The Pyramid of Senusret II

Senusret II’s pyramid complex from Mark Lehner, The Complete Pyramids.

Data was published on some of the measurements of the interior of Senusret II’s pyramid at Illahun in Illahun, Kahun and Gurob, by WMF Petrie at al, Chapter 1, pages 1-4 and see Plate II for a diagram of the passage system. https://archive.org/details/cu31924086199514

An overview and diagrams are provided by Keith Hamilton in The Pyramid of El-Lahun A Layman’s Guide, https://www.academia.edu/39001967/The_Pyramid_of_El-Lahun_A_Laymans_guide – one of many excellent guides by Keith available here: https://independent.academia.edu/KeithHamilton?swp=tc-au-39001967

The diagram from Petrie et al shows first an entrance passage, which continues through an entrace chamber, then another strethc of passage leading to the passage chamber, with the passage continuing on to the east end of the limestone chamber. A passage leads from the south side of the limestone chamber to the granite chamber or sepulchre, first branching off to the left then turning right four times to lead back to the granite chamber.

I should say that after spending an evening pouring over the data, that what I see is sort of a strange mix of the self-explanatory, and of the less self-explanatory, with plenty of risk of confusion between the metrological values used and straightforward expressions of Royal Cubits or common subdivisions thereof. Some of it seems simple, and some of it deceptively simple.

This pyramid and data represent a welcome opportunity becase there are rounded ceilings in a number of the passages and both the height of the walls and the height of the arches are given along with the passage width.

It represents an opportunity to further explore various hypotheses, the main one at present being that both of these height values will be meaningful, not only in relation to the width, but to each other as well.

The numbers sometimes look rounded, but the results of inquiry may nonetheless be recognizable.

The passage from the entrace chamber to the passage chamber is described as being 684 inches long, 64 inches wide, 74 inches high on the sides, and 80 inches high in the center, to the the top of the arch.

684 inches / 12 = 57.00000000 ft
80 inches / 12 = 6.666666666 ft
74 inches / 12 = 6.166666666 ft
64 inches / 12 = 5.333333333 ft

Several measures here may already be self-explanatory.

6.166666666 = ~6.164045669 ft = 10 Inverse Assyrian Cubits = 7.5 Inverse Remens
5.333333333 = ~5.337190795 ft = 5 Hashimi Cubits = 5 x 1.067438159

74 / 64 = 1.156250000
80 / 74 = 1.081081081
80 / 64 = 1.25

It’s not certain what these mean. 1.156250000 looks like a slightly large figure for the Egyptian Royal Foot in modern feet, while 1.25 looks a little less like data than it could.
.125 is the reciprocal of 4, and 2^n is something we were probably expected to use as an obvious probe just plying simple reasoning, so there’s relatively little need to actually write 4 or 125 as data.

1.081081081 typically might mean 1.080000000, 1.081540977 (10.67438159 / (Pi^2)), or 1.082323231. I’ve mentioned these have been important to expressions of the Mayan calendar and the same would go for ancient Egyptian calendars.

1.082323231 x 4000 = 4329.292924, so far considered a likely and rather useful approximation of the Jupiter Orbital Period of 4332.59

What is interesting is that we will see more of numbers in close range to these figures in the passage system, including that

684 inches / 74 = (1 / 1.081871345) / 10, so we actually seem to be able to find a number that looks like this twice in the very same part of the passage system.

The passage leading from the passage chamber to the limestone chamber is described as being 894 inches long, 76 inches wide, 69 inches high on the walls and 79 inches high in the center.

894 inches / 12 = 74.54 ft
76 inches / 12 = 6.3333333333 ft
69 inches / 12 = 5.75 ft
79 inches / 12 = 6.5833333333 ft

I don’t have much here in the way of predictions, but I should note that 6.5833333333 much resembles the 6585.322 day Saros Cycle / 10^n. 5.75 could be 5.741903085, although it’s only several 100ths of a foot away from 5.729577951 ft.

I believe 74.54 feet may mean 74.54995935.ft, or 125 Inverse units of 1.676727943, a mysterious unit of measure whose existence is supported by a number of important relationships to known and established units. It may be a variation on the Royal Cubit, since as Jim Wakefield has pointed out, 1.676727943 could be essentially derived from the Indus / Northern Foot (which can in turn be directly related to the Royal Cubit).

110.0078966 ft = 100 Indus Feet = 64 Royal Cubits.

76 / 69 = 1.101449275
79 / 69 = 1.144927536 = ~Double Radian / 100 = (57.29577941 x 2) / 100 = 1.145915590
79 / 76 = 1.039473684 = ~1.039030303 or 1.040913798

1.145915590, which may likely be a shorter version of the Egyptian Royal Foot, when taken literally as 2/3 of one Royal Cubit of 1.718873385

1.039030303 is 1/100 of the outer sarsen circle diameter of Stonehenge, and the square of the ratio between the height of the Cheops and Chephren pyramids in Munck’s models. 1.040913798 is 1/100 of the outer lintel circle diameter of Stonehenge. Both may have functions as approximations of twice the important calendar number 52 (as in 52 weeks in a year, or 52 years in the Half Venus Cycle).

