refraction of short radio waves in the upper atmosphere

$: Refraction of Short Radio Waves in the Upper Atmosphere$
Refraction of Short Radio Wavesin the Upper Atmosphere

BY WILLIAM G. BAKER, and CHESTER W. RLCE2Non-member Associate, A I. E. E.

Synopsis. The paper shows that the striking phenomena of distributed over the surface of the earth. A focussing of energy justshort-wave radio transmission (i. e., below 60 meters) can be quanti- beyond the skip distance, and again just inside the point wheretatively accounted for on a simnple electron refraction theory in which the ray tangent to the ground at the transmitter comes back to earth,the effect of the earth's magnetic field and electron collisions may be is clearly shown. The reflection of waves at the sulrface of theneglected as a first approximation. The distribution and number earth is also considered.of electrons per unit volume in the upper atmosphere required on The results of these calculations make it possible to estimate thethis theory to account for the meager experimental data appear most sutitable wave lengths for night and day conmunication betweento be in general accord wvith the values required in the explanation any two points on the earth's surface. It is also pointed out thatof the diurnal variations of the earth's magnetic field, aurorae and there will be a minimum ivave length, in the vicinity of 10 meters,long-wave radio transmnission. below which long distance communication becomes impossible. It is

The paths taken by the waves from an antenna to distant points on .shown thatfrom th.e point of view of long distance communication lowthe surface of the earth are calculated. The path calculations give a angle radiation is most effective. The ray paths and energy fluxdefinite picture of the noow familiar skip distance effects. Ideal density in the wave front of the sky waves are independent of thesignal intensity curves (i. e., neglecting absorption and scattering) plane of polarization of the transmitter. The effects of polarizationare given, which show how the energy sent outt by a transmitter is on the reception problem are not discussed.

I. INTRODUCTION assume full daylight or night conditions at both theSHORT-wave (i. e., 60 to 15 meters) transmission transmitter and receiver in order to eliminate sunset and

experiments during the past two years have sunrise effects as well as peculiarities which arise whendefinitely brought to light many peculiarities which one station is in darkness and the other in daylight.

were entirely unexpected as extrapolations from our This, of course, limits the diagram to practically northmany years of long-wave experience. Until recently and south transmission; nevertheless, it may be used toany announcement of long-distance short-wave trans- estimate the general trend for other conditions.mission was put down as an unexplained freak l-y theaverage radio man, and dismissed from his mind. As O"m 0 aOz Q Q Zothe number of such reports increased, we could no longer gOo

be content to dismiss them as freaks and were forced to 60 0

abandon our preconceived notions as to what normal 40 0 tshort-wave transmission should be. and extend ourtheory in such a way as to give these remarkable results a _ L 'Ha definite place in the new scheme of things.

In Fig. 1 we have attempted to summarize the avail- t ° _ 'able data on short-wave transmission characteristics. 0

Most of the data are from the valuable papers by 404 ~~~~~~~~~~~~~~~30Taylor3 and Taylor and Hulburt4 with a few check points

kindly supplied by Young.' We have also obtainedconsiderable help, in drawing the smooth curvesthrough the few scattered points, from the valuablework published by many amateurs.' The curves FIG. 1-APPROXIMATE TRANSMISSION CHARACTERISTICS

umethat 5 kw. are being supplied to an average MAINLY FROM DATA BY A. H. TAYLOR FOR 5 KW. IN ANTENNAassume that 5 kw. are being supplied to an average AND LIMIT OF RECEPTION, 10 MICROVOLTS PER METER. NORTHantenna and that the practical limit of reception is AND SOUTH TRANSMISSIONreached at 10 microvolts per meter. The curves further

1. W. and E. Hall Fellow of the University of Sydney, New As a typical example of the peculiarities of short-South Wales, Australia. Research Laboratory, General Electric wave transmission, let us describe the experience ob-Co., Schenectady, N. Y. tained with a 5-kw., 30-meter transmitter. Here the

2. Research Laboratory, GeneralElectricCo., Schenectady, signal rapidly decreases as we leave the transmitter and3. A. Hoyt Taylor, Inst. Radio Eng., Vol. 13, p. 677, 1925. reaches the lower useful limit of 10 microvolts per4. A. Hoyt Taylor and E. 0. Hulburt, Q. S. T., p. 12, Oct. meter at about 70 miles. This short range is what

1925. might be called the expected value as viewed from our5. C. J. Young, Unpublished Reports on Short-Wave Trans- long-wave experience and is represented in Fig. 1 by

mission Tests by the General Electric Co. at Schenectady, N. Y .pasn to th ih fteln mre iifGon6. See for example Q. S. T., 1924 and 1925. ' pslgt h lh fteln akdLmto r?nPresented at the Midwinter Convention of the A. I. E. E., Wae Ifw no cntueogrtrdiacsth

at New York, N. Y., February 8-11, 19W?. signal remains out until we reach approximately 450302

Feb. 1926 BAKER AND RICE: REFRACTION OF SHORT RADIO WAVES 303

miles where the day signal unexpectedly becomes strong sively nearer the transmitter. Finally a critical angleagain. This is represented in the figure by crossing to is reached for which the refracted ray comes down at thethe right of the curve marked Minimum Range of Day nearest distance to the transmitter. For higher anglesSky Wave. Continuing to greater distances we find the the points of return recede from the transmitter untilsignal gradually falling off in intensity, reaching the finally a second critical angle is reached where the rayuseful limit of 10 microvolts per meter in the vicinity of does not return to earth, but instead goes out into4500 miles by day. This is represented in Fig. 1 by space and is lost.passing to the right of the curve marked Maximum Day The distance from the transmitter to the nearestRange. On a summer night the signal does not reappear point at which the refracted sky wave returns to earthafter the 70-mile extinction until we are approximately has been called the "skip distance." For a given2000 miles from the transmitter, after which the wave length the skip distance is a minimum in thesignal falls off gradually to a very low value at 7500 miles. middle of the day and a maximum on a winter night,The present explanation of the above peculiar the summer night value being somewhat less than the

phenomena is as follows: Assume, for simplicity, that winter night skip. This variation is theoreticallyenergy is radiated equally in all directions by the trans- accounted for by a change in the height, thickness andmitter. As we go away from the source, the signal maximum value of the electron density in the upperstrength will decrease in the usual manner due to spread- atmosphere.ing and energy absorption by the ground, with the The experimental summary given in Fig. 1 shows thatresult that the 30-meter signal practically vanishes in the skip distance for a given time of day or nightthe vicinity of 70 miles. In other words, the ground decreases with increasing wave length. This observa-wave component of the 30-meter signal behaves as we tion is in agreement with the increase in refraction onshould expect from our long-wave experience, i. e., the longer wave lengths. If we neglect the effect ofis rapidly attenuated. The unexpected thing happens collisions between the electrons and gas molecules whichwhen we go out to 450 miles and find that the signal by prevent the refractive index from going to zero, weday has reappeared. This reappearance of signal is obtain a sharp upper limit in wave length above whichaccounted for by electronic refraction of a portion of the skip distances fall to zero. For the ionization valuesenergy which is radiated towards the sky. A reflection assumed in the paper, this occurs for a wave length oftheory of this effect has been proposed by Reinartz.7 60 meters on a winter night as shown in the calculatedMore recently a refraction theory has been de- skip distance curves of Fig. 16.veloped by Taylor and Hulburt.4 The calculations If the effect of collision frequencieswere taken intoin the present paper are based on the electron account, this sharp upper limit would disappear and wetheory of optical dispersion in metallic media as would obtain skip distances on wave lengths greaterdeveloped by Lorentz,' Drude,9 etc. We are also greatly than 60 meters for the assumed winter night ionizationindebted to Eccles,1' Larmor,11 Appleton12 and Nichols values. The effect of the earth's magnetic field willand Schelleng"3 who have worked out many important also require consideration in the vicinity of the upperconclusions which follow from an application of the limiting wave length. The experimental determinationoptical theory to various phases of the radio transmis- of skip distances on the longer wave lengths will besion problem. difficult owing to the masking effect of a relativelyThe present calculations show that by making certain strong ground wave.

reasonable assumptions as to the number and distribu- On the present theory we may expect severe fadingtion of the free electrons in the upper atmosphere the near the transmnitter under certain circumstances.main characteristics of the relatively meager experi- For example, consider the case of 60-meter transmissionmental results can be accounted for. on a winter night. Here a refracted or sky wave willThe calculated paths for rays going out from the fall inside of the ground wave limit and at a certain

transmitter at different angles to the horizontal show distance may be approximately equal in magnitude tothe following general characteristics: A ray starting out the ground wave value. Under these conditionsat a low angle will be only slightly refracted and come severe interference effects between the two waves willto earth again at a great distance from the transmitter. result. If the ground absorption is high this effectFor higher angles the rays will return to earth progres- may be found quite close to the transmitter. Over

salt water the effect should occur at a greater distance.7. John L. Reinartz, Q. S. T., p. 9,April 192Z5. is

8. H. A. Lorentz, The Theory of Electrons, Teubner, 1909. O hre ae hr h kpdsac swl9. Paul Drude, The Theory of Optics (Engl. Trans. by Ma,nn beyond the ground wave limit, severe fading iS expected

and Millikan) LongmaUns, 1917. in the region just beyond the skip distance, where the10. W. H. Eccles, Proc. Roy. Soc., Lond., Vol. 87, p. 79, 1912. two sky waves of approximately equal intensity overlap.11. Joseph Larmor, Phil. Mag., Vol. 48, p. 1025, 1924. Appleton'2and a little later Nichols and Schelleng" in-12. E. V. Appleton, Proc. Phy. Soc., Lould., Vol. 37, Part 2, P. dependently pointed out that the earth's magnetic

13. W. H. Nichols and J. C. Sehelleng, The Bell System field should produce some very interesting effects onTechnical Journal, Vol. IV, p. 215, 1925. radio transmission, especially in the vicinity of 214

304 BAKER AND RICE: REFRACTION OF SHORT RADIO WAVES Transactions A. I. E. E.

meters and above. It is interesting to note, in this Differentiating equation (1) with respect to t we haveconnection, that Fig. 1 shows a marked absorption in d Ethe 214 meter region. A very interesting study of d = wEmCos W ttransmission phenomena in the broadcast wave length t (6)band has recently been reported by Bown, Martin and Substituting this relation in (5) we obtain for thePotter14. The present paper is confined to the propa- charging current densitygation phenomena on the short-wave side of 214 meters 1where the effective electron restoring force due to the i2= 4 XE,, cos w t (7)earth's magnetic field will cease to be important com-pared with the electron inertia force, and may, there- The total current, i, is obviously the sum of the twofore, be neglected as a first approximation. currents i, and i2.

II. REFRACTIVE INDEX OF A MEDIUM CONTAINING 1 { 4 7r Ne2 EFREE ELECTRONS 4 7r m co2

When an electromagnetic wave passes through By substituting equation (6) in (8) the total currentan electron atmosphere it sets the electrons in motion, density becomeswith the result that the apparent or phase velocity of 1 rthe wave is increased above that in free space. Follow- I_ 2_1- ___ d (9)ing Larmor", let the electric field intensity of the wave 4 7r m c2 dtbe given by It will be observed that this becomes the ordinary

E = Em sin co t (1) expression for the charging current of our unit condenserThe force exerted on the electron of charge e is e E if we let the bracketed quantity stand for the effective

and since for a free electron no elastic restoring or dis- dielectric constant of the material between the plates.sipative forces exist, the total applied force is con- In other words, the effective dielectric constant of ansumed in accelerating the electron mass m, giving electron atmosphere becomes

4drNwNe2m = e E,,, sincw t (2) k m 2 (10)d t

The quantities in the above expression are all in c. g. s.Integrating equation (2) and putting the constant of electrostatic units.

integration equal to zero, since we are not concerned N = number of electrons per cu. cm.with the random velocities, we obtain e = electron charge = 4.77 X 10 -lo e. s. u.

e E m = 8.97 x1-28gramsv m cos X t (3) w = 2 7r f

f = frequency in cycles per secondwhere v = electron velocity due to electric field. 4 wr e2/m = 3.2 XI10' c. g. s. (e. s. u.)

If there are N of these electrons per cu. cm. moving We see from equation (10) that the effect of electronswith the velocity v they will constitute a current ii per in reducing the apparent dielectric constant of space is,square cm. equal to N e v. We thus have from (3) for a given frequency, proportional to the total number

N present per cu. cm., to the square of the chargeN e2 Em and therefore independent of the sign of the charges,el = _- eos X t (4) and inversely proportional to their mass. The

lightest gas ion is that of hydrogen having a mass ofIn addition to this current we have the ordinary con- approximately 1800 times that of an electron. It isdenser charging current flowing across the unit cube. evident, therefore, that for anything like equal numbersThe electric intensity or potential gradient in the wave the effect of gas ions on the effective dielectric constantis E and, therefore, the potential difference between can be neglected.two planes one cm. apart is E times one cm. The A comparison of the signs of equations (4) and (7)capacity of our imaginary condenser which has two shows that the el-ectron current is opposite in phase toplates of one sq. cm. area and one cm. apart is the condenser charging current. In other words, theC = 1/4 wr since the dielectric constant is unity. The inertia due to the to-and-fro motion of the electrons is.charging current for this unit condenser is equivalent to shunting the capacity of each unit cube of

1 dE space by an inductance equal to rn/N e'.2 4 wr d t 5 h velocity of anelectromagnetic waein amedium

is inversely proportional to the square root of the14. Ralph Bown, De Loss K. Martin and Ra,lph K. Potter, permeability times the dielectric constant of the

The Bell Syst. Tech. Jour., Vol. V, No. 1, p. 143, 1926. medium. For our case the permeability is unity and


the effective dielectric constant that given by equation III. GENERAL EQUATION FOR THE PATH OF A RAY(10). We may, therefore, write for the velocity IN A MEDIUM OF VARYING REFRACTIVE INDEX

c The distribution of ionization in the upper atmos-c = cm. per sec. (11) phere will, in general, increase with height, reaching a

maximum, and will then decrease again to a small valuewhere c = t3X 10i cm. per sec. velocity of light, at very great heights. We have seen that the apparentThe refractivelindext of amedium is defined as the refractive index of the medium is reduced by the

ratio of the velocity of light in a vacuum to that in the presence of the free electrons which are present in anmedium. 12 ionized gas. This means that the energy, which is

= c/c' (12) radiated up into this region from our antenna, will beFrom (10), (11) and (12) we obtain bent around, and return to earth at some distant point,

1 4 7r N e2 instead of vanishing into space. We picture the energyM = 1 m N2 (13) leaving the antenna as consisting of an infinite number

of rays just as is commonly done when dealing withas the refractive index of our electron atmosphere. problems in light. We can then follow the paths of the

Thisexpression for the refractve index dependsupn various rays, which leave the antenna in differentthe assumptions that the average time between the directions, up through the medium and back to earthcollision of an electron with molecules is sufficiently again. In thisuwa weobtin acp pictureogreat to allow the full effect of the accelerating action of what happentsthe electric field to be realized. When an electroncollides with a massive particle, the directed velocity In Fig. 2 we have drawn two rays at a distance d n

given to it by the wave is turned at random and cannot apart. Let the wave front be along A B at a time t

be returned to the wave. Thus the effect of a col- and along C D at a time t + d t. It then takes thelision between an electron and a molecule is to convertthe energy given to the electron by the wave into ran- (+ d)tdom motion or heat energy. The effect of a collision C,between two electrons does not withdraw energy fromthe wave, because a collision between two particles, ofequal charge and mass, results in such an interchange ofvelocities that the total charge movement is unchangedby the collision.

