breakthroughinenergy.combreakthroughinenergy.com/sitefiles/media/PDF Documents... · Table of Contents Paper A Numerical evidence for repulsion-force positioning of landmasses on

- --- - - ·-- --- --~ ------~-- -==--==

Underground Geophysics for Geoelectric Energy

Michael Csuzdi

Core Publishing December, 1994 Kingston, Ont. CANADA

Table of Contents

Paper A Numerical evidence for repulsion-force positioning of landmasses on the Earth, Mars, and Moon

Appendix Al

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii Acknowledgement . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi Distorted Image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Indoctrination to One View . . . . . . . . . . . . . . . . . . . . . . 3 Applying Another View . . . . . . . . . . . . . . . . . . . . . . . . . 4 "Blueprint" Projection . . . . . . . . . . . . . . . . . . . . . . . . . 10 Area Centers ................................ 13 Zigzagging of Continents . . . . . . . . . . . . . . . . . . . . . . . 15 Pentagonal Pyramid ........................... 19 Electrical Force .............................. 23 Physical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Mathematical Model . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Characteristic Patterns . . . . . . . . . . . . . . . . . . . . . . . . . 42 Continents of Mars . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Continents of Moon . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

User's Guide to Electroglobe

AppendixA2

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Installation . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . 67 Operations ................................. 68 Pattern examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Numerical results of Mars and the Moon . . . . . . . . . . . . 80 Program Listing of Electroglobe . . . . . . . . . . . . . . . . . . 82

Calculating Centers of Areas of continents on a Spherical Surface

Africa ................................... 103 Input of coordinate points . . . . . . . . . . . . . . . . . . . . . . 103 Designate subcontinents . . . . . . . . . . . . . . . . . . . . . . . 105 Perform the calculations . . . . . . . . . . . . . . . . . . . . . . . 106 Eurasia ................................... 112 North America . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 South America . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Antarctica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 Australia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Mars-A ................................... 118 Mars-B ................................... 119 Mars-C .................................. 120

ii Underground Geophysics

Moon-A 121 Moon-B ........................ , . . . . . . . . . . 122 Moon-C ..................... , ............ 123 Summary of Calculations . . . . . . . . . . . . . . . . . . . . . . 124

AppendixA3 Derivation of Equation A-4 of PaJJer A

AppendixA4 Measuring Charge Densities with a Pithball Electroscope

.........•..............................

In preparation:

Paper B . . . . . Earth Currents arul Auroras Paper C . . . . . Generation arul Reversals of the Geomagnetic Field Paper D . . . . . Sudden Oceanic Storms Paper E . . . . . Cyclones arul Anticyclones Paper F . . . . . The Geopower Plant Paper G ..... Electricity, Ether, Gravity, arul Inertia Paper H . . . . . Explosion of the Space Shuttle Challenger Paper J . . . . . Interior Structure of the ·Earth arul the Sun

125

129

Introduction

Geophysics is a jigsaw puzzle. It consists of a myriad of observations, and a theory to put them together into a meaningful picture. In fact, there were several such theories in the past, sometimes many of them simultaneously active and vigorously debated. But now we have only one theory that is widely accepted by the entire scientific community. There is no more vigorous debate. There is only agreement, and mutual support among scientists. Have they found the final truth? Or are they simply in a cul-de-sac? I'm afraid, they are. They all have failed to observe a great message of the Earth about its internal workings. And this has serious implications for their current theory.

I am a Professional Electrical Engineer. In 1974, while studying an Earth globe, I noticed a certain regularity in the positions of the continents. Please take your own globe, and view it from above Africa. Take a marker pen, and mark the area centers of all six continents. Then connect Africa's center to those of the other continents. You will see a pentagonal star.

soutb ""'e,ica

I wondered if this was known to geophysicists. It was not. Thus, I decided to investigate it myself. My ideas started taking shape in about a year, and in 1976 I gave a presentation at the National Research Council, in Ottawa.

As time went by, my theory kept expanding. At one point I realized that a major new energy source may become available from the Earth, and in 1980, and again in 1984, I published a book, "Breakthrough in Energy". I gave a paper at a meeting of the American Geophysical Union under the title of "Electrical Plate Tectonics", EOS Transactions, Vol. 62, No. 45, November 10, 1981, p. 1035, and another one, "Continents of Mars", EOS Transactions, Vol. 66, No. 18, April 30, 1985, p. 378. However, my papers were not well received ( ... you are not a geophysicist to talk to us about continents). A private journal in England published my paper

iii

iv Underground Geophysics

"Electrostatic forces in the earth's crust", Open Earth, No. 18, 1982-C, pp. 38-42. However, it generated no response. Over the years, I submitted several papers for publication in recognized geophysical journals that were all rejected by the editors. Geophysicists have decided not to pay attention.

Apparently, I have opened a door on a wide field. As the Table of Contents shows, almost all major areas of geophysics are involved. Most importantly, I have found the equivalent signs to the Earth's pentagonal star on Mars and on the Moon. Consequently, the sign is not accidental. Furthermore, there is also a close connection to the physics of electricity. The Earth shows us how to generate energy. The Earth operates its entire weather system from an internal energy source; it is a total misconception that this energy is solar. The energy is coming from the thermionic electric charge emission of the Earth's internal magma surface. However, this huge energy is converted into several different forms during its transition through the crust, the oceans, and the atmosphere. Thus, cursory observers may fail to see the link between them. The energy takes the shape of the geomagnetic field, earthquakes, cyclones, hurricanes, tomados, lightning, the rain and wind, and finally the polar light, before it joins the solar wind. The pentagonal arrangement of the continents is also a sign of that energy as it positions the continents according to a fu.ndamental law of electricity.

One possible reason for geophysicists' inattention is that the necessary knowledge of electricity is not readily available to them. True enough, that part of physics, although elementary, is also esoteric and obscure outside of a very small group of people. Thermionic charge emission is studied and applied only by those who design and develop the cathode of electron tubes. Not many people do this today since electron tubes have been replaced by transistors, with the exception of TV picture tubes. I have met many, otherwise properly trained engineers and physicists, especially the younger generation, who barely knew the concept, let alone its working mechanism. Why would geophysicists know it?

Physicists of electricity would be the best candidates to observe the Earth's charge emission. However, there is the turf problem, or specialization; physics and geophysics are two totally separate departments, they are even located in separate buildings at universities. I have candidly interviewed many physicists about their information on the structure of the continents and the Earth's interior. They matched those of geophysicists about charge emission.

The price of this gap is the expensive delay of accessing a totally harmless, inexpensive, and perfectly renewable large scale energy source. A geoelectric power plant would run entirely on the Earth's electric energy and, would produce no dangerous radiation or any other kind of pollution. It could be constructed anywhere on the Earth, and there would be practically no limit to the number of such power plants that could be

Introduction V

constructed. Once you know how the Earth's own power plant operates, you should be able to model it in a smaller, but optimized and controlled environment. This would be a tall, wide, and empty concrete tower where a continuous water cycle would operate the same way it does in Nature. An amount of water would be continuously lifted by the atmospheric electric field, then it would fall back to the ground in pipes, and a conventional hydroelectric generator would convert the water's kinetic energy into electricity for consumers. Then the water would be recycled back into the tower for endless circulation. The amount of energy available from geoelectric power plants would be comparable to that of nuclear generators.

The amount of energy available from a geoelectric generator rapidly increases with its height. However, it works with any small height, although its energy yield is small. Unbelievable as it may sound, we have such equipment already in operation by the millions; all chimneys operate on the same principle, contrary to current explanations. We can also observe hundreds of very tall natural chimneys operating every day. In a tropical cumulonimbus cloud a million tons of water rises to an altitude of 20 kilometers in 30 minutes, its upward speed sometimes reaches 100 km/hour. Ask any scientist about this mechanism, he will not know how it operates. The best current explanation is that "a certain atmospheric imbalance sets in when dry air and moisture are mixing". We could now model this operation, and use the obtained energy for the benefit of mankind.

The word underground in the book's title has a dual meaning by design. The main objective of my work is to propose the constructi9n of geoelectric power plants. The source of this energy is in the Earth's interior, that is, underground. The discipline which should deal with this subject is 'underground geophysics'. The other meaning comes from the fact that geophysicists have consistently refused the discussion of my fundamental observation about the continents, and my proposed explanation of it. Thus, they have forced me underground, so to speak, in the political sense. Consequently, the theory described in this book is outside of mainstream thinking, it is decidedly revolutionary. Its (future) adherents will practice 'underground geophysics' until the old system is overturned.

Although geophysicists' inattention had distressed me for a while, eventually I realized that I have remained totally free in my research. Not many scientists can say that. Tough and anonymous peer reviews of all published material, amounting to total censorship, assures that. For the price of time (twenty years so far), nobody has interfered with my thinking, writings, and mathematics. Nobody has denied me research grants. And nobody has fired me for heretic ideas. I only had to keep working hard as an engineer to earn a good salary to finance my

vi Underground Geophysics

geophysical research. Then I did that research, sometimes intermittently, whenever I found time for it. My theory's rejection has been, in fact, a blessing in disguise. But that rejection will come to an enforced end; when an energy-concerned organization decides to build the first Geopower Plant, geophysicists will take their globe into their hand, and will look at it from above Africa.

I am addressing this book first to people who are interested and concerned about large scale energy generation, its cost, and its pollution. They could bring geophysicists and electro-physicists together under a well-financed feasibility project that would include the construction of an experimental energy generator.

I also address this book to undergraduate physics and geophysics students. To those who are willing to consider an alternative and new interpretation of the Law of Nature that has seldom happened since the Ether Controversy in physics in the 1880's, and the Granite Controversy in geophysics in the 1920's.

Acknowledgment

I am indebted for help, support and contribution to this study only to one person: David C. Gilmore, M.A., Professional Engineer in Mechanics. He worked out the three-dimensional geometry necessary for ca\culating the movement of objects on the spherical surface for forces acting in 30 space. He also prepared the necessary computer codes for the mechanics in a user-friendly and interactive way. Without this vital tool my theory would have remained only a qualitative discussion and a vague proposal. I probably would have never arrived at the general conclusions about characteristic patterns of continental masses. He also gave me invaluable help on the effective use of computers. This took place at the time when computers all too often displayed messages like 'not enough memory', 'disk full', or the required calculations tended to take forever.

I met Dave shortly after the first presentation of my theory in 1976. My paper was not well received at that meeting. None of the six geophysicists in my audience could deny the pentagonal arrangement of the continents around Africa ( and they all admitted that they had never before noticed this pattern, or heard about it). But they unanimously declared that the pattern was coincidental, like constellations in the sky, and it was without any merit to investigate it further. Moreover, they also declared that my proposed electrical positioning was "just impossible".

That verdict came as a total and unmitigated surprise to me. When I first noticed the pentagonal pattern myself, I had two possibilities in my mind. One was coincidence of course, and the other that it may be the

Introduction vii

result of some yet undiscovered forces. What would be its scientific value if it were the latter? Perhaps enormous. Consequently, an off-hand dismissal was out of the question for me, and I felt that an investigation was morally compulsory. I knew very well that my knowledge in geophysics was little, but I thought I could make up for the shortfall by reading all the necessary material, even though that might take some time. Then, I decided to keep my discovery a secret until I found it out. I thought that if a geophysicist learned about it, he would run away with it, would solve it overnight, and publish it at once. Then I would be left behind, and forgotten.

However, I had bad conscience about this secrecy, for causing a delay in making the discovery public. Because of my vanity, scientists would be prevented from doing something significant with it. I knew the story of the French scientist who discovered insulin years before Banting and Best, but he kept it secret. He deposited its sealed documentation with the French Academy of Science. Then, whert the Canadian discovery became known, he asked that the document be opened to prove his own priority. In response, the Academy publicly disgraced and condemned him for committing a crime against humanity for the delayed availability of insulin. The only difference was, perhaps, that I was willing to make my discovery public, as soon as I could also propose an explanation, formally at a recognized institute. That is exactly what I did at the presentation of my theory. Curiously enough, while I failed to make the discovery public because of the resistance of those scientists, I did achieve an implied official acknowledgment of my priority in their letter of rejection of my proposal.

I soon recovered from my bitter disappointment because I realized that the scientific jury did not offer any scientific evidence that my proposal was wrong. Evidently, they did not have any. I worked out a plan in my mind to expand my evidence to a higher level. I decided to make a computer model in which electrically charged objects would be allowed to move on a spherical surface under Coulomb's law of electrical repulsion forces. If six mathematical continents would move to the same relative positions on the computer screen as the real continents do on the Earth's surface, then this would be irrefutable evidence.

However, the task appeared to me too difficult, beyond my mathematical and computer skills. Furthermore, I was also at the end of my money. I had quit my job a year before, to work exclusively on the theory in preparation for the presentation, (and I expected a job offer or a research grant after it that never came). Thus, I joined another engineering company. On my very first day I was sitting, a little intimidated, in the middle of a huge and busy engineering office, among some 50 new faces, and a new post of duty on my hand. That was the time when a tall, thin,

viii Underground Geophysics

well-dressed, and quiet gentleman walked slowly towards my desk, and greeted me as a newcomer. That was Dave Gilmore.

Later we talked more and more often, and then I asked him how he would solve certain geometric problems. To my surprise, he always had an elegant, short, and exact answer. Sometimes I showed him pages of my computer programs, and he would say "why don't you do it this way?", and he did it in just a few lines. Eventually we spent hours at a time discussing these problems. I always found these discussions spectacular in a certain way; how two introvert people would exchange ideas. Always in total patience and attention to each other, and never yielding without being totally convinced on purely technical merits. And never trying to force an issue on any other grounds. No urgency, or convenience, or amount of effort would count; only a perfectly satisfactory technical solution. Then, after years of struggle, when perfect results were obtained, Dave would say "Yes, Mike. Any other problem?"

In time, we became friends. We discussed also many other subjects, personal aspirations, art, philosophy, religions. I never saw the end of Dave's patience and understanding. Sometimes, when the subject tended to go out of control, Dave would say "this is hot vine for me", but nevertheless continued the discussion and offered his honest opinion on the subject in a highly civilized manner.

But most importantly, he had eternal patience with the frequently ambiguous, obscure, and sometimes totally wrong results, and with my ever changing plans on how to tackle the problems, and how to formulate the question to which the computer was expected to answer. His endurance with this task is without parallel in my life experience. He unfailingly supported me for more than 14 years without a shade of compensation for his untold hours of hard work. Or better to say, until I obtained the computer's answer to my question; yes, the mathematical continents move to the very same relative positions where the real continents are. And not only on the Earth, but also on Mars, and on the Moon, too. Thus the force is real.

Then our ways parted. He accepted engineering consulting positions in the Far East where he is also doing exquisitely beautiful watercolors on ancient and natural objects. A few years later I retired from engineering to work full time on the finishing touches of my theory. I always feel extremely privileged for having such a friend and a godsend at the same time.

Paper A

Numerical evidence for repulsion-force positioning of landmasses on the Earth, Mars, and Moon

Distorted Image

The Earth globe, as a sphere, is the simplest body in Nature. Only one dimension, its diameter, is needed to determine its size. Still, it is a sophisticated object. Its surface has no boundary, you can move on it without ever running out, bumping into an edge, or getting into a comer. Still it is finite, you are never too far from home after any length of traveling. The sphere is a favorite product of Mother Nature, she has created all stars and planets in this shape.

But mankind has certain difficulties with the globe ever since.· the ancient Greeks first attempted to measure the circumference of the Earth. One problem is that you cannot see its entire surface simultaneously.

90

N

60

30

0

30

s 90

180W 120 60

u

a 0 60 120 E 180

Figure A-1. Mercator projection of the entire surface of the Earth into a single map.

Whenever you are studying one area and you want to relate it to another, you must rely on your memory. Even if you make a transparent globe, it does not help. You will see greatly distorted, projected views on the horizons, and reversed shapes on the far side. In addition, the entire view will be cluttered.

1

2 Underground Geophysics

Another difficulty is that a globe is not suitable for storing information in the co11venient way as books and papers do. Its bulk just takes up valuable space but it carries no information about the globe's surface. If you want more than just minimum surface details, then very large globes are needed.

Mankind prefers information written and drawn on thin sheets of paper. Thus, the map was invented. The map solves the information storage and the surface details problems, but only for a price: distortion. It is caused by the projection of the three-dimensional spherical surface on the two-

Figure A-2. Homolographic or Mollweide's Equal Area Projection within an Ellipse, as if all continents occupied one side of an elliptical globe.

dimensional paper sheet. A piece of information that represented the third dimension, is lost. This information is the correct distance between surface points. As the distance gets distorted, and so does the square of distance: the area.

Gerardus Mercator, in 1569, invented a projection to simultaneously view all areas of the Earth. He changed the length of the longitudes at all latitudes to be equal to that of the Equator. Thus he converted the spherical surface to a cylindrical one, and rolled out the cylinder into a rectangle. The Mercator projection is still in use today. Its merit is that its distortion is controlled in a strict and simple mathematical way, by dividing distances by a factor depending on the latitude. This is fine as long as you are a skilled scientist or navigator who knows what to do. To ordinary people the map shows continents completely out of proportions. On a Mercator projection Greenland is greater than South America, and Antarctica is as large as all continents and oceans at the Equator, with South America, Africa, Oceania, the Pacific, the Indian Ocean, and the Atlantic, all added together (Figure A-1).

The Mercator map has its hazards for scientists, too. When you look at a nice person always in a distorting mirror, eventually you will remember an ugly person. When you do quick unconscious comparisons between people, you do not multiply that person's nose length by a numerical factor

Paper A 3

to establish its proper size. Your mental image becomes permanently altered. A repeated view of a distorted image is a form of brainwashing.

Mercator maps are still used today in geographic publications, in newspapers, magazines, and books. They also can be found as wall maps in travel bureaus, and in elementary schools where they establish distorted views of the Earth in children during their most vulnerable years, including those who event;ually become geophysicists.

Ever since Mercator's time, mapmakers have been trying to overcome the problem of area distortion. In a recently published Atlas I have counted

Figure A-3. Goode's Homolosine Equal-Area projection, as if a model globe was cut up and spread out on a flat surface.

13 different area projections such as Zenithal Equidistant, Oblique Conical Orthomorphic, Cylindrical Equidistant, Oblique Zenithal Equal Area, ·etc. · The two most often used formats in newspapers and magazines are illustrated in Figures A-2 and A-3. Note that besides area distortion the correct comparative relations between continents are completely lost.

Indoctrination to One View

If we want to establish a clear view about the Earth's surface features, perhaps we should return to the model globe. However, there is a problem here as well. All globes are made with the North Pole up. This may be called a "convention", but since no alternative is ever presented this view has become unconsciously indoctrinated into people's mind as the "proper" orientation of the globe. In reality, of course, this is completely wrong. In space there are no ups and downs, and the plane of the ecliptic (the plane of the Earth's orbit) is not "horizontal". When a spacecraft travels in the solar system, the floor of the pilot's cabin is not necessarily set in this plane.

You can make an experiment in "involuntary brainwashing in society". In your room or office, suspend a terrestrial globe from the ceiling, on a


thin thread attached, say, at the Equator. Then count the percentage of your visitors who say that "you hang the globe wrong", or make any other critical remark.

There is no easy way to escape from the North-Is-Up concept. While there are now available free standing globes that can sit in their cup-like stand in any orientation, the lettering on their surface is still in the northup direction. But what is wrong with this indoctrination? Is not this a kind of law and order? Yes, it is. But chances are that you might ignore weak signs of yet hidden features of the globe if they contradict "well established and widely accepted" views.

How many people in the world have looked at the terrestrial globe during the past 300 years? And how many scientists, geophysicists, mapmakers, ship and airplane captains were among them? Perhaps hundreds of thousands. They all missed something to observe because of the strength of indoctrination to one view that is only one of I!lany others possible.

Applying Another View.

In the early 1970's I worked on a satellite transmitter antenna as an electrical development engineer. The satellite was to orbit the Earth at the geostationary altitude of 35,900 km, and to "illuminate" certain continents with a transmitted radio beam. Thus I frequently had to consult a terrestrial globe, and study the locations of continents. These had been the pioneer years of satellite operations around the Earth. The Earth globe attracted my attention beyond my professional interest. At the same time I acquired the knowledge of scientific programming of computers. The computer's capacity for solving equations, its speed, and accuracy in mathematical calculations, greatly fascinated me. Thus I decided to start a private project, in which I apply my academic training in electricity to the Earth, through a computer.

At that time I had only a vague concept about the scientific view of the Earth, that is, geophysics. But I recalled reading in the popular literature that scientists had difficulties explaining the physics behind the geomagnetic field. I started reading the related literature, old and new, and I found the subject suitable for my project. Since nobody supported my project financially, I regarded myself as an independent scientist not responsible to anyone about ideas and progress.

My first impression from the literature was that authors spent more time quoting each other than proposing a working model. Furthermore, authors often freely referred to processes unheard of in physics, like "the continuous conversion of gravitational energy into electric current and magnetic field". There is no such a thing. On the other hand, many well-known potential processes were never mentioned. Arrogantly perhaps,

Paper A 5

but I felt that the authors had never been properly trained in the science of electricity.

By 1974 I had become deeply involved in this field. I noticed that the geomagnetic poles were offset from the rotational poles in both Hemispheres in a systematic way. Both poles were displaced in the direction away from continents. The continents happen to concentrate on the African side of the globe; the "away" direction is toward the Pacific. Furthermore, both the continents and the magnetic poles are known to be drifting slowly. I suspected a correlation between the positions of the continents and the magnetic poles.

This assumed correlation gave me the idea that the geomagnetic field could be generated by the rotation of the crust, and mostly by the continental crust. In fact, the generation of a magnetic field by physical rotation of electric charges was demonstrated by H. A. Rowland, an American scientist in 1875. He placed charges on an ebonite plate that he rotated like a gramophone disk, and observed the formation of a magnetic pole at the center of the disk.

Based on Rowland's idea, W. Sutherland suggested in 1900 that the geomagnetic field could be generated by charges co-rotating with the Earth (the concept of "crustal layer" had not been known yet). He said that if the upper part of the Earth was negatively charged, and an equal amount of positive charges was positioned anywhere below it, then the observed polarity of geomagnetic field would be generated by the co-rotation of the · charges with the Earth. Opposite charges rotating in the same direction generate two magnetic fields of opposite polarities that subtract from. each other. However, the negative charges rotate at a greater radius;, they generate a stronger magnetic field than that of the positive charges rotating at a shorter radius. Therefore, their algebraic difference is non zero, and it carries the polarity that of the negative charges. But this idea has been abandoned because nobody could suggest a source for the charges, let alone to explain why those charges would be separated in such a specific way.

I did have ideas for both. The source of the charges could be thermionic electric charge emission that always takes place on the surface of red-hot objects, like on the cathode of an electron tube. This is the Earth's internal magma surface that is positioned right below the crust. Such an emission always carries a kinetic energy; the charges would be "pumped" into the crust from below. The crust could be viewed as a semiconductor that is a conductor for the physically very small free electrons (negative charges), but it is an insulator for the much larger positive ions. This would be charge separation in the correct direction.

This mechanism also obeys Gauss's Law of electrostatics. If equal amounts of opposite polarity charges are carried by a spherical structure,


then no electric field is generated above the surface even if the charges are positioned at different spherical layers.

The beauty of this mechanism is that it can be numerically calculated. The dimensions, the rotational speed, and the rotational radius of the crust are all known. It is not difficult to calculate the amount of charges required to generate the observed magnitude of the geomagnetic field ( see my detailed calculations in Paper C). The calculations indicate that about 1016 free electrons per m3 in the crust would generate the observed geomagnetic field. This is a relatively extremely small number of electrons. In the crustal rock there are about 1028 atoms per m3 (Avogadro's number), thus the electron density is only one part per 1012 atoms. In comparison, in current carrying copper wires there is one free electron for each atom (as measured by the Hull effect).

The existence of free electrons, a net charge in the crust reminded me Coulomb's electrostatic repulsion force that would arise between blocks of charged objects. On the Earth, the charged blocks would be the continents. Just how much repulsion force would be generated by the above calculated free electrons? Take, for instance, two Africa-size continents with 40 km crustal thickness, and place them 10,000 km apart (in space, to avoid the complications of the spherical surface of the globe). They would be repelling each other by an unbelievable 2x1025 kg· force. They would accelerate apart at 70,000 (seventy thousand) m/sec2

• By comparison, in free fall bodies accelerate at 10 m/sec2

, and the bullet in the barrel of a shotgun accelerates at 200,000 m/sec2.

I called this result an impasse and deadlock. Under this force the continents should have blown up themselves long time ago. Perhaps they do not have any net charge after all, and my co-rotation theory of the geomagnetic field is all wrong. I spent long months pondering about this problem, trying to find evidence one way or another.

One day, when my desk was cluttered with model globes and model continents, my little daughter, 8, walked into my room. She had just received her copy of World Magazine of the National Geographic Society. Two sheets of cardboard came with it as a supplement that could be assembled into a 24-sided toy globe. Properly called, it was a Physical Globe, an Icositetrahedron with a Gnomonic Projection. "Daddy, would you please assemble this globe for me, you have been playing with globes anyway" she said. Of course I dropped everything and started to work on it at once.

When it was ready, I decided to combine the demonstration of the globe with some geographic education. When I worked with my own cutout continents I myself had difficulties sometimes to recognize them by their shape, thus I had to write their names on. I wanted my daughter to know them by their shape, as the price for the assembly work. To prepare the demonstration I hung the new globe from the ceiling on a thin thread

Paper A 7

attached to its North Pole. The hanging substituted for the stand of the globe, and I could easily rotate it. Then in the pre-demonstration test runs I tried to adjust the height of the globe for optimum visibility from her eye level. This turned out to be difficult. If I set the height below eye level, Antarctica could not be seen. If I set it above eye level, Antarctica came

OS

Figure A-4. An Icositetr edron Physical Globe for chi~dren, to be assembled at home from a printed cardboard.

properly into view, but Alaska, Siberia, and Greenland disappeared. In either case, when the Pacific side rotated into view, nothing interesting was to be seen.

To improve the visibility of continents, I considered changing the point of rotation. In fact, the globe offered this possibility by having 25 pointed comers and 24 sharp edges while having no prepared facilities for rotation at the traditional North Pole. I bent a needle into a hook, pierced through the cardboard, and attached the thread to the hook. Then I tested the different peaks and edges as centers of rotation.

I wanted to see less of the Pacific, and more of the dry land, to trade in one for the other. When I first tried it, I immediately noticed another benefit. The globe looked unexpected and funny, just like when someone puts on his hat backward. I thought my little daughter will like it, and she will pay more attention to it.

Where should I put the point of rotation? After a short experiment I found two possible locations. I could place it either at the center of the Pacific, and position the globe above her eye level. Or, I could place it at the center of Africa, and position the globe below eye level. This latter solution appeared to be more convenient since both of us could look downward at the globe. And indeed, all continents came into view in one rotation, and no time was wasted on viewing the Pacific. The demonstration went on very well. When my daughter came into the room she immediately said "Daddy, you hung the globe wrong . . . "


After the lecture in geography, the globe remained in my room for a while, and I frequently looked at it simply because it was unusual and interesting. It easily rotated on its long and thin thread when I gave it a light spin, and it kept rotating for several turns. One day I viewed the rotating globe from the plane of its new rotational equator, and tried to prepare a new lecture. "Imagine that we are in a satellite, in orbit around the Earth. The astronauts do not feel that they are moving as we don't feel

Figure A-5. The Icositetrahedron Globe is suspended and rotated at the center of Africa to improve visibility of the continents.

it on an airplane. They see the Earth rotating outside (not below). But this is not the Earth's real rotation that is much slower, once in 24 hours, and around another axis. This is an apparent rotation caused by our orbital movement. You can see now that in space there are no ups and downs, and the North Pole does not have to be 'up'. The globe can be positioned in any direction in space".

"We are now looking out of the window and we see the continents coming and going, one after the other. This must be a special orbit that makes this lineup possible. Which continents are these? This is Eurasia, North America, South America, Antarctica, Australia, then Eurasia again. But wait a second, there is a certain regularity here. The continents appear to be located on alternating sides of the new equator. If we consider Africa to be at the top of the globe, then Eurasia is above, North America is below, and South America is above, the new equator. They follow a regular up-down-up-down sequence, or zigzagging" (Figure A-6).

At first, I did not take zigzagging seriously. Just as anyone else, I thought it was coincidence. Furthermore, looking at negative evidence, only four of the six continents took part in it. Antarctica was not zigzagging although it was located along the new equator. It was just sitting

· between South America and Australia.

Paper A 9

Then there was Africa, not located along the new equator and not zigzagging. I thought that if there was a mysterious force causing zigzagging it would not have made exception with two of the six continents. Perhaps.

However, I soon realized that the continents came into view regularly. That meant that they were evenly spaced from each other along the new rotational equator. This was a new positive evidence of something at work, not randomness. Then there was Africa. While it did not take part in the

Antarctica .'!'............ ; South. .., .,_, ,, ~;

cnca .., ~ ..

North America Australia

Australia

LEFf FRONT RIGHT

Figure A-6. Zigzagging of the continents is illustrated by the thick dashed line interconnecting their area centers. The thin dashed line is the new equator.

tnAr"-r,d.ca

Figure A-7. The area center of Africa is5:terconnected with those of the other five continents. The lines radiating from Africa form a visibly near perfect pentagon.

zigzagging, it was definitely doing something special. It was rotating around its own center of area. This was not randomness. True enough, I placed the point of rotation there myself, but not because it was the area center. My first choice of the axis of the rotation was the middle of Africa, on the Equator. But when I noticed the zigzagging, I adjusted the axis point to optimize zigzagging, to make its amplitude symmetrical on the two sides of the new rotational equator. During this optimization the axis point shifted northward from the Equator, and settled at a point that eventually turned out to be the area center. Then came the real surprise: I

BACK


realized that the lines interconnecting Africa's area center with those of the other five continents formed a near-perfect pentagonal star (Figure A-7).

Blueprint projection

Based on these quite impressive first observations, I started studying the zigzagging and the pentagonal formation of the continents. First, I selected a method of illustrating the globe on paper. I realized the distorting effects of the mapmakers' geographical projections, and I turned to the standard engineering six-view projection, the method used in industrial and construction blueprints. This is orthogonal projection, (similar to the popular perspective view of small foreshortening). See the example of a simple house in Figure A-8(a). This projection is similar to the one created by the

(a) (b) Figure A-8. Orthogonal projection (a), and its extended version the wireframe in modem CAD drafting (b).

human eye as it projects a 30 object into a 20 image on the retina. Then the brain, long accustomed to this method, generates a 30 "feeling" of the object in one's mind. This is not happening with the usual map projections: the brain does not generate a 30 feeling from a Mercator or any of the other 12 different projections I am aware of. On the contrary, these projections try to convince one's brain that the Earth is a flat object, by eliminating the third dimension.

The problem with the orthogonal view is that it hides information that is on the back side of the object. One solution is the transparent orthogonal view called "wireframe" in Computer Aided Drafting (CAD), Figure A-8(b). But the view can become cluttered with too many lines. Furthermore, the reversed view of the rear side, with the visible hidden lines, inhibits the brain's ability to generate a correct 30 feeling. When I view that house in wireframe I either see it from above with the door on the front, or from below with the door at the rear.

A good practical solution to this problem is the engineering projection that is a collection of six plane views, each rotated 90° from the other. One view is arbitrarily named front, then the right side view of the object is placed to the right of it, and the back side view is placed further to the

Paper A 11

right of it. The object's left side view is placed to the left of the front view, the top view goes above it, and the bottom view is placed below it. No vital information is lost, and there is no cluttering.

The meaning of some views may not be immediately clear. For instance, the roof on the front view looks like a flat vertical rectangle. But this is not distortion. You can immediately see the right or left side view that conveys the slanting nature of the roof to your brain. This is, in fact, how you see the real house directly from the front when you close one of your eyes. The brain needs only a little hint, a small amount of a side view, to generate the 3D feeling. Indeed, Mother Nature has equipped us

ODD

CJ

TOP

BOTTOM

RIGHT

00 BACK 00000

Figure A-9. A "blueprint", or six-view engineering projection to provide two side views to any of the views.

with two eyes always to supply an amount of side view. An engineering drawing does this by providing two side views to any of the views, Figure A-9. But the most important feature of this projection is that it preserves the correct comparative relations between all views of the object.

Figure A-10 illustrates the Earth globe as the object, and its six side views (of which three are hidden). I project each curved spherical side, with the continents, to the flat surface of the enclosing (imaginary) box. The contour line of the globe is a circle. I always display continents in featureless solid black color, and I do not write their names into this area. This leaves me free to rotate their views as necessary, without reference to the North-South direction that is irrelevant for the present purposes.

In my earlier studies I also projected the spherical coordinate grid because it further conveyed the spherical shape of the globe on flat paper. However, I rotated the grid away from the irrelevant North and South


Poles, to relevant locations my study suggested. The original purpose of the grid is to make location of surface points possible. For this purpose it does not have to be tied to the rotational axis at the Poles. The grid's poles

Figure A-10. A six-view engineering projection of the Earth's surface.

could have been placed anywhere, to a city for example, like the zero longitude tied to the city of Greenwich, England, quite arbitrarily. Of course, it is convenient and practical to use the Poles because the grid becomes also the dial of the Earth as a timepiece. This feature is also irrelevant for my study. In fact, the tying the two independent poles together, that of the rotation, and that of the coordinate grid, has become indoctrinated in people's mind as a single entity. For reasons I give further

Figure A-11. Africa is rotated to the top in the six-view projection.

down, I always placed the pole of the grid at the area center of Africa, and rotated the globe to have this point at the top of the view. Then people usually asked the question "why did Africa move to the North Pole?". To avoid this misunderstanding I am not using the grid system anymore.

