Ancient Calendars ...

Let us suppose that Egyptian artisans made circular mats in the way described here. To obtain a mat having T rings, what will be the length of string L needed? As before, let us take the width of the string as the unit of measurement. Table II gives rounded-off experimental results. The smallest values of T and d (d = 2T + 11 for which L becomes equal to a square, n?, are T = 4 and d = 9, respectively (see Fig. 12). with n = 8. In this case, we have n = (8/9)d and, as a consequence, we arrive once more at theancient Egyptian formula for the area of a circle:

 

 

Similar to Monte Albán for the scale of modifications to the natural hilltop, Xochicalco is best known for its Temple of the Feathered Serpent. Large feathered snakes curl around the walls of the pyramid base, with Maya-style lords seated within the curvesof the serpent's body

 

 

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A reference guide to various constants by wendy.krieger

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rodz.akitu : schema;diagram, template ...

8 / 9 * 320=284,4444444444444
284,4444444444444/ 0,2666666666666666=1066,666666666667
1066,666666666667 / 3,333333333333333=320
284,4444444444444/ 0,2666666666666666/1,6=666,6666666666667
(284,4444444444444 / 0,2666666666666666 / 1,6) * 1,0125=675
(284,4444444444444/ 0,2666666666666666/1,65)*1,1=711,1111111111112
711,1111111111112/2=355,5555555555556
284,4444444444444 / 0,8888888888888888=320
284,4444444444444 / 0,9=324

28,44444444444444*12,5=355,5555555555555
28,44444444444444*12,6=358,4***
28,44444444444444 * 12,8=364,0888888888888
28,44444444444444 * 12,96=368,64
28,44444444444444 * 13,122=373,248
28,44444444444444 * 13,125=373,3333333333333
28,44444444444444 * 13,2=375,4666666666666
28,44444444444444 * 13,23=376,32
28,44444444444444 * 13,33333333333333=379,2592592592591
28,44444444444444 * 13,5=384


28,44444444444444 * 14,4=409,6
28,44444444444444 * 14,58=414,72

324/675=0,48
324/666,6666666666667=0,486
7,111111111111111 * 48=341,3333333333333
7,111111111111111 * 48,6=345,6

7,111111111111111 * 49,5=352

7,111111111111111 *53,15625=378
53,15625 / 1,008=52,734375
52,734375 / 1,0125=52,08333333333333
7,111111111111111 *53,26=378,7377777777778

(52,08333333333333 * 10 / 9) *10,125=585,9375 (Thales sun eclipse)

etc.......etc.........etc.......etc....etc.......etc.........etc.......etc....etc.......etc.........etc.......etc


8 / 9 * 325=288,8888888888889
288,8888888888889/0,2666666666666666=1083,333333333334
1083,333333333334 / 3,333333333333333=325
288,8888888888889 / 0,2666666666666666/1,6=677,0833333333335
(288,8888888888889 / 0,2666666666666666/1,6)*1,0125=685,5468750000002


8 / 9 * 324=288
28,8*12,5=360
360/0,625=576
576/1,6=360
576/1,62=355,5555555555556
etc....etc...etc ...apply a schema;diagram, template ....



28,88888888888889*12,5=361,1111111111111(Mesoamerican calendars)
361,1111111111111(Mesoamerican calendars)/0,625=577,7777777777778
[361,1111111111111(Mesoamerican calendars)/0,625]*1,0125=585
585/1,6=365,625
585/1,62=361,1111111111111
etc...etc..etc 




8 / 9 * 315,9=280,8






8 / 9 * 315=280


8/9*316,8=281,6



8/9*307,2=273,0666666666667
273,0666666666667 / 0,2666666666666666=1024







********************************************************************************************************************************************

Excerpt from the session at academia.edu: The Great Pyramid, Egypt and the Ecliptic Plan by John Neal




