Ceta Files

7/25/2019 Ceta Files

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C Files

Ian Beardsley

Copyright 2016 by Ian Beardsley

ISBN: 978-1-365-26289-0


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Climate Science 4

Model Future 14

The Source Code 19

Running The Models 38

Modeling In Python And Java 47

Paul Levinson And Manuel Heredia 55

Amarjit 70

Manuel 74

The Bronze Age 81

The Mystery In Our Units of Measurement 86

The Sequences 104

The Wow! Signal 109

Aquila 115

Manuelss Integrals 121


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Theories


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Climate Science


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Albedo

Albedo is a function of surface reflectivity and atmospheric reflectivity. Atmospheric albedo

seems to play the primary role in the overall albedo of a planet. Albedo is the percent of light

incident to a surface that is reflected back into space. It has a value ranging from zero to one

inclusive. Zero is a black surface absorbing all incident light and one is a white surfacereflecting all incident light back into space. Albedo plays a dominant role in the climate of a

planet. Let us see if we can find a relationship between composition of a planet and its albedo if

not in its distance from the star it orbits and its albedo, even a relationship between its albedo and

orbital number, in that albedo could be a function of distance from the star a planet orbits

because composition seems to be a function of distance of a planet from the star it orbits. As in

the inner planets are solid, or terrestrial, and the outer planets are gas giants. There may be an

analogue to the Titius-Bode rule for planetary distribution, but for albedo with respect to

planetary number. The inner planets are dominantly CO2, Nitrogen, Oxygen, and water vapor,

the outer planets, hydrogen and helium.

1. Mercury albedo of 0.06 composition 95% CO22. Venus albedo of 0.75 composition clouds of sulfuric acid

3. Earth albedo of 0.30 composition Nitrogen, Oxygen, H20 or water vapor

4. Mars albedo of 0.29 composition CO2

5. Asteroids

6. Jupiter albedo of 0.53 composition hydrogen and helium

7. Saturn albedo of 0.47 composition hydrogen and helium

8. Uranus albedo of 0.51 composition hydrogen, helium, methane

9. Neptune albedo of 0.41 composition of hydrogen and helium

We see the outer gas giant, which are composed chiefly of hydrogen and helium have albedosaround 50%. Earth and Mars, the two planets in the habitable zone, are about the same (30%).

Go to the next page for a graph of albedo to planetary number.


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The average for the albedo of the inner planets is:(0.06+0.75+0.3+0.29)/4 = 0.35This is close to the albedo of the habitable planets Earth and Mars.

The average for the albedo of the outer planets is:(0.52+0.47+0.51+0.41)/4 + 0.4775 ~0.48This says the outer planets are all close to 0.48~0.5

mercury 0.06

venus 0.75

earth 0.3

mars 0.29

asteroids

jupiter 0.52

saturn 0.47

uranus 0.51

neptune 0.41


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All this also says, if the planet is solid and habitable it probably has an albedo of around0.3, otherwise it is an outer gaseous planet and probably has an albedo of around 0.5.


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The luminosity of the sun is:

The separation between the earth and the sun is:

The solar luminosity at the earth is reduced by the inverse square law, sothe solar constant is:

That is the effective energy hitting the earth per second per square meter.

This radiation is equal to the temperature, , to the fourth power by the

steffan-bolzmann constant, sigma . can be called the effectivetemperature, the temperature entering the earth.

intercepts the earth disc, , and distributes itself over the entireearth surface, , while 30% is reflected back into space due to theearths albedo, a, which is equal to 0.3, so

But, just as the same amount of radiation that enters the system, leaves it,to have radiative equilibrium, the atmosphere radiates back to the surface

so that the radiation from the atmosphere, plus the radiation entering

the earth, is the radiation at the surface of the earth, . However,

and we have:

L0= 3.9 "10

26J/

1.5"1011

S0 =3.9"1026

4# (1.5 "1011)2=1,370Watts/meter

2

T

(" )

T

S0

"r2

4"r2

"Te

4=

S0

4(1# a)

(1# a)S0$r

2

4$r2

"Ta

4

"Te4

"Ts4

"Ta

4

="Te

4


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So, for the temperature at the surface of the Earth:

Lets convert that to degrees centigrade:

Degrees Centigrade = 303 - 273 = 30 degrees centigrade

And, lets convert that to Fahrenheit:

Degrees Fahrenheit = 30(9/5)+32=86 Degrees Fahrenheit

In reality this is warmer than the average annual temperature at the surfaceof the earth, but, in this model, we only considered radiative heat transferand not convective heat transfer. In other words, there is cooling due tovaporization of water (the formation of clouds) and due to the condensationof water vapor into rain droplets (precipitation or the formation of rain).

"Ts

4="T

a

4+"T

e

4= 2"T

e

4

Ts= 2

1

4Te

"Te

4=

S0

4(1# a)

" = 5.67 $10#8

S0 =1,370

a= 0.3

1,370

4(0.7) = 239.75

Te

4=

239.75

5.67 $10#8

=4.228 $109

Te= 255Kelvin

Ts=2

1

4Te=1.189(255) =303Kelvin


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The incoming radiation from the sun is about 1370 watts per square meteras determined by the energy per second emitted by the sun reduced by theinverse square law at earth orbit. We calculate the total absorbed energyintercepted by the Earth's disc (pi)r^2, its distribution over its surface area4(pi)r^2 and take into account that about 30% of that is reflected back intospace, so the effective radiation hitting the Earth's surface is about 70% ofthe incoming radiation reduced by four. Radiative energy is equal totemperature to the fourth power by the Stefan-boltzmann constant.However, the effective incoming radiation is also trapped by greenhousegases and emitted down towards the surface of the earth (as well asemitted up towards space from this lower atmosphere called thetroposphere), the most powerful greenhouse gas being CO2 (CarbonDioxide) and most abundant and important is water vapour. This doublesthe radiation warming the surface of the planet. The atmosphere is

predominately Nitrogen gas (N2) and Oxygen gas (O2), about 95 percent.These gases, however, are not greenhouse gases. The greenhouse gasCO2, though only exists in trace amounts, and water vapour, bring thetemperature of the Earth up from minus 18 degrees centigrade (18 belowfreezing) to an observed average of plus 15 degrees centigrade (15degrees above freezing). Without these crucial greenhouse gases, theEarth would be frozen. They have this enormous effect on warming theplanet even with CO2 existing only at 400 parts per million. It occursnaturally and makes life on Earth possible. However, too much of it and theEarth can be too warm, and we are now seeing amounts beyond thenatural levels through anthropogenic sources, that are making the Earthwarmer than is favorable for the conditions best for life to be maximallysustainable. We see this increase in CO2 beginning with the industrial era.The sectors most responsible for the increase are power, industry, andtransportation. Looking at records of CO2 amounts we see that it was 315parts per million in 1958 and rose to 390 parts per million in 2010. It roseabove 40s in radiative equilibrium, that is, it loses as much radiation as itreceives. Currently we are slightly out of radiative balance, the Earthabsorbs about one watt per square meter more than it loses. That means

its temperature is not steady, but increasing.


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Equilibrium: The Crux of Climate Science

Let us say the Earth is cold, absolute zero, then suddenly the sun blinks on. The Earth willreceive radiation and start to get warmer. As it gets warmer, it starts to lose some of the heat itreceives, and warms slower and slower. There are various mechanisms by which the Earth can

lose heat; which ones kick in and by how much they draw heat off the planet, are varynumerous and, vary in a wide spectrum as to the amount of heat, or energy in other words, thatthey can draw off the planet and, at what rates. We have discussed two mechanisms: radiativeheat transfer, and convective heat transfer. An example is the vaporization of the ocean, whichis water becoming a gas, or clouds in other words. The heat required to raise its temperature tothe point that it vaporizes is one calorie per gram degree centigrade. This represents a loss ofheat, or energy, from the sun, that would have gone into warming the planet. As well, when thevaporized water, or what are called clouds, condenses into liquid, this represents another loss ofheat-energy that would have gone into warming the planet, for the same reason it takes energyto make a refrigerator cold. This condensation of water vapor to its liquid form, is calledprecipitation, the formation of water droplets, or what we commonly call rain. The amount ofwater that is vaporized from the ocean must equal the amount that precipitates, rains back upon

the earth, in other words. If these two were not equal, then the oceans would dry up. Back tothe warming earth: as it warms, it does so slower and slower as the cooling mechanisms kick in.Eventually the rate at which the earth warms will slow down to zero. At this point the amount ofenergy it receives equals the amount of energy it loses and the earth is at a constanttemperature. This is called an equilibrium state. For the earth, this should be about 15 degreescentigrade in the annual average temperature. If the earth goes out of equilibrium, that is growswarmer or colder with time, then there can be a great deal of causes for this to happen, andmany complex factors must be considered to calculate how long it will take the earth to return toa stable temperature (equilibrium state) and to determine what the temperature of the earth willbe when it is back in equilibrium.

From a purely mathematical perspective, equilibrium states can be described by placing a ball in

a dish and displacing it to either the left or right: it will roll back and forth until by friction it settlesat the bottom of the dish motionless (in an equilibrium state). There can be two types ofequilibrium states. One, like we just described, a valley, or two, the reverse: a peak where wehave a ball balanced at the apex of a mountain. In this scenario, if I displace the ball to the leftor right, it will go out of equilibrium, but never return to equilibrium, like it did in the previousexample of a trough: but rather roll down the mountain, never to return.

The earth is currently out of equilibrium, that is, it receives more energy per second than it losesby one watt per square meter. This means the earth is warming. The reason for this is mostlybecause human activity is putting more CO2 into the atmosphere than should be there, whichmeans the earth retains more heat than it can lose.

