C Files

1 of 126 C Files Ian Beardsley Copyright © 2016 by Ian Beardsley ISBN: 978-1-365-26289-0

2 of 126 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

Manuels’s Integrals

121

3 of 126 Theories

4 of 126 Climate Science 


5 of 126 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 surface reflecting 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.

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7.

8.

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Mercury albedo of 0.06 composition 95% CO2 Venus albedo of 0.75 composition clouds of sulfuric acid Earth albedo of 0.30 composition Nitrogen, Oxygen, H20 or water vapor Mars albedo of 0.29 composition CO2 Asteroids Jupiter albedo of 0.53 composition hydrogen and helium Saturn albedo of 0.47 composition hydrogen and helium Uranus albedo of 0.51 composition hydrogen, helium, methane 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 albedos around 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.

6 of 126

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

The average for the albedo of the inner planets is: (0.06+0.75+0.3+0.29)/4 = 0.35 This 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 ~0.48 This says the outer planets are all close to 0.48~0.5

7 of 126 All this also says, if the planet is solid and habitable it probably has an albedo of around 0.3, otherwise it is an outer gaseous planet and probably has an albedo of around 0.5.

8 of 126

The luminosity of the sun is: 26 ! L0 = 3.9 ×10 J /s

The separation between the earth and the sun is: €

! 1.5 ×1011 m The solar luminosity at the earth is reduced by the inverse square law, so the solar constant is:

€

!

S0 =

3.9 ×10 26 = 1,370Watts /meter 2 11 2 4π (1.5 ×10 )

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

This radiation is equal to the temperature, ! Te, to the fourth power by the steffan-bolzmann constant, sigma ! (σ). ! Te can be called the effective temperature, the temperature entering the earth.

€

€ 2 S ! 0 intercepts the earth disc, € €! πr , and distributes itself over the entire 2 earth surface, ! 4πr , while 30% is reflected back into space due to the

earth’s albedo, a, which is equal to 0.3, so €

S € a) σTe = 0 (1− 4 πr 2 (1− a)S0 4πr 2 !

€

4

€

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 4 so that the radiation from the atmosphere, ! σTa plus the radiation entering 4 4 the earth, ! σTe is the radiation at the surface of the earth, ! σTs . However, 4 4 ! σTa = σTe

€

and we have: €

€ €

9 of 126 4

4

4

σTs = σTa + σTe = 2σTe

4

1 4

Ts = 2 Te S 4 σTe = 0 (1− a) 4 σ = 5.67 ×10−8 S0 = 1,370 a = 0.3 1,370 (0.7) = 239.75 4 239.75 4 Te = = 4.228 ×10 9 5.67 ×10−8 ! Te = 255Kelvin

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

1 4

! Ts = 2 Te = 1.189(255) = 303Kelvin Let’s convert that to degrees centigrade: €

Degrees Centigrade = 303 - 273 = 30 degrees centigrade And, let’s 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 surface of the earth, but, in this model, we only considered radiative heat transfer and not convective heat transfer. In other words, there is cooling due to vaporization of water (the formation of clouds) and due to the condensation of water vapor into rain droplets (precipitation or the formation of rain).

10 of 126

The incoming radiation from the sun is about 1370 watts per square meter as determined by the energy per second emitted by the sun reduced by the inverse square law at earth orbit. We calculate the total absorbed energy intercepted by the Earth's disc (pi)r^2, its distribution over its surface area 4(pi)r^2 and take into account that about 30% of that is reflected back into space, so the effective radiation hitting the Earth's surface is about 70% of the incoming radiation reduced by four. Radiative energy is equal to temperature to the fourth power by the Stefan-boltzmann constant.

However, the effective incoming radiation is also trapped by greenhouse gases and emitted down towards the surface of the earth (as well as emitted up towards space from this lower atmosphere called the troposphere), the most powerful greenhouse gas being CO2 (Carbon Dioxide) and most abundant and important is water vapour. This doubles the 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 gas CO2, though only exists in trace amounts, and water vapour, bring the temperature of the Earth up from minus 18 degrees centigrade (18 below freezing) to an observed average of plus 15 degrees centigrade (15 degrees above freezing). Without these crucial greenhouse gases, the Earth would be frozen. They have this enormous effect on warming the planet even with CO2 existing only at 400 parts per million. It occurs naturally and makes life on Earth possible. However, too much of it and the Earth can be too warm, and we are now seeing amounts beyond the natural levels through anthropogenic sources, that are making the Earth warmer than is favorable for the conditions best for life to be maximally sustainable. We see this increase in CO2 beginning with the industrial era.

The sectors most responsible for the increase are power, industry, and transportation. Looking at records of CO2 amounts we see that it was 315 parts per million in 1958 and rose to 390 parts per million in 2010. It rose above 40s in radiative equilibrium, that is, it loses as much radiation as it receives. Currently we are slightly out of radiative balance, the Earth absorbs about one watt per square meter more than it loses. That means its temperature is not steady, but increasing. 

11 of 126 Equilibrium: The Crux of Climate Science Let us say the Earth is cold, absolute zero, then suddenly the sun blinks on. The Earth will receive radiation and start to get warmer. As it gets warmer, it starts to lose some of the heat it receives, 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 vary numerous and, vary in a wide spectrum as to the amount of heat, or energy in other words, that they can draw off the planet and, at what rates. We have discussed two mechanisms: radiative heat transfer, and convective heat transfer. An example is the vaporization of the ocean, which is water becoming a gas, or clouds in other words. The heat required to raise its temperature to the point that it vaporizes is one calorie per gram degree centigrade. This represents a loss of heat, or energy, from the sun, that would have gone into warming the planet. As well, when the vaporized water, or what are called clouds, condenses into liquid, this represents another loss of heat-energy that would have gone into warming the planet, for the same reason it takes energy to make a refrigerator cold. This condensation of water vapor to its liquid form, is called precipitation, the formation of water droplets, or what we commonly call rain. The amount of water 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 to the 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 of energy it receives equals the amount of energy it loses and the earth is at a constant temperature. This is called an equilibrium state. For the earth, this should be about 15 degrees centigrade in the annual average temperature. If the earth goes out of equilibrium, that is grows warmer or colder with time, then there can be a great deal of causes for this to happen, and many complex factors must be considered to calculate how long it will take the earth to return to a stable temperature (equilibrium state) and to determine what the temperature of the earth will be 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 settles at the bottom of the dish motionless (in an equilibrium state). There can be two types of equilibrium states. One, like we just described, a valley, or two, the reverse: a peak where we have a ball balanced at the apex of a mountain. In this scenario, if I displace the ball to the left or right, it will go out of equilibrium, but never return to equilibrium, like it did in the previous example 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 loses by one watt per square meter. This means the earth is warming. The reason for this is mostly because human activity is putting more CO2 into the atmosphere than should be there, which means the earth retains more heat than it can lose.

Ian Beardsley March 25, 2016 




12 of 126

13 of 126

Climate Modeling with Radiative Heat Transfer And Convection

14 of 126

Model Future


15 of 126 Star System: Alpha Centauri Spectral Class: Same As The Sun Proximity: Nearest Star System Value For Projecting Human Trajectory: Ideal


16 of 126

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): 

17 of 126 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 the universe. 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.

18 of 126 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. 

19 of 126 The Source Code


20 of 126 bioplanet.c #include<stdio.h> #include<math.h> int main(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"); float s, 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"); float joules; joules=(3.9E26*s); printf("The luminosity of the star in joules per second is: %.