Suppose, for example, that on the average an electronmakes ten collisions with molecules during one cycle ofthe wave then it is evident that the wave would scarcelyhave begun to accelerate the electron before a collision 0

would interfere with the process. It seems clear that FIG. 2where a great number of collisions occur per cycle ofthe wave the amount of energy which goes into electronacceleration during the short interval between col- wave the same time to go from B to D as from A to C.lisions is relatively small, and consequently the energy The distance from B to D is v d t and the distanceabstracted from the wave is small. It is also evident from A to Cils (v + d v) d t, where v and (v ± d v)that under these conditions the effect of the electrons are respectively the velocities along the lines B D andin reducing the refractive index from unity is slight. A C.From a qualitative point of view, we can see that the If 0 B is the radius of curvature of the ray B D and

maximum energy dissipation will occur when there is equal to R, we must haveare about four electron collisions per cycle of the wave A C/A 0 = B D/B 0 (14)since the full acceleration is obtained in a quarter cycle. since each is equal to the angle B 0 D in radians.Calculation shows that actually the maximum effect Thusoccurs for 1 /2 r collisions per cycle. This may readilybe understood since the greater number of collisions (v + d v) d t v d twill more than make up for the slightly lower velocity R + d n R (15)acquired by the electrons during the shorter intervalof time. With fewer collisions per cycle, the energy From this we obtainloss will rapidly decrease and the full effect of the elec- v ± d v R ± d ntrons on the refractive index will result. v R (16)A rigorous solution of the problem taking into ac-

count the average frequency of collision between an and upon subtracting one from each sideelectron and a molecule is given in Appendix I. dyv/v = dn/R (17)


It follows that Substituting (24) in (21) gives1 1 dv d zld n = -1/V/1 + (d z/d x)2 (25)R v d n 18 We now put (25) in (20) and obtain

But we have from the definition of the refractive index d MA _ 1 d A (26)that v = c/,ll where c is the velocity of light. On d n d z \2 d zsubstituting this relation in (18) we obtain15 A + dx J