Ian BeardsleyMarch 25, 2016


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Climate Modeling with Radiative Heat Transfer And Convection


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Model Future


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Star System: Alpha Centauri

Spectral Class: Same As The Sun

Proximity: Nearest Star System

Value For Projecting Human Trajectory: Ideal


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The probability of landing at four light years from earth at Alpha Centauri in 10

random leaps of one light year each (to left or right) is given by the equation of a

random walk:

{ W }_{ n }({ n }_{ 1 })=\frac { N! }{ { n }_{ 1 }!{ n }_{ 2 }! } { p }^{ n1 }{ q }^{ n2 }\\

N={ n }_{ 1 }+{ n }_{ 2 }\\ q+p=1

To land at plus four we must jump 3 to the left, 7 to the right (n1=3, n2 = 7: 7+3=10):


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Using our equation:

We would be, by this reasoning 12% along in the development towards hyperdrive.

Having calculated that we are 12% along in developing the hyperdrive, we can use the equation

for natural growth to estimate when we will have hyperdrive. It is of the form:

t is time and k is a growth rate constant which we must determine to solve the equation. In 1969

Neil Armstrong became the first man to walk on the moon. In 2009 the European Space Agency

launched the Herschel and Planck telescopes that will see back to near the beginning of theuniverse. 2009-1969 is 40 years. This allows us to write:

log 12 = 40k log 2.718

0.026979531 = 0.4342 k

k=0.0621

We now can write:

log 100 = (0.0621) t log e

t = 74 years

1969 + 74 years = 2043

Our reasoning would indicate that we will have hyperdrive in the year 2043.


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Study summary:

1. We have a 70% chance of developing hyperdrive without destroying ourselves first.

2. We are 12% along the way in development of hyperdrive.

3. We will have hyperdrive in the year 2043, plus or minus.

Sierra Waters was handed the newly discovered document in 2042.


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The Source Code


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bioplanet.c

#include#includeintmain(void){printf("\n");printf("\n");printf("Here we use a single atomospheric layer with no\n");printf("convection for the planet to be in an equilibrium\n");printf("state. That is to say, the temperature stays\n");printf("steady by heat gain and loss with radiative\n");printf("heat transfer alone.\n");printf("The habitable zone is calculated using the idea\n");printf("that the earth is in the habitable zone for a\n");printf("star like the Sun. That is, if a star is 100\n");printf("times brighter than the Sun, then the habitable\n");

printf("zone for that star is ten times further from\n");printf("it than the Earth is from the Sun because ten\n");printf("squared is 100\n");printf("\n");

floats, a, l, b, r, AU, N, root, number, answer, C, F;printf("We determine the surface temperature of a planet.\n");printf("What is the luminosity of the star in solar luminosities? ");scanf("%f", &s);printf("What is the albedo of the planet (0-1)?");scanf("%f", &a);printf("What is the distance from the star in AU? ");

scanf("%f", &AU);r=1.5E11*AU;l=3.9E26*s;b=l/(4*3.141*r*r);

N=(1-a)*b/(4*(5.67E-8));root=sqrt(N);number=sqrt(root);answer=1.189*(number);printf("\n");printf("\n");printf("The surface temperature of the planet is: %f K\n", answer);

C=answer-273;F=(C*1.8)+32;printf("That is %f C, or %f F", C, F);printf("\n");floatjoules;joules=(3.9E26*s);printf("The luminosity of the star in joules per second is: %.2fE25\n", joules/1E25);


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floatHZ;HZ=sqrt(joules/3.9E26);printf("The habitable zone of the star in AU is: %f\n", HZ);printf("Flux at planet is %.2f times that at earth.\n", b/1370);printf("That is %.2f Watts per square meter\n", b);

printf("\n");printf("\n");

printf("In this simulation we use a two layer atmospheric model\n");printf("where equilibrium is maintained by both radiative heat\n");printf("transfer and convection,\n");printf("\n");

printf("This program finds the temperature of a planet\n");

floatL0,sun,S0,r0,R,S,A,sigma,TE,delta,sTe4,sTs4;floatresult, answer2, c, f, x;printf("Luminosity of the star in solar luminosities? ");scanf("%f", &L0);printf("Planet distance from the star in AU? ");scanf("%f", &r0);printf("What is the albedo of the planet (0-1)? ");scanf("%f", &A);printf("What is the temp dif between layers in kelvin? ");scanf("%f", &delta);sun=3.9E26;S0=L0*sun;

R=(1.5E11)*r0;S=(S0)/((4)*(3.141)*R*R);sigma=5.67E-8;TE=(sqrt(sqrt(((1-A)*S*(0.25))/sigma)));x=delta/TE;sTe4=(1-A)*S/4;sTs4=3*(sTe4)-(sTe4)*(2-(1+x)*(1+x)*(1+x)*(1+x))-(sTe4)*(1+(1+x)*(1+x)*(1+x)*(1+x)-(1+2*x)*(1+2*x)*(1+2*x)*(1+2*x));result=(sTs4)/(sigma);answer2=sqrt((sqrt(result)));printf("\n");printf("\n");

printf("planet surface temp is: %f K\n", answer2);c=answer2-273;f=(1.8)*c+32;printf("That is %f C, or %f F\n", c, f);printf("flux at planet is %f watts per square meter\n", S);printf("\n");printf("\n");}


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cipher.c

#include #include #include intmain (intargc, string argv[1]){intk = atoi(argv[1]);if(argc>2|| argc


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climate.c


printf("zone for that star is ten times further from\n");printf("it than the Earth is from the Sun because ten\n");printf("squared is 100\n");floats, a, l, b, r, AU, N, root, number, answer, C, F;printf("We determine the surface temperature of a planet.\n");printf("What is the luminosity of the star in solar luminosities? ");scanf("%f", &s);printf("What is the albedo of the planet (0-1)?");scanf("%f", &a);printf("What is the distance from the star in AU? ");scanf("%f", &AU);r=1.5E11*AU;

l=3.9E26*s;b=l/(4*3.141*r*r);

N=(1-a)*b/(4*(5.67E-8));root=sqrt(N);number=sqrt(root);answer=1.189*(number);printf("\n");printf("\n");printf("The surface temperature of the planet is: %f K\n", answer);C=answer-273;F=(C*1.8)+32;

printf("That is %f C, or %f F", C, F);printf("\n");floatjoules;joules=(3.9E26*s);printf("The luminosity of the star in joules per second is: %.2fE25\n", joules/1E25);floatHZ;HZ=sqrt(joules/3.9E26);


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printf("The habitable zone of the star in AU is: %f\n", HZ);printf("Flux at planet is %.2f times that at earth.\n", b/1370);printf("That is %.2f Watts per square meter\n", b);printf("\n");printf("\n");}


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convection.c

#include#includeintmain (void){printf("\n");printf("\n");printf("This program finds the temperature of a planet\n");floatL0,sun,S0,r0,r,S,a,sigma,TE,delta,sTe4,sTs4;floatresult, answer, C, F, x;printf("Luminosity of the star in solar luminosities? ");scanf("%f", &L0);printf("Planet distance from the star in AU? ");scanf("%f", &r0);printf("What is the albedo of the planet (0-1)? ");scanf("%f", &a);

printf("What is the temp dif between layers in kelvin? ");scanf("%f", &delta);sun=3.9E26;S0=L0*sun;r=(1.5E11)*r0;S=(S0)/((4)*(3.142)*r*r);sigma=5.67E-8;TE=(sqrt(sqrt(((1-a)*S*(0.25))/sigma)));x=delta/TE;sTe4=(1-a)*S/4;sTs4=3*(sTe4)-(sTe4)*(2-(1+x)*(1+x)*(1+x)*(1+x))-(sTe4)*(1+(1+x)*(1+x)*(1+x)*(1+x)-(1+2*x)*(1+2*x)*(1+2*x));

result=(sTs4)/(sigma);answer=sqrt((sqrt(result)));printf("\n");printf("\n");printf("planet surface temp is: %f K\n", answer);C=answer-273;F=(1.8)*C+32;printf("That is %f C, or %f F\n", C, F);printf("flux at planet is %f watts per square meter\n", S);printf("\n");printf("\n");}


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modefuture.c

#include #include intmain (void){printf("\n");intN, r;doubleu, v, y, z;doublet,loga, ratio;intn1, n2;charname[15];floatW,fact=1,fact2=1,fact3=1,a,g,rate,T,T1;doublex,W2;printf("(p^n1)(q^n2)[W=N!/(n1!)(n2!)]");printf("\n");printf("x=e^(c*t)");

printf("\n");printf("W is the probability of landing on the star in N jumps.\n");printf("N=n1+n2, n1=number of one light year jumps left,\n");printf("n2=number of one light year jumps right.\n");printf("What is 1, the nearest whole number of light years to thestar, and\n");printf("2, what is the star's name?\n");printf("Enter 1: ");scanf("%i", &r);printf("Enter 2: ");scanf("%s", name);printf("Star name: %s\n", name);

printf("Distance: %i\n", r);printf("What is n1? ");scanf("%i", &n1);printf("What is n2? ");scanf("%i", &n2);printf("Since N=n1+n2, N=%i\n", n1+n2);N=n1+n2;printf("What is the probability, p(u), of jumping to the left? ");scanf("%lf", &u);printf("What is the probability, p(v), of jumpint to the left? ");scanf("%lf", &v);printf("What is the probability, q(y), of jumping to the right? ");

scanf("%lf", &y);printf("What is the probability, q(z), of jumping to the right? ");scanf("%lf", &z);printf("p=u:v");printf("\n");printf("q=y:z");printf("\n");for(inti=1; i


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{fact = fact*i;printf("N factorial = %f\n", fact);

a=pow(u/v,n1)*pow(y/z,n2);}for(intj=1; j


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modelocean.c

#include intmain (void){intoption;printf("\n");printf("The surface area of the earth is 510E6 square km.\n");printf("About three quarters of that is ocean.\n");printf("Half the surface area of the earth is receiving sunlight atany given moment.\n");printf("0.75*510E6/2 = 200E6 square km recieving light from the sun.\n");printf("There is about one gram of water per cubic cm.\n");printf("\n");printf("Is the section of water you are considering on the order of:\n");

printf("1 a waterhole\n");printf("2 a pond \n");printf("3 the ocean\n");scanf("%d", &option);