In Figure A-11, I rotate the globe into the position I have used with the Icositetrahedron globe; the area center of Africa is at the top. This position is nearly on the old Equator, it is at 18.36°E, 7.14°N. Thus the conventional Equator runs nearly vertically, and the North-South axis is nearly in

Paper A 13

the plane of the new rotational equator that I eventually named E-equator (for Electrical Equator). The front view must be chosen arbitrarily; I selected the side with North and South America.

Area Centers

Area centers play an important role in my study. This is a subset of Center of Gravity, a household word in mechanical engineering where it is frequently called "cg". This concept has come into use in the 1770-s in

up

H ...

down

Figure A-12. Eurasia is above the -equator (up), and North America is below it (down). This illustration is the left side of the globe in the definition of Figure A-11.

America

I I

up

Figure A-13. On the front side of the go e North America is below (down), and South America is above (up) the E-equator.

studies of the motion of projectiles in flight, or Ballistics. The motion of large three-dimensional bodies, even if their shape is irregular, can be described mathematically as the motion of a single point, their center of gravity.

The concept can be also applied to two-dimensional bodies like a sheet of paper where it is called area center. For this purpose, for example, a


large city can be viewed as two dimensional on the surface of the ground. When you are driving on a highway that goes toward such city, but does

\

not go into it, where is the exact point on the highway where you are 'at' the city (ignoring exits)? It is the point where the city's area center is the nearest to the highway. Thus when we traveled in our imaginary spacecraft

---•-= down

Antarctica

Australia

"B. Figure A-14. On this right side of the go e South America is above (up), and Australia is below ( down) the E-equator. Antarctica is on the straight line which interconnects the area centers of South America and Australia, and it is at half-way beween them.

up

down

Figure A-15. On this back-side view of the go e Australia is below (down), and Eurasia is above (up) the E-equator. This view completes the sequence of sideways rotations because another 90° rotation returns to the left side of the globe.

along the E-equator of the lcositetrahedron globe, we were 'at' the continents when their area centers were nearest to us.

Apart from guessing and estimating, the easiest way to determine a continent's area center is to cut out its shape from cardboard, and to rotate it in the vertical plane around certain different points. Pierce a needle through it somewhere in the middle, and hold the needle horizontally as an axis. Start rotating the cutout by a small jolt, and then leave it alone.

Paper A 15

If it comes to a rest always at the same position, then the needle is not at its area center. Change the needle's place until the rotation stops always at different positions. (Tire balancing at your garage is done for a similar purpose: to move the tire's center of gravity to the middle of the shaft by adding extra weight to the tire at different locations). The cardboard method is completely adequate for the present purposes to find the area

Continent Longitude Latitude

Africa 18. 36° 7 .14°

Eurasia 76.31° 50.66°

North America -98.90° 50 .40°

South America -60,65° -14.24°

Antarctica 77. 70° -·86.11°

Australia 134.37° -25 .80°

Table A-1. Calculated positions of area centers of continents.

centers of continents. Nevertheless, I have developed a computer program of a mathematical method (see Appendix A2) to do it more exactly. Divide the surface area into several triangles of arbitrary sizes, and enter the coordinates of the triangles. The program calculates the area centers of such spherical triangles, and sums all centers into a single point. Table A-1 lists these values for all six continents.

Zigzagging of Continents

Figures A-12, A-13, A-14, and A-15 show four sides of the globe along the E-equator as the globe is rotated around the area center of Africa. The small cross in the white circle on the continents are the calculated positions of their area centers. The E-equator is drawn in dashed line.

The four side views suggest that neighboring continents along the E-equator are always on the opposite side of the E-equator, except Antarctica. This also can be expressed as zigzagging when we observe the globe rotating, or when we orbit the Earth in a spacecraft in the plane of the E-equator. Figure A-16 illustrates the four views placed side by side, as if you rotated the globe from right to left.

In Figure A-17 I display the zigzag lines extracted from Figure A-16. The circles with the cross represent the area centers. The rule appears to be that two adjacent continents must not stay in the same hemisphere. Even Antarctica obeys this rule by taking no sides. What kind of force would cause this? My conclusion is that a repulsion force is acting among the continents. Under this force each continent would move as far away from its neighbor as possible; thus they would move to a great circle of


the globe, to the E-equator. Then, if the force is very great, it would cause "buckling" along this great circle.

Buckling is another engineering expression that means to bend, heave, warp, or kink usually under the influence of an external force, or collapse, give way, yield. Buckling sometimes affects very tall and thin radio towers. These towers are designed to stand perfectly vertically thus their weight causes only a compressional force on their steel structure. Steel

Antarctica

Australia North America

..:·-..Ali.~-------~:---.-~-;&~, once~.. ,.

Australia

LEFT FRONT

Figure A-16. Zigzagging of the continents is illustrated by the thick dashed line interconnecting their area centers. The thin dashed line is the new equator. (This is the same illustration as Figure A-6).

Australia

Figure A-17. The zigzag lines of the continents are extracted from the spherical surface of Figure A-16, and redrawn in rectangular plane coordinates.

bars are the strongest against compression. Usually several guy wires are attached to the tower to keep it vertical under windy conditions (Figure A-18). However, if very strong winds attack the tower between two guide points, it may cause the steel bars to bend. Then the weight of the structure above this point will cause buckling, a series of left-right-left-right bending centered at the guide points (thick lines in Figure A-18). This could cause the tower to collapse.

Repulsion is a compressional force between objects on the closed surface of a spherical body. But what about Antarctica? Why does not take part in it? The special nature of the spherical surface is responsible. On the spherical surface there are no start and end of the zigzag, it is an endless ring. Thus, no two points can exist side by side in the same hemisphere. Figure A-19 illustrates the buckling map of two, three, four, and five bodies on a spherical surface. Of two bodies of A and B, A is located in the upper, and Bis in the lower hemisphere. For three bodies there is no

BACK

(

Paper A 17

such simple solution: 3 cannot be divided evenly into two whole numbers. If we place one body in the upper hemisphere, then two would go to the lower hemisphere, ending up side by side. The solution is that A goes to

Figure A-18, A tall and thin radio tower, supported by guy wires, can buckle in a non-uniform strong wind. The buckled tower forms a zigzag line (for a short time before collapsing).

the upper hemisphere, B to the lower hemisphere, and C stays on the equator, halfway between the other two. This could be called the "odd-number solution". For four bodies the solution is simple again: two in each hemisphere. For five bodies the odd-number solution applies again.

A A

Psz1 B

A A

E2=J B

A C A

E0ZSiJ B D

Number of bodies

2 (even)

3 (odd)

4 (even)

~I 5/oddJ

B D Figure A-19. Repulsion forces on a spherical surface keep one of the zigzagging bodies on the equator when the number of bodies is odd.

Two bodies go to the upper, and two to the lower hemisphere, and one body stays on the equator. On the Earth, Antarctica plays the role of staying on the £-equator.


In a radio tower only a very small amplitude of zigzag state can exist, and this can be observed only by sensitive instruments. When the amplitude reaches a critical magnitude, the tower collapses. However, on a spherical surface it is much different. Here buckling is the only stable condition. Let's ignore Africa for the moment. If the continents are positioned by a mutual repulsion force then they move as far apart as

j

Figure A-20. On a sphere, mutually repelling adjacent bodies can buckle only into the opposite hemispheres.

possible. Say, there are five of them (as on the Earth). Then they would move to a 72° separation strictly on the E-equator. In an ideally accurate mathematical model this would really happen. However, if there is any misalignment, if any of the continents is slightly out of the equatorial plane, then a sideways component of the force develops. This forcecomponent causes the buckling. In fact, there is no opposing force against moving sideways because there is no repelling continent in this direction, unlike in the equatorial plane where there are repelling continents only at 72° angular distances away (Figure A-20a).

Thus the continents start moving in directions perpendicular to the equator, that is, toward a pole. These directions are the meridians, and all meridians converge together at the pole. However, if two adjacent continents start out toward the same pole, then they also move closer and closer together as they approach the pole (b). Their mutual repulsion force would increase with decreasing distances. This feature defeats the original reason to go in the direction of decreasing forces. This prevents them to move into the same hemisphere right from the beginning. Instead, they move to opposite hemispheres where the force decreases as their mutual distances increase. In fact, they help each other moving into the opposite hemispheres by repelling each other in that direction. Furthermore, with their increasing distance their mutual repulsion force decreases (c).

However, if two bodies move long enough in opposite directions on a spherical surface, eventually they will approach each other again, after moving more than 180° apart. This means an arising force from the forward direction. The two forces, the decreasing one from behind, and the

Paper A 19

increasing one from ahead, will be equal at one point. These are the positions where the buckling continents stop moving polewards. These are their stable and force-balanced positions that can be observed on the Earth globe.

Pentagonal Pyramid

The idea that buckling is caused by repulsion force implies that the bodies should move as far apart from each other as possible along a great circle that is the E-equator. That is, they should be evenly positioned along ihe E-equaior. Is this observable? Apart from measuring their angular separations, visual evidence is available by viewing the globe from above Africa. This view constitutes the correct top view of the standard six-view rotation as defined in Figure A-11.

In Figure A-21 I interconnected the area center of Africa with those of the other five continents located along the E-equator. I included in this view those parts of the other continents that are above the E-equator, thus visible from above Africa. I also included Australia in dotted lines, to show its position, as it is completely below the E-equator. Numerical values related to their angular positions around Africa are listed in Table A-2. The bottom view completes the six-view rotation of the globe, Figure A-22. Figure A-23 contains all my new observations about the positions

1l

...... "'··· J

sout\l ,\11\edca Figure A-21. The area center of Africa is interconnected with those of the five other continents. The lines radiating out of Africa form a visibly near-perfect pentagon.

of the continents; the area centers, the E-equator, the zigzag lines, and the pentagon lines. One possible conclusion from this observation is that the zigzagging is not a random feature, it has its own mechanism that causes it. It is like a line in a complex spectrum; it can be separated or eliminated from the rest without upsetting the other features left behind.


Angle Measured Between Angle0 Deviation from 72°

Continents Degree I Percent

Eurasia - North America

I 71.6

I -0.4

I -0.56

North America - South America 70.3 -1. 7 -2.36

South America - Antarctica 77.4 +5.4 +7.50

Antarctica - Australia 61. 7 -10,3 -14,30

Australia - Eurasia 79.0 +7,0 +9.70

Table A-2. Numerical values associated with the pentagonal positioning of continents around Africa. Deviations are calculated from an ideally perfect pentagon's 72° angle.

area center of

Antarctica

'g

l! "' G .q .;:a ..... ts 9 ,~

Figure A-22. The model's bottom view is almost completely surrounded by evenly distributed continents. The area center of the Pacific coincides with the antipode of Africa's area center.

I

Pacific

EB

~ • i·

:r. Q

I. Q

~

' I \ .,g·~.. ' I \

~ ~ -------•-_'!'_ - )(

._ .. > Antarctica !a e!. .;·

I

~ I .:i E-equator

! ,· ,' .. .. , y._

Australia

Figure A-23. In an Africa - Pacific centered rendering of the Earth, five continents zigzag in an ordered system along the E-equator. They also form a near-perfect pentagon around Africa. Furthermore, the area center of the Pacific coincides with the antipode of Africa's area center.

N , ~

"'Cl ~ ~ ::i,..

~


Figure a-24. After reducing zigzagging to zero, the five continents along the E-equator move to the plane of that equator where their area centers form a plane pentagonal pattern.

Figure A-25, When the area center of Africa is interconnected with those of the continents along the E-equator, and zigzagging is removed, a pentagonal pyramid pattern results. ·

Figure A-26. When the area centers of the continents along the E-equator are interconnected with that of the Pacific, a double pentagonal pyramid results.

(

Paper A 23

The purpose of the separation is to simplify the structure left behind, that is, the pentagon-like structure, and to find out what is causing it. When the zigzagging is eliminated, the five continents along the E-equator move to that equator. Their area centers will be located on a single plane, on the equatorial plane. The area centers can be interconnected by straight lines (through the bulk of the globe), and they form a simple geometrical pattern; a plane pentagon (Figure A-24). Then the comers of the plane pentagon can be connected to the center of Africa, resulting in a threedimensional geometrical pattern; a pentagonal pyramid (Figure A-25). Finally, the comers of the plane pentagon can also be interconnected to the area center of the Pacific. The resulting object is a double pentagonal pyramid, Figure A-26.

Electric Force

During my geomagnetic studies I was surprised to learn from the geophysical literature that the Earth's solid upp~r surface, the crust, was only 20 km thick on the average, SO km on the continents, and only 5 km under the oceans. Considering the 12,000-km diameter of the Earth, the crust is relatively as thick as the skin of an apple. Below the crust the temperature is at the melting point of granite. If the crust was removed, the entire Earth would shine red hot.

I have studied red-hot surfaces in my own profession of electrical engineering; the cathodes of electron tubes. I imagined that we live on a huge thermionic cathode. Is this an operational cathode? Is it working as a cathode? Obviously, there is no anode plate or anode voltage around it, as in an electron tube. Normally, if you tum off the anode voltage of an electron tube in a radio, it stops working even if the cathode is kept red hot by the heating voltage.

By implication I realized the Sun was also a thermionic cathode, and it certainly operated as a cathode. Astronomers report that a constant flow of electric charges leaves its surface, that they call "solar wind". It consists of an equal amount of positive and negative charges as protons and electrons. These are generated by the thermal breakup of atomic hydrogen, the very matter of the Sun's surface. However, for some reasons astronomers do not identify the solar wind with thermionic electric charge emission in the sense as it is used in studies of electron tubes.

The Earth's red hot inner surface and the solar wind reminded me of the principles of electron tubes I studied at the university in the 1950's. An electron tube may stop operating when the anode voltage is turned off in normal everyday use in a radio, but it does not cease its charge emission. The "zero-voltage anode current" is an important measure of the cathode's

efficiency in converting its thermal energy into electrical energy. This had been an important problem just before the transistors came into use, in


battery operated portable radios. This property is measured in a test circuit of Figure A-27.

In the diagram D is the electron tube, a diode, B is an adjustable power supply, A is an ammeter to measure the anode current Ia, and V is a voltmeter to measure the anode voltage Ua. The tube's normal operating range is in the P region where a small change in the anode voitage results in a relatively large and linear change in the anode current. In the S region saturation occurs because the cathode is unable to supply enough charges for a linear increase of the current. However, at the Z point in the diagram, at zero anode voltage, the anode current is not zero. Furthermore, even at negative anode voltages the current flows, against the anode voltage.

The standard explanation of this mechanism is that at red-hot temperatures atoms of the cathode ( of whatever material) break up into their constituent particles of positive ions or protons, and free electrons. Another name of these particles is "electric charges". Then collisions with thermally fast vibrating particles accelerate them in random directions. Those

V

B

A

Figure A-27. Test circuit to measure the zero voltage anode current of an electron tube. The current is generated by the thermal energy of the cathode.

particles that happen to be accelerated toward the cathode's surface, depart the cathode material and continue traveling away from it. They carry an energy that originates in the cathode's thermal energy. The flow of charges away from the cathode is properly called thermionic electric charge emission (not "wind"). The amount of material emitted by the cathode is determined by the type of the cathode's material, its surface temperature, and its surface area. Most notably, the thickness of the cathode, and the inner temperature of the cathode, do not come into account.

If the cathode is located in the vacuum of free space, both types of charges are emitted in equal quantities. This equality is maintained in digital accuracy because of the nature of charge generation. Any neutral atom consists of a number of positively charged protons in its nucleus, and the same amount of negatively charged electrons in orbits around the

. (

Paper A 25

nucleus. The equal number of opposite charges negate each other; the atom does not exhibit an external electric charge. When thermal collision removes one electron from orbit, the electron becomes a free negative charge. The loss of one electron means the loss of one negating counterpart for one proton, and the nucleus will exhibit one external positive charge. A general expression for this process is ionization of the atom, and the non-negated atom becomes a positive ion.

To understand the operation of the electron tube ( and the Earth's cathode) it is necessary to discuss two concepts. One is the penetration or infusion of matter with electric charges, and the other is the two types of energy the charges carry.

In free space the emitted charges travel together to indefinite distances since nothing removes their energy. However, the opposite charges behave . differently if there is an impediment in their way. The difference is caused by the significantly different physical size of the charges. The negative charge or free electron is several orders of magnitude smaller than a positive ion. It is so small that it can easily infuse solid matter and propagate in it, including insulators (semiconductors) and metals, while a positive ion cannot infuse or penetrate either.

Emitted charges simultaneously carry energy in two different forms. One is a kinetic energy determined by the velocity and the mass of the involved particles. This is the same type of energy that is carried by, say,

1-------the-bullet-shot-by-a-gun-;-This-energy-is-active-only-when-the-partidesmove at some velocity. The other type of energy is charge separation. When opposite charges are separated, an electric force develops between them that tends to reunite the charges in the original form of a neutral atom. This energy is active even if the particles are not moving. Its related force is an electric force, called electrostatic force, Coulomb's force, or any combination of these names.

In the electron tube at zero anode voltage emitted charges propagate from the cathode toward the anode plate initially unhindered, just like the solar wind. In this form there is no significant charge separation, thus the charges collide with the solid-metal anode plate by their kinetic energy. Free electrons easily move in, they infuse it, and continue moving inside. However, positive ions cannot do this. Instead, they stay in the space between the cathode and the anode plate. This constitutes a charge separation, free electrons in the metal and ions outside. An electrical energy builds up in which the positive ions tend to attract the departing free electrons back.

However, this attraction extends not only toward the anode plate. It is omnidirectional, it acts also toward the cathode. Indeed, electrons traveling through the circuit from the anode to the cathode, reappear on the cathode. They have lost their kinetic energy in the resistive wire, slowed down, and they are capable to recombine with the waiting and attracting positive ions


into their original neutral atom. Depending on heat loss, resistance, charge separation, and particle velocity, a balance sets up at some value of the current that remains constant as long as the cathode's temperature is maintained. No anode voltage is needed for the electrons to infuse the solid-matter anode plate and wiring, and propagate in them.

The above experiment can be extended by a further simplification. If no anode voltage is needed, then the anode plate can also be removed, together with its external circuitry. The emitted charges still keep moving away from the cathode by their kinetic energy, and they will collide with the next piece of solid matter in their way. This is the glass envelope of the electron tube. Again, free electrons infuse it and propagate in it ( although at smaller velocity than in metals), and the positive ions stay behind. This is charge separation again, in which the ions exert an electrical attraction force on the free electrons that entered into the glass wall, and perhaps exited at its outer surface.

However, there is no return circuitry to the cathode. A completely balanced situation arises in which the space in the tube is filled to some density of both charges that prevents further emission from the cathode (in the industry this is called "space charge"). In the balance there are more positive ions than electrons inside, and the missing electrons are partly in the glass, partly in the air just outside the glass. This is easily observable on any operating TV screen when you approach it with your fingers. The electrons cause a tingling sensation as they move into your skin, and even you may experience electrical discharge, or sparks.

Free electrons outside the thick glass screen all have moved through the glass, and there are still more free electrons in the glass wall as long as the cathode remains operating. (This is a very important mechanism toward understanding the Earth's electrical operation). In this TV tube the anode voltage is not zero. However, the presence of an anode voltage causes only quantitative changes, not qualitative ones. We can vary the elements of the experiment to maintain a significant amount of free electrons in the glass wall. We can reduce the anode voltage, increase the size of the cathode, and decrease the relative thickness of the glass wall. Eventually we arrive at the dimensions of the Earth, and we will have a significant amount of free electrons in the glass-like granite envelope of the magma cathode, in the crust.

Electrically charged bodies exhibit a force between themselves. In fact, this is the most ancient observation of electricity. Thales of Miletus reported in 600 B.C. that a piece of amber, having been rubbed, would attract small bits of straw and feathers. This phenomenon remained only a curiosity until the dawn of modern age. In 1785 a French scientist, Charles-Augustin de Coulomb, determined the numerical relationship between the quantity of electricity that two bodies carry, Q1, Q21 the distance between the bodies, R, and the force they develop, F.

Paper A

F=k Q1Q2 R2

27

(A-1)

The measure of electricity is named after him: the unit is one coulomb. After J. J. Thomson, an English physicist, discovered the electron in 1897, it was determined that one coulomb electrical charge contains 6.24x10 18

electrons. The word "electron" is a reference to ancient observations of electricity, it is the Greek word for amber.

Coulomb's law, or equation, states that the electrical force that develops between two charged bodies is proportional to the product of the charges, and to the inverse square of their distance. In addition, there is a constant

p u .,,

Figure A-28. The Absolute Electrometer measures the quantity of electricity by mechanical means, by measuring the electrical force in a weight balance.,

in the equation that carries the numerical description of the electrical environment in which the force propagates between the bodies, and a factor depending on the system of units of charges, forces, and distances. For example, if one body carries 2 coulombs, and the other 3tcoulombs of charges, their distance is 230 meters, and their environment is air, then the electrical force between the bodies is 1 million kg force. As this very large force indicates the unit of coulomb is a very large charge. In laboratory practice much smaller charges are used, and the resulting forces are in the range of grams. The direction of the force is an attraction if the signs of the charges are opposite, and it is repulsion if the signs are identical.

The amount of electrical charges can be directly measured by mechanical means, by the Absolute Electrometer (Figure A-28). This is a laboratory balance in which mechanical and electrical forces are directly compared, or calibrated, against each other. The left arm of the balance is a flat metal plate above a fixed plate of the same size (P), working together as a capacitor. When opposite electric charges are placed on the plates, that is, the capacitor is charged, the plates attract each other, and the upper plate moves downward. This is balanced by a weight in the right arm (W). From the dimensions and weight the magnitude of the applied electric charge can be calculated. This is an absolute and direct conversion between electrostatic and mechanical forces.


Another instrument of electrical laboratories is the Gold Leaf Electroscope, Figure A-29. Its active component, unlike the two-piece capacitor in the Absolute Electrometer, is a single piece of metal rod, (M), sitting in a glass jar. Two very flexible gold leaves are hanging freely, (A), (B), at the bottom of the solid brass rod. All three pieces form a single metal electrode. When any polarity of charges is placed on this rod, its attached gold leaves deflect (C), (D), although they are electrically "short-circuited" at their base.

For this property I always found it a mysterious instrument. As everybody else, I grew up in a world where electricity is always represented by two wires and two terminals. The gold leaf electroscope appears to be out of this world because of its singleness. Its operation is completely self-contained, it does not depend on an "electrical ground", or on

p I

111 11) I r I D

AB

Figure A-29. The Gold Leaf Electroscope is a sensitive instrument to demonstrate the repulsion nature of electrical like-charges within a single body.

another electrode. Left alone in deep space it would work just as well. In this context it has a fundamental similarity with the continents of a live planet as it will become evident during my study.

The deflection within the very same material means that if large enough charges were applied (and discharge did not prevent this application), then the entire metal rod would split apart along its axis (p-p), and the two halves would fly apart. This instrument is a clear demonstration that internal like-charges within a single body can break up that body (which may be a continent). The Gold Leaf Electroscope works only as an indicator, and it is not suitable for numerical measurements.

A popular high school experiment is to demonstrate fundamental electricity with the Pithball Electroscope (Figure A-30). This is made of two pithballs, or plastic foam balls, which are suspended side by side on thin threads. Uncharged, the balls sit side by side, (K, L). When the balls are very lightly rubbed with cat's fur, they acquire a charge, start repelling each other and deflect, (P, R). This is poor man's inexpensive Absolute

Paper A 29

Electrometer because the applied electric charge can be numerically determined from its dimensions.

The balls move apart along the radius of the chords, upward against gravity to a height h, and to the distance d from each other. From the dimensions and from the weight of the balls the acting electrical charges can be calculated. The charge density of volume, expressed in coulombs per meter, is a practical measure of charges because it is independent of the dimensions of the tested objects. It makes comparison easy between different size of objects.

I built a Pithball Electroscope for myself to get a feeling for charge densities. I wanted to charge it with some easily attainable value, and to compare this number with that of the crust I calculated earlier, the one that is capable of generating the geomagnetic field. It is fundamental human nature that one has an expectation from any planned action. I expected that

R

K L

d

Figure A-30. The Pith Ball Electroscope :is a very simple instrument that is suitable to measure charge densities numerically.

in comparison the crust's calculated charge density would be enormously, orders of magnitude greater. After all, I make the pithballs deflect by very lightly brushing them with cat's fur. Obviously, the energy involved is not more than that of brushing, it must be minuscule. On the other hand, in the Earth's red-hot magma ball of 12,000 km diameter acting energies are immeasurably greater, and they produce the geomagnetic field.

However, I was wrong. After a few experiments and adjustments I succeeded making an electroscope with the same charge density as that of the crust (in my calculations). The dimensions were: diameter of the balls 1.24 cm, thread lengths 10 cm, deflection 3.33 cm. (See detailed account in Appendix A4). I did the experiment in a warm room on a cold winter day to have low humidity in the room. In higher humidity it is difficult to impart charges to the balls. Furthermore, I used rabbit's fur instead of cat's.


My conclusion is that the crust's charge density is extremely small since the same value can be easily achieved in an electroscope by very lightly brushing the plastic balls with fur. Now, here is a cardinal question. If the electroscope is deflected by such a small charge density, would the continents be repelling one another by the same small charge density? Again I had my own expectations; that they would not be positioned to the pentagonal pattern by this charge density. Thus, I resisted the necessary calculations as worthless. I expected that a few grams of force would not make any difference. But the idea persisted, and finally I decided to do it 'for the fun of it' as the saying goes. I was new to computers, and I enjoyed doing calculations for whatever purposes. (At first I did not trust the simple form of Coulomb's equation in using for this geometry. I numerically integrated the force of a large number of small cubic volumes on the computer). If for nothing else, I thought the result would be a good piece of conversation. "Did you know that America is pushing Europe by 32 grams over the Atlantic?"

Let's calculate first the force between two Africa-size continents. Africa's surface size is about the same as that of a square plate of 5000 km by 5000 km, and its average crustal thickness is about 40 km. This is a volume of 109 km3 or 1018 m 3

• In a cubic shape the same volume has an edge length of 1000 km or 106 m. Imagine a large pithball-type electroscope with two of these granite cubes hanging on it. Placed side by side their distance between their volume centers is 1000 km. 1016 electrons per meter 3 is equivalent to 1.6x10· 3 coulomb charge, thus the charge of one cube is 1018x1.6x10· 3 = 1.6x10 15 coulombs. The arising force in Coulomb's equation is F = k Q1 Q2 I R

2, where k = 1/4Jte0, and Eo = 8.85x10· 12 (the

electrical permittivity of air), thus k = 9x10 9• Q1 and Q2 are the charges in

coulombs, and R is their distance, in meters. Thus, the resulting force F = 2.3x10 28 newtons, or 2.3x10 27 kgforce. However, this appeared to me an unbelievably large number, thus I immediately rejected it as wrong. First quickly, then slowly looking through the calculations for an obvious error, I found nothing. Then I spent several months analyzing all possible inputs to this result, from the Earth's dimensions to the charge of an electron in the SI system of dimensioning, using a different computer, and doing the calculations longhand. All inputs had been correct. In a final effort to reduce this number I placed the two blocks at 10,000 km apart as a "more reasonable distance for continents". The result was, of course, only 1/100 times smaller, 2.3x10 25 kgforce, still a huge number.

Eventually, I found the explanation. Coulomb's force equation has an implied steep nonlinearity when the volume charge of the bodies are involved. The total charge Q of a body, when it is charged uniformly in its volume, can be expressed as the product of the cube of its linear dimension and the charge density. If the body is a cubic block with edge length b, then its volume is b3

, and !ts total charge is Cd x b3 where Cd is·

Paper A 31

the charge density. If the two repelling bodies are identical, then the nominator of Coulomb's equation becomes Cd x b3 x Cd x b3 or Cj b6

•

The charge density is constant, it can be included in k, thus the equation becomes

b6 F=k-

R2 (A-2)

The electric force increases with the 6th power of the linear dimensions of the blocks if they carry uniform volume charges. This appears to be an original conclusion of great importance, although Coulomb's equation has been known for 200 years.

If no error is evident in the calculations the only thing one can do is to study this force and learn as much as possible about it. There are many potential problems still looming about it, like electrical surface discharges caused by large total charges, but I take only one step a time.

First, let's build a bridge between the small and the large electroscopes. Table A-3 lists the force, weight, and the force/weight ratios for cubic blocks at touching distances apart. The edge length b of the blocks covers the range from 0.01 m to 1000 km in nine logarithmic steps. The smallest

edge length b force F, weight W F.IW kg-force kg

0.01 m 2x10-s 3x10- 3 0,007

0.1 m 2x10- 1 3xl0° 0,07 1 m 2xl0 3 3xl0 3 0.7

10 m 2x10 7 3xl0 6 7 100 m 2xl0 11 3xl0 9 70

1 km 2x10 15 3xl0 12 700 10 km 2xl0 19 3xl0 15 7,000

100 km 2xl0 23 3xl0 18 70,000 1000 km 2x10 27 3xl0 21 700,000

1.5 m 2x10 4 3xl0 4 1

Table A-3. The force/weight ratios of two charged granite blocks. The ratio increases with the size, thus the force between very large blocks are enormously greater than their weight.

block is about the same as my experimental electroscope, and the largest one is equivalent to Africa. The force/weight ratio has a value of unity somewhere around the 1-m edge length size. The exact numbers appear at the bottom of Table A-3, when b=l.5 meters the electrical repulsion force between the blocks is equal to the weight of one block.

Above the 1.5 meters size the electrical force is greater, or more specifically, it is increasingly greater than the gravitational weight of the involved blocks. If two blocks of 1.5 meters size were placed on top of each other the upper block would float weightlessly in the air. Larger blocks would accelerate upwards against gravity, the larger the block the

I


more actively they would fly. However, they would all stop accelerating at some altitudes because the gravity, measured from the center of the Earth, decreases slower than the electrical force that is measured from the other block at the surface.

Repulsion Force kg•force

Size Distance meters m 100 101 102 1()3 104 10s 106 107

10-2 2x10-' 2x10-u 2x10-u 2x10-1J 2x10-n 2x10-" 2x10-•1 2x10-"'

10-1 2x10-3 2x10-5 2x10-7 2x10-9 2x10-11 2x10-13 2xl0- 15 2x10-11

100 2xl0 3 2xl0 1 2x10-1 2x10-3 2x10-5 2x10-7 2x10-9 2x10-11

101 - 2xl0 7 2x10S 2xl0 3 2xl0 1 2x10-1 2x10-3 2x10-5

1D2 - - 2x1011 2x109 2x107 2x10S 2x1D3 2x101·

1D3 - - - 2x1D15 2xl0 13 2xl0 11 2xl0 9 2x107

104 - - - - 2xl0 19 2xl0 17 2x1015 2xl0 13

10S - - - - - 2x1D23 2x1D21 2xl0 19

106 I - - - - - - 2x1D27 2x10 25

Table A-4. The repulsion force decreases between blocks by the inverse square of the distance. However, the force between large blocks remains very large even at great distances.

At this point feasibility becomes an issue. With small blocks, as in the pithball electroscope, the experiment is certainly feasible. But at much larger sizes it may not be possible to infuse the blocks with the same charge density. Electric discharge on the surface of the blocks would prevent (shortcircuit) the buildup of the charge in this simple construct. But this is not sufficient reason to abandon investigating the basic idea because in other constructs it still may be feasible. These ideas must be discussed, labeled, and put aside for possible future application. Geophysicists have already observed that in certain 'gravity anomalies' the greater the mass the larger the anomaly is. Furthermore, if you discard evidence all too soon, on the grounds that 'it does not make sense', you will never be able to put together even a simple jigsaw puzzle because at the start of the game not one piece makes sense.

Repulsion forces between charged blocks can be further studied when the blocks are at varying distances from one another (Table A-4). The force decreases with the 2nd power of the distance. When the blocks move apart 10 times their original distance, their force drops to the one-percent value. However, the magnitude of the force of larger blocks is very large. Their force remains significant even at great distances. For example, 1-km size blocks at a distance of 1000 km still repel each other by 2x109

kg-force (two million metric tons of thrust). The 6th power increase of force with the size (Equation A-2) is demonstrated by the ten-times larger

I

Paper A 33

10-km size blocks at the same 1000 km distance; it is one million times larger, 2x1015 kg-force. Continent-size blocks of 1000-km edge length at 10,000 km distance repel each other by 2x1025 kg-force.

These large numbers are beyond the limit where one can still visualize their magnitude, or compare the number to something already experienced. However, the concept of acceleration could help. Acceleration is described by the equation a=Flm, where a is in m/sec2

, Fis in newtons, and m is in kg (Table A-5). The same numbers are also illustrated in graphics form in

Acceleration m/sec

Size Distance meters m 10° 10 1 10 2 10 3 10 4 10 5 10 6 10 7

10- 2 7x10- 6 7x10-s 7x10- 10 7x10- 12 7x10- 14 7x10- 16 7x10- 18 7x10- 20

10- 1 7x10- 3 7x10- 5 7x10- 7 7x10- 9 7x10- 11 7x10- 13 7x10- 15 7x10- 11

10° 7xl0° 7x10- 2 1x10- 4 7x10- 6 7x10-s 7x10- 10 7x10- 12 7x10- 14

10 1 - 7x10 1 7x10- 1 7x10- 3 7x10- 5 7x10- 7 7x10- 9 7x10- 11

10 2 ·- - 7xl0 2 7xlo 0 7x10- 2 7x10- 4 - 7xro~ 6 7x10-s

10 3 - - - 7xl0 3 7x10 1 7x10- 1 , 7x10- 3 7x10- 5

10 4 - - - - 7xl0 4 7xl0 2 7xl0° 7x10- 2

10 5 - - - - - 7x10 5 7xl0 3 7xl0 1

10 6 - - - - - - 7xl0 6 7x10 4

Table A-5. When repelling bodies are free to move, they accelerat~ because the force continuously acts between them. Acceleration values are easier related to experience like acceleration of a car, train,~or a bullet shot.