1 1 d y Substituting (26) in (19) we haveR ,u dn (19) 1 1 d 1 (27)

- ~~~R,u d z ad z '2If we know the value of ,u everywhere this equation d 1+ ( d x

enables the path to be calculated. The calculationsare much simplified, however, by making the following It remains to express the value of 1/R in terms oftransformations. z and x.

Let z denote the height above the ground, and let The required relation for the curvature in rectangularus assume that the electron density is a function of z coordinates is given in any standard book on theonly. In this case , is a function of z alone since the calculus ascollision frequency can also be considered as being d2 za function of z. We may write 1 d X2

dA,u dg,u dz (20) R =l 1 + ( d z 213/2 (28)d_ d__ dz 1+(

dn dz dn (20) dx/On equating the two expressions (27) and (28) for the

curvature we obtain:

,//o~~~~~~~~~~~~ d 2zak (2 1 d_ dx2 (29)

FIG. 3 This is the required differential equation for thepath of a ray since we have assumed that AL is a func-

To evaluate d z/'d n, observe that we must take an tion of z only. It is convenient to obtain a first integralelement d n of the normal to the ray path drawn in- of this equation before expressing ,u as a function of z.wards, and find the corresponding change d z in z. d zReference to Fig. 3 will bring out the construction. To do this we multiply (29) by 2 d dxand reduce toSince the inward normal has a downward component,xit is evident that z will decrease and, therefore, d z will d z 2be negative. It follows from Fig. 3 that 2 d (d x )I dAt1 (30)

dz/dn=-cos f (21) 1+( d z \2 At

where q$ is the acute angle between d z and d n. But kd x/the angle X is also the angle between the tangent to The two sides of the equation may now be integratedthe path and the horizontal so that directly to

dz/dx =tan f= sin fcos f (22) /d z iFurthermore _ ____log{ 1 dx ) ==2log,u+ C1 (31)

sin = dzV d x) + ( z)2(23)Taking the anti logarithms we haveCombining (22) and (23) we find 2______ cos¢ = 1/V/1+ (dz/dX)2 (24) 1 + ( z J) C2 (32)

13. We must be careful in which direction we draw the normalto perform the differentiation d At/d n. If the refractive index In order to determine the value of C2, observe that JUStat A is less than thaUt at B, so tha.t the velocity at A is greaUter before the ray enters the ionized medium we havethan that at B, the path going through A will bend over towards At = 1, while the left hand side is the square of theB. As dra.wn it is a convention that the radius of curvature secant of the angle between the ray and the horizontal.shall be considered negative. Hence d At Id n must be consideredpositive, a,nd we must differentiate along the inward drawn 16. For example, Granville, Differential and Integral Cal-normal to the path. culus, p. 161.


This can readily be seen by reference to Fig. 4 as follows: mass of gas is subject when moved from one level tod z!d x = tan 3 another would automatically bring its temperature to

I + (d zld x)2 = 1 + tan2' = sec2 / that of its surroundings. Actually, however, the heatTherefore when A = 1, C2 = sec2 ,B. Substituting this exchanges by radiation and conduction reduce thevalue of C2 in equation (32) and putting 1 /cos2' for observed gradient to approximately one-half of thesec2' we obtain adiabatic gradient. The height of the troposphere in

cos d z different places may vary from 10 to 15 kilometers.d x = 2r__ (33) The outer shell is called the stratosphere and is

V,u -Ccos2 distinguished by the absence of a temperature gradientThis is the general differential equation for the ray and consequently any large scale mass motion. It is

paths, applicable to any distribution of ionization, which sometimes referred to as the isothermal layer. Theboundary between the two layers is, of course, notsharp, and the thickness of the transition layer, in which

W.un-8D.t.fz, of Uthe mass motion practically dies out, is not at presentkoniz,d Nfediu.M.dx/& precisely known. We will assume, however, for the

present discussion that convection ceases at a heightFIG. 4 of 20 kilometers.

In Fig. 5 we have reproduced the proportional compo-allows us to express the resulting refractive index A sition of the atmosphere, by mass, as calculated byas a function of the height z alone. In order to plot Chapman and Milne.'7 This is based on the assump-the path of a ray we require the value of the horizontal tions that convection ceases at 20 kilometers, and thatdistance x for any given value of the vertical height Z. no hydrogen is present. The temperature of theThis required relation between x and Z is obtained by stratosphere is taken at-54 deg. cent. In Table I theperforming the integration indicated below: pressure, number of molecules, approximate molecular

Cos r3 d z mean free path, and the average collision frequencyx = . (34) between an electron and gas molecule, are given for

Z=0 V.Y2- cos2Here: is a constant for any given ray as will be seen 210

from Fig. 4.The above expression for the path of a ray is for a 70-f0

plane earth. The corresponding equation in polar X50

coordinates which takes into account the curvature of 3Ethe medium is given in Appendix II. O

Before we can actually perform the integration .0 IIindicated in (34) we must express A as a function of z GO

and substitute it into the equation. In the following 40

section we shall review the present status of our Hirgon~knowledge concerning the constitution and probable °11'o,1,I405 6D

state of ionization in the upper atmosphere, in order to FIG. 5-PROPORTIONAL COMPOSITION OF AIR BY MASSselect a reasonable, and at the same time analyticallymanageable, assumption as to the variation of refrac- TABLE Itive index with height above the earth. Electron

Pressure Number of Molecular CollisionIV. CONSTITUTION AND DISTRIBUTION OF IONIZATION Height in Dynes per molecules M. F. P. frequency

IN THE UPPER ATMOSPHERE kilometers sq. cm. per cu. cm. cm. per sec.

A. Constitution. The earth's atmosphere is com- 0 1.01 X 106 2.7 X 101| 9. X lo- 9.5 X 10'112 1.92 X 105 6.5 X 101 4. X 105 2.1 XlO"1

posed of two concentric spherical shells. The inner one 20 5-53 X 104 1.9 X 1018 1. X 1o-4 8.5 X 1010is called the troposphere, and is distinguished by the 40 2.55 X 10- 8.6 X 1016 3. X 10- 2.8 X 109

60 1.24 X 102 4.2 X 1015 6. X 10-2 1.4 X 108existence of a temperature gradient, which decreases 80 6.27 2.1 x 1014 1.0 8.5 X 1o0from the ground upwards, the mean gradient being in 100 0.363 1.2 X 1013 20. 4.3 X i05

150 1.49 X 10 5.0 X 1011 500. 1.7 X 104the neighborhood of 6 deg. cent. per kilometer. In 200 5.62 X 10-3 1.8 X 1011 1000. 8.5 X 103this region, masses of air are continually being raised or 300 6.99 X< i0~4 2.4 X i01' 1 X lO 8.5 X 102lowered, and thus moved to places where the pressure is 600 2159 X< 10-6 8.8 X< 107 3 X 106 2.8different. 800 7.97 X 10-8 2.7 X i06 9 X i07 0.95

In the absence of heat exchanges by conduction and 100 29X10 9.Xi0 3Xio 28X10radiation, the resulting temperature distribution would* 1.l . ~~~~17.5. Chapman and E. A. Milne: Quarterly Jour. of the Roybe that of an atmosphere in adiabatic equilibrium. Meteorological Soc., Vol. 46, p. 357, 1920. This is a veryconIn other words, the temperature distribution would be plete discussion of the present status of our knowledge of thesuch that the adiabatic heating or cooling to which a upper atmosphere.


different heights. The approximate values of the above 70 kilometers, the radio waves will not be sub-molecular free paths were calculated by Chapman and jected to large absorption.Milne"6 from the relation Recent work by Lindemann and Dobson"8 based on

1 the observation of falling meteors leads them to a higherV 2 7r MU2 cm. (35) estimate of the density above 60 kilometers than that

corresponding to a temperature of 219 deg. K. for thewhere isothermal layer. Their calculations indicate a valueM = number of molecules per cu. cm. in the neighborhood of 300 deg. K. They have alsoUf = mean molecular diameter assumed 3 X 10 -x cm. suggested reasons why we may expect this higherThus no account is taken of the fact that the air is a temperature. It has occurred to the writers that themixture. The approximate values of the electron free heating effect of the eddy currents induced in thepaths are obtained from the relation ionized upper atmosphere required in the explanation

1 of the diurnal variations of the earth's magnetic field,l'=- l e Cm. (36) may be a contributing factor.M r Ul2<1 + - The effect of a higher temperature and consequently

m higher densities at a given height in the isothermalwhere layer will be to increase the height at which the electronM - number of molecules per cu. cm. assumed large collisions cease to be important and will require us,

compared with the number of electrons. therefore to put the refracting medium at a greaterUf + U' height if we wish to avoid high absorption of the radio

al 2 = average of molecule and electron waves

diameters B. Ionization. The theories of the aurorae,"9 di-

M' = mass of electron urnal and semi-diurnal variations in the earth's mag-netic field as well as magnetic storms20 are based on them = mass of molecule

For our purposes we may neglect the mass and exist.nce of ionization in the upper atmosphere.Again in the early days of radio, Kennelly2' and in-diameter of the electron in comparison to those for the d H

molecule so that the electron free path iS dpnetyHaiie ugse h edo o

ducting layer to account for the observed long distance1' - 4 transmission. Recently Watson23 has shown that an

7w M U2 equation of the Austin-Cohen type can be obtainedtheoretically if a conducting upper atmosphere isCornparison with (35) shows that the electron free path ased.

is 4 / 2 times the molecular free path. Thed.The average electron collision frequency is The present views appear to favor the hypothesis

v - c/l' (38) that the ionization in the upper atmosphere is main-where c is the average electro8 velocity as given by tained principally by the action of high velocity elec-kherecistheaveragelectroveloctyasgtrons which reach the earth's atmosphere from the sun.

= 8 km cm. per sec. (39) When they approach the earth, the magnetic field willwM concentrate them at the poles and bend them around

in which the earth into the dark hemisphere. On this view, thek = 1.37 X 10-16 universal gas constant maximum ionization intensity will be expected in theT = electron temperature deg. K. polar regions and on the sunlit hemisphere. Ray-mI = 9 X 10t-2 the electron mass in grams. leigh24 has shown definitely that the green auroral line

In making the calculations we have assumed an elec- is always present in the night sky and in this waytron temperature of 6000 deg. K., i. e., the electrons proved the existence of strong ionization on the darkhave not had sufficient time since entering the side of the earth.stratosphere to acquire the mean temperature of 18. F. A. Lindemann and G. M. B. Dobson, Proc. Roy. Soc.219 deg. K. This gives Lond., Vol. 102, p. 411, 1923.

c = 4.85 X 107 cm. per sec. (40) 19. For a summary of theories and references see W. J.and consequently Humphrey's Physics of the Air, p. 422, Lippincott, 1920.

20. See excellent summary by S. Chapman in Glazebrook,4.8 X 107 8.6 X 106 Dictionary of applied Physics. Vol. II, p. 543, MacMvillan, 1922.4 V21 (41) 21. A. E. Kennlelly: Electrical World and Engineer, Maurch 15,

+/ ~~~~~~~~~~~1902.Inspection of the values in Table I shows that for 22. Oliver Heaviside: Encyclopedia Britannica, tenth edition,

a wave length of 60 meters we will obtain the maximum Vol. 33, Dec. 1902.1~~ ~~~ ~7 kioetr sic at this 23. G. N. Watson: Proc. Roy. Soc. Lond., Vol. 95, p. 562,absorption at approximately 0 loerssn atn 1919.

point v = ct. Consequently if we assume that the 24. Lord Rayleigh, Proc. Roy. Soc. Lond., Vol. 109, p. 428,lower boundary of the ionized medium is considerably 1925.


Chapman and Milne16 have calculated the distribu- lisions between electrons and molecules on the refrac-tion of ionization which would result from the absorp- tive index of the ionized upper atmosphere can betion of high velocity electrons in the earth's atmosphere. neglected. We therefore substitute the expression forWe have reproduced their curve in Fig. 6. From this N given in equation (42) in equation (13) and obtainwe see that from the ground to nearly 40 kilometers the following expression for the square of the refractiveheight the relative ionization remains practically zero. index as a function of the height z above the earth:It then begins to increase slowly at first, and then moreand more rapidly with the height. As we approach the .2 = 1 - sin2 2 - b) (31j,(A2-in2 (B-b) (43)

4 7re2so where = m

= 3.2 X 109 c. g. s. (e. s. u.) for electrons.This expression is only assumed to hold in the ionized

N

eigA Atw4 &o

FIG. 6(-)

maximum, the rate of increase decreases again, and after 4OIRzlvil Ion=uon . (4 t 1| / ~~~~~~~~~~~~~~~~~~~~~~~Ionfiztori-Ass.

passing the maximum at 54 kilometers, it falls off tob.zix.more gradually to a small value at 90 kilometers. Ourpresent interest is mainly in the general shape of thecurve, since for a different absorption coefficient the (z-6}Tabsolute values calculated would be changed, but the N t4 2(8-b)general shape characteristics would remain the same. FIG. 7

In calculating the refraction of radio rays which riseinto an ionized medium of this type and are refracted Partbof the medium from z = b to z = B. Belowback to earth, we are only concerned with the shape ofthe curve from the ground to the maximum since any equal to unity down to the surface of the ground, i. e.,

z = 0. As previously pointed out, we are not concernedrays which riseabovthe maximum wil with the behavior of the refractive index after passingspace and be lost. With this in mind, it will be evident

that a good approximation for our purposes may be hadby writing assume that the refractive index returns graduallyby writing to unity at a great height.

N = N sin2 (z-b)(42)2 (B- b) (4

whereb = height of the lower boundary of the ionized -chitM 7aximum

medium above the earth. 8-6)B = height of the maximum ionization above the I Ih At w

earth. y|oin.No = maximum number of electrons per cu. cm. 6'N = number of electrons per cu. cm. at height z

o- &z...X%ZZZZZZzZ/XM%Xw%zzzze z ax_above the ground. -x)Here z is restricted to values between b and B. By KX-b

varying the values No, b and B we can represent a thick FIG. 8or thin, high or low medium with any desired value of SubsI gvnaoeb ( imaximum ionization. The general shape Of the curveeqain(4weotniS shown in Fig. 7. -

V. CALCULATION OF THE PATH OFA RAY INA MEDIUM (X- X) = NJ co -3dIN WHICH THE ELECTRON DENSITY IS A SINE A!b 1- arN sin2('z b) -rcos%@

SQUARE FUNCTION OF THE HEIGHT \ a2 *2(B-b)A. Case of Flat Earth. We have seen in the pre- (44)

vious section that for short waves the effect of col- as the horizontal distance (x -X) traveled by the ray


in the ionized medium, while rising a distance (Z - b) we havein the ionized medium. Inspection of Fig. 8 will make cos u d u = sin a cos w d w (55)this clear. and the integral (53) becomesThe trick involved in the integration of this equation

consists in finding substitutions which will transform (x-X) =it into a known form. In what follows it is shown that f sin 7r(Z=b)lit can be reduced to an elliptic integral. w=ssnlsin a J2 (B-b)cot/3sinacoswdw

First of all let us put r (56)u = 7r (z-b) /2 (B-b) (45) J=o 7r COS U cos w

and consequentlyd u = 7r d z/2 (B - b) (46) Let us now define on angle 4 such that

also wr(Z-b)1- COs2/ = sin2f, ' sin a sinq5 = sin2(Bb) (57)

in equation (44) and obtainsr(Z -b) and substitute it in (56); we obtain20) 2 (B- b) cos/3du w=O

(X XX) =2 (B-b) eos du ( f47\ 2 (B- b)cot/3sinadww

(x- \!sin2/ - No sin2 u (x- X) = wo wcosu (58)

Squaring equation (54) and subtracting each sideWe will now divide numerator and denominator by from unity we have

sin ,3, and getsr(Z-b) 1- sin2 - 1- sin" asin2w (59)

2(BX-b) 2 (B-b) cot 3du and hence(x~ ~0 =F No_ 48 os2 ut = 1- Sin2ao sin2 w (60)7r! 1- sin2u

w2 sin22 Substituting equation (60) in (58) we obtainIt can be shown by the following argument that the U.( 2-(B)- b)cotsisin a d w

expression o- NO/c 2 sin2 /under the radical sign is (x - X) 14 w (61)greater than unity. Going back to equation (32)we have If we now put the value of sin a from the relation

1 + (d z 'd x)2 = /2/cos2/ (49) W2 sin2,/lo- No = sin2 a (62)and if the ray becomes horizontal and returns to earth, in the numerator of (61) we obtainwe must have d z,/d x = 0 at the summit, and conse-quently (x-X)=2 (B-b) X cos d w

/.t2 = coS2A (50) Xwv No wN0 1- sin2 a sin2wSubtracting each side of equation (50) from unity we (63)have The integral in the above equation is called the

1 - 2 = sin2/ (51) elliptic integral of the first kind. For brevity it isat the summit of the path. customary to writeBut we have from equation (13) d

1- y2 = o N/W2 (52) F - = F (sin a, 4) (64)and No is greater than N. Hence from equations W=0 \/1-sin 2a sin2 w(51) and (52) we have o- Nol W2 greater than or equal It iS also customary to put k = sin cx and write itto sin2 3 5so that the expression o- No/ W2 sin2 /3 is greaterthan or equal to unity. We may, therefore, put it F(k, The numerical value of the integral for anythanor eualto ulty We ay,therfor, pu ltvalue of ax and 4) has been calculated and recorded inequal to one over the square of the sine of some angle a talue of e integras Forc examl wecfindfosince the cosecant square of an angle is always equal the tables thatto or greater than one. Hence

X (-)2 (B-b) cot/ du (53) F'((sna 1 -sin2a sin2w =.48(65)n== W+\ 1- in when 4) = 25deg. and ae = 75deg.

si2K The numerical values of 4) and ag for any valueA further substitution will be required since the 25Foeaml,seASrtTbef ngas,yB..coefficient ~ofsn*sgetr hnoe fw u Pierce, p. 118. A more complete table will be found in Jahnke

sin u = sin ae sin w (54) und Emde, Funktionentafeln, p. 53-68.


z = Z on a given ray may be obtained from (62)and foot of the perpendicular which passes through the(57), as follows: summit of the path and Z, is the height of the summit

a = sin-' { co sin /-V No0 (66) of the path above the ground.7r (Z- b) The integral in equation (75) is called the Complete

a// No sin 2 Elliptic Integral of the First Kind. For brevity it is4 = sin-' (B- b) (67) customary to write it

co sin3 j 7r2 dw

The corresponding numerical value of F (sin a, 4)) 1 V1-sin2 sin2w = K (sin a) (76)is looked up in the tables and the numerical values ofthehorizontal component (x - X), traveled by the ray It is also customary to put k = sin a and write itin the ionized medium while rising a distance (Z - b) K (k).in the ionized medium as Numerical values of this integral for any value of a