{floatarea, depth, cubic, density=0.000, mass=0.000;printf("How many square meters of water are warmed? ");scanf("%f", &area);printf("How many meters deep is the water warmed? ");scanf("%f", &depth);cubic=area*depth;

density=100*100*100; //grams per cubic meter//mass=(density)*(cubic);if(option==2){printf("That is %.3f E3 cubic meters of water. \n", cubic/1E3);printf("%.3f cubic meters of water has a mass of about %.3f E6 grams.\n", cubic, mass/1E6);printf("\n");printf("\n");}if(option==1){

printf("That is %.3f cubic meters of water.\n", cubic);printf("%.3f cubic meters of water has a mass of about %.3f E3 grams.\n", cubic, mass/1E3);}if(option==3){printf("That is %.3f E3 cubic meters of water.\n", cubic/1E3);


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printf("%.3f E3 m^3 of water has a mass of about %.3f E12 g\n", cubic/1E3, mass/1E12);printf("\n");}}printf("\n");floatreduction, incident, energy, watts, square, deep, volume, vol,densiti, matter;floattemp, increase, temperature;printf("The specific heat of water is one gram per calorie-degreecentigrade.\n");printf("One calorie is 4.8400 Joules.\n");printf("The light entering the earth is 1,370 Joules per second persquare meter.\n");printf("That is 1,370 watts per square meter.\n");printf("By what percent is the light entering reduced by clouds? (0-1)");

scanf("%f", &reduction);incident=reduction*1370;printf("Incident radiation is: %.3f watts per square meter.\n",incident);printf("\n");printf("The body of water is exposed to the sunlight from 10:00 AM to2:00 PM.\n");printf("That is four hours which are 14,400 seconds.\n");watts=14400*incident;printf("How many square meters of water are to be considered? ");scanf("%f", &square);printf("How deep is the water heated (in meters)? ");

scanf("%f", &deep);volume=deep*square; //volume in cubic meters//vol=volume*100*100*100; //volume in cubic centimeters//printf("The volume of water in cubic meters is: %.3f\n", volume);printf("That is %.3f E3 cubic centimeters.\n", vol/1E3);densiti=1.00; //density in grams per cubic cm//matter=densiti*vol; //grams of water//printf("That is %.3f E3 grams of water in %.3f cubic meters of water.\n", matter/1E3, volume);energy=watts*square/4.84;printf("That is %.3f cubic meters heated by %.3f calories\n", volume,energy);

printf("What is the intitial temperature of the body of water? ");scanf("%f", &temp);increase=energy/(matter*temp);temperature=increase+temp;printf("The temperature of the body of water has increased; %.3fdegrees C.\n", increase);printf("That means the temperature of the body of water is: %.3fdegrees C.\n", temperature);


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}

modelplanet.c

#include #include intmain(void){printf("\n");printf("We input the radii of the layers of a planet,...\n");printf("and their corresponding densities,...\n");printf("to determine the planet's composition.\n");printf("Iron Core Density Fe=7.87 g/cm^3\n");printf("Lithosphere Density Ni = 8.91 g/cm^3\n");printf("Mantle Density Si=2.33 g/cm^3\n");printf("Earth Radius = 6,371 km\n");

printf("Earth Mass = 5.972E24 Kg\n");printf("\n");floatr1=0.00, r2=0.00, r3=0.00, p1=0.00, p2=0.00, p3=0.00;printf("what is r1, the radius of the core in km? ");scanf("%f", &r1);printf("what is p1, its density in g/cm^3? ");scanf("%f", &p1);printf("what is r2, outer edge of layer two in km? ");scanf("%f", &r2);printf("what is p2, density of layer two in g/cm^3? ");scanf("%f", &p2);printf("what is r3, the radius of layer 3 in km? ");

scanf("%f", &r3);printf("what is p3, density of layer three in g/cm^3? ");scanf("%f", &p3);printf("\n");printf("\n");printf("r1=%.2f, r2=%.2f, r3=%.2f, p1=%.2f, p2=%.2f, p3=%.2f \n",r1,r2,r3,p1,p2,p3);printf("\n");

floatR1, v1, m1, M1;{R1=(r1)*(1000.00)*(100.00);

v1=(3.141)*(R1)*(R1)*(R1)*(4.00)/(3.00);m1=(p1)*(v1);M1=m1/1000.00;printf("the core has a mass of %.2f E23 Kg\n", M1/1E23);printf("thickness of core is %.2f \n", r1);}floatR2, v2, m2, M2;{


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R2=(r2)*(1000.00)*(100.00);v2=(3.141)*(R2*R2*R2-R1*R1*R1)*(4.00)/(3.00);m2=(p2)*(v2);M2=m2/1000.00;printf("layer two has a mass of %.2f E23 Kg\n", M2/1E23);printf("layer two thickness is %.2f \n", r2-r1);}floatR3, v3, m3, M3;{R3=(r3)*(1000.00)*(100.00);v3=(3.141)*(R3*R3*R3-R2*R2*R2)*(4.00)/(3.00);m3=(p3)*(v3);M3=m3/1000.00;printf("layer three has a mass of %.2f E23 Kg\n", M3/1E23);printf("layer three thickness is %.2f \n", r3-r2);}printf("\n");

printf("\n");printf("the mass of the planet is %.2f E24 Kg\n", (M1+M2+M3)/1E24);}


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starsystem.c


printf("zone for that star is ten times further from\n");printf("it than the Earth is from the Sun because ten\n");printf("squared is 100\n");printf("\n");

floats, a, l, b, r, AU, N, root, number, answer, C, F;printf("We determine the surface temperature of a planet.\n");printf("What is the luminosity of the star in solar luminosities? ");scanf("%f", &s);printf("What is the albedo of the planet (0-1)?");scanf("%f", &a);printf("What is the distance from the star in AU? ");

scanf("%f", &AU);r=1.5E11*AU;l=3.9E26*s;b=l/(4*3.141*r*r);

N=(1-a)*b/(4*(5.67E-8));root=sqrt(N);number=sqrt(root);answer=1.189*(number);printf("\n");printf("\n");printf("The surface temperature of the planet is: %f K\n", answer);

C=answer-273;F=(C*1.8)+32;printf("That is %f C, or %f F", C, F);printf("\n");floatjoules;joules=(3.9E26*s);printf("The luminosity of the star in joules per second is: %.2fE25\n", joules/1E25);


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floatHZ;HZ=sqrt(joules/3.9E26);printf("The habitable zone of the star in AU is: %f\n", HZ);printf("Flux at planet is %.2f times that at earth.\n", b/1370);printf("That is %.2f Watts per square meter\n", b);


printf("In this simulation we use a two layer atmospheric model\n");printf("where equilibrium is maintained by both radiative heat\n");printf("transfer and convection,\n");printf("\n");

printf("This program finds the temperature of a planet\n");

floatL0,sun,S0,r0,R,S,A,sigma,TE,delta,sTe4,sTs4;floatresult, answer2, c, f, x;printf("Luminosity of the star in solar luminosities? ");scanf("%f", &L0);printf("Planet distance from the star in AU? ");scanf("%f", &r0);printf("What is the albedo of the planet (0-1)? ");scanf("%f", &A);printf("What is the temp dif between layers in kelvin? ");scanf("%f", &delta);sun=3.9E26;S0=L0*sun;

R=(1.5E11)*r0;S=(S0)/((4)*(3.141)*R*R);sigma=5.67E-8;TE=(sqrt(sqrt(((1-A)*S*(0.25))/sigma)));x=delta/TE;sTe4=(1-A)*S/4;sTs4=3*(sTe4)-(sTe4)*(2-(1+x)*(1+x)*(1+x)*(1+x))-(sTe4)*(1+(1+x)*(1+x)*(1+x)*(1+x)-(1+2*x)*(1+2*x)*(1+2*x)*(1+2*x));result=(sTs4)/(sigma);answer2=sqrt((sqrt(result)));printf("\n");printf("\n");

printf("planet surface temp is: %f K\n", answer2);c=answer2-273;f=(1.8)*c+32;printf("That is %f C, or %f F\n", c, f);printf("flux at planet is %f watts per square meter\n", S);printf("\n");printf("\n");


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printf("\n");printf("We input the radii of the layers of a planet,...\n");printf("and their corresponding densities,...\n");printf("to determine the planet's composition.\n");printf("Iron Core Density Fe=7.87 g/cm^3\n");printf("Lithosphere Density Ni = 8.91 g/cm^3\n");printf("Mantle Density Si=2.33 g/cm^3\n");printf("Earth Radius = 6,371 km\n");printf("Earth Mass = 5.972E24 Kg\n");printf("\n");floatr1=0.00, r2=0.00, r3=0.00, p1=0.00, p2=0.00, p3=0.00;printf("what is r1, the radius of the core in km? ");scanf("%f", &r1);printf("what is p1, its density in g/cm^3? ");scanf("%f", &p1);

printf("what is r2, outer edge of layer two in km? ");scanf("%f", &r2);printf("what is p2, density of layer two in g/cm^3? ");scanf("%f", &p2);printf("what is r3, the radius of layer 3 in km? ");scanf("%f", &r3);printf("what is p3, density of layer three in g/cm^3? ");scanf("%f", &p3);printf("\n");printf("\n");printf("r1=%.2f, r2=%.2f, r3=%.2f, p1=%.2f, p2=%.2f, p3=%.2f \n",

r1,r2,r3,p1,p2,p3);

printf("\n");

floatR1, v1, m1, M1;{

R1=(r1)*(1000.00)*(100.00);v1=(3.141)*(R1)*(R1)*(R1)*(4.00)/(3.00);m1=(p1)*(v1);M1=m1/1000.00;printf("the core has a mass of %.2f E23 Kg\n", M1/1E23);printf("thickness of core is %.2f \n", r1);

