The Great Pyramid: metamorphose of the architecture
The Great Pyramid: metamorphose of the architecture
METAMORPHOSE OF THE PYRAMID'S ARCHITECTURE
Angle C-B - E: 26,3026897°
A-B = 365,242 Sacred Cubits (SC)
C-D = 232,5204754 SC
H-K = 206,065819 SC
Radius of the circle: 116,2602377 SC
a) Volume of the sphere (radius 116, 2602377 SC): 6.582.363,505 cubic Sacred Cubits,
Volume of the Gr. pyramid: 10.339.543,67 cub. Sacred Cubits:
10.339.543,67 : 6.582.363,505 = 1,570795 ( 1/2 Pi )
Sacred Cubit
The English have a measurement of length, which they call skein : 109,728 meters.
If a man was walk one day, with his every step 1 Sacred Cubit (63,5 cm) and with the speed of two steps in a second (2 Sacred Cubits or 127 cm), in 24 hours he would travel 109,728 km (1.000 skeins). For one year of 365,242 days, he would travel around the planet Earth. That is, he would travel the length of the Equator: 40.077,27418 km:
The original ancient base-breadth was 760,9208333 present feet = 9.131,05 inches = 23.192,867 centimeters = 365,242 Sacred Cubits. The differences between the originally and the present measurements were caused by the earthquakes and by the meteorological reasons: the Sun’s heat and the night’s coldness: the most sun exposed south side of the present Pyramid’s base is the longest side, and the north side, mostly in a shadow, is shortest.
THE GREAT PYRAMID AND THE EARTH
One side of the Great Pyramid's base is 23.192,867 cm long. Two sides together have 46.385,734 cm: this is the length of speed of Earth's turning on the Equator in one second. For the amount of time of one minute one point on the Equator moves by 2.783.144,04 cm or 27,8314404 km. For one hour this is the length of 1.669,886424 km. For the amount of 24 hours this is 40.077,27418 km. This is, according to the Great Pyramid, the length of Earth's Equator:
The height of the Great Pyramid is 14.765,05019 cm.
If a certain object was to travel on with a speed of 14.765,05019 cm in a second, it would travel a distance of 53.154.180,68 cm in one hour. For the amount of time of one day this distance would be 12.757,00336 km. This, according to the Great Pyramid, is the length of Earth's Equator.
14,765,05019 x 3,14159 = 46.385,734 cm = two sides of the Pyramid’s base.
Angle C-B - E: 26,3026897°
A-B = 365,242 Sacred Cubits (SC)
C-D = 232,5204754 SC
H-K = 206,065819 SC
Radius of the circle: 116,2602377 SC
a) Volume of the sphere (radius 116, 2602377 SC): 6.582.363,505 cubic Sacred Cubits,
Volume of the Gr. pyramid: 10.339.543,67 cub. Sacred Cubits:
10.339.543,67 : 6.582.363,505 = 1,570795 ( 1/2 Pi )
Sacred Cubit
The English have a measurement of length, which they call skein : 109,728 meters.
If a man was walk one day, with his every step 1 Sacred Cubit (63,5 cm) and with the speed of two steps in a second (2 Sacred Cubits or 127 cm), in 24 hours he would travel 109,728 km (1.000 skeins). For one year of 365,242 days, he would travel around the planet Earth. That is, he would travel the length of the Equator: 40.077,27418 km:
The original ancient base-breadth was 760,9208333 present feet = 9.131,05 inches = 23.192,867 centimeters = 365,242 Sacred Cubits. The differences between the originally and the present measurements were caused by the earthquakes and by the meteorological reasons: the Sun’s heat and the night’s coldness: the most sun exposed south side of the present Pyramid’s base is the longest side, and the north side, mostly in a shadow, is shortest.
THE GREAT PYRAMID AND THE EARTH
One side of the Great Pyramid's base is 23.192,867 cm long. Two sides together have 46.385,734 cm: this is the length of speed of Earth's turning on the Equator in one second. For the amount of time of one minute one point on the Equator moves by 2.783.144,04 cm or 27,8314404 km. For one hour this is the length of 1.669,886424 km. For the amount of 24 hours this is 40.077,27418 km. This is, according to the Great Pyramid, the length of Earth's Equator:
The height of the Great Pyramid is 14.765,05019 cm.
If a certain object was to travel on with a speed of 14.765,05019 cm in a second, it would travel a distance of 53.154.180,68 cm in one hour. For the amount of time of one day this distance would be 12.757,00336 km. This, according to the Great Pyramid, is the length of Earth's Equator.
14,765,05019 x 3,14159 = 46.385,734 cm = two sides of the Pyramid’s base.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
THE GREAT PYRAMID AND EARTH'S MEASUREMENTS
- [B]β [/b] Pyramid's angle = 51,85399754˚ (tangent = 1,273240621).