will be found in the tables of elliptic integrals.25(x - X) =2 (B-b) w cos

- F (sin a, 4)) (68) We may now write the expression for the total hori-ri/rV No zontal distance between transmitter and the point at

On the assumption of a flat earth, inspection of Fig. 8 which a given ray will return to ground for the Caseshows that we must add the horizontal projection of the of a Flat Earth as follows.straight part of the path 2 x, = 2 b cot f

X = b cot ,B (69) ~~4(B-b) X cos0X = b cot 03 (69) + 4 (B - b K (sin a) kilometers (77)to equation (68) to obtain the total horizontal distance r Va/o Nox to the point (x, Z) on the ray path. The complete In whichexpression for the case of a flat earth then becomes r Q

2 (B- b) w cos a = sin-' sin A3 in degreesx=bcot 3+ F(sin a,4) (70) VaNo

b - height above the ground at which ionization isIf we wish to know wher te ray will return to earth assumed to begin; measured in kilometers.

we must takhe tw the value of x corresponding to thet = angle between ray and horizontal at the lowersummit of the path since the path is clearly symmetrical, boundary of the medium in degrees.At the summit of the path we have by equations (51) B = heightabove the ground at which the ionization isand (52) assumed to reach its maximum value, in kilo-

sin2l =a¢N/~CA)2 (71) meters.But co = 2 r f and f = frequency of the radiation in

2(Z - b) w cycles per second.

N = N-sin- 2 (B-b) (72) af = 4 re21/m = 3.2 X 109c. g. s. (e. s. u.)where e is the charge and m the mass of an

from which z electron.o- No (Z - b) 7r No number of electrons per cu. cm. at the height B of

sin2 = @2 sin2 2 (B - b) (73) maximum ionization.K (sin a) = K (k) = complete elliptic integral of the

Substituting the value of sin: from (73) in equation first kind to be looked up in tables. For(67) for 4) we obtain example, when a =10 deg., K(sin a) = 1.5828.

s0B. Approximate Allowance Made for the Effect of the

) = sin- 1 = 90 deg. = 2(74) Curvature of the Earth on the Ray Paths. When con-sidering long distance transmission, the effect of the

at the summit of the path. earth's curvature cannot be entirely neglected. TheSubstituting this value of ) in equation (63) we analytical difficulties, however, become considerably

obtain increased when we attempt to take into account the(x,-X) = W= urvature of the ionized medium and use equation (15)2 (B - b) c cos 23 d w of Appendix II instead of equation (34). Happily the

J (75) most important efect produced by the earth's curva-w V aN,~~ V 1 - si2 a sin2wture iS to increase the angle s3 at which a ray of given

as the equation for the horizontal component initial inclination to the horizontal 6, strikes the lower(x - X) traveled by the ray in the ionized medium boundary of the ionized medium. This effect can bewhile rising to the summit of the path in the ionized readily taken into account as follows:medium. Or, in other words, while rising a distance Let r, be the radius of the earth and let r, be the(Z - b) in the ionized medium. In the above x, is radius from the center of the earth to the lower bound-the horizontal distance from the transmitter to the ary of the ionized medium. Then from Fig. 9 the


angle 0 A B is equal to (90 deg. + 0) while the angle tance 0 6 or range of the ray on the assumption of a0 B A is equal to (90 deg. - 0). It follows that the flat earth. Radii are then drawn from the center ofangle A 0 B is equal to (o -0) since the sum of the the earth through the points 1', 2', 3', 4', 5', and 6' andangles of a triangle is equal to 180 deg. From the sine continued above the lower boundary of the ionizedrule for triangles we have medium. Distances equal to the ordinates at 1, 2, 3,

ro/sin (90 deg. - 3) = ri/sin (90 deg. + 0) (78) etc., of the full line curve are then measured off on thefrom which corresponding radii 1', 2', 3', etc., from the concentric

cos 6 = (ri/ro) cos B (79) circle representing the lower boundary of the ionizedand sincer, = (ro + b) we can write (79) medium. These distances become the ordinates of the

cos , = cos 0/j1 + b/ro) (80) dotted curve which represents the approximate path ofthe ray in the ionized medium. At E the dotted ray

From equation (80) we can calculate the value of : forany assigned value of initial angle 0 and height of (x x)Zmedium b. Knowing : we can calculate (/ - 0) andthe distance A C measured along the surface of theearth since the arc length

27w roAC 360 3) (81)

where (,-) is in degrees.The method of approximating the portion of the path

in the ionized medium for the case of a curved earthwill be understood by reference to Fig. 10 in whichthe effects of curvature are greatly exaggerated forthe sake of clearness. Assume a ray starting out FIG. 10from the transmitter at A which makes an angle 0 withthe horizon. This ray is assumed to take a straight emerges from the ionized medium and is assumed topath through the lower atmosphere and meet the take a straight path, E to e,back to the surface of thelower boundary of the ionized medium, vertically earth, theacuteangentplaneatEover the point marked C. The angle between the ray andtheraybeingovdand the tangent plane at the point of intersection being It will be observed that the above method of con-,B. The value of 3 for any value of 0 is obtained from structing the approximate path for a curved mediumequation (80) from the calculated path for a flat medium increases

the length of the transformed path. This effect is inthe right direction since in the actual problem themedium would be bending around with the ray andconsequently retain the ray in the medium for a greaterdistance.

6 8 On the above assumptions the total range of a ray forA Se the case of a curved earth, i. e., the arc length A D in

Fig. lOis given by

¾X'')WT Range 460 (A3-)360

°0 4 co (B- b)cos3FIG. 9 ± 4( )os K (sin a) (82)w V - No

whereThe distance over the surface of the earth A C is

obtained from equation (81). We now calculate the ro = 6300 kilometers, the approximate radius of thepath which a rav with this value of A would take in a earthFlat Medium by equation (68). This path is illustrated 0 = angle between ray and horizontal at the trans-by the full line curvze. The value of (x -X) from 0 to mitter in degrees.2 is indicated as (x-X) 2. Similarly the ordinate cos 6of the path at the point 2 is indicated as (Z -b)2. /3 cos'/r) = acute angle betweenThe distance 0 to 6 is 2 (x - X) as given by equation{(1 br)(75). The distances 0' 1', 0' 2', etc., equal to 0 1, 0 2, the ray and the tangent to the lower boundary of theetc., are then laid off on the surface Qf the earth as ionized medium at the point where the ray of initialshown. Thus the arc length 0' 6' is equal to the dis- -angle 6, enters the medium; measured in degrees.


The other factors are as given for the case of a flat amplitude of the integral, is determined from equationearth under equation (77). (57) which becomes (see also equation 45)

VI. TYPiCAL PATH CALCULATIONS . . *a

9 (Z- 200) lIn order to make clear the method used in calculating s 8

ray paths, we give an example which is representative and consequently varies along the path.of summer day conditions. The values used for the At the summit of the path k becomes a right angle,constants appearing in the formulas were selected with and F (sin a, b) becomes K (sin a).the idea of fitting the experimental data concerning The path is symmetrical about a radius, (i. e.,skip distances given in Fig. 1. For summer day con- vertical) drawn through its summit.ditions we assume the maximum electron density to The distance X appearing in the expression is thebe Nn = 6 X 105 per cu. cm., the lower edge of the horizontal distance to where the ray enters the ionizedionized medium at a height b = 200 km., and the medium, and is given by 110(the- 6), where (d3 6) isdistance from this lower edge to the height of maximum expressed in degrees.intensity, (B - b) = 80 km. Putting these values All distances are expressed in kilometers.in the equation (42), we have The calculations are set out in the accompanying

9 (z - 200) tables.N = 6 X 105sin2 t 8 (83) Table II gives the values of 1, (,- 6) and X, for

We shall calculate the paths for a wave length of 21 TABLE II*

meters. On referring to the formula (68) developed Line Quantity 1 2 3 4 5 6 7 8

in Section V we find that if (x - X) denotes the distance I 0 15 20 23 25 25.7to the right of the point where the ray enters the II lo 14.25 15.04 17.3 20.7 24.5 26.9 28.55 29.2

ionized medium,III (,B-0)" 14.25 10.04 7.3 5.7 4.5 3.9 3.55 3.5IV X,km. 1568 1104 803 627 495 429 390 385

2.12 X 104 *The values of d for different values of 0 were unfortunately read

(x - X) = - (B- b) cos ,B F (sin a, 4) (84) from a curve instead of being directly calculated from equation (88).A. X/ No The parts of the paths in the ionized medium are calculated correctly

for the values of,B used, but some small errors made in reading the curve

and is therefore in this case given by (,8 vs. 0) mean that the paths are for slightly different values of 0. These*x -n X) =104.5 co (si a,errors do not amoLnt to more than a small fraction of a degree in 0, and

(x- X) = 104.25 cos 3 F (sin ac, c) (85) therefore do not appreciably affect the results. In the same way small

The angle a is determined from the equation (66) which inaccuracies exist in the si-nal intensity calculations of Tables Xl andXII.

becomes105 various values of 0. These values depend only on b.

sin a - sin (86) Table III gives the calculations of a and 104.25 cos 13 X -,~/No by means of sin 1, sin a and cos1. These quantities

and in this case reduces to are constant for a given ray path.43 Table IV gives the calculations for the points corre-

sin a = 21 sin (87) sponding to 4 = 18 deg., 36 deg.,54 deg.,72deg.,and90deg. The quantities determined for each point are

The angle : is determined by the equation (80), which is z and (x - X).Table V gives some additional points for the curve

roCos1 = Cos 0 (88) with 0 = 25.7 deg. This curve goes to infinity, and

ro + b is the ray path which separates the rays which return

in which ro = 6300 km. = earth's radius to earth from those which go out into space.and b = height of lower edge of medium. Table VI gives the values of the co-ordinate x for

The angle 0 is the angle made by the ray with the the points calculated. The downward curve is filledhorizontal at the transmitting station, 1 is the angle at in by symmetry, the corresponding points being tabu-which the ray cuts the ionized medium, and a is an lated against supplementary values of ¢. The pathsangle which is required in the calculation of the path. obtained from the calculations in the above tables are

The elliptic integral F (sin a, 4) is given in the tables21 plotted in Fig. 11.for various values of a and p6. The angle 0, called the It is of some interest to consider the paths of those

TABLE III

Line Quantity 1 2 3 4 5 6 7 8

I 0° 0 5 10 15 20 23 25 25.7

II sine 08O.246 0.2595 0.2974 0.3535 0.4147 0.4524 0.478 0.4879III sina 0xO..503 0.531 0.608 0.724 0.849 0.927 0.979 1

IV oa 30.2 32.1 37.4 46.4 58.1 68 78.2 90 °V cos3 0O.9692 0.9657 0.9548 0.9354 0.9100 0.8918 0.8784 0.8729VI 104. 25Scos.f km. 101 100.5 99.4 97.4 94.7 93 91.5 91


TABLE IVLine Quantity 0 1 2 3 4 5 6 7 8

I 0 0. 50 10° 15° 200 23° 250 25.70II sin 4 sin a 18° 0.1565 0.164 0.188 0.2235 0.2625 0.2865 0.308 0.309III U 9. 9.440 10.840 12.9° 15.2° 16.6° 17.93° 18°IV z 208 208.4 209.6 211.5 213.5 214.8 215.9 216V F (sin a, 4) 0.3154 0.3156 0.3161 0.3169 0.318 0.3187 0.3192 0.3195VI (x - X) 31.8 31.7 31.5 30.9 30.2 29.6 29.2 29.1VII sin i sin a 360 0.296 0.312 0.358 0.4255 0.499 0.5445 0.575 0.5878VIII U 17.3° 18.2° 21° 25.2° 29.9° 33° 35.10 360IX z 215.4 216.2 218.7 224 226.6 229.3 231.2 239X F (sii a, 0) 0.6384 0.6397 0.6433 0.6502 0.6596 0.6666 0.6719 0.6743XI (x - X) 64.5 64.3 63.9 63.4 62.5 62 61.5 - 61.3XII sin 0 sin a 540 0.407 0.43 0.492 0.585 0.687 0.75 0.792 0.809XIII U 24° 25.5° 29.5 ° 35.8° 43.4° 48.60 52.1 ° 54°XIV z 221.4 222.7 226.3 231.8 238.6 243.2 246.4 248XV F (sin a, 0) 0.9746 0.9769 0.9946 1.017 1.054 1.086 1.111 1.124XVI (x - X) 98.5 97.5 94 98.5 100 101 101.5 102XVII sin 0 sin a 720 0.478 0.505 0.578 0.689 0.807 0.881 0.93 0.9511XVIII U 28.6° 30.3° 35.3° 43.6° 53. 8X 61.8° 68.4° 72XIX z 225.4 226.9 231.2 238.8 247.8 2.54.9 260.8 272XX F (sin a, 01 1.326 1.336 1.365 1.418 1.546 1.648 1.769 1.843XXI (x - X) 137 134 136 138 146.5 154 162 167.5XXII z 90, 228.4 228.5 233.3 241.3 251.7 260.4 269.5 280XXIII K(sin a) 1.687 1.704 1.757 1.875 2.117 2.419 2.995 ooXXIV (x - X) 170.5 171.5 174.5 183 200.5 225 274 00

The values of z and (x - X) are in kilometers.

rays which do not return to earth. We can only < Ncalculate these paths as far as the height of maximum the expression 2 No becomes less than unity. Weelectron density, because we have made no assumptionsconcerning the electron distribution at greater heights. proved this expression to be greater than unity for allFor this purpose we require a different formula, since rays that return to earth, but in the case of rays going

o- NoTABLEV into space we can prove that we must have 2 sn2TABLE V CO S1 2

When 0 = 25.7° a = 90° and a = U

Line Quantity 1 2 3 4 5 6

I or U 750 78° 81° 840 87° 890 <II z 267 269.3 272 274.7 277.3 279.13-M m -TIII F (sin a, 4) 2.028 2.253 2.542 2.949 3.642 4.741-8IV (x .-X) 184.5 205 231.5 268.5 331.5 432 -) 0Km

V x 570 590 617 654 716 817 IME

XIII l Ground I | t78 | 2552 | 1956 620|1390|1308|1328|F1Ttz-fF*<-ti_0 41

____ ~~~TABLE VILine f Values of x in k. m. bENaGeTCsDITarNe

If we go back a fewsteps in the derivation of the that used previously. We mINy'LhNniput

I x-X) 0 5 10 15 200 230 250 25.7((B-b) 430dKuII 0 x 1568 1104 803 627 495 429 390 385

III= 18 1600 1136 835 658 525 459 419 414IV 36 1633 1168 867 690 558 491 460 475V 54 1666 1202 897 725 595 530 492 487VI 72 1705 1238 939 765 642 583 552 553VII 90 1739 1276 978 810 695 654 664 05ViII 108 1773 1314 1017 855 748 725 776IX 126 1812 1350 1059 895 795 778 836X 144 1845 1384 1089 930 832 817 868xi 162 1878 1416 1121 962 865 849 909 4`XII 180 1910 1448 1153 993 895 879 938 FXIII Gxround 3478 2552 1956. 1620. 1390. 13081 1328.

FIG.1equation (68) breaks down when 03 exceeds the limitingvalue for return of the ray to earth. less than unity, by a somewhat similar argument to

If e go-I--bac a fe stp in th deiato of th thatf usedar npreviously. We may-v then putf


The above integral then becomes an elliptic integral, ray, and the angle of incidence is equal to the angle ofand its value is reflection. Clearly in this case the second path

2 (B - b) described will be a repetition of the first, provided that(x - X) = cot 3 F( sin a', U) (93) the reflection occurs at a horizontal plane, for example,

at sea.

In the case considered, we have In Fig. 12 we show an example of some of these21

sin a' =94-43 sin ~ (4

160(x- X) = cot /3F (sin a', U) (95)

In table VII are given calculations of paths whichdo not return to ground.The curves given in Figs. 11 and 12 which correspond

to the ionization conditions assumed for winter nights,FIG. 12 REFLECTION OF TANGENT AND SKIP DISTANCE RAYS

TABLE VII FOR WINTER NIGHT CONDITIONS

Line Quantity q or U z 1 2

1 0 40° 70' multiple paths, corresponding to the winter night curvesII d+0 42.070 70. 65° given above in Fig. 11.III - 0) 2.07' 0.65'IV x 200 227 143 VII. DISCUSSION OF SKIP DISTANCE CALCULATIONSV sin d 0.6700 0.9435VI sin a' 0.73 0 519 We have shown in the preceding section that underVII lcoll 46.9 31.3 the conditions assumed we obtain a skip distance.

160 Some experimental data are available concerning these,IX - cotB 56.3 17.9 so that the parameters NC, b and (B - b), appearingX F (sin a', U) 300 226.7 0.5365 0.5299 in the path equations, can be chosen to fit the observedXi (z - X) 30.2 9.4 values for skip distances.XII x 257 152XIII F (sin a', U) 600 253.3 1.151 1.096 As a preliminary, it is necessary to find the effect ofXIV (x - X) 64.9 19.6 each of the three parameters. This is done by holdingXV x 292 163 prmtr.bXVI K (sin a') 90' 280 1.883 1.705 two constant, varying the third, and calculating theXVII (x stances fr variouXVIII[ (z - X)) 106 130.5 skip distances r various wave-lengths.xviii x 333 174

The effect of varying N,, the maximum electronwere calculated in a similar manner, for a wave length density, can easily be calculated. We have the funda-of 25 meters. The assumed constants are as shown mental equationin the figure. _- N

It is seen from an inspection of these curves that we 2= 1- 2 (96)have a skip distance in both cases. As we increase 0from zero, the distance to the point where the ray 4 7r e2returns to ground decreases at first, and afterwards in = 3.2 X 109 for electrons. Inincreases, having passed through a minimum.

It is evidently not necessary to calculate the complete the above expression collisions are entirely neglected.paths in order to determine the skip distances. For we Suppose now that we multiply N by 4, and co by 2,are only interested in the points where the different rays or, what is the same thing, halve the wave length X.reach the ground, and these are twice the distances to On putting the new values in the above equation wethe summits of the various paths. We must, however, obtain the same value for ,2. But if we multiply Nocalculate the total range for several values of 0, and if by 4, we multiply the value of N at every point by 4we plot a curve giving the relation between this range also. Hence if we obtain a skip distance S for a waveand the angle 0, we shall be able to determine the skip length X with maximum electron density No we shalldistance by reading the minimum from the curve, obtain the same skip distances for a wave lengthThese curves are given in Figs. 23 and 26 for the cases A/2 with maximum electron density 4 No.already calculated. We can, in this manner, make a In general, if we multiply No by n2 and divide Xspecial study of the skip distances apart from the paths. by n, we shall not alter ,u~2 at any point, and thereforeWe must admit the possiblity of a wave being re- we cannot alter the skip distance.

flected at the earth's surface when it comes down. For convenience we show the relation between wave-We shall assume that the reflection is specular, that is length and skip distance by plotting log (skcip distance)to say, the reflected ray is in the plane of the incident as abscissa and log (wave length) as ordinate. On such


