Luke Piwalker

Reaching For The Sky

Well, there you have it with that last one – looking at my math isn’t so bad after all after having seen some of my philosophy, eh?

Silly me, when I got back into this I thought if it calendars were what it was all about, it should be simple.

Nonetheless, the past week has seen what appears to be the identification of the correct way to assign figures for the Anomalistic Month, the Eclipse Year, the Lunar Year and the Venus Synodic Period. If these have been assessed correctly now, more progress has been made with them in the past few days than in the past few years, and if one dwells on that, it should be tremendously encouraging.

I was hoping that one of the new numbers Stonehenge just taught us – the perimeter of the bluestone “oval with corners” – being an unfamiliar number might optimistically serve as some sort of master key to the missing planetary values, but this may not be the case.

It may largely be a very clever expression regarding the missing values already recently solved.

Still, if we reach for it and stretch a bit, we might just be able to make a little more progress. At least for the data on Mars, Jupiter and Saturn still being missing, thankfully the Saturn Synodic Period seems like it may be solved some ways back with the assignment of 10758.28707 ((1 / sqrt 8640) x 10^n) to represent its 10,759.22 day Synodic Period.

That’s a little bit of departure, but this is calendars we’re talking about, and it’s big number. The error by percentage may be that much more tolerable.

At present I’m wondering if the correct ratio for the Mars Synodic Period might be either “2 Remens to the Builder” or “2 ‘Thoth Remens’ to the corresponding half Venus Cycle”.

I’m still uncertain about 2160 / Pi = 687.5493542 as a representative of the Mars Orbital Period of 686.971 days, but among other things going for it, it has the potential to make the ratio between the Venus Synodic Period and the Orbital Period into 1.17745771, which would be quite welcome and appropriate.

The raw value is estimated at 686.971 / 583.92 = 1.176481367, so 1.177245771 is probably exactly what that should suggest and how we can we stand it being anything else?

This being the case, now that we know that the Venus Synodic Period value of 584.0321292 apparently belongs to a “B” set of calendar values, we would know that 584.0321292 x 1.177245771 = 687.5493542 also belongs to the “B” set if that will hold up under further scrutiny.

Speaking of numbers so important that they are very hard to resist, it’s very hard to resist the idea that Lunar Year / Saturn Synodic Period should equal 1.067438159. Have I written a post yet that doesn’t mention this very important number?

The raw data says 378.09 / 354.367 = 1.066944721, so that one isn’t really much of a stretch either, and already our planet ratio table is beginning to fill up with illustrious figures, if we have them figured right.

Shall we call it that for now, so we can begin looking at how it works out to do so?

There’s yet another possible clue in the form of the ratio between Lunar Year and Jupiter Synodic Period being about 1/2 of the Venus Orbital Period / 100.

Here the raw data says 398.88 / 354.367 = 1.125612712 = 225.1225424 / 2 / 100, but there are a number of possible forms of the Venus Orbital Period, so it’s not entirely certain what was intended here.

If it were the standard Venus Orbital Period figure that’s halved to formed the correct ratio, the Jupiter Synodic Period that corresponds to it would be 353.9334578 / (224.8373808 / 2) = 397.8873581, but that may be the first time that number’s come up in this context and I’m not sure what to think of it, and the difference between 397.8873581 and 398.88 might be too large of one for working with such relatively small numbers.

Up until now the candidacy has been dominated by the possibility of 399.1413898 being the correct value to represent 398.88.

If that isn’t what the right ratio is, though, it may be very close, and indeed may well be half of some figure for the Venus Orbital Period.

Likewise, the ratio between Mars Synodic Period and Earth Year could be 2.134876318 – that’s 1.067438159 x 2.

The raw data gives 365.243 / 779.96 = 2.135455026 = 1.067727513 x 2

The raw ratio for Saturn Orbital Period / Lunar Year is 10759.22 / 354.367 = 30.36180005, resembling 300 / Pi^2 = 30.39635509, which seems to be part of at least one Aztec calendar stone, and indeed the ratio between 10758.28707 and 353.9334578 is

10758.28707 / 353.9334578 = 300 / Pi^2 = 30.39635509

Now that we seem to be able to tell through deduction that 353.9334578 seems to belong to the “B” calendar set, we would also know what 10758.28707 belongs to the same set if this ratio has been correctly identified.

While I’m thinking of, I recently mentioned a number that reminded me of a number in some work being done by others (our metrologist friends on GHMB), and it looks as if it may appear in the planet ratios as “Venus Orbital Period / Lunar Year”:

353.9334578 / 224.8373808 = 1.574175329

The ratio between Jupiter Orbital Period and the Mars Orbital Period could be 6.307546992 – it calculates from raw data at 4332.59 / 686.971 = 6.306801888.

6.307546992 is an important number I’ve discussed before on a number of occasions.

My unfinished planet ratio tables haven’t been quite as much help as they could have been sometimes for not being quite able to sort planetary value candidates into their respective groups properly, but a possible ratio of 2.062648062 between Mars Synodic Period and Saturn Synodic Period.

The raw value is 779.96 / 378.09 = 2.062895078

If the Mars Synodic Period were 779.27277283,

779.27277283 / 2.062648062 = 377.8020801, another vote for the half side length of the Great Pyramid unpaved perimeter as a possible representative of the 378.09 day Saturn Synodic Period, as is 353.9334592 x 1.067438159 = 377.8020801

One other item of note for now: Both the Saturn Synodic Period / Mercury Orbital Period ratio (10759.22 / 87.9691 = 122.3068100) and the Jupiter Orbital Period / Lunar Year ratio (4332.59 / 354.367 = 12.22627954) could be 12.23194029 = Great Pyramid ideal apothem 611.5970145 x 2 = Stonehenge inner sarcen circle circumference 305.7985073 x 4.

The full consequences of these proposals remain to be seen along with the overall sensibility thereof, but it’s possible that more of the missing pieces of a full calendar set are already herein.

–Luke Piwalker

Rock Flattens Paper…

It may be time for a break from Stonehenge, I’ve done a lot with it recently and it’s assisted us in having what look to be like the right formulas for Anomalistic Month and Eclipse Year, and it’s even helped us to understand a “C” set of calendar values and the correct placement of the Venus Synodic Period within its respective calendar set, for the first time since this inquiry into the possibility of ancient calendrical architecture began three years ago.

As I said, I really wouldn’t mind getting back to Mayan studies, and the first place my mind tries to wander back to is still Rio Bec. I’m still in awe of the architect that designed at least one of the main structures partly described by Andrews, who seemed to have an unusual gift for making use of neglected numbers.

Something else that presumably the same architect achieved was making rooms into mathematical formulas on multiple levels. It’s still too soon for me to declare that since I can’t offer all the data involved and indeed cannot count the structure in question as being successfully solved, but I invite the reader to consider this carefully for a moment.

What we are talking about here is most likely not only rooms designed as self-referential geometric formulas, but at the same time the usual type of data display is likely riding “piggy-back” on top of that.

Imagine for just a moment just how mathematically demanding a mandate like that has to be. As I said, I am literally in awe of whoever designed that. It seems to make a relatively small and modest Mayan building into something to in ways rival the Great Pyramid, and we barely have any data on Rio Bec to speak of yet can still point this out.

Anyway, I thought it might make for a good occasion to go over some of the basic premises I work with. I expect every room I encounter in ancient architecture to have carefully chosen length, width, and height that assemble into meaningful mathematical formulas, and years of working with the data have taught me that I should expect that, even if I still haven’t solved half of what I’m looking at.