Vladimir (Владимир) Belobrov (Белобров)
7 days ago
Hi John I also thought for a long time that the Egyptians used this approximation (22/7), but recently discovered that in the Egyptian papyri around 2000 BC there is a different formula for PI: 4*(8/9)^2 (3.160493827), and 22/7 is attributed only to Archimedes (which does not exclude the fact that he got this information somewhere in the bowels of the Library of Alexandria). Thus, we have no documentary grounds to attribute knowledge of the 22/7 proportion to the Egyptians.
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John Neal
7 days ago
Hello Vladimir, When you have the Great Pyramid, that every investigator claims was built a pi ratio of 22/7, who needs texts?
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Vladimir (Владимир) Belobrov (Белобров)
7 days ago
Tell me please, John, who is interested in the opinions of modern researchers when there are ancient texts?
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David Kenworthy
7 days ago
hello Vladimir yes the texts give 16/9 x 16/9 as Pi but this is a measure of area, it is a square. 22/7 replaced it when grain measures paid out were not balancing to the amounts in the silos. Measures were being overstated. Milo Gardener posts on Academia, look up his papers, he has interpreted the Rhind papyrii. He confirms this happened and that is who i got it from. He is a top class American code breaker. 44/7 x 14/11 = 8 sides to the Giza pyramid. The GP is designed against a rise run of 11 horizontal to 14 as height. So the on the ground base to height relationship is 44 to 7 or 1760 to 280 and this is the imperial system. Within this framework the cubit can only function as a variable unit. The GP can be reconciled against a fixed length cubit but within this framework the only cubit length that will work is Berriman's 20.625 imperial inches. This is the imperial rod divided by 9.6. This reconciles to John's 7920,1080 diagram Dave
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Vladimir (Владимир) Belobrov (Белобров)
7 days ago
Hi David There are modern interpretations and there are ancient texts. What should I prefer? - When there are no clear texts, of course, I should rely on interpretations. But when there are clear texts, interpretations are no longer needed. There is no Egyptian text that says that in the Pyramid of Cheops encrypted the number PI and that it is 22/7. There aren't such texts! But "4*(8/9)^2" is written distinctly and clearly. Why interpret this record? Vladimir
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David Kenworthy
7 days ago
Because it relates to 800/81 and you should read the work of Algernon Berriman 3.125 x 16/9 x 16/9 = 800/81 and this is what the Giza Pyramid is designed against and it is a million miles away from John's analysis, even though John's analysis is correct D
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David Kenworthy
7 days ago
Milo uses the ancient texts , look up his work John's mathematics to a large part is decimal but Milo's is based on Egyptian fractional series
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Vladimir (Владимир) Belobrov (Белобров)
6 days ago
O'k! But where is "Milo Gardener"?
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Terence Hogan
6 days ago
Hi David, Re Berriman's cubit of 20.625 imperial inches, which is also Petrie's cubit of 1.7181818 Imp. Ft. Nobody ever appears to give the source of this cubit, it just gets related to other measures. So here is my suggestion ~ One centisecond-of-longitudinal-arc at Giza, with Pi at 25/8 equals 0.88 of the equatorial value (1). It's reciprocal value is 1.13636363r Multiply by 1.008 to convert to an arc-of-meridian value. Results in 1.145454545r x 1.5 (for a STANDARD cubit) = 1.71818181818r The 'cubit' of the GP, which is based on it's latitudinal location. This works for many ancient measures and is precise.
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David Kenworthy
2 days ago
Berriman hs this 20.1618181818r cubit using the remen as 20 digits with the digit being 0.729 and 729 is 3x3x3x3x3x3 0.729 x 20 = 14.58 x root 2 being 140/99 equals 20.16181818r exactly I think Berriman has the relationship between the quadrant arc and chord as 10/9 working with 22/7 and this is a huge relationship in terms of base 10 being the natural ancient number in every circle so 4 x 2.5 per Thom. this means if the drew a line one meter long 3.3 indus valley feet the the arc would be 3.666r and x 4 = 14.666r the eclipse unit.the double suare would be 29.333r. cheers dave
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Harry Sivertsen
15 hrs ago
Regarding the source of various cubits, there are extent cubit rods in museums that give some of the mayor values including the royal value of Egypt. However, care needs taking in the interpretation of the values as numerous close but erroneous metric values have been applied. Please see Stonehenge Metrology on my page here pp 61-71 for details.
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Rodz 123
1 day ago
Hi Vladimir... О происхождении так называемого «короткого стадия Эратосфена by Vladimir Belobrov (...)Стадий Марина Тирского: 40 007,860 км / 360 / 500 =222,266 м (...) 
___________________________________________ 
40007,52 km/360/500=222,264(Eratosthenes' experiment) 222,264(Eratosthenes' experiment)/420*1000=529,2m 529,2*10/9=588 _______________________
 (588 / 35,55555555555555)/0,0375=441 (588 / 0,8888888888888888)*10/18=367,5 367,5 / 441=0,8333333333333333