a plot, if we multiply No by 4, we must slide the curve curves for various values of (B - b) is given in Fig. 14.down so that every point will have for its ordinate log When we vary b, keeping No and (B - b) constant,X/2 instead of log X. But log X/2 = log X- log 2, there appears to be practically no effect for wave lengthshence we have moved every point of the curve through near that which makes the skip distance zero. Thethe same vertical distance, log 2, and the shape is effect becomes larger as we move away from this wave-unaltered. This rule applies also to any path at a length, so that a family of curves for various values of bgiven angle 0. is spread out into a fan shape at the lower wave lengths.A family of curves for various values of No is given The effect of successive equal increments of b becomes

in Fig. 13. less and less. A family of these curves is given inFor the other parameters empirical results only can Fig. 15.

be obtained. An interesting point is shown by all these curves. AsWhen we vary (B - b), keeping No and b constant, we decrease the wave length, the maximum permissible

value of f, in order that the ray may return, wille I 3,I decrease. It may happen that the ray tangent to the

I<q ILi:rj 4 [ [ t l^earth cuts the ionized medium at an angle greater thanthe critical angle for return. In this case no ray would

40EE/ErRLWfit1 -106E 6 2= 50 K. 0

XOrMcE

MAIOWER (oGroNAX. 8- 6A1z be able to return at all.

When the ray tangent to the earth cuts the ionizedmedium at the critical angle, it is readily shown that

LA+LLALA ! j 1'- LL w the skip distance is infinite, since the ray at the critical

L=1etX Rr -E oo

500H3.00/sos Q

500/So. e

FIG. 13-EFFECT OF INTENSITY OF IONlZATION ON SKIP 1

DISTANCE _|0

the result appears to be to move the curve towards the K -right by a constant amount all along the curve. Suc- l-cessive equal increments of (B - b) appear to produce I4 <Ve~~~~~~~~~~&ee ta41:Ti

S,,slllsP ,0Os/,SEA/sEE /d/ /SE s -

I i00 SIVId -1--1A1'V-016Heo0t FIG. 15-EFFECT OF HEIGHT OF MEDIUM ON SKIP DISTANCE

P3005/59 (05304 EER3I7)~ P.. 5'/$0ON,AX^/NU/1 rrAooezvsf^yo x/v.zIl-w II0o0

DIJANC FRz.OWC;F"CC TO 3 angle goes off to infinity. The curves have, therefore,(5b!1/00^'- o 20 a horizontal asymptote, at the wave length for which the

*00 !K |3| X wo^ 0 xoray tangent to the earth cuts the medium at the- <4qw08 critical angle.

-t70 rioWhen we vary (B-b) we do not affect the criticalangle, which depends only on No. The angle at which

t1X|fr0 "a z the ray tangent to the earth cuts the medium depends.I +__X_- only on b. It is therefore evident that when we vary

(B- b), we do not alter the lower asymptoticwave3II° I R§g' °00 length, but a variation of either N0 or b will evidently085ZACE IJV IrLOEf.S 10change its value.

FIG. 14-EFFECT OF THICKNESS OF MEDIUM ON SKIP DISTANCE The critical angle is always given byco No = sin2 (97)

slightly diminishing displacements of the curve on the /o

logarithmic plot. The shape is again unaltered. It and cosA = ( b) cos 6, so that the ray tangentmust be noticed that at a certain wave length the skip rO+distance tends to zero and the curve becomes asym-- rptotic in the logarithmic plot. A displacement to the to the earth will have cos a3 = ( + b ) and hence,right can obviously have no effect on this asymptote. r0+The limiting wave length is that which makes the value b2 + 2 r bo 98of 1u2 at the maximum ionization equal to zero, hence sin2 : = (r02+(b))we have o No = w2 from equation (96). A family of r+)


Substituting (98) in (97) we find that the greatest We shall also neglect energy absorption in the skyvalue of cw for which a ray can return to earth is given by wave path. We shall further assume that the energy

b2 + 2 ro b radiated by the transmitter at an angle 0 above theo- No = W2 (99) horizontal is the same in all directions, i. e., the radia-

(ro + b)2 tion curve is symmetrical about the vertical axis.Or Suppose that a ray starting out from the transmitter

co = (rO + b) /2N0 (100) at an angle 0 above the horizontal comes down to2 rOb + b2 earth again at a distance x (0). We write this distanee

which evidently depends on both N0 and ~ as x (0) to keep before us the fact that this distance is awhich evidently depends on both No and b.function of the initial angle 0 at which the ray leaves theCalculations show that there is a sudden bend in the tasitr najcn a triga h nltransmitter. An adjacent ray starting at the anglecurve very close to the lower asymptotic wave length. ((d + d 0) will come down at a distance x (O + d 0)Knowing the effect of varying each of the parameters,

it is fairly easy to choose values to fit curves drawn to ( dx (0)which may also be written x (0) + d d 0 The

o _> in e 2 o:t8°t; °°to t |at8' ,power emitted by the transmitter between the angles 0--1--1-_*L.; IlX _ r_t 80 and (6 + d 0) is a function of the distribution curve

C iiX t 29 5 of the antenna, which is here assumed to be a functionof 0 only. We will therefore write the power emitted

_t xr-XX f 0 >Xf 9 t e 8 by the transmitter through this zone as p (0) d 0.lH H L f X< '-_A',, ,, _ This power must come down distributed over the

1 019 Y]11A IV ~~~~area lying between the two circles oni the earth's surface

5tf# <8ofo<<_ { dx(

___

11 i".,LI....-0 4 iatdistancesx ( and x(0) + d6( d j. from

- - +X rt 4 s ll Il Ie the transmitting station. This of course assumes thatCVI7o^D<g.o.,a:7alt-L,2, " these distances are equal in all directions.

If ro be the radius of the earth, the angle A 0 B inFIG. 16 Fig. 17 will be x (0) /ro radians. The radius r of the

zone will then be ro times the sine of x (0) /rO and henceindicate the results of experimental observations. the area will beIn this way we find that the assumed experimental x (0) d x (6)values given in Fig. 1 and redrawn in Fig. 16, require: 2 7r rO sin - d ° (101)For summer days No = 6 X 100 electrons per cu. cm., Dividing the power sent out from the transmitter

b = 200 km., (B - b) = 80 km. For summer nightsNo = 4.83 X 105 electrons per cu. cm., with b = 500 between the angles 0 and (6 + d 0) by the area of thekm. and (B - b) = 430 km. For winter nights the zone on the earth's surface over which it is spread wecurve is nearly parallel to that for summer nights, so obtainwe take N0 = 3.08 X 105 electrons per cu. cm., b = 500 p (0)1km., (B-b) = 430 km., and let the change in No w = d x (0) . x() (102)take up the whole of the difference in the curves. The 2 sdata concerning skip distances, especially at night, areas yet very meager so that our assumed values may be for the power per unit area at the distance x (0) fromfar from accurate. As further data are obtained the transmitter.the curves may be moved in anv desired direction by For the Case of a Flat Earth the angle 0 is equal to ,3changing No or (B - b), and we may change their and we therefore writeshape by changing b. p (,B)

w= dx ()- (103)VIII. CALCULATION OF THE POWER RECEIVED AT THE 2 7r X(x)SURFACE OF THE EARTH FROM A DISTANT SHORT d ,BWAVE TRANSMITTER, NEGLECTING ENERGY for the power received per unit area at a distance x (,B)

ABSORPTION from the transmitting station. In this equationIn the present discussion we will confine our attention x (,s) = 2 xs as given in equation (77).

to the power which comes down to the earth from the It is custosmary to represent the field intensity of thesky due to refraction in the upper atmosphere. The wave radiated by an antenna in different directions byground wave, which is only of importance near the a polar diagram, in which the lengths of the radii drawntransmitter, can be considered in the usual manner, in various directions are made proportional to the field


intensities emitted in the directions indicated by the I = 1o cos2 0 (107)radii. where

Polar diagrams are also frequently drawn in which the Io = horizontal source intensity in watts per unitlengths of the radii are made proportional to the power solid angle.intensities emitted by the source in the different direc- Substituting (107) in (106) and performing thetions. The radii for a power intensity curve of a given integration we havesource are then proportional to the squares of the radii 4 Tfor the corresponding field intensity curve. P = Io (108)

If I be the amount of radiant power which falls onunit surface at unit distance from the antenna system If we put the value of Io in terms of P in equationin the direction 6, the source is said to have a power (107) and substitute the resulting value of I in (105)intensity I for the given direction. In other words the we obtainsource radiates I watts per unit solid angle in the 3direction 0. Then the amount of power which will p (0) = - P cos3 0 (109)fall on unit surface at a distance r from the antennasystem in the direction 0 will be I/r2, since for a source This may be used in equation (102) when calculatingof given intensity I the radiation density in a given the received power from a short vertical antenna.direction falls off inversely as the square of the distance. If the power per unit area radiated by the antennaThe power radiated by the antenna through the zone system is equal for all values of 0, i. e., if the system

lying between 0 and (6 + d 0) which we have called constituted a point source of radiation, we would havep (0) d 0,will be equal to the area of the zone times'the I = Io (110)

and from (106). P=2wrIo (111)

Thereforep (0) = Pcos 0 (112)

-- - - _ B-<A de4@)@ for use in equation (102) on the assumption of a pointsource radiator. Actually of course none of our simpleantenna systems radiate equally in all directions,nevertheless, an assumption of this kind is interestingwhen we wish to study the limitations imposed on short

o /4An5ex ) wave transmission by the properties and configurationFIG. 17 of the medium, irrespective of the directive properties

of our antenna system. When this ideal case has beenamount of radiation falling on the zone per unit area. studied, and the directions of useful radiation determinedFrom Fig. 18 this is seen to be for the particular problem in hand, the most suitable

2 7r r cos 0 r d 0 X I/r2 = p (0) d 0 (104) type of antenna directive curve can be chosen readily.where d 0 is in radians, Before we can calculate the radiation density at

or different distances from the transmitter from equationp (0) = 2w7rIcos 0 (105)

The total power radiated by the antenna, assumingthat it all goes out through the hemisphere above thesurface of the earth, will be obtained by integrating(104) over the surface of the hemisphere.This gives

7r/2P =f 2w IcosOdO (106)

0

For a simple vertical antenna in which the height is FIG. 18a small fraction of a quarter wave length we have theusual expression26 (102) we require the value of d x (0) d 0. This is

26. Curves of this kind were first given by A. Blondel,otie ydfeetain qain(2 hc xFrench Association for the Advancement of Science a,t Angers in presses the range x (6) of a ray as a function of the1903. See J. A. Fleming, Principles of Elect. Wav$ Tel. and Tel>., initial angle 6 at which the ray left the transmitter.Fourth Edition, 1919, p. 362. The directive curves for vertical For the Case of a CurvedG Earth we have from equationantennas excited in various harmonics were first given by Baith.- (82) after expressing the angles in radiansvan der Pol, Jr., in Proc. Phy. Soc. Lond., Vol. 29, p. 269, 1917.Further reference may be made to Stuart Ballantine, I. R. E., 4 wS (B-b)Vol. 12, No. 6, p. 823, 1924, and Balth. van der Pol, Jr., I. R. E., X (6) = 2 r0 (3 -06) ± COS /3Kf (sin a) (113)Vol. 13, No. 2, p. 251, 1925. r+vu No


in which 1, the acute angle at which the ray enters the the transmitter at various values of the angle 0 fromionized medium, is given by equation (82). We have seen from the sample ray

cos 0 paths calculated in section VI and given in Figs. 11cosfl = (1±b) (114) and 12 and further illustrated in Figs. 19 and 20 that

for the case of a flat earth there are in general two directand sin a, which is called the modulus of the elliptic rays with two corresponding values of 0 which traverseintegral, K (sin a), is given by different paths but strike the earth beyond the skip

distance at the same distance x from the transmitter.co

sin a = sin 1 (115) For the case of a curved earth illustrated in Fig. 20V o No we again have no sky rays until the skip distance has

The differentiation of equation (113) involves a good been passed. Beyond this there is a region of two directmany steps and is therefore set aside in Appendix III. rays, until we reach the point where the limiting ray,The final value obtained in equation (17) of AppendixIII gives for the Case of a Curved Earth, -dx 2r tan__ 4 (B-b) sinao -iii

d 0 tan 713 + tan tan 0 X

F E (sin a) K (sin a) ] (Ltan' 13 cos2 a sin' 3 ] 16

The new expression E (sin a) in the above equationdenotes the complete elliptic integral of the secondkind and its value for any value of a. will be found inTables.24 FIG. 20

For the Case of a Flat Earth we require the value ofd x (3) Id 1 from equation27 (77) for use in the power which starts tangent to the earth's surface at the trans-density equation (103). This gives mitter, comes to earth. Outside of this region we againd x (,8) 2 b 4 (B- b) .find only one direct ray. We shall postpone the dis-

(13)- 2b 4(B+-b sin ax cussion of reflected rays for the moment, and confined 1 sin' 13 W our attention to the direct rays.

E (sin ae) K (sin az) Except near the skip distance we can usually neglect[_E(si sin a the ray which has the largest value of initial angle 6,tan213os2 a sin213 since the power density which it produces at x is

(117) relatively small. Where both rays are important weWe have now completed all of the expressions re- add the two values of w to obtain the resultant power

quired by equation (102) for the calculation of the density at the point. This assumes that the relativepower per unit area which a ray leaving the antenna at phases of the two rays are distributed at random when

they arrive at the ground.If everything were constant in the medium traversed

,-'"\by the rays, there would of course be a definite phaserelation between the two rays at the ground which

As,L \\ \ would be a function of the time of propagation of thetwo rays over their respective paths and the wave

// \' \ \ \ ', length. The phenomena of fading indicate irregulari-ties in the refracting medium, which means that at oneNo 9.ys - > rwo DixePtRaysI - ,Vo Rays - T~oDi/ecORays~~ instant of time the amplitudes of the two rays may add

FIG. 19 directly in phase and a moment later the phases willshift so that they will oppose each other. Under thesean angle 0 yields when it again reaches the earth, conditions the resultant power density at the point,

neglecting absorption in the medium. taken over a considerable length of time, is equal to theFor example let us assume that we have a short sum of the power densities of the two rays.

vertical antenna which radiates a total power of P We shall now assume that we have determined thewatts into space, and desire to know the number of initial angles of the two rays which strike the groundwatts per square kilometer w received at a distance x at the required distance x from the transmitter, andkilometers from the transmitter, forthe caseofacurved that we are so far from the skip distance that

earth. we can neglect the power density contributed by theWe first have to calculate ranges of the rays leaving ray having the larger initial angle. Equation (102)27. For convenience we have put 2 x8 = x (13) in this equation. then states that

320 BAKER AND RICE: REFRACTION OF SHORT RADIO WAVES Transactions A. 1. E. E.

p (0) the sine we require the angle in degrees, which isw =(6 wattspersq.km. (118) 180 x(O)

27rro d0 sin (0)

dO ro Wr ro

The value of p (0) for our case of a short vertical We have thus obtained the power density in wattsantenna radiating P watts into space is given by per square kilometer of the earth's surface fallingeantiona(09raslatlng P watts Into space ls gl byat the distance x from the transmitter. To find theequation (109) as

power density on a surface normal to the ray we haveP (0) =

3P cos3 0 watts (119) to divide by sin 0, since the ray comes down at the anglep 2(6)=P cos3 6 watts (119)L6 to the horizontal

where 0 is the initial angle which we have calculated for Let wo w/sin 6 (123)the importantraystrikingthe groundaWe can calculate the received electric field intensity

theirompttasintgodthe distanceter. in microvolts per meter by means of Poynting'sx from the transmitter.Theorem which states that the energy flux den-

We next calculate the value oDfd

from equation sity wo, in ergs per second per square cm. ofd 0 c

an electromagnetic wvave is equal to -~timestheprod-(116). Here ro = 6300 kilometers, the approximate 4 7rradius of the earth; 0 is our calculated initial angle;: uct of the electric field intensity E in c. g. s. (e. s. u.)is calculated from equation (80) in which b equals the and the magnetic field intensity H in c. g. s. (e. m. u.)height of the lower boundary of the ionized medium for the case of an isotropic medium of unit permeabilityabove the earth in kilometers. The angle a is given in and dielectric constant, that isequation (66) as

Cco ~~~~~~~~~~ ~~~wo=- [E Hi (124)sina = sin6 (120) 4 [(

A/ ¢Nowhere

this may be written as c 3 X 1010 cm. per sec. the velocity of light.2 ir c For an electromagnetic wave in space we have

sin = sing (121) Einc.g.s. (e.s.u.) = Hinc.g.s. (e.m.u.)X x/a No This can be seen easily by comparing the expressionsin which for the energy per unit volume stored in the electro-

c = velocity of light in meters per sec. static and magnetic fields, and remembering that bothand forms carry equal amounts of energy. We may there-

X = wave length in meters fore write-= 4 ir e2/m = 3.2 X 109c. g. s. (e. s. u.) for elec- c

trons Sn = 4 7r ~~~~~~~~~~E2ergs per sq. cm. per sec.trons w" 4w7No = maximum number of electrons per cu. cm. where E is in c. g. s. (e. s. u.)

which occurs at the height B above thewhichd.occurs at the height B abovethe If we wish to translate our values of energy fluxground. density, which are given in watts per square kilometer,

Hence into microvolts per meter we have

in = 100 in (122) c1 . g. s. (e. s. u.) = 300 volts3 X -\INo 1 erg per sec. = 10-v watts

The value of (B - b) is the thickness of the ionized E2medium from the lower boundary to the maximum in wo = 120 r watts per sq. cm.kilometers. The values of the complete elliptic inte- . .grals of the first and second kinds, K (sin a) and whereE iS involtspercm.E (sin a), respectively, are found in tables of elliptic Expressing E in microvolts per meter we haveintegrals for any value of a. E = 19.4 X 108 V/WoThe final quantity required is the sine of the angle where w, is still in watts per sq. cm.

