Advanced Flat Earth Theory

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #180 on: December 13, 2013, 01:54:06 AM »
More details concerning the Tibetan Acoustic Levitation (original message posted here: http://theflatearthsociety.org/forum/index.php?topic=30499.msg1388219#msg1388219 )

https://web.archive.org/web/20110304035318/http://www.nilsolof.se/ljudkraft.htm






http://web.archive.org/web/20060306044903/http://www.ianlawton.com/sl2.htm

Rudolf von Linauer, tibetan levitation:

http://ufoarchives.blogspot.ro/2013/07/rudolf-von-linauer-and-tibetan-mystery.html

UPDATE ON R. V. LINAUER:

http://ufoarchives.blogspot.ro/2013/10/rudolf-von-linauer-and-tibet.html

http://ufoarchives.blogspot.ro/2013/09/new-data-on-rudolf-von-linauer.html




On the meaning of the crescent (raif) and the bindu:

http://www.mandalayoga.net/pretty_print.php?rub=what&p=mantra_om&lang=en


An equivalent symbol is the Thule swastika: red = laevorotatory ether, white = dextrorotatory ether, black = shadow swastika/aether swastika (see http://theflatearthsociety.org/forum/index.php?topic=58472.msg1487905#msg1487905 )

Analysis of the acoustic levitation (D. Davidson):

If we assume that each monk with his instrument produced one half this much sound energy (which is highly unlikely) and we make the further gross assumption that this is the amount if power that reaches the stone (actually sound dissipates rapidly over distance), we would have about 0.04 watts (i.e., (19 instruments + 19 x 4 monks) x 0.000094) hitting the huge stone block.

This is an astoundingly small amount of energy actually hitting the 1.5 cubic meter stone to produce the effect.

To lift the stone 250 meters takes a prodigious amount of energy. Rocks such as granite and limestone have weights in the neighborhood of 150-175 pounds per cubic foot.

If we assume a nominal value of 160 pounds per cubic foot then the 1.5 cubic meter stones weighed around 8475 pounds (i.e., over 4 tons!!!). To lift the 8475 pounds 250 meters would require about 7 million ft-pounds of work (i.e., 8475 pounds X 250 meters / 0.30408 meters/foot = 6,968,035).

Since this was done over a 3 minute period then about 70 horsepower was produced (i.e., 7 x 106 foot-pounds / 180 seconds / 550 horsepower/foot-pound/second = 70.384). This is equivalent to 52 kilowatts (i.e., 70.384 X 0.74570 kilowatts/horsepower = 52.5).

The over unity power factor we obtain is 5,250,000 over unity (i.e., 52,500 watts/0.01 watts).





The density of ether and aether where humans are present is markedly different (photographs taken near a Black Sea resort in the 1970s and 1980s showing the telluric currents: no flash reflection or fake streamers; other examples of real streamers here, http://www.pinellaspascoparanormal.com/aboutorbsandstreamers.htm )

"Nikola Tesla -- the literal inventor of modern civilization (via the now worldwide technology of "alternating current") -- experimentally anticipated the ether waves by finding them in nature; from massive experimental radio transmitters he had built on a mountain top in Colorado, he was broadcasting and receiving (by his own assertion) "longitudinal stresses" (as opposed to conventional EM "transverse waves") through the vacuum. This he was accomplishing with his own, hand-engineered equipment (produced according to Maxwell's original, quaternion equations), when he detected an interference "return" from a passing line of thunderstorms. Tesla termed the phenomenon a "standing columnar wave," and tracked it electromagnetically for hours as the cold front moved across the West."


Thus Tesla was able to change the initial data for the path of the ball lightning: the destination of the trajectory, based on a spherical earth hypothesis, was wrong, and had to be modified to reach an unhabitated area (the desired location of Tunguska). His equipment detected a different density of ether and aether, signaling the presence of human habitation (city of Kezhma).

http://olkhov.narod.ru/tunguska_trajectory.gif

The initial path approached Kezhma from the south, then it abruptly changed course to the east. Two hundred and fifty miles later, while above Preobrazhenka, it reversed its heading toward the west. It exploded above the taiga at 60ş55' N, 101ş57' E (LeMaire 1980).

The same opinion was reached by Felix Zigel, who as an aerodynamics professor at the Moscow Institute of Aviation has been involved in the training of many Soviet cosmonauts. His latest study of all the eyewitness and physical data convinced him that "before the blast the Tunguska body described in the atmosphere a tremendous arc of about 375 miles in extent (in azimuth)" - that is, it "carried out a maneuver." No natural object is capable of such a feat.

Felix Zigel, professor of aerodynamics (Moscow Aviation Institute) and other space experts agree that, prior to exploding, the object changed from an eastward to a westward direction over the Stony Tunguska region.


It was the research done by professor Giuseppe Longo (University of Bologna) which uncovered the forgotten photographs taken in Irkutsk on June 30 and July 1, 1908.



He also was able to find the map drawn by Vasilyev (of course, Vasilyev's research ONLY reached as far east as the city of Krasnoyarsk).

There were other expeditions which went much further to the east, for example the I.M. Suslov voyage...

Evenki tribe account.

http://www.vurdalak.com/tunguska/witness/lyuchetkana_a.htm

A bright summer night fell, the fire began to diminish. In place of the heat, it grew cold. We decided to move toward the Katanga [river]. By the time we got to the Chambe river, we were already totally weak, all around we saw marvels, terrible marvels. It wasn’t our forest [any more]. I never saw a forest like that. It was strange somehow. Where we lived there had been dense forest, an old forest. But now in many places there was no forest at all. On the mountains all the trees lay flat, and it was bright, and everything was visible for a far distance.

(translation by Bill DeSmedt)
« Last Edit: May 30, 2018, 04:41:53 AM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #181 on: December 27, 2013, 06:26:46 AM »
“The universe is more like a giant thought than a giant machine and the substance of the great thought is consciousness which pervades all space.”
Sir James Jeans

Spirit – transcendent verb
Thought/Emotion – intentional/visualized verb (desire)
Imagination – potential verb (word for the idea/emotion) (knowledge)
Sound – manifested verb (action)

http://www.eso-garden.com/specials/the_secret_life_of_nature.pdf (takes the reader inside a subquark: the structure of a boson/antiboson) walnut = subquark/Anu/UPA

Chapters 7, 8 and 9

Taking a closer look at one walnut, Ron saw that two threads came out of it, one of which appeared fainter than the other.The clearer one looked like a tangled, twisted piece of string, which could be pulled out into a straight line with little effort and which, on being relaxed, resumed its tangled state.

Thinking he would see a spiral within one of these strings, Ron magnified it. Instead he saw a stream of bubbles flowing back and forth so quickly he could not observe the moment they reversed direction.

As the bubbles came out of the walnut in single file to move along what looked like a tube, some form of energy appeared to expand them to their maximum over a distance of up to ten bubble diameters.
Then the current reversed.

Fastening his attention onto a single bubble, Ron saw that as it moved through the tube the tube rotated one instant in one direction, next in the opposite, clockwise as the bubbles moved away, counterclockwise as they moved toward him, though again he could not distinguish the actual instant of transition.

Estimating the distance between successive bubbles as about six times the width of a bubble, Ron noted that as each bubble passed, the tube seemed to collapse very slightly, its edges no sharper than the boundary between two liquids. Managing to move along with a bubble-obviously not moving his physical body but his viewpoint-Ron saw that it was shaped like a fat doughnut, with an indented sort of cap that led the bubble's motion and trailed a tail.Wanting to see what was happening close up to one of the walnuts, Ron approached a thread that appeared to link two walnuts.

On closer inspection, the bubbles seemed to Ron to be created in the corkscrew spiral near the exit because there was no sign of bubbles at the start of this spiral. As the bubbles flowed back into the walnut, instead of forming a puff like those entering from the other thread, they simply shrank down to nothing. Whenever bubbles reversed direction, the tail would fade away, to reappear on the opposite side.

Bubbles in what to Ron was thread number two started out as mere squiggles of energy, pointed at both ends. Then the squiggle got fatter, turning into the stable tadpole shape.


Therefore, it is the tail (Ron’s own description) itself which forms both the bosons and antibosons.


What is the structure of the tail, so far the smallest particle known to science?

The Gizeh Pyramid is a large scale model of the particles which do make up the tail (which itself becomes a boson or an antiboson).

One of the most mysterious features of this Pyramid is the scale of its measurements: certainly the pyramid could have been built larger or smaller (keeping all of the proportions equal, of course), but it could only function if and only if it was built to its present scale.

Let us imagine our Universe (http://www.freewebs.com/raacoz/enclosure3%5B1%5D4.jpg ) to be the size of a subquark (http://www.alliancesforhumanity.com/matter/matter_files/image010.jpg ). We know that a subquark has some 14 million bosons (and many more antibosons) inside its structure, and that a boson consists of two inverted pyramids which exchange aether and ether. Then, roughly, the Gizeh Pyramid would correspond to the size of such a boson’s interior pyramids.


Where are the “secret” tunnels/passages of the Gizeh Pyramid located?

Chinese five notes cycle

jiao (1)
shi (2)
gong (3)
shang (4)
yi (5)

1-3
3-5
5-2
2-4
4-1

(this, by the way, is also the origin of the Chinese five elements creation cycle)

1 – will designate the area from the base of the Gizeh pyramid to the top of the queen’s chamber
2 – from the bottom of the queen’s chamber, to the bottom of the king’s chamber
3 – from the bottom of the king’s chamber, to the top of the djed apex (just before the block separating the apex from its lower chambers)
4 – the djed itself
5 -  from the top of the djed apex, all to the way to the top of the Gizeh pyramid

I predict, therefore, that there is a narrow passage from the top of the djed apex all to the way to the top of the pyramid. The “secret” passage from the top of the pyramid which leads to the queen chamber has already been discovered a few years ago.

Two more secret passages will be discovered: leading from the queen’s chamber to the djed apex, and the other one descending from the djed apex to the base.

The most mysterious feature of the pyramid is the groove (FA-MI interval) inside the Grand Gallery:




In the center of the boson we have the two apexes (called parabindu) which rotate as follows:

http://www.eaglespiritministry.com/pd/howto/images/mt_01.gif

The virtual (thought-like) pyramid is facing downwards: this is called the aparabindu particle in vedic physics. It produces aether, the medium needed for the sound to propagate.

The upward facing pyramid (imagination) produces sound, which activates the shadow/thought pyramid.

The Gizeh pyramid has a virtual twin pyramid: it faces downwards, like in the following images (posted in a different context here: http://www.aloha.net/~johnboy/Reticulum.htg/Great_Pyramid.jpg ):





Examples of the virtual component of matter (electrophotography/kirlian images):

http://webspace.webring.com/people/gl/lemagicien/kfpage/kf.html

http://webspace.webring.com/people/gl/lemagicien/kfpage/kfgalery/gal.html (plants)

http://webspace.webring.com/people/gl/lemagicien/kfpage/kfjava/kfjava.html (plants)





More photographs here: http://www.crystalinks.com/kirlian.html

The Secret Life of Plants, Tompkins and Bird, 1973


ETHERIC REGION OF THE PHYSICAL WORLD (BARYONS, MESONS, QUARKS AND SUBQUARKS):

http://www.rosicrucian.com/rcc/rcceng01.htm#part2


Who actually built the Gizeh pyramid?

Zecharia Sitchin - The Wars of Gods and Men (3rd Book of Earth Chronicles)

Chapter 7, figure 42 (a and b)

In chapter 7 we will also find a very good demonstration that the Gizeh pyramid could not have been built by any pharaohs (sixth or first dynasty).

In chapter 10 (figures 66 and 71), we can see the four meters of masonry at the base of the pyramid which means that the actual height measures 136.1 meters.

In figure 42 beings are shown with a crystal in front of their foreheads. Using this crystal, it was possible to build the Gizeh pyramid in less than two months, utilizing acoustic levitation and drilling.

In all humans, this virtual crystal (actually made up of baryons, mesons and quarks) is latent: only a few have been able to activate it on a very limited scale, Nikola Tesla was one of them.

Tesla Mind Lab:

http://www.creativethinkingwith.com/Nikola-Tesla-Creative-Thinking-Secrets.html


The original configuration of the Gizeh pyramid included many pairs of stones which actually made it possible for the inverted virtual pyramid to be activated.

Ninurta (Nimrod) found inside the pyramid...

"....Escorted by the Chief Mineralmaster, Ninurta inspected the array of "stones" and instruments. As he stopped by each one of them, he determined its destiny - to be smashed up and destroyed, to be taken away for display, or to be installed as instruments elsewhere. We know of these "destinies" and of the order in which Ninurta had stopped by the stones, from the text inscribed on tablets 10-13 of the epic poem Lugal-e. It is by following and correctly interpreting this text that the mystery of the purpose and functions of many features of the pyramid’s inner structure can be finally understood.