Figure A-31. This is a log-log graph with the distance on the horizontal axis, and a on the vertical axis. For reference, the acceleration of free fall, a=9.81 m/sec2

, and the acceleration of the bullet in a shotgun, 200 km/sec2

, are represented by dashed lines. Also displayed is the line of "blocks touching".

The floating granite blocks calculated in Table A-3, are on the graph where the blocks touching, the free fall, and the 1.5 m lines intersect. The acceleration can be enormous if the blocks are large enough. When the continent-size blocks of 1000 km edge length are set free to move from a side-by-side position, they start moving apart at an acceleration greater than that of a bullet shot. At the 7x106 m/sec2 acceleration they would travel 7000 km in the first second. Their repulsion force is very high even at 10,000 km distance, causing the blocks to accelerate at 70 km/sec2

•

.§


Physical Model

The near-perfect geometrical pentagonal pattern I obtained after removing the zigzag component, suggests that the acting force between the continents is very powerful. It moves the landmasses into those positions against inertia, friction and other possible obstacles. If the force is really an electrical one, then the first question is how can such force "grip" a continent?

The Earth's surface is about 70% ocean, and 30% continental landmass. At the bottom of the ocean there is a very thin, solid layer of rock, the "ocean floor", of only 5-km thickness, which separates the red-hot fluid

I I I acceleration of a bullet 200 km/sec2 -~1~-~r~-~-~---Blocks touching _.- _.---103 -l------l-~------b--=-------1-"""""----l------'"'-'---~--"~---I

'E 10 -3 -'--~--'---..,,.,_~ _ ___,__

]

10 -15---1----1----"'-l------l-~"-------l'-----""'-l-----l--.3.-- Distance 1 m 10m 100m 1km 10 km 100 km 1000 km 10,000 km

Figure A-31. Graphical illustration of acceleration v. distance, as in Table A-5. For comparison with everyday experience, the acceleration of free fall, and a bullet shot, are shown in dashed lines.

magma surface from the water. The ocean floor does not support itself, it is floating in the magma. The continental landmasses are similarly floating, and they are submerged to about 90% of their 50-km thickness. This is similar to a large frozen lake covered by a thin sheet of ice. The continents are represented by ships embedded in the ice sheet.

The red-hot magma surface continuously operates as a thermionic cathode by emitting electrical particles. The particles carry kinetic energy that, in free space, would carry them to indefinite distances. However, the

Paper A 35

ocean floor and the continents form a solid-layer envelope, the crust, that prevents the particles from freely escaping into outer space. Thus the particles are forced into the crust by their own kinetic energy and also by those particles that are following them. There is also a filtering action at work in which only free-electrons penetrate the crust. Then they propagate upward in the crust and infuse it in its volume.

The continental landmasses are much thicker than the ocean floor, by about ten times. However, on global scales, both of them are very thin with respect to their lateral expanse. For example, the Atlantic Floor is about 5000 km wide, thus its relative thickness is 1: 1000, or 0.1 %. Similarly, the continent of North America is also about 5000 km wide, thus its relative thickness is 1:100, or 1 %. From the magma cathode's point of view the crust is only a thin film with some minor variation in its thickness. Thus, the cathode's rate of emission is independent of the local thickness of the crust. Consequently, free electrons infuse both the ocean floor and the continents at the same rate; the same number of electrons enters them per meter2 per second.

The different thickness becomes significant when we compare conti- -nents to the ocean floor, because the solid mass of rock acts as an accu- , mutator of free electrons. Consequently, electron accumulation is much greater in continents than it is in the floor. Electrons travel relatively slowly in the crust, it may take several days or weeks to move through (while in metals it would be only a few microseconds). Thus, while they are still in the crust they constitute a volume charge. Consequently, the total charge per unit surface area of crust is ten times higher when it is a continent than when it is an ocean floor.

Therefore, in the first approximation, the electrical. role of the ocean floor can be ignored. In this view the physical model of the Earth is that there are six continents floating in a liquid on a spherical surface. The continents are charged in their volume, thus they exert Coulomb's electrical force on each other. This is very similar to a pith ball electroscope in which the pith balls are moving along the radius of their suspending threads, thus they are like moving on a spherical surface. But the earth's gravity also acts on the balls, from an outside direction, and it makes this electroscope a little complicated. For the continents the role of the thread is played by gravity, thus no force acts on them from the outside. This is a clear and simple gravity centered terrestrial electroscope.

One can visualize a real pithball electroscope with six balls suspended on six fine threads, and placed in a gravity-free environment like on a Space Shuttle. Then, when charged, the balls would move apart along a spherical surface as each ball exerts a repulsion force on all other balls .. In fact, I was considering building such a multi-ball electroscope myself to

· operate in my room. If the applied charges were large enough, then the outside effect of gravity would become negligible. A high-voltage power


supply, like a Van de Graaff generator, could be the source of charges. However, I decided to go ahead instead with a computer based mathematical model. A science museum may wish to build such a physical model in the future.

Since gravity does not effect the surface movements of the continents, it can be ignored. What is left is the simplest possible model: an all-ocean planet on which ships are floating. The ocean's water represents the liquid magma surface, and the ships are representing the area centers of continents. Then electric charges are placed on the ships that would freely move wherever the repelling forces direct them.

The electrical force grips the continents the same way as it grips the pith balls: by the fact that the propagation speed of electrical volume charges in semiconductor (insulator) bodies is slow. Imagine, that there were only two electrons involved: one inside the body, and one outside. The force acts only between them. However, one is confined in an electrically neutral body (for the duration of its transit). Thus when a force is exerted on it, the electron tries to move accordingly. For a small electron there is always space in solid matter, to move. As it moves it collides with the nearest atom in its way. This is a mechanical transfer of kinetic energy; the electron makes the entire body move. A sail boat moves by the same mechanism; the kinetic energy of moving air molecules is transferred to the entire boat when air molecules collide with the sail.

Where would two ships move on an all-ocean planet? I made my first physical model by taking a ping-pong ball, and I drew two small dots on it, side by side, in pencil. The dots, or ships, represented the area centers of two continents. Then I applied electric like-charges on them, in thought, and I watched them moving. Where would they go? They certainly would not move forever. They finally would come to a certain balanced position where they stop moving. It was not very difficult to find these points. They were the antipodes, two points at ±90° from the starting point, 180° angular distance apart.

Then I placed three ships side by side on the ping-pong ball, and watched them move. It took only a little longer to visualize that they moved to 120° angular distances apart on a great circle, and·stopped there permanently.

The advantages of this modeling can already be seen. The model is independent of the "absolute values" of the applied charges, and also independent of any "constants" in the equations. When the ships carry twice or ten times as much charges, they regardless move to the same relative positions. In mathematical terms "the constants drop out of the equations", as multipliers they take the value of one. The main constant is kin Equation (A-1), in Coulomb's electrical force. Another constant is the

\

(

Paper A 37

radius of the model. A triangle is a triangle on any size of spherical body, planets included.

With the three-ship system the easy time came to an end with the pingpong model. It took me a great effort to tell where four ships were at their final positions (you may wish to try it yourself now, without the benefit of hindsight). I felt that their positions somehow should be related to multiple antipodes. First I held the ball at two antipodes with two fingers of my right hand, and I placed two fingers of my left hand at the other two antipodes at right angles to the first pair. This resulted in four points at 90° apart on a great circle. This arrangement met the force balance

Figure A-32 Geometric obJect formed when the positions of four mutually repelling bodies on a spherical surface are interconnected by straight line~.

requirement, but I realized that it would "buckle" since there was plenty., of empty space available in the two hemispheres. Thus, I let them buckle. I rotated the two pairs away from each other into opposite hemispheres. As I rotated them, the distance between the ships in each pair decreased as they moved on meridians, and the distance between the two pairs increased as they moved deeper into opposite hemispheres. Then, when the distances between all four ships were about equal, I stopped. In this form they were again in force balance, and they were buckle-proof as well.

How can I identify this arrangement? I felt that this must be some fundamental form in Nature since it obeys a very simple rule (balanced forces) on a very simple object (sphere). First, I tried to put their four pairs of coordinate numbers into tables like a mathematical matrix set, and find some characteristic pattern between the rows and columns. Then I tried a geometrical form by interconnecting the points with lines on the ball's surface. Then finally I interconnected the points with straight lines through the bulk of the ball. This looked promising; I made a model of it from thin wooden sticks. Indeed, I obtained the simple object of Nature I have been looking for (Figure A-32). I thought it was beautiful. It used the minimum number of uniform elements to define a three-dimensional space.


However, I could not name it, I did not remember seeing this object before. It reminded me of the children's swing set we had in our back yard, except that this would tip over for having its legs closed. I even tried to coin a name for it, although I expected that all elementary geometrical objects have already been discovered and named by ancient Greeks.

Mathematical Model

Next, I placed five dots on the ping-pong model to represent five repelling ships on the all-ocean planet. But I could not possibly find out where their force balanced positions would be. Thus, I constructed a mathematical model (but that took years). There are two main forms of mathematical modeling. One is a set of equations that gives the final positions of the ships when the equations are solved for force balance (like Equ. A-4 further down). This method has one great disadvantage if more than three objects are involved. It is difficult to know whether the result is correct or not, especially when one does not know in advance what the result should be. Imagine that you have entered the positions of, say, six ships, that is, 12 coordinate points. Then after days or weeks of strenuous work (in which there could be errors), another 12 coordinate points are obtained, and they are scattered all over the spherical surface as the final positions. How do you identify their arrangement? Furthermore, the output points will have lost their identification with the input points, thus you would not know which ship has moved where. This method is for teaching the subject in hindsight.

I decided on a dynamic, computer based mathematical model which would give me intermediate positions of the moving ships any time during their travel. The ships are initially placed somewhere on the surface, side by side, for example. Then repulsion forces are applied, and the ships start moving in the direction of the force. After a short travel their positions are displayed on the computer screen in a graphical form, in the 6-view engineering projection. Then another short traveling starts, and another display. This is called "animation", and if the steps are short (and the computer is fast enough), then the movements appear to be continuous.

The ships leave behind a trace of their path on the screen. This allows the movement of each ship to be followed, and to observe any anomalous move. In a research project such anomaly could lead to discoveries of unknown behavior of objects or mechanism. But more often it reveals programming errors like division by zero, a numeric overflow, or an incorrect trigonometric sign after crossing the 90° -point, especially in the tangent function. On the other hand, the absence of anomaly gives a great comfort and confidence in the proper operation of the model when a reasonable result slowly unfolds.

Paper A 39

The continuous display also allows the comparison of the model with the real movements of continents. If the principal idea of the model is correct (the repelling nature of the continents), then the model would properly describe the "drifting" of continents (as geophysicists refer to continental movements). There is a number of observed behavior of the continents in their movements that should also be reproduced by the model. For example, it has been shown that certain continents once were part of a common landmass, and they broke up since then. Such a breakup mechanism should be an implicit nature of the model that operates only when some special conditions are met.

The model also should not do certain things. For example, when two ships move apart from adjacent initial positions, they would be approaching each other without ever turning back, when they have passed the antipodal positions. This should not take place in a system where repulsion forces are acting, unless the program's logic is faulty.

Eventually I developed a model to run up to eight ships simultaneously. First I applied two or three ships for which I knew the results by simple reasoning. Then, when my confidence was high enough, I added only one more ship at a time, and watched for new results. The model is based on Coulomb's equation,

F= 01 ~ (A-3) d2

without the constant k, and it is applied on bodies whose movements are confined to a spherical surface. An individual repulsion force is generated between each two bodies along the straight line interconnecting the bodies, in three-dimensional space. Then all forces on each body are vectorially summed, and the bodies are allowed to move in the direction of the local horizontal force components.

My model ignores friction between the ships and the fluid medium, and it also ignores inertia. The justification is that the real forces in the Earth are expected to be so great (see Table A-4) that the effect of friction must be negligible. Furthermore, fluid friction always approaches zero at decreasing velocities anyway. My model investigates final force balanced positions reached at velocities approaching zero. Inertia causes overshooting and oscillation around the final positions, but it does not prevent the continents from reaching these positions eventually. The diagrams of Figures A-33(a) and (b) describe the movements of two ships and their force vectors. In ( a) the two ships are placed initially at positions A and E. When like-charges are applied, Q1, Q21 the ships start repelling each other in the local horizontal directions, represented by the vectors Fh. The magnitudes of the forces, represented by the length of the arrows, are very high because dis very small in Equ. A-3. The ships move with respect to


each other, and at equal accelerations and velocities. The acting force does not affect the globe itself, it does not impart any momentum to it.

As the ships move apart, their driving force decreases for two reasons. For example, at the midway position of B and G the ships are at d distance apart, thus one reason for a smaller force is the greater linear distance at which the force is down by a factor of 1/d2. The other reason is that the direction of the force on the ships are not horizontal anymore, they come from underneath at an angle. The absolute value of the force is expressed by the vector Fa which points somewhere skyward. This vector can be broken into local vertical and horizontal components, Fv and Fh. Only the horizontal components drive the ships. The driving force is down by another factor, by cos(a/2).

Figure A-34 illustrates the force factors as the function of a angular distance between the ships. These forces vary differently with the angular distance. The linear distance factor decreases rapidly first, then slowly, while the angular factor decreases slowly first, then more and more

(a) (b) Figure A-33. The absolute positions of two ships can be calculated by the symmetrical diagram (a). When only their relative positions are investigated, one ship can be kept fixed as in the simpler diagram (b).

rapidly. They are both multipliers of the absolute force vector, thus if either of them becomes zero, then the force itself becomes zero as well. Indeed, the angular factor becomes zero at a=180°. At this point the driving force totally disappears from the ships, even though the vertical force may remain significantly high. This is the mathematical form of the simple reasoning that two ships move to the antipodal positions and stop there.

Further inspection of the diagram reveals that the angular factor not only decreases sharply to zero, but it switches sign at crossing the 180° point. When the absolute value of this factor is viewed, (the minus-sign side is mirrored into the plus-sign side in dashed line in Figure A-34), its ditch-like shape becomes very visible (see the wheel at the bottom of the ditch). In fact, this is how it works. When a ship (or continent) arrives at

Paper A 41

the 180° position the driving force ceases to act on it. However, when an external force (its inertia, wind, or a deliberate action) attempts to remove it from this position in any direction, then an electrical force arises which tends to move it back to the 180° force-balance position. But no force or energy is needed to keep the continent in the force-balance position. This is like when your car gets into a ditch, no energy is needed to keep it there (see the small wheel drawn symbolically at the bottom of the ditch). Energy is needed only when you want to remove it. Then the ditch resists the removal. The deeper the ditch the greater the energy is which removes it.

What controls the depth of the electrical ditch for a continent on a planet? The absolute value of the repulsion force. The greater the force the deeper the ditch is. The significance of the enormous forces calculated for

10

--- -as 'I... ..J

'\ ' ,., / /

'\. ' I/ ,I a,

1' '- / , ... Q4

d2 '

' ,. , Q2

-' '-

cos0 '-

' 2 '

-0.4

.Q6

" ...... ~ -as

-10 .....

0 fl !is iii!;!~~ !i! !! ~ ~ ~ ~ ~ !;l !.'l ~ ;i!O

Figure A-34. The two multiplier factors of the repulsion force vary differently with a. The 1/d2 factor rolls over smoothly at 180° with a non-zero value, while the cos a/2 factor drops sharply to zero, and changes sign after.

the Earth's continents in Table A-4 is that the continents sit in very deep electrical ditches. The slope of the ditch is not uniform; it, follows the cosine function. The closer you get to its node the steeper it is. Thus it is more accurate to say that very great repulsion forces result in very steep ditches.

The electrical ditch does not exist without the continents. In fact, the continents themselves dig their own electrical ditches. Nevertheless, it is easy to calculate the force that drives the continent down into it, or the external force needed to lift the continent out of it.

Consider two Africa-size continents (as in the example on Page 30) volume = 1018 m3, mass =

3x10 21 kg, that are very near to their antipodes on an Earth-size globe, d = 1.27x107 m. The absolute value of the force between them

F. = 9x109 x (1.6x10 15 )2 I (1.27x107)2 = 1.43x1026 newtons

Let's suppose that the ships are still at 1500 m away from the antipode positions. Then


sin ~ = 1500 I 6.37xl0 6 = 2.35xl0 4

where ~ is the angle between B and C subtended from the center in Figure A-33.

~ = asin 2.35x10 4 = 0.0135°

a= 180°-2~ = 179.973°

cos(a/2) = 2.35x10· 4.

Fh = F. x 2.35x1Q·4 = 3.37x10 22

a = Fh I mass= 3.33x1022 I 3xl0 21 = 11.2 m/sec2

If you pulled these continents away from their 180° positions by 1500 m, and then released them, they would accelerate back to their balanced position at 11.2 m/sec2

, about the same rate as if they were free-falling under gravity (9.8 m/sec2

). This is a very great restoring force only at 1500 m from their force balance position. The example suggests that effective restoring forces may exist within meters of the actual force balanced positions of continents.

The steepness of the electrical ditch defines the stability of the continents against drifting. Further down I will analyze the observable drifting of continents in terms that are compatible with this high stability value. For larger number of continents, like those of the Earth, the stability values are significantly tighter, down to a few centimeters.

The ships in Figure A-33( a) arrive at their force-balance positions at 180° angular distance apart at points C and H. The horizontal · force component completely disappears, and the vertical component becomes the same as the absolute component. The ships are firmly held here by an electrical force that arises only when an external force tries to displace the ships in any direction.

The two ships behave exactly identically, and all forces are the mirror images on the two sides. It is redundant to deal with both sides for the present purposes. In Figure A-33(b) one ship is fixed at position K, while the other ship is let to move freely. All linear and angular distances and the forces are the same as in ( a). I use the (b) format in my computer model because it cuts down the number of calculations by half. Since mutual repulsion forces do not impart momentum to the globe, the relative positions of the ships are "floating" on the globe, they are not tied to it. Their absolute positions depend only on the ships' initial positions.

Characteristic Patterns

I repeated all my previous experiments on the ping-pong ball, with the computer model. I equipped the program with a graphics subroutine to interconnect all adjacent final positions with thick straight lines to give them a geometrical interpretation. This feature has become vital in classifying and drawing general conclusions of the final, force-balance positions, thus I named them characteristic patterns or patterns for short.

Paper A 43

Figures A-35 and 36 illustrate the two and three-ship systems with their characteristic patterns of straight line, and triangle, respectively.

In the two-ship system the ships can carry either equal or different amount of charges, but they nevertheless move to the 180° positions. The introduction of three ships reveals a new property of the model. The ratio

Figure A-JS. In a two-ship system repelling bodies move 180° apart. . .. _

of charges becomes relevant. The ships move to equal distances oh the great circle only when they carry equal amount of charges, that is, their charge ratio is 1, or more correctly 1:1:1, since there are three possible ratios in the same system. At other charge ratios their angular distances become a function of the charge ratios. For the next analyses I fix all ratios to unity (but I return to the non-uniform ratios in a later study).

Next, I placed four ships on the model. The fourth and more ships introduced three-dimensionality and uncharted waters. As usual, I tried to predict results. It is often said in science that researchers should be

Figure A-36. In a three-ship system equally charged repelling bodies move 120° apart on a great circle; they form a plane triangle.

unbiased about future test results. This is not really possible even if one wanted it, because the test equipment must be prepared for a predetermined range of results both qualitatively and numerically. You must expect something, and prepare your equipment for that. The most you can


do for non-bias is to watch for "anomaly", a deviation from the expected, and not to dismiss or explain it away.

I prepared myself for two possible scenarios. One, that a given number of ships always move to different final positions from different initial positions. This would be a random mechanism, and not many conclusions

Figure A-37. Four ships always form a triangular pyramid (tetrahedron).

could be drawn from it. In this pessimistic or skeptical scenario my already observed pentagonal pyramid pattern would be a chance configuration. Zigzagging would not really be an independent and removable feature; it would be part of inherent randomness. That would be similar to constellations where you cannot find a force that would move stars to form them.

The other scenario is that a given number of ships move always to the same relative positions from any initial positions, and would always form the same pattern. This would be a highly predictable scenario since the number of continents would determine the planet's characteristic pattern. My computer model was to tell me which scenario is at work.

My first results with four ships were difficult to interpret. A line drawing of a 3D object on a 2D computer screen does not say which line is in front or behind another line, thus the object does not appear to be 3D. I solved this problem by rotating the viewed object around its 3D center, thus when points in front moved to the right, points in the rear moved to the left. This helped perception to interpret the 2D line-drawing as a 3D object. (I applied the technique of 'hidden line removal', used in the present illustrations, only much later. You may decide to use some new technique now Figure A-38. An example for

available, like solid modeling). switching visual perception.

I

'

Paper A 45

I noticed that I sometimes got different patterns although I always used four ships. It appeared that the ships form one of a few possible patterns. To my disappointment this implied the possibility that with further tests

Figure A-39. Five ships form a double triangular pyramid.

the 'few' would grow greater and greater, and eventually the constellation scenario would win. The point in question actually centered on my favorite 'swing-stand' model, which showed up occasionally, and I noticed that it depended on the selection of initial positions of the ships. Normally, the display indicated another simple (and well known) geometrical object: a triangular pyramid or tetrahedron. Eventually I solved the problem by

Figure A-40. Six ships form a double rectangular pyramid.

making a physical model; both displays were the same tetrahedron. I unintentionally rotated the object on the screen by certain selection of initial positions. Then viewing the result was a case of "switching perception", like in the old example of "two faces or one vase" (Figure A-38). The brain attempts to give meaningful interpretation to a poorly defined picture, and it switches from one interpretation to another. The same object looked to me a triangular pyramid in some views, and a


swing-stand in other views, while in fact Figures A-32 and A-37 illustrate the very same object.

The swing-stand affair may look like a minor fiasco, but in fact it was a blessing in disguise. By the time I solved it I had been experimenting with larger number of ships, 5, 6, and 7. However, I only rarely could

Figure A-41. Seven ships form a double pentagonal pyramid.

interpret the displays. There were dots all over the screen which did not make much sense even if I interconnected adjacent points. The swing-stand affair asked for the simple solution to designate one point on the object that would always be on the top of the display, regardless of rotation of the initial positions. I designated one comer of the pyramid for the fixed position. Eventually I restricted this rule to the top comer (apex) of the pyramids, thus they stood always upright, regardless of initial positions,

# of points Characteristic Pattern

1 point

2 line

3 triangle

4 triangular pyramid

5 double triangular pyramid

6 double rectangular pyramid

7 double pentagonal pyramid

Table A-6. All objects formed by repelling bodies on a · spherical surface belong to this simple geometrical set.

Paper A 47

and the swing-set has never showed up again. This procedure does not violate the purpose of the model which is to find and interpret the relative positions of the force balance state. No rotation of the entire object changes that.

The fixing of the object's orientation allowed me to realize that all objects are pyramids when the number of ships is four or more. As Table A-6 illustrates, all objects fit into a single system of elementary geometrical patterns. Figures A-39, 40, and 41 illustrate the rest of the patterns, the double-pyramids. These are the practical objects that occur under realistic conditions where stability must be considered. My computer model, being numerically very accurate (16-decimal places), also produced the non-stable patterns: the plane rectangle, pentagon, hexagon, and heptagon, and all single pyramids above the triangular one. These all buckle into those in Table A-6, upon the slightest misalignment.

Thus the "single pentagonal pyramid" (Figure A-25), which has led me originally to the idea of repulsion forces, has also fallen victim to my computer analysis. I did not let it die easily as a viable explanation, but I had to. My computer model strictly rejected it as totally unstable. The six ships, which were supposed to form it, never stopped at the required positions. They kept moving on and always ended up forming a double rectangular pyramid. It was a tough problem. I spent years in searching for an explanation that was acceptable both from the geophysical and from the electrical point of view, as to why the computer model does not approve it while the continents do appear in that form. Here it is: the Pacific Floor acts electrically as the seventh continent, thus the complete characteristic pattern of the Earth's continents is a double pentagonal pyramid, Figure A-41. This may not be obvious at this point. I will return to it later in this study.

Continents of Mars

So far I have described my models from the basic two-ship system to the seven-ship dynamic computer model. Between these extremes I found a good reason to have a closer look at the three-ship system. When I realized that all repelling systems belong to the "simple geometrical set" listed in Table A-6, I speculated that a planet other than the Earth could be an independent evidence if it fell into one of the categories. These were the years of the mid 1970's, with the Viking mission to Mars, but to my knowledge at that time no such map had been published yet. Then, during a casual visit to a geological library, I found one by chance. "Martian topography derived from occultation, radar, spectral, and optical measurements" by E. J. Christensen, in the Journal of Geophysical Research, Volume 80, pages 2909-2913, in 1975. The article contained a map showing a vertical range of surface elevations of 13 km, and contour

.g ~-

~

-60 180

,.--- .~ /

150 240 210

LONGITUDE, deg

Figure A-42. Martian surface topographical map redrawn from a published paper (see text).

180

~ ,'"'-,

" "' .. a

&

[ Cl)

~

i !:I.

~ ,§ ~ "' ~-

I

229.24 ° ( observed)

f3=234.33° ( calculated)

144.19° (observed)

w '

so

40

30

20

10

0

-10

-20

-70--ll-~---,-~--90 --00 -30 30 w 90

-105.42° 38.77° 123.82°

A B C

150 180

LONGITIJDE

"ti ~ ~ ::t,.

Figure A-42. Martian surface topographical map redrawn from a published paper (see text). I t


lines were drawn in 1 km resolution (Figure A-42). On the Earth the vertical range is about 20 km filled halfway up with the water of the ocean. The water does not control the size and position of the continents, but it conveniently delineates them. The halfway delineation gives the most accurate results. If the level was too low the entire surface would form a single dry land with a few lakes. A too high a level would show a single ocean with a few small islands. On the map the highest elevation is at + 10 km, the lowest one is at -3 km, thus the halfway mark is at +3.5 km.

I hypothetically inundated the Martian surface with water near to the halfway mark, at the +3 km contour lines. I expected that ancient Martian continents could be found by such inundation. Indeed, three large areas remained above the water (Figure A-43). Their positions revealed for the first glance that mutual repulsion forces had positioned them. All three area centers were on the same great circle as in the three-ship model. By this time I had evidence (from my geomagnetic studies) that the total charge in a continental plate is linearly proportional to its surface area. This was also evident on the map: the two largest continents were farther apart than the two smallest ones because large continents repel one another more than small continents according to Coulomb's law (Equation A-3). In details, however, I had to slightly modify the original map. A terrestrial example for this need is shown by the connection between the continents of North America and South America. Some maps show a dry-land connection, others show that the area is a swamp. In fact, some years ago a British military unit tried to cross it by dry-land vehicles (Landrovers) as an exercise. But they ran into difficulties because the water level was a few meters higher than expected. Obviously, lowering or elevating water levels does not change the status of continents. My other example is the boundary between Africa and Eurasia between the Red Sea and the Mediterranean Sea. This is a strip of dry land about 150 km wide. Thus, continents can be in dry-land contact. Therefore, I decided that the landmass between the 280° and 190° longitudes, -5° to -45° latitudes, is a separate continent, and I cut its two isthmuses.

I also added arbitrarily an area to the largest landmass below the limit of the map at -60° latitude. The 3-km line goes below the -60° latitude at 92° and 175° longitudes, thus an extra area belongs to this continent. The map does not show it, but it would be a greater error ignoring this area. The area in question is small, only Mercator projected maps show it large in this arctic region. I also converted the 0° to 360° scaling of longitudes into 0° to ±180° to conform with my Earth's longitudes, thus the same equations and computer programs could work on both.

In my first attempt to verify numerically the applicability of Coulomb's equation to Mars, I re-plotted the Mercator map onto a blank globe, and graphically measured the outlined areas by overlaying a fine square grid,

Paper A 51

and counting the number of squares. Then I determined the area centers by paper cutouts, and I measured the angular distances between the area centers. These were the "observed" positions. Next, I ran the computer model with the measured area ratios. I found that the model virtually did not move, indicating that the continents were already in their force balanced positions. Conversely, when I placed the bodies at much different initial positions, then they moved back to these positions.

FcA Figure A-44. Vector diagram of a three-body system to prepar~ ,a set of equations for the force-balanced positions. -· ·

QA Qccos (/3/2) QA Q9 cos (a/2) __ 0 ~~~~~~ + ~~~~~~

2 (1-cos /3) 2 (1-cos a)

QA Q9 cos (a/2)

2(1-cos a)

Q9 Qccos ((/3-a)/2) = 0

2(1-cos (/3-a))

;''c''

(A-4)

I considered this result an independent evidence, besides that of the Earth, for the general planetary validity of repelling-force positioning of continents. However, I soon realized that for such a bold proposal an 1800-line long computer program is not suitable as 'proof'. In other words, nobody would believe that the program is showing the truth, and

52 , Underground Geophysics

furthermore, nobody would take the time ( at this stage) to verify the program's integrity.

Therefore, I decided to work out a simpler proof of mathematics. The three-body system lends itself for such simplicity because the three positions at the force-balanced state define a single plane. Thus, very simple planar trigonometry can be used to calculate the final forcebalanced positions. It would be a set of two simultaneous equations for two unknowns, an easy undergraduate task. The equations could be easily verified by anyone with basic mathematical background.

The calculations are based on the diagram in Figure A-44. The results are similar to that of Figure A-33, except that three bodies are used. The bodies are A, B, and C, which carry Q» QB> and Qc charges. As I already pointed out only the ratios of the charges are needed, thus all three values are normalized to the smallest one. The bodies exert repelling forces on each other in the straight directions over DAB, DBc, and DCA distances, and generate F AB=F BA , FBc=F cB , and FAc=F CA forces, proportional to the product of the charges and to the inverse square of distances, as in Coulomb's equation. The bodies are allowed to move only along the circular path of unity radius, thus only their local horizontal vectorcomponents, designated by the subscript h, come into account. There are two such force components on each body, acting in opposite directions, thus their algebraic difference becomes the effective driving force. The bodies move until these forces all become zero, FAch+F ABh=O, FBch+FBAh=O, and FcAh+FcBh=O.

The final positions are expressed by two angular distances, a and f3, measured from A, thus we will have a set of two simultaneous equations (Equation A-4, see its derivation in Appendix A3). Because of the trigonometric functions involved, this is a transcendental set, thus a method of successive approximations should be used to solve it. Several commercial programs are available for this purpose. As it turned out, Equation A-4 has revealed several very important features of the repelling system related to its nonlinearity.

To visualize the operation of the three-body system I have calculated five test cases shown numerically in Table A-7, and Figure A-45 illustrates the resulting characteristic patterns. #1 is an equilateral triangle produced by equal charges, the same as in Figure A-36. In #2A is twice as powerful as B or C, thus it repels them farther away from itself. Consequently, B and C move closer to each other in spite of their own repelling charges that remain the same. A cursory observer may notice two continents moving toward each other and would consider it a "contradiction" for a repelling system.

Since only the ratios of the charges are relevant, the same effect occurs if the charges of B and C decrease simultaneously while the charge of A remains the same. Furthermore, instead of being pointlike objects as in the

Paper A 53

model, the bodies may have lateral expanse as real continents have, which may touch each other at the edges. Here the point-like objects represent the area centers ( or center of gravity) of the objects. Geophysicists call this event a "collision of continents", and may mistakenly consider it as

#1 #2

240.0"

#3 #4 112.s·

C

#5

Figure A-45, Graphical illustration of the characteristic patterns of five test cases.

counter-evidence. As this example illustrates, small continents can be pushed together from behind by one or more larger continents through repulsion forces acting at a distance.

In #3 all three charges are different; the distance between the two strongest ones is the greatest, and between the two weakest ones is the shortest. This is the feature that is observable on Mars at the first glance.

#4 is an extreme case where one body is much stronger ( a thousand times) than the other two, and it repels them to the opposite side of the circle. However, the nonlinearity of the system comes into effect in which


the distance between B and C cannot become zero even if the charge of A becomes infinitely large. (For non-pointlike bodies Band C represent their area centers, thus they still can collide at the edges).

The nonlinearity is caused by two different factors. One is the cosine function that approaches zero near to 90°. For example, in Figure A-44 the local horizontal driving force FBAh on Bis proportional to FBA x cos(a/2) where a is approaching 180°, or a/2 is approaching 90°, thus cos(a/2) and

# QA QB Qc a /3 1 1 1 1 120.0° 240, 0°

2 2 1 1 12 9. 2° 230, 8°

3 4 2 1 145. 6° 238. 9°

4 1000 1 1 172.2° 187,2°

5 1000 1000 1 179, 8° 269. 9°

Table A-7. Numerical values for the five test cases, as in Figure A-45. Because of nonlinearity, charge ratios as small as 2:1, or as large as 1000:1, result in substantial displacement.

also FBAh would become zero at 180° regardless of the value of FBA. The other factor is that as the angular distances of B and C from A are nearing 180°, their own distances from each other are nearing zero. Thus, however small their. own charge is, their mutual repulsion force is approaching infinity according to Coulomb's law of F=QB Qcf tf , where d is their distance that is approaching zero. These two tactors counteract and negate the repelling force from A at an observable distance from the exact 180° position.

#5 further tests the nonlinearity effect to see how much a small object (a small continent) can influence the positioning of the entire system. In this example both A and B are extremely large, a thousand times larger than C. In a linear system one would say that the effect of such a small body is negligible, and the system defaults to a two-body system in which the large bodies position themselves at 180° angular distance from each other. Furthermore, the hapless small body is simply pushed aside as far as possible, to the halfway point between the large ones. But not in a nonlinear system. As the numerical -example shows the tiny small body

· noticeably repels the large ones, it forces them to be at 179.8° (179.8382859°) angular distance apart on the far side, and at 180.2° (180.1617141 °) on the near side instead of the expected exact 180° distance. This is not inaccuracy in the calculations.

There are three factors involved in the enhanced capability of the small body. The first is that the large ones have rapidly decreasing local horizontal force-components for each other as they approach the 180°

. (

(

Paper A 55

positions. The second is that the small body is positioned (by the large ones) at half-way between the large ones. Thus, each of the large ones are at an effective 45° angle from it; its own repulsion force is not much reduced by the cosine of the angle, only by 1/v'2=0.7 times. Similarly, its distance from the large ones is shorter, by-V2, which translates into a force factor of 2 because of the inverse square term of the distance in the equation.