The premise here is that “sacred” mathematics must have been the domain of both astronomers and architects, with who knows how much overlap between the two in any given instance.

Ancient architecture was used as if “paper”, and ancient architects used it to try to immortalize it to “write down” important formulas that they knew (particular those concerning astronomy, is what it looks like), and important building commissions may not come along every day, so the “paper” they used was still in limited supply, so they let nothing go to waste.

They used stone as “paper” to the best of their ability with no feature being simply by chance, with every measurable quantity having a deliberate and well-chosen meaning.

The idea of nothing being by chance and nothing going to waste here is premise you will encounter as far back in this work as Carl Munck’s own efforts and writings.

Beyond this, the durability of the medium – the permanence of stone compared to that of paper – and the advantages of using a more durable medium to “write on”, are hopefully self-explanatory.

I recently raised the possibility with a skeptic who asks for written proof of this ancient mathematics – show us a papyrus where it’s written down that the ancient Egyptians could conceive of a decimal point or long division – that maybe ancient people simply weren’t in the custom of writing it down because once you start committing these equations to papyrus you’re at risk of depleting the national supply of the stuff.

I’ve filled whole shelves with my limited calculations, and Carl Munck is said to have filled whole rooms with his, the sheer volume of his notes being owed in part to many of them having been taken prior to the general availability of pocket calculators, and we are still barely scratching the surface of ancient mathematics.

I want to tell the reader the point of all this, why all this is so important to consider, and I’ve probably never been good at articulating it. For my own participation in it, I’ve felt like it was important to prove that ancient people were capable of understanding highly sophisticated mathematics, as if it would actually prove that ancient customs were something that should matter to highly sophisticated people and not a bunch of superstitious rot.

The most common thread running through any account of why this matters is something of an object lesson in prejudice – sometimes I think I’d like to approach any member of the 22/7 squad who thinks that 22/7 was the best ancient people could do for Pi as to grip them by the lapels and say, “Look, would it kill you to try to give them the benefit of a doubt that they were smart enough to have figured out decimal points and long division long before a fragmented and devastated historical record gives them credit for?”

What I am saying in my work is what the ancient Egyptians commemorated as metrological units are some of the oldest numbers and math to mankind that demonstrably, easily derive from trying to work out the very first calendars, and that the AEs as we call them, had worked that notion out for themselves, with their architects and mathematics taking it into consideration in their choice of metric units.

That would have given mankind hundreds or more likely thousands of years to perfect mathematics.

What on earth is so implausible or unthinkable about that unless one actually relishes the idea that they are so much smarter than their ancestors?

If I could find my copy of John Michell’s Secret of the Stones, I’d quote the passages on urdummheit, the theory of “original stupidity”, that the early back you go in time, the dumber people were, until finally way back there, they were too stupid to come in out of the cold and became extinct before they could give rise to the likes of us.

What’s wrong with that picture? Often plenty, obviously.

I’ve no wish to become an example of Godwin’s Law here, and that certainly isn’t the point or intention, but there is a reason after all that a German word is being used for the phenomenon of urdummheit, and much more innocently many of have simply spent a lifetime having it drummed into our heads by advertising that what we have is bigger, better, faster, stronger and smarter than anyone has ever had before and that anything that came before it including last year’s model is junk.

Better yet, I could recommend some of U. Utah Phillips’ astute comments on how this is done with with own contemporary culture, the dismissive bundling of popular music or culture by decade – “Oh, that’s that fifties stuff” or “sixties stuff” or what have you.

By this time, having urdummheit creep its way into our view of ancient history risks going so far as coloring what we are able to even perceive about our own past as humans. Remarks may arise like “the ancients couldn’t possibly have known something like that” – because? Because they were ancient? As if a single individual cannot acquire remarkable skill at something with a lifetime of practice, no matter what century it may be?

I am not even going to bother asking what could happen next if prejudice gets a foot in the door, but that isn’t the point. I’m sure it’s how much that decent conscientious people still outnumber raging bigots that keeps us safe before it’s knowing that many ancient mathematicians, like many of us in our vocations, were awesome at what they did.

The ancients, unlike us, may have a very different mindset than the “bigger, better, faster, stronger and smarter” that we’ve become so accustomed to ourselves, and instead honored and upheld traditional ideas and standards to the point that we may look back and mistake that for cultural stagnation.

The tragic part of this is that if we rest a little more faith in them, and their skills and their abilities, the ancients may still have something to offer us, if only reassurance, even if we haven’t really any use for all this fancy math I try to demonstrate that ancient people were doing.

Imagine someone being convinced that the Y2K Bug was going to set civilization back a thousand years, but not looking to people a thousand years ago and their writings for tips and advice on how to live at that level of civilization.

Those people wrote the book on how to survive without the last 1000 years of technology, and they may have some helpful hints because of it, and a phenomenal wealth of ideas may survive as folklore and mythology if we sometimes dig just a little deeper than face value.

That’s really what we’re talking about here, is the possibility of cheating ourselves out of seeing something truly amazing (or even something useful) by reflexively looking down our noses at our ancestors, that’s all.

In case that sounds a little preachy in parts, if I’m preaching it might only be the choir, but I already know I’m not entitled to a soapbox anyway if the sermon is still as much for my own benefit as anyone else’s.

My recent work with Stonehenge has probably taught me already that even the likes o’ me still has at least a nagging touch of urdummheit myself. If I’m not actually a little surprised they were even capable of what I see, I’m still a little too surprised that they summoned the skill to actually pull it off.

I’m rightfully in awe of what I see, but I shouldn’t be surprised by it.

The best part of it is perhaps that with Stonehenge, I have a sense of my own ancestry being involved that I’m not going to get from studying ancient architecture in Egypt or the Americas, and the level of skill I’m seeing in it lately makes me pretty proud to think that this is what I come from, as opposed to thinking they were primitive, superstitious idiots.

As far as I can see, the view seems to be worth the climb.

So, this post being about what mathematicians wrote on back in those days, and how paper was precious, I think it might be safe to say that at Stonehenge, they were definitely not wasting paper.

–Luke Piwalker

Still More Half-Baked Stonehengery

Still in pursuit of an understanding of how to group candidate figures for the Lunar Year value(s), I took up the question with four different figures for the half Venus Cycle, and four of the most familiar versions of the terrestrial year. I have asked Stonehenge, and it think it may have answered.

The answer thus far is that regardless of what figure for the earth year may belong to a particular calendar set (sets A, B, C…), the ratio between the particular half Venus Cycle and the Lunar Year may always be 2 / 365.0200808.

That may or may not withstand additional testing, but at the moment it looks rather promising.

The relatively limited success in generating more accurate figures for the Eclipse Year (if there is any hope of keeping these figure accurate enough to actually predict an eclipse or what-have-you, we should certainly aspire to achieve this) may automatically group these figures into the calendar sets that the equation associates them with, thus we have values of

A = 346.4769011 B = 346.7274999 C = 346.5939367

If this is correct, my favorite version of this which is C, has come along just as we are recognizing how other aspects of Stonehenge talk about half Venus Cycle C and Venus Orbital Period C.

I probably can’t overemphasize that I obtained a value for the perimeter of the bluestone “oval with corners” (Thom: perimeter = 51.06 MY = ~138.88832 of 138.6375748 then lo and behold 138.6375748 / 4 = 34.65939370.

138.6375748 also turns out to belong to a prodigious series formed from 1.622311470 all the way to the 9th power at least, that also just so happens to display multiple prominent figures from the C set:

138.6375748 x (1.622311470^1) = 224.9133278 Venus Orbital Period C

138.6375748 x (1.622311470^2) = 364.8794714 Earth Year C

Also in this curious but important series is an advertisement for half Venus Cycle A

138.6375748 / (1.622311470^1) = 1.709136345 / 2 (see Teti’s pyramid ?)