2fE25\n", joules/1E25);

21 of 126 float HZ; 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"); float L0,sun,S0,r0,R,S,A,sigma,TE,delta,sTe4,sTs4; float result, 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"); }

22 of 126 cipher.c #include <stdio.h> #include <cs50.h> #include <string.h> int main (int argc, string argv[1]) { int k = atoi(argv[1]); if (argc>2 || argc<2) { printf("Give me a single string: "); } else { printf("Give me a word: "); } string s = GetString(); for (int i=0, n=strlen(s); i<n; i++) { printf("%c", s[i]+k); } printf("\n"); }â&#x20AC;Š

23 of 126 climate.c #include<stdio.h> #include<math.h> int main(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"); float s, 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"); float joules; joules=(3.9E26*s); printf("The luminosity of the star in joules per second is: %.

2fE25\n", joules/1E25); float HZ; HZ=sqrt(joules/3.9E26);

24 of 126 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"); }

25 of 126 convection.c #include<stdio.h> #include<math.h> int main (void) { printf("\n"); printf("\n"); printf("This program finds the temperature of a planet\n"); float L0,sun,S0,r0,r,S,a,sigma,TE,delta,sTe4,sTs4; float result, 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"); }

26 of 126 modefuture.c #include <stdio.h> #include <math.h> int main (void) { printf("\n"); int N, r; double u, v, y, z; double t,loga, ratio; int n1, n2; char name[15]; float W,fact=1,fact2=1,fact3=1,a,g,rate,T,T1; double x,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 the star, 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 (int i=1; i<=N; i++)

27 of 126 { fact = fact*i; printf("N factorial = %f\n", fact); a=pow(u/v,n1)*pow(y/z,n2); } for (int j=1; j<=n1; j++) { fact2 = fact2*j; printf("n1 factorial = %f\n", fact2); } for (int k=1; k<=n2; k++) { fact3 = fact3*k; printf("n2 factorial = %f\n", fact3); x=2.718*2.718*2.718*2.718*2.718; g=sqrt(x); W=a*fact/(fact2*fact3); printf("W=%f percent\n", W*100); W2=100*W; printf("W=%.2f percent rounded to nearest integral\n", round(W2)); } { printf("What is t in years, the time over which the growth occurs? "); scanf("%lf", &t); loga=log10(round(W*100)); printf("log(W)=%lf\n", loga); ratio=loga/t; printf("loga/t=%lf\n", ratio); rate=ratio/0.4342; //0.4342 = log e// printf("growthrate constant=%lf\n", rate); printf("log 100 = 2, log e = 0.4342, therfore\n"); printf("T=2/[(0.4342)(growthrate)]\n"); T=2/((0.4342)*(rate)); printf("T=%.2f years\n", T); printf("What was the begin year for the period of growth? "); scanf("%f", &T1); printf("Object achieved in %.2f\n", T+T1); } }

28 of 126 modelocean.c #include <stdio.h> int main (void) { int option; 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); { float area, 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);

29 of 126 printf("%.3f E3 m^3 of water has a mass of about %.3f E12 g\n", cubic/ 1E3, mass/1E12); printf("\n"); } } printf("\n"); float reduction, incident, energy, watts, square, deep, volume, vol, densiti, matter; float temp, 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); 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);

30 of 126 } modelplanet.c #include <stdio.h> #include <math.h> int main(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"); float r1=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"); float R1, 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); } float R2, v2, m2, M2; {

31 of 126 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); } float R3, 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); }

32 of 126 starsystem.c #include<stdio.h> #include<math.h> int main(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"); float s, 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"); float joules; joules=(3.9E26*s); printf("The luminosity of the star in joules per second is: %.

2fE25\n", joules/1E25);

33 of 126 float HZ; 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"); float L0,sun,S0,r0,R,S,A,sigma,TE,delta,sTe4,sTs4; float result, 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"); float r1=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"); float R1, 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); } float R2, 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); }

35 of 126 float R3, 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); int option; 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); { float area, 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");

36 of 126 } 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 E12 g\n", cubic/1E3, mass/1E12); printf("\n"); } } printf("\n"); float reduction, incident, energy, watts, square, deep, volume, vol, densiti, matter; float temp, 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);

37 of 126 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); }â&#x20AC;Š

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Running The Modelsâ&#x20AC;©

39 of 126 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 no convection for the planet to be in an equilibrium state. That is to say, the temperature stays steady by heat gain and loss with radiative heat transfer alone.

The habitable zone is calculated using the idea that the earth is in the habitable zone for a star like the Sun. That is, if a star is 100 times brighter than the Sun, then the habitable zone for that star is ten times further from it than the Earth is from the Sun because ten squared is 100 We determine the surface temperature of a planet.

What is the luminosity of the star in solar luminosities? 3 What is the albedo of the planet (0-1)?0.5 What 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 F The luminosity of the star in joules per second is: 117.00E25 The habitable zone of the star in AU is: 1.732051 Flux 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 model where equilibrium is maintained by both radiative heat transfer and convection, This program finds the temperature of a planet Luminosity of the star in solar luminosities? â&#x20AC;Š

40 of 126 running modelocean.c Last login: Wed Jun 29 22:10:37 on ttys000 Claires-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 given moment.

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 waterhole 2 a pond 3 the ocean 1 How many square meters of water are warmed? 5 How many meters deep is the water warmed? .3 That 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 square meter.

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? 5 How deep is the water heated (in meters)? .5 The volume of water in cubic meters is: 2.500 That 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 calories What is the intitial temperature of the body of water? 75 The 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]â&#x20AC;Š

41 of 126 running modelplanet.c Last login: Wed Jun 29 22:13:02 on ttys000 Claires-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^3 Lithosphere Density Ni = 8.91 g/cm^3 Mantle Density Si=2.33 g/cm^3 Earth Radius = 6,371 km Earth Mass = 5.972E24 Kg what what what what what what

is is is is is is

r1, p1, r2, p2, r3, p3,

the radius of the core in km? 500 its density in g/cm^3? 7.87 outer edge of layer two in km? 3000 density of layer two in g/cm^3? 8.91 the radius of layer 3 in km? 6371 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 Kg thickness of core is 500.00 layer two has a mass of 1.00 E23 Kg layer two thickness is 2500.00 layer three has a mass of 2.26 E23 Kg layer three thickness is 3371.00 the mass of the planet is 0.33 E24 Kg 2016-06-29 22:17:16.672 modelplanet[77374:6559658] Hello, World! logout [Process completed]â&#x20AC;Š

42 of 126 running starsystem.c Last login: Wed Jun 29 22:18:39 on ttys000 Claires-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 no convection for the planet to be in an equilibrium state. That is to say, the temperature stays steady by heat gain and loss with radiative heat transfer alone.

The habitable zone is calculated using the idea that the earth is in the habitable zone for a star like the Sun. That is, if a star is 100 times brighter than the Sun, then the habitable zone for that star is ten times further from it than the Earth is from the Sun because ten squared is 100 We determine the surface temperature of a planet.

What is the luminosity of the star in solar luminosities? 1 What is the albedo of the planet (0-1)?.3 What 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 F The luminosity of the star in joules per second is: 39.00E25 The habitable zone of the star in AU is: 1.000000 Flux 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 model where equilibrium is maintained by both radiative heat transfer and convection, This program finds the temperature of a planet Luminosity of the star in solar luminosities? 1 Planet distance from the star in AU? 1 What is the albedo of the planet (0-1)? .3 What is the temp dif between layers in kelvin? 2 planet surface temp is: 259.447876 K That is -13.552124 C, or 7.606177 F

43 of 126 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^3 Lithosphere Density Ni = 8.91 g/cm^3 Mantle Density Si=2.33 g/cm^3 Earth Radius = 6,371 km Earth Mass = 5.972E24 Kg what what what what what what

is is is is is is

r1, p1, r2, p2, r3, p3,

the radius of the core in km? 200 its density in g/cm^3? 8 outer edge of layer two in km? 4000 density of layer two in g/cm^3? 9 the radius of layer 3 in km? 6371 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 Kg thickness of core is 200.00 layer two has a mass of 24.12 E23 Kg layer two thickness is 3800.00 layer three has a mass of 20.37 E23 Kg layer 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 given moment.

