Another work in progress, some internal debate continues over the exact original specifications of the Bent Pyramid of Dashur – the parts seem identifiable but exactly how they work together seems uncertain.
Here are some recent notes on the Bent Pyramid from a GHMB post
I’m in general agreement that some sort of circumference figure is being expressed here, and I guess I do seem to have some possible preference for Dorner’s data as it is expressed here. I’d usually want to stick with Petrie when in doubt, but Dorner’s data for this seems pretty remarkable and difficult to ignore.
I think it’s a tough one though – this is where I really get a sense of pyramids having been designed somewhat irregularly for a purpose
(1 / .3048)
x N 189.41 = 24856.95538 / 40 Polar = ~24860 (189.4332000 m)
x E 189.75 = 24901.57480 / 40 Equatorial = ~24901.5
x S 189.71 = 24896.32546 / 40 “” ?
x W 189.57 = 24877.94276 / 40 Mean = ~24883.2 (189.6099840 m)
I work with a pair of equatorial circumference figures, 24901.19742 and 24903.442232, and possibly both were accommodated here. There may be a little tolerance in the data since this is another case where there’s a lack of consensus between sources which sides are longest and shortest.
I might have to guess that Dorner nailed the East side but came up a little short on the South side (24903.442232 / 40 / (1 / .3048) = 189.7642305 m vs 189.71)
That would give a mean of 189.6393536 m = 622.1763570 ft x 4 = 24887.05428 (24883.2 to .9998451291, which is within my usual >.9995 tolerance when adding and subtracting.
The mean of Petrie’s three figures shown here of course gives 24863.51706, not far from ~24860, so even if we dismissed Dorner we seem to be looking at the same thing essentially.
An interesting thing about these sidelengths is that with the numbers I use, at least two of them are equivalent to “Palestinian Cubit” x (Royal Cubit^2) using familiar values.
2.107038476 x (1.718873385^2) = 622.5299356 = 24901.19742 / 4
2.105515607 x (1.718873385^2) = 622.0800000 = 24883.20000 / 4
I wonder what “the rules” are for a pyramid like this? Are we allowed to take the different sides and construct abstract equal-sided pyramids from them?
If permitted an abstract pyramid of perimeter 24901.19742 / 4 = side 622.5299355 ft = 360 units of 1.729249821 ft – or just one side of 622.5299355 suffices, I guess. I don’t know if I want to call these cubits or even metrological units, but 200 such hypothetical units make up a side of the Mycerinus as I reckon it (and 1000 of them the perimeter of Silbury Hill).
I also find it interesting that the mean value in meters 189.6393536 seems suspiciously similar to 360 / Half Venus Cycle ~18980 — ((360 x 10^n) / 18980 = 189.6733404 while the figure I work with is 360 / 18983.99126 = 189.6334626), and on another note perhaps implying some ancient meter value at work here, Petrie gives the upper height of the Bent Pyramid at 2277 in = 189.75 feet.
I have the upper perimeter at probably 1622.311470 ft, which may actually work with 24883.2 to give a meaningful upper/lower perimeter ratio (this gives the average of 405.5778675 ft per side to Petrie’s 4873.4 in = 406.11666666 ft), although of course I can’t stop anyone from making 1618.033989 out it or whatever if they prefer, and that may be correct also.
My sketchy notes say that Dorner gave 405.446194 ft for the upper sidelength.
Either way, it would seem a rather meaningful point for a course correction on the slope angle.
People have of course debated over whether the Great Pyramid expresses Pi or Phi, but I wouldn’t doubt it was tweaked to accommodate both, since sqrt (1.61803398 x 10) = ((1 / 24860.28939) / 10^n) and sqrt (2 Pi x 10) = 24902.31984 / Pi (10^n)
We might be able to even better than 360 / Half Venus Cycle, namely the Half Venus Cycle directly as discussed in some even more recent notes appearing for the first time here.
How much of this can all be present and working together at the same time?