"Going up the Ascending Passage, Ninurta reached its junction with the imposing Grand Gallery and a Horizontal Passage. Ninurta followed the Horizontal Passage first, reaching a large chamber with a corbelled roof. Called "vulva" in the Ninharsag poem, this chamber’s axis lay exactly on the east-west center line of the pyramid. Its emission ("an outpouring which is like a lion whom no one dares attack") came from a stone fitted into a niche that was hollowed out in the east wall. It was the SHAM ("Destiny") Stone. Emitting a red radiance which Ninurta "saw in the darkness," it was the pulsating heart of the pyramid. But it was anathema to Ninurta, for during the battle, when he was aloft, this stone’s "strong power" was used " to grab to kill me, with a tracking which kills to seize me." He ordered it "pulled out... be taken apart... and to obliteration be destroyed."

Among other features, Ninurta encountered:

"....Whereas in the narrow passages only " a deem green light glowed," the Gallery glittered in multicolored lights - "its vault is like a rainbow, the darkness ends there." The many-hued glows were emitted by twenty-seven pairs of diverse crystal stones that were evenly spaced along the whole length of each side of the Gallery.... each crystal stone emitted a different radiance, giving the place its rainbow effect....

Ninurta’s priority was the uppermost Grand Chamber and its pulsating stone.... he reached the Antichamber of unique design...."There three portcullises - "the bolt, the bar and the lock" of the Sumerian poem - elaborately fitted into grooves in the walls and floor, hermetically sealed off the uppermost Grand Chamber: "to foe it is not opened...." But now, by pulling some cords, the portcullises were raised, and Ninurta passed through.

"He was now in the pyramid’s most restricted ("sacred") chamber, from which the guiding "Net" (radar?) was "spread out" to "survey Heaven and Earth...." It responded to vibrations with bell-like resonance. The heart of the guidance unit was the GUG Stone ("Direction Determining").... Ninurta ordered this stone destroyed: "Then, by the fate-determining Ninurta, on that day was the Gug stone from its hollow taken out and smashed."

Finally there was the Apex Stone of the Pyramid, the UL ("High As The Sky") Stone: "Let the mother’s offspring see it no more," he ordered. And, as the stone was sent crashing down, "let everyone distance himself," he shouted. The "Stones," which were "anathema" to Ninurta, were no more.


There were several attempts to build a similar pyramid on a much smaller scale. They all ended in failure: the corners of the structure could not be aligned perfectly.



http://www.mysticalconspiracy.info/2013/12/16/the-1978-nippon-pyramid-project/

http://sacredsites.com/africa/egypt/the_great_pyramid_of_giza.html


The structure of a magnet is larger scale version of the boson itself:

http://theflatearthsociety.org/forum/index.php?topic=60367.msg1563059#msg1563059


Comments on antigravitons:

http://dougvanvenrooij.wordpress.com/2013/03/15/anti-gravitons-may-explain-dark-matter-dark-energy-and-the-universe-we-observe-today-2/



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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #182 on: January 06, 2014, 02:43:58 AM »
Gizeh Pyramid Advanced Calculus

The concept of radian measure, as opposed to the degree of an angle, is normally credited to Roger Cotes in 1714. He had the radian in everything but name, and he recognized its naturalness as a unit of angular measure

The first mention of the natural logarithm was by Nicholas Mercator in his work Logarithmotechnia published in 1668, although the mathematics teacher John Speidell had already in 1619 compiled a table on the natural logarithm.

The problem of extending the factorial to non-integer arguments was apparently first considered by Daniel Bernoulli and Christian Goldbach in the 1720s, and was solved at the end of the same decade by Leonhard Euler.



Basically, all the previous attempts to discover the hidden mathematical properties in the Gizeh Pyramid used only π and the golden section.

However, the most important figure of all is the ratio 136.1/53.33 = one hundred sacred inches. Then if we multiply this number by 25 we get the sacred cubit, or 0.63566 m. One sacred inch = 0.025424 m.

As I will demonstrate, the constructors of the pyramid had at their disposal all the details needed from advanced calculus: radian measure, Taylor series expansion, natural logarithm, gamma function, Stirling series (complete with realistic error bounds), and much more.



The sacred cubit is designated in the form of a horseshoe projection, known as the "Boss" on the face of the Granite Leaf in the Ante-Chamber of the Pyramid. By application of this unit of measurement it was discovered to be subdivided into 25 equal parts known now as: Pyramid inches.


http://guardians.net/egypt/gp2.htm

http://books.google.ro/books?id=8LZCAAAAYAAJ&pg=PA270&lpg=PA270&dq=vyse+operation+gizeh+1837+queen+chamber+niche&source=bl&ots=4fJ-tNxlTs&sig=akFC7UCDV6SBqW87gq9VcokwMGU&hl=ro&sa=X&ei=OrNAUKzXE4TAhAfshYGwDA&redir_esc=y#v=onepage&q=vyse%20operation%20gizeh%201837%20queen%20chamber%20niche&f=false (page 112)

https://web.archive.org/web/20120314235117/http://www.aiwaz.net/queen-chamber/a25

http://www.world-mysteries.com/mpl_2.htm

http://thegreatpyramidofgiza.ca

http://www.samuellaboy.com/New_Folder/Special_Topics/Advanced_s.htm

http://www.ancientegyptonline.co.uk/great-pyramid.html


100 – 36.43 = 100 sc (sacred cubits) = 360/π2

0.63566 radians = 36.4206 degrees

sin 136.12 = ln 2

ln 20sc = 2.5424 (=1si)

140.7 (total height without the apex) / 203 (steps of the pyramid )~= ln 2

sin 72.7 degrees = 1.5 sc

7.2738 = 286.1 si = displacement factor

tan 51.8554 degrees = 2 sc

(51.87 degrees = angle formed by the height of the pyramid, 5813 si, and the base, 9131 si)

Apex plateau triangle: base 224.6 si, height 286.1 si

Angle 1 = 51.87 degrees
Angle 2 = 38.13 degrees

sin 38.13 degrees  = 0.618

sin 51.87 degrees = 1/2sc

51.87/38.13 = 1.3603

3.813 = 6 sc

sin 5 x 286.1 sc = sin 909.31163 degrees = -1.618/10

286.1 si x 0.4 = 1.361 x 4 x 0.534

Triangle with sides 309.5si, 286.1si and 118.1si and angles of 67.57 and 22.43 degrees

sin 22.43 degrees = 0.381562

Queen chamber niche measurements

First step – w 1.568m / l 1.0414 m / h 1.743 m
Second step – w 1.34 m / l 1.0414 m / h 0.87266 m

π/360= 0.0087266

2.618/20 = 0.1309
0.1309/1.5 = 0.087266

1.743 = 0.87266 x 2

Third step – w 1.062 m / l 1.0414 m / h 0.69733 m

0.1309 x 5.34 = 0.699

0.69733 x 1.25 = 0.87266

0.69733 = 40π/360

https://web.archive.org/web/20170605173236/http://reocities.com/CapeCanaveral/Hall/3324/nefersschooloflearning.htm

https://web.archive.org/web/20170605171457/http://reocities.com/CapeCanaveral/Hall/3324/nefershouse.htm

https://web.archive.org/web/20120802231648/http://reocities.com/CapeCanaveral/Hall/3324/pan3.gif

There are several values to be used for the sacred cubit depending on the color of the light spectrum: starting from 0.62832 all the way to 0.64 – the most important value is of course 0.63566, the sacred cubit.

The authors of the work even express each and every value of the Gizeh pyramid using a very special type of circle:

https://web.archive.org/web/20120802231648/http://reocities.com/CapeCanaveral/Hall/3324/pan5.gif

However, I will use a different and more interesting representation in order to uncover the hidden mathematical symbols of the Gizeh pyramid, where the radius is equal to 68.05 (68.05 x 2 = 136.1 = the diameter of the circle).

On such a circle, using s = r x @ (@ measured in radians), we will obtain some very special values:

Degrees   -   Arclength

22.5 – 26.66
45 – 53.4
90 – 106.68 (exactly the frequencies used by the Tibetan monks)
72.9 – 86.5 (=136.1 sc)
2.142 – 2.542 (1si x 100)
136.1 – 161.8

FULL PYRAMID VOLUME

2,658,672.883 m3

We divide three times by 1si and we get,

1.6178314 x 1011






I will use all the terms up to and including 1/12x.


Then,

Γ(15.23065) = 1.6178314 x 1011

15.23065 x 180/π = 872.652 degrees (let us remember that π/360= 0.0087266).

15.23065 - 4π = 2.66428

Now let us use the special circle described above (r = 68.05) –

2.66428 x 68.05 = 181.3 (286.1 sc = 181.6)

FULL PYRAMID LATERAL AREA

87,326.424 m2

If we divide once by 1si we get 3,434,802.71

If we divide twice we get 135,100,798.7

Γ(10.966) = 3,434,802.71

15.23065/10.966 = 1.3889 = 25/18

Γ(12.495) = 135,100,798.7 

12.495 x 180/π = 715.91

12.495 - 3π = 3.0702

3.0702 x 68.05 = 208.93


208.93 = 72.83 + 136.1


FULL PYRAMID VOLUME/TOTAL LATERAL AREA = 30.4452

But 15.23065 x 2 = 30.4452


APEX TOTAL VOLUME

78.636 m3

Again, we divide three times by 1si to get,

4,785,112.6

Γ(11.116) = 4,785,112.6

11.116 x 180/π = 636.9

11.116 - 3π = 1.69122

1.69122 x 68.05 = 115.087

115.087 = 181.1 x 1sc


APEX TOTAL LATERAL AREA

88.93 m2

88.93 = 3,497.9 si

If we divide 88.93 twice by 1si we get

137,582


Γ(7.815) = 3497.9

7.815 x 180/π = 447.7665

447.7665 /5 = 140.8 x 1 sc (140.8 total height of the pyramid without the apex)

447.7665 – 360 = 87.766

7.815 -2π = 1.5318

153.18 + 26.18 = 180

1.5318 x 68.05 = 104.24

286.1/104.24 = 1/(1 – 1sc)



11.116/7.815 = 1.4224 = 64/45

VOLUME OF THE APEX/TOTAL APEX LATERAL AREA = 0.8842

1.4224 x 0.618 = 0.8842



Γ(9.564) = 137,582

9.564 x 180/π = 547.97 (= 447.76 + 100)

9.564 - 3π = 0.139222

0.139222 x 68.05 = 9.474

9.474/5 = 1.8948 = 1.361 + 5.34


It should be noted that we obtain these figures only for the Gizeh pyramid.

As an example, for the following values used for another pyramids, no correspondence can be obtained from the same kind of calculations.

1.   Side of the pyramid = 8, height =3, volume = 64 cubic meters
2.   Side of the pyramid = 6.18034, height = π, volume = 40 cubic meters


Let us proceed further with the Gizeh pyramid.

KING CHAMBER VOLUME

305.258 cubic meters

Dividing three times by 1si we get,

18,575,281.96

Γ(11.685) = 18,575,281.96

11.685 x 180/π = 669.5

11.685 - 3π = 2.26

2.26 x 68.05 = 153.81 (same figure as in the apex total lateral area)

153.81 + 26.18 = 180

KING CHAMBER SARCOPHAGUS VOLUME

77.56si x 21.77si x 33.46si = 69288.7

Γ(9.251) = 69,288.7

9.251 x 180/π = 530

9.251 - 2π = 2.9678

2.9678 x 68.05 = 201.96

201.96/π = 450/7

King chamber volume/201.96 = 1.511 exactly the distance from the queen chamber niche to the apex of the queen chamber itself.


