There is a typo in the following statement that you made in your post to Tim:

"The right-angle formula stipulates the square of the two sides equaling the square of the hypotenuse, and we know the hypotenuse/base ratio is 1.618, therefore the ratio of these squares would be (1.618x1.618)/(1x1) = 2.164:1.
Returning to the joints...
330.9/2.164 = 126.6”...the distance to the first joint...three down!"

1.618 x 1.618 = 2.618.
divided by 1 x 1 still equals 2.618.

I know this is what you meant because 330.9/2.618 = 126.4 inches.

However, I disagree with your discounting the implication of the golden section. Since the total length of the floor is equal to the limestone portion of the floor times 2.618, this also means that the limestone portion of the floor times 1.618 is equal to the granite portion of the floor. This is a classic expression of the golden section, with the total length of the floor being divided into the golden section by the division of the limestone portion and the granite portion. ie:

Granite/limestone = total/granite = 1.618 = phi.

Phi is also implicated since phi is derived from the square root of five and the square root of five is present in the height of the KC (sq rt 5 x 5, or 1/2 of the diagonal length of the floor of the KC), and in the height of the antechamber (sq rt 5, plus 5).

The granite portion of the floor of the antechamber is 5 cubits, and the height to the top of the porticullis is 5 cubits, giving a square with an area of 25 cubits. The length of the limestone portion of the floor from the south wall of the grand gallery to the joint where the granite begins is 64.9 inches, or pi cubits (3.14 cubits). The length of the ceiling of the passage between the south wall of the grand gallery to the north wall of the antechamber is 2.5 cubits. Thus, the length of the antechamber is pi cubits, plus 5 cubits, minus 2.5 cubits, or pi cubits plus 2.5 cubits, or 5.64 cubits. A circle with a diameter of 5.64 cubits also has an area of 25 cubits. This same circle has a circumference of 17.72 cubits or sq rt pi x 10 cubits, the same distance as the length from the midline of the pyramid to the south wall of the subterranean chamber.

The height of the Kc passage between the chambers is generally given as 2 cubits, but according to the surveys of Petrie and Smyth, the height is actually about an inch more than 2 cubits. This is significant because it is the required height for the walls of the KC to express both the pi and phi proportions. The pi proportion requires a height for the five courses of 11.416 cubits (pi x 20, minus 40, divided by 2) for the ratio between the total perimeter of the wall, divided by the length of the wall (20 cubits), to equal pi.

Given that the KC walls are comprised of five courses of equal height, and given that the height from the raised floor of the KC to the ceiling of the KC is 11.18 (sq rt 5 x 5) cubits, and given that the height of the ceiling of the KC passages is the same as the top of the first course/bottom of the second course of the walls of the KC, The height from the raised floor of the KC to the top of the first course must be 2.048 cubits for the chamber to express both the pi and phi proportions. This is in accordance with the surveys by Petrie and Smyth for the height of the KC passages.

Best,

Jim

PS - I think the slope of the gallery is slightly less than 2/1, however, if the intended slope was 2/1 and the perpendicular height of the gallery is twice the height of the antechamber (sq rt 5 plus 5) x 2, then the vertical height of the gallery would be 16.18 cubits, or phi x 10.




Edited 1 time(s). Last edit at 07/14/2009 09:55PM by Jim Alison.