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 waterhole 2 a pond 3 the ocean 2 How many square meters of water are warmed? 500 How many meters deep is the water warmed? .1 That is 0.050 E3 cubic meters of water.

50.000 cubic meters of water has a mass of about 50.000 E6 grams.

44 of 126

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 square meter.

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? 500 How deep is the water heated (in meters)? .1 The volume of water in cubic meters is: 50.000 That 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 calories What is the intitial temperature of the body of water? 72 The 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]â&#x20AC;Š

45 of 126

running modelfuture.c Last login: Wed Jun 29 22:20:17 on ttys000 Claires-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, and 2, what is the star's name?

Enter 1: 4 Enter 2: alphacentauri Star name: alphacentauri Distance: 4 What is n1? 3 What is n2? 7 Since N=n1+n2, N=10 What is the probability, p(u), of jumping to the left? 1 What is the probability, p(v), of jumpint to the left? 2 What is the probability, q(y), of jumping to the right? 1 What is the probability, q(z), of jumping to the right? 2 p=u:v q=y:z N factorial = 1.000000 N factorial = 2.000000 N factorial = 6.000000 N factorial = 24.000000 N factorial = 120.000000 N factorial = 720.000000 N factorial = 5040.000000 N factorial = 40320.000000 N factorial = 362880.000000 N factorial = 3628800.000000 n1 factorial = 1.000000 n1 factorial = 2.000000 n1 factorial = 6.000000 n2 factorial = 1.000000 W=59062.500000 percent W=59063.00 percent rounded to nearest integral n2 factorial = 2.000000 W=29531.250000 percent W=29531.00 percent rounded to nearest integral n2 factorial = 6.000000 W=9843.750000 percent W=9844.00 percent rounded to nearest integral

46 of 126 n2 factorial = 24.000000 W=2460.937500 percent W=2461.00 percent rounded to nearest integral n2 factorial = 120.000000 W=492.187500 percent W=492.00 percent rounded to nearest integral n2 factorial = 720.000000 W=82.031250 percent W=82.00 percent rounded to nearest integral n2 factorial = 5040.000000 W=11.718750 percent W=12.00 percent rounded to nearest integral What is t in years, the time over which the growth occurs? 40 log(W)=1.079181 loga/t=0.026980 growthrate constant=0.062136 log 100 = 2, log e = 0.4342, therfore T=2/[(0.4342)(growthrate)] T=74.13 years What was the begin year for the period of growth? 1969 Object achieved in 2043.13 logout [Process completed]

47 of 126 Modeling In Python And Java Ian Beardsley 2016â&#x20AC;©

48 of 126 bioplanet.java import comp102x.IO; /** * Here we write a program in java that models the temperature of a planet for a star * of given luminosity. * @author (Ian Beardsley) * @version (Version 01 March 2016) */ public class bioplanet { public static void bioplanet() { System.out.print("Enter the luminosity of the star in solar luminosities: "); double lum = IO.inputDouble(); System.out.print("Enter the distance of the planet from the star in AU: "); double r=IO.inputDouble(); System.out.print("Enter albedo of the planet (0-1): "); double a=IO.inputDouble(); double R=(1.5E11)*r; double S=(3.9E26)*lum; double b=S/(4*3.141*R*R); double N = (1-a)*b/(4*(5.67E-8)); double root = Math.sqrt(N); double number = Math.sqrt(root); double answer = 1.189*number; IO.outputln("The surface temperature of the planet is: "+answer+ " K"); double C = answer - 273; double F = 1.8*C + 32; IO.outputln("That is: " +C+ " degrees centigrade"); IO.outputln("Which is: " + F + " degrees Fahrenheit"); } }

49 of 126 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)*au l=(3.9)*(10**26)*s b=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*number print("The surface temperature of the planet is: "+str(answer)+"K") C=answer-273 F=(9.0/5.0)*C + 32 print("That is " +str(C)+"C") print("Which is " +str(F)+"F") joules=3.9*(10**26)*s/1E25 lum=(3.9E26)*s print("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/1370.0 print("Flux at planet is "+str(flux)+" times that at earth") print("That is " +str(b)+ " watts per square meter")

50 of 126 double.py print("This program finds the temperature of a planet.")

L0=float(raw_input("Luminosity of the star in solar luminosities? ")) sun=3.9E26 S0=L0*sun r0=float(raw_input("planet distance from star in AU? ")) r=(1.5E11)*r0 S=S0/((4)*(3.141)*(r**2)) a=float(raw_input("What is the albedo of the planet (1-0)?: ")) sigma=5.67E-8 TE=((1-a)*S*(0.25)/(sigma))**(1.0/4.0) delta=float(raw_input("temp dif between two layers in Kelvin: ")) x=delta/TE sTe4=(1-a)*S/4 sTs4=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-273 F=(1.8)*C+32 print("That is "+str(C)+" C, or "+str(F)+" F") print("flux at planet is "+ str(S)+" watts per square meter")â&#x20AC;Š

51 of 126 objective.py import math object=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));â&#x20AC;Š

52 of 126 input.c #include <stdio.h> #include <math.h> 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 in 2042.\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 = 4.6 achievements,\n"); printf("Tzero=1969 when we landed on the moon, which at 2009 is 0.552=.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<=0); for (int i=1; i<=n; i++) { do {

53 of 126 printf("What is percent development towards objective(0-100)? "); scanf("%f", &object); } while (n<0 && n>100); 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"); }â&#x20AC;Š

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object.c #include <stdio.h> 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 = 4.6 achievements,\n"); printf("Tzero=1969 when we landed on the moon, which at 2009 is 0.552=.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<=0); for (int i=1; i<=n; i++) { do { printf("What is distance to object(0-4.6?) "); scanf("%f", &object); } while (n<=0 && n>=4.6); printf("What is the starting point (year) ?"); scanf("%f", &Tzero); T= object/(0.0621); time= Tzero+T; printf("Time to object=%f years.\n", T); printf("That is the year: %f\n", time); printf("\n"); } }

55 of 126 Paul Levinson And Manuel Herediaâ&#x20AC;Š

56 of 126 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 Astronomer’s 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 in the 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 developed sophisticated 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 wasn’t it, either”…”Oh, 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 wasn’t 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 Beardsley March 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 â&#x20AC;&#x153;Timewave Calculator Version 1.0â&#x20AC;?. 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 the Universe 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, â&#x20AC;&#x153;The Questionâ&#x20AC;?, 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 the Egyptian 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.

62 of 126 LEVMAN When we consider the 9/5 of five-fold symmetry (the biological) and the 5/3 and 11/6, of six-fold symmetry (the physical), we can make three equations and therefore find a place 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 we let the parameter, t, be eliminated we have a place in space that points to the constellation Aquila. The former relationship came to me via a Gypsy Shaman called Manuel in Granada, Spain. The latter relationship came to me via a Fordham University professor and science fiction author called Paul Levinson, in New York. The former and latter relationships stemmed from separate and independent research about two totally different topics, the former dealing with what I call the Yin and Yang of the Universe, the latter Fiction-Reality Entanglement, or what could be called the unfolding of the McKenna time-wave. Now we find the two concepts are part of one theory and are bound to one another by the standard reference for concert pitch, A440; that tone which the 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 to something even deeper than what its makers know themselves.