0 ~~~~~~~~~~~~~~Ifwe express the power in watts per square kilometerx() Here x (6) is the range of the ray which strikes we have

rO ~~~~~~~~~~~~E= 19.4 X 103 V/wO microvolts per meter (125)

the ground at the distancezx kilometers from the trans- Since in our equations the power is expressed inmitter, and r0 is the radius of the earth in kilometers. watts per square kilometer of earth surface we haveThis gives the angle in radians, and for looking up E = 19.4 X 103 V w/sin 6 microvolts per meter (126)


This is the total value of potential gradient and lies Here - is the fraction of the energy which is assumed toin the wave front. If it lies in a vertical plane, the be specularly reflected.resultant field obtained by combining with the reflected Similarly for the Case of a Curved Earth we have forray, assuming perfect metallic reflection at the surface the second zone from Fig. 22of the earth, will be -"p (0) d 0

2 E cos 1 (127) 2 Xrosin2: (0) X2dx(6)da (131)

If the plane containing the direction of the ray and the rO d 6

RAYS ( < D S ^y4 7r ro sin 2x (0) dx (6) (132)ro dO0

The quantities in the above expressions are the same asx (/3) ++-- (p3)-+ 2those already used in calculating the power density

z(13+d3)- P~ Xo1 +dtl) > received in the first zone except that in (132) we haveto look up the sine of the angle 2 x (0) /ro instead of

dxa3;r3 (< x (0)/r. Thus the labor involved in taking accountdN 3 2 do of repeated reflections is not serious.

FIG. 21 To calculate the energy flux density at a pointdistant x from the transmitter we must take first thatwhich comes by the direct ray, then that which has

vector E makes an angle 4 with the vertical, the been reflected from rays at x/2, x/3, x/4, etc. Absorp-resultant field at the earth's surface will be tion in the atmosphere will tend to reduce the intensity

2 E cos 0 cos A = 38.8 X 10" cos 0 cos 1 V/w/sin 0 (128) of the direct and subsequently reflected rays, also theIn the above calculations we have determined the reflection efficiency factor X will be reintroduced for

energy flux density received at the surface of the earth every reflection so that we may expect that most of thewhen a ray of given initial angle first strikes the ground. energy will come from paths with few reflections. ItObviously a portion of this incident energy will be may of course happen that some or all of the multiplereflected toward the sky and return to earth again at reflected rays will be absent at the distance x undertwice the initial distance from the transmitter. If we consideration: For example we cannot have a rayassume that the irregularities of the earth's surface, at coming down after two sky paths at less than twice thethe point of incidence, are small compared with the skip distance, etc.wave length, the angle of reflection will be equal to theangle of.incidence and the path taken by the reflected IX. INTENSITY CALCULATIONSray will be a repetition of the first. The direct and To make clear the method of calculating the energyreflected rays are illustrated in Fig. 21 for the Case of aFlat Earth. Here the power emitted by the transmitter dbetween the angles / and (B + d /3) first strikes the dg f 2 doearth in the ring between x (d) and x (/3 + d /), and thereflected portion next strikes the earth in the ringbetween 2 x (/) and 2 x ( +d). This process of +\successive reflections will go on until all of the energywhich was originally radiated from the antenna throughthe zone d J is dissipated by absorption, scattering due X 2

to ground irregularities, etc. The power densityreceived in the second zone from the power radiated by FIG. 22the antenna between the angles / and (/ + d /3) will be

W2

p (/3) do 1flux density at distant points, we give an example

ws = d x (A) (129) worked out in full. The case chosen is that for which2 7F X 2 X (/) X 2 d x d/3the path calculations were given. The constants are

d3 lN = 6 X 10" electrons per cc. =maximum density;or b = height of lower edge of medium above ground

Ti p (/3) = 200 kilometers; (B -b) = distance from lower edgeW2 = ~dx (/3) (130) to maximum = 80 km. The wave length chosen is 21

8 ii x (/3) dmeters. If we consider only the end point of the path,d and denote the distance x for this point by x (6), we

with similar expressions for the third, fourth, etc., zones. can draw a curve giving the relation between x (6) and


the initial angle 6. This curve is given in Fig. 23. p (6)If we wish to calculate the energy flux density at a w d x (0) (0) (133)

distance x from the transmitting station, we must find 2 X rd

sin xthe corresponding value of 0 from this curve, since the d (6)formulas used give the energy flux density for rays for a point source p (0) = P cos 0, where P is thestarting out at the angle 0.The energy flux per square kilometer per second is . . . d x (0)

givenbyequatio(102) hich ispower radiated into the sky. The quantity d 6]given by equation (102) which iS df 0- ________________ is the change of x (0) per radian change in 0. In°l|tb 4 + 43*144|'ttjorder to get more convenient units we put3e00.i{~___

ov48 s 6 Dlo' '. .

(8)9000{18

w0 x6. f (). 14Line Quantity1 2 3 4 5 61d70si

180~~

0 4 lb10°15°20° 23° x2(6)5is°alsoto equivalent,

radians chaged its 11

I 6 0~~~~~---5whee10i0exr15e0i 20gres 23an 25ots0 h

II 14.25° 15.04° 17.3° 20,7° 24.5° 26.9° 28.55°III a 30.20 32.1° 37.40 46.40 58.1° 68° 78.20IV sin A 0.246 0.2595 0.2974 0.3535 0.4147 0.4524 0.478V sin a 0.503 0.531 0.608 0.724 0.849 0.927 0.979VI x (e) 3478 2552 1956 1620 1390 1308 1328VII K (sin a) 1.687 1.704 1.757 1.875 2.117 2.419 2.995

_~~~~~~~~~~~~~~~ _

TABLE INLine Quantity 1 2 3 4 5 6 7

I e 00 50 100 150 200 230 250II tan 0 0,0875 0.1763 0.2679 0.3640 0.4245 0.4663III tan d 0.254 0.2685 0.3115 0.3779 0.4557 0.5073 0.5441IV cOS a 0.8643 0.8471 0.7944 0.6896 0.5284 0.3746 0.2045V E(sin a) 1.688 1.453 1.414 1.338 1.229 1.136 1.052

TABLE XLine Quantity 1 2 3 4 5 6 7

I 0 0. 50 100 15° 200 230 250

IV sitan 0 0.326 0.566 0.709 0.799 0.837 0.857III (IT - 1) -1 -0.674 -0.434 -.291 -0.201 -0.163 -0.143IV sina X0I 0 0.1731 0.3445 .513 0.688 0.775 0.839

V XIV 0 0.308 0.612 0.912 1.223 1.379 1.49

VI tan d 0.0721 0.0971 .1428 0.208 0.2575 0.296VII cos2 a 0.717 0.63 0.4759 0.2795 0.1402 0.0417VIII VI X0VII 0.0517 0.0612 0.0679 0.0581 0.0361 0.01236IX E (sin a! III 28.1 23.1 19.7 21.12 31.45 85.2X Cs52 a 0.0674 0.0884 0.1136 0.172 0.205 0.2285XI K (sin a) IX 25.3 19.87 16.5 12.29 11.79 13.10XII IX -XI 2.8 3.233.2 8.83 19.66 72.1XIII 220X(III -220 -4-1.274 -9.53 -. -044.2 -035.9 -31.5XIV XII X V 0 0.86 1.98 2.92 10.8 27.1 107.5

XVdT -220 -147.5 -93. -61.1 -33.4 - 8.8 +76


d x (6) A curve is given in Fig. 24 showing the relationdegrees. The value of is given by d x ( ad6

between d ()and ()d x (0) _ tan Od 220 Itan: In the actual intensity calculation we are not con-

dx (6) _+ (B - b) sin ae tan (6) F (sin a) K (sin cx) cerned with the sign of d 06 but only with its

45- )si tan,B ) (sin2 A)Kos (sin a)45 tan /3 [tanl /3 cos2 a sin' 0 magnitude. This can be seen from the way that thekilometers per degree (135) expression is derived. We now have

which is obtained from the formula (116) by multiplying

by ~~and putting in the value 6300 for ro, the e4rth's

radius.In Table VIII are given the values of quantities A

already determined in the path calculations, viz.,,B, a, sin /3, sin a, x (0), and K (sin a).

In Table IX are given the values of tan 0, tan /,cos a, and E (sin a), as read from appropriate tables.

In Table X are given the calculations to determine FIG. 25-CALCULATED INTEN-SITY OF SIGNAL

d x (0)d (O) *W180 P cos 0

16W~~~~~~~~~~~~~~8W 27rr,,dx (0) . ( x(06) (136)

5r-,WI+A1 d-onputtind0 110

col_ ._T=1.13/sin6(137)I ITTA.-, -1--- dx () x____2228 sin x()

d600 110

812-/'^ < ~~~~~~~inwhlicbh E is in microvolts per meter and w is in watts>t' , /; 1 | t s W I t-per square kilometer.

In T ableXI a regiven the calculations to determine-24o # 4{JIZ tt ,620 , i rE and w from the above formulas.

Aact zvv OEG8CE5-The valueti ofil stfrengthios gvalen byx()ar ie

FIG. 24-RATE OF CHANGE OF RANGE WITH RESPECT TO ThvausoEfrvrisvlesfx(6aegvn

TABLE I*

Line Quantity 1 2 3 4E 5 p 6 7

I e 0O 5 10° 15° 20° 23° 25°

II ~~~110 31.6° 23.2° 17.8° 14.7° 12.6° 11.9° 12.07°

III 3in~x (e) 0.524 0.3939 0.3957 0.2538 0.2181 0.2062 | 0.2091110l

IV XII- 115.1 58. 28.6 15.5 | 7.3 | 1.81 | 15.9d 0 XI deg. Ia

V cos a 1 0.9962 .9848 0.9659 0.9397 0.9205 0.9063VI V .÷IV 0.00868 0.01717 .0344 0.0623 0.1288 0.509 0.057VII 106 XVI .2228 =106 w 3.79 7.69 15.4 27.9 57.7 lr 228. 25.6VIII 01+OIs./VII 1.946 2.77 3.92 5.28 7.6 ! 15.1 | 5.06

19.4 XVIII~~TALEXI

IX E 182.4 182.4 221.2 2452 469.5 151.4

X x (o) 3478 2552 1956 1620 1390 11308 1328*See foot note to Table II.


The calculations were also made for a wave length of : 0

25 meters under winter night conditions, and the results d2x(0) }sin 2 x (0)are given in Figs. 25, 26 and 27. Fig. 26 gives the WI _ d 6 110 (139)relation between x (0) and 0; Fig. 27 gives the value of w2 d x (0) sinx ()

dx (6) ~~~ ~~~~~~~~~~~~~~~~~dO0 110dO for different values of 0, while in Fig. 25 are i w d r a w odO in which w , refers to the direct ray and W2 to the

reflected ray. This simplifies to

* 1'ii_ w2= ",1fW (140)x (6)4 coskk~~ ~ ~ ~ ~~* 110

t~~~~~~~~~'Es WA5 =61 = 015 AX _.Wr e.t6 Sr 1 # s__-:ffso that the value of E2 will be

14~~~~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~~~F* J$-lt *et-i-DD:t-f E2 (141)

cos2CFIG. 26-RANGE OF RAYS WITH DIFFERENT INITIAL ANGLES N 110

plotted the signal strength values against the distance The calculations for the intensity of the receivedx (0). signal after reflection are given in Table XII. WhenTo show the effect of multiple reflections, we have there are several rays coming down at a point, the most

calculated the signal strength of rays coming down a probable signal strength would be given by the squareroot of the sum of the squares of the separate values,

woo VX-A--ti!; 4KtfX lt0t0-=S l Xsince the energies may be assumed to be additive.WttiL+l4-tl003.'0t0 Actually there may be fluctuation all the way from the

4Lo0XXI<t!AXI'DELt:tt:l4-0t LI1sum to the difference of the intensities of the rays, soV0OK, it seems best to leave the curves in the form of the

tT ii fth>n 0-01:separate signal strengths due to the different rays.v>eai600 __L_ e - XWe can also see in this way where severe fading is

1t4 likely to occur. The signal strengths for the directV00 t L 4 W XX and once reflected rays are plotted in Fig. 25.

Et- L, WIii - lX. SUMMARY AND CONCLUSIONS

tILAjiX_l,[iIX.:005ll It00:=u:Ltt:ta]~Wehave seen that the striking phenomena of shortwave radio transmission (i. e., below 60 meters) can

FIG. 27-CHANGE OF RANGE PER DEGREE CHANGE OF INITIAL be quantitatively accounted for on a simple electronANGLE refraction theory in which the effects of the earth's

magnetic field and collisions of electrons with moleculessecond time after being reflected once at the earth's mav be neglected as a first approximation. The distri-surface. If the direct ray comes down at x (6), the bution and number of electrons per unit volume inreflected ray comes down at 2 x (6), and the value of the upper atmosphere required on this theory to accountd x (0) . for the meager experimental data appear to be in generaldO will also be doubled. accord with the values required in the explanation of

the diurnal variations of the earth's magnetic field,The ratio of the two values of w is aurorae and long wave radio transmission28.

TABLE XII*Line Quantity 1 2 3 4 5 6 7 8

I x (0) 3478 2552 1956 1620 1390 1308 1328 140011 E (x) X 182.4 182.4 221.2 252.3 469.5 151.4 80

III~110 31.6 |23.2 17.8 14.7 |12.6 |11.9 |12.07 |12.72

IV cos 110 0.8517 0.9191 0.9521 0.9673 0.9759 0.9785 0.9779 0.9754

V V IV 0.923 0.957 0.976 0.984 0.988 0.989 0.989 0.987.5VI E (2 x) X |95.2 93.5 113 |128. |237 76.5 |41VII 2 x (0) 6956 5104 3912 3240 l2780 12616 12656 12800*See foot note to Table II.

28. H. J. }Round, T. L. Eckersley, K. Tremellenand F. C. Lunnon. Journal. Inst. Elec. Enqg, Vol. 63, NTo. 346, p. 933,Oot. 1925.


Thus the large increase in the skip distance on a given Under these conditions we would, of course, face thewave length at night compared with that by day is a possibility of considerable fading due to interferencenatural consequence of the greater ionization produced between the weak direct ray and the reflected ray.on the sunlit hemisphere by the streamers issuing from Another method would be to use a still longer wavethe sun. The ideal field intensity calculations given length, for example around 25 meters, and depend uponin Fig. 25 show that we may make an ample allowance the reflection of the energy falling at 2500 km. which isfor scattering and absorption and still account for the already beginning to pile up towards the focus at 3500strong signals observed at great distances. The high km.field intensities indicated in the ideal curves of Fig. 25 In this connection it is interesting to note that theat the, skip distance and where the tangent ray strikes reflected 25-meter signal should be much stronger atthe earth are due to the focussing effect discussed in 6400 km. than at 5000 km. since the focus is in theAppendix IV. The intensities at the two foci will, of vicinity of 3200 km. In other words, it is apparentlycourse, Le limited bv the finite size of the source as well easier to get a good day signal at 6400 km. than at 5000as by absorption and scattering. km. This point of view indicates that in short wave

Let us now discuss the problem of maintaining night transmission problems, there will be certain favoredand day communication between two points 5000 kilo- distances. Another point of interest is that a 28-metermeters apart from the point of view of the present wave length should give good day and night communi-theory. For full winter night conditions (i. e., night cation between two points about 6400 km. apart due toat both transmitter and receiver) Fig. 16 indicates that the direct ray by night and the reflection from thethe selection of a 30-meter wave length would put the tangent ray focus by day.receiver right at the skip distance. Such a wave The numerical values deduced for this example are, oflength selection would result in the arrival of two course, very uncertain since the ionization constants forintense sky waves and consequently severe fading. the upper atmosphere which are required before a setInspection of the field intensity calculations of Fig. of radio transmission characteristics like Fig. 16 can25 shows that the field intensit produced by the be calculated were figured backward from the veryray leaving the transmitter at the higher initial angle meager radio data. In other words, it is probably fairdies out very rapidly compared with that produced by to say at the present time that short wave radio trans-the ray having the lower initial angle. Therefore, if we mission experiments are the most direct method wepick a somewhat longer wave length than 30 meters we have of estimating the ionization conditions in the upperwill receive a reasonably strong signal from one ray atmosphere. We should not lose sight of the fact thatonly and consequently be less likely to find severe and the skip distances etc., which depend upon the ioniza-rapid fading effects. We therefore would probably tion conditions in the upper atmosphere are probablychoose a wave length in the vicinity of 32 meters for not constant but will vary from year to year followingwinter night operation. the 11-year sun spot period, the last minimum of whichAn alternative would be to select a wave length occurred in 1922.

in the vicinity of 55 meters, with the idea of taking When both the transmitter and receiver are not inadvantage of the focussing which occurs just inside the sunlit or darkened hemisphere the ray paths will noof the distance at which the ray leaving the transmitter longer be symmetrical about the middle point, and duetangent to the ground comes back to earth. For full allowance will have to be made for the variation ofday conditions this wave length would be weak since ionization conditions between the two stations. Sunset5000 kilometers is too far from the day skip distance. If and sunrise effects will also require special treatment.we applied the above reasoning in selecting the best It is also probable that transmission to or from the polarvalue for full day operation from Fig. 16 we would regions will require special study of this kind, dueprobab]y be led to select a wave length in the vicinity of to the high concentration of ionization over the polar11 meters. This, however, would bring us down very regions as compared with that over the middle belt ofclose to the lower wave length limit which would the earth. We can conclude in a general way thatmean a very weak signal. The reason for the weak transmission from the sunlit into the darkened hemi-signal near the low wave length limit will be readily sphere will result in longer skip distances than wouldappreciated when we remember that under these result if daylight extended over the whole path. Forconditions we are working close to the value at which example, a ray entering the ionized medium from thea ray leaving the transmitter tangent to the earth's sunlit side will at first meet the normal day refraction,surface strikes the lower boundary of the ionized which starts bending the ray back towards the earthmedium at the second critical angle. Under these and, as it moves into the darkened hemisphere, theconditions there is only a small fraction of the emitted bending by refraction will become less and less untilradiation which returns to earth. normal night conditions exist. Thus the ray will strike

It would be better, therefore, to select a wave length the ground at a greater distance from the transmitterin the vicinity of 15 meters and operate on the ray which than it would have, if full day conditions extended overis reflected from 2500 kilometers to 5000 kilometers. the entire path. Here the general conditions for


reciprocity29 are satisfied, so that if a ray were started desirable to place the transmitter on a hill or mountain,back along the same path it would retrace the entire so as to obtain an unobstructed path to the horizonpath back to the starting point. in the desired direction. Raising the antenna system

