What Discover Is Finding

What Discover Is Finding By Ian Beardsley Copyright Š 2015 by Ian Beardsley

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In an effort to find patterns in mathematics, nature, and the universe, I have decided to undertake writing a computer program, called Discover, that searches for them in the hopes that from these patterns a theory of everything can be established.

In a vast sea of billions of grains of sand on the Earth alone, and billions of stars in the Galaxy alone, it takes a program like Discover to find its hidden nuances.

I have already had some success. I have found a relationship pertaining to silicon circuitry, perhaps even AI. I wrote: If the golden ratio conjugate is to be found in Artificial Intelligence, it should be in silicon, phosphorus, and boron, since doping silicon with phosphorus and boron makes transistors.

We take the geometric mean between phosphorus (P) and Boron (B), then divide by silicon (Si), then take the harmonic mean between phosphorus and boron divided by silicon:

Arithmetic mean of these two numbers: (0.65 + 0.57)/2 = 0.61. 0.61 is the first two digits of the golden ratio conjugate.

The code for some of the programs, which I wrote in the language C, can be found in the appendices, for which I provide the theories behind them.

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Notice near fahrenheit-centigrade equivalence for the golden ratio and its conjugate when applied to the sun (solar luminosity) and the earth (AU) and the albedo of the planet: Running Discover 08 jharvard@appliance (~): cd Dropbox/descubrir jharvard@appliance (~/Dropbox/descubrir): ./discover8 bash: ./discover8: No such file or directory jharvard@appliance (~/Dropbox/descubrir): ./discover08 This program calculates the temperature of a planet,...

Given the luminosity of the star and the albedo of the planet,..

What is brightness of the star in solar luminosities? 1.618 What is the albedo of the planet (0-1)? 0.618 What is the distance of the planet 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 jharvard@appliance (~/Dropbox/descubrir):

Feb 23, 2015

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running star.c (discover 03) for HD 26961 and NGC 6738 jharvard@appliance (~): cd Dropbox/discover jharvard@appliance (~/Dropbox/discover): make star clang -ggdb3 -O0 -std=c99 -Wall -Werror star.c -lcs50 -lm -o star jharvard@appliance (~/Dropbox/discover): ./star 0 enter an int 4 1 enter an int 18 2 enter an int 15 3 enter an int 50 4 enter an int 17 5 enter an int 44 4.00 hours 18.00 minutes 15.00 seconds 50.00 deg 17.00 min 44.00 sec

RA For Star One = 60.30 deg Dec For Star One = 50.30 deg 0 enter an int 19 1 enter an int 2 2 enter an int 1 3 enter an int 11 4 enter an int 37 5 enter an int 28 19.00 hours 2.00 minutes 1.00 degrees

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11.00 deg 37.00 min 28.000 sec

RA For Star Two =285.03 deg Dec For Star Two = 11.62 deg The separation between star one and star two is 228.03 The ratio of separation to 360 is 0.63 jharvard@appliance (~/Dropbox/discover):

The result is close to the golden ratio conjugate and levinson’s number.

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Appendix 1 The Theory Behind discover 04 and discover 08 (Goldilocks.c and Discover 08) As climate science is a new science, there are many models for the climate and I learned my climate science at MIT in a free online edX course. One can generate a basic model for climate with nothing more than high school algebra using nothing more than the temperature of the sun, the distance of the earth from the sun, and the earth’s albedo, the percent of light it reflects back into space.

The luminosity of the sun is:

L0 = 3.9 "10 26 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 4 # (1.5 "1011 ) 2

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 (" ) . Te can be called the effective temperature, the temperature entering the earth. ! the earth disc, "r 2 , and distributes itself over the entire earth surface, S0 intercepts ! !

4 "r 2 , while 30% is reflected back into space due to the earth’s albedo, a, which is equal to 0.3, so !

S "Te 4 = 0 (1# a) 4 $r 2 (1# a)S0 4 $r 2

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But, just as the same amount of radiation that enters the system, leaves it, to have radiative equilibrium, the atmosphere radiates back to the surface so that the radiation 4 4 from the atmosphere, "Ta plus the radiation entering the earth, "Te is the radiation at 4

the surface of the earth, "Ts . However,

"Ta 4 = "Te 4

!