Ian Beardsley December 1, 2014â&#x20AC;Š

â&#x20AC;©

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64 of 126 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 Levinsonâ&#x20AC;&#x2122;s number is the growth rate for human progress. (See my work, ET to AI). The mathematical trick used to get the new value for manuelâ&#x20AC;&#x2122;s number was using an equation 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.â&#x20AC;Š

â&#x20AC;©

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â&#x20AC;©

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67 of 126 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=Levinsonâ&#x20AC;&#x2122;s Number Ga=69.72 As=74.92 Ga=Gallium As=Arsenic Ga and As are doping agents for making diodes, transistors, integrated circuitry,â&#x20AC;Śthe operational components of AI. a) b) c) d) e)

2(69.72)(74.92)=10446.8448 (69.72)+(74.92)=144.64 a/b=72.2265~72.23 (69.72)(74.92)=5223.4224 sqrt(5223.4224)=72.27324816

9/5 connects pi to phi: 3.141+1.618=4.759 7=(9+5)/2

68 of 126 360/5=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 = 300 300/360 = 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

=> <5/36, -10/33, -1/9> (Here we have inverted the coefficients of the equation of the plane) sqrt( ((5/36)^2) + ((20/33)^2)) = 0.0621 = Levinson’s 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 = manuel’s number 621=L= levinson’s number ML=440 440=standard concert pitch 23, 27 = Manuel Numbers 23X27=621=L

69 of 126 0.0621, Levinsonâ&#x20AC;&#x2122;s 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.â&#x20AC;Š

70 of 126 Amarjitâ&#x20AC;©

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An Indian Tabla Set

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I found myself in the medioambiente (atmosphere) of an Indian ethnomusicologist from the University 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 first things he told me was that in India there are many false gurus, that will not really teach you, but that this was not his consciousness, that he would really teach me. He said he would put me on a program of learning to play Bahjrans and Kirtans, rythms of 6 and 7 which fall under the category of Guzals, or Indian romantic music, but that he would be playing and composing temple music, called tin tal, which was the cycle of 16 considered the highest and most spiritual form of North Indian Classical music.

The training began with the history of the tabla which has its origins in the mridangam. It was the Muslim King in India, Amir Kusuro, who took the mridangam, which was closed on both ends, the left side played with the left hand and the right side played with the right hand, and broke 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 of dry ink that allows the drums to be tuned to precise pitches. The ink is rubbed into the tabla, as was explained to me, with a stone that floats on water and glows like a cat’s eye and only exists in 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 of seven and one half and a cycle of 13 1/2. I find that interesting. If a person considers each beat 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 can’t help but think in reference to that, the earth rotates through 15 degrees in an hour, and the most abundant element in the earth’s atmosphere is nitrogen which is in chemical group 15 in the periodic table. Let us multiply Amarjit’s 7 1/2 by the 16 of his tin tal. It is 120. 120 are the degrees 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 that same 360 and then add the result to one: 360-120 = 240 240/360 = 2/3 2/3+1 = 5/3 This is the value that represents the yang of the cosmic yin and yang that came to us from the Gypsy 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 = 72 360-72 = 288 288/360 = 4/5 4/5 + 1 = 9/5 = 1.8 Let us divide Amarjit’s stressed cycle 13.5 by 7.5. We find it is 1.8, which equals the yin of 9/5 that is representative of the organic aspects of nature to which the Gypsy Shaman, Manuel guided us in my story Gypsy Shamanism and the Universe, which I will present following the story 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 were coming into the house. I noticed in the living room was lots of clothing and art from India. I was introduced to these people, who obviously ran a store, and they told me they were just coming back 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 name tags 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 buying instruments. On our way back, with his student driving, me in the front seat, Amarjit laid stretched out on the back back seat telling me that the store owner’s refusal of our price offer for a crude guitar indicated that he was “A very greedy man and would not get far in life”. At some point I told Amarjit that I had dreams of him giving me tabla lessons. He told me he could communicate with me in this way.

Upon learning that God told me the Gypsy Shaman, Manuel, always second guesses him, and Manuel telling me that because of this, he goes out into the world to do God’s work for him at his request, Amarjit and his students were going to change their course from one of merging with God, to one of merging with Manuel.

Ian Beardsley May 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, 2012â&#x20AC;Š

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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 Observatory—to 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, as these 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 didn’t roll up the hose and put it away, so it got stolen, and that I had to buy 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 wasn’t 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 the man 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

77 of 126 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.â&#x20AC;Š

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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 Shaman’s 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 Kubrick’s “2001: A Space Odyssey”. The Shaman tells me, telepathically, that this time the mission won’t 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”. I understand 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

79 of 126 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 â&#x20AC;&#x153;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.â&#x20AC;?

I know ratios, in mathematics are the most interesting subject, like pi, the ratio of the circumference 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 a mole 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 Shamanâ&#x20AC;&#x2122;s 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. Nine fifths 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.

80 of 126 Nine-fifths was in the elements gold to silver, Saturn to Jupiter, Sun to moon. Where was fivethirds? Salt of course. “The Salt Of The Earth” is that which is good, just read Shakespeare’s “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, and the 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 Manuelâ&#x20AC;&#x2122;s and Amarjitâ&#x20AC;&#x2122;s 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? â&#x20AC;Š

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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, for example, evolved out of a rough estimate of the approximate time it took to walk a horse around a track of no precise length, in order to exercise it. A kilometer was defined in modern times as one ten thousandth of the distance from the pole of the earth to its equator. Yet it is a curious fact that there 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 two over two is equal to A440, which is standard concert pitch, the cycles per second of the frequency the oboe sounds for the orchestra to tune all of its instruments to the same pitch before performing a work. I first began to discover how these randomly evolved units of measurement were connected to the Universe, Nature, and each other back around 2012. It all began with the observation: R=solar radius r=lunar orbital radius Au=molar mass of gold Ag=molar mass of silver R/r = Au/Ag =9/5 Which lead to: Five-fold Symmetry: The Biological 360 288 8 8 9 = 72;360 â&#x2C6;&#x2019; 72 = 288; = ; +1 = 360 10 10 5 ! 5

Six-fold Symmetry: The Physical 360 240 2 2 5 = 60;360 â&#x2C6;&#x2019; 60 â&#x2C6;&#x2019; 60 = 240; = ; +1 = 360 3 3 3 ! 6

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Alternate Six-fold: The Physical 360 300 5 5 11 = 60;360 − 60 = 300; = ; +1 = 360 6 6 6 ! 6

9/5: 5, 14, 23, 32,… and 1.8,3.6, 5.4, 7.2,… ! an = 7.2n− 4

5/3: 8, 13, 18, 23,… and 1.7, 3.3, 5, 6.7,… ! an = 3.3n+ 3 11/6: 6, 17, 28, 39,.. and 11/6, 11/3, 11/2, 22/3,… π + φ = 3.141+1.618 = 4.759; 7 = (5 + 9) / 2

! π + e= 3.141+ 2.718 = 5.859;9 / 5 = 1.8 This lead me to consider the following integral: (v) = 3 + 3.3t (v)=at=(33/10)t where v is velocity, a is acceleration, and t is time.

3=(33/10)t (t) = 30/33 980 cm/s/s = g = the surface gravity of the earth to nearest 10 (980 cm/s/s)(3.3)=3,234 cm/s/s (3,234 cm/s/s)(30/33 s) = 2,940 cm/s = v_0 v_0 is the initial velocity Thus we can write the equation as: (v) = 2,940 cm/s + (3,234 cm/s/s)t This is the differential equation: (dx) = (2,940 cm/s)dt + (3,234 cm/s/s)t dt

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The only thing we lack in solving this is a time for which we can derive a distance. The Earth rotates through 15 degrees in one hour, so we consider 15 seconds. The result is (x_0) = (2,940 cm/s)15 + ((1/2)(3,234) cm/s/s)15^2 = 44,100 +363,825 = 407925 cm 407925 cm/100/1000 = 4.07925 km This is nearly four kilometers. If a kilometer is defined as one ten thousandth of the distance from the pole to the equator, then 4 kilometers is one ten thousandth the circumference of the Earth.