I’d like to note at this point that the first passage entered, the south passage, is stated to be 734 inches long. Compared to the 894 inches given for this passage

894 inches / 734 inches = 1.217983651, the Remen as ratio once again.

The passage between the granite chamber and adjacent chamber is said to be 42.5 inches wide, and in height, 62.8 to the top of the walls and 72.8 at the center.

Taking the numbers first in inches, 62.8 much resembles 2 Pi (a relevant figure in Royal Cubits, metrologically speaking) and 73.00401616, or 1/5 of a 365.0200808 day year.

The height to the arch is therefore 73.00401616 / 12 = 6.083668013 = 5 Remens

42.5 / 12 = 354.1666666. Lunar Year = ~354 days.

Adjusting to the customary lunar value, the width is therefore (57.29577951 / 1.618829140) = 3.539334578 and 3.539334578 x 12 = 42.47201498 in (5 / 1.177245771 = 4.247201498), and ratio between 73.00401616 and 42.47201498 is

73.00401616 / 42.47201498 = 1.718873385 – now the Royal Cubit as ratio once again.

(73.00401616 / (42.47201498^2)) = 1.618829140 / 4 / 10^n.

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The width of the limestone chamber is given as 123.7 inches to the East end and 122.8 West. The mean is (123.7 + 122.8) / = 123.25 inches.

In inches this looks much like 100 times 2 / 1.622311470 = 1.232808888 — for some time now, one of the two most powerful mathematical probes (data retrieval tools) known, something that ancient architects like to build into their architecture in order to help maximize data retrieval capacity.

Something that I hadn’t noticed for there being so much else to notice is that 123.2808888 inches / 12 = 1.027340740.

That in itself may explain a great deal, since 1.027340740 seems to be very popular in ancient architecture, particularly Giza and Egypt. It was presumed that this is largely because this would be the value in feet of an inverse Roman / Egyptian foot of 1 / 1.027340740 = 0.973396992 (1/100 the inner sarsen circle diameter of Stonehenge = 1/100 of 80 Remens of 1.216733603 (this relationship between the Roman / Egyptian foot and the Remen is why the Roman / Egyptian foot has been largely ignored in my work).

It would appear that there was still more to 1.027340740 than meets the eye.

The length of the limestone chamber is given as 196.7 N and 195.3 S. The mean is
(196.7 + 195.3) / 2 = 196 inches / 12 = 16.33333333, which looks like 10 of the straightfoward form of the Assyrian Cubit of 1.622311470 ft = .75 Remens.

If this is correct so far, the length / width ratio would be 16.22311470 / 10.27340740 = 1.579136703 = 16 x (Pi^2) / 10.

The metrological value of 1.579136703 isn’t necessarily obvious, but interestingly it is the reciprocal of 1 / 1.579136703 = 6.332583981, while the width of the passage to the limestone chamber 76 inches maybe 75.99088777 inches / 12 = 6.33573981 ft.

6.332583981 is 1/8 of the fourth root of

(((6.332583981 x 8)^2)^2) = 6586.941099 x 1000, 6586.941099 being very good representation of the 6585.322 day Saros Cycle.

The slope of the limestone chamber calculates at about 128.7705421 inches. If there is a way for this to come out 129.0994449 (100 x (1/ sqrt 60)), the granite chamber would embody both of the two most powerful mathematical probes known, 2 / 1.62231147 and 1 / sqrt 60.

On their W ends, the granite chamber and adjacent chamber measure 123.7 and 129.7 inches respectively; on their E ends, 123.1 and 126.4 respectively.

129.7 / 123.7 =1.048504446, resembling 1.047197551 (Pi / 3), another very powerful data retrieval tool.
126.4 / 123.1 = 1.026780747, looking like 1.027340740 again, this time as a ratio rather than a measure.

On their South sides the granite chamber and adjacent chamber measure 104.2 and 206.9 inches respectively, and on the North 206.2 and 105.4

206.9 / 104.2 = 1.985604607, which might be 1/10 the value of the Pole/Rod in inches
1.981574329
206.2 / 105.4 = 1.956356736, which might be 1.962076285 = 10 x (1 / 1.622311470) / Pi)

The mean measures for the granite chamber are (123.7 + 123.1) / 2 = 123.4 inches and (206.9 + 206.2) / 2 = 206.55

As 123.2808888 / 12 = 10.27340740 ft and 206.2648062 / 12 = 17.18873885, their ratio would be

17.18873885 / 10.27340740 = 1.673128805 and their product would be

17.18873885 x 10.27340740 = 176.5868655 = 150 x 1.177245771.

1.673128805 – which I tend to think of as something like “the next best thing to 1.676727943 in that range” – is a venerated number that has been around for some time in Munck’s publications, along with some of its simple multiples or fractions.

Numbers like these do also appear in my own work. Mean earth diameter in feet 4.182822013 x (10^n) x 4 = 1.673128805.

1.673128805 may also appear in data for a surviving Great Pyramid casing block. (I will have to find that material again but many thanks to Glass Jigsaw for sharing the link).

At any rate, the work here is largely unfinished, but still also rewarding. I encourage others to have at the data and see what they can find.

Cheers!

–Luke Piwalker