I’m going to try to find a way around the apparent error in the data for the cistern of Hadrian’s library in the diagrams by Athanasios Angelopoulos.
The specific issue is that if all of the proportions shown describe its measures inside the rim, 57.82 – (5.706 x 2) = 46.408 m, whereas it is labelled 46.16 m.
I presume that these are all interior measures of the cistern because taking pixel measures of both diagrams shows the ratio between inner width of the cistern and edge around the cistern to be quite close to equal especially considering the method. By projection then the cistern edge would have a width of very roughly 2.5 feet, which is about 4 times the error of 46.61 – 46.408 = 0.202 m = ~.6627 feet, so that this later difference does not seem to reflect the cistern’s edge width, but rather seems to present a smaller error of some kind.
Hopefully we can rest some extra confidence in the figure of 57.82 m since it is given twice as the length of the cistern and also as the width of the courtyard, and Angelopoulos specifically states that these two measures are equal.
If we can rest some confidence in the measures for the semicircles and their remarkable interpretive value, then the actual inner length figure for the cistern may be more likely 46.408 rather than 46.61.
46.408 m = 152.2572178 ft which is quite reminiscent of the Greek Cubit of 1.5 Greek Feet. There is some question about accuracy of approximation here which implies that perhaps they were using a different “360 / Half Venus Cycle” figure for the total length than the “A” version. 1.5 of the Greek Foot value I use customarily would be 1.013944669 x 1.5 = 1.520917004 ft.
If we re-work the equations, it rather looks as if out of necessity they may have meant the “C” version of the Half Venus Cycle that was first taught to us by Stonehenge, but I’m not certain of this and it might interfere with the parallels between the semicircular ends and the calculations for the Aztec Sun Stone unless perhaps the Sun Stonehenge somehow also incorporates the C value for the “HVC” as well as the A and B values.
Interestingly, the mean of the A and B values, 18983.99126 and 18997.72194 respectively, is (18983.99126 + 18997.72194) / 2 = 18990.85660; the HVC C value is 18990.40328 (accuracy of approximation for addition/subtraction = 18990.40328 / 18990.85660 = a very acceptable .9999761582), so it may indeed be possible that the Aztec Sun Stone and its apparent Mayan counterpart at Yaxchilan are even more clever than previously thought, and that the architect referenced the HVA C value of necessity without departing from the “Aztec Sun Stone” theme.
At this point, however, the use of the Greek Cubit is only speculative, and I don’t have the highest confidence level that “1.5” is the true ratio between Greek Foot and Greek Cubit either. I believe this is about the point where we may encounter some turbulence in the metrological schemes of Michell. We have seen the relationship between Egyptian Royal Foot and Egyptian Royal Cubit given as an equally questionable “2/3” even when it is demonstrably NOT the relationship between primary values of these two units.
The inner dimensions of the courtyard outside the columns are given as 73.468 m = 241.0367454 ft and 95.352 m = 312.8346457. The first expression, the width, does not appear to be in Greek Feet although perhaps it’s possible that the second expression, the length, could be 1/324th a Greek Foot x 10^n ; 1.013944669 / 324 = 312.9458855 / 10^n. This is somehow an unfamiliar number so I don’t know yet know what to make of it. The ratio between the two raw data figures is 95.352 / 73.468 = 1.297871182, which may be 51.95151515 / 4 / 10 = 1.298787879. I went though several phases where I was specifically looking out for this number and raw values that looked like this often seemed to turn out to be exactly that.
Assuming symmetry in design, we can obtain the width outside the columns as 73.486 – (7.47 x 2) = 58.546 m = 192.0800525 ft; the length outside the columns is given as 84.454 m which appears to be adequately consistent with the determination (95.352 – 5.45 x 2) = 84.452. 84.454 m = 277.0734908 ft. I am not sure offhand what either of these values is supposed to represent. Neither figure appears to be in Greek Feet in the Inductive Metrology sense.
277.0734908 could represent the Sidereal Month and/or reciprocal of the Solar Leap Year; intriguingly 192.0800525 in Greek Feet is approximately the HVC “B” Builder Figure. The ratio between the two raw data figures is 84.454 / 58.546 = 1.442523827. One of the more attractive figures in this range might be 1.441041519 = (1 / Metonic Cycle 6939.425799) x 10^n, although a fraction of the Eclipse Year (ideally 346.5939351 / 240 = 1.444141396) might also be possible.