In the next message, full calculations for the queen chamber niche, apex cone, and the queen chamber itself.
« Last Edit: July 26, 2018, 08:38:06 PM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #183 on: January 08, 2014, 01:27:12 AM »
Gizeh Pyramid Advanced Calculus (II)


VOLUME OF THE CONE GENERATED BY THE APEX PYRAMID

123.523 cubic meters

Dividing the volume three times by 1si, we get

7,516,502.3

Γ(11.308) = 7,516,502.3

11.308 x 180/Π = 647.9

11.308 - 3Π = 1.8832

1.8832 x 68.05 = 128.153


128.153 - 90 = 38.153 = 60sc

128.153/53.4=2.4

VOLUME OF THE APEX/VOLUME OF THE APEX CONE = 0.63657


VOLUME OF THE CONE GENERATED BY THE PYRAMID

4,176,223.827 cubic meters

2.5412838 x 1011

Γ(15.398) = 2.5412838 x 1011

15.398 x 180/Π = 882.24

15.398 - 4Π = 2.83163

2.83163 x 68.05 = 192.7

192.7 = 7.2738 x 26.5


Γ(11.06) = 4,176,223.827

11.06 x 180/Π = 633.691

11.06 - 3Π = 1.635222

11.06 x68.05 = 111.277

111.277/1.618 = 68.774

111.277 + 68.774 = 180

 FIBONACCI NUMBER #106

F106 = 6,356,306,993,006,846,248,183 = 6.3563 x 1021

F106/VOLUME OF THE PYRAMID = 1/2.5412838 x 10-11 (VOLUME OF THE PYRAMID CONE)





QUEEN’S CHAMBER VOLUME

VOLUME OF THE RECTANGULAR PRISM

140.9 cubic meters

8,574,194.2

Γ(11.363) = 8,574,194.2

11.363 x 180/Π = 651.05

11.363 - 3Π = 1.938

1.938 x 68.05 = 131.896

131.896/0.618 = 213.4 (=4 x 53.4)


VOLUME OF THE TRIANGULAR PRISM

45.57 cubic meters

2,773,032.4

Γ(10.884) = 2,773,032.4

10.884 x 180/Π = 623.61

10.884 - 3Π = 1.4592

1.4592 x 68.05 = 99.3

68.05Π – 99.3 = 11.448 (11.444 = 4 x 286.1si)


TOTAL VOLUME OF THE QUEEN’S CHAMBER

186.47

11,346,902.7

Γ(11.4803) = 11,346,902.7

11.4803 x 180/Π = 657.8

11.4803 - 3Π = 2.0555

2.0555 x 68.05 = 139.88

139.88/99.3 = 1.408 (height of pyramid = 140.7 meters)

139.88/2.618 = 53.43


QUEEN’S CHAMBER NICHE, FIRST SECTION

w = 1.568 m
h = 1.743 m
l = 1.0414 m

V = 2.8462 cubic meters

173,192.62

Γ(9.67) = 173,192.62

9.67 x 180Π = 554

9.67 - 3Π = 0.24522

0.24522 x 68.05 = 16.687

16.687 = 27 x 0.618

QUEEN’S CHAMBER NICHE, SECOND SECTION

w = 1.34 m
h = 0.87266 m (Π/360 = 0.0087266)
l = 1.0414

V = 1.217776 cubic meters

74,103

Γ(9.28) = 74103

9.28 x 180/Π = 531.7

9.28 - 2Π = 2.9968

2.9968 x 68.05 = 203.93 = 6sc x 53.4 = 136 x 1.5

QUEEN’S CHAMBER NICHE, THIRD SECTION

w = 1.062 m
h = 0.69733 m
l = 1.0414 m

V = 0.77122 cubic meters

46,929.81

Γ(9.07) = 46,929.81

9.07 x 180/Π = 519.67

9.07 - 2Π = 2.7868

2.7868 x 68.05 = 189.64

189.64 = 136.1 + 53.4

QUEEN’S CHAMBER NICHE, FOURTH SECTION

w = 0.773
h = 0.69733
l = 1.0414

V = 0.56135 cubic meters

34,158.9

Γ(8.924) = 34,158.9

8.924 x 180/Π = 511.3

8.924 - 2Π = 2.6408

2.6408 x 68.05 = 179.707

179.707/286.1 = 2Π/10

QUEEN’S CHAMBER NICHE, FIFTH SECTION

w = 0.5156
h = 0.69733
l = 1.0414

V = 0.37443 cubic meters

22,784.38

Γ(8.725) = 22,784.38

8.725 x 180/Π = 500

8.725 - 2Π = 2.4418

2.4418 x 68.05 = 166.166

166.166 ~= 10 x 16.687 (value obtained for the first section)

554 (value from the first section) x 3 = 10 x 166.2


CONCLUSIONS

The builders of the Gizeh Pyramid had at their disposal a deep knowledge of differential and integral calculus.

The notions of the radian, Taylor series expansion (used to calculate the decimal/fractional values of the trigonometric functions), tables of sine/cosine/tangent values, error estimates for the Taylor series (which uses the Extended Mean-Value Theorem, attributed to A. Cauchy), natural logarithm, Gamma function, Stirling series were well known to these builders as the foregoing calculations have shown in great detail.

Furthermore, the Stirling series could not have been used without an adequate error analysis which would provide realistic error bounds. Such an error analysis would involve knowledge of the Euler integral, Euler limit form, Euler constant, the notion of an asymptotic expansion, the Zeta function.

For proofs see Asymptotics and Special Functions, F.W.J. Olver, chapter 2 - sections 1.3, 1.4, 11.1-5, chapter 8 - section 4.1


It is inconceivable to state that the Gizeh pyramid was built some 5,000 years ago: no matter which hypothesis is used (extraterrestrials coming from the 12th planet [Z. Sitchin] or from some other galaxy - which would need negative energy/tachyon technology, thus debunking/disproving the infinite universe/multiple galaxy conjecture, Atlantis, or any other) there simply is not enough time for a round planet to have formed itself, given the official chronology used for any solar system evolution physics.

The Gizeh pyramid must have been built just a few hundreds of years ago: some decades later, the mathematics and physics used in its construction were infused into the Western scientific mainstream.


L. Euler: a fictional character invented at the end of the 18th century:

http://theflatearthsociety.org/forum/index.php?topic=30499.msg1483598#msg1483598

http://theflatearthsociety.org/forum/index.php?topic=30499.msg1483917#msg1483917
« Last Edit: January 10, 2014, 06:26:00 AM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #184 on: January 10, 2014, 06:20:00 AM »
Origin of Calculus: How Mathematical Analysis Was Imported to India, Italy, France and England

http://www.hinduwisdom.info/Yuktibhasa.pdf

A relevant epistemological question is this: did Newton at all understand the result he is alleged to have invented? Did Newton have the wherewithal, the necessary mathematical resources, to understand infinite series? As is well known, Cavalieri in 1635 stated the above formula (the infinite series expansion for the sine function) as what was later termed a conjecture. Wallis, too, simply stated the above result, without any proof. Fermat tried to derive the key result above from a result on figurate numbers, while Pascal used the famous “Pascal’s” triangle long known in India and China. Though Newton followed Wallis, he had no proof either, and neither did Leibniz who followed Pascal. Neither Newton nor any other mathematician in Europe had the mathematical wherewithal to understand the calculus for another two centuries, until the development of the real number system by Dedekind.

The next question naturally is this: if Newton and Leibniz did not quite understand the calculus, how did they invent it? In the amplified version of the usual narrative, how did Galileo, Cavalieri, Fermat, Pascal, and Roberval etc. all contribute to the invention of a mathematical procedure they couldn’t quite have understood? The frontiers of a discipline are usually foggy, but here we are talking of a gap which is typically 250 years.

Clearly a more natural hypothesis to adopt is that the calculus was not invented in Europe, but was imported, and that the calculus took nearly as long to assimilate as did zero.


Dr. Joseph George Gheverghese from the University of Manchester said there was strong circumstantial evidence that the Indians passed on their discoveries to mathematically knowledgeable Jesuit missionaries who visited India during the 15th century.

That knowledge may have eventually been passed on to Newton himself, he said.



A key development of pre-calculus Europe, that of generalisation on the basis of induction, has deep methodological similarities with the corresponding Kerala development (200 years before). There is further evidence that John Wallis (1665) gave a recurrence relation and proof of the Pythagorean theorem exactly as Bhaskara II did.

Although it was believed that Keralese calculus remained localised until its discovery by Charles Whish in 1832, Kerala had in fact been in contact with Europe ever since Vasco da Gama first arrived there in 1499 and trade routes were established between Kerala and Europe. Along with European traders, Jesuit missionaries from Europe were also present in Kerala during the 16th century. Many of them were mathematicians and astronomers, and were able to speak local languages such as Malayalam, and were thus able to comprehend Keralese mathematics. Indian mathematical manuscripts may have been brought to Europe by the Jesuit priests and scholars that were present in Kerala.


Other pieces of circumstantial evidence include:

James Gregory, who first stated the infinite series expansion of the arctangent (the Madhava-Gregory series) in Europe, never gave any derivation of his result, or any indication as to how he derived it, suggesting that this series was imported into Europe.


Kerala's established trade links with the British East India Company, which began trading with India sometime between 1600 and 1608, not too long before Europe's scientific revolution began.

There was some controversy in the late 17th century between Newton and Leibniz, over how they independently 'invented' calculus almost simultaneously, which sometimes leads to the suggestion that they both may have acquired the relevant ideas indirectly from Keralese calculus.


Some of Bhaskara's contributions to mathematics include the following:

Integer solutions of linear and quadratic indeterminate equations (Kuttaka). The rules he gives are (in effect) the same as those given by the renaissance European mathematicians of the 17th Century.

A cyclic, Chakravala method for solving indeterminate equations of the form ax2 + bx + c = y. The solution to this equation was traditionally attributed to William Brouncker in 1657, though his method was more difficult than the chakravala method.

Solutions of Diophantine equations of the second order, such as 61x^2 + 1 = y^2. This very equation was posed as a problem in 1657 by the French mathematician Pierre de Fermat.

Preliminary concept of mathematical analysis.

Preliminary concept of infinitesimal calculus, along with notable contributions towards integral calculus.

He conceived differential calculus, after discovering the derivative and differential coefficient.

Stated Rolle's theorem, a special case of one of the most important theorems in analysis, the mean value theorem.

Traces of the general mean value theorem are also found in his works.

Calculated the derivatives of trigonometric functions and formulae.


The calculus has played a key role in the development of the sciences, starting from the “Newtonian Revolution”. According to the “standard” story, the calculus was invented independently by Leibniz and Newton. This story of indigenous development, ab initio, is now beginning to totter, like the story of the “Copernican Revolution”. The English-speaking world has known for over one and a half centuries that “Taylor” series expansions for sine, cosine and arctangent functions were found in Indian mathe-matics/astronomy/timekeeping (jyotisa) texts, and specifically in the works of Madhava,Neelkantha (Tantrasangraha, 1501CE), Jyeshtadeva (Yuktibhâsâ, c. 1530 CE) etc. No one else, however, has so far studied the connection of these Indian developments to European mathematics.


The force on a body is the resultant of gravity and the work done against it. V.S 5.1.13

In the absence of all other forces gravity exists. V.S 5.1.7

Action is opposed by an equivalent opposite reaction - V.S 5.1.16-18

Newton's laws of motion copied from the Naya Vaiseshika Sutra.

Suppose that the mass of an object is 'm' and in time interval 't', the velocity of the object changes from 'u' to 'v' due to the force acting on it. Then,

Initial momentum = mu
Final momentum = mv
Change in momentum = m(v-u)

Therefore, the rate of change of momentum = m(v-u)/t = ma (from Kanada's first law)

From Kandas second law,
force is proportional to the rate of change of momentum.
Or, p k ma
Or, p = kma (where k is a constant)

If m=1 and a=1, then
1 = k*1*1 or k = 1
Or, p = ma

Therefore, unit force is the one that produces unit acceleration in an object of unit mass.

Prashastpada



ISAAC NEWTON, THE CALCULUS THIEF: (some excerpts)

He copied his laws of gravity from "Surya Sidhanta" the great Sanskrit astronomical work written in the Vedic age . Reproduced in another written text by Bhaskara , 1200 years before Newton it clearly explains gravity without an apple. However Vedic gravity was a push ( after observing the solar eclipse ) and NOT a pull. 

They took Calculus to Europe , from where the likes of Gottfried Wilhelm Von Liebniz , Isaac Newton and Robert Hooke raced with each other to translate , re-invent and market it in their own names, in a acrimonious manner.

It was John Wallis , while he was the keeper of Oxford Univeristy archives who first started pondering over translated Mathematics stolen from India. 

John Wallis patented Vedic Math infinity and infinitesimal in his own name.  Rest he could NOT understand .  Whatever he could make head or tail of, he included in his Arithmatica Infinitorum and Treatise on Algebra.

His baton was taken over by Isaac Barrow, who tutored Isaac Newton in Kerala Calculus.




http://ckraju.net/IndianCalculus/Bangalore.pdf

The Infinitesimal Calculus: How and Why it Was Imported into Europe


https://web.archive.org/web/20130713214810/http://indianrealist.com/2009/01/26/how-jesuits-took-calculus-from-india-to-europe/

‘Calculus is India’s Gift to Europe’

In his speech at ICIH 2009, Professor C.K. Raju revealed that calculus was an Indian invention that was transmitted by Jesuit priests to Europe from Cochin in the second half of 16th century. “Indian infinite series has been known to British scholars since at least 1832, but no scholar tried to establish the connection with the calculus attributed to Newton and Leibnitz,” he said.

Dr. Raju’s 10-year research that included archival work in Kerala and Rome was published in a book “Cultural Foundations of Mathematics.” It established that the Jesuit priests took trigonometric tables and planetary models from the Kerala mathematicians of the Aryabhata school and exported them to Europe starting around 1560 in connection with the European navigational problem.

“When the Europeans received the Indian calculus, they couldn’t understand it properly because the Indian philosophy of mathematics is different from the Western philosophy of mathematics. It took them about 300 years to fully comprehend its working. The calculus was used by Newton to develop his laws of physics,” Dr. Raju added. Ironically, some British scholars claimed credit for this research despite being warned against plagiarizing Professor Raju’s work.



However, what Dr. C.K. Raju does not realize is that the same science of calculus was also imported to India, in order to create the false impression of an ancient indian history.

http://madhesi.wordpress.com/2008/09/24/did-ashoka-exist/ (how Emperor Ashoka, India's greatest historical figure, is a fictional character invented in the 19th century)

Not so ancient India:

http://breakfornews.com/forum/viewtopic.php?p=27888#27888


Ptolemy's Almagest: it was created at least after 1350 AD, here are the complete proofs:


When was Ptolemy's star catalogue in Almagest compiled in reality?

https://web.archive.org/web/20160306134153/http://hbar.phys.msu.ru/gorm/fomenko/fomenko3.pdf



The dating of Ptolemy's Almagest based on the coverings of the stars and on lunar eclipses

https://web.archive.org/web/20141008085625/http://hbar.phys.msu.ru/gorm/fomenko/fomenko4.pdf


Both works appeared in the Acta Applicandae Mathematicae (17 - 1989 and 29 - 1992).