138.6375748 / (1.622311470^2) = 1 / 18983.99124, half Venus Cycle A

We can take this novel series at least as far down as

138.6375748 / (1.622311470^7) = 46875 without departing from recognizable territory.

If then the equation 18990.40381 (half Venus Cycle C) / (2 / 365.0200808) = 346.5939367 has correctly grouped 346.5939367 into calendar set C, then we can determine from this group that

346.5939367 / 224.9133278 = 1.541011111

and 364.8794714 / 346.5939367 = 1.0527578031

And hence these should be the ratios between corresponding figures for all currently suggested sets, which now includes A, B, and C.

About these numbers:

1.541011111: 1.541011111 x 2 = 3.082222222. I’ve identified this with Thom’s “Pi'” = 3.0840 value for the Type D flattened ring. This would be the one and only time that the “Giza standard of accuracy” for approximation of >.9995 hasn’t quite been met, but it may be due to the particular choice of geometry used by Thom; actual examples of Type D rings might help shed light on this but they are rather scarce.

(See: Alexander Thom, Megalithic Sites in Britain, pgs 28-29, 51)

All of the other Perimeter/MN and AB/MN values for Thom’s flattened rings were readily recognizable, including uncanny resemblance to numbers already associated with Stonehenge, namely the resemblance between Thom’s Type A P/MN ratio = 3.0591 and the 305.7985077 ft sarcen circle inner perimeter, and between Thom’s Type B P/MN ratio of 2.5972 and 1/2 of the outer sarcen circle radius, 51.95151515 / 2 = 25.97575758

1.0527578031: awhile back this came up and I was finding interesting properties that it has, not the least of which is that it turned out that it can be seen as simply 2 / half Venus Cycle B.

2 / 18997.72194 = 1.0527578031 / 10^n

If the ratios generated from these figures and others will continue to hold up, Stonehenge’s vision of a Calendar Set C may continue to make itself manifest, and the place of the Anomalistic Month and Eclipse Year in at least our three primary calendar sets may have finally been understood.

Postscript: I should be so lucky, but I was looking at some of my data tables from various calendar experiments in the course of this, and I happened to randomly notice where I had that the Venus Synodic Period relates to the 1/2 Venus Cycle Builder figures, but at the time I didn’t realize it would be a fixed ratio so it looks at first glance like a failed experiment for lack of consistent results.

I believe I may have just taken that hint and ran with it where we’re supposed to go. This makes for a rather strange direct ratio between half Venus Cycle and Venus Synodic Period (Morton’s Apex Displacement Ratio / 32) – not wonder that was hard to figure out without flipping it backwards by bouncing it off 360.

Apex Ratio 1.040913798 / 32 = 3.25285519 = 1/2 Venus Cycle / Venus Synodic Period.

Reminder, the “Builder” figures used for 360 / x = half Venus Cycle y are largely courtesy of the fact that the canonical half Venus Cycle 18980 likes to think it’s the square root of 3600), and the “Builder” will be the smaller of the two

If this too is correct, we now know at long last that the standard Venus Synodic Period value of 584.0321292 is the value belonging to the B set of calendar numbers, that the A value for the VSP is 584.4545456 and the C value for the VSP is 584.2571915.

If the rest of the possible internal ratios prove pleasing for these so far, some new understanding of ancient calendar systems, at least those which may have contributed to the design of ancient architecture, seems to have been achieved.

–Luke Piwalker

Uncharted Skies?

Stonehenge seems to have pointed us in the direction of a third value for representing the half Venus Cycle (it’s nearly twice as accurate as the standard “B” value for the half Venus Cycle, adding about 10 days as opposed to about 17), and it looks like it may be mathematically compatible with its environment.

I’ve recently noticed and pointed out that this “C” value for the half Venus Cycle might have unsuspected geodetic value, and I’ve recently talked about some prospects of using internal ratios to reconstruct calendar number groups.

I’m still looking for some of the formulas that might be needed to expand those calendar sets, as well as other possible formulas that offer guidance with such efforts.

The internal ratios that have been identified or proposed thus far have been fairly simple and straightforward. Essentially, the ratio between earth calendar year and Venus Orbital Period is 1.622311470, and the ratio between the earth calendar year and Lunar Year is 1.03132401, or 1 / 20 of one Royal Cubit in inches – exactly, if we use the formula that derives the Royal Cubit from circular geometry.

These internal ratio formulas also help to sort out potential calendar numbers into their respective groups in ways that other, sometimes more elaborate, formulas may not.

We know that 225 and 365.0200808 must belong to the same set since 225 / 1.622311470, and we know that 365.0200808 and 353.9334582 must belong to the same set since 365.0200808 / 353.9334582 = 1.031324031.

If this much has been established, then the ratio between the VOP and the Lunar Year would be expected to be 225 / 353.9334582 = 3.178563580 x 2, which might help explain how 3.178563580 found its way into both Mayan mathematics in spite of the very strong showings of a very similar number, and it might also help explain how 3.178563580 found its way into the mathematics of Silbury Hill.

Taking the half Venus Cycle / Venus Orbital Period ratio to be 1 / (12 x (Pi^2)), we then know that 225 x (1 / (12 x (Pi^2))) = 18997.72193 / 10^n, so we know that all three of these numbers – 225, 365.0200808, 353.9334582 – must therefore belong to the “B” set of calendar numbers along with 18997.72193, half Venus Cycle “B”, and so forth.

If we can find additional ratios, then we can likely group additional calendar values into their respective sets correctly.

Sadly, finding additional ratios might not be quite so easy. Let’s look at a few possibilities here…

Calendar Year / Eclipse Cycle

365 / 346.62 = 2.106042738 / 2. That obviously looks like half of some kind of Palestinian Cubit or Sacred Cubit, but which one? There are numerous candidates for Palestinian Cubits.

Venus Synodic Period / Anomalistic Month

584 / 27.55 = 2.119782214, which looks a lot like (360 / 200) x 1.177245771, and if we use my now customary figure of 584.0321292 (in feet, equal to 480 Remens of 1.216733603),

584.0321292 / (360 / 200) x 1.177245771 = 27.56113481, which may be an acceptable approximation of the ~27.55 day Anomalistic Month, but a challenge here is that 584.0321292 itself doesn’t seem to have been successfully placed into a particular set of calendar numbers.

Venus Synodic Period / Lunar Year

584.0321292 / 353.9334582 = 1.650118449 = 16 x 1.031324031, and

Venus Synodic Period / Calendar Year

584.0321292 / 365.0200808 = 1.6

But in either one of those formulas, 16 / 10 or 16 could mean 16 x 1.000723277.

Half Venus Cycle / Calendar Year

To use raw canonical numbers, 18980 / 365 = 52. We might see 52.04568991 ((Pi / (paved Perimeter Great Pyramid) x 2) in some of these equations, but it may not be certain if it really belongs there, and we’d expect this should be another fixed internal ratio, but Stonehenge’s 51.95151515 may have other ideas about this.

Lunar Year / Eclipse Year

Canonically, 354 / 346.62 = 1.021291328 – it’s very tempting to want to take that to be the “Mayan” wonder number 1.021521079, it would certainly help explain why that came out of some seemingly very calendar-related workings at Tikal, but it’s still not certain if 1.021521079 truly works in this role.