- 45˚ = the angle of the south shannel of the King's Chamber.
- 32,48165854˚ = the angle of the north channel of the King's Chamber (tangent = 0,63662031).
Circle G = Earth
C-E = Earth's radius = 6.378,501681 km:
6.378,501681 x 0,63662031 = 4.060,6837 km = E-L
4.060,6837 x 3,14159 = 12.757,00336 km = Eart's equatorial diameter = W-E = A-B =B-D =D-F = F=A
A-B =B-D =D-F = F-A = 151.028,01344 km = circle R
Radius of the circle R = 8.121,367435 km = C-N:
8.121,367435 : 1,273240621 = 6.378,501681 km = W-C = Earth's radius.
If a certain object was to travel with a speed of 2 Sacred Cubits ( 50 inches = 127 cm) in a second, for 24 hours (1 day) it would travel a distance of 109,728 km = 1.000 skeins:
51.028,01344 : 109,728 = 465,0409507 days = 1,273240621 years = tangent β
Great Pyramid's base = 365,242 Sacred Cubits (SC):
365,242 x 8 = 2.921,936 SC = 185.542,936 cm = 1,85542936 km = 1 ' of the Earth's length on the Equato: 1˚ = 60' = 111,3257616 km.
- [B]β [/b] Pyramid's angle = 51,85399754˚ (tangent = 1,273240621).
- 45˚ = the angle of the south shannel of the King's Chamber.
- 32,48165854˚ = the angle of the north channel of the King's Chamber (tangent = 0,63662031).
Circle G = Earth
C-E = Earth's radius = 6.378,501681 km:
6.378,501681 x 0,63662031 = 4.060,6837 km = E-L
4.060,6837 x 3,14159 = 12.757,00336 km = Eart's equatorial diameter = W-E = A-B =B-D =D-F = F=A
A-B =B-D =D-F = F-A = 151.028,01344 km = circle R
Radius of the circle R = 8.121,367435 km = C-N:
8.121,367435 : 1,273240621 = 6.378,501681 km = W-C = Earth's radius.
If a certain object was to travel with a speed of 2 Sacred Cubits ( 50 inches = 127 cm) in a second, for 24 hours (1 day) it would travel a distance of 109,728 km = 1.000 skeins:
51.028,01344 : 109,728 = 465,0409507 days = 1,273240621 years = tangent β
Great Pyramid's base = 365,242 Sacred Cubits (SC):
365,242 x 8 = 2.921,936 SC = 185.542,936 cm = 1,85542936 km = 1 ' of the Earth's length on the Equato: 1˚ = 60' = 111,3257616 km.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
ANGLE OF THE NORTH CHANNEL OF THE KING'S CHAMBER
The angle ( α ) of the Northern Channel of the King's Chamber and the Pyramid's proporcion
The ascending direction of angle R-C (N-C) to the base of the Pyramid (alpha, α) is the angle of north channel of King's Chamber = 32, 48165854˚.
Length in Sacred Cubits:
R-S (N-S) = 182,621
S-C (C-A) = 116,2602377
V-X (Z-Y) = 91,3105
P-O-T-U-P = 597,7624754
The angle ( α ) of the Northern Channel of the King's Chamber and the Pyramid's proporcion
The ascending direction of angle R-C (N-C) to the base of the Pyramid (alpha, α) is the angle of north channel of King's Chamber = 32, 48165854˚.
Length in Sacred Cubits:
R-S (N-S) = 182,621
S-C (C-A) = 116,2602377
V-X (Z-Y) = 91,3105
P-O-T-U-P = 597,7624754
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
THE PYRAMID AND THE N. CHANNEL OF THE KING'S CHAMBER (II)
β = 32,48165854˚ (North channel of the King's chamber)
tang. β = 0,63662031
sin. β = 0,537029596
E-N = 91,3105 Sacred Cubits (SC) = E-K
r = 58,13011882 (SC) = 1453,25297 inches = 3691,262545 cm
Circle N = 0,115964335 km = circle M
If a certain object was to travel with a speed of 0,115964335 km/sec., for 24 hours (1 day) it would travel a distance of 10.019,31854 km = 1/4 of the Earth's Equator.
Earth's equatorial radius = 6378,501681 km:
6378,501681 x 0,63662031 (tang. β) = 4060,683721 km:
4060,683721 km x 3,14159 = 12757,00336 km =Eart's diameter.
(4060,683721 x 4) x 3,14159 = 51.028,01348 km
The great Pyramid - Greenwich meridian ( distance) = 31,13513514˚
1˚ on the Earth's curved surface = 111 km: 31,13513514˚ = 3456 km:
51.028,01348 : 3456 = 14,7650502 km; the height of the Great pyramid = 0,1476505019 km.