}floatR2, v2, m2, M2;

{R2=(r2)*(1000.00)*(100.00);v2=(3.141)*(R2*R2*R2-R1*R1*R1)*(4.00)/(3.00);m2=(p2)*(v2);M2=m2/1000.00;printf("layer two has a mass of %.2f E23 Kg\n", M2/1E23);printf("layer two thickness is %.2f \n", r2-r1);

}


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floatR3, v3, m3, M3;{

R3=(r3)*(1000.00)*(100.00);v3=(3.141)*(R3*R3*R3-R2*R2*R2)*(4.00)/(3.00);m3=(p3)*(v3);M3=m3/1000.00;printf("layer three has a mass of %.2f E23 Kg\n", M3/1E23);printf("layer three thickness is %.2f \n", r3-r2);

}printf("\n");printf("\n");printf("the mass of the planet is %.2f E24 Kg\n", (M1+M2+M3)/

1E24);

intoption;printf("\n");

printf("The surface area of the earth is 510E6 square km.\n");printf("About three quarters of that is ocean.\n");printf("Half the surface area of the earth is receiving sunlight

at any given moment.\n");printf("0.75*510E6/2 = 200E6 square km recieving light from the

sun.\n");printf("There is about one gram of water per cubic cm.\n");printf("\n");printf("Is the section of water you are considering on the order

of: \n");printf("1 a waterhole\n");printf("2 a pond \n");

printf("3 the ocean\n");scanf("%d", &option);

{floatarea, depth, cubic, density=0.000, mass=0.000;printf("How many square meters of water are warmed? ");scanf("%f", &area);printf("How many meters deep is the water warmed? ");scanf("%f", &depth);cubic=area*depth;density=100*100*100; //grams per cubic meter//mass=(density)*(cubic);

if(option==2){

printf("That is %.3f E3 cubic meters of water. \n", cubic/1E3);

printf("%.3f cubic meters of water has a mass of about %.3f E6 grams.\n", cubic, mass/1E6);



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}if(option==1){

printf("That is %.3f cubic meters of water.\n", cubic);printf("%.3f cubic meters of water has a mass of about %.

3f E3 grams.\n", cubic, mass/1E3);}if(option==3){

printf("That is %.3f E3 cubic meters of water.\n", cubic/1E3);

printf("%.3f E3 m^3 of water has a mass of about %.3f E12g\n", cubic/1E3, mass/1E12);

printf("\n");}

}printf("\n");

floatreduction, incident, energy, watts, square, deep, volume,vol, densiti, matter;floattemp, increase, temperature;printf("The specific heat of water is one gram per calorie-degree

centigrade.\n");printf("One calorie is 4.8400 Joules.\n");printf("The light entering the earth is 1,370 Joules per second

per square meter.\n");printf("That is 1,370 watts per square meter.\n");printf("By what percent is the light entering reduced by clouds?

(0-1) ");scanf("%f", &reduction);

incident=reduction*1370;printf("Incident radiation is: %.3f watts per square meter.\n",incident);

printf("\n");printf("The body of water is exposed to the sunlight from 10:00 AM

to 2:00 PM.\n");printf("That is four hours which are 14,400 seconds.\n");watts=14400*incident;printf("How many square meters of water are to be considered? ");scanf("%f", &square);printf("How deep is the water heated (in meters)? ");scanf("%f", &deep);

volume=deep*square; //volume in cubic meters//vol=volume*100*100*100; //volume in cubic centimeters//printf("The volume of water in cubic meters is: %.3f\n", volume);printf("That is %.3f E3 cubic centimeters.\n", vol/1E3);densiti=1.00; //density in grams per cubic cm//matter=densiti*vol; //grams of water//printf("That is %.3f E3 grams of water in %.3f cubic meters of

water.\n", matter/1E3, volume);


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energy=watts*square/4.84;printf("That is %.3f cubic meters heated by %.3f calories\n",

volume, energy);printf("What is the intitial temperature of the body of water? ");scanf("%f", &temp);increase=energy/(matter*temp);temperature=increase+temp;printf("The temperature of the body of water has increased; %.3f

degrees C.\n", increase);printf("That means the temperature of the body of water is: %.3f

degrees C.\n", temperature);}


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Running The Models


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running bioplanet.c

Last login: Wed Jun 29 18:57:29 on ttys000/Users/ianbeardsley/Desktop/c\ files/modelsystems\ execs/bioplanet\copy ; exit;Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/c\ files/modelsystems\ execs/bioplanet\ copy ; exit;

Here we use a single atomospheric layer with noconvection for the planet to be in an equilibriumstate. That is to say, the temperature stayssteady by heat gain and loss with radiativeheat transfer alone.The habitable zone is calculated using the ideathat the earth is in the habitable zone for astar like the Sun. That is, if a star is 100

times brighter than the Sun, then the habitablezone for that star is ten times further fromit than the Earth is from the Sun because tensquared is 100

We determine the surface temperature of a planet.What is the luminosity of the star in solar luminosities? 3What is the albedo of the planet (0-1)?0.5What is the distance from the star in AU? 2

The surface temperature of the planet is: 259.846832 K

That is -13.153168 C, or 8.324298 FThe luminosity of the star in joules per second is: 117.00E25The habitable zone of the star in AU is: 1.732051Flux at planet is 0.76 times that at earth.That is 1034.70 Watts per square meter

In this simulation we use a two layer atmospheric modelwhere equilibrium is maintained by both radiative heattransfer and convection,

This program finds the temperature of a planet

Luminosity of the star in solar luminosities?


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running modelocean.c

Last login: Wed Jun 29 22:10:37 on ttys000Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/c\ files/modelsystems\ execs/modelocean ; exit;

The surface area of the earth is 510E6 square km.About three quarters of that is ocean.Half the surface area of the earth is receiving sunlight at any givenmoment.0.75*510E6/2 = 200E6 square km recieving light from the sun.There is about one gram of water per cubic cm.

Is the section of water you are considering on the order of:1 a waterhole2 a pond3 the ocean

1How many square meters of water are warmed? 5How many meters deep is the water warmed? .3That is 1.500 cubic meters of water.1.500 cubic meters of water has a mass of about 1500.000 E3 grams.

The specific heat of water is one gram per calorie-degree centigrade.One calorie is 4.8400 Joules.The light entering the earth is 1,370 Joules per second per squaremeter.That is 1,370 watts per square meter.By what percent is the light entering reduced by clouds? (0-1) 1

Incident radiation is: 1370.000 watts per square meter.

The body of water is exposed to the sunlight from 10:00 AM to 2:00 PM.That is four hours which are 14,400 seconds.How many square meters of water are to be considered? 5How deep is the water heated (in meters)? .5The volume of water in cubic meters is: 2.500That is 2500.000 E3 cubic centimeters.That is 2500.000 E3 grams of water in 2.500 cubic meters of water.That is 2.500 cubic meters heated by 20380166.000 caloriesWhat is the intitial temperature of the body of water? 75The temperature of the body of water has increased; 0.109 degrees C.

That means the temperature of the body of water is: 75.109 degrees C.logout

[Process completed]


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running modelplanet.c

Last login: Wed Jun 29 22:13:02 on ttys000Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/c\ files/modelsystems\ execs/modelplanet ; exit;

We input the raddi of the layers of a planet,...and their corresponding densities,...to determine the planet's composition.Iron Core Density Fe=7.87 g/cm^3Lithosphere Density Ni = 8.91 g/cm^3Mantle Density Si=2.33 g/cm^3Earth Radius = 6,371 kmEarth Mass = 5.972E24 Kg

what is r1, the radius of the core in km? 500what is p1, its density in g/cm^3? 7.87

what is r2, outer edge of layer two in km? 3000what is p2, density of layer two in g/cm^3? 8.91what is r3, the radius of layer 3 in km? 6371what is p3, density of layer three in g/cm^3? 2.33

r1=500.00, r2=3000.00, r3=6371.00, p1=7.87, p2=8.91, p3=2.33

the core has a mass of 0.00 E23 Kgthickness of core is 500.00layer two has a mass of 1.00 E23 Kglayer two thickness is 2500.00

layer three has a mass of 2.26 E23 Kglayer three thickness is 3371.00

the mass of the planet is 0.33 E24 Kg2016-06-29 22:17:16.672 modelplanet[77374:6559658] Hello, World!logout

[Process completed]


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running starsystem.c

Last login: Wed Jun 29 22:18:39 on ttys000Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/c\ files/modelsystems\ execs/starsystem\ copy ; exit;You have chosen to run starsytem by Ian Beardsley!.It is a simulator that models a habitable starsystem.