An area ratio of 1000:1 is well over any practical consideration, it only serves to show the existence of the effect. However, the same effect becomes stronger for smaller ratios. For example, at the area ratio of 100: 1 the A to B distance on the far side is 178.41 °, and for 10:1 it is 166.53°. As I will point out further down, this effect is clearly observable on the surface of the Moon.

These results warrant a closer look at the delineated Martian landmasses. I have re-measured the area ratios and area centers as accurately as I could, using high resolution digitizer equipment. Table A-8 lists the values I measured on the map, and the result of subsequent calculations from these measured values. I also numerically compare observed and calculated values. I call this number the degree of agreement between theory ( calculation) and practice ( observation). This degree is frequently used in science to evaluate theories according their ability to "predict" future observations in physics. Simply put, my theory says: tell me the area ratios of continents, and I will tell you the relative positions of those continents. For the Martian continents the accuracy of my prediction is better than 90%. Figure A-46 is a visual illustration of the comparison.

The concept of "continents of Mars" is a new proposal in science; In other words, my theory's ability to predict accurately the positions of large landmasses, amounts to the discovery of Martian continents. Furthermore, the existence of Martian continents is an independent planetary evidence, besides the special positioning Earth's continents, for electrical repulsion forces in large landmasses. This independence potentially expands the scope of the discovery into planetary environment in general. That is, it should be applicable to all planets.

To further explain the idea of Martian continents, I have applied the six-view projection technique (Figure A-47) according to the definition in Figure A-10. Thus the center line of the Front View is the 0° longitude (at the west-end of Continent B, where Martian Greenwich could have been located), and the Northern Hemisphere is in the Top View. I included the spherical grid because on Mars the E-equator almost coincides with the rotational Equator, thus it does not cause any misunderstanding (as it would in Figure A-23 for the Earth).

The Martian globe gave me the first hint about the role of the Pacific Floor as the electrically acting seventh continent of the Earth (see Figure A-41). In a three-ship computer model the E-equator always coincides with

56 Urulerground Geophysics

Martian Data

Observed Values

Area QA:Q 8 :Qc = 4,74sl.83sls00 ratios

Area ALON= -105.42° ALAT = -13,88° centers BLON = 38. 77° BLAT = -13 .19°

CLON = 123,82° CLAT = -24. 99°

Angular A to B = 144 .19° distances B to C = 85. 05°

C to A = 130. 76°

Calculated Values

Angles a= 146,42° p = 234.33°

Angular A to B = 146.42° distances B to C = 87.91°

C to A= 121. 67°

Comparison Agreement

Angular A to B, 144 .19° v. 146.42° 98% distances B to c, 85 ,05° v. 87,91° 97%

C to A, 130. 76° v. 121. 67° 93%

Table A-8. Observed and calculated values from the Martian map, and their comparison. Note the high degree of agreement between calculated and observed values.

(

(

(

(

Paper A

OBSERVED POSITIONS

. 0

~ ~ 00 R & ~ ~ 0 ,... .... ~ '£; .....

~ "i ~ lib ".;; '-1.;,o ..... ,,~

'-l..?o .... '),~ C 123.82 ° -1.1.

-105.42°.A o -100 100

-90 90

-80 80

:10 7o ,f.)~ ./</</

b~ ·./.90

~ ' ,:, 'b 75 fl ~ 0 ....

7 0 0

CALCULATED POSITIONS

~ 0 R 00 & ~ 0 .... ....

..... ~ ~ 1::- "i

~ lib '-1.;,o ".;; ~,,~ C 128.91 °

'-l..?o .... '),~

-10sA2°A ·J.1.0 \\0

-100 100

-90 90

-80 80

.10 7o

-'"~ 60

0 .... 0

Figure A-46. The equatorial plane of Mars from above the North Pole. In the floating CALCULATED POSITIONS diagram the area center of Continent A is pegged to its Observed Position to make visual comparison easier.

~ a

57

,,.--- ,~ -~ ,,--.. ~- ,.--._ ,,.--- ,--._ ,,--.\ ;- /~

Continents of Mars

Figure A-47. The complete six-view projection of Mars to illustrate the shapes and positions of its three continents named A, B, and C.

~

[ II>

~

i ;:,.

~ .g ~ c..,

~a

~ ,-----..._ ..._ ~ ,-...., ..._,

Paper A 59

a great circle of the globe. But on Mars (and on the Moon) this equator is definitely shifted to one of the hemispheres. The shift amounts to more than 10°, therefore, it cannot be an inaccuracy in the topographical mapping. My explanation is that the ocean floor acts as an active and powerfal source of repulsion force like a real continen~ but with one significant difference.

First, the similarities. Both are part of the magma cathode's solid-matter envelope, therefore both are infused with a net electric charge, and both exert repulsion forces on all other infused bodies (continents). The ocean floor is much thinner than the continents are, but its lateral extent is very large: it occupies an entire hemisphere on the Earth. Thus its volume and its charge are large, they are second only to those of the largest continent, Eurasia.

Now, the difference. The continents are embedded in the ocean floor, but they can move in it (when they break it up temporarily). Thus the floor is a background, like 'space'. When an object moves in space, space disappears at the object's new location, and reappears at its old location. But you cannot identify moving space, you cannot say that 'this space was there before'. The most you can do is to identify the center of a contiguous volume of space, or ocean floor, with respect to some objects nearby. If objects move in it, then the center may move, too.

On Mars there are two such ocean floors, the Northern Floor and the Southern Floor. Let's place the three Martian continents initially exactly on the equator between the two Floors. In this expanded view Mars is now a five-ship system, forming a double triangular pyramid. There are four different forces on each continent, and two of the forces , are from the ocean floors. All forces on each continent are in balance; none of the continents move. Now, break off a large piece from one continent (and disintegrate it by melting or corrosion) to deliberately upset the force balance. The continents immediately start moving to seek out their new force-balanced positions. In a perfectly symmetrical computer model all movements would take place in the plane of the E-equator. But on a real planet the ocean floor would not be symmetrically strong everywhere, it would be stronger in one region than in another. It would crack and break easier in one direction than in another. Therefore, as the continents move in the direction of the force in the equatorial plane, they may swerve ever so little from their course. They may slip slightly into one of the Hemispheres.

At the moment a continent slips into one hemisphere, the floor of that hemisphere becomes smaller and the floor of the other hemisphere becomes larger. The repulsion force from a larger floor is larger, and from a smaller floor it is smaller. Thus, the larger floor pushes the slipping continents deeper into the hemisphere of the smaller floor only to make it