Half Venus Cycle / Anomalistic Month

Canonically, 18980 / 27.55 = 6.889292106, which may be likely to be 6.890283797. I’ll spare most of the history of this number but Munck placed a fair amount of importance on it and most likely with good reason, and in my work, 6.890283797 / Pi = sqrt of height of Great Pyramid, unpaved

(6.890283797 / Pi)^2 = 481.0325483 ft / 10^n

The reciprocal of 6.890283797 is 1 / 6.890283797 = 145.131904 / 10^n

Which is the circumference of the Stonehenge inner bluestone circle if 17 MY is reckoned as (2 / 1.177245771) x 10 and the “Incidental Meg Yard” of 2.719256444 is used.

6.890283797 / 4 = 2.756113483, by the way, which may be a fortuitous shortcut therefore in the internal workings of calendar number groups.

Perhaps interestingly, this would group 2.756113483 into the “C” group of calendar figures that Stonehenge seems to have some enthusiasm about.

Hopefully the correct ratio here has now been identified and can be used to associate candidate values for the Anomalistic Month with the correct set of calendar figures via their relationships to the half Venus Cycle.

Also, not that it’s of additional avail in grouping new numbers into correct sets, but at present the ratio between half Venus Cycle and Lunar Year appears to be (1.073519411 / 2) x 10^n, and I’m happy to see that because 1.073519411 is closely linked to at least some of the “wonder numbers” from Tikal, leading to the prediction that this figure might be of some importance to calendars.

Does this help to explain why after three years, not one full calendar set has been solved, not to mention the difficulties of trying to integrate some of this with geodetics?

For the Draconic Month (fingers crossed), I might have gotten a lead from an unexpected place.

For at least several years, I’ve been advising people of the formula

360 / (Royal Cubit^2) = ~1 Remen

It may not always even work, and yet it might have been influential in shaping ancient views of the most appropriate metrological values, and the most literal use of the formula along with other circumstantial evidence may help support the existence of a so-called “Thoth Remen”.

What hadn’t occurred to me concerning this formula is that because of the geometric relationship between Remen and Megalithic Yard, the formula

360 / (Megalithic Yard^2) = ~4 Remens may also be valid whenever possible.

From here, we can apply another possible formula that I stumbled over of

120 Megalithic Yards / Royal Cubit = half Venus Cycle

This seems to work rather well in actual practice

MY = 2.720174976 = half Venus Cycle 18990.40381 (“C” value)

MY = 2.719256444 = half Venus Cycle 18983.99120 (“A” value)

and the big surprise,

MY = 2.721223210 = half Venus Cycle 18997.72194 (“B” value)

That’s quite an unexpected way to be able to incorporate a MY or MY-like value of 2.721223210 that so closely approximates the “textbook” Draconic Month figure of 27.212220815 days.

I’ll keep hoping that the pieces keep falling into place, because they seem to continue to do so.

Postscript: Well I didn’t mean to break my own rule again about putting all my cards on the table if someone’s going to be kind enough to sit down to that table and read this junk, but I did forget something here, speaking of geodesy.

We can see in the formula above where 27.212220815 is working at the second power. Never in all these years did I suspect that squaring this figure was going to be good for anything, it’s already a preposterous and difficult enough figure at the first power, and one would think that squaring that kind of number is only going to make things worse.

Having seen this in action, however, next I was inspired to try to pay heed to the formula that revealed this surprising fact, again

360 / (Megalithic Yard^2) = ~4 Remens may also be valid whenever possible

360 / (2.72122321^2) = 4 x 1.215385852, which rather looks like a geodetic Remen, and 1.215385852 / 64 = 18990.40393, the half Venus Cycle “C” value that Stonehenge seems to working at trying to introduce us to.

I wrote about this back in Part 8 of this series – this sums up some of it

“Next there is a note about a meter-like figure of 3.281541775 and that it associates with a polar circumference figure of ~24860. A Remen-like figure of 1.215385842 follows, with the notation that this figure divided by 2^n equals 18990.40378. “n” must be unusually high here, but 18990.40381 is the figure suspected of being an additional version of the Long Count. n=6, as a matter of fact. 18990.40381 x 64 = 1.215385842

Okay, since that was mentioned in the context of polar circumference, we’ll estimate with 24860 x 5280 = 131260800.0, and 131260800.0 / 1.215385842 = 107999283.4, near to the ideal of 108000000.0.”

…but I am absolutely amazed at ending up back there again via this

360 / (2.72122321^2) = 1.215385852

If we plug AEMY 2.720174976 into this equation,

360 / (2.721074976^2) = 1.216322743 x 4

I am again very surprised, I really wouldn’t have thought these “linear” Megalithic Yards AEMY 2.720174976 and IMY 2.719256444 would be useful at higher powers either. I’ve seen extremely little evidence of that if any.

Yet 1.216322743 is indeed what we’d think it is, it’s the square of the very Stonehegey 1.177245771 / 1.067438159 = 1.102870233

1.102870233^2 = 1.216322751

That such a thing might have some actual role to play, and that 1.216322751 is thus related to another of Stonehenge’s favorite topics, 2.720174976 is remarkable – 2.721074976 squared, no less.

I’m not quite sure exactly what all this means, but Stonehenge may know even more than I’d realized.

By the way, is everyone aware what happens when we go back and use the Munck Megalithic Yard in that equation?

360 / (2.71971567^2) = 1.216733693, the classic.

There is a fly in the ointment, however, which is that the “Remen / 64 = half Venus Cycle” formula we saw does not seem to work for this

1.216322743 / 64 = 19005.04298, and I have not been authorized by the ancients or by anyone else to declare anything as high as 19000.00000 to be a legitimate representation of 18980, and there may still be too many candidates for half Venus Cycle values as it is, not to mention that there are possible “D” and “E” values where 1/2VC = ~360/(1/2VC) more accurately describes mathematical truth so that these possible “D” and “E” values are really two half Venus Cycles each, pushing the number of possible Venus Cycles up to at least eight.

I’m sure that’s a complication nobody wants, and if nobody needs to, maybe we’ll never have to go there, who knows?

Maybe the correct ratio for Remen / x = half Venus Cycle is the revised Mycerinus’ perimeter/height ratio of 64 x 1.000723277 = 6.404628973 rather than 64, but if that too only seems to end up working part of the time, I’m sure I’m going to end up confused by any lack of a fixed ratio here.

–Luke Piwalker

Rio Bec, Slight Return

Aside from Tikal, I’m not sure there any ancient American archaeological sites that have given me a sense of wonder and inspiration quite the way that Rio Bec, the namesake of the Rio Bec archaeological style, has. That was all any archaeologists had to say to me was to propose that the stately false tower facades of the Rio Bec style might be meant to suggest the lofty pyramid temples of Tikal, and I was on the job, calculator in in hand.

To date, though, I have relatively little finished work on Rio Bec. There’s something about the Rio Bec site proper and I can’t quite put my finger on it, but I always end up feeling like I’m looking at the work of a particularly proficient ancient mathematician, and it’s sometimes as confusing as it is humbling. It’s definitely a place to look, though, if you’re eager to learn new things about math, just as is Tikal.

One of the things that Rio Bec provided inspiration for is a series of “Rio Bec Equations” that were hoped would help coordinate identification and correct grouping of calendar values into respective sets.

Rio Bec Equation #1 is a very simple one that may still actually hold up. Rio Bec was where I first learned that 12 x (Pi^2) could be very important to calendars, and Rio Bec Equation 1 reads,

Half Venus Cycle / reciprocal of (12 x (Pi^2)) = Venus Orbital Period.