Here we use a single atomospheric layer with noconvection for the planet to be in an equilibriumstate. That is to say, the temperature stayssteady by heat gain and loss with radiativeheat transfer alone.The habitable zone is calculated using the ideathat the earth is in the habitable zone for astar like the Sun. That is, if a star is 100

times brighter than the Sun, then the habitablezone for that star is ten times further fromit than the Earth is from the Sun because tensquared is 100

We determine the surface temperature of a planet.What is the luminosity of the star in solar luminosities? 1What is the albedo of the planet (0-1)?.3What is the distance from the star in AU? 1

The surface temperature of the planet is: 303.727509 K

That is 30.727509 C, or 87.309517 FThe luminosity of the star in joules per second is: 39.00E25The habitable zone of the star in AU is: 1.000000Flux at planet is 1.01 times that at earth.That is 1379.60 Watts per square meter

In this simulation we use a two layer atmospheric modelwhere equilibrium is maintained by both radiative heattransfer and convection,

This program finds the temperature of a planet

Luminosity of the star in solar luminosities? 1Planet distance from the star in AU? 1What is the albedo of the planet (0-1)? .3What is the temp dif between layers in kelvin? 2

planet surface temp is: 259.447876 KThat is -13.552124 C, or 7.606177 F


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flux at planet is 1379.603149 watts per square meter

We input the radii of the layers of a planet,...and their corresponding densities,...to determine the planet's composition.Iron Core Density Fe=7.87 g/cm^3Lithosphere Density Ni = 8.91 g/cm^3Mantle Density Si=2.33 g/cm^3Earth Radius = 6,371 kmEarth Mass = 5.972E24 Kg

what is r1, the radius of the core in km? 200what is p1, its density in g/cm^3? 8what is r2, outer edge of layer two in km? 4000what is p2, density of layer two in g/cm^3? 9

what is r3, the radius of layer 3 in km? 6371what is p3, density of layer three in g/cm^3? 2.5

r1=200.00, r2=4000.00, r3=6371.00, p1=8.00, p2=9.00, p3=2.50

the core has a mass of 0.00 E23 Kgthickness of core is 200.00layer two has a mass of 24.12 E23 Kglayer two thickness is 3800.00layer three has a mass of 20.37 E23 Kglayer three thickness is 2371.00

the mass of the planet is 4.45 E24 Kg

The surface area of the earth is 510E6 square km.About three quarters of that is ocean.Half the surface area of the earth is receiving sunlight at any givenmoment.0.75*510E6/2 = 200E6 square km recieving light from the sun.There is about one gram of water per cubic cm.

Is the section of water you are considering on the order of:

1 a waterhole2 a pond3 the ocean2How many square meters of water are warmed? 500How many meters deep is the water warmed? .1That is 0.050 E3 cubic meters of water.50.000 cubic meters of water has a mass of about 50.000 E6 grams.


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The specific heat of water is one gram per calorie-degree centigrade.One calorie is 4.8400 Joules.The light entering the earth is 1,370 Joules per second per squaremeter.That is 1,370 watts per square meter.By what percent is the light entering reduced by clouds? (0-1) 1Incident radiation is: 1370.000 watts per square meter.

The body of water is exposed to the sunlight from 10:00 AM to 2:00 PM.That is four hours which are 14,400 seconds.How many square meters of water are to be considered? 500How deep is the water heated (in meters)? .1The volume of water in cubic meters is: 50.000That is 50000.000 E3 cubic centimeters.

That is 50000.000 E3 grams of water in 50.000 cubic meters of water.That is 50.000 cubic meters heated by 2038016384.000 caloriesWhat is the intitial temperature of the body of water? 72The temperature of the body of water has increased; 0.566 degrees C.That means the temperature of the body of water is: 72.566 degrees C.logout

[Process completed]


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running modelfuture.c

Last login: Wed Jun 29 22:20:17 on ttys000Claires-MBP:~ ianbeardsley$ /Users/ianbeardsley/Desktop/c\ files/modelfutre\ execs/modelfuture\ copy ; exit;

(p^n1)(q^n2)[W=N!/(n1!)(n2!)]x=e^(c*t)W is the probability of landing on the star in N jumps.N=n1+n2, n1=number of one light year jumps left,n2=number of one light year jumps right.What is 1, the nearest whole number of light years to the star, and2, what is the star's name?Enter 1: 4Enter 2: alphacentauriStar name: alphacentauri

Distance: 4What is n1? 3What is n2? 7Since N=n1+n2, N=10What is the probability, p(u), of jumping to the left? 1What is the probability, p(v), of jumpint to the left? 2What is the probability, q(y), of jumping to the right? 1What is the probability, q(z), of jumping to the right? 2p=u:vq=y:zN factorial = 1.000000N factorial = 2.000000

N factorial = 6.000000N factorial = 24.000000N factorial = 120.000000N factorial = 720.000000N factorial = 5040.000000N factorial = 40320.000000N factorial = 362880.000000N factorial = 3628800.000000n1 factorial = 1.000000n1 factorial = 2.000000n1 factorial = 6.000000n2 factorial = 1.000000

W=59062.500000 percentW=59063.00 percent rounded to nearest integraln2 factorial = 2.000000W=29531.250000 percentW=29531.00 percent rounded to nearest integraln2 factorial = 6.000000W=9843.750000 percentW=9844.00 percent rounded to nearest integral


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n2 factorial = 24.000000W=2460.937500 percentW=2461.00 percent rounded to nearest integraln2 factorial = 120.000000W=492.187500 percentW=492.00 percent rounded to nearest integraln2 factorial = 720.000000W=82.031250 percentW=82.00 percent rounded to nearest integraln2 factorial = 5040.000000W=11.718750 percentW=12.00 percent rounded to nearest integralWhat is t in years, the time over which the growth occurs? 40log(W)=1.079181loga/t=0.026980growthrate constant=0.062136log 100 = 2, log e = 0.4342, therfore

T=2/[(0.4342)(growthrate)]T=74.13 yearsWhat was the begin year for the period of growth? 1969Object achieved in 2043.13logout

[Process completed]


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Modeling In Python And Java

Ian Beardsley

2016


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bioplanet.java

importcomp102x.IO;/*** Here we write a program in java that models the temperature of aplanet for a star* of given luminosity.* @author(Ian Beardsley)* @version(Version 01 March 2016)*/publicclassbioplanet{

publicstaticvoidbioplanet(){

System.out.print("Enter the luminosity of the star in solar

luminosities: ");doublelum = IO.inputDouble();System.out.print("Enter the distance of the planet from the

star in AU: ");doubler=IO.inputDouble();System.out.print("Enter albedo of the planet (0-1): ");doublea=IO.inputDouble();doubleR=(1.5E11)*r;doubleS=(3.9E26)*lum;doubleb=S/(4*3.141*R*R);doubleN = (1-a)*b/(4*(5.67E-8));doubleroot = Math.sqrt(N);

doublenumber = Math.sqrt(root);doubleanswer = 1.189*number;IO.outputln("The surface temperature of the planet is:

"+answer+ " K");doubleC = answer - 273;doubleF = 1.8*C + 32;IO.outputln("That is: "+C+ " degrees centigrade");IO.outputln("Which is: "+ F + " degrees Fahrenheit");

}}


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stellar.py

print("We determine the surface temperature of a planet.")s=float(raw_input("Enter stellar luminosity in solar luminosities: "))a=float(raw_input("What is planet albedo (0-1)?: "))au=float(raw_input("What is the distance from star in AU?: "))r=(1.5)*(10**11)*aul=(3.9)*(10**26)*sb=l/((4.0)*(3.141)*(r**2))N=((1-a)*b)/(4.0*((5.67)*(10**(-8))))root=N**(1.0/2.0)number=root**(1.0/2.0)answer=1.189*numberprint("The surface temperature of the planet is: "+str(answer)+"K")C=answer-273F=(9.0/5.0)*C + 32print("That is "+str(C)+"C")

print("Which is "+str(F)+"F")joules=3.9*(10**26)*s/1E25lum=(3.9E26)*sprint("luminosity of star in joules per sec: "+str(joules)+"E25")HZ=((lum/(3.9*10**26)))**(1.0/2.0)print("The habitable zone is: "+str(HZ))flux=b/1370.0print("Flux at planet is "+str(flux)+" times that at earth")print("That is "+str(b)+ " watts per square meter")


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double.py

print("This program finds the temperature of a planet.")L0=float(raw_input("Luminosity of the star in solar luminosities? "))sun=3.9E26S0=L0*sunr0=float(raw_input("planet distance from star in AU? "))r=(1.5E11)*r0S=S0/((4)*(3.141)*(r**2))a=float(raw_input("What is the albedo of the planet (1-0)?: "))sigma=5.67E-8TE=((1-a)*S*(0.25)/(sigma))**(1.0/4.0)delta=float(raw_input("temp dif between two layers in Kelvin: "))x=delta/TEsTe4=(1-a)*S/4sTs4=3*(sTe4)-(sTe4)*(2-(1+x)**4)-(sTe4)*(1+((1+x)**4)-(1+2*x)**4)result=(sTs4)/(sigma)

answer=(result)**(1.0/4.0)print("planet surface temp is: "+ str(answer)+" K")C=answer-273F=(1.8)*C+32print("That is "+str(C)+" C, or "+str(F)+" F")print("flux at planet is "+ str(S)+" watts per square meter")


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objective.py

importmathobject=float(raw_input("Enter percent development towards objective:"));Tzero=float(raw_input("Enter the starting point (enter 1969): "));L=math.log10(object)/math.log10(2.718);T=L/(0.0621);Time=Tzero+T;print("Time to objective is: "+ str(T) + "years");print("That is the year: "+ str(Time));


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input.c

#include #include int main (void){printf("\n");char s[15], w[15], t[5], b[10];printf("Sierra Waters\n");printf("The Brain\n");printf("Enter Last Name: ");scanf("%s", w);printf("Enter First Name: ");scanf("%s", s);printf("Enter Name: ");scanf("%s", b);printf("Enter Definite Article: ");

scanf("%s", t);printf("%s, %s: She was handed the newly discovered document in2042.\n", w, s);printf("%s, %s: He designed hyperdrive in 2044.\n", b, t);printf("Between 2042 and 2044 is 2043.\n");printf("\n");printf("\n");float object, Tzero, T, time, L;int n;printf("If we use alphacentauri as the key to our model,\n");printf("for modeling the future, then our task has been reduced,\n");printf("through the work I have done, to quite a simple one.\n");

printf("growthrate=k=0.0621, objective=log 100/log e = 4.6achievements,\n");printf("Tzero=1969 when we landed on the moon, which at 2009 is0.552=.0.12(4.6)\n");printf("1/0.55 = 1.8=9/5 = R/r = Au/Ag, putting us in the age ofgold:silver\n");printf("Our equation is then, Time=(Object Achieved)/(Achievements/year)\n");printf("\n");

do{

printf("How many simulations would you like to run (10 max)? ");scanf("%d", &n);}while (n>10&& n


