The Great Pyramid: metamorphose of the architecture

[Vidusa]

ASCENDING PASSAGE AND THE EARTH


Figure 1

North Celestial Pole (NCP)= 90º
Angle of the Ascending passage = 26.3026897º = β

90º - 26.3026897º = 63.6973103º = the angle of the Ascending passage doesn't point to the star Thuban.

Thuban (Alpha Draconis) declination (DEC) = +64.37583333º

90º- 64.37583333º = 25.62416667º

90º - 26.3026897º = 63.6973103º = the angle of the Ascending passage doesn't point to the star Thuban.

The Ascending passage (Descending passage) of the Great Pyramid is perfectly aligned with the center of the Ecliptic Pole (EP).

Angle of the Ascending passage = 26.3026897º
Tangent = 0.494289195
Sinus = 0.443113275
Cosinus = 0.896465629

SINUS 26.3026897º = 0.443113275


Figure 2

Earth:
Equatorial radius = 6378.50168 km = d/2
Area of the circle A = 127,816,488.4 km2 = area of the square BC = 11,305.59509 km

6378.50168 x 0.443113275 = 2826.398773 km

2826.398773 x 4 = 11,305.59509 km = C (Figure 2).

Antechamber:
length = 116.2602377 inches

116.2602377 x 0.443113275 = 51.51645468 inches

51.51645468 x 4 = 206.0658187 inches = width of the King's Chamber.

The Great Pyramid:
height = 14,765.05019 cm

14,765.05019 x 0.443113275 = 6542.589745 cm

If a certain object was to travel with a speed of 6542.589745 cm in one second, for 24 hours it would travel a distance of 5652.79754 km:

5652.79754 x 2 = 11,305.59508 km = C (Figure 2).

The Great Pyramid:
slope angle = 51.85399754
tangent = 1.273240621


Figure 3

1.273240621 x 0.443113275 = 0.564189821 = C (Figure 3).

d = 0.636620309 = tangent of the angle of King's Chamber North channel = 1.273240621/2

[Vidusa]

GEOMETRY AND THE ASCENDING PASSAGE


Figure 1

BCD = 90º
Angle of the Ascending passage = 26.3026897º = β = BC-EP = to Ecliptic Pole
γ = 63.6973103º

tan β = 0.494289195
sin β = 0.443113275
cos β = 0.896465629


Figure 2

Height of the Great Pyramid (CD) = 147.6505019m = CEP

147.6505019 x 0.443113275 = 65.42589745m = CL (Figure 1) = C (Figure 2)

Area of the square B = Area of the circle A

Diameter (d) of the circle A = 73.82525085m = 147.6505019/2

[Pot Noodle]

Scientists exploring the Great Pyramid in Egypt sent a robot into the northern shaft in the past few days, discovering another blocking stone. The "door" appears to be identical to the one in the southern shaft that was already known. The doors are equidistant (65 meters/208 feet) from the queen's chamber. It is the third such block discovered within the shafts of the pyramid.



The announcement of the discovery was made Monday by Farouk Hosni, Egypt's minister of culture, and Zahi Hawass, secretary general of Egypt's Supreme Council of Antiquities and a National Geographic explorer-in-residence.

A specially developed combination of robotics, camera, and lighting technology developed by iRobot of Boston, yielded the new information. Until this discovery, no one knew that the northern shaft extended to the north as far as the southern shaft goes to the south.

Prior explorations of the northern shaft have failed because, unlike the southern shaft, the northern shaft has a number of bends and sharp corners. Hawass suggested that the layout of the northern shaft may have been designed to avoid intersection with the pyramid's grand gallery.

"This find in the northern shaft, coupled with last week's discovery of a second 'door' behind the blocking stone in the southern shaft, represents the first major new information about the Great Pyramid in more than a century. We will now carefully study the data and plan out further investigation of the two shafts in order to accurately map and interpret the find," Hawass said.

The newly discovered northern shaft door appears to be very similar to the one in the southern shaft, including the presence of a pair of copper "pins" or "handles." The southern shaft "door" was discovered in a 1993 investigation conducted under the auspices of the German Archaeological Institute.