The Integral From 0 To 15 Seconds: !∫

15 0

(2,940cm /s)dt +

∫

15 0

(3,234cm /s/s)tdt = 4.07925km

Thus we can write the equation as:

€

(v) = 2,940 cm/s + (3,234 cm/s/s)t This is the differential equation: (dx) = (2,940 cm/s)dt + (3,234 cm/s/s)t dt Integral From 0 To 15 Seconds

∫ ! €

15 0

(2,940cm /s)dt +

∫

15 0

(3,234cm /s/s)tdt = 4.07925km

Mach 1 = 768 mph =1,235 km/hour That is mach 1 in dry air at 20 degrees C (68 degrees F, or room temperature) at sea level.

If we write, where 1,235 km/hr (mach 1) = 0.343 km/s, then: 34,300 cm/s =2,940 cm/s + (3234 cm/s/s)t and t=9.696969697 seconds = 9 23/33 s = 320/33 seconds ~ 9.7 seconds

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So, the integral is a time of 9.7 seconds to reach mach 1. Putting that time in the integral: (x) = (2,940)(320/33) + 1/2(3234(320/33)^2 = 180,557 cm 1.80557 km ~ 1.8km Thus, with the integral we reach mach one in about 9.7 seconds after traveling a distance of 1.8 kilometers.

1.8=9/5=R/r=Au/Ag

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I have taken upon myself to search for some basic unit of energy that is connected to Nature and The Universe. Here is how I have done it: Electron Volt: A unit of energy equal to the work done on an electron in accelerating it through a potential of one volt. It is 1.6E10-19 Joules (Google Search Engine) Volt: Potential energy that will impart on joule of energy per coulomb of charge that passes through it. (Wikipedia) Coulomb: The charge of 6.242E18 protons or 6.242E18 electrons.

Forward Bias: A diode (silicon) must have 0.7 volts across it to turn it on, 0.3 volts (Germanium). This is called forward voltage. The forward voltage threshold is 0.6 volts.

(0.6 volts)(1.6E-19)=9.6E-20 Joules This is the energy to turn on a diode, or the threshold of life for artificial intelligence.

Aerobic respiration requires oxygen (O2) in order to generate ATP. Although carbohydrates, fats, and proteins are consumed as reactants, it is the preferred method of pyruvate breakdown in glycolysis and requires that pyruvate enter the mitochondria in order to be fully oxidized by the Krebs cycle. The products of this process are carbon dioxide and water, but the energy transferred is used to break strong bonds in ADP as the third phosphate group is added to form ATP (adenosine triphosphate), by substratelevel phosphorylation, NADH and FADH2 Simplified reaction: C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) + heat ΔG = −2880 kJ per mol of C6H12O6 (From

Wikipedia)

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(2,880,000 J)/(6.02E23 C6H12O6) =4.784E-18 J = basic unit of biological life (4.784E-18 J)/(9.6E-20 J)=49.8~50

This says the basic energy unit of organic, or biological life, is about 50 times greater than the basic energy unit of electronic life, or artificial intelligence.

That is 0.6(50)=30 electron volts = basic unit of energy for biological life.

So, we see the visible spectrum for one photon of light begins where the energy of the photon is 2 “bue” electronic which is 100 “bue” biological and that that photon has a wavelength of 1.0 micrometers.

This is all about vision in a robot or AI.

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+1.2eV —————|>—————————> out (9.6E-20 J, or 0.6 eV) | | —————|>———| | | R | | —————————

A photon has to have a minimum energy of 1.2 electron volts to impart to an electron for it to turn on the simplest of logic gates; a one on, one off, OR GATE, for there to be an output of one “bue” (basic unit of energy electronic) , which 9.6E-20 Joules, as I have calculated it.

Use Planck’s Equation: E=hv where h= 6.626E-34 Joule seconds v=2(9.6E-20)/(6.626E-34)=3.067E14 cycles per second wavelength = lambda = c/v where c is the speed of light equal to 3E8 m/s lambda = (3E8)/(3.067E14) = 9.78E-7 meters 1 micrometer = 1E-6 meters lambda ~ 1 micrometer (This is where the visible spectrum begins) So we see the visible spectrum for one photon of light begins where the energy is 2 bue. This is an output of 1 “bue”. Thus units fall in sync with Nature here as well,

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We can see that the mile has further connection to Nature than just the golden ratio, and square root of two over two, which I should point out is the sin of 45 degrees and the cosine of 45 degrees, 45 degrees the angle of maximum distance for the trajectory of a projectile.

We ask: How far is far?

I could say the distance from my house to the village (about a mile) is close. Yet, if I consider the distance from my bedroom to the front door, the distance to the village is far. Everything is relative. Therefore, what can we say is close and what can we say is far? Perhaps the answer to that is embedded in Nature. Let us consider something on the smallest scale we know, the distance of an electron from a proton in an atom of hydrogen and call the distance of one from the other as close. It is about 0.053 nanometers. That is, point zero five three billionths of a meter (0.053E-9 m). Let us consider that which is closest to us on the largest scale we know, the distance to the nearest star, alpha centauri and call it far. It is 4.367 light years away (one ly is 9.56E15 meters) putting alpha cenatauri about 25.6 trillion miles way. We will take the geometric mean of of the electron-proton separation in a hydrogen atom with the earth-alpha centauri separation and consider the result an average manageable distance.

One light year is 9.46E15 meters.

(9.46E15 m/ly)(4.367 ly)=4.13E16 m sqrt[(0.053E-9 m)(4.13E16 m)]=sqrt(2189526 square meters)=1,480 meters (1,480 m)(1 km/1000 m) = 1.480 kilometers (1.480 km)(one mile/1.60934 kilometers)=0.9196 miles ~ 1 mile Therefore, when humans chose the unit of a mile to measure distance, they may have been in tune with the cosmos (atoms of hydrogen and the closest star).

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What We Now Have We have an energy of 2 bue, where the energy of a photon begins for the visible spectrum of an electronic eye receiving 1 “bue” at one micrometer.

We have an acceleration, the acceleration at the surface of the earth connected to kilometers and the circumference of the earth through our integral. We have the unit of a mile, shown to be significant. From these three we can determine a mass: (1 mile)(1 km/0.621 mi) =161,000 cm 1.92E-12 ergs = 2 bue 1.92E-12 ergs = (161,000 cm)(981cm/s/s)m m =1.2E-20 grams (The Mystery is in whatever this mass means) mass of a proton: 1.67 E-27

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Suffice to say the real search is in finding energies connected to the bue, basic unit of energy. So, I have written a program for computing energies:â&#x20AC;Š

97 of 126 #include <stdio.h> int main (void) { printf("\n"); int n, number; float mass, distance, acceleration, bue, bue_bio, joules, work=0.00; float h, v, c, lambda, ergs; printf("We calculate the energy for 1. work done, or 2. kinetic energy \n"); printf("How many calculations would you like to make? "); scanf("%i", &n); printf("\n"); for (int i=1; i<=n; i++) { printf("Would you like the bue (basics units of energy) for: \n"); printf("1. Input of work in centimeters-grams-seconds (dyn), or: \n"); printf("2. Input of work in kilograms-meters-seconds? (Newtons) Or, \n"); printf("3. Input of kinetic energy in centimeters-grams-seconds, or \n"); printf("4. Input of kinetic energy in kilograms-meters-seconds: \n"); printf("5. Or simply input an energy in ergs: \n"); scanf("%i", &number); if (number == 1) { printf("What is the mass? "); scanf("%f", &mass); printf("What is the acceleration? "); scanf("%f", &acceleration); printf("What is the distance? "); scanf("%f", &distance); work=mass*acceleration*distance; printf("\n"); bue = work*((1E-5)/(9.6E-20)); bue_bio = work*((1E-5)/(9.6E-20))*49.8; joules=work*(1E-5); printf("That is %.2f ergs\n", work); printf("Which is %.2f joules\n", joules); printf("Or, that is %.2f calories\n", work*(1E-5)/4.184); printf("Or, %.2f E18 eV (electron volts)\n", (work*(1E-5)/ (1.602E-19))/1E18); printf("basic unit of energy (bue electronic): %.2f E16\n", bue/1E16); printf("bue biological: %.2f E18", bue_bio/1E18); printf("\n"); printf("\n"); }

