Michael Lawrence Morton
Excellent .. and it relates to *Current* March Equinox Node
Thu Oct 31 09:28:06 2002
64.12.97.7

Chronos wrote ..
<< 54* 44' 08.19713832" = Grid Latitude 54 x 44 x 08.19713832 = 19476.40065... so we are 
looking at a Hexahedral Constant which is a decimal harmonic of the Tetrahedral Constant, 
geometrically c.a. 19471.22603 and for the Matrix, approximated as the alternate 
19467.73670 N / S, which in hexigesimal mode is virtually indistinguishable on any map
of the scale that has always had to serve us, big as these major nodes might be... 
(the Bermuda Triangle ain't no dinky pyramid, it's freaking big...) I don't know what 
else it could be... but 19467.73670 N / S

It's got it's mother's eyes, you might say... >>

Yes .. very good !! .. The ("HD" ?) Hexahedral Grid LAT ..

Hey !! .. 

8760.48194 * 2.222222222 = 19467.73764 ..
= 54(deg) * 44(min) * 8.193492274(sec) N/S

I wonder if there's anything in Scotland at such a
latitude ? Or .. in Ireland ?

Here's something I noticed about the 2.222222222 ratio
between HD-TET & HD-HEX Grid LATs ..
IF you apply the 7.5 "Incoming Solar-ecliptic-crossing-
point" {re: Nibiru Orbit-path, as-observed-from-
Earth} ..
and then apply the "162" light-speed (B.Cathie) ..
(2.222222222 * 7.5 * 162) = 2700 .. which is the
Ecliptic_Grid LONG_(W.ALNITAK) of the "Incoming Solar-
ecliptic-crossing-point" (of Nibiru's orbit-path). 

Again; keep in mind that this 7.5 Node .. (2700 / 360) ..
is the result of an alignment at Circa 2000 A.D. which
involves an exact "mesh" (sync) between Earth's
precession cycle_and_Earth's main Obliquity-to-the-
ecliptic cycle .. as viewed_against_the sky back-drop
(from Earth, of course). 

Apparently, Nibiru always crosses the Solar-ecliptic
(as viewed from Earth) against the sky_back-drop_of
the Galactic Cross-Nodes. But it is rare, indeed, 
when the Galactic Cross-Nodes *themselves* are precisely
aligned with our Equinoxial points.
Again; indication of a Grand Display.

When the subject of the sync between Earth precession
and obliquity comes up .. I think "83.33333333" ..
= (50 / 0.6) .. ratio of the mean (or "ideal") annual
arc-distance movements of those 2 cycles, in arc-seconds.

(83.33333333 / 2.222222222) = 37.5 .. an important
decimal-harmonic .. and also the DIAMETER of the "Miami
Circle" in regular feet.

Then; you find the Celtic-Unit {Brennan} hanging-around ..
(3.75 / 2.222222222) = 1.6875 .. regular feet.

The "Miami Circle" location .. the NE corner of "Miami
Square" .. I think is "symbolizing" the "UPWELLING"
node of the Galactic Pole, if you will. This would be
like a "fountain-head". This would be akin to Earth's
geographic "North Pole" .. in terms of an "outflow"
of magnetic force or polarity.

Grid Point Value of "Miami Circle" .. is .. Sqrt 15.
(3.872983346 / 1.6875) = Sqrt 3.75 = AVEBURY CIRCLE
Grid Point Value, as revised by Robert Carl in 2002. 

-- MLM 

(c) 2002 by mailto:Milamo@aol.com Michael Lawrence Morton ~ Archeocryptographer.

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