On September 17, 2002, a National Geographic robot, specially designed to traverse the southern shaft to the blocking stone, inserted a miniature fiber-optic camera into a three-quarters-of-an-inch hole to reveal the rough-hewn blocking stone lying seven inches beyond the original southern shaft door. That earlier portion of the expedition was broadcast live in an international television event carried on National Geographic Channel and on Fox in the U.S.

"The mystery of the Great Pyramid becomes all the more compelling with each new discovery coming from the queen's chamber and the Supreme Council of Antiquities/National Geographic expedition," Terry Garcia, executive vice president of mission programs at the National Geographic Society said. "This continuation of our century-long involvement in archaeological breakthroughs in Egypt is an exciting extension of the National Geographic mission."

Portions of the northern shaft have been previously explored. In 1872 Waynman Dixon found a small bronze hook and granite ball. In the 1920s a pyramid enthusiast, Morton Edgar, attempted to learn more about the queen's chamber shafts by using flexible metal rods. In the southern shaft he was stopped, presumably by the blocking door. In the northern shaft, which appears to bend and curve around the grand gallery, Edgar's flexible rods broke and remain there to this day. The SCA/NG robot "rover" had to navigate around the metal rods to reach the end of the northern shaft.

In the course of the German Archaeological Institute's 1993 investigation, Rudolf Gantenbrink's robot traversed part of the shaft but only succeeded in covering 19 meters (63.3 feet).

[Vidusa]

THE SIGN OF THE SOLOMON'S TEMPLE


Thomas Ustick Walter: Solomon's temple*

Solomon’s temple (measurements in sacred cubits "cubit of the old standard"):

- height (portico) = 120
- length = 60
- height (temple) = 30
- breadth = 20

The numbers of the Temple are the Code:
- number 120
- number 60
- number 30
- number 20

Equatorial diameter of the Earth = 12,757.00336 km = 20,089,769.07 sacred cubits**:

20,089,769.07 : 120 = 167,414.7423 sacred cubits

167,414.7423 : 60 = 2790.245704 sacred cubits

2790.245704 : 30 = 93.00819014 sacred cubits

93.00819014 : 20 = 4.650409507 sacred cubits = 2 x 2.325204754 sacred cubits = 2 x 147.6505019 cm = 1/100 of the height of the Great Pyramid.

The question is: who was the architect of the Great Pyramid?
-------------------------
*Solomon's temple:
http://www.philaathenaeum.org/tuw/ancient.html

**Earth:
http://www.google.ca/search?hl=en&q=...e+Search&meta=

[Pot Noodle]

To much sun metinks

[Vidusa]

THE TEMPLE OF METROLOGY


Figure 1, The Great Pyramid and Temple (model).

On the east side of the Great Pyramid is the Temple of metrology with its basalt base, a rectangular building measuring about 171 feet from north to south and about 132 feet from east to west. Patches of the black basalt paved courtyard remain along with the sockets which held huge fifty granite pillars that formed a colonnade around the courtyard. From the Temple a causeway leads off in a straight line at an angle 14º North of due East:


Figure 2, The Temple of metrology

Basalt base (pavement) = basalt!? Why was basalt used?
Basalt = lava

The answer is here: when basaltic lava cools, it shrinks. In thick sheets of basaltic lava, this shrinking can produce shrinkage cracks that often occur in a hexagonal pattern and create hexagonal columns of rock, a process known as columnar jointing:


Figure 3, Basalt hexagonal patterns, Devils Postpile National Monument, California


Figure 4, Hexagon

Temple = 50 pillars (Figure 2)

Base of the Great Pyramid (four sides) = 927.71468 meters:

927.71468 : 50 = 18.5542936m = 1/100 of one minute (1') of Longitude.

Angle of the Temple's causeway = 14º
Tangent 14º = 0.249328002

(927.71468 : 2) : 0.249328002 = 1860.430181 m

1860.430181 : 4 = 465.1075454 = AB (Figure 6)

Hexagon = 6 triangles
1 triangle = 3 sides
6 triangles = 18 sides

465.1075454 : 18 = 25.83930808 = r (Figure 4) = radius of the circle A (Figure 5).


