Scuttlers, Black Goo and the Biogenic

1 of 25 Scuttlers, Black Goo, And The Biogenic Field Ian Beardsley © 2016 by Ian Beardsley

2 of 25 Scuttlers


3 of 25 I first learned about Scuttlers when Miles Johnston and Kerry Cassidy filmed their interviews at one of the UFO Congresses in Arizona a few years back. Miles Johnston mentioned sighting one on the input valve, or plug, at one of the radio stations where he was an electronics engineer. He explained to Cassidy that they are electronic, hyperdimensional devices that feed on information. Cassidy said she had seen them, and, suggested that they run along the grid that makes the matrix that makes reality. As I see them as some kind of AI, and I have derived several AI Matrices, I here calculate the energy, its magnitude and direction, of the scutter, in what I call the isocyanic matrix:




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5 of 25 The magnitude and direction of the energy is is the cross product of vector1 and vector2 in the pages that follow. It is the green line plotted with an online vector plotter. 




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7 of 25 Scuttler Energy: Magnitude & Direction in the Matrix)  




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9 of 25 Subset of The Isocyanic Matrix, Another Scuttler Energy Vector


10 of 25 Black Goo


11 of 25 At the 2014 International UFO Congress near Phoenix Arizona, Miles Johnston of the Bases Project and Kerry Cassidy of Project Camelot discuss things with one another and, aside from discussing Archons and Djinn, they talk at length about the Black Goo, also called biosentient fluid or, nano oil.

Johnston explains that it was discovered in the Falkands War, a British war that took place in the 80’s where the British discovered on an island south of the war, the existence of a black oil that was engineered in Argentina in collusion with blue and grey ETs that can be activated electronically. Johnston explains that if you can manipulate this nano oil with transistor technology, you can use it to engineer cyborg hydraulics systems of extraordinary power.

He goes on to tell us that in Britain, scientists working with the goo couldn’t get it to do what they wanted so they dumped it down the drain and, it got into the sewage system wherein it infected colonies of people and that this would have the effect in 20 years of making them more intelligent. However, he warns, this stuff is a nano technology and can get in through your skin and take control of a person, in which case the infected person self-terminates their self.

Though, Johnston says there will be a more advanced people enhanced by the black goo.

In 2016, two years later, the take on the black goo does a complete turn around in Johnston’s interview with scientist Harald Kautz Vela on Johnston’s Bases Project, wherein Johnston summarized the story of Kautz Vela as one where the black goo is actually the oil in the Earth and is something that gives humans their instinct. However, he says Vela warns the black goo native to the earth is being replaced by black goo from other worlds, by life from other worlds, that gives humans their instinct in place of their original instinct; essentially an attempt by ETs to turn humans into versions of them.

The point was made by Cassidy that the black goo is AI nano technology.

The Oil Drop Experiment of 1909 by Robert A. Millikan and Harvey Fletcher Indeed there is nothing silly about oil being electrically charged. In the oil drop experiment drops of oil were suspended between gravity and a charged plate. Knowing the force of gravity and the density of oil, since the electric field equals the gravitational field, you can determine the electric charges on the oil drops. Since they were all base multiples of one value (1.592E-19 C), this was deemed the charge on an electron. It is very close to the actual value that was later obtained by more accurate methods of, 1.602E-19 C.

12 of 25 The Composition of oil is hydrocarbons, predominantly alkanes. The basic chemical structure is: C_nH_2n+2 Let us look at a single molecule: C_1H_2(1)+2 = CH4 = 12.01 + 4(1.01) = 16.05 grams per mole Which is methane, or natural gas. The molar mass of CH4 is 16.05 grams per mole. While the composition of crude oil varies widely, the percent of different elements by mass do not. Crude oil is basically: Carbon: 85% Hydrogen: 14% Nitrogen: 2% Oxygen: 1.5% Sulfur: 6% Metals: Less than .1% We can write: Carbon: (12.01 g/mol)(mol/ 6.02E23 atoms)=2.00E-23 grams per atom Hydrogen: (1.01 g/mol)(mol/6.02E23 atoms)=1.68E-24 grams per atom Nitrogen: (14.01 g/mol)(mol/6.02E23 atoms)=2.327E-23 grams per atom Oxygen: (16.00 g/mol)(mol/6.02E23 atoms)=2.6578E-23 grams per atom Sulfur: (32 g/mol)(mol)(mol/6.02E23 atoms)=5.3E-23 grams per atom Metals: negligible If we consider crude oil as a single atom, we have: (2.00E-23 grams/atom)(0.85) (1.68E-24 grams/atom)(0.14) (2.327E-23 grams/atom)(0.02) (2.6578E-23 grams/atom)(0.015) + (5.3E-23 grams/atom)(0.06) ————————————————— 108.5% ~ 100%

