I’ve been working on this awhile, but have only just now gotten back to the subject since the initial effort, so I will try to recount the entire history of it here as best as possible.
Having been informed by reliable sources that the width of the circles formed the horizontal lintels of the Stonehenge sarcen circle differs from the width of the circles formed by the upright sarcens themselves, I set about to try to deduce what the proportions of these lintel circles might be.
Jan 21, 2020
I noticed some remarks in one of Harry Sivertsen’s works that I fished out from the Google cache, The Metrology of Stonehenge including dimensional explanations and supporting example:
“Petries survey, published in his 1880 work Stonehenge: Plans, Description and Theories p 237 made the inner diameter of the Sarsen Circle 1167.9 inches or 97.325 feet and he claimed that this was within a maximum tolerance of 0.72 of an inch. This is much closer in reality than the survey by Alexander Thom. 100 long Roman feet equate with 97.32096 feet which is within 0.04848 of an inch of Petrie’s conclusion. The lintel width according to Atkinson in the 1950s in agreement with Petrie, was ‘about’ 3.5 feet.8 Note that the Sarsen Circle was here being defined by the lintel circle dimensions and NOT via the stones supporting the lintels. Chippindale states that, Obviously some unit of measure must have been used and the megalithic yard and rod seem to work better than any of Petries measures.9 So here we have another little piece of nonsense, how does one set of measures ‘work better’ than another? Either they fit what is being surveyed or they do not.”
Indeed, I have the sarcen thickness at 3.282806350 ft = 57.29577951^2 based on the interpretation of the data from Petrie and Thom. The difference here between the thickness of the uprights and the thickness of the lintels to take this at face value, is by ratio about 3.5 / 3.282806350 = 1.066160969, much resembling 1.067438159 or possibly 1.066666666.
Assuming the lintels are centered on the uprights, to take the difference 3.5 – 3.282806350 = 0.21719365 ft and add this to the outer diameter, 103.9030304 + 0.21719365 = 104.0312241 ft, and to subtract it from the inner diameter, 97.33868822 – 0.21719365 = 97.12149457 ft, for estimates of the possible intended lintel circle radius, diameter and circumference values.
If we take 104.0312241 to be Pi / Great Pyramid Perimeter, paved = Pi / 3.018110298 = 104.0913798 (.999422087, raw data) and 97.12149457 to be 9.712974840 = 6 x 1.618829140 (.999915022, raw data)
D 104.0913798 x Pi = C 327.0127141 ft
D 97.12974840 x Pi = C 305.1421040 ft
Years ago, when I worked on the outer value for the Sarcen Circle, I spent a lot of time working on 327.0127141.
1 / 327.0127141 = 305.7985078 = 360 / 1.177245771, the inner sarcen circle circumference I work with, and it was a very attractive proposition that the outer circumference might be the reciprocal of the inner circumference and vice versa. Munck insisted they knew their numbers forwards and backwards and I agree.
However, because of the some of the consequences of the proposition including a possible 2.725105951 ft Meg Yard = 327.0127141 / 120 and how well it didn’t and didn’t measure things at the time, I kept looking and found 2.720174976.
It’s surprising, since I have many kind words for 1 / 2.725105951 = 3.669582093, which still comes back from inquires as the possible answer to “366.6666666” that the numbers I use requires.
Note the 305.1421040 ft value = 360 / 1.179778193
The thickness mandated by (57.29577951^2 x 1.067438159) / 10^n = 3.504192766 ft = 360 / (1 / 9.733868822) = 2.880000000 Remens = 1.019328360 x 2 Royal Cubits.
1.019328360 = ht Cheops / ht Chephren (Munck) = mean sarcen circle diameter 100.6036766 / (Pi^2)
Some of the consequences of this proposition are still a bit strange, but they may start to make more sense a little later as so many things often do. This might have actually been been part of the design?