That truth has held up for both “A” and “B” calendar sets, and is predicted to hold true for any additional calendar sets that may be eventually confirmed, and it exemplifies the way we may be able to “lock” internal ratios of calendar sets. and learn to construct them from these internal formulas.

Notably, the hugely important ratio 1.622311470 is another we should probably expect to see preserved internally within calendar sets.

Looking at the tables for Rio Bec Equation #1, though, I’m increasingly amazed that this VOP-like number that Stonehenge has been showing us, 224.9133273, never managed to get onto the radar at the time that the table was created.

Stonehenge has also been trying to talk to us about 18990.40378 as a viable half Venus cycle apparently, an idea that received even more emphasis today when setting out the first version of the Stonehenge data tables.

Do you know what the 120 Meg Yard = 326.4209971 ft outer sarcen circle circumference is in standard Royal Cubits? Don’t feel bad if you don’t know the answer, I didn’t know it – but it’s 189.9040383

326.4209971 / 1.718873385 = 189.9040383

We can use Rio Bec Equation #1 to see that this and the Venus Orbital Period figure appear to belong to the same set of calendar numbers

189.9040383 / (1 / (12 Pi^2)) = 224.9133273

It might be interesting if we start feeding different Megalithic Yards into that equation. Hopefully Stonehenge’s teaching about representing the Venus Cycle don’t end with this set, which is currently becoming the default “C” set of calendar numbers which hopefully upholds the last attempt at assigning the “C” designation to a calendar number group.

In fact, if we use 120 x Incidental Meg Yard in place of 120 AE Meg Yards, we get the “A” value for the half Venus Cycle – not a new teaching but a nice confirmation of an old one.

Anyway, if I ever get business at Stonehenge tidied up, I’d like to get back to some Mesoamerican studies, and there’s still nothing that intrigues me quite like Rio Bec. I still salute its architect for giving us such a run for our money.

–Luke Piwalker

Scrutinizing Stonehenge 2

I don’t imagine I yet have enough additional material on Stonehenge for another part in the series yet, but in case I have anything new to say on the subject in the meantime, let’s see what I can make out of another post.

With the rapid influx of Stonehenge data lately, I’ve begun building a table to keep it organized. I’ve made tables of Stonehenge data before which haven’t outlived their usefulness but a lot of this data is being tabled for the first time.

Even with the very few metrological columns I’m adding in order to get a bitter idea of what the proposed values are in various metrological units, I think some interesting things have appeared.

Partly because Petrie’s Stonehenge unit is probably more outstanding when expressed in inches as 224.8373808 inches, I’m looking at some of Stonehenge’s proportions in inches for probably the first time. Normally things work perfectly well if I simply work in feet, but Petrie’s unit does put focus on inches.

Something that’s already come out of that is looking at the Great Pyramid in inches a little bit for the first time. It’s really not necessary, we could (and should) use 12 or multiples of 12 as mathematical probes, but expressing things in inches can lend an expanded perspective that perhaps begins to shed light on what the mathematical advantages were of having so many metrological units that can be linked together very simply by the number 12.

That would seem to be a bit redundant, but maybe there’s wisdom in it?

I recently mentioned a raw figure from Thom’s data that might have suggested the original “Mayan Annoyance”, although 24 Remens may be more promising, and yet the “Mayan Annoyance” is there. The mean sarcen circle diameter of 50.30183830 ft is 29.26442327 standard Royal Cubits.

After a surprising amount of material generated on the presence of the Venus Orbital Period / Petrie Unit 224.8373808, I’m reminded by the unfinished tables that the minimum sarcen circle radius of 48.66934411 ft = 584.0321293 inches, 584.0321293 days being the standard Venus Synodic Period.

Having the data laid out for an overview also allows spotting metrological patterns. For example, the minimum (inner) sarcen circle dimeter of 97.33868822 feet is equal to 1168.064259 inches, and the suggested Aubrey Circle maximum (outer) radius of 142.1223025 is equal to 116.8064252 standard Remens of 1.216733603 ft each.

The suggested outer bluestone circle circumference value of 244.8157478 ft is equal to 201.2073532 Remens, twice the number of feet for the mean sarcen circle diameter, and having just pointed out where “Mayan” wonder number 1.424280287 occurs with using the Aubrey Center to re-apportion the inner bluestone circle in Part 10 of the “More About Stonehenge” series, I’m very pleased to learn that 244.8157478 ft in standard Royal Cubits is

244.8157478 / 1.718873385 = 142.4280287

The maximum sarcen circle radius value of 51.95151522 ft is in Royal Cubits

51.95151522 / 1.718873385 = 30.22416641, 1/100 of my base perimeter (unpaved) for the Great Pyramid.

These are some of the sort of things that can come to light with a more organized overview of the data.

More mysteries also come to light as well – there are some numbers I don’t recognize although I have reason to expect that they are there for good reason.

The logic of the Aubrey Circle is somewhat hard for me to grasp, though, even with a better view of it. The values proposed again:

Circumference, Max 892.9807632 ft

Circumference, Mean 891.0639963 ft

Circumference, Min 889.5317998 ft

The last one is 2 / 224.8373808, the VOP/Petrie unit again, but the same trick may not work with the other two. 2 / 892.9807632 = 2.239689904, which is in false square root 5 territory rather than VOP territory, so they don’t all have to be about Venus per se, but 2 / 891.0639963 = 2 / 224.44507699, which brings me back to the question of whether it was ever used to represent the Venus Orbital Period. I might believe it, but I want to see more evidence first.

I almost feel like I’m cheating readers with such a short post, so here’s a little more on the subject that I hope will be of interest.

When we convert it to inches, we can observe that the 305.7985077 min (inner) circumference of the sarcen circle is 305.7985077 x 12 = 3669.582093. That’s a number that’s come up again twice just in the past several days. I was scrambling for an example of where it may exist in real life (it may be involved in Hawass’ find of a Giza subsidiary pyramid, G1d, depending on what we think it the slope angle?) but it’s also posted thus at Stonehenge already, something one may tend to forget when heavily focused on measuring in feet.

That being the case, more background on this number is probably due soon.

Most recently, one of the reasons I had occasion to write about 3669.582093 is because its reciprocal may suggest some kind of Megalithic Yard to some of us (this may not include me, or at least not yet) in case three Megalithic Yards aren’t enough already.

1 / 3.669582093 = 2.725105951

What I hope to do, is round out my Stonehenge data table with more conversions of its proportions into other metrological units. Looking at things in putative Indus feet is something I’ve never really done in the past, as focusing on the Indus foot is something rather new to me that happened only after encouragement from Jim Wakefield who’s something of a champion of the Indus foot.

My favorite Indus foot candidate may well show metrological significance at Silbury Hill, which may still have the potential to be something of a metrological metropolis.

Naturally, when looking at additional components of Stonehenge, it would definitely be a good idea to try measuring things out in Squared Munck Megalithic Yards.

In my humble opinion, it was very astute of Munck to associate this unit with Stonehenge, because just as

360 / min sarcen circle circumference 305.7985077 = 1.177245771

360 / min sarcen circle radius 48.669334410 = 7.396853331, the Squared Munck Megalithic Yard (“SMMY”) in feet. Since 48.669334410 = 40 Remens, we see more about how the Remen and “SMMY” are directly related to one another.

At least as far as the minumum sarcen circle values go, 7.396853331 is able to put on quite a display, operating at least as high as the sixth power, SMMY^6. At or near the bottom of the series is another occurrence of the “Mayan” wonder number 1.02152107. On the way there, the sidelength of the Great Pyramid (unpaved) is also found, along with a figure equal to the sarcen circle’s mean radius in Remens, and a number that might be the main thing we’re supposed to think the Aubrey Circle’s 56 holes refer to, 55.8903914.