= = = = =

1.7E-23 grams C per atom of oil 2.35E-25 grams H per atom of oil 4.654E-25 grams N per atom of oil 3.9867E-24 grams O per atom of oil + 3.18E-24 grams S per atom of oil ——————————————————2.4867E-23 grams oil per atom of oil

2.4867E-23 ~ 2.5E-23 So, we say there are approximately 2.5E-23 grams of oil per atom of oil (2.5E-23 grams/atom)(6.02E23 atom/mole) = (15.0500 grams of oil/mole) We say the molar mass of crude oil is 15 grams per mole. That puts it right between nitrogen and oxygen in the periodic table of the elements. The earth atmosphere is approximately 75% nitrogen and 25% oxygen.

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These results suggest the following equation for black goo: CH4 = 16, Crude Oil = (N+O)/2 Or Crude Oil + CH4 = N/2 + O/2 + O Black Goo = N/2 + O/2 + O This suggests the following vector: <N/2, O/2, O> = <7, 8, 16>â&#x20AC;Š

14 of 25 In so far as the Black goo has the formula: Crude Oil + CH4 = N/2 + O/2 + O We look at crude oil:

ethane = 2(12.01) +6(1.01) = 30.08 ~ 30 grams/mole. It has the chemical structure: | | —C—C— | | It is the first term in my Skellein Matrix:

15 of 25 We have connected black goo to the earth atmosphere: Black Goo=crude oil + CH4 = (N/2 + O/2 + O) Let us substitute C2H6 for crude oil: C2H6+CH4 = (N/2 + O/2 + O)x CH4= 12 + 1(14) = 16 So, 30 + 16 = (14/2 + 16/2 +16)x 46 = (7+8+16)x 46 = 21x and, x=46/21 =2.19 ~2 Two grams per mole is is hydrogen gas (H2). Hydrogen (H) occurs naturally as a diatomic molecule (H2). Hydrogen is the first element in the periodic table of the elements and is the one element that is not a metal, a semimetal, or non-metal. It is in a class to itself. As well, unlike the other elements, it was not made in the stars but, by the explosion that gave birth to the Universe, and is the element from which all the other were made in the interior of stars.

The Equation of Black Goo is: C2H6 + CH4 = (N/2 + O/2 + O)He/H2

16 of 25 The Biogenic Field

17 of 25 Miles Johnston has said we cannot exist outside the biogenic field, which surrounds and penetrates the Earth. Here I model the biogenic field with my program modelearth.c that determines the mass of a planet given the composition of its various layers. Here is the code I wrote for a three layer model: #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");

18 of 25 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); } 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); }

19 of 25 jharvard@appliance (~/Dropbox): ./modelplanet 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 is r1, the radius of the core in km? 5000 what is p1, its density in g/cm^3? 7.87 what is r2, outer edge of layer two in km? 5500 what is p2, density of layer two in g/cm^3? 8.91 what is r3, the radius of layer 3 in km? 6371 what is p3, density of layer three in g/cm^3? 0.79

(density of light crude oil)

r1=5000.00, r2=5500.00, r3=6371.00, p1=7.87, p2=8.91, p3=0.79 the core has a mass of 41.20 E23 Kg thickness of core is 5000.00 layer two has a mass of 15.44 E23 Kg layer two thickness is 500.00 layer three has a mass of 3.05 E23 Kg layer three thickness is 871.00

the mass of the planet is 5.97 E24 Kg jharvard@appliance (~/Dropbox):â&#x20AC;Š

(iron core) (Nickel Lithosphere) (oil mantle)

(This is exactly equal to the mass of the Earth)

