a) (26)

Similarly

H - ii 2kr i(kr -r/4) sin(kh sina) (27)

The outward power flow is

Ez, Qc H sin (kh sino) W-m (28)

Thus the current and its image produce a far-field interference pattern that changes

with kh, but always has a null along the earth's surface. The power flowing through

a 1 m length of a large cylinder centered on the Z-axis of the XYZ coordinate sys-

tem is

11

-T--T-Tir-T -T- I I 1 I ffli I I , I11111 1T 1 - if IW | / I

2 -- -- -- - -- -- -FREE- SPACE VALUE ____-

I

0

10 I-

i--._

"'J- 10"3t-

1 10-4,

10 . 4 1 - 3 10 2 I0 " IHEIGHT OF WIRE IN WAVELENGTHS .h/\

Figure 2. Radiation Resistivity of aWire Carrying a Uniform Current as a Function

of HJeight Above a Perfectly Conducting Flat Earth

oa [2 r o 2 /2

2, f sin ks~n~o :2,, f c " os (2kh sin a)) do00O

0 1 - Jol2kh)] (29)

where use is made of a standard integral form. This is, of course, the same as

the power expended in driving the current, as discussed above.

12

- d ' -... ' -. . . w , J

5. SUPPLFMENTARY FIELDS AND FORMAL SOLUTION

Inside the earth the Maxwell-Ampere equation is

VXH = (0-ii.' )i (30)

instead of Eq. (2), while Eq. (1) remains unchanged. Symmetry considerations

show that the only rectangular components of the fields are E z , Hx and Hy, and

these do not change with z. The Maxwell equations then reduce to

' aEH I - o z (31)

DE

H =-i z (32)y WM ax

aH aHY x iic) E(33)ax- - (a - i zE) E

It may be shown by direct substitution that these equations are satisfied by the

following expressions:

2C 2y -

E (X, y) G(s) e cos xs ds (34)

0

2 2s"-k2(xy) f -/,o- k G(s) e cos xs ds (35)x 0 . f S ~se S~k

0

2r 2

H (x, y) i s G(s) e - sin xs ds (36)

where

k 9 2 2 0 + ioP (37)

13

4

and G(s) is an arbitrary function of the variable of integration, to be chosen later

to satisfy the boundary conditions. In accordance with symmetry requirements

E and H are e ven functions of x, while H yis an odd function. If a~ > 0 the

quatits - k-is always in the lower half of the complex plane, and hence2-k2is either in the second or fourth quadrant depending on the choice of

sign. For Eqs. (34), (35). and (36) to represent physically realistic fields, the

real rtof must be positive so that the fields vanish as y - - -. Thus

/s - -kg is taken to be in the fourth quadrant.

At points above the ground, Eqs. (31). (32), and (33) with E = E and a 0 are

satisfied by the fields

Ez(xy) f A(s) e y /sk cos xs ds + Ez, 0 (38)

0

IH (x, y) =-. A5s e -y k 2,x Ji Vs -k 2 Ase cos xs ds + H X c(9

0

0 c/ -.

0

where Ez, 0,' Hx, cc and H are the fields for the case of perfect ground conduc-

tivity already discussed in Section 4. The parts of the fields containing the arbi-

trary function A(s) individually satisfy the Maxwell equations if k w. I as

before. When s 2 > k , the positive real value for /s 2 - k must be chosen so2 ~ 2

that the fields decrease with increasing y. If s < k2 , the negative value for'2 - k2 must be chosen to have the solution correspond to outgoing waves rather

than physically unrealistic incoming waves. As a reminder of this sign choice, an

asterisk (") is added to the square root sign in subsequent expressions.

At the earth's surface (y = 0) the tangential components of the electric and

magnetic fields must be continuous. Continuity of Ez requires

F: (X, o) f A(s) cos xs ds JG(s) cos -s ds (41)0 0

since E'z, vanishes at y = 0. Evidently A and G are essentially Fourier trans-

forms of the same even function Ez(x, 0), so that

14

A (s) G G(s) (42)

Simillarly, continuity of H x requires

Vj S -K (s) cos xs ds HX - f 5 2 k 2 G(s) cos xsds

0 0(43)

On using Eqs. (22) and (42) this condition becomes

WM 1:0 +X/vsi A(s) cosxs ds 2 2li 2~)k4 T ~

00

I.f F(h, s) cos xsds

0

(44)

where by the Fourier integral theorem the (dimensionless) function

F(h, s) = k ,h H1 () (k Vx 2 + 112 )cos xs dx (45)

It follows that

A(s) " M0 w I F (h, s) (46)2 2,) , * /2-.2

154

On consolidating the results, the fields above the earth are:

E 2 (x, Y) 0 ~Zf e V iI F(h, s) cos2xs ds-,/ kg2 + Vsk

0

- WI2i [H (1)k rJ H (1)(k r+)j (47)

k *2 rI2*/Hl i /j e -y V - k F(hs) cos xs ds(x,y) 2 + 2- 2

0

+ W -hy HI(1)(kr_) + h+y H ()(kr+) (48)

H (x, y) e -y V s - k sF(h, s) cos xs dsY - 2.2 +* s2 k2

ikI1 (1(k r_ 2L H :(k r+)] (49)s - 1 r+-I

Below the surface of the earth the fields are:

E e F(h,s) cos xs ds (50)Ez~xy - J , 2/_2 +'' 2 _2

2 2iif e \ kg F (h, s) cos x5 s (H (x,y) = iJ ds (51)

" Isy 2 k : +*/sk 2

0

16

........ ..... = . .. , ... . ;, _ , -. Je' '. . : ' -:.., ' ' ' J

*1 - ,'= ...- .

