Hi all,

I was doing some thinking about anthonys' comments which is probably a dengerous thing
NB the following assumptions will not make much of a difference to the overall scheme and generally accepted data of the great pyramid. They are for the purpose of illustration.

Assumption 1.
4 : pi gradient

Assumption 2.
9069 inch side length.

Assumption 3
10,000 / sqrt 3 height.

Assumption 4
280 cubit heigtht

Assumption 5.
Max height of 1st course 58.6
Min height of 1st course 57.6
Mean of 1st course 58.1 inches
[www.ronaldbirdsall.com]

The distance from the base to the apex = 10,000 / sqrt 3 inches = 5773.5 inches
The distance from the 1st course to the apex = as above less 58.1 = 5715.4 inches

5715.4 / 5773.5 = 0.9899....

I have come across this ratio before with the work on herodotus and the cubit definitions.
Face area = 80,000 setjat
Height = sqrt 80,000 = 282.8427 cubits
280 / 282.8427 = 0.9899.....
28 / 20 x sqrt 2 (remen diagonal) = 0.9899....

Questions.
1. Is the base to apex 280 cubits of 20 x sqrt 2 fingers (remen diagonals) ?
2. Is the 1st course to apex 280 x 28 finger cubits ?
3. Is base to apex therfore sqrt 80,000 28 finger cubits ?

This would then tie in excellently with the 1/100th idea of Anthonys' and Herodotus comments stated to originate from an Egyptian priest.
using 99:100 for the 28 finger and remen diagonal cubits as Khet and short Khet with the 140 / 99 / 70 ratios of sqrt 2. A square with 70 remen cubit sides has a diagonal of 100x28 finger cubits or 99 remen cubits.
A 100 x 100 28 finger cubit square has twice the area of a 70 x 70 remen diagonal square.
Therefore 8....100x100 28 finger squares = 16....70x70 remen diagonals = 32....70x70 remen



Edited 1 time(s). Last edit at 10/11/2005 09:11AM by Mick.