Figure 5

Figure 5:

Radius of the circle A = 25.83930808 m

Area of the circle A = 2097.544899 m² = area of the square B = 45.79896176² = 52.12661939 x 40.23941939 m = 171.0190925 x 132.0190925 feet = dimensions of the Temple of metrology.


Figure 6, April 27: ephemeris of the Sun +14º

Figure 6:

β = causeway's angle = 14º

AB = 465.1075454 m

BC = 115.964335 m = 1/2 of the Pyramid's base = position of the Temple = C.

April 26-28th*: ephemeris of the Sun = +14º

Gregorian calendar: April 27th = 116.2602377th day of a year = diameter of a year:


Figure 7, Diameter of the circle A = 116.2602377 days

Figure 7:

Circle A (circumference) = 365.242 days = 1 year

Diameter of the circle A = 116.2602377 days

Area of the circle A = area of the square B
Every side of the square B = 103.0329095 days

April 27 of the Gregorian calendar is April 14 of the Julian calendar = 103.0329095-th day of a year.


Figure 8, Diameter of the circle = 103.0329095 days

Figure 8:

d = 103.0329095 days

Area of the circle = area of the square

AB = 91.3105 days = BC = CD = DA = Seasons


Figure 9

Figure 9:

Radius (r) of the circle B = 103.0329095 days (1/2 of the King's Chamber width = 103.0329095 inches).

Area of the circle B = 33,350.42967 days = area of the square A

One side of the square A = 182.621 "days" = 1/2 of a year

Angle β = 32.48165854º = north "air-channel" of the King's Chamber (ascend).
Tangent 32.48165854º = 0.63662031:

103.0329095 x 0.63662031 = 65.59284278 days = radius of the circle C:

(2 x 65.59284278) x 3.14159 = 412.131638 days = circumference of the circle C = length of the King's Chamber (in inches).

412.131638 days = 1.128379644 years:

1.128379644² = 1.273240621 = tangent of the Pyramids angle of rise.

Circle C (area) = 13,516.44286 "days" = area of the square A1.

One side of the square A1 = 116.2602377 days

Earth's equatorial diameter = 12.757.00336 (km):

(12.757.00336 : 10) - 465.1075454 = 810.5927906 = length of the Causeway (in meters):

"The total length of the causeway from the east face of the Great Pyramid to the site of the lower temple is not less than 810m"**
---------------------------------

*Ephemeris for the Sun: http://eclipse.gsfc.nasa.gov/TYPE/TYPE.html

**Length of the causeway: http://www.zahihawass.com/egyptian_hist_dev_khufu.htm

[Vidusa]

THE GREAT PYRAMID: TEMPLE OF KNOWLEDGE


The Great Pyramid and the obliquity of the ecliptic.

Pi = 3.14159
10Pi = 31.4159 = one side of the square F

Area of the square F = 986.9587728m2 = area of the circle E

Radius (r) of the circle E = 17.72453102m
Diameter of the circle E = 35.44906205m = ABBase of the Great Pyramid = 231.92867m = BC
1/2 of the Pyramid's base =115.964335m = CD

BC + AB = 267.3777321m

115.964335 : 267.3777321 = 0.433709771 = tangent of 23.44684885º = β = obliquity of the ecliptic.

King's Chamber:

- length = 412.1316378 (inches)

- width = 206.0658189 (inches)

a) AB = 35.4490625 m:

412.1316378 : 35.4490625 = 11.62602362 (x 31.4159 = 365.2419953)

206.0658189 : 35.4490625 = 5.813011808 (the Pyramid's height = 5813.011885 inches).

b) AB = 35.4490625m = 1395.632382 inches:

412.1316378 : 1395.632382 = 0.295300999 = 2 x 0.147650499 (the Pyramid's height = 0.1476505019km).