90W 60W 30W O 30E 60E 90E 120E 150E 180 150W 120W 90W

SON . ·-A--. SON

30N 30N

0 0

30S 30S

SOS 1 >. -C::::::::: ·,.c: :::::;.,, I .C: Z L C C .LJ.?::bD ·::::,..., ., :\ > ',,. :::,..,· I SOS

90W 60W 30W O 30E 60E 90E 120E 150E 180 150W 120W 90W

Figure A-48. Lunar surface topographical map redrawn from a published paper (see text).

~~~~~~~~~~~~ ~~~~ ~-, ~ ~ ~ ,~.

. ..

~

[ <II

~ ~ 5.. ~ ~ ~ ;:!. r;i

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

250.65° ( observed)

~=260.08" (calculated)

162.70° (observed)_ a:162.97° (calculated)

~ 60 ---, ........ 0 ....

~ 50

. ...

~ :s 40

30

20

10

0

-10

-20

-30

-40

-50

-90 -60 -30 0 30 60 90 120 150 180 -150 -120 -90

-9.76° 78.19° -172.46° LONGITUDE

B C A Figure A-49. Areas above the 1738.05-km contour lines are delineated and shaded as continents. The separations between area centers are read from the map as observed angular distances. For comparison, distances calculated from area ratios are also shown.

~ {; ~ :i,..

~


Lunar Data

Observed Values

Area QA:QB:QC = 7.7817.2511100 -ratios

Area ALON = -172.46° ALAT = -13. 88° centers BLON = -9. 76° BLAT = 10 .47°

CLON = 7.0.19° CLAT = 5 .98°

Angular A to B = 162. 70° distances B to C = 87.95°

C to A = 109.35°

Calculated Values

Angles a= 162.97° 13 = 260. 08°

Angular A to B = 162.97° distances B to C = 97 .11 °

C to A= 99 .92°

Comparison Agreement

Angular A to B, 162. 70° v. 162.97° 99% distances B to c, 87.95° v. 97 .11 ° 91%

C to A, 109 .35° v .. 99. 92° 90%

Table A·9, Observed and calculated values from· the Lunar map, and their comparison. Note the high degree of agreement between calculated and observed values.

(

(

(

(

(

(

(

(

C

Paper A

OBSERVED POSITIONS

A-112.46 °

~ ~ ~ ~ ~ ~ 0 ........

~17, 0 '-I ~ J'o

'l<o -110

·100

-90

.so .10

.<:iO

0 '5

B-9.76 °

90

8° C 18.19° 7o

CALCULATED POSITIONS

~ ~ 0 1'b

'-1.;,o

''<o -110

-100

-90

.so .10

_<:,0

A-112.46 °

r----r--1~:. 9° C 81.62 °

0 '5

B-9.49 °

80

.. 0

a

Figure A-50. The equatorial plane of the Moon from above the North Pole. In the floating CALCULATED POSITIONS diagram the area center of Continent A is pegged to its Observed Position to make visual comparison easier.

63


even smaller. The imbalance between the two floors increases more and more.

However, when two or more continents are being pushed into the same hemisphere then they are necessarily converging closer to each other as they move toward the E-pole along meridians (E-pole of the E-equator). Decreasing distances mean rapidly increasing mutual force between them, and at one point this force becomes equal to the one that is pushing them toward the E-pole. At this point a new force balance arises and the continents stop moving. But the circle passing through their area centers is not a great circle anymore. The Martian continents are no:w in these positions. Their E-equator is about 10° off from its parallel great circle. The shift of the Martian E-equator deep into a hemisphere is evidence that a large ocean floor acts as a powerful repelling landmass, or "electrical" continent. This mechanism is applicable also to the Earth (and to the Moon). The Pacific Floor acts as the electrical seventh continent whose area center is at the antipode of Africa's area center, and it makes the characteristic pattern of the Earth a double pentagonal pyramid as illustrated in Figure A-41.

Continents of the Moon

Eight years after my first study of Martian continents, a Moon map was published: "Topographic Mapping of the Moon" by Sherman S. C. Wu, in the journal of Earth, Moon, and Planets, Volume 32, No. 2, page 169, April 1985. This time I immediately saw that the Moon also follows electrical repelling positioning: in Figure A-48 all the mountain peaks are lined up along a great circle that almost exactly coincides with the Equator just like on Mars. On the Moon the range of surface elevations is only 920 meters between the lowest point at 1737.51 km (in the North Polar region, not shown in my illustration), and the highest point at 1738.43 km, at the top of the largest continent (the contour lines are given as radii from the center of the Moon). Contour lines are drawn in every 50 meters. The half-way elevation is at 1737.91 km, and the nearest line drawn is at 1738.00 km. This line delineates only one large landmass going all around the Moon; it is obviously too low. Nevertheless, I was ready to cut the isthmuses at 50°E, 105°E, and at 90°W to separate the landmasses like I did with continent C of Mars. But I found the next higher contour line at 1838.50 km an attractive choice since it already neatly delineated three landmasses, or continents, and I could avoid arbitrary delineations altogether.

The results are shown in Figure A-49, and Table A-9 lists the main · numerical values. The agreement between the observed and calculated figures is better than 90% in the worst case in which the very small

(

(

Continents of the Moon

Figure A-51. The complete six-view projection of the Moon to illustrate the shapes and positions of its three continents named A, B, and C.

~ l ),.

~


continent C is involved, and it is better than 99% between the two large continents A and B.

All comments I made with respect to the Martian Data are valid for the Lunar Data. For example, the two large conwnents, A and B, with respect to the very small C, nearly position themselves at 180° apart. However, I call your attention to Figure A-50, predicted by my computer model in Figure A-45, #5, several years before the publication of the Lunar map. That very small landmass, Continent C, keeps the two large ones, A and B, from reaching the 180° positions exactly the predicted way: it makes them to be closer together on the far side, at 162.70°, and farther from each other on the near side, at 109.35+87.95=197.30°. And all these to come out so accurately, in spite of the obvious difficulties at the original surveying the Moon from an orbiting satellite, determining its radius and measuring the shape and height of that flat hill. You would not even notice it there; the land rises only 120 meters over 600 kilometers, the entire length of Florida. Still, it obsevably obeys Coulomb's electrical repulsion force to more than 90% accuracy. Obviously, the electrical force is a very sensitive mechanism. This is because the acting force on each continent is the difference of two very large forces.

Appendix Al

User's Guide to Electroglobe.

Introduction

Electroglobe is an interactive computer program to demonstrate that electrically charged objects freely floating on an all-ocean planet, move under their mutual repulsion forces to the same relative positions where the real continents of the Earth are now. Such objects move away from each other on the spherical surface until all repulsion forces become balanced on each object. The program can also be used to numerically demonstrate that landmasses on Mars and on the Moon are also in such force balanced positions.

The program is based on rigorous mathematical equations to describe movements on a spherical surface when repulsion forces are acting in three-dimensional space. At the start the program calculates the local horizontal force on each object, allows the object to move in the direction of that force for a short period. Each such move is a 'step'. After all objects have moved one step, the new force values and· new force directions are calculated, and a new sequence of stepping is done. Since all objects move away from each other under their mutual repulsion forces, the force values decrease after each step. This is not only because of the increasing distances, but also for force balancing. In the three-dimensional space between the repelling objects all forces vectorially sum to . a net single value on each object. The objects move only until the net forces become zero ( or negligibly small).

Input parameters are user accessible for editing. Electroglobe is written in the Fortran language (Microsoft® Fortran Version 5.1, by Microsoft Corporation). The program codes are printed in this Appendvc, except for specific graphics 'INCLUDE' subroutines that are part of the Fortran package. The Main Program of Electroglobe is called EGLOBE.FOR. A complete compiled execute-only program on a diskette may be accompanying this paper, or it might be available for a limited time from the publisher. This program is ready to run on most current Personal Computers. The program codes can be modified by the user, for example, to print out intermediate values. Then the modified program must be recompiled by a licensed Compiler which also comes with the Fortran package.

Installation Required hardware is a 386 or higher PC with a graphics card, and a

VGA monitor. Electroglobe can be run directly from the diskette, or from the hard disk, after copying all files to a new directory.

67


Operations

1. Electroglobe has two parts: a visual introduction, and the mathe-matical model. To start the program, type EGLOBE. A header screen appears, and an input line. Follow instructions on the screen. Type I for Introduction and go to Step 2 of this Guide. Or press ENTER to run the model, and go to Step 3.

2. The Introduction consists of 14 screens. After adequately studying each screen, press ENTER to continue with the next one. After the Introduction the first screen of the Mathematical Model appears.

3. The Model starts with a file list and an input line. The list shows all different setups available for the Model. Any file can be selected by typing its number portion. But for first time use, to see all the features available, run the seven-ship model. Type 700. This is a prepared input file IN700.FIL for seven ships.

4. The screen shows now: -To edit input file, press "I", to start the Model, press "Enter"-. To edit, go to Step 13. To run the prepared input file IN700.FIL, press ENTER. A side view of the globe appears. A color dot on the equator represents the 0° Longitude 0° Latitude position. If the color is green, the dot is on the near side of the globe. If it is yellow, it is on the far side. The grid lines are at 10° apart in Latitudes, and 5° apart in Longitudes. Bright red dots are the small ships on the near side, and pale red dots are ships on the far side. Single-character letters on blue background to the right of the ships are the names of the ships. In the first line at the top left corner of the screen the name of the input file number is displayed, and the current view number. IP is the current view-number, 1 is the top view, 6 is the bottom view, and 2 to 5 are four side views in the equatorial plane. In IP = 3 the 0° Longitude 0° Latitude position is in the center on the near side. The second line is menu selection. G is for 'go', to start stepping. Numbers 1-6 are for the six views of the globe, E is for 'end', C is for 'connect', and R is for 'rotate'. On the third line LM is the 'limit' of the number of steps the ships will take. The next number on this line, initially zero, shows the momentary step-number while the ships are moving. The table below this line, called the Position Table, shows the momentary coordinates of all ships at each step.

5. Type any number between 1 and 6. The selected view appears with the initial positions of the ships. You can repeat typing numbers in. This could be useful to familiarize yourself with the positions of the ships in the spherical coordinate system in connection with the values in the input file.

6. Press ENTER once. All ships take one step in the direction of the mqmentary repulsion forces. The Position Table is updated. At each subsequent pressing of ENTER one more step is taken.

(

(

(

(

Appendix Al 69

7. When you do not want to do manual stepping any more, type G or g. The ships start moving automatically until the number of steps equals LM. Press PAUSE to temporarily halt stepping, and press SPACE BAR to continue. Ship A is always anchored at its initial position to make comparisons easier between different patterns, since only the relative positions are relevant. Usually it is located at, or near to the top pole. Note the change of color of the ships as they move over the horizon, from bright red to pale red, or the other way round. Steps near to LM are too small to see, but the Position Table shows their positions at higher resolution. The ships are reasonably close to their theoretical final positions when the Longitude and Latitude values do not change more than 0.01 °, equal to 1 km on the Earth. (normally, the ships are stepping 5° or more at start). Accuracy of approach of 10-10 can be achieved, 0.1 millimeter on the Earth. Calculations are always done at 16 decimal digits accuracy, but the numerical display shows only a maximum of six.

8. When the number of steps have reached LM, the names of the ships appear on a yellow background to the right of the ships .. positions. This allows to identify the paths the ships have followed: Some paths are sometimes twisted when a ship moves over the hodzon with respect to another one, and its repulsion force-component switches from coming from behind to coming from ahead. You can· type now either a number from 1 to 6, or the letter C, or E. If you have typed a number, the related view appears with the ships back at their initial positions. They start stepping immediately, and they go until LM is reached again. You are back at Point 8 of this Guide. '

9. If you type C, for 'connect', adjacent final positions ofthe1

ships are interconnected by straight green lines to form the characteristic pattern for the given number of ships according to this list:

# of ships characteristic pattern

2 line 3 triangle 4 triangular pyramid 5 double triangular pyramid 6 double rectangular pyramid 7 double pentagonal pyramid 8 double hexagonal pyramid

Seven ships form a double pentagonal pyramid, the same as the area centers of the Earth's continents form. The ships (and continents) always arrive at these patterns from any initial positions. A certain amount of distortion or zigzagging may appear on the patterns when four or more ships are involved. This is not an inaccuracy of calculation, it is part of the acting mechanism as discussed in Paper


A. On the display of the Mathematical Model the Earth's rotation around the north - south axis is ignored, and the coordinate grid system is not tied to the North - South Poles. Instead, the axis of the pyramidal pattern is positioned upright for easy comparison with other patterns. When the Earth is modeled, Africa is at the top of the pyramid, Eurasia, North America, South America, Antarctica, and Australia, are at the model's equatorial plane (with a certain degree of regular zigzagging), and the area center of the Pacific is at the bottom apex of the pyramid.

10. The Chords Table appears in the left bottom comer of the screen to show the chord or edge lengths between the apexes of the pyramid that is enclosed in a sphere of radius 100. The names of the ships at those apexes are printed on the top line and in the left column.

11. If it is not clear what is the geometrical object you see on the display, then check first that the ships have arrived at their forcebalanced positions by not moving anymore while the program is still stepping. If they have, then it is possible that the Connect function is not interpreting the apex positions properly (see. Step 20). However, the Chords Table will always help you to make a three-dimensional physical model of the pattern, from thin rods or wooden sticks. From the physical model modify the related GRxxx.FIL file (see Step 20) to display the proper shape.

12. To end the program, Type E . The program performs a normal exit to the DOS prompt. CNTR-C is an emergency end and exit, used under error conditions (for example, when the program keeps stepping too long). This leaves the screen in graphics mode that can be restored by typing M.

13. After selecting a file in Step 4, you can type I to edit it. On the next screen type A to edit file parameters, and go to Step 14. Or type P to edit initial positions of the ships, and go to Step 18.

14. Do not modify DT and IT. ED is electron density. With the given other default parameters its useful range is between .1E19 and .1E20, corresponding to repulsion force factors 1 to 100. With smaller ED numbers the ships make shorter steps, thus the paths are more detailed near to the initial positions. But the ships require larger number of steps to get closer to the true final positions. With larger ED numbers the ships get closer to the true final position, but they may take very large steps immediately after start. Also, the forces are very large if the ships are too close together initially. This may result in the "explosion" of the system in which ships irregularly jump all over the surface of the globe. In this case either ED should be reduced, or the ships should be initially placed farther apart.·

15. DE is 'delay' in seconds, inserted between steps. A practical minimum value is 0.1 sec. This slows down automatic stepping (animation)

Appendb:Al 71

if you want to observe changes in the values of the Position Table. When DE is set to zero, the ships start moving automatically after a G (go) instruction, and the Position Table is not shown until the final positions are reached. This is the fastest mode of displacement used for visual study of the paths the ships follow, or, when only the final positions are investigated.

16. LM is the limit of the nuniber of steps the ships take before their positions are declared final. Practical numbers are between 100 and 10,000.

17. IP is the default view number which is displayed when the globe first appears.

18. Any item in the Edit Parameters screen can be changed according to the instructions on the screen. Take note of the values before any change, in case the new ones upset the system.

19. If you have typed P to edit position parameters, a menu appears with the names and positions of all ships. The names should remain in alphabetical order. New values of the coordinates are in effect only for the current run unless the Parameters Menu is also edited, in which case the changes remain permanent for subsequent runs. Th~se ate the initial positions of the ships. Do only small changes (a few degrees), because the 'connect' process (Step 20) is not yet smart enough to always meaningfully interconnect the final positions, thus the pattern might be confusing. However, the Chords Table will be always correct, and the pattern can be realized by making a physical model. When you have finished editing, press ENTER, and you are back at Step 5 of this Guide.

20. The Program uses the GRxxx.FIL files to interconnect adjacent points. These files must be written manually by an editor, after studying the new physical model (it is not always obvious what pattern is formed by several ships that are apparently 'scattered' on the globe's surface). Look at an example on the diskette, for example GR400.FIL. The first line is only an identification of the file number. From the second line down, the character in the first column is either an M, or an L. M moves the beam in pen-up state to the coordinate position of the ship whose one-character name is in the second column. An L in the next line draws the beam (pen-down) to the position of the ship whose name is in the second column. The third column is optional to keep track of lines drawn. Any number of lines can be drawn by this file, in any order.

21. New files of INxxx.FIL and GRxxx.FIL can be created (in DOS) by copying existing files, and changing the third and/or second digits of their number to assign the file a serial number (the first digit indicates the number of ships involved). Then you can modify the parameters either in the DOS editor, or during the startup of Electroglobe. Thus


you can make 99 versions of any configuration. The File Menu on the first screen will always display all available input files, new or old.

P;ttem examples

In Paper A there are six illustrations to show the system of characteristic patterns of two to seven objects; Figures A-35, -36, -37, -39, -40, -41, and listed in Table A-6. These patterns can be generated with Electroglobe by entering prepared input files as follows.

IN200 illustrates the principal pattern; the straight line between two ships. Initially both are placed on the equator, thus they remain on it during the displacement. A is fixed at the 0° Longitude, and B starts at 20°. Since the repulsion force is the greatest while their distance is the shortest, B starts moving very fast. After 100 steps it already arrives at the 146.66466° position. As their distance increases, and their local horizontal force components decrease, B rapidly slows down. At Step 1000 it is at 179.99295°. Nevertheless, it closely approaches its theoretical final position of the exact 180° at Step 1500, when its position is 179.99993°.

IN200.FIL 2

.lOOE+Ol

.500E+l9

.lOOE-01

A

1500 1

100

.o

.o • 0

40000.0 DIS01.CRD

B 20.0 o.o

• 0 40000.0

DIS01.CRD

filename number of continents duration of step, sec electron density delay stepping stepping limit view number

name of continent longitude latitude rotation thickness shape file


GR200.FIL MA L B 1

IN300 produces an equilateral triangle in the plane of the equator. A is fixed at 0° Longitude.Bis at 20°, C is at -20°. Their perfectly balanced state is at ±120°. They approach these positions after 1000 steps, at the ±120.00000 longitudes. The numbers in the first five decimal places are already zero as a result of rounding. If more digits were displayed they would be all 9-s, and subsequent numbers would be between O and 9.

Appendix Al

IN300.FIL 3

.lOOE+Ol

.280E+19

.lOOE-01

A

1000 5

100

• 0 • 0 • 0

40000.0 DIS01.CRD

B 20.0

.o • 0

40000.0 DIS01.CRD

C -20.0

• 0 .o

40000.0 DISOl.CRD





GR300.FIL MA L B 1 L C 2 LA 3

73

IN302 calculates pattern #2 of Figure A-45 in which the charge ratios are listed in Table A-7, both in Paper A. B goes to 129.18579° whose rounded value is the same as a. in the Table. C goes to -129.18579°, thus ~=360+(-129.18579)=230.81421, or 230.8°, rounded. Note the significantly different 'thickness' values used, to directly correspond to Q» QIP and Qc For relative displacement only their ratios are relevant. However, for a reasonably fast display, a greater ED value should be used.

IN302.FIL 3

.lOOE+Ol

.500E+21

.lOOE-01

A

1000 1

100

.o

.o

.o 2.0

DIS01.CRD

B 20.0

.o • 0

1. 0 DIS01.CRD

C -20.0

• 0




name of continent longitude latitude


74

.o 1. 0

DIS01.CRD

rotation thickness shape file

Underground Geophysics

IN303 calculates pattern #3 of Figure A-45 in which the charge ratios are listed in Table A-7, both in Paper A. B goes to 145.57808° whose rounded value is the same as a in the Table. C goes to -121.09040°, thus P=360+(-121.09040)=238.90960, or 238.9°, rounded.

IN303.FIL 3

,lOOE+Ol ,500E+21 ,lOOE-01

A

1000 1

100

.o

.o

.o 4,0

DIS01.CRD

B 20.0

.o

.o 2.0

DIS01.CRD

C -20.0

.o • 0

1.0 DIS01.CRD





GR303.FIL MA L B. 1 L C 2 LA 3

IN304 calculates pattern #4 of Figure A-45 in which the charge ratios are listed in TableA-7, both in Paper A. B goes to 172.79549° whose rounded value is the same as a in the Table. C goes to -172.79549°, thus P=360+(-172.79549)=187.20451, or 187.2°, rounded.

IN304.FIL 3

,lOOE+Ol ,300E+20 .lOOE-01

A

1000 1

100

• 0 .o .o

1000.0 DIS01.CRD

B 20.0

.o

.o



name of continent longitude latitude rotation


Appendix Al

1.0 DIS01.CRD

C -20,0

• 0 • 0

1. 0 DIS01.CRD

thickness shape file


75

IN305 calculates pattern #5 of Figure A-45 in which the charge ratios are listed in Table A-7, both in Paper A. B goes to 179.83826° whose rounded value is the same as a in the Table. C goes to -90.08087°, thus ~=360+(-90.08087)=269.91913, or 269.9°, rounded.

IN305,FIL 3

.lOOE+Ol

.400E+20 ,lOOE-01

A

1000 1

100

.o

.o

.o 1000.0

DIS01.CRD

B 20.0

• 0 • 0

1000.0 DIS01.CRD

C -20.0

• 0 • 0

1. 0 DIS01.CRD





GR305.FIL MA L B 1 L C 2 L A 3

IN400 produces a three-dimensional pattern; an equilateral triangular pyramid. In order to have an upright pyramid, ship A is placed at the 90° latitude. The other initial positions are arbitrarily set. After 1500 steps the ships are adequately in force balance as the latitudes are final in the first three decimal places (within 10 meters on an Earth-size model). Note Ship B as it starts out southward, but as the positions of the other ships change, it turns northward on a large radius. Note also that the chord lengths in the Table are all equal, (163 units long).

IN400,FIL 4

,lOOE+Ol .300E+l9 .lOOE-01

1500

filename number of continents duration of step, sec electron density delay stepping stepping limit

GR400.FIL MA L B 1 L D 2 L A 3 L C 4

76

3 view number

A

100

.o 90.0

.o 40000.0

DIS01.CRD

B 60.0

-30,0 .o

40000.0 DIS01.CRD

C .o

45 •. 0 .o

40000.0 DIS01.CRD

D 15.0 35.0

.o 40000.0

DIS01. CRD


name of continent ·longitude latitude rotation thickness shape file




L D 5 MB L C 6

IN500 generates a double triangular pyramid. Note that three ships settles exactly on the equator, and the fourth one goes to the south pole, and that Ships B and E significantly alter their course during displacement. Besides the coordinate values, rotation of the pattern as well as displaying it in the top or bottom view, shows that the equatorial triangle is equilateral.

IN500.FIL 5

.lOOE+Ol

.400E+19

.lOOE-01 1500

2 100

A • 0

90.0 .o

40000.0 DIS01.CRD

B -50.0 -20.0

.o 40000.0

DIS01.CRD

C 100.0

40.0 .o

40000.0




name of continent longitude latitude rotation thickness

GR500.FIL MA L B 1 L D 2 L A 3 L C 4 L D 5 L E 6 L C 7 L B 8 L E 9

Appendix Al

DIS01.CRD

D 300.0 -45.0

• 0 40000,0

DIS01.CRD

E 290.0 -80.0

• 0 40000.0

DIS01.CRD

shape file



77

IN600 illustrates the fast displacement by printing the coordinates only at the end of the run. Ship A is fixed at a slightly offset position from the usual north pole to allow a more vivid 3D view while rotating the double rectangular pyramid.

IN600,FIL 6

.lOOE+Ol ,500E+19 .OOOE+OO

A

1500 3

100

• 0 80.0

• 0 40000.0

DIS01.CRD

B 60,0 20.0

• 0 40000.0

DIS01.CRD

C 120.0

15.0 .o

40000.0 DIS01.CRD

D 200.0 -35.0

.o 40000,0

DIS01.CRD

E • 0

45,0 .o

40000.0 DISOl.CRD

F 250.0 -85.0

• 0








GR600.FIL MA L B 1 L F 2 L C 3 L B 4 LE 5 L A 6 L D 7 L E 8 L F 9 L D 10 L C 11 LA 12

78

40000.0 DIS01.CRD



IN700 generates a double pentagonal pyramid shown in the top view. This is the Earth's characteristic pattern. However, the Earth's continents are of different volumes, while this input file specifies equal volumes. (Electroglobe uses pointlike objects, but different volumes (in effect, electron content) can be simulated by assigning different values to 'Thickness', as used in IN800). Note the appearance of zigzagging of ship positions along the equator when you switch to a side view.

IN700.FIL 7

.lOOE+Ol

.600E+19

.OOOE-01 1500

1 100

A .o

90.0 • 0

40000.0 DIS01.CRD

B 40.0 40.0

• 0 40000.0

DIS01.CRD

C 90.0 40.0

.o 40000.0

DISOl.CRD

D 140.0 -20.0

• 0 40000.0

DIS01.CRD

E 260.0 -50.0

.o 40000.0

DIS01.CRD

F 60.0

-30.0 .o

40000.0 DIS01.CRD

G 70.0

-80.0 • 0






name of continent longitude · latitude rotation thickness shape file



GR700.FIL MA L B 1 L G 2 LE 3 L B 4 L F 5 L A 6 L E 7 L D 8 L G 9 LC 10 L D 11 L A 12 LC 13 L F 14 L G 15

l.

l.

Appendix Al

40000.0 DIS01.CRD


79

INBOO is an attempt to predict the characteristic pattern the Earth may take up when the next breakup occurs in which one of its existing continents breaks into two. One candidate is North America whose crust in the central region is the thinnest of all, 20 km only, while in the east and west coasts it is 40 km or more. The relative volume values ( or area values, since the continental plates are large sheets), are encoded in the 'Thickness' parameters. The relative volumes are: Eurasia (E) = 58600, Pacific Floor (H) = 39400, Africa (A) = 37200, North America = 22800, South America (B) = 19700, Antarctica (C) = 14400, and Australia (D) = 10000, as the reference. In the breakup North America would split into NA1 (F) = 11400, and NA2 (G) =11400 similarly. The resulting pattern is a greatly zigzagged double hexagonal pyramid, as shown on the display.

INBOO,FIL 8

.lOOE+Ol

.200E+19

.OOOE+OO

A

5000 1

100

.o 90.0

.o 37200.0

DIS01.CRD

B • 0

10.0 • 0

19700.0 DIS01.CRD

C 72.0

.o

.o 14400.0

DIS01.CRD

D 144.0 -15.0

.o 10000.0

DIS01.CRD

E 216,0

25.0 .o

58600.0 DIS01.CRD

F 288.0







name of continent longitude

GRBOO,FIL M A 1 L B 2 L H 3 L C 4 L B 5 L G 6 L A 7 L F 8 L G 9 L H 10 L F 11 L E 12 LA 13 L D 14 L E 15 L H 16 L D 17 LC 18 L A 19

80

-15.0 .o

11400.0 DIS01.CRD

G 288.0

-3.0 .o

11400.0 DIS01.CRD

H 90.0

-90.0 .o

39400.0 DIS01.CRD

latitude rotation thickness shape file



Numerical results of Mars and the Moon


IN310 contains the numerical coding of the continents of Mars; their area ratios are in the 'thickness' lines. A = 47400, B = 18300, and C = 10000 as the reference. All three area centers are initially placed on the equator since on the map they are very close to it. For easy comparison with the analytically calculated separations, as in Table A-8 and Figure A-46 in Paper A, Continent A is fixed (pegged) at its observed location of -105.42°. B and C are arbitrarily set at about ±15° from A, at -90° and -120°, respectively. After 1500 steps, B settles at 40.99665°, and C settles at 128.91227°. After rounding to two decimal places, 41.00° and 128.91 °, these are the same values as those in Paper A which were calculated by a different method, namely, by solving a nonlinear analog set of equations, Equ. A-4.

IN310.FIL 3

.lOOE+Ol

.400E+l9

.lOOE-01 1500

1 100

A -105.42

.oo

.oo 47400.0

DIS01.CRD

B -90.00

.oo

.oo 18300.0

DIS01.CRD

C -120.00

.oo




name of continent longitude latitude

GR310.FIL MA LB L C LA

Appendix Al

.oo 10000.0

DIS01.CRD

rotation thickness shape file

81

IN320 contains the numerical coding of the continents of the Moon; their area ratios are in the 'thickness' lines. A = 77800, B = 72500, and C = 10000 as the reference. All three area centers are initially placed on the equator since on the map they are very close to it. For easy comparison with the analytically calculated separations, as in Table A-9 and Figure A-50 in Paper A, Continent A is fixed (pegged) at its observed location of -172.46°. B and C are arbitrarily set at about ±15° from A, at -160° and 170°, respectively. After 1500 steps, B settles at -9.49159°, and C settles at 87.62246°. After rounding to two decimal places, -9.49° and 87.62°, these are the same values as those in Paper A which were calculated by a different method, namely, by solving a nonlinear analog set of equations, Equ. A-4.