–Luke Piwalker

“Not A Cubit”

Well, I don’t really know if it’s a cubit or not. Nobody actually put me in charge of ancient metric standards, and that’s not something I want to be in charge of anyway. Had I been recruited by some ancient agency in charge of metric standards, I presume my job would be try to keep the number of possible variations Cubits, Remens, and what-nots approved for actual use down to a minimum…

Which is the same thing one can accomplish just be being hesitant to “endorse” any values that don’t have some obvious and compelling standard such as the geometric origins of precise values for Cubits, Remens, and etc.

I know lots of possible Royal Cubit values, including that sqrt 3 is ostensibly similar enough to the consensus Royal Cubit value (1.732050808 and ~1.718+ respectively, or 1.718873358 if we apply a precise geometric standard).

However, I can tell the difference between these two numbers, and I’d like to at least give the benefit of a doubt to ancient mathematicians that they were able to do the same at at least as easily, and quite possibly more so for having actual careers in mathematics.

I’ve nominated a few possible Royal Cubits in my time – 1.722570927, 1.721325932, 1.712234567, 1.720116607 (and perhaps 1.731717175, which might be better off considered as an ancient version of sqrt 3), but I’m still not sure if any ancient persons considered them to be Cubits any more than I do, I’m still not sure that any register as the fundamental metrological unit of any ancient architecture, although 1.720116607 might have been used conditionally with Chephren’s pyramid and its subsidiary.

I often attempt to indulge the many who insist the Great Pyramid is “280 x 440” cubits, although it isn’t as far as I can tell. In my models, 438.9663491 Royal Cubits is the correct figure for the Great Pyramid after placing of final pavement, and 439.5926811 Royal Cubits, and keep in mind with that last figure I am being as conservative as I possibly can with consensus data.

I’ve also considered some larger values. The Meg Yard x a 2 Pi pyramid’s overt version of Phi is about 440 (literally, 2.72 x 1.618033989 = 440.105245 and my school of though might make of this 2.721074976 x 1.618829140 = 440.3498517, but given the fixed parameters of the model, this would produce Cubits of 1.713472874 and 1.715917826, which are several things I really don’t know enough about to have any business whatsoever going around calling them cubits.

I should note that the Megalithic Yard may be the true unit of the Great Pyramid’s baseline as it exists without the hypothetical missing pavement, rather than the Royal Cubit, and nothing odd about that since the Megalithic Yard would have been a very obvious unit to the ancient Egyptian through direct geometric relationship to some of their major units such as the Royal Cubit and the Remen.

Anyway, as with “Not A Remen” which could be a Remen for all I know (and I need to get back to Moyer’s “Happiness Interrupted” page about these matters), there’s a figure I’ve run into once too often and I don’t know what to do to help keep it from falling through the cracks but slap the dumbest name imaginable on it so it has a working title.

As a matter of fact, we just saw it – “Not A Cubit” is what I’m calling 1.715917826. It’s very hard for me to take it seriously as Royal Cubit or any metrological value, but as with all of my nominations for possible Royal Cubits, it’s something I have to take very seriously as a mathematical constant.

“Not A Cubit” 1.715917826 has a very interesting pedigree – it’s (1 / 1.618829140) / 360, so we would assume right there that it’s something important and useful, but it yet it may be something that defies general utility, making its true purpose something that remains to be seen.

I’ve found a use for “Not A Cubit” – linking the secondary figure for Earth’s equatorial circumference in miles to what I presume to be the (or one of the) intended value

24903.44232 / 1.715917826 = 145.1319051

Which may help to reinforce the idea of 1.715917826 being fairly secular in nature, and having specific but very limited functional value.

On the other hand, once we realize that 1.715917826 may generally reside near the bottom rather than the top of certain important series, that view could still quickly change.

1.715917826 x (360^1) = 1 / 1.618849140

1.715917826 x (360^2) = 5 / 224.8373808

1.715917826 x (360^3) = 8.005786209

1.715917826 x (360^4) = 2882.083035

1.715917826 x (360^5) = 10375.49893

That’s not a bad series! Those are all nice numbers.

1.715917826 may be a poor responder to 2 Pi (?) but 2 Pi is able to bring forth 168.0334823 from this “Not A Cubit”, which may be important.

As long as a Leap Year figure for these calendar systems hasn’t been figured out yet, it could be important that 2 Pi / 1.715917826 = 366.1705247

Who knows? Again it’s just a largely unexplored number that’s come up once too often and has thereby aroused curiosity.

This just a general advisory – be aware that “Not A Cubit” could be in your neighborhood, approach it with caution, don’t fire until you see the whites of its eyes, etc.

–Luke Piwalker

More About Stonehenge, Pt 11

MORE AUBREY CENTER DISPLACEMENT EXPERIMENTS

In Part 9 of this series, we looked the effect of the Aubrey Center displacement on the proportioning of the mean Trilithon “horseshoe” ellipse along its major axis (major diameter), and the we took a look at is effect on the proportioning of the outer bluestone circle.

For the mean ellipse, we observed that (28.5 / 2) – 1.5 = 12.75 MY = ~34.68 ft

And that this is suspiciously close to the “textbook” (Wikipedia) figure of 346.62 days for the Eclipse Year.

Now let’s look at the inner bluestone circle, with a diameter of ~17 MY, according to Professor Thom.

(17 / 2) + 1.5 = 10 MY x 2.72 = ~27.20 ft

(17 / 2) – 1.5 = 7 MY x 2.72 = ~19.04 ft

There is a remarkable number I first encountered at Tikal. I ended up writing a post specifically devoted to it.

http://grahamhancock.com/phorum/read.php?1,1198957,1203919#msg-1203919

Being this magical mystery number is 1.424280287, would it surprise the reader to know that 27.20174976 / 19.09859317 (60/Pi) = 1.424280287?

I think that one might actually be more less solved already? The main components, at least.

Next let’s try something particularly impetuous. Surely it’s not a surprise that the mean of the two Trilithon ellipses that contain the Trilithon “horseshoe” might turn out to have an intentional-looking mean value, but how about the mean of the inner Trilithon ellipse and the bluestone ellipse?

Surely that must be stretching things too far, right?

Well, this would be the mean of two ellipses described by Professor Thom as 27 x 17 MY and 22 x 14 MY, so

(27 + 22) / 2 = 24.5 MY x 2.72 = ~66.64 ft

(17 + 14) / 2 = 15.5 MY x 2.72 = ~42.16 ft

24.5 / 15.5 = 1.580645161

I don’t think I want to try to solve that right now, but I would like to note that 42.16 ft in Remens is about 42.16 / 1.2167 = 346.5110545, since we’ve been talking about how Stonehenge is starting to throw figures at use that are reminiscent of the “346.62” day Eclipse Year.

There’s another one – so yes, this almost looks like something we were intended to do in our interpretation.

What I’d like to do here is just see how the Aubrey Center offset affects the proportions here

(24.5 / 2) + 1.5 = 13.75 MY x 2.72 = ~37.4 ft

(24.5 / 2) – 1.5 = 10.75 MY x 2.72 = ~29.24 ft

Yet these figures look less mysterious. The first figure looks like “Alternate Pi” / Pi, or 1.177245771 / Pi = the marvelous number 3.747289674.

