In the last post, I pointed out how Maler’s data for the El Castillo pyramid at Chichen Itza gives a horizontal distance of 2000 cm from the edge of the temple platform to the edge of the base on any side, and recounted how, as pointed out by Munck, a recurring figure of 200 cm is prominent in the data for the Tikal pyramid temples whether our source is Teobert Maler or George F. Andrews.
These could be blatant displays of a “Mayan Meter”, but whether or not they represent ancient American use of metric, in these cases the correct interpretation of the value in use may be 57.29577951^2 / 1000 = (360 / (2 Pi))^2 / 1000 = 3.282806350.
This is the same value that is obtained at Stonehenge by simply honoring the work of WMF Petrie and Alexander Thom as much as humanly possible, accepting Petrie’s conceptualization of the inner sarsen circle as equivalent to 80 Remens diameter (with a Remen “fine tuned to 1.216733603 ft) and Thom’s conceptualization of the outer measures of the sarsen circle as 120 Megalithic Yards circumference (with a Megalithic Yard “fine tuned” from 2.72 to 2.720174976 ft).
The equation looks like this (120 x 2.720174976 / Pi) – (80 x 1.216733603) = 6.564342201 total difference, divided across the two opposite sides of the circle equals 6.564342201 / 2 = 3.282171101.
Obviously addition and subtraction take a toll on the absolute accuracy that is possible here, which is why minute allowances are made in such cases, but the figure is accurate to 3.282171101 / 3.282806350 = .999806492 = ~99.98%
The reason I say this in regard to the Maya even though there are other possible candidates for an ancient meter, is because with the Maya we are in the heart of “Venus worship territory” where there is little getting around an astronomical, iconographic, and artistic preoccupation with the planet Venus.
I was experimenting again with something entirely different, namely the interactions between Megalithic Yard and Megalithic Foot. Thus far,
2.720174976 / 1.177245771 = 2.310626246; 2.310626246 x 150 = 346.5939368, our best value for approximating the Eclipse Year of about 346.62 days.
2.310626246 / 2 = 1.155313123 = ~Tzolkin / Venus Orbital Period = ~260 / 225 = 1.1555555555.
Correct decimal placement notwithstanding, invoking the standard premiere value for the Venus Orbital Period, 224.8373808 = ~”225 days” gives us quite a lovely series of astonomical constants when applied exponentially
1.155313123 x (224.8373808^1) = 259.7575766; 1/4 outer sarsen circle diam Stonehenge = 25.97575766 = an apparently valid version of the 260 day “Mayan” Tzolkin
1.155313123 x (224.8373808^2) = 584.0321316 x 10^n, the standard Venus Synodic Period
1.155313123 x (224.8373808^3) = 131.3122548 = 32.82806369 x 4
1.155313123 x (224.8373808^4) = 29.52390343 “best value for approximating the Lunar Month” x 10^n
The calculation for the first entry is (circumference 120 x 2.720174976) / Pi = outer diameter 103.9030304 ft; 103.9030304 / 2 = outer radius 51.95151515 ft, and 51.95151515 / 2 = 25.975757575.
Thus we can see that the value 3.282806350 (or its simple multiples like 32.82806369 x 2 or 32.82806369 x 4 etc) is a wonderful thing to include in architecture so that we can do the obvious with it, namely expose it to the Venus Orbital Period.
These considerations also hold true for Stonehenge, where 3.282806350 is present as the sarsen circle thickness, and 224.8373808 is present as the apparent correct fine-tuned value of Petrie’s Stonehenge Unit of ~224.8 inches
https://books.google.com/books/about/Stonehenge_Plans_Description_and_Theorie.html?id=rUUIAQAAMAAJ
Note that both the 6800 inches mentioned here (2.72 x 2500 = 6800) and the inner sarsen circle circumference when converted to Megalithic Yards — 80 Remens x Pi = circumference = 80 x 1.216733603 = 305.7985079 ft / 2.720174976 = 224.8373808 / 2 — represent where it’s plausible, or even rather likely, that if Petrie would have simply spent a little more time pouring over his data, he would have discovered the Megalithic Yard well in advance of Thom, and quite possibly the Megalithic Foot as well (360 / 305.7985079 ft = 1.1772435771).
Whatever reason that either culture might have been so preoccupied with astronomy, there seems to be much more to ancient archaeoastronomy and mathematics than simply the moon.
–Luke Piwalker
Luke Piwalker 