IN320.FIL 3

.lOOE+Ol

.300E+19

.lOOE-01 1500

1 100

A -172.46

.oo

.oo 77800.0

DIS01.CRD

B -160.00

.oo

.oo 72500.0

DIS01.CRD

C -190.00

.oo

.oo 10000.0

DIS01.CRD





GR320.FIL MA L B 1 L C 2 L A 3


Program Listing of Electroglobe

If the program diskette is missing, the program steps can be typed in ( although it may be laborious). The line numbers are not part of the programs. The complete program consists of ten files, the INCLUDE files are part of the Microsoft® Fortran package.

1 EGLOBE.FOR 2 AREA.FOR 3 CGTRI.FOR 4 CGSUB.FOR 5 CGCONS.FOR

6 BASICS.FOR 7 BASIS.FOR S SHS.FOR 9 NEWPOS.FOR

10 SETS.FOR

All these programs should be compiled and linked.

PROGRAM EGLOBE.FOR

1 C 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 1 27 28 29 30 31 32 2 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5 55 56

PROGRAM EGLOBE INCLUDE 'FGRAPH.FI' IMPLICIT REAL*8 (A-H,0-Z) REAL*8 MS,MLM(8),ILM(8),M0(8) INTEGER*2 JXT(8),JYT(8),kxt(8),kyt(8) COMMON /PDR/ PI,DRC,RR COMMON /ITEMS/ CNA$(8),SHP$(8) COMMON /TEX/ ICNT,LM,NM$,NN,JXT,JYT COMMON /TRN/ JROTA,kxt,kyt COMMON /LST/ TH(8),PH(8),GG(8),TCK(8) COMMON /FNAME/ INFILENAM$ COMMON /GGC/ IP,JG,JFL,LL CHARACTER INFIL*5,INFILENAM$*9,FILENAME$*9,SHP$*9,

+CNA$*1,NM$*3,ED$*1,I$*1,J$*1,K$*1,FNA$*9,str$*10 DIMENSION AASC(8,9),ATR(8),BE(8,9,7),BCG(8,9),CCG(8,9),

+CE(8,9,7),FI(8),PHSC(9),PPH(9,7), +PX(9),QC(8),RG(8,9),SFX(8),SFY(8),SFZ(8),SX2(8),SY2(8), +SZ2 (8)',THSC(9) ,TTH(9, 7) ,TX( 8) 1

+NNS(8) ,NP(8, 9) ,XC(8, 9) 1 YC(8, 9), ZC(8,9) ,XCG( 8), YCG(8) 1 ZCG(8) RR=6.37E6 PI=DATAN(1)*4.0 I 3.141592653589793 MS=3.0E3 ECH=l.610-19 POT=PI/2. DRC=PI/180. ICNT=O JGOFLG=O JREPEAT=O JROTA=O THX=O.O PHX=O.O CALL SETGLB ( ) PRINT ' (/III/) ' CALL settextwindow (INT2(6),INT2(65),INT2(29),INT2(74))

WRITE (str$,'(1X,A)') 'Available' CALL outtext (str$) WRITE (str$,'(1X,A)') 'input' CALL outtext (str$) WRITE (str$,'(1X,A)') 'files:' CALL outtext (str$) WRITE (str$,' (lX,A)') CALL outtext (str$) DO I=0,9 I$=CHAR(I+48) DO J=0,9 J$=CHAR(J+48)

DO K=0,9 K$=CHAR(K+48)

FNA$='IN'//I$//J$//K$//' .FIL' OPEN (l,FILE=FNA$,STATUS='OLD' ,MODE='READ',ERR=5) WRITE (str$, '(lX,A)') FNA$ CALL outtext (str$)

CLOSE (1) END DO I K

END DO I J END DO II

Appendix Al

57 58 59 60 61 62 63 10 64 12 65 66 67 68 69 70 71 72 20 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 80 95 96 97 98 99

100 101 102 100 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 120 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 140 136 137 138

CALL settextwindow (INT2(1),INT2(1),INT2(29),INT2(79)) CALL SETPRINT (INT4(18),INT4(1)) WRITE(*, '(5X,A\)') 'Enter input file number: READ (*,'(A3)') NM$ INFIL='IN'//NM$ INFILENAM$=INFIL//'.FIL' OPEN (2,FILE=INFILENAM$,STATUS='OLD',MODE='READ') READ (2,' (5X,A9)') FILENAME$ READ (2,'(1X,I6)') NN READ (2,'(lX,FB.3)') OT READ (2,' (lX,EB.3)') ED READ (2,'(lX,FB.3)') DE READ (2,'(1X,I6)') LM READ (2,' (1X,I6)') IP READ (2,'(1X,I6)') IT

DO I=l,NN READ (2,*) READ (2,'(lX,Al)') READ (2,'(1X,F7.2)') READ (2,'(1X,F7.2)') READ (2,'(1X,F7.2)') READ (2,' (lX,FB.l)') READ (2,'(1X,A9)')

ENDDO I I CLOSE (2)

CNA$(I) TH(I) PH(I) GG(I) TCK(I) SHP$(I)

IF (JREPEAT .EQ. 1) GOTO 80 CALL clearscreen ($GCLEARSCREEN) CALL SETPRINT (INT4(18),INT4(1))

number of continents duration of step, sec electron density delay stepping stepping limit view number (not used)


WRITE (*,'(5X,A)') 'To edit input file, WRITE (*,'(5X,A)') 'to start the Model, WRITE (*, '(5X,A\)')'

press "I" press "ENTER"

>>> : • READ (*,'(A)') ED$ IF (ED$ .EQ. 'I' .OR. ED$ .EQ. 'i') THEN CALL SETIN (NN,TH,PH,TCK) ELSE NOED=l END IF

IF (JREPEAT .EQ. 1) THEN IP=IPNEW GOTO 170 END IF

DO NCT=l,NN OPEN (UNIT=3,FILE=SHP$(NCT),STATUS='OLD',MODE='READ') READ (3,*) READ (3,' (F5.1)') SMP IF (.NOT. EOF(3)) THEN

I size multiplier

READ (3,'(I2,I2,1X,F5.1,2X,F5.l)') NSC,NPT,RPH,RTH PPH(NSC,NPT)=RPH*SMP

TTH(NSC,NPT)=RTH*SMP NP(NCT,NSC)=NPT GOTO 100 END IF

CLOSE (3) NNS(NCT)=NSC

DO J=l,NSC NPTS=NP(NCT,J) NTRI=NPTS-2 ASUM=O.O THSC(J)=O.O PHSC(J)=O.O

DO K=l,NTRI Pl=PPH(J,1) Tl=TTH(J,1) P2=PPH(J,K+l) T2=TTH(J,K+l) P3=PPH(J,K+2) T3=TTH(J,K+2) ATR(K)=AREA(Tl,T2,T3,Pl,P2,P3) ASUM=ASUM+ATR(K) CALL CGTRI (Tl,T2,T3,Pl,P2,P3,TC,PC) TX(K)=TC PX(K)=PC ENDDO I K

CALL CGSUB (NTRI,ATR,Tl,Pl,TX,PX,TC,PC) THSC(J)=TC PHSC(J)=PC AASC(NCT,J)=ASUM ENDDO I J CALL CGCONB (NCT,NSC,AASC,Tl,Pl,THSC,PHSC,TC,PC) CALL BASICS (NCT,NSC,THSC,PHSC,TC,PC,BCG,CCG,RG) CALL BASIS (NCT,NSC,NP,TTH,PPH,TC,PC,BE,CE) ENDDO I NCT

83


139 CALL NEWPOB (PH,TH,GG,NN,NNS,BCG,CCG,XC,YC,ZC) 140 IF (NOED .EQ. 1) GOTO 150 141 IF (JREPEAT .EQ. 0) THEN 142 CALL clearscreen ($GCLEARSCREEN) 143 CALL SETPRINT (INT4(14),INT4(1)) 144 CALL SETS (DT,ED,DE,LM,IP,IT,NN) 145 END IF 146 150 DO I=l,NN 147 DO J=l,NNS(I) 148 MLM(I)=MLM(I)+MS*TCK(I)*AASC(I,J) 149 ILM(I)=ILM(I)+MS*TCK(I)*AASC(I,J)*RG(I,J)**2 150 ENDDO I J 151 ENDDO I I 152 DO I=l,NN 153 QC(I)=TCK(I)*ECH*ED 154 ENDDO 155 170 CALL GLOBE (THX,THY) 156 ICNT=O 157 CALL GLB (DBLE(O.O),DBLE(O.O),THX,THY,3) I mark zero deg. 158 CALL SH8 (TH,PH,SFX,SFY,SFZ,MO,FI,QC,AASC,NN,XC,YC,ZC,NNS) 159 CALL PATH (THX,THY,DE) I first points 160 C----------------------------------------------------------------161 16 ICNT=ICNT+l 162 DO I=l,NN 163 IF (ICNT .GT. 1 .AND. I .EQ. 1) GOTO 200 164 IF (I .EQ. 8) GOTO 200 165 DK=DT 166 IF (I .EQ. 1) DK=O.O 167 SX2(I)=SFX(I)/MLM(I)*DK*DK/2 168 SY2(I)=SFY(I)/MLM(I)*DK*DK/2 169 SZ2(I)=SFZ(I)/MLM(I)*DK*DK/2 170 PHD=PH(I)*DRC 171 THD=TH(I)*DRC 172 XCG(I)=DCOS(PHD)*DCOS(THD) 173 YCG(I)=DCOS(PHD)*DSIN(THD) 174 ZCG(I)=DSIN(PHD) 175 RX=RR*XCG(I) 176 RY=RR*YCG(I) 177 RZ=RR*ZCG(I) 178 SXR=RX+SX2(I) 179 SYR=RY+SY2(I) 180 SZR=RZ+SZ2(I) 181 SSR=DSQRT(SXR**2+SYR**2) 182 TH(I)=DATAN2(SYR,SXR)/DRC 183 PH(I)=DATAN2(SZR,SSR)/DRC 184 200 ENDDO I I 185 CALL NEWP08 (PH,TH,GG,NN,NNS,BCG,CCG,XC,YC,ZC) 186 CALL SH8 (TH,PH,SFX,SFY,SFZ,MO,FF,QC,AASC,NN,XC,YC,ZC,NNS) 187 IF (ICNT .GT. 1) GOTO 250 188 GOTO 300 189 250 CONTINUE 190 CALL PATH (THX,THY,DE) 191 300 IF (ICNT .GE. LM .AND. JGOFLG .EQ. 1) GOTO 500 192 IF (ICNT .GE. LM) GOTO 600 I finish step-by-step 193 IF (JGOFLG .EQ. 1) GOTO 16 194 500 IF (JGOFLG .EQ. 0) GOTO 600 195 600 CALL RC16EN (IGT,IP) 196 620 IF (!GT .EQ. 16) THEN I rotate 197 CALL CIRCLE() 198 630 CALL GRAPH () 199 CALL PATH (THX,THY,DE) 200 JC=JC+l 201 IF (JC .GE. 720) THEN 202 JC=O 203 JROTA=O 204 GOTO 600 205 END IF 206 GOTO 630 207 END IF 208 IF (IGT .EQ. 15) THEN I draw chords 209 CALL GRAPH () 210 GOTO 600 211 END IF 212 IF (IGT .LT. 10) THEN I new IP 213 DO I=l,NN 214 kxt(I)=O 215 kyt(I)=O 216 END DO 217 IPNEW=IP 218 JREPEAT="l 219 GOTO 10 220 END IF

Appendix Al 85

221 IF (!GT .EQ. 11) GOTO 999 222 IF (!GT .EQ. 13) THEN I go 223 CC CALL TIMER (1,DE,JREPEAT) 224 JGOFLG=l 225 GOTO 16 226 END IF 227 IF (!GT .EQ. 14) GOTO 16 continue step-by-step 228 READ(*,*) wait for <CR> 229 999 .CALL RESETGLB () 230 END 231 C======~~~~~~~============~========~=========== 232 SUBROUTINE SETGLB () 233 IMPLICIT REAL*8 (A-H,0-Z) 234 INCLUDE 'FGRAPH.FD' 235 INTEGER*2 st,sta,xaf(86),xeu(82),xsa(30) 236 INTEGER*4 RGB,tmp 237 CHARACTER str$*80,ANS$*1 238 RECORD /xycoord/ xy 239 RECORD /xycoord/ poly [FAR] (5) 240 RECORD /xycoord/ bout [FAR] (5) 241 RECORD /xycoord/ blenter [FAR] (5) 242 RECORD /xycoord/ afr [FAR] (43) 243 RECORD /xycoord/ eur [FAR] (41) 244 RECORD /xycoord/ sam [FAR] ( 15) 245 DATA xaf/280,230, 300,230, 303,242, 330,248, 332,242, 356,242, 16 246 +400,302, 402,304, 415,300, 420,301, 418,320, 408,331, 396,348, 113 247 +395,366, 398,378, 396,390, 380,398, 378,410, 372,418, 361,430, 120 248 +352,440, 341,442, 330,432, 318,406, 310,400, 312,390, 311,370, 127 249 +310,358, 300,350, 300,330, 290,323, 270,321, 256,330, 239,324, 134 250 +222,310, 221,298, 221,281, 222,270, 240,258, 250,242, 261,240, 141 251 +272,236, 280,230/ 143 252 DATA xeu/360,168, 386,178, 406,188, 421,200, 436,212, 447,228, 16 253 +456,240, 463,252, 462,290, 450,261, 436,250, 420,249, 430,262, 113 254 +421,278, 416,288, 400,291, 382,261, 370,248, 364,231, 358,230, -120 255 +348,226, 341,220, 335,219, 337,226, 330,224, 320,218, 308,212, 127 256 +294,214, 280,215, 266,230, 258,230, 259,220, 270,212, 269,203, 134 257 +302,190, 298,180, 310,170, 330,170, 340,172, 350,168, 360,168/ 141 258 DATA xsa/163,320, 170,335, 173,352, 185,360, 197,370, 195,380, 16 259 +192,380, 192,390, 191,408, 190,412, 184,400, 174,380, 168,360, 113 260 +163, 340 I 163, 320/ 115 26i st=setvideomoderows ( $VRES16COLOR, 30) 262 tmp=RGB(INT4(31),INT4(31),INT4(31)) 0 background 263 st=remappalette (INT2(0),tmp) 0 background 264 tmp=RGB(INT4(0),INT4(60),INT4(63)) 1 1st lettering bckgr 265 st=remappalette (INT2(1),tmp) 1 1st lettering bckgr 266 tmp=RGB(INT4(40),INT4(40),INT4(40)) 2 grid rear 267 st=remappalette (INT2(2),tmp) 2 grid rear 268 tmp=RGB(INT4(0),INT4(63),INT4(0)) 3 chords 269 st=remappalette (INT2(3),tmp) 3 chords 270 tmp=RGB(INT4(0),INT4(0),INT4(0)) 4 gr letters 271 st=remappalette ( INT2 ( 4), tmp) 4 gr letters 272 tmp=RGB(INT4(63),INT4(63),INT4(0)) 5 last lettering bckgr 27 3 st=remappalette ( INT2 ( 5), tmp) 5 last lettering bckgr 274 tmp=RGB(INT4(0),INT4(0),INT4(0)) 6 grid front 275 st=remappalette (INT2(6),tmp) 6 grid front 276 tmp=RGB(INT4(63),INT4(63),INT4(63)) 7 text color 277 st=remappalette (INT2(7),tmp) 7 text color 278 tmp=RGB(INT4(63),INT4(63),INT4(0)) 8 title tex (yellow) 279 st=remappalette (INT2(8),tmp) 8 title tex (yellow) 280 tmp=RGB(INT4(10),INT4(10),INT4(63)) 9 title disk (blue) 281 st=remappalette (INT2(9),tmp) 9 title disk (blue) 282 tmp=RGB(INT4(63),INT4(63),!NT4(63)) 10 white (dialog) 283 st=remappalette (INT2(10),tmp) 10 white (dialog) 284 tmp=RGB(INT4(0),INT4(63),INT4(0)) 11 285 st=remappalette (INT2(11),tmp) 11 286 tmp=RGB(INT4(62),INT4(63),INT4(0)) 12 zero deg. near side 287 st=remappalette (INT2(12),tmp) 12 zero deg. near side 288 tmp=RGB(INT4(0),INT4(63),INT4(0)) 13 zero deg. far side 289 st=remappalette (INT2(13),tmp) 13 zero deg. far side 290 tmp=RGB(INT4(62),INT4(2),INT4(11)) 14 path near side 291 st=remappalette (INT2(14),tmp) 14 path near side 292 tmp=RGB(INT4(60),INT4(30),INT4(40)) 15 path far side 293 st=remappalette (INT2(15),tmp) I 15 path far side 294 C title text-----------------------------------------------------295 st=registerfonts ('\FLIB\*.FON') 296 tmp=RGB(INT4(0),INT4(0),INT4(0)) I set bckgrd black 297 st=remappalette(INT2(0),tmp) I set bckgrd black 298 CALL TITLE() 299 st=setcolor (int2(10)) 300 CALL moveto (INT2(10),INT2(460),xy) 301 st=setfont ("t'courier' "//'hl2w9') 302 CALL SETPRINT (INT4(20),INT4(23))

303 304 305 306 307 308

86

309 550 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375, 376 377 378 379 380 381 382 383 384

WRITE ( *, ' ( lX,A, I ,23X,A\)') +' Type I for Introduction,', +'or press ENTER to start the Mathematical Model

READ(*,' (A)') ANS$ IF (ANS$ .EQ. 'I' .OR. ANS$ .EQ. 'i') GOTO 550 GOTO 600 st=setcolor(O) CALL clearscreen ($GCLEARSCREEN) CALL TITLE ( ) st=setfont ("t'helv' "// 'hl8w9b') I dialog text st=setcolor ( INT2 (0)) I blackout text blenter(l).xcoord=INT2( 0) blenter(l).ycoord=INT2(450) blenter(2).xcoord=INT2(225) blenter(2).ycoord=INT2(450) blenter(3).xcoord=INT2(225) blenter(3).ycoord=INT2(480) blenter(4).xcoord=INT2( 0) blenter(4).ycqord=INT2(480) blenter(5).xcoord=INT2( 0) blenter(5).ycoord=INT2(450)


st=polygon ($GFILLINTERIOR,blenter,5) st=setcolor (9) st=ellipse ($GFILLINTERIOR,

blue globe draw the globe

+INT2(162),INT2(162),INT2(478),INT2(478)) st=setcolor(INT2(8)) , DO i=l,86,2 j=j+l afr(j).xcoord=xaf(i) afr(j).ycoord=xaf(i+l) END DO st=polygon ($GFILLINTERIOR,afr,43)

st=setcolor (int2(10)) CALL moveto (INT2(50),INT2(80),xy) WRITE (str$,'(A)') 'Introduction in 14 steps.' CALL outgtext (str$)

CALL moveto (INT2(300),INT2(80),xy) WRITE (str$,'(A)') "l. This is the Earth's globe" CALL outgtext (str$) CALL moveto (INT2(320),INT2(100),xy) WRITE (str$,'(A)') 'as seen from above Africa.' CALL outgtext (str$)

sta=setcqlor (10) CALL moveto (I.NT2(10),INT2(465),xy) WRITE (str$,'(A\)') 'Press ENTER to continue' CALL outgtext .( str$) READ (*,*) st=setcolor(O) st=polygon ($GFILLINTERIOR,blenter,5)

st=setcolor (INT2(0)) I blackout text bout(l).xcoord=INT2( 0) bout(l).ycoord=INT2( 80) bout(2).xcoord=INT2(639) bout(2).ycoord=INT2( 80) bout(3).xcoord=INT2(639) bout(3).ycoord=INT2(150) bout(4).xcoord=INT2( 0) bout(4).ycoord=INT2(150) bout(5).xcoord=INT2( 0) bout(5).ycoord=INT2( 80) st=polygon ($GFILLINTERIOR,bout,5) CALL moveto (INT2(300),INT2(80),xy) st=setcolor(INT2(10)) WRITE (str$,' (A)') "2. This is Africa's area center." CALL outgtext (str$) DO I=l, 15 st=setcolor(INT2(10)) st=ellipse ($GFILLINTERIOR,

+INT2(315),INT2(315),INT2(325),INT2(325)) CALL DELAY (DBLE(.2)) st=setcolor(INT2(0)) st=ellipse ($GFILLINTERIOR,

+INT2(315),INT2(~15),INT2(325),INT2(325)) CALL DELAY (DBLE(.2)) END DO sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE(2))

I area center of Africa

I area center of Africa

WRITE (str$,'(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ (*,*)

I '·

(

\ (

Appendix Al

385 386 C 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 C--450 100 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466

st=polygon ($GFILLINTERIOR,blenter,5) Eurasia ----------------------------------------------------------j=O

DO i=l,82,2 j=j+l eur(j).xcoord=xeu(i) eur(j).ycoord=xeu(i+l) END DO st=setcolor (INT2(8)) ·st=polygon ($GFILLINTERIOR,eur,41) eurasia st=setcolor (INT2(0)) blackout text st=polygon ($GFILLINTERIOR,bout,5) st=setcolor(INT2(10)) CALL moveto (INT2(300),INT2(80),xy) WRITE ( str$, ' (A)' ) • 3. This is Eurasia. Its area center is" CALL outgtext (str$) CALL moveto (INT2(320),INT2(100),xy) WRITE ( str$,' (A)') "only slightly this side of the horizon." CALL outgtext (str$) DO I=l, 15 st=setcolor(INT2(10)) st=ellipse ( $GFILLINTERIOR, I area center of Eurasia

+INT2(400),INT2(182),INT2(410),INT2(192)) CALL DELAY (DBLE(.2)) st=setcolor(INT2(0)) st=ellipse ($GFILLINTERIOR, I area center of Eurasia

+INT2(400),INT2(182),INT2(410),INT2(192)) CALL DELAY (DBLE(.2)) END DO sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE(2)) WRITE (str$,'(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ (*,*) st=polygon ($GFILLINTERIOR,blenter,5) st=setcolor (INT2(0)) I blackout text st=polygon ($GFILLINTERIOR,bout,5) st=setcolor(INT2(10)) CALL moveto (INT2(300),INT2(80),xy) WRITE (str$,'(A)') "4. Interconnect the two area centers." CALL outgtext (str$) CALL DELAY (DBLE(4)) st=setcolor (14) poly(l).xcoord=INT2(320) 1 A poly(l).ycoord=INT2(316) 1 poly(2).xcoord=INT2(324) 2 poly(2).ycoord=INT2(318) 2 poly(3).xcoord=INT2(407) 12 poly(3).ycoord=INT2(188) 12 poly(4).xcoord=INT2(403) 11 poly(4).ycoord=INT2(186) 11 poly(5).xcoord=INT2(320) 1 poly(5).ycoord=INT2(316) I 1 st=polygon ($GFILLINTERIOR,poly,5)

sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE(2)) WRITE (str$, '(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ(*,*) st=polygon ($GFILLINTERIOR,blenter,5)

North America-----------------------------------------st=setcolor (INT2(0)) I blackout text st=polygon ($GFILLINTERIOR,bout,5) st=setcolor(INT2(10)) CALL moveto (INT2(300),INT2(80),xy) WRITE ( str$,' (A)' ) "5. North America and its area center" CALL outgtext (str$) CALL moveto (INT2(320),INT2(100),xy) WRITE ( str$, ' (A)' ) "is slightly beyond the horizon.• CALL outgtext (str$) DO I=l, 15 st=setcolor(INT2(10)) I area center of N.A st=ellipse ($GFILLINTERIOR,INT2(203),INT2(206),

+INT2(213),INT2(216)) CALL DELAY (DBLE(.2)) st=setcolor(INT2(0)) I area center of N.A st=ellipse ($GFILLINTERIOR,INT2(203),INT2(206),

+INT2(213),INT2(216))

87

467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548

88

CALL DELAY (DBLE(.2)) END DO

sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE ( 2 )) WRITE (str$,'(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ (*,*) st=polygon ($GFILLINTERIOR,blenter,5)


st=setcolor ( INT2 (0)) I blackout text st=polygon ($GFILLINTERIOR,bout,5) st=setcolor(INT2(10)) CALL moveto (INT2(300),INT2(80),xy) WRITE ( str$,' (A)' ) "6. Interconnect with Africa's area center." CALL outgtext (str$) .CALL DELAY (DBLE(4)) st=setcolor (14) poly(l).xcoord=INT2(316) 9 E poly(l).ycoord=INT2(319) 9 poly(2).xcoord=INT2(319) 10 poly(2).ycoord=INT2(316) 10 poly(3).xcoord=INT2(208) 20 poly(3).ycoord=INT2(209) 20 poly(4).xcoord=INT2(205) 19 poly(4).ycoord=INT2(212) 19 poly(5).xcoord=INT2(316) 9 poly(5).ycoord=INT2(319) I 9 st=polygon ($GFILLINTERIOR,poly,5)

sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE(2)) WRITE (str$,'(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ (*,*) st=polygon ($GFILLINTERIOR,blenter,5)

c-- South America-------------------------------------------j=O DO i=l,30,2 j=j+l sam(j).xcoord=xsa(i) sam(j).ycoord=xsa(i+l) END DO st=setcolor (INT2(8)) st=polygon ($GFILLINTERIOR,sam,15) I South America st=setcolor (INT2(0)) blackout text st=polygon ($GFILLINTERIOR,bout,5) st=setcolor(INT2(10)) CALL moveto (INT2(300),INT2(80),xy) WRITE (str$,'(A)') "7. South America's area center is" CALL outgtext (str$) CALL moveto (INT2(320),INT2(100),xy) WRITE ( str$, ' (A)' ) "only slightly this side of the horizon." CALL outgtext (str$) DO I=l, 15 st=setcolor(INT2(10)) I area center of S.A st=ellipse ($GFILLINTERIOR,INT2(168),INT2(370),

+INT2(178),INT2(380)) CALL DELAY (DBLE(.2)) st=setcolor(INT2(0)) I area center of S.A st=ellipse ($GFILLINTERIOR,INT2(168),INT2(370),

+INT2(178),INT2(380)) CALL DELAY (DBLE(.2)) END DO

sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE(2)) WRITE (str$, '(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ(*,*) st=polygon ($GFILLINTERIOR,blenter,5)

st=setcolor (INT2(0)) I blackout text st=polygon ($GFILLINTERIOR,bout,5) st=setcolor(INT2(10)) CALL moveto (INT2(300),INT2(80),xy) WRITE (str$,'(A)') "8. Interconnect with Africa's area center." CALL outgtext (str$) CALL DELAY (DBLE(4)) st=setcolor (14) poly(l).xcoord=INT2(318) I 7 D

(

Appendix Al 89

549 poly(l).ycoord=INT2(323) 7 550 poly(2).xcoord=INT2(316) 8 551 poly(2) .ycoord=INT2(319) 8 552 poly(3).xcoord=INT2(171) 18 553 poly(3) .ycoord=INT2(372) 18 554 poly(4) .xcoord=INT2(173) 17 555 poly(4).ycoord=INT2(377) 17 556 poly(5) .xcoord=INT2 (318) 7 557 poly(5).ycoord=INT2(323) I 7 558 st=polygon ($GFILLINTERIOR,poly,5) 559 sta=setcolor (10) 560 CALL moveto (INT2(10),INT2(465),xy) 561 CALL DELAY (DBLE(2)) 562 WRITE (str$,'(A\)') 'Press ENTER to continue' 563 CALL outgtext (str$) 564 st=setcolor(O)

. 565 READ (*,*) 566 st=polygon ($GFILLINTERIOR,blenter,5) 567 C-- Antarctica-------------------------------------------568 200 st=setcolor (INT2(0)) I blackout text 569 st=polygon ($GFILLINTERIOR,bout,5) 570 st=setcolor(INT2( 10)) 571 CALL moveto (INT2(300),INT2(80),xy) 572 WRITE (str$,'(A)') "9. Antarctica's area center is" 573 CALL outgtext (str$) 574 CALL moveto (INT2(320),INT2(100),xy) 575 WRITE (str$,'(A)') "slightly beyond the horizon." 576 CALL outgtext (str$) 577 DO I=l, 15 578 st=setcolor(INT2(10)) I area center of Antarctica 579 st=ellipse ($GFILLINTERIOR,INT2(330),INT2(470), 580 +INT2(340),INT2(480)) 581 CALL DELAY (DBLE( .2)) 582 st=setcolor(INT2(0)) I area center of Antarctica 583 st=ellipse ($GFILLINTERIOR,INT2(330),INT2(470), 584 +INT2(340),INT2(480)) 585 CALL DELAY (DBLE(.2)) 586 END DO 587 sta=setcolor (10) 588 CALL moveto (INT2(10),INT2(465),xy) 589 CALL DELAY (DBLE(2)) 590 WRITE (str$, '(A\)') 'Press ENTER to continue' 591 CALL outgtext (str$) 592 st=setcolor(O) 593 READ (*,*) 594 st=polygon ($GFILLINTERIOR,blenter,5) 595 st=setcolor(sta) 596 st=setcolor (INT2(0)) blackout text 597 st=polygon ($GFILLINTERIOR,bout,5) 598 st=setcolor(INT2(10)) 599 CALL moveto (INT2(300),INT2(80),xy) 600 WRITE (str$,'(A)') "10. Interconnect with Africa's area center." 601 CALL outgtext (str$) 602 CALL DELAY (DBLE(4)) 603 st=setcolor (14) 604 poly(l).xcoord=INT2(323) 5 C 605 poly(l).ycoord=INT2(323) 5 606 poly(2).xcoord=INT2(318) 6 607 poly(2).ycoord=INT2(324) 6 608 poly(3).xcoord=INT2(335) 16 609 poly(3).ycoord=INT2(477) 16 610 poly(4).xcoord=INT2(339) 15 611 poly(4).ycoord=INT2(477) 15 612 poly(5).xcoord=INT2(323) 5 613 poly(5).ycoord=INT2(323) I 5 614 st=polygon ($GFILLINTERIOR,poly,5) 615 sta=setcolor (10) 616 CALL moveto (INT2(10),INT2(465),xy) 617 CALL DELAY (DBLE(2)) 618 WRITE (str$,'(A\)') 'Press ENTER to continue' 619 CALL outgtext (str$) 620 st=setcolor(O) 621 READ (*,*) 622 st=polygon ($GFILLINTERIOR,blenter,5) 623 C-- Australia-------------------------------------------624 300 st=setcolor (INT2(0)) I blackout text 625 st=polygon ($GFILLINTERIOR,bout,5) 626 st=setcolor(INT2(10)) 627 CALL moveto (INT2(300),INT2(80),xy) 628 WRITE (str$,'(A)') "11. Australia's area center is" 629 CALL outgtext (str$) 630 CALL moveto (INT2(320),INT2(100),xy)

631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 712

90

WRITE (str$,'(A)') "a little beyond the horizon." CALL outgtext (str$)


DO I=l, 15 st=setcolor(INT2(10)) I area center of Antarctica st=ellipse ($GFILLINTERIOR,INT2(461),INT2(375),

+INT2(471),INT2(385)) CALL DELAY (DBLE(.2)) st=setcolor(INT2(0)) I area center of Antarctica st=ellipse ($GFILLINTERIOR,INT2(461),INT2(375),

+INT2(471),INT2(385)) CALL DELAY (DBLE(.2)) END DO

sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE(2)) WRITE (str$,'(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ(*,*) st=polygon ($GFILLINTERIOR,blenter,5)

401 st=setcolor (INT2(0)) I blackout text st=polygon ($GFILLINTERIOR,bout,5) st=setcolor(INT2(10)) CALL moveto (INT2(300),INT2(80),xy) WRITE (str$,' (A)') "12. Interconnect with Africa's area center." CALL outgtext (str$) CALL DELAY (DBLE(4)) st=setcolor (14) poly(l).xcoord=INT2(324) 3 B poly(l).ycoord=INT2(319) 3 poly(2).xcoord=INT2(322) 4 poly(2).ycoord=INT2(323) 4 poly(3).xcoord=INT2(466) 14 poly(3).ycoord=INT2(380) 14 poly(4).xcoord=INT2(468) 13 poly(4).ycoord=INT2(376) 13 poly(5).xcoord=INT2(324) 3 poly(5).ycoord=INT2(319) I 3 st=polygon ($GFILLINTERIOR,poly,5)

sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE(2)) WRITE (str$, '(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ (*,*) st=polygon ($GFILLINTERIOR,blenter,5)

C------------------------------------------------------------st=setcolor ( INT2 ( 0)) I blackout text st=polygon ($GFILLINTERIOR,bout,5) st=setcolor(INT2(10)) CALL moveto (INT2(290),INT2(80),xy) WRITE (str$,'(A)') "13. Interconnect all adjacent area centers." CALL outgtext (str$) CALL DELAY (DBLE(4)) st=setcolor (INT2(14)) poly(l).xcoord=INT2(203) 29 J poly(l).ycoord=INT2(214) 29 poly(2).xcoord=INT2(208) 20 poly(2).ycoord=INT2(209) 20 poly(3).xcoord=INT2(402) 21 poly(3).ycoord=INT2(185) 21 poly(4).xcoord=INT2(408) 22 poly(4).ycoord=INT2(189) 22 poly(5).xcoord=INT2(203) 29 poly(5).ycoord=INT2(214) I 29 st=polygon ($GFILLINTERIOR,poly,5) CALL DELAY (DBLE(2)) poly(l).xcoord=INT2(174) I 27 I poly(l).ycoord=INT2(381) I 27 poly(2).xcoord=INT2(171) I 18 poly(2).ycoord=INT2(372) I 18 poly(3).xcoord=INT2(205) I 19 poly(3).ycoord=INT2(212) I 19 poly(4).xcoord=INT2(211) I 28 poly(4).ycoord=INT2(206) I 28 poly(5).xcoord=INT2(174) 1· 27 poly(5).ycoord=INT2(381) I 27 st=polygon ($GFILLINTERIOR,poly,5) CALL DELAY (DBLE(2)) poly(l).xcoord=INT2(342) I 25 H poly(l).ycoord=INT2(476) I 25

Appendix Al

poly(2).xcoord=INT2(335) 16 poly(2).ycoord=INT2(477) 16 poly(3).xcoord=INT2(173) 17 poly(3).ycoord=INT2(377) 17 poly(4).xcoord=INT2(170) 26 poly(4).ycoord=INT2(370) 26 poly(5).xcoord=INT2(342) 25 poly(5).ycoord=INT2(476) I 25 st=polygon ($GFILLINTERIOR,poly,5) CALL DELAY (DBLE(2)) poly(l).xcoord=INT2(469) 23 G poly(l).ycoord=INT2(372) 23 poly(2).xcoord=INT2(466) 14 poly(2).ycoord=INT2(380) 14 poly(3) .xcoord=INT2 ( 339) 15 poly(3).ycoord=INT2(477) 15 poly(4).xcoord=INT2(330) 24 poly(4).ycoord=INT2(478) 24 poly(5).xcoord=INT2(469) 23 poly(5).ycoord=INT2(372) I 23 st=polygon ($GFILLINTERIOR,poly,5) CALL DELAY (DBLE(2)) poly(l).xcoord=INT2(402) 21 F poly(l).ycoord=INT2(185) 21 poly(2).xcoord=INT2(408) 22 poly(2).ycoord=INT2(189) 22 poly(3).xcoord=INT2(468) 13 poly(3).ycoord=INT2(376) 13 poly(4).xcoord=INT2(465) 24 poly(4).ycoord=INT2(383) 24 poly(5).xcoord=INT2(402) 21 poly(5).ycoord=INT2(185) I 21 st=polygon ($GFILLINTERIOR,poly,5) CALL DELAY (DBLE(2))

sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE(2)) WRITE (str$, '(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ (*,*) st=polygon ($GFILLINTERIOR,blenter,5)

st=setcolor (INT2(0)) I blackout text st=polygon ($GFILLINTERIOR,bout,5) st=setcolor(INT2(10)) CALL moveto (INT2(270),INT2(75),xy) WRITE (str$,' (A)') "14. When the red lines run straight through" CALL outgtext (str$) CALL moveto (INT2(320),INT2(95),xy) WRITE ( str$, ' (A)' ) "the body of the globe they form a" CALL outgtext (str$) CALL moveto (INT2(320),INT2(120),xy) st=setcolor (15) WRITE ( str$' ' (A)' ) "PENTAGONAL PYRAMID." CALL outgtext (str$)

sta=setcolor (10) CALL moveto (INT2(10),INT2(465),xy) CALL DELAY (DBLE(2)) WRITE (str$,'(A\)') 'Press ENTER to continue' CALL outgtext (str$) st=setcolor(O) READ(*,*) st=polygon ($GFILLINTERIOR,blenter,5)

713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 C 792 793 794

c .. text ••••••••••••••••••••••••••••••••••••••••••••••••••• 501 st=setcolor (10)

CALL clearscreen (GCLEARSCREEN) tmp=RGB(INT4(63),INT4(63),INT4(63)) st=remappalette (INT2(7),tmp) PRINT '(A,/ ,A,// ,A,/ ,A,/ ,A,// ,A,/A, / ,A,/ ,A,/)',

+' When objects take up ordered (non-random) positions', +' it is reasonable to assume the presence of directed forces.', +' When the area centers of the continents are at the corners of', +' an equilateral pyramid they are as far away from each other', +' as possible on the spherical surface enclosing the pyramid.', +' This arrangement suggests that mutual repulsion forces are at', +' work between the continental landmasses. Electrical repulsion', +' forces can fulfill this role if the landmasses carry a net', +' electric charge.'

PRINT '(A, I ,A, I ,A, I ,A, I ,A, I ,A, /A,// ,A, I ,A, I ,A, I ,A, I ,A, I ,A, I ,A)', +' Imagine an all-ocean planet.', +' Place two small ships on the surface near to each other,',

91

795 796 797 798 799


+' and place like-charges on them. Then let them move under', +' the force. Where will they move? They certainly will not move', +' forever. They will move to some positions, and will stay there', +' even if the electric charges and the repulsion forces remain', +' fully active and totally powerful.',

+' Of course, two ships move exactly 180 degrees apart.', +' Three ships move 120 degrees apart on a great-circle.', +' However, it is not very easy to intuitively find out where', +' four ships move. (They move to form an equilateral triangular', +' pyramid). But for five or more ships you definitely need the', +' help of mathematics to find it out. For example, seven ships', +' move to form a double pentagonal pyramid, the same pattern', +" the Earth's continents form."

CALL moveto (INT2(10),INT2(465),xy) WRITE (str$,'(A\)') 'Press ENTER to continue' CALL outgtext (str$) READ (*,*) CALL clearscreen (GCLEARSCREEN) PRINT '(////////////,9X,A,/,9X,A,/,9X,A,/,9X,A,/,9X,A)',

+' Program ELECTROGLOBE is a rigorous and high precision', +' mathematical model which investigates how electrically', +' charged bodies move on a spherical surface under mutual', +' repulsion forces, and what geometrical pattern they form', +' in their final positions.'

WRITE (str$,'(A\)') 'Press ENTER to continue' CALL moveto (INT2(10),INT2(465),xy) CALL outgtext (str$) READ (*,*)

c •..•.•..•.•..•...•••••...............•........••......•.•••

800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 818 819 820 821 822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848 849 850 851 852 853 854 855 856 857 858 859 10 860 861 862 863 864 865 866 867 868 869 870 871 872 873 874 875

600 st=settextcolor (l) tmp=RGB(INT4(31),INT4(31),INT4(31)) st=remappalette (INT2(0),tmp) CALL clearscreen ($GCLEARSCREEN) CALL TITLE() st=setfont ( "t' courier' "/I' hl2w9' ) END

I O bckgrd grey

C=========================================================== SUBROUTINE RESETGLB ( ) IMPLICIT REAL*B (A-H,0-Z) INTEGER*2 st INCLUDE 'FGRAPH.FD' CALL unregisterfonts() st=setvideomode ($DEFAULTMODE) END

C====================================================== SUBROUTINE SETPRINT (jrow,jcol) IMPLICIT REAL*B (A-H,O-Z) INCLUDE 'FGRAPH.FD' RECORD /rccoord/ curpos CALL settextposition (INT2(jrow),INT2(jcol),curpos) END

c======================================================== SUBROUTINE GLOBE (THX,THY) IMPLICIT REAL*8 (A-H,0-Z) INCLUDE 'FGRAPH.FD' COMMON /GGC/ IP,JG,JFL,LL COMMON /PDR/ PI,DRC,RR COMMON /TRN/ JROTA,kxt,kyt INTEGER*2 st,kxt(8),kyt(8) N=36 I no. of longitudes M=72 I no. of line increments K=lB I no. of parallels st=setwritemode ($GOR) CALL clearscreen ($GCLEARSCREEN)

SELECT CASE (IP) CASE (1)

THX=O.O THY=270.0

CASE (2) THX=90.0 THY=270.0

CASE (3) THX=O.O THY=270.0

CASE (4) THX=90.0 THY=90.0

CASE (5) THX=lBO.O THY=90.0

CASE (6)

Appendix Al

876 877 878 879 880 881 20 882 883 884 885 886 887 888 889 30 890 891 892 893 894 99'5

THX=O.O THY=180.0

END SELECT THX=THX*DRC THY=THY*DRC

DO IG=l,N+l TH=PI/N*IG TH=TH+(5.0*DRC)

DO JG=l,M+l PH=(2.0*PI/M)*JG CALL GLB (PH,TH,THX,THY,l) I Meridians END DO I JG

END DO I IG DO IG=l,K+l PH=PI/K*(IG-1)-PI/2.0 DO JG=l,M+l TH=2.0*PI/M*JG

CALL GLB (PH,TH,THX,THY,l) I Latitudes END DO I JG

END DO I IG END

c==============================. =================== 896 897 898 899 900 901 902 903 20 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957

503

504

600

SUBROUTINE GLB (PH,TH,THX,THY,LFL) IMPLICIT REAL*S (A-H,0-Z) INCLUDE 'FGRAPH.FD' COMMON /GGC/ IP,JG,JFL,LL RECORD /xycoord/ xy X=DCOS(PH)*DCOS(TH) Y=DCOS(PH)*DSIN(TH) Z=DSIN(PH) Xl=X Yl=Y IF (THX .EQ. 0.0) GOTO 503 Y=Yl*DCOS(THX)-Z*DSIN(THX) Z=Yl*DSIN(THX)+Z*DCOS(THX) SELECT CASE (IP) CASE (1)

CC=Z DYP=400.0 DZP=240.0

CASE (2) CC=Z DZP=400.0 DYP=240.0

CASE (3) CC=X DYP=400.0 DXP=240.0

CASE (4) CC=Z DZP=400.0 DYP=240.0

CASE (5) CC=X DYP=400.0 DXP=240.0

CASE (6) CC=Z DYP=400.0 DXP=240.0

END SELECT IF (THY .EQ. 0.0) GOTO 504 X=Xl*DCOS(THY)+Z*DSIN(THY) Z=-Xl*DSIN(THY)+Z*DCOS(THY) X=X*230.0+DXP Y=Y*230.0+DYP Z=Z*230.0+DZP

IF (LFL .NE. 1) GOTO 510 SELECT CASE (IP) CASE (1)

display planes

IF (JG .EQ. 1 .OR. JFL .EQ. 1) THEN CALL moveto(Y,Z,xy) ELSE

IF (CC .LT. 0.0) THEN st=setcolor(2) st=lineto(Y,Z) ELSE st=setcolor(6) st=lineto(Y,Z) END IF

END IF CASE (2)

93

958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033 1034 1035 1036 1037 1038 1039

94

IF (JG .EQ. 1 .OR. JFL .EQ. 1) THEN CALL moveto(Z,Y,xy) ELSE