Speaking of the possibility of Stonehenge being a profoundly well designed calendrical calculator mathematically speaking (in addition to who knows what all else), the same may be said of the Aztec (and Mayan) calendar stones (“circular altars”), and 1.177245771 / Pi = 3.747289674 seems to feature prominently in the design of what may be the best known of them all, the Aztec Sun Stone

http://grahamhancock.com/phorum/read.php?1,1198957,1203987#msg-1203987

Interpreted as measuring

Radius 1.873644837 ft – Diameter 3.747289674 ft – Circumference 11.77245771 ft – Thickness 1 = 32.27486122 = (1 / sqrt 960) = 1 / 30.98386677 ft – Thickness 2 = 3.225153443 = (1 / (Pi^3)) = (1 / 31.00627668) ft

~29.24 ft might want to tempt us with exotic possibilities like “The Mayan Annoyance” 29.26442327, but the safer bet might be 24 Remens = 29.20160646, a potent geodetic modelling figure.

37.47289674 / 29.20160646 = 1 / 779.2727272, 779.2727272 being a fairly well established figure now for the diameter of the outer bluestone ring, and still striving for acceptance as one valid representation of the 779.96 day Mars Synodic Period.

I have to say, I’m rather amazed at the way Stonehenge seems to balance redundancy (backup) with the incorporation of new numbers. I’m very impressed, and I’m not very easily so impressed.

What then are the hopefully now “fine-tuned” figure 37.47289674 and 29.20160646 in Megalithic Yards?

29.20160646 / 2.720160646 = 10.73519414.

Always happy to see that figure. If it’s going to turn up at Stonehenge, someone should put (Pi / 3) near to it so we can set off the same fireworks display as at Tikal when 10.73519414 and (Pi / 3) are combined.

Oh, wait, someone did put (Pi / 3) near to it at Stonehenge, as if they knew precisely what they were doing.

37.47289674 / 2.720160646 = 1.377591407

37.47289674 / 2.719256444 = 1.378056742

And what are those?

Well, I’ve already mentioned how Stonehenge seems to be flinging some unexpected calendrical figures at us – and not just a ~346.62 day Eclipse Year, but also some numbers that look suspiciously like the lunar Anomalistic Month of 27.554549886 days (Wikipedia)

So, quite intriguingly,

1.377591407 x 2 = 27.55182814

1.378056742 x 2 = 27.56113483

We may get this figured out yet if Stonehenge keeps giving us so much help.

–Luke Piwalker

More About Stonehenge, Pt 10

Yeah, ten parts and we haven’t covered the half of it yet… What does that suggest about what Stonehenge is?

What now? I bet we’re all very curious now what the Aubrey Center offset does to the message of the sarcen circle, I know I certainly am – but we can get back to that because it’s on the table now at least.

Well, I’ve been doing my best to reveal that Stonehenge seems to be just about as drenched in Petrie’s Stonehenge unit (or the Venus Orbital Period, if you prefer) as it is in the magic number 1.067438159…

While at the same time I’ve had to get into the subject of how Stonehenge may be expressing data on more than two sets of calendar numbers, even though it may be very focused on two primary sets of calendar numbers labelled as “A” and “B”.

Just when we think we’re going to find 224.8373808 occupying a niche in every conceivable interstice of Stonehenge’s design, guess what? It might have brought some company with it.

Mean diameter sarcen circle 316.0557714 / 224.8373808 = 224.9133273 / 16, and there are a number of other signs that the possible alternate representation of the ~225 day Venus Orbital Period, 224.9133273, may also be on Stonehenge’s agenda.

Since the VOP “A” and “B” values are already accounted for, if 224.9133273 truly does belong to a likely set of calendar numbers, Stonehenge would have to be showing us the “C” or “D” value here, and believe me, working out those sets has been hard enough that if Stonehenge wants to help us out, it would probably not be wise to turn it down on the offer.

One problem is that I’ve already found enough possible Venus Orbital Period values for eight or ten sets of calendar numbers, even while no more than five corresponding sets of Venus Cycle figures have ever surfaced with any consistency, and the fifth (very similar to the first) has already come and gone so quickly as to cast doubt on whether it’s worth pursuit.

It’s gotten to look a lot like something you can’t find unless you’re looking for it, and that seems quite contrary to the purposes of a system with the resonance to jump out and grab the unsuspecting and make them wonder why they are seeing the same numbers over and over again when they start looking at ancient architectural ratios.

For that, it helps if numbers volunteer themselves often enough from equations and experiments.

So, basically still four Venus Cycles and as many as eight Venus Orbital Periods, and I still don’t know a way to organize and embrace the lot. It may be that a good part of the time they used symbolic numbers for the VOP, at least in an incidental sense, that don’t even have calendar sets of their own – I really couldn’t tell you, not yet anyway.

I’ll post more of the equations concerning this curious character 224.9133273 when I get the chance to look into it more. For what it’s worth, it does have a history, but again its been part of attempts at “master formulas”. Thus far they reveal important calendar numbers but don’t sort out into respective sets as needs to be done.

On second, though I should never be coy with someone who’s had the decency and fortitude to actually read any of my voluminous writings, here is what I’m talking about that I have so far.

Inner circumference of sarcen circle in feet: 305.7985078 equals, depending on Megalithic Yard used

305.7985078 / 2.720174976 = 224.8379803 / 2

305.7985078 / 2.719256444 = 224.91332772

Outer circumference sarcen circle in feet: 326.4209971 ft (120 x Meg Yard of 2.720174976 ft)

326.4209971 / 224.8373808 = 1.451809285

326.4209971 / 224.9133272 = 1.451319053

Again, 145.1319053 ft appears to the primary intended value for the max diameter of the inner bluestone circle, and the VOP/Petrie unit links it to the the sarcen circle maximum vale, and links the sarcen circle minimum directly to the Megalithic Yard.

It’s a great pity that Petrie didn’t chop up his Stonehenge data just a little more and realize that his unit was so ubiquitous as a ratio (or for that matter, how often the ancient Egyptians might have used the Remen and the Royal Cubit as a ratio, possibly a very good sign they used our modern foot precisely and that the true origins of the foot have been lost to history).

We could plot out the alternate calendar sets with a complete inventory of internal ratios belong to any one them, but there isn’t such a complete inventory yet, and attempts to advance the cause of having one continue to stumble over the fact of dealing with as as many as four different sets of numbers at a time with no clear way of sorting them, so it’s a bit of swimming upstream, isn’t it?

There’s a question that haunts me, though, and that’s whether Stonehenge will be able to sort additional calendar numbers into their respective sets with its brilliant equations any better than the not-so-brilliant-but-rather-obvious equations I tried to come up with.

That’s a question that can only be answered, no doubt, by plunging even deeper into Stonehenge’s mysteries.

So, for the mystery presently at hand, let us say this – we should want more proof that this near-VOP figure (canonically, anyway) 224.9133272 really has any business at Stonehenge. Is this number really on the same wavelength with Stonehenge’s agenda of serving as glorious advertisement for 1.177245771 and 1.067438159?

We should expect this number 224.9133272 to answer to 1.177245771 and 1.067438159, should we not?

Hopefully then it will be of no small interesting that upon exploring this critical question, we quickly find that whereas 240 / 1.06738159 = 224.8373810, 224.9133272’s rapport with these numbers involves a somewhat different conversation

224.9133272 / (1.067438159^1) = 2.017038476, my favorite Palestine Cubit

224.9133272 / (1.067438159^2) = 19.73920886 = 2 x (Pi^2)

224.9133272 / (1.067438159^3) = 184.912337 = Squared Munck Megalithic Yard / 4

We can use 1.067438159 here at the fourth power also, but I think I’ll skip that because it could turn into a long story that’s probably beside the point right now. Suffice it that’s probably rare enough to get 1.067438159 to work at the third power generally.