γ = 57,51834146 ( tang. γ = 1,570795 = 1/2 Pi
A-B = 0,185542936 km = 10-th part of 1' (minute) on the curved Earth's surface .
β = 32,48165854˚ (North channel of the King's chamber)
tang. β = 0,63662031
sin. β = 0,537029596
E-N = 91,3105 Sacred Cubits (SC) = E-K
r = 58,13011882 (SC) = 1453,25297 inches = 3691,262545 cm
Circle N = 0,115964335 km = circle M
If a certain object was to travel with a speed of 0,115964335 km/sec., for 24 hours (1 day) it would travel a distance of 10.019,31854 km = 1/4 of the Earth's Equator.
Earth's equatorial radius = 6378,501681 km:
6378,501681 x 0,63662031 (tang. β) = 4060,683721 km:
4060,683721 km x 3,14159 = 12757,00336 km =Eart's diameter.
(4060,683721 x 4) x 3,14159 = 51.028,01348 km
The great Pyramid - Greenwich meridian ( distance) = 31,13513514˚
1˚ on the Earth's curved surface = 111 km: 31,13513514˚ = 3456 km:
51.028,01348 : 3456 = 14,7650502 km; the height of the Great pyramid = 0,1476505019 km.
γ = 57,51834146 ( tang. γ = 1,570795 = 1/2 Pi
A-B = 0,185542936 km = 10-th part of 1' (minute) on the curved Earth's surface .
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
GEOMETRY OF THE KING'S CHAMBER
King's Chamber:
- length = 412.1316378 inches
- width = 206.0658189 "
- height = 230.3800057 "
The Chamber's width = 206.0658189 inches = R
Circle C = 33.350.42964 squared inches = square S
One side of the square S = 182.621 inches = two sides of the G. Pyramid's base.
If a certain object was to trawel with a speed of 182.621 inches in one seconnd. for 24 hours (1 day) it would travel a distance of 15.778.454.4 inches = 100-th part of the Earth's Equator.
King's Chamber:
- length = 412.1316378 inches
- width = 206.0658189 "
- height = 230.3800057 "
The Chamber's width = 206.0658189 inches = R
Circle C = 33.350.42964 squared inches = square S
One side of the square S = 182.621 inches = two sides of the G. Pyramid's base.
If a certain object was to trawel with a speed of 182.621 inches in one seconnd. for 24 hours (1 day) it would travel a distance of 15.778.454.4 inches = 100-th part of the Earth's Equator.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
KING'S CHAMBER
King's Chamber (plan)
King's Chamber:
- length = 412.1316378 inches
- width = 206.0658189 "
- height = 230.3800057 "
Te volume of King's Chamber is 19.565.308.45 cubic inches = 320.618.1934 liters.
The ceiling is constructed with nine granite blocks (stones). The two last blocks in the Chamber are only 1/2 of the size of other blocks: 1/2 + 1/2 = 1
Therefore, there are "eight" granite blocks in the ceiling. This signals us that the volume of the King's Chamber needs to be divided in eight parts:
320.618.1934 : 8 = 40.077.27418 = the length of the Earth's Equator (1 liter = 1 km).
King's Chamber (plan)
King's Chamber:
- length = 412.1316378 inches
- width = 206.0658189 "
- height = 230.3800057 "
Te volume of King's Chamber is 19.565.308.45 cubic inches = 320.618.1934 liters.
The ceiling is constructed with nine granite blocks (stones). The two last blocks in the Chamber are only 1/2 of the size of other blocks: 1/2 + 1/2 = 1
Therefore, there are "eight" granite blocks in the ceiling. This signals us that the volume of the King's Chamber needs to be divided in eight parts:
320.618.1934 : 8 = 40.077.27418 = the length of the Earth's Equator (1 liter = 1 km).

-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
SUN – EARTH
Sun and Earth
149.597.870 = in kilometers the Sun’s mean distance from the Earth = 1 Astronomical unit (1 AU).
Earth’s orbit = 939.950.345 km
Earth’s equatorial diameter = 12.757,00336 km:
939.950.345 : 12.757,00336 = 73.681,12389 km
Mean solar tropical year = 365,242 days ( 1 mean day = 24 hours):
a) 73.681,12389 : 365,242 = 201,7323415 km.
The height of the Great Pyramid = 0,1476505019 km:
201,7323415 x 0,1476505019 = 29,78588148 km/sec. = Earth’s orbital velocity.
12.757,00336 : 365,242 = 34,9275367 km: if a certain object was to travel with a speed of 40,4253897 cm in one second, for 24 hours (1 day) it would travel a distance 34,9275367 km:
40,4253897 cm = 0,63662031 Sacred Cubits (tangent of the angle of the King’s Chamber North channel).