It does increase my confidence in the idea of 327.0127141 that 360 / 327.0127141 = 1.100874628, there’s a meaningful output from the 360 / x test, my favorite Indus foot candidate – and it looks like 327.0127141 / ~56 = Venus Synodic Period ~584. Let’s see how well that actually works
327.0127141 / 584.0321292 = 55.99224730. Having proposed 892.9807632 for the max on the sarcen circle based on Thom’s later figures, 5 / 55.99224730 = 892.9807632 / 10^n, although it’s easy enough to see why 55.89093144 would have also been intended, or mainly intended, or whatever.
55.89093144 would also be needed because it links the outer sarcen perimeter established by the uprights to the VSP (120 x 2.720174976 = 326.4209969) / 55.89093144 = 584.0321292 / 10^n.
326.4209969 / (55.89093144^2) = 1.044949716 which is likely also meaningful. 1.044949716 / 2 = 5.224748579 = (57.29577951^2) / (2 Pi) = 20.62648063 / ((2 Pi)^2) = 360^2 / ((2 Pi)^3).
Feb 04, 2020
I’ve encountered a more accurate value for the Earth Year of 365.2437808 (1.177245771 / (12^8)) that I mentioned in the Harris and Stockdale Megalithic Foot thread.
I think it’s interesting that I encountered it while trying to work with 1/120 of the possible value for the sarcen lintel circle, (1 / 305.7985078) = 327.0127141; / 12 = 2.725105950. Somehow I got from there to 365.2437808 using 12s and 36s, according to my notes.
So far if I go to explore this strange number 365.2437808, I start to see more familiar versions of the Earth year in the equations instead
(Note: (1.177245771 / (12^8)) x 10^n = 365.2437808)
February 08, 2020
I’m not sure if this belongs under Stonehenge, or the Megalithic Yard, but I may not have gotten much into the sarcen circle mean in some aspects, and of course on aspect would be what it measures in Megalithic Yards rather than feet.
There’s that disparity between measures between ~1.006 and ~1.008 and perhaps I’m finally starting to understand that a little better?
I don’t know how others are going to approach this problem, but in terms of Megalithic Yards, I see the base unit for the sarcen circle mean as half of the Squared Munck Megalithic Yard
100.6036766 / AEMY 2.720174976 = 36.98426663 = SMMY 7.396853331 / 2
36.98426663 x AEMY 2.720174976 = 100.6036766
36.98426663 x IMY 2.71925644 = 100.5697054
36.98426663 x DMMY 2.721223218 = 100.6424451
So it’s kind of like a possible advertisement for different interpretations of Michell’s long/short unit ratio.
If the outer lintel circumference is 327.0127141, to divide it by 120 as we do the outer upright circumference of 120 MY,
327.0127141 / 120 = 2.725105950
To treat it as if it were a Megalithic Yard
36.98426663 x MY ? 2.725105950 = 1.007860451, or 1.008 to .9998615582
The ratio between 2.725105950 and 2.720174976 is 2.725105950 / 2.720174976 = 1.001812741, the “New 2 Pi Root Ratio” posted to my blog, which would make it a sensible pairing in that respect too.
April 10, 2020
I decided last night that it was perhaps time to revisit my interpretation of the Stonehenge lintel circle, because it’s never been adequately explored.
I have yet to explore the suggestion of the lintel width being 2 Royal Cubits, I’m working with the figure of 3.5 feet with this, just to see what the possible merits of that might be.
I obtained an inner diameter of 97.12974854 ft for the lintel circle based on the figure of 3.5 feet, and an outer circumference of 327.0127141, which is the reciprocal of the inner circumference of the sarcen circle.
The inner circumference of the lintel circle is then 97.12974854 x Pi = 305.1421048 ft, and its outer diameter 327.0127141 / Pi = 104.0913798 ft
360 / 305.1421048 = 1.179778193, the “next best thing” to 360 / 305.7985077 = 1.177245771.