Inspection of the day curve of Fig. 16 shows that well above the ground will also assist by reducing groundit will be impossible to maintain communication be- losses and lowering the horizon.tween two distant stations on less than a 10-meter wave For long-distance work the plane containing the elec-length. This limitation is due to the fact that the tric vector at the transmitter may make any angletangent ray will strike the lower boundary of the whatever with respect to the ground without appreci-ionized medium at an angle greater than the second ably affecting the ray paths or energy flux densitycritical angle and will therefore not return to the earth, in the wave front.but be refracted out into empty space. In any case the earth's magnetic field will produceThe precise value of this minimum wave length is, enough rotation of the plane of polarization to make the

of course, not 10.6 meters, as indicated in Fig. 16, angle of polarization of the received ray at the surfacebecause of the meager data upon which the figure is of the ground purely a question of chance. The bestbased. There is a similar limit for full winter night type and orientation of receiving system (loop orconditions given as 22.5 meters in Fig. 16. Near sunset antenna) will depend upon the direction and polariza-or sunrise, however, we can use a wave length less than tion of the arriving wave as well as upon the con-the above full winter night limiting value, when trans- ductivity of the ground and height of the receivingmitting from the dark into the light hemisphere, since system above the earth. Some interesting work onthe refraction is increasing in the direction of travel of determining the direction of arrival of signal waves hasthe ray and may be sufficient to bend the rav back to recently been done by Appleton and Barnett30.earth. 9 The best polarization of the transmitter can therefore

It is now interesting to see what type of antenna be considered from the point of view of ground losses,directive curve will be most effective for long distance mechanical construction and such questions as nearbycommunication from the point of view of the present interference, due to the ground wave, etc.theory. We have seen that all of the energy which It should be pointed out in closing that electron

strikes the lower boundary of the ionized medium collisions and the effect of the earth's magnetic field willstrikes the lower boundary of the ionized medium modify the shape of the skip distance curves in theabove the second critical angle is refracted out into vicinity of the upper asymptotic wave length. Here

space and is lost. The initial angle at the transmitter absorption and double refraction (". e., splitting of a raywhich corresponds to this condition is obtained from into two components having different velocities ofequations (80) and (97) aspropagation) will require consideration in a complete6 = cos-' { (1 -F- b sr,) V/ 1- y NTo, t'W2 } theorv of short wave transmission.

which may be written for convenience as The above theory is based on continuous wave theory0 = cos-' { (1 + b/6300) /1- 9 X 10--.9 No X2 and will not apply directly tovery highly damped waves.

In highly damped spark transmission we are dealing withThe summer day and winter nighteondltlons on a wide band of frequencies and therefore skip distancelong wave lengths yield the largest useful values of the effects, etc., will be considerably blurred. Hereinitial angle. For our assumed ionization condition methods similar to those used by Eckersley3' in hiswe obtain0 = 67.8 deg. for 40 meters on a summer day very interesting explanation of the familiar "trolleyand 6 = 64.5 deg. for 55 meters on a winter night. car noise" often heard by radio men, would have to beFor shorter wave lengths the critical values for the applied.initial angles will be much less as will be seen from Figs. The relatively small effect of molecular refraction,23 and 26. Here 0 = 25.9 deg. for 21 meters on a duet ensity san tempertur grdents in athowe

summedayand 11. deg for25 mterson adue to density and temperature gradients in the lowersummer day and 6 = 11.2 deg. for 25 meters on a atmosphere, discussed by Fleming32 and Larmor,"winter night. Thus we may conclude that on these has been neglected.short wave lengths all of the useful radiation is emitted It should also be kept in mind that only an approxi-between the horizontal and approximately 70 deg., and mate allowance has been made for the efFect of thethe greatest distances are reached by the low-angle curvature of the ionized medium on the ray paths.radiation. We therefore conclude that for long-distanceAwork, on short waves, maximum efficieney is obtained T ACKNOgraLEDGMENTby low-angle radiation. This also means thiat nearb th ayilmntn ugetoswihh otiobstructions which cut off the low angle radiation willthmaylumntn uesoswhche nr-

bedermna toln.itnewrig ti hrfr buted during the preparation of the paper.-- ~~~~~~~~~~~~~~~30.E. V. Appleton and NVI. A. F. Barnett, Proc. Roy. Soc. A,X29. Cases where two-wvay communication on the same wave Vol. 109. p. 621, Dec. 1925.

length will not hold, due to the effect of the earth's magnetic 31. T. L. Eckersley, A Note on Musical Atmospheric Dis-field, or hecause of an electron drift velocity, have heen discussed turhance, Phil. MUag., Vol. 49, p. 1250, June, 1925.respectively hy E. V. Appleton, Nature, p. 382, MIarch 7, 1925, 32. J. A. Fleming, Proc. Phy. Soc. Lond., Vol. 26, p. 318,T. L. Eckersley, Natture, p. 466, Sept. 26, 1925. 1913-1914.


Appendix I diEFFECT OF ELECTRON COLLISIONS WITH MOLECULES E Le d + Ra i (12)ON THE REFRACTIVE INDEX OF AN IONIZED MEDIUM

Let N = number of electrons per c. c. which is the familiar equation for a voltage impressedv = collision frequency between an electron on an inductance and resistance in series.33 Thusequa-

and molecule (per sec.) tions (10) and (11) give the effective inductance andv = average instantaneous velocity in the direc- resistance of a unit cube of space due to the presence of

tion of the electric field. the electrons and molecules.The equation of motion is Let us now imagine an infinitely long square section

of one sq. cm. area cut out of space through which ad v e E (1) plane electromagnetic wave is being propagated.dt m

where $-------e = electron chargeE = instantaneous value of electric field intensity in FIG. 1

the wavem = electron mass The length is taken in the direction of propagation andIf we multiply (1) byN e, we obtain the rate of change the sides of the square are taken parallel to the electric

of current density in the absence of collisions intensity E, and magnetic intensity H, respectively.The resulting slab or beam is pictured in Fig. 1. Here

d v d i N e2 E the electric intensitv produces a potential differenceNe t= d t = (2) between the top and bottom surfaces of the beam and

we are therefore concerned with the capacity betweenThe total number of collisions which occur per c. c. these two faces per unit length of the beam. This will

in the time interval d t will beN vd t (3) 1

be c. g. s. (e. s. u.) assuming unit dielectric con-and in the absence of an accelerating field these col- 4 wlisions would cause the current density to decrease in tatthe interval d t by the amount stant.tir dbtaNt d4 If an imaginary current i, per cm. of width, flowed

d i = -N v d t e v (4) from right to left on the top of the slab and an equalor and opposite current flowed along the bottom of the

di slab, we would obtain a magnetic field H in the directiond t =- v e v (5) shown which would be equal to 4 r i, if i is in c. g. s.

(e. m. u.). The flux from a cm.-length of slab would beThe net rate of change of current considering the 4 7r i lines and therefore the inductance per unit length

effect of collisions will then be the sum of (2) and (5), of slab will be 4 7r in c. g. s. (e. m. u.) since the induc-that is tance is defined as the flux linkage per unit current.

d i N e2 E The inductance per unit length in c. g. s. (e. s. u.) willdt=- -N vev (6)d t m

But the total charge movement or current density is 4always

i = Nev (7) FIG. 2Substituting (7) in the second term on the right handside of (6) we obtain

di N e2 E then be -,where c is the velocity of light in cm.

d t m (8)per second.

Transposing and rewriting (8) we obtain We have already seen that the effect of the electron

m d i mvcurrent is to shunt each cu. cm. of space by a resistance.E=N e9 d t + N e2 ()and inductance in series of the values given in equations

(11) and (10). Hence we may represent the equivalentIf we put circuit of our wave beam in an ionized medium as a

Le = m/~Ne2 ....... C. G. S. (E. S. U.) (10) transmission line of the form shown in Fig. 2. TheRe = m v/N e2 .. . C. G. S. (E. S. U.) (11) 33. The above simple deduction proposed by E. XV. KVellogg

in (9) we have replaces a rather intricate proof by the authors.

328 BAKER AND RICE: REFRACTION OF SHORT RADIO WAVES Transactions A. I. E. E

constants per unit length in c. g. s. (e. s. u.) are as a2 +1\2 /=XV(b-b,)2+g2 (28)follows: Adding (26) and (28) we obtain

Series Inductance L0 = 4 7r/C2 (13) a x°t =':2{ /(-)2 + g2 + (-c

Shunt Capacity C0 = 1/4 r (14)(29)

Shunt Inductance Le 2 N (15) - (bc + g (bcb)

(30)____ Subtracting we haveShunt Resistance Re e2N)

The propagation constant of such a line iS34 24i { V(b-b2)2 + g2- (b-b,)(31)

eriesImpedanceShunt Impedance = { V(bc-b)2 +Vg2+ (be-b)

= V/ Series Impedance X Shunt Admittance (18) (32)For our line we have a is called the attenuation constant per unit distance and

is the reciprocal of the distance in which the amplitudeP = a113=i.T 1 ~ of the wave will decrease to 1/e of its value. A3is called

P +=JO( R+ L +J Cj) CO) the wave length constant and is the phase change,measured in radians, in the wave per unit distance of

(19) travel. In other words, a wave length on the trans-

We shall substitute the following expressions for mission line is 2 r units long. If the frequency of theconvenience

x = Lo co (20) wave is f, the phase velocity or apparent velocity ofpropagation c' will be

Re9 = R 2 + (h c4) = Conductance (21) c' = f = 2wf - (33)

Le c Hence the refractive index of optical theory for theb = Re2 + (e )2 = Susceptance (22) ionized medium through which the light beam was

drawn isb& = Co @ Suseeptance (23)

c _c13(4Then 8 c' =- (34)

P = a +j,B = V/x (g-j b + j bc) (24) From equations (20), (21), (22), (23), and (13), (14),Squaring (24) we have

(16)equavea2 + 2 a( 13-32 =j x g + x b-x bc (25)

Equating the real and imaginary parts we have 4 c,) gX = 2 C. g. S. (e. s. u.) (5

a 2 -2 = x(b-bc) c2 ~~~~(26)12ao =jxg

(6

jze2N vEquating the absolute value of these vectors we have g = m (V2 + w2) (36)V(a2-_132)2+ (2 a1A)2 =Vx2(b- bG)2 + X2g2 (27)

from which b = eN.V C,+OJ2 (37)34. See for example J. A. Fleming, The Propagation of Electric

Currents in Tel. and Tel. Conductors, D. Van Nostrand Co., orK. S. Johnson, Transmission Circuits forTelephonic Communica- b 4 c (38)tion, D. Van Nostrand Co.


Substituting these values in (30) and (32) weobtain (t =c>g2 4 /\ L 1- 4e7r2N 2 47r e2N V 2 4 7re2Nobtain a = +vi___ 4w2N2)F1- 4

CV2 M (V2 +co2) L mwco(PI +w2) JL m(p2+ )i(39)

X,/ _ 4 ir e2N _2 4 e2 2 4 7r e2N ]= cV2 L,/ m(p2 + W) JL Mm (p2 +c I2 p +W(0

From equation (34) we obtain for the refractive

index,u = m([1_)7r ] V[i_ 4 7r e2 ]2+ 2 [ ]2 (41)

If we assume that v is small compared with w, we may v2G / 1 \ vG vG2neglect the second term under the inner radical sign a = 2 2 = 2 c + 4 c (52)of equation (40) and obtain the approximation

If now G is small compared with unity we can neglectco= 1 ( 7r+e2 (42) its square and write the simpler formula

and by equation (34) a= 2c m (2 + C2) (53)

_ 4w e2 N (4)Appendix IIGENERAL EQUATION FOR THE PATH OF A RAY IN A

For the same conditions an approximation for a may MEDIUM OF VARYING REFRACTIVE INDEX INbe obtained as follows, POLAR COORDINATESLet Transforming equation (27) into the polar coordi-

G = 4 r e2 N/rm (V2 + Co2) (44) nates r and 6 we write,Then 1 1 dA r (1)

a=~~4V(1~~~G )2+(PG)2 ( ~ 4)R A dr Vdr+ 2

c -\/2 (1 G (1 -o G) 45 r2 + (d 0)in which the sign has been reversed because of the

w / f {v G 82 | difference in the respective conventions concerning= 2+4\'(1-G)2 { 1+ )

2 -(1-G) (46) the sign of the curvature when using rectangular and

c+\/2 (1- G) I polar coordinates. For rectangular coordinates apositive curvature is such that the convexity is towards

co I s1 ( X )the X-axis while in polar coordinates a positive curva-=___2_i_(1-G) \W1 (1- G) ture has its concavity towards the origin. The general

V2(1G)expression for the radius of curvature in polar coordi-(47) nates is given in text books of the calculus35, as

(7G2 d 2 r

cV,2 \I(1 G) 'L1++ ')t;(1-G)(48) rR+ ( d r d 2vG ) 1+ (1-G)2 J R{2 ( )2 3 (2)

(x 2 c \/1-G (49) Equating expressions (1) and (2) we have

Using equation (43) we may write it 2 dr 2 r d2 r2eN dO) d 62 r d___c~tm(P2± W2) (0r2 +(dr g dr(3

The value given by Nichols and Schelleng,13 page 229,for the absorption coefficient yields the following value 1 drof a, in our notation, Multiplying both sides by- - d 6 we obtain7 y I~~~~~~~~~~~rdO0

ct=m (p2 + z5)25) (r dOr +72 ( d6)3 dd 62 d 6 )d6_0 d ,v_When ,u is nearly equal to unity both expressions will ( d1r\2=_give substantially the same result. r2 + , d60 J(4A further approximation may be obtained from equa -____

tion (49) by expanding the denominator in a binomial 35. See for example Granville: Differential and Integralseries and taking the first term only. This gives Calculus, Ginn and Co., p. 162.


For purposes of integration we split up the left-hand so that the equation for the path in polar coordinatesside as follows: becomes

ri cos ,B d r2 d r d r d2r d6= r cos (14)

2L dO +2 ~ d2 Jd6 r Vr\/ 2- r1csg(4dr )2 and

02 = J rVr2it2-r12cos2 (15)

2 d r 3 d2 r d rri

LT2I ( dO0 ) -2 dd2 dd JS d0 DF N T Appendix III+ / d 2- DIFFERENTIATION OF THE RANGE EQUATION WITH

r ± d O RESPECT TO 0 FOR THE CASE OF A CURVED EARTH(5) The equation for the range of a ray for the case of a

On dividing the numerator and denominator of the curved earth is given in equation (113) aslast term on the left-hand side by r9 we have 4 c (B-b)-[i2r dr} + 2dr -dS2 1_ x (0) = 2 ro(3-6) + V 3NcosfK (sin a) (1)

dO ~~~~~ Jd6~~~in which

d )2 Cose3= cos =A cos 0 (2)

2_ dr 2 d2 r dr dO

+ +( d d) sina=a N sin ,3=C sin (3)r2 dO8(6) Differentiating (2) we have

We may now integrate and obtain -sin d f =-A sin 0 d 02 so that1 ~~~dr1 dr2-1g 2 lg'0 +- C-Clog IA2 d 2(r2 d r-lgI d =- A sin (4)

(7) dO0 sin~i 4which reduces to Differentiating (3) we obtain

1 F ~~1 d r \1d a Cosflogr- ( d l ) J = C-log I (8) d =C cosa (5)or Multiplying (4) by (5) we have

1 r- 2~~ da df da _ C sinO (6-~log -_C_-_ g_IA_2 1+ I d =C-2log (9) d O' = dO' dd° A C osatan (6)The differentiation ofK (sin a) is given by WhittakerThe denominator on the left-hand side of this ex- and Watson, Modern Analysis, 3rd edition p. 521 in the

pression iS the square of the cosecant of the angle formwhich the ray path makes with the radius vector. Atthe lower boundary of the ionized medium, i. e., d K (k) 1 E (k) K (k) (7)r = r1, we have IA = 1 and the angle between the ray d k k 1 k2

and the radius vector is - in which k sin a2 from which

Hence d K (sin a) cos a E (sin a)1 { sna

C = 2 log (r 2cos2I,) (10) d. sina 1-sin2a K= log (r1 cos 3) (11) cot a E (sin a) _ K (sin a) (8)

Substituting this value in (9) we obtain COS2 a

r2 - r12 cos2 g In the above the new expression E (sin a) denotes the1± 1 ( dr\82 - (12) complete elliptic integral of the second kind and its1__r2 kdoJ values will be found in Tables36.

InvertinandtrnsposingwehaveFor convenience we shall simplify equation (1) byInerlg n taspsngw2hv writing(iL'd [.622Jr2c=2f (13) 36i. For example, Jalinke andEmde, Funktionentafeln, p. 68,\dO/r12 cos ~~~~~~~~~ B. 0). Pierce, Short Table of Integrals, p. 118.


D = 2 rO (9) as the final expression for the rate of change of theand range with respect to the initial angle 6.

4 w (B -b)-F =-7r (B Nb (10) Appendix IV

On differentiating (1) we then obtain FOCUSING EFFECTS AT THE SKIP DISTANCE AND INSIDEOF THE TANGENT RAY

d x -D ( do - + F COS

d K (sin a) d a It will be observed from Fig. 25 that there is a focus-d- dO .ing of the rays in the vicinity of the skip distance, and

d A also where the ray which starts out tangent to the- F sin3 K (sinae) d(11) ground returns to the earth's surface. The reason for

the focusing effect at the skip distance may be readilyOn substitutingathevvaluesefound in (4), (8) and (6) seen from the following consideration. As we increase

the angle 6 from zero to larger values, the range of thed x (0)

- D A sin ° 1 + F cos cot a X rays decreases rapidly at first and then more slowlyd 0 sin until the minimum range (i. e., skip distance) is reached

E(sin o) . {qt sinGe X

for a certain value of O. For higher values of 0, thecosin2a) K (sin a) A C sin range increases rapidly until the second critical angle iscos'aJcos a tan reached. Near the skip distance the rate of change ofsin 6 the range with respect to 0 is very small and conse-

- F sin ,B K (sin a) A sin quently the energy emitted between the angles 6 and

(12) (6 + d 6) must come down in a small zone, whichWe may simplify to

d ) ( AsiA _ 1 ) cos2 A sin 0dxOD Asn6 1+FACdO s.sin /3 snasin/A

{ cos2a >-K (sin a) FFA sin 6K (sin a) (13) XNi

Substituting the value of C given by equation (3) wereadily obtain /35d x () ( A sin -i

[ D, in/3 - K (FAsin6 /e

DO(siin01)+ 0Fa[cotn (E(sina)1cos2a-K (sin a) -K(sin a)j

(14)Asn 1)lsinin6 FIG.1

means a high energy flux density. This effect is shown[ E (sin a) _ K (sin a) inFig. 26 between the angles 0 = 70 and 0 = 9°.L tan2 / cos2 a sin2 g The focusing effect where the tangent ray returns to

(15) earth will be made apparent from the following geo-Putting back the values of A, D, and F we obtain metrical explanation. The tops of the ray paths are

d x (6) =2 r0'tan 0i±4(B-b) c curved while below the ionized medium the paths aredO = 2 ro tan /3 + V/a NO cos / tan 0 straight. If we produce the straight portions of any

path they will meet above the summit of the path, andE (sin a) _ K (sin a) I the result is as if sharp reflection had occurred at this

tan2 / COS2 a sin2 / point instead of a gradual bending at lower heights.(16) Referring to Fig. 1A, let 0 A B D be the actual path of

a ray starting out tangent to the earth. Produce 0 APutting in the value of N from (3) we have and D B to meet in H. Then H is the point at which

+/° ~~~~sharp reflection may be imagined to occur. Thed x (6) - 'tan 6 -1 ajdacent ray, 0 E F G, has the very small initial angledO kr tan:/3 ) d6Owith its sharp reflection point atK.

The arc length, A B, may be shown to be equal to the+4(B-b)sin atan 6 F E (sin a) K (sin a) J arc length, E F, as follows. The difference betweeniv tan /3 Ltan2 /3 cos2 ae sin2, /3 A nd E F measured along the circle E A FB is

