Apologies again for that last post getting somewhat out of hand. It’s good to revisit some of those related subjects and to explore possibilities, but I’m sure it’s understandably confusing, especially for anyone who may be “just tuning in”.
As with the subjects of other Egyptian studies, including the Great Pyramid, we could try to get in touch with the basic design first. We presume that these designs begin with a simple, regular structure before various adjustments or “tweaks” are made to the design to accomodate a higher volume of stored data.
The basic design of the Sepulchral Chamber would seem to be rectangle 10 Royal Cubits long by 6 Royal Cubits wide. The height may be 6 ft Imperial; the varying values for the height of the walls tends to be in the neighborhood of 72 inches = 6 feet.
Such a choice of height figure would make the volume of the room to be, in Imperial feet, (10 x 1.718873385) x (6 x 1.1718873385) x 6 = 1063.629257 cubit feet.
This is intriguing because this is a long known if rarely seen value; in a manner of speaking it represents the squaring of the circle in the sense that that we achieve a generic assessment of the circle by substituting 360* for the actual perimeter, or the Radian value of 57.29577951 for the actual radius, so that the area of such a circle is 57.29577951^2 x Pi = 10313.24031, generic area of any circle.
This generic circular area squared = 10313.24031^2 = 106362925.7
We can see it more simply as Royal Cubit (1.718873385)^2 x 360 = 1063.629257.
I really don’t know a lot yet about why we would be seeing it here, but we may wish to note that relative to the King’s Chamber of the Great Pyramid (length 20 Royal Cubits by width 10 Royal Cubits, height 18 Hashimi Cubits = (1.718873385 x 20) x (1.718873385 x 10) x (18 x 1.067438859) = 113535.8459 cubit feet)
Volume King’s Chamber 113535.8459 / Volume Senwosret II’s Chamber 1063.629257 = 10.67438159 = 10 Hashimi Cubits, in feet.
This seems quite interesting as it seemingly implies that architects retained access to blueprints and /or data from the interiors of 4th Dynasty pyramids like the Great Pyramid, at very least up until the 12th or 13th Dynasty.
This of course only concerns the lower part of the Lahun Sepulchral Chamber which can be considered as a rectangular box. A more complete volume figure may be obtained by adding the volume that occurs above the top of the walls to the volume figure below this point of 1063.629257 cubic feet.
We can also look to nearby features for possible guidance, such as the Offering Chamber.
Petrie’s description of the Offering Chamber in the Lahun pyramid, from “Illahun, Kahun, and Gurob”
Diagram of the Sepulchral Chamber plus Offering Chamber with Petrie’s data for the Offering Chamber labelled onto it.
Let’s attempt some initial observations:
1. Measurements. 62.8 inches; 2 Pi = 6.283185307. 72.8 / 12 = 6.06666666 ft; 5 Remens = 1.216733603 x 5 = 6.083668015 ft. 45.2 is near to 45.23893421, but also 45.2 / 12 = 3.766666666 and Munck Perimeter Great Pyramid 3018.110298 / 8 = 377.2637873; Saturn Synodic Period 378.09 days.
89 / 12 = 7.416666667; both 109.6622711 and 741.2764993 have both been reported in the layout of Giza (the later figure in the enclosure wall of Mycerinus’ pyramid)
4450 inches / 12 = 741.6666666 / 2. Hashimi Cubit 1.0673438159 / 12 / 12 = 741.2764993 / 10^n
Concerning the possibility that 45.2 / 12 = 3.766666666 references Saturn’s Synodic Period of 378.09 days: If we obtain a Saturn Synodic Period value from dividing the Great Pyramid’s perimeter by 8 as measured from the base,
3022.416640 / 8 = 377.8020800
377.8020800 / 7.412764993 = 50.96641811 – recognize that from the previous posts on the Sepulchral Chamber? – and
377.8020800 / (7.412764993^2) = 6.875493579 = 4 Royal Cubits in Feet = Mars Orbital Period / 100 (textbook value 686.971 days).
So 7.412764993 is an astronomical constant whose square is able to link Saturn with Mars.
I don’t want to try to grapple with it just yet, but we can again tentatively ask the question of whether the irregularities, this time in the Offering Chamber, may be intentional.
129.7 / 126.4 = 1.026107595 = ~1 Inverse Roman / Egyptian Foot = 1 / .9733868822 = 1.027340740
105.4 / 104.2 = 1.011516315 = 16.18426104 / 16 = ~16.18829140 / 16 = 1.011768213.
Let’s look at the mean values here: (129.7 + 126.4) / 2 = 128.05 and (105.4 + 104.2) / 2 = 104.8.
The are approximately 128.0925795 and 104.7197551 ((Pi / 3) x 100) respectively, which have a product of 128.0925795 x 104.7197551 = 134.1382356 – there’s that again – and a ratio of 128.0925795 / 104.7197551 = 1.223194032. Combining 128.0925795 with 1/3 Pi actually generates a significant if modest series which includes the diameter of the Stonehenge Lintel Circle.
Let’s convert those into inches now: 104.7197551 / 12 = 8.726646260 ft = 15 Inverse Royal Cubits = 15 / 1.718873385, and 128.0925795 / 12 = 10.67438159, 10 Hashimi Cubits length, whereas the Sepulchre Chamber just outside the Offering Chamber is 10 Royal Cubits in length.
So, we may have already recovered the original basic design specifications and the extended design specifications for the floor plan of the Offering Chamber.