IF (CC .GT. 0.0) THEN st=setcolor(2) st=lineto(Z,Y) ELSE st=setcolor(6) st=lineto(Z,Y) END IF

END IF CASE (3)

IF (JG .EQ. 1 .OR. JFL .EQ. 1) THEN CALL moveto(Y,X,xy) ELSE

IF (CC .LT. 0.0) THEN st=setcolor(2) st=lineto(Y,X) ELSE st=setcolor(6) st=lineto(Y,X) END IF

END IF CASE (4)

IF (JG .EQ. 1 .OR. JFL .EQ. 1) THEN CALL moveto(Z,Y,xy) END IF IF (CC .LT. 0.0) THEN st=setcolor(2) st=lineto(Z,Y) ELSE st=setcolor(6) st=lineto(Z,Y) END IF

CASE (5) IF (JG .EQ. 1 .OR. JFL .EQ. 1) THEN CALL moveto(Y,X,xy) ELSE

IF (CC .GT. 0.0) THEN st=setcolor(2) st=lineto(Y,X) ELSE st=setcolor(6) st=lineto(Y,X) END IF

END IF CASE (6)

IF (JG .EQ. 1 .OR. JFL .EQ. 1) THEN CALL moveto(Y,X,xy) ELSE

IF (CC .GT. 0.0) THEN st=setcolor(6) st=lineto(Y,X) ELSE st=setcolor(2) st=lineto(Y,X) END IF

END IF END SELECT


510 IF (LFL .EQ. 3) THEN I mark zero degree latitude & longitude SELECT CASE (IP)

CASE (1) CALL ELLI (Y,Z,4,I) bright green

CASE (2) CALL ELLI (Z,Y,4,I) bright green

CASE (3) CALL ELLI (Y,X,4,I) bright green

CASE (4) CALL ELLI (Z,Y,4,I) bright green

CASE (5) CALL ELLI (Y,X,3,I) dim green

CASE (6) CALL ELLI (Y,X,4,I) bright green

END SELECT END IF END

C================================================== SUBROUTINE PATH (THX,THY,DE) IMPLICIT REAL*8 (A-H,O~Z) INCLUDE 'FGRAPH.FD' COMMON /PDR/ PI,DRC,RR COMMON /GGC/ IP,JG,JFL,LL

(

(

1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 1050 1051 1052 1053 1054 1055 1056 1057 1058 1059 1060 1061 1062 1063 1064 1065 1066 1067 1068 1069 1070 1071 1072 1073 1074 1075 1076 1077 1078 1079 1080 1081 1082 1083 1084 1085 1086 1087 1088 1089 1090 1091 1092 1093 1094 1095 1096 1097 1098 1099 1100 1101 1102 1103 1104 1105 1106 1107 1108 1109 1110 1111 1112 1113 1114 1115 1116 1117 1118 1119 1120 1121

Appendix Al 95

18

20

503

504

COMMON /ITEMS/ CNA$(8),SHP$(8) COMMON /LST/ TH(8),PH(8),GG(8),TCK(8) COMMON /TRN/ JROTA,kxt,kyt COMMON /TEX/ ICNT,LM,NM$,NN,JXT,JYT INTEGER*2 JXT(8),JYT(8),kxt(8),kyt(8) CHARACTER CNA$*1,SHP$*9,NM$*3 RECORD /rccoord/ curpos IF (JROTA .EQ. 1) GOTO 18 I don't print coordinates IF (DE .EQ. 0.0 .AND. ICNT .LT. LM) GOTO 18 CALL settextposition (INT2(5),INT2(1),curpos) .PRINT '(9X,A, I4,2X, I4)', 'LM =' ,LM, ICNT PRINT '(4X,A)', 'LONGITUDE LATITUDE' CALL DELAY (DE)

DO I=l,NN IF (JROTA .EQ. 1) GOTO 20 IF (DE .EQ. 0.0 .AND. ICNT .LT. LM) GOTO 20 PRINT '(1X,A,1X,Fl0.5,1X,F9.5)' ,CNA$(I),TH(I),PH(I)

IF (JROTA .EQ. 1) TH(I)=TH(I)+0.5 I rotate by 0.5 degree X=DCOS(PH(I)*DRC)*DCOS(TH(I)*DRC) I plot the path

Y=DCOS(PH(I)*DRC)*DSIN(TH(I)*DRC) Z=DSIN(PH(I)*DRC)

Xl=X Yl=Y IF (THX .EQ. 0.0) GOTO 503 Y=Yl*DCOS(THX)-Z*DSIN(THX) Z=Yl*DSIN(THX)+Z*DCOS(THX) SELECT CASE (IP)

CASE (1) cc=z DYP=400.0 DZP=240.0

CASE (2) CC=Z DZP=400.0 DYP=240.0

CASE (3) CC=X DYP=400.0 DXP=240.0

CASE (4) CC=Z DZP=400.0 DYP=240.0

CASE (5) CC=X DYP=400.0 DXP=240.0

CASE (6) CC=Z DYP=400.0 DXP=240.0

END SELECT IF (THY .EQ. 0.0) GOTO 504 X=Xl*DCOS(THY)+Z*DSIN(THY) Z=-Xl*DSIN(THY)+Z*DCOS(THY) X=X*230.0+DXP Y=Y*230.0+DYP Z=Z*230.0+DZP IF (JROTA .EQ. 1) GOTO 520 SELECT CASE (IP) CASE (1)

IF (CC .GE. 0.01) K=l IF (CC .LT. 0.01) K=2 CALL ELLI (Y,Z,K,I)

CASE (2) IF (CC .LT. 0.01) K=l IF (CC .GE. 0.01) K=2 CALL ELLI (Z,Y,K,I)

CASE (3) IF (CC .GE. 0.01) K=l IF (CC .LT. 0.01) K=2 CALL ELLI (Y,X,K,I)

CASE (4) IF (CC .GE. 0.01) K=l IF (CC .LT. 0.01) K=2 CALL ELLI (Z,Y,K,I)

CASE (5) IF (CC .LT. 0.01) K=l IF (CC .GE. 0.01) K=2 CALL ELLI (Y,X,K,I)

CASE (6) IF (CC .LT. 0.01) K=l

1122 1123 1124 1125 1126 1127 1128 1129 1130 1131 1132 1133 1134 1135 1136 1137 1138 1139 1140 1141 1142 1143 1144 1145 1146 1147 1148 1149 1150 1151 1152 1153 1154 1155 1156 1157 1158 1159 1160 1161 1162 1163 1164 1165 1166 1167 1168 1169 1170 1171 1172 1173 1174 1175 1176 1177 1178 1179 1180 1181 1182 1183 1184 1185 1186 1187 1188 1189 1190 1191 1192 1193 1194 1195 1196 1197 1198 1199 1200 1201 1202 1203

96

IF (CC .GE.' 0.01) K=2 CALL ELLI (Y,X,K,I)

END SELECT CYCLE


520 SELECT CASE (IP) CASE (1)

jxt(i)=Y jyt(i)=Z

CASE (2) jxt(i)=Z jyt(i)=Y

CASE (3) jxt(i)=Y jyt(i)=X

CASE (4) jxt(i)=Z jyt(i)=Y



END SELECT END DO I I END

C===================================== SUBROUTINE ELLI (xxc,yyc,K,I) IMPLICIT REAL*8 (A-H,O-Z) INCLUDE 'FGRAPH.FD' REAL*8 xxc,yyc INTEGER*2 st,xu,yu,xl,yl,JXT(8),JYT(8),kxt(8),kyt(8) CHARACTER SHP$*9,CNA$*1,NM$*3 COMMON /PDR/ PI,DRC,RR COMMON /ITEMS/ CNA$(8),SHP$(8) COMMON /TEX/ ICNT,LM,NM$,NN,JXT,JYT COMMON /LST/ TH(8),PH(8),GG(8),TCK(8) COMMON /TRN/ JROTA,kxt,kyt RECORD /xycoord/ xy IF (K .EQ. 1) st=setcolor (14) IF (K .EQ. 2) st=setcolor (15) IF (K .EQ. 3) st=setcolor (12) IF (K .EQ. 4) st=setcolor (13) xc=INT2(xxc) I for path yc=INT2(yyc) size=INT2(2) xu=xc-size yu=yc-size xl=xc+size yl=yc+size st=ellipse ($GFILLINTERIOR,xu,yu,xl,yl) I path

IF (ICNT .EQ. 0 .AND. K .LE. 2) THEN I 1st bckgr of lettering xcl=xc+l2 ycl=yc sizel=6 xul=xcl-sizel yul=ycl-sizel xll=xcl+sizel yll=ycl+sizel st=setcolor (INT2(1)) st=ellipse ($GFILLINTERIOR,xul,yul,xll,yll) CALL moveto (INT2(xxc)+8,INT2(yyc)-6,xy) I letters st=setcolor (INT2(4)) CALL outgtext (CNA$(I))

END IF IF (ICNT .EQ. LM .AND. K .LE. 2) THEN I last bckgr of lettering xcl=xc+l2 ycl=yc sizel=6 xul=xcl-sizel yul=ycl-sizel xll=xcl+sizel yll=ycl +sizel

st=setcolor (INT2(5)) st=ellipse ($GFILLINTERIOR,xul,yul,xll,yll) CALL moveto (INT2(xxc)+8,INT2(yyc)-6,xy) I letters st=setcolor (INT2(4)) CALL outgtext (CNA$(I))

END IF IF (ICNT .EQ. LM) THEN

JXT(I)=INT2(xxc) I chords JYT(I)=INT2(yyc) I chords

(

(

(

(

Appendix Al

,END IF END

C===========================================

1204 1205 1206 1207 1208 1209 1210 1211 1212 1213 1214 1215 1216 1217 1218 1219 1220 12 1221 1222 1223 1224 1225 1226 1227 1228 1229 1230 24 1231 1232 1233 1234 1235 1236 1237 1238 1239 1240 1241 1242 1243 1244 1245 1246 1247 1248 1249 1250 1251 1252 1253 1254 1255 1256 1257 1258 1259 1260 1261 1262 1263 1264 1265 1266 1267 1268 1269 1270 1271 1272 1273 1274 1275 1276 1277 1278 1279 1280 1281 1282 1283 1284 1285

SUBROUTINE GRAPH ( ) IMPLICIT REAL*8 (A-H,0-Z) INCLUDE 'FGRAPH.FD' COMMON /TEX/ ICNT,LM,NM$,NN,JXT,JYT COMMON /TRN/ JROTA,kxt,kyt INTEGER*2 jxt(8),jyt(8),kxt(8),kyt(8),st CHARACTER NM$*3,GRFILE$*9,T$*1,R$*1 RECORD /xycoord/ xy st=setwritemode ($GPSET) GRFILE$='GR'//NM$//'.FIL' OPEN (4,FILE=GRFILE$,STATUS='OLD' ,MODE='READ') READ (4,*) IF (JROTA .EQ. 1) st=setcolor (0)

IF (.NOT. EOF(4)) THEN READ (4,'(A,1X,A)')T$,R$ ind=ICHAR(R$)-64 IF (T$ .EQ. 'M') IF (T$ .EQ. 'L') GOTO 12

END IF st=setcolor (3) REWIND 4 READ (4,*)

CALL moveto (kxt(ind),kyt(ind),xy) st=lineto (kxt(ind~,kyt(ind))

IF (.NOT. EOF(4)) THEN READ (4,' (A,lX,A)' )T$,R$ ind=ICHAR(R$)-64 IF (T$ .EQ. 'M') IF (T$ .EQ. 'L') GOTO 24

END IF

CALL moveto (jxt(ind),jyt(ind),xy) st=lineto (jxt(ind),jyt(ind))

IF (JROTA .EQ. 0) CALL CHORD() REWIND 4 DO i=l,NN kxt(i)=jxt(i) kyt(i)=jyt(i) END DO END

C================================================== SUBROUTINE CHORD ( ) IMPLICIT REAL*8 (A-H,0-Z) INTEGER*2 JXT(8),JYT(8),st,kxt(8),kyt(8) INCLUDE 'FGRAPH.FD' COMMON /LST/ TH(8),PH(8),GG(8),TCK(8) COMMON /PDR/ PI,DRC,RR COMMON /TEX/ ICNT,LM,NM$,NN,JXT,JYT COMMON /TRN/ JROTA,kxt,kyt COMMON /ITEMS/ CNA$(8),SHP$(8) CHARACTER NM$*3,CNA$*1,SHP$*9,

+str2*6,str3*10,str4*14,str5*18,str6*22,str7*26 DIMENSION L(7,7) RECORD /rccoord/ curpos R=lOO DO I=l,NN

DO J=l,NN IF (J .GE. I) CYCLE T2I=TH(I) T2J=TH(J) P2I=PH(I) P2J=PH(J)

IF (T2I .EQ. T2J) IF (P2I .EQ. P2J) IF (T2I .EQ. 0.0) IF (P2I .EQ. 0.0)

AD=90.0-P2I BD=90.0-P2J CD=T2I-T2J AR=AD*DRC BR=BD*DRC CR=CD*DRC

T2I=T2I+l.OE-06 P2I=P2I+l.OE-06 T2I=l.OE-06 P2I=l.OE-06

Bl=DCOS((BR-AR)/2.0) B2=1/DCOS((BR+AR)/2.0) B3=1/DTAN(CR/2.0) BPA=2*DATAN(Bl*B2*B3) B4=DSIN((BR-AR)/2.0) B5=1.0/DSIN((BR+AR)/2.0) BMA=2.0*DATAN(B4*B5*B3) B6=DTAN((BR-AR)/2.0) B7=DSIN((BPA)/2.0) BBB=DSIN(BMA/2.0)

97

98

IF (BBB .EQ. 0.0) BBB=l.OE-06 B8=1.0/BBB CC=2.0*DATAN(B6*B7*B8) . L(I,J)=IDNINT((DABS(2.0*DSIN(CC/2.0))*l00.0))

ENDDOIJ END DO I I st=settextcolor (7)


CALL settextwindow (INT2(0),INT2(0),INT2(30),INT2(30)) CALL settextposition (INT2(23),INT2(1),curpos) WRITE (*,'(5X,6(A,3X))') (CNA$(K),K=l,NN-l)

CALL settextposition (INT2(24),INT2(1),curpos) WRITE (str2,' ( lX,A, 2 ( 1X,I3))') CNA$ (2), (L(2 ,J) ,J=l, 1) CALL outtext (str2) IF (NN .EQ. 2) RETURN

CALL settextposition (INT2(25),INT2(1),curpos) WRITE ( str3,' ( lX,A, 3( lX, I3))') CNA$ (3), (L(3,J) ,J=l, 2) CALL outtext (str3) IF (NN ;EQ. 3) RETURN

CALL settextposition (INT2(26),INT2(1),curpos) WRITE (str4,'(1X,A,4(1X,I3))') CNA$(4),(L(4,J),J=l,3) CALL outtext (str4) IF (NN .EQ. 4) .RETURN

CALL settextposition (INT2(27),INT2(1),curpos) WRITE (str5,' ( lX,A, 5( 1X,I3))') CNA$ (5), (L(5 ,J) ,J=l, 4) CALL outtext (str5) IF (NN .EQ. 5) RETURN

CALL settextposition (INT2(28),INT2(1),curpos) WRITE (str6,'(1X,A,6(1X,I3))') CNA$(6),(L(6,J),J=l,5) CALL outtext (str6) IF (NN .EQ. 6) RETURN

CALL settextposition (INT2(29),INT2(1),curpos) WRITE (str7,'(1X,A,7(1X,I3))') CNA$(7),(L(7,J),J=l,6) CALL outtext (str7)

END C========================================================

1286 1287 1288 1289 1290 1291 1292 1293 1294 1295 1296 1297 1298 1299 1300 1301 1302 1303 1304 1305 1306 1307 1308 1309 1310 1311 1312 1313 1314 1315 1316 1317 1318 1319 1320 1321 1322 1323 1324 10 1325 1326 1327 1328 1329 1330 1331 1332 1333 1334 1335 1336 1337 1338 1339 1340 1341 1342 1343 1344 1345 1346 1347 1348 1349 1350

·1351 1352 1353 1354 1355 1356 1357 1358 1359 1360 1361 1362 1363 1364 1365 1366 1367

SUBROUTINE DELAY (DT) IMPLICIT REAL*8 (A-Z) begin time=secnds() end_time=secnds() diff_time=end_time-begin_time IF (diff time .LT. DT) GOTO 10 END -

C=========================================== REAL*8 FUNCTION secnds() INTEGER*2 hour,minute,second,hundredth CALL GETTIM(hour,minute,second,hundredth) secnds=((DBLE(hour)*3600.0)+

+(DBLE(minute)*60.0)+DBLE(second)+ +(DBLE(hundredth)/100.0))

END C===================================================

SUBROUTINE CIRCLE () IMPLICIT REAL*8 (A-H,O-Z) INCLUDE 'FGRAPH.FD' INTEGER*2 st st=setcolor(6) st=ellipse ($GBORDER,INT2(168),INT2(8),INT2(632),INT2(472)) END

C==================================================== SUBROUTINE RC16EN (IGT,IP) IMPLICIT REAL*8 (A-H,O-Z) INCLUDE-'FGRAPH.FD' COMMON /TEX/ ICNT,LM,NM$,NN,JXT,JYT COMMON /TRN/ JROTA,kxt,kyt COMMON /FNAME/ INFILENAM$ INTEGER*2 JXT(8),JYT(8),kxt(8),kyt(8) CHARACTER R$*1,NM$*3,INFILENAM$*9 RECORD /rccoord/ curpos CALL settextposition (INT2(1),INT2(1),curpos) WRITE (*, '(1X,A,2X,A,Il)') INFILENAM$,'IP=',IP IGT=O WRITE (*,'(lX,A) ')'' WRITE (*,'(lX,A\)')'G, 1-6, E, C, R ? : ' READ (*,'(A)') R$

IF (R$ .EQ. 'R' .OR. R$ .EQ. 'r') THEN IGT=16 JROTA=l GOTO 9 END IF

rotate

IF (R$ .EQ. 'C' .OR. R$ .EQ. 'c') THEN I draw chords IGT=15 GOTO 15

(

(

Appendix Al

END IF IF (R$ .EQ. 'G' .OR. R$ .EQ. 'g') THEN 1 go IGT=13 GOTO 15 END IF

IF (R$ .EQ. 'E' .OR. R$ .EQ. 'e') THEN IGT=ll GOTO 9 END IF

1 end

IF (R$ .EQ. '1' .OR. R$ .EQ. '2' .OR. R$ .EQ. '3' .OR. + R$ .EQ. '4' .OR. R$ .EQ. '5' .OR. R$ .EQ. '6') THEN

IGT=ICHAR(R$)-48 IP=IGT GOTO 9 END IF IGT=14 GOTO 15

1 continue step-by-step

CALL clearscreen ($GCLEARSCREEN) CONTINUE END

1 new IP

c===================================================

1368 1369 1370 1371 1372 1373 1374 1375 1376 1377 1378 1379 1380 1381 1382 1383 1384 1385 9 1386 15 1387 1388 1389 1390 1391 1392 1393 1394 1395 1396 1397 1398 1399 1400 1401 1402 1403 1404 1405 1406 1407 1408 1409 1410 1411 1412 25 1413 1414 1415 1416 1417 1418 1419 1420 1421 1422 1423 1424 1425 1426 1427 1428 1429 1430 1431 1432 1433 1434 1435 1436 1437 1438 1439 1440 1441 1442 1443 1444 1445 1446

INTEGER*4 FUNCTION RGB (r,g,b) INTEGER*4 r,g,b RGB=ISHL (ISHL(b,8) .OR. g,8) .OR. r RETURN END

C===================================================

144 7 35 1448 1449

SUBROUTINE SETIN (NN,TH,PH,TCK) IMPLICIT REAL*8 (A-H,O-Z) INCLUDE 'FGRAPH.FD' DIMENSION TH(8),PH(8),TCK(8) CHARACTER ED$*1 CALL clearscreen ($GCLEARSCREEN) CALL SETPRINT (INT4(18),INT4(1)) WRITE (*,'(5X,A)') 'To Edit Positions, enter "P"' PRINT '(lX)' WRITE (*,'(5X,A)') 'To Edit Parameters, enter "A"' PRINT '(lX)' WRITE (*, '(5X,A)') 'else, press "ENTER" PRINT '(lX)' WRITE (*, '(5X,A\)')' READ (*,'(A)') ED$ IF (ED$ .EQ. 'P' .OR. ED$ .EQ. 'p') GOTO 25 IF (ED$ .EQ. 'A' .OR. ED$ .EQ. 'a') GOTO 40 CALL clearscreen ($GCLEARSCREEN) CALL SETPRINT (INT4(1),INT4(1)) PRINT '(5X,A)',' - EDIT POSITIONS -PRINT '(lX)' PRINT '(5X,A,F8.l)','A PRINT '(5X,A,F7.2)','A PRINT '(5X,A,F7.2)','A PRINT '(5X,A,F8.l)' ,'B PRINT '(5X,A,F7.2)' ,'B PRINT '(5X,A,F7.2)' ,'B

IF (NN .EQ. 2) GOTO 35 PRINT '(5X,A,F8.1)','C PRINT '(5X,A,F7.2)','C PRINT '(5X,A,F7.2)','C

IF (NN .EQ. 3) GOTO 35 PRINT '(5X,A,F8.l)','D PRINT '(5X,A,F7.2)' ,'D PRINT '(5X,A,F7.2)' ,'D

IF (NN .EQ. 4) GOTO 35 PRINT '(5X,A,F8.l)','E PRINT '(5X,A,F7.2)','E PRINT '(5X,A,F7.2)','E

IF (NN .EQ. 5) GOTO 35 PRINT '(5X,A,F8.l)' ,'F PRINT '(5X,A,F7.2)','F PRINT '(5X,A,F7.2)','F

IF (NN .EQ. 6) GOTO 35 PRINT '(5X,A,F8.1)','G PRINT '(5X,A,F7.2)' ,'G PRINT '(5X,A,F7.2)' ,'G

IF (NN .EQ. 7) GOTO 35 PRINT '(5X,A,F8.1)','H PRINT '(5X,A,F7.2)','H PRINT '(5X,A,F7.2)','H PRINT '(5X)' WRITE (*,'(5X,A\)')

1 thickness 2 longitude 3 latitude 4 thickness 5 longitude 6 latitude

7 thickness 8 longitude 9 latitude






+'Enter item number [99 to end] READ (*,*) ITM

',TCK(l) ',TH(l) ',PH(l) ',TCK(2) ',TH(2) ',PH(2)

',TCK(3) ',TH(3) ',PH(3)

',TCK(4) ',TH(4) ',PH(4)

',TCK(5) ',TH(5) ',PH(5)

',TCK(6) ',TH(6) ',PH(6)

',TCK(7) ',TH(?) ',PH(?)

',TCK(8) ',TH(8) ',PH(8)

>>

99

100

1450 1451 1452 1453 1454 1455 1456 1457 1458 1 1459 1460 1461 2 1462 1463 1464 3 1465 1466 1467 4 1468 1469 1470 5 1471 1472 1473 6 1474 1475 1476 7 1477 1478 1479 8 1480 1481 1482 9 1483 1484 1485 10 1486 1487 1488 11 1489 1490 1491 12 1492 1493 1494 13 1495 1496 1497 14 1498 1499 1500 15 1501 1502 1503 16 1504 1505 1506 17 1507 1508 1509 18 1510 1511 1512 19 1513 1514 1515 20 1516 1517 1518 21 1519 1520 1521 22 1522 1523 1524 23 1525 1526 1527 24 1528 1529 1530 40 1531

IF (ITM .EQ. 99) GOTO 40 IF (ITM .GT. NN*3) THEN PRINT '(5X,A)', 'Wrong item number (press ENTER)' READ (*,*) GOTO 25 END IF

GOTO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17, +18,19,20,21,22,23,24) ITM

.WRITE (*, '(5X,A\)') ' New thickness READ (*,*) TCK(l) GOTO 25 WRITE(*, '(5X,A\)') New longitud~ READ (*,*) TH(l) GOTO 25 WRITE(*, '(5X,A\)') New latitude READ (*,*) PH(l) GOTO 25 WRITE(*, '(5X,A\)') ' New thickness READ (*,'(A)') TCK(2) GOTO 25 WRITE (*, '(5X,A\)') New longitude READ (*,*) TH(2) GOTO 25 WRITE(*, '(5X,A\)') ' New latitude READ (*,*) PH(2) GOTO 25 WRITE (*,'(5X,A\)') New thickness READ (*,*) TCK(3) GOTO 25 WRITE(*, '(5X,A\)') New longitude READ (*,*) TH(3) GOTO 25 WRITE (*,'(5X,A\)') New latitude READ (*,*) PH(3) GOTO 25 WRITE(*, '(5X,A\)') ' New thickness READ (*,*) TCK(4) GOTO 25 WRITE (*, '(5X,A\)') New longitude READ (*,*) TH(4) GOTO 25 WRITE(*, '(5X,A\)') ' New latitude READ (*,*) PH(4) GOTO 25 WRITE(*, '(5X,A\)') New thickness READ (*,*) TCK(5) GOTO 25 WRITE (*,'(5X,A\)') ' New longitude READ (*,*) TH(5) GOTO 25 WRITE(*, '(5X,A\)') New latitude READ (*,*) PH(5) GOTO 25 WRITE (*,'(5X,A\)') ' New thickness READ (*,*) TCK(6) GOTO 25 WRITE (*, '(5X,A\)') New longitude READ (*,*) TH(6) GOTO 25 WRITE (*, '(5X,A\)') New latitude READ (*,*) PH(6) GOTO 25 WRITE (*, '(5X,A\)') ' New thickness READ (*,*) TCK(7) GOTO 25 WRITE (*, '(5X,A\)') ' New longitude READ (*,*) TH(7) GOTO 25 WRITE (*, '(5X,A\)') ' New latitude READ (*,*) PH(7) GOTO 25 WRITE (*, '(5X,A\)') New thickness READ (*,*) TCK(8) GOTO 25 WRITE (*,'(5X,A\)') New longitude READ (*,*) TH(8) GOTO 25 WRITE (*, '(5X,A\)') ' New latitude READ (*,*) PH(8) GOTO 25 CALL clearscreen ($GCLEARSCREEN) END


Appendix Al 101

1532 C======~====~================~=================== 1533 SUBROUTINE TITLE() 1534 IMPLICIT REAL*8 (A-H,0-Z) 1535 INCLUDE 'FGRAPH.FD' 1536 INTEGER*2 st 1537 CHARACTER str$*30 1538 RECORD /xycoord/ xy 1539 st=setcolor (INT2(2)) 1540 st=setfont ("t'helv"'//'h18w9b') I title line 1 1541 CALL moveto ( INT2 (285), INT2 (5) ,xy) 1542 WRITE ( str$, ' (A)' ) 'Program' 1543 CALL outgtext (str$) 1544 st=setfont ("t'helv"'//'h28w16b') I title line 2 1545 st=setcolor (13) 1546 CALL moveto (INT2(155),INT2(30),xy) 1547 WRITE (str$,' (A)') 'ELECTRO GLOBE' 1548 CALL outgtext (str$) 1549 END

AppendixA2

Calculating Center of Areas of Continents on a Spherical Surface.

Necessary equipment is a computer (PC), a Fortran compiler ( e.g. MS-Fortran® 5.1 by Microsoft Corporation), and a Globe or a World Atlas. There are three steps in this procedure. One is to read the coordinates of a number of points of each continent's boundary, and enter them into a data file using Program LONGLAT. Two is to designate subcontinent areas in the data file by using an Editor. Three is to run Program CGCONT which calculates the surface area of each continent and the position of its area center. The programs are prepared to work on 12 continents, six of the Earth, three of Mars, and three of the Moon. I discuss the procedure in details for Africa, but I present all numerical values for the other continents as well. All programs and data files are on the diskette that may come with this book, in directory \APDX_A2.

Africa

Input of coordinate points

The first step is to input coordinate points of the continent's boundary. Ideally, the boundary at the continental shelf would be taken, but for the present purposes the water lines are perfectly adequate. Run LONGLAT which displays the names of the continents in association with a number:

1 Africa 2 Eurasia 3 North America 4 South America 5 Antarctica 6 Australia 7 Mars-A 8 Mars-B 9 Mars-C

10 Moon-A 11 Moon-B 12 Moon-c

Enter the required number, say, 1. The program searches for Africa's data file (afric.dat). If it finds, it counts the number of data points already written, (say, 87) and displays the next input line:

Enter# 88 "LON,LAT" [999.9 to end]

If no such file exists yet, it opens a new one, and the input line will be Enter # 1 •••. The purpose of this arrangement is to facilitate repeated

103


return to the same data file which may be needed when a large number of points are to be entered. If you make an error in the entry, use an Editor to delete or correct the line, and continue with LONGIAT. (The Editor could be used to create the entire file, but LONGLAT is more convenient). Enter longitude first, then latitude, each with one decimal number, and a comma to separate the two numbers. LONGLAT accepts up to 999 entry points. However, after about 50 points for small continents or 150 point for large ones, the results do not improve significantly.

C LONGLAT.FOR to enter coordinates from a globe or map. CHARACTER names$*13,cnt$*9 DIMENSION names$(12),cnt$(12) DATA names$/'Africa', 'Eurasia', 'North America',

+ 'South America','Antarctica', 'Australia', + 'Mars-A' , 'Mars-B' , 'Mars-C' , + 'Moon-A', 'Moon-B', 'Moon-C' I

DATA cnt$ /'afric.dat', 'euras.dat', 'namer.dat', + 'samer.dat', 'antar.dat', 'austr.dat', + 'm.arsa.dat', 'marsb.dat', 'marsc.dat', + 'moona.dat', 'm.oonb.dat', 'moonc.dat'/

PRINT '(////////////)' DO i=l, 12

PRINT '(1X,I2,2X,A)',i,names$(i) END DO PRINT '(//,lX,A\)', 'Enter number of continent : ' READ (*,'(I2)') nn OPEN (1,FILE=cnt$(nn),STATUS='UNKNOWN') OPEN (2,FILE= cnt$(nn)(:5)//'z',STATUS='NEW')

4 IF (.NOT. EOF(l)) THEN READ (1,' (Tl,I3,TS,F6.1,Tl2,F6.l)') kk,rlon,rlat WRITE (2,'(Tl,I3,TS,F6.1,Tl2,F6.l)') kk,rlon,rlat GOTO 4

END IF CLOSE (1) OPEN (l,FILE=cnt$(nn),STATUS='OLD',ACCESS='APPEND')

6 kk=kk+l PRINT '(1X,A,I3,A\)'·,

+'Enter #' ,kk, ', "LON,LAT" [999.9 to end] - ' READ (*,' (F6.l,F6.l)') rlon,rlat IF (rlon .EQ. 999.9) GOTO 8 WRITE (1, '(Tl,I3,TS,F6.l,Tl2,F6.l)' )kk,rlon,rlat GOTO 6

8 CLOSE (1) END

My inputs for Africa are as follows:

afric.dat # LON LAT # LON LAT 1 -5.9 35.8 23 9.0 4.5 2 -7.0 34.0 24 9.5 3.0 3 -9.0 33.0 25 9.0 ·o.o 4 -10.0 30.0 26 10.0 -2.S 5 -12.0 29.0 27 11.0 -4.0 6 -13.0 28.0 28 13.0 -6.0 7 -15.0 26.0 29 13.S -8.0 8 -14.S 24.0 30 14.2 -11.0 9 -16.0 23.5 31 14.0 -12.S

10 -17.0 23.0 32 12.5 -15.0 11 -17.0 22.0 33 11.5 -17.0 12 -16.0 19.0 34 12.0 -18.0 13 -17.0 16.0 35 14.0 -22.0 14 -16.0 12.0 36 15.0 -26.0 15 -15.0 11.0 37 16.0 -29.0 16 -13.0 8.0 38 18.0 -32.0 17 -8.0 5.0 39 18.1 -33.0 18 -4.0 s.s 40 20.0 -35.0 19 -2.0 5.0 41 23.0 -34.0 20 3.0 6.5 42 25.0 -34.5 21 5.0 6.0 43 29.0 -32.S 22 6.0 4.0 44 31.0 -30.0

# LON LAT 45 33.0 -25.0 46 35.S -24.0 47 35 .o -20.0 48 37.5 -17 .s 49 41.0 -15.0 so 40.8 -11.0 51 39.0 -8.0 52 38.8 -6.0 53 41.0 -2.0 54 43.0 o.o 55 45.5 2.0 56 48.S s.o 57 so.o 8.0 58 51.0 11.0 59 50.8 12 .o 60 48.0 11.0 61 45.0 10.3 62 43.5 11.1 63 43.6 12.2 64 41.2 14.6 65 39.0 15.8 66 38.8 17 .8

# 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87

LON LAT 38.0 19.2 37.0 22.0 35.S 23.5 35 .o 25.0 32.S 30.0 32.4 32.4 30.0 31.6 25.0 31.7 22.s 33.0 20.0 32 .o 19.S 30.2 16.7 31.0 15.0 31.2 14.9 32 .2 11.0 33.5 10.8 37.0 s.o 36.8 1.4 36.7

-1.0 35 .2 -4.0 35.1 -4.9 35 .8

AppendixA2 105

Designate subcontinents

The main program CGCONT to calculate the surface area and the center of area of each continent is based on the mathematical principle that the surface area and the area center of a spherical triangle can be exactly calculated if the coordinates of its vertices are known. Accordingly, the irregular surface area of the continent is divided into an arbitrary but adequately large number of perfectly adjacent spherical triangles.

Figure A2-1. · The Center of Area (CA) of a continent is calculated by dividing the area into subcontinents, dividing each into touching spherical triangles, finding the CA of each triangle, and then summing them.

For each triangle two or sometimes all three of its vertices are chosen from the boundary coordinate points. When only two are chosen, then the coordinates of the third vertex point are arbitrarily selected somewhere inland. This third point is a common point for a larger number of triangles. Consider the layout of Africa (Figure A2-1).

87 points determine the boundary as listed in "afric.dat", and all adjacent points are interconnected by straight lines. Two vertices of the first triangle are Points 1 and 2. The only requirement for the third vertex point is that the entire area of the generated triangle must belong to the continent. For


example, 15°E, -10° cannot be chosen because Points 23 and 24 would cut into the triangle, and the area to the west from those points do not belong to Africa. All other considerations are for convenience only. For example, the third vertex point should be common to as many triangles as possible to reduce the manual labor of entering data points. I have chosen 5°E, 20°N, marked on the map as #1. From this strategic location 60 points can be reached on the boundary, and 60 clean and adjacent triangles can be formed. I call the sum of these areas a "subcontinent". I have covered Africa with four such subcontinents. Two large ones, #1 and #2 cover large areas while #3 and #4 are for the nooks and crannies.

You have to prepare a list to designate which boundary point belongs to which subcontinent. The list for Africa is as follows:

CGAF,TBL Cl 20.0 5.0 RA 1 23 RA 47 56 RA 61 74 RA 77 87 FI 1 C2 -10,0 20.0 RA 23 47 FI 23 C3 9.0 47.5 RA 56 61 FI 56 C4 32 .o 22.5 RA 74 77 FI 74

Cl, C2, etc. are the coordinates of the common vertex point in subcontinent 1, 2, etc. RA indicates the starting and ending points in each unbroken sequence of boundary points. Thus the first sequence runs from Point 1 to 23, it jumps here to 47, and runs to 56, etc. Program CGCONT which reads this file, will generate all triangles, including the large one between #1, Point 23, and Point 47. After one or more RA sequences there is a FI line which ensures that the triangle between Point 47 and Point 1 is taken into account, and the subcontinent is one contiguous area. More than one FI point can be in the list, one after the other, providing that the main rule is followed in which the last side of a triangle forms the first side of the next triangle.

Perform the calculations

Run Program CGCONT to obtain the coordinates of the area center, and the area size of a continent. Two specific data files must be available for each continent: for Africa these are "afric.dat" and "cgaf.tbl". The program displays the same input screen as that of LONGLAT, the names of twelve continents, each associated with a number. Type in the number. The results are displayed as follows:

AppendixA2

Africa LON= 18.37 LAT= 7 .15 AREA= .293E+14

107

LON and LAT are the coordinates of the area center in angular degrees of the same system as that of "afric.dat". AREA is in square meters.

Program CGCONT also produces a file with information on all triangles formed in the given continent. This file can be displayed with an Editor, or printed. It can be useful for trouble shooting or for confirmation of data integrity, but otherwise it remains transparent for the user. For Africa the file is this:

cgaf.crd 1 1 20. 0 1 2 35.8 1 3 34.0 1 4 33.0 1 5 30.0 1 6 29.0 1 7 28.0 1 8 26. 0 1 9 24. 0 1 10 23.5 1 11 23.0 1 12 22.0 1 13 19.0 1 14 16.0 1 15 12.0 1 16 11.0 1 17 8.0 1 18 5. 0 1 19 5 .5 1 20 5. 0 1 21 6.5 1 22 6.0 1 23 4.0 1 24 4.5 1 25 -20.0 1 26 -17.5 1 27 -15.0 1 28 -11.0 1 29 -8.0 1 30 -6.0 1 31 -2.0 1 32 .o 1 33 2.0 1 34 5 .o

5.0 -5.9 -7.0 -9.0

-10.0 -12.0 -13.0 -15 .o -14.5 -16.0 -17 .o -17 .o -16.0 -17 .o -16.0 -15.0 -13.0

-8.0 -4.0 -2.0

3.0 5.0 6.0 9.0

35.0 37.5 41.0 40.8 39.0 38.8 41.0 43.0 45.5 48.5

1 35 10.3 1 36 11.1 1 37 12.2 1 38 14.6 1 39 15.8 1 40 17.8 1 41 19.2 1 42 22.0 1 43 23.5 1 44 25.0 1 45 30.0 1 46 32.4 1 47 31.6 1 48 31. 7 1 49 30.2 1 50 31.0 1 51 31.2 1 52 32.2 1 53 33.5 1 54 37.0 1 55 36.8 1 56 36.7 1 57 35.2 1 58 35.1 1 59 35.8 1 60 35.8 2 1 -10.0 2 2 4.5 2 3 3.0 2 4 .o 2 5 -2.5 2 6 -4.0 2 7 -6.0 2 8 -8. 0

45 .o 43.5 43.6 41.2 39.0 38.8 38.0 37.0 35.S 35 .o 32.5 32 .4 30.0 25.0 19.5 16.7 15 .o 14.9 11.0 10.8 5.0 1.4

-1.0 -4 .o -4.9 -5.9 20.0

9.0 9.5 9.0

10.0 11.0 13.0 13.5

2 9 -11.0 2 10 -12.5 2 11 -15.0 2 12 -17 .o 2 13 -18.0 2 14 -22 .o 2 15 -26.0 2 16 -29 .o 2 17 -32.0 2 18 -33.0 2 19 -35 .o 2 20 -34.0 2 21 -34.5 2 22 -32.S 2 23 -30.0 2 24 -25.0 2 25 -24.0 2 26 -20.0 2 27 4 .s 3 1 9.0 3 2 5 .o 3 3 s.o·· 3 4 11.0 3 5 12 .o 3 6 11.0 3 7 10. 3 · 3 8 5.0 4 1 32.0··· 4 2 31. 7 4 3 33.0 4 4 32 .o 4 5 30.2 4 6 31. 7

14.2 14.0 12.5 11.s 12.0 14.0 15.0 16.0 18.0 18.1 20.0 23.0 25.0 29.0 31.0 33.0 35.5 35.0

9.0 47.5 48.5 50.0 51.0 so.a 48.0 45.0 48.S 22.S 25.0 22.S 20.0 19.5 25.0

Column 1 is the subcontinent's number, 2 is the vertex number, 3 is the latitude of the vertex, and 4 is the longitude of the vertex. Vertex No. 1 in each subcontinent is the common point for all triangles in that subcontinent. Considering the rule of triangle formation in which "the last side of a triangle is the first side of the next triangle", any triangle can be identified in this table. For example the first triangle is determined by Vertex Nos 1, 2, 3, the second triangle is by 1, 3, 4, etc.


The program printout of CGCONT is as follows:

c CGCONT,FOR to calculate cg-points and areas, c .dat and .tbl files are required, .crd is produced.

IMPLICIT REAL*8 (A-H,0-Z) CHARACTER b$*2,c$*2,names$*13,cnt$*9,crd$*12,tbl$*8,crdfile$*8 DIMENSION dla(200),dlo(200),names$(12),cnt$(12),crd$(12),

+tbl$ ( 12) ,R( 12) COMMON PI,DRC,RR DATA names$/'Africa', 'Eurasia', 'North America',

+ 'South America','Antarctica','Australia', + 'Mars-A', 'Mars-B', 'Mars-C', + 'Moon-A' , 'Moon-B' , 'Moon-C' I

DATA R / 6370000, 637.0000, 6370000, + 6370000, 6370000, 6370000, + 3395000, 3395000, 3395000, + 1738000, 1738000, 1738000/

DATA cot$ I 'afric.dat', 'euras.dat', 'namer.dat', + 'samer.dat', 'antar.dat', 'austr.dat', + 'marsa.dat', 'marsb,dat', 'marsc.dat', + 'moona.dat', 'moonb,dat', 'moonc.dat'/

DATA crd$ /'cgaf,crd', 'cgeu.crd', 'cgna.crd', + 'cgsa.crd', 'cgan.crd', 'cgau.crd', + 'cmaa.crd', 'cmab.crd', 'cmac.crd', + 'cmoa.crd', 'cmob.crd', 'cmoc.crd'/

DATA tbl$ /'cgaf.tbl', 'cgeu.tbl', 'cgna.tbl', + 'cgsa.tbl', 'cgan.tbl', 'cgau.tbl', + 'cmaa.tbl', 'cmab, tbl', 'cmac.tbl', + 'cmoa. tbl', 'cmob, tbl', 'cmoc, tbl' /

PRINT '(///////////,lX)' DO i=l,12 PRINT '(1X,I2,1X,A)' ,i,names$(i) END DO PRINT '(lX,A\)','Enter number of continent READ(*,*) nn RR=R(nn) crdfile$=crd$(nn) OPEN (l,FILE=tbl$(nn),STATUS='OLD',MODE='READ') OPEN (2,FILE=crdfile$,STATUS='UNKNOWN',MODE='WRITE') OPEN (3,FILE=cnt$(nn),STATUS='OLD',MODE='READ') READ (1, '(lX) ') WRITE (2, '(A)') crd$(nn)

2 IF (,NOT, EOF(3)) THEN READ (3, '(Tl,I3,T5,F6,l,T12,F6.1)') i,dlo(i),dla(i) GOTO 2 END IF

4 IF (,NOT, EOF(l)) THEN READ (1, '(A)') b$

SELECT CASE (b$(1:l)) CASE ( 'C')

kk=O BACKSPACE 1 READ (1, '(A,T7,F5,l,T13,F6.l)') c$,rla,rlo kk=kk+l NSC=ICHAR(b$(2:2))-48 WRITE (2, '(Tl,I2,T5,I2,T8,F5,l,Tl4,F6.l)') NSC,kk,rla,rlo

CASE ( 'R') BACKSPACE 1 READ (1,'(A,T7,I3,T13,I4)') c$,Jl,J2 DO i=Jl,J2 kk=kk+l WRITE (2, '(Tl,I2,T5,I2,T8,F5,1,Tl4,F6,l)') NSC,kk,dla(i),dlo(i) END DO

CASE ( 'F') BACKSPACE 1 READ (l,'(A,T7,I4)') c$,n kk=kk+l WRITE (2, '(Tl,I2,T5,I2,T8,F5,1,Tl4,F6.l)') NSC,kk,dla(n),dlo(n)

END SELECT GOTO 4

END IF

AppendixA2

PRINT '(lX)' CLOSE (1) CLOSE (2) CLOSE (3) OPEN (2,FILE=crdfile$,STATUS='OLD',MODE='READ') PRINT '(///////////////////,1X,A)',nallles$(nn) CALL CGCONT ( ) END

109

C================================================================= SUBROUTINE CGCONT ()

IMPLICIT REAL*8 (A-H,O-Z) COMMON PI, DRC, RR CHARACTER CARD$*8 DIMENSION PPH(l0,100), TTH(l0,100), NP(S,100), AASC(S,100),

+THSC(lOO), PHSC(lOO),ATR(lOO),TX(lOO),PX(lOO),NNS(8) NCT=l PI=DATAN(l)*4 DRC=PI/180 READ (2, '(A)') CARD$

12 IF (.NOT. EOF(2)) THEN READ (2, '(Tl,I3,T5,I3,T8,F5.l,Tl4,F6.l)') NSC,NPT,RPH,RTH PPH(NSC,NPT)=RPH TTH(NSC,NPT)=RTH NP(NCT, NSC)=NPT

GOTO 12 END IF CLOSE (2) NNS(NCT)=NSC DO J=l,NSC NPTS=NP(NCT,J) NTRI=NPTS-2 ASUM=O THSC(J)=O PHSC(J)=O DO K=l,NTRI Pl=PPH(J,l) Tl=TTH(J,l) P2=PPH(J,K+l) T2=TTH(J,K+l) P3=PPH(J,K+2) T3=TTH(J,K+2) IF ((Tl .EQ. T2 .AND. Pl .EQ. P2)

+ .OR. (T2 .EQ. T3 .AND. P2 .EQ. P3) + .OR. (T3 .EQ. Tl .AND. P3 .EQ. Pl)) THEN

PRINT '(A, I2 ,A, I2) ', +' ERROR: Faulty triangle at NSC=',J,' NTRI=',K

STOP END IF ATR(K)=AREA (Tl,T2,T3,Pl,P2,P3) ASUM=ASUM+ATR(K) CALL CGTRI (Tl,T2,T3,Pl,P2,P3,TC,PC) TX(K)=TC PX(K)=PC END DO IK CALL CGSUB (NTRI,ATR,Tl,Pl,TX,PX,TC,PC) THSC(J)=TC PHSC(J)=PC AASC(NCT,J)=ASUM ARSUM=ARSUM+ASUM END DO IJ CALL CGCON8 (NCT,NSC,AASC,Tl,Pl,THSC,PHSC,TC,PC) PRINT '(A,F7.2)',' LON=',TC PRINT '(A,F6.2)' ,' LAT=',PC PRINT '(A,E9.3)',' AREA=',ARSUM END

C=========================================== FUNCTION AREA (Tl,T2,T3,Pl,P2,P3) IMPLICIT REAL*8 (A-H,O-Z) COMMON PI,DRC,RR Xl=RR*DCOS(Pl*DRC)*DCOS(Tl*DRC) Yl=RR*DCOS(Pl*DRC)*DSIN(Tl*DRC) Zl=RR*DSIN(Pl*DRC)

110