98 of 126 else if (number==5) { printf("bue in ergs: 9.6e-13\n"); printf("How many ergs would you like to consider (1.92e-12)? "); scanf("%f", &ergs); h=6.626E-27; //erg-seconds// v=ergs/h; c=3E10; lambda=(c/v); printf("That is a wavlength of: %.4f meters\n", lambda); printf("Which is: %.4f micrometers", lambda*1E4); printf("\n"); } } }â&#x20AC;Š

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jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): make bue clang -ggdb3 -O0 -std=c99 -Wall -Werror bue.c -lcs50 -lm -o bue jharvard@appliance (~/Dropbox): ./bue We calculate the energy for 1. work done, or 2. kinetic energy How many calculations would you like to make? 1 Would you like the bue (basics units of energy) for: 1. Input of work in centimeters-grams-seconds (dyn), or: 2. Input of work in kilograms-meters-seconds? (Newtons) Or, 3. Input of kinetic energy in centimeters-grams-seconds, or 4. Input of kinetic energy in kilograms-meters-seconds: 5. Or simply input an energy in ergs: 5 bue in ergs: 9.6e-13 How many ergs would you like to consider (1.92e-12)? 1.92e-12 That is a wavlength of: 0.0001 meters Which is: 1.0353 micrometers jharvard@appliance (~/Dropbox): â&#x20AC;Š

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jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): ./bue We calculate the energy for 1. work done, or 2. kinetic energy How many calculations would you like to make? 3 Would you like the bue (basics units of energy) for: 1. Input of work in centimeters-grams-seconds (dyn), or: 2. Input of work in kilograms-meters-seconds? (Newtons) Or, 3. Input of kinetic energy in centimeters-grams-seconds, or 4. Input of kinetic energy in kilograms-meters-seconds: 5. Or simply input an energy in ergs: 1 What is the mass? 25 What is the acceleration? 981 What is the distance? 150 That is 3678750.00 ergs Which is 36.79 joules Or, that is 8.79 calories Or, 229.63 E18 eV (electron volts) basic unit of energy (bue electronic): 38320.31 E16 bue biological: 19083.51 E18 Would you like the bue (basics units of energy) for: 1. Input of work in centimeters-grams-seconds (dyn), or: 2. Input of work in kilograms-meters-seconds? (Newtons) Or, 3. Input of kinetic energy in centimeters-grams-seconds, or 4. Input of kinetic energy in kilograms-meters-seconds: 5. Or simply input an energy in ergs: 1 What is the mass? 5 What is the acceleration? 150 What is the distance? 25 That is 18750.00 ergs Which is 0.19 joules Or, that is 0.04 calories Or, 1.17 E18 eV (electron volts) basic unit of energy (bue electronic): 195.31 E16 bue biological: 97.27 E18

101 of 126 Would you like the bue (basics units of energy) for: 1. Input of work in centimeters-grams-seconds (dyn), or: 2. Input of work in kilograms-meters-seconds? (Newtons) Or, 3. Input of kinetic energy in centimeters-grams-seconds, or 4. Input of kinetic energy in kilograms-meters-seconds: 5. Or simply input an energy in ergs: 5 bue in ergs: 9.6e-13 How many ergs would you like to consider (1.92e-12)? 5 That is a wavlength of: 0.0000 meters Which is: 0.0000 micrometers jharvard@appliance (~/Dropbox): â&#x20AC;Š

102 of 126 stellar.c #include<stdio.h> #include<math.h> int main(void) { float s, 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("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"); float joules; joules=(3.9E26*s); printf("The luminosity of the star in joules per second is: %.

2fE25\n", joules/1E25); float HZ; 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/3.9E26); printf("That is %f Watts per square meter\n", b); }

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Notice running stellar.c for the golden ratio and its conjugate in terms of solar luminosities, earth orbit (AU) and albedo of the hypothetical planet gives near equivalence between the fahrenheit and centigrade scales for its surface temperature. jharvard@appliance (~): cd Dropbox/descubrir jharvard@appliance (~/Dropbox/descubrir): ./stellar We determine the surface temperature of a planet.

What is the luminosity of the star in solar luminosities? 1.618 What is the albedo of the planet (0-1)?0.618 What is the distance from the star in AU? 1.618 The surface temperature of the planet is: 231.462616 K That is -41.537384 C, or -42.767292 F The luminosity of the star in joules per second is: 63.10E25 The habitable zone of the star in AU is: 1.272006 jharvard@appliance (~/Dropbox/descubrir):

To the matter at hand. The foot-pound system was not derived from any relationship to nature that we know of. The Metric system was: one gram is the mass of a cube of water at STP one centimeter on each side. A centimeter is a hundredth of a meter, and a meter is a thousandth of a kilometer. One kilometer is one ten thousandth of the distance from the pole to the equator.

Centigrade is derived such that there are 100 units between freezing and boiling points of water, freezing being zero and boiling, 100. Fahrenheit is not derived from anything natural: It stars at freezing of water at 32 degrees and who knows why? We do know the foot-pound system has earth gravity at 32 feet per second per second. Let us find the temperature where centigrade and Fahrenheit are the same: F=(9/5)C + 32 F=C C=(9/5)C=32 (25/25)C-(45/25)C=32 (-20/25)C=32 C=-32(25)/20=-800/20=-40 Degrees C -40 degrees C = -40 degrees F In earlier work, I have suggested that our different systems of measurement are connected not just to one another, but, to nature, even when the evolution of these systems had no such intent to do so.â&#x20AC;Š

104 of 126 The Sequencesâ&#x20AC;©

105 of 126 We considered the ratio nine to five, then the proportion and found it in Saturn Orbit to Jupiter orbit, Solar Radius to Lunar Orbit, Gold to Silver and if flower petal arrangements. It is left then to consider the whole number multiples of nine-fifths (1.8) or the sequence: 1.8, 3.6, 5.4, 7.2,â&#x20AC;Ś in other words, and we look to see if it is in the solar system and find it is in the following ways: 1.8 Saturn Orbit/Jupiter Orbit Solar Radius/Lunar Orbit Gold/Silver 3.6 (10)Mercury Radius/Earth Radius (10)Mercury Orbit/Earth Orbit (earth radius)/(moon radius)= 4(degrees in a circle)(moon distance)/(sun distance) = 3.7 ~ 3.6 There are about as many days in a year as degrees in a circle.

(Volume of Saturn/Volume Of Jupiter)(Volume Of Mars) = 0.37 cubic earth radii ~ 3.6 The latter can be converted to 3.6 by multiplying it by (Earth Mass/Mars Mass) because Earth is about ten times as massive as Mars.

5.4 Jupiter Orbit/Earth Orbit Saturn Mass/Neptune Mass 7.2 10(Venus Orbit/Earth Orbit)

106 of 126 The Neptune Equation If we consider as well the sequence where we begin with five and add nine to each successive term: 5, 14, 23, 32…Then, the structure of the solar system and dynamic elements of the Universe and Nature in general are tied up in the two sequences: 5, 14, 23, 32,… and 1.8, 3.6, 5.4, 7.2,…

How do we find the connection between the two to localize the pivotal point of the solar system? We take their difference, subtracting respective terms in the second sequence from those in the first sequence to obtain the new sequence: 3.2, 10.4, 17.6, 24.8,…

Which is an arithmetic sequence with common difference of 7.2 meaning it is written 7.2n – 4 = a_n The a_n is the nth term of the sequence, n is the number of the term in the sequence.