HI (x. Y) = ies F~h. s) sin xs ds (52)f 2_ 2 * 2s. -Sk. + s -k

0

On converting to the dimensionless parameters:

u = s/k

1,=k (53)

Y,= ky

h 1= kh

Eq. (45) becomes

0 x1 +h 1

and, for instance, with y > 0,

-Yi' &- 1fe F (h. u) Cos ux

E z(x, Y) -?01 du2u k 9 k + * 24

0

0~w [(1)(k - 11)(k r+)](5

where

kg2/ = + ia/ue 0 (56)

6. PLAUSIBLE APPROXIMATIONS

Before proceeding with a numerical investigation of a particular case of theexact solution, it may be of interest to mention, without justification, some

17

plausible approximations. If the real part of the Hankel function H I Q) is ignored,

and its small-argument approximation

Ir

[see Eq. (10)] is used in the integral for F(h. u) even though the range of integra-

tion extends to infinity. then

Fh1 ) 2 i fh-Cos WCI (58)

0 1 1

This integral is a standard form, giving

-h u

F(h1 .u) _-ie 1(59)

2 l -

Next f Vu' - 1 is replaced by u, Eq. (55) gives

E (x, Y) 1_- L 0 wI f e -(h 1+y,)u Co u du 'U 0 [1H (1) (k r. H H(D(k r+

o 2 k9/ 2 + (60)

Finally, if the Hankel functions TI:(1) are replaced by small argument approxima-0

tions:

H ( 1)(k r I - In -2

[see Eq. (10)), the result is

E (X, Y) 2 0C - (h 1 )u Csu u+In\ -

INj~ 77-k 9 /k2 + x, + (y, -h 1

(61)

18

This is similar to a form used by Wait3 .ar highly conducting earths and low

frequencies. t

7. NUMERICAL EXAMPLE

Using the exact solution [Eq. (55)], the real part of the electric field at the

line current itself (x = 0; y = h) will be calculated numerically to within a few per-

cent for the following case:

f = 3.5 X 104 Hz

) = 8571.42857 m

k = 7.33(38287 X 10 - 4 m " 1

-3(Y = 10- mho/m-

e/E° = 10

a/lwe = 514.28571430

h = lom

Yl = hI = 7. 33038287 X 10 3

kr+ = 2kh = 0.014660764

kr- = 0

Then

Re [H 1)(kr -H (1)(kr+) = J(0)- Jo(0. 014660764)

= 5.38 X 10 5

by interpolation of tables, 4 and

J Ao W I 0 " o W I -Re(E - 0 Re (Q Q) F(hl, u)du - 0 (5.38 X 10") (62)z Vr f r i 1'4

0

tWaitis formulation evidently contains a typographical error in which his sym -bols x and y were interchanged in three places.

3. Wait, J. R. (196 1) On the impedance of a lcng wire suspended over the ground,Proc. IRE, 49, October, p 1576.

4. Cambi, E. (1948) Eleven and Fifteen-Place Tables of Bessel Functions of theFirst Kind to all Significant Orders, Dover, New York.

19He[E z ]

where

Q= Qr + iQi e (63)2_V u . E/Eo - ia/w o + *u -1

The functionsQ andQ in the interval 0 <u < 400 are illustrated in Figure 3.Then if Fr and F i are, respectively, the real and imaginary parts of F(h1 , u),

)A 0W1 0 001. 08 X 10-4]Re8EJrr d -du QFidu+ (64)

Now

F = h f - cosux 1 dx (65)0

with

22+ h, x1, +i h (66)

Figure 4 shows a graph of Jl(p)/p for 0 <x 1 <20. For large values of xl,Jl(p)/p fapproaches /2-/ cos (x1 - 3r/4)/x 13/2 so that

f J1 ( p ) - d

[ cos ux dx I I< / dx 2 36J I f 1 <2 mJ 32 V/ 0 __

20 20 1

for all values of u. The portion of the integral from 0 to 20 is not larger than about0. 5, so it is concluded that F is less than something of the order of h1 or 7 X 10 3

rIfor all values of u, and will be much less than h at large values of u. Inspection

1of Figure 3 shows that th? area under the curve of Qr(U) is about 1o so that thefirst term inside the large bracket in Eq. (64) is not larger than about 7 X 10

20

- -

• - .. . ... .... " •. . . . . . . .. . . - -- ,- -.-. <l ..= -' . . . . '. .. . , ',. ... : . + ,.. .