Sun and Earth
149.597.870 = in kilometers the Sun’s mean distance from the Earth = 1 Astronomical unit (1 AU).
Earth’s orbit = 939.950.345 km
Earth’s equatorial diameter = 12.757,00336 km:
939.950.345 : 12.757,00336 = 73.681,12389 km
Mean solar tropical year = 365,242 days ( 1 mean day = 24 hours):
a) 73.681,12389 : 365,242 = 201,7323415 km.
The height of the Great Pyramid = 0,1476505019 km:
201,7323415 x 0,1476505019 = 29,78588148 km/sec. = Earth’s orbital velocity.
12.757,00336 : 365,242 = 34,9275367 km: if a certain object was to travel with a speed of 40,4253897 cm in one second, for 24 hours (1 day) it would travel a distance 34,9275367 km:
40,4253897 cm = 0,63662031 Sacred Cubits (tangent of the angle of the King’s Chamber North channel).
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
THE GREAT PYRAMID AND THE SATELLITE MAP OF EARTH
The Grat Pyramid and the Earth's satellite map
Earth's Equator = 40.077,27418 km.
Equatorial diameter = 12.757,00336 km
Pyramid's angle: 51,85399754 degrees (tangent = 1,273240621)
12.757,00336 : 1,273240621 = 10.019.31854 km = 1/2 of the Pyramid's base or 1/4 of the Equator.
The Grat Pyramid and the Earth's satellite map
Earth's Equator = 40.077,27418 km.
Equatorial diameter = 12.757,00336 km
Pyramid's angle: 51,85399754 degrees (tangent = 1,273240621)
12.757,00336 : 1,273240621 = 10.019.31854 km = 1/2 of the Pyramid's base or 1/4 of the Equator.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
GEOMETRI OF THE KING'S CHAMBER (II)
King's Chamber:
- length = 412,1316378 inches
- width (R) = 206,0658189 inches
Circle C = 33.350.42964 squared inches = square S
One side of the square S = 182,621 inches.
1 sacred cubit = 25 inches: if a certain object was travel with a speed of 25 inches/sec.,for 24 hours (1 day) it would travel a distance of 2.160.000 inches.
Equatorial diameter = 12.757,00336 km = 502.244.226,8 inches:
502.244.226,8 : 2.160.000 = 232,5204754 inches;
- one side of the square S = 182,621 inches;
- tangent of the Pyramid's angle ascend = 1.273240621
182,621 x 1,273240621 = 232,5204754 inches.
King's Chamber:
- length = 412,1316378 inches
- width (R) = 206,0658189 inches
Circle C = 33.350.42964 squared inches = square S
One side of the square S = 182,621 inches.
1 sacred cubit = 25 inches: if a certain object was travel with a speed of 25 inches/sec.,for 24 hours (1 day) it would travel a distance of 2.160.000 inches.
Equatorial diameter = 12.757,00336 km = 502.244.226,8 inches:
502.244.226,8 : 2.160.000 = 232,5204754 inches;
- one side of the square S = 182,621 inches;
- tangent of the Pyramid's angle ascend = 1.273240621
182,621 x 1,273240621 = 232,5204754 inches.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
PASSAGE INTO THE KING’S CHAMBER
Figure 30
The measurements of the King’s Chamber passage (the square S) in inches (Figure 30):
· Length = 100,5951944
· Width (a) = 41,2131638
· Height (h) = 41,2131638
Square S = circle C =1.698,52487 squared inches
Diameter of the circle Z = 23,25204754 = 59,06020075 cm = R
R = 59,06020075 cm:
59,06020075 x 3,14159 = 185,5429361 cm = circle Z (Figure 30):
185,5429361 cm x 1.000 = 1,855429361 km = the length of one minute (1’) on the curved Earth’s surface around the Equator.
If a certain object was to travel with a speed of 59,06020075 cm in one second, for 24 hours it would travel a distance of 51,02801345 km:
Earth’s equatorial diameter = 12.757,00336 km:
12.757,00336 : 51,02801345 = 250 parts of the Equator.
Figure 30
The measurements of the King’s Chamber passage (the square S) in inches (Figure 30):
· Length = 100,5951944
· Width (a) = 41,2131638
· Height (h) = 41,2131638
Square S = circle C =1.698,52487 squared inches
Diameter of the circle Z = 23,25204754 = 59,06020075 cm = R
R = 59,06020075 cm:
59,06020075 x 3,14159 = 185,5429361 cm = circle Z (Figure 30):
185,5429361 cm x 1.000 = 1,855429361 km = the length of one minute (1’) on the curved Earth’s surface around the Equator.