- ~~~~~(17)equal to r1 d 6jr0, times the second term on the right


hand side of equation (116), and as the appropriate subtended by a given chord at any point on the circle,value of 0 is zero, the difference vanishes. and on the same side of the chord, is constant. TheHence converse is obviously true. Hence it follows that the

are A B = arc E F (1) five points K, H, D', C and 0 lie on one circle. Again,and since the angles K D C and HO C are right angles, the

L A C B = Z E C F (2) points H, 0, C and D lie on one circle. It followsWe also have that K, H, 0, C, D and D' lie on one circle. The straight

1 line H D meets this circle in H, D and D' which canZ OCH = Z OCA + Z A CB (3) onlybe true if D' is the same point as D.

Thus D is a focus for rays leaving 0 at sufficientlyand 1 small initial angles.

Z O C K = Z OC E + -2 Z E C F (4) The point D may be regarded as the mirror image of0 in a nearly spherical mirror with center at C and going

Taking the difference we have through H. This image is, of course, real; that is, theZ K C H = Z 0 C A - Z 0 C E (5) rays would go through the point D if not intercepted

But the angle 0 C A is equal to the entrance angle, ,B, as by the earth. On the other hand, the mirror image O'may be readily seen by reference to Fig. 9 or 10. in Fig. 1B is virtual. In this case the rays would di-We may, therefore, write (5) as verge by the angle d 0, appearing to come from the

L K C H = -Z 0 C E (6) point O'. The focussing effect is due, therefore, to theThe angle 0 C E is equal to curvature of the imaginary reflecting surface H K.

W-t d 0) (7)where f' is the entrance angle for the ray 0 E F G Discussionand is, therefore, given by A. E. Kennelly: The paper is very timely and interesting

d o because so much attention has been drawn in recent months,A =-dO (8) not to say years, to the marvelous properties of very short waves.d 0 The paper makes a definite and very reasonable attempt to

where *is the value for 6 = 0. From equation (4) of explain some of those properties. The direct wave dies out ata relatively short distance from the sending station and then

Appendix III we find that for 6 = 0, d /3, d 0 = 0, so that nothing more is heard of it or received from it until it has' 1=,B(9) traveled a relatively great distance. That phenomenon repeats

Substituting (9) in (7) we have itself at least once.Z 0 C E = (j- d 0) (10) Here we are between certain rival theories of refraction and

(10) in (6) we have reflection, from effects produced in the upper atmosphere at aPutting

d 0distance at which we can only guess. It is very remarkable

/ K C H = d 6 (11) thatwe know so little. We are necessarily ill-informed concerningBy construction the reflecting surface at H is perpen- the conditions that exist in the atmosphere at a distance of, let

dicular to H C and that at K is perpendicular to K C. us say, fifty kilometers above our heads, while fifty kilometersFrom equation (11) we see that the mirror at K is along the ground we can cover in an automobile in an hour or less.turned through an-angle d 6 with respect to H. If the The wonderful thing is that we are able already to form

mirror aK m ep lopinions, such as are expressed in the paper, as to what does takemirror at K remained parallel to that at H as shown in place in that region above our heads which is so near and yet soFig. 1B, a ray 0 K inclined at an angle d 0 to 0 H would far. The promise is very brilliant that we shall be able to learncontinue to diverge from H D along K G as illustrated. from radio observations, in the not far-distant future, the elee-On the other hand, if the mirror is turned through an trical properties of the atmosphere at distances from ten to 100angle d 6 the reflected ray will be turned through twice or more kilometers above the earth, and it will be very surprising

the anled ndthreorcnvrtthdvegeced0if the information thereby gained does not have a marked in-the angle d 6, and therefore convert the divergence d 6 fluence upon weather forecasts at least in the region up to twentyinto a convergence d 0. Hence the ray K G will inter- or thirty kilometers above the surface.sect the ray H D and therefore give a focusing effect. Personally, I think it is early yet to form hard and fastThe problem may be further analyzed as follows: opinions or to make very definite conclusions as to just whatSuppose thatblem mayebetsHDfurther aalyzetas fthese phenomena are. We know that there must be refraction.

Suppose that K G meets H D at D'. Let H 0 meet We know that there must be rotary polarization and we alsoK C in L and K G in M. Then in the triangles 0 K L know that there must be reflection. Probably all three of theseand C H L we have the angles at 0 and C equal, also things occur simultaneously. I think we must therefore retainthe angles at L equal, and consequently an open mind for the present-until more information can beOKnLe CHL (12) securedobyeobservation.

W. B. Kouwenhoven: The author's conclusions are similarand to those of Dr. J. Zenncek who published a paper on this subject

/ 0 K D' = / 0 H D' (13) in "Elektrotechnik und Maschinenbau", Vol. 43, p. 593 and 612,In the triangles 0 K M and D' H M we have the angles 1925. Dr. Zenneck concludes from observations that radioat K and H equal and the angles at M equal, whence it waves enter the upper atmosphere and that these waves mayfo1llw that come back to earth at some distant point because of refraction.

Z K D'H-Z H O K 14 ~~Radio transmission takes place by means of ground waves and/ KD' =/H0- (14) by means of waves thaJt pass through the upper atmosphere.It is a well-known property of circles that the angle In the case of short waaves the ground radiation is rapidly ab-


sorbed as pointed out by the authors, while the absorption is earth's magnetic field on the electron motion will require con-much less for long-wave transmission. sideration in a complete theory of the propagation of radio wavesThe electric field produced by a radio wave in the upper atmos- as Dr. Kennelly has pointed out. We have evaded that phase of

sphere sets the ionized particles in motion. In the case of long the question as well as we could by dealing only with the case ofwaves the mean free path of the ionized particle is short com- short waves where the earth's magnetic field cannot have muchpared to the wave length, and collisions occur. The energy is effect. There is a certain resonance frequency of the electronstherefore absorbed and very little if any of the wave is refracted produced by the earth's magnetic field which corresponds to aback to earth. In the case of short waves the absorption is wave length of about 214 meters. If we are working well abovemuch less in the upper atmosphere, and the wave is refracted that frequency the effect of the earth's magnetic field on theand reaches the earth again at some point distant from the motion of the electrons can be neglected without serious error.transmitter. We wish to thank Dr. Kouwenhoven for calling our attentionM. I. Pupin: Mr. Einstein would probably say that you are to Dr. Zenneck's paper, and we are interested to hear that his

wrong when you say that the wave can be propagated by a ve- conclusions concerning the propagation of short radio waves arelocity greater than the velocity of light wave. What would you similar to ours.answer if he did object? Recently there has been a good deal of surmising, on the partW. G. Baker: I think it is one of the cases of distinction be- of radio men, as to the height of the ionized medium. Obv.iouslywe cannot speak of the height of the actual mnedium since we do

tween the velocity of a group of waves and the phase velocity. notkw ere o say itbi whe the maxium is,o werenot know where to say it begins, wrhere the maximum iS, or whereWith reference to the question of reflection versus refraction it ends. The effective height of the layer as judged by a sharp

raised by Dr. Kennelly, I believe that most of the people who reflection theory will vary, of course, with the wave length,talk about reflection from the Kennelly-Heaviside layer use it distance between transmitter and receiver, etc. On shortas an approximation; they do not wish to bother with the com- waves we require a relatively large amount of ionization to bendplexity introduced by refraetion. Actually the transition from the ray back to earth. Therefore the ray will penetrate deepthe neutral to the ionized medium must be gradual, and will into the ionized medium before it returns to earth. Here thetherefore not be sufficiently abrupt to produce appreciable apparent height of the layer as judged on the reflection theoryreflection except on the very long wave lengths. Unless the may be very great. On longer waves a smaller ionization densityionization changes by a large amount in a distance comparable is sufficient to bend the same ray back to earth so that the depthwith the wave length of the radio wave no appreciable reflection of penetration will be less and the apparent height of the mediumcan occur, but electron refraction may easily bend the wave will be lower than that estimated from the short-wave experi-back to earth. ments. A further difference is brought in by the effect of the

Rotary polarization and other effects due to the action of the earth's magnetic field.

refraction of short radio waves in the upper atmosphere

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