In case I haven’t done so previously, I should point out what these dimeters consist of
Inner diameter sarcen circle = 97.33868819 ft = 60 x 1.622311470
Inner diameter lintel circle = 97.12974854 ft = 60 x 1.618829140
Hopefully, that should remind us of this, where the architects appear equally eager to pay homage to both of these very important “Phi-like” figures
The ratio between the two is 97.33868819 / 97.12974854 = 1.002151140, the “Polar 2 Pi Root” ratio: 1.002151140 x ((2 Pi)^3) = 24858.38047 / 100
1.002151140 = (1 / 72) / (1.177245771^2)
The metrology of the lintel circle appears to be different than the sarcen circle. The sarcen circle can be seen as having diameter of 80 Remens
97.33868819 / 80 = 1.216733603
Trying to make the same of 97.12974854 seems to give either unwanted forms of the Remen, or unwanted forms of 80.
The actual base unit may be Petrie’s Stonehenge Unit
97.12974854 / (224.8373808 / 12) = 5184 / 100
51840 / 2 = 25920, 25920 x 1.177245771 = 305.1421048, inner circumference of lintel circle.
25920 / (1.177245771^2) = 187.0254547 = 22443.05456 / 12, data that is also in the Aubrey Circle in my model
However, we can force the issue to obtain
97.12974854 / 1.216733603 = 7.982827818 = 399.143509 / 200, with 399.1413901 being the current A value for the 398.88 day Jupiter Synodic Period.
97.12974854 can also be reckoned in Morton Cubits to give the Lunar Calendar Month
97.12974854 / 1.718873385 = 56.50779705 = 2 / 353.9334577
305.1421048 / 1.718873385 = 177.5244805 = Lunar Leap Year 355.0489611 / 2, whose corresponding Lunar Leap Month is 355.0489611 / 12 = 29.58741342 days = Squared Munck Megalithic Yard x 4 (7.396853356 x 4 = 29.58741342)
Also, 97.12974854 / 1.177245771 = 82.50592267 = 1.718873385 x 48
305.1421048 can also be seen as (57.29577951^3) x 1.622311470
Still, I was treated to an interesting surprise when applying the Radian value 57.29577951 as a mathematical probe
97.12974854 / 57.29577951 = 1.695233914, which is my nomination for a Nippur Cubit in feet that is distinctive from a Karnak Cubit of 2.721074976 / 1.6 = 1.700109360 ft
What this Nippur Cubit is, is my flavor of the HSMF, gone through the Berriman / Michell 12^n metrology check.
1.177245771 x 12 x 12 = 169.5233910 = 1.177245771 x 144
This is to the Morton Cubit, as the Imperial Foot is to the Greek Foot
1.718873385 / 1.695233914 = 1.013944668 = 365.0200808 x 360
This unit value is still largely untested, but you can why I am just as happy for Peter Harris’ sake that this turned out to be the winning suggestion based on sheer mathematics, rather than 2 / 1.177245771 = 1.698880598.
However, I should point, however obvious, that 1.698880598 does also have the potential to be a metrological unit, and in a directly analogous manner.
Inner diameter sarcen circle = 97.33868819 ft
97.33868819 / 1.698880598 = 57.29577951
We should perhaps not let it escape us that these values then link actual radius and diameter, to generic radius and diameter.
In terms of whether the putative ancient pied du roi was applied metrologically to Stonehenge, we can observe that for the inner circumference sarcen circle (360 / 1.177245771 = 305.7985077 ft)
305.7985077 / 1.067438159 = 57.29577951 / 2
To continue with the Radian probe,
97.12974854 / (57.29577951^1) = 1.695233910
97.12974854 / (57.29577951^2) = 29.59741335
97.12974854 / (57.29577951^3) = 5.163977801 = (1 / sqrt 60) x 4 = 1 / sqrt 3.75
97.12974854 / (57.29577951^4) = 1.802568302 x 2
97.12974854 / (57.29577951^5) = 1 / (54 x 1.177245771)
97.12974854 / (57.29577951^6) = 1.372734261 x 2 = 1 / 364.2365563, which may yet prove to belong to an unexplored calendar group representing 364.