If a certain object was to travel with a speed of 59,06020075 cm in one second, for 24 hours it would travel a distance of 51,02801345 km:
Earth’s equatorial diameter = 12.757,00336 km:
12.757,00336 : 51,02801345 = 250 parts of the Equator.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
THE ANTECHAMBER
Tropical years = 365,242 days:
365,242 : 3,14159 = 116,2602377
Antechamber
Length of the Antechamber D = 116,2602377 inches.
Circle Y = 10.615,78044 squared inches = square T = 103,0329095 x 103,0329095 inches:
2 x 103,0329095 = 206,065819 inches = two side of the square T = widths of the King’s and Queen’s Chamber.
4 x 103,0329095 = 412,131638 inches = four sides of the square T = length of the King's Chamber.
Tropical years = 365,242 days:
365,242 : 3,14159 = 116,2602377
Antechamber
Length of the Antechamber D = 116,2602377 inches.
Circle Y = 10.615,78044 squared inches = square T = 103,0329095 x 103,0329095 inches:
2 x 103,0329095 = 206,065819 inches = two side of the square T = widths of the King’s and Queen’s Chamber.
4 x 103,0329095 = 412,131638 inches = four sides of the square T = length of the King's Chamber.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
MILLENIUM DOME
The O2 - Millennium Dome, Greenwich
(http://en.wikipedia.org/wiki/Millennium_Dome)
Location: Greenwich, UK
Longitude: O˚
Latitude: 51˚ 30' 8'' N
Diameter: 365 m (1 m = 1 day)
Height: 49,98259186 m (50 m)
Height: 49,98259186 m = radius of the circle A
Area of the circle A = 7.848.507028 m2 = area of the square B
Circumference of the square B = 354.3672 m = Lunar year = 354,3672 days
Latitude: 51˚ 30' 8'' N
1˚ on te curved Earth's surface = 111 km:
51˚ 30' 8'' = 5.716.746667 km:
5.716.746667 x 7 = 40.017,22667 km = Earth's mean size
The O2 - Millennium Dome, Greenwich
(http://en.wikipedia.org/wiki/Millennium_Dome)
Location: Greenwich, UK
Longitude: O˚
Latitude: 51˚ 30' 8'' N
Diameter: 365 m (1 m = 1 day)
Height: 49,98259186 m (50 m)
Height: 49,98259186 m = radius of the circle A
Area of the circle A = 7.848.507028 m2 = area of the square B
Circumference of the square B = 354.3672 m = Lunar year = 354,3672 days
Latitude: 51˚ 30' 8'' N
1˚ on te curved Earth's surface = 111 km:
51˚ 30' 8'' = 5.716.746667 km:
5.716.746667 x 7 = 40.017,22667 km = Earth's mean size
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
GREAT PYRAMID ABOVE SEA LEVEL
The Great Pyramid: 198 feet above sea level (A-B).
Square C-D + D-E + E-F + F-C = 597,7624754 Sacred Cubits = Pyramid’s base and Pyramid’s height ( 365,242 + 232,5204754 SC).
A-C = A-F = 1.868,007736 inches
Angle beta = 51,85399754° (Pyramid’s angle of ascent)
Tangent beta = 1,273240621
1.868,007736 x 1,273240621 = 2.378,42333 inches = A-B = = 198,2019442 feet above sea level.
Square H-I + I-J + J-K + K-H = 4 x 365,242 = 1.460,968 Sacred Cubits:
a) Square C-D + D-E + E-F + F-C = 597,7624754 Sacred Cubits
b) Square H-I + I-J + J-K + K-H = 1.460,968 Sacred Cubits:
1.460,968 : 597,7624754 = 2,444061078
A) Earth’s Equator = 40.077,27418 km:
40.077,27418 : 2,444061078 = 16.397,82023 km
Circumference of the circle S = 18.978,9586 cm = 0,189789586 km:
If a certain object was to travel with a speed of 0,189789586 km in one second, for 24 hours it would travel a distance of 16.397,82023 km. The Great Pyramid stands on the northern edge of the Giza Plateau,
55
B) Equatorial diameter = 12.757,00336 km:
12.757,00336 : 2,444061078 = 5.219,5927 km
Diameter of the circle S = 2.378,423329 inches = 6.041,195256 cm:
If a certain object was to travel with a speed of 6.041,195256 cm in one second, for 24 hours it would travel a distance of 5.219,5927 km.
C) Pyramid’s base = 365,242 Sacred Cubits = H-K =H-I = I-J = J-K
365,242 : 2,444061078 = 149,4406188 SC = C-F = C-D = D-E = E-F
The Great Pyramid: 198 feet above sea level (A-B).
Square C-D + D-E + E-F + F-C = 597,7624754 Sacred Cubits = Pyramid’s base and Pyramid’s height ( 365,242 + 232,5204754 SC).
