Hi RLH,

RLH wrote: but can you expand a little more on the “ zero point energy....” part.

Not at this time, there is enough confusion regarding Keplers conjecture and as you yourself said: Not sure I completely understood all that. While there are hints that coincide with some of Tesla's works within the mathematics of the Ancient Egyptians demonstrating they were at least cognizant of zero point energy there is nothing at this time I am willing to release towards that end. I have found in the past too much information muddles the works. It is first necessary to identify precisely when, how and where Kepler's concepts, principles and elements occur prior to any speculation on the actual basis for their being there beyond plain old coincidence. It has crossed my mind hundreds of times, maybe the Ancient Egyptians had mathematical abilities far beyond what orthodoxy has been willing to recognize.

It could be a coincidence, but in rational numbers the Kepler conjecture (pi/3*sqrt2) times the Royal Egyptian cubit 189/110 equals the rise-run of G1. (14/11) i.e. (20/27 * 189/110 = 14/11) for me that seems like a pretty bold hint. inversely the rise-run of G1 divided by the Royal Egyptian cubit equals the Kepler conjecture.

It could be a coincidence, but the square root of 2 divided by pi/3 multiplied by the run-rise of G1 = 99/70 / 22/21 * 14/11 = 189/110 feet equals the Royal Egyptian cubit, the inverse of Royal Egyptian cubit is the Kepler conjecture divided by the rise-run of G1, (22/7) / (3 * 99/70) = (20/27)…0.740740740…(20/27) / (14/11) = 110/189

From what I have been able to ascertain the square is no different from the circle, they were considered interchangeable as were the sphere and cube. The Royal Egyptian cubit of 20 34/55 inches is the cubit of the sphere. There are other length cubits derived from different shapes. Demonstrated by the Kings Chamber length, 1/22 the base of G1 and 1/7 the length of the sphinx. That is 34 4/11 feet the surface area square footage of a 36" circumference sphere.

But Making a 36" circumference from a 9-inch square:
9 * 9 = 81 area of square perimeter of square 4*9 = 36
81 * 14/11 = 103 1/11 inches area of circle with a 36 inch circumference with a diameter of 11 5/11 inches with a surface area of 412 4/11 inches (20 cubits)

Making a 36" sphere from a 9 inch cube
9 * 9 = 81 area of square
81 * 6 = 486 square inches = surface area of a 9 inch cube
(412 4/11 / 486 = 28/33) is always the ratio between the surface areas of a sphere and cube, i.e. (14^2) * 22/7 = 616…616 / 28/33 = 726…726 / 6 = 121 (11^2)

14/11 * 2/3 = 28/33 equals the ratio between the surface area of a sphere and cube from the AE perspective. Using their system reduced complicated mathematical equations to simple fractional math problems.

Relating to your grid, I find it extremely interesting that the ventilation shafts exit at at the 102 course the exact point of 28/22 grid, (14/11) and drawing a correlation of which there are many, but 28 * 22 equals an area grid with 616 squares same as the surface area of a 14 inch diameter sphere (14^2) * 22/7 = 616…616 / 28/33 = 726…726 / 6 = 121 (11^2) equals an 11 inch cube.

Regards