I made a few comment posts yesterday on molders other topic that introduced David Kenworthy's book entitled Stonehenge, 'The Eclipse'. So while I am waiting on some sort of response from this author I would just like to take that discussion back to my topic once more since we seem to be discussing GP dimensions relationship to the Earth at the moment. The comment I made on the other topic is found here:
<Your calculation of the height of the GP is using the AE RC unit of 20.625 ins. which is comparable to the old Saxon Ft. of 13.2 ins as 25/16 ratio. But there are a few AE RC units to consider in this matter also. The best one in this case, is the .525 Meter RC unit of 20.671875 ins. which divided by 7/4 gives an Egyptian Ft. of 11.8125 ins., which is in ratio to the Imp.Ft as 63/64 = 12 ins. The difference between this AE RC and the one you used being the ratio 441/440. This larger AE RC unit can then be expanded by another ratio of some familiarity around here of 176/175 to obtain a unit of 20.79 ins. And what makes this unit so intriguing is that it is in ratio to the Imp.Ft. as 1:1.7325 ins., which is a very ancient value of the sq.rt. of 3 which is also found in the grid layout of the Giza Plateau, defined by the rectangle of the three main pyramids distance length in Cubits of 20.625 ins. x 1732.5 units. While you have been focusing on the Lunar Saros cycle, the layout of the Giza plateau width factor of 1417.5 cubits in this rectangular grid gives a product of 99 Lunar cycles of 29.53125 days, which is the Metonic 8 year cycle of the Moon.>
This is more or less correct as I see it, in that the Giza Grid rectangle of 9/11 units does have a correlation to the Metonic cycle, but not as I described above. The dimensions of this Giza plateau grid being the stated 1417.5 cubits by 1732.5 cubits has a perimeter of 6300 cubits, without specifying which cubit value is involved yet. So if I divide this perimeter into 99 sections of 636.363 cubit units, and then specify an AE RC value to those sections such as 20.625 ins., the product is 13125 ins. per section. Already I am seeing a decimal fraction of the Earth Circ. in Ft., as per the model I have been studying for many years now when multiplied by 10^4. But more interesting to me than that observation, is to go ahead and state the entire perimeter in that 20.625 in. cubit x 6300 units perimeter = 129,937.5 ins. This perimeter can then be converted to a square of / 4 = 32,484.375 ins. per side. This is equivalent to 825 Old Meters when divided by 39.375 in. units, or 1100 Lunar units of 3/4 ratio to the Meter of 29.53125 ins. Another way of stating this square is to divide 6300 cubits / 4 = 1575 per side. This is also another of those values I have been discussing over the years now in various contexts, such as the number of inches in this Earth Circ. model x 10^6, or as 157.5 barley weight grains per cubic inch. This side of 1575 cubits has an area of when squared = 2,480,625 sq. cubits. Such that I can now divide this area into the aforesaid 99 sections of the grid and multiply by 100 = 2,505,681.818 cyc, expanded area sq. units. What does this larger figure indicate? How about an Earth Radius decimal fraction, when divided by 12 ins. / 5280 Ft. = 39.54674587 x 2 = 79.09349174 x 22/7 Pi, = 248.579545 x 100 Mi. Earth Circ. I don't know why it has taken me this long now to do this calculation, LOL!
Best Holiday regards,
Edited 2 time(s). Last edit at 26-Dec-17 20:53 by magisterchessmutt.