A-C = A-F = 1.868,007736 inches
Angle beta = 51,85399754° (Pyramid’s angle of ascent)
Tangent beta = 1,273240621
1.868,007736 x 1,273240621 = 2.378,42333 inches = A-B = = 198,2019442 feet above sea level.
Square H-I + I-J + J-K + K-H = 4 x 365,242 = 1.460,968 Sacred Cubits:
a) Square C-D + D-E + E-F + F-C = 597,7624754 Sacred Cubits
b) Square H-I + I-J + J-K + K-H = 1.460,968 Sacred Cubits:
1.460,968 : 597,7624754 = 2,444061078
A) Earth’s Equator = 40.077,27418 km:
40.077,27418 : 2,444061078 = 16.397,82023 km
Circumference of the circle S = 18.978,9586 cm = 0,189789586 km:
If a certain object was to travel with a speed of 0,189789586 km in one second, for 24 hours it would travel a distance of 16.397,82023 km. The Great Pyramid stands on the northern edge of the Giza Plateau,
55
B) Equatorial diameter = 12.757,00336 km:
12.757,00336 : 2,444061078 = 5.219,5927 km
Diameter of the circle S = 2.378,423329 inches = 6.041,195256 cm:
If a certain object was to travel with a speed of 6.041,195256 cm in one second, for 24 hours it would travel a distance of 5.219,5927 km.
C) Pyramid’s base = 365,242 Sacred Cubits = H-K =H-I = I-J = J-K
365,242 : 2,444061078 = 149,4406188 SC = C-F = C-D = D-E = E-F
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
NUMBER 1 AND THE NUMBERS OF THE PYRAMID
Number 1
1 : 3,14159 = 0,318310155
0,318310155 x 2 = 0,63662031 = tangent of the angle 32.48165854º (north air-channel of the King's Chamber).
0,63662031 x 2 = 1,27324062 = tangent 51,85399754º (the G. Pyramid's angle of ascend).
√1,27324062 = 1,128379644 {1,128379644 year = 412,1316379 days = length of the King's Chamber (in inches)}.
412,1316379 : 2 = 206,065819 {width of the King's Chamber (in inches)}.
1,128379644 x 206,065819 = 232,5204754 {heigth of the G. Pyramid (in Sacred Cubits)}.
232,5204754 x 3,14159 = 730,4840003 (length of speed of Earth's turning on the Equator in one second: 730,4840003 x 60" x 60' x 24 h = 63.113.817,64 Sacred Cubits = 40.077,27419 km = Earth's Equator).
NUMBER 1 = 1 INCH
Antechamber passage, Granite Leaf and the Boss (Seal)
C-D = 1 inch
A) 1 : 3,14159 = 0,318310155 inches
(0,318310155 : 2) x 100 = 15,91550077 inches:
If a certain object was to travel with a speed of 15,91550077 inches/sec., for the amount of time of one day (24 hours) this distance would be 502.244.226,8 inches = Earth's equatorial Diameter (12.757,00336 km).
232,5204754 inches:
(232,5204754 X 100) X 2,54 = 59.060,20075 cm = 0,590602008 km:
If a certain object was to travel with a speed of 0,590602008 km/sec., for 24 hours (1 day) it would travel a distance of 51.028,01345 km.
Earth's Equator= 40.077,27418 km:
51.028,01345 : 40.077,27418 = 1,273240621 days of Earth's rotation = tangent 51,85399754º (the G. Pyramid's angle of ascend).
C) 730,484003 inches:
730,484003 x 100 = 73.048,4003 inchies = 1,855429368 km = 1' (minute) on the curved Earth's surface around the Equator.
D) 206,0658189 inches
206,0658189 x 100 = 20.606,58189 inches:
If a certain object was to travel with a speed of 20.606,58189 inches/sec., for 1 year (365,242 days) it would travel a distance of 650.280.025.400 inches:
Earth's Equator = 40.077,27418 km = 1.577.845.440 inches:
650.280.025.400 : 1.577.845.440 = 412,1316378 days of the Earth' rotation = 412,1316378 inches = length of the King's Chamber.
.
Number 1
1 : 3,14159 = 0,318310155
0,318310155 x 2 = 0,63662031 = tangent of the angle 32.48165854º (north air-channel of the King's Chamber).
0,63662031 x 2 = 1,27324062 = tangent 51,85399754º (the G. Pyramid's angle of ascend).
√1,27324062 = 1,128379644 {1,128379644 year = 412,1316379 days = length of the King's Chamber (in inches)}.
412,1316379 : 2 = 206,065819 {width of the King's Chamber (in inches)}.
1,128379644 x 206,065819 = 232,5204754 {heigth of the G. Pyramid (in Sacred Cubits)}.
232,5204754 x 3,14159 = 730,4840003 (length of speed of Earth's turning on the Equator in one second: 730,4840003 x 60" x 60' x 24 h = 63.113.817,64 Sacred Cubits = 40.077,27419 km = Earth's Equator).