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Rodz 123
1 day ago
Hello Vladimir. You wrote: I also thought for a long time that the Egyptians used this approximation (22/7), but recently discovered that in the Egyptian papyri around 2000 BC there is a different formula for PI: 4*(8/9)^2 (3.160493827) . 
______________________________ 
8/9*400=355,5555555555556
 8/9*405=360
 0,8888888888888889 * 0,8888888888888889=0,7901234567901235
 (4*0,7901234567901235)*1,0125*1,0125*100=324( Jim Wakefield)
 4 * 0,7901234567901235 * 1,0125 * 1,0125 * 96 * 1000 * 10=3110400 
3110400/8640/24/15=1 
(3110400/324/324)*12=355,5555555555556
 (355,5555555555556 * 0,8888888888888888)/1000=0,3160493827160494
 0,8888888888888889 / 0,3160493827160494=2,8125 
28,125 / 1,0125=27,77777777777778 
355,5555555555555 / 27,77777777777778=12,8 
28,125*12,8=360

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Harvey Price
8 days ago
Dear John, Thank you for allowing me to join your session. Long time follower here. Bought your Done With Mirrors book from you personally several years ago. All being said, a quick technical question for you. Where do you get the lengths of degree for various latitudes from? I have calculated the lengths of degree using various online calculators and got different results from what you state the lengths to be. For instance, for 66 degrees latitude, I have got from one online calculator, 365837.16ft (2 dp) rather than the 365783.04ft that you have stated on page 4 of your document (I believe this is also stated the same in your Done with Mirrors book). I understand the numbers are almost identical (but I'm a stickler for these things, and little differences can mount up etc ...) and I have often wondered the source for your data (maybe you have already mentioned which ellipsoid you are using, and I have missed it?). Anyway, if you let me know, I would be extremely grateful. Having just scanned the document, I will probably have a few more questions before the session is over ... Best wishes, Harvey Price (Palmerston North, New Zealand)
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John Neal
7 days ago
You have rather answered your own question inasmuch "which ellipsoid you are using?" There is no hard and fast agreement. I believe, not "know", that the count begins right on the equator. That is a half degree either side of 0º this would maintain a difference in every degree, otherwise there are two adjacent the same at the pole and equator.
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Harvey Price
7 days ago
Thank you for that, John. As I said, I am a stickler for precision when it comes to these things. Maybe too much, possibly, but I have seen a lot of people state things without any references for years. It is good to know how you approach such things as it helps put your work in a greater context, as it were. Anyhow, I believe there is still a lot in the Great Pyramid that we haven't even considered as of yet. For instance, I believe that the Pyramid is at a latitude where the circumference of the Earth ('longitudinally" speaking) is almost 3960 miles (radius) using conventional Pi. I believe it to be about 3958 miles, roughly (haven't got the number to hand at the moment). Could be important (???) as it is exceedingly close to the 3960 number over the circumference of the earth. I'm sure there will be more as time progresses. Many thanks for your reply. Kind regards, Harvey P.
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John Neal
7 days ago
Try 864/275
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John Neal
7 days ago
in fact 3.15 on the polar (Vitruvius), 22/7 on the mean and 864/275 on the sphere all produce the same result.
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David Kenworthy
7 days ago
hello Harvey and John i used to work as an auditor and systems analyst and in that world everything had to balance to the penny so Harvey i understand where you are coming from. 20.736 x 4 = 99.5328 20.736 / 0.995328 = 20.8333r This I think, is the reason why the base 100 cubit is 20.8333r 20.736 is a fraction short.
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Harvey Price
6 days ago