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printf("What is percent development towards objective(0-100)? ");scanf("%f", &object);}while (n100);printf("What is the starting point (year:enter 1969) ?");scanf("%f", &Tzero);L= ((log10 (object))/((log10 (2.718))));T=L/(0.0621);time= Tzero+T;printf("Time to object=%f years.\n", T);printf("That is the year: %f\n", time);}printf("\n");printf("If you chose tzero as moon landing (1969), then you found\n");printf("obect acheived 2043 between Sierra Waters and The Brain.\n");printf("That time being reached in 74 years after time zero.\n");printf("If you ran a second simulation again with t zero at 1969, and

\n");printf("ran the program for the 74 years to hyperdrive reduced by\n");printf("a factor of ten (that is input 7.4 percent development.)\n");printf("Then, you found object achieved in 2001, the year of Kubrick's\n");printf("Starchild\n");printf("\n");}


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object.c

#include int main (void){printf("\n");float object, Tzero, T, time;int n;printf("If we use alphacentauri as the key to our model,\n");printf("for modeling the future, then our task has been reduced,\n");printf("through the work I have done, to quite a simple one.\n");printf("growthrate=k=0.0621, objective=log 100/log e = 4.6achievements,\n");printf("Tzero=1969 when we landed on the moon, which at 2009 is0.552=.0.12(4.6)\n");printf("1/0.55 = 1.8=9/5 = R/r = Au/Ag, putting us in the age of

gold:silver\n");printf("Our equation is then, Time=(Object Achieved)/(Achievements/year)\n");printf("\n");

do{printf("How many simulations would you like to run (10 max)? ");scanf("%d", &n);}while (n>10&& n


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Paul Levinson And Manuel Heredia


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I have written three papers on the anomaly of how my scientific investigation shows the

Universe related to the science fiction of Paul Levinson, Isaac Asimov, and Arthur C. Clarke. In

my last paper, The Levinson-Asimov-Clarke Equation part of the comprehensive work The

Levinson, Asimov, Clarke Triptic, I suggest these three authors should be taken together to make

some kind of a whole, that they are intertwined and at the heart of science fiction. I have now

realized a fourth paper is warranted, and it is just the breakthrough I have been looking for to put

myself on solid ground with the claim that fiction is related to reality in a mathematical way

pertaining to the Laws of Nature. I call it Fiction-Reality Entanglement.

In my paper Paul Levinson, Isaac Asimov, Arthur C. Clarke Intertwined With An Astronomers

Research, I make the mathematical prediction that humans have a 70% chance of developing

Hyperdrive in the year 2043 to word it as Paul Levinson worded it, and I point out that this is

only a year after the character Sierra Waters is handed a newly discovered document that sets in

motion the novel by Paul Levinson, The Plot To Save Socrates.

I now find that Isaac Asimov puts such a development in his science fiction at a similar time inthe future, precisely in 2044, only a year after my prediction and two years after Sierra Waters is

handed the newly discovered document that initiates her adventure. So, we have my prediction,

which is related to the structure of the universe in a mystical way right in between the dates of

Levinson and Asimov, their dates only being a year less and a year greater than mine.

Asimov places hyperdrive in the year 2044 in his short story Evidence which is part of his

science fiction collection of short stories called, I, Robot.

This is a collection of short stories where Robot Psychologist Dr. Susan Calvin is interviewed by

a writer about her experience with the company on earth in the future that first developedsophisticated robots. In this book, the laws of robotics are created and the idea of the positronic

brain introduced, and the nature of the impact robots would have on human civilization is

explored. Following this collection of stories Asimov wrote three more novels, which comprise

the robot series, The Caves of Steel, The Naked Sun, and The Robots of Dawn.

I, Robot is Earth in the future just before Humanity settles the more nearby stars. The novels

comprising The Robot Series are when humanity has colonized the nearby star systems, The

Foundation Trilogy, and its prequels and sequels are about the time humanity has spread

throughout the entire galaxy and made an Empire of it. All of these books can be taken together

as one story, with characters and events in some, occurring in others.


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Hyperdrive is invented in I, Robot by a robot called The Brain, owned by the company for which

Dr. Susan Calvin works when it is fed the mathematical logistical problems of making

hyperdrive, and asked to solve them. It does solve them and it offers the specs on building an

interstellar ship, for which two engineers follow in its construction. They are humorously sent

across the galaxy by The Brain, not expecting it, and brought back to earth in the ship after they

constructed it. This was in the story in I, Robot titled Escape!.

But Dr. Susan Calvin states in the following short story, that I mentioned, Evidence:

But that wasnt it, eitherOh, eventually, the ship and others like it became government

property; the Jump through hyperspace was perfected, and now we actually have human colonies

on the planets of some of the nearer stars, but that wasnt it.

It was what happened to the people here on Earth in last fifty years that really counts.

And, what happened to people on Earth? The answer is in the same story Evidence from which

that quote is at the beginning. It was when the Regions of the Earth formed The Federation. Dr.

Susan Calvin says at the end of the story Evidence:

He was a very good mayor; five years later he did become Regional Co-ordinator. And when

the Regions of Earth formed their Federation in 2044, he became the first World Co-ordinator.

It is from that statement that I get my date of 2044 as the year Asimov projects for hyperdrive.

Ian BeardsleyMarch 17, 2011


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I watched a video on youtube about Terence McKenna where he lectured on his timewave zero

theory. I found there was not an equation for his timewave zero graph but that a computer

algorithm generated the graph of the wave. The next day I did a search on the internet to see if a

person could download timewave software for free. As it turned out one could, for both Mac and

pc. It is called Timewave Calculator Version 1.0. I downloaded the software and found you had

to download it every time after you quit the application and that you could not save the graph of

your results or print them out. So I did a one-time calculation. It works like this: you input the

range of time over which you want see the timewave and you cannot calculate past 2012,

because that is when the timewave ends. You also put in a target date, the time when you want to

get a rating for the novelty of the event that occurred on that day. You can also click on any point

in the graph to get the novelty rating for that time. I put in:


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Input:

Begin Date: December 27 1968 18 hours 5 minutes 37 seconds

End Date: December 2 2011 0 hours 28 minutes 7 seconds

McKenna said in the video on youtube that the dips, or valleys, in the timewave graph represent

novelties. So, I clicked on the first valley after 1969 since that is the year we went to the moon,

and the program gave its novelty as:

Sheliak Timewave Value For Target:

0.0621

On Target Date: August 4, 1969 9 hours 53 minutes 38 seconds

I was happy to see this because, I determined that the growth rate constant, k, that rate at which

we progress towards hyperdrive, in my calculation in my work Asimovian Prediction For

Hyperdrive, that gave the date 2043, a year after Sierra Waters was handed the newly discovered

document that started her adventure in The Plot To Save Socrates, by Paul Levinson, and a year

before Isaac Asimov had placed the invention of hyperdrive in his book I, Robot, was:

(k=0.0621)

The very same number!!!

What does that mean? I have no idea; I will find out after I buy The Invisible Landscape by

Terence McKenna, Second Edition, and buy a more sophisticated timewave software than that

which is offered for free on the net.

Ian Beardsley

March 19, 2011


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There is a common thread running through the Science Fiction works of Paul Levinson, Isaac

Asimov, and Arthur C. Clarke.

In the case of Isaac Asimov, we are far in the future of humanity. In his Robot Series, Asimov has

man making robots whose programming only allows them to do that which is good for humanity.

As a result, these robots, artificial intelligence (AI), take actions that propel humanity into

settling the Galaxy, in the robot series, and ultimately save humanity after they have settled the

Galaxy and made an empire of it (In the Foundation Series).

In the case of Paul Levinson, scholars in the future travel through time and use cloning, a

concept related to artificial intelligence (it is the creating of human replicas as well, but

biological, not electronic), and the goal is to save great ancient thinkers from Greece, and to

manipulate events in the past for a positive outcome for the future of humanity, just as the robots

try to do in the work of Asimov.

In the case of Arthur C. Clarke, man undergoes a transformation due to a monolith placed on the

moon and earth by extraterrestrials who have created life on earth. The monolith is a computer. It

takes humans on a voyage to other planets in the solar system, and in their trials, humanity goes

through trials that result in a transformation for the ending of their dependence on their

technology and for becoming adapted to life in the Universe beyond Earth. That is, the character

Dave Bowman becomes the Starchild in his mission to Jupiter. The artificial intelligence is the

ship computer called HAL.

So, the thread is the salvation of man through technology, and their transformation to a new

human paradigm, where they can end their dependence on Earth and adapt to the nature of theUniverse as a whole.

At the time I was reading these novels, I was doing astronomical research, and, to my utter

astonishment, my relationships I was discovering pertaining to the Universe were turning up

times and values pivotal to these works of Levinson, Asimov, and Clarke. Further, I was

interpreting much of my discoveries by developing them in the context of short fictional stories.

In my story, The Question, we find Artificial Intelligence is in sync with the phases of the first

appearance of the brightest star Sirius for the year, and the flooding of the Nile river, which

brings in the Egyptian agricultural season. It is presumed by some scholars that because theEgyptian calendar is in sync with the Nile-Sirius cycle, theirs began four such cycles ago.

I then relate that synchronization to another calculation that turns up the time when the key

figure of the Foundation Series of Asimov begins his program to found a civilization that will

save the galaxy. We later find his actions were manipulated into being by robots, in order to save

intelligent life in the galaxy by creating a viable society for it called Galaxia.


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In the case of Paul Levinson, I was making a calculation to predict when man would develop

hyperdrive, that engine which could take us to the stars, and end our dependence on an Earth that

cannot take care of humans forever. That time turned out to be when the key scholar in the work

by Paul Levinson, began her quest to help humanity by traveling into the past and using cloning,

in part, to change history for the better. I can now only feel her quest to save humanity is going to

be through changing history to bring about the development of hyperdrive, so humanity will no

longer depend on Earth alone, which, as I have said, cannot take care of life forever.