I think 1.802568302 has surfaced in discussions lately, in the context of the similar number 1.803872059 = 1.80256302 x 1.000723277 = 2.920160646 / 1.6188219140 that was part of the original study of Avebury that I did.
1.80256302 / 2 = 8 / SRVS 887.6223994
360 / 97.12974854 = 7.412764996 / 2, with 7.412764996 x 10^n seemingly appearing at least once in the enclosure wall of the Mycerinus Pyramid according to Flinders Petrie’s data.
If we use 1.177245771 more as a mathematical probe than a metrological unit,
97.12974854 / (1.177245771^1) = 82.50592251 = 48 x 1.718873385
97.12974854 / (1.177245771^2) = 7.008385550, one of the candidates for what this system of numbers has to offer for the number 7, which does not belong to it.
97.12974854 / (1.177245771^3) = unknown
97.12974854 / (1.177245771^4) = 505.6892338 = Venus Orbital Period A 224.8373808 x Venus Orbital Period C 224.9133276
97.12974854 / (1.177245771^5) = unknown
97.12974854 / (1.177245771^6) = 364.8794717 = Earth Calendar Year C value
We can also observe that for the lintel circle model in question
outer circumference 327.0127141 x inner radius 97.12974854 = 317.6266269, which appeared early on in my work on Silbury, and repeatedly surfaced in the vicinity of Tikal, no doubt for having its own significance to calendars even if I don’t have that information handy.
(I still have very little of Tikal area data logged, the most interesting thing listed is still 5400 / 170.0109360 = 3.176266269)
The interaction between 97.12974854 and the Squared Munck Megalithic Yard 7.396853364 ft includes
97.12974854 / (7.396853364^1) = 4 x (57.29577951^2)
97.12974854 / (7.396853364^2) = 177.5244798 = Lunar Leap Year 355.0489596 / 2, as previously discussed as 305.1421048 / 1.718873385 = 177.5244805
97.12974854 / (7.396853364^3) = 240.000000
97.12974854 / (7.396853364^4) = 3.244622937 = 1.622311470 x 2
97.12974854 / (7.396853364^5) = 4.386490839 = (sqrt height Great Pyramid unpaved) x 2: 4.386490839 / 2 = 21.93245419 = sqrt 481.0325470
If we measure this in experimental Indus Foot A, 1.100874628 (1/500 Diameter of Silbury Hill),
97.12974854 / 1.100874628 = 88.22961855 = 4.411480921 x 2 = 240 / 2.720174976
In experimental Indus Foot B, 1.100874628 / 1.000723277 = 1.100078966
97.12974854 / 1.100078966 = 88.29343304 = 75 x 1.177245771
I think I mentioned before that for the outer lintel circle,
360 / outer lintel circumference 327.0127141 = 1.100874628, situated in a prominent position analogous to 360 / inner sarcen circumference 305.7985077 = 1.177245771 as if indeed they were fond of the Indus Foot.
A word about my planetary tables – I have as many as 12 or 13 sets of planetary values, possibly more, that are derived from keeping track of and logging values near to 18980, and then attempting to extrapolate the remaining planetary values using a set of formulas with fixed ratios, which often turn out to be notable figures in their own right.
The main goal is often to identify the A and B versions of a particular value, and to spot their appropriate placement in the A and B groups according the telltale ratio between them of 1.000723277, which is the ratio between A and B groups in general.
The A and B (and C) groups are generally the most important, and the ones we might expect to see most often.
However, this would be such an ambitious goal on the part of the ancients that everything may not quite work out as ideally as we might hope, and in fact the larger spread of groups A-L (or more) might be required as a “net” to “catch” some of the more ideal or more realistic values.
As with any system, there may be compromises and trade-offs.