Hi David, Thank you for that. I used to work with Actuaries at the Government Actuary's Department in London ... so numbers and accuracy were very important to them, and I think I may have picked up some bad habits from them! But accuracy in these matters is I believe reasonably important since measures have to be accurate otherwise there is no point in using them. And if there was a system of measure going back to deep antiquity, then accuracy would have been almost sacred to them ... or am I reading too much into the mind of ancient man?? Nevertheless, I am still learning, and am still very green around the gills with this study. I feel that I am very much the infant compared to the people in this session, and that I cannot add too much to the conversation. Anyway, thank you for your reply. It's greatly appreciated. Best wishes, Harvey P.
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Terence Hogan
6 days ago
Harvey, more power to you, keep going! David, here is what I believe is the probable derivation sequence for 20.736" cubit. It works on many others. One centisecond-of-arc of longitude, at the latitude of Giza, with Pi at 22/7 equals 0.875 of the Equatorial value (1). Reciprocal of 0.875 = 1.142857r 1.142857r x 1.008 (converts to arc-of-meridian measure) = 1.152 1.152 x 1.5 (for a STANDARD cubit) = 1.728 Imp. Ft. 1.728 x 12 = 20.736" Query ~ What is the 'base 100 cubit'? I don't remember coming across it before.
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Harvey Price
6 days ago
Thank you, David. And thanks for posting those links to your friend's YT videos. I've been playing around with Google Earth for a few years, and it might give me impetus to do more with it. It is certainly interesting to try to show how John Michell was right regarding the distance between Silbury Hill and Avebury (does Avebury look like an octagon rather than a circle to anybody else???) ... so seeing other people do such things is definitely inspiring. A tangent: Also a few years ago, I started plotting Albert Watkins leys from his Old Straight Track book, and was fascinated to see that the leys weren't actually straight in a two dimensional sense ... because hills and valleys are not actually 2D structures. Well I never! Strange, that! Might have to dig out the old files and carry on from there. The great problem with that little project was that the names of certain points on the leys were really difficult to discern from the illustrations in the book, and some were even harder to find on online maps. Very tricky business. Regardless ... another question for people to mull with. Did the measures come first, or did they "measure the earth" and go from there. A sort of chicken and egg situation which may be rather dumb to think about but one that may give insight into other areas of the measures we are talking about. Just some random thoughts as always. Best wishes HP
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David Kenworthy
6 days ago
Bill and i have been doing this for a while now https://www.megalithic.co.uk/modules.ph ... 15&forum=4 thanks for the feedabck HP
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Terence Hogan
6 days ago
Hi John, This is in relation to your reply to David Kenworthy re 20736. I wonder if things are not getting a little muddled here? 1) David refers to 20736 as being a Royal Egyptian cubit, which means he is taking 20736 as 20.736 inches, which is the cubit of 1.728 Eng. Ft. 2) I agree that there is a Sacred Cubit of 2.0736 Eng. Ft. which is 24.8832 inches ( note that these are the same digits that express the meridian circumference, in miles), so David's 20736 is an Egyptian cubit of 1.728 Eng. Ft. But 20736 is not a Sacred Cubit if we are still taking it as inches. Which, as noted above, is 24.8832" for the Sacred Cubit of 2.0736 Eng. Ft. 3) We can, of course, divide a Sacred Cubit by two and obtain the measure of 1.0368, known as the common Greek foot. I definitely don't have a problem with that. However, just because the digits comprising a number value can be related mathematically, in this case as the ratio 6 : 5, this should not necessarily be assumed to be a causal relationship. In fact, in this case, it is a mathematical mirage as the two values have completely independent derivations, to my understanding. As follows ~ My understanding is that all genuine so-called 'Sacred Cubits' are, specifically, a 1/10,000,000 part of any of the three radii of the Earth, at both 25/8 Pi and 22/7 Pi values. As such they are not, to my understanding, 'Cubits' of any sort, even if they have been termed that for millennia. I understand, I think with you John, that a 'cubit' is an equivalent term to 'covid', and basically means 'a measure', but one that is related to a 'base' foot measure. The true 'Sacred Cubits' are a specific fraction of Earth radii, and