206.0658189 : 1395.632382 = 0.14765050499

[b]c) [/B]AB = 35.4490625m = 55.82529528 sacred cubits:

412.1316378 : 55.82529528 = 7.382524996 = 14.76504999 : 2

[Vidusa]

QUESTION WITHOUT ANSWER





Hot molten basalt lava in the Temple of metrology was placed on the limestone surface. When this lava cooled down it was cut into blocks.

On the above pictures it perfectly shows the connection between the basalt lava and the limstone*. Are there any egiptologysts, arheologists or architects who will explain all these methods of construction? No, they are quiet and mute because they do not have the answer. When all of them are silent nobody else will even start to think about all this**.
--------------------------------------
*Photo: http://www.ancient-wisdom.co.uk/egyptghizapage.htm

http://www.gizapower.com/pma/index.htm

** Basalt pavement (connection between the basalt lava and the limstone, photos): http://www.solomonseries.com/freedownloads1.htm

[missarchi]

I knew there would be another pyramid scheme...
impei and co...

http://www.bdonline.co.uk/story.asp?...0000000182eb6b

[Vidusa]

WHY GREENWICH?

The Greenwich Meridian, based at the Royal Observatory, Greenwich, was established by Sir George Airy in 1851.


Earth's Prime Meridian and the Pyramid's Meridian (blue line).


Measurements:

Earth's Equator = 40,077.27418km*

Base of the Great Pyramid (one side) = 231.92867m**

Greenwich Prime Meridian - Pyramid's Meridian = 31.13513514º E (31º 8' 6.486504" E).

1º = 111 km***:

31.13513514º = 3456km

40,077.27418 : 3456 = 11.5964335km

Base of the Great Pyramid (one side) = 231.92867m = 2 x 115.964335m

-----------------------
*Equator:

http://www.google.ca/search?hl=en&sa...,077km&spell=1

http://www.google.ca/search?hl=en&q=...e+Search&meta=

** Real length of the Pyramid's base (PYRAMIDISTS VS PYRAMIDOLOGITS):

http://www.archiseek.com/content/sho...?t=5399&page=4

**1º = 111km:

http://eesc.columbia.edu/courses/ees...labs/lab1.html
.

[Vidusa]

WHY GREENWICH?

Length od the Pyramid's base:

W. M. Flinders Petrie: Socket Sides:
....................Length in inches:
N......................9129.8
E......................9130.8
S......................9123.9
W.....................9119.2
(The Pyramids and Temples of Gizeh by W. M. Flinders Petrie, p. 38:
http://www.ronaldbirdsall.com/gizeh/petrie/c6.html )

Pyramidologists: original architectural dimension of the Gr. Pyramid's socket sides: 9131.5 inches:

9131.05 - 9130.8 = 0.25 inches = 0.635cm

[Vidusa]

WHY GREENWICH?


Earth's Prime Meridian and the Pyramid's Meridian (blue line).

Earth's Equator = 12,757.00336 km

Greenwich Meridian - Pyramid's Meridian = 31.13513514º = 3456 km

12,757.00336 : 3456 = 3.691262546 km

3.691262546 x 4 = 14.76505019 km

Height of the Great Pyramid = 0.1476505019 km
.

[Vidusa]

4320 km


House of the Temple, Washington DC

House of the Temple, 1733 16th - S St. NW, Washington, DC (the map: http://maps.google.ca/maps?hl=en&q=w...-8&sa=N&tab=wl )
Located in Dupont Circle, the House of the Temple is considered one of the most beautiful monuments in the city:
Latitude: 77.036538º W
Longitude: 38.91891892º N

1º = 111 km

38.91891892º = 4320 km

Earth's Equator = 40,077.27418 km

40,077.27418 : 4320 = 9.277146801 km

9.277146801 : 4 = 2.3192867 km

Base of the Great Pyramid (one side) = 0.23192867 km

[Vidusa]

POSITION OF THE R.F.KENNEDY MEMORIAL STADIUM


R.F. Kennedy Stadium, Washington DC (the white circle right)*

Position**:
38.889897º N
76.97387387º W

Circle = 360º

360º - 76.97387387º = 283.0261261º = 31,415.9km= 10,000Pi

-------------------------------------------------------
* Field dimensions:
Left Field: 335 ft (102 m)