HISTORY: FICTION OR SCIENCE? VOLUME 3, DATING PTOLEMY'S ALMAGEST

https://archive.org/stream/AnatolyFomenkoBooks/History-FictionOrScienceByAnatolyFomenkoVol.3#mode/2up

Pg. 209 - 214

Tycho Brahe = N. Copernicus


Pg. 248 - 259

Who actually wrote the works attributed to Hipparchus, T. Brahe and C. Ptolemy


Pg. 302 - 327

J. Kepler = N. Copernic = T. Brahe = C. Ptolemy the most extraordinary analysis

The other pages include one of the best ever discussion on the new chronology of the times of J. Kepler, C. Ptolemy, T. Brahe, N. Copernicus, who were actually one and the same person.


Dating Ptolemy's Almagest (a more technical work):

https://www.jaks.sk/dokumenty/fomenko/Fomenko-%20Kalashnikov-%20Nosovsky%20-%20Dating%20Ptolemy-s%20Almagest%20-1993-.pdf

The coverings of the stars, and the lunar eclipses described in Almagest, could have occurred ONLY during the period 800 - 1350 a.d. and not one thousand years earlier. Archimedes' Palimpsest was also forged after 1750 AD.
« Last Edit: December 06, 2019, 08:55:12 AM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #185 on: January 28, 2014, 12:47:36 AM »
Therefore, the truncated Maxwell equations refer ONLY to the temporary hertzian ripples in the ether sea, and NOT to the scalar/ether waves themselves:

http://www.theflatearthsociety.org/forum/index.php/topic,58190.msg1489693.html#msg1489693

http://www.theflatearthsociety.org/forum/index.php/topic,58190.msg1489785.html#msg1489785


Now, a closed-form formula for the natural logarithm:

LN v = ((-2 +{2+[2+...(2+ 1/v + v)^1/2]^1/2}^1/2))^1/2 x 2^n

n+1 parantheses to evaluate - in the last parenthesis we substract 2 and take the square root one last time (n+1), before we multiply the result by 2^n

For v very large, we can omit the term 1/v


By summing the nested square root function, we obtain the final result:

LN v = 2n x ( v(1/2^n) - 2 + v-(2^-n) )1/2

Of course, we can use the first formula for computation utilizing only a pocket calculator with only the four basic arithmetic operations (since a square root function is essentially a continued fraction).


And there are more formulas to be derived from the logarithm continued function:

COS @ = 1/2 X (({[(2 - @^2/2^n)^2 -2)^2...]-2}^2 -2)) (n/2+1 evaluations)

COS^-1 @ = ((2- {2+ [2+ (2+ 2@)^1/2]...^1/2}))^1/2 x 2^n (n+1 evaluations)

 
COSH v = 1/2 x (({[( ( (2 + v2/2^n)^2) -2)^2] -2)^2 ...-2}^2 -2)) (n/2 +1 evaluations)

TAN-1 v = ((2- {2+ [2+ (2+ 2{1/(1+ v2)1/2})^1/2]...^1/2}))^1/2 x 2^n (n+1 evaluations)


ln v = 2n x ( v1/2n+1 - 1/v1/2n+1 )

This is the correct formula for the natural logarithm function, linking algebraic functions with elementary and higher transcendental functions, providing the fastest way to calculate the value for any logarithm, and at the same time help to evaluate any integral containing lnx terms.


For a first approximation:

ln v = 2n x ( v1/2n - 1 )

First results appear for n = 8 to 12, all the remaining digits for n = 19 and greater...

Example: x = 100,000 ; lnx = 11.5129255

with n=20, the first approximation is lnx = 11.512445 (e11.512445 = 100001.958 )


For the function 2n x v1/2n, there is a certain pattern for the succesive approximations, (fk+1 - fk)/(fk - fk-1); as an example for 1x108 after the first four evaluations, the ratios approach 2; for 1x105, after the first four calculations, for 5.23 x 1012, after the first five evaluations.

Now, for the first time, we can evaluate and obtain estimates for the logarithmic integral (li(x)):



Perhaps we will encounter integrals of the form ( u2n+1/(u-1) x du ) or some form of partial fraction decompositions containing 2n+1 factors, but there is an excellent chance to obtain a formula or even some kind of an estimate which will settle the matter:



li(x) - Π(x) = O(x1/2lnx) (number of primes not exceeding x and a verification of Riemann's hypothesis at the same time - I recommend H.R. Edwards' Riemann's Zeta Function for further information)

We all know that the integral of lnx = xlnx - x; however, the real beauty and significance of this formula is revealed only when we use the correct logarithm expression derived above for the first time:

2n x ((1/(1 + 1/2n+1)) x v1 + 1/2n+1  -  (1/(1 - 1/2n+1)) x v1 - 1/2n+1 )



Einstein, 1905:

"The principle of the constancy of the velocity of light is of course contained in Maxwell's equations”


ALBERT IN RELATIVITYLAND

http://www.gsjournal.net/old/ntham/amesbury.pdf

However, space-time as a fourth dimension is nothing more than the product of professor Minkowski's cerebral and mathematical imagination.


On Physical Lines of Force, the original set of Maxwell's equations:

http://vacuum-physics.com/Maxwell/maxwell_oplf.pdf


The best work done on the original set of Maxwell's equations belongs to Dr. Frederick Tombe.

http://web.archive.org/web/20071006083222/http://www.wbabin.net/science/tombe4.pdf

The correct demonstration of the rotating aethereal substance within Maxwell’s vortex cells.

Dr. Tombe's paper demonstrates quite clearly the fallacy of Einstein's statement.


Dr. Tombe went even further with his paper: Gravity and Light -

http://www.gsjournal.net/old/science/tombe18.pdf

Abstract. Gravity and light are two different manifestations of aether flow.

Another classic by Dr. Tombe:

http://www.gsjournal.net/old/science/tombe5.pdf

Gravitation and the Gyroscopic Force



Double helix theory of the Magnetic field:

http://www.gsjournal.net/old/science/tombe.pdf


There is no such thing as the theory of relativity:

http://theflatearthsociety.org/forum/index.php?topic=60367.msg1563056#msg1563056


Dayton Miller ether drift experiments:

http://theflatearthsociety.org/forum/index.php?topic=60367.msg1563058#msg1563058
« Last Edit: January 31, 2014, 02:17:20 AM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #186 on: February 12, 2014, 12:34:18 AM »
Circle arclength formula:

A = 2n x (2c2 - c(2c2 + c(2c2 ... +c(2c2 - 2bc)1/2)1/2)1/2...)1/2

c = radius, b = side of the right triangle which lies on the x-axis


Then, we can certainly factor out the term c1/2n+1, which means that the logarithm of the radius of a circle is definitely linked to the trigonometric functions.


Sacred cubit radians

Degrees to sacred cubit radians - divide by 36.4206 = (100 - 100 sc)

Sacred cubit radians (scr) to degrees - multiply by 36.4206

50 degrees = 1/0.7284 scr (7.28 = displacement factor of the Gizeh Pyramid)

100 degrees = 2.7457 scr = 1/0.364206


sc = sacred cubit = 0.63566 (there are several sacred cubit measurements, ranging from 0.625 to 0.64)


b12 + a12 = c12

b1 = d1 x d2 (divisors of b1)

a1 = (d12 - d22)/2

c1 = (d12 +d22/2


b1, a1, c1 in the natural numbers set

If b1 is prime, then b12 + a22 = c22 (where c2 = (b12 +1)/2 )



Using sacred cubit radians, we finally understand the importance of the arclength for the b12 + a12 = c12 formula.


b1 = 33
c1 = 65

arccos 33/65 = 1.0385 radians = 59.4897 degrees, no symmetry can be detected

However, using scr, we get 1.6334 scr.

Multiplying the arclength for the given angle by one sc we get 106.1785, and then by the scr value: 167.025

65sc2 = 26.264

167.025/26.264 = 6.36 (10 sc)

Similarly, for 5177 = b1 (c1 = 14425), we get the value 2.5424 x 1/0.2861

for 3173 = b1 (c1 = 14125), we get (2 + 1/sc2)/sc, and so on.


The sequence 2sc - 1/2sc x N (N = 1,2,3 ... ) will give the values: 5.34, 7.287, 13.6034, 63.65 and much more.


The Fibonacci numbers are actually sacred cubit numbers.

1,618034 = 4sc2 (sc = 0.636009827)

Then Fn = 1/(8sc2 -1) x 22n x sc2n



Then we get:

(b12)sc + (a12)sc =~ ([(b1 + a1 + c1)/2]2)sc

b11/sc + a11/sc =~ c11/sc


b1sc + a1sc = (k x c1)sc , 1 < k <~2




Let p(n) = partitions of a natural number




I am going to derive the asymptotic formula for p(n) using just the sacred cubit as a guide; using the above formula I have calculated p(n) up to n = 96, we also know that p(243) = 133,978,259,344,888

For starters let n!/([lnn]! x (n - [lnn])!) = f(n), where [ x ] is the integer part of x, then

f(83)/p(83) = 1/sc

f(91)/p(91) = 1/12


Using various functions to approximate ln p(n) such as (n/sc2)1/2, and n1/2/sc2, we get some special values:

ln p(33) = 331/2 x 1.6195

[(1 + 4sc2) x 33 )]1/2/sc = ln p(33) x 1/sc

For h1 = ((1 + 4sc2) x lnn)1/2/sc we get

h1(33) - (3sc + lnn) = 1sc
h1(51) - (3sc + lnn) = 1/2sc

2431/(1 - sc) = 3sc + ln 243


Finally, without using complex analysis or Ramanujan sums, we get:

ln p(n) = [(1 + 4sc2) x n )]1/2/sc  - (lnn + 1 + 1/sc2 - 1/sc) - a very good approximation



Next, I am going to attempt to solve the most difficult known problem in number theory: large number factorization of semiprimes (product of two very large prime numbers), at least for a semiprime which has ten digits, using just the sacred cubit: a new formula (the leading asymptotic term) which solves the problem for numbers with ten digits or less, and a new algorithm featuring Fibonacci numbers remainders.

There is a wealth of information which can be obtained from the b1 term, using sacred cubits, and which can be the starting point to a whole new approach to factoring semiprimes.

« Last Edit: January 06, 2018, 10:39:19 PM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #187 on: February 14, 2014, 12:53:54 AM »
(b12)sc + (a12)sc =~ ([(b1 + a1 + c1)/2]2)sc

b11/sc + a11/sc =~ c11/sc



a1 + c1 = d12

List of 4sc2 sequence numbers (Fibonacci numbers):

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html


With a reasonable approximation for a1, we can obtain a very good estimate for d1 from the first formula, presented for the first time in my previous message.

8141 x 131071 = 1073602561

1073602561 = 286572 + 252378911 = 28657 x 46368 - 255165215  (where 28657 = F23, and 46368 = F24)

If b1<a1, then the a1 term will be of the form F242 - ..., or F24 x F25 - ..., F25 x F26 - ..., or F252 - ... ; if b1>a1, then a1 will equal F23 x F24 - ..., F222 - ... , that is, only 4-6 possible choices.

In order to get a very good estimate for d1, we will use the first remainder (and a few subsequent remainders if needed, more explanation below) obtained from the b1 for each of the above choices .

For the a1 =  F25 x F26 - ... choice, using a10 = 255165215, and substituting in the first formula, we get:

d1 = 132578.957, an excellent approximation.

Actually, a1 = 8556257280 = 750252 + 2927506655 = 75025 x 121393 - 551252545


65537 x 131071 = 8590000127 = 750252 + 2961249502 = 75025 x 121393 - 517509698

Using the same reasoning and the same formula, we get a first estimate for d1, d1 = 130095.707


It is only by using the power of the sacred cubit that we can actually get these estimates, impossible to obtain otherwise by any other method.


But we can actually accomplish much more, by using the 4sc2 sequence to reveal the sacred cubit structure of the natural number system.


821 x 941 = 772561

772561 = 610 x 987 + 170491 = 9872 - 201608

170491 = 3772 + 28362 = 377 x 610 - 59479

The sum of the remainders obtained by dividing the numbers by Fn will equal a product of Fn numbers.

201608 = 377 x 610 - 28362 = 3772 + 59479

28362 = 144 x 233 - 5190 = 1442 + 7899

59479 = 2332 + 5190 = 233 x 377 - 28362

7899 = 892 - 22 = 89 x 55 + 3004

5190 = 89 x 55 + 295 = 892 - 2731


3004 = 552 - 21 = 55 x 34 + 1134

2731 = 552 - 294 = 55 x 34 + 861


1134 = 342 - 22 = 34 x 21 + 420

861 = 34 x 21 + 147 = 342 - 295

420 = 212 - 20 = 21 x 13 + 147

294 = 21 x 13 + 21 = 212 - 147 ; 147 + 21 = 168

147 = 21 x 8 - 21 = 13 x 8 + 43

43 = 8 x 5 + 3 = 82 - 21

21 = 5 x 3 + 6 = 52 - 4


Of course, we can immediately obtain a first approximation for d1, d1 = 918; by summing the remainders of b1 in their corresponding order, we can obtain even better estimate for d1.