Relative to DavidK’s system, which might be somewhat more inflexible in places if there are unifying values involved, although more useful and consistent, in reference to 346.6666666 (difference 346.6666666 – 346.62 = 0.046666666).
The most prominent value in my tables is the A value of 346.5939367 (difference 346.62 – 346.5939367 = 0.026063300), but due to the nature of the tables, less accurate versions may also appear.
The most accurate value I have in my tables for the Eclipse Year is
346.6251893 but the trade-off so far is obscurity. This value is all the way out in the H or I group, depending on which experimental version of the Eclipse Year we are talking about, and seems accordingly very difficult to reference with equations.
I would also like to point out that so far, nothing is known about 97.12974854 having exponential value. Frankly, I would be skeptical.
So let me show you just what level of intelligence and mathematical skill I think we are talking about.
(Maybe I should stop calling them “stone age Einsteins” and call them something else instead like “stone age Ramanujans” since they seem to be mathematicians rather than rocket scientists, based on a conspicuous absence of Megalithic rockets. Perhaps Aubrey Burl and other skeptics of Thom would find that notion slightly easier to swallow?)
Ready?
Outer lintel circle circumference 327.0127141 / inner lintel circle diameter 97.12974854 = 3366.251810 = unknown, it’s not the expected 3366.458185 = ((2 Pi) x 2.921060646)^2…
Outer lintel circle circumference 327.0127141 / inner lintel circle diameter squared (97.12974854^2) = 346.6251810
That is brilliant as in (expletive deleted) brilliant. Absolutely (expletive deleted) brilliant. I wouldn’t have dreamed that was even possible.
I will eventually go back and try to develop a model for the lintel circle based on the proposed width of 2 Royal Cubits, but I’d like to explore this model a bit more, and I’m very glad I started.
April 10, 2020
The outer / inner sarcen circle ratio according to my model is the quintessential 1.067438159. Using the outer and inner circumference values to demonstrate,
326.4209748 (120 x 2.720174976) / 305.7985077 (80 x 1.216733603 x Pi = 360 / 1.177245771) = 1.067438159
For the lintel circle, the figure ends up rather unusual, not so quintessential, and in fact, downright secular. Even for being made of familiar parts, the constitution of the outer / inner ratio of the proposed lintel circle model is suprising
327.0127142 (1 / 305.7985077) / 305.1421048 = 1.071673522
1.071673522 = (1 / 72) / (360^2) = (1 / 360) / 25920 = 72 / (25920^2) = 1 / 933120
However, if we combine the two ratios in an equation, we obtain
1.071673522 / 1.067438159 = 1.003967783 = 2 Pi Root of alternate Equatorial Circumference
1.003967783 x ((2 Pi)^3) = 24903.44229 / 100
I’m greatly reminded of both the way Tikal embodies both 24901.19742 and 24903.44229 atop the very same temple pyramid, and the secular equations started by Munck and advanced and refined by Michael Morton, that led to my discovery of the figure 24901.19742.
Morton had only to solve the equation he’d cued up, and I would be talking how he discovered the figure, and Munck had only to fine tune his own equation or I would be telling you how he discovered 24901.19742.
I also discovered that the cube root of 24901.19742 = 24 Remens of 1.216733603 ft each, modelling the earth a ratio of 1 foot = 10^n miles.
If the lintel circle outer / inner ratio is only going to know one trick, it certainly picked a very good one.
Also, I frequently talk about the most important Stonehenge numbers, and to me they are
1.216733603, 1.177245771, 1.067438159, and 224.8373808
I noticed yesterday that
97.12974854 / 224.8373808 / 1.216733603 / 1.177245771 / 1.067438159 = 2.825389857 = 1 / 353.9334572, the reciprocal of A value for the Lunar Calendar Year.
That too strikes me as a bit on the (expletive deleted) brilliant side.
Cheers!
–Luke Piwalker
Luke Piwalker 