X2=RR*DCOS{P2*DRC)*DCOS(T2*DRC) Y2=RR*DCOS(P2*DRC)*DSIN(T2*DRC) Z2=RR*DSIN(P2*DRC) X3=RR*DCOS(P3*DRC)*DCOS(T3*DRC) Y3=RR*DCOS(P3*DRC)*DSIN(T3*DRC) Z3=RR*DSIN(P3*DRC) XXl=DSQRT{Xl*Xl+Yl*Yl+Zl*Zl) XX2=DSQRT(X2*X2+Y2*Y2+Z2*Z2) XX3=DSQRT{X3*X3+Y3*Y3+Z3*Z3) A=DACOS{(X2*X3+Y2*Y3+Z2*Z3 )/ (XX2*XX3))


I B=DACOS({Xl*X2+Yl*Y2+Zl*Z2)/(XXl*XX2)) C=DACOS({Xl*X3+Yl*Y3+Zl*Z3)/(XXl*XX3)) S=0,5*(A+B+C) TR=DSQRT((DSIN(S-A)*DSIN(S-B)*DSIN(S-C))/(DSIN(S))) AA=2,0*DATAN(TR/DSIN(S-A)) BB=2,0*DATAN(TR/DSIN(S-B)) CC=2,0*DATAN(TR/DSIN(S-C)) E={AA+BB+CC)/DRC-180.0 AREA=PI*E*RR*RR/180.0 END

C=========================================== SUBROUTINE CGTRI {Tl,T2,T3,Pl,P2,P3,TC,PC) IMPLICIT REAL*8 (A-H,O-Z) COMMON PI,DRC,RR Xl=RR*DCOS(Pl*DRC)*DCOS(Tl*DRC) Yl=RR*DCOS(Pl*DRC)*DSIN(Tl*DRC) Zl=RR*DSIN{Pl*DRC) X2=RR*DCOS(P2*DRC)*DCOS(T2*DRC) Y2=RR*DCOS{P2*DRC)*DSIN(T2*DRC) Z2=RR*DSIN(P2*DRC) X3=RR*DCOS(P3*DRC)*DCOS(T3*DRC) Y3=RR*DCOS(P3*DRC)*DSIN(T3*DRC) Z3=RR*DSIN(P3*DRC) XM=(X2+X3)/2,0 YM=(Y2+Y3)/2,0 ZM=(Z2+Z3)/2,0 XC=2,0/3,0*(XM-Xl)+Xl YC=2,0/3,0*(YM-Yl)+Yl ZC=2,0/3,0*{ZM-Zl)+Zl ZS=DSQRT (XC**2+YC**2) TC=DATAN2(YC,XC)/DRC PC=DATAN2{ZC,ZS)/DRC END

C=========================================== SUBROUTINE CGSUB (NTRI,ATR,Tl,Pl,TX,PX,TC,PC) IMPLICIT REAL*8 (A-H,O-Z) COMMON PI,DRC,RR DIMENSION ATR(8),TX(8),PX(8) Xl=RR*DCOS(Pl*DRC)*DCOS(Tl*DRC) Yl=RR*DCOS{Pl*DRC)*DSIN(Tl*DRC) Zl=RR*DSIN(Pl*DRC) XMSUM=0,0 YMSUM=0,0 ZMSUM=0,0 ASUM=O,O

DO I=l,NTRI XX=RR*DCOS(PX(I)*DRC)*DCOS(TX{I)*DRC) XMSUM=XMSUM+ATR(I)*(XX-Xl) YY=RR*DCOS{PX(I)*DRC)*DSIN(TX(I)*DRC) YMSUM=YMSUM+ATR(I)*(YY-Yl) ZZ=RR*DSIN(PX(I)*DRC) ZMSUM=ZMSUM+ATR(I)*(ZZ-Zl) ASUM=ASUM+ATR{I) ENDDO I I

XC=Xl+XMSUM/ASUM YC=Yl+YMSUM/ASUM ZC=Zl+ZMSUM/ASUM ZW=DSQRT(XC**2+YC**2) TC=DATAN2 (YC,.XC) /DRC PC=DATAN2(ZC,ZW)/DRC END

C=============================================================

AppendixA2

SUBROUTINE CGCON8 (NCT,NSUB,AASC,Tl,Pl,THSC,PHSC,TC,PC) IMPLICIT REAL*8 (A-H,0-Z) COMMON PI,DRC,RR DIMENSION AASC(8,9),PHSC(9),THSC(9) Xl=RR*DCOS(Pl*DRC)*DCOS(Tl*DRC) Yl=RR*DCOS(Pl*DRC)*DSIN(Tl*DRC) Zl=RR*DSIN(Pl*DRC) XMSUM=O.O YMSUM=O.O ZMSUM=O.O ASUM= 0.0

DO I=l,NSUB X=RR*DCOS(PHSC(I)*DRC)*DCOS(THSC(I)*DRC) XMSUM=XMSUM+AASC(NCT,I)*(X-Xl) Y=RR*DCOS(PHSC(I)*DRC)*DSIN(THSC(I)*DRC) YMSUM=YMSUM+AASC(NCT,I)*(Y-Yl) Z=RR*DSIN(PHSC(I)*DRC) ZMSUM=ZMSUM+AASC(NCT,I)*(Z-Zl) ASUM=ASUM+AASC(NCT,I) ENDDO ! I

IF (ASUM) 10,20,30 10 STOP 'error in CGCON8' 20 ASUM=l.OE-10 30 XC=Xl+XMSUM/ASUM

YC=Yl+YMSUM/ASUM ZC=Zl+ZMSUM/ASUM ZW=DSQRT(XC**2+YC**2) TC=DATAN2(YC,XC)/DRC PC=DATAN2(ZC,ZW)/DRC END

111


Eurasia

Figure A2-2. Eurasia

euras.dat # LON LAT # LON LAT # LON LAT # LON LAT 1 25.0 72.0 31 -4,5 36,0 61 74.5 13.5 91 126.5 37.5 2 20.0 70.0 32 -4.0 36,5 62 77 .o 9.0 92 126.5 35 .o 3 11.0 65.0 33 -1.0 36.5 63 79.0 8.0 93 128.0 34.8 4 9.0 63.5 34 o.o 38.0 64 79.5 9.0 94 129 .5 36.0 5 7.0 63.0 35 -0.3 39.0 65 80.5 13.0 95 130.0 37.0 6 5.0 62.0 36 1.5 41.0 66 80.0 15 .o 96 129.0 38.0 7 5.0 60.0 37 3.0 42.0 67 82.0 17 .o 97 127.0 40.0 8 7.0 58.5 38 5.0 43.0 68 85 .o 19.0 98 130.0 42 .o 9 8.0 57.0 39 10.0 44.0 69 87.0 21.0 99 135 .o 43.0

10 5.0 56.0 40 15.0 40.0 70 90.0 22.0 100 138.0 46.0 11 1.0 57.5 41 20.0 40.0 71 92.0 21.0 101 140.0 50.0 12 -2.5 59.0 42 25.0 36.0 72 95.0 16.0 102 142.0 54.0 13 -5.0 58.0 43 33.0 36.0 73 97.5 13.0 103 138.0 55.0 14 -8.0 57.5 44 35.0 34.5 74 102.0 11.5 104 140.0 58.5 15 -10.0 55.0 45 35.0 33.5 75 105 .o 10.0 105 145 .o 60.0 16 -11.0 53.0 46 32.4 32.4 76 108.0 11.0 106 152.0 58.0 17 -9.0 51.0 47 35.0 28.0 77 109.0 13.0 107 155.0 55.0 18 -5.0 50.0 48 38.0 25.0 78 107.0 17.0 108 157.0 51.0 19 -2.5 48.0 49 40.0 21.0 79 106.0 19.0 109 160.0 53.0 20 -1.0 46.0 50 43.8 17,0 80 107.0 22.0 110 163.0 58.0 21 -1.0 44.5 51 45.0 13.0 81 112.0 22.5 111 172 .o 61.0 22 -2.0 44.2 52 50.0 15.0 82 117 .o 23.0 112 180.0 62 .o 23 -8.5 44.0 53 55.0 17.0 83 120.0 26.0 113 -174.0 64.0 24 -9.0 43.5 54 59.0 21.0 84 122.0 32 .o 114 -170.0 66.0 25 -8.5 41.5 55 60.0 23,5 85 121.5 33.5 115 180.0 69.0 26 -9.5 39.0 56 63.0 25.0 86 121.0 35 .o 116 170.0 70.0 27 -9.0 37.5 57 67.0 25.0 87 119.0 37.0 117 160.0 71.0 28 -8.5 37.0 58 69.0 22.0 88 120.0 39.0 118 140.0 73.0 29 -6.5 37.0 59 73.0 20.0 89 124.0 40.0 119 135.0 71.0 30 -5.5 36.0 60 73.5 16.5 90 125.0 38.0 120 120.0 73.0

AppendixA2 113

121 120.0 75.0 125 80.0 72.5 129 65 .o 70.0 133 47.0 67.0 122 113.0 77.0 126 74.0 71.5 130 60.0 10.0 134 40.0 68.0 123 105.0 78.0 127 72.0 73.0 131 55.0 69.0 135 32.0 71.0 124 85.0 75.0 128 70.0 73.0 132 50.0 68.0

cgeu.tbl RA 45 56 Cl 50.0 70.0 FI 45 RA 4 10 cs 15 .o 100.0 RA 39 45 RA 71 78 RA 56 71 FI 71 RA 78 89 C6 43.0 128.0 RA 97 105 RA 89 97 RA 119 134 FI 89 FI 4 C7 65 .o 160.0 C2 50.0 0.0 RA 105 119 RA 10 21 FI 105 RA 36 39 CB 68.0 25.0 FI 10 RA 1 4 C3 Ann -5.0 FI 134 -.v.v

RA 21 36 FI 135 FI 21 FI 1 C4 25.0 50.0

North America

namer.dat

# LON LAT # LON LAT # LON LAT #' LON LAT 1 -155.0 72.0 28 -105.0, 19.0 55 -96.0 28.0 81 -58.0 55.0 2 -160.0 71.5 29 -100.0 16.5 56 -95 .o 29.0 82 -62 .o 58.0 3 -163.0 70.0 30 -96.0 15.5 57 -90.0 29.0 83 -65 .o 60.0 4 -165. 0 68.0 31 -95.0 16.5 58 -85 .o 29.5 84 -65 .o 63.0 5 -164. 0 67.0 32 -93.0 15.5 59 -83.5 29.0 85 -62 .o 65.0 6 -165. 0 66.5 33 -92.0 14.0 60 -82.5 27.0 86 -64.0 68.0 7 -163.0 65.8 34 -90.0 13.5 61 -81.5 25.5 87 -75.0 72.5 8 -165.0 64.5 35 -88.0 12.5 62 -80.0 26.0 88 -78.0 75.0 9 -165 .o 63.0 36 -85.0 10.0 63 -80.5 29.0 89 -80.0 77.0

10 -163.0 61.3 37 -84.0 8.0 64 -81.0 30.0 90 -70.0 80.0 11 -162 .o 58.0 38 -81.0 7.0 65 -81.0 31.5 91 -65 .o 81.0 12 -158.0 57.5 39 -80.0 8.0 66 -80.0 32.5 92 -67.0 82.0 13 -153. 0 59.0 40 -80.0 9.0 67 -76.0 35.0 93 -80.0 82.0 14 -145.0 6q.o 41 -82.0 9.0 68 -75.5 36.0 94 -90.0 81.5 15 -138.0 58.5 42 -84.0 10.0 69 -76.0 37.5 95 -100.0 80.0 16 -134.0 55.0 43 -84.5 12.0 70 -74.0 40.5 96 -110.0 78.0 17 -130.0 52.0 44 -84.0 15.0 71 -70.0 42 .o 97 -120.0 77 .o 18 -127. 0 50.0 45 -85.0 16.0 72 -70.0 43.0 98 -123.0 76.0 19 -125. 0 48.0 46 -88.0 16.0 73 -68.5 44.0 99 -125.0 75.0 20 -124. 0 45.0 47 -87.5 17.5 74 -65 .o 43.5 100 -125.0 72.0 21 -124.0 40.0 48 -87.0 22.0 75 -60.0 46.0 101 -127.0 70.5 22 -123.0 38.0 49 -89.0 22.0 76 -55.0 46.5 102 -130.0 70.0 23 -120.0 35.0 50 -91.0 21.0 77 -53.0 47.0 103 -135.0 69.0 24 -118.0 33.0 51 -92.0 19.0 78 -52.5 48.5 104 -145.0 70.0 25 -116. 0 30.0 52 -95.0 19.0 79 -54.0 50.0 105 -150.0 72.0 26 -113.0 26.0 53 -96.5 20.0 80 -55.0 53.0 27 -110.0 23.0 54 -98.0 23.5


Figure A2-3. North America

cgna.tbl FI 29 cs 14.0 -87,0 Cl 60.0 -100.0 FI 30 RA 35 36 RA 14 29 FI 31 RA 42 46 RA 54 58 FI 52 FI 35 RA 65 103 FI 53 C6 8.0 -82,0 FI 14 FI 54 RA 36 42 C2 65.0 -150,0 FI 29 FI 36 RA l 14 C4 17.0 -90.0 C7 30.0 -82.0 RA 103 105 RA 31 35 RA 58 65 FI l RA 46 52 FI 58 C3 18.0 -97.0 FI 31

AppendaA2 115

South America

Figure A2-4. South America

samer,dat # LON LAT # LON LAT # LON LAT # LON LAT 1 -65,0 10.0 10 -78,0 7.0 27 -65.0 -55.0 43 -35.0 -6.5 2 -68.0 10.5 11 -77.5 5.0 28 -68.0 -53.0 44 -36.0 -5.0

12 -80.0 1.0 29 -69.0 -50.0 45 -37.5 -5.0 3 -70,0 12 .o 13 -81.0 -5.0 30 -66,0 -47,0 46 -40,0 -3.0

14 -78,0 -10.0 31 -65 .o -43.0 47 -43,0 -2.5 15 -76.0 -15 .o 32 -60.0 -39.0 48 -45.0 -2.0 16 -72.0 -17 .o 33 -55.0 -35.0 49 -50.0 0.0

4 -73.0 11.5 17 -71.0 -18.0 34 -50.0 -30.0 50 -51.0 2.0 5 -75.0 11.0 18 -70.5 -20,0 35 -48.0 -28,0 51 -52.0 4,0

19 -71.0 -25 .o 36 -48.0 -26.0 52 -53.0 6.0 6 -77.0 8,0 20 -72.0 -30,0 37 -45 .o -24.0 53 -55.0 6.5

21 -73.0 -35 .o 38 -43.0 -23.0 54 -57.5 6.5 7 -78.0 9.0 22 -74.0 -40.0 39 -40.0 -20,0 55 -60.0 8.0

23 -74.5 -45.0 40 -39.0 -15.0 56 -62.0 10.0 8 -80,0 9.0 24 -75.0 -50.0 41 -37.0 -11.0 9 -80.0 8,0 25 -74.0 -54.0 42 -35 .o -8.0

26 -70,0 -55.0

cgsa.tbl Cl -15 .o -60.0 RA 1 6 RA 10 24 RA 31 56 FI 1 C2 8,0 -78.0 RA 6 10 FI 6 C3 -52.0 -73,0 RA 24 31 FI 24


Antarctica

I?

~ ,::; 0

0 ~

~ ,!. cl

'ti

" • 08· 06· ooi· Figure A2-5. Antarctica

antar.dat

# LON LAT # LON LAT # LON LAT # LON LAT 1 o.o -71.0 14 -100.0 -72.0 26 125.0 -66.0 38 40.0 -69.0 2 -1.0 -70.0 15 -105 .o -75.0 27 120.0 -67.0 39 38.0 -70.0 3 -10.0 -71.0 16 -125.0 -73.0 28 110.0 -67.0 40 35 .o -69.0 4 -20.0 -73.0 17 -145 .o -75.0 29 105.0 -65.0 41 30.0 -69.0 5 -36.0 -77 .o 18 -165. 0 -79.0 30 95 .o -65.0 42 28.0 -70.0 6 -50.0 -77.0 19 170.0 -77 .o 31 90.0 -67.0 43 25.0 -70.0 7 -60.0 -75.0 20 160.0 -78.0 32 80.0 -68.0 44 18.0 -70.0 8 -60.0 -66.0 21 165.0 -75 .o 33 75.0 -70.0 45 14.0 -69.0 9 -57.0 -63.0 22 170.0 -72.0 34 70.0 -68.0 46 13.0 -70.0

10 -65.0 -66.0 23 160.0 -70.0 35 60.0 -68.0 47 6.0 -70.0 11 -70.0 -69.0 24 145.0 -67 .o 36 55.0 -66.0 48 1.0 -70.0 12 -75.0 -72.0 25 135.0 -66.0 37 50.0 -67.0 13 -80.0 -73.0

cgan.tbl Cl -90.0 0.0 RA 1 7 RA 12 20 RA 24 48 FI 1 C2 -67.0 -70.0 RA 7 12 FI 7 C3 -72.0 155.0 RA 20 24 FI 20

I (

AppendixA2

Australia

120 125 130 135 140

Figure A2-6. Australia

austr,dat # LON LAT # LON LAT # LON 1 130.0 -14.0 16 114.5 -26.0 31 136.0 2 129.0 -15.0 17 114.0 -27.0 32 138.0 3 128.0 -15 .o 18 114.5 -28.5 33 138.0 4 127.0 -14.0 19 115.0 -29.5 34 139.0 5 126.0 -14.0 20 116.0 -32.0 35 142.0 6 125.0 -15 .o 21 116.0 -33.0 36 143.5 7 124.0 -16.0 22 115.0 -34.5 37 145 .o 8 123.0 -17 .o 23 117 .s -35 .o 38 146,5 9 122.0 -18.0 24 118.0 -35 .o 39 150.0

10 121.0 -20.0 25 120.0 -34.0 40 151.0 11 120.0 -20.0 26 123.5 -34.0 41 153.0 12 117,0 -21.0 27 124.5 -33.0 42 153.5 13 115.0 -22.0 28 130.0 -32 .o 43 153.0 14 , 114. 0 -24.0 29 133.0 -32 .s 44 148.0 15 114.0 -25.0 30 134.5 -33.S 45 146.0

cgau.tbl Cl -20.0 130.0 RA 55 60 RA 1 32 FI 55 C2 -25.0 145 .o RA 32 55 FI 32

145 150

LAT # -35.0 46 -33.0 47 -36.0 48 -37.0 49 -38.5 so -39.0 51 -38.0 52 -39.0 53 -37.5 54 -35.0 55 -32.0 56 -28.5 57 -26.0 58 -20.0 59 -19.0 60

155

·10

·15

-20

·25

LON 146.0 145.0 144,5 144,0 143.5 142.5 141.5 141.5 141.0 140.0 139.0 135 .s 136.5 132 .s 131.0

117

LAT -17.0 -15.0 -14.0 -14.0 -12.5 -11.0 -12.5 -15.0 -17,5 -18.0 -17,0 -15.0 -12.3 -12.0 -12,5

118

Mars-A

50

40

30

20

10

0

-10

-20

-30

-40

-50

-60

marsa.dat # LON 1 -145, 7 2 -149.0 3 -151. 3 4 -155,5 5 -159. 6 6 -164.7 7 -169.l 8 -173.0 9 -176.9

10 -182. 3 11 -186.1 12 -189.3 13 -190 .2 14 -188.2 15 -184.3 16 -181.l 17 -178. 7 18 -177.6 19 -177.9 20 -178.1 21 -174.8 22 -168.7


"'""I"'''' ''l'''''''''P''''''"l'"''''''l'"''"''l'"'""'"''''""I' ............. l""''"'"'"''"'l""''"T'"'""I 11111111

-180 -150 -120 -90 -60

Figure A2-7. Continent A of Mars

LAT # LON LAT # LON LAT # LON LAT -24.2 23 -162 ,2 -69.2 45 -45 ,4 -36.0 66 -103.5 51.5 -26.0 24 -155. 8 -69.5 46 -47.7 -31.2 67 -108.6 51.3 -26.0 25 -149.5 -69.8 47 -49.8 -26.0 68 -113.6 51.l -25.0 26 -143,1 -69 .s 48 -50,9 -20.6 69 -118.7 51.0 -23.8 27 -136. 7 -69.0 49 -51.6 -14.0 70 -124.1 51.0 -22.8 28 -130.0 -68.6 50 -52.4 -6.2 71 -129.8 50.3 -22.9 29 -122.7 -68.0 51 -55.4 1.4 72 -135.8 49.0 -23.5 30 -114. 8 -66.8 52 -61.2 6.2 73 -140.9 46.8 -25.0 31 -107.6 -65.4 53 -67.3 10.0 74 -145.1 43.5 -27.0 32 -101.1 -63.5 54 -73.1 13.2 75 -146.3 39.0 -29.8 33 -96.0 -62 .o 55 -78.9 17,5 76 -142.6 34.0 -34.0 34 -91.3 -60.0 56 -77.0 22.0 77 -137.4 29.8 -39.0 35 -87.2 -58.5 57 -71.1 26.8 78 -133.1 26.5 -44.0 36 -83.2 -57.0 58 -66.2 32.0 79 -130.3 24.0 -48 .• 6 37 -79.3 -56.0 59 -63.0 38.0 80 -129.2 21.s -51.8 38 -75.4 -54.5 60 -63.5 44.0 81 -131.5 14.0 -54.6 39 -7L3 -53.0 61 -68.4 48.1 82 -134.7 6.5 -57.2 40 -67.0 -51.5 62 -75 .9 so.a 83 -137.4 -1.0 -60.5 41 -62.5 -49,7 63 -84.0 51.8 84 -139.6 -8.5 -64.5 42 -56.9 -47.2 64 -91.8 52.0 85 -141.3 -14.5 -67 .3 43 -51.0 -44.8 65 -97.8 51.8 86 -143.1 -20.0 -68.8 44 -46,3 -41.0

cmaa.tbl RA 81 86 Cl -50.0 -150 .o FI 1 RA 1 30 FI 30 FI 1 C3 30.0 -110.0 C2 -35.0 -100.0 RA 55 80 RA 30 55 FI 55 FI 80

AppendixA2 119

Mars-B 40

30

20

10

-10

-20

-30

-40

-50

-30 0 30 60 90

Figure A2-8. Continent B of Mars

marsb.dat # LON LAT # LON LAT # LON LAT # LON LAT 1 15.6 -6.8 14 29.6 -49.5 27 76.7 -14.l 39 66.8 36.l 2 10.5 -15.0 15 34.l -46.6 28 78.5 -10.4 40 65 .s 37.0 3 5.3 -22.0 16 38.7 -41.0 29 78.9 -6.2 41 62.8 37.3 4 -l.8 -28.0 17 42.9 -34.2 30 78.5 -3.8 42 61.0 37.2 5 -14.l -33.8 18 45.6 -29.2 31 77. 7 -1.l 43 58.8 36.8 6 -21. 7 -37.0 19 48.9 -24.8 32 75.6 l.8 44 54.8 33.8 7 -23.3 -39.9 20 so. 7 -22.8 33 72.7 5.2 45 50.8 28.7 8 -21.3 -42.6 21 52.0 -21.5 34 69.6 10.4 46 47.0 23.2 9 -14.8 -44.8 22 53.3 -20.6 35 67.6 15. l 47 42. 7 17.8

10 -2.6 -48.0 23 56.5 -19.3 36 67.l 21.1 48 37.8 12.l 11 8.1 -50.0 24 65.5 -17 .2 37 67.5 28.2 49 31.2 5.4 12 17.6 -51.0 25 71.2 -16.2 38 67.6 33.5 50 23.2 -0.l 13 24.5 -51.0 26 74.3 -15 .8

cmab.tbl Cl -30.0 20.0 RA l 18 FI l C2 -5,0 50.0 RA 18 50 FI l FI 18


Mars-C

-10

-20

-30

-40

-SO tlljililllllijlllljlllijlllljlllljliilliii)liiiillllljlllljlllijlillllllijiillllliifllllllliijllitJ

90 120 1:50 180

Figure A2-9. Continent C of Mars

marsc.dat # LON LAT # LON LAT # LON LAT # LON LAT 1 -235,0 -11.0 10 -259.4 -2,5 25 -247.0 -34,2 40 -195.4 -40.0

11 -262 .6 -2.5 26 -243.7 -36.9 41 -193.7 -36.1 12 -266.9 -3.0 27 -241.5 -39.0 42 -193.6 -32 .1 13 -271.2 -5.1 28 -239.3 -41.0 43 -196.9 -28.1 14 -273.8 -7.8 29 -236.8 -42.2 44 -200.9 -25.0

2 -237 .s -10.0 15 -275.3 -10.9 30 -234.2 -43.8 45 -204.0 -22.1 16 -275.5 -14.2 31 -231.2 -44.8 46 -206.9 -20.0 17 -274.2 -17.8 32 -228.8 -45.4 47 -209.8 -18.0

3 -240.1 -8.6 18 -270.9 -20.8 33 -224.5 -45.8 48 -212.6 -16.2 19 -265 .9 -23.4 34 -221. 0 -45.5 49 -215.4 -14.8

4 -242 .4 -7.5 20 -262 .3 -25.2 35 -217.5 -45 .o so -218.4 -13.S 5 -244.6 -6,2 21 -259.0 -27,0 36 -213.9 -44.5 51 -221.3 -12 .8 6 -247 .o -5.1 22 -255.5 -28.7 37 -210.1 -43.9 52 -224.4 -12 .o 7 -249.2 -4.2 23 -251.9 -30.4 38 -205.7 -43.2 53 -227.6 -11.9 8 -252,8 -3.0 24 -249 .s -31.8 39 -199.8 -42 ,2 54 -231. 0 -11.8 9 -25,5. 7 -2.5

cmac.tbl Cl -20.0 -250.0 RA 1 24 FI 1 C2 -30.0 -225,0 RA 24 54 FI 1 FI 24

AppendixA2 121

Moon-A

I l""1""l"" 1""l""'""l""'""l'""""l"""'"l''" 1""l"''1""l"''"1"l""1""l"""'''l"" 1""l"'""''l'""""l"" 1""l''''1"''l"''"'''I

a m m ~ ~ ~ • m oo = = = = = = = = •

Figure A2-10. Continent A of Moon

moona.dat # LON LAT # LON LAT # LON LAT # LON LAT 1 165.0 21.9 18 116.3 -25.7 35 -147. 7 -27.4 52 -117 .2 25.2 2 161.2 20.5 19 120.1 -30,6 36 -143.1 -27,3 53 -123.3 27.0 3 157. 7 20.0 20 125.0 -34.0 37 -138,2 -27,3 54 -128.4 28.3 4 154,2 20.0 21 131.0 -36.6 38 -132.4 -27,9 55 -132.9 29.5 5 150.1 20.9 22 137,4 -37.8 39 -125.2 -29.0 56 -137,0 31.0 6 144.8 22.4 23 144.0 -37,5 40 -115. 7 -30.1 57 -140.9 32.2 7 139.0 23.0 24 150.2 -36.1 41 -109.1 -29,9 58 -144.9 33.9 8 132.5 22.8 25 155.4 -34.8 42 -103.8 -27,8 59 -149.2 35.8 9 125.6 21.4 26 160.0 -33.4 43 -100.7 -24.0 60 -154,3 37,9

10 118,9 18.5 27 165.7 -31.9 44 -98.2 -17 .o 61 -160,2 40.0 11 112,0 14.3 28 172.4 -30.3 45 -97.2 -10,0 62 -166.9 41.1 12 108.3 11.2 29 178.1 -28.9 46 -97.2 -2.0 63 -172,1 41.4 13 105,7 7.1 30 -173.9 -27.9 47 -97.4 6.3 64 -177,0 40.8 14 106.1 0.6 31 -167.3 -27.7 48 -98,2 11.8 65 179.7 39.0 15 107,8 -7.5 32 -161. 8 -27,5 49 -100.0 16.3 66 175.5 34.9 16 110.2 -14.0 33 -156,9 -27.5 50 -104.3 20.2 67 172 .5 30.1 17 113.1 -20.0 34 -152.3 -27.5 51 -109.7 23.0 68 169.6 26.0

cmoa.tbl Cl -5.0 160.0 RA 1 29 FI 1 C2 5.0 -150.0 RA 29 68 FI 1 FI 29

122

Moon-B

moonb.dat # LON 1 59.6 2 59.4 3 59.6 4 61.0 5 63;3 6 66.3 7 69.6 8 72.9 9 76.1

10 78.9


10 20

40

2 . . 3 . -10

6 7

jj I II I I I I I I I r I I I I I I I I I I II f r I I I I I I I I I I I II ii I I I I I I I I I I I I I

50 60

LAT # LON 3.1 11 81.4

-0.5 12 83.7 -4.4 13 85.9 -7.9 14 87.9

-10.3 15 89.9 -11.8 16 92.0 -12 .8 17 94.2 -12 .9 18 96.5 -12.3 19 98.6 -11.0 20 99.8

70 80 90 Figure A2-11. Continent B of Moon

LAT # LON -9.9 21 99.1 -8.8 22 97.7 -7.5 23 96.0 -6.0 24 93.6 -4.4 25 90.9 -2.7 26 88.1 -0.4 27 85.4

2.0 28 82.7 5.1 29 79.9 8.8 30 77.0

cmob.tbl Cl 7.4 78.0

RA FI

1 1

40

LAT 11.9 14.8 17 .3 19.5 21.0 22.1 23.2 24.1 24.7 24.2

100

# LON LAT 31 74.3 23.6 32 71. 7 22.7 33 69.4 21.3 34 67.3 19.8 35 65 .6 17.9 36 64.4 15.8 37 63.3 13.7 38 62 .2 11.5 39 61.1 9.1 40 60.1 6.3

(

(

AppendixA2 123

Moon-C

38 37 36

23

7 8

111 1""'""1""' 111 11" 1 • 't 111 • 'I """ I" """l'""""l""'""I '"'""l""'""I ""'"'I

-70 -60 -SO -40 -30 -20 -10 10 20 30 40 so

Figure A2-12. Continent C of Moon

moonc.dat # LON LAT # LON LAT # LON LAT # LON LAT

12 -14.3 -31.8 25 41. 7 8.0 38 -11.5 58.8 13 -8.2 -31.8 26 40.0 14.0 39 -18.8 57.8

l -57.3 5.9 14 -3.4 -32.2 27 38.7 19.9 40 -25.8 55.9 2 -57.8 -1.l 15 2.6 -33.2 28 37.3 25.4 41 -32.4 53.6 3 -59.2 -8.8 16 9.5 -34.2 29 35.8 30.9 42 -38.7 51.0 4 -61.2 -16.6 17 17.2 -35 .o 30 33.6 35 .9 43 -44.7 48.l 5 -63.9 -26.l 18 25.5 -34.8 31 30.7 40.8 44 -49.9 44.5 6 -63.6 -34.9 19 33.7 -33.0 32 27.0 45.2 45 -54.8 40.6 7 -56.5 -39.l 20 40.2 -29.0 33 22.5 49.4 46 -58.6 35.8 8 -46.6 -40.0 21 44.4 -23.0 34 17.2 53.2 47 -61.4 30.6 9 -36.0 -37.8 22 47.l -16.0 35 10.7 56.3 48 -61.8 24.2

10 -26.8 -34.0 23 47.0 -8.3 36 3.5 58.l 49 -59.8 17.7 11 -20.0 -32 .3 24 44.3 0.5 37 -3.9 58.9 50 -58.0 11.8

cmoc.tbl

Cl 10.0 -10.0 RA l 50 FI l


The results of calculations are in Table A2-1.

Planet Continent Longitude Latitude Area, m2

Earth Africa 18 .37° 7.15° .293e14

Eurasia 76.31° 50.66° .540e14

North America -98.90° 50. 38° .251e14

South America -60. 65° -14.24° , 181el4

Antarctica 77. 70° -86.11° .140e14

Australia 134 .37° -25 .80° .762e13

Mars A -105.42° -13. 88° ,309e14

B 38. 76° -13 .19° , 119el4

C 123. 82° -24. 99° ,652e13

Moon A -172.46 -1.14° • 774e13

B 78, 19° 5 .98° ,995e12

C -9. 76° 6 .10° • 721e13

Table A2-1. Calculated positions of Center of Areas and area sizes.

(

(

AppendixA3

Derivation of Equation A-4 of Paper A

The fundamental law of the electrical repulsion force between two charged bodies is expressed by the equation

F=k Q1 Q2 R2

(1)

where Q1 and Q2 are the charges the bodies carry, R is the distance between the bodies, k is a constant for the units of measurement and for the electrical environment ( dielectric constant), and F is the mutual electrical force that arises between the bodies. The force is repulsion when the bodies carry like-charges. When more than two bodies are involved,

~c

FCA Figure A3-1. Vector diagram of a three-body system to prepare a set of equations for the force-balanced positions.

the above equation is independently valid between any two bodies. The resulting force acting on each body is the vectorial sum of the individual forces.

When three small bodies are involved and they are located on a spherical surface, the bodies define a circle on a plane. On this plane planar trigonometry can be used to determine the force components. When

125


the bodies are free to move on the spherical surface (like ships on an allocean planet), the bodies move to positions which are on a great circle of the spherical surface.

Figure A3-1 illustrates the geometry of three pointlike bodies, A, B, and C, on a circle. They carry Q1 , Q2 , and Q3 amount of charges, respectively. The angular distance between A and B is a, and between A and C (in the direction of B) is p. The line distances between the three bodies are:

the squares of the distances:

DAB= 2 sin(a/2)

DBc = 2 sin((P-a)/2)

DCA =2sin(P/2)

Dla =4 sin2 (a/2)

D;c = 4 sin2((13-a)/2)

D~ = 4sin2(fl/2)

a simpler form is obtained by using this type of identity:

sin2( q,) = 1/2-1/2 cos (24>)

thus (5), (6), and (7) become 2 DAB =2(1-cos(a))

D;c = 2(1-cos(J3-a))

D~ = 2(1-cos(/3))

The forces on the bodies are:

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

(11)

(12)

(13)

(14)

AppendixA3 127

The horizontal force components are:

(15)

(16)

(17)

There are three more horizontal force components, but they are the same as (15), (16), and (17) because the same charges generate them over the same distances. These are:

(18)

(19)

(20)

The ultimate driving force on each body is the difference of the two oppositely directed horizontal force components:

(21)

(22)

(23)

The identities of (18), (19), and (20) can be substituted into (21), (22), and (23). The results are:

(24)

(25)

(26)


The bodies are in force balance when their horizontal forces cancel out to zero

(27)

(28)

(29)

Since there are only two unknowns, a and /3, only two of the three simultaneous equations are required to solve. Let's select the first two. The final set of equations is the same as (A-4) of Paper A.

QA QC COS (/3/2) + QA QB COS (a/2) = Q

2(1-cos /3) 2(1-cos a)

QA QB COS (a/2) 2(1-cos a)

_Q_0 _Q_c co_s (_(/3_-_a)_/2_) = 0 2(1-cos (/3-a))

(30)

The equations can be solved by a program of Successive Approximations (available in most commercial mathematical softwares). Table A3-1 lists the examples used in Paper A, to 12-digit precision. Case #1 to #5 are those of Table A-7, #6 is that of Table A-8 (Mars), and #7 is that of Table A-9 (the Moon).

# QA QB a 13 error-1 error-2

1 1 1 120.000000000 240.000000000 .17E-15 .OOE+OO

2 2 1 129.185787974 230.814212026 .98E-16 . 25E-11

3 4 2 145.578128604 238.909641321 -.93E-12 - .76E-11

4 1000 1 172. 795486900 187.204513100 .31E-13 -.23E-08

5 1000 1000 179.838285936 269.919142968 - .92E-06 - .. 92E- 06

6 4.74 1.83 146. 416656994 234.332280628 .16E-07 -.33E-07

7 7.78 7.25 162.968806517 260.082704685 -.17E-10 -.58E-07

Table A3-1. Examples for the solution of Equation (30).

AppendixA4

Measuring Charge Densities with a Pithball Electroscope.

Electrical like-charges exert repulsion forces on each other. The force is numerically expressed by Coulomb's law of the electrostatic force:

F =k QA QB " R2

(1)

where QA and QB are the charges expressed in coulombs, R is the distance between them. k is a constant of the environment (dielectric), and the measurement system employed. F. is the generated force.

While the force exists strictly between the charges, those charges can be carried by electrically neutral matter, like metals and insulators; the electrical force will appear to exist between the charge carriers. In metals the charges move very fast, in microseconds per meter, and they spread out as far apart as possible under their own mutual repulsion forces. Thus, they usually occupy the external surface of the metal object. In insulators the speed of propagation is slow, seconds, minutes, or hours per meter. They appear virtually 'static' when their speed is very slow (thus the expression static charge. However, sooner or later all charged objects regain their neutral state. They lose their charge, discharge, under all circumstances.

The details of Coulomb's law are as follows:

1. In a homogeneous environment the electrical force is directed along a straight line between the charges.

2. The force is proportional to the product of the charges, and it is inversely proportional to the square of the distance between them.

3. The force is repulsion if the charges have the same sign of polarity, and it is an attraction if the signs are opposite.

4. The forces obey the principle of linear superposition. When more than two separate charges are present, the net force on any one charge is the linear vector sum of the electrical forces exerted by all other charges.

5. The electric force on a charge carrier combines with nonelectric forces by simple vector addition.

The physical unit of an electrical charge is the charge of the electron. In the SJ system of measurement the electron's charge is e=1.6x10- 19

coulomb, and k=8.99x10 9, in a vacuum environment. For all practical

purposes, the same value is used for air. Thus, on two electrons at 1 meter distance apart the electrical force


.newton (2)

As textbooks of electricity frequently point out, the electric force between two electrons is much greater than their gravitational force. This latter can be calculated by

F=GmAmB g R2

(3)

here G=6.67x10-11 is the gravitational constant, mA = mn = 9.1x10-31 kg is the mass of the electron, and R = 1 meter, thus the gravitational force

~= 6.67x10-11 (9.lxl0- 31) 2 = s.sx10- 11 newton (4)

Accordingly, the ratio between the electrical and gravitational forces, between two electrons

F,,/~=2.3x10- 28/S.Sx1Q-7l =4.2xl0 42 (5)

The electrical force is inconceivably greater than the gravitational force. The same is true for protons. While the protons charge and the resulting force are the same as those of the electrons, their larger mass results in greater gravitational attraction. The proton's mass is m = 1.67x10·27 kg, thus the gravitational attraction between two protons at 1 meter distance

newton (6)

Thus the ratio between the electrical and gravitational forces between two protons is still very large:

(7)

The popular belief, that "the electrostatic force is a weak force" is totally incorrect. What people do is that they compare a small electric force with a large gravitational force, and they say "do you see how small the electric force is?". The fact is that it is difficult to create large electrical forces between small objects only because it is difficult to maintain a large charge density in objects. But as I point it out in Paper A, Equ. A-2, the electrical force is proportional to the 6th power of the uniformly charged object's linear dimensions. Consequently, even a very small, thus easily maintained charge density generates an enormously large electrical force in large objects.

The common usage that "a charged object carries an electric charge" is also not quite correct. An electrically neutral object always carries all the charges it can ever have, both the positive and the negative charges. However, in a neutral state the opposite charges negate each other inter-

AppendixA4 131

nally, thus the object does not show any external electrical effect. But when its number of protons and electrons become unequal, then the nonnegated charges become observable, they generate an external electrical effect.

Positive charges are represented by protons that are located in the atomic nucleus, and negative charges by electrons that orbit the nucleus. Normally, and most easily, an object becomes charged when it loses one or more of its orbital electrons. Because of the loss of electrons, the object becomes positively charged. Those atoms which lost electrons are called 'positive ions', and the removed electrons become 'free electrons'. This action is called ionization.

The outermost orbital electrons can be easily removed from atoms, by simple mechanical means like touching, striking, or rubbing. This is mechanical ionization. In thermal ionization the object is exposed to heat in which high velocity 'thermal' electrons collide with orbital electrons and dislodge them from orbit. In both the mechanical and thermal processes orbital electrons are removed. Individual atoms can not be forced to take extra electrons into orbit, more than the number of protons they have. Thus atoms can be made only positively charged (contrary 'to chemists). However, when several atoms form molecules, or objects, including liquids, extra electrons can be forced and (temporarily kept) embedded in the inter-molecular space. For example, when such an object'Surrounds an electron source, like the glass envelope of a thermal cathode, or the water envelope of an all-ocean planet, energetic free electrons can enter those envelopes from inside. These electrons make the envelopes negatively charged. · ·

In rubbing or touching two different materials together;electrons can be removed from one material and injected into the other, making the former positively, and the latter negatively charged. For example, when a glass rod is rubbed with silk, the glass becomes positively charged, and the silk negatively. However, when amber is rubbed with silk, the amber becomes negative, and the silk positive.

A simple means of numerically measuring the charged state of an insulator material is the pithball electroscope. In this very simple equipment two balls of pith (the soft, spongy central cylinder of parenchymatous tissue in the stems of dicotyledonous plants), or plastic foam balls (like styrofoam, an expanded plastic made from polystyrene), are suspended from a fixed point P side by side on fine non-conducting threads. Then the balls are gently rubbed with cat's fur. The balls acquire like-charges in the rubbing, start repelling each other, and deflect to a certain distance from each other (Figure A4-1).

There are two forces acting on each ball: the force of gravity Fg generates a vertically downward force, and the electrical repulsion force between the balls generates a horizontal force, Fe . As a result, the balls


deflect from their original positions, and move apart along the radius of the threads. During their movement the distance R increases, thus the horizontal force decreases on the balls. At one point the two forces come into balance, and the balls stop moving any further apart. The balance

Fig1.1re A4-1. The pithball electroscope to numerically measure the electron density of charged insulator objects.

position is when the direction of the resulting force vector Fa becomes lined up with the thread.

In my experiment each ball is of 1 cm3 volume (1.24 cm diameter), its mass ism= 1.36x10·5 kg, thread length L = 10 cm, and the gravitational acceleration is taken as g = 9.81 m/sec2

• After rubbing the balls deflect to R = 3.33 cm distance. At this point the following equations are established:

.,,,j

''° I

f •

c•.· R sma=-

2L (8)

(9)

(10)

These variables are substituted into Coulomb's equation, Equ. (1), with the simplification that QA = QB = Q. The equations are transcendental because· of the trigonometric functions involved, thus I used the method of successive approximations by a commercial mathematical program. The input equations and the solutions are listed in Table A4-1.

(

(

Appendbc A4 133

Input K=8,99e9 Coulomb's constant

L=0,1 thread length, m FG=9.81*0.0136e-03 mass of foam ball, newton

R=O. 033 deflection, m

EC=l.60219e-19 electron's charge, coulomb SIN(AL)=(R/2)/L

TAN(AL)=FE/FG FE=K*QA2/RA2 electric force, newton

QE=Q/EC number of electron!! ED=QE/EC electron density, m~3

', "-.-.-, ..

Solution Variable Value

EC l.6021900e-19 ED l.0262730e+l6

FE 0,000022319562

FG 0.0001334160q K 8,9900000e+09

L

Q

QE R

AL

Maximum error

0.10000000

1.6442844e-09 1,0262730e+10 0.033000000 0.16575801

2,7755576e-17

Table A4-1. Solution of the deflected pithball electroscope.

i. ·1·.,,

),,., d

',0:. .• i . ·.:· :~i

In this experiment it is not important what material has been used for the balls or for rubbing, and whether the balls become positively or negatively charged. The important fact is that a certain amount of charge units have been imparted to the insulator balls which deflect under the arising electrical force. In reality, electrons have been moved. I express the charge units, besides coulombs, in electrons and in electron density to underscore the real physics behind the charged state. Accordingly, the final result of the pithball experiment is that the achieved electron density in the insulator balls is

ED= 1016 electrons per meter 3