Now, we can actually get the remainders of the a1 term either by noticing that 6 and 4 (remainders obtained by dividing 21 by F5 and F4) can be used to start the a1 sequence of remainders starting from the bottom up, or by using a very interesting shortcut involving b1sc.

Actually, a1 = 105720 = 3772 - 36409 = 377 x 233 + 17879

Using the same scheme as above, we get finally:

40 = 82 - 24 = 8 x 3 + 16

16 = 52 - 9 = 32 + 7

9 = 3 x 5 - 6 = 2 x 3 - 3


65 = 82 + 1 = 8 x 13 - 39

39 = 8 x 5 - 1 = 52 + 14

14 = 3 x 5 -1 = 2 x 5 + 4


Knowing that 6 and 4 are the remainders of a1, we can see that from the possible choices we eventually get (11, 19, 9 and 14) only 9 and 14 will make any sense, given the fact that the sum of the remainders at each stage of the calculation will equal a product of Fn numbers.


One of the remainders of a1 will be 2857.


3004 - 2857 = 147



772561sc = 5530


5530 - 5063 = 2 x 233

5530 - 2857 = 89 x 30

(5063, another a1 remainder)

That is, there is a certain symmetry and relationship between b1sc and some of the a1 remainders.


The same reasoning can be used for any b1 = d1 x d2.

For 1000009 = 3413 x 293, we get a first estimate of 3486, and by summing the remainders of b1 (576230 + 204130 + 62001 + 25840 + 5104 + 2817 + 947 ...) we get an estimate of 3400, which is amazing, because we only use the remainders from b1 and very simple approximations.


For 1000009, b1sc = 6515.72

9368 - 6515.72 = 610 x 4.66 = 987 x 2.88  (4.66 = 2 x 2.33 , and 2.88 = 2 x 1.44, both 233 and 144 are Fn)

9368 is one of the a1 remainders

Another a1 remainder is 3448

6515.72 - 3448 =~ 552 = 233 x 13


The algorithm uses only Fn numbers, and is proportional to the number of the digits of b1 and not to any divisor of b1.


I believe that this formula is just the first leading term of a certain asymptotic approximation to d1, and we have seen the extraordinary approximations which can be obtained effortlessly:

(b12)sc + (a12)sc =~ ([(b1 + a1 + c1)/2]2)sc


« Last Edit: February 28, 2017, 12:37:35 AM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #188 on: February 16, 2014, 07:47:45 AM »
2123 = b1 = 11 x 193
18564 = a1

2123sc = 130.26

301 - 130.26 = 34 x 5

854 - 130.26 =~ 144 x 5

301 and 854 remainders of 18564 upon division by the corresponding Fn numbers.


21232sc = 16967.50815

a1 - 21232sc = 1596.4918  (1597 = F17)


18564sc = 516.92

902 - 516.92 = 7 x 55

902 remainder of 2123


15617 = b1 = 97 x 161
8256 = a1


8256sc = 308.84

15617sc = 463.13

463.13 - 106 = 89 x 4   (106 remainder of 8256)

932 - 308.84 = 7 x 89   (932 remainder of 15617)


1000009 = b1 = 3413 x 293
5781360 = a1

10000092sc = 293627 x 144.5  (293627 remainder of 5781360)
10000092sc - (293627 x 144.5) = 25519.6
25840 - 25519 = 233 + 89 (25840 remainder of 1000009)

1000009sc = 6515.72

6515.72 - 3448 =~ 552


231 - 1 = 2147483647

261 - 1 = 2.3059 x 1018

b1 = (231 - 1) x (261 - 1) = 4.951760152 x 1027
a1 = 2.658455989 x 1036

In a situation like this b12sc can be used to find useful relationships between the remainders of b1 and a1, and even estimates.

In fact, with a1 trial function 4 x 1035, we get an estimate for d1 = 2.353 x 1018.


I would need access to a computer which can handle division/multiplication of integers with 50 digits+, and then use the b1, a1 remainders to discover the hidden sacred cubit symmetries:

-the following powers of b1 also would be very useful to discover further formulas: 2sc + 1/2sc, 1/sc, 2sc, 1, 1/2sc, 2 - 2sc, sc, sc/2, 2sc - 1/2sc (in fact it would cover the range of all possible values of a1)

-ln (b1/a1 + a1/b1) leads to the conclusion that the remainders of 2c1 and b1 + a1 also do contain useful information


For a 200 digit number (semiprime), the required computational time (1990) for the methods then used in integer factorization will take 4 x 1015 years.

For a 300 digit number, we would need 5 x 1021 years

For a 500 digit number, the figure would rise to 4.2 x 1032 years.

An elegant method would not resort to "needle in the haystack algorithms", but would make full use of the very interesting mathematical relationships which do exist between the remainders of b1 and a1 upon division by the corresponding Fn numbers - an algorithm which is proportional to the number of the digits of b1.


Even the universal constants of bifurcation theory are related to the sacred cubit.

136/18 x 1/4sc2 = 4.66933 (Feigenbaum constant)

136/48 x 1/4sc2 = 2.8333 (-2.8333 is the Shenker-Rand constant)

4.6692 + 2.618034 = 7.2872 (displacement factor of the Gizeh pyramid, 286.1 sacred inches = 7.28)

2 x 2.8333 = 4.66933 x 3sc2


Albert Einstein,Relativity, The special and the general theory, 11th ed., 1936, p.64:

“In contrast to electric and magnetic fields, the gravitational field exhibits a most remarkable property, which is of fundamental importance ... Bodies which are moving under the sole influence of a gravitational field receive an acceleration, which does not in the least depend either on the material or the physical state of the body.”


PROJECT MONTGOLFIER - Dr. Thomas Townsend Brown proves the fallacy of Einstein's statetment; also the Biefeld-Brown effect shows that terrestrial gravity and antigravity are electrical forces of opposite spin.


http://theflatearthsociety.org/forum/index.php?topic=59837.msg1540161#msg1540161

http://theflatearthsociety.org/forum/index.php?topic=59837.msg1552735#msg1552735

http://theflatearthsociety.org/forum/index.php?topic=59837.msg1554378#msg1554378

http://theflatearthsociety.org/forum/index.php?topic=59837.msg1553607#msg1553607
« Last Edit: February 18, 2014, 12:23:01 AM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #189 on: February 28, 2014, 07:55:42 AM »
Riemann’s Hypothesis – Sacred cubit structure of the zeros of the Zeta function



The zeros of the Zeta function of Riemann are angles (expressed in radians); by finding the corresponding angles in the first four quadrants, and multiplying by 68.05 (radius of the circle), we obtain a certain arclength.

Then, an intricate and extraordinary system/network of sacred cubit equations will become apparent, showing that the structure of the zeros of the Zeta function cannot be understood without employing the notion of the  sacred cubit.

Furthermore, each group of five consecutive zeros of the Zeta function will form a five element cycle (as described earlier), with striking sacred cubit mathematical relationships becoming evident.


The zeros of the Zeta function can be viewed as sort of a Poincare map: the iteration of an initial point (of a periodic differential equation on a cylinder, for our case, for visualization purposes) under the Poincare map is the succesive interaction of the spiral with a vertical line on the cylinder (the sequence of Poincare maps fit together to make a spiral on the cylinder).


http://mathworld.wolfram.com/RiemannZetaFunctionZeros.html

http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/physics1.htm

http://member.melbpc.org.au/~tmajlath/Riemann.html

http://www.dtc.umn.edu/~odlyzko/zeta_tables/index.html



The Stieltjes constants are actually sacred cubit constants.



A0 = 0.5772156649
A1 = 072815845

7.28158 is one of the values of the displacement factor of the Gizeh pyramid.

In fact,

0.078158455 – 0.07273942 = 7.6425 x 10-5

7.6425 x sc = 4.858 = 10 x (2sc – 1/2sc)

Then, the Euler constant can be expressed as:

= 0.07273942 x 18sc/(18sc – 10)



We can express Catalan’s constant, G, in terms of sacred cubits.



G x 0.2861 = 1/6sc

G/1.361 = 1/[1 + (2sc – 1/2sc)]

The Feigenbaum constant can be expressed in terms of G.

4.6693 x 0.72738 = 5.34 x sc

(18sc – 10)/18sc x sc x 53.4 = 4.27678

4.27678/4.6693 = G


The sum of the negative integer powers of the nontrivial zeros is again a sacred cubit constant.






The sum will equal 0.02309571

2.309571/sc = 7.2738/2 – 3.5567 x 10-3

355.555 = 2/3 x 533.33

http://mathworld.wolfram.com/RiemannZetaFunctionZeros.html (for all the sums of the negative integer powers of the nontrivial zeros, from Z(1) to Z(6) )


FIRST FIVE ZEROS OF THE ZETA FUNCTION (see http://www.dtc.umn.edu/~odlyzko/zeta_tables/index.html )

We reduce the value of the zero itself (angle expressed in radians) to the first few quadrants, as needed, then multiply this value by 68.05 to find the value of the arclength.

14.134725 - 4π = 1.568354 (value of the width of the first section from the queen chamber niche)

1.568354 x 68.05 = 106.726

Let us now find further sacred cubit equations featuring 14.134725

Ln 14.134725 = 14.134725/5.3366

14.134725 x 0.618 = 1/0.11447 (11.444 is one of the most important constants of the Gizeh pyramid, and equals 4 x 2.861)

14.134725/2.5424 = 1/180

14.134725 – 21/2 = 12.7205 = 20sc

14.134725 - 5π = -1/sc

26.666 x 4 = 106.666

106.726sc = 14.134725 + 5.332

Let us now express 45 degrees as sacred cubit radians:

45/36.4206 = 1.23556

1.23541 x 180 = 222.37

Dividing 14.134725 by one sacred cubit we obtain:

14.134725/sc = 22.22363

Since 106.666 is the radian value for 90 degrees multiplied by68.05, we can see that the first zero of the Zeta function is absolutely related to the value of 90 degrees.


21.02204 - 6π 2.172476

2.172476 x 68.05 = 147.837


25.0108 - 7π = 3.0196

3.0196 x 68.05 = 205.487


30.424 - 8π = 5.29125

5.29125 x 68.05 = 360.07


32.935 - 2π = 26.666
32.935 - 8π = 7.8022

7.8022 x 68.05 = 530.94

37.586

by substracting 9π and multiplying by 68.05 we get 633.658
by substracting 8π and multiplying by 68.05 we get 847.444

Now, the crucial observation is made: each set of five consecutive zeros of the Zeta function, expressed as arclength (and at this early stage also as the zero divided by one sc) will obey the rules of the five elements cycle defined earlied on this page.


Therefore,

360.07 – 106.726 = 253.444
253.42 = 156.6 x 1.618

360.07 – 205.487 = 154.583

154.583 x 0.618 = 100sc x 1.5

530.94 – 360.07 = 170.874 (Gizeh pyramid value)

633.659 – 530.94 = 102.719
102.719/sc = 161.594 (136.1 expressed in radians and multplied by 68.05)


Since the values of the zeros at this early stage are actually spaced out quite nicely, we can divide their values by one sc to obtain further equations.

Dividing the previous values of the first five zeros of the Zeta function by one sc we get:

22.22363

33.07

39.346 (1/1si)

47.862

51.81

33.07 = (the Euler constant x 180)/π

39.346/22.22363 = 1.77

e1/1.77 = 1.759398 = 1 + 39.346/51.81

e0.75943 = 4 x 53.4

0.75943 = 1/1.31678

51.81/33.07 = 1.5666

47.862/33.07 = 1.4473

1.4487 = 52.762 degrees/36.4206

140.6/52.762 = 2.666

With an angle of 5.29125 and a radius of 68.05 the triangle formed will have sides measuring: 37.2285, 56.9635, 68.05 – their sum will be 99.884

If we subtract π/2 from the 3.0196, we get again 1.4488 – the corresponding triangle will have sides measuring 8.2768, 67.54, 68.05, and a sum of 143.872

143.872/99.884 = 1.4462

47.862/22.223 = 2.154

2.154/5.34 = sc2

2.154/1.5666 = 1/0.727296

2.154/1.316777 = 1 + sc

1.316777 + 0.25 = 1.56666

32.935 – 14.134725 = 18.8

Volume of Gizeh pyramid/Lateral area = 30.4452

30.4452 x 0.618 = 18.815

30.4452 x π/180 =1/1.882

Volume of apex/Lateral area of apex = 0.8842


5.29125 – 1.4488 = 3.84246

3.84246 x 68.05 = 261.474


205.487/261.479 = 1/2sc

261.479/360.07 = 0.7263 (one of the values for the displacement factor)

360.07 x 2sc = 458.01

530.94 – 458.01 = 72.931


ZEROS 6 THROUGH 10

37.586

40.9187

43.32707

48.00515

49.774

Using the same procedure (1. Deriving the value of the arclength and 2. Dividing each zero by one sc) we will get again equations involving sc.