.0,,

.02

.o01- Qre- '.

CQi -I- -*- '-e-h'u

o00 200 300 400 500

U -*

Figure 3. Graphs of the Functions Qr and Qi

Now,

F. = h 0 cos ux dx (67)

0

may be written in terms of the small-argument approximation mentioned in Sec-tion 6, plus a remainder D which accounts for the difference between the exactand approximated function. This gives the exact equation

OCF. _f j2 Cos ux dx + D(u) (6 8)

21

.5

.4-

COSX,-3ir/4)

Figure 4. Graph of the Function J1 (p)/p

where

2h~ 0 os dD(U) 6 Cosf ux1 (69

0

6 .r y (P) + 1(70)

Evaluating the integral of Eq. (68) as in Section 6,

-h uF.=- 1 + D(u) (71)

A graph of a portion of the function 6 is shown in Figure 5. Inspection shows that

the portion of .6 for x< 10 contributes a magnitude less than about 0. 3(2h /ir)

22

2.0,

SINUk,-37T/4) + I1

VT v§ -31 2 2

-10

-20

-2.8-301

0 1 2 3 4 5 6 7 8 9 10X,

Figure 5. A Portion of the Function 6(x

to the function D, for all values of u. For values of x 1 greater than 10, the

limiting form

x sin (x 3r/4) 1(72)nA(x x 3/2 +-2

represents A with an error less than about 10 - 3 The contribution to D from

x > 10 is then less than about

+ 1 dx = o. 9 (73)

7 10 1 1 l

for all values of u. Thus for 0 u u - 100, D(u) !S 2h /77 = 0. 0o5, or less than-hlu 1

1 percent of e which is between 1 and 0.480 in this range. At larger values

of u, the entire factor F. becomes quite unimportant because the multiplying fac-

tor Qi(u) in Eq. (64) is then much less than 1 percent of its initial value, as shown

in Figure 3.

The product function -Q F. in Eq. (64) is shown in Figure 6 for the important

part of its range. A numerical integration gives

23

MI

f -Qi (u) Fi(u) du 0. 33 (74)

0

and hence

Re[Ez - (0. 33) )i 1 + 1. 1 X 10 -

= -0.84 (75)8

The corresponding resistivity of the line is 0.84 Mo /8 0. 029 £P/m.t

TBased on the previously mentioned approximate formulation (Section 6),Wait 1, gave two terms of a series for the resistivity:

i ~ ~ 0 -- -3-vr o h +..

For the above numerical example, the quantity (8,//3) h = 0.200, andthe corresponding line resistivity is 0.80 mo.jT8 il/m.

5. Wait, J. R. (1974) Comments on 'The horizontal wire antenna over a conductingor dielectric half space: current and admittance,' Radio Science 9:1165.

24

-mom

02-

0''-

00 10 20 30 40 50

Figure f Graph of -Qi F.

25

References

I. ( arson, J. H . (102 f) WVa ve propagation in ov( rhead %k it-es ith Lround l.tu Iri,Bell System Technical Journal 5:539-5-4.

2. Abramo( itz, A. , and Stegun, 1. A. (19f;4) Handbook of M athem atical F un, tions,National Bureau of Standards, Applied Mathematics Series n, 1". (.oven-ment Printing Office, Washington, 1. C.

3. Wait, J. R. (19VI) On the impedance of a long wire suspended over the ground,-roc. IRE, 49, October, p 1-7,7.

4. (ambi, E. (1948) Eleven and Fifteen-Place Tables of Besse.l I Unctions of thtF'irst Kind to all Significant Ordez s, Dover, Net York.

5. Wait, J. P. (1974) (omments on 'The horizontal %% ire antenna over - (odt, tingor dielectri,- half spac,: current and aim ittant, ) , adio S itncr , I' .

27

hPAZ 'N-O II.oW

MISSION* Of

Rom Air Development CenterRAVC ptans and execute6 k'e6eoA~ch, devetopment, teAt andaetected acquiezition pLogLam in suppott o4 Command, ControfCormirncation6 and Intettiqence, (C31) acauitiei&. Technicateand engZneekiung 4uppo~C wiLthin oA'ea4 o4 technictco competencei.6 p~ovided to ESP P~ogutam O66ice.6 (POt6) and ote' ESVeteenit6. The, p'tincpaL tehncat mL5-3s.on a~eau Wte.cominuniZcationa, etec~t~omagneic guidance and ctontko, 65U&-vei~tance o6 q'Lound and ae,pace. ob ect6, intettigence da-tacottction and handting, indo'na-tion sy~tem technotogy,iono6pheAic pkopagation, sotid state 4cienceA6, mic'wkzephyq6cA and etect'onic 'tetiabi"~, ma-intainabJitty andcompatib-LCiq.

Printed byUnited States Air ForceHanscom AFB, Massn. 01731