! !

and we have:

!

"Ts4 = "Ta 4 + "Te 4 = 2"Te 4 1 4

Ts = 2 Te S0 (1# a) 4 " = 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 #8 5.67 $10 Te = 255Kelvin

"Te 4 =

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

!

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Ts = 2 4 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

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clouds) and due to the condensation of water vapor into rain droplets (precipitation or the formation of rain).

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Discover 04 (Goldilocks.c) #include <stdio.h> #include <math.h> int main (void) { printf("This program finds the habitable zone of a star,...\n"); printf("And the surface temperature of the planet in the habitable zone\n"); float LC, r, L, HZ, AU, a, root, number, N, answer, C, F; printf("What is the luminosity of the star in Joules per second? \n"); scanf("%f", &L); AU=L/3.9E26; HZ=sqrt(L/3.9E26); printf("The luminosity of the star in solar luminosities is: %f\n", AU); printf("The habitable zone of the star is in AU: %f\n", HZ); r = HZ*1.5E11; LC=L/(4*3.141*r*r); printf("luminosity constant of star in watts per square meter: %f\n", LC); printf("What is the albedo of the planet? (between 0 and 1): "); scanf("%f", &a); N = (1-a)*LC/(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 \n", C, F); }

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Discover 08 #include <stdio.h> #include <math.h> int main(void) { float s, a, l, b, r, AU, N, root, number, answer, C, F; printf("This program calculates the temperature of a planet,...\n"); printf("Given the luminosity of the star and the albedo of the planet,..\n"); printf("What is brightness 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 of the planet 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"); }

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Appendix 2 Theory Behind Discover 01 The arithmetic mean is the midpoint, c, between two extremes a, and c:

b=

!

The harmonic mean is not necessarily the midpoint between two extremes but is the value that occurs most frequently:

b=

!

a+c 2

2ac a+c

The geometric mean, b, between a and c, is the side of a square that has the same area as a rectangle with sides a and c:

b = ac The following relationship holds:

! a:

a + c 2ac :: :c 2 a+c

!

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Appendix 3 discover 03 (star.c) #include <stdio.h> #include <math.h> int main(void) { float num[6], a, b, c, d, e, A, B, D, E, F; for (int i=0; i<=5; i++) { printf("%d enter an int \n", i); scanf("%f", &num[i]); } printf("%.2f hours\n", num[0]); printf("%.2f minutes\n", num[1]); printf("%.2f seconds\n", num[2]); printf("%.2f deg\n", num[3]); printf("%.2f min\n", num[4]); printf("%.2f sec\n", num[5]); printf("\n"); printf("\n"); a=num[0]*15; b=num[2]/60; c=b+num[1]; d=c/60; e=a+d; printf("RA For Star One = %.2f deg\n", e); A=num[3]; B=num[5]/60; D=B+num[4]; E=D/60; F=E+A; printf("Dec For Star One = %.2f deg\n", F); float dig[6], h, i, j, k, l, H, I , K, L, M; for (int j=0; j<=5; j++) { printf("%d enter an int \n", j); scanf("%f", &dig[j]); } printf("%.2f hours\n", dig[0]); printf("%.2f minutes\n", dig[1]); printf("%.2f degrees\n", dig[2]);

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printf("%.2f deg\n", dig[3]); printf("%.2f min\n", dig[4]); printf("%.3f sec\n", dig[5]); printf("\n"); printf("\n"); h=dig[0]*15; i=dig[2]/60; j=i+dig[1]; k=j/60; l=h+k; printf("RA For Star Two =%.2f deg\n", l); H=dig[3]; I=dig[5]/60; K=I+dig[4]; L=K/60; M=L+H; printf("Dec For Star Two = %.2f deg\n", M); float dif, ratio; dif=sqrt(((e-l)*(e-l))+((F-M)*(F-M))); ratio=dif/360; printf("The separation between star one and star two is %.2f\n", dif); printf("The ratio of separation to 360 is %.2f\n", ratio); }

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The Author

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