Finally, where Arthur C. Clarke is concerned, I find values in the solar system and nature that are

in his monolith, and I connect it to artificial intelligence of a sort, that kind which would be

based on silicon.


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LEVMAN

When we consider the 9/5 of five-fold symmetry (the biological) and the 5/3 and 11/6, ofsix-fold symmetry (the physical), we can make three equations and therefore find aplace in space. If we let the parameter, t, be zero we have a place in space that is near

the SETI Wow! Signal (extraterrestrial message) in the constellation Sagittarius. If welet the parameter, t, be eliminated we have a place in space that points to theconstellation Aquila. The former relationship came to me via a Gypsy Shaman calledManuel in Granada, Spain. The latter relationship came to me via a Fordham Universityprofessor and science fiction author called Paul Levinson, in New York. The former andlatter relationships stemmed from separate and independent research about two totallydifferent topics, the former dealing with what I call the Yin and Yang of the Universe, thelatter Fiction-Reality Entanglement, or what could be called the unfolding of theMcKenna time-wave. Now we find the two concepts are part of one theory and arebound to one another by the standard reference for concert pitch, A440; that tone whichthe oboe sounds before the symphony plays so that all the instruments can be tuned to

it. Threaded through it is the discovery that AI (artificial intelligence) is connected tosomething even deeper than what its makers know themselves.

Ian BeardsleyDecember 1, 2014


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So, M could be also, the square root of two over two, which we immediately recognize is an

important number as well in that it is the the sine of 45 degrees which is equal to the cosine of 45

degrees, which is the not only derived from the important 45-45-90 triangle, but is the angle for

maximum range in projectile physics, it is steep enough that it allows a lot of time in the air for a

projectile and shallow enough that the projectile has a lot of horizontal motion. So, we can also

write another expression for Manuel and Levinson which is:

ML = 440

Let us look at this. This says the product of Levinson and Manuel is A440. That is, taken

separately Manuel and Levinson cannot put the earth in tune, but taken together they can. This is

interesting because Manuel and Levinson come from very different places, but one can see

clearly that their different talents working together, would produce an Earth in tune, that has

maximum range if we consider Manuels number is the 45 degrees for maximum range of a

projectile and Levinsons number is the growth rate for human progress. (See my work, ET to

AI). The mathematical trick used to get the new value for manuels number was using anequation like a template. That is the numbers in the equation are merely place holders for which

you can substitute other values that make sense in terms of them, like when using a template to

design a website or blog. The template is an idea, but you can change the content. Whether or

not this is an acceptable approach or not does not matter because, through it we discovered the

above relationship.


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R = Solar Radius

r= lunar orbital radius

Au = gold

Ag = silver

R/r = Au/Ag = 9/5

(9/5)(4) = 7.2

Mars = 4

The earth precesses through one degree in 72 years

0.72 = Venus orbital radius in Astronomical Units (AU)

harmonic mean: Ga; As = 72.23

geometric mean: Ga; As = 72.27

23=Manuel Number

27=Manuel Number

23X27=621=L=Levinsons Number

Ga=69.72

As=74.92

Ga=Gallium

As=Arsenic

Ga and As are doping agents for making diodes, transistors, integrated circuitry,the operational

components of AI.

a) 2(69.72)(74.92)=10446.8448

b) (69.72)+(74.92)=144.64

c) a/b=72.2265~72.23

d) (69.72)(74.92)=5223.4224

e) sqrt(5223.4224)=72.27324816

9/5 connects pi to phi: 3.141+1.618=4.759

7=(9+5)/2


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360/5=72 360-72=288

288/360 = 8/10 (8/10)+1 = 9/5

360/6 = 60 360-60-60=240

240/360 = 2/3 (2/3)+1 = 5/3

360/6=60 360-60 = 300300/360 = 5/6 (5/6)+1 = 11/6

9/5, 5/3, 11/6

9/5: 5, 14, 23, 32,

1.8, 3.6, 5.4, 7.2,

5/3: 8, 13, 18, 23,

1.7, 3.3, 5, 6.7

11/6: 6, 17, 28, 39

11/6, 11/3, 11/2, 22/3

9/5: a_n=7.2n-4

5/3: a_n=3.3n+3

11/6: a_n=9n-5

=> (Here we have inverted the coefficients of the equation of the plane)

sqrt( ((5/36)^2) + ((20/33)^2)) = 0.0621 = Levinsons number~phi=0.618~0.62

Here we have eliminated n and taken the gradient to find the normal to the plane. The figure in

the square root is the right ascension vector pointing to the constellation Aquila.

sin 45 = (sqrt(2))/2

(sqrt(2))/2= M = manuels number

621=L= levinsons number

ML=440

440=standard concert pitch

23, 27 = Manuel Numbers

23X27=621=L


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0.0621, Levinsons number is a growth rate for progress, There are 0.621 miles in a kilometer

(km) and 1 km=1/10,000 of the distance from the pole to the equator. 0.0621 is the novelty rating

in the McKenna Timewave for the year humans first set foot on the moon.


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Amarjit


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An Indian Tabla Set


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I found myself in the medioambiente (atmosphere) of an Indian ethnomusicologist from theUniversity of Delhi, taking tabla lessons. First he explained to me that he was not an idle man,that he had many students and taught from five in the morning to 6 in the evening, and that

during that time he prepared a special soup for all those that were his students. One of the firstthings he told me was that in India there are many false gurus, that will not really teach you, butthat this was not his consciousness, that he would really teach me. He said he would put me ona program of learning to play Bahjrans and Kirtans, rythms of 6 and 7 which fall under thecategory of Guzals, or Indian romantic music, but that he would be playing and composingtemple music, called tin tal, which was the cycle of 16 considered the highest and most spiritualform of North Indian Classical music.

The training began with the history of the tabla which has its origins in the mridangam. It wasthe Muslim King in India, Amir Kusuro, who took the mridangam, which was closed on bothends, the left side played with the left hand and the right side played with the right hand, andbroke it into two, the Dayan and Bayan, with the Dyan being the high tones and the Bayan being

the low tones played with right and left hands respectively. In the center of each is a circle ofdry ink that allows the drums to be tuned to precise pitches. The ink is rubbed into the tabla, aswas explained to me, with a stone that floats on water and glows like a cats eye and only existsin a few secret, undisclosed locations, only known to tabla makers. Amarjit, that was his name,had made it a point of telling me that among the rhythms I would be learning was a cycle ofseven and one half and a cycle of 13 1/2. I find that interesting. If a person considers eachbeat of one half a beat of one, then that is a cycle of 15. It was the Gypsy Shaman, Manuel,who first pointed out to me that 15 was of primary importance, and as a scientist, I cant help butthink in reference to that, the earth rotates through 15 degrees in an hour, and the mostabundant element in the earths atmosphere is nitrogen which is in chemical group 15 in theperiodic table. Let us multiply Amarjits 7 1/2 by the 16 of his tin tal. It is 120. 120 are thedegrees in the angles of a regular hexagon, an equal angled, equal sided polygon with six

sides. Let us subtract 120 from the 360 degrees that are in a circle and divide the result by thatsame 360 and then add the result to one:

360-120 = 240240/360 = 2/32/3+1 = 5/3

This is the value that represents the yang of the cosmic yin and yang that came to us from theGypsy Shaman, Manuel, that represents six-fold symmetry, or the physical aspects of nature,like snowflakes. The biological aspects are in five-fold symmetry, derived as above:

360/5 = 72360-72 = 288288/360 = 4/54/5 + 1 = 9/5 = 1.8

Let us divide Amarjits stressed cycle 13.5 by 7.5. We find it is 1.8, which equals the yin of 9/5that is representative of the organic aspects of nature to which the Gypsy Shaman, Manuelguided us in my story Gypsy Shamanism and the Universe, which I will present following thestory we are telling now.


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After my tabla lesson, I left the room and just as I came out, several people from India werecoming into the house. I noticed in the living room was lots of clothing and art from India. I wasintroduced to these people, who obviously ran a store, and they told me they were just comingback from an interactive convention between Indians and Mexicans. The interchange was one

between ideas in the cooking of Indian food and Mexican food. They were all wearing nametags that said on them, Friendly Amigo.

Later I met with Amarjit and he took me to a music store to give me a lesson in buyinginstruments. On our way back, with his student driving, me in the front seat, Amarjit laidstretched out on the back back seat telling me that the store owners refusal of our price offer fora crude guitar indicated that he was A very greedy man and would not get far in life. At somepoint I told Amarjit that I had dreams of him giving me tabla lessons. He told me he couldcommunicate with me in this way.

Upon learning that God told me the Gypsy Shaman, Manuel, always second guesses him, andManuel telling me that because of this, he goes out into the world to do Gods work for him at

his request, Amarjit and his students were going to change their course from one of mergingwith God, to one of merging with Manuel.

Ian BeardsleyMay 15, 2015


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Manuel


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Gypsy Shamanism And The Universe

I wrote a short story last night, called Gypsy Shamanism and the Universe about the AE-35 unit,

which is the unit in the movie and book 2001: A Space Odyssey that HAL reports will fail and

discontinue communication to Earth. I decided to read the passage dealing with the event in

2001 and HAL, the ship computer, reports it will fail in within 72 hours. Strange, because Venus

is the source of 7.2 in my Neptune equation and represents failure, where Mars represents

success.

Ian Beardsley

August 5, 2012


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Chapter One

It must have been 1989 or 1990 when I took a leave of absence from The University Of Oregon,

studying Spanish, Physics, and working at the state observatory in Oregon -- Pine Mountain

Observatoryto pursue flamenco in Spain.