It appears as if the inner courtyard length is supposed to be 80.664 meters with 80.479 m as a speculative alternative, as it appears may also be the case with the figure of 57.82 m where 57.76 is seemingly offered as a speculative alternative. The figure of 80.479 as an actual label would seem to be be misplaced as it appears over a longer rather than a shorter part than is labelled 80.664 m. 80.664 m = 264.6456693 ft, which is roughly ten times the value of the width of the Propylon (8.05 m = 26.41076115 ft).
These do not appear to be in Greek Feet unless perhaps 1/384 of a Greek Foot times 10^n, but rather may be measures in Indus Feet. The ratio between the two raw data figures for length and width is 80.664 / 57.82 = 1.395088205, which might be either 1.3941274004 (a simple fraction of Munck’s “Giza Vector Grid Point”) or 1.397273286 (a simple fraction of the proposed primary meaning of “56” at the Aubrey Circle).
For the two figures for the span outside the columns to the inner perimeter of the outer courtyard, 5.45 m = 17.88057743 and 7.47 m = 24.50787402. One thing this second figure in feet might be is (1/48th of a Megalithic Foot) x 10^2, or perhaps possibly 24.46388062, a simple multiple of the inner circumference of the Stonehenge sarsen circle (46.61 m taken literally probably would have meant 1/2 the inner sarsen circle circumference). 17.88057743 does not appear to be in Greek Feet. It may be the reciprocal of the suggested primary approximation of “56” at Stonehenge (1.789199025 x 10^n), or perhaps alternately, the product of Egyptian Mystery Unit (“LSR”) x Hashimi Cubit (1.789803389 x 10^n).
The ratio betwen these two figures is 7.47 / 5.45 = 1.370642202, a range that sometimes requires some thought as to what we are seeing. For example (Pi / 365.0200808) / 2 Pi = 1.369787654 = 1 / 73.00401616) x 10^n, while Michael Morton had a notable fondness for 1.370778390 = (1 / (72 x (Pi^2))) x 10.
That is most of the readily available data on the courtyard given a cursory treatment then; the ratio 97.24 / 95.352 = 1.019800319, which surprisingly might also be valid data and could represent numbers like 1.019328359 or perhaps even 1.021521078, which may be the two best candidates in this particular range.
Two of the next steps we might take are to consider any possible good alternatives to some of these suggestions, and then to begin to see how well any of the suggested component values actually fit together into a cohesive whole.
The relationships between successive courtyard lengths are 95.352 / 84.454 = 1.129040661 = ~(1 / Lunar Year) x 4 x 10^n, and 84.454 / 80.664 = 1.046985024, quite likely to prove to be (Pi / 3) = 1.047197551 (probably a nice Tikal-style style complement to some of these numbers that may be effective at mining additional data from them).
The corresponding relationships between the courtyard widths are 73.468 / 58.528 = 1.255262438 (4 Pi = 1.256637061) and 58.528 / 57.82 = 1.012244898, which may be the Greek Foot in Imperial as ratio, and presumably the intended value is a bit closer to the traditional long Greek Foot value 1.013944669 although the short value of (10 / (Pi^2)) ft may also be possible. 1/16th of at least one form of Phi is probably also in this range (1.618829140 / (16/10) = 1.011768213)
The projected diagonals are: for 73.468 and 95.352 m, d = 120.3725506 m = 394.9230661 ft; for 58.528 and 84.454 m, d= 102.7521528 m = 337.1133621 ft; and for 80.664 and 57.82 m, d = 99.24632636 m = 325.6113070 ft. These do not appear to be in Greek Feet with the possible exception of 337.1133621 (Greek Foot / 3 = 337.9815564 / 10^n).
The diagonals across the corner of the cistern after correction would be about sqrt ((12.49^2 m) + (46.408^2 m)) = 48.05936500 m = 157.6750820 ft, which may be 1296/10 Remens = 157.6886749, or in Greek Feet, 15552 / 10, with the value as expressed in Remens probably being the more economical whole number. 157.6886749 x 2 x 10^n = 31537734.98, the number of seconds in a year of 365.0200808 days.
The length between two circles projected from the semicircles at the ends would be about 57.82 – (5.706 x 2) = 34.996 m = 114.8162730. This may be 114.83806170, which was found at Tikal as its reciprocal, Squared Munck Megalithic Yard x 1.177245771.
The width across the squarish alcoves of 9.72 m = 31.89976378 ft may not be written in Greek Feet per se, but may represent the Apsidal Precession Cycle converted into Greek Feet: 31.89976378 x 1.0139 = 3233.303150.
Thus we conclude for now the first phase of the analysis of the courtyard of Hadrian’s Library.
–Luke Piwalker
Luke Piwalker 