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Hi David,

Pardon the long post, but if by chance you in any way comprehend the methods of the Ancient Egyptians, you might also understand each side length of G1 defines a different cubit length established for a particular purpose. I am of the opinion each different cubit length is a one off unit employed much like the slide rules of old giving directions for use in particular lengths.

The cubit of 20 34/55 inches is derived geometrically from the arc second circle and in the process proving both the length of the Royal Egyptian cubit and the inch as valid units of measure employed by the Ancient Egyptians. Derived as follows: 360x60x60 = 1296000 arc seconds.

If i arc second equals one inch then 1296000 equal to 1000 perimeters of the North Wall in the Kings Chamber multiplied by the rise run of G1's 5 1/2 seked, being 7 1/2 / 5 = 1649454 6/11 = equal to 1000 times the horizontal length of the Grand Gallery (80 cubits). That divided by 8 = 2061818 9/11 equal to 1000 times the width of the Kings Chamber (20 cubits) or 10,000 times the Royal Egyptian Cubit of 20 34/55 inches (1 cubit). (We simply say 1296000 / 44/7 = 206181.818181...), but 8 /(14/11) = 44/7
From the same arc second circle using Ancient Egyptian methods, it is simple mathematics to compute the rise run of the Grand Gallery is found along with the width and other factors all relating to G1. Using their methods it is also possible to mathematically link the Grand Gallery dimensions to G2 base.

Without mountains of supporting information or a cubit rod specimen to confirm the length. i.e. 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 = 481/280 ft cubit length. 481/280 x 12 = 1443/70 inches (20.6142857142857..inches). 481/280 x 440 = 755 6/7 feet,
it is amazing how this exactly matches Specimen #3: Turin Museum Catalog #6347:
The Amenemope Rod.
Scale A (bevel face):
Length = 20.614 with subdivisions of the 15 right-most digits on the front. There are no palm divisions
144/7 x 481/480 = 1443/70

If you want to talk about cubit rods this site is most helpful: [www.egyptorigins.org]
or maybe the dimensions of G1 and G2 Sarcophagi will provide a cubit length.

G2 sarcophagus has a mean length of 2592/25 (103.68) inches, divided by 5 gives a cubit length of 2592/125 or 20.736 inches as a cubit length. 2592/125 multiplied by 440 equals the length of the South side socket base of G1 9123 21/25 inches (9123.84 inches with in 6/1000 inch of Petrie's measured length 9123.9 inches.
G1 sarcophagus with a length of 90 inches divided by 5 = 18 inches. 18 / (8/9) = 81/4 or 20 1/4 inches within 3/1000 inch of:
Specimen #4: Turin Museum catalog #6349. Scale B divided into 27 parts:
Useful Length = 20.247
No Palm divisions
(27) Digit mean length = 0.750 if there were 28 digits the rod would be exactly 21 inches long.
[www.egyptorigins.org]

Whatever Encyclopedia Britannica states as the value of a cubit is really irrelevant unless there is physical evidence or an actual cubit rod specimen to support the claimed length. After reading some of the information on your posted website, according to my research a better value for the cubit for Stonehenge would be 20.736 G2's sarcophagus cubit multiplied by (176/175) for a value of 20.8544914285714... inches or inches equal to 1/1,000,000 earth polar radius in feet and also equal to 1/2 hunab, the STU found by Hugh Harrelston Jr at Teotihuacan. Inner diameter of the Sarsen Circle is 56 units (like the 56 Aubrey Holes) and outer diameter equal to 60 units. Fractional unit in feet is 38016/21875 feet. Where did it come from. Earth Sun dimensional ratios aka As the Ancient Egyptian geographic cubit.
i.e. Suns diameter 864000 x 5280 = 4561920000 feet
4561920000 / 21875 = 20854491 3/7 feet Earth Polar Radius.
20854491 3/7 / 12,000,000 = 38016/21875 feet. Of which 38016 is 1/2 G1 South Side socket base in feet, 38016 divided by 100 = 380.16 feet.
Earth Circumference 24883.2 x 5280 = 131383296 feet
131383296 / 38016 = 3456
3456 times (twice G1 casing perimeter 3024) 6048 feet = 20901888 feet equals Earth Mean Radius.

Stonehenge Sarsen Circle outer radius 30 x 38016/21875 = 52.13622857142857...x 7000 = 365953.6 feet or one degree of arc.
Since Stonehenge predicts eclipses it is only fitting the moons dimensions should show up somewhere.
52.13622857142857..x 21875 = 1140480 or about 1/10 the moons diameter. 2160 x 5280 = 11404800 feet.

Proportional inch value list of cubit progressions in a nutshell:
G1 Sarcophagus Cubit 18 x (9/8) = 81/4 Specimen #4: Turin Museum catalog #6349:
Specimen #4: Turin Museum catalog #6349: 81/4 x (64/63) = 144/7 Standard Cubit
Standard Cubit: 144/7 x (441/440) = 1134/55 Royal Egyptian Cubit
Royal Egyptian Cubit 1134/55 x (176/175) = 2592/125 = G2 Sarcophagus Cubit
2592/125 x (175/176) = 456192/21875 = Geographic Cubit of Giza, Stonehenge and Teotihuacan.

There appears to be a time stamp contained within the different length cubits, but my time is limited so I will stop here.

Regards,
Jacob

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Re: Let there be Phos :) 251 magisterchessmutt 22-Dec-17 10:44
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Re: Let there be Phos :) 288 magisterchessmutt 24-Dec-17 04:07
Re: Let there be Phos :) 225 drew 18-Dec-17 00:51
Re: Let there be Phos :) 221 magisterchessmutt 18-Dec-17 02:11
Re: Let there be Phos :) 317 mmstud 22-Dec-17 14:44
Re: Let there be Phos :)Doubling a cube 298 molder 22-Dec-17 21:21
Re: Let there be Phos :)Doubling a cube 316 molder 23-Dec-17 22:27
Re: Let there be Phos :)Doubling a cube 247 magisterchessmutt 24-Dec-17 01:55
Re: Let there be Phos :)Doubling a cube 259 magisterchessmutt 26-Dec-17 14:53
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Re: Let there be Phos :)Doubling a cube 189 molder 27-Dec-17 05:02
Re: Let there be Phos :)Doubling a cube 256 magisterchessmutt 27-Dec-17 11:07
Re: Let there be Phos :)Doubling a cube 216 rodz111 29-Dec-17 22:44
Re: Let there be Phos :)Doubling a cube 186 Sirfiroth 28-Dec-17 18:28
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