If I take Petrie’s 2277 inch / 12 = 189.75 ft height for the upper pyramid to mean 189.8399126 ft (~18980 / 10) and the perimeter to be 1622.311470 (~365 / 225), I get
Base a = 405.5778675 / 2 = 202.7889338, height h = 189.8399126 (Half Venus Cycle A value = 18983.99126)
hypotenuse b 277.78146822226
angle ? 43.111052082443
area S 19,248.71673442 = (18997.72194 x 10) / (Pi^2) (18997.72194 is Half Venus Cycle B, exactly = (5 / (1.622311470^2) x 10^n)
277.78146822226 = 1 / 360 = 277.7777777 to .9999867146
Height as 200 standard Royal Cubits of 1.718873385 ft = 343.7746770 ft
Total height 343.7746770 ft – upper height 189.8399126 = 153.9347644 ft (Petrie 154.75)
(153.9347644 / 8.0) x (Pi^2) = 18990.94035 = ~18990.04038 (18990.04038 is Half Venus Cycle C = perimeter Great Pyramid 3022.416640 ft x (2 Pi))
153.9274695 ft = (height Great Pyramid x 32) / 10^n
However some of these Half Venus Cycle figures might be rearranged somewhat to permit the lower height to be something like 154.1011111 — i.e., about Eclipse Year 346.62 / Venus Orbital Period 225 = 153.8942222.
Not at all certain how interchangeable 153.9274695 and 154.1011111 might be or why exactly.
An alternate way of generating Eclipse Year figures seems to be Half Venus Cycle / ((2 / 1.622311470)^2).
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I’d like a better look at the possible irregularity, for one thing
Flinders Petrie, A Season in Egypt pg 30
43. The height of the pyramid may be approximately calculated from the angles. The place of the N.E. corner, where the change of slope occurs, was triangulated, and it is 1301.1 from the edge of the N. face, and 1288.7 from the edge of the E. face, measuring horizontally. This difference shows a still larger variation than do the angles, as observed from below; the vertical height by the N. at 55″ 2′ appearing as 1860.5, and by the E. at 55° 12′, 1854.2, mean 1857 inches. Assuming the line of change in the face to be level all round, the height of the upper part will be 2277 inches : the whole, therefore, 4134 inches.
1301.1 / 12 = 108.425 ft, 1288.7 / 12 = 107.3916667 ft
Assuming symmetry for a rectangular shape and a lower base of essentially 24883.2 / 4 = 622.08 ft,
622.08 – (108.425 x 2) = 405.23 ft
622.08 – (107.3916667 x 2) = 407.2966666 ft
Mean (108.425 + 107.39166666) / 2 = 107.908333333
622.08 – (107.908333333 x 2) = 406.263333333
407.2966666 x 4 – 1629.1866667 = ~1631.553867 (1.177245771^3 x 10^n)?
Using Petrie’s mean figure 1857 in / 12 = 154.75 ft for the lower height, the lower slopes would be
sqrt ((154.75^2) + (108.425^2) = 188.9538121 = ~Half Venus Cycle
sqrt ((154.75^2) + (107.3916667^2) = 188.3627685 = ~sqrt Lunar Year, as at El Castillo at Chichen Itza
mean
(188.9538121 + 188.3627685) / 2 = 188.6582903 = ~6 Pi
(405.23 x 2) + (407.2966666 x 2) = 1625.053333 = ~Solar Year / Venus Orbital Period = ~1625.801322 height of Cheops pyramidion in standard Royal Cubits.
Just how many of those proposals might fit together or not?
Maybe it wouldn’t hurt to try and have another look at where they might have been headed with this?
Data from Dorner via John Romer, attributed to Dorner, Form und Ausmasse der Knickpyramide MDAIK 47 pp 81-92
Source – Pyramid – Base – Height
Dorner 1986 Bent Pyramid 1 (first scheme) 156.99 123.43
Dorner 1986 Bent Pyramid 2 (second scheme) 189.43 136.00
Dorner 1986 Bent Pyramid 3 (upper part, as completed) 123.58 57.56
(1 / .3048) x 156.99 = 515.0590551 ft = ~300 cubits of ~1.72
(1 / .3048) x 123.43 = 404.9540682 ft = (117.7192059 x 2) cubits of ~1.72
—> (117.72445771 x 2) Royal Cubits of 1.718873385 = 404.7072841 = 1618.829139 / 4
—> Perimeter / height ratio (1200 x 1.718873385) / 404.7072841 = 5.096641803 = 162.2311470 x Pi
(189.43 x 4) / 136 = 5.571470588
(1 / .3048) x 189.43 = 621.4895013 = 24859.58005 / 4
(1 / .3048) x 136.00 = 446.1942257 = ~5577.096019 x 8 = 446.1676815 = 24883.2 / 5577.096019 or ~89298.07632 / 2 = 446.4903816 = 24901.19742 / 5577.096019
Munck’s 5577.096019 “Giza Vector Grid Point” value certainly seems to want to fit in there somewhere.