Most importantly, again, the consecutive five zeros will obey the rules of the five elements cycle defined earlier.

In fact 1463.0528/1074.2345 = 1.361 (1463.0528 is the arclength for 49.772, and 1074.2345 is the arclength for 40.9187)

Moreover, we meet again with the constant 2.534


ZEROS 125 THROUGH 129

278.2507

279.229251

282.465

283.2111851

284.875964

Again the same precise mathematical relationships involving sc and the five elements cycle.




From the list posted earlier, let us pick at random the following five zeros:

361574.0875
361574.94195
361575.08737
361575.76592
361576.1480

The angles reduced to the first two quadrants will measure:

1.90582 (3sc)
2.76022
2.90562
3.584213
3.9663

Multiplying by 68.05 we get:

129.69105
187.833
197.727
243.906
269.91

I invite the reader to discover the precise equations involving sc and the five element cycle of values expressed as various formulas involving sc.






At random, again, let us pick another set of five consecutive zeros.

1132486.2441
1132486.3922
1132486.87488
1132487.57186
1132487.951

Angles:

1.7827
1.9307
2.9127
3.1107
3.4897

Arclengths:

121.3127
131.384
164.184
211.683
237.474

We obtain again mathematical equations involving sc, and the precise five element cycle values expressed as various formulas involving sc.


Let us now put the sacred cubit to the ultimate test: the Lehmer phenomenon.

http://mathworld.wolfram.com/LehmersPhenomenon.html

7005.0606918
7005.10055

Angles:

2.450667
2.4905252

Arclengths:

166.7679
169.48024

169.48024 – 166.7679 = 2.71234

2.71234 x euler’s constant = 1.5656

2.71234/2sc = 4 x 0.533

2.71234/8 = 0.339 = 0.53333 x sc

2.71234 – 0.1695 = 2.54284  (2 x 0.1695 = 3.39)


Conclusions:

The builders of the Gizeh pyramid must have had among their ranks the equivalent of G.F.B. Riemann and must have had at their disposal the Riemann-Siegel formula:

http://mathworld.wolfram.com/Riemann-SiegelFormula.html

Either all zeros lie on the x=1/2 line, or none of them do (each set of five consecutive zeros are part of a five element cycle, and of an intricate system of equations involving sc: if a single zero is to be found on different x = @ line, then the previous four zeros will no longer be part of a five element cycle, and so on right back to the very first zero.

The zeros of the Zeta function are intersections of the helical sound wave travelling from the pyramid to the apex (at the quantum physics infinitesimal level, see my previous message on this page for a complete explanation) with the outer surface of the cylinder (or some similar shape) which links the pyramid with its apex, then the values themselves are projections onto the central axis (in our case the x = ˝ line).
« Last Edit: January 06, 2018, 10:44:23 PM by sandokhan »

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Ichimaru Gin :]

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Re: Alternative Flat Earth Theory
« Reply #190 on: March 11, 2014, 05:02:17 PM »
So if I get the gist of it, Newton merely adopted what was known earlier (and aided in the Gizeh construction), as evidenced by the surprising finding that he did not fully comprehend his own supposed invention? Very interesting!!! I also found the Wolfram links quite helpful. Thank you Levee.
I saw a slight haze in the hotel bathroom this morning after I took a shower, have I discovered a new planet?

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #191 on: March 13, 2014, 07:06:20 AM »
Here is a work on Riemann's Zeta function which explains further the findings of C.L. Siegel in the 1930s: the discovery of the asymptotic formula of the Zeta function in Riemann's archives, and of course, much more on the Riemann's hypothesis, if you are interested in this problem:

http://books.google.ro/books?id=ruVmGFPwNhQC&pg=PA298&lpg=PA298&dq=edwards+riemann+zeta+function&source=bl&ots=P4HbjJaNFk&sig=ySMxAy_uEaOaKAkKCo6gbN_w-CQ&hl=ro&sa=X&ei=LbEhU8jmKafF0QHukIHoBg&ved=0CGoQ6AEwBjgK#v=onepage&q=edwards%20riemann%20zeta%20function&f=false


Believe or not, the value of the first zero of the Zeta function on the x = 1/2 line has everything to do with the value of the frequency of the musical instrument (drum) used by the tibetan monks, as described earlier - the single drum of 534.4 Hz.

Height of apex of pyramid: 7.2738 units (286.1 si)

Two apexes in merkabah formation: 9.245 units total height (http://www.absoluteempowerment.com/attachments/Image/Merkabah/002.gif )

Pyramid total height (including subterranean chamber): 174.6 units (140.6 + 30 + 4)

Each and every boson/antiboson has in its center two truncated pyramids facing each other, with their two apexes rotating in opposite direction right in the center of the distance between the truncated summits.

The importance of the value of 534.4 is as follows: 340 + 9.245 + 170 = 534 - 2x7.27

The distance between the pyramids is 186 (actually 185.58625)

In the center lie the two apexes in a merkabah mathematical figure.


Then, 185.586 - 1/0.0063566 = 2 x 14.134725 (where 14.134725 is the value of the first zero of the Zeta function on the x = 1/2 line).


9.245 - 7.2738 = 1.9712

9.245 - 2x1.9712 = 5.3026

2x14.134725 - 5.3026 = 2x11.483425

11.483425 - 11.444 = 0.039425   -   where 0.039425 = 2x0.0197125

In order to discover the true significance of the value of 14.134725 we must make use of the one of the most interesting geometrical figures of the Gizeh pyramid: the sothic triangle (http://www.subtleenergies.com/ormus/tw/SothicTri.jpg ).

That is, we embed the apex itself into a larger triangle as follows:

the side angles of the sothic triangle will measure, of course, 51.87 degrees

its height will measure 16.786 units

the base of the sothic triangle will measure 26.35224 (2x13.17612)


For the apex there are only two possible side angle measures: 68.57 degrees and 51.87 degrees.

68.57 degrees

Substracting 1/2 the value of the base of the apex from 1/2 the value of the base of the sothic triangle we get:

10.32135

14.134725 - 10.32135 = 3.813375 = 6 sacred cubits


51.87 degrees

Substracting 1/2 the value of the base of the apex from 1/2 the value of the base of the sothic triangle we get:

7.4666

14.134725 - 7.4666 = 6.666 = 1/0.15


Now things get really interesting.

tan 51.87 degrees x 13.17612 = 16.786

16.786 - 14.134725 = 2.65129 = 5.3026/2

e2.65129 - 14.134725 = 1/26.6

2.65129 - 2.648634 = 0.00266

We remember that ln 14.134725 = 2.648634


Then, we understand precisely the significance of this figure:

http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/zeros.jpg

The distance from the very center of the boson to the apex of the sothic triangle will measure exactly 14.134725

Then the sound wave vortex manifests itself and each intersection of this helix with the outer surface of the cylinder (in the shape of a very complex sine wave) will appear on the center axis (its projection) with a certain value: the exact values of the zeros of the Zeta function on the x= 1/2 line).

As the a new wave starts from the pyramid and travels towards the its apex in the center, more values of the zeros of the Zeta function will be plotted on the x = 1/2 line, and so on.


"Although the Riemann zeta-function is an analytic function with [a] deceptively simple definition, it keeps bouncing around almost randomly without settling down to some regular asymptotic pattern. The Riemann zeta-function displays the essence of chaos in quantum mechanics, analytically smooth, and yet seemingly unpredictable."

M.C. Gutzwiller, Chaos in Classical and Quantum Mechanics (Springer, 1990), p. 377

"One idea for proving the Riemann hypothesis is to give a spectral interpretation of the zeros. That is, if the zeros can be interpreted as the eigenvalues of 1/2 + iT, where T is a Hermitian operator on some Hilbert space, then since the eigenvalues of a Hermitian operator are real, the Riemann hypothesis follows. This idea was originally put forth by Pólya and Hilbert, and serious support for this idea was found in the resemblance between the "explicit formulae" of prime number theory, which go back to Riemann and von Mangoldt, but which were formalized as a duality principle by Weil, on the one hand, and the Selberg trace formula on the other.

The best evidence for the spectral interpretation comes from the theory of the Gaussian Unitary Ensemble (GUE), which show that the local behavior of the zeros mimics that of a random Hamiltonian. The link gives a more extended discussion of this topic."

"Gutzwiller gave a trace formula in the setting of quantum chaos which relates the classical and quantum mechanical pictures. Given a chaotic (classical) dynamical system, there will exist a dense set of periodic orbits, and one side of the trace formula will be a sum over the lengths of these orbits. On the other side will be a sum over the eigenvalues of the Hamiltonian in the quantum-mechanical analog of the given classical dynamical system.

This setup resembles the explicit formulas of prime number theory. In this analogy, the lengths of the prime periodic orbits play the role of the rational primes, while the eigenvalues of the Hamiltonian play the role of the zeros of the zeta function. Based on this analogy and pearls mined from Odlyzko's numerical evidence, Sir Michael Berry proposes that there exists a classical dynamical system, asymmetric with respect to time reversal, the lengths of whose periodic orbits correspond to the rational primes, and whose quantum-mechanical analog has a Hamiltonian with zeros equal to the imaginary parts of the nontrivial zeros of the zeta function. The search for such a dynamical system is one approach to proving the Riemann hypothesis."   (Daniel Bump)


However, these distinguished mathematicians will NEVER be able to understand the true significance of Riemann's hypothesis outside of the subject of my previous messages: the Gizeh pyramid and the sacred cubit.


The Gizeh pyramid is the architectural equivalent of Riemann's Zeta function.


There were several people who wrote the works attributed to Newton: one wrote the Principia, another the texts on alchemy, another the treatises on chemistry, and others who compiled the works on optics and Newton's private letters.

Newton merely adopted what was known earlier (and aided in the Gizeh construction), as evidenced by the surprising finding that he did not fully comprehend his own supposed invention? Very interesting!!!

Indeed. The Gizeh pyramid was constructed only a few hundreds of years ago: calculus was then slowly infused into the Western cultural/scientific mainstream during the Renaissance which occurred during the 18th century.
« Last Edit: January 06, 2018, 10:45:14 PM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #192 on: March 15, 2014, 03:09:56 AM »
Ukraine/Crimeea/Moldavia - Flat Surface of the Earth

In the official historical chronology, Dimitrie Cantemir is recognized as one of the greatest geniuses ever produced by Eastern Europe.

Dimitrie Cantemir (1673–1723) was twice Prince of Moldavia (in March–April 1693 and in 1710–1711). He was also a prolific man of letters – philosopher, historian, composer, musicologist, linguist, ethnographer, and geographer.



In 1714 Cantemir became a member of the Royal Academy of Berlin. Between 1711 and 1719 he wrote his most important creations. Cantemir was known as one of the greatest linguists of his time, speaking and writing eleven languages, and being well versed in Oriental scholarship. His oeuvre is voluminous, diverse, and original; although some of his scientific writings contain unconfirmed theories and inaccuracies, his expertise, sagacity, and groundbreaking researches are widely acknowledged.


http://www.dacii.go.ro/materiale/dacii/spiritualitate/pagini/monument_megalitice_ceahlau.htm

From the classic work signed Dimitrie Cantemir, Descriptio Moldaviae (http://www.educatlaiasi.ro/uploadpoze/2%20martie%20semnificatii.jpg ), we have the following quote:

"Cel mai inalt multe al Moldovei este Ceahlaul si daca acest munte ar fi fost cunoscut poetilor vechi, el ar fi fost tot atat de celebru ca si Olimpul, Pindul sau Pelia. De alta parte, cat de inalt este muntele acesta se poate conchide din imprejurarea ca in timpul cat cerul este senin si soarele se inclina spre apus, acest munte se poate vedea intreg si asa de curat de la orasul Acherman (Tyras, Cetatea Alba), departe de 60 de ore, ca si cand ar fi in apropiere. Iar pe dealurile din jur se vad urme de cai, de caini si de pasari, imprimate in stanci, in numar asa de mare ca si cand ar fi trecut pe acolo o oaste imensa de calareti.”

Acherman = Cetatea Alba = White Fortress

Here is the map:





Carpathian Mountains, Ceahlaul Range, Toaca Peak:

http://upload.wikimedia.org/wikipedia/commons/thumb/b/ba/Csalh%C3%B3.jpg/800px-Csalh%C3%B3.jpg

Exact location on map: http://ro.wikipedia.org/wiki/V%C3%A2rful_Toaca,_Masivul_Ceahl%C4%83u


TOACA PEAK MEASURES 1907 METERS IN ALTITUDE

DISTANCE CETATEA ALBA/WHITE FORTRESS TO TOACA PEAK: OVER 300 KM


Dimitrie Cantemir is saying that he could see the highest peak of the Ceahlaul Range all the way from the White Fortress very clearly.

The visual obstacle, for a distance of 300 km, altitude of observer some 20 meters above sea level (White Fortress) measures some 4 km.


Of course, now we know that the works attributed to Cantemir were invented at the beginning of the 19th century...however, the description is extraordinary.


Here is a historical figure which did exist: Gheorghe Asachi.

