The Great Pyramid: metamorphose of the architecture
Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Sat Aug 16, 2008 6:43 pm
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
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
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Mon Aug 25, 2008 9:21 pm
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
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
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Re: The Great Pyramid: metamorphose of the architecture
by Pot Noodle » Tue Aug 26, 2008 11:19 am
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).
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).
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by Vidusa » Wed Aug 27, 2008 5:09 pm
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=earth+40%2C077km&btnG=Google+Search&meta=
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=earth+40%2C077km&btnG=Google+Search&meta=
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Re: The Great Pyramid: metamorphose of the architecture
by Pot Noodle » Thu Aug 28, 2008 2:49 pm
To much sun metinks
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Fri Aug 29, 2008 3:32 pm
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
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
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Tue Sep 09, 2008 1:07 am
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
c) [/B[B]]AB = 35.4490625m = 55.82529528 sacred cubits:
412.1316378 : 55.82529528 = 7.382524996 = 14.76504999 : 2
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
c) [/B[B]]AB = 35.4490625m = 55.82529528 sacred cubits:
412.1316378 : 55.82529528 = 7.382524996 = 14.76504999 : 2
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Tue Sep 23, 2008 10:28 pm
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
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
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Re: The Great Pyramid: metamorphose of the architecture
by missarchi » Mon Sep 29, 2008 9:59 am
I knew there would be another pyramid scheme...
impei and co...
http://www.bdonline.co.uk/story.asp?sectioncode=725&storycode=3123600&c=2&encCode=000000000182eb6b
impei and co...
http://www.bdonline.co.uk/story.asp?sectioncode=725&storycode=3123600&c=2&encCode=000000000182eb6b
- missarchi
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Fri Oct 17, 2008 4:24 pm
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=X&oi=spell&resnum=0&ct=result&cd=1&q=Earth%27s+equator+40,077km&spell=1
http://www.google.ca/search?hl=en&q=Eart%27s+equator+40%2C077km&btnG=Google+Search&meta=
** Real length of the Pyramid's base ([B]PYRAMIDISTS VS PYRAMIDOLOGITS):[/B]
http://www.archiseek.com/content/showthread.php?t=5399&page=4
**1º = 111km:
http://eesc.columbia.edu/courses/ees/lithosphere/labs/lab1.html
.
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=X&oi=spell&resnum=0&ct=result&cd=1&q=Earth%27s+equator+40,077km&spell=1
http://www.google.ca/search?hl=en&q=Eart%27s+equator+40%2C077km&btnG=Google+Search&meta=
** Real length of the Pyramid's base ([B]PYRAMIDISTS VS PYRAMIDOLOGITS):[/B]
http://www.archiseek.com/content/showthread.php?t=5399&page=4
**1º = 111km:
http://eesc.columbia.edu/courses/ees/lithosphere/labs/lab1.html
.
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Sun Oct 19, 2008 1:46 am
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
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
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Tue Oct 21, 2008 8:27 pm
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
.
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
.
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Wed Oct 22, 2008 4:46 pm
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=washington%20DC&um=1&ie=UTF-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
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=washington%20DC&um=1&ie=UTF-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
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Vidusa - Member
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- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Wed Oct 22, 2008 11:26 pm
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/encyclopedia/Robert-F.-Kennedy-Memorial-Stadium
*http://maps.google.ca/maps?hl=en&q=washington%20DC%20map&um=1&ie=UTF-8&sa=N&tab=wl
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/encyclopedia/Robert-F.-Kennedy-Memorial-Stadium
*http://maps.google.ca/maps?hl=en&q=washington%20DC%20map&um=1&ie=UTF-8&sa=N&tab=wl
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Vidusa - Member
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- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Thu Oct 30, 2008 4:32 pm
Vidusa wrote: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/encyclopedia/Robert-F.-Kennedy-Memorial-Stadium
*http://maps.google.ca/maps?hl=en&q=washington%20DC%20map&um=1&ie=UTF-8&sa=N&tab=wl
Position 76.97387387º W (click on the maps)
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Pot Noodle » Thu Oct 30, 2008 5:08 pm
Vidusa wrote:
Position 76.97387387º W (click on the maps)
You really are barking
- Pot Noodle
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Fri Oct 31, 2008 11:55 pm
Pot Noodle wrote:You really are barking
Hi, brother!
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Sat Nov 08, 2008 4:23 pm
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.
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.
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Fri Nov 14, 2008 1:23 am
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
.
(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
.
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Wed Nov 19, 2008 6:16 pm
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
.
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
.
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Wed Nov 19, 2008 6:17 pm
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
-------------------------
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
-------------------------
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Sat Nov 22, 2008 8:13 pm
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
.
(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
.
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Tue Nov 25, 2008 1:07 am
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=sxxJAAAAMAAJ&pg=PA99&lpg=PA99&dq=james+Fergusson+great+pyramid&source=bl&ots=EQIfFzEwAN&sig=eJqTeSmPky-vaqtF7swQA0UWVOY&hl=en&sa=X&oi=book_result&resnum=1&ct=result
.
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=sxxJAAAAMAAJ&pg=PA99&lpg=PA99&dq=james+Fergusson+great+pyramid&source=bl&ots=EQIfFzEwAN&sig=eJqTeSmPky-vaqtF7swQA0UWVOY&hl=en&sa=X&oi=book_result&resnum=1&ct=result
.
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Vidusa - Member
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Re: The Great Pyramid: metamorphose of the architecture
by Vidusa » Fri Nov 28, 2008 1:00 am
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
.
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
.
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