Hi Mark,

Mark Heaton wrote: There is a slight error in Petrie's calculation of the height of the groove. It is reported as 166.2 + 5.9 = 172.1 inches, where 5.9 inches is the mean of 4 measurements taken at NE, NW, SE, ans SW. In reality he should have used the measurements near the north wall which were 5.4 and 5.7 inches perpendicular (6.0 and 6.4 vertical), so the height is between 172.2 and 172.6 from Petrie's measurements with a mean of 172.4.

I think that the height at the higher end of the gallery was intended to be less, but explaining why is another topic.

[www.ronaldbirdsall.com]

The roof stones are set each at a steeper slope than the passage, in order that the lower edge of each stone should hitch like a paul into a ratchet-cut in the top of the walls; hence no stone can press on the one below it, so as to cause a cumulative pressure all down the roof; and each stone is separately upheld by the side walls across which it lies. The depth of two of these ratchet-cuts, at the S. end, I measured as 1.0 and 1.9 to 2.0; and the angles of the two [p. 73] slabs there 28º 0' to 28º 18', and 27º 56' to 28º 30', mean 28º 11'; which on a mean slab 52.2 from N. to S., would differ 1.74 inches from the passage slope.

The remarkable groove in the lower part of the third lap, along the whole length of the sides, was measured thus, perpendicularly: At the S.W. it is cut to a depth of .8 inch, at the S.E. to .6 (?); the upper edge of it is often ill-defined and sloping. According to Prof. Smyth the mean [p. 74] height of this lap above the gallery floor is 166.2 ± .8 vertically; hence the groove is at 172.1 to 179.0 vertically over the floor, and its lower edge is therefore at half the height of the gallery, that varying from 167 to 172.

Let me attempt to explain your subject for another thread. I don’t think there was an error, if there was an error it was in the cutting of the groove. According to Petrie, adding all components as he stated gives values of 336 to 344 +/- 2 as the variation of the height for the Grand Gallery if the grove is the center line of the wall height. But he did say the lower edge of the groove, does not run parallel to the floor, is half the height of the gallery.

The apparent spherical anomalies are you detect in the Grand Gallery are not anomalies at all, but are based on spherical aspects of the design for the entire Pyramid. Now, If you would bear with me for a moment and forget cubits and we will return to them in a moment, instead use 360 * (1/(4*22/7)) = 28 7/11...28 7/11 * 12 = 343 7/11 inches, (50/3 cubits) which times hexagon arc to chord ratio (22/21), 343 7/11 * 22/21 = 360 inches (1100/63 cubits) circumference. Therefore the hexagons arc to chord ratio of (22/21) times 12 = 88/7 or 4pi, therefore 28 7/11 * 22/21 = 30 = 1/12 of a 360 foot circumference circle or sphere. 30 is the arc and 28 7/11 is the chord.

This gives as the height of the Grand Gallery a value of 343 7/11 the same number of inches/10 as the number of feet in the length of the Kings Chamber 34 4/11 feet.

Supporting Information:
28 7/11 / (5/6) = 34 4/11 width of the Kings chamber, the length of the Kings Chamber represents 1/6 the radius or chord of a circle with a circumference of 36 units. 1/10 a 360 unit circle, the height of the Kings Chamber is 4/7 the length. 34 4/11 * 4/7 = 19 7/11
19 7/11 / 189/110 = 11 3/7 cubits, 11 3/7 * 28 = 320 digits from the sub floor

The Grand Gallery height 28 7/11 is 6/5 the height of the Kings chamber: 28 7/11 * 22/21 = 30 or 1/12 the circle of 360 unit circle.

Cubit: 9! / (189/110) = 211200
Series representation: (9!)/(189/110) = 110/189 sum_(k=0)^infinity((9-n_0)^k Gamma^(k)(1+n_0))/(k!) for (((script n)_0(not element)Z or (script n)_0>=0) and (script n)_0->9)

Integral representation: (9!)/(189/110) = 110/189 integral_0^infinity t^9/e^t dt

(9!)/(189/110) = 110/189 integral_0^1 log^9(1/t) dt

(9!)/(189/110) = 110/189 integral_1^infinity t^9/e^t dt+110/189 sum_(k=0)^infinity(-1)^k/((10+k) k!)