Gheorghe Asachi, surname also spelled Asaki; March 1, 1788 – November 12, 1869) was a Moldavian-born Romanian prose writer, poet, painter, historian, dramatist and translator. An Enlightenment-educated polymath and polyglot, he was one of the most influential people of his generation. Asachi was a respected journalist and political figure, as well as active in technical fields such as civil engineering and pedagogy, and, for long, the civil servant charged with overseeing all Moldavian schools. Among his leading achievements were the issuing of Albina Românească, a highly influential magazine, and the creation of Academia Mihăileană, which replaced Greek-language education with teaching in Romanian. His literary works combined a taste for Classicism with Romantic tenets, while his version of the literary language relied on archaisms and borrowings from the Moldavian dialect.

http://www.mlnar.ro/system/files/images/gheorghe_asachi.thumbnail.jpg


http://romaniapress-misterelelumii.blogspot.ro/2011/01/misterele-ceahlaului-fenomene.html

Ceahlaul nu se ridica, nici pe departe, la altitudinea altor piscuri muntoase din România sau din tarile vecine. Cu toate acestea, in mod paradoxal, el este singurul masiv care poate fi vazut de la sute de kilometri departare. in anumite conditii atmosferice si de luminozitate solara, piscurile Ceahlaului se zaresc cu o deosebita claritate de pe tarmul Marii Negre si de pe malurile Nistrului. Gheorghe Asachi scria despre acest fenomen inca din anul 1859: „Corabierul de pe Marea Neagra vede piscul cel inalt al acestui munte, de la Capul Mangaliei si pâna la Cetatea Alba. Locuitorul de pe tarmul Nistrului vede soarele apunând dupa masa acestui munte, iar pastorul nomad, dupa ce si-a iernat turmele sale pe câmpiile Bugeacului, se intoarce catre casa având in vedere vârful Pionului, sau Ceahlaului”.


Translation: The ship owner (corabier denotes actually each member of the crew, from captain to sailor) sailing the Black Sea can see the highest peak of the Ceahlaul range, starting from Mangalia all the way to the White Fortress.

The local inhabitants located on the banks of the river Nistrul can see the sunset, as the sun disappears behind the Ceahlaul range.

Mangalia on the map:

http://romeonet.ro/imagini/forum.romeonet.ro_litoral.jpg

River Nistru:

http://romaniancoins.org/harti/entransnistria.jpg

Banks of river Nistru:

http://upload.wikimedia.org/wikipedia/commons/4/4e/Ua_river_dnestr_piliptsche.jpg

THE VISUAL OBSTACLE FROM MANGALIA TO THE TOACA PEAK MEASURES OVER 12 KM.

THE VISUAL OBSTACLE FROM RIVER NISTRU TO TOACA IS 2.7 KM.


Mountain climbers say that they can see the Black Sea all the way from the Carpathian mountains, Bucegi Range, Omu Peak (2505 m in altitude):

https://web.archive.org/web/20090422052937/https://www.infomontan.ro/Galerie/Turism%20Diverse/Files/009.html


It is absolutely impossible to see the Black Sea from Peak Omu on a spherical earth: the visual obstacle measures 800 meters (Peak Omu on the map):

http://ro.wikipedia.org/wiki/V%C3%A2rful_Omu,_Mun%C8%9Bii_Bucegi


Photographs of Peak Omu taken from Bucharest, residential neighborhood, building measuring 20 meters in height:


http://forum.softpedia.com/lofiversion/index.php/t21996.html










Each and every geographical detail can be seen over the 150 km distance: no curvature whatsoever.

http://theflatearthsociety.org/forum/index.php?topic=28196.msg674444#msg674444 (curvature, visual obstacle formulas)



« Last Edit: March 26, 2018, 09:32:13 AM by sandokhan »

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sandokhan

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Re: Alternative Flat Earth Theory
« Reply #193 on: March 19, 2014, 03:43:27 AM »
Here is another clear proof that the surface of the Earth is actually flat:

 
Grand Haven Daily Tribune   April 3, 1925

COAST GUARDS SEE MILWAUKEE LIGHTS GLEAM

Captain Wm. J. Preston and Crew See Lights of Milwaukee

and Racine Clearly From Surf Boat

ANSWER TO FLARE

Crew Runs Into Lake in Search For Flashing Torch

Grand Haven Daily Tribune   April 3, 1925

Captain Wm. J. Preston and his U. S. Coast Guard crew at Grand Haven harbor witnessed a strange natural phenomenon last night, when they saw clearly the lights of both Milwaukee and Racine, shining across the lake.  As far as known this is the first time that such a freak condition has prevailed here.

 The phenomena was first noticed at shortly after seven o’clock last night, when the lookout called the keeper’s attention to what seemed to be a light flaring out on the lake.  Captain Preston examined the light, and was of the impression that some ship out in the lake was “torching” for assistance.

Launch Power Boat

   He ordered the big power boat launched and with the crew started on a cruise into the lake to locate, if possible, the cause of the light.  The power boat was headed due west and after running a distance of six or seven miles the light became clearer, but seemed to be but little nearer.  The crew kept on going, however, and at a distance of about ten and twelve miles out, a beautiful panorama of light unfolded before the eyes of the coast guards.

 Captain Preston decided that the flare came from the government lighthouse at Windy Point at Racine.  Being familiar with the Racine lights the keeper was able to identify several of the short lights at Racine, Wis.

Saw Milwaukee Also

   A little further north another set of lights were plainly visible.  Captain Preston knowing the Milwaukee lights well, easily distinguished them and identified them as the Milwaukee lights.  The lights along Juneau Park water front, the illumination of the buildings near the park and the Northwestern Railway station were clearly visible from the Coast Guard boat.  So clearly did the lights stand out that it seemed as though the boat was within a few miles of Milwaukee harbor. 

   Convinced that the phenomenon was a mirage, or a condition due to some peculiarity of the atmosphere, the keeper ordered the boat back to the station.  The lights remained visible for the greater part of the run, and the flare of the Windy Point light house could be seen after the crew reached the station here.


DISTANCE GRAND HAVEN TO MILWAUKEE: OVER 80 MILES (128 KM).

http://www.coastwatch.msu.edu/images/twomichigans2a.gif


Windy Point Lighthouse:

http://upload.wikimedia.org/wikipedia/commons/thumb/5/5f/Wind_Point_Lighthouse_071104_edit2.jpg/800px-Wind_Point_Lighthouse_071104_edit2.jpg

The lighthouse stands 108 feet (33 m) tall

THE CURVATURE FOR 128 KM IS 321 METERS.

Using the well known formula for the visual obstacle, let us calculate its value:

h = 3 meters BD = 1163 METERS

h = 5 meters BD = 1129 METERS

h = 10 meters BD = 1068 METERS

h = 20 meters BD = 984 METERS

h = 50 meters BD = 827.6 METERS

h = 100 meters BD = 667.6 METERS


No terrestrial refraction formula/looming formula can account for this extraordinary proof that the surface across lake Michigan is flat.

In fact: http://ireland.iol.ie/~geniet/eng/refract.htm#

If we use h = 50 for the observer, and 140 for the distant object height, we get a negative answer: no way it could be seen over a 128 km distance; while the actual data for the account is h = 5 m, and d = 40 m.


Looming/modified lapse rate:

http://mintaka.sdsu.edu/GF/explain/atmos_refr/altitudes.html

The formula used here does not recognize the change in the range of temperature values, nor do we know if it takes into consideration the very basic formula I posted earlier for the visual obstacle: http://theflatearthsociety.org/forum/index.php?topic=28196.msg674444#msg674444 - however, it is an excellent place to start and to explore the effect of looming/ducting on the visual target being observed.

Let us use several values, starting with the value of 15 C for that day (Milwaukee/Racine/Holland/Grand Haven) and increasing the value for the target by 1-3 degrees.

For a value of 15 C overall we get of course a negative altitude value of the target.

For a value of 16 C (for the target) we get, again, a negative altitude value for the target (−0.317 degrees of arc) - target is hidden by horizon

For a value of 17 C (for the target) we get: −0.207 degrees of arc, target is hidden by horizon

For a value of 18 C (for the target) we get: −0.098 degrees of arc, target is hidden by horizon


Let us decrease the value to 12 C.

Increasing the value for the target to 15 C degrees, again, we get negative values. This would also correspond to a huge k = 0.613 value.

From the textbook on atmospheric science:

 "So the ray curvature for an arbitrary lapse rate  γ K/m will be

k  = ( 0.034 − γ ) / 0.154

where we take γ to be positive if the temperature decreases with height, and a positive curvature means a ray concave toward the Earth.

Example 1: the Standard Atmosphere:

In the Standard Atmosphere, the lapse rate is 6.5°/km or  γ = 0.0065 K/m. The numerator of the formula above becomes .034 − .0065 = .0275, so the ratio k is about 1/5.6 or 0.179. In other words, the ray curvature is not quite 18% that of the Earth; the radius of curvature of the ray is about 5.6 times the Earth's radius.

Example 2: free convection:

In free convection, the (adiabatic) lapse rate is about 10.6°/km or  γ = 0.0106 K/m. The numerator of the formula above becomes .034 − .0106 = .0234, so the ratio k is about 1/6.6 or 0.152. In other words, the ray curvature is about 15% that of the Earth; the radius of curvature of the ray is about 6.6 times the Earth's radius. This is close to the condition of the atmosphere near the ground in the middle of the day, when most surveying is done; the value calculated is close to the values found in practical survey work."


Moreover, as we have seen, the light from Windy Point was continuously observed, during the approach, and during the return to the station:

The power boat was headed due west and after running a distance of six or seven miles the light became clearer, but seemed to be but little nearer.  The crew kept on going, however, and at a distance of about ten and twelve miles out, a beautiful panorama of light unfolded before the eyes of the coast guards.

The keeper ordered the boat back to the station.  The lights remained visible for the greater part of the run, and the flare of the Windy Point light house could be seen after the crew reached the station here.



Now, the calculation for the most pronounced form of looming: ducting.

However, ducting requires the value for the ray curvature, k, to be greater than or equal to 1.

This amounts to at least a five degree difference in temperature.

With 10C in Grand Haven (or Holland) and 15C in Racine, we get k = 1.182.


For the very same geographical/hydrographical conditions, for the same latitude in question, for cities located on the opposite shores of Lake Michigan, it is absolutely impossible to have a five degree difference, at the very same instant of time - moreover, looming/ducting do not apply to the two cases presented here:

FURTHERMORE, as we have seen, the light from the lighthouse located in Racine was seen all of the time.

For the second case exemplifed here, see below, Mr. Kanis did see the very shape of the buildings: in the case of ducting/looming a very distorted image would appear making it instantly recognizable:

http://upload.wikimedia.org/wikipedia/commons/f/f4/Superopr_mirage_sequence.jpg
http://3sky.de/Div/Luftspieg/Summary.html
http://finland.fi/public/default.aspx?contentid=160069&contentlan=2&culture=en-US







'As twilight deepened, there were more and more lights.'

Bringing out a pair of binoculars, Kanis said he was able to make out the shape of some buildings.

'With the binoculars we could make out three different communities,' Kanis said.

According to one Coast Guard crewman, it is possible to see city lights across the lake at very specific times.

Currently a Coast Guard crewman stationed in Holland, Todd Reed has worked on the east side of Lake Michigan for 30 years and said he's been able to see lights across the lake at least a dozen times.

The highest building in Milwaukee has a height of 183 meters, the difference from h = 5 meters in altitude being 946 meters, and those residents saw the buildings from THREE DIFFERENT COMMUNITIES, two of which have buildings whose heights measure way under 183 meters.

Therefore, the only way those buildings could be seen, given the 128 km distance, would be if the surface of Lake Michigan is completely flat.

THE TALLEST BUILDING IN RACINE IS THE COUNTY COURTHOUSE, 40 METERS; IT WOULD BE ABSOLUTELY IMPOSSIBLE TO SEE THIS COURTHOUSE FROM 128 KM DISTANCE, FROM HOLLAND.


On Memorial Day, it was 60 F degrees (15 C) in Milwaukee on that day.


Black Sun photographs, the Moon and the Sun are disks and not spheres:


http://theflatearthsociety.org/forum/index.php?topic=58190.msg1490160#msg1490160

http://theflatearthsociety.org/forum/index.php?topic=58190.msg1490183#msg1490183 (solar/lunar ISS transit videos)

http://theflatearthsociety.org/forum/index.php?topic=58190.msg1490299#msg1490299 (Thierry Legault photographs)
« Last Edit: June 05, 2018, 06:33:45 AM by sandokhan »

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #194 on: March 31, 2014, 02:46:09 AM »
Earth is not a Globe, chapter IX, Cause of Sunrise and Sunset:

ALTHOUGH the sun is at all times above the earth's surface, it appears in the morning to ascend from the north-east to the noonday position, and thence to descend and disappear, or set, in the north-west. This phenomenon arises from the operation of a simple and everywhere visible law of perspective.

It does not only appear to ascend (rise), it actually DOES so.



http://www.moonglow.net/eclipse/2003nov23/


Official science information:

The sun "travels " between the Tropic of Cancer (23.5 N) on the Northern Solstice (first day of summer in the Northern Hemisphere) to the Tropic of Capricorn (23.5 S) on the Southern Solstice and back to the Tropic of Cancer in one year (365 days). Thus the sun travels 47 degrees of latitude * 2 (to compensate for the round trip) = 94 degrees of latitude during a solar year.