This we notice can be written: [(Venus-orbit)/(Earth-orbit)][(Earth-mass)/(Mars-mass)]n – (Mars orbital #) = a_n We have an equation for a sequence that shows the Earth straddled between Venus and Mars. Venus is a failed Earth. Mars promises to be New Earth.

The Mars orbital number is 4. If we want to know what planet in the solar system holds the key to the success of Earth, or to the success of humans, we let n =3 since the Earth is the third planet out from the Sun, in the equation and the result is a_n = 17.6.

This means the planet that holds the key is Neptune. It has a mass of 17.23 earth masses, a number very close to our 17.6.

Not only is Neptune the indicated planet, we find it has nearly the same surface gravity as earth and nearly the same inclination to its orbit as earth. Though it is much more massive than earth, it is much larger and therefore less dense. That was why it comes out to have the same surface gravity.

107 of 126 The Uranus Equation I asked what needs to be done to solve My Neptune Equation, by going deep with the guitar in Solea Por Buleras. I found the answer was that I didn’t have enough information to solve it.

Then I realized I could create the complement of the Neptune equation by looking at the Yang of 5/3, since the Neptune equation came from the Yin of 9/5.

We use the same method as for the Neptune equation: Start with 8 and add 5 to each additional term (we throw a twist by not starting with 5) 5/3 => 8, 13, 18, 23,…

List the numbers that are whole number multiples of 5/3: 5/3n = 1.7, 3.3, 5, 6.7,…

Subtract respective terms in the second sequence from those in the first: 6.3, 9.7, 13, 16.3,…

This is an arithmetic sequence with common difference 3.3. It can be written: (a_n) = 3 + 3.3n This can be wrttten: Earth Orbital # + (Jupiter Mass/Saturn Mass)n = a_n Letting n = 3 we find a_n = 13 The closest to this is the mass of Uranus, which is 14.54 earth masses. If Neptune is the Yin planet, then Uranus is the Yang planet. This is interesting because I had found that Uranus and Neptune were different manifestations of the same thing. I had written: I calculate that though Neptune is more massive than Uranus, its volume is less such that their products are close to equivalent. In math: N_v = volume of Neptune N_m = mass of Neptune U_v = volume of Uranus U_m = mass of Uranus (N_v)(N_m) = (U_v)(U_m)

108 of 126 The Earth Equation We then sought the Yang of six-fold symmetry because it is typical to physical nature, like snowflakes. We said it was 5/3 since it represents the 120 degree measure of angles in a regular hexagon and we built our universe from there, resulting in the Uranus Integral, which was quite fruitful. Let us, however, think of Yang not as 5/3, but look at the angles between radii of a regular hexagon. We have: 360 – 60 = 300 300 + 360 = 660 660/360 = 11/6 We say Yin is 9/5 and Yang is 11/6 and stick with The Gypsy Shaman’s 15 (See An Extraterrestrial Analysis, chapter titled “Gypsy Shamanism And The Universe”) and build our Cosmology from there.

We already built The Neptune Equation from 9/5 and used it with 5/3 to derive the planet Europia, but let us apply 11/6 in place of 5/3: 11/6 => 11/6, 11/3, 11/2, 22/3,… = 1.833, 3.667, 5.5, 7.333,…

11/6 => 6, 6+11 = 17, 17+11=28, 28+11=39, … = 6, 17, 28, 39,…

Subtract the second sequence from the first: 4.167, 13.333, 22.5, 31.667,…

Now we find the common difference between terms in the latter: 9.166, 9.167, 9.167,… (a_n) = a + (n-1)d = 4.167+(n-1)9.167 = 4.167 + 9.167n – 9.167 = 9.167n-5 Try n=3: 9.167(3) – 5 = 27.501 – 5 = 22.501 (works) Our equation is: (a_n) = 9.167n –5 We notice this can be written: [(Saturn Orbit)/(Earth Orbit)]n – (Jupiter Orbital #) = (a_n) The Neptune Equation for n=3 gave Neptune masses, the Uranus equation for n=3 gave Uranus masses. This equation for n=3 gives close to the tilt of the Earth (23.5 degrees) in a form that is exactly half of the 45 degrees in a square with its diagonal drawn in. In the spirit of our first cosmology built upon 9/5, 5/3, and 15, we will call this equation The Earth Equation.

109 of 126 The Wow! Signal

110 of 126 We have the Neptune Equation: 7.2x â&#x20AC;&#x201C;4 We have the Uranus Equation 3.3x + 3 And now with our alternate cosmology we have The Earth Equation: 9x-5 With three equations we can write the parameterized equations in 3-dimensional space, parameterized in terms of t, for x, y, and z. We can write from that f(x,y,z) and find the gradient vector, or normal to the equation of a plane in other words, and from that a region in space.

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! a=5/36 b=-10/33 2

2

! c = (5 /36) + (10 /33) = 0.0918 + 0.019 = 0.3328 d=-1/9

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tanα = b/a α = −65.358 ! tan β = d /c ! ! β = −18.46 ! !

-65.358 degrees/15 degrees/hour =-4.3572 hours

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24 00 00 – 4.3572 = 19.6428 hours RA: 19h 38m 34s Dec: -18 degrees 27 minutes 36 seconds

!

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!

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114 of 126 The projection by my calculation through my cosmology of yin, yang, and 15 for the origin of my message from extraterrestrials was somewhere in the easternmost part of the constellation Sagittarius. This happens to be the same place where the one possible alien signal was detected in the Search For Extraterrestrial Intelligence (SETI). It was called “The Wow Signal” because on August 15, 1977 the big ear antenna received something that seemed not like star noise, but exactly what they were looking for in an extraterrestrial signal. Its name is what it is because the astronomer on duty, Jerry R. Ehman wrote “Wow!” next to the numbers when they came in. Incredibly, it lasted the full 72 seconds that the Big Ear antenna listened for it. I say incredible because I have mentioned the importance of 72, not just in my Neptune equation – for which my location in space was derived in part – but because of its connection to the Gypsy Shaman’s AE-35 antenna and its relation to 72 in the movie “2001: A Space Odyssey”. The estimation of the coordinates for the origin of the Wow signal are two: 19h22m24.64s 19h25m17.01s With declination of: -26 Degrees 25 minutes 17.01 s That is about 2.5 degrees from the star group Chi Sagittarri It is very close to my calculation for an extraterrestrial civilization that I feel hid a message in our physics, which I calculate to be near HD 184835 and exactly at: 19h 38m 34s -18 Degrees 27 minutes 36 seconds The telescope that detected the Wow Signal was at Ohio Wesleyan University Delaware, Ohio called The Perkins Observatory.

Ian Beardsley May 6, 2013

115 of 126 Aquila

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Sagittarius And Aquila For the Sagittarius vector, we took the parameter, t, to be zero. If you instead eliminate t in the equations the vector points to an open cluster in Aquila, NGC 6738.