211200 * 1/12 = 17600 /10 = 1760…1760 (base perimeterG1) * 7/88 = 140 cubits Sphinx base
211200 * 1/7 * 1/12 = 2514 2/7…2154 2/7 / 10 = 251 3/7 = 80 pi cubits
21120 0 * 1/28 * 1/12 = 7542 6/7…7542 6/7 / 10 = 754 2/7 = 240 pi cubits
And as already shown: 211200 cubits / 9 / 8 / 7 / 6 / 5 / 4 / 3 / 2 / 1 = 110/189 which is the inverse of the Royal Egyptian Cubit. 110/189 * 14/11 = 20/27 (.740740740...) In 1611, Kepler proposed that close packing (either cubic or hexagonal close packing, both of which have maximum densities of pi/(3sqrt(2)) approx 74.048%) is the densest possible sphere packing, and this assertion is known as the Kepler conjecture. Which in rational form would read, (22/7)/(3 99/70) = 20/27 = (0 .704704704…) I know there is much doubt the ancient Egyptians were aware of this, but the truth is they were quite adept at hyperbolic geometry so far there are only a handful of individuals that really understand the mathematics and how the pyramids were designed and you evidently see at least part of it.
If you have enough patients and can devote enough time to the mathematics, it will demonstrate, through geometry not only the dimensions of the pyramids, but the natural evolution, origin and source of the cubit, the order and reason behind their use of only rational values unfold from the lowly inch into the into the magnificent wonder that is G1 spelling out in the process out the natural progressions from line to square to circle, cube sphere, hypercube etc.

14/11 * 64 = 81 5/11…81 5/11 / 81 = 896/891…896/891 * 22/7 = 256/81

9 * 9 = 81 area of square perimeter 36

81 * 14/11 = 103 1/11 inches area of circle circumference 36

103 1/11 * 4 = 412 4/11 square inches = surface area of a 36-inch circumference sphere.

412 4/11 * 88/7 = 5184...5184 * 7 = (1/10 * 9!) or the base perimeter of G1.

Area of circle:
103 1/11 / 5 = 20 34/55 if inches is the cubit, described as (14/11) * (99/70) / (22/21)
103 1/11 / 6 = 17 2/11 if feet is the width of the Kings Chamber

103 1/11 * 6 = 618 6/11 the number of feet in Bent Pyramid base
103 1/11 * 7 = 721 7/11 the number of feet in Red Pyramid base
103 1/11 * 7 1/3 = 756 the number of feet in G1 base (7 1/3 = (11 * 2/3) = (22/3)
103 1/11 * 28 = 2886 6/11 if feet the perimeter base of Red Pyramid, if inches the base of the Sphinx.
103 1/11 * (121 / 3) = 4158 if inches equals the base of G3 (11^2 / 3)
103 1/11 * 99/70 = 729/5 that is (9^3 / 5)
Deriving the cubit:
Ancient Egyptian:
(14/11) * (99/70) / (22/21) = 189/110 feet
(14/11) * (99/70) / (126/11) = 20 34/55 inches, 126/11 is the diameter of a circle with a circumference of 36 inches.
You may assume the 12 is inches, but it is also the kissing number!
[en.wikipedia.org]
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Our current system states:
(14/11 * sqrt(2))/(pi/3)×12 = 20.625428333780975411628199648518...

(504 sqrt(2))/(11 pi) is a transcendental number

Series representation: (12 (14 sqrt(2)))/((11 pi)/3) = (504 (sqrt(z_0) sum_(k=0)^infinity((-1)^k (-1/2)_k (2-z_0)^k)/(k! z_0^k)))/(11 pi)

(12 (14 sqrt(2)))/((11 pi)/3) = (504 (exp(i pi floor((arg(2-x))/(2 pi))) sqrt(x) sum_(k=0)^infinity((-1)^k (2-x)^k (-1/2)_k)/(x^k k!)))/(11 pi) for (x element R and x<0)

(12 (14 sqrt(2)))/((11 pi)/3) = (504 ((1/z_0)^(1/2 floor((arg(2-z_0))/(2 pi))) z_0^(1/2 (1+floor((arg(2-z_0))/(2 pi)))) sum_(k=0)^infinity((-1)^k (-1/2)_k (2-z_0)^k)/(k! z_0^k)))/(11 pi)

Is it any wonder we can’t see anything the Ancient Egyptians wrote, they used numbers, but in a context different from our own.
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Regards