Left-Center: 380 ft (116 m)

Center Field: 410 ft (125 m)

Right-Center: 380 ft (116 m) : 116,2602377 x 3.14159 = 365.242

Right Field: 335 ft (102 m)

http://www.nationmaster.com/encyclop...morial-Stadium

*http://maps.google.ca/maps?hl=en&q=w...-8&sa=N&tab=wl

[Vidusa]

Originally Posted by Vidusa

POSITION OF THE R.F.KENNEDY MEMORIAL STADIUM


R.F. Kennedy Stadium, Washington DC (the white circle right)*

Position**:
38.889897º N
76.97387387º W

Circle = 360º

360º - 76.97387387º = 283.0261261º = 31,415.9km= 10,000Pi

-------------------------------------------------------
* Field dimensions:
Left Field: 335 ft (102 m)

Left-Center: 380 ft (116 m)

Center Field: 410 ft (125 m)

Right-Center: 380 ft (116 m) : 116,2602377 x 3.14159 = 365.242

Right Field: 335 ft (102 m)

http://www.nationmaster.com/encyclop...morial-Stadium

*http://maps.google.ca/maps?hl=en&q=w...-8&sa=N&tab=wl

Position 76.97387387º W (click on the maps)

[Pot Noodle]

Originally Posted by Vidusa

Position 76.97387387º W (click on the maps)

You really are barking

[Vidusa]

[Vidusa]

Originally Posted by Pot Noodle

You really are barking

Hi, brother!

[Vidusa]

WHY GREENWICH (III)


Earth's Prime Meridian and the Pyramid's Meridian (blue line).

Greenwich Meridian - Pyramid's Meridian = 31.13513514º = 3456 km

Pi = 3.14159

100Pi = 314.14159

3456 : 314.159 = 11.00079896 km

1 day = 86,400" (seconds)

11.00079896km = 11,000.79896 meters = 110,007.9896 decimeters (dcm)

110,007.9896dcm/86,400" = 1.273249621 = tangent of the Pyramid's angle of rise.

[Vidusa]

THE SOURCE OF THE GREAT PYRAMID
(Geometry's law)



Circumference of the Circle A = 3.14159 = infinity

Diameter of the Circle A = 1 = a single entity, first

Area of the Circle A = 0.7853975squared units* = Area of the Square B

One side of the Square B = 0.886226551 = C

Pyramid's angle (ascent) = 51.85399754º
Tangent 51.85399754º = 1.273240621

0.7853975 x 1.273240621 = 1
----------------------------------------
*0.7853975 x 4 = 3.14159
.

[Vidusa]

SQUARING THE CIRCLE (I)


Figure I


Figure II

CD = 0.5
CB = 0.564189822
Scale, proportion CD-CB* = 1.128379644 = √1.273240621 = tangent of the Great Pyramid's angle of arise (51.85399754˚).


Figure III

Angle α = 26.302689˚ = angle of the Pyramid's passages.
CE = 1.141416458
BE = 1.273240621 = tangent of the Pyramid's angle.


Figure IV

AB = 1.128379644 = √1.273240621


Figure V

CD = 0.5
DF = 1


Figure VI

GI = 1


Figure VII

IH = 1


Figure VIII

HJ = 1


Figure IX

JG = 1


Figure X

Diameter of the circle K = 1.128379644
Area of the circle K = 1

One side of the square L = 1
Area of the square L = 1

Area of the circle K = area of the square L

-------------------------------------------------------------
* CD = 0.5
CB = 0.5 x 1.128379644 = 0.564189822 = r
.