In the Flat Earth Theory, the diameter of the Sun is only some 600 meters - therefore, as it travels between the two tropics, rising and setting each and every day, there will be a limited amount of space/distance to account for the yearly solar precession phenomenon.

Due to precession, the Sun's apparent position relative to the vernal equinox slowly regresses about 50 arc seconds (0.014 degree) every year, or approximately one degree every 72.2 years. This results in the difference of 20 minutes between the sidereal and tropical years.

http://img585.imageshack.us/img585/8711/bun1copy.jpg

No matter whether the other FE will accept the six gates/180 windows theory or not (http://theflatearthsociety.org/forum/index.php?topic=30499.msg1566598#msg1566598 ) we can understand that there will be an upper bound for the distance/space alloted for the precession of the Sun, as it orbits above the flat Earth.

Works on precession:

http://www.aloha.net/~johnboy/Preces.htg/precession.htm

http://www.crystalinks.com/precession.html

This means that our entire history is only some hundreds of years old - confirming the comet tail paradox/new radical chronology proofs. Moreover, if the other FE are not happy with a 354 year old history, they can calculate themselves an upper bound: it cannot exceed some 500 years, because of the very distance between the two tropics.


It is not nearly enough to state that the surface of the Earth is flat: the shape of the Earth is part of a much larger question - has our entire history up to at least 1825 been forged/falsified?

Let us remember that currently the FE cannot answer at all the axial precession question (with my exception), as the RE can immediately state that the historical/astronomical records from Hipparchus up to B. Franklin are true, and thus prove that the Earth has been rotating around its own axis for at least the past 2000 years.

The solar precession is the best key to answer not only the question about the correct chronology of history, but also it provides a direct way to prove that the Earth is actually flat.

http://theflatearthsociety.org/forum/index.php?topic=30499.msg1488947#msg1488947

One of the most important canonical books of the orthodox church, in the official history, is the Syntagma by Matthew Vlastar.

Matthew Vlastar’s Collection of Rules Devised by Holy Fathers, or The Alphabet Syntagma. This rather voluminous book represents the rendition of the rules formulated by the Ecclesial and local Councils of the Orthodox Church.

Matthew Vlastar is considered to have been a Holy Hierarch from Thessalonica, and written his tractate in the XIV century. Today’s copies are of a much later date, of course. A large part of Vlastar’s Collection of Rules Devised by Holy Fathers contains the rules for celebrating Easter. Among other things, it says the following:


“The Easter Rules makes the two following restrictions: it should not be celebrated together with the Judaists, and it can only be celebrated after the spring equinox. Two more had to be added later, namely: celebrate after the first full moon after the equinox, but not any day – it should be celebrated on the first Sunday after the equinox. All of these restrictions, except for the last one, are still valid (in times of Matthew Vlastar – the XIV century – Auth.), although nowadays we often celebrate on the Sunday that comes later. Namely, we always count two days after the Lawful Easter (that is, the Passover, or the full moon – Auth.) and end up with the subsequent Sunday. This didn’t happen out of ignorance or lack of skill on the part of the Elders, but due to lunar motion”

Let us emphasize that the quoted Collection of Rules Devised by Holy Fathers is a canonical mediaeval clerical volume, which gives it all the more authority, since we know that up until the XVII century, the Orthodox Church was very meticulous about the immutability of canonical literature and kept the texts exactly the way they were; with any alteration a complicated and widely discussed issue that would not have passed unnoticed.

So, by approximately 1330 AD, when Vlastar wrote his account, the last condition of Easter was violated: if the first Sunday happened to be within two days after the full moon, the celebration of Easter was postponed until the next weekend. This change was necessary because of the difference between the real full moon and the one computed in the Easter Book. The error, of which Vlastar was aware, is twenty-four hours in 304 years.

Therefore the Easter Book must have been written around AD 722 (722 = 1330 - 2 x 304). Had Vlastar known of the Easter Book’s 325 AD canonization, he would have noticed the three-day gap that had accumulated between the dates of the computed and the real full moon in more than a thousand years. So he either was unaware of the Easter Book or knew the correct date when it was written, which could not be near 325 AD.


G. Nosovsky: So, why the astronomical context of the Paschalia contradicts Scaliger’s dating (alleged 325 AD) of the Nicaean Council where the Paschalia was canonized?

This contradiction can easily be seen from the roughest of calculations.

1) The difference between the Paschalian full moons and the real ones grows at the rate of one day in 300 years.

2) A two-day difference had accumulated by the time of Vlastar, which is roughly dated 1330 AD.

3) Ergo, the Paschalia was compiled somewhere around 730 AD, since

1330 – (300 x 2) = 730.

It is understood that the Paschalia could only be canonized by the Council sometime later. But this fails to correspond to Scaliger’s dating of its canonization as 325 AD in any way at all!

Let us emphasize, that Matthew Vlastar himself, doesn’t see any contradiction here, since he is apparently unaware of the Nicaean Council’s dating as the alleged year 325 AD. A natural hypothesis: this traditional dating was introduced much later than Vlastar’s age. Most probably, it was first calculated in Scaliger’s time.

With the Easter formula derived by C.F. Gauss in 1800, Nosovsky calculated the Julian dates of all spring full moons from the first century AD up to his own time and compared them with the Easter dates obtained from the Easter Book. He reached a surprising conclusion: three of the four conditions imposed by the First Council of Nicaea were violated until 784, whereas Vlastar had noted that “all the restrictions except the last one have been kept firmly until now.” When proposing the year 325, Scaliger had no way of detecting this fault, because in the sixteenth century the full-moon calculations for the distant past couldn’t be performed with precision.


Few scientists remember the original debate between Immanuel Velikovsky and Carl Sagan, where the creator of the original Cosmos series was debunked on each and every statement on cosmology made later in the series:

http://www.varchive.org/lec/aaas/transcripts.htm (original debate)

http://www.varchive.org/lec/aaas/afterword.htm (Velikovsky's final response and notes)
http://www.varchive.org/lec/aaas/challenge.htm
« Last Edit: April 03, 2014, 05:23:17 AM by sandokhan »

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #195 on: April 04, 2014, 06:20:26 AM »
Flat Earth Solar Precession Facts





The real orbit of the Sun above the Flat Earth:



Distance between Tropic of Cancer and Tropic of Capricorn: 3,234.64 miles (center of first map)

Arclength: 6356.621 KM (side of figure/map) - this is the absolute uppermost bound/limit

Polar axis length (official numbers):  6356.7519 km = actual radius of the flat earth map (http://img836.imageshack.us/img836/7857/africabrazil.gif )


Precession = a slow westward shift of the equinoxes along the plane of the ecliptic


RE maps of the solar precession:



The orange axis was the Earth's rotation axis 5000 years ago when it pointed to the star Thuban. The yellow axis, pointing to Polaris is the situation now. (official astronomical information)



The orange axis was the Earth's rotation axis 5000 years ago when it pointed to the star Thuban. The yellow axis, pointing to Polaris is the situation now.




Annual solar precession: 50 arcseconds

1 full degree every 72 years


TIME CALCULATION

The time difference between the mean solar day and "precession"(sidereal "what's real" time) is 3.141592654 (Pi) seconds (of time) per day. Pi seconds per day accumulates to 22.94884425 hours in 72 years or approximately 1 degree (day) of arc. (Precession is approximately 1 degree in 72 years.)

The 3 . 14 seconds is the mean deviation.  The actual deviation, over time, is from 3.11 seconds to 3.17 seconds daily. In other words, the TIME required for the Earth to complete one orbit around the Sun (a year) decreases daily by the rate of 3.14 seconds (of time) and accumulates, annually, to approximately 18 minutes (+-).

http://www.aloha.net/~johnboy/Preces.htg/precession.htm

Every (approx.) 72 years the annual 18 (+ -) minutes has accumulated to almost one full day, and has done so despite the addition of 'leap years', 'leap seconds', etc.  So, after 72 years we actually 'show up early' for the 'Vernal Equinox' and decide that the entire outer visible universe has 'mysteriously' backed-up one whole degree (day) in the last 72 years!  We call this 'backing up' The Precession of The Equinoxes.


π = 3.1416 = 2/sc per day

1146.68132 seconds per year

As I have argued before, the year really has 364 days (of a slightly different length).

1146.68132/364 = π + 1.361sc/100

For a 354 year old history, we obtain 404778.5054 seconds

404778.5054 = (1000sc)2

That is, the total precessional time for a 354 year old history amounts to exactly 1,000,000sc2 seconds or 636,619.7723sc seconds.


The reader can calculate a similar total precessional time of his/her own choosing (the choice rests upon the figure used for the total duration of history in years).


DEGREE/ARCLENGTH CALCULATION

Full arclength between the tropics: 6356.621 km

The calculation for the six gates/180 windows configuration has already been done (the figures used in the Bundahisn).

Now, the calculation for the six gates/12 window used in the Book of Enoch.

http://www.johnpratt.com/items/docs/enoch.html#Enoch_71

And in the fourth gate, through which the sun with the moon proceeds, in the first part of it, there are twelve open windows


Solar precession, 50 arcseconds per year

1 degree per 72 years

For a 354 year old history, we get 4.91666 degrees (0.02861 x 3 radians)

Total arclength of precession (using a radius of 6356.621 km) = (534 + 11.444) km


In the Flat Earth model, the effect of the precession is cumulative: that is, as the Sun moves between the summer/winter solstices, we add 4.233 meters for each day of the year.

For a 354 year old history, there will be 1.5408 km/year of arclength precessional distance.

1.5408/364 = 0.004233 km


For the 12 window configuration, there will be 2.5 days assigned for each window.

One gate arclength = 1060.86 km

One window arclength = 88.405 km (actually we could use 87.266 km and account for the difference by assuming that there is a certain distance between the gates themselves).

545.4 (total solar precessional arclength) km/6.18034 = 88.405 km

That is, in a 354 year old history, if we divide by approximately six, the total figure of 545.4 km, we get the exact arclength of a single window.


87.266/12 = 7.272166 (= the actual displacement factor the Gizeh Pyramid!)


The reader can choose a certain total duration of history and play with the figures to calculate the effect of the solar precession on any FE model.

He/she will discover that already for a 354 year old history, the total effect of the solar precession will account for exactly 1/2 the arclength of a single gate: thus we can see that our history is very short, not more than some 500 years old for its total duration.


Since the orbit of the Sun is bounded/limited between the two tropics, we can see that the solar precession can last only for a very precise amount of time (see the TIME CALCULATION section): in that very year where the solar precession will exceed the arclength of single window, it would mean that at either the summer or winter solstice, the Sun would rise BEYOND either the Tropic of Cancer or the Tropic of Capricorn, which is impossible.



« Last Edit: July 02, 2017, 04:35:40 AM by sandokhan »

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #196 on: April 11, 2014, 05:21:18 AM »
There is one remaining possibility for the flat earth solar precessional hypothesis: the tropics and the north/south poles move also, along with the equinoxial points.

For a very nice estimate concerning this case, I will make use of the information contained here:

http://www.bibliotecapleyades.net/tierra_hueca/tierrahueca/Chapter5.htm

http://www.bibliotecapleyades.net/tierra_hueca/tierrahueca/Chapter2.htm

"Starting at 70 to 75 degrees north and south latitude the Earth starts to curve IN. The Pole is simply the outer rim of a magnetic circle around the polar opening. The North Magnetic Pole, once thought to be a point in the Arctic Archipelago, has been lately shown by Soviet Arctic explorers to be a line approximately 1000 miles long. However, as we stated above, instead of being a straight line it is really a circular line constituting the rim of the polar opening. When an explorer reaches this rim, he has reached the North Magnetic Pole; and though the compass will always point to it after one passes it, it is really not the North Pole even if one is deluded into thinking it is, or that he discovered the Pole due to having been misled by his compass. When one reaches this magnetic circle (the rim of the polar opening), the magnetic needle of the compass points straight down. This has been observed by many Arctic explorers who, after reaching high latitudes, near to 90 degrees, were dumbfounded by the inexplicable action of the compass and its tendency to point vertically upward. (They were then inside the polar opening and the compass pointed to the Earth's North Magnetic Pole which was along the rim of this opening."

For a circumference of approximately 1600-1800 km, the diameter can be calculated (the reader can increase the circumference by actually following the Peary/Cook expeditions as they are described in chapter 5 of Hollow Earth)


The region the HE take as the entrance to the inner earth, is actually the region on the Flat Earth which cannot be accessed by either land, sea or air, as the expeditions of both Peary and Cook proved clearly: neither could discover the North Pole at all (in Antarctica, we have already seen how R. Scott was assasinated in an earlier discussion). The curvature paradox also applied equally well to the HE hypothesis: since there is no curvature at the surface of the Earth, the Hollow Earth cannot be true.


Antarctica Maps: http://www.bibliotecapleyades.net/esp_antartica.htm
« Last Edit: April 30, 2014, 05:50:22 AM by sandokhan »