As I continue to search for previous work I’ve done on ancient Greek architecture, I discover that apparently I’ve never really posted about the Parthenon although I’ve done a little bit of previous work with it.
I’ve been using Angelopouolos’s diagrams for that. Of course their figures aren’t identical but two other sources provide general corroboration of his figures and help establish the data he gathered himself as sufficiently accurate and reliable.
It’s possible that the Parthenon makes a very good case in point for the problem with Greek Metrology. For some of its most basic proportions, we can see the use of the Greek Foot more clearly than ever (another probable vote of confidence for quality of the data used), but even for these most fundamental measures we do not seem to see it used consistently, which probably means an importation of units that is beyond the understanding of Wikipedia’s source on ancient Greek Metrology.
These basic proportions are
69.589 m = 228.3103675 ft = ~225 Greek Feet
59.087 m = 193.8549869 ft = not Greek Feet?
30.935 m = 101.4927882 ft = ~100 Greek Feet
21.75 m = 71.35826772 ft = not Greek Feet?
228.3103675 is suggestive of 225, this is one of the classic sources for my ongoing remarks that the Greeks seem to have an inclination toward the use of 225 Greek Feet in Imperial.
71.35826772 does not seem to be related to the Greek Foot in the Inductive Metrology sense. Tolerating an error of all of about 8 inches would allow 193.8549869 to be 192 Greek Feet, and about 7 inches in the event of the short Greek Foot; thus this figure may not be in Greek Feet either.
One of the most dramatic things to have emerged from earlier work is that
“69.589 / 59.087 = 1.17773791189 = 11.77245771 I presume.” (Ancient Measures forum, Feb 15, 2020).
I also commented that “On the end the ratio is 30.935 / 21.715 = 1.424591296 – Don’t tell me it’s the Tikal wonder number, 1.424280268? I’ll have to expect (Pi/3) here to decompress it…”
It might turn out to the Faiyum Wonder Number (3 / Palestinian Cubit 2.107038476 ft = 1.423799344) as well, but whatever it is it may be an important key.
It was also suggested that
“Dimensions given at end inside 2nd row of columns
21.715 m / 1.68 m = 1.292559524 = 1.290994449?”
These equations were also given as possible suggestions:
59.087 / 21.715 = 2.720122335 = 2.720174976?
69.589 / 30.935 = 2.249523193793 = 224.8373803?
69.589 / 21.715 = 3.204651162790 = 3.202314486 = 64.04628973 / 2?
72.59 / 69.589 = 1.043124632 = 1.047197551 = (Pi / 3)?
72.59 / 59.087 = 1.228527426 = 1.2328088888 = 2 / 1.622311470?
72.59 / 30.935 = 2.346533053 = 2.354491542 = 1.177245771 x 2?
72.59 / 21.715 = 3.342850564 = 3.346257611 = 1.673128806 x 2?
72.59 / 59.89 = 1.212055435 = 1.216733603??
At least half of that may be correct, and we would have (Pi / 3) where we expected it to work with a Wonder Number. The Faiyum Wonder number has an important but limited run with a series from from it and (Pi /3) to about (Pi / 3)^6. I might be inclined to keep giving equal credence to the Tikal Wonder Number for now.
I was going to suggest that maybe the confusion over the true meaning of “225 Greek Feet” owed to a preference for a different Venus Orbital Period (about 225 days) value such as possibly VOP C, but VOP C doesn’t seem to go well with the long Greek Foot, and perhaps not well with the short Greek foot either, which may narrow it down to VOP A 224.8373808 or VOP B 225.
So far I am working with the premise that the preferred Greek Foot is the long one until forced to concede otherwise; if this bears out it may also help to clarify exactly what was intended.
It was noted that 227.8078955 is able to tap into the “Best Lunar Month” value.
227.8078955 x 360 = 1 / 1.219350967; 227.8078955 x (360^2) = 29.52390326
That astronomical gesture might be something else that could have made some of Ranieri’s models come out somewhat strange?
I”m not sure how pertinent they are, but these observations from the original proceedings were interesting (SMMY = Squared Munck Megalithic Yard = 9 / standard Remen 1.216733603)
“1.873644839 / SMMY = 2.533029594 = (sqrt (sqrt 6586.899509)) / 2
(1.177245771 / Pi) / SMMY = (sqrt (sqrt 6586.899509))”
These are reminiscent of some recent observations about other Greek Temples and the Saros Cycle of 6585.3211 conveyed through square roots or other root functions like root 4 (~square root of the square root of the Saros) as seen here.
There was also the intriguing suggestion that “1.415733832 x 4 = 5.662935328 for the Temple of Olympian Zeus” might have been the number that the “Bat Palace” at Tikal was trying so desperately to convey. I don’t know if I’ve ever followed up on that, although right in the middle of trying to learn a new architecture may not be a great time to do it.
The diagonals of these two rectangles would be approximately
sqrt (69.589^2 + 30.935^2) = 76.15512554 m = 249.8527741 ft, a bit too large for a geodetic figure (?) and
sqrt (59.078^2 + 21.75^2) = 62.95452791 m = 206.5437287 ft, just a bit large to be the standard Egyptian Royal Cubit, but could be the rarely seen long Egyptian Royal Cubit?
206.5437287 / 12 = 1.721197739; long Egyptian Royal Cubit = 1.718873385 x 1.000723277 = 1.720116607 ft
So we are seeing 76.15512554 again only this time in meters, and in fact there are a few more figures involved that can catch attention even when in meters, so this may also be an opportunity to explore ancient meters and how the Greeks used them, although I have few expectations that any ancient persons were able to use a single value for the meter consistently.
In context, although it’s very tempting indeed to take these 7.6-something figures and make 7.599088770 of them, metrologically this involves the short Greek Foot, but because of the prevalence of the long Remen (1.2 long Greek Feet) in Egypt and because I am going to use the long Greek Foot in Greece for as long as I can away with it, the opening bid should probably be 1/16th of 100 long Remens, or (100 x 12.16733603) / 16 = 7.604585019 ft, or 6.25 Remens if Petrie isn’t looking over our shoulders to make sure we are hunting for whole numbers of metrological quanta.
I think I will leave it at that for now until I can find some deeper insights on the subject to share.
–Luke Piwalker
Luke Piwalker 