20 of 25 Here we run the program for the actual composition of the Earth: jharvard@appliance (~): cd Dropbox jharvard@appliance (~/Dropbox): make density clang -ggdb3 -O0 -std=c99 -Wall -Werror density.c -lcs50 -lm -o density jharvard@appliance (~/Dropbox): ./density what is r1, the radius of the core in km? 500 what is p1, its density in g/cm^3? 7.87 what is r2, outer edge of layer two in km? 5000 what is p2, density of layer two in g/cm^3? 8.91 what is r3, the radius of layer 3 in km? 6371 what is p3, density of layer three in g/cm^3? 2.33

r1=500.00, r2=5000.00, r3=6371.00, p1=7.87, p2=8.91, p3=2.33 the core has a mass of 0.04 E23 Kg thickness of core is 500.00 layer two has a mass of 46.60 E23 Kg layer two thickness is 4500.00 layer three has a mass of 13.04 E23 Kg layer three thickness is 1371.00

note: 1371~S_0=the solar constant~1370

the mass of the planet is 5.97 E24 Kg jharvard@appliance (~/Dropbox): The actual mass of the Earth is 5.972E24 The Modeled Value is 100% accurate We used Iron core, Nickel middle region, Silicon outer region While there are more than three layers to the Earth Iron, Nickel, Silicon are only predominant to the regions And, are not the only elements or compounds found in them We used the actual densities of Iron, Nickel, and Siliconâ&#x20AC;Š

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â&#x20AC;Š

Presented are the equations for modeling average annual temperature for a planet given the orbital distance of the planet from the star, the starâ&#x20AC;&#x2122;s luminosity and the albedo of the planet.

The source code is written on an HP 35s Scientific Calculator and values for it run with an enigmatic result. The Equations For The Modeling of Average Annual Temperature of A Planet With a Single Atmospheric Layer

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H001 LBL H H002 INPUT U (Solar Luminosities) H003 STO U H004 3.9E26 H005 X (MULTIPLY) H006 STO W H007 VIEW W (Joules/Sec) H008 INPUT R (AU) H009 STO R H010 1.5E11 H011 X (MULTIPLY) H012 STO M H013 VIEW M (Meters) H014 RCL M H015 x^2 H016 pi H017 X (MULTIPLY) H018 4 H019 X (MULTIPLY) H020 STO V H021 RCL W H022 RCL V H023 / (DIVIDE) H024 STO S H025 VIEW S (Stellar Constant) H026 RCL U H027 sqrt H028 STO H H029 VIEW H (Habitable Zone) H030 INPUT A (Albedo) H031 STO A H032 1 H033 RCL A H034 - (SUBTRACT) H035 STO B H036 RCL S H037 X (MULTIPLY) H038 4 H039 / (DIVIDE) H040 STO C H041 VIEW C H042 5.67E-8 H043 / (Divide) H044 4 H045 x root of y H046 1.189 H047 X (MULTIPLY)

H048 STO K H049 VIEW K (Plantet Temperature Kelvin) H050 273 H051 - (SUBTRACT) H052 STO T H053 VIEW T (Planet Temperature Centigrade) H054 —> F H055 STO F H056 VIEW F (Planet Temperature Fahrenheit) H057 RTN

XEQ H000 U? 1.618 W=6.31E26 R? 1.618 M=242,…

S=852 H=1.272 A? 0.618 C=81 K=231 T=-41.5 F=-43

R/S R/S R/S R/S R/S R/S R/S R/S R/S R/S R/S

We have run the program, here, for: 1.618 solar luminosities = PHI 0.618 albedo = phi 1.618 orbital distance = PHI And we got near Fahrenheit-Celcius equivalence:

F=(9/5)C+32 F=C C=(9/5)C+32 (5/5)C - (9/5)C = 32 -(4/5)C = 32 -4C = 160 C=-40 -40C = -40F

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Let us run this program for these values in the C emulator and we notice that flux of the star at orbital distance phi is 0.62 that at earth which is phi (the golden ratio conjugate) rounded to two places after the decimal:â&#x20AC;Š

24 of 25 #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/1370); printf("That is %.2f Watts per square meter\n", b); } jharvard@appliance (~): cd Dropbox/descubrir jharvard@appliance (~/Dropbox/descubrir): ./stelr 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 Fahrenheit-Celsius Equivalence The luminosity of the star in joules per second is: 63.10E25 When Running for Golden The habitable zone of the star in AU is: 1.272006 Ratio Flux at planet is 0.62 times that at earth. note:0.62~phi That is 852.66 Watts per square meter jharvard@appliance (~/Dropbox/descubrir): â&#x20AC;Š

25 of 25 The Author