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We have: <5/36,-20/33,1/9> a=5/36, b=-20/33 d=1/9 c=sqrt((5/36)^2 +(20/33)^2)=sqrt(0.01929 + 0.3672)=sqrt(0.38659) =0.62176 tan(alpha)=b/a=-48/11=-4.363636 alpha=-77 degrees (-77 degrees)/(15 degrees)/hour =-5 hours 24 hours - 5 hours = 19 hours tan(beta)=d/c=(1/9)/0.62176=0.1787 beta = arctan(0.1787) = +10 degrees This is an object with right ascension 19 hours And declination 10 degrees This is in the constellation Aquila and is around the open cluster NGC 6738

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Next Contact

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Another Projection For Extraterrestrial Contact There is another way of calculating when the possible message from extraterrestrials in Sagittarius, the SETI Wow! Signal, will repeat it self. Our discovery of another message from the same place began with the Gypsy Shaman’s hose collection of 15 hoses making us realize that 15 was important because the Earth rotates through 15 degrees in an hour, and 15 seconds lead to the dynamic integral, Manuel’s Integral. The first message was on August 15, 1977. We noted that that 15 of August pointed to the Shaman’s 15 hoses and the two sevens in 1977 add up to 14, which when added to the one in 19 is 15 as well, while the 9 is the nine of nine-fifths that we found in Nature from which we calculated a place in Sagittarius where the SETI Wow! Signal is. We decoded another message on around May 5 of 2013. The next message should then be on August 15, 2015, to line up the Shaman’s fifteens. August is a good time to view the constellation Sagittarius from where I am in Southern California. Sagittarius has always been my favorite constellation, because it is in the center of the Galaxy, rich with globular, and open clusters that can be viewed with binoculars. Summer is when stargazing becomes exciting because you get both a rich sky and warm, uplifting weather. Also, August is the eighth month and our nine-fifths divided out is one point eight. The one takes you around a circle once, leaving point eight.

Ian Beardsley Dec 24, 2013

HoHoary 15, 2015 Guess For Second Contact It is my feeling that first contact with ET's was in Sagittarius (The Wow! Signal) because that is where my equations point when the parameter t, is zero. However, it is my feeling that second contact will be in Aquila, around NGC 6738, because that is where my equations point when I eliminate the parameter, t. I have indicated that one possible time for this second contact would be around August 15, 2015. Perhaps even on August 14, 15, and 16 of that same year.

Ian Beardsley February 15, 2015

121 of 126 Manuel’s Integrals

122 of 126 It has always been thought that an extraterrestrial message would be mathematical. Are the anomalous Fast Radio Bursts (FRBs) an extraterrestrial beacon encoding Manuel’s Third Integral that I discovered when doing research that lead me to the other theoretical ET signal, the Wow Signal in the constellation Sagittarius? Here I present the integral and how it is connected to the FRBs. To learn of all the research, read my work “SETI: Another Signal in Sagittarius” by Ian Beardsley or my work “ET Conjecture” (Red 01) with the title page “This Is 440” at the opening.

As well read my work “The Program Discover” and “The Genesis Project”.

Ian Beardsley April 2, 2015

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FRBs and Manuel’s Third Integral The occurrence of FRBs (Fast Radio Bursts) from outside the Galaxy that can be explained by no known natural process I find have their source to be in an integral that was brought to me by a Gypsy Shaman, Manuel. There are three Manuel Integrals, and my first one has been part of an overall theory that does not just propose extraterrestrial contact in 2015, but tells us where to look. Manuel’s third integral is derived from the nine fifths that is the ratio of the molar masses of gold to silver and is in the ratio of the solar radius to the lunar orbital radius, the sun gold in color, the moon silver. It also uses earth gravity rounded to the nearest ten, and the 15 that describes the amount of degrees through which the earth passes in one hour. The New Scientist reports on March 31, 2015 that the time between the beginning of the first burst and the end of the last burst is a multiple of 187.5. This happens to be the same as Manuel’s Third Integral, the decimal part beyond 9 kilometers: We consider the earth equation: (y) = 9x-5 v=9t-5 g=980 cm/s/s 9(980)=8,820 cm/s/s 5=9t (t)=5/9 (v_0) = (8,820 cm/s/s)(5/9) =4,900 cm/s dx = (8,820 cm/s/s)t dt – (4,900 cm/s) dt 15

∫ 8,820cm /s/stdt − ∫ 4,900cm /sdt = 9.1875km !

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15

0

0

The Integral has the 9 of nine-fifths, the five, the seven, which is the average of 9 and 5 and the one and eight of the 1.8 that is 9/5 divided out. We call this Manuel’s Third Integral.

Ian Beardsley April 1, 2014

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Once you realize nine-fifths is not just at the crux of Gold and Silver, Pi and the Golden Ratio, Pi and Euler’s Number e, the five-fold symmetry that is typical of life, Jupiter and Saturn, Sun and Moon, it is not long before you realize its compliment is 5/3 and that you form the sequences: (For 9/5) 5, 14, 23, 32,… and 1.8, 3.6, 5.4, 7.2,…

(For 5/3) 8, 13, 18, 23,… and 1.7, 3.3, 5, 6.7,…

For which you get: 7.2n – 4 = a_n and

(a_n) = 3 + 3.3n respectively.

In the latter, letting the 3.3 be Earth Gravities rounded to the nearest ten (980), we have: (v) = 2,940 cm/s + (3,234 cm/s/s)t This is the differential equation: (dx) = (2,940 cm/s)dt + (3,234 cm/s/s)t dt !∫

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15 0

(2,940cm /s)dt +

∫

15 0

(3,234cm /s/s)tdt = 4.07925km

(Manuel’s Integral)

15 seconds because there are 15 degrees in an hour of right ascension. The factor of one fourth enters because the kilometer is defined by the distance from the pole to the equator, not by the circumference of the Earth. Notice the 0.07925 has the nine and five of nine-fifths, the average of nine-fifths and the 2 used to make it.

Mach 1 = 768 mph =1,235 km/hour That is mach 1 in dry air at 20 degrees C (68 degrees F) at sea level.

If we write, where 1,235 km/hr (mach 1) = 0.343 km/s, then: 34,300 cm/s =2,940 cm/s + (3234 cm/s/s)t and t=9.696969697 seconds = 9 23/33 s = 320/33 seconds ~ 9.7 seconds So, the Uranus equation is a time of 9.7 seconds to reach mach 1. Putting that time in the integral: (x) = (2,940)(320/33) + 1/2(3234(320/33)^2 = 180,557 cm 1.80557 km ~ 1.8km Thus we see Manuel’s Integral reaches mach one in about 9.7 seconds after traveling a distance of about 1.8 kilometers. Let’s convert that to miles: (1.8km)(0.621371mi/km)=1.118478 miles Manuel’s Integral reaches mach one in 1.8 kilometers, which is the amount of kilometers in a mile and is the 9/5 that occurs in Nature and the Universe, not to mention that it unifies pi and golden ratio and pi and Euler’s number e. It is one compact statement that embodies everything and connects it to Earth Gravity. – Ian Beardsley, August 15, 2013

125 of 126 Manuel’s Second Integral Earth gravity (g) is 9.81 m/s/s This is close to 9.80 m/s/s Indeed if rounded it to one place after the decimal, it would be 9.8 m/s/s This value when converting to cm/s/s gives g = 980 cm/s/s There may be good reason to write it like this (which is rounding it to the nearest ten) because we see in our research that it is fruitful not mention that it provides a nice form for the value if we want to create a new system of units both with a zero at the end for the value and that is connected to nature, which it is, in Manuel’s integral. Also the nine is the nine in the nine-fifths connected to nature and mathematical constants, as we have shown in our research, and the eight is the 0.8 in the 1.8 that is nine fifths, the fraction around a circumference of a circle that is ninefifths of a circumference.

Let us consider the Neptune Equation: 7.2x –4 = y Let 7.2 be Earth Gravities: (v) = 7.2t –4 (dx/dt) = 7.2t –4 (dx) = 7.2t dt – 4dt (7.2)(980 cm/s/s) = 7,056 cm/s/s v=4=7.2t t=(5/9) (7,056)(5/9)=v_0 = 3,920 so, (dx) = 7,056 cm/s/s t dt –3,920 cm/s dt 15

15

!

∫ 7,056cm /s/stdt − ∫ 3,920cm /sdt = 7.35km = 0

0

We call this Manuel’s Second Integral

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Ian Beardsley November 26, 2013

147 km 20

126 of 126 The Author