Numerical Code of its Architecture Hiding in the Great Pyramid’s Summit?

The Summit of the Great Pyramid. The Tripod was placed in 1874 by David Gill during the Transit of Venus Party, but it was removed in 2019.

If you could climb up to the summit of the Great Pyramid, you would arrive on a platform at the 201st course. This platform measures 11.7 (N-S) x 11.9 (E-W) meters (Kawae et al.). The summit floor is composed of 86 irregular limestone blocks. In the center are the remnants of two additional, presumed courses, #202 and #203.

This is Figure 2 from the above paper.

However, it is generally believed that the Great Pyramid's summit was originally a peak made of a pyramidion cap stone and not a flat summit or a sphere as some have proposed. On a New Kingdom relief, the Stele of Mentuhor, the artist showed the Great Pyramid with a pointed peak. It seems the artist would not have left out a flat top or sphere if it had been there. This ancient 3500 year-old relief is the closest to a photo we have.

The Stele of mntw-hr, Montuhor.

Is it possible that the pyramidion fell off and that what we see today was the platform supporting it, or is what we see today the result of stone looting. In other words, is the current platform an artifact of later history and not part of the original architectural design?

The Great Pyramid and its architectural geometry based on a Kepler Triangle.

If you look at the dimensions of the Great Pyramid as they may have been designed by the architect in Royal Egyptian Cubits ("Cubits"; 0.5236 m; 20.62 inch), you could reasonably reconstruct a base of 440, a half-base of 220, and a height of 280 Cubits. If you use Pythagoras to derive the length of the apothem (the side-slope), it would be 356.09 Cubits, or better 356 Cubits and 2 1/2 fingers. These figures give us an apothem/half-base ratio of 356.09/220=1.619, which is near the Golden Ratio φ (1.618). There is no unequivocal proof that the architect of the Great Pyramid knew this ratio for it could be incidental to choosing the 280 and 220 Cubit dimensions in the design. However, φ eerily appears in other places (see my previous blog post, for example) when the architecture and Giza pyramids' lay-out is closely examined (for another example see Jim Penniston and Gary Osborn's 2019 The Rendlesham Enigma) and so the question is justified: Did the architect know this ratio, perhaps as the fraction of 34 and 21, two numbers in the Fibonacci Sequence?

The dimensions of the Great Pyramid in Royal Egyptian Cubits.

Well, let's take a look at the current summit: 11.7 x 11.9 meters is 139.23 square meters. If you divide this surface area by the number of blocks used to make the core masonry, 86, you get 1.619! How could this be? This presumes knowledge of the meter. Didn't the Egyptians think in Cubits? The meter was not known until it was invented by the French at the end of the 18th century, was it?

This is where master architect Jean-Paul Bauval has a word to say. Jean-Paul has proposed that the indentation of the core masonry at the base can only mean one thing at the top: A virtual summit concealed by the casing stones. Think of a holographic projection, a hidden architectural signature so to speak imbued into the stony peak. And that virtual summit creates a virtual space into which you can exactly fit a sphere with radius diameter...one meter. The paper is published and can be viewed here: https://www.academia.edu/27553148/THE_CONCAVITY_OF_THE_GREAT_PYRAMID_a_design_feature_Did_the_designer_know_the_Meter_Unit

We are left with a nagging question, now nagging even more: Did the architect know the meter and did s/he know φ? How would you prove it? That is for all you to decide.

Update (5-3-2020): Further studying this topic, an interesting ratio appears when you divide the surface of the Bauval virtual sphere by the reconstructed surface of the cased summit (Kawae: 13.64 m). The ratio looks a lot like a cycle time and our sidereal orbit number in Earth revolutions. The relationship between squaring a circumference of a circle and the circle's surface area is x 4π. If you reverse-compute the orbit radius which corresponds to 366² annual Earth revolutions squared using revolutions as an orbit distance measure, you get 3.545 Astronomical Units (1 A.U. = Earth orbital radius average). Where is this? In a very sparse zone at outmost Asteroid belt. Not too promising yet. Maybe something else will come to mind.

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