5577.096019 = 1729.249824 / (Pi^3)
446.1676815 / 343.7746770 = 1.297849177
446.4903816 / 343.7746770 = 1.298787873 <– favored?
446.1676815 – 343.7746770 = 1.023930045 = ?
446.4903816 – 343.7746770 = 1.021757046 = 1.0215135539 <– favored??
1.0215135539 x (Pi^3) = 107.3519416 see 107.3916667 from Petrie’s data
1.021757046 = 1.0215135539 to .999761692
Postscript: Some additional notes from another GHMB post so that as much as possible on the Bent Pyramid is here in one place for consideration. You can see here much about where the model may be headed, and why isn’t already there yet. The Bent Pyramid may still present us with some tough interpretive choices.
Note that for some possible scenarios, the Half Venus Cycle itself / 10^n may be even more amenable than the alternative, (360 / Half Venus Cycle) x 10^n. This is a number that the Bent Pyramid may be capable of being fairly saturated in, depending on the specifics.
Also, speculative projection of a rectangular based upper pyramid from Petrie’s data may result in an upper base perimeter different from the projection from the data of 1622.311470 for an equal-sided pyramid.
(405.23 x 2) + (407.2966666 x 2) = 1625.053333 = ~Solar Year / Venus Orbital Period = ~1625.801322 height of Cheops pyramidion in standard Royal Cubits x 10^n, accuracy 1625.053333 / 1625.801322 = >.9995 = .9995288724
One of the curious things about this number 1625.801322 is that “divided by what we could have hoped it was”,
1625.801322 / 1.622311470 = 1.002151160, a very familiar ratio from working with pyramids that has geodetic utility (1.002151160 x ((2 Pi)^3) = 24858.38047 / 10^n), and
1625.801322 / (1.622311470^2) = (1 / 1.618829104) / 10^n
1.618829104 thus far being the next most important Phi-like number in my vocabulary after 1.622311470, 1.618829104 also having geodetic functions when paired with 2 Pi.
An interesting thing about the value 1625.801322 is that it would work to give a pleasing lower / upper perimeter ratio for the Bent Pyramid of 152.8992514 / 10^n.
Besides the fact that this is 1/2 of the standard value given for the inner sarcen circumference of Stonehenge in feet, this may be an interesting number because of recent efforts at geodesy.
In terms of qualifying the Remen for geodetic functions, we can say that ~108 x (10^n) x Remen = Earth Circumference in feet, but what is the equivalent for the Royal Cubit?
For the Circumference values in miles, it’s been proposed that ~((12^2) = 144) x 10^n Royal Cubits can be considered as making up the circumference at a cubits:miles ratio, but what about the actual number of Royal Cubits in the Circumference value in feet?
In lieu of a nice round number for this, an interesting number is possible, which is
76.44962559 x 10^n x 1.718873385 = 131407226.7 = 24887.73233 x 5280 = ~24883.2 x 5280
131407226.7 is the answer I get using any of the metrological units I use, for John Michell’s 12^n metrological formulas = 108 x 10^n x 1.216733603
76.44962559 x 2 = 152.8992514 and 76.44962559 x 4 = 305.7985024 = 360 / 1.177245771 and 76.44962559 x 8 = 611.5970047, Great Pyramid apothem length
This may also raise questions about whether the mean of three attributed to Petrie of 189.46 m = 24863.51706 inches might not have validity to indicate a correct mean for the Bent Pyramid of the Polar Circumference, rather than the Mean Circumference.
It’s very difficult getting away from the idea of terrestrial circumference here, but there are choices to be made when it comes to circumference.
Again, just how many splendid proposals can the Bent Pyramid data accommodate?
I wish I had a complete picture to offer, but at least we might find some pointers for now?
Are we finally beginning to understand this curious and very novel Egyptian pyramid?
–Luke Piwalker
Luke Piwalker 