as such are exactly what they are, i.e. 2.0736 Eng.Ft is precisely 1/10,000,000th of an Earth radii. That is it's derivation, end of story! Yet, 20.736 inches is an Egyptian cubit, STANDARD class, ( 24 digits). As follows ~ It is derived from the Egyptian foot of 1.142857142857 Eng. Ft. 1.142857 x 1.008 (converts to meridian value) = 1.152 1.152 x 1.5 (for a STANDARD cubit) = 1.728 Eng. Ft. or 20.736" So we have 2.0736 Eng.Ft for a Sacred Cubit, and 20.736 inches for an Egyptian cubit. BUT, I consider that they have totally different derivation sequences and are geodetically quite seperate. I really think that a more comprehensive understanding of the full theoretical and geodetic framework that the ancients developed would help to sort out many of these mathematical mirages ( false ratios), and simplify our understanding of the overall structure which, as I now understand it, is a theoretical framework that was applied to the Earth, with one physical datum to initiate the whole structure. Hope the above helps, all the very best John. Query ~ Do you have any idea of a release date for Vol III of Ancient Metrology?? Cheers, Terry
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David Kenworthy
6 days ago
hello terry thanks for your excellent response John has hit 23.5 within the GP design 23.5 x 4 = 94 x real pi equals something hardly believable 94 3.141592654 295.3097094 it speaks fo itself this is courtesy of metrologist Jim Wakefield from Australia http://www.dozenalsociety.org.uk/pdfs/Rollrights08.pdf fantastic paper D
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David Kenworthy
6 days ago
can't resist a couple of liks from someone who i have worked with for a few years  and another 
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David Kenworthy
6 days ago
Bill is really trashing it 
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David Kenworthy
6 days ago
Thom's flattened northern circle is brought into 3 d life by the brilliant Bill Wilkinson this is straight from Thom's books Bill knows https://www.megalithic.co.uk/modules.ph ... 15&forum=4
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David Kenworthy
6 days ago
Would you Adam and Eve it https://www.academia.edu/5384849/The_Me ... olar_Stars
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PETUR HALLDORSSON
6 days ago
In every cosmology (Cosmic image, CI), as recently discovered in Iceland, there was a center and two southwest-northeast, and southeast-northwest axis, which have their “omega point” of departure from a specific burial ground in its center. In Egypt they extended from Memphis with a 45° ancle on latitudes, to the shores of the Mediterranian and the furthest reaches of the Nile desert; Madinet Maadi in the southwest to Tanis/Pelusum in the northeast: And from the monastery of Saint Anthony in the Der al Memum mountains in the southeast, to Alexandria in the northwest. The immediate “center area” seems to have been designed as a square with the axis crossing at Memphis (See: The World of the Sphinx. Academia.edu). With the Egyptian concepts of birth and death, Heliopolis and Memphis, on the south and north end of its 60.000 royal cubits north-south axis, which bisect the 6 x 6 square down the middle (See my papers and books). What is specialy intrigguing about this Egyptian square is the idea of Giza – meaning BORDER, and located on the center off its west side. There the diagonal axis; Saint Anthony Monastery - Alexandria, crosses its border (as in Memphis), and there the Sphinx bisects the square, east-west with its gaze. Thus we see an amazing “concept of the world of the Sphinx,” – the Square, bisected north-south-east-west. And futher more, thus it seems that the center area is composed of two worlds (geographical squares) paralelled with each other, with one tilting 45°, making each of its 4 corners protrude penatrate the square, as the idea of a “Pyramid” with its top protruding through the squares west site (Fig 1 in John Neal's paper), lending more weight to the idea of "The world beond" -burrial grounds all along the sacred square’s west side. Maybe I will be so lucky that my friends; Neil, Harry and Dave, come up with an exact measure of these concepts. Best regards. Petur: petur@po.is
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John Neal
9 days ago
Why has this never been noticed before; how did they calculate the duration of the cycle of the plane of the ecliptic?
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David Kenworthy