So how about 1.177245771 (Munck’s “Alternate Pi”)?

Oh, well – I said I shouldn’t be coy with the reader, right? We already hit 1.177245771 — 224.9133272 / (1.067438159^4) = (1.177245771^2) / 8.

To apply 1.177245771 in a more straightforward manner,

224.9133272 x (1.177245771^1) = 264.7782633 = 132.3891316 / 2

I talk about 132.3891316 often enough, remember those circumscription and inscription experiments with the Great Pyramid?

224.9133272 x (1.177245771^2) = 3117.090907, the surface area value that the Great Pyramid’s theoretical capstone would have been happy to have

224.9133272 x (1.177245771^3) = 366.9582088

I was just telling people about this number a couple of hours ago on the GHMB

http://grahamhancock.com/phorum/read.php?1,1215095,1217302#msg-1217302

224.9133272 x (1.177245771^4) = 432.0000000

224.9133272 x (1.177245771^5) = 508.5701724 = 1.017140345 / 2

As some readers may recall, 360 / 1.017140345 = 353.9334585, the main representative of the ~354 day Lunar Year

So yes, 224.9133272 does answer to 1.067438159 and 1.177245771 (even thought its responses to the standard Remen are at least outwardly perplexing, which is quite surprising).

224.9133272 does look to be something that belongs at Stonehenge in that sense, it’s certainly passed two important tests with some flying colors.

–Luke Piwalker

More About Stonehenge, Pt 9

EXCERPTS FROM THE AUBREY EXPERIMENTS

In case it had something to offer us, yesterday I experimented with trying to figure out what the point might be of the Aubrey Center being off-centered. I don’t know if I’ve ever seen anyone give an accounting for that and it seems like a very strange thing without some accounting.

Sort of reminds one of “outlandish” ideas like apex displacement in the pyramids of Cheops and Chephren, that these pyramids are “off-centered from their centers”, doesn’t it?

Tell you what, why don’t I just start with the experimental mean values for the Trilithon horseshoe? Maybe we will shed more light on that in the process of looking at in the context of the Aubrey Center displacement.

A note on Thom’s diagram – he shows the Aubrey Center as slightly off-center from both axes, but that may be a liberty for illustrative purposes that isn’t supported as reality by the text descriptions.

The raw mean figures for the Trilithon ellipses again being

…mean values would be about (30 + 27) / 2 = 28.5 MY as the mean of the major diameters, and (20 + 17) / 2 = 18.5 MY as the mean of the minor diameters.

Thom gives the distance from the center of Stonehenge to the Aubrey Center as 1.5 Megalithic Yards, so if we displace the center of the mean ellipse by this value,

(28.5 / 2) + 1.5 = 15.75 MY = ~42.84 ft

(28.5 / 2) – 1.5 = 12.75 MY = ~34.68 ft

I’m not sure offhand what the first figure might represent, but does the second figure 34.68 remind anyone else of the 346.62 (day) Eclipse Year figure?

15.75 / 12.75 = 1.235294118, which could be at least several things, 2 / 1.618829140 and 1.234567901 being among them.

Let’s try another one, let’s try dividing up the outer bluestone circle along its major axis at the Aubrey Center rather than it’s actual center.

Thom gives the diameter as 28.65 MY and that’s already been interpreted as 57.29577951 / 2 = 28.64788976 MY x 2.270174976 = 77.29727283 = .8 x (Pi^4).

We’ll use 28.65 for the estimate, half it, add 1.5 MY to half of it, and subtract 1.5 from the other half.

(28.65 / 2) + 1.5 = 15.825 MY x 2.72 = ~43.044 ft

(28.65 / 2) – 1.5 = 12.825 MY x 2.72 = ~34.884 ft

15.825 / 12.825 = 1.23391812865

Are any of these actually sensible statements?

Well, 43.044 suggests to me that someone finally got around to addressing 43.03314828, which can be seen as 2 / (sqrt 2160).

(I hope I don’t under-appreciate sqrt 2160 myself since it’s of considerable importance – important enough to merit frequent expression in ancient architecture, and it’s prominent in Munck’s own work as well).

How about 1.23391812865? It could possibly be 1.223194031, for starters.

1223.194031 / 2 = 305.7985078, inner sarcen circle circumference in feet = 611.5970155 (unpaved apothem) value for Great Pyramid / 2.

Already again it looks like we could be seeing sensible statements rather than mere random consequences of the Aubrey Center’s displacement, but it does remain to be seen how the pieces really work together when engaged in equations.

That would be something I couldn’t emphasize enough for would-be detractors, but there is no reasonable expectation in all this of being able to haphazardly slap nice numbers together and having them work together in ways that speak of intent. The case seem much more to be that they need to be well chosen for that if our own well chosen numbers are really going to stick.

34.884 I’m not so sure about, numbers in this range can be challenging (Harleston’s “hunab” being one of them) and confusing (the “hunab” has been taken for 2 Royal Cubits before), and that’s just for starters. It’s probably going to overshoot the 34.662 range successfully given accuracy standards…

Perhaps the best thing I can do with this is to take it for sqrt (1215) = 34.85685012

So the ratio between parts along the major axis because of the Aubrey Circle center offset would be

43.03314828 / 34.85685012 = 1.23456790

So that could be what it is, although more alternatives might be inspired by looking at those suggestions in action.

I’m going to want to see some overarching reason for 1.234567901 to be thrown into the mix at Stonehenge. It’s not enough that it’s a very interesting number, there are lots of those.

What’s troubling here is I don’t think I’ve ever seen that come out of Stonehenge before, and I’m not sure whether it’s about time we saw that, or whether that suggests an interpretive mistake.

I may have gotten things wrong here on the first and second tries, and 2 / 1.622311470 = 1.232808889 might really be what was intended here. This helps coordinate various parts of Stonehenge with the Great Pyramid’s unpaved perimeter value via exponential application of 1.177245771, and 1.177245771 is one of the things that Stonehenge seems to be demonstrably very preoccupied with.

That might well be the most rewarding interpretation of that raw ratio – 2 / 1.622311470 is able to operate at impressively high powers, even more so than 1.622311470 itself, whereas by the time it’s squared, 1.234567901 may typically already be worn out as a data retrieval tool.

Just to show what that looks like, the unpaved perimeter Great Pyramid is modeled as 3022.416640 ft, and one side theoretically measures 3022.416640 / 4 = 755.6041600 ft.

755.6041600 x (1.177245771^1) = 889.5318019 = 2 / 224.8373808, which 224.8373808 appears to be WMF Petrie’s Stonehenge unit in inches, with which Stonehenge’s design seems to be heavily infused, not forgetting that 224.8373808 effectively represents the ~225 day Venus Orbital Period

755.6041600 x (1.177245771^2) = 104.7197552, (Pi / 3) which we’ve seen as a probably ratio in the Stonehenge ellipses already in this series

755.6041600 x (1.177245771^3) = 1.232808888

755.6041600 x (1.177245771^4) = 145.1390152, and we’ve just seen that at Stonehenge too in this series.

So you see what I mean, not only is 1.232808888 a very powerful data retrieval tool, but it appears to be another case of a missing piece belonging to a series already running at Stonehenge.

The question now would be, which of these two lovely numbers is in error if it’s that kind of situation?

43.03314828 or 34.85685012

At any rate, hopefully enough order is on already display here to safely speculate that whatever other purposes the Aubrey Circle or the Aubrey Center may have then, boosting the data storage and retrieval capacity of Stonehenge may be one of them – a very important purpose when data storage and retrieval is the very name of the game.

–Luke Piwalker