The Moors, who carved caves into the hills for residence when they were building the Alhambra

Castle on the hill facing them, abandoned them before the Gypsies, or Roma, had arrived there

in Granada Spain. The Gypsies were resourceful enough to stucco and tile the abandoned

caves, and take them up for homes.

Living in one such cave owned by a gypsy shaman, was really not a down and out situation, asthese homes had plumbing and gas cooking units that ran off bottles of propane. It was really

comparable to living in a Native American adobe home in New Mexico.

Of course living in such a place came with responsibilities, and that included watering its

gardens. The Shaman told me: Water the flowers, and, when you are done, roll up the hose

and put it in the cave, or it will get stolen. I had studied Castilian Spanish in college and as

such a hose is una manguera, but the Shaman called it una goma and goma translates as

rubber. Roll up the hose and put it away when you are done with it: good advice!

So, I water the flowers, rollup the hose and put it away. The Shaman comes to the cave the

next day and tells me I didnt roll up the hose and put it away, so it got stolen, and that I had tobuy him a new one.

He comes by the cave a few days later, wakes me up asks me to accompany him out of The

Sacromonte, to some place between there and the old Arabic city, Albaicin, to buy him a new

hose.

It wasnt a far walk at all, the equivalent of a few city blocks from the caves. We get to the store,

which was a counter facing the street, not one that you could enter. He says to the man behind

the counter, give me 5 meters of hose. The man behind the counter pulled off five meters of

hose from the spindle, and cut the hose to that length. He stated a value in pesetas, maybe

800, or so, (about eight dollars at the time) and the Shaman told me to give that amount to theman behind the counter, who was Spanish. I paid the man, and we left.

I carried the hose, and the Shaman walked along side me until we arrived at his cave where I

was staying. We entered the cave stopped at the walk way between living room and kitchen,

and he said: follow me. We went through a tunnel that had about three chambers in the cave,

and entered one on our right as we were heading in, and we stopped and before me was a

collection of what I estimated to be fifteen rubber hoses sitting on ground. The Shaman told me


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to set the one I had just bought him on the floor with the others. I did, and we left the chamber,

and he left the cave, and I retreated to a couch in the cave living room.


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Chapter Two

Gypsies have a way of knowing things about a person, whether or not one discloses it to them

in words, and The Shaman was aware that I not only worked in Astronomy, but that my work in

astronomy involved knowing and doing electronics.

So, maybe a week or two after I had bought him a hose, he came to his cave where I was

staying, and asked me if I would be able to install an antenna for television at an apartment

where his nephew lived.

So this time I was not carrying a hose through The Sacromonte, but an antenna.

There were several of us on the patio, on a hill adjacent to the apartment of The Shaman s

Nephew, installing an antenna for television reception.

Chapter Three

I am now in Southern California, at the house of my mother, it is late at night, she is a asleep,

and I am about 24 years old and I decide to look out the window, east, across The Atlantic, to

Spain. Immediately I see the Shaman, in his living room, where I had eaten a bowl of the Gypsy

soup called Puchero, and I hear the word Antenna. I now realize when I installed the antenna, I

had become one, and was receiving messages from the Shaman.

The Shamans Children were flamenco guitarists, and I learned from them, to play the guitar. I

am now playing flamenco, with instructions from the shaman to put the gypsy space program

into my music. I realize I am not just any antenna, but the AE35 that malfunctioned aboard The

Discovery just before it arrived at the planet Jupiter in Arthur C. Clarke s and Stanley Kubricks

2001: A Space Odyssey. The Shaman tells me, telepathically, that this time the mission wont

fail.

Chapter Four

I am watching Star Wars and see a spaceship, which is two oblong capsules flying connected in

tandem. The Gypsy Shaman says to me telepathically: Dios es una idea: son dos. Iunderstand that to mean God is an idea: there are two elements. So I go through life basing

my life on the number two.

Chapter Five

Once one has tasted Spain, that person longs to return. I land in Madrid, Northern Spain, The

Capitol. The Spaniards know my destination is Granada, Southern Spain, The Gypsy


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Neighborhood called The Sacromonte, the caves, and immediately recognize I am under the

spell of a Gypsy Shaman, and what is more that I am The AE35 Antenna for The Gypsy Space

Program. Flamenco being flamenco, the Spaniards do not undo the spell, but reprogram the

instructions for me, the AE35 Antenna, so that when I arrive back in the United States, my

flamenco will now state their idea of a space program. It was of course, flamenco being

flamenco, an attempt to out-do the Gypsy space program.

Chapter Six

I am back in the United States and I am at the house of my mother, it is night time again, she is

asleep, and I look out the window east, across the Atlantic, to Spain, and this time I do not see

the living room of the gypsy shaman, but the streets of Madrid at night, and all the people, and

the word Jupiter comes to mind and I am about to say of course, Jupiter, and The Spanish

interrupt and say Yes, you are right it is the largest planet in the solar system, you are right to

consider it, all else will flow from it.

I know ratios, in mathematics are the most interesting subject, like pi, the ratio of thecircumference of a circle to its diameter, and the golden ratio, so I consider the ratio of the orbit

of Saturn (the second largest planet in the solar system) to the orbit of Jupiter at their closest

approaches to The Sun, and find it is nine-fifths (nine compared to five) which divided out is one

point eight (1.8).

I then proceed to the next logical step: not ratios, but proportions. A ratio is this compared to

that, but a proportion is this is to that as this is to that. So the question is: Saturn is to Jupiter

as what is to what? Of course the answer is as Gold is to Silver. Gold is divine; silver is next

down on the list. Of course one does not compare a dozen oranges to a half dozen apples, but

a dozen of one to a dozen of the other, if one wants to extract any kind of meaning. But atoms

of gold and silver are not measured in dozens, but in moles. So I compared a mole of gold to amole of silver, and I said no way, it is nine-fifths, and Saturn is indeed to Jupiter as Gold is to

Silver.

I said to myself: How far does this go? The Shamans son once told me he was in love with the

moon. So I compared the radius of the sun, the distance from its center to its surface to the

lunar orbital radius, the distance from the center of the earth to the center of the moon. It was

Nine compared to Five again!

Chapter Seven

I had found 9/5 was at the crux of the Universe, but for every yin there had to be a yang. Ninefifths was one and eight-tenths of the way around a circle. The one took you back to the

beginning which left you with 8 tenths. Now go to eight tenths in the other direction, it is 72

degrees of the 360 degrees in a circle. That is the separation between petals on a five-petaled

flower, a most popular arrangement. Indeed life is known to have five-fold symmetry, the

physical, like snowflakes, six-fold. Do the algorithm of five-fold symmetry in reverse for six-fold

symmetry, and you get the yang to the yin of nine-fifths is five-thirds.


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Nine-fifths was in the elements gold to silver, Saturn to Jupiter, Sun to moon. Where was five-

thirds? Salt of course. The Salt Of The Earth is that which is good, just read Shakespeares

King Lear. Sodium is the metal component to table salt, Potassium is, aside from being an

important fertilizer, the substitute for Sodium, as a metal component to make salt substitute.

The molar mass of potassium to sodium is five to three, the yang to the yin of nine-fifths, which

is gold to silver. But multiply yin with yang, that is nine-fifths with five-thirds, and you get 3, andthe earth is the third planet from the sun.

I thought the crux of the universe must be the difference between nine-fifths and five-thirds. I

subtracted the two and got two-fifteenths! Two compared to fifteen! I had bought the Shaman

his fifteenth rubber hose, and after he made me into the AE35 Antenna one of his first

transmissions to me was: God Is An Idea: There Are Two Elements.

It is so obvious, the most abundant gas in the Earth Atmosphere is Nitrogen, chemical

group 15 and the Earth rotates through 15 degrees in one hour.


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The Bronze Age

Often the one thing you are looking for is the one thing that was left out of the story.

If you are an archaeologist you understand that gold and silver were important to early civilizations,

especially to be used for ceremonial jewelries. But, you would also know that copper was used earlier

and more as it is a soft and malleable metal that can be worked without being heated, pounded out into

flat sheets.

Copper (Cu) used tin (Sn) as an alloying metal to make bronze, which was the beginning of the Bronze

Age in Mesopotamia around 3500 BC.

These elements are the elements left out of Manuels and Amarjits stories, and so are just what are being

suggested. Today the alloying metal for bronze is zinc (Zn). Let us look at the ratio of the molar masses

of tin to zinc:

Sn/Zn = 118.71/65.39 = 1.8154 ~ 1.8 = 9/5

It is the nine-fifths around which our stories have been centered.

Ian Beardsley

May 15, 2015


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Hand pounded copper ashtray demonstrating its malleability.


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Two works in silver, one in gold, demonstrating its use for ceremonial and spiritual purposes.


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Five-fold Symmetry: The Biological

Six-fold Symmetry: The Physical

Alternate Six-fold: The Physical

9/5: 5, 14, 23, 32,and 1.8,3.6, 5.4, 7.2,

5/3: 8, 13, 18, 23,and 1.7, 3.3, 5, 6.7,

11/6: 6, 17, 28, 39,.. and 11/6, 11/3, 11/2, 22/3,

We Have Three Equations


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Eliminating t In Our Three Equations

0.62176 is the magnitude of the right ascension vector that points to the constellation Aquila.

Where have we seen this?


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The Mystery In Our Units of Measurement


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Our units of measurement evolved out of a complex history. The mile, forexample, evolved out of a rough estimate of the approximate time it took towalk a horse around a track of no precise length, in order to exercise it. Akilometer was defined in modern times as one ten thousandth of thedistance from the pole of the earth to its equator. Yet it is a curious fact thatthere are 0.621 miles in a kilometer, which is close to the golden ratio(0.618). More interesting is that 0.621 multiplied with the square root of twoover two is equal to A440, which is standard concert pitch, the cycles persecond of the frequency the oboe sounds for the orchestra to tune all of itsinstruments to the same pitch before performing a work. I first began

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