An initial look at volume:
For the entrance way into the Offering Chamber, the volume of a rectangular box below the roof arc in cubit feet estimates at about length (89 / 12) x width (45.2 / 12) = floor area ~27.93611111 sq ft x height (62.8 / 12) = 146.1989815
This seems to refer back to data from the Great Pyramid’s missing section which is also reiterated in the King and/or Queen’s Chambers of the Great Pyramid.
That is, floor area ~27.9361 square feet and projected height of Great Pyramid missing section = 27.9454657 ft.
27.9454657 x (2 Pi / 12) = 146.3221163 / 10 and this is a number we’ve run into several times recently exploring the pyramidal architecture of the Faiyum Oasis.
For the Offering Chamber itself, we’ve suggested mean length and width of 128.0925795 and 104.7197551 inches respectively.
The meaning of 70 inches height of the Offering Chamber walls isn’t certain, but we can explore: 70 inches / 12 = 5.83333333 ft, which looks a bit like half a Royal Cubit, but 5.83333333 x 2 = 11.6666666 and 11.6666666 / 12 = .9722222222, about 1 Egyptian / Roman foot of .9733868822 modern feet = .8 Remens, so (.9733868822 x 12) / 2 = 584.0321293 Venus Synodic Period / 10^n.
In the course of building the volume figure, we note that 5.840321293 x ((Pi / 3) / 12) x 100 = 50.96641797; 5.096641797 x 10.67438159 = 544.0349937 cubic feet.
In linear feet, 544.0349937 = 200 Megalithic Yards of 2.720174976.
Let’s go back to this diagram for a moment because I’d like to point something out
Little if any work on diagonals has been done yet for the Sepulchral Chamber and Offering Chamber. The West door seems like an inviting place to start. We have a door 81.9 inches high and 61.1 inches wide.
sqrt ((81.9^2) + (61.1^2)) = sqrt 1044.082 = 102.1803308
One of our original “Wonder Numbers” from Tikal in Guatemala which was later found at Stonehenge and at Giza, is 102.1521078. It was found at Giza because it is the perimeter of the missing apex section of the Great Pyramid when expressed in Royal Cubits (something that no one thought to do with the model because the value in feet is already factored across the Royal Cubit as ratio, wherein 1 Royal Cubit = 1.718873385, and the size of the missing apex section is the total size of the pyramid from the pavement, divided by 17.18873385).
In feet this would be 102.1521078 / 12 = 8.512675648, another one that “sort of” looks like half the standard Royal Cubit. This is the reciprocal of 1 / 8.512675648 = 1.174718786.
This number got its own post for being flagged as a “Megalithic Wonder Number” and appears on the list of Wonder Numbers.
Note that the mean value estimate for the Sepulchral Chamber of 102.5326195 waxes rather close to 102.5135530, which at Tikal is the “companion” Wonder Number to 102.1521078 (I had to work at not getting the two confused). At Tikal, they are linked by the powerful series former (Pi / 3) which we appear to have already found in the Lahun Pyramid. These are based on 1/2 of the Venus Orbital Period written as 224.8373808 (it’s Mayan, of course it has Venus in it):
(224.8373808 / 2) / ((Pi / 3)^2 = 102.5135530 and 224.8373808 / 2) / ((Pi / 3)^3 = 1 / (102.1521078 / 10^n).
Finally, if I didn’t do it previously, I should like to point out the possibility that one or more of the equatorial “Pi Root” may have also been built into the Sepulchral Chamber by means of the variable wall height. Petrie gives height of 71.7 and 72.0 for the height of the walls on the South and North sides of the West door respectively.
72.0 / 71.7 = 1.004184100 = ~1.003877283; 100.3877283 x ((2 Pi)^3) = 24901.1974 = ~equatorial circumference in miles (textbook value ~24901.55)
Edit: I’m going to add a couple more comments on this rather than kick them forward to another post.
The difference in height between the entrance way and the Offering Chamber is estimated at 109.6 / 72.8 = 1.505494502. I’m not certain what this is but if the numbers it’s built out of here are what I suspect they are, is would be one of our more enigmatic metrological values / tools / building blocks and one which may have been used in association with one of the surviving Egyptian pyramid capstones.
109.6622711 / 73.00401618 = 1.502140250
The difference between height of the walls of the two chambers would be about 70 / 62.8 = 1.11464968, for which my first guess would be 1.115419204. For fellow students of Munck, that’s the Double Giza Vector / 10^n. In feet it is seen as a measure in Sacred Cubits (Inverse OR Forward!); it is the square root of 1/2 of the consensus mean Earth circumference 24883.2 miles; it is also geodetic in that (1.115419204 x 360) / 10^n = 1 / (24903.44229); it is the square root of 12 x 25920; the Maya were apparently fond of it and perhaps especially so in the form of Munck’s Giza Vector; it can also be fashioned from Royal Cubits and Megalithic Feet (Munck’s original context for it recontextualized metrologically) etc etc.
In short, it’s an important number and we should expect to see it fairly often accordingly.
Lastly, I was looking near the door to the Offering chamber. The door width is given as 41.1 inches and just inside it widens to 45.2 for the entrance way. I’m still not certain if 45.23893421 was meant, however likely, but we note the ratio between 45.2 and 41.1 is 45.2 / 41.1 = 11.099756691, nearly the Indus Foot.
If we fill in the pieces as 45.23893421 / Long Indus Foot 1.100874628 ft = 41.09362961 inches, which is probably the first thing I should have guessed for 41.1 is Inverse Remens converted to inches. (5 / 41.09362961 = 1.216733603 ft, standard Remen).
–Luke Piwalker
Luke Piwalker 