NUMBER 1 = 1 INCH
Antechamber passage, Granite Leaf and the Boss (Seal)
C-D = 1 inch
A) 1 : 3,14159 = 0,318310155 inches
(0,318310155 : 2) x 100 = 15,91550077 inches:
If a certain object was to travel with a speed of 15,91550077 inches/sec., for the amount of time of one day (24 hours) this distance would be 502.244.226,8 inches = Earth's equatorial Diameter (12.757,00336 km).
232,5204754 inches:
(232,5204754 X 100) X 2,54 = 59.060,20075 cm = 0,590602008 km:
If a certain object was to travel with a speed of 0,590602008 km/sec., for 24 hours (1 day) it would travel a distance of 51.028,01345 km.
Earth's Equator= 40.077,27418 km:
51.028,01345 : 40.077,27418 = 1,273240621 days of Earth's rotation = tangent 51,85399754º (the G. Pyramid's angle of ascend).
C) 730,484003 inches:
730,484003 x 100 = 73.048,4003 inchies = 1,855429368 km = 1' (minute) on the curved Earth's surface around the Equator.
D) 206,0658189 inches
206,0658189 x 100 = 20.606,58189 inches:
If a certain object was to travel with a speed of 20.606,58189 inches/sec., for 1 year (365,242 days) it would travel a distance of 650.280.025.400 inches:
Earth's Equator = 40.077,27418 km = 1.577.845.440 inches:
650.280.025.400 : 1.577.845.440 = 412,1316378 days of the Earth' rotation = 412,1316378 inches = length of the King's Chamber.
.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
Sloping walls are a hoor to seal and drain properly. I'd say if they were making it today they'd go with a cube.
- helloinsane
- Member
- Posts: 175
- Joined: Mon Jul 22, 2002 3:22 pm
- Location: Vancouver, Canada
Re: The Great Pyramid: metamorphose of the architecture
helloinsane wrote:Sloping walls are a hoor to seal and drain properly. I'd say if they were making it today they'd go with a cube.

-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
ANTECHAMBER AND THE BOSS
Antechamber
Length of the Antechamber = 116,2602377 inches:
If a certain object was to travel with a speed of 116,2602377inches/sec., for 24 hours (1 day) it would travel a distance of 10.044.884,54 inches:
10.044.884,54 x 3,14159 = 31.556.908,8 inches = 50-th part of the Earth's Equator.
The length of the entrance passage into the King's Chamber = 100,5951944 inches. The width and the heigth are same: 41,2131638 inches. The cubic diagonal of the passage is 116,2602377 inches = 4.320.000-th part of Earth's equatorial diameter.
The Boss
The length of the boss on the Granite Leaf = 5 inches:
If a certain object was to travel with a speed of 5 inches in one second, for 365,242 days (1 year) it would travel a distance of 15.778.454,4 inches = 40.077,27418 km = Earth's Equator.
Antechamber
Length of the Antechamber = 116,2602377 inches:
If a certain object was to travel with a speed of 116,2602377inches/sec., for 24 hours (1 day) it would travel a distance of 10.044.884,54 inches:
10.044.884,54 x 3,14159 = 31.556.908,8 inches = 50-th part of the Earth's Equator.
The length of the entrance passage into the King's Chamber = 100,5951944 inches. The width and the heigth are same: 41,2131638 inches. The cubic diagonal of the passage is 116,2602377 inches = 4.320.000-th part of Earth's equatorial diameter.
The Boss
The length of the boss on the Granite Leaf = 5 inches:
If a certain object was to travel with a speed of 5 inches in one second, for 365,242 days (1 year) it would travel a distance of 15.778.454,4 inches = 40.077,27418 km = Earth's Equator.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
I think if you clad it in siberian larch it could look quite good and also help break up the bulk.
- alan d
- Senior Member
- Posts: 746
- Joined: Wed Mar 19, 2003 1:07 pm
- Location: glasgow
Re: The Great Pyramid: metamorphose of the architecture
alan d wrote:I think if you clad it in siberian larch it could look quite good and also help break up the bulk.
Off topic!

-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
The Pyramids and Temples of Gizeh by W. M. Flinders Petrie
Length of sides of casing
Socket Sides:
9129.8 inches
9130.8 inches
9123.9 inches
9119.2 inches
-------------------------------------------
Pyramid's original length of sides of casing: 9131,05 inches:
9131,05 - 9130,8 = 0,25 inches = 6,35 millimeters
Length of sides of casing
Socket Sides:
9129.8 inches
9130.8 inches
9123.9 inches
9119.2 inches
-------------------------------------------
Pyramid's original length of sides of casing: 9131,05 inches:
9131,05 - 9130,8 = 0,25 inches = 6,35 millimeters
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
ANTECHAMB. PASSAGE, GRANITE LEAF AND THE BOSS
Antechamber passage, Granite Leaf and the Boss (Seal)
A-B = 41,21316378 inches = one side of the Square S1 = diameter of the circle Z.