8 days ago
hello John I hope you are well, obviously in fine form if this paper is anything to go by. When i first read about 20736 being 12 x 12 x 12 x 12 and an Egyptian Royal cubit to boot my initial thoughts were. What am I wasting my time for, surely this is spot on. Had a good look at this number and noticed it was 36 x 36 x 16 so wondered if it was of Babylonian origin. Well is has to be. then realised it is 1296 x 16 then 1296 /16 x 9 = 729 being 3 x 3 x 3 x 3 x 3 x 3 Berriman's Roman digit and... Read More
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John Neal
8 days ago
Hello David, just because it is compatible with the sexagesimal numbers does not mean it is Babylonian. In fact the Babylonian methods were little different to all of the others, they are not truly sexagesimal because they ascend by 10 times 6 so are really a decimal mix which is universal. Five for the merchants and military and six for the builders and sciences. 20736 is not royal Egyptian, it is the sacred cubit at six to five of the Egyptian 1.728; because it begins with 2 then divide it by 2 and its foot is 1.0368 this is a common Greek foot; at 12.4416 inches it was identified by Petrie as used in Athens. 20736 is the length of the schoinos at 176 to 175 of the value used by Eratosthenes in the text. It has many applications and sub-divisions. At Eratosthenes value it is identical to the Spanish marine league of 7,500 varas and the common Greek foot value was used all over Europe particularly Scotland and Rhineland.
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David Kenworthy
8 days ago
thanks John, interesting that it has been traced to Scotland where the Ell is 37 inches x 3 = 111. i know for sure 12 x 12 x 12 x 12 x 12 x 100 = 24883.20 Also 6.6 x 3,14181818r = 20.736 and 6,6 is 1/10th of a chain being 79.2 inches the point I want to make is that i take these measures as proved beyond reasonable doubt and 20.736 is coming out of the Babylonian clay tablet Plimpton 322. The shorter value as 20.736 x 175/176 also is proved. i apologise for getting the names of the cubits wrong. What i am really interested in is this 17.28 miles that Michell had from Stoneehege to Avebury because I have checked it on GE and it works like clockwork and obviously you did not have that luxury. Are there any other values to go with this value because i am using 17.333r miles and have been fot some time. i am looking for values above 17.28 so they would be expressed as cubits in feet. Do you have one as 1.736111r in your workigs. I have Opus 2 Thanks for writing it it What I have looked at is the ancient metre and based on your findings I have it as 39.375 short x 176/175 = 39.60 The Indus valley yard/metre in inches i got 39.375 from Berriman Now Berriman has the Royal Cubit at 20.625 and this lies between your two values as 20.6181818r and 20.736 an I have reconciled 20.6181818r to Berriman's analysis. By my reckoning this makes Berriman's work line up with yours in respect of this cubit but the method of calculation is completely different. i was looking to try to align the cubit with the metre as a fixed relationship and I think i have it. It means that cubits and metres are variable lengths and the relationship between them is fixed and complicated by the fact that there are long and short versions. But the Berriman calculation agreeing 20.6181818r was something of a Eureka moment. I have Michell's work and the Heaths ' i also have the GP framework as 1760/280 or 2 x 22/7 this gives the 5 and 1/2 sekhed as 14/11 an can be taken as read. I wonder what you think? It works for 176 but for 175? did they use 3.125? Also i am into Ccrowhursts work at Carnac especially his root 5 calculation using Thom's meg yard and he also goes with root 2 and 3. I know your position on this but wonder if his findings may have had any impact on it? If they used 22/7f or pi why could they not approximate the other constants? Finally I agree with Thom that the evidence suggests they looked on base 10 as 'universally sacred' Also , the evidence suggests they used multiple versions of pi and certainly the Roman water men did. To line their pockets perhaps. finally not to ignore the article you have posted, when you talk about the variable ecliptic does this mean they knew the mean version of the eclipse month? Would there be a value in days and what would it be? Does this mean they knew the Saros and could predict eclipses, something the Babylonians were famous for. Sorry i have gone on a bit cheers Dave