[Vidusa]

SQUARING THE CIRCLE (II)


Figure XI

Angle β = 51.85399754˚
Tangent 51.85399754˚ = 1.273240621

MF = 0.785395 = 1/4Pi


Figure XII

MN = 1.570795 = 1/2Pi


Figure XIII

Angle γ = 32.48165854˚ = angle of the north "air" channel of the King's Chamber.
Tangent 32.48165854˚ = 0.63662031 = 1.273240621/2

Angle δ = 38.14600246˚
Tangent 38.14600246˚ = 0.785395 = 1/4Pi

-------------------------

[Vidusa]

SQUARING OF THE CIRCLE
(conclusion)



δ = 26.3026897˚ = angle of the Pyramid's passages.
Tangent = 0.494289195
Sinus = 0.443113275

β = 51.85399754˚ = Gr. Pyramid's angle.
Tangent = 1.273240621
Sinus = 0.786439353

γ = 32.48165854˚ = angle of the north "air" channel of the King's Chamber.
Tangent = 0.63662031 = 1.273240621/2

α = 38.14600246˚
Tangent = 0.7853975 = Pi/4

ω = 48.4517858˚
Tangent = 1.128379644 = ω = √1.273240621

ζ = 41.5482142˚
Tangent = 0.886226551 = √0.7853975 = √Pi/4

HP = 0.443113275 = sinus 26.3026897˚

CB = 0.564189822 = radius of the Circle K = r

0.443113275 : 0.564189822 = 0.7853975 = Pi/4

CD = 0.5 = a/2

Constant for squaring the circle: ω (omega)

ω = 1.128379644

r = a/2 x ω = 0.5 x 1.128379644

a/2 = r/ω = r/1.128379644

.

[Vidusa]

PERFECTION OF THE GEOMETRY



γ = 32.48165854˚ = angle of the north "air" channel of the King's Chamber.
Tangent 32.48165854˚ = 0.63662031 = 1.273240621/2

β = 51.85399754˚ = Gr. Pyramid's angle
Tangent 51.85399754˚ = 1.273240621
Sinus 51.85399754˚ = 0.786439353

AB = dijameter of the circle K = 1.128379644 = √1.273240621

1.273240621 years = 412.1316378 days = length of the King's Chamber (in inches)

GF = 1/2 GI = 0.5

δ = angle of the Pyramid's passages = 26.3026897˚
Sinus 26.3026897˚ = 0.443113275

0.5 : 1.128379644 = 0.443113275

0.5 x 1.1283279644 = 0.564189822 = radius CB

0.564189822 : 0.443113275 = 1.273240621

√3.14159 = 1.772453102

1.772453102 : 0.443113275 = 4

4 : 3.14159 = 1.27324062

1,772453102 : 1.273240621 = 1.392080234

1.392080234 : 3.14159 = 0.443113275

James Fergusson, in his great work, the History of Architecture, describes the Great Pyramid as
"the most perfect and gigantic specimen of masonry that the world has yet sin." (James Fergusson, History of Architecture)

http://books.google.ca/books?id=sxxJ...um=1&ct=result

.

[Vidusa]

THE HARMONIC NUMBERS OF CREATION



Circle = 360º

Length of the Earth's equator = 40,077.27418 km

Equatorial diameter of the Earth = 12,757.00336 km

Esoteric code number = 4320

Code numbers of the Solomon's Temple (in cubits):
120 (Portal's height)
..60 (length of the Temple)
..30 (height of the Temple)
..20 (width of the Temple)

Height of the Great Pyramid = 147.6505019 = 0.1476505019 km

2 x 0.1476505019 = 0.2953010038 km

Number Pi = 3.14159

1 meter (m) = 10 decimeters (dcm)

Code measurement = 3.14159 dcm (decimeter)

3.14159 x 120 x60 x 30 x 20 = 13,571,668.8 dcm = 1357,166.88 m = 1357.16688 km

1357.16688 : 4320 = 0.314159

In order to travel the distance of 1357.16688 km in one day (24 hours), a certain object would have to travel at a speed of 15.70795 m/sec. = 5Pi.

40,077.27418 : 1357.16688 = 29.53010037 km

12,757.00336 : 29.53010037 = 432.0

1º of the equatorial latitude = 40,077.27418 : 360º = 111.3257616 km

1' of the equatorial latitude = 1.85542936 km

29.53010037 km on the Equator = 0.265258462º = 15' 54.95967683" = Sun's parallax and Semi-diameter of the Sun on the sky (summer, July)*.

360º : 1357.16688 = 0.2652585462º = 29.53010036 km
.