Circle Z = 1.334,017186 squared inches = Area of the square S2
Square S2:
F-G = 36,5242 inches
50
If a certain object was to travel with a speed of 36,5242 inches in one second, for 24 hours it would travel a distance of 3.155.690,88 inches = 500-th part of the Earth’s Equator:
3.155.690,88 x 500 = 1.577.845.440 inches = 40.077,27418 km.
C-E = Level of bottom of boss at base = 5 inches above horizontal joint between upper and lower slabs of the Leaf:
If a certain object was to travel with a speed of 5 inches in one second, for 24 hours it would travel a distance of 432.000 inches:
a) Earth’s equatorial diameter = 12.757,00336 km = 502.244.226,8 inches:
502.244.226,8 : 432.000 = 1.162,602377 inches = A-B
b) Earth’s Equator = 40.077,27418 km = 1.577.845.440 inches:
1.577.845.440 : 432.000 = 3.652,42 inches = circle Z
C-D = 1 inch = Position of the center of boss is 1 inch to right (west) of center of Granite Leaf:
If a certain object was to travel with a speed of 1 inch in one second, for 24 hours it would travel a distance of 86.400 inches:
Earth’s equatorial diameter = 502.244.226,8 inches:
502.244.226,8 : 86.400 = 5.813,011884 inches = the height of the Great Pyramid.
Antechamber passage, Granite Leaf and the Boss (Seal)
A-B = 41,21316378 inches = one side of the Square S1 = diameter of the circle Z.
Circle Z = 1.334,017186 squared inches = Area of the square S2
Square S2:
F-G = 36,5242 inches
50
If a certain object was to travel with a speed of 36,5242 inches in one second, for 24 hours it would travel a distance of 3.155.690,88 inches = 500-th part of the Earth’s Equator:
3.155.690,88 x 500 = 1.577.845.440 inches = 40.077,27418 km.
C-E = Level of bottom of boss at base = 5 inches above horizontal joint between upper and lower slabs of the Leaf:
If a certain object was to travel with a speed of 5 inches in one second, for 24 hours it would travel a distance of 432.000 inches:
a) Earth’s equatorial diameter = 12.757,00336 km = 502.244.226,8 inches:
502.244.226,8 : 432.000 = 1.162,602377 inches = A-B
b) Earth’s Equator = 40.077,27418 km = 1.577.845.440 inches:
1.577.845.440 : 432.000 = 3.652,42 inches = circle Z
C-D = 1 inch = Position of the center of boss is 1 inch to right (west) of center of Granite Leaf:
If a certain object was to travel with a speed of 1 inch in one second, for 24 hours it would travel a distance of 86.400 inches:
Earth’s equatorial diameter = 502.244.226,8 inches:
502.244.226,8 : 86.400 = 5.813,011884 inches = the height of the Great Pyramid.
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
It's amazing how fluent the pyramid builders were in both inches *and* kilometres. They must have been Canadian.
- helloinsane
- Member
- Posts: 175
- Joined: Mon Jul 22, 2002 3:22 pm
- Location: Vancouver, Canada
Re: The Great Pyramid: metamorphose of the architecture
It's amazing how fluent the pyramid builders were in both inches *and* kilometres. They must have been Canadian: Vancouver, Canada! 

-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada
Re: The Great Pyramid: metamorphose of the architecture
THE CIRCLE
1 : 3,14159 = 0,318310155
0,318310155 = 1/2 of 0,63662031 (tangent of the angle of north King's Chamber "air-channel" = 32,48165854° ).
Diameter (of Earth) = 12.757,00336 km:
12.757,00336 x 0,318310155 = 4.060,683717 km
4.060,683717 x 3,14159 = 12.757,00336 km = diameter
12.757,00336 x 3,14159 = 40.077,27418 km
40.077,27418 x 0,318310155 = 12.757,00336 km
1 : 3,14159 = 0,318310155
0,318310155 = 1/2 of 0,63662031 (tangent of the angle of north King's Chamber "air-channel" = 32,48165854° ).
Diameter (of Earth) = 12.757,00336 km:
12.757,00336 x 0,318310155 = 4.060,683717 km
4.060,683717 x 3,14159 = 12.757,00336 km = diameter
12.757,00336 x 3,14159 = 40.077,27418 km
40.077,27418 x 0,318310155 = 12.757,00336 km
-

Vidusa - Member
- Posts: 141
- Joined: Sat Oct 21, 2006 12